topology
function
integration
calculus
derivative
logarithms
complex numbers

Let $f$ be analytic in the annulus $A= \{z;\ 1 \leq |z| \leq 2 \}$ such that $|f|=5$ on the frontier of $A$. Show that $f$ has at least $2$ zeroes in $A$.

It's boundary, not frontier.

A silly counterexample: $f(x)=5$.
I suppose the problem really says "nonconstant". I've got it up to details, but the topology is really annoying.

It's not trivial to show such functions exist.

Yeah nonconstant. Well I found that such a function has at least one zero (this is rather trivial via the maximum modulus principle), but I still cannot see (even intuitively) where the second zero comes from.

Also, that should be "counting multiplicity"- I can construct an example that has one zero of multiplicity two.

Here's the nice argument:
Let $C_1$ be the inner circle and $C_2$ be the outer circle. By the argument principle, the number of zeros of $f$ (counting multiplicity) is $\int_{C_2}\frac{f'(z)}{2\pi i f(z)}\,dz-\int_{C_1}\frac{f'(z)}{2\pi i f(z)}\,dz$. Each of those integrals must be an integer. Since $f$ is not constant, $\frac{f'}{f}$ is not uniformly zero on either curve.

Now, by geometric considerations, $z\frac{f'(z)}{f(z)}$ is pure real on both circles. In addition, since $|f(z)|<5$ inside the annulus by the maximum modulus principle, $z\frac{f'(z)}{f(z)}$ has positive (nonnegative, really) real part on $C_2$ and negative real part on $C_1$ (look at what happens to $f$ along the curve $tz$). This makes $\int_{C_2}\frac{f'(z)}{2\pi i f(z)}\,dz$ a positive integer and $\int_{C_1}\frac{f'(z)}{2\pi i f(z)}\,dz$ a negative integer, so the difference is at least 2.

I stumbled upon this problem recently and couldn't figure it out. I am happy to see someone solved it   :)
Could you elaborate on it. Why is zf'/f real? Why is it important that the f'/f is not uniformly zero; even if it is not zero, then still the integrals could be. Any help would be appreciated!

The argument isn't just that $z\frac{f'(z)}{f(z)}$ is real- it's that $z\frac{f'(z)}{f(z)}\ge 0$ on the outer circle and $z\frac{f'(z)}{f(z)}\le0$ on the inner circle. With that, the only way either integral can be zero is if $z\frac{f'(z)}{f(z)}$ is identically zero on the circle.

Why is $z\frac{f'(z)}{f(z)}$ purely real? Differentiate $\ln(f(re^{i\theta}))$, where $r$ is the radius of one of the circles. The derivative is $ire^{i\theta}\frac{f'(re^{i\theta})}{f(re^{i\theta})}$; this is purely imaginary since the real part of $\ln(f(e^{i\theta}))$ is identically $\ln 5$. Divide by $i$ and set $z=re^{i\theta}$: $z\frac{f'(z)}{f(z)}$ is real when $z$ is on one of the circles.

There are other, more geometric ways of seeing this- look at the definition of the derivative along certain curves. To get the sign result, we use a radial line instead of the circle.

Hi,
Thank you for your reply! I have a few more questions. How do you change your argument if f(z) is a negative number? Do you change the branch cut for the log function? Also, how do you know exactly that zf'/f is positive/negative? I am not able to see this. Also, how do you go from zf'/f to dz f'/f? Do you just use the fact that z~i dz?
Thank you very much, and sorry to bother you with all the details!

Actually, I think I got it! Thank you for your help!

In one of your previous posts you mention that "the topology is really annoying" Is there a general approach to this kind of questions? Could you recommend a book?

Also, can you give a function with a double root? How do you construct one?

Once again, thank you for your help. THis question really intrigued me and that's why I have so many questions.


geometry
3D geometry
prism
Pythagorean Theorem

Three Cubes are stacked as shown. If the cubes have edge lengths $1$, $2$, and $3$ as shown, what is the length of the portion of the segment $\overline{AB}$ that is contained in the center cube?

[img]http://i96.photobucket.com/albums/l167/adasarathy/cubesstacked.gif[/img]

Yes, that's correct, I believe, I remember this problem. Here's my soluiton...

[hide]$B$ is 6 units down, 3 units forward, and 3 units to the right of $A$. Thus, by the pythagorean theorem, $\overline{AB}=\sqrt{54}=3\sqrt{6}$. $\frac{1}{3}$ of the line segment is in the middle cube, so divide by 3 to get $\sqrt{6}$.[/hide]

[color=blue]Kyyuanmathcount: Previous post deleted - Answers without solutions will now be deleted.  [/color]

i havn't read any solutions so here it goes
[hide]
$3\sqrt{5}$???????[/hide]

-jorian

[color=red]o, wait, crap, i did it wrong[/color]

[color=blue]kyyuanmathcount: please no color.  And, posts like this one will ALSO be deleted.  if you haven't read any solutions, you should read them, then only post yours (your solution i.e.) if it is different.  An different answer w/out a solution contributes nothing to a thread.  Thanks.  [/color]

[quote="bpms"][quote="espark52"]is it
[hide="probably wrong..."]$\sqrt{6}$?[/hide][/quote]
THat is what I got.
[hide="hint"]Lookj at it like a 2-D problem.[/hide]
[hide="bigger hint"]Construct a right triangle[/hide]
[hide="biggestest hint"] Make a right triangle with height from A to the point six beneath it one leg and the distance from that point to B the other leg.[/hide][/quote]

[hide="summing up the hints"]
Make a rectangular prism, then find the space diagonal.[/hide]

i don't mean to sound very stupid, but how do u find the diagonal of a rectangular prism

isn't a cube's diagonal: $edge*\sqrt{3}$???

-jorian

If the sides of a rectangular prism are a, b, and c and you want the diagonal d, use the equation :

$a^{2}+b^{2}+c^{2}=d^{2}$

so its sort of like pythagoream

thank you sooooo much mr. frost (now i sound stupid...wait, i already did :) )

thank you,
jorian hoover

[quote="Aequilipse"]Not posting a new thread as I only have a question.

How does one know that $\overline{\text{AB}}$ is trisected through the cubes? As in in the first cube there is $\sqrt6$ of the line, in the center there's $\sqrt6$ and in the last there's $\sqrt6$...[/quote]

Its not trisected... but we can pretend it is since the average of 1 and 3 is 2. Just ignore everything else and focus on that 1/3 of the prism.

Actually it's quite easy to show with similar triangles.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that if a group of order 28 has a normal subgroup of order 4,then G is abelian. (without quoting Sylow Theorem)

[hide="Outline"]
By Cauchy's Theorem, it has a subgroup M of order 7, and by the material developed in section 2.9, that group is normal. Let N be the subgroup of order 4. Then elements of M and N commute with each other, and MN is the whole group, and you can pretty much show from there that the group is abelian.

[/hide]


Let $ a, b, c$ be the side-lengths of a triangle, then

$ \sum\frac{a^2(b\plus{}c)}{b^2\minus{}bc\plus{}c^2}\geq a\plus{}b\plus{}c$

[quote="arqady"][quote="Ligouras"]Let $ a, b, c$ be the side-lengths of a triangle, then

$ \sum\frac {a^2(b \plus{} c)}{b^2 \minus{} bc \plus{} c^2}\geq a \plus{} b \plus{} c$[/quote]
$ \sum\frac {a^2(b \plus{} c)}{b^2 \minus{} bc \plus{} c^2} \minus{} ( a \plus{} b \plus{} c) \equal{} \sum_{cyc}\left(\frac {a^2(b \plus{} c)}{b^2 \minus{} bc \plus{} c^2} \minus{} \frac {b \plus{} c}{2}\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {(b \plus{} c)(a^2 \minus{} (b \minus{} c)^2 \plus{} a^2 \minus{} bc)}{2(b^2 \minus{} bc \plus{} c^2)} > \sum_{cyc}\frac {(b \plus{} c)(a^2 \minus{} bc)}{2(b^2 \minus{} bc \plus{} c^2)} \equal{}$
$ \equal{} \sum_{cyc}\frac {(b \plus{} c)((a \minus{} b)(a \plus{} c) \minus{} (c \minus{} a)(a \plus{} b))}{4(b^2 \minus{} bc \plus{} c^2)} \equal{}$
$ \equal{} \sum_{cyc}(a \minus{} b)\left(\frac {(b \plus{} c)(a \plus{} c)}{4(b^2 \minus{} bc \plus{} c^2)} \minus{} \frac {(a \plus{} c)(b \plus{} c)}{4(a^2 \minus{} ac \plus{} c^2)}\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {(a \minus{} b)^2(ab(a \plus{} b) \plus{} c(a^2 \plus{} ab \plus{} b^2) \minus{} c^3)}{4(a^2 \minus{} ac \plus{} c^2)(b^2 \minus{} bc \plus{} c^2)}\geq$
$ \geq\sum_{cyc}\frac {(a \minus{} b)^2ab(a \plus{} b \minus{} c)}{4(a^2 \minus{} ac \plus{} c^2)(b^2 \minus{} bc \plus{} c^2)}\geq0.$[/quote]

Very nice.... thanks :)   :wink:


Especially for Hell(pun intended)

Q.Reaction of 1 molecule of HBr with 1 molecule of 1,3-batadiene at 40C gives predominantly

A. 3-bromobutene under thermodyanamically controlled conditions
B. 1-bromo 2-butene under thermodyanamically controlled conditions
C. 3-bromobutene under kinetically controlled conditions
D. 1-bromo 2-butene under kinetically controlled conditions

[quote="chemrock"][quote="rituraj007"]1,3-batadiene [/quote]
I didn't know such wonderful "SHOE" compounds existed. :rotfl:  :rotfl:[/quote]
dei rakesh its just a spelling mistake tell the answeer am i rite??

[quote="Albert Einstien"]i think the answer the answer is D da as 1,2 addition is the product feasible at 40 c and it is a kinetic product not thermodynamic one only 1,4, addition is thermodynamic!!! i think so plz correct me if i am wrong!!!! i am just a layman!!! :D[/quote]
yep. ur rite. here kinetic product is favored, as low temp are used. but the answer is C i think.

i don't kno right answer but sol says B--sai is rite

actually, sai is rite.  :)  40c seems to a high enuf temp for the thermodynamic product to dominate. but it also depends on how much time the reaction mixture is kept at that temp, as the thermodynamic product is favored by quilibrium.

[quote="achu_182"]actually high and low are relative terms

for this particular reaction, anything above about -80 C is high enough, so 40 is decently high, for thermodynamic products to dominate.[/quote]
'anything above -80 C is high enough'. i dont think so. the percentages of kinetic and thermodynamic products are as follows: -80C: kinetic 80 thermodynamic 20
            40C: thermodynamic 80 kinetiic 20
so anything btw these temp, youll get a mixture of both and at -40, in roughly equal amounts.

[quote="rituraj007"]so kinetic product will be 3-bromobutene?[/quote]
yes,so b and c are the only viable options.

where do u read things???????? :blush:
 and cud u suggest a suitable book 4 organic chemistry????????


AMC
AIME
Euler
modular arithmetic
number theory
relatively prime
geometric sequence

mock aime: http://www.tjhsst.edu/~tmildorf/math/PracticeAIME.pdf

solutions: http://www.tjhsst.edu/~tmildorf/math/PracticeAIMESolutions.pdf

1.
#4 first solution, what is the motivation for for the first factorization?

2.
#5 first solution, are there the other factorizations similar to the one used with a and b cubed terms in this problem for higher powers? How can you tell when this factorization or similar ones is useful?

3.
#7 solution, Why do we have k-1 B's and 5-k C's to replace the 4 A's?
Where does that summation come from?

1. Well i think the motivation is just the $ 7^{4k}$ works out very nice mod 10. But it seems like you could do the problem just as easily with no thought by using geometric sum formula and fermat-euler ($ \frac {7^5 \minus{} 1}{6}$ is not very hard to compute).

2. Well if you are given $ a \plus{} b$ and $ a^3 \plus{} b^3$ it seems natural to use an identity involving $ (a \plus{} b)^3$. There's a similar identity for 7th powers I think.

3. For example, think about how you were to count the number of permutations of AABBB. It would be $ 5C2$ or $ 5C3$, both are equivalent. The same concept is applied to the problem, because suppose 6 C's were replaced completely with k A's and 6-k B's, the amount of permutations of that would be $ 6Ck$. Apply that for each of the three original strings of A's, B's, and C's and multiply to get the summation. As for $ k \minus{} 1$ B's and $ 5 \minus{} k$ C's to replace A's, think of it this way: After all the C's are replaced, you have like $ 6$ C's kinda floating around, but then $ 1 \plus{} k$ of these C's have to be used to replace B's, and then u have $ 6 \minus{} 1 \minus{} k \equal{} 5 \minus{} k$ C's left to replace A's. Then $ 4 \minus{} 5 \plus{} k \equal{} k \minus{} 1$ B's.

[quote="mathemonster"]what is fermat-euler?
is it $ a^p \equal{} a(\pmod p)$[/quote][url]http://en.wikipedia.org/wiki/Euler%27s_theorem[/url]

[quote="mathemonster"]what is fermat-euler?
is it $ a^p \equal{} a(\pmod p)$[/quote]

Define phi(n) to be the number of positive integers a less than n such a and n are relatively prime.

If b is an integer relatively prime to n, then b^(phi(n)) is congruent to 1 modulo n.

A generalization of Fermat's Little Thereom. Becaues when n is prime, phi(n) = n-1. (Because every positive integer less than n is relatively rpime to n.

[quote="mathemonster"]??? u just posted nothing wawb..[/quote]Typo fixed.

so how is fermat euler $ (b^{phi(n)} \equal{} 1\pmod n)$ useful in problem 4, as aznlord suggessted?

Another way of doing it is to think of it as a geometric sequence

$ \frac{7^{2005}\minus{}1}{6}$

so $ 7^{400} \equal{} 1 (\mod{1000})$

so $ 7^{2005} \equiv 7^5 (\mod{1000})$

but remember so $ \frac{7^{2005}\minus{}1}{6} \equiv \frac{7^5 \minus{} 1}{6} \equiv \boxed{801}$. in modulo 1000.

if this helps... :(

yes it does, but how did you get
$ 7^{400} \equal{} 1\pmod {1000}$ ?
if your using $ 7^{phi(1000)} \equal{} 1\pmod {1000}$ how do you know there are $ 400$ integers less than $ 1000$ that are relatively prime to it?

Your hint is that there are formulas to calculate $ \phi (n)$ based on the prime factorization of $ n$. Rather than look those formulas up, I highly suggest you try to come up with them and prove them; you'll be much more likely to recall them quickly if you need them on a contest.

in other words:

$ 1000 \equal{} 2^x \times 5^y$

so use PIE, there are 500 multiples of 2, 200 multiples of 5, and 100 multiples of both, so there are 500 + 200 - 100 = 600 numbers that are not relatively prime to 1000. so there are 400 that are.


(Russia 1195) If x, y > 0, prove that

x/(x^4 + y^2) + y/(y^4 + x^2) <= 1/(xy)

Thoughts?

By the way, this was in Kiran Kedlaya's MOP notes and said to be from Russia 1995, but John Scholes' Kalva archive doesn't have this problem under Russia 1995. Odd.

Hint:

[hide]This was in the Power Mean Inequality section in the lecture notes.[/hide]

I have a rather unorthodox solution to this:

[hide]By substituting a = 1/x and b = 1/y into the inequality, we see that the equality is equivalent to a / (a4 + b2) + b / (b4 + a2) :le: 1/(ab). This is identical to our original inequality. Thus we can assume WLOG that xy = 1. Then,

x / (x4 + y2) + y / (y4 + x2) :le: 1/(xy)

x / (x4 + 1 / y2) + (1 / x) / (1 / x4 + x2) :le: 1

2x3 / (1 + x6) :le: 1

2x3 :le: 1 + x6

This is trivially true.[/hide]

Thanks for the solutions.

[quote]I think I've seen this before though... was it from MOP 2002 ?[/quote]

Like I said, its in Kedlaya's MOP notes, and so I guess they have been used at MOP before.

[quote="belenos"]Yet an other solution :

x^4 + y >= x^3y + xy = xy(x+1) (rearrangement or (x^3-y)(x-y)>= 0)

So ineq becomes x/(x+1) + y/(y+1) <= 1 which is true (x+1 >= 2x).[/quote]

Except the first part isnt quite right...

take x = 2; y = 3 ;

then LHS = 16 + 9 = 25
        RHS = 24 + 6 = 30

and LHS < RHS . => <=

(Rearrangement only works when x and y (or w/e variables are in the inequality) are symmetric with one another. Since its x^4 + y^2, swapping x and y does NOT give you the same value so its not symmetric. Furthermore, in order to get the contradiction case, you wrote (x^3-y)(x-y) >= 0. Thats also not true, since its easy to get a case where x^3 > y and x < y. (like 2, 3)  )

Good try though :)

[quote="Nukular"]Except the first part isnt quite right... [/quote]

Oooops  :cry:   You're right.

It's alright we've all seen a nice result before and assumed it worked without checking (I definitely have on MULTIPLE occasions)

But the thing on this question I like is that we didnt multi-use AMGM too much, since AM-GM tends to make inequalities weaker and weaker the more it is used :(

[quote]But the thing on this question I like is that we didnt multi-use AMGM too much, since AM-GM tends to make inequalities weaker and weaker the more it is used [/quote]

How do you mean? Equality is always possible with AM-GM, so I don't understand how it would become weaker.

fyty - just because there is equality at one place does not mean the inequality cannot be made sharper at other places. For example, someone brought up taylor series. If you take just the first term of the taylor series, you have a sorta good approximation to the function. This approximation can be made a lot sharper by taking the first two terms. However, in both approximations, there is still equality at one point (the point at which the taylor series was evaluated). I think this is what's wrong with the "there's still equality" thought. I had trouble with it for a long time and this is the best explanation I could come up with. Hope it's right.

[quote="sqing"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=299899&p=1645181&hilit=Russia+1995#p1645181]Russia1995:[/url]For positive $ x,y$ prove that\[\frac1{xy}\geq \frac x{x^4 \plus{} y^2} \plus{} \frac y{x^2 \plus{} y^4}.\][/quote]
[hide="Solution"]
Expansion gives $x^6 + y^6 + x^2y^2 + x^4y^4 >=  x^4y+x^2y^5+x^5y^2+xy^4$. This is simple because \[\dfrac12 x^6 + \dfrac12 x^2y^2 \ge x^4y\]
\[\dfrac12 y^6 + \dfrac12 x^2y^2 \ge xy^4\]
 \[\dfrac12 x^6 + \dfrac12 x^4y^4 \ge x^5y^2\]
 \[\dfrac12 y^6 + \dfrac12 x^4y^4 \ge x^2y^5\]
[/hide]

[quote="sqing"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=299899&p=1645181&hilit=Russia+1995#p1645181]Russia1995:[/url]
For positive $ x,y$ prove that\[\frac1{xy}\geq \frac x{x^4 \plus{} y^2} \plus{} \frac y{x^2 \plus{} y^4}.\][/quote]

Hi,

also refer to this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=595705


Ahmet and Betül play a game on $ n\times n$ $ \left(n\ge 7\right)$ board. Ahmet places his only piece on one of the $ n^{2}$ squares. Then Betül places her two pieces on two of the squares at the border of the board. If two squares have a common edge, we call them adjacent squares. When it is Ahmet's turn, Ahmet moves his piece either to one of the empty adjacent squares or to the out of the board if it is on one of the squares at the border of the board. When it is Betül's turn, she moves all her two pieces to the adjacent squares. If  Ahmet's piece is already on one of the two squares that Betül has just moved to, Betül attacks to his piece and wins the game. If Ahmet manages to go out of the board, he wins the game. If Ahmet begins to move, he guarantees to win the game putting his piece on one of  the $\dots$  squares at the beginning of the game.

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  n^{2} \qquad\textbf{(C)}\  \left(n\minus{}2\right)^{2} \qquad\textbf{(D)}\  4\left(n\minus{}1\right) \qquad\textbf{(E)}\  2n\minus{}1$

If Ahmet places his piece to the border squares, he always wins because he is the first player.
If he places his piece to the inner squares (the squares that are not at the border of the board), Betül will place her two pieces diagonally such that Ahmet's pieces will be between her pieces. Ahmet moves his pieces to upward,downward, left or right. Betül always eats the piece. So there are are $4(n-1)$ safe places for Ahmet. The answer is $\boxed{D}$.

[quote=xeroxia]... Ahmet moves his pieces to upward,downward, left or right. Betül always eats the piece.[/quote] Is it very obvious?I will be happy if someone help me about this part.

That's not a game if you can just put your piece at the border of the board and win


Have any of you seen [url=http://shygypsy.com/az]this page yet[/url]?
(Hint: the first answer is "b")

Can you give me some hints about 'harrison'

went till giraffe
but what is the logic behind it?

Me too!
I think it should be a name, but i've tried almost all starting with I

Everyone got past giraffe? My IQ must be below average.

I tried everything: hyena, horse, hog, hitchhiker, hosepipe, honululu, house, hostel, hall, halloween, houston, happiness, horoscope, horror, hussar, hattrick, harpoon - but nothing works.

But don't tell me. I will exhaust my voacabulary.

Are you dumb because you got accition, or because you can't get dad?

P.S.: Still stuck

I'm pretty sure he meant he's stuck on accition...
So am I  :blush:

Come on. This is a [b]math[/b] forum. What math term could be similar to accition?  :D

Oh, and I'm stuck on the next one "dad".  :blush:

ah, right.
Well, I didn't even know was accition was...  :oops:

According to Merriam-Webster Online, it's not even a real word.  :maybe:

dad was tricky, but english was the trickiest. It had a hint. For dad, forget that it is a maths forum.

P.S.: Still stuck.

Err, I don't get it.  I go on the page and see the letter "a" with a blank below it, what are you supposed to do?  :?:

Bleh, I'm miserable at this game.  I'm stuck on dad too.

EDIT:  Yay!! I did it!

There are kind-of rules--the answer has to do with the clue, as well as the next letter of the alphabet.

I finally got india; now I'm stuck on Jeremiah.

EDIT: Now I'm on ampersand.

Still stuck on india. :wallbash:

But I think I might know the answer to jeremiah.

[quote="bubka"]Is it the name of a movie by harrison? Harrison Ford or Greg Harrison or Brett Harrison or Rex Harrison? I am not into current movies much.[/quote]

the actor that i referred to is one of the three there

[quote="nutz_for2.718281828"]I give up on off.[/quote]
Same here, lol

i'm still stuck on dad :(

edit: whoops nevermind! :)

[quote="Xantos C. Guin"]Off was totally random. I typed in one thing, then another thing, and it worked.[/quote]
Can you please please give me a hint?

[quote="wobster109"]off was strange.  it worked a lot like dad.[/quote]
Yes it did :D
thanks
Edit: got giraffe in 2 tries lol

Could someone give me a hint on india? :help:

Just a minor hint, please--nothing that gives it away.

Thanks. :)

I don't get the logic on the answer of Dad... Can anybody explain it to me? :blush:
I just tried things randomly and get it...

EDIT: Got English at 3rd time and Off at first :D


trigonometry
trig identities
Law of Cosines

Given a quadrilateral ABCD with sides AB = 4 cm, BC = 8 cm, CD = 6 cm , DA = 7 cm and BD = 10 cm.
Find the length of AC. ( Answer should be expressed in surd form.)

I got a really ugly number with two square roots in it, one of which is sqrt(1911)... basically, I used the law of cosines on triangle ABC and found cosine and sine of angles DBC and ABC to get the cosine of the whole thing.  Any better way?


Katrina and Abbie start a game with one pile of 40 pennies. They take turns. On each turn, a player must take 1, 2, 3, 4 or 5 pennies from the pile. The player who takes the last penny from the pile of 40 pennies wins the game. If Abbie plays first, what number of pennies must she take from the pile on her first turn in order to guarantee that she can win the game?

[hide]Well, working backwards, we know that whoever takes the 7th penny (leaving the 6th) wins, because then, no matter how many pennies the other girl takes, Katrina will be able to take the last penny. And since you can take up to 5 pennies, we'll count up multiples of 6, and the closest multiple of 6 is 36 (meaing if she doesn't take the 36th penny, she'll win). So if Katrina takes 4 pennies in the first turn, she'll win.[/hide]

[hide]she wants the 7th penny, so she should go for multiples of six.  closest is 36, so [color=red]4[/color][/hide]


functional equation
number theory
prime numbers
algebra
IMO
IMO 1998
IMO Shortlist

Determine the least possible value of $f(1998),$ where $f:\Bbb{N}\to \Bbb{N}$ is a function such that for all $m,n\in {\Bbb N}$,

\[f\left( n^{2}f(m)\right) =m\left( f(n)\right) ^{2}. \]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Do the brackets indiciate floor() or just another layer of parentheses?

when $f: \mathbb{N}\rightarrow \mathbb{N}$
it's easy to see that it's parentheses ;)

13 year bump?!

We claim the minimum value of $f(1998)$ is $\boxed{120}$.

Let $P(m, n)$ denote the given assertion. Additionally, let $f^k(x)$ denote $f$ applied $k$ times on $x$, let $f(x)^k$ denote $f(x)$ to the $k$'th power, and let $c = f(1)$. Suppose $f$ is such that $f(1998)$ is minimal. \\

From $P(m, 1)$, we get
\begin{align*}
	f^2(m) &= mc^2.
\end{align*}
From $P(1, m)$, we get
\begin{align*}
	f(cm^2) &= f(m)^2.
\end{align*}
From $P(f(cm^2), n)$, we get
\begin{align*}
	f(n^2f(f(cm^2))) &= f(cm^2)f(n)^2 \\
	f(c^3m^2n^2) &= f(m)^2f(n)^2 \\
	f(cmn)^2 &= f(m)^2f(n)^2 \\
	f(cmn) &= f(m)f(n).
\end{align*}
Setting $n = 1$ in the above, we obtain
\begin{align*}
	f(cm) &= cf(m).
\end{align*}
Hence, for all $m, n$ we have
\begin{align*}
	cf(mn) &= f(m)f(n).
\end{align*}
In particular, setting $m = n$ here yields $c \mid f(m)^2$ for all $m$. Suppose $c > 1$, and let $p$ be a prime dividing $c$. Note that, for all $m$, we must have $\nu_p(f(m)) \geq \frac{\nu_p(c)}{2}$. Suppose there exists $m$ for which $\nu_p(f(m)) < \nu_p(c)$. Then, we have
\begin{align*}
	cf(m^2) &= f(m)^2 \\
	\nu_p(c) + \nu_p(f(m^2)) &= 2\nu_p(f(m)) \\
	\nu_p(f(m^2)) &= 2\nu_p(f(m)) - \nu_p(c) \\
	&< \nu_p(f(m)).
\end{align*}
Iterating this, we find that for large enough $k$ we have $\nu_p(f(m^{2^k})) < \frac{\nu_p(f(m))}{2}$, a contradiction. Therefore, we have $\nu_p(f(m)) = \nu_p(c)$ for all primes $p \mid c$, implying that $c \mid f(m)$ for all $m$. Now, let $g(m) = \frac{f(m)}{c}$. We rewrite $P(m, n)$ as follows:
\begin{align*}
	cf\left(n^2\cdot \frac{f(m)}{c}\right) &= mc^2\left(\frac{f(n)}{c}\right)^2 \\
	c^2g(n^2g(m)) &= mc^2g(n)^2 \\
	g(n^2g(m)) &= mg(n)^2,
\end{align*}
so $g$ satisfies $P$ as well. In the case that $c > 1$, we have that $g(1998) < f(1998)$, contradicting the fact that $f$ minimizes $f(1998)$. Therefore, we have $c = 1$, so that
\begin{align*}
	f(mn) &= f(m)f(n)
\end{align*}
for all $m, n$, and $f^2(m) = m$. \\

We are now nearly done. Let $p$ be a prime, and let $f(p) = \prod_{i = 1}^k p_i^{e_i}$ for distinct primes $p_i$ and positive integers $e_i$. We have
\begin{align*}
	f^2(p) &= f\left(\prod_{i = 1}^k p_i^{e_i}\right) \\
	p &= \prod_{i = 1}^k f(p_i)^{e_i}.
\end{align*}
If $k > 1$, it follows that there exists a prime $q$ for which $f(q) = 1$. In this case, we have $f^2(q) = f(1) = 1$, contradicting the fact that $f^2(q) = q$. Hence, $k = 1$, and we have $p = f(q^r) = f(q)^r$ for some prime $q$ and positive integer $r$. Clearly, this implies that $r = 1$, so $f(p) = q$. Thus, $f$ is a multiplicative involution which maps primes to primes. We claim that all such functions $f$ satisfy the given functional equation. Indeed, for any such $f$, we have
\begin{align*}
	f(n^2f(m)) &= f(n^2)f^2(m) \\
	f(n^2f(m)) &= mf(n)^2.
\end{align*}

To finish, we have
\begin{align*}
	f(1998) &= f(2 \cdot 3^3 \cdot 37) \\
	&= f(3)^3f(2)f(37) \\
	&\leq 2^3 \cdot 3 \cdot 5 \\
	&= 120,
\end{align*}
as claimed. $\Box$

Let $P(m,n)$ denote the FE. The key identities we'll use are
[list]
    [*] $P(m,1)$ gives $f(f(m))=mf(1)^2$.
    [*] $P(1,n)$ gives $f(f(1)n^2)=f(n)^2$.
[/list]
Consider $S=f\big(a^2f(f(1)b^2)\big)$. Note $f(f(1)b^2)=f(b)^2$, so $S=f(a^2f(b)^2)$. On the other hand, $S=f(1)b^2f(a)^2$ by using the FE. Hence,
\[ f(a^2f(b)^2)=f(1)b^2f(a)^2. \]
Taking $f$ of both sides gives and using the two bulleted identities gives
\[ a^2f(b)^2 f(1)^2 = f(bf(a))^2 \implies af(b)f(1)=f(bf(a)). \]
Substitute $f(b)$ for $b$ above to get $af(f(b))f(1)=f(f(b)f(a))$, i.e. $abf(1)^3=f(f(b)f(a))$. Taking $f$ of both sides gives
\[ f(abf(1)^3) = f(b)f(a)f(1)^2.\]
Plug in $b=1$ above to get $f(af(1)^3)=f(a)f(1)^3$. So $f(abf(1)^3)=f(ab)f(1)^3$, which means
\[ f(ab)f(1)^3 = f(b)f(a)f(1)^2 \implies \boxed{f(1)f(ab)=f(a)f(b)}. \]
---------------------------------------------------------------------
Now we claim $f(1)\mid f(x)$ for all $x$. We have $f(a^k)=f(a)^k/f(1)^{k-1}$ by iterating the above. Consider some $p\mid f(1)$, then $k\nu_p(f(a)) \ge (k-1)\nu_p(f(1))$, i.e. $\nu_p(f(a)) \ge \tfrac{k-1}{k}\nu_p(f(1))$. Since this holds for all $k\ge 1$, actually $\nu_p(f(a)) \ge \nu_p(f(1))$. Therefore $f(1)\mid f(a)$, as claimed.

Now let $g(x)=f(x)/f(1)$ be an integer function. Then $g(ab)=g(a)g(b)$. We claim $g(g(m))=m$. Indeed,
\[ mf(1)^2=f(f(m))=f(1)g(f(m))=f(1)g(f(1)g(m))=f(1)g(f(1))g(g(m)). \]
But $g(f(1))=f(f(1))/f(1)$, and since $P(1,1)$ gives $f(f(1))=f(1)^2$, actually $g(f(1))=f(1)$. Hence, the above equalities give $mf(1)^2=f(1)^2g(g(m))$, i.e. $g(g(m))=m$.

Now, we simply pair up primes $(p,q)$ and set $f(p)=q$, $f(q)=p$, and then since $g$ is multiplicative, all other outputs are determined. It is easy to check that $g$ is involutive in general in this construction (each prime in the prime factorization involutes). This describes the general function $g$.

Finally, to minimize $f(1998)$, set $f(1)=1$, and then
\begin{align*}
    f(1998) &= g(1998) = g(2\cdot 3^3\cdot 37) = g(2)g(3)^3g(37) \ge 3\cdot 2^3 \cdot 5 = 120,
\end{align*}
where we set $g(2)=3,g(3)=2$ and $g(37)=5$.

[hide=Remarks]
The main part of the problem is to reduce to the ``almost-multiplicative''ness of $f$. The FE manipulations are actually very tricky to come up with; the main ideas in it were (1) compute $S=f(a^2f(f(1)b^2))$ in two different ways, and later (2) Once we have $f(abf(1)^3)=blah$, then set $b=1$ to get a formula for $f(af(1)^3)$, and then plug in $ab$ to get a new equation.

The idea behind (1) was to think of the FE very structurally; I set $f(1)=1$ to get something without $f(1)$'s floating around everywhere, and then I had $f$ was an involution satisfying the FE. Now $f(2)$ could not be pinned down, so I set $f(2)=3$ to see what happens. Then $f(3)=2,f(4)=9$. For $f(6)$, I found you can use $f(36)=f(6)^2$ and $f(36)=f(2^2\cdot f(4))$ to compute $f(6)$. This trick generalized nicely.
[/hide]

We first provide a construction for all functions $f$, then prove all functions $f$ are of this form. Let $d$ be an arbitrary positive integer and $g$ be an involutive multiplicative function that takes primes to primes and fixes $d$. Then, define $f(n)=dg(n)$. It is not too hard to check $f$ works.

Let $P(m,n)$ denote the condition for $m,n$. Let $d=f(1)$. Observe that
[list]
    [*] By $P(1,1)$, $f(d)=d^2$.
    [*] By $P(n,1)$, $f(dn^2)=f(n)^2$.
    [*] By $P(1,n)$, $f(f(n))=d^2n$.
[/list]
Now, note that
\[f(a)^2f(b)^2=f(b^2f(f(a)^2))=f(b^2f(f(da^2)))=f(d^3a^2b^2)=f(dab)^2,\]
hence
\[f(a)f(b)=f(dab)\qquad (\diamondsuit).\]
Take $a=1$ in $\diamondsuit$ to yield $df(n)=f(dn)$, so $\diamondsuit$ rewrites as
\[f(a)f(b)=df(ab)\qquad (\heartsuit).\]
Note that the information we have extracted implies the given equation because
\[f(n^2f(m))=\frac{1}{d}f(n^2)f(f(m))=mf(n)^2,\]
so we ignore it now.

[b][color=#f00]Claim:[/color][/b] Every $f(n)$ is divisible by $d$.

[i]Solution:[/i] Note
\[f(n^k)=\frac{1}{d}f(n)f(n^{k-1})=\cdots = \frac{1}{d^{k-1}}f(n)^k,\]
which implies $d\mid f(n)$ by taking large $k$ and analyzing the $p$-adic valuation of $f(n),d$. $\fbox{}$

Now, let $g(n)=f(n)/d$ be a function from $\mathbb{N}\to\mathbb{N}$, which exists by the previous claim. Our information rewrites as $g(d)=d$, $g(a)g(b)=g(ab)$, $g(g(n))=n$. By multiplicativity, $g$ preserves the number of divisors of a number, hence $g$ takes primes to primes, and we are done by taking $g$ and $d$ in our construction to be as extracted.
-------

Now, we actually solve the problem; prime factorize $1998$ as $2\cdot3^3\cdot37$, so $g(1998)\ge 2^3\cdot3\cdot5=120$.

[quote=abvk1718][quote=pohoatza]Denote $f(1)=a$, and put $m=n=1$, therefore $f(f(k))=a^{2}k$ and $f(ak^{2})=f^{2}(k)$, $\forall k \in \mathbb{N}$.

Thus now, we have: $f^{2}(x) f^{2}(y)=f^{2}(x)f(ay^{2})=f(x^{2}f(f(ay^{2})))=$
$=f(x^{2}a^{3}y^{2})=f(a(axy)^{2})=f^{2}(axy)$

$\iff f(axy)=f(x)f(y) \Rightarrow f(ax)=af(x)$

$\iff af(xy)=f(x)f(y) , \forall x,y \in \mathbb{N}$.

Now we can easily prove that $f(x)$ is divisible by $a$ for each $x$, more likely we have that $f^{k}(x)=a^{k-1}\cdot f(x^{k})$ is divisible by $a^{k-1}$.

For proving the above asertion we consider $p^{\alpha}$ and $p^{\beta}$ the exact powers of a prime $p$ that tivide $f(x)$ and $a$ respectively, therefore $k\alpha \geq (k-1)\beta , \forall k \in \mathbb{N}$, therefore $\alpha\geq \beta$, so $f(x)$ is divisible by $a$.

Now we just consider the function $g(x)=\frac{f(x)}{a}$.
Thus: $g(1)=1, g(xy)=g(x)g(y), g(g(x))=x$.
Since $g(x)$ respects the initial condition of the problem and $g(x)\leq f(x)$, we claim that it is enough to find the least value of $g(1998)$.

Since $g(1998)=g(2 \cdot 3^{3}\cdot 37) =g(2) \cdot g^{3}(3)\cdot g(37)$, and $g(2),g(3),g(37)$ are disting prime numbers (the proof follows easily), we have that $g(1998)$, is not smaller than $2^{3}\cdot 3 \cdot 5=120$. But $g$ beeing a bijection, the value $120$, is obtained for any $g$, so we have that $g(2)=3, g(3)=2, g(5)=37, g(37)=5$, therefore the answer is $120$, and thus the problem is solved![/quote]

Can anyone tell HOW $\alpha ≥ \beta  $?[/quote]

We know that $\frac{\alpha}{\beta} \geq \frac{k-1}{k}$. Let $X=\left\{ \frac{k-1}{k} | k \in \mathbb{N} \right\}$. Then $\sup X=1$. But $\frac{\alpha}{\beta}$ is a upper bound of $X$. Therefore, $\frac{\alpha}{\beta} \geq 1$.

Let $c=f(1).$ Then, $$m=n=1\implies  f(f(m))=c^2m \text{ and } f(cn^2)=f^2(n).$$ It follows that \begin{align*} f^2(m)f^2(n)=f^2(m)f(cn^2)\\=f(m^2f(f(cn)))=f(m^2c^3n^2)\\=f^2(cmn) \implies f(m)f(n)=f(cmn).\end{align*}  Also $$f(cm)=cf(m)\implies cf(mn)=f(m)f(n).~~~~~(*)$$
---------

[b][u]Claim 1: $c|f(n)~~ \forall n\in \mathbb{N}.$[/u][/b]
[i]Proof.[/i] Let $p$ be any prime and let $s\geq 0$ and $r\geq 0$ be the degrees of $p$ in the prime representation of $c$ and $f(n),$ respectively. By induction from $(*)$, $$f^t(n)=c^{t-1}f(n^t) ~~\forall t \in \mathbb{N}.$$ It follows that $r\geq s,$ which provides the claim. $\blacksquare$

Let $g(n)=\frac{f(n)}{c},$ then \begin{align*} g(mn)=g(m)g(n) ~~~~~(\alpha)\\ g(g(m))=m ~~~~~(\beta) \\ \forall m,n \in \mathbb{N}.\end{align*} Setting $m=n=1$ in $(\alpha) \implies g(1)=1 ~~~~~(\gamma).$
-----------

[b][u]Claim 2: $g(p)$ is a prime whenever $p$ is a prime.[/u][/b]
[i]Proof.[/i] Let $g(p)=ab.$ From $(\alpha)$ and $(\beta)$, $$p=g(g(p))=g(ab)=g(a)g(b).$$ So we can assume, $g(u)=1 \implies u=1.$ As desired. $\blacksquare$

Now let $n\geq 2$ and $n=p_1^{t_1}\dots p_k^{t_k}$ be the prime representation of $n.$ Then from $(\alpha)$, $$g(n)=g^{t_1}(p_1)\dots g^{t_k}(p_k).~~~~~(**)$$ It follows that the function $f$ is determined solely by $c=f(1)$ and the function $g$ mapping from the set of primes to the set of primes s.t. $g(g(p))=p.$
-------------

[b][u]Claim 3: $f(1998)=120.$[/u][/b]
[i]Proof.[/i] Notice that $$f(1998)=f(2\cdot 3^3\cdot 37)=f(1)g(2)g^3(3)g(37).$$ Because $g(2),g(3),g(37)$ are distinct primes we have $$g(2)g^3(3)g(37)\geq 3\cdot 2^3 \cdot 5=120\implies f(1998)\geq 120.$$ It is easy to check that the inequality is achievable when $f(1)=1, g(2)=3,g(3)=2,g(37)=5$ and $g(p)=p~~\forall p \neq 2,3,37.$ Thus the proof is complete. $\blacksquare$

The answer is $f(n)  = cg(n)$, such that $g\colon \mathbb{N}\to \mathbb{N}$ maps primes to primes, $g$ is an involution, $g$ is multiplicative, and $g(c) = c$. These can be checked to work.

Let $P(m,n)$ denote the given assertion and $f(1) = c$.

$P(m,1): f(f(m)) = m\cdot c^2$. From here we see that $f$ is injective.

$P(1,m): f(m^2 \cdot c) = f(m)^2$.

$P(1,1): f(c) = c^2$.

$P(m,c): f(f(m)c^2) = m c^4$

$P(c,m): f(m^2c^2) = c f(m)^2$.

$P(f(m),n): f(m c^2 n^2 ) = f(m) f(n)^2$.

$P(m^2c, f(n)): f(f(m)^2 f(n)^2) = m^2 n^2 c^5$.

$P(f(m),c): f(mc^4)  = c^4f(m)$.

So \[f(m^2n^2c^5) = f(c^4 ((mn)^2 c)) = c^4f(mn)^2 \]

Thus $f(m)^2 f(n)^2 c^2 = c^4 f(mn)^2$, which implies \[f(m)f(n) = cf(mn)\]

By induction we find \[f(m^k) = \frac{f(m)^k}{c^{k-1}}\] for any $k$.

Suppose we had a prime $p$ for which $\nu_p(c) > \nu_p(f(m))$. If $\nu_p(c) = a, \nu_p(f(m)) = b$, we have \[bk\ge a(k-1) = ak-a,\]so $(a-b)k \le a $, which implies $k \le \frac{a}{a-b}$. However, we can set $k$ arbitrarily large, contradiction.

Thus $c\mid f(m)$ for all $m$. Let $g(m) = \frac{f(m)}{c}$. The equations \[f(m)f(n) = cf(mn) \text{ and } f(f(m)) = c^2 m\] imply that $g$ is multiplicative and involutive. Since $f(c) = c^2$, we have $g(c) = c$. It remains to show $g$ maps primes to primes.

Note that if $g(m) = 1$ then $m=1$. So for any $x,y>1$, we have $g(xy) = g(x)g(y)$, which implies $g(n) $ is composite for all $n$. Since $g(g(p)) = p$, $g(p)$ can't be composite, so $g(p)$ is prime.

Denote by $P(m,n)$ the assertion. Let $f(1)=c$. Then, we have that
\begin{enumerate}[(i)]
\item $P(1,1)$ giving $f(c)=c^2$
\item $P(m,1)$ giving $f(f(m))=mc^2=mf(c)$.
\item $P(1,n)$ giving $f(n^2c)=f(n)^2$
\end{enumerate}
Note that $P(f(m),n)$ gives $f(m)f(n)^2=f(n^2f(f(m)))=f(n^2mc^2)$. Substituting $m\to m^2c$, we get
\[f(m)^2f(n)^2=f(m^2c)f(n)^2=f(m^2n^2c^3)=f((mnc)^2c)=f(mnc)^2\]
Thus, we have $f(m)f(n)=f(mnc)$. Letting $m=1$ in that, we have $f(cn)=cf(n)$, so $f(m)f(n)=cf(mn)$. Let $f(n)=cg(n)$ then $g(m)g(n)=g(mn)$, $g(1)=1$, and a manipulation of $P(m,1)$ gives $c^2g(g(n))=cg(cg(n))=c^2n$ so $g(g(n))=n$. Additionally, $g$ is bijective.
\\$~$\\
We claim that $g(p)$ is prime for primes $p$. If not, then let $g(p)=xy$ where $x\ge y>1$, then
\[p=g(g(p)=g(xy)=g(x)g(y)\]
so $g(x)$ or $g(y)$ is one, a contradiction because $g(x)=1\iff x=1$.
\\$~$\\
We have $f(1998)=cg(2)g(3)^3g(37)\ge 1\cdot 3\cdot 2^3\cdot 5 = 120$, achievable when $g$ just involutes the primes so that $2\to 3, 37\to 5$, and $c=1$.

Denote the assertion with $P(m, n)$.
Let $d = f(1)$ and let $c = n^2f\left(\frac{1}{f(n)^2}\right)$ such that $f(c) = 1$.
Then, $P(1, 1)$ gives that $f(d) = d^2$. Likewise, $P(n, 1)$ gives that $f(f(n)) = d^2n$ so $f$ is injective.
Finally $P(1, n)$ means that $f(dn^2) = f(n)^2$
As such, \[ f(a)^2f(b)^2 = f(da^2)f(b)^2 = f(b^2f(f(da^2))) = f(b^2a^2d^3) = f(dab)^2 \] so $f(a)f(b) = f(dab)$ and thus $df(a) = f(da)$. Thus, this becomes $f(a)f(b) = df(ab)$.
Thus, $d = \frac{f(a)f(b)}{f(ab)}$ so \[ \nu_p(f(a)) + \nu_p(f(b)) - \nu_p(f(ab)) = 1 \]
If there is some $f(a)$ such $d \nmid f(a)$, then \[ \nu_p(f(b)) - \nu_p(f(ab)) = 1 \]
However, then the sequence $f(b), f(ab), f(a^2b), \dots$ has ever decreasing $\nu_p$, contradiction.
Thus, $d \mid f(n)$ for all $n$.
Define $g(n) = \frac{f(n)}{d}$. Then we have that $d^2g(a)g(b) = d^2g(a)g(b)$ so $g$ is multiplicative.
Furthermore, $g(dg(n)) = dn$ implies that $g(d)g(g(n)) = dn$ so $g(g(n)) = n$.
We now characterize all multiplicative involutions.
Note that $g(1) = 1$. Let $p_1, p_2, \dots$ be the primes and have $g(p_i) = q_i$, and note that $q_i \ne 1$ by injectivity.
Then, $g(q_i) = p_i$ can only hold if  $q_i$ is also a prime.
Thus, for any such pair $(g, d)$, $f(n) = d \cdot g(n)$ satisfies the original equation.
Thus $f(1998)$ is minimized when $d = 1$ and since $1998 = 2 \cdot 3^3 \cdot 37$, the minimal value is $2^3 \cdot 3 \cdot 5 = 120$.


The equation of the line that passes through the points $ (\minus{}2,0)$ and $ (0,2)$ can be expressed in the form $ y\equal{}mx\plus{}b$. What is the value of $ m\plus{}b$?

Drawing a line from (-2,0) to (0,2) on a graph, you would see that for the increase in 2 on the x-axis (from -2 to 0), there is also an increase of 2 on the y-axis (from 0 to 2).

Thus gradient = m = 1.

b is the value of y when x = 0, so it is the y-intercept = -2 (as given from the first coordinate point).

Then m + b = 1 + (-2) = -1

The slope of a line, given 2 points is $ \frac {y' \minus{} y}{x' \minus{} x}$, where $ (x, y)$ and $ (x', y')$ are the two points. Plugging in the points gives us $ \frac {2 \minus{} 0}{0 \minus{} ( \minus{} 2)} \equal{} 1$. The second point given for the problem displays the y intercept, or b. Thus, m+b=1+2=3.

A bit more on finding the value of $ b$:

To find slope, we use the formula $ \frac{y_2 \minus{} y_1}{x_2 \minus{} x_1}$. Plugging in our values, we find that the slope, or $ m$ is equal to $ 1$.

So we know our equation looks something like:

$ y \equal{} x \plus{} b$

We can substitute $ y$ and $ x$ for one of the points. Let's use $ (0, 2)$. When we plug these in, we get:

$ 2 \equal{} 0 \plus{} b$

$ 2 \equal{} b$

So our equation is $ y \equal{} x \plus{} 2$.

We want to find the sum of $ m$ and $ b$, so our answer is $ \boxed{3}$.

Is $ \frac{y_2\minus{}y_1}{x_2\minus{}x_1}$ known as Point slope format or something else?

Well, that's the formula to find slope.

Point-Slope form is another way of writing equations. The way we used in this problem is called slope-intercept form. The other way is called Standard Form.

Point-Slope Form: $ y \minus{} y_1 \equal{} m(x \minus{} x_1)$

Slope-Intercept Form: $ y \equal{} mx \plus{} b$

Standard Form: $ Ax \plus{} By \equal{} C$


geometry
3D geometry
ratio
real analysis
real analysis unsolved

On page 44 of Pugh's "Real Mathematical Analysis,"

[hide]A rational number $ p/q$ is [b]dyadic[/b] if $ q$ is a power of $ 2$, $ q=2^{k}$ for some nonnegative integer $ k$.  For example, $ 0$, $ 3/8$, $ 3/1$, $ -3/256$ are dyadic rationals, but $ 1/3$, $ 5/12$ are not.  A dyadic interval is $ [a,b]$ where $ a=p/2^{k}$ and $ b=(p+1)/2^{k}$.  For example, $ [.75,1]$ is a dyadic interval but $ [1,\pi]$, $ [0,2]$, and $ [.25,.75]$ are not.  A dyadic cube is the product of dyadic intervals having equal length.  The set of dyadic rationals may be denoted as $ \mathbb{Q}_{2}$ and the dyadic lattice as $ \mathbb{Q}_{2}^{m}$.[/hide]
21. (d) Given $ \epsilon>0$, prove that finitely many disjoint discs can be drawn inside the unit square so that the total area of the discs exceeds $ 1-\epsilon$.

At first, I wanted to prove that in the diagram attached, that any sequence of subregions (outside the circles) in the square obtained through such repeated iterations will have a subsequence consisting of regions "thick" enough that the inscribed circle represents an appreciable percentage.  Namely, that the subsequence will have inscribed-circle-area to region-area ratios bounded below by a positive number.  This I no longer believe to be true.  So, here's my new approach:

Call the ratio of the area of a circle to it's circumscribed regular hexagon $ r$, which is bigger than $ 0.9$.  We can finely pack the plane with circles so that drawing any predetermined (and well-behaved) shape on the plane and considering only the circles that lie completely inside the shape yields a total circular area close to $ r$ of the shape.  More precisely, we can fill more than 90% of any region with congruent circles.  Each iteration in this scheme multiplies the area of the square that is uncovered by something less than $ 0.1$.  This seems like proof enough to me, but I have a feeling it wouldn't fly with modern mathematicians (just as emphasized in Pugh's preliminary disapproval of convincing argument vs. mathematical proof) simply because I assumed the borders become negligible as the circles shrink to fit the curve.  Is this sort of argument at all valid?  Maybe it just needs to be purged of hand-waving?

Thanks!

That is a hole- you need to prove that the boundary becomes "small enough".

The key lemma you need: Let $ U$ be an open, bounded set in $ \mathbb{R}^{2}$, and let its area be $ A$. For each $ \delta>0$, let $ U_{\delta}$ be the set of all points in $ U$ which are at distance at least $ \delta$ from the complement of $ U$.
Now, for any $ \epsilon>0$, there is some $ \delta$ such that the area of $ U_{\delta}$ is at least $ A-\epsilon$.

Once you have this, you can fill $ U$ with a mesh of (closed) disks with radius less than $ \delta$, and thus cover a large fraction of $ U$.

The lemma does require proof, and I haven't put together all the details. It should also work in higher dimensions.

Okay, I mistook your lemma for something other than the border, so this post is irrelevant.  It's as if we fill up the square with small hexagons and consider the hexagons that intersect with both the square and the square's complement (i.e. the border).  We let the radii of the disks, $ \delta$, be small enough to where the area of "$ U-U_\delta<\epsilon$".


What would be the optimal finishing time for an old national sprint round? What about an old state sprint round? Thanks.

[quote="hoshattack"]what do you mean?  time yourself for the 40 minutes for the sprint.[/quote]

I think he means how long it takes you to finish an old nats or state test...finish as in completed all the problems with no double checking, etc...but still shouldn't you use all the 40 min you're given?

considering the difficulty of this year's chapter, I'm speculating that you will need to be done with sprint in ~35 min


There are 10 girls and 15 boys in a junior class, and 4 girls and 10 boys in a senior class. A committee of 7 members is to be formed from these two classes Find the number of ways this can be done if the committee must have exactly 4 senior students and exactly 5 boys.

These are the three groups of possible combinations:

2 senior boys + 2 senior girls + 3 junior boys + 0 junior girls
3 senior boys + 1 senior girls + 2 junior boys + 1 junior girls
4 senior boys + 0 senior girls + 1 junior boys + 2 junior girls

These are the number of ways for each group of possible combinations:
$ \dbinom{10}{2}*\dbinom{4}{2}*\dbinom{15}{3}*\dbinom{10}{0}=122850$
$ \dbinom{10}{3}*\dbinom{4}{1}*\dbinom{15}{2}*\dbinom{10}{1}=504000$
$ \dbinom{10}{4}*\dbinom{4}{0}*\dbinom{15}{1}*\dbinom{10}{2}=141750$

Therefore the total number of ways is 768600.


What is the energy of a photon? and can a photon actually have energy ?(I think I am thinking about this wrong but the math is good).

so anyway:

$E=MC^{2}$ plug in the variable for the mass and we get

\[E=0C^{2}\] and anything times 0 is zero, so

\[E=0C^{2}\Longrightarrow E=0\]

so can a photon have no mass, but still have energy?

or does the photon's energy increase by its mass?

First off, a photon is said to have zero [i]rest mass[/i].  The thing is, it's never in rest, so $E=mc^{2}$ itself won't work and you have to use an alternate form.

ok I see now, I thought I was missing something,

Thanks

[quote="Derek"]$E^{2}= c^{2}p^{2}+m^{2}c^{4}$
Since $p = \frac{hv}c$
$E = hv$[/quote]

This is the general form of the equation. For most objects, we assume that the momentum is zero, so that the $c^{2}p^{2}$ term drops out and we have

$E^{2}=m^{2}c^{4}$

By taking the positive square root of both sides (of course, we could have antimatter and negative energy) we get the familiar equation

$E=mc^{2}$

However, as quantum leap has stated, this doesn't work for photons.


I have the choice between these two courses and wonder which one will be better to put on resume. Also, which one will be more math? Thanks for the advice.

I'd go with Industrial because that usually has high pay, and I think it has more math involved.

This question is impossible to answer in the abstract, because:

a) game theory can be taught in more or less rigorous ways

and

b) industrial organization can be taught more or less mathematically--in fact, game theory is the basis for many economic models used in industrial organization.

However, as someone who has studied and taught both game theory and industrial organization, I disagree with hunter34---industrial organization does not, in general, have more math involved than game theory.

A highly theoretical graduate industrial organization class might develop and use a lot of mathematical game theory.  On the other hand, a typical  undergraduate industrial organization class might use only very simple models of game theory and mostly cover descriptive economics (e.g., defining concentration ratios to measure the degree of monopoly power in different industries, discussing barriers to entry, product differentiation, antitrust laws and regulations, methods of collusion, etc.)

Game theory can also be more or less mathematical.  Sometimes game theory is taught as an upper level elective for math majors (in which case it's likely to be quite mathematical.)   Sometimes game theory is a course offered with no prerequisites as a way that non-mathy students can meet a general education distribution requirement (in which case it's not going to be very challenging mathematically.)

Take a look at the syllabus and readings used for both courses, consider recommendations from friends who have taken both classes, and then decide which will be a better learning experience for you.

Ideally, you attend a college that will let you sit in on the first week of both classes before you make a decision.  In that case, you should ask yourself which professor will inspire and challenge you to learn the most.

As to which course "will look more impressive on a resume," I don't think it will make a lot of difference, since savvy interviewers will know that either course could be more or less rigorous.  Unless you are interviewing with a recent alum of your institution, they are unlikely to know a whole lot about the particular courses offered at your classes.  The course title, by itself, means little.

Also depends upon what you want to do.
Some banks and hedge funds really like game theory and explicitly ask for it.


[quote="guyinPA"]can't show pic but it's an L shape of 5 equal sized boxes up and down and then next to bottom to boxes are 3 adjacent boxes so a total of 11 same sized boxes -- Question: The shaded region shown consists of 11 unit squares and rests along the x-axis and y-ais.  The shaded region is rotated about the x-asis to form a solid.  In cubic units, what is the volume of the resulting solid? Please express answer in simplest form in terms of pi.[/quote]
next to bottom [size=200][b]two[/b][/size] boxes?

eidt: o reading this for real, i actually remember this question
[hide="The Answer"]i wanna eat 35 pies! [/hide][/hide]

[quote="guyinPA"]can't show pic but it's an L shape of 5 equal sized boxes up and down and then next to bottom to boxes are 3 adjacent boxes so a total of 11 same sized boxes -- Question: The shaded region shown consists of 11 unit squares and rests along the x-axis and y-ais.  The shaded region is rotated about the x-asis to form a solid.  In cubic units, what is the volume of the resulting solid? Please express answer in simplest form in terms of pi.[/quote]
[hide]$\pi \cdot 5^{2}\cdot 1+\pi \cdot 2^{2}\cdot 3 = 37\pi$[/hide]

[quote="guyinPA"]can't show pic but it's an L shape of 5 equal sized boxes up and down and then next to bottom to boxes are 3 adjacent boxes so a total of 11 same sized boxes -- Question: The shaded region shown consists of 11 unit squares and rests along the x-axis and y-ais.  The shaded region is rotated about the x-asis to form a solid.  In cubic units, what is the volume of the resulting solid? Please express answer in simplest form in terms of pi.[/quote]

My solution:
[hide]Divide the rotated figure into 2 cylinders:
One on left with height 1 and raidus 5: $V=\pi \cdot 1 \cdot 5^{2}= 25\pi$
One on right with height 3 and radius 2: $V=\pi \cdot 3 \cdot 2^{2}= 12\pi$

Total it up, and it's $37\pi$[/hide]
EDIT: Oops, didn't see that $35\pi$ isn't the answer, I'll go back and rework it.  :oops_sign:
EDIT again: Oh, rotated about the X-AXIS.  I've changed my answer.

Um dude don't revive month-old threads please.

Sorry, someone posted a link to this thread in another thread :huuh:

Sorry about that. I posted the wrong link. I edited it now.

The answer's 35, but I'm too lazy to explain in depth.  You split it into two cylinders. :D

[hide]split it into two cylinders
25pi + 12pi = 37pi[/hide]

THis problem basically involves a 2- dimentional object rotated into a three dimensional object. Therefore, resulting in a spherical section of the previously 2-d object.  :)

Ugh I still am really bad at visualizing this. Can somebody hopefully good with asy/geogebra diagram it? How does one rotate about the x axis? When I hear that I think about flipping it, rotating it 180 degrees, stuff like that. But that keeps it 2D..


Yes, this is necessary to post here, because:
1) This is directed only toward logic812 and MustItAlwaysBeMe.
2) My PM system reacts oddly to them; it seems that I can't communicate with them by means of a PM, only through posts like these.
3) The Indiana Forum is the best location to try to get them to talk to me, and it won't bother TOO many people (compared to other parts of the AoPS forum).

The rest of you can just ignore this if you want.

Anyways, my PM system, as most of you know, has been strange lately.  However, I realized that the strangest part of it has been occurring because I am a fool.  As levans and Valentin Vornicu have pointed out to me, deleting messages in the Outbox right after you have sent them notifies the receiver of the message, but doesn't actually deliver the message (since I deleted the message in the Outbox).  So I've been blindly sending messages, then deleting them in the Outbox, and wondering what the heck was happening.

However, one problem remains, which I propose a solution to (again, this is only for MustIt and logic).

I'm not sure if I'm able to send PMs to MustIt and logic, but I know that they sent me PMs, but I haven't received them.

My solution is that you two fell into the same lapses of judgment I have: perhaps you also have been deleting the messages in your Outbox after sending them, which is why I am notified of them, but do not receive them.

Or, my solution may not work and the whole PM system may just be screwed up between us three.  However, I have consistently been receiving PMs from other people; so it's only you two that I have a problem with.

Try sending a PM to me again; and if you've been deleting Outbox messages, don't do it again!

I hope this matter can be resolved.

ooh that's interesting.. I never knew that you could sort of cancel sent messages. I'm guessing the message stays in the outbox until the recipient reads it, then it gets moved to the sentbox. Wondered why they were separate :P

The PM was there, and I replied.  So I guess this topic has no more use (since MustIt doesn't seem to want to PM me).  Let it burn!

1. I don't know when I PMed you.  Or when I wanted to.  So why is this targeted to me?

2. No offense (possibly), but the fact that you kept deleting messages in your outbox is just funny. :D

Well, cognos599 told me that you also complained about not being able to PM me, but I guess he actually meant logic812.  Oh well then.  (BTW, you did PM me before.)

Now that all is resolved about my PM system, this topic can [i]officially[/i] die.

you did PM me once but i didn't get it. i know this because, when i registered, i said "e-mail me whenever i receive a private message", so when i checked my e-mail, i saw that i had gotten a PM from MathAndKnowledge, but when i checked my inbox, i had no PM's from him. so can you PM me and see if it works, MathAndKnowledge?


function
trigonometry
real analysis
real analysis unsolved

Let $f$ be a $\mathcal{C}^1$ function and $f(0,0)=0$. Show that any solution of differential equation $x''=f(x,x')$ has at most one zero.

[quote="Kent Merryfield"]Either I misunderstand what you mean, or it isn't true. Let $f(u,v)=-u.$ That's a smooth function and $f(0,0)=0.$ But the differential equation $x''=f(x,x')$ is $x''+x=0$, which has solutions such as $x(t)=\sin t$ which have many zeros.

Is there perhaps another condition?[/quote]
Sorry, I made a mistake. The problem should say to prove that every nonzero solution (which has a zero) has simple zero.

Then it's a direct consequence of the "only theorem in the whole Differential Equations course": the existence-uniqueness theorem.

Note that this is an autonomous differential equation; that simplifyies things.

Consider this initial value problem:

$x''=f(x,x');\, x(t_0)=0,\, x'(t_0)=0.$

The function $x(t)=0$ is a solution. (That's what we get from $f(0,0)=0.$) By the uniqueness side of the existence-uniqueness theorem, there are no other solutions.

A simple zero is a time $t_0$ at which $x(t_0)=0$ but $x'(t_0)\ne 0.$ We can certainly have many of these; my example above shows that.

A zero of higher order would be a time $t_0$ at which $x(t_0)=0$ and $x'(t_0).$ This cannot happen for any solution which is not identically zero.


prove that any positive integer greater than 7 can be written as a sum of 3s and 5s, i.e. in the form 3a+5b with non-negative integers a and b

[hide]Using the "Chicken McNugget Theorem":

The largest integer that can't be written as $ 5a \plus{} 3b$ is $ 5\cdot3 \minus{} 5 \minus{} 3 \equal{} 7$. Therefore, all greater integers can be expressed as $ 5a \plus{} 3b$.[/hide]

[hide]

$ 8\equal{}5\plus{}3$
$ 9\equal{}3\plus{}3\plus{}3$
$ 10\equal{}5\plus{}5$

Now we have formed 3 consecutive integers as a sum of 5's and 3's and we can form the rest by repeatedly adding 3's to these base integers.
[/hide]


linear algebra
matrix
inequalities
induction
geometry
3D geometry
sphere

here is a very interesting problem:
Find the convex hull of $O_n(R)$. That is the set of
matrices $M$ such that:
$M=\sum_{i}{a_iP_i}$ ,$\sum_{i}{a_i}=1$,$a_i \geq 0$,$P_i \in O_n(R)$

I think those are real matrices $ M $, with norm $ || M || \leqslant 1 $.
First of all, all matrices in this convex hull have norm $ \leqslant 1 $, because
orthogonal matrices always have norm $ 1 $. Now, if we denote by $ K $ the set of
all matrices of norm $ \leqslant 1 $, then it is a closed convex set, so
by the Krein-Milman theorem, it is closure of $ conv(E) $, where $ E $ is
a set of matrices which are extreme points of $ K $. If we will show that the
set $ E $ of extreme points of $ K $ are exactly the orthogonal matrices
$ O_{n}( \mathbb{R} ) $, then $ E $ is closed and since we are in finite
dimensional situation, $ K = \overline{conv(E)} = conv( \overline{E} ) = conv(E) $.
Now suppose that the matrix $ S \in K $ lies in $ E $. We know that every matrix
can be written as product of a non-negative symmetric matrix and an orthogonal
matrix. So we can write $ S = AU $, where $ A $ is nonnegative symmetric
and $ U $ is orthogonal. Of course, $ || A || \leqslant 1 $, so all eigenvalues
of $ A $ are in the interval $ [0,1] $. Now, if some eigenvalue $ \lambda $ is strictly
less than $ 1 $, then we can write that $ A = V^{-1}DV $, where $ V $ is orthogonal
and $ D = diag( \lambda_{1}, \lambda_{2}, \ldots , \lambda_{n} ) $, where
$ \lambda_{1} = \lambda $, so if we take some $ \lambda_{1}', \lambda_{1}'' \in
( -1,1 ) $, such that $ \lambda_{1} = \lambda = \frac{1}{2} ( \lambda_{1}' + \lambda_{1}'' ) $
and define $ A_{1} = V^{-1} diag( \lambda_{1}', \lambda_{2}, \ldots , \lambda_{n} ) V $,
$ A_{2} = V^{-1} diag( \lambda_{1}'', \lambda_{2}, \ldots , \lambda_{n} ) V $,
$ S_{1} = A_{1}U $, $ S_{2} = A_{2}U $, then of course $ S_{1} \neq S_{2} $,
$ ||S_{1}|| , ||S_{2}|| \leqslant 1 $ and $ S = \frac{1}{2} (S_{1} + S_{2}) $,
so $ S $ will not lie in $ E $. So we get that all eigenvalues of $ A $ are equal
to $ 1 $, so $ A = I $, so $ S = U \in O_{n}( \mathbb{R} ) $.
Now, if we have some $ S \in O_{n}( \mathbb{R} ) $, such that
$ S = \mu A + (1- \mu )B $, where $ 0 < \mu < 1 $ and $ || A || , || B || \leqslant 1 $,
then of course $ || A || = || B || = 1 $,
then for every $ x \in \mathbb{R}^{n} $ we have by triangle inequality
$ || x || = || Sx || = || \mu Ax + (1- \mu )Bx || \leqslant
\mu || Ax || + (1- \mu )|| Bx || \leqslant \mu || x || + (1- \mu )|| x || = || x || $,
so we have an equality in the triangle inequality, so for every $ x $, we have that
$ Ax $ and $ Bx $ are linearly dependent. Therefore $ A,B $ are proportional, so there
exists some $ a \in \mathbb{R} $, such that $ A = aB $, then because of
$ || A || = || B || = 1 $ we have $ a = 1 $ or $ -1 $, and because of $ S \neq 0 $,
$ a = 1 $, so $ A = B $, so $ A = B = S $.
So we see that every point $ S \in O_{n}( \mathbb{R} ) \subset K $ lies in
$ E $.

Wow! Great solution, Leva! Alekk, do you have a different approach? Don't be lazy and say it's the same!  ;)

great solution Leva ;-)
another solution without Krein Millman theorem,
(but I prefers yours):

We will show the result by induction on $n$. Here are the steps of
the proof:
$n=1$, OK since $O_1(R)=\{I,-I\}$
$n\geq 2$.
Let $M \neq 0$ be a matrix such that $|M| \leq 1$, $v_0 \in S$
such that $m=|M(v_0)|=max_{v \in S} \{ |M(v)| \}$ where $S$ is the unit sphere of $R^n$.
Let $P \in O_n(R)$ be an orthogonal operator such that $P(v_0)=(\frac{1}{m})u(v_0)$.
Show that the hyperplan orthogonal to $v_0$ is stable by $P^{-1}M$ (because $v_0$ is
an eigenvector of $M^{t}M$).
Conclude by induction.

Tell if you have problem, I don't have time
to write all the details.

(and what is the convex hull of $SO_n(R)$ ? I don't know ..)


hameye 3 tayihaye (a,b,c) az adade sahihe Na manfi ra biabid ke \[ 2^c\minus{}1|2^a\plus{}2^b\plus{}1\]

[hide]
Hint : avval sabet konin ke age $ a\equal{}ct\plus{}r$ va $ b\equal{}ct'\plus{}r'$ va $ (a,b)$ javab bashe oon moghe $ (r,r')$ ham javaban . bad ba namosavi hameye javabaro bedast biarin
[/hide]


By this i'm assuming we can still discuss problems from the 2007-2008 Handbook?

Yes, it is only the School test problems (sprint/target/team/countdown) that you cannot discuss.

[quote="Nerd_of_the_Ages"]Oh whoops. i have a blog entry from before the winter break saying i got certain score on the target round (but not on the problems themselves). i'll go delete it right now.[/quote]

Yes please do; no scores or any evaluation of its difficulty.  Best to just forget about it.

Because some schools might not have taken the test yet and we might be giving them answers or letting them know how hard the test is.

[quote="Jongao"]Do you mean scores or specific questions?[/quote]Rule of thumb is DO NOT DISCUSS ANYTHING ABOUT THE COMPETITION. Is that clear enough?

DO NOT DISCUSS ANYTHING RELATING THE 2008 MATHCOUNTS COMPETITION.

Do not mention your score. Do not mention the problems. If you see any discussion of the problems or a discussion of scores on the 2008 School Round before February 29, 2008, then email or PM us a link to the topic in which the discussion takes place.

why can't we dicuss our scores? It's not like we are giving answers.

it discusses difficulty.  Which can influence how much people prepare.

Another question can we mention what place we got into.

If you mean what place in your school, I doubt that many people would know about which school you go to anyways, it doesn't really matter much I guess....
I dunno...
 :wink:

When can we discuss it?
im assuming february, as that is when chapter competitions begin to take place?

Read the first post...2/29/08. Some schools are still choosing their team right up until their chapter competition, which can tke place up until the end of February.

Guys, it's like this for all mathcounts tests...the only thing you may say for school is if you made the team...but then again, we don't really need to know that. And for chapter and state, you can say what place you got in and what place your team got, but no mention of scores or the type of the test, or anything. Hope that clarifies things,
mamathbug

THE 2008 MATHCOUNTS SCHOOL ROUND IS A SCHOOL ROUND :P

but yeah, I was getting annoyed at how many people go like "the problem with <stuff> was <more stuff>"

Would people still be taking the school round still in February? I thought everyone should be done with those rounds by the end of January. Because according to the MATHCOUNTS site, it said that the chapter competitions take place in February. That means, by the start of February, all schools should have finished the 2008 School Competition, right?

Nope.  It all depends when the chapter competition is in February.  It could be at the very end, and if there is a big snowfall or something, it could actually get delayed past that, as it happened several years ago.  We'll let you know when it can be discussed.

Is the time that we can talk about our chapter round scores at the same time as when we can talk about our school round scores?

btw, i didn't do school round.

Yes, you're generally allowed to start talking about the school and chapter rounds at the same time. Thus, the school round is often eclipsed by the more challenging and important chapter round.

Look guys, sillyness like "My score was between -1 and 47" and arguing about dates is not necessary. Every year people argue about the dates, and [i]every[/i] year the answer is always the same: Never till we tell you can you discuss any Mathcounts round.

As the discussion on the FTW problems appearing on the real tests, and vice versa, I saw a question from the school round in FTW. although the school round is over...Just warning you. If an administrator would like to remove the problem for any reason, even though there is only 1 day left, you can pm me details on the question. Im also not sure if anyone else is still taking the school round..

remember: mathcounts has made many, many problems.  it's certainly possible that the same question on a previous test was in the school round from this year.  however, if you see a problem from the test, keep quiet about it if some people on this site try cheating.

are we allowed to discuss the school round yet???


The sides of the base of a triangular pyramid are 3, 4 and 5 feet and the altitude is 6 feet. What is the number of cubic feet in the volume of the pyramid?

Since the base of the pyramid is a right triangle, we have $ \frac{3(4)}{2}\equal{}6$ as the area of the base. Then, by using $ V\equal{}\frac13Bh$, we have $ \frac{6(6)}{3}\equal{}\boxed{12}$.


Find all polynomials $ P(x)$ with coefficients in $ R$ satisfy:

$ P(a\plus{}3b\plus{}5c)\equal{}P(a)\plus{}3P(b)\plus{}5P(c), \forall a,b, c \in R$ and $ ab(a\plus{}b\plus{}c)\equal{}3b^3$

$ a\equal{}b\equal{}c\equal{}0 \Rightarrow P(0)\equal{}0$
$ a\equal{}b\equal{}c \ne 0 \Rightarrow P(9x)\equal{}9P(x)$
$ \Rightarrow P(x)\equal{}Cx$

... what about the condition $ ab(a\plus{}b\plus{}c)\equal{}3b^3$ (strange, $ b$ not canceled from both sides) ... ?


induction
pigeonhole principle
combinatorics unsolved
combinatorics

49 student solve three problem with mark in {0,1,...,7}.Prove that there exist two studentA and B such mark of each problem A $\leq $ mark of B

Suppose by absurd that $\forall A, B$ exist marks number $i,j \in \{1,2,3\}$ such that: ${mark_{i}(A)}>{mark_{j}(B)}$

Under this hypotesis, let $C$ be the student with higher marks in all three problems, if we choose another student $A$ we have that do exist marks number $i,j \in \{1,2,3\}$ such that: ${mark_{i}(A)}>{mark_{j}(C)}$. But this is absurd, it follows that your thesis it's true.

Seems wrong....you don't use anywhere the fact how many students there are.

I don't know where is the mistake in this.
The problem is about finding the maximum possible cardinality of a set $X\subset \left(\mathbb{N}^{*}_{7}\right)^{3}$, such that $\forall\;\; (a,b,c),(a',b',c')\in X,\;\;\neg (a\geq a'\wedge b\geq b' \wedge c\geq c')$, to which we refer as suitable set; of course, the definition can be extended to subsets of $({N}^{*}_{n})^{m}$ for any $m,n\in \mathbb{N}$.
First of all we will prove the following fact: given a suitable set $Y\subset \left({N}^{*}_{7}\right)^{2}$, with $|Y|\geq 8-m$, $m\in \mathbb{N}^{*}_{7}$, $\forall (a,b)\in Y\;\;\;d(a,b)=|a+b-7|\leq m$. Suppose, on the contrary, that it existed an $(a,b)\in Y$ such that $d(a,b)>m$ and suppose without, loss of generality (due to mapping $(a,b)\mapsto(7-a,7-b)$),   $a+b\leq 7$. $\forall (c,d)\in Y,\;\;(c,d)\neq (a,b)\;\;\; (c<a\wedge d>b)\vee(c>a\wedge d<b)$; but since $\forall (x,y),(w,z)\in Y\;\;x\neq w\wedge y\neq z$, by the pdgeonhole principle there are at most $min \{ a, 7-b \}$ elements of $Y$ which satisfy the first of the two conditions and $min\{ b, 7-a \}$ for the second; we get a total: $|Y|\leq 1+min\{ b,\; 7-a \}+min \{ a,\; 7-b \}=$
$=1+min\{ b,\; m+b \}+min \{ a,\; m+a \}=1+a+b<8-m$.
Now, returning to our previous $X$, see $\forall m\in \mathbb{N}^{*}_{7}\;\;D(m)=\{(a,b)\in \left({N}^{*}_{7}\right)^{2}\;|\;(a,b,m)\in X \}$ is a suitable set; in addition $D(i)\cap D(j)=\emptyset \forall i,j\;\;i\neq j$. We will prove by induction that given a family $F$, of disjoint suitable subsets of $Y\subset \left({N}^{*}_{7}\right)^{2}$, $|\bigcup F|\leq \sum_{i=1}^{|F|}a_{i}$ where $a_{1}=8,\;\;a_{2}=7,\;\;a_{3}=7\;\;a_{4}=6\;\;a_{5}=6 \ldots$.  The base case $|F|=1$, $|\bigcup F|\leq 8$ follows from the pidgeonhole principle and the definition of suitable set. Now suppose we had proved the statement $\forall F'\;/\;|F'|=|F|-1$. Take $s= min \{ |S|\;/\;S\in F \}$, then $\bigcup F \subset \{(a,b)\;/\;d(a,b)\leq 8-s \}=K(s)$ and $|\bigcup F|\leq |K(s)|$. If it was $s>a_{|F|}$, then letting $l=|F|-1$ if $|F|\equiv 0 \;\;(mod\;2)$ and $l=|F|-2$ otherwise, we have $|X|\leq |K(a_{|F|}+1)|=\sum_{i=0}^{l}a_{i}<\sum_{i=0}^{|F|}a_{i}$ and we are done. On the other hand if $s\leq a_{|F|}$, by applying the induction hypothesis to $F-\{S'\}$, where $S'$ is such that $|S|=a_{|F|}$, we get that $|\bigcup F|\leq a_{|F|}+\sum_{i=1}^{|F|-1}a_{i}=\sum_{i=1}^{|F|-1}a_{i}$, and that's all.
With the last result and the observation at the beginning of the previous paragraph (in this case $|F|=|{N}^{*}_{7}|=8$) we can conclude $|X|\leq \sum_{i=1}^{8}a_{i}=8+7+7+6+6+5+5+4=48<49$.
After seesing the proof is esay to construct the equality case (well, actually for me it has happened exactly the opposite way).
Sorry in advance.


calculus
derivative
limit
calculus computations

If the derivative of $ y\equal{}k(x)$ equals $ 2$ when $ x\equal{}1$, what is the derivative of

a) $ k(2x)$ when $ x\equal{}\frac{1}{2}$?
b) $ k(x\plus{}1)$ when $ x\equal{}0$?
c) $ k(\frac{1}{4}x)$ when $ x\equal{}4$?

[hide]I have no idea how to start with this.
$ \frac{dk(x)}{x}$
$ \frac{dk(1)}{x} \equal{} 2$[/hide]

Use
\[ f'\left( {x_0 } \right) \equal{} \mathop {\lim }\limits_{x \to 0} \frac{{f\left( {x \plus{} x_0 } \right) \minus{} f\left( {x_0 } \right)}}
{x}
\]
or chain rule, e.g.,
\[ \begin{gathered}
  g\left( x \right) \equal{} f\left( {2x} \right) \hfill \\
  g'\left( x \right) \equal{} 2f'\left( {2x} \right) \hfill \\
\end{gathered}
\]
this yields
\[ g'\left( {\frac{1}
{2}} \right) \equal{} 2f'\left( 1 \right) \equal{} 4.
\]

Applying that to $ b$ and $ c$ I got $ 2$ and $ 1/2$ as my answers.
Thanks!


calculus
integration
inequalities
triangle inequality

[hide] this uses the triangle inequality, that in a triangle, the sum of any two sides is greater than the third.  so the smallest integral value of the third side is 52, and the greatest is 130 so the average is 91[/hide]

[quote="mr. math"]Shang is making a triangle that has two side lengths of 91 and 40. what is the mean of all integral values that are possible for the third side?

how would u do this?

and can someone explain the property involved in this problem?

thanks.[/quote]

[hide] $x$ is the 3º side of the triangle. Then, by the triangle inequality, we have that $91+40>x$ and $91-40<x$. Now, we have that $51< x < 131$. Then, you can finish the problem.

The property involved here says that, for any triangle with sides $a, b$ and $c$m with $c\ge a \ge b$, you have $a+b>c$, $a-b<c$[/hide]

This problem uses the triangle inequality.

[hide="read this if you want to try to figure it out for yourself"]If you have 3 sticks of length 1, 1, and 1000, can you make a triangle with them?

What about if you had sticks of length 1, 2, and 3?[/hide]

[hide="more..."]
If two of the sides of a triangle aren't as long as the third, can they reach all the way across to make a triangle?[/hide]

[hide="more general..."]What about if the sides where a, a, and b?

a, b, and c?[/hide]

[hide="if not..."]For a triangle with sides of length $a$, $b$, and $c$, the sum of the lengths of any two sides of the triangle must be at least the length of the longest side, or the sides won't be able to connect.  Thus,
$a+b>c$
$a+c>b$
$b+c>a$[/hide]


trigonometry
search
inequalities unsolved
inequalities

First, denote $21ab=xy,2bc=yz,8ca=zx$, where $x,y,z$ are positive reals. Second, denote $x=\cot{A},$ etc,..., where $ABC$ is acute triangle.

And final we need minimum of $m\tan{A}+n\tan{B}+p\tan{C}$, where $m,n,p$ are positive reals.

search in posts of mine and topic of gemath, please  :wink:


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

show that for each $n$ integer there exists a polynomial$P_n\in \mathbb{C}[x_1,\dots,x_n]$ such that for every $n*n$ matrix $A$ over $\mathbb{C}$ such that $det(A)=P_n[Tr(A),\dots,Tr(A^n)]$.

My idea is that it all comes to solving a linear system of $n$ equations with $n$ unknown terms.

follows from Newton's formula about symetric polynomials.


We have $16\times 16$-square grid divided into 16 smaller squares, each of which has side-length 4 squares. In every grid we can put one number between 1 and 16. In every row, every coloum and every small square we can not have two or more same numbers?

Two players will mark at each turn some number between 1 to 16 to some non-marked square. The loser is the one who can not mark a square according to the rules of the game. Which one, first or second player, will win if both players are playing optimally?

The question was easier than I thought :blush:. We have even number of small squares. If the first player plays his move anywhere, the second player can play a move to the square which is symmetric to the middle point of the board and put there the same number as first player. No two such a moves will be on a same square nor the same line. Because the total number of small squares is even, the second player will put the last move in the case that the whole board will be covered. Therefore the second player will win the game.


Determine all the pairs of positive integers $a$, $b$ such that $8a+1$ is divisible by $b$, and $8b+1$ is divisible by $a$.

There are more, like (9,73).
[hide="sketch"]
Observe that $a,b$ are relatively prime (if $d$ divides $a$ and $b$, then $d$ divides $8b+1$ and $b$, so $d$ divides $8b+1-8b=1$). Also, $a$ and $b$ both divide $8a+8b+1$. It follows that $ab$ divides $8a+8b+1$.

Now we must have $8a+8b+1=kab$ for positive integer $k$: clearly $k,a,b$ are odd and $k$ is no greater than 17 (if $a\geq b$, then $8a+8b+1\leq 17a\leq 17ab$), and $k^{2}ab=8k(a+b)+k$ so that
\[(ka-8)(kb-8)=k+64\]
The rest is casework... also a sharper bound for $k$ when $a\geq 3$ would probably help.
[/hide]


geometry
perimeter
analytic geometry
rotation
percent

Equiangular triangle ABC has a perimeter of 60 inches. Point Q is placed in or on the triangle, and the lengths AQ, BQ and CQ are added together. What is the minimum value of the sum of the three lengths?

What is the next number in the following sequence: 2,6, 30, 210, 2310, 30030...?

Points A(-4,1), B (-1,4), C (-1,1) are the vertices of the triangle ABC. What will be the coordinates of the image of point A if triangle ABC is rotated 90 degrees clockwise about the origin? ( I really don't get what it really means!)

Bill scores exactly 28 percent correct on the first part of a true/false test. On the second part of the True/ False test he was correct on exactly four questions. If he correctly answered one-third of the questions on both parts of the tests combined, what is the fewest possible number of questions on the second part of the test?

In the sequence 1,_,9,_25,_,... the terms in the even positions are missing while the terms in the odd position shave the property that the nth term of the sequence is the square of n. Let each missing term be the arithmetic mean of the two terms on either side. What is the 50th term of this sequence?

1.
[hide]Draw ABC and label it... Q should be the middle of the circle. Using 30-60-90 triangles and given their sides $ s, s\sqrt3,$ and $ 2s$, Let any distance (AQ, BQ, CQ) be the longest side $ 2s$, and half of any side of the triangle be $ s\sqrt3$.
$ 10 \equal{} s\cdot\sqrt3$
$ \frac {10} {\sqrt3} \equal{} s$
Rationalize
$ s \equal{} \frac {10\sqrt3} {3}$
Times three for the sum of all of them,
$ \boxed {10\sqrt3}$[/hide]

2.
[hide]You're just multiplying by consecutive primes, so multiply $ 30030$ by the next prime to continue the sequence.
$ 30030 \cdot 17 \equal{} \boxed {510510}$[/hide]

3.
[hide]Draw ABC and... rotate it, which I'm not sure how to explain. I havent been taught this either, but my guess is that C goes to $ (1,1)$, B goes to $ (4,1)$ and A goes to $ \boxed {(1, 4)}$. I'm not sure, though...[/hide]

4.
[hide]If Bill scored $ 28$ percent exactly, then the smallest number of questions possible on the first part would be the denominatior of $ \frac {28} {100}$ simplified. Since $ \frac {7} {25}$ is its simplification. Then we set $ \frac {7\plus{}4} {25\plus{}x}$ equal to $ \frac {1} {3}$.
$ \frac {11} {25\plus{}x} \equal{} \frac {1} {3}$
We see $ 25\plus{}x \equal{} 33$, so $ \boxed {33}$ is our answer. We could easily find x, but it isn't necessary.  [/hide]

5.
[hide]So, any odd $ nth$ term of the sequence is $ n^2$. Since the evens are the arithmetic mean of the two surrounding squares, we just need to compute $ \frac {49^2 \plus{} 51^2} {2} \equal{} \boxed {2501}$.[/hide]

Just for 4, you need to find x. That's the answer to the question.

Ohh, sorry, misread. Our answer is [hide]8.[/hide]


[hide]$\cot x+\cot y=\frac{\tan x+\tan y}{\tan x\tan y}=\frac{25}{\tan x\tan y}=30\implies \tan x\tan y=\frac{5}{6}$

$\tan\left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{25}{1-\frac{5}{6}}=6\cdot 25 = \boxed{150}$[/hide]

[quote="pianoforte"]If $\tan x+\tan y=25$ and $\cot x+\cot y=30$, what is $\tan(x+y)$?[/quote]

L[hide]et $\tan x=a$, and $\tan y=b$.

This means that $a+b=25$, and $\frac{1}{a}+\frac{1}{b}=30$.

Hence $ab=\frac{5}{6}$.

We are looking for $\frac{a+b}{1-ab}$, which is $150$

$150$[/hide]


Show that there exist 8 consecutive positive intergers such that none of them can not be expressed as \[ |7x^2\plus{}9xy\minus{}5y^2|\] for \[ x, y\] are intergers