summer program
MathPath
MATHCOUNTS
AMC
AIME
trigonometry
induction

I read the entire website and all, but I have some questions:

1. According to the Math Jam, the courses are rated with stars.

Can someone give me examples of different star levels, and their difficulty?

2. I see that there is Contest preperation during camp, for contests such as

MathCounts, AMC, AIME, USAMO. How is this training done?

3. The schedule says "Evenings will be planned around camper's academic and physical

needs". Exactly what is an example of an evening activity?

4. To previous campers: How was the experience? On a scale of 1-10, how much did

you learn? How much did you like it?

Thanks!

-BOGTRO.

to add to that:
do they teach trig at mathpath?

These are the best answers I can give, although they might not be accurate now as I was a camper in '06:

1. 1-star/2-star: The Mathcounts problem solving class was one of these.

1-star Intro to Combinatorics: we did (according to class outline) Permutations, The Mississippi formula, meanings of nCr, Comdinatorical arguments, Pascal's triangle, Binomial Theorem, "Balls and urns" formula, Inclusion-exclusion.

2-star Problem Solving: Included the induction method. I didn't take the whole class (it probably did other stuff besides induction), I switched to 4-star problem solving, but here are a couple problems from the homework:
State and prove a formula for 2+4+6+...+2n.
Prove by induction the standard formula for the sum of the first n+1 terms of a geometric series with r<>1.

3-star Problem Solving: Discussed lots of different problem solving methods. Problems:
Can every positive integer be written as a sum of consecutive positive integers? If so, prove it. If not, show with proof which integers can and which cannot.
Determine the largest number which is the product of positive integers whose sum is 2006.
Consider the following two-player game. Each player takes turn placing a penny on the surface of a rectangular table. No penny can touch a penny already on the table. [not going to type this whole problem..] Does the first player have a winning strategy?

4-star Problem solving: With Dr. V, we did geometry, sequences, fibonacci, and a couple other days I missed. Sample Problem:
Dennis's scoop of ice cream is a perfect sphere with radius 2 cm. It rests in a cone which has a radius of sqrt3 cm at the widest part and is tangent to the cone at the rim. He eats some of the ice cream and find the remainder can be pushed down to fill the cone exactly. How much ice cream did he eat?
With Dr. Z:
Prove that x^2+y^2+z^2=2xyz has no integer solutions.
Suppose each camper at MathPath plays a single chess game with every other camper, and there are no draws. Prove the no matter who wins what games, it is possible to order all the campers in a "pecking order," where the first student beats the second, the second beats the third, etc. Must this pecking order be unique?

3-star: Hyperbolic Geometry and Spherical geometry from Dr. T, where we proved theorems from postulates, such as how many parallels, and did other random stuff like triangles in the different geometries. There was also Analytical geometry by Dr. T.

4-star Number theory or 4-Star Problem Solving by Willy. Really hard. From a problem set:
Competition Problems (easier): When a two-digit number is multiplied by 5, the result is a three-digit number. When the digit 7 is written after the resulting three-digit number, the four-digit number is 1281 great than the original. What is the original?
Theoretical Problems:
Prove that if x is represented by a terminating decimal, then x can be written as a quotient p/q where p and q are integers and q is of the form 2^a 5^b for some integers a and b.
Prove that any rational number p/q can be represented as a recurring decimal.
Prove the Euclidean algorithm indeed produces the GCD of two numbers a and b.

4-star Advanced Combinatorics: Don't have any problem sets, but we did a lot of graph theory and its relation to combinatorics.

That's mostly just samples of homework (given for in class or outside of class), but there's a lot more. Classes vary from a short-ish lecture on a topic then solving problems on that topic, to just solving problems, to mostly just strategies and examples (some solved by students) with homework, to no homework but lots of proofs in class with students solving parts of the proofs or giving suggestions.

Of course, there are many other topics for classes, these are just some of the ones I took.

And this also answers your second question partly. Training is done in the problem solving classes, by various methods with the different teachers. Usually strategy is discussed, and homework is given or done in class, and students usually get to present problems they solve.

3. Some of the more organized evening activities the year I went were the soccer tournament, the chess tournament, and the table tennis tournament. I actually did none of these, but did play some chess and soccer outside of the tournament. Other activities were going to the pool during midday break or going to the library at midday break. Also, students played basketball and there was a morning running club. There was a practice schedule for piano (everyone interested signed up for a time spot in the morning, afternoon, or evening to play. But they didn't have to go every day or anything if they signed up.) I'm sure they'll arrange something if a camper needs to practice a different instrument. There were study groups after dinner and classes, supervised by counselors. And the only mandatory activity besides classes was watching these half-hour long statistics videos each evening right after dinner. Plus there were plenty of other games (mental or physical) or spontaneous activities that went on during evenings. I also worked on the yearbook crew, which was really enjoyable (understatement).

4. What was my experience? 9 out of 10. It wasn't perfect, of course, but it was totally awesome. I met tons of cool kids around my age also interested in math. Not to mention all the wonderful staff (counselors, teachers, and directors).
How much did I learn? I'm not sure how to rate it, but I learned more than I expected, and enjoyed learning it. The classes were very clear and engaging (with one exception. The teacher's no longer there, so don't worry), and I also learned from working on homework with other campers.
How much did I like it.... 10 out of 10. It was the greatest summer I'd ever had. (Mathcamp the next summer tied it--I really can't decide which was better.)

And to answer abacadaea, I believe the first week of Willy's number theory had trig in it, and there might have been a class called Intro to Trig. I'm not sure, though. But it was also taught in various other classes, including ones which the whole camp went to.

Thanks a lot!

There is a chess tournament? YES!!!

The kids set up the chess tournaments. So, yes, there's one if someone sets it up :) Maybe you could do it!

Oh, well then nice... I WILL WIN!!! YAY!!!

i am terrible at chess....i can get checkmated in <10 by some of my friends almost every game...

my mom wants to know....my mom wants to know everything.

I wasn't aware of any rules against it. :P

Hmm, while I was there it was never a problem. So, gum is probably allowed - but I probably wouldn't suggest chewing gum during class. Do it during free times.

Hmm. That's an interesting question...

LadyKnight, thanks for all the detail about your experience!


Call a set of integers [i]spacy[/i] if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\}$, including the empty set, are spacy?

[b](A)[/b] 121      [b](B)[/b] 123      [b](C)[/b] 125     [b](D)[/b] 127    [b](E)[/b] 129

[hide]

Consider the case where we want a spacy subset of size $n$. There are $\binom{12-2(n-1)}{n}$ ways to do this, because there is a bijection between this problem and choosing a (arbitrary) subset of size $n$ from a set of size $12-2(n-1)$. If you do not see the bijection, I would suggest trying to find it on your own, because this is a really good problem, and the technique generalizes immensely.

Anyways, adding up, you get

$1+\binom{12}{1}+\binom{10}{2}+\binom{8}{3}+\binom{6}{4}= 129$, so the answer is $(E)$.

[/hide]

[hide="general case"]

If we don't want any more than $1$ out of $m$ consecutive integers, then it is simply the coefficient of $x^{n}$ in the power series expansion of
\[\frac{1-x^{m}}{(1-x)(1-x-x^{m})}\]
Proof will be left as an excercise to the reader =).

(If you don't believe me, and you are willing to wait a long time, you can do $nDeriv((1+x+x^{2})/(1-x-x^{3}),x,12)|x=0$ on the TI-89 and you'll get $129$).

We can actually solve this when $m=2$. It becomes
\[\frac{1+x}{1-x-x^{2}}= 1+2x+3x^{2}+5x^{3}+\ldots \]
or in particular the $n+2$nd Finobacci number is the coefficient of $x^{n}$.
[/hide]

[hide="Recursive Solution"]

Let $E_{n}$ denote the number of spacy subsets of the integers ${\{1,...,n}$.

We can see that $E_{0}=1$ (the empty set) and $E_{1}= 2$ (the empty set and the set containing 1).

We now wish to show that $E_{n+1}=E_{n}+E_{n-2}$:

Notice that for each spacy subset of $n+1$ that contains $n+1$, we can take out $n+1$, and because that means that it does not contain $n$ or $n-1$, this forms a spacy subset of $n-2$ (and equivalently for each spacy subset of $n-2$, we can form one for $n+1$).

As well, if the spacy subset does not contain $n+1$, it automatically forms a spacy subset for $n$, and similarly we can see that each spacy subset of $n$ is also a spacy subset of $n+1$.

Thus, we have shown that the number of spacy subsets of $n+1$ corresponds to the sum of the number of spacy subsets of $n$ and $n-2$, producing $E_{n+1}= E_{n}+E_{n-2}$.

We now use the recursion and initial conditions to calculate $E_{12}$ and get $E_{12}=\boxed{129}$.[/hide]

JSteinhardt, I think you wanted 6 choose 4, not 7 choose 5 on the last line.

Isn't this just like an easier version of the #25 on the 2006 12A?

[quote="mysmartmouth"]Yeah, and there was a similar AIME problem not too long ago, something with mailboxes I believe...[/quote]

Oh yeah...

I love these questions with non consecutiveness.

calc rulez: How did you think of that? I took one look at the problem and had no idea.

[quote="lowfatsourcreme"]What's a bijection?[/quote]

I think it's like a one-to-one relationship or something.  Smart people use it to mean something like "these are identical problems."  Or something.

You know, if you didn't see the bijection and simply brute forced it, it takes less than five minutes. The cases are very straightforward - kind of odd for a number 25.

A bijection is a function

\[f : S \to T\]

such that if $f(s_{1})=f(s_{2})$, then $s_{1}=a_{s}$, and, furthermore, if $t \in T$, then there exists $s$ such that $f(s) = t$. Note that the function need not be a function in the sense that we usually used, where we map reals to reals. Most often it maps the elements of one set to those of another set. The most important thing to note about a bijection is that the existence of such a function implies that if $S$ and $T$ are finite then they have the same number of elements. This is because, if we think of it in terms of those arrow diagrams you see in algebra 2, then each element in $T$ gets "pointed to" exactly once.

[hide]
${6\choose 4}+{8\choose 3}+{10\choose 2}+{12\choose 1}+{14\choose 0}=129$.

The diagram illustrates some elements of the ${6 \choose 4}$ case.
[/hide]


[quote="Fang-jh"]Let F be the Feuerbach point of triangle ABC, Let I be incenter of triangle ABC, Incircle of triangle ABC touches AB,AC at D,E.respectively.M is projection C on AB. N is projection B on AC. DE intersects MN at P. if P is inside triangle ABC, then F,P, and I are collinear.[/quote]
I think this is wrong.


limit
logarithms
number theory unsolved
number theory

Show $\lim_{n\to\infty}\frac{p_{n}}{n\ln n}=\lim_{n\to\infty}\frac{n\ln p_{n}}{p_{n}}=1$

Where $p_{n}$ is the $n$th prime

This is a corollary of the prime number theorem.  :wink:

In case you were looking for more details...
[hide]We use the asymptotic notation $f(x) \sim g(x)$ if $\lim_{x\rightarrow \infty}f(x)/g(x)=\lim_{x\rightarrow \infty}g(x)/f(x)=1$.  Note that the PNT says $\pi(x) \sim x/\log x$.  So, $\pi(x) \cdot \log x \sim x$.  Letting $x=p_{n}$ yields that $n \cdot \log p_{n}\sim p_{n}$ (i.e. $\lim_{n\rightarrow \infty}p_{n}/\big(n \cdot \log p_{n}\big)=1 )$.

To derive the other limit, notice that from PNT we get $\log \pi(x) \sim \log x$.  So, it follows that $\pi(x) \log \pi(x) \sim x$.  Again letting $x=p_{n}$ yields that $p_{n}\sim n\log n$ (i.e. $\lim_{n\rightarrow \infty}p_{n}/\big(n\log n\big) =1$.) [/hide]

Both statements \[\lim_{n\to\infty}\frac{p_{n}}{n\ln n}=1\] and \[\lim_{n\to\infty}\frac{n\ln p_{n}}{p_{n}}=1\] are obviously equivalent to the prime number theorem.

[quote="sylow_theory"]Are you sure we need the prime number theorem here?
Because I happen to know a somewhat elementary proof that shows that:
$\lim \frac{p_{n}}{n}= 0$
And the prime number theorem seems to involve $\pi (n)$ which isnt mentioned here.[/quote]

I think you mean: \[\lim \frac{p_{n}}{n}= \infty.\] Anyways, just post what you are thinking and others will tell you their opinions.


vector function trigonometry algebra domain

find a vector valuied function to represent the graph of the intersection between the semiellipsoid x^2/12 + y^2/24 + z^2/4 = 1 and the parabolic cylinder y=x^2

i tried solving for x^2 in the first function and setting
y = 12 - y^2/2 - 3z^2
y^2 + 2y + 6z^2 - 24 = 0
(y+1)^2 - 1 - 12 + 6z^2 = 0
(y+1)^2 + 6z^2 = 13
(y+1)^2/6 + z^2 = 13/6

would this be the right way to go about this problem?(and then just parameterize with trig functions and write in vector notation)

oops i was thinking off for a second itd b simplified as...

(y+1)^2/13 + 6z^2/13 = 1

then again as Ci + (sqrt(13)sinT-1)j + (sqrt(13/6)cosT)k

btw if i am right how would i find C?

[quote="joml88"]Can't you just let $x=t$ so then $y=t^{2}$ and $z = \sqrt{\frac{24-2t^{2}-t^{4}}6}.$  In other words the vector is given by
\[\left\langle t, t^{2}, \sqrt{\frac{24-2t^{2}-t^{4}}6}\right\rangle. \]
[/quote]

two things:
1) z can be negative
2) not all real t work, the domain must be restricted


geometry
Pythagorean Theorem
power of a point

Please solve this problem step by step, and writting all theorems you use. Can someone give me some tips for solving this kind of exercises?

Find the radius of the given circle:

[img]http://i204.photobucket.com/albums/bb126/agerador/figure4.jpg[/img]

PD:I would like to thank 1000th user for his help in geometry.  :lol:

[hide="Solution"]

$ PT \equal{} PS$ since they're both tangent to the circle.
Now, using Power of a Point, $ QR(QT) \equal{} QS^2$.
Let $ RT \equal{} x$, and perform Pythagorean Theorem:
\[ (1 \plus{} x)^2 \plus{} 9 \equal{} (3 \plus{} \sqrt {1 \plus{} x})^2\]

\[ 1 \plus{} 2x \plus{} x^2 \equal{} 1 \plus{} x \plus{} 6\sqrt {1 \plus{} x}\]

\[ (x^2 \plus{} x)^2 \equal{} 36(1 \plus{} x)\]

\[ x^2(x \plus{} 1) \equal{} 36 \implies x\equal{}3 \implies \text{radius} \equal{} \boxed{\frac{3}2}\]
.

(Note the 3-4-5 triangle  :wink: )

[/hide]

EDIT: Thanks to TalchyonPulse.

It isn´t clear to me why PT equals PS. Why if both points are tangent to the circle  their distances to a certain point are equal?

Isn't the radius $ \frac{3}{2}$ because $ x \equal{} RT \equal{} 2r$?

I don´t understand the last part of the pythagorean theorem.

Why? $ (3\plus{}\sqrt{1\plus{}x})^{2}$ at the end

Because $ QP \equal{} QS \plus{} SP \equal{} QS \plus{} 3$

and $ QS^{2} \equal{} QR\times QT \equal{} 1 \times (1\plus{}x) \equal{} 1\plus{}x$, so $ QS \equal{} \sqrt{1\plus{}x}$

and so $ QP \equal{} (3 \plus{} \sqrt{1\plus{}x})$

and so $ QT^{2} \plus{} TP^{2} \equal{} QP^{2}$ becomes $ (1\plus{}x)^{2} \plus{} 9 \equal{} (3 \plus{} \sqrt{1\plus{}x})^{2}$.

Hope this helps you.

[quote="Chalao"]PD:I would like to thank 1000th user for his help in geometry.  :lol:[/quote]Thanks, I'm glad to help you any time I'm on here. :) Geometry is one of my favorite math subjects and I'm one of those that would favor a synthetic approach over a trigonometric approach whenever I can unless it gets ugly computationally :P

Anyways, I hope that Caelestor's solution and AndrewTom's explanation are clear to you. If there are any questions, let me know :)


:rotfl:  :roll: prove or disprove the inequality

It's easy if you think tangent of $ f$ such that $ f(x)\equal{}\frac{a^{2}}{(a^{2}\plus{}1)^{2}}$
then, tangent of $ f$ at the point $ f(\frac{1}{2})$ is $ y\equal{}\frac{48}{125}x\minus{}\frac{4}{125}$
therefore $ LHS \le \frac{48}{125}(a\plus{}b\plus{}c\plus{}d)\minus{}4\frac{4}{125} \equal{}\frac{16}{25}$


geometry calculus HMMT Harvard college MIT Princeton

Maybe you guys can make teams for some of these competitions here.

===============================

[b][url=http://www.theitest.com]iTest[/url][/b]
[url=http://74.125.95.132/search?q=cache:http://www.theitest.com/]mirror[/url]
teams of 5, must be from the same state
online
free

===============================

Most of the University math competitions have Team, Power (proof-based), and Individual components, and the individs are broken into subject tests (Algebra, Geometry, Number Theory, Combinatorics, Calculus, Advanced Topics)

[b][url=http://web.mit.edu/hmmt/www/]HMMT[/url][/b]
Harvard-MIT Math Tournament
teams of 8, must be from either the same school or within a 100-mile radius, and same state
held at Harvard and at MIT, switching every other year
10 dollars per student if you register on time

[b][url=https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition]PUMaC[/url][/b]
Princeton University Mathematics Competition
teams of 8 (no geographical restrictions)
held at Princeton University
10 dollars per student if you register on time

[b][url=http://sumo.stanford.edu/smt/]SMT[/url][/b]
Stanford Math Tournament
I don't know anything about this, and couldn't find much. Similar format though.

[b][url=http://www.ruf.rice.edu/~eulers/RMT.html]RMT[/url][/b]
Rice Math Tournament
teams of 24 at max, from the same school
power, team, individs/subjects

many of these contests have geographical limits in order to prevent students from forming "dream teams". nobody really wants to compete if it's like, hypothetically, the IMO team versus all.

I wonder who wants to be on an iTest team for 2009 IL...

Who was on it in 2008?

I don't know who was on it in 2008.. do you want to make a team for 2009 guass?

there were several teams from IL

Team Chicago was
Kevin Hu (aryth)
Kate Rudolph (k8reindeer)
Kiaran Dave
Arjun Puranik

i think Josh Dement had one? i'm not sure who was on it.

As a hint, if you guys tryout for and make the Chicago ARML teams, that's a good way to find buddies for geography-bound contests.

@ guass : That's true.. it's a long way lol.

@ Aryth : Can 8th graders make the team?

I think 8th graders are allowed on teams. For example, Bobby shen was on an itest team and on an ARML team. :D

Oh ok I hope so.  When I looked at the D team info it said you must be a 9th or 10th grader in bold though... ?

Errr.......suma do u do usamts btw?

But...there's a problem...

*suma_milli gets kicked off the ernie tourney mathcounts team  :D because of her lack of a  :D avatar*

No I was going to do but I never really got around to it... :(  I  now realize I should have.  I'll do round 4 for sure!

What ernie tourney mathcounts team? *goes to check middle school tournaments* OH. By the way I now have TWO :Ds in my avatar so it still gives my avatar the Mr.Green terifficness regardless of the razzes :P.  Well not really if you think that each razz cancels the :D out.  But at least I've still got the :D spirit! So therefore I can still be on Team :D.  (Wait.. I never was in the first place but w/e)  I'll probably fail epiclly but it can't do any harm I will sign up to be on Team :D!

Wow.. this was a long and confusing post  :wacko:

Btw does anyone know when ARML tryouts are?

[url=http://www.mathkangaroo.org/]Math Kangaroo[/url]

Registration is still open for this one. The problems seem rather easy, probably mid AMC12 level at hardest, though I can't really tell.

i guess...?

btw, how did u fit so many smilies in your post?

 :P  :D  :D

Well then I might sign up for math kangaroo..

I don't know I just did?

I'd be interested in joining the team for the 2009 iTest. When does it start?

[quote="Aryth"]ARML tryouts are February 26 (late this year).
Ugh, math kangaroo D:[/quote]

What's the Ugh, math kangaroo for  :?:  :P

What's the Ugh, math kangaroo for :?:  :D

Eh, it's some international math contest for grades 2-12. This year's contest is hosted by IMSA!!!!

And registration deadline is TOMORROW. :(

Darn. Not again! Too many smilies in your post: 4.
You can have at most 3 smilies in this message. Please remove some smilies before submitting this message.

i missed a stupid question last year. it was like, you have the digit 2 on your digital calculator and you flip it horizontally. what digit do you get?

HINT: NOT TWO.  :blush:

[quote="suma_milli"]LOL  :rotfl:  :rotfl:  :rotfl:[/quote]
Oh snap, that reminds me of the anonymous rotfl'er who spams all the blogs..... :o

I don't know. It's a mystery...ANd I'm afraid that anonymous kid is gonna get to my blog...

*shudders*


The probability of rain on any given day in Atlanta is 20%.  after how many days would you expect it to have rained on 30 days

Is this some kind of trick question?
wouldn't you just go 1/5*30=6?

Actually it isn't: [hide]you expect it to rain 1 day in every 5, so you expect it to rain 30 days in 150.[/hide]

leeyuhang, you solved it for how many days would it rain in 30 days. but you got the concept

:oops: i thought it had said how many times would you expect it to rain in 30 days.
gotta pay more attention to the directions.


Ένα πεπερασμένο σύνολο Α θετικών ακεραίων θα το λέμε καλό αν για κάθε στοιχείο χ του Α ισχύει ότι ο χ διαιρεί το άθροισμα όλων των στοιχείων του Α.
Να αποδειχθεί ότι κάθε πεπερασμένο σύνολο θετικών ακεραίων είναι υποσύνολο ενός καλού συνόλου.

Ωραιο προβλημα Παυλο.

[hide]Ας το δουμε κατασκευαστικα.

Αρκει να βρουμε ενα "καλο" συνολο φυσικων που να περιεχει το $ \{1,2,..,n\}$ για τον τυχαιο $ n$.

Παρατηρουμε ομως οτι ισχυει
$ \sum_{k \equal{} 3}^{n}( (n \plus{} 1)_{k \minus{} 1}) \equal{} \sum_{k \equal{} 2}^{n \minus{} 1}( (n \plus{} 1)_{k})$
$ \leftrightarrow \sum_{k \equal{} 3}^{n}( (n \plus{} 1)_{k} \minus{} (n \plus{} 1)_{k\minus{}1}) \equal{} (n \plus{} 1)! \minus{} n(n \plus{} 1)$ με $ (n)_k \equal{} \frac {n!}{(n \minus{} k)!}$

Η παραπανω ομως ισοτητα σε συνδυασμο με το οτι ισχυει $ n(n \plus{} 1) \equal{} \sum_{i \equal{} 1}^n (i) \plus{} \frac {n(n \plus{} 1)}{2}$
μας δινει την δυνατοτητα να θεωρησουμε το συνολο,

$ \{ c_3,..c_{n \minus{} 1},1,2,..,n,\frac {n(n \plus{} 1)}{2} \}$, οπου $ c_i \equal{} (n \plus{} 1)_{i } \minus{} (n \plus{} 1)_{i\minus{}1}$, το οποιο πληρει το ζητουμενο αφου ισχυει $ c_i \equal{} (n \plus{} 1)_{i\minus{}1}(n\plus{}1\minus{}i)|(n \plus{} 1)!$[/hide]

Μια διαφορετική προσέγγιση:

Θεωρούμε το σύνολο $ \{a_1,a_2,...,a_n\}$ θετικών ακεραίων.
Έστω $ S\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n$ και $ P\equal{}EK\Pi[a_1,a_2,...,a_n]$. Γράφουμε τον P στο δυαδικό σύστημα αρίθμησης οπότε ο P γράφεται στη μορφή
$ \displaystyle P\equal{}2^{\beta_1}\plus{}2^{\beta_2}\plus{}...\plus{}2^{\beta_k}$ με $ \beta_1<\beta_2<...<\beta_k$.
Θεωρούμε το σύνολο $ \{a_1,a_2,...,a_n,S,2S,...,2^{\beta_1\minus{}1}S,2^{\beta_2}S,...,2^{\beta_k}S,PS,2PS,2^2PS,...,2^{\beta_k\minus{}1}PS\}$.
Τότε το άθροισμα των στοιχείων του συνόλου αυτού είναι $ 2^{\beta_k}PS$ και κάθε στοιχείο του συνόλου αυτού διαιρεί τον αριθμό αυτό. Άρα είναι καλό σύνολο που περιέχει το $ \{a_1,a_2,...,a_n\}$


For which positive integers is $ {m}^2\equal{}{n}^2\plus{}80$ true?

$ m^2\equal{}n^2\plus{}80\implies m^2\minus{}n^2\equal{}(m\minus{}n)(m\plus{}n)\equal{}80$, giving us

$ (m,n)\equal{}\boxed{\{(21,19),(12,8),(9,1)\}}$.

Note that $ m\equal{}\frac{(m\minus{}n)\plus{}(m\plus{}n)}2$ and $ n\equal{}\frac{(m\plus{}n)\minus{}(m\minus{}n)}2$, so both numerators in the equation/factors of $ 80$ must be even.

[quote="math154"]$ m^2 \equal{} n^2 \plus{} 80\implies m^2 \minus{} n^2 \equal{} (m \minus{} n)(m \plus{} n) \equal{} 80$, giving us

[/quote]

I got this step myself and but how do get $ (m,n) \equal{} \boxed{\{(21,19),(12,8),(9,1)\}}$, just trial and error? And how do you know this is the finite set?

Since $ m,n\in\mathbb Z^ \plus{}$, $ m \minus{} n,m \plus{} n\in\mathbb Z^ \plus{}$ as well.

Let $ x \equal{} m \minus{} n$ and $ y \equal{} m \plus{} n$ such that $ xy \equal{} 80$ ($ x,y\in\mathbb Z^ \plus{}$, obviously), and

$ (m,n) \equal{} \left(\frac {x \plus{} y}2,\frac{y \minus{} x}2\right)$.

Thus, both $ x \plus{} y$ and $ y \minus{} x$ must be even, and since $ 80$ is even, at least one of $ x,y$ must be even.

However, if the other is not even, then both $ x \plus{} y,y \minus{} x$ are odd, which gives a contradiction, so both $ x,y$ must be even.

The rest is just trial and error--the working pairs $ (x,y)$ are $ (2,40),(4,20),(8,10)$.


Had posted this in the Intermediate section but found no response.

Looking for some approaches to this problem  :-)

Vineet likes to talk only in integer numbers. So much that he rounds off everything including his course average points to the nearest integer. For example, 89.34 is 89 and 99.54 is 100, and 115.5 is 116. Vineet always calculates the average (real) on the cumulative points so far. After his 75 points in Finance, his rounded average drops by 1 point. Next, after 83 points in strategy paper, his rounded average further plummets down by 2 points. Which among the following is not true?

(a) The minimum possible number of courses is less than 15
(b) The maximum possible number of courses is not more than 50
(c) The minimum possible current rounded average is 95
(d) Either of 126 or 127 can be the current rounded average
(e) none of the foregoing

By current average, the question means after Finance and Strategy marks.

Agreed this is tough but it doesn't need any advanced concepts of Mathematics.
Isn't there anyone who can attempt this?


Prove that for any triangle each exradius is less than four times the circumradius.

It's well known that $4R+r=r_A + r_B + r_C$, where $R,r,r_A,r_B,r_C$ are, respectively, the circumradius, inraduis and the three exradius of a triangle $ABC$. Then, since $r_A,r_B,r_C>r$, we have $4R+r>2r+r_X\Rightarrow r_X <4R-r<4R$, for any $X \in \{ A,B,C \}$.


function
floor function
inequalities
modular arithmetic
algebra proposed
algebra

Find all real numbers $x$ such that there is a positive integer $n$ with $15\lfloor x \rfloor ^2 + 10\lfloor x \rfloor x + 18\lfloor x \rfloor + 2x + 6 - 4^n = 0$. ($\lfloor x \rfloor$ denotes the greatest integer which is not more than $x$.)

This is one of the ugliest problems I've ever seen.

[hide="Idea from an stupid ignorant freshman."]group the terms, prove that x is an even integer. [/hide] :) :D

I'll give you a clue: the solution is that $x$ is an element of one of two negative infinite sequences which are pretty horrendous.

The solution is not elegant, I'll tell you that.

Its not a particularly nasty solution, it just doesn't involve any clever ideas, and requires accuracy in algebra, a facility I seem to lack!

Oh well, if one of you feels like posting a solution, go for it :D

Ok, I'll give it a try :) :

[hide="attempted solution"]
So, grouping the terms by creative ways give us $2x$ must be an integer and $[x]$ must be even and $2x$ must be even as well.  Thus, x is an even integer.  Now, the original equation becomes: $15x^2+10x^2+18x+2x+6=25x^2+20x+6=4^n$
Note that $(5x+2)^2=25x^2+20x+4$, so the equation becomes: $(5x+2)^2+2=4^n$  Now, the left side must be $2\mod4$, since $5x+2$ is even, while the right side must be $0\mod4$, so no solution.
[/hide]

Point out the mistake if one exists in Z.  :D

There are no solutions for $x \in \mathbb{Z}$. But $x$ is not necessarily an integer, $(10\lfloor x \rfloor + 2)x$ is necessarily an integer.

Huh?  :huh:

If (10[x]+2)x is an integer, then wouldn't x be an integer by default(isn't 10[x]+2 an integer)?

No, $x$ must be rational but its denominator may be a factor of $10\lfloor x \rfloor + 2$

But $(3[x]+2x)$ must be even, and [x] must be even, thus $2x$ must be an even integer, so x is an integer.

No, that's wrong. Take $x = \frac{1}{2}$, then $(10[x] + 2)x = 1$ is an integer.

If you still don't believe me, try putting $x = -\frac{1}{8}$.

Wait, Im confused... :huh:
I was saying that we can group: $(2x+3[x])(5[x]+1)=4^n-6-15[x]$ and then [x] must be even, 2x must be an even integer, etc...... :roll:

$2x$ is not an integer. $2x(5[x] + 1)$ must be an integer...but this does not mean $2x$ is an integer.

Eg $\frac{1}{8}(16)$ is an integer but, $\frac{1}{8}$ isn't an integer.

Argh  :oops:  :wallbash_red:
Darn, oh well...

As a start, I'll prove that there are no nonnegative values of $x$ that work.

Let $y = \lfloor x \rfloor$.  By usual properties of floor, we have $y \le x < y + 1$.  The given equation becomes
\[ 15 y^2 + 10xy + 18y + 2x + 6 = 4^n \, .  \]

Let $x$ be nonnegative.  Then $y$ is nonnegative too.  We will show that $5y + 2 < 2^n < 5y + 3$.  But that's a contradiction because $5y + 2$ and $5y + 3$ are consecutive integers.

First, let's show that $5y + 2 < 2^n$.  We have
\begin{eqnarray*} (5y + 2)^2 & = & 25 y^2 + 20y + 4 \\ & < & 25y^2 + 20y + 6 \\ & = & 15y^2 + 10y^2 + 18y + 2y + 6 \\ & \le & 15y^2 + 10xy + 18y + 2x + 6 \\ & = & 4^n \, . \end{eqnarray*}
Taking the square root of both sides gives $5y + 2 < 2^n$.

Next, let's show that $5y + 3 > 2^n$.  We have
\begin{eqnarray*} (5y + 3)^2 & = & 25y^2 + 30y + 9 \\ & > & 25y^2 + 30y + 8 \\ & = & 15y^2 + 10y(y+1) + 18y + 2(y+1) + 6 \\ & > & 15y^2 + 10xy + 18y + 2x + 6 \\ & = & 4^n \, . \end{eqnarray*}
Taking the square root of both sides gives $5y + 3 > 2^n$.

Putting both inequalities together gives $5y + 2 < 2^n < 5y+ 3$, which is the contradiction we sought.

I'm pretty sure the method above can determine the negative values of $x$ too, but I'll leave that case for other people...

This is weird.... isn't $4^n$ arbitary?.....oh well... :huh:  :(

I'm not sure what you are asking.  Maybe you are pointing out that the only property of $4^n$ used in the previous post is that it is a square, which is true.

Also, as I had suspected, the method above does find the negative solutions too.  Recall that the given equation is
\[ 15 y^2 + 10xy + 18y + 2x + 6 = 4^n \, .  \]
Solving for $x$ gives
\[ x = \frac{4^n - 15y^2 - 18y - 6}{10y + 2} \, .  \]
So once we know $y$ (and $n$), $x$ is determined.

Let $x$ be negative.  Then its floor satisfies $y \le -1$.  Using the method above, we can prove that $(5y + 4)^2 < 4^n < (5y + 1)^2$.  Taking square roots gives $-5y - 4 < 2^n < -5y - 1$.  Thus either $2^n = -5y - 3$ or $2^n = -5y -2$.  We will handle the two cases separately.

Suppose that $2^n = -5y - 3$.  To make $2^n \equiv -3 \pmod 5$, we must have $n \equiv 1 \pmod 4$.  Thus we know $y$ (in terms of $n$), and so by the equation above, we can solve for $x$:
\[ x = - \frac{2 \cdot 4^n - 3}{10(2^n + 2)} \, .  \]
We can check that the floor of this value is indeed $- \frac{2^n + 3}{5}$, which is our value of $y$.  So we have found an infinite sequence of solutions.

Next, suppose that $2^n = -5y - 2$.  To make $2^n \equiv -2 \pmod 5$, we must have $n \equiv 3 \pmod{4}$.  Plugging in the value for $y$ into our equation for $x$, we get
\[ x = - \frac{4^n + 3 \cdot 2^n - 3}{5(2^n + 1)} \, .  \]
Again, we can check that the floor of this value is indeed $- \frac{2^n+2}{5}$, our value of $y$.  Thus we have found the second infinite sequence of solutions, as Post #4 hinted at.

Oh, nvm, I understand it now. :)
Nice proof btw.

That solution was far nicer than my solution and involved far less horrific algebra bashing.

However, the question is still not a particularly elegant question in the sense it really was just algebra slogging throughout.


AMC
AIME
geometry
perimeter
parallelogram
trigonometry
3D geometry

As the 2004 AIMEA is this tuesday, let's do something to prepare: the AIME Chain.

I'll post an AIME problem.  The person who solves it gets to post another problem.  The person who solves this problem gets to post one, and so on.

[b](1987/11) Find the largest k such that 3^11 is the sum of k consecutive positive integers.[/b]

[hide]The sum of n consecutive integers with first term a is n(a+n-1)/2. So, a=2*3^11/n - n + 1. So, n is a divisor of 2*3^11. Furthermore, since all terms are positive integers, a >= 1, i.e. 2*3^11>=n^2. The largest factor of 2*3^11 less than the square root of this number is 2*3^5=****486****[/hide]

Triangle ABC is isosceles with AB=AC, and a point D lies on median AM such that AD=10 and DM=1. Furthermore, <BDC = 3 <BAC. Find the perimeter of triangle ABC. (Its from Kalva, so the answer might not be a 3 digit integer. Based on 1995/9)

Zabelman, the sum of n consecutive integers with first term a is n(2a+n-1)/2 not  n(a+n-1)/2.  Your answer is still correct though.

zabelman wrote:Triangle ABC is isosceles with AB=AC, and a point D lies on median AM such that AD=10 and DM=1. Furthermore, <BDC = 3 <BAC. Find the perimeter of triangle ABC. (Its from Kalva, so the answer might not be a 3 digit integer. Based on 1995/9)

[hide]call AM=b, angle BAC= :theta: . Tan :theta: =b/11. With some algebra we have tan(3 :theta:)=(363b-b^3)/(1331-33b^2)=b.

Solve for b, we have b=11/2. AB^2=(11/2)^2+11^2, so AB=11  :sqrt: (5) /2. the total perimeter =11 :sqrt: (5)+11  [/hide]

Circles radii 5, 5, 8, k are mutually externally tangent, where k=m/n in lowest term, find m+n.

I changed the wording of Kalva's problem to make a integer answer

beta wrote:Circles radii 5, 5, 8, k are mutually externally tangent, where k=m/n in lowest term, find m+n.

[hide]Let circle A, B have radius 5, circle C radius 8, circle D radius k. Call P the pt. of tangency between circles A, B. CP is the altitude of :tri: ABC. BC = 13, BP = 5, therefore CP = 12. Call E the pt. of tangency between circles C, D. CE = 8, thus EP = 4. BD = 5+k and DP = 4-k. By Pythagoras, DP2 + BP2 = BD2. Solve to get k = 8/9. m+n = 17. [/hide]

The graph of y2 + 2xy + 40|x| = 400 partitions the plane into several regions. What is the area of the bounded region?

wyra wrote:
The graph of y2 + 2xy + 40|x| = 400 partitions the plane into several regions. What is the area of the bounded region?

[hide]the equation=y :^2: +2xy=400 :pm: 40x. Add x :^2: to both side we have (x+y) :^2: =(x :pm: 20) :^2: . so the lines are y=20, y=-20, y=-2x-20, y=-2x+20. The lines form a parallelogram with base 20 and height 40. the area=20*40=800[/hide]

n is the smallest positive integer which is a multiple of 75 and has exactly 75 positive divisors. Find n/75.

Quote:n is the smallest positive integer which is a multiple of 75 and has exactly 75 positive divisors. Find n/75.

[hide]The number must be of the form n = 2^a * 3^b * 5^c *...., where (a+1)(b+1)(c+1).....=75. Since n has an odd number of divisors, it is a perfect square, and so a, b, c, ..... are all odd. We already know that n has 75 as a factor, or 5^2 * 3. n, therefore, has at least two factors of 5 and one factor of 3. We find that the least possible n will be 2^4 * 3^4 * 5^2, and n/75 = 2^4 * 3^3 = 432.[/hide]

(1988/7) In the triangle ABC, the foot of the perpendicular from A divides the opposite side into parts length 3 and 17, and tan A = 22/7. Find area ABC.

Quote:In the triangle ABC, the foot of the perpendicular from A divides the opposite side into parts length 3 and 17, and tan A = 22/7. Find area ABC.

[hide]Wow, my first AIME problem using trig! Yay!

length of altitude=a

two angles formed by altitude=x and y (x+y=A)

tan(x+y)=(tan x +tan y)/(1 - tan x tan y)=(17/a + 3/a)/(1 - 51/a^2)=22/7

20/a=22/7(1-51/a^2)

70/11a=1-51/a^2

70a/11=a^2-51

70a=11a^2-561

11a^2-70h-561

factorizing

(11h+51)(h-11)

take the postive root or 11

[/hide]

1992/7

ABCD is a tetrahedron. Area ABC = 120, area BCD = 80. BC = 10 and the faces ABC and BCD meet at an angle of 30o. What is the volume of ABCD?

Quote:ABCD is a tetrahedron. Area ABC = 120, area BCD = 80. BC = 10 and the faces ABC and BCD meet at an angle of 30o. What is the volume of ABCD?

Solution:

[hide]Make the base ABC and one of the sides BCD. The altitude of BCD is 80*2/10 = 16. Draw the altitude of the tetrahedron. It forms a triangle with opposite angle 30 with the altitude of BCD. Then the tetrahedron's altitude is 16*sin 30 = 8. The volume of the tetrahedron is 120*8*1/3=320.[/hide]

11/2003 A

The angle x is chosen at random from the interval 0o < x < 90o. Find the probability that there is no triangle with side lengths sin^2x, cos^2x and sin x cos x.

Tetrahedron, i misread the problem as sin (2x) instead of sin :^2: (x). Could you edit that?

[hide] we must have cos :^2: x > sin :^2: x + sin x cos x.

                          or cos :^2: -sin :^2: >sinxcosx.

                                  cos(2x)>(1/2)sin(2x)

                                 2(cos(2x)>sin(2x)

                                 4cos :^2:(2x)>sin :^2: (2x)

                                 4cos :^2: (2x)>1-cos :^2: (2x)

                                  5cos :^2: (2x)>1

                                     cos (2x)>  :rt5:/5

it could be expressed nicely with tangent. cos(2x)=:rt5:/5 , then tan(2x)=2. 2x=tan^(-1) 2. x=(1/2)tan^(-1) 2. x must be less than 45 or we would have cos(2x)<0. so the answer is ((1/2)tan^(-1)(2))/45=(tan^(-1)(2))/90.[/hide]

  Find the largest positive integer n for which there is a unique integer k such that 8/15 < n/(n+k) < 7/13.

For Beta's problem:

[hide]We can cross multiply and simplify to obtain the inequalities 7n>8k and 6n<7k, or we can simplify it to (6/7)n<k<(7/8)n. Thus, for k to be unique, (7/8)n-(6/7)n < 2, and {(6/7)n} + q < 2, where q is some number between 1 and 2 that the two fractions actually differ by. Simply testing numbers below 111, we obtain 105 as our largest solution.

[/hide]

Here's one of my personal favorites from geometry:

ABCDEF is a convex hexagon with all sides equal and opposite sides parallel. Angle FAB = 120o. The y-coordinates of A, B are 0, 2 respectively, and the y-coordinates of the other vertices are 4, 6, 8, 10 in some order. Find its area.

(AIME 2003 B, 14)


Find the nonzero digits $ a,b,c$ such that: $ \sqrt {a} \plus{} \sqrt {ab} \plus{} \sqrt {abc} \plus{} \sqrt {a \plus{} b \plus{} c} \equal{} cc \minus{} bb \minus{} aa$

[quote]whats are nonzero bytes?[/quote]

I've corrected  :D


inequalities
trigonometry
algebra
polynomial
inequalities proposed

x, y, z ∈ R
prove that  x^4 + y^4 + z^4 + xy^3+ yz^3 + zx^3 ≥ 2(yx^3 + zy^3 + xz^3)

[quote="eunice914"]$ x, y, z \in \mathbb R$
prove that  $ x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3\plus{} yz^3 \plus{} zx^3 \geq2(yx^3 \plus{} zy^3 \plus{} xz^3)$[/quote]
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3\plus{} yz^3 \plus{} zx^3 \minus{}2(yx^3 \plus{} zy^3 \plus{} xz^3)\equal{}\frac{1}{2}\sum_{cyc}(x^2\minus{}y^2\minus{}xy\plus{}yz)^2.$ :wink:

[quote="eunice914"]x, y, z ∈ R
prove that  $ x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3\plus{} yz^3 \plus{} zx^3 \ge\ 2(yx^3 \plus{} zy^3 \plus{} xz^3)$[/quote]
Assum $ c\equal{}max${$ a;b;c$}
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3\plus{} yz^3 \plus{} zx^3 \ge\ 2(yx^3 \plus{} zy^3 \plus{} xz^3)$
$ \leftrightarrow [(a\plus{}b)^2\plus{}c^2\plus{}ab\plus{}c^2\minus{}2ac\minus{}2bc](a\minus{}b)^2\plus{}[ab\plus{}(a\plus{}c)(b\plus{}c)\plus{}2ab\plus{}bc\minus{}a^2\minus{}2a^2\minus{}2ac](a\minus{}c)(b\minus{}c) \ge\ 0$
We have done! :wink:

[quote="arqady"][quote="onlylove_math"][quote="eunice914"]x, y, z ∈ R
prove that  $ x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3 \plus{} yz^3 \plus{} zx^3 \ge\ 2(yx^3 \plus{} zy^3 \plus{} xz^3)$[/quote]
Assum $ c \equal{} max${$ a;b;c$}
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3 \plus{} yz^3 \plus{} zx^3 \ge\ 2(yx^3 \plus{} zy^3 \plus{} xz^3)$
$ \leftrightarrow [(a \plus{} b)^2 \plus{} c^2 \plus{} ab \plus{} c^2 \minus{} 2ac \minus{} 2bc](a \minus{} b)^2 \plus{} [ab \plus{} (a \plus{} c)(b \plus{} c) \plus{} 2ab \plus{} bc \minus{} a^2 \minus{} 2a^2 \minus{} 2ac](a \minus{} c)(b \minus{} c) \ge\ 0$
We have done! :wink:[/quote]
Why done? I don't see it. :maybe: Explain please. Thx![/quote]
Sorry arqady, With first proof i don't sure $ ab \plus{} (a \plus{} c)(b \plus{} c) \plus{} 2ab \plus{} bc \minus{} a^2 \minus{} 2a^2 \minus{} 2ac \ge\ 0$
I think second proof 'll true  :)
Assum $ c \equal{} min${$ a;b;c$}
$ Ineq \leftrightarrow [(a \plus{} b)^2 \plus{} c^2 \plus{} ac \plus{} bc \minus{} 2ab \minus{} 2c^2](a \minus{} b)^2 \plus{} [ab \plus{} (a \plus{} c)(b \plus{} c) \plus{} b^2 \plus{} ab \minus{} 2(ab \plus{} bc \plus{} ca \minus{} b^2)](a \minus{} c)(b \minus{} c) \ge\ 0$
$ S_a \equal{} (a \plus{} b)^2 \plus{} c^2 \plus{} ac \plus{} bc \minus{} 2ab \minus{} 2c^2 \equal{} a^2 \plus{} b^2 \plus{} ac \plus{} bc \minus{} 2c^2 \ge\ 0$ by $ c \equal{} min${$ a;b;c$}
$ S_b \equal{} ab \plus{} (a \plus{} c)(b \plus{} c) \plus{} b^2 \plus{} ab \minus{} 2(ab \plus{} bc \plus{} ca \minus{} b^2) \equal{} ab \plus{} c^2 \plus{} 3b^2 \minus{} bc \minus{} ca \ge\ 0$by $ c \equal{} min${$ a;b;c$}
We have done!

[quote="eunice914"]$ x, y, z \in \mathbb R$
prove that  $ x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3 \plus{} yz^3 \plus{} zx^3 \geq2\,(yx^3 \plus{} zy^3 \plus{} xz^3)$[/quote]

\[ {x^4 \plus{} y^4 \plus{} z^4 \plus{} xy^3 \plus{} yz^3 \plus{} zx^3 \minus{} 2\,(yx^3 \plus{} zy^3 \plus{} xz^3) \equal{} \frac {1}{4}\left( {x}^{2}\minus{}xy\plus{}2\,xz\plus{}{y}^{2}\minus{}yz\minus{}2\,{z}^{2} \right) ^{2}\plus{}\frac{3}{4}\left( {x}^{2}\minus{}xy\minus{}{y}^{2}\plus{}yz \right) ^{2}}\]

Inequality becomes equality if $ x\equal{}y\equal{}z$ or if $ x: y: z\equal{}x_0: y_0: z_0$, where

\[ x_0\equal{}1\plus{}2\cos{\frac{2\pi}{9}}\]
\[ y_0\equal{}1\plus{}2\cos{\frac{4\pi}{9}}\]
\[ z_0\equal{}1\plus{}2\cos{\frac{8\pi}{9}}\]

See also: http://www.mathlinks.ro/viewtopic.php?p=1167938#1167938 and other messages in this thread.

Sergey Markelov

[quote="eunice914"]For $ x, y, z \in\mathbb R$ prove that  $ x^4 \plus{} y^4 \plus{} z^4 \plus{} yz^3 \plus{} zx^3 \plus{} xy^3 \geq 2\left(y^3z \plus{} z^3x \plus{} x^3y\right).$[/quote]Very nice Vasc type problem.  :!:

$ x^6 \plus{} y^6 \plus{} z^6 \plus{} k\left(yz^5 \plus{} zx^5 \plus{} xy^5\right)\geq (k \plus{} 1)\left( y^5z \plus{} z^5x \plus{} x^5y\right)$

holds for all real $ x,y,z$ if and only if $ \minus{} 1 \minus{} k_1 \equal{} k_2\leq k\leq k_1 \equal{} 0.54210\cdots,$

where $ k_1$ and $ k_2$ are two real roots (the greatest and the least respectively) of the following irreducible polynomial over $ \mathbb{Q}:$

$ 12800k^{16} \plus{} 102400k^{15} \plus{} 404900k^{14} \plus{} 1042300k^{13} \plus{} 2016545k^{12} \plus{} 3208570k^{11}$

$ \plus{} 4056892k^{10} \plus{} 3511385k^9 \plus{} 1362730k^8 \minus{} 943720k^7 \minus{} 1614990k^6 \minus{} 669733k^5$

$ \plus{} 425390k^4 \plus{} 700260k^3 \plus{} 209650k^2 \minus{} 77545k \minus{} 121358;$

with equality if and only if $ k \equal{} k_1,x \equal{} t_1 \equal{} 0.856455\cdots,y \equal{} t_3 \equal{} 0.537388\cdots,z \equal{} t_4 \equal{} \minus{} 0.393843\cdots,$

and in the cyclic permutations of this case, where $ t_1,t_3$ and $ t_4$ are three real roots (the greatest, the third and the least respectively)

of the following irreducible polynomial over $ \mathbb{Q}:$

$ 21952t^{24} \minus{} 175616t^{23} \plus{} 685188t^{22} \minus{} 1747269t^{21} \plus{} 3285247t^{20} \minus{} 4893476t^{19}$

$ \plus{} 6109383t^{18} \minus{} 6699849t^{17} \plus{} 6653722t^{16} \minus{} 6040686t^{15} \plus{} 4988323t^{14} \minus{} 3717434t^{13}$

$ \plus{} 2484152t^{12} \minus{} 1469086t^{11} \plus{} 740898t^{10} \minus{} 292700t^{9} \plus{} 71470t^{8} \plus{} 4820t^{7} \minus{} 15290t^{6}$

$ \plus{} 8725t^{5} \minus{} 3136t^{4} \plus{} 798t^{3} \minus{} 144t^{2} \plus{} 17t \minus{} 1.$


LaTeX
ratio
geometry
rectangle
3D geometry
pyramid
quadratics

[quote="mysmartmouth"]I am making a latex solutions document for a team member, and I need nice solutions to these.  Any help would be appreciated.

[u]1994 Sprint[/u]
6) The numbers from 1 through 8 are seperated into two sets, A and B. The numbers in A are multiplied together to get A. The numbers in B are multiplied together to get b. The larger of the two numbers a and b is written down. What is the smallest possible number that can be written down using this procedure?

10) The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length. If the width is 2, how many units are in the length of the diagonal?

[u]1994 Team[/u]
5) Four different digits are selected at random from the digits 1 through 9.  $S$ is the sum of all possible four-digit numbers that can be created by using these four digits.  $F$ is the greatest common factor of all such terms.  Find F.

6) A geometric solid is created by rotating the segment $2x+3y=6$ in the first quadrant around the x-axis.  What is the ratio of the surface area to volume of this solid?  Express your answer as a common fraction in simplest radical form.

10) A plane parallel to the base of a square pyramid divides the volume into two equal amounts.  If the dimensions of the pyramid are six inches for an edge of the base and six inches for the height, how many inches are in the distance from the plane to the base of the pyramid?  Express your answer in simplest radical form.[/quote]
[hide="10"] 2, 2x and 2x^2 are the sides. pythagorean gets 4+4x^2=4x^4
x^4-x^2-1=0. sub x^2 with u
u^2-u-1=0
u=(1+sq(5))/2=x^2
Double that to get the hypotenuse
1+sq(5)[/hide]

1994 Sprint #4
[hide]This a summary from the book:
You want the closest number to sqrt(40320) which would be around 200.
The book says for you to check the prime factorization for each number starting with 201 (because the squareroot is a little larger than 200).

Edit: forgot to write answer:210

It might not be much help...[/hide]

1994 Team #5
[hide]First look at the number of possibilities there are if the first number is one:
There are 3*2*1=6 possibilities.  So each number of the four numbers will show up 6 times in thousands place, the hundreds place, the tens place, and the ones place.  That is 1111*6=6666.[/hide]

1994 Team #6
[hide]3y=-2x+6
y=-2/3x+2

So you can see the y intercept is 2 and the x intercept is 3.  The solid is a cone.  The volume is 2^2*3/3=4pi.  The surface area is 4pi+(pi)2sqrt(13).  The formula for surface area is (pi)r^2+(pi)r(slant length) (I hope).  Since pi can be factored out of both sides we divide 4/(4+2srt(13))=
4(4-2sqrt(13))/(16-52)=(16-8srt(13))/(-36)=(2(sqrt(13)-2))/9

I think that is the general idea as long as I didn't mess up somewhere.[/hide]

1994 Team 10
[hide]you can see how 1/sqrt^3(2) to the third power is 1/2.  So that means the height, width, and length of the piece that still looks like a pyramid would have all its sides multiplied by 1/sqrt^3(2).  How far it is away from the base is (1-1/sqrt^3(2))times 6(the height.  That is 6-6/sqrt^3(2)=6-3sqrt^2/3(2)

edit: 6-3sqrt^3/2(2)

I think that is the general idea as long as I didn't mess up somewhere.[/hide]

[quote="ehehheehee"]1994 Sprint #4
[hide]This a summary from the book:
You want the closest number to sqrt(40320) which would be around 200.
The book says for you to check the prime factorization for each number starting with 201 (because the squareroot is a little larger than 200).

Edit: forgot to write answer:210

It might not be much help...[/hide]

1994 Team #5
[hide]First look at the number of possibilities there are if the first number is one:
There are $3*2*1=6$ possibilities.  So each number of the four numbers will show up 6 times in thousands place, the hundreds place, the tens place, and the ones place.  That is $1111*6=6666$.[/hide]

1994 Team #6
[hide]$3y=-2x+6$
$y=-2/3x+2$

So you can see the y intercept is 2 and the x intercept is 3.  The solid is a cone.  The volume is $2^2*3/3=4\pi$.  The surface area is $4\pi+(\pi)2sqrt(13)$.  The formula for surface area is $(\pi)r^2+(\pi)r(slant length)$ (I hope).  Since pi can be factored out of both sides we divide $4/(4+2srt(13))= 4(4-2sqrt(13))/(16-52)=(16-8srt(13))/(-36)=(2(sqrt(13)-2))/9$

I think that is the general idea as long as I didn't mess up somewhere.[/hide]

1994 Team 10
[hide]you can see how 1/sqrt^3(2) to the third power is 1/2.  So that means the height, width, and length of the piece that still looks like a pyramid would have all its sides multiplied by 1/sqrt^3(2).  How far it is away from the base is (1-1/sqrt^3(2))times 6(the height.  That is 6-6/sqrt^3(2)=6-3sqrt^2/3(2)

edit: 6-3sqrt^3/2(2)

I think that is the general idea as long as I didn't mess up somewhere.[/hide][/quote]

Hahaha you see why I seemingly quoted nothing.  I meant to only hit preview, but I guess I hit reply.  Latex would be nice.  I was converting to latex, but got lazy with the square roots and then quit...

[hide="6 sprint"]We want A and B to be as close to each other as possible. $8!=40320$ implies that A and B are each approximately 200. Prime factorizing the numbers starting from 201:
201=3*67
202=2*101
203=
204=2^2*3*17
205=5*41
206=2*103
207=3^2*23
208=2^4*13
209=
$\boxed{210}$=2*3*5*7[/hide]
[hide="10 sprint"]Letting x be the length of the diagonal, we have
$\frac{x}{\sqrt{x^2-4}} =\frac{\sqrt{x^2-4}}{2}$

$\implies 2x=x^2-4\implies x^2-2x-4=0$.
Using the quadratic formula, and taking the positive root, we get $x=\boxed{1+\sqrt{5}}$.[/hide]
[hide=" 5 team"]Taking any 4 distinct numbers a,b,c,d, and for all of the possible 4-digit number arrangements of them, each letter appears 6 times in each place value, so the sum of all the 4-digit number arrangements for a,b,c,d is $(a+b+c+d)(6000+600+60+6)=(a+b+c+d)(6666)$ and we see that $\boxed{6666}$ is our answer.[/hide]
[hide=" 6 team"]The line has y-intercept 2, and x-intercept 3. The figure is a cone.
The volume=$4\pi$.
The slant height is $\sqrt{13}$, and the surface area is $4\pi+2\pi\sqrt{13}$.
The ratio is $\frac{4\pi+2\pi\sqrt{13}}{4\pi}=\boxed{\frac{2+\sqrt{13}}{2}}$(?)[/hide]
[hide=" 10 team"]The volume of the pyramid is $6^3/3=72$, thus each of the resulting figures after the cut have a volume of 36. The cut off pyramid is similar to the original one, and has a height equal to its base edges: it has a volume of $x^3/3=36$, where x is the height. Solving for x, and subtracting it from the original pyramid's height, we get $\boxed{6-3\sqrt[3]{4}}$ as our distance from the base to the plane.[/hide]

[quote="mysmartmouth"][quote="ehehheehee"]1994 Sprint #4
[hide]This a summary from the book:
You want the closest number to $sqrt(40320)$ which would be around $200$.
The book says for you to check the prime factorization for each number starting with $201$ (because the squareroot is a little larger than $200$).

Edit: forgot to write answer:$\displaystyle{\boxed{210}}$.

It might not be much help...[/hide]

1994 Team #5
[hide]First look at the number of possibilities there are if the first number is one:
There are $3*2*1=6$ possibilities.  So each number of the four numbers will show up 6 times in thousands place, the hundreds place, the tens place, and the ones place.  That is $1111*6=6666$.[/hide]

1994 Team #6
[hide]$3y=-2x+6$
$y=-2/3x+2$

So you can see the y intercept is 2 and the x intercept is 3.  The solid is a cone.  The volume is $2^2*3/3=4\pi$.  The surface area is $4\pi+(\pi)2sqrt(13)$.  The formula for surface area is $(\pi)r^2+(\pi)r(slant length)$ (I hope).  Since pi can be factored out of both sides we divide $4/(4+2srt(13))= 4(4-2sqrt(13))/(16-52)=(16-8srt(13))/(-36)=(2(sqrt(13)-2))/9$

I think that is the general idea as long as I didn't mess up somewhere.[/hide]

1994 Team 10
[hide]you can see how $1/sqrt^3(2)$ to the third power is 1/2.  So that means the height, width, and length of the piece that still looks like a pyramid would have all its sides multiplied by $1/sqrt^3(2)$.  How far it is away from the base is $(1-1/sqrt^3(2))times 6(the height)$.  That is $6-6/sqrt^3(2)=6-3sqrt^(2/3)(2)$

edit: $6-3sqrt^(3/2)(2)$

I think that is the general idea as long as I didn't mess up somewhere.[/hide][/quote][/quote]

Easiest way to learn latex: quote people who use it, I can't find the way to to squareroot by fractional powers oh well, just find somebody that made a fraction

Third roots are made by \sqrt[3]{x}->$\sqrt[3]{x}$.


1. Bridge is played with 4 players and an ordinary deck of 52 cards. Each player begins with a hand of 13 cards. In how many ways can a bridge game start?

2. There are n identical sticks lined up in a row and k of them are chosen.
a) how many choices are there?
b) how many choices are there if no two of the chosen sticks can be consecutive?
c) how many choices are there if there must be at least j sticks between each pair of chosen sticks?

For the first one I assume that it will be some variation on the n!, but its the setup of the question that confuses me. As for the second one any suggestions at how to solve it would help.

1:

52 cards are dealt out in order, then split into 4 partitions of 13. But each partition has 13! symettries.

N = (52!)/(13!13!13!13!)

2:

Let F(n,k,j) be the number of ways of choosing k sticks from n sticks in a line, so that there are at least j sticks between each stick chosen.

If the first stick chosen is stick 1, the next j sticks cannot be chosen, so k-1 sticks must be chosen from the remaining n-j-1 sticks. Similarly, if the first stick chosen is stick 2, k-1 sticks must be chosen from the n-j-2 sticks to the right of it, etc...

So we get a recursive relationship: F(n,k,j) = F(n-j-2,k-1,j) + F(n-j-3,k-1,j) + F(n-j-4,k-1,j) + F(n-j-5,k-1,j) ... 1
But we know that F(n,1,j) = n.

These conditions are satisfied by $F(n,k,j)\ =\ \binom{n+(1-k)j}{k}$ (consider pascals triangle).

The first two parts of the question are special cases of this result.

I think it may be possible to show the result more neatly by demonstrating a direct equivalence between the choice $\binom{n+(1-k)j}{k}$ and the initial problem, but I can't see a good way to express it.

This question can be 'solved' by just working out values for different n,k when j = 1, then again when j = 2, and looking at the sequences you get - they are recognisable as those from pascals triangle. But this isn't a reliable method!  :D


Triangle ABC is circumscribed within a circle of radius R. Prove that the distance between the circumcentre and orthocentre of triangle ABC is always less than 3R.

try to prove that:

$ OH^2\equal{}9R^2\minus{}(a^2\plus{}b^2\plus{}c^2)$

sorry,if i find time,i wil post the proof :wink:

Call G the centroid of ABC and K the projection of O on BC, then for the stewart theorem we have:

$ AP \cdot (GO^2 \plus{} AG \cdot GP) \equal{} AG \cdot PO^2 \plus{} GP \cdot OA^2$

from wich we can find $ GO^2$ and then $ OH^2$.


If a midpoint convex function is defined at all real numbers, is it convex? And if it is, is it differentiable at all real numbers?

Thanks in advance.

in general, no, a midpoint convex function need not be convex. however, if the function is continuous, then midpoint convexity implies convexity.


trigonometry
LaTeX
inequalities proposed
inequalities

Let a,b,c be positive numbers such that:a^2+b^2+c^2+abc=4.Prove that;
ab+bc+ca+4≥abc+2(a+b+c).
Good luck. :)

Here is my solution:
We know $ abc \le 1$ from AM-GM.
We have to show
$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)^2 \ge 4(a \plus{} b \plus{} c)^2$
From the given, $ (a \plus{} b \plus{} c)^2 \equal{} 4 \minus{} abc \plus{} 2ab \plus{} 2bc \plus{} 2ca$ and $ 4 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} abc$
Substituting and expanding, we get
$ \sum a^4 \plus{} 3 \sum a^2b^2 \plus{} 2 \sum a^2bc \plus{} \sum a^3bc \plus{}a^2b^2c^2 \ge 4 \sum a^2 \plus{} 2 \sum ab^2c^2 \plus{} 4abc$
Subtract $ 4\sum a^2 \equal{} (a^2 \plus{} b^2 \plus{} c^2 \plus{} abc)(a^2 \plus{} b^2 \plus{} c^2)$ from both sides.
$ \sum a^2b^2 \plus{} 2 \sum a^2bc \plus{} a^2b^2c^2 \ge 4abc \plus{} 2 \sum ab^2c^2$
$ \sum a^2b^2 \plus{} 2 \sum a^2bc \plus{} 3a^2b^2c^2 \minus{} 2 a^2b^2c^2 \ge 4abc \plus{} 2 \sum ab^2c^2$
By AM-GM, $ a^2b^2 \plus{} a^2b^2c^2 \ge 2a^2b^2c$, so we have to show
$ 2ab^2c \plus{} 2a^2bc \plus{} 2abc^2 \ge 4abc \plus{} 2a^2b^2c^2$
$ a \plus{} b \plus{} c \ge 2 \plus{} abc$
Squaring, we have to show
$ 2ab \plus{} 2bc \plus{} 2ca \ge 5abc \plus{} a^2b^2c^2$
Since $ abc \le 1$, $ 6a^{2/3}b^{2/3}c^{2/3} \ge 5abc \plus{} a^2b^2c^2$
Then, it is true by Muirhead.

It is very long. Is there a shorter solution?

[quote="quykhtn-qa1"][quote="dgreenb801"]After I long bash, I have
$ a \plus{} b \plus{} c \ge abc \plus{} 2$
Is this true?[/quote].
It must be :(a+b+c-2)^2≥abc.
But i think your way is  maletic :)
i have a nice solution :lol:[/quote]
Where is the error in my solution, and can you show yours?

[quote="cosinator"][quote="quykhtn-qa1"](1-cos A)(1-cosB)(1-cos C)≥cos A cosB cosC
[/quote]
Could you please prove this?[/quote]
OK. use cosin this inequaliti become:
$ (a^2 - (b - c)^2)(b^2 - (c - a)^2)(c^2 - (a - b)^2)\geq \(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(c^2 + a^2 - b^2)$ (*).
I will prove (*) when $ a^2,b^2,c^2$ are the three lengths of sides of a triangle.
I will prove
 $ ((a^2 - (b - c)^2))^2\geq \(a^2 + b^2 - c^2)(a^2 - b^2 + c^2) < = > (b - c)^2 (b^2 + c^2 - a^2)\geq\ 0$ .
Which is obvious true.:)

[quote="quykhtn-qa1"][quote="dgreenb801"]Here is my solution:
We know $ abc \le 1$ from AM-GM.
We have to show
$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)^2 \ge 4(a \plus{} b \plus{} c)^2$
From the given, $ (a \plus{} b \plus{} c)^2 \equal{} 4 \minus{} abc \plus{} 2ab \plus{} 2bc \plus{} 2ca$ and $ 4 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} abc$
Substituting and expanding, we get
$ \sum a^4 \plus{} 3 \sum a^2b^2 \plus{} 2 \sum a^2bc \plus{} \sum a^3bc \plus{} a^2b^2c^2 \ge 4 \sum a^2 \plus{} 2 \sum ab^2c^2 \plus{} 4abc$
Subtract $ 4\sum a^2 \equal{} (a^2 \plus{} b^2 \plus{} c^2 \plus{} abc)(a^2 \plus{} b^2 \plus{} c^2)$ from both sides.
$ \sum a^2b^2 \plus{} 2 \sum a^2bc \plus{} a^2b^2c^2 \ge 4abc \plus{} 2 \sum ab^2c^2$
$ \sum a^2b^2 \plus{} 2 \sum a^2bc \plus{} 3a^2b^2c^2 \minus{} 2 a^2b^2c^2 \ge 4abc \plus{} 2 \sum ab^2c^2$
By AM-GM, $ a^2b^2 \plus{} a^2b^2c^2 \ge 2a^2b^2c$, so we have to show
$ 2ab^2c \plus{} 2a^2bc \plus{} 2abc^2 \ge 4abc \plus{} 2a^2b^2c^2$
$ a \plus{} b \plus{} c \ge 2 \plus{} abc$
Squaring, we have to show
$ 2ab \plus{} 2bc \plus{} 2ca \ge 5abc \plus{} a^2b^2c^2$
Since $ abc \le 1$, $ 6a^{2/3}b^{2/3}c^{2/3} \ge 5abc \plus{} a^2b^2c^2$
Then, it is true by Muirhead.

It is very long. Is there a shorter solution?[/quote]
 I have a nice solution. :)
since a^2+b^2+c^2+abc=4.Let a=2cos A,b=2cos B,c=2cos C.
This inequaliti become:
(1-cos A)(1-cosB)(1-cos C)≥cos A cosB cosC
It very easy and nice. :lol:[/quote]
i think my solution is shorter than dgreenb801

$ \sum\frac{4}{4\minus{}a^2}\ge5\minus{}abc$

$ \Leftrightarrow\sum\frac{4}{4\minus{}a^2}\ge1\plus{}\sum a^2$

Making substitution $ a\equal{}2\cos A$, etc yields

$ \sum\frac{1}{\sin^2A}\ge1\plus{}\sum 4\cos^2A$

$ \Leftrightarrow\sum\left(\frac{1}{\sin^2A}\minus{}4\cos^2A\right)\ge1$

which is Jensen's.

[quote="cosinator"]$ \sum\frac {4}{4 \minus{} a^2}\ge5 \minus{} abc$

$ \Leftrightarrow\sum\frac {4}{4 \minus{} a^2}\ge1 \plus{} \sum a^2$

Making substitution $ a \equal{} 2\cos A$, etc yields

$ \sum\frac {1}{\sin^2A}\ge1 \plus{} \sum 4\cos^2A$

$ \Leftrightarrow\sum\left(\frac {1}{\sin^2A} \minus{} 4\cos^2A\right)\ge1$

which is Jensen's.[/quote]
You can write more appearly?

[quote="cosinator"]What do you need more clarification on?  Please isolate the step you don't understand so that I don't have to overexplain everything.[/quote]
Sorry :oops: your proof can solve this problems :blush: ???
Find the best k constant such that;
$ \frac{4}{4\minus{}a^2}\plus{}\frac{4}{4\minus{}b^2}\plus{}\frac{4}{4\minus{}c^2}\plus{}kabc\geq\ 4\plus{}k$
 :P

cosinator :you can solve this problem? :)  :)  :)

[quote="quykhtn-qa1"][quote="cosinator"]What do you need more clarification on?  Please isolate the step you don't understand so that I don't have to overexplain everything.[/quote]
Sorry :oops: your proof can solve this problems :blush: ???
Find the best k constant such that;
$ \frac {4}{4 \minus{} a^2} \plus{} \frac {4}{4 \minus{} b^2} \plus{} \frac {4}{4 \minus{} c^2} \plus{} kabc\geq\ 4 \plus{} k$
 :P[/quote]
Nobody interested in this problem?

[quote="nguoivn"][quote="quykhtn-qa1"]
Nobody interested in this problem?[/quote]
See solution here: http://www.mathlinks.ro/Forum/viewtopic.php?t=258051[/quote]
I know.but i asked cosinator :)

[quote=quykhtn-qa1]With$ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4,a,b,c>0$.prove that:
$ \frac{4}{4\minus{}a^2}\plus{}\frac{4}{4\minus{}b^2}\plus{}\frac{4}{4\minus{}c^2}\plus{}abc\geq\ 5$[/quote]

[img]http://s12.sinaimg.cn/middle/0018zOAxgy71zVzaTOz1b&690[/img]


"Revisited" because this was discussed in a previous thread but that thread died.

The Mathnerd314 Triangle Function for a positive integer x is defined as:
Suppose you have infinitely many tiles shaped as equilateral triangles. (Who would make that type of tile, I don't know.) A side of a tile is defined as one unit. You decide to use these tiles to make a tesselation in the shape of an equilateral triangle, with a side of x units. The Mathnerd314 Triangle Function for x is the total number of triangles formed by the edges of these tiles in the tesselation. That is, one counts all of the triangles of side 1 unit, all triangles formed by 4 tiles with a side length of 2 units, etc.

What is a polynomial representation for this function? [b]PROVE[/b] your answer.

I posted this in Intermediate Topics because I didn't know where it should go, and I didn't want to be blamed for posting problems that were too hard repeatedly.

I have been working on this problem for almost two years. I must be doing something wrong.

Pleasee explain the function in some easier way. I was not able to understand what you want to say.

You draw a big equilateral triangle with a bunch of little equilateral triangles in it. How many total equilateral triangles are formed?

For example, if you have a triangle formed by four small triangles (that is, two triangles to a side), there are the four small triangles plus the big triangle. That makes five triangles.

mathnerd314, you were being forgotten by everyone these days.
Well I was trying your problem all these days in the school. And gosch I found it today. So, here is the answer to your problem:
For [b]even n[/b],
No. of triangles = $\frac{n(n+2)(2n+1)}{8}$
For [b]odd n,[/b]
No. of triangles = $\frac{(n+1)(2n^2+3n-1)}{8}$

I'm sorry I am not providing the solution now because it would take time to write it. Also if you can send me a picture file containing you equilateral triangle with 6 or 5 sides than I could answer quickly.
I'll try to post the solution within next 2 days.

I'd like to know why anyone didn't try this problem?
Am I a fool to try it alone. Please comment on this.

I posted this while I was away at camp. When I got back to the forum, I didn't read very carefully because everyone else was waiting for the computer (cough, impatiently, cough).

[quote="Altheman"]think about a way to do it with a recursive statement, then find the explicit form, i.e. let f(n) be the number for n little triangles, then if you add 1 row, how many do you add, then take this from recursive to explicit and prove it with induction[/quote]

Well, Altheman, nice trick. I spent two days solving that problem and you will just use induction to prove that it is correct and say that you have done it.
Its not at all done.
If you really want to do post your solution on the net. :mad:


geometry
inequalities
algebra unsolved
algebra

let the $x,x_i,a_i$s ($1\leq  i \leq n$) be positive numbers.
prove:
1. $\prod_{i=1}^n (1+a_i) \geq (1+\sqrt[n] {\prod_{i=1}^n a_i})^n $
prove or disporve
2.$\prod_{i=1}^n (x+a_i) \geq (x+\sqrt[n]{ \prod_{i=1}^n a_i})^n $
3.${\prod_{i=1}^n (x_i+a_i) \geq (\sqrt[n]{ \prod_{i=1}^n x_i}+\sqrt[n]{ \prod_{i=1}^n a_i})^n} $

$\prod_i (1+a_i)=\sum 1+\sum_i a_i+\sum_{i<j}a_ia_j+\ldots+\prod_i a_i$

By AM-GM

$\sum_{a<b<\ldots<n}a_i\geq \binom{k}{n}\sqrt[\binom{k}{n}]{\prod_i a_i^{\binom {k}{k-1}}}=\binom{k}{n}\prod_i a_i^{\frac{n}{k-n+1}}$

Sorry, I have to run, I'll post this and come back to it later.

If the first problem is true, then it is easily seen that the 2nd and 3rd problems are true. Since x, x_i, a_i are all positive and no more conditions are needed, divide both side by x^n (in 2nd problem) or x_1*x_2*...x_n (in 3rd problem), then change the problems to the first one.

could you post your solution[b]RAFAL[/b], by alternatively form the generalised Holder's inequality?

and anyone have any idea about my suggestion?

zhaoli , as you have said we can  change the right side of the second and the third one to the first one (by deviding it by your way) , but what about the left sides ?
dividing them dont give the needed result


If the equation:

$f(\frac{x-3}{x+1}) + f(\frac{3+x}{1-x}) = x$

holds true for all real x but $\pm 1$, find $f(x)$.

This one is from Korea 1999.
You may find it, with its solution, as exercice #13 here :
http://www.animath.fr/cours/eqfonc.pdf

Pierre.

[quote="pbornsztein"] This one is from Korea 1999.
You may find it, with its solution, as exercice #13 here :
http://www.animath.fr/cours/eqfonc.pdf
[/quote]

 :cool:  :P That's false , because of the french exception  :P  :cool:


[quote="BlnGcc"]Happy New Year! Good luck to everyone![/quote]

_________
dear BlnGcc has given a good contribution to the evolution of the inequalities...
 :)

indeed very nice identity, thanks @BlnGcc for uploading  :)


function
limit
algebra
domain
calculus
calculus computations

Show that the function is continuous on the given interval.

f(x) = x(sqrt{16 - x^2}), [-4, 4]

The function is not continuous when $16-x^2<0\rightarrow x<-4 $ and $x>4$, so the function is only defined in $[-4,4]$.

... so now we know it is continuous at the ends of its domain ...

As an expression of elementary functions, it is continuous on its definition domain. Period.


$ x^{2}\equal{}AABBCC$, x is an integer and A,B, and C are digits with A being non-zero.  If S is the sum of all possible $ x$, and $ B\equal{}4,$ what is $ S$?

I know the really slow way to do this is to actually right the number like 100000A+10000A+1000B.....so forth and you find that the number is divisible by 121..and then there are like 10 cases to consider for (A,C) but using huge numbers, that takes forever to do...since you also have to consider that when you divide that by 121, itll have to be a perfect square.

Is there an easier way to do this?

Well, I don't know if this is a good way, but it's slightly faster.
If you divide AA44CC by 121, it's like dividing by 11 twice.
$ \frac{AA44CC}{11}\equal{}A040C$
So, by the divisibility rule for 11, A+C+4=11 or 22, or A+C=7 or 18.
If A+C=7, there are only 7 cases, so you can check them all using long division, since it's only a 5-digit number divided by 11.
Also, you can eliminate (4,3) and (5,2) for (A,C) since no perfect squares end in 3 or 2.
It turns out that the only pair for (A,C) that works is (7,0), giving 70400/11=6400, which is 80 squared.
Since we divided by 121 to get 80 squared, we multiply the 80 by 11 to get 880, which is the answer (I think).

I guess this is slightly faster, but there probably is a better way.


haha i love these easy problems
they're like warm-ups

[hide]101[/hide]

lol are these even math questions; they seem to test calculation more than logic  :D

Remember, this website is called [i]The Art of Problem Solving[/i], not the art of speedy calculations,  :P

When I see these kinds of easy problems, I think there are some tricks involved. I can't see any in this one, and it's labeled an easy problem.

when i saw the title i thought it might be a little harder . I mean this is so easy its funny :rotfl:  :wow:  :wow:  :rotfl:  :rotfl:  :wow: [hide]101[/hide]

[hide="Answer"]$101$

EDiT: Sorry forgot LaTeX[/hide]

[quote="math92"]what number is 900 less than 1001?[/quote]

[hide]$1001-900=101$[/hide]

[quote="math92"]what number is 900 less than 1001?[/quote]

[hide="Hidden Text!!!"]
[hide="Hidden Text!!!"]
[hide="Hidden Text!!!"]

[color=white]1001-900=101(my ssiter can answer this and she is in first grade!!!)[/color]
[/hide][/hide][/hide]

that gets annoying after a while.  and my sister is in 6th grade and can solve it. :D

[hide]Stupid me it has to be even so it's 20...[/hide]

[quote="math92"]try this one:
The average of four consecutive even integers is 17.
Find the largest of the four integers.[/quote]

[hide=""Solution""]Let $n$ be the largest integer

$\frac{(n-6)+(n-4)+(n-2)+n}{4}=17$
$n-3=17$
$n=20$
[/hide]

[hide]101[/hide]
thats funny...
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:   :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

please don't put all those smileys in a post.

[hide]101[/hide] that was easy :lol: yet fun!

enough warmups lets get to the good stuff  :D

[hide]7+4x/2x=9^2
ok so maybe i was jkin  :P [/hide]
[/hide]

:wow: [hide]20 :wow: . The average is 17 so the third is 18 and the fourth is ${\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{20}}}}}}}}}}}}}}}}}$ :D [/hide]

[quote="nonie"][quote="math92"]okay that problem was too easy how bout this one (it's not hard either)

solve for x:

$4x-4=6x+2$[/quote]
[quote="math92"]okay that problem was too easy how bout this one (it's not hard either)

[hide]-4=2x+2
-6=2x x=-3[/hide][/quote][/quote][hide]

[hide]-3[/hide][/hide]

i really don't see how these problems are funny. :huh:

daermon: I have noticed this in almost all of your 40 posts, please provide a solution and please do not bump up topics that are almost a year old that have had many answers already.


summer program
Mathcamp
email
probability
geometry
3D geometry
function

Yay. Acceptance emails (or at least some of them) have been sent out.

Who's going to MathCamp or is at least potentially going, with decision pending from scholarships?

*raise hand*  :lol:

do they send to people who are rejected???

I'm in! :):)

And yes, I think they will send emails to people who are rejected.

1:30 AM my time--so tired, and then I get this e-mail saying "Congratulations on your acceptance!"

Now I'm happy :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)

Oh my gosh, that qualifying quiz tingy drained the heck out of me to get about 6 problems cause I only found out in mid-April about this and crammed on all the problems, and for me they were so so so so so hard--well, I must've written a good personal statement or had good recs or something.

I'm very happy :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)

Yes! I'm in! OK I can finally sleep now.

*I'm in!* Looking forward to seeing you all.  :)

It seems that I'll be also passing by after the IMO ;) :P

[quote="bubala"]
Oh my gosh, that qualifying quiz tingy drained the heck out of me to get about 6 problems cause I only found out in mid-April about this and crammed on all the problems, and for me they were so so so so so hard--well, I must've written a good personal statement or had good recs or something.  [/quote]

I think 6 is probably about enough to qualify, especially considering you're an 8th grader.  Some of the six easiest (in my opinion) were rather clever.  I thought the EGG one was really quite nice, but my solution for it was pretty poorly explained :(.  I also crammed hehe.  I think I might have gotten in on personal statement/recommendations, though, considering the low quality of some of my solutions and my poor LaTeXing skills.  Well I'm just glad I made it!

Shameless advertisement: If you're coming to Mathcamp, come to Wednesdays there!

I'm on the waiting list... which means... what exactly?

I think that means they wait until some acceptees have decided not to attend Mathcamp.  Then they consider you among the other waitlisted and late folks.

[quote="bubala"]1:30 AM my time--so tired, and then I get this e-mail saying "Congratulations on your acceptance!"

Now I'm happy :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :) :)[/quote]
Congratulations!!! Very happy for you!!! :lol:

Addendum to the shameless advertisement: Come on tuesdays and fridays too :)

[quote="juggle7"]I'm in.  Need money.  Trying to fundraise and get a job.  If I go, I'll bring my juggling stuff.[/quote]
I hope you can make it!  :)

Yeah, that would be really great to meet everybody in person, not to mention all the math that goes on.

Could I know how I did on the qualifying quiz?  I answered everything and was fairly confident about my answers.

if you PM ztbb, he might tell you.  I haven't tried to figure out what I got, though.

Hey, first post, got my acceptance letter a couple weeks ago... think it'll be the first time I get to be with "those of my kind"... even private school doesn't have enough nerds...

But yea, I got the invitation from my AIME scores, and I had registered for a program at Yale, so I had to get the application in REALLY early, spend 15 hours on Saturday, told myself I wouldn't' sleep til I finished the test... let's just say the next day even trig homework made my head hurt...

I got 1-5, 6a, 7a, and 8, but 9&10 were giving me problems so I didn't finish solving them; I just submitted a few ideas.

So, anyone else flying from the northeast?  Looking forward to the five hour flight?

Congrats! Go Northeast--the Number 1 part of the country by far, except maybe Silicon Valley.  Where in the Northeast are you from?

And actually, it's a six hour flight w/ about 5.5 hours flight time going there.
It's 5 hours back.

It's always longer flying west than east--I think it's because of the jet stream or something like that...

I'm probably repeating myself now, but if so, so be it... Central Jersey, Princeton to be exact.  I'm probably the only one around here who immediately thinks Nash and Conway when talking about Princeton U, others think a good shopping center and some cafés...

My dad works in Jersey City, right across the river, but he was born and raised in Phili, so we're a fully fledged Philies-Eagles-76ers-Flyers family... How off topic am I now?  Oh, yea, travel...

Yup, it's 6 hours there, 5 back.  I hope to be constructive and get some reading done on the flights, but I have a feeling that isn't going to happen...

But a big sigh of relief that I have a connecting flight to where we are having a family reunion!  I almost had to miss the last 4 days!  Luckily I was able to convey to my parents the importance of the last days, though I might have confused them by being upset that I couldn't stay up all night and do math.  Well... at least you guys understand...

I'm going to Mathcamp too, and I'm from New York City...  I figure I'll find something to do during a 6-hour flight, I always end up passing 6 hours without much difficulty anyway...
I have to miss the last 2 days, though, because I'll have to leave on the morning of August 6.

I'll be going as well (one day after MOP ends!) and I'll probably be leaving slightly less than a week early.

Dad finally approved my going!  I need to by Conway's book on Dots and Boxes so I can compete with him.

I got Conway's book a few weeks ago. I hope I'll be able to compete with him for Dots and Boxes and possibly Phutball. I'm really looking forward to meeting him. (By the way, if you didn't already know, you can play Dots on Yahoo! which is the exact same as Dots and Boxes.)

Anyone care to enlighten those of us uninitiated to productive gaming about Conway's other games?

Hmm... Who are you? You got your username near the end of camp, so I don't think you're in the yearbook...


probability
limit
number theory
prime numbers

A real number is chosen randomly from the interval $[0,1]$.
What is the probability that there is a $3$ in its decimal expansion?

It looks like $\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+.....$, but there are probability overlaps (0.33333....)

Probability that there is no $3$ is that the first digit is not $3$ (9/10), the second digit is not $3$ (9/10), ...
So probability is $\lim_{n\rightarrow +\infty} \Big(\frac{9}{10}\Big)^n = 0$
Then probability that there is at least one $3$ in the decimal expansion of $x$ is $1$ !!! Looks strange but I'm OK with that. Is it true ?!

nope, 0.1, 0.11, 0.111, etc. have no three. :dry:

isn't it just 1/10?

intuitively you think that taking a real $x$ in $[0;1]$ there is only probability of $\frac{1}{10}$ that it has a $3$ in its decimal expansion ?!
We can't use here the formula number of possible cases divided by the number of total cases since we are not in a finite set. For an example, density of prime numbers in natural numbers is $0$ ...

Isn't it the same case with all digits 0 through 9, so 3 is just one of those digits, thus probability 1/10?

Rodolphe2005 is correct: With probability 1, a random number has a 3 in it.  Remember that in infinite probability spaces, an event can have probability 1 without always being true.

More is true: With probability 1, a random number has all 10 digits, each appearing an infinite number of times.

[quote="Ravi B"]  Remember that in infinite probability spaces, an event can have probability 1 without always being true.
[/quote]

Really? I havet o read more probability books  :roll:


hi, im having trouble understading how to do these problems, could someone help me start these off, thanks

1.)determine whether the series is convergent or divergent

sigma (((3/(n^3)^(1/6))+(1/(n^6))) n=1 to infinty

if convergent, enter the 4th partial sum to estimate the sum of series

2.)if the following sum converges, comupte its sum. if divergent find whether its inf, minus inf, or div.

sigma (3+4^n)/(4^n) n=1 to infinity

any help would be great, thanks again

1) Split up into two series which may or may not converge (you may know them as p-series, that is, series of the form $\textstyle \sum_n n^{-p}$)

2) Does the nth term tend to 0?

1.) im still having trouble in the process of finding the sum, when i try to find r they are never constant for both.

2.)what do you mean by the n term tending ti zero? do you mean n=0?

is this your question?
i)$\sum_{n=1}^{\infty} \frac{3}{(n^3)^{\frac{1}{6}}} + \frac{1}{n^6}$

ii) $\sum_{n=1}^{\infty} \frac{3+4^n}{4^n}$

The nth term tend to 0 mean that $\lim_{n \rightarrow \infty} a_n = 0$.
do you recognize that theorem ? ;)

yes thoese are the question, for the first one (n^3)^(1/6) is the sixth root of n^3 on the problem, but its all the same, so it shouldnt matter.

for the second one would i use l'hopitals rule? getting (n4^(n-1))/(n4^(n-1))? if so where would i go from there?

thanks

as you can see that $\lim_{n \rightarrow \infty} \frac{3+4^n}{4^n} = \lim_{n \rightarrow \infty} \frac{3}{4^n} +1$. Now compute it.
 if the $\sum_{n=0}^{\infty} a_n$ converges then $\lim_{n \rightarrow \infty} a_n = 0$.   :)

so for the second one would it be -infinity +1? so it diverges?

The limit of the term is used to know wether the series converges or diverges. I think you know that the limit is $1 \not= 0$. so by the last proposition in my post. we have
if $\lim_{n \rightarrow \infty } a_n \not=0$ then the series divergent (why?). So the second series  diverges.
To know the sum noticed that

$\sum_{i=0}^{\infty} 2^n$ goes to $+ \infty$,
$\sum_{i=0}^{\infty} (1-2n)3^n$ goes to $- \infty$, and
$\sum_{i=0}^{\infty} (-1)^n$ is $-1$ for odd $n$ and $0$ for even $n$.

No.  You are erronously thinking that something of the form $1/\infty=-\infty$.

In fact, $\lim_{n\to\infty} \frac{3}{4^n}=0$.

ok so the limit for it equals 1? so the sum equals 1 as well?

No, the nth term does not go to zero, so the sum diverges.  Since it is a series of positive terms, you could well say that 'it diverges to $\infty$'.

i see, i just looked over my book again and saw that theorem, thanks for your help.


linear algebra
matrix
abstract algebra
linear algebra unsolved

$A,B \in Mn(C)$ such that $Rg(AB-BA) \leq 1$.
Prove that one can trigonalise $A$ and $B$ in a same basis.

I think the more general problem was solved by Leva 1980 in a smart smart way about finite family of matrice. I am trying to find it, since it seems to me this is a particular case (of course $AB-BA$ is nilpotent).

Sorry, I was wrong, thre was supplimentary strong condition in that problem.

Phew, what a tough problem. (I think it's from Ulm. ;))

Well, it is plain that we have to separate two cases :
(1) : $Rg(AB-BA) = 0$
and
(2) : $Rg(AB-BA) = 1$.

In the first case, we have that the matrices commute, and it is not hard to derive the desired conclusion (look at the common eigenvectors).
The second case is harder, but, since I don't have much time now, I will post my solution in a few days.
(If you really want the solution for the first case too, just ask. :))

sorry for being stupid,but what's Rg? :blush:

i saw this problem as a theorem in prasolov's book,of course it is not that hard with same idea that Leva used to prove there exists common eigenvector.

hi all,
I was checking the proof of leva 1980 mentioned by harazi,then i came across a post by grobber,actually i have no prior knowledge in lie algebras,but i just tried on web and read some related material,but i got totally confused when i tried to understand his proof,sorry if this is something away from the present discussions,but i hope someone of u can clear my misconception:
------------------------------------------------------------------------------------------------------------------
grobber:
I was looking through some older topics and I found this problem. There seem to be a lot of extra conditions here Confused.

The non-trivial part of the double implication asks for a proof that if $A_iA_j,\ A_rA_s$ commute for all $i,j,r,s$ and $[A_i,A_j]$is nilpotent for all$i,j,$then$A_i$ can be trigonalized simultaneously. In fact, we can prove that the conclusion holds if$[A_i,A_j],\ [A_r,A_s]$ commute for all $i,j,r,s$:

Let $\mathcal A$be the Lie algebra generated by $A_i,\ i=\overline{1,k}$, and let
$\mathcal D\mathcal A$ be the derived ideal of$\mathcal A$, i.e. the subalgebra generated by all the commutators $[A_i,A_j],\ i,j\in\overline{1,k}$. By our hypothesis, the Lie algebra $\mathcal D\mathcal A$ is abelian and thus solvable. On the other hand,
$\mathcal A/\mathcal D\mathcal A$ is also abelian (solvable). This means that$\mathcal A$ itself is solvable, so, by Lie's Theorem, it's triangulable.
-----------------------------------------------------------------------------------------------------------------
my doubt is if as above if i take any two matrices $A,B$ and consider the lie algebra generated
then the derived algebra as per grobber is the subalgebra generated by $AB-BA$(so abelian always) and following the above reasoning,wo'nt it imply that always the subalgebra generated by $A,B$ is solvable and hence triangulable,definitely i know i am totally stumped  :?

I deleted my post in that topic. It doesn't work that way (I don't know what I was thinking :)):

You'd have to know that $A_i$ are actually a system of generators for the Lie algebra, i.e. every $[A_i,A_j]$ would have to be a linear combination of $A_k$'s, and this is not given as a hypothesis. In order for the argument to work in your problem, with two matrices $A,B$, you'd have to have $[A,B]=\alpha A+\beta B$ for some scalars $\alpha,\beta$. With this hypothesis, $A$ and $B$ can indeed be simultaneously triangulated (I believe sam-n posted this little problem some time ago, I gave a direct but longer proof, and amfulger killed it immeidately with Lie's Theorem).

ooof thanks grobber,actually i became totally confused and was not able to think anything else :)

Our teacher told us today a generatlization. Suppose $\mathcal F$ is a family of trigonlisable operators which for each $A,B\in\mathcal F$ we know $Rank (AB-BA)\leq 1$. Then all of these operators can be trigonalised in a basis.


Let $ z_1\ ,\ z_2\in \mathbb{C}^*$ such that $ |z_1\plus{}z_2|\equal{}|z_1|\plus{}|z_2|$ . Prove that $ (\exists)\alpha \in [0,\infty)$ such that $ z_1\equal{}\alpha z_2$

On one hand, $ (|z_1| \plus{} |z_2|)^2 \equal{} |z_1|^2 \plus{} |z_2|^2 \plus{} 2|z_1z_2|$. On the other hand $ |z_1\plus{}z_2|^2 \equal{} (z_1\plus{}z_2)(\overline{z_1\plus{}z_2}) \equal{}  (z_1\plus{}z_2)(\overline{z_1}\plus{}\overline{z_2}) \equal{} |z_1|^2 \plus{} |z_2|^2 \plus{} z_1\overline{z_2} \plus{} \overline{z_1}z_2$.

Therefore $ z_1\overline{z_2} \plus{} \overline{z_1}z_2 \equal{} 2|z_1z_2|$, or $ (z_1|z_2| \minus{} |z_1|z_2)^2 \equal{} 0$, i.e. $ z_1 \equal{} \alpha z_2$ with $ \alpha \equal{} \left | \frac {z_1} {z_2} \right |$. By the way, $ \alpha \in (0, \infty)$.

The equality condition for the triangle inequality states these points must be colinear with the origin.


function
limit
real analysis
real analysis unsolved

If $\ f: \mathbb{R}\to\mathbb{R}$ is a continuous at the origin function such that the limit $\ \lim_{x\to0}\frac{f(\lambda x)-f(\mu x)}{x}$ exists ($\lambda,\mu>0,\lambda\neq\mu$), prove that f is differentiable at the origin.

$0<a<b$, $\frac{f(bx)-f(ax)}{x}=L+e(x)$ where $e(x)$ tends towards $0$ with $x$.
$a=bc$ where $c=\frac{a}{b}$ and one wants to look at $\frac{f(x)-f(0)}{x}$.
By continuity of $f$ at $0$:
$f(ax)-f(0)=\sum_{k=0}^{k=\infty} f(a(c^kx))-f(a(c^{k+1}x))=\sum_{k=0}^{k=\infty} f(a(c^kx))-f(b(c^{k}x))$
Take $\epsilon>0$ and $x_0$ such that $|e(x)|<\epsilon$ when $|x|<|x_0|$. then, for such $x$,
$\frac{f(ax)-f(0)}{x}=\sum_{k=0}^{k=\infty} \frac{f(a(c^kx))-f(b(c^{k}x)}{c^kx}c^k=\sum_{k=0}^{k=\infty} (L+e(c^kx))c^k$.

If you call $D=\sum_{k=0}^{k=\infty} Lc^k$, then show that $\frac{f(ax)-f(0)}{x}-D$ tends towards $0$. So $f'(0)=\frac{L}{b-a}$, as one could have guessed.


What is the greatest common factor of 68 and 92?

ok well breaking up into primes we have 2^2 * 17 = 68
and 2^2 *23 = 92
the largest common term is 2 squared or 4.

Or you could use the Euclidean Algorithm.

$ 92 \minus{} 68 \equal{} 24$, $ 68 \minus{} 24 \equal{} 44$, $ 44 \minus{} 24 \equal{} 20$, $ 24 \minus{} 20 \equal{} 4$. So we narrowed it down to the greatest common factor of 24 and 4, which is $ \boxed{4}$.

[quote="isabella2296"]Or you could use the Euclidean Algorithm.
.[/quote]
what is that?

It's better answered in Intro To NT, last section of chapter 3, I believe. It's really easy, but hard to explain.


inequalities
calculus
rearrangement inequality

There is something I can't understand:

Let a>=c .It doesn't matter the way x,y,z are, but for some of the three configurations we will have that   ap +  cq >=aq + cp, for p,q from x,y,z
p<>q. If this doesn't happen, then the inequality is not true. if it is true, then it is obvious that one of the A,B,C will have it's numerator>=its denominator, so >=1.
Where did I use a+c=2b I don't know. :) :) :) :)

I think I have a solution.
   Let supose that max(A,B,C)<1
   That means that A<1;B<1;C<1
    So if A<1 we have ax+by+cz/az+by+cx<1
    We also know that a,b,c,x,y,z>0
    After some more calculs we will have (a-c)(x-z)<0-for A
    For B and C we will have (a-c)(y-x)<0 (a-c)(z-y)<0. Now we will have the cases 1.a>c that means that a-c>0 So in this case we shall have
    x-z<0 so x<z ;y-x<0 so y<x and z-y<0 so z<y
    From all the conditions we shall have z<y<x<z But this relation it is false. So max(A,B,C)>1
               2.a<c In this case you will have the same contradiction
               3.a=c In this case you will have max(A,B,C)=1
      I did not use the fact that a+c=2b so iam not sure that the solution is good. Hope it is!

Well both of you didn't use the fact that  a+c=2b, I think that you proved the inequality, but it seems strange, perhaps this is an extra thing, I don't know, or the author of the problem when he solved it, he had to use this, I don't know!  :?

Anyway this was given in 2001, for the 9th grade in Arges .
tschuess :D

[quote="Lagrangia"]Let $ a,b,c,x,y,z$ be positive real numbers and

$ A\equal{}\frac{ax\plus{}by\plus{}cz}{az\plus{}by\plus{}cx},B\equal{}\frac{ay\plus{}bz\plus{}cx}{ax\plus{}bz\plus{}cy},C\equal{}\frac{az\plus{}bx\plus{}cy}{ay\plus{}bx\plus{}cz}$.

Show that $ \max\{A,B,C\}\geq 1$.[/quote]
$ \max\{A,B,C\}\geq \frac{(az\plus{}by\plus{}cx)A\plus{}(ax\plus{}bz\plus{}cy)B\plus{}(ay\plus{}bx\plus{}cz)C}{(az\plus{}by\plus{}cx)\plus{}(ax\plus{}bz\plus{}cy)\plus{}(ay\plus{}bx\plus{}cz)}\equal{}1$.

[quote="Lagrangia"]Let a,b,c,x,y,z be pozitive real numbers so that a+c=2b and
A=(ax+by+cz)/(az+by+cx)
B=(ay+bz+cx)/(ax+bz+cy)
C=(az+bx+cy)/(ay+bx+cz).
Show that max(A,B,C)>=1.

cheers! :D[/quote]

If I am correct, this is quite an easy problem.

WLOG, let x>=y>=z; a>=b>=c.
Now, with rearrangement inequality, ax+by+cz >= az+by+cx, which means A >=1.

Prove that $ \forall a,b,c \geq 0$

$ \sum{\frac{{a}^{3}}{bc}}\geq\sum{\frac{{a}^{2}\plus{}{b}^{2}}{2c}}\geq\sum{a}$

thank's :oops:

[quote="cristinarm30"]Prove that $ \forall a,b,c \geq 0$

$ \sum{\frac {{a}^{3}}{bc}}\geq\sum{\frac {{a}^{2} \plus{} {b}^{2}}{2c}}\geq\sum{a}$

thank's :oops:[/quote]

This is not related to Post #1 :wink: And $ a,b,c$ can't be zero or $ \sum{\frac {{a}^{3}}{bc}}$ is not defined.

The Right inequality $ \Leftrightarrow (2c \plus{} 2c \plus{} 2b \plus{} 2b \plus{} 2a \plus{} 2a)(\sum{\frac {{a}^{2} \plus{} {b}^{2}}{2c}}) \ge (a \plus{} b \plus{} c \plus{} a \plus{} b \plus{} c)^2$, which is true by Cauchy-Schwarz.

The Left inequality $ \Leftrightarrow 8abc \sum{\frac {{a}^{3}}{bc}}\geq 8abc \sum{\frac {{a}^{2} \plus{} {b}^{2}}{2c}}$
$ \Leftrightarrow 8(a^4 \plus{} b^4 \plus{} c^4) \ge 4(a^3b \plus{} ab^3 \plus{} b^3c \plus{} bc^3 \plus{} c^3a \plus{} a^3c)$
$ \Leftrightarrow (3a^4 \plus{} b^4) \plus{} (3b^4 \plus{} a^4) \plus{} (3b^4 \plus{} c^4) \plus{} (3c^4 \plus{} b^4) \plus{} (3c^4 \plus{} a^4) \plus{} (3a^4 \plus{} c^4)$
$ \ge 4(a^3b \plus{} ab^3 \plus{} b^3c \plus{} bc^3 \plus{} c^3a \plus{} a^3c)$, which is true by AM-GM.

$ LHS \minus{} MHS$ :
$ \sum(\frac {a^3}{bc} \plus{} \frac {b^3}{ac})\ge\sum(\frac {a^2 \plus{} b^2}{c})$
$ \sum(a^4 \plus{} b^4)\ge \sum(a^3b \plus{} ab^3)$
$ \sum(a^4 \plus{} b^4) \equal{} \sum\left[^{a^3 b^3}_{a\ b}\right]\ge\sum\left[^{a^3 b^3}_{b\ a}\right] \equal{} \sum a^3b \plus{} ab^3$

$ MHS \minus{} RHS$
$ \frac12 (\sum \frac {a^2}{c} \plus{} \frac {c^2}{a})\ge\sum \sqrt {ac}$
$ \sqrt {ac} \plus{} \sqrt {ba} \plus{} \sqrt {cb} \equal{} \left[^{\sqrt a\ \sqrt b\ \sqrt c}_{\sqrt c\ \sqrt a\ \sqrt b}\right]\ge\left[^{\sqrt a\ \sqrt b\ \sqrt c}_{\sqrt a\ \sqrt b\ \sqrt c}\right] \equal{} a \plus{} b \plus{} c$

[quote="cristinarm30"]Can we prove that?

$ \forall a_{i} > 0$

$ \sum{\frac {{a_{i}}^{3}}{a_{1}a_{2}...a_{i \minus{} 1}a_{i \plus{} 1}...a_{n}}}\geq\frac {1}{2}(\sum{a_{i}})(\sum{{a_{i}}^{2} \minus{} 1})\geq\sum{a_{i}}$

Thank's!  :oops:[/quote]

No if there are no other constraints, as the three expressions are of different degrees.


There are $ n \geq 4$ cards in a circle, with numbers 1, -1 in each card. Heli wants to know the product of all $ n$ cards, and to do that he can ask about the product of every subset of three cards. Which is the minimum number of question Heli can ask to be sure to know the product of all $ n$ cards?

It was posted before (here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202660&search_id=1804069133 ), but it's not clear whether that thread contains a correct solution  :maybe:


If you are in 10th grade, are you allowed to take both the amc10 and the amc12 this year, if on different testing dates? Thanks.

Basically they have the AMC 10/12 A and AMC 10/12 B.

You need to take one on the "A" date, and the other on the "B" date.

Frequently Asked Question AMC 10/12 #12
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, Mathematics Competitions

Sure
You can either take 10A followed by 12B or 12A followed by 10B.
I personally suggest take 10A first then 12B


Are we allowed to post [b]current [/b]MATHCOUNTS questions? (The questions in the 2004 MATHCOUNTS School Handbook)

Well I was afraid it might be copyrighted...

they have it for free on their site

so no


superior algebra
superior algebra unsolved

Any help with the following problem??

Let G be a finite group, and suppose there exist representatives g_1,...,g_r of the r distinct conjugacy classes in G, such that for all i and j we have g_ig_j=g_jg_i.  Prove that G is commutative.

Thanks!

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282859]here[/url].

Steve


conics parabola calculus derivative quadratics

I need help in finding the equations of two parabola's. One has a turning point at (60,10). The second crosses through the origin (0,0). They both need to join smoothly and have the points (24,5). Differentiation needs to be used to ensure that the gradient of the curve meet smoothly. The graph look something like this to give you the  whole picture. The graph needs to be considered as two parabolas.

[url=http://img468.imageshack.us/my.php?image=graphwl8.png][img]http://img468.imageshack.us/img468/8946/graphwl8.th.png[/img][/url]

Is anyone able to help me please. Try and show the basic steps of how you came to solve it.

Thanks

P.S. [b]DiscreetFourierTransform[/b] you were a legend and helped with my last problems. Thanks heaps and I hope you are able to help with this one as well. Your working out was tops!

The graph cannot look the way you have drawn it.

The unique solution consists of the two parabolas

$f(x) = \frac{5}{1728}x(x+48)$,

$g(x) =-\frac{5}{1296}(x^{2}-120x+1008)$.

Note that the quadratic $f$ has roots at x = 0, and x = -48; which means that it cannot intersect the positive x-axis, as you have drawn it.

Yep. Please continue discussion of this problem in the High School Basics thread of it, since the author posted it there also, and more thorough solutions of the problem have already been posted there.

It's right here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154509[/url]

Thanks

Dear atomicwedgie,

How did you get those two equations? If you click the link
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154509
you will see someone else's solution to the problem. How did you get two different equation to that persons?

The first equation that Chickendude boxed is equivalent to $g(x)$ here.
The second equation that Chickendude boxed is equivalent to $f(x)$ here.

I don't see how it's different.

FYI, for the future, calculus computations should go in:
$\boxed{\text{Forum}\; \to \; \text{College Playground}\; \to \; \text{Calculus Computations and Tutorials}}$


Let $ \alpha ,\ \beta$ be the roots of a quadratic equation $ x^2\plus{}px\plus{}q\equal{}0\ (p,\ q\in\mathbb{R})$ such that $ |\alpha \plus{}p|\plus{}|\beta \plus{}p|\leq 1$.
Find the maximum and minimum values of $ p^2\plus{}2q$.

The condition can be rewritten as $ |\alpha| \plus{} |\beta| \le 1$

And we need to find the max. and min. of $ (\alpha \plus{} \beta)^2 \plus{} 2\alpha\beta$

For the max, $ (\alpha \plus{} \beta)^2 \plus{} 2\alpha\beta \le 1 \plus{} 2\alpha\beta \le 1 \plus{} 2|\alpha\beta|\le$
$ 1 \plus{} 2(\frac {|\alpha| \plus{} |\beta|}{2})^2\le 1 \plus{} 2(1/4) \equal{} 3/2$ Which can be attained when $ \alpha \equal{} \beta \equal{} 1/2$

For the min, Note that $ 2|\alpha\beta| \le 1$ which follows from AM-GM, so $ \minus{} 1/2 \le \alpha\beta$. Therefore,
$ \minus{} 1/2 \le (\alpha \plus{} \beta)^2 \minus{} 1/2\le (\alpha \plus{} \beta)^2 \plus{} 2\alpha\beta$ We get get equality when $ \alpha \equal{} 1/2$ and $ \beta \equal{} \minus{} 1/2$. Therefore, max is 3/2, min. is -1/2.


inequalities
real analysis
real analysis unsolved

Prove that if the series $\displaystyle\sum_{k=1}^{+\infty}a_k^{1+\frac{1}{k}}$ converges then the series $\displaystyle\sum_{k=1}^{+\infty}a_k$ also converges

Put $ a_k^{1+1/k}=b_{k+1}$. The problem reduces to proving that $\sum{b_k}$ converges imply $\sum {b_k^{1-1/k}}$ converges. This follows from the well-discussed inequality $\sum_{k=1,n}{a_k^{1-1/k}}<(1+\sqrt{\sum_{k=2,n}{a_k}})^2$.

I proved this inequality while solving this problem but my proof wasn't quite short. Maybe you can give the link where this one was discussed

I think Myth solved it some not long time ago and I also gave a proof. I do not remember exactly the topic, but I think it is proposed problems inequalities, something like Kolmogorov inequality...

[quote="eugene"] I proved this inequality while solving this problem but my proof wasn't quite short.[/quote] There is also a simple general approach to eugene's problem that works in many other situations. The idea is that the extra factor $a_k^{1/k}$ matters only if $a_k<e^{-k}$ but $\sum_k e^{-k}$ converges. So, you just split the series $\sum_k a_k$ into $\sum_{k:a_k>e^{-k}}a_k\leqslant e\sum_k a_k^{1+\frac 1k}<+\infty$ and $\sum_{k:a_k\leqslant e^{-k}}a_k\leqslant \sum_k e^{-k}<+\infty$.

Following the proof of fedja we have this version

----------------------------------------------------------------------------------
Let $a_n,q_n>0$ reals sequences such that

$\sum_{n\geq 1}a_n^{1+\frac{1}{q_n}}$ converge
and
$\sum_{n\geq 1}e^{-q_n}$ converge

Then $\sum_{n\geq 1}a_n$ converge

I tried, but did not manage to solve the very nice problem: what is the necessary and sufficient condition for a sequence $(b_n)$ of positive numbers to verify the property: if $\sum_{n}{a_n}$ converges and $a_n>0$ then $\sum_{n}{a_n^{b_n}}$ also converges. Any idea?

WLOG, $b_n<1$ for all $n$. Then the condition is "There exists $C>0$ such that $\sum_n\exp\left\{-\frac C{1-b_n}\right\}<+\infty$".  :)