geometry
circumcircle
ratio
geometry proposed

In the plane is given a convex quadrilateral $ABCD$. The point $A',B',C',D'$ are orthocenters  of  the triangles $BCD,CDA,DAB,ABC$. Prove that in the quadrilaterals $A'B'C'D'$ and $ABCD$ the corresponding diagonals in the points of intersection are divides in the same relation.

We denote as Ha, Hb, Hc, Hd, the orthocenters of the triangles BCD, CDA, DAB, ABC respectively, instead of A', B', C', D'.

     We denote also as Ga, Gb, Gc, Gd, the centroids and as Oa, Ob, Oc, Od, the circumcenters of the above triangles, respectively.

      Let as P, H, G, O be, the intersection points of diagonals of the quadrilaterals ABCD, HaHbHcHd, GaGbGcGd, OaObOcOd respectively and we will prove that

PA / PC = HHa / HHc and PB / PD = HHb / HHd, based on the follows proposition.

       LEMMA. - Let ABCD be an arbitrary quadrilateral. Consider the points E, F, on sides AD, BC, respectively such as AE / ED = BF / FC = m.
Let M, N, P be points, on AB, EF, DC respectively, such as AM / MB = EN / NF = DP / PC = n
Prove that the points M, N, P, are collinear.

      I will post here next time, two simple proofs I have in mind about this proposition which has many nice applications and it is also true for non convex quadrilateral and also when the vertices A, B, C, D, are not on the same plane.

1 - In our configuration now, the quadrilaterals  GaGbGcGd,  ABCD  are similar, because of
     their  sides are parallel  to each  other  ( one per one ) and also with the same ratio ( = 1/3 ). So,  we  have  that

     GGa / GGc = PA / PC,   (1) and similarly we have that GGb / GGd = PB / PD,  (2).

2 - Because  of  the  perpendicularities  of  the sides of the  quadrilateral  OaObOcOd,  to the
     sides of  ABCD,  we have that        <OOaOb = < PDC,          <OObOa = <PCD.

     So the triangles  OOaOb,  PCD  are similar  and  then         OOa / PD = OOb / PC         (3)

     Also  we  have       <OObOc = < PAD,       <OOcOb = <PDA.     So  the  triangles  OObOc,

     PAD,  are  similar  and  then      OOc / PD = OOb / PA       (4)

     From  (3), (4)  =>                     OOa / OOc = PA / PC          (5)

     By  the  same  way  we  may  prove  that          OOb / OOd = PB / PD        (6)

     From  (1), (5)  =>      OOa / OOc = GGa / GGc      (7).

     From  (2), (6)  =>      OOb / OOd = GGb / GGd      (8).

3 - Let H', H'' be the points at which the line OG intersects the lines HaHc,HbHd  respectively.

     We consider the non convex quadrilaterals OaHaHcOc,  ObHbHdOd and based on the

     Lemma, from (7), (8), we have that OG / GH' = 1 / 2, (9)  and also OG / GH'' = 1 / 2, (10)

     ( because of OaGa / GaHa = OcGc / GcHc = 1/2 )

     From  (9), (10)  =>  H' ≡ H'' ≡ H,  that  is  the  points  O, G, H are collinear. Hence we

     conclude that HHa / HHc = GGa / GGc = OOa / OOc = PA / PC

     and HHb / HHd = GGb / GGd = OOb / OOd = PB /PD and the proof is completed.

     In  an  arbitrary  Cartesian  system  the quadrilateral ABCD in my  drawing  formed  by
     the  points  A(3, 12), B(12, 13),  C(20, 0),  D(0, 0).

     Kostas Vittas.

Another interesting result in this configuration, is that the areas of the quadrilaterals ABCD, HaHbHcHd (A'B'C'D'), are equal.

But dear Ticks, I have a question. Do you have a different solution of this problem? I would like to know your approach, if it is possible.

Kostas Vittas.

[quote="vittasko"]LEMMA. - Let ABCD be an arbitrary quadrilateral. Consider the points E, F, on sides AD, BC, respectively such as AE / ED = BF / FC = m.
Let M, N, P be points, on AB, EF, DC respectively, such as AM / MB = EN / NF = DP / PC = n
Prove that the points M, N, P, are collinear.

I will post here next time, two simple proofs I have in mind about this proposition which has many nice applications and it is also true for non convex quadrilateral and also when the vertices A, B, C, D, are not on the same plane.[/quote]

You can see at http://www.mathlinks.ro/Forum/viewtopic.php?t=97493

kostas Vittas.

Dear Vittas congratulations with a nice solution.My solution is mostly the same with yours,in fact, it is also the official solution :wink:.I wasn't able to solve this problem during the contest(actually,it was unsolved in the contest) but I did solve it in an hour after the contest finished.Anyway I got first place in the ARGO,even without solving this problem :) .


Mary drove from Amity to Belleville at a speed of 50mi/h. On the way back, she drove at 60mi/hThe total trip took 4.4 hours of driving time. Find the distance between these two cities

[hide="simple hint "]use the formula distance = rate * time[/hide]

[quote="math92"][quote="venkat3848"]120 miles[/quote]
you should try and show some work :)[/quote] AND HIDE your answer! :)

I'm fairly new at this but...

I'll explain it since the answer was given already   :|

I'm not so good at latex, but this problem doesn't require it.

as nonie hinted earlier, use D=RT

on the way to Bellville she drives at 50m/h, so let's assign a variable for miles per hr:

50x

on the way back, it's 60(4.4-x) because the total time is 4.4 hours and the time that is excluded from the previous trip goes into the trip back to Aville.

In these problems one variable is always constant, in this case the distance between the two cities cannot change so  D=50x, D=60(4.4-x)

set those equal, plug x back into either equation to find the distance (120)


email
real analysis
real analysis theorems

OK, I am reading a real analysis book.  And this is the preliminary section.  It talks about  ring, delta-ring, algebra, and delta-algebra.  OK, I know about rings, but what is delta-ring?  I can't even find it in wikipedia.

if you have expertise on this subject, please email me at misogyny1987@gmail.com.  thank you

Are you sure that's a delta? I bet it was really a sigma, that is, $\sigma$-ring and $\sigma$ algebra. Right?

We're in the set of subsets of a given set. There are two main operations: union and intersection. (And union distributes over intersection, and intersection distributes over union.) A ring and/or algebra of sets is some collection of the subsets of a given set that are closed under finite unions and intersections and maybe complements. (Look again at your textbook for the exact definitions.) The prefix $\sigma$- attached to the word ring or algebra extends the definition to being closed under countably infinite unions and countably infinite intersections.

In general, the prefix $\sigma$- (sigma) has the meaning of "countable." For instance, a $\sigma$-finite measure space is the countable union of sets of finite measure; a $\sigma$-compact topological space is the countable union of compact sets.

thanks alot man.  YEs that was this circle with a tail over it.  Sigma!  That's it.  your reply is very much appreciated.

prop:  IF X is any set, the class of countable subsets of X is a sigma-algebra iff X is countable.

my understanding:

  <==

  if X is countable, then all of its subsets are countable.  If this particular class of sets is algebra, then it automatically is sigma algebra.

  ==>

  if a class of countable subsets of X is sigma-algebra, then "everything" not in this class of countable subsets is also in algebra, and consequently in sigma-algebra.  Since it's in sigma algebra, then it is also countable.

Everything not in this class of countable subsets + class of countable subsets of X = X, i.e., the union of two contable sets is countable.

Am I right?


let $ (u_n)$ a sequence such that :
$ u_1\equal{}1$ and $ u_2\equal{}c$  ($ c$ is natural integer)
and $ u_n\equal{}(2n\plus{}1)u_{n\minus{}1}\minus{}(n^2\minus{}1)u_{n\minus{}2}$  $ (n>\equal{}3)$
find the values of $ c$  such that : $ u_i$ divides $ u_j$  for all  $ i\equal{}<j$


The Monge-d'Alembert Theorem states the following:

The external center of similitude of three circles are collinear.

(i) What is an external center of similitude? How do we construct it? (it's the point of intersection of the two external tangents, but what if there aren't external tangents(i.e. one of the circle contains another)?)

(ii) How do we prove the Monge-d'Alembert Theorem using Euclidean geometry only?

Thanks.

Let $ \mathcal{C}_{1}(O_{1},\ r_{1})$ and $ \mathcal{C}_{2}(O_{2},\ r_{2})$ be two non-intersecting circles such that $ r_{1} \geq r_{2}$. The external/internal center of similtude $ E$/$ I$ of $ \mathcal{C}_{1}$ and $ \mathcal{C}_{2}$ is the intersection point of the common external/internal tangents of the two circles. Thus, $ EO_{1}/EO_{2} \equal{} IO_{1}/IO_{2} \equal{} r_{1}/r_{2}$ (the segments are not directed). In this case, the Monge-D'Alembert theorem and its 1-2 variant (see the beggining of my post from [url]http://www.mathlinks.ro/viewtopic.php?p=1186805#1186805[/url] for these "definitions") follow from Menelaus' theorem.

Applying menelaus to the triangle with its vertice the centers of three circles ...

I always see only 3D proofs for Monge. But the above post should be right. Menelaus may solve it, since it's collinear

How do we prove the Monge-d'Alembert Theorem using  projetive geometry ?

I do not know how to use the Desargues theorem in this demonstration.


inequalities
number theory proposed
number theory

Let a possible integer $ n$. Denote S is the sum of all divisors of n, k is the number of divisors of n. Prove that $ k\sqrt {n} < S < n\sqrt {2k}$

let $ p$ prime and $ a>0$
$ \frac{1\plus{}p\plus{}...\plus{}p^{a} }{p^a}\le 1\plus{}\frac{1}{2}\plus{}...\plus{}\frac{1}{2}<2\le \sqrt{2(a\plus{}1)}$
you prove that $ \frac{1\plus{}p\plus{}...\plus{}p^{a}}{p^{a/2}}>a\plus{}1$ (use for exemple $ \inf_{x\ge 2}(\frac{x^{a\plus{}1}}{(x\minus{}1)\sqrt{x}^a})$ )

then $ \boxed{p^{a/2}(a\plus{}1)<1\plus{}p\plus{}...\plus{}p^a< p^{a}\sqrt{2(a\plus{}1)}}$
so if $ n\equal{}\prod_{i\equal{}1}^{m}p_i^{a_i}$ then
$ \boxed{k\sqrt{n}\equal{}\prod_{i}p^{a_i/2}(a_i\plus{}1)<\prod_{i}(1\plus{}p_i\plus{}...\plus{}p_i^{a_k})\equal{}S< \prod_{i}p_i^{a_i}\sqrt{2(a_i\plus{}1)}\equal{}n\sqrt{2k}}$

[quote="aviateurpilot"]${ \prod_{i}p_i^{a_i}\sqrt {2(a_i + 1)} = n\sqrt {2k}}$[/quote]
I think $ \prod_{i}p_i^{a_i}\sqrt {2(a_i + 1)} = n\sqrt {2^sk}$
where $ s$ is the number of prime divisors of $ n$...

another solution for left inequality:
$ S = \sum_{d|n} d\Rightarrow 2S = \sum_{d|n} (d + \frac {n}{d}) \geq \sum_{d|n} 2\sqrt {d\cdot\frac {n}{d}} = \sum_{d|n} 2\sqrt {n} = 2k\sqrt {n} \\
\Longrightarrow S \geq k\sqrt {n}$
and obviously for $ n>1$ the equality not happen...(because  $ 1+n>2\sqrt{n}$...)


Do you think iTunes should lower their prices? Why or why not?
With songs costing $0.99 and videos costing $1.99, the average American teeanger spends $180 on iTunes (i dont know for sure if the source i used to find this is correct).

A lot of my friends download songs for free from other websites and I think that if Apple reduces iTunes's prices, their profit won't go down because people will be motivated to buy more songs.

Kind of a useless topic, but what ever

ehh...even if their like 20 cents why wouldnt people still go for the free options...and just fyi, i thought of this but definetely dont do it(im very law abiding) you could just get one of those microphones that you sometimes use to record youself that save your recording directly onto the computer(i have one they're like 20 bucks) but y couldnt someone just go on you tube and record the songs and save them to their ipod(not the best quality) but it would probably work...like i said i havent tried it and i dont suggest that people try it(probably illegal) but i will say that the movies are kinda pricy...14 dollars for a new release...WOW!!!maybe 5 or 6!!! and wuts up with these rentalls...kinda useless in my mind!!!

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

Grammar, its a good thing.

Anyways, I believe that paying one dollar isn't expensive, if the music they provided WAS NOT JACKED UP BECAUSE OF DRM.

This is why I only purchase my music (I have about $ \$$6000 of it) from Amazon, since they have better quality .mp3's and no strings attached.

Also (directed towards Zolojetto), why use a microphone when you could simply use a program like Audacity to directly record from your sound card.

[quote="happyme"]Also (directed towards Zolojetto), why use a microphone when you could simply use a program like Audacity to directly record from your sound card.[/quote]

Audacity is so useful. Although, I do remember having some video editing freeware that could strip the audio file from a video, as well as some other minor editing tasks.

I've wondered about the ethics of this: Can't you go to the public library, check out some CDs, and rip the files?

[quote="Kalle"]iTunes really has no natural market since you can get better quality for free of charge. Their market is an artificial one, created by laws pushed on by lobbying. It's pure protectionism, trying to save a traditional market that doesn't exist anymore because of new technology. By standard economic principles, it's inefficient. "Oh no save us from the free music!" really rings the same tone "oh no save us from the cheap shoes produced elsewhere" or "oh no save us from the cheap labour".

Plain, stupid-as-always, protectionism.[/quote]

Cheap $ \neq$ Free

Also, by saying that you want to be saved by "cheap shoes" equates to music - then you inherently mean you are fine listening to crap music produced by tone-deaf kids in some province of China.

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

Cheap $ \neq$ Free

Also, by saying that you want to be saved by "cheap shoes" equates to music - then you inherently mean you are fine listening to crap music produced by tone-deaf kids in some province of China.[/quote]
I don't know what you're trying to say, and I suspect you haven't added anything to the discussion. You've just taken words with some connotation and put them into a different context. I did not imply what you said.

By cheap I simply meant low price, not a bad product in any way.

Should Itunes lower their prices? Well of course, because they're obviously competing with cheaper song download sites, and not to mention Bittorent and other filesharing programs  :P  But come to think of it, when Ipods are sold, they say they can hold 6000+ songs and for computer-not-so-savvy people, they will probably go with itunes, and if they intend to use most of their space, that will be extremely expensive, which is all a part of their marketing strategy.

1 dollar for a SINGLE SONG is way too expensive. I do not agree with zolojetto, if the price was less I think more people would be encouraged to buy music, than download it from the n (where ${ n \to \infty}$) available pirated sources. But as the above user says, if it were not the not-so-computer savvy users, iTunes store might not have been a hit though a lot of tech-savvy guys do use iTunes but still......

P.S On a sidenote, I prefer Mediamonkey over the iTunes Software anyday. :D

iTunes makes profit because it is relatively free of overhead, convenient, and very mainstream. The demographic it caters to is not the hey-look-at-me-I-can-rip-audio-files-from-YouTube group. Just because there are people smart and amoral enough to realize that torrenting albums is better does not mean that everybody else is.

iTunes should offer a program like Verizon Wireless programs. Your friend can recommend x number of songs for you and you can get those songs for free, like unlimited texting or 1000 minutes on your cell phone. I do not recommend that they actually lower their prices, but offer a program that affordable for their customers' sakes and the company's.

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

That would mean about around $ \$1,000$ lost in free songs. Maybe you could pay 50 cents, instead of getting free songs almost every other album you purchase.

[quote="Ignite168"][quote="Cyber-Shreyas"]1 dollar for a SINGLE SONG is way too expensive. I do not agree with zolojetto, if the price was less I think more people would be encouraged to buy music, than download it from the n (where ${ n \to \infty}$) available pirated sources. But as the above user says, if it were not the not-so-computer savvy users, iTunes store might not have been a hit though a lot of tech-savvy guys do use iTunes but still...... [/quote]

Yeah right; if you think the prices are set too high, then why wouldn't you buy from its competitors?

It would be an idiotic move for iTunes to lower their prices. Their consumer demographic is primarily buying songs based on either: a) impulse b) convenience c) not knowledgeable about alternative sources d) marketing. Impulse is the reason you can buy songs with a couple of touches on the iPhone; people don't usually gauge price when making impulse decisions. Convieniencewise, why would people take the hassle of pirating something when with one click they can buy a song from iTunes? c) is obvious, for marketing, Apple has marketed themselves as trendy and upscale. I bet one of their main sources of revenue is gift cards judging by the fact that iTunes gift cards are now everywhere. People perceive iTunes gift cards as something desirable and rightly so, and then people buy music on iTunes.

iTunes' business model does not have a focus on competing pricewise and they have no reason to lower their prices judging by the success of iTunes and its dominance over Napster, Google's brief attempt at selling music and video online, etc.[/quote]

this is a really accurate post.  i remember hearing that they sold some obscene number of songs on christmas day last year, something like 30 million, so they obviously are selling a lot of gift cards.

[quote="D3m0n Shad0w"][quote="aanjohri"]and maybe once in a while they could do sales like "if you buy 10 songs you get the 11th one free" or something[/quote]

That would mean about around $ \$1,000$ lost in free songs. Maybe you could pay 50 cents, instead of getting free songs almost every other album you purchase.[/quote]

I am guessing that neither of you actually read Ignite's post.

Well...It's not too expensive, I don't mind paying $ .99 a song or up to$15 a movie. This is what you would have to pay at the store anyway. Also if you think its too expensive try Limewire?

Sound is not copyrighted, so anything you hear on the radio could be kept by your storage media. If you do not sell it, there will be absolutely no problem of being arrested. Besides, how many of us borrowed a movie from a friend or have our friends give us a CD of their favorite tracks?


Two perpendicular chords intersect in a circle. The segments of one chord are 3 and 4; the segments of the other are 6 and 2. Then the diameter o the circle is:

a. $ \sqrt{89}$
b. $ \sqrt{56}$
c. $ \sqrt{61}$
d. $ \sqrt{75}$

For some reason I keep getting an answer that's not listed:

[hide]
(I don't know Asymptote, so I'll just try to get along without it:)
Ok, so I constructed the circle with the 3 and 4 chord vertically on the right, and the 6 and 2 horizontally. I then drew the radius meeting the 6 and 2 chord on where it meets the circle. Therefore, we get a right triangle with hypotenuse $ r$, and leg$ 4$, since the vertical line that passes through the center bisects the chord. Therefore, the remaining side has length $ \sqrt{r^{2}-16}$ Moreover, this must be equal to $ \frac{7}{2}-3=\frac{1}{2}$(Using the same properties on the other chord), so solve:
$ \sqrt{r^{2}-16}=\frac{1}{2}$
$ r^{2}-16=\frac{1}{4}$
$ r^{2}=\frac{65}{4}$
$ r=\sqrt{\frac{65}{4}}$
$ r=\frac{\sqrt{65}{2}}$
, so diameter= $ \boxed{\sqrt{65}}$...
which is not listed :(
[/hide]

Here is a diagram:[asy]pair A=(0,0), B=(-2,4), C=(0,7), D=(6,4), E=(4,7), P, Q, O; P = intersectionpoint(A--C,B--D); Q = foot(E,B,D); O = midpoint(O--E); draw(circumcircle(A, B, C)); draw(Q--E--A--C--B--D--E--C); dot(A^^B^^C^^D^^E^^P^^O^^Q); label("A",A,SW);label("B",B,W);label("C",C,NW);label("D",D,NE);label("E",E,NE); label("P",P,SE);label("O",O,SE);label("Q",Q,S);[/asy]Let $ AP=4$, $ PC=3$, $ BP=2$, $ PD=4$. Then draw a line from $ C$ parallel to line $ BD$ to intersect the circle again at $ E$. Since $ CE\parallel BD$, $ BC=ED$, and $ CP=EQ$, so $ BP=QD=2$. Then $ PQ=CE=4$, so the diameter is $ \sqrt{(4+3)^{2}+4^{2}}=\sqrt{65}$.
I get the same answer as erdogankerem123.. :huh:

In general, we have the formula: $ D^{2}\equal{}AP^{2}\plus{}BP^{2}\plus{}CP^{2}\plus{}DP^{2}$ [according to minsoens' diagram].

This also produces $ \sqrt{65}$.

@minsoens: you have awesome asymptote diagrams.

[quote="Altheman"]In general, we have the formula: $ D^{2}\equal{} AP^{2}\plus{}BP^{2}\plus{}CP^{2}\plus{}DP^{2}$ [according to minsoens' diagram].

This also produces $ \sqrt{65}$.

@minsoens: you have awesome asymptote diagrams.[/quote]thank you :D i always like to include a diagram with my solutions because it's difficult for me to explain without one. you should learn asy too, since you frequently post in geometry topics


This comes from the 1st AIME.

What is the remainder on dividing  by 49?

Seems easy, but I think I must have taken the long route to answer it.

Here's what I did:

[hide]Change 6=7-1 and 8=7+1.  Then you expand it and get the following:

(xCy) means x!/y!(x-y)!, since I don't know the TeX for combinations yet.  And I couldn't find some easy way to simplify this.  I'm sure there is some easy way.  The answer has to be a multiple of 7 between 0 and 42.  I checked the answer and it is 35.[/hide]

PS Could you please tell me how to get combinations in TeX?

TeX : {n \choose k} gives \[ {n \choose k} \]

In your expansion,
$2{83 \choose 0}7^{83}, 2{83 \choose 2}7^{81}, ... , 2{83 \choose 80}7^3$ are multiple of $7^2=49$,
so $x \equiv 2{83\choose 1}7^1 \pmod {49}$.

Thanks, I realized that shortly after posting it... I should give myself more time.

[quote="WarpedKlown1335"]Could've done it using (7-1)83+(7+1)83 too. ;-)[/quote]

I think that's what he did!!


Let $ n$ and $ k$ be natural numbers ,prove that $ n^{kn} \geq (n^{k} \plus{}n^{k\minus{}1} \plus{}...\plus{}1)^{n\minus{}1}$

[b]Full Solution[/b]
We can first use the fact that $ n^{k} \plus{} n^{k \minus{} 1} \plus{} ... \plus{} 1 \equal{} \frac {n^{k \plus{} 1} \minus{} 1}{n \minus{} 1}$, so we have to show
$ n^{kn} \cdot (n \minus{} 1)^{n \minus{} 1} \ge (n^{k \plus{} 1} \minus{} 1)^{n \minus{} 1}$
Since $ n \ge n \minus{} 1$, we have $ LHS \ge n^{kn \minus{} k} \cdot (n \minus{} 1)^{n \minus{} 1 \plus{} k} \equal{} n^{k(n \minus{} 1)} \cdot (n \minus{} 1)^{n \minus{} 1 \plus{} k}$
Case 1: $ (n \minus{} 1)^{n \minus{} 1 \plus{} k} \ge (n \minus{} 1 \plus{} k)^{n \minus{} 1}$ is true
Then we have to show
$ n^{k(n \minus{} 1)} \cdot (n \minus{} 1 \plus{} k)^{n \minus{} 1} \ge (n^{k \plus{} 1} \minus{} 1)^{n \minus{} 1}$
Taking the n-1th root of both sides, we have to show
$ n^k(n \minus{} 1 \plus{} k) \ge n^{k \plus{} 1} \minus{} 1 \iff (k \minus{} 1)n^k \ge \minus{} 1$, which is obvious.
Case 1 is usually true as $ a^b \ge b^a$ is true for $ a < b$ if $ a > e \equal{} 2.71828...$ since by taking ln of both sides, it is equivalent to $ \frac {ln a}{a} > \frac {ln b}{b}$, and $ \frac {ln x}{x}$ is strictly decreasing for $ x > e$. So the proof is true for $ n \ge 4$. It is obviously true for n=1. If n=2 it is equivalent to $ 2^{2k} \ge 2^{k \plus{} 1} \minus{} 1$ which is obvious. If n=3 it is equivalent to $ 4 \cdot 3^{3k} \ge (3^{k \plus{} 1} \minus{} 1)^2$ which is also pretty obvious. So we are done!


A MATHCOUNTS committee contains exactly three women and four men. what fraction of the committee is men? Express your answer as a common fraction.

[hide="Solution"]There are $ 4$ men. There a total of $ 7$ people. Thus, $ \frac {\text{success}}{\text{possible}} \equal{} \boxed{\frac {4}{7}}$.[/hide]


In how many ways can U.S. coins be combined to total $ 26$ cents?

$ 25a\plus{}10b\plus{}5c\plus{}d\equal{}26$
If a=1, one way
a=0

$ 10b\plus{}5c\plus{}d\equal{}26$
b=2
$ 5c\plus{}d\equal{}6$ 2 ways
b=1
$ 5c\plus{}d\equal{}16$ 4 ways
b=0
$ 5c\plus{}d\equal{}26$ 6 ways

Total: 13 ways

Alternatively, use the formula for $ 10\le n\le49$: in this range, there are $ m\equal{}3\plus{}\frac{\lfloor\frac n5\rfloor(\lfloor\frac n5\rfloor \minus{} 1)}2$ ways.

[hide="Proof"]
Let $ k\equal{}\left\lfloor\frac n5\right\rfloor$.

[b]Lemma:[/b] If $ 5k$ coins gives us $ m$ ways to do this, then $ 5n\plus{}1,5n\plus{}2,\ldots,5n\plus{}4$ also have $ k$ ways.
[b]Proof:[/b] In each original arrangement of the $ 5k$ coins, say there are $ q$ quarters, $ d$ dimes, $ n$ nickels, and thus $ 5k\minus{}(q\plus{}d\plus{}n)$ pennies. This means that once $ q,d,n$ are chosen, there is exactly one possible value for the number of pennies, so if we add any number to $ 5k$ (a multiple of $ 5$!) from $ 0$ to $ 4$, we will still have the same number of possible arrangements of $ q,d,n$, since we need at least $ 5$ to get another nickel, hence boosting the the number of combinations for $ q,d,n$.

This reduces our casework tremendously (we can just go by $ 5$'s instead of one by one).

[hide="Casework by $ k$"]
$ k\equal{}2\implies10\le n\le14\implies m\equal{}4$
$ k\equal{}3\implies15\le n\le19\implies m\equal{}6$
$ k\equal{}4\implies20\le n\le24\implies m\equal{}9$
$ k\equal{}5\implies25\le n\le29\implies m\equal{}13$
$ k\equal{}6\implies30\le n\le34\implies m\equal{}18$
$ k\equal{}7\implies35\le n\le39\implies m\equal{}24$
$ k\equal{}8\implies40\le n\le44\implies m\equal{}31$
$ k\equal{}9\implies45\le n\le49\implies m\equal{}39$
[/hide]

Now subtract $ 3$ from each of the cases above to get the sequence $ 1,3,6,\ldots,28,36$, which are the triangular numbers shifted once to the right!

Finally, this means that for $ 2\le k\le9$, $ m\equal{}3\plus{}\frac{k(k\minus{}1)}2$, or equivalently, for $ 10\le n\le49$, $ m\equal{}3\plus{}\frac{\lfloor\frac n5\rfloor(\lfloor\frac n5\rfloor \minus{} 1)}2$, as desired.
[/hide]

Very nice formula! Is there a general case?

I don't think so. As you can see, I didn't really "derive" the formula, it was more or less experimentation and just a way to "condense the results."

The "real way" to solve this is to use generating functions...if we can use all types of US coins, then the number of ways to get a value of $ n$ cents is the coefficient of $ x^n$ in $ \sum_{k\equal{}0}^\infty x^k\sum_{k\equal{}0}^\infty x^{5k}\sum_{k\equal{}0}^\infty x^{10k}\sum_{k\equal{}0}^\infty x^{25k}\sum_{k\equal{}0}^\infty x^{50k}$ (see if you can figure out the reason why this works).

So that only works for 10-49 cents? btw very nice. :!: Are there other formulas for bigger ranges?

Maybe. I haven't looked at it much though. But I doubt a "universal formula."

[b]Edit:[/b] There's no polynomial formula for $ 10\le k$ in pennies, nickels, dimes, and quarters. I'm pretty sure the formula above is just a coincidence. There's also no formula for $ 10\le k$ in pennies, nickels, dimes, quarters, and half dollars.


If $ A\ \spadesuit\ B$ is defined as $ A\ \spadesuit\ B \equal{} A \plus{} B \plus{} 4$, what is the value of $ A$ for which $ A\ \spadesuit\ 2 \equal{} 19$?

We plug in $ 2$ for $ B$. $ A\plus{}2\plus{}4\equal{}19 \Rightarrow A\equal{}19\equal{}6 \Rightarrow A\equal{}13$

let's just plug in the values

$ A\plus{}2\plus{}4\equal{}19$
$ A\equal{}13$

edit:beaten to it....


[b]the two ends of a long cylinder containing gas are maintained at different temperatures.which of the following quantities will have the same value throughout the cylinder?

(a)pressure
(b)density
(c)ratio of pressure to density
(d) the no. of molecules per unit volume. [/b]

c is definitely not true because P/density = RT.

b and d imply same thing, since only 1 answer so a. :p

Srry for stupid reply.

$ \text{Obviously, pressure will not be same throughout, but not density, since,}\; \rho\propto \frac {1}{V}, \; \; V\propto T. \\
\therefore \rho \propto T \\
 \\
 \\
P: \rho \equal{} \frac {RT}{M}, \; \; \text{where M is the molecular mass of the gas} \\
 \\
 \\
\;\text{a is the answer.}$

[quote="Rajiv"]$ \text{Obviously, pressure will not be same throughout, but not density, since,}\; \rho\propto \frac {1}{V}, \; \; V\propto T. \\
\therefore \rho \propto T \\
 \\
 \\
P\rho \equal{} \frac {RT}{M}, \; \; \text{where M is the molecular mass of the gas} \\
 \\
 \\
P: \rho \equal{} \frac {RT}{M\rho} \\
 \\
 \\
\;\text{Hence, only Pressure is constant. a is the answer.}$[/quote]

Example of anxiety induced by fiitjee.... :P  :P

[quote="hell_ever"]
Example of anxiety induced by fiitjee....

[/quote]

I'd rather say it was hunger...!  :D  :lol:

[hide]
The ideal gas law is
\[ PV \equal{} NRT\]
where $ P$ is the pressure of the gas, $ N$ is the number of moles of gas, $ R$ is the universal gas constant, and $ T$ is the absolute temperature.

$ V$ equal to the volume of the cylinder, is constant.

$ N$ is constant, assuming that no gas is added to or released from the cylinder.

The number of moles per unit volume is $ n \equal{}\frac{N}{V}$ is constant because $ N$ and $ V$ are both constant.

The number of molecules per unit volume is $ 6.02\times 10^{23}\cdot n$ is constant.

The density, $ \rho \equal{} M n$, where $ M$ is the molecular weight of the gas, is constant.

Now, the ideal gas law can be rewritten as $ P \equal{} nRT$.
\[ P \equal{} nRT \implies \Delta P \equal{} nR\Delta T\].

Since $ T$ is not constant, $ P$ is not constant.

One can also rewrite the gas law as $ \frac{P}{\rho} \equal{}\frac{R}{M}T$.

\[ \frac{P}{\rho} \equal{}\frac{R}{M}T \implies \Delta \left(\frac{P}{\rho}\right) \equal{}\frac{R}{M}\Delta T\]

Likewise, since $ T$ is not constant, $ \frac{P}{\rho}$ is not constant.

Therefore the answer to your question is (b), and (d).[/hide]

[quote="Conjurer"][quote="Dr. No"]

The number of moles per unit volume is $ n \equal{} \frac {N}{V}$ is constant because $ N$ and $ V$ are both constant.

[/quote]

This is just,saying due to lack of words, average density. But actually there maybe more molecules in one part of the volume and less in the other. So I dont think we can comment anything surely.[/quote]
[hide]Essentially you're saying that $ \frac {\partial N}{\partial V} \neq n$.

Consider $ \frac {\partial n}{\partial V} \equal{} 0 \equal{} \frac {\partial}{\partial V}\left(\frac {N}{V}\right)$.
\[ \frac {\partial}{\partial V}\left(\frac {N}{V}\right) \equal{} \frac {\frac {\partial N}{\partial V} V \minus{} N \frac {\partial V}{\partial V}}{V^2} \equal{} 0 \\
 \\
\implies \frac {\partial N}{\partial V} V \minus{} N \equal{} 0 \\
 \\
\implies \frac {\partial N}{\partial V} \equal{} \frac {N}{V} \equal{} n
\]
.

Therefore we have by conjecture that $ \frac {\partial N}{\partial V} \neq n$, and we have by calculation that $ \frac {\partial N}{\partial V} \equal{} n$. One of these is wrong. I'll leave it to you to decide which one. [/hide]


Could an administrator make the source field available for all the forums for me. It is very helpful to have a small caption indicating where you got the problem from. Thanks.

[quote="Peter VDD"]Maybe it could be helpful to indicate in which forums they are still forgotten. ;)[/quote]

A lot. That's all I can tell you. I only remember the source field in the Mathcounts forum.


$Question 1$ This one has three parts...
(Part A): There are $100$ people in a ballroom. Every person knows at least $67$ other people (and if I know you, then you know me). Prove that there is a set of four people in the room such that every two from the four know each other. (We will call such a set a "clique".)

(Part B): Two people in the room are Joe and Grace, who know
each other. Is there a clique of four people which includes Joe
and Grace?

(Part C): Oops, I miscounted. There are actually $101$ people in the room, But it's still true that each knows at least $67$ others. That can't make a difference, can it?

$Question 2$ This is related to Prime numbers... (lifted from some where....)
After doing the necessary calculations, I found myself looking at a staggering $6$-digit number on my pocket calculator. I observed that it was a prime number that had non-zero and distinct digits when the hawk-eyed officer sitting across the table remarked that he also saw a prime with the same characteristics from his side. “An astounding number!!” said the officer; little realizing that he was looking at the number upside down,” Moreover, as you also might see the last four digits of the number make a prime number, too.

Actually, the number that I saw had the last four and the last three digits also making up a prime number but I didn’t tell the officer.

Can you tell me what numbers we were looking at?

uhhh i think i might have #1?

[hide]

a) Consider four groups of people, A,B,C, and D, such that there are 1,33,33, and 33 people in those groups, respectively. Then let's say that everyone in group A, B, C, or D, doesn't know any of the people in their group. So, laying this out, we would have everyone in group B knows everyone in groups A, C, and D, and likewise for groups C and D. For Group A, however, let's just say that person knows everyone in groups B and C but only one person in group D (the idea behind this though is just that A must know at least one person in each group). Hence, connecting the lines it is clear that there is at least one clique in this room, since we have a person from A who knows someone from B,C, and D, and likewise a person from B who knows those people in A,C, and D, and etc.

b) For this question, I think it depends on whether or not the person in group A knows the two people... Since, let's just wlog and say that Joe is in group B and Grace is in group D, then they know each other, but let's say that the one person in group D that the person in group A knows is NOT Grace. Then we have a problem, as there cannot be a clique of four people. But, if that one person in group D that the person in Group A knows IS Grace, then we have a clique which includes both Joe and Grace.

c) The fact that there are 101 people in the room now does affect the outcome of part A. We can wlog and put that extra person in group A, so now there are 2 people. But, since each person still needs to know just 67 others, the people in groups B,C, and D now have the "option" of knowing either one of the people in group A (whereas before there was only one person so they had to know that one person). Because of this new second option, there no longer has to be a clique in the room.

[/hide]


Solve $n^{4}\equiv n\pmod{10000}$ for all integers.

Is there any way to use Fermat's theorem, $a^{p-1}\equiv1\pmod{p}$, and/or Euler's generalization to Fermat's theorem to do this?

Thanks to SnowStorm for the tip.

Here's a latex side note anyway:  use \pmod{} instead of just \mod it will put the parentheses around the mod automatically.

[code]a \bmod n[/code]
also works, giving $a \bmod n$.  :)

[hide]Well, I got 0, 1, and 625 so far.[/hide]

[hide="Solution"]Subtract $n$ from both sides and factor.  We get $n(n-1)(n^{2}+n+1)\equiv 0\pmod{10000}$.

By manually plugging in values for $n$, we find that $n^{2}+n+1$ is not divisible by 2 or 5.  Therefore, we have $n(n-1)\equiv 0\pmod{10000}$.  Note that $n$ and $n-1$ can't both be divisible by 2 and can't both be divisible by 5.  Now we do some casework:

1)  $n$ is divisible by $2^{4}$ and $5^{4}$.  Then $n\equiv 0\pmod{10000}$.

2)  $n-1$ is divisible by $2^{4}$ and $5^{4}$.  This gives $n\equiv 1\pmod{10000}$.

3)  $n$ is divisible by $2^{4}$ and $n-1$ is divisible by $5^{4}$.  Let $n-1=m\cdot 5^{4}$.  We have $625m+1\equiv 0\pmod{16}$.  Subtracting $624m$ (which is allowed because 624 is divisible by 16), we get $m\equiv 15\pmod{16}$.  Therefore $n=(16k+15)5^{4}+1=10000k+9376$, so $n\equiv 9376\pmod{10000}$.

4)  $n-1$ is divisible by $2^{4}$ and $n$ is divisible by $5^{4}$.  Let $n=m\cdot 5^{4}$.  We have $625m-1\equiv 0\pmod{16}$.  Again we subtract $624m$ to get $m\equiv 1\pmod{16}$.  Now we have $n=(16k+1)5^{4}=10000k+625$, so $n\equiv 625\pmod{10000}$.

The answers are $n\equiv 0,1,625,9376\pmod{10000}$.[/hide]

Can anybody clarify the above solution?

Why did you only consider when $n$ is divisible by $2^{4}$ and $5^{4}$ (in Case 1). Should there be other cases where $n$ is only divisible by $2^{3}$, $5^{3}$, etc? This question also goes towards Case 2, 3, and 4.

[quote="1/(ln x)"]Can anybody clarify the above solution?

Why did you only consider when n is divisible by $2^{4}$ and $5^{4}$  (in Case 1). Should there be other cases where $n$ is only divisible by $2^{3}, 5^{3}$, etc? This question also goes towards Case 2, 3, and 4.[/quote]

[hide="hint"]Um, if $n-1$ is divisible by 5, then is $n$ divisible by 5?[/hide]


I had quite an adventure doing this round of USAMTS.  In particular, I was in Hong Kong with my school's Jazz Band and Orchestra when the round was due.  They decided to fill our days completely from 7 AM to 9 PM, so I was working out my solutions in the wee hours of the morning.  With no access to a computer, I handwrote all of my solutions, hoping to fax them on the 13th (which would be the 12th in the states).  Unfortunately, someone (either our hotel or the Telecoms) charges an exorbitant rate for faxing (10 USD per page), so I ended up not submitting anything (Bronze winner works well enough for me).

That having been said, late nights seem to have hampered my reading comprehension skills, since I worked a completely different problem for #5.  I read condition (1) as:

$x_{i}\not= x_{j}$ for all $i \not= j$

I had a good bit of fun with this problem, so I figured I'd toss it out for others to try.  I haven't solved it yet, but I'm fairly sure that there's something simple that I'm missing.  Anyway, here's what I have so far:

I was able to prove that the following sequence is the sequence with minimal sum satisfying the given conditions (where, again, the given conditions are my modified version of condition (1).)

$a_{n}= A_{n-1}$ if $A_{n-1}$ has not yet appeared in the sequence
$a_{n}= A_{n-1}+n$ otherwise.

I'd post the method for showing minimality here, but it's somewhat tedious analysis, and, again, my solutions are handwritten, so I can't just copy-paste it over (and I don't have access to a scanner at the moment).

Unfortunately, I wasn't able to go too far past that sequence, but I did notice something strange.  Below are the first 21 values of this sequence.
\begin{tabular}{| c | c | c | }\hline n & A_{n}& a_{n}\\ \hline 1 & 1 & 1 \\ \hline 2 & 2 & 3 \\ \hline 3 & 2 & 2 \\ \hline 4 & 3 & 6 \\ \hline 5 & 4 & 8 \\ \hline 6 & 4 & 4 \\ \hline 7 & 5 & 11 \\ \hline 8 & 5 & 5 \\ \hline 9 & 6 & 14 \\ \hline 10 & 7 & 16 \\ \hline 11 & 7 & 7 \\ \hline 12 & 8 & 19 \\ \hline 13 & 9 & 21 \\ \hline 14 & 9 & 9 \\ \hline 15 & 10 & 24 \\ \hline 16 & 10 & 10 \\ \hline 17 & 11 & 27 \\ \hline 18 & 12 & 29 \\ \hline 19 & 12 & 12 \\ \hline 20 & 13 & 42 \\ \hline 21 & 13 & 13 \\ \hline \end{tabular}

Notice that, in the above $a_{a_{n}}= n$ and $A_{A_{n-1}+n}= n.$  Using either condition, we can "leapfrog" our way up the sequence.  For instance

$A_{6}= 4$, so $A_{4+7}= A_{11}= 7,$ so $A_{7+12}= A_{19}= 12,$ so $A_{12+20}= A_{32}= 20,$ and so on, and so forth.

Unfortunately, I wasn't able to proceed from this point.  Anyone else want to take a crack at it?  To be honest, I'm more interested in why $a_{a_{n}}= n$ than anything else.  Help would be appreciated.

That's exactly what i did which confused me. I got the same sequence as you but had trouble calculating $A_{2007}$


limit
calculus
derivative
integration
algebra
binomial theorem
calculus computations

Use L'Hopital for indeterminations $0/0$ (you derivate the numerator and denominator by separate and then you divide them) The limit is $-5/96$

You will have to nicely factor it, and I don't know how.  Does anyone?

Let $f(x)=\sqrt[4]{x+15}-\sqrt[3]{x+7},$ we have $f(1)=0.$ Thus from the difinition of differential coefficient our limit is

$\lim_{x\rightarrow 1} \frac{f(x)-f(1)}{x-1}=f'(1)=\frac{1}{24}.$

[color=blue]Last edited[/color]: The right answer is $-\frac{5}{96}.$

kunny

How about no calculus? Although it looks like your solutions differ.

I guess now that this has been moved to a calculus forum we're allowed calculus.

I'm going to use the first stage of the binomial theorem:

$(1+u)^{\alpha}=1+\alpha u+O(u^2).$

The first step for the problem is to make a substitution:

$y=x-1$, so $x=y+1$ and $y\to0$ as $x\to1.$

We have

$\lim_{y\to0}\,\frac{\sqrt[4]{y+16}-\sqrt[3]{y+8}}y= \lim_{y\to0}\frac2y\left[\left(1+\frac y{16}\right)^{\frac14} -\left(1+\frac y8\right)^{\frac13}\right]$

$=\lim_{y\to0}\,\frac2y\left[\left(1+\frac y{64}+O(y^2)\right)-\left(1 +\frac y{24}+O(y^2)\right)\right]$

$=\lim_{y\to0}\,2\left(\frac1{64}-\frac1{24}\right)+O(y)=-\frac5{96}.$

SOL's answer is correct.

SOL and Dr. Merryfield's answer is correct, but kunny's method is valid and leads to the correct answer if you differentiate properly.  It is a bit like pulling a rabbit out of a hat, but it is a nice idea because it allows you to evaluate many 'contrived' limits.

Another limit:

$\displaystyle\lim_{x\to\infty}6x^2(\frac{pi}{4} - atan\frac{x^2}{x^2-1})$

that would be interesting, but i dont think it can be nicer than the L'hopital one  ;)

[quote="Fermat -Euler"][quote="Fermat -Euler"]Find
$\lim_{x\rightarrow\ 1}{\frac{\sqrt[4]{x+15}-\sqrt[3]{x+7}}{x-1}}$[/quote]
[quote="Fermat -Euler"]withowth using L'hopital ;)[/quote]
[quote="SOL"]i cannot think in a nicer solution  :oops:
what about you?[/quote]
I have a solution and i want see ather solution .
do you want know mine :)[/quote]

   It is very easy and I think we can use caculus with Kunny have already done. It is the fastest and easy way to do.
  By the way , I can prove it without caculus but I like Kunny's method.
If we observed that  $\sqrt[4]{1+15}=2$ and $\sqrt[3]{1+7}=2$ so on the numerator we can manipulate it like:
$A= \sqrt[4]{x+15} -2 +2- \sqrt[3]{x+7}$
$A1=\sqrt[4]{x+15} -2$ can be factor by multiplying with rational conjugate $\sqrt[4]{x+15}+2$ and $\sqrt{x+15}+4$
$A2=2- \sqrt[3]{x+7}$ . Time it with $4+4 \sqrt[3]{x+7}+ \sqrt[3]{(x+7)^2}$

  Sorry . I am lazy at doing computation like this. I think you can do it easily .

[quote="Gibbenergy"] It is very easy
...
 Sorry . I am lazy at doing computation like this. I think you can do it easily .[/quote]
I think that is it were truly "very easy" you would have finished it yourself. The notion of "rational congugate" with roots other than square roots, is, well, messy. Oh, the method is perfectly correct, and following the path you have started will lead to the right answer.

But this is the point at which I'll say that I think the binomial series (see my post above) is simpler than this.

Hi Sir Kent Merryfield!
 I am sorry that I says "It is very easy" and don't go further on my sols. I think you and SOL have given the correct answer. And also I am in the right track. So, that would be nice if I don't post my sols because it is complicated after factoring and I am not good at Latex.
   About your sols, I think it is nifty and also gives me a big surprise. It is very helpful through solving any limit. But in order to use it, you must be in the high level of college and  I didn't reach that level. :lol:

I've been stumping for people to think that things like big-O, little-o notation or the binomial series aren't "really high-level stuff" - they're calculus. And calculus understood right is not some purely formal game with "derivative" and "integral" as mystery black-box operations; at its core, it's about comparing the sizes of things.

Let me make one more stab at explaining this problem. I'll start with the same substitution I had earlier, so that we have

$\lim_{y\to0}\frac{\sqrt[4]{16+y}-\sqrt[3]{8+y}}y$

The next thing to do is to notice that the two terms of the numerator are essentially unrelated to each other; we're not going to wrap them up into one nice package. That being the case, let's use a "put and take" to pull them apart. The "put and take" is adding and subtracting 2 because both roots tend to 2.

$=\lim_{y\to0}\frac{\sqrt[4]{16+y}-2+2-\sqrt[3]{8+y}}y$

$=\lim_{y\to0}\frac{\sqrt[4]{16+y}-2}y-\lim_{y\to0}\frac{\sqrt[3] {8+y}-2}y$

Both limits are now the definition of a derivative:

$f'(x)=\lim_{y\to0}\frac{f(x+y)-f(x)}y.$

In the first case, $f(x)=\sqrt[4]{x},$ so $f'(x)=\frac1{4x^{3/4}}$ so that $f'(16)=\frac1{32}.$

In the second case, $f(x)=\sqrt[3]{x},$ so $f'(x)=\frac1{3x^{2/3}}$ so that $f'(8)=\frac1{12}.$

And the answer is $\frac1{32}-\frac1{12}=-\frac5{96}.$

Here is most elementaly My solution, like Colombus egg!:lol:

$\lim_{x\rightarrow 1} \frac{\sqrt[4]{x+15}-2}{x-1}=\lim_{a\rightarrow 2} \frac{a-2}{a^4-16}=\frac{1}{32}.$

$\lim_{x\rightarrow 1} \frac{\sqrt[3]{x+7}-2}{x-1}=\lim_{b\rightarrow 2} \frac{b-2}{b^3-8}=\frac{1}{12}.$

Thus the desired limit is $\frac{1}{32}-\frac{1}{12}=-\frac{5}{96}.$

Japanese high school students are not allowed to use l'Hopital's theorem in case of university entrance examination.

Note that kunny in #18 used essentially the same "put and take" as in my #17. Yes, that's the most elementary solution. Note also that kunny used a substitution (actually two slightly different substitutions on the two separated pieces of a problem.) Always remember that substitutions are extremely powerful, even if sometimes their power is mearly in directing your perceptions. The original variables aren't a straightjacket - always feel free to substitute.

Thanks a lot for the people that wrote the solutions and i find them very useful for myself. My teacher in the University doesn't like when we use some methods that u used and prefers the Dr. Merryfie's method with the Big O's buti dont understand why? and  what is the L' Hopital ?

[quote="rustam"]  what is the L' Hopital ?[/quote]
see  [url=http://mathworld.wolfram.com/LHospitalsRule.html]L'Hospital's Rule[/url]

[quote="Kent Merryfield"]
$\lim_{y\to0}\,\frac{\sqrt[4]{y+16}-\sqrt[3]{y+8}}y= \lim_{y\to0}\frac2y\left[\left(1+\frac y{16}\right)^{\frac14} -\left(1+\frac y8\right)^{\frac13}\right]$
[/quote]

I don't follow this one. How did you arrive at the RHS side?

I really like this method [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=368509#p368509]#17[/url]. :D

[quote="Hazel"]I don't follow this one. How did you arrive at the RHS side? [/quote]
The operative principle behind all practical uses of the binomial series: "factor out the big thing." In the first root, $16$ is much bigger than $y$ and in the second root, $8$ is much bigger than $y$. So we factor $16$ out of the first root and $8$ out of the second root. They both become $2$ outside the root.


geometry
3D geometry
pyramid
perimeter
trigonometry
floor function
trig identities

What is the measure of the smallest angle in a triangle with sides of length 8, 9, and 10?
	a. 18.21°
	b. 40.54°
	c. 49.46°
	d. 58.75°
	e. 71.79°

If [img]http://petersons.ilrn.com/ilrn/books/ptsnSATIIM2Cbank/img803.png[/img] and [img]http://petersons.ilrn.com/ilrn/books/ptsnSATIIM2Cbank/img804.png[/img], then [img]http://petersons.ilrn.com/ilrn/books/ptsnSATIIM2Cbank/img805.png[/img]
	a. w^2x
	b. wx^–2
	c. w^–3x^2
	d. w^3x^6
	e. w^6x^–3

A pyramid has a height of 10 and a square base with a perimeter of 24. If one gram of sand has a volume of 8, then how many grams of sand can this pyramid hold?
	a. 1.5
	b. 3
	c. 8
	d. 12
	e. 15

[quote="kuteypie123"]What is the measure of the smallest angle in a triangle with sides of length 8, 9, and 10?
	a. 18.21°
	b. 40.54°
	c. 49.46°
	d. 58.75°
	e. 71.79°
[/quote]
[hide]The smallest angle in a triangle will be opposite the smallest side (In this case, the angle is opposite the side with length $ 8$). Let $ \theta$ be angle between sides of length $ 9$ and $ 10$.

By using the law of cosines we get,

$ 8^2 \equal{} 9^2 \plus{} 10^2 \minus{} 180\cos\theta$

$ 64 \equal{} 181 \minus{} 180\cos\theta$

$ \minus{} 180\cos\theta \equal{} \minus{} 117$

$ \cos\theta \equal{} 117/180$

$ \theta \equal{} \boxed{49.46^\circ}$c.[/hide]

[hide="#2"]
$ y^{1/3} \equal{} (w^{2/3})^{3/2} \equal{} w$

$ z^{1/3} \equal{} (x^{3/2})^{4/3} \equal{} x^2$

$ (y/z)^{1/3} \equal{} y^{1/3}/z^{1/3} \equal{} w/x^2 \equal{} \boxed{wx^{\minus{}2}}$ b.[/hide]

[hide="#1"]
Maybe not the best way, but...
Heron's formula gives the area of the triangle as $ 34.197$. Use the formula $ \text{Area} \equal{} \frac {1}{2}\cdot a\cdot b\cdot \sin C$ to get the equation $ 34.197 \equal{} \frac {1}{2}\cdot 10\cdot 9\cdot \sin\theta$ (where $ \theta$ is our angle).
Then, use a calculator to find $ \arcsin 0.760$ and you get $ \theta \equal{} 49.46\implies \boxed{\text{C}}$.
[/hide]

[hide="#2"]
$ \left(w^\frac {2}{3}\right)^\frac {3}{2} \equal{} \left(y^\frac {2}{9}\right)^\frac {3}{2}\implies w \equal{} y^\frac {1}{3}$
$ \left(x^\frac {3}{2}\right)^\frac {4}{3} \equal{} \left(z^\frac {1}{4}\right)^\frac {4}{3}\implies x^2 \equal{} z^\frac {1}{3}$
$ \frac {y^\frac {1}{3}}{z^\frac{1}{3}} \equal{} \frac {w}{x^2} \equal{} wx^{ \minus{} 2}\implies\boxed{\text{B}}$
[/hide]

[hide="#3"]
The volume of the pyramid is $ \frac {1}{3}\cdot\left(\frac {24}{4}\right)^2\cdot 10 \equal{} 120$.
And $ \left\lfloor\frac {120}{8}\right\rfloor \equal{} 15\implies\boxed{\text{E}}$
[/hide]

For the first problem, the law of cosines is probably the most efficient way (on a test, where time is important). By using it, you avoid having to find the area of the triangle, which saves a great deal of time. Despite this, using Heron's formula is also important for more difficult problems and the particular method used by grn_turtle is good.


Let $ A,B,C,D$ be four points on a circle (occurring in clockwise order), with $ AB<AD$ and $ BC>CD$. The bisectors of angles $ BAD$ and $ BCD$ meet the circle at $ X$ and $ Y$, respectively. Consider the hexagon formed by these six points on the circle. If four of the six sides of the hexagon have equal length, prove that $ BD$ must be a diameter of the circle.

The conditions $AB < AD$ and $BC > CD$ imply that the points appear in the order $A,B,X,C,D,Y$ along the circle.
Note that $X$ is the midpoint of the arc $BD$ containing $A$ and $Y$ is the midpoint of the arc $BC$ containing $C$ and so $YD=YB>YA,AB$ and $XB=XD>XC,CD$. So if four sides are equal then $YA=AB=XC=CD$. This implies $XD=BY$ and so $\angle XAD=\angle BCY$. But $\angle XAD=\angle BAX$ and $\angle BCY=\angle DCY$ so $2\angle BAD=\angle BAD+\angle BCD=180^{\circ}$ means $BD$ subtends an angle of $90^{\circ}$ so is a diameter.


algebra
polynomial
binomial theorem
algebra unsolved

Let a be an integer not being divisible by 5. Prove that x^5-x+a cannot be written as the product of two polynomials with integer coefficients whose degree is positive !

This problem can be generalize:

For prime p, (a,p) = 1 the polynomial  P(x) = x^p - x - a is inreducible in Zp[x] (=> inreducible in Z[x]).

Hints: Use the following lemma - If f(x) from Zp[x] then f(x+1), ..., f(x+p-1) are either all distinct either all equal (on Zp!) and the fact that P(x+i) = P(x) for all i.

Try to finish it.

Source: Exercises in Algebra /Ed. A.I.Kostrikin, Gordon and Breach Publ., 1996.

Okay .. after quite some time I just came up with this solution:

Lets prove that
f(x) cannot be written as (x-b)g(x), nor (x^2-bx-c)h(x). with b,c integer coeficients.

Lets say that f(x) can be represented like (x-b)g(x).
we have that f(b)=0 which is equivalent to: b^5-b=-a. from Fermat we get that b^5-b==0(mod 5) which gives a==0(mod 5). Contradiction.

Lets say that f(x) can be represented like (x^2-bx-c)h(x).
we divede f(x) to (x^2-bx-c)h(x) and get that:
(b^4+3b^2c+c^2-1)X+(a+2bc^2+cb^3)=0
so b^4+3b^2c+c^2-1=0 and also a+2bc^2+cb^3=0
we get that b^5-5bc^2-b=3a and we have that b^5-5bc^2-b==0(mod 5) so we need 3a==0(mod 5). Contradiction again!!

cheers! hope it's okay! :D

you have not solved the problem lagrangia. a polynomial can still be reductible in Z[X] while not having any divisors of degree less or equal  than 2 in z[X].

for example (x^4+x^3+x^2+x+1)(x^6+x^5+x^4+x^3+x^2+x+1) (or for that matter any product of irreducible polynomials of degree greater than 2).

this solutions' problem follows like this:

reduce this polynomial f to Z/pZ [x]. By the reduction criterion if this polynomial is irreducible then also f is irreducible in Z[X].

It is obvious from Fermat's theorem that f has no roots in Z/pZ[x], so let us consider an extension of this field to another field F (which still has characteristic p) and in which f has a root call it d.

Now it is easy to check that if the char. of a field is a prime p, then by Newton's Binomial theorem (a+b)^p=a^p+b^p because pCk is divisible by p (easy exercise) and therefore it is easy to check that if d is a root then also is d+k where k=0,1,2,...,p-1 and because F is a finite field, by Wedderburn it is commutative (actually it has exactly p^2 elements) so it has at most p roots in F.

But we already found p roots all in F, thus f(x)=(x-d)(x-d-1)...(x-d-p+1). Suppose f(x)=g(x)h(x) in Z/pZ[x] thus g(x)=(x-d-i_1)(x-d-i_2)...(x-d-i_s) where i_1,i_2,....,i_s are amongst 0,1,2,..,k. But g(x) is a polynomial in Z/pZ so by Viette sd+i_1+i_2+...+i_s is in Z/pZ, and as i_j is in Z/pZ for all j=1,2,...,s we have sd \in Z/pZ and as s is in Z/pZ we have that d \in Z/pZ which is absurde.

Thus our reducibility supposition is false, so f is irreducible in Z/pZ[x] and also in Z[x].

Just another atomic bomb on this problem is Dum's criteria.

[i]Edit[/i]
I have just realize that I proved the following result:
If $x^{n}+ux+v$ ($u\neq 0$) has not integer root, then it is irreducible.

Is it true???

Newton diagram of $p(x)=x^{n}+ux+v$ modulo $p$ (where $p\nmid u,v$) consist of two edges. It means that if $p(x)=f(x)g(x)$ then $f(x)=x^{k}+p\cdot a(x)+b$ ($\deg a<k$) and $g(x)=x^{n-k}+p\cdot c(x)+d$ ($\deg c<n-k$). It follows that $k=1$ or $n-k=1$.


Let $ ABCD$ be a quadrilateral with non-parallel opposite sides, and let $ O$ be the point of intersection of its diagonals $ AC$ and $ BD$. Denote by $ \left(XYZ\right)$ the circumcircle of any triangle $ XYZ$. Let $ A^{\prime}$ be the point of intersection of $ \left(OAB\right)$ and $ \left(OCD\right)$ apart from $ O$. Let $ B^{\prime}$ be the point of intersection of $ \left(OBC\right)$ and $ \left(ODA\right)$ apart from $ O$. Let the circles with diameters $ AC$ and $ BD$ intersect at $ C^{\prime}$ and $ D^{\prime}$. Prove that the points $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$, $ D^{\prime}$ are collinear or lie on one circle.

First of all, perform inversion centered at $O$ to get an equivalent problem: in a quadrilateral $ABCD$ diagonals intersect at point $O$, pairs of opposite sides intersect at $A'$ and $B'$. Circles with diameters $AC$ and $BD$ intersect at points $X$ and $Y$. Prove that $X$, $Y$, $A'$, $B'$ are either concyclic or lie on a straight line. However, this is trivial since it is well known that in a complete quadrilateral circles built on diagonals as diameters are coaxial, i.e. circle with diameter $A'B'$ passes through $X$ and $Y$ as desired. By the way, it is easy to see why this fact holds - just prove that all 4 orthocenters of the triangles determined by the 4 sides of a complete quadrilateral have equal powers with respect to these circles.


Here's one cool prb (also known as "the butterfly's problem"):

We have a circle C and a chord. Lety A be the middle of that chord. B and C are pts on this chord s.t. AB=AC. PQ and MN are chords through B and C respectively. PM and NQ intersect the initial chord at R and S respectively. Prove that AR=AS.

http://www.cut-the-knot.org/pythagoras/Butterfly.shtml

Enjoy these ten proofs

Moubinool Omarjee

I have just uploaded a note called [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i] on my website (which is mirrored on [url]http://de.geocities.com/darij_grinberg/[/url]).

  darij


In triangle $ABC$, altitude $AA'$ is drawn, such that $2(BA')=A'C$. Also, cevian $BB'$ is drawn, intersecting $AA'$ at $P$ such that $2(AP)=PA'$. Find the ratio of the area of triangle $BAP$ to the area of quadrilateral $B'PA'C$.

[hide="hint"]

Use the fact that the ratio of the area = ratio of the bases if two triangles have the same height.

[/hide]

[hide="solution"]
Let $[ABP] = z$

$2 [ABP] = [A'BP] = 2z$ because both share the same height ($AB$)
$2 [A'AB] = [A'AC] = 6z$ because both share the same height ($AA'$)

If we connect $B'A'$, notice that $2[B'PA] = [B'PA']$
So, let $[B'PA'] = y$

$[A'B'B] = [A'B'P]+[PA'B] = 2y+2z$

Notice $2 [A'B'B] = [A'B'C] = 4y+4z$ because both share the same height (make $CB$ the base)

$[A'AC] = [A'B'C]+[B'PA]+[B'PA']$

$6z = 4y+4z+y+2y$

So, $y = \frac{2z}{7}$

We want the ratio of $\frac{[CB'A']+[A'B'P]}{[APB]}= \boxed{\frac{40}{7}}$

Someone should check my arithmetic >.>.

EDIT: Silly me, forgot the symbol for area :P
[/hide]


algebra
polynomial
function
binomial coefficients
number theory unsolved
number theory

Is there an infinite set of natural numbers such that all sufficiently large integer can be represented in the same number of ways as the sum of two elements of the set?

I think it's something like that.
Is there a set $ A$ of natural numbers and $ M\in \mathbb{N}$ such that if $ X_n\equal{}\{(a_i,b_i)\in A^2|a_i\plus{}b_i\equal{}n\}$ then $ |X_M|\equal{}|X_n|$ for all $ n\geq M$.

Let $ A(x) \equal{} \sum_{a \in A} x^a$.  The problem statement is equivalent to $ A(x)^2 \equal{} p(x) \plus{} \frac {C}{1 \minus{} x}$ for some polynomial $ p(x)$ with integer coefficients and some positive integer constant $ C$.  But this implies that $ A(x)$ has a square-root singularity at $ x \equal{} 1$, and standard results imply that $ [x^n] A(x) \equal{} \Theta(n^{ \minus{} \frac {3}{2} })$; contradiction.

Could you explain second part of your solution? Thank you.

The standard example of such a singularity is the generating function $ \frac{1}{ \sqrt{1 \minus{} 4x} }$ for the [url=http://mathworld.wolfram.com/CentralBinomialCoefficient.html]central binomial coefficients[/url].  By Stirling's formula this generating function has coefficients asymptotically equal to $ 4^n n^{\minus{} \frac{3}{2} }$, and it turns out that this is typical of square-root singularities in general.  A good reference is [url=http://algo.inria.fr/flajolet/Publications/book.pdf]Flajolet and Sedgewick[/url].


P1: Let \[ a_n\equal{}(1\plus{}(\minus{}1)^n)\frac{n\plus{}1}{n}\plus{}(1\plus{}(\minus{}1)^{n\plus{}1})\frac{3n\plus{}2}{n\plus{}1}\]
Find supA and infA, with \[ A\equal{}\{a_n|n\in\mathbb{N},\ \ n\geq 1\}\]

P2: Let set \[ A\equal{}\{\frac{3n\minus{}5}{2n\plus{}10} | n\in\mathbb{N} \}\]
Find supA and infA

P1. $ a_{2n} \equal{} 2\frac {2n\plus{}1} {2n} \equal{} 2 \plus{} \frac {1} {n}$; $ a_{2n\minus{}1} \equal{} 2\frac {6n\minus{}1} {2n} \equal{} 6 \minus{} \frac {1} {n}$. The first sub-sequence is decreasing from $ 3$ to the limit $ 2$, while the second is increasing from $ 5$ to the limit $ 6$. Therefore $ \sup A \equal{} 6$ and $ \inf A \equal{} 2$.

P2. $ \frac {3n\minus{}5} {2n \plus{} 10} \equal{} \frac {3} {2} \minus{} \frac {10} {n \plus{} 5}$ is a sequence increasing from $ \minus{}\frac {1} {2}$ (for $ n\equal{}0$) to the limit $ \frac {3} {2}$.


Solve for $ n$: $ 5^{2n \plus{} 1} \equal{} \frac{1}{25}$. Express your answer as a common fraction.

$ \frac{1}{25}\equal{}5^{\minus{}2}$, so $ 5^{2n\plus{}1}\equal{}5^{\minus{}2} \implies 2n\plus{}1\equal{}\minus{}2 \implies n\equal{}\boxed{\frac{\minus{}3}{2}}$.


Suppose that $ a; b; c$ are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$ prove that
$ \frac {a^2 \plus{} b^2c}{b \plus{} c} \plus{} \frac {b^2 \plus{} c^2a}{c \plus{} a} \plus{} \frac {c^2 \plus{} a^2b}{a \plus{} b}\ge \frac {2}{3}$

Νομίζω ότι έχω μία λύση αλλά είναι αρκετά άσχημη(expanding),γι'αυτό περιμένω τις δικές σας :)

Edited now

Το κατω φραγμα ειναι $ \frac {2}{3}$ και οχι $ \frac {3}{2}$
Δεν ειναι και πολυ δυσκολη αλλα μου πηρε αρκετη ωρα διοτι δεν ειμαι καθολου γρηγορος με τις πολλες πραξεις.

Η Αποδειξη μου:

Η Παρασταση Ισουτε με:

$ \sum{\left(\frac{a^2 + c^3}{b + c} + 3bc\right)} - 1 = \sum{\left({\frac{a^2}{6(b + c)} + \frac {5}{8}(a(b + c)) + \frac {5a^2}{6(b + c)} + \frac {5}{36} + \frac {c^3}{b + c} + \frac {7}{4}bc - \frac {5}{36}\right)} - 1 \geq \frac{a + b + c}{12} + \sum{\left(3\sqrt [3]{(\frac {5}{12})^3a^3}\right)} + \sum{\left(\frac {c^3}{b + c} + \frac {7}{4}bc}\right)} - \frac{5}{12} - 1 = \frac{1}{12} + \frac{5}{4}(a + b + c) + \sum{\left(\frac {c^3}{b + c} + \frac {7}{4}bc \right)} - \frac {17}{12} = \sum{\left(\frac {c^3}{b + c} + \frac {7}{4}bc \right)} + \frac{2}{3} - \frac {3}{4}$

Αρκει να δειχθει λοιπον οτι:

$ \sum{\left(\frac{c^3}{b + c} + \frac {7}{4}bc \right)} \geq \frac {3}{4}$

Πραγματι απο AM-GM:

$ \sum{\left(\frac {4c^3}{b + c} + c(b + c) \right)} \geq 4\sum{c^2} = 4 - 8(ab + bc + ca) \Leftrightarrow \sum{\left(\frac {4c^3}{b + c} + 9ab + a^2 \right)} = \sum{\left(\frac {4c^3}{b + c} + 7ab \right)} + (a + b + c)^2 = \sum{\left(\frac {4c^3}{b + c} + 7ab \right)} + 1 \geq 4 \Rightarrow \sum{\left(\frac {c^3}{b + c} + \frac {7}{4}bc \right)} \geq \frac {3}{4}$


[quote="xzlbq"]$ x,y,z > 0$,prove that:
$ x \plus{} y \plus{} z\geq \frac {4xyz}{(z \plus{} x)(x \plus{} y)} \plus{} \frac {(z \plus{} x)^2 \plus{} (x \plus{} y)^2}{2x \plus{} y \plus{} z}.$[/quote]

The inequality is equivalent to

$ (x^2\plus{}2y^2\minus{}xy)z^2\plus{}(x^3\minus{}xy^2\minus{}4x^2y)z\plus{}yx^3\plus{}y^2x^2\geq 0$

Note

$ x^2\plus{}2y^2\minus{}xy>0$

$ yx^3\plus{}y^2x^2>0$

if  $ x^3\minus{}xy^2\minus{}4x^2y\ge 0$

then the inequality holds.

if  $ x^3\minus{}xy^2\minus{}4x^2y< 0$

$ \Delta\equal{}\minus{}x^2(7y^2\plus{}10xy\minus{}x^2)(\minus{}x\plus{}y)^2$

we only prove while $ x^3\minus{}xy^2\minus{}4x^2y< 0$(we have $ 0 <\frac{x}{y}\le \sqrt{5}\plus{}2$)

$ 7y^2\plus{}10xy\minus{}x^2\geq 0$

easy prove it.

 :lol:

[quote="xzlbq"][quote="can_hang2007"][quote="xzlbq"]$ x,y,z > 0$,prove that:
$ x \plus{} y \plus{} z\geq \frac {4xyz}{(z \plus{} x)(x \plus{} y)} \plus{} \frac {(z \plus{} x)^2 \plus{} (x \plus{} y)^2}{2x \plus{} y \plus{} z}.$[/quote]
See here: http://can-hang2007.blogspot.com/2009/12/inequality-11-b-q-liu.html[/quote]

I have not see,post to me:
wshr987@163.com.
thank you.
BQ[/quote]
You can download my file [url=http://cid-b2deb869e6bb1093.skydrive.live.com/self.aspx/Inequality%20Problems/Inequality%2011.rar]here.[/url]

[quote="xzlbq"][quote="can_hang2007"][quote="xzlbq"]$ x,y,z > 0$,prove that:
$ x \plus{} y \plus{} z\geq \frac {4xyz}{(z \plus{} x)(x \plus{} y)} \plus{} \frac {(z \plus{} x)^2 \plus{} (x \plus{} y)^2}{2x \plus{} y \plus{} z}.$[/quote]
See here: http://can-hang2007.blogspot.com/2009/12/inequality-11-b-q-liu.html[/quote]

I have not see,post to me:
wshr987@163.com.
thank you.
BQ[/quote]

see here


Let $ D$ and $ C$ be arbitrary points on a circle,which touches the segment $ AB$ in its midpoint.$ AD$ and $ BC$ intersect
with circle at points $ X$ and $ Y$,respectively,and $ CX$ and $ DY$ intersect with $ AB$ at points $ M$ and $ N$,respectively.
Prove that $ AM\equal{}BN$.

Let $ T$ be tangency point. Consider a degenerated chord  $ TT$, and $ T$ is also its midpoint. Suppose that $ C,D,X,Y$ are arbitrary points on  the circle. Than by applying Butterfly theorem we obtain that $ T$ is midpoint of $ MN$. Because $ AT \equal{} TB$ we obtain that $ AM \equal{} BN$, q.e.d.

I don't understand your solution Ahiles please do you have another solution.More simple or could you explain to me.


geometry
geometric transformation
rotation

Alright, I'm still confused on this...dispite having looked at it a year ago. Can anyone explain Polya's couting theory to me?
'n here's a problem:
How many ways are there to seat 2 mathematicians, 2 chemists, and two biologists at a table with 6 seats such that two people of the same occupation are not distinct and rotation doesn't matter?

I don't know about Pólya's counting theory, but if the table is circular,[hide="Solution"]we have $\displaystyle \frac{\left(6-1\right)!}{2!2!2!}=15$[/hide]

Here's an interesting paper about Polya's theory : http://www.geometer.org/mathcircles/polya.pdf :)

Yeah, that's where I'm reading about his theory. The problem is the notation and all that is confusing the heck out of me.


HE MOVED TO MICHIGAN!!!!!!!
lucky michigan.... they'll win arml 4 sur  :(

IM ALEX SONG'S FRIEND'S FRIEND!!!!!
beat THAT!!!! :P

He's my friend :) . (Well kinda...he's in my math class)

of the two posts he ever made on aops, one of them was in reply to my question about vectors.
Beat that!

I met a person who found him extremely annoying.

BEAT THAT!

Oh yea?

I got beasted by him in mathcounts practice
beat that

I heard of him a few times on some math forum.
Beat that.

[quote="athunder"]Yeah he is in Michigan now. His username is Alex Song. I saw him yesterday, he is cool.[/quote]

Tell him i'm Edward's friend (he'll know what that means). :lol: Thx SAY ITS FROM TIM

I threw a paper airplane towards him at ARML, and when it had landed in the vicinity of the Ontario team, they cheered.

Beat that.

I could say something inappropriate.

Beat that?

:censored: maybe?

well, I DONT KNOW HIM!! OWN!


algebra
polynomial
ratio
trigonometry
limit
geometric sequence
calculus

hi, im need help finding whether these series are absolutly convergent, coverges but not absolutley convergen, or if its divergent.

sigma n=1 to infinity  1.) (sin(6n))/(n^2)
                                  2.) ((-1)^n)/(5n+6)
                                  3.) ((n+1)((10^2)-1)^n)/(10^(2n))
                                  4.) ((-1)^n)*((sqrt(n))/(n+8))
                                  5.) ((-6)^n)/(n^2)

any help would be great. thanks.

Settle absolute convergence first. Take absolute values (ignore $\pm$ signs), and "de-clutter." The size is what matters, which is to say that all the tests that matter are comparison tests. Sharpen your eyes to look only at the most important size feature as $n$ tends to $\infty.$ (You can usually rigorously justify this with the limit comparison, if necessary. Don't worry too much about that.)

One issue that #1 brings up - how big should we say $\sin(6n)$ (or anything else like it) is as $n\to\infty$? I'd prefer to say that it's size is "one, only wigglier."

So what does the "decluttered" view of the absolute values see in each of these cases?

#1: The terms are comparable to $\frac1{n^2}.$

#2: The terms are comparable to (a constant times) $\frac1n.$

#3: The terms look like (mere polynomial)$\cdot \left(\frac{99}{100}\right)^n.$ The geometric sequence with ratio less than $1$ is vastly more important than the polynomial factor.

#4: The terms are comparable to $\frac1{\sqrt{n}}.$

#5: The terms look like $\frac{6^2}{n^2},$ which goes to $\infty$ since geometric sequences grow faster than polynomial sequences.

So, right away from that: #1 and #3 converge absolutely. #2 and #4 do not converge absolutely, although the terms to at least tend to zero. #5 clearly diverges because the terms do not tend to zero.

If you happen to like using the ratio test or the root test, go ahead and apply them to #3 and #5, which include geometric sequences. These tests are not magic; they are merely comparisons - comparisons to geometric series. #1, 2, and 4 do not usefully compare to anything geometric; if you attempt to use the ratio test or root test on them, you are merely wasting your time.

#2 and #4 are worth revisiting with the $\pm$ signs included. In each case, we ask whether the criteria of the alternation series test are met. Since the answer is yes, we conclude that both converge. Since we had previously ruled out absolute convergence for either, we conclude that they converge conditionally.

1) as you know $|\sin 6n| \leq 1|$ then you can use basic comparasion test $\left| \frac{\sin 6n}{n^2}\right| \leq \left|\frac{1}{n^2}\right|$. it is well-known that $\frac{1}{n^2}$ is converges.
2) $\left| \frac{(-1)^n}{5n+6} \right| = \frac{1}{5n+6}$, the series  $\sum \frac{1}{5n+6}$ is diverges, but $\sum \frac{(-1)^n}{5n+6}$ converges, so it is conditionally convergen.
3) by ratio test $\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right | = \lim_{n \rightarrow \infty} \frac{99}{100} < 1$, converges absolutely.
4)  $\left | (-1)^n \frac{\sqrt{n}}{n+8} \right |= \frac{\sqrt{n}}{n+8}$, you can check the convergen of $\frac{\sqrt{n}}{n+8}$ .
5) use ratio test.


superior algebra
superior algebra unsolved

Can anyone help me to solve this simultaneous equation.

(1) : a = -2(q 2  - 2pr + 2) +(p 2 -2q) 2
(2) : b = (q 2  - 2pr + 2) 2 + 2 - 2(p 2 -2q)(r 2 -2q)
(3) : c = -2(q 2  - 2pr + 2) + (r 2 -2q) 2

where p, q, r be variable and a, b, c = constant

I simplified this problem to

(4) : a = -2x + y 2
(5) : b = x 2 + 2 - 2yz
(6) : c = -2x + z 2

where : x = q 2  - 2pr + 2, y = p 2 -2q , z = r 2 -2q

I want to get parametrice solution
Ex.
(?)
p 4 = a + b + 4t + 8 (I guess)
q4 = b + c + 4t + 8 (I guess)
r4 = c + a + 4t + 8 (I guess)
where t must come from auxiliary equation
ex. t 4 - (a + b +c)t - 6abc = 0

from (1),(2),(3) we will get

(7)
a + b + c + 2
= p 4 + q4 + r4 - 2p 2 r 2 - 4prq 2
= (q 2 -2pr) 2 + (p 2 - r 2 ) 2

if anyone can solve this problem. one of value t from auxiliary equation must sastisfy with equation (?)
one of solution of eq (7) is (a, b, c, p, q, r) = (4,6,4,4,6,4)

however anyone that try to solve this problem. Please feel free to use your intuition. Thanks for everyone that try to solve this problem.

Note. I " m not good at in english. Please don't care much about my gramma.

At last I solved this problem completely. So I don't need help any more.


[quote="smdsmd"]Can you choose to get questions from a particular subject? Like counting, or Venn diagrams? If not, you should add that. Then you could work on getting a particular badge.
Also--Do you get karma from rating posts?[/quote]
Future versions of Alcumus will likely add the ability to choose from a certain topic.

There is no karma for rating posts.

Additionally, as is the case with most situations like this, if you are rating stuff simply to gain karma, there is absolutely no point, so rate only if you feel your rating would be contributive (if that's a word).


find all soluation  from set ={1,2,3,4,..............}

satesfay this equathon

 $x{}^{y}{}^{2}=y^{x}$

If $p|y$ then $p|x$. It give $x=c^{n},y=c^{m}$. Therefore $m|n$ and $n=mc^{k}$. It give $k=m(c^{k}-2)$.


algebra
polynomial
Rational Root Theorem

Show that polynom $f=X^3+pX^2+pX+p$ (p is a prime number) can't be writen as a two integer polynoms product  :? (Eisenstein prime :? what's that?!  :blush: )

Any polynomial with integer coefficients that factors over the rationals factors over the integers.  Eisenstein's criterion says that if you have a polynomial, all of whose coefficients otehr than that of highest degree are divisible by a prime p and whose constant coefficient is not divisible by p^2, it does not factor over the integers.  In your case, for a cubic, we could also easily use the rational root theorem.  Eisenstein can be proven by supposing that such a polynomial has a factorization over the integers and deriving a contradiction.

[hide]If it can be written as a product of at least two polynomials, one must be linear; thus the polynomial has one integer root.  By the Rational Root theorem, the only possible rational roots are +/- p, +/- 1, none of which are roots.[/hide]


$ x$, $ y$ and $ z$ are positive real numbers such that $ xyz\equal{}1$
prove that
$ \frac{x\minus{}1}{y\plus{}1}\plus{}\frac{y\minus{}1}{z\plus{}1}\plus{}\frac{z\minus{}1}{x\plus{}1} \geq 0$

[quote="greatestmaths"]$ x$, $ y$ and $ z$ are positive real numbers such that $ xyz \equal{} 1$
prove that
$ \frac {x \minus{} 1}{y \plus{} 1} \plus{} \frac {y \minus{} 1}{z \plus{} 1} \plus{} \frac {z \minus{} 1}{x \plus{} 1} \geq 0$[/quote]
$ \sum_{cyc}\frac {x \minus{} 1}{y \plus{} 1}\geq0\Leftrightarrow\sum_{cyc}(x^2z \minus{} z \plus{} x^2 \minus{} 1)\geq0,$ which true because
 $ \sum_{cyc}(x^2z \minus{} 1) \equal{} \sum_{cyc}(x^2z \minus{} xyz)\geq0$ is true by AM-GM and
$ \sum_{cyc}(x^2 \minus{} z) \equal{} \sum_{cyc}(x^2 \minus{} x) \equal{} \sum_{cyc}(x^2 \minus{} x^{\frac {4}{3}}y^{\frac {1}{3}}z^{\frac {1}{3}})\geq0,$ which true by Muirhead, for example.

The following inequality is true too:
$ x$, $ y$ and $ z$ are positive real numbers such that $ xyz \equal{} 1.$
Prove that
\[ \frac {x \minus{} 1}{2y \plus{} 1} \plus{} \frac {y \minus{} 1}{2z \plus{} 1} \plus{} \frac {z \minus{} 1}{2x \plus{} 1} \geq 0\]
 :wink:


TAN (60) TAN (40) TAN(20) TAN(80)=3
 PROVE IT IN A SHORT WAY

[quote="ayush_2008"]good shot [b]sir[/b][/quote] :rotfl:  :rotfl:  :P  :rotfl:


Source: ARML 1991
T10. Compute the smallest positive integer that can not be the difference between a square and a prime, if the square is greater than the prime.

[hide]

$\fbox{16}$

we have

$a^{2}-(a^{2}-16) = a^{2}-(a-4)(a+4) = 16$

we see that $(a-4)(a+4)$ will never be prime because when $a=5$, it is still equal to $9$ which is a perfect square

now it suffices to show that all the values up to $16$ can be obtained, and the hardest ones were $81-71 = 10$ and $196-181 = 15$, otherwise the rest are easy

[/hide]

[quote="srulikbd"]can't we use mods?[/quote]

How exactly would you plan on using mods?  The prime numbers are not very nicely behaved in modular arithmetic.


USAMTS
summer program
Mathcamp
email
ARML
geometry
inequalities

What prizes are we going to get for the USAMTS? I think along with other prizes, everybody should get a free AOPS T-shirt for participating in the USAMTS, or at lease for winning a medal  :D .

hey I already have AoPS t-shirt already.

Those who do well with be able to choose from a list of options for prizes.  We will also be putting together a USAMTS T-Shirt for winners, and Mathcamp is giving 200 dollar scholarships to the top 30 women and top 30 men in the USAMTS (provided they earn acceptance to Mathcamp - winning a scholarship is no guarantee of being admitted).

Do you mean winners as in people who win gold medals or just anybody who wins at lease a honorable mention?

And we get the a medal itself, of course?

[quote="DPatrick"]There are no medals.

Full prize information will be announced in the coming weeks.  Please be patient![/quote]
That WOULD be pretty cool if we actually got medals for achieving gold, silver, or bronze winner status. :)

umm, how do we find out the prize we won, is it just besed on the score and the cutoffs that are on the website?

An email was sent out this morning regarding the prizes:

[quote]This is a reminder to visit the USAMTS website (http://www.usamts.org) to select your prizes.  The prizes will be allocated on a first come, first serve basis.  The dates on which winners will be able to select their prizes are listed below (you can select your prize on or after the dates listed):

Gold: April 19
Silver: April 23
Bronze: April 26
Honorable Mention: April 29

If you do not select a prize by May 5, we will select one randomly for you.[/quote]

Wow. Cool books to choose from! I'm looking forward to some interesting summer reading :lol:

The prizes are a list of books to choose from.  Also note that the top 30 females and top 30 males will receive 200 dollar scholarships to Mathcamp if they earn admission to Mathcamp.

[quote="Geehoon"]and a t-shirt.[/quote]

 :blush: I forgot about that.

[quote="Geehoon"][hide="problem of the week"]
it is multiple of 3 and 4.
so 2 times 2 times 3 times a big number.
and I know that the big number is not a prime; why four integers?
[/hide][/quote]
[hide="problem of the week"]It is only a multiple of 2 and 3 . . . and two others -- but it can't be a multiple of four, because the last two digits are "22" => not divisible by 4.[/hide]

[quote="anhnguyen"]What should we expect the prizes to arrive?[/quote]
I think it says that they will arrive "sometime in mid-June" near the top of the "My USAMTS" page.

[hide="problem of the week"]
wow, for some reason, i thought it was 11...1122...2233...3344...44.
2*3*10000...0001*the rest, there will be 18 zero's.
[/hide]

[quote="rrusczyk"]The prizes are a list of books to choose from.[/quote]
For those who didn't win (and for those who didn't participate), could you please be a little more explicit about the actual prizes?

I understand the winners had a menu of choices, but what was on the menu when the restaurant opened?  Did the Gold medal winners get to choose more than one item?  Etc.  TY

Aside from the Mathcamp scholarships, which went to the first 30 boys and first 30 girls, and the T-Shirts, which everyone from HM and above will receive, and certificates, which all HM and above will receive, the students had a list of books and one software package (Geometer's Sketchpad) to choose from.  Gold and Silver were able to choose 2 prizes, and Bronze + HM choose 1.  Prizes were up for Gold winners first, then Silver, and so on.  Therefore, the Gold medal winners have the best shot at the scarce prizes (we only have a limited supply of each).

The books generally are problem books and general interest books.  The books include ARML books, The Code Book by Simon Singh, Music of the Primes, 360 Problems for Mathematical Contests, Challenging Problems in Geometry, Old and New Inequalities, C2K2: Century 2 of Komal, and An Introduction to Diophantine Equations.

Suggestions for book prizes for next year are most welcome.

Just wondering, how much were the Mathcamp scholarships?

i heard it was something like 500 dolllars.

EDIT: paladins got the right amount... :D


Each interior angle of a regular polygon measures $ 140^\circ$. How many sides does the polygon have?

The exterior angle is $ 180^\circ\minus{}140^\circ\equal{}40^\circ$.

Since for a regular $ n$-sided polygon, the exterior angle is $ \frac{360^\circ}n$, we get

$ \frac{360^\circ}n\equal{}40^\circ\implies n\equal{}\boxed9$.


Espero que suban aquí sus soluciones.. sobre todo de los problemas 3 y 4 de cada nivel, pronto subiré las soluciones oficiales... :D

$Tipe$

Ok ..

  muy bien Tipe
...
siempre es bueno hacer referencia a las soluciones oficiales
pero siempre estara abierta la posibilidad de enviar otras ...
haber si se animan y envian ideas  para estos problemas ..

Carlos Bravo :)
Lima - PERU

buno tare editamndo un pocedimiento q segui...ahi  estare poniendolo cualquier momento

jeje por fin encontre peruanos, todo este tiempo e estado en Olimpiad section pero aora revisare aqui  :D

PD. Tipe, ablamos casi todos los dias pq no me dijiste que estabas aqui plop

Bueno hucht

Bienvenido entonces a la Comunidad ....

espero que tambien de tu parte ... recibamos aportes importantes para que esta comunidad cresca....

Carlos Bravo ;)
Peru - Lima


hey,

should I take both the IC and IIC, or would that be a waste of time? Will colleges assume your math is fine as long as you only take IIC and get a good score (750+)?

Also, which prep books are the best for the following SAT II's?

Physics
Chem
Bio
US History

Thanks

Our plan here is to take only the SAT II Math Level 2 (which is the new name for the IIC). There appears to be little benefit to also taking the Level 1 (old IC) for someone who is able to take the Level 2.

taking both math 1 and math 2 would be a waste of time AND money.
I can't tell you but for what I used for Physics: Barron's and PR. Barron's was a lttle more difficult..and their practice tests are not like the real test, but PR seemed quite realistic to me. also heard about Sparknotes being good, but I haven't tried it :)
good luck ;)


$ x^3\minus{}px^2\plus{}qx\minus{}r\equal{}0$

If $ r\equal{}\frac{p^3}{r^3}$ show that $ \frac{p}{r}$ is a root.

:)

Your problem puts no condition on $ q$.  Is it written correctly?


Determine the smallest and the greatest value of the expression

$ T\equal{}(xy\minus{}x\plus{}1)^2\plus{}(yz\minus{}y\plus{}1)^2\plus{}(zx\minus{}z\plus{}1)^2  \\  (x, \ y, \ z \in [0;1])$

[quote="mcdat"]Determine the smallest and the greatest value of the expression

$ T \equal{} (xy \minus{} x \plus{} 1)^2 \plus{} (yz \minus{} y \plus{} 1)^2 \plus{} (zx \minus{} z \plus{} 1)^2 \\
(x, \ y, \ z \in [0;1])$[/quote]

All surrendered  to it  :?:  :?:

I think you shouldn't expect that any problem will be solved just  in a few hours.  :)

[quote="mcdat"]Determine the smallest and the greatest value of the expression

$ T \equal{} (xy \minus{} x \plus{} 1)^2 \plus{} (yz \minus{} y \plus{} 1)^2 \plus{} (zx \minus{} z \plus{} 1)^2 \\
(x, \ y, \ z \in [0;1])$[/quote]

Observe that $ xy\minus{}x\plus{}1\equal{}1\minus{}(x)(1\minus{}y)$ since $ x,1\minus{}y\in[0,1]$. It is easy to see that $ 1\ge 1\minus{}(x)(1\minus{}y)\ge 0$ so $ T\le 3$. An example of equality is when $ x\equal{}y\equal{}z\equal{}1$.

I do not know about the lower bound but it occurs for $ (x,y,z)\equal{}(0.5,0,1)$

[quote="Altheman"][quote="mcdat"]Determine the smallest and the greatest value of the expression

$ T \equal{} (xy \minus{} x \plus{} 1)^2 \plus{} (yz \minus{} y \plus{} 1)^2 \plus{} (zx \minus{} z \plus{} 1)^2 \\
(x, \ y, \ z \in [0;1])$[/quote]

Observe that $ xy \minus{} x \plus{} 1 \equal{} 1 \minus{} (x)(1 \minus{} y)$ since $ x,1 \minus{} y\in[0,1]$. It is easy to see that $ 1\ge 1 \minus{} (x)(1 \minus{} y)\ge 0$ so $ T\le 3$. An example of equality is when $ x \equal{} y \equal{} z \equal{} 1$.

I do not know about the lower bound but it occurs for $ (x,y,z) \equal{} (0.5,0,1)$[/quote]

Thankyou . I think so but that is very difficult  :(  :(

[quote="mcdat"][quote="kunny"]$ T \equal{} (1 \minus{} x)^2(1 \minus{} y)^2 \plus{} (1 \minus{} y)^2(1 \minus{} z)^2 \plus{} (1 \minus{} z)^2(1 \minus{} x)^2$
$ \plus{} 2\{y(1 \minus{} x)(1 \minus{} y) \plus{} z(1 \minus{} y)(1 \minus{} z) \plus{} x(1 \minus{} z)(1 \minus{} x)\} \plus{} x^2 \plus{} y^2 \plus{} z^2.$[/quote]

Hiz. I can't know how to do . Can you guide again ?

Thank  :)[/quote]
$ T \equal{} (1 \minus{} x)^2(1 \minus{} y)^2 \plus{} (1 \minus{} y)^2(1 \minus{} z)^2 \plus{} (1 \minus{} z)^2(1 \minus{} x)^2$
$ \plus{} 2\{y(1 \minus{} x)(1 \minus{} y) \plus{} z(1 \minus{} y)(1 \minus{} z) \plus{} x(1 \minus{} z)(1 \minus{} x)\} \plus{} x^2 \plus{} y^2 \plus{} z^2\equal{}\left( \sum x(1\minus{}y)(1\minus{}z) \right)^2\ge 0.$ So $ T_{min}\equal{}0$ .

I think this is what you mean , am I right, kunny-san? :blush:

[quote="Potla"][quote="mcdat"][quote="kunny"]$ T \equal{} (1 \minus{} x)^2(1 \minus{} y)^2 \plus{} (1 \minus{} y)^2(1 \minus{} z)^2 \plus{} (1 \minus{} z)^2(1 \minus{} x)^2$
$ \plus{} 2\{y(1 \minus{} x)(1 \minus{} y) \plus{} z(1 \minus{} y)(1 \minus{} z) \plus{} x(1 \minus{} z)(1 \minus{} x)\} \plus{} x^2 \plus{} y^2 \plus{} z^2.$[/quote]

Hiz. I can't know how to do . Can you guide again ?

Thank  :)[/quote]
$ T \equal{} (1 \minus{} x)^2(1 \minus{} y)^2 \plus{} (1 \minus{} y)^2(1 \minus{} z)^2 \plus{} (1 \minus{} z)^2(1 \minus{} x)^2$
$ \plus{} 2\{y(1 \minus{} x)(1 \minus{} y) \plus{} z(1 \minus{} y)(1 \minus{} z) \plus{} x(1 \minus{} z)(1 \minus{} x)\} \plus{} x^2 \plus{} y^2 \plus{} z^2 \equal{} \left( \sum x(1 \minus{} y)(1 \minus{} z) \right)^2\ge 0.$ So $ T_{min} \equal{} 0$ .

I think this is what you mean , am I right, kunny-san? :blush:[/quote]
How did you obtain that from what kunny wrote? If $ x\equal{}y\equal{}z\equal{}1$, then $ T\equal{}3$ and not $ T\equal{}0$..

[quote="Potla"][quote="mcdat"][quote="kunny"]$ T \equal{} (1 \minus{} x)^2(1 \minus{} y)^2 \plus{} (1 \minus{} y)^2(1 \minus{} z)^2 \plus{} (1 \minus{} z)^2(1 \minus{} x)^2$
$ \plus{} 2\{y(1 \minus{} x)(1 \minus{} y) \plus{} z(1 \minus{} y)(1 \minus{} z) \plus{} x(1 \minus{} z)(1 \minus{} x)\} \plus{} x^2 \plus{} y^2 \plus{} z^2.$[/quote]

Hiz. I can't know how to do . Can you guide again ?

Thank  :)[/quote]
$ T \equal{} (1 \minus{} x)^2(1 \minus{} y)^2 \plus{} (1 \minus{} y)^2(1 \minus{} z)^2 \plus{} (1 \minus{} z)^2(1 \minus{} x)^2$
$ \plus{} 2\{y(1 \minus{} x)(1 \minus{} y) \plus{} z(1 \minus{} y)(1 \minus{} z) \plus{} x(1 \minus{} z)(1 \minus{} x)\} \plus{} x^2 \plus{} y^2 \plus{} z^2 \equal{} \left( \sum x(1 \minus{} y)(1 \minus{} z) \right)^2\ge 0.$ So $ T_{min} \equal{} 0$ .

I think this is what you mean , am I right, kunny-san? :blush:[/quote]

Oh no , $ T > 0 \ \forall \ x, \ y, \ z \in [0,1]$   :!:  :!:


http://www.qoos.com/bbs/attachments/forumid_69/pbypUqZX_RMCPcoD0yk17.jpg

Thank you..

I've seen it used to show that a sequence majorizes an other (see http://planetmath.org/encyclopedia/Majorization.html)


It seems pretty difficult...

Bashing it out, we see it's equivalent to:

$(ab+bc+ca)(a+b+c)+6 \geq 5(a+b+c)$

Now, we make the subsitution $x^3=a$, etc ($xyz=1$), so this becomes

$(x^3y^3+y^3z^3+z^3x^3)(x^3+y^3+z^3)+6x^3y^3z^3 \geq 5x^2y^2z^2(x^3+y^3+z^3)$

$ \iff \sum_{\text{sym}} 2x^6y^3+3x^3y^3z^3-5x^5y^2z^2 \geq 0.$

However, this is quite hard to prove (i.e. standard inequalities like Schur, Muirhead, etc don't work).  I tried Schur on the numbers $x^2y, y^2z, z^2x$, etc and that got a part of the inequalitity--so this seems to be much stronger.  Anyone have some ideas?

Well, rest assured that the problem is true. There is this incredibly bad and objectionable "solution":

[hide]WLOG $x \le y \le z$. Let $x = a$, $y = a+b$, $z = a + b + c$ with $a > 0;\,b,c\ge 0$. Then compute that

$\sum_{sym}2x^6y^3 + 3x^3y^3z^3 - 5x^5y^2z^2 = $

$6 a^7 b^2 + 40 a^6 b^3 + 124 a^5 b^4 + 216 a^4 b^5 + 218 a^3 b^6 + 124 a^2 b^7 + 36 a b^8 + 4 b^9 + 6 a^7
    b c + 60 a^6 b^2 c + 248 a^5 b^3 c + 540 a^4 b^4 c + 654a^3 b^5 c + 434 a^2 b^6 c + 144 a b^7 c + 18 b^8 c + 6 a^7 c^2 + 24 a^6 b c^2 + 150 a^5 b^2 c^2 + 496 a^4 b^3 c^2 + 794a^3b^4 c^2 + 646 a^2 b^5 c^2 + 252 a b^6 c^2 + 36 b^7 c^2 + 2 a^6 c^3 + 26 a^5 b c^3 + 204 a^4 b^2 c^3 + 498 a^3 b^3 c^3 + 530 a^2 b^4 c^3 + 252 a
      b^5 c^3 + 42 b^6 c^3 + 10 a^5 c^4 + 60 a^4 b c^4 + 180 a^3 b^2 c^4 + 250 a^2 b^3 c^4 + 150a b^4c^4 + 30 b^5 c^4 + 14 a^4 c^5 + 40 a^3 b c^5 + 62 a^2 b^2 c^5 + 48 a b^3 c^5 + 12b^4 c^5 + 4 a^3 c^6 + 6 a^2 b c^6 + 6 a b^2 c^6 + 2 b^3 c^6$

(No animals were harmed in the calculation!)

Since all terms are non-negative, the expression is at least 0. Moreover, no term consists solely of a power of $a$, thus, it is zero iff $b = c = 0$.[/hide]

I don't yet have progress on a real solution, although perhaps checking weird squares might help, since for example you can get

$(x^2+y^2+z^2-xy-yz-zx)^2 = \sum_{sym} \frac{1}{2}x^4 - 2 x^3 y + \frac{3}{2}x^2y^2 \ge 0.$

Haha eww...

Anyways, here's another random inequality that might help? (from Topics in Inequalities by [url=http://ultrametric.googlepages.com/tin2006new.pdf]Hojoo Lee[/url]):

$\sum_{\text{sym}}5x^{6}+30x^{4}y^{2}+4x^{3}y^{2}z+x^{2}y^{2}z^{2}-16x^{5}y-8x^{4}yz-16x^{3}y^{3}$

$= \frac{1}{6}\sum_{\text{cyc}}(y-z)^{4}(x^{2}+15y^{2}+15z^{2}+8xy+4yz+8zx) \geq 0$

WLOG assume $a \ge b \ge c$. Let $f(a,b,c)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{6}{a+b+c}$.
\[f(a,b,c) \ge f(a,\sqrt{bc},\sqrt{bc}) \Leftrightarrow\]
\[\frac{(\sqrt{b}-\sqrt{c})^2}{bc(a+b+c)(a+2\sqrt{bc})}((a+b+c)(a+2\sqrt{bc})-6bc) \ge 0 \Leftrightarrow\]
\[(a+b+c)(a+2\sqrt{bc}) \ge 6bc.\]
As $a \ge \frac{b+c}{2} \ge \sqrt{bc}$, so $(a+b+c)(a+2\sqrt{bc}) \ge 9bc \ge 6bc$, hence the above inequality is true.
\[f(\frac{1}{x^2},x,x) \ge 5 \Leftrightarrow\]
\[(x-1)^2(2x^4+4x^3-4x^2-x+2) \ge 0.\]
As $2x^4+4x^3-4x^2-x+2 >0$ if $x>0$, so the above inequality is true. Therefore \[f(a,b,c) \ge f(a,\sqrt{bc},\sqrt{bc}) = f(\frac{1}{bc},\sqrt{bc},\sqrt{bc}) \ge 5.\]

[quote="xtar"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {6}{a \plus{} b \plus{} c}\ge5$.[/quote]

Another solution:

From Vasc's inequality in http://www.mathlinks.ro/Forum/viewtopic.php?t=23651 post #13, we have

$ \left(a \plus{} b\right)\left(b \plus{} c\right)\left(c \plus{} a\right) \plus{} 7\geq 5\left(a \plus{} b \plus{} c\right)$.

But

$ \left(a \plus{} b\right)\left(b \plus{} c\right)\left(c \plus{} a\right) \equal{} a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b \plus{} 2abc$
$ \equal{} \left(a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b \plus{} 3abc\right) \minus{} abc$
$ \equal{} \left(a \plus{} b \plus{} c\right)\left(bc \plus{} ca \plus{} ab\right) \minus{} abc \equal{} \left(a \plus{} b \plus{} c\right)\left(bc \plus{} ca \plus{} ab\right) \minus{} 1$,

where we have used that abc = 1 in the last step of our calculation. Thus, we have

$ \left(\left(a \plus{} b \plus{} c\right)\left(bc \plus{} ca \plus{} ab\right) \minus{} 1\right) \plus{} 7\geq 5\left(a \plus{} b \plus{} c\right)$;

in other words,

$ \left(a \plus{} b \plus{} c\right)\left(bc \plus{} ca \plus{} ab\right) \plus{} 6\geq 5\left(a \plus{} b \plus{} c\right)$.

Upon division by a + b + c, this becomes

$ \left(bc \plus{} ca \plus{} ab\right) \plus{} \frac {6}{a \plus{} b \plus{} c}\geq 5$.

Finally, since abc = 1, we have $ bc \equal{} \frac {1}{a}$, $ ca \equal{} \frac {1}{b}$ and $ ab \equal{} \frac {1}{c}$, and thus we get

$ \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \plus{} \frac {6}{a \plus{} b \plus{} c}\geq 5$,

what completes the proof.

  Darij


Consider the equation $x^3 = 3x + p$ and define $f(p)$ as follows:

*if the equation has 3 real roots, $f(p)$ is the product of the greatest and smallest roots.
*if the equation has 1 real root, $f(p)$ is the square of this root.

Determine the minimum of $f(p)$ as $p$ ranges over all real numbers.

This is Tokyo University Entrance exam 1991 (1).

A start:
Make the substitution $\displaystyle x=w+\frac{1}{w}$.  Then the equation simplifies to
\[w^6-pw^3+1=0\]
which is a quadratic in $w^3$.  Solve to get
\[\displaystyle w^3=\frac{p \pm \sqrt{p^2-4}}{2}\]
If the equation has only one real root then $p=\pm 2$ which implies $w=\pm 1$.  Either way $f(p)=x^2=4$.  Still thinking about the other part.


combinatorics solved
combinatorics
Coloring
duplicate

Every integer is colored red, blue, green or yellow. $m$ and $n$ are distinct odd integers such that $m + n$ is not zero. Show that we can find two integers $a$ and $b$ with the same color such that $a - b = m$, $n$, $m + n$, or $m - n$.

[color=#cc0000][mod says: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=19789& ][/color]

http://www.kalva.demon.co.uk/short/soln/sh99c6.html

tschuess! :D


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 10
complex numbers

My goal this year is to qualify for USAMO. I will be a sophmore so all I need to do is reach the AIME floor. I can get about a 2-4 on the AIME and I am on the borderline of qualifying for the AIME through the AMC 12 (about 100). About how many hours a day should I try working on problem solving to reach USAMO? Thanks for you advice.

i think you should start with practicing for AMC...and i think the best way is AoPS Vol one.
and hours is based upon you, and how much you can handle.

also AoPS Vol I will help you to get 5-9 on AIME.

you have all the AIME's?

good

you will know you are doing enough AIME problems when you do not have any free time.

good hourage is as much as you can, without making yourself hate the practice.

(4 hours a day if you can)

do the contest
read the solutions
make sure you understand the first 12 problems at least
work out the ones you didn't get right

also checking up aops 1&2 on topics you don't know.

if you keep this up til feb you'll be in excellent shape.

Don't forget HW either

I'd say follow amirhtlusa's advice, do some AMC's first

If you're only getting slightly above 100 on AMC's, theres not that high of a chance you can get 7 or 8 on the AIME.

do AIMEs

they're just like AMCs, but harder

so in effect, you'll be improving both your aime and amc score.

trust me.  i was in a situation mostly identical to yours just a year ago.

Do a lot of past AIMEs off Kalva, and also use AoPS book 2, but book 1 if you have not finished that. practice for AMCs also i guess, but if u are sure u can make it to AIME, focus more on AIME, as it will be more important than AMC. Otherwise, good luck.

[quote="amirhtlusa"]
also AoPS Vol I will help you to get 5-9 on AIME.[/quote]

Maybe I'm not using it right....  :P

Yeah, I want to make USAMO this year.  I'm gonna be working really hard on the AoPS Forum and reading AoPS Volume 2 (and re-reading AoPS Volume 1).  I'll probably wind up taking 3-6 AoPS Classes between now and AIME time.  I reallllly wish I'd taken classes this summer, mostly Intro to Counting/Probability and Intermediate Algebra.  Is there anyway that we can at least find out what they did in the class?

My AMC/AIME goals...

140+ on the AMC 12, 11+ on the AIME.

And USAMO, at least a 20?  =).

same for me as WarpedKlown. except i would expect a lower AMC score (around high 130s) and higher USAMO  :D

[quote="no_one_2000"]My goal is also to make USAMO.  Last year I got a 6 on the AIME and I've improved a lot since then, so I think I have a chance.  Unfortunately, it's my last year... :(:([/quote]
same here except i got a 4. should have started before my senior year.
i'll be hoping for a 125 AMC and a 11 AIME. need to be doing more math though to reach this goal.

I want to make USAMO this year, and since I'm a freshman, I just need to get the floor.

[quote="JesusFreak197"]I'm hoping for about 140-something AMC-10 and 120-something AMC-12, 10 on the AIME, and at least a 10 on the USAMO, more like a 15. :)

Ideally, I would have a 150 AMC-10, 130-something AMC-12, 13 or 14 on the AIME, and at least 30 on the USAMO. :D[/quote]

i'm rather confident you will reach your goal :)

[quote="JesusFreak197"]Thank you for the confidence. :) It would be nice to reach USAMO as a sophomore, and hopefully MOP by the time I'm a senior, at least. :) Don't want to stretch too far and say as a junior. ;)[/quote]Isn't the score to make it as a junior lower than for seniors?

These are my goals:
AMC12: 130, more likely I'll get 120-129
AIME: 10 (or anything above the floor value), more likely I think I'll get around 8...

My goals:

Qualify for USAMO: I got an 8 this year without practicing, or knowing anything about AIME, and as I'll be a freshman, all I want is the floor.

And then qualify for MOP: ideally with a 15-22, but I'd be satisfied with whatever qualified me for MOP.

[quote="eryaman"]Isn't the score to make it as a junior lower than for seniors?[/quote]

Yes, but I'll have more experience with the USAMO by senior year. I dunno, maybe I'll have an even chance both years.

136 AMC 12
10/11 AIME

Really  far away, though: averaging 130 on AMC's and only 6-7 on AIME. I'm still stuck in AoPS V 2

Goals

140-150 AMC 10
10+ on AIME
Qualify for MOP (I'm in 9th grade)

Current Reality

130-140 AMC 10
2-6 on AIME :(

[quote="chess64"]Goals

140-150 AMC 10
10+ on AIME
Qualify for MOP (I'm in 9th grade)

Current Reality

130-140 AMC 10
2-6 on AIME :[/quote]

If I can get an 8 this year, you can get a ten next year!

Goal for AMC-12: Enough to Make AIME
Goal for AIME: Enough to Make USAMO

I'm not actually joking by my goals because I didn't make AIME on stupid mistake on AMC-10 last year and this year, I'll be taking AMC-12 (that seems to be easier to qualify and I'm better off with the test with log, complex numbers, and trig). As for AIME, I'm not sure how well I'll do. I'm positive that I'll get at least 1 right but I'm not sure.

Maybe you guys can guess what I might get :D

Silverfalcon, I got a 98.5 last year on the AMC 12. You seem to be more adept than me at solving these problems so I don't think making AIME will be too difficult for you.

Jesusfreak, and chess64, and all the other freshmen have 2 opportunities to make AIME floor, compared to 0 for me. :(

I feel your pain.  This is the first year I'm getting serious about making USAMO, so I blew my two chances to utilize the floor value for freshmen and sophomores  :(  I didn't even make it to the AIME as a freshman because I wasn't really interested in math contests at the time.

No, I'm not a freshman, I'm going into 10th grade, but I still do have two opportunities. ;) Unless you weren't including me in the "freshmen" thing...

[quote="JesusFreak197"]I got a 134.5 on AMC-10 and 3 on AIME. If you barely didn't qualify, I'm guessing you got just under 120, which means you might have gotten a 2 on AIME last year. So... I don't think you'll have a big problem qualifying for USAMO this year, with all the practice you've been doing. ;)[/quote]

I got a 3 last year, and I just barely got to take the AIME (a 121.5 on the AMC 10).  But that was the AIME I, which was supossed to be really easy, so I dunno.

[quote="DPopov"]Hmm, goals, I guess I'll try for 150 on AMC 12 and 15 on AIME  :lol:  :P Always best to aim high, that way you end up somewhere in the middle![/quote]

They are realistic for you..... you got a13 this year! :lol:

I still don't know how I managed that one, I just hope I can do it again! :)

If I could get 5, that would be great!

Im guessing I will get around 2 though.

[quote="Derek"]Same penguin, I got a 5 last year because I was extremely lucky and the aime itself was easy.[/quote]

If you already qualifies for the AIME, then how come your goal is to qualify for the AIME?

I'm still aiming for a 12 although a 15 is nice(I solved every single problem on 2003 AIME I and II, but I am too slow, I can only do about 13 problems in the time limit+careless mistakes). Maybe I'll get a little better by March.

My problem is also slowness. I am aiming for a 12. I think that's plenty reasonable... though not sure I can do it. Only time will tell.

[quote="mathclass"]As an eigth (SP?? i never get this right) grader I got a 8 on the AIME because of a careless mistake. :( the floor was 9 I believe...[/quote]

eh i got a 8 as a 9th grader, missed my MOP chance... GL next year tho

[quote="DPopov"]Let's hope that all those complaints about the AIME I being too easy this year (which it wasn't, though some people such as myself did well on it) have not angered the test writers and caused them to write a cut-throat test this year.[/quote]

I'm sure it'll be [i]well designed.[/i]

[quote="DPopov"]Let's hope that all those complaints about the AIME I being too easy this year (which it wasn't, though some people such as myself did well on it) have not angered the test writers and caused them to write a cut-throat test this year.[/quote]

It would be funny if the floor was a 5 this year. :P

I'd like it to be pretty difficult... last year I walked out confidently with 13 answered only to make 2 stupid mistakes and find myself 2 points above the cutoff. Kinda frustrating because it's nice to know ABOUT what the cutoff is before the test (I had predicted 7-8). Nonetheless, I did like some of the questions.

Well I guess my AIME goal for this year is 15 (duh... lol) but realistically I'm shooting for 13+. Then again I'm kind of out of AIME practice so I don't know. As long as I get past it's good enough.