Sorry about posting that question twice. I have edited the post with another question.

[quote="mr. math"]Could anyone tell me the formula for the sum of the first n perfect squares? Its come up on a couple of problems I've done recently.[/quote]
$ 1^2 \plus{} 2^2 \plus{} \ldots \plus{} (n \minus{} 1)^2 \plus{} n^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$

The proof is in the [url=http://www.usamts.org/About/U_Gallery.php]proofs without words gallery[/url].

Another proof. Let
$ a_k\equal{}\frac{(2k\minus{}1)^3}{24}$
We see that $ a_{k\plus{}1}\minus{}a_k\equal{}k^2$ and now it suffices to consider
$ \sum_{i\equal{}1}^n (a_{i\plus{}1}\minus{}a_{i})\equal{}a_{n\plus{}1}\minus{}a_1\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$ ;)

Also see [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]here[/url] for a proof by Hockeystick.


[quote="joejia"]A man digs a hole 6 inches deep for a post to hold his mailbox. In the square base of the post is 4 in. $\times$ 4 in., how many cubic inches of dirt will be displaced by the post?[/quote]
[hide]
$4*4*6=16*6=96$[/hide]

[quote="IntrepidMath"][hide]
6*4*4=96 Cm^{3}$
[/hide][/quote]
do you mean
[hide="this?"]
$6*4*4=96 Cm^{3}$
[/hide]

[hide] $ 4 \cdot 4 \cdot 6 = \boxed{96in^3}  $[/hide]


Consider the points $ A(a) , B (b) , C(c)$ , where $ a,b,c \in \mathbb{C}$ and $ |a|\equal{}|b|\equal{}|c| \neq 0$ . We have that $ OB \bot OC$ and there exist $ \alpha \in \left(0 ; \frac{\pi}{2} \right)$ such that $ a \equal{} b \cdot \cos{\alpha} \plus{} c \cdot \sin{\alpha}$ .
It is true that $ A \in \stackrel{\huge{\frown}}{BC}$ ? Please explain your answer :)

What is supposed to be the midpoint of this arch? if it's $ O$, I think we need the additional constraint $ |b| \equal{} |c|$.

Edit: Yeah, I do have to learn to read...

$ O$ is the origin of the complex plane , also center of the circumcenter of $ \triangle{ABC}$ .

[hide="Solution"]
$ OB\perp OC \implies c \equal{} bi \implies a \equal{} b(\cos \alpha \plus{} i\sin \alpha) \equal{} be^{i\alpha}$ and since $ \alpha\in\left[0,\frac{\pi}{2}\right]$ and $ \angle AOB \equal{} \alpha$ the result follows immediately.
[/hide]


Prove or disprove that 7 is expressible as the sum of the squares of 3 rational numbers.

Assume of sake of contradiction that we have solutions. Let the three rational numbers be $ \frac{a}{d}$, $ \frac{b}{d}$, and $ \frac{c}{d}$ where $ a$, $ b$, $ c$, and $ d$ are integers and $ ((a,b,c),d)\equal{}1$. (If that common factor is greater than 1, divide it out of all of a,b,c which does not change the fractions). If we clear the fractions, we get
\[ 7d^2\equal{}a^2\plus{}b^2\plus{}c^2\]

Note that the squares modulo $ 7$ are $ 0$,$ 1$,$ 4$. We have that the LHS is $ 0\pmod{7}$, so the right side is  as well. So then the only combination of those three mods that gives zero mod 7 is 0,0,0. But then if $ 7|a^2$, then $ 7|a$, so $ 7^2|a^2$, and $ 7^2|(a^2\plus{}b^2\plus{}c^2)$, so $ 7^2|(7d^2)$, so $ 7|d^2$, so $ 7|d$, which is a contradiction.

Now $ 2$ is in fact a square mod $ 7$, since $ 3^2 \equiv 2$. So your argument does not rule out the possibility that $ a \equiv 4$, $ b \equiv 2$, $ c \equiv 1$ mod $ 7$.

(This corresponds nicely to the fact that modulo an odd prime, exactly half of the non-zero residues are quadratic).

Instead, write the equation, as you do, but consider it modulo $ 8$ instead. Then

$ a^2\plus{}b^2\plus{}c^2\plus{}d^2 \equiv 0,$

and since the only squares mod $ 8$ are $ 0,1,4$, then none of $ a^2,b^2,c^2,d^2$ can be $ 1$ mod $ 8$. Thus $ a,b,c,d$ are all even, a contradiction to the fact, that we canceled all common factors at the beginning of the argument.

[quote="ghjk"]The answer is no. To prove it, we use the following lemma:
"A natural number is the sum of the squares of three rational numbers if and only if it is the sum of the squares of three integers"
And 7 can't be expressed as the sum of the squares of three integers[/quote]
How would we prove that?

Well, the square of any integer is either positive or zero.  The only integer squares less than 7 are 0,1, and 4.  It is pretty clear that no combination of 3 of these numbers will add up to 7.

(Unless you were asking how to prove the lemma.)

Altheman used squares modulo 7, not just smaller than 7. That includes 2.

I thought diophantient was referring to ghjk's comment:
[quote="ghjk"]And 7 can't be expressed as the sum of the squares of three integers[/quote]

The solution for the lemma I mentioned above:
Suppose that a natural number n is the sum of the squares of three rational numbers.Reducing all the three rational numbers to the same denominator, we may write:m^2n=a^2+b^2+c^2 where a,b,c are integers.If n=4^l(8k+7) where k,l are integers >=0, let's putting m=2^r(2s+1)(s,r are non-negative integers) and m^2n=4^(k+r)(8t+7) where t is also a non-negative number. However, according to the Gauss's theorem, this is impossible because m^2n=a^2+b^2+c^2. So m^2n must be the sum of the squares of three integers.Thus we see that this condition is necessary and also sufficient as well :)
PS: Altheman's proof is exactly the same as the proof of Gauss's theorem! :wink:


I have this image, which I made in Word and took a screen shot of (yes, I was that desperate =\):
[img]http://img100.imageshack.us/img100/9489/39thcmoquestion1ty9.gif[/img]
I was wondering if there's any way this could be done in LaTeX easily, and furthermore if it could be done with the math tags. So far, the only solution I could find was to use arrays and "X" to denote a filled square, which, as you may imagine, is less than ideal.

Obviously it can be done using packages but you want it to work on the forum. I suggest using the picture environment to get the squares, see [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]The Not So Short Introduction to LaTeX2e[/url] [i]5.2.6 \multiput and \linethickness[/i] to do this. That won't produce shading but try filling the squares to be shaded with lots of lines or dots to simulate shading.


Notice that there are 94 terms in your product and

$ \big( 1\minus{}\frac{1}{10^6}\big)\big(1\minus{}\frac{1}{10^7}\big) \ldots \big(1\minus{}\frac{1}{10^{99}}\big)$

$ \equal{} 1\minus{}\sum_{i\equal{}6}^{99} \frac{1}{10^i} \plus{} \sum_{i\equal{}6}^{98}\frac{1}{10^i 10^{i\plus{}1}} \minus{} \sum_{i\equal{}6}^{97}  \frac{1}{10^i 10^{i\plus{}1}10^{i\plus{}2}} \plus{} \cdots \plus{} \frac{1}{10^610^7\cdots 10^{99}}$

This should help you finish the problem.


Let $k$ be an integer and let $n=\sqrt[3]{k+\sqrt{k^{2}+1}}+\sqrt[3]{k-\sqrt{k^{2}+1}}+1$. Prove that $n^{3}-3n^{2}\in\mathbb{Z}$.

[quote="JBL"]I think it is just wrong.[/quote]
I edited  :oops:

[quote="cincodemayo5590"]I still think it's wrong; I'm pretty sure it should be:
\[n=\sqrt[3]{ k+\sqrt{k^{2}-1}}+\sqrt[3]{ k-\sqrt{k^{2}-1}}+1 \]
[hide="Since it works out nicely:"]
Let $a=\sqrt[3]{ k+\sqrt{k^{2}-1}}$ and $b=\sqrt[3]{k-\sqrt{k^{2}-1}}$. Then, we have:
\[(n-1)^{3}=(a+b)^{3}=2k+3ab(a+b)=2k+3(n-1) \implies n^{3}-3n^{2}=2k-2 \in \mathbb{Z}\]
[/hide][/quote]

What is the actual problem? :P  I'm wondering if my guess was correct, but I'm sure the problem is still incorrect as stated. (Test $k=3$.)

[quote="Altheman"]OMG THIS HAS A STRIKING RESEMBLANCE TO THE ARML HW PROBLEM

ooo nooo

they just used this:

$\sqrt[3]{k+\sqrt{k^{2}-n^{3}}}-\sqrt[3]{k-\sqrt{k^{2}-n^{3}}}$, n=-2, k=6

ONO[/quote]

Rofl, DUH.


[hide]
Factor $1560$ to get $2^{3}\times 3 \times 5 \times 13$. Obviously, $13$ is the smallest integer whose factorial is divisible by this.[/hide]

:!:

What's the point getting through all that factoring? How do you find an easy way? Shortcut?

[quote="ckck"]What is the least positiver integer n such that 1560 divides n!?[/quote]
[hide]
1560 prime factors into 2^3*3*5*13
this needs at least 13!
13! has 8, 3, and 5
so $n=13$
[/hide]

ProtestanT: Thanks for copying my exact solution.

MiG29: There is no faster way. Factoring is very fast, especially in cases like this where all of the factors are really small, so you can factor it down to a number like 65 really fast.

Thank you for explaining. This is a good problem.

[quote=ariedel][quote=Dude156]I know that you probably don't know this, but please refrain from reviving old threads. Thanks![/quote]

No only revive if you have an important question or something.  Just saying "this is cool" or posting another solution to an already resolved problem is not good.[/quote]

Wait posting another solution is fine as long as the same solution hasn't been posted, there are a lot of different solutions to a problem, many people want to know all of them.


Both my teachers that I want a recommendation letter from are leaving next year. (Math and History) What should I do?

Is it possible that I could approach them in two years (I'm a sophomore right now) when they are at another school and ask them to write a recommendation? In other words, do the recommendations have to come from teachers that are currently teaching?

Another one of my teachers is retiring, am I allowed to ask her in two years and ask her to write me a recommendation?

It would make more sense just to ask them to write it now, and then file it away somewhere safe (or ask them to do so).  Of course, it's quite possible that over the next 2 years you'll find a teacher who you'd rather have write your recommendation.

I think its better in general to have letters from teachers you had your junior or senior year.  These two might have been great teachers who could write you outstanding letters, but I'm sure you'll be able to find 2 more within the next 2 years.


you'd probably get a quicker response if you post in the high school forum.  but anyway...

$\frac {(-3^{-1} x^2 y)^{-3}}{6x^{-9}} = \frac{-3^3 x^{-6} y^{-3}}{6x^{-9}} = \frac{-9x^{-6+9}}{2y^3} = \frac{-9x^3}{2y^3}$

Really, it is for the Intermediate Section.

oh, there's an intermediate section?  i forgot about that...

[quote="evoss001"]Simplify:

           (-3^-1 x^2 y)^-3/6x^-9[/quote]
It's kind of hard to tell what expression you mean.  It could be $-(3^{-1})$ or $(-3)^{-1}$, and whether or not there are parenthesis around the exponents makes a difference too.

yeah, though in this case (-3)^-1 and -3^-1 amount to the same thing


How can I show that the function $f(x,y,z) = (1-x^2-y^2+z)e^{x+y-z}$ has a largest value on $x^2 + y^2 - z \le 1$ ?

I have to find a good upper bound on the function (smaller than $e^{1/2}$) that is true in the given region as long as we are far away enough from the origin. But I can't find such an upper bound.

Let $u=z-x^2-y^2\ge -1$. You just have to show that $(1+u)e^{x+y-x^2-y^2-u}=(1+u)e^{-u}e^{1/2}e^{-(x-\frac 12)^2}e^{-(y-\frac12)^2}$ tends to $0$ as $(x,y,u)\to\infty$, which is not hard since each factor is bounded and tends to $0$ as the corresponding variable tends to infinity ($+\infty$ in the case of $u$). Of course, once you wrote the formula this way, it becomes also obvious that to maximize the function, it is necessary and sufficient to maximize each factor and it is quite clear how to do that.

[quote="Kalle"]How did you see what to do so easily?[/quote]
That's a much harder question, not to say that it is not exactly calculus. :) The general idea is that it almost always makes sense to play with variables a bit to simplify the expression. In this particular case it was relatively clear that the combination $z-x^2-y^2$ is of importance, so I just took it for a new variable and looked at what resulted from such substitution.


algebra
polynomial
number theory unsolved
number theory

prove that $ \sqrt[3]{2}\plus{}\sqrt[3]{3}\plus{}\sqrt[3]{5}$ is rationa.

Did you mean to say "irrational"? This number is an algebraic integer, so if it had been rational it would have been an ordinary integer, which clearly it isn't (it's greater than 4 and smaller than 5).

Proving that the sum of two algebraic integers (= roots of monic polynomials with integer coefficients) is another algebraic integer is not quite trivial but it's not too hard, either. Proving that a rational algebraic integer must be a rational integer (namely, belong to $ \mathbb{Z}$) is even easier. Here's a quick introduction covering this material:

http://modular.fas.harvard.edu/papers/ant/html/node14.html

yes i take a mistake in writing. you right i didnt type irrational. :maybe:


modular arithmetic
number theory proposed
number theory

If $n$ is is an integer such that $4n+3$ is divisible by $11,$ find the form of $n$ and the remainder of $n^{4}$ upon division by $11$.


How come the banner at the top of  the screen says that there is a Game Theory Math Jam today, but the Math Jam page doesn't have it listed?  Is there or isn't there a game theory Math Jam today?

The Game Theory Math Jam has been postponed.  I simply forgot to remove the banner until now.

[quote="LynnelleYe"]I was sooo disappointed when I thought I'd missed it, and then I saw your post and jumped for joy (mentally, not literally)...so when will it be held now?[/quote]

Joel, the Cal Tech grad student who will be running it, needs to find the right date in his schedule.  This time he was pulled away by school duties.  We'll have to see when he is able.

[quote="Wumbate"]I noticed that it has been completely pulled from the list.  Are you planning to have it in the near future or at a date in the far future (after October, for example)?[/quote]

It will be back on the list when there is a date for it.


1. Which acid should be stored in plastic containers rather than in glass ones?

(A) Hydrofluoric acid
(B) Nitric acid
(C) Phosphoric acid
(D) Sulfuric acid

2. Which gas is odorless?

(A) $\text{CH}_{4}$
(B) $\text{HCl}$
(C) $\text{NH}_{3}$
(D) $\text{O}_{3}$

#1 is $\boxed{A}$.  Fluoride breaks the bonds in $SiO_{2}$.

#2 is $\boxed{D}$.  Ozone is odorless.  This is one of those things you either know or don't.  Methane (edit: oops) and ammonia definitely have smells.  Apparently hydrogen chloride has a "pungent odor."

[quote="t0rajir0u"]#1 is $\boxed{A}$.  Fluoride breaks the bonds in $SiO_{2}$.

#2 is $\boxed{D}$.  Ozone is odorless.  This is one of those things you either know or don't.  Methane and ammonia definitely have smells.  Apparently hydrogen chloride has a "pungent odor."[/quote]
The answer for #2 was given as $\boxed{A}$.

Hmm.  Apparently what people smell as methane is usually other products that are released along with methane either from bacteria or from natural gas.

But according to two sources, what most people think is the smell of ozone, say, in a lightning storm is actually just a result of chemical changes, not of the ozone itself.

If natural gas were delivered as produced (mostly methane, with smaller quantities of ethane, propane, and other very light hydrocarbons), it would indeed be odorless. That would be a bad thing from a safety point of view. Hence commercially delivered natural gas is intentionally spiked with a tiny amount of an intensely odoriferous gas (a mercaptan, meaning a compound with a free -SH group). If methane is produced in some biological rotting process, then there would be other biological side products, including some with their own -SH groups. But growing plants actually give off methane (see the latest issue of [i]Scientific American[/i]) in ways that do not involve anaerobic bacteria.

I'm not quite sure that "pungent odor" quite captures the effect a small amount HCl in the air would have on you. Your eyes would water, your nose would sting - HCl is in fact intensely toxic. Just rips up your eyes, your lung lining, and any other sensitive surfaces. I heard about someone in an organic chemistry labs who tossed some excess anhydrous aluminum chloride into a trash can. That cleared the room - and the offending student was invited not to return. (Anhydrous aluminum chloride reacts with small amounts of water to produce HCl gas.)


$P$ is a point located in $\triangle ABC$ such that $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=\overrightarrow{0}$. Prove that $P$ is the centroid of $\triangle ABC$.

You mean, if we define the centroid as a meeting point of the medians? Since what you've written is general definition of the centroid of a system of points.

[quote="Farenhajt"]You mean, if we define the centroid as a meeting point of the medians?[/quote]
Yes.

Let $M$ be the mid point of $BC,$ we have $\overrightarrow{PA}+2\overrightarrow {PM}=\overrightarrow{0}.$

[hide="Proof"]Denote the midpoint of $BC$ by $A_1$. It is quite obvious that $\overrightarrow{AA_1}={1\over 2}(\overrightarrow{AB}+\overrightarrow{AC})$ (by the rule of the parallelogram).

(Or if it isn't obvious, here's a quick proof: Adding the identities $\overrightarrow{AA_1}=\overrightarrow{AB}+\overrightarrow{BA_1}$ and $\overrightarrow{AA_1}=\overrightarrow{AC}+\overrightarrow{CA_1}$ we get $2\overrightarrow{AA_1}=\overrightarrow{AB}+\overrightarrow{BA_1}+\overrightarrow{AC}+\overrightarrow{CA_1}$. But as $\overrightarrow{BA_1}+\overrightarrow{CA_1}=\overrightarrow{0}$ - collinear vectors of the same length and opposite directions - the last identity simplifies to $2\overrightarrow{AA_1}=\overrightarrow{AB}+\overrightarrow{AC}.$)

Now, since $P$ divides $\overrightarrow{AA_1}$ in the ratio $1: 2$, it follows that $\overrightarrow{PA}=-{2\over 3}\overrightarrow{AA_1}=-{1\over 3}(\overrightarrow{AB}+\overrightarrow{AC}).$ Similarly, we have $\overrightarrow{PB}=-{1\over 3}(\overrightarrow{BA}+\overrightarrow{BC})$ and $\overrightarrow{PC}=-{1\over 3}(\overrightarrow{CA}+\overrightarrow{CB})$. Therefore, $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=-{1\over 3}(\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{AC}+\overrightarrow{CA}+\overrightarrow{BC}+\overrightarrow{CB})=\overrightarrow{0}.$[/hide]


calculus
integration
real analysis
real analysis unsolved

thank you ,fedja

fixed:

improper integal $ \int_{ \minus{} 1}^{1}\frac {1}{\sqrt {1 \minus{} t^{2}}}$

1.Is $ \frac {1}{\sqrt {1 \minus{} t^{2}}}$  legesgue measure on $ \left[ 1, \minus{} 1\right]$?

2.Dose the  formula of integration by parts hold over legesgue integral ?

Do you mean

1) Is $ \frac 1{\sqrt{1\minus{}t^2}}$ Lebesgue integrable on $ [\minus{}1,1]$?

2) Under what assumptions does the integration by parts formula hold for the Lebesgue integral?

? (I couldn't really interpret the second question as a "yes/no" one meaningfully, so, perhaps, you had in mind something else there :?)


Estw a,b,c,d mh arnitikoi pragmatikoi arithmoi tetoioi wste : $ a\plus{}b\plus{}c\plus{}d\equal{}2$ .Na apodeixtei oti:

        $ ab(a^2\plus{}b^2\plus{}c^2)\plus{}bc(b^2\plus{}c^2\plus{}d^2)\plus{}cd(c^2\plus{}d^2\plus{}a^2)\plus{}da(d^2\plus{}a^2\plus{}b^2) \leq 2$

Ειναι
$ ab(a^2 \plus{} b^2 \plus{} c^2) \leq ab(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$,
$ bc(b^2 \plus{} c^2 \plus{} d^2) \leq bc(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$,
$ cd(c^2 \plus{} d^2 \plus{} a^2) \leq cd(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$,
$ ad(d^2 \plus{} a^2 \plus{} b^2) \leq ad(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$,
$ 0 \leq ac(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$,
$ 0 \leq bd(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)$.

Σε ολα τα παραπανω εχουμε ισοτητα χωρις να ειναι ολα τα ανα 2 γινομενα 0, αν: $ a \equal{} c \equal{} 0$, η $ a \equal{} d \equal{} 0$, η $ c \equal{} b \equal{} 0$, η $ c \equal{} d \equal{} 0$.

Προσθεση των παραπανω κατα μελη δινει:

$ LHS \leq (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)(ab \plus{} bc \plus{} ca \plus{} ad \plus{} bd \plus{} cd) \equal{} \frac {1}{8}4(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)(2(ab \plus{} bc \plus{} ca \plus{} ad \plus{} bd \plus{} cd)) \leq \frac {1}{8}[a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 2(ab \plus{} bc \plus{} ca \plus{} ad \plus{} bd \plus{} cd)]^2 \equal{} \frac {1}{8}[(a \plus{} b \plus{} c \plus{} d)^2]^2 \equal{} \frac {16}{8} \equal{} 2$.

Ισοτητα εχουμε για $ (a,b,c,d) \equal{} (1,1,0,0)$ και ολες τις μεταθεσεις αυτων.

ΥΓ: Ειναι σαν του Αρχιμηδη 2009

μια ευκολη λυση γινεται και με κυρτοτητα ως προς καθε μεταβλητη :wink:

Αν κανεις αντικατασταση και μετα παραγωγηση και προκειψει κυρτη συναρτηση τοτε εχεις το δικαιομα να πεις οτι το μεγιστο ειναι στις κορυφες του 4-διαστατου κυβου, δεν μπορεις ομως να πεις το ιδιο αν δεν γινει πρωτα αντικατασταση απο τη συνθηκη.


geometry
geometric transformation
reflection
rotation

Let F be the midpoint of side BC of triangle ABC.
Construct isosceles right triangles ABD and ACE externally on sides AB and AC,
with right angles at D and E respectively. Show that DEF is an isosceles right triangle.

[i]Ireland 1996 [/i]

Let points D' and E' be mid-points of sides AB and AC respectively then
triangles DD'F and triangles EE'F are congruent.

therefore DF=EF.

Triangles DAE,DD'F are similar triangles.
Therefore <ADE = <D'DF which implies that <ADD'=<EDF=45 degrees

Hence the result.

Nice geometric solution from kalyanram.

Here is an algebraic solution, using standard notation :

By Law of cosines/Al-Hashi in triangle EBF :

EF² = c²/2 + a²/4 - ac/sqrt(2) cos(B+pi/4) = .. = b²/4 + c²/4 + acsinB/2 = b²/4 + c²/4 + abc/4R

it is symetric ./. b and c so EF=DF

By Law of cosines/Al-Hashi in triangle AED :

ED² = c²/2 + b²/2 -2bc cos (A+pi/2) = 2 EF²

So triangle EFD is isosceles right.

Take $M$ reflection of $B$ about $D$ and $N$ - reflection of $C$ about $E$. A $90^\circ$ rotation about $A$ will swap triangles $MAC$ and $BAN$, hence $BN=CM, BN\bot CM$; $DF, EF$ being midlines in triangles $BMC, BCN$, we are done.

Best regards,
sunken rock

Nice proof, [b]sunken rock[/b] :) Here's another approach.

See the attached diagram.

Reflect the entire image about $F$, as shown. Then it's practically obvious that $\triangle ECD'$ is the rotational image of $\triangle EAD$ about $E$ for $90^\circ$ ($AD\parallel A'D'\iff AD\perp CD')$. Hence $ED\perp ED'\land ED=ED'$. Continuing the argument, we find that $ED'E'D$ is a square. The conclusion follows.

[quote="fmasroor"]If you say that ABD and ACE are similar right triangles then the result is the same minus the part that says 'right/[/quote]
... and the proof is similar with the one I gave above: if $\angle ABD=\angle ACE=\alpha$, then a $2\alpha$ rotation about $A$ will swap the two triangles, thus $DE=DF$ and $\angle EDF=180^\circ-2\alpha$.

Best regards,
sunken rock

Dear Mathlinkers,
this result comes fromJoseph Neuberg...

See for example
http://perso.orange.fr/jl.ayme  vol. 16  Deux triangles... p. 31-32
Sincerely
Jean-Louis


Find all positive reals $a$ such that $a^{x}\geq ax\forall x\in (0,+\infty)$.

This is trivial.
$f(x)=a^{x-1}-x\geq 0$ for all $x>0$ and $f(1)=0$,
then $f'(1)=0$ and $a=e$.
Conversely for $a=e$ we have $f'(x)\leq 0$ for $x\in (0,1]$ and $f'(x)\geq 0$ for $x\in [1,\infty)$.
So $f(x)\geq f(1)=0$ for all $x>0$.


Okay, I don't understand how to do this problem

There are eight circles in the picture below. The shaded circles are congruent. If each circle in the picture is tangent to each of its neighbor circles, what is the ratio of the total area of the seven shaded regions to the area of the largest circle? express your answer as a common fraction.

I don't understand it. All I know is that you have to find the ratio of the circles to the circle they are in. But how do I find that?! :huh:

Let the radius of one of the small circles be $ r$.  Then, the radius of the largest circle is $ 3r$, and the ratio of the areas is
\[ \frac {7\left (r^2\pi \right)}{\left( 3r\right)^2 \pi } \equal{} \boxed{\frac {7}{9}}
\]

Oh... I see... Obviously I'm not thinking today...

You could use variables, like <insert number>x did, but just try having 1 for the radius of the smaller circles, or any other number.  The answer will always be the same though.


Consider the sequence of numbers: $ 4, 7, 1, 8, 9, 7, 6, \ldots .$ For $ n > 2$, the $ n$th term of the sequence is the units digit of the sum of the two previous terms. Let $ S_n$ denote the sum of the first $ n$ terms of this sequence. The smallest value of $ n$ for which $ S_n > 10,000$ is:

$ \textbf{(A)}\ 1992 \qquad \textbf{(B)}\ 1999 \qquad \textbf{(C)}\ 2001 \qquad \textbf{(D)}\ 2002 \qquad \textbf{(E)}\ 2004$

[hide]
The sequence repeats: $4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3...$
The sum of these $12$ terms is $60$.
To find the number of times that the $12$ terms must repeat:
${\displaystyle\frac{10,000}{60}=166 \displaystyle\frac\{2}{3}$
After repeating $166$ times, the sum is: $166 \ast 60=9960$
Add $7$ more terms for $S_n$ to be over $10,000$:
$9960+4+7+1+8+9+7+6=10,002$
$n=166 \ast 12+7=1999$
$(B)$
[/hide]

[hide]The series is cyclic $\{4,7,1,8,9,7,6,3,9,2,1,3\}$ this has a sum of $60$.

$\lfloor\frac{10000}{60}\rfloor = 166$, but then we have 40 left over, to get a sum of at least 40, we must add the first 7 terms, so we need a total of $166(12) + 7 = 1999$ terms.  So the answer is $B$.[/hide]

sorry for the revive, but I just wanted to point out that I was kind of shocked by this problem because I thought it had to be 2002 (the year) because why else would they put this problem this year?

If the answer on a 2011 test was always 2011, then it would making guessing too easy, no?

Often, you'll find a problem from 2005 that involves subtracting 1 and working with 2004 or something along those lines.

[b]Solution:[/b] Since this sequence is just based on the two previous ones, it is plausible to presume that it will repeat eventually. We see that it repeats every $12$ numbers (the sum of those every $12$ numbers is $60$). When we take $\lfloor{\frac{10000}{60}}\rfloor$, we see that $166$ full groups of $12$ numbers will be needed to add up to $10000$. However, in addition, we will need a sum of $40$ more to reach at least $10000$, which can happen with the first $7$ terms of the next cycle of $12$ numbers. Hence our answer is $166 \cdot 12 + 7$, or B.


geometry
similar triangles
cyclic quadrilateral
geometry unsolved

Given is a triangle $ ABC$ and let $ H$ be it’s orthocenter. An arbitrary point $ P$ is given inside triangle. Let $ HL$ and $ HM$ be perpendiculars from $ H$ to lines $ AP$ and $ BP$, respectively, with $ L$ on $ AP$ and $ M$ on $ BP$. Further, denote by $ X$ intersection of $ LH$ with $ BC$ and by $ Y$ intersection of $ MH$ and $ AC$. Prove that lines $ PH$ and $ XY$ are perpendicular.

Let $ PH \cap XY \equal{} Z, AH \cap BC \equal{} A_1$, denote $ B_1 , C_1$ similarly
Thus $ \angle ALX \equal{} \angle AA_1 X$, $ A, L, A_1, X$ are concyclic. So $ AH.A_1 H \equal{} LH.XH$
similarly, we can know that $ BH.B_1 H \equal{} YH.MH$
because $ BH.B_1 H \equal{} AH.A_1 H$, $ LH.XH \equal{} YH.MH$
So $ X, Y, L, M$ are concyclic and it is true that $ L, P, M, H$ are concyclic.
then $ \angle LPH \equal{} \angle LMH \equal{} \angle LXY$.
so $ L, P, X, Z$ are concyclic.
$ \angle PLX \equal{} \angle PZX$ and $ \angle PLX \equal{} 90^\circ$
this is the end of the proof

Very nice and simple solution Thank you!

We denote the points $ X'\equiv LH\cap BP$ and $ Y'\equiv MH\cap AP$ and it is enough to prove that $ X'Y'\parallel XY,$ because of $ PH\perp X'Y'$
from the point $ H,$ as the orthocenter of the triangle $ \bigtriangleup PX'Y',$ also $ ($ except $ \bigtriangleup ABC$ $ ).$

From the similar triangles $ \bigtriangleup BXX'$ and $ \bigtriangleup HAY',$ because of $ \angle BX'X \equal{} \angle HY'A$ $ ,(1)$ and $ \angle XBX' \equal{} \angle A_{1}BM \equal{} \angle A_{1}HM \equal{} \angle AHY'$ $ ,(2)$ $ ($ from the cyclic quadrilateral $ A_{1}BHM$ $ ),$

 we have that $ \frac {XX'}{AY'} \equal{} \frac {BX'}{HY'}$ $ \Longrightarrow$ $ (XX')\cdot (HY') \equal{} (BX')\cdot (AY')$ $ ,(3)$

From the similar triangles $ \bigtriangleup BHX'$ and $ \bigtriangleup YAY',$ because of $ (1)$ and $ \angle BHX' \equal{} \angle B_{1}HL \equal{} \angle B_{1}AL \equal{} \angle YAY'$ $ ,(4)$ $ ($ from the cyclic quadrilateral $ AB_{1}LH$ $ ),$

we have that $ \frac {HX'}{AY'} \equal{} \frac {BX'}{YY'}$ $ \Longrightarrow$ $ (HX')\cdot(YY') \equal{} (BX')\cdot(AY')$ $ ,(5)$

From $ (3),$ $ (5)$ $ \Longrightarrow$ $ (XX')\cdot (HY') \equal{} (HX')\cdot (YY')$ $ \Longrightarrow$ $ \frac {HX'}{XX'} \equal{} \frac {HY'}{YY'}$ $ \Longrightarrow$ $ X'Y'\parallel XY$
 and the proof is completed.

Kostas Vittas.


This was the last question in a Canadian contest last tuesday (some of you might have seen this question in my blog). It's a contest for grade 12's and anyone can write it, so I'm guessing it is like AMC 12 level:

There are lockers numbered 1 to n in a school (All the lockers are initially opened). A person walks down the row of lockers closing every locker, i.e., where the case is n=10, he closes lockers 2, 4, 6, 8, 10:

1 2 3 4 5 6 7 8 9 10
1 X 3 X 5 X 7 X 9 X
Now the person, now at the right end of the line walks back to left end of the line of lockers, closing every second open locker he sees:

1 X 3 X 5 X 7 X 9 X
1 X X X 5 X X X 9 X

The person, now at the left hand side of the line, goes again and closes every second open locker he sees,

1 X X X 5 X X X 9 X
1 X X X X X X X 9 X

He does this once more,
1 X X X X X X X 9 X
X X X X X X X X 9 X
And 9 is the last open locker remaining. Thus, we let f(n) = the last locker open, i.e. f(10)=9.

Now, the question:
a) Find f(50)
b) Prove that there is no n where f(n)=2005.
c) Prove that there are an infinite number of n where f(n)=f(2005).

an interesting problem.
i've seen problems similar to this one before.
there are some things that are obvious about the $f(n)$
that $f(n)$ can not be an even number and for every odd number such as $z$, $f(z+1)=f(z)$

the key to this problem is just what said before.
hint:
[hide]$f(50)=f(49)=1$
[/hide]
helpfull numbers for this problem:
[hide] $1,3,1,3,9,11,9,11,1,3,1,3,9,11,9,11...$
the numbers above are the $f(n)$ for odd $n$. ($f(n+1)=f(n)$)[/hide]

Good, but now about proving the sequence goes on forever  :)

to be or not to be... that's the question :D
i'm working on it  :lol:

this is my aproach for proving this:
if you can prove that for every odd $n$ the following is true then your problem is solved.
$f(n)=f(n+4)=f(n+16)=1$
and for the [b]same $n$[/b]
$f(n+2)=f(n+6)=f(n+18)=3$
and so on. you are finished.

but i haven't been so succesfull yet.

[quote="amirhtlusa"]do you know the proof?[/quote]

Not for this question.

[quote="amirhtlusa"]it's a very interesting problem but it's bugging me so much cause i haven't been able to prove that the sequence goes for ever. :?[/quote]
Lol, same here. [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=165]my blog[/url]

[quote="amirhtlusa"]all right i might have a solution for this that just poped into my head. but it's still incomplete. so whatever.
lets say you have the number $n$. (it doesn't matter if it's odd or even) you can write the numbers one through n, (or n+1 if, n is odd), in a $2*\frac{n}{2}$ table. then you can write these numbers in such a way that each time you go through the lockers you eliminate exactly on colum of this table. when ever a colum of the table is eliminated a new table is made with 2*\frac {n}{4} elements in it. it follows that based on this you can logically show that if the number is in a form of $4K+1$, $f(n)$ is either $1$, or $9$.  if $k$ is in a form of $4m$ or $4m+1$ $f(n)$ is one, and if $k$ is in a form of $4m+2$ or $4m+3$, $f(n)=9$.
it also follows that if $n$ is in a form of $4k+3, f(n)$ is either $3$ or $11$. this follows that if $k$ is in a form $4m+1$ $f(n)$ is 3, and if $k$ is in a form of $4m+2$ or $4m+3$, $f(n)=11$.
the reason i'm saying these numbers is that if you draw the table after a while you will see why i got these numbers, because $1,3,9,11$ are the numbers that will remain on the bottom of the table and always one of them is going to be the $f(n)$.
by the way $f(2005)=1$[/quote]

Looks good. I guess if you were to be completely meticulous, then you'd have to prove that 1, 3, 9, 11, actually individually repeat, (so that i don't get a series, 1, 3, 1, 3, 9, 11, 9, 11, 11, 11 ,11 ,11 ,11 ,11 ,11 ,11 ,11 ,11 ,11... with only 11's repeating.). But ya, other than that, your proof is good. I'd probably never think of that myself  :( . I sorta tried another approach, but I didn't get very far yet, and your approach seems a LOT easier.

well i said that that if the number is in a form of $4k+1,f(n)$  is either $1$ , or$9$ . if  is in a form of $4m$  or $4m+1$   is one, and if  is in a form of $4m+2$ or$4m+3$ , $f(n)=9$.
it also follows that if  is in a form of $4k+3, f(n)$ is either  $3$ or $11$ . this follows that if $f(n)$ is in a form of $4m$ or $4m+1, f(n)$ is $3$, and if $f(n)$ is in a form of $4m+2$  or $4m+3, f(n)=11$ .  thus it would be impossible for you to get a sequence like: $1, 3, 1, 3, 9, 11, 9, 11, 11,11 ,11 ,11 ,11 ,11 ,11 ,11 ,11 ,11...$ with only $11$'s repeating


calculus
integration
limit
calculus computations

Let $ F(t) \equal{} \int_0^t \frac {dx}{1 \minus{} x^\alpha}$, $ (\alpha \neq 0)$. Find $ lim _{t \rightarrow 0} F(t)$

Hint: For |x| < 1, We have:
$ \lim_{h \to 0} \frac{F(t \plus{}h) \minus{} F(t)}{h} \equal{} \lim_{h \to 0} \frac{1}{h} \int_{t}^{t \plus{} h} \frac{1}{1 \minus{} x^a}dx \equal{} \lim_{h \to 0}\frac{1}{h} \int_{t} ^{t \plus{} h} \sum_{k}^{\infty} x^{ak}dx$

That seems like overkill for a limit that's just zero. Note that the integral in Plurain's problem isn't divided by $ t$ or anything like that.

what is difference between phhj and Pluraln ???

http://www.mathlinks.ro/viewtopic.php?t=267801


Hi.

Couldnt solve the following exercise:

$\liminf_{n \to +\infty}Bn=1 ,$

$\lim_{n \to +\infty}An=1$

$B_{n+1} \leq \frac{A_nB_n}{A_{n+1}}$

Prove that   $\lim_{n \to +\infty}Bn=1.$

Sorry I didnt know how to mark lower limit symbol for Bn.

Thanks.

RealTime, what kinds of things did you try when you were trying to prove this?

I tried

$B_{n+1} A_{n+1} \leq A_nB_n$

And making a New series that is defined

$C_n = A_nB_n$

and $C_n > C_{n+1}$

Meaning $C_n$ is decreasing. Then im clueless.  :(

Is this the right forum for this question or should I have posted in the other one?

Sounds like a promising start.

Since $A_n$ is getting really really close to $1$, and $\liminf_{n \to \infty} B_n = 1$, it seems plausible that $\liminf_{n \to \infty} A_n B_n = 1$.  Have you tried proving that?

(Oh, yes, I think this is the correct forum.)


This is not ture, for an counter-example,

$a=1,\, b=80,\, c=20,\, d=160\,$.

oops! I confused this with another inequality which has $(ac+bd)(ad+bc)$ instead of $(ad+bc)^{2}$ in the last term.

This problem is very interesting. It'sn't simple as you think. I work very hard to find this inequality. Can you post the proof

[quote="Albanian Eagle"]I confused this with another inequality which has $(ac+bd)(ad+bc)$ instead of $(ad+bc)^{2}$ in the last term.[/quote]
The following inequality is true and ugly :
$2 \sqrt{ \frac{ (a-b)^{2}}{2( a^{2}+b^{2})}}+6 \sqrt{ \frac{ (c-d)^{2}}{2( c^{2}+d^{2})}}+3 \sqrt{ \frac{(ac+bd)(ad+bc)}{( a^{2}+b^{2})( c^{2}+d^{2})}}$
$\leq 3\sqrt 2+\sqrt{\frac{13}{2}}=6.7921504439156775614191\cdots$,
with equality if $a=9,b=13+2\sqrt{22},c=0,d=1$.


Let $D,E, F$ be points on the sides $BC, CA,AB$ of a triangle $ABC$, respectively. Suppose that the inradii of the triangles $AEF,BFD, CDE$ are all equal to $r_1$. If $r_2$ and $r$ are the inradii of triangles $DEF$ and $ABC$ respectively, prove that $r_1 + r_2 = r$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=49374 .

  darij


1) consider a regular pentagon, 5 vertices, 2 colors by pigeonhole 3 must be the same color.   Now these three must be an iscoseles triangle.

2)no color half the circle black and the other white

[quote="cancer"]Each point of a circle is painted black or white.

(a) Prove that there exists an isosceles triangle inscribed in this circle such that all its vertices are painted the same color.

(b) Does there always exist an equilateral triangle inscribed in this circle such that all its vertices are painted the same color?[/quote]

[hide="(a)..."]Inscribe a regular pentagon in the circle, with vertices $ P_1, P_2, \ldots , P_5$. Then two vertices next to each other, say $ P_1, P_2$ have to be the same color, say white. Then the other three vertices have to be black or else we could form an isosceles triangle with all vertices white, but then with $ P_3, P_4, P_5$ we can create an isosceles triangle with all vertices black. [/hide]

[hide="(b)..."]If you take one point, the point forms a certain equilateral triangle that no other point, besides the three vertices, can create. Then just color those, say black black white, and do that for all the points on a $ 120^\circ$ arc...? And there is no equilateral triangle inscribed in the circle with all vertices painted the same color?[/hide]

Edit: Blargh...late


integration
trigonometry
real analysis
real analysis unsolved

I'd like to show that $ \int_0^\infty \frac{|Sin(x)|}{x} dx \equal{} \infty$, can someone give me a hint?


geometry
geometric transformation
reflection
projective geometry

Hrm...I've been wondering about this problem for a while; it's on the WOOT page.
[img]http://www.artofproblemsolving.com/Classes/Images/wootgeoprob.GIF[/img]

Any ideas?

Extend BO to meet the circle again at B'
Similarly, extend CO to meet the circle again at C'
Let B'N meet the circle again at K
By Pascal's Theorem applied to cyclic hexagon C'AB'CKB we have that N,O, and C'K$\cap$AB are collinear.  It follows that C'K$\cap$AB is M.
By the definitions of S and R, SO is parallel to B'N and RO is parallel to C'M.  Thus the angle SOR is the same as angle C'KB', which is 1/2 the measure of arc C'B'.  Now, <BAC is 1/2 the measure of arc CB.  As arc CB is just the reflection of arc C'B' about O, we are done.


I've tried this very long but still cannot find the answer :(
$ \sum\limits_{cyc}^{a,b,c} {\left( {a \plus{} \frac{1}{b} \minus{} 1} \right)} \left( {b \plus{} \frac{1}{c} \minus{} 1} \right) \ge 3(a,b,c > 0)$

I assume that
$ S \equal{} \minus{} 3 \plus{} \sum^{a,b,c}_{cyc} (a \plus{} \frac {1}{b} \minus{} 1)(b \plus{} \frac {1}{c} \minus{} 1) \\
\equal{} \minus{} 3 \plus{} (a \plus{} \frac {1}{b} \minus{} 1)(b \plus{} \frac {1}{c} \minus{} 1) \plus{} (b \plus{} \frac {1}{c} \minus{} 1)(c \plus{} \frac {1}{a} \minus{} 1) \plus{} (c \plus{} \frac {1}{a} \minus{} 1)(a \plus{} \frac {1}{b} \minus{} 1)$
Am I right?

We make some substitutions:

$ a \plus{} \frac {1}{b} \equal{} x \\
b \plus{} \frac {1}{c} \equal{} y \\
c \plus{} \frac {1}{a} \equal{} z$

This gives us that

$ S \equal{} (x \minus{} 1)(y \minus{} 1) \plus{} (y \minus{} 1)(z \minus{} 1) \plus{} (z \minus{} 1)(x \minus{} 1) \minus{} 3 \equal{} (xy \plus{} yz \plus{} xz) \minus{} 2(x \plus{} y \plus{} z)$

Which gives us by substituting back

$ (ab \plus{} bc \plus{} ac) \plus{} 3 \plus{} (\frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b}) \plus{} (\frac {1}{bc} \plus{} \frac {1}{ac} \plus{} \frac {1}{ab}) \minus{} 2(a \plus{} b \plus{} c) \minus{} 2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$.

By using the AM-GM inequality on each term  we achieve that $ S \geq T$ where

$ T \equal{} 3(abc)^{\frac {2}{3}} \minus{} 6(abc)^{\frac {1}{3}} \plus{} 6 \minus{} \frac {6}{(abc)^{\frac {1}{3}}} \plus{} \frac {3}{(abc)^{\frac {2}{3}}}$.
Substituting $ (abc)^{\frac {1}{3}} \equal{} p , \ ( \Rightarrow p > 0)$ gives us

$ 3p^2 \minus{} 6p \plus{} 6 \minus{} 6p^{ \minus{} 1} \plus{} 3p^{ \minus{} 2}$

Substituting $ u \equal{} p \plus{} \frac {1}{p}$ gives us $ 3u^2 \minus{} 6u$, where $ u \geq 2$ by AM-GM which obviously suggest that this expression is larger than 0. Since this is less than or equal the original expression we have proved that it is positive.

Please let me know if I did anything wrong!

:D By using AM-GM we can't achieve $ \ S\geq\ T$

You cannot use AM-GM with
$ \ \minus{} 2(a \plus{} b \plus{} c) \minus{} 2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$.It is convert. :ninja:


AMC
AIME
geometry
3D geometry
pyramid
octahedron
probability

3 hours, no calculators, infinite scratch paper, answers are integers from 000 to 999 inclusive, etc.  Happy holidays, good luck, and have fun! [b]This contest has ended.[/b]

1. For how many positive integers n > 1 is it possible to express 2005 as the sum of n distinct positive integers?

2. $a_1, a_2, \dots$ is a sequence of real numbers where $a_n$ is the arithmetic mean of the previous $n-1$ terms for $n > 3$ and $a_{2004} = 7$. $b_1, b_2, \dots$ is a sequence of real numbers in which $b_n$ is the geometric mean of the previous $n-1$ terms for $n > 3$ and $b_{2005} = 6$. If $a_i = b_i$ for $i = 1, 2, 3$ and $a_1 = 3$, then compute the value of $a_2^2 + a_3^2$.

3. Compute the largest integer $n$ such that $2005^{2^{100}} - 2003^{2^{100}}$ is divisible by $2^n$.

4. ABCDEFG is a regular heptagon, and P is a point in its interior such that triangle ABP is equilateral. There exists a unique pair $\{m,n\}$ of relatively prime positive integers such that the measure of angle CPE is $\left(\frac{m}{n}\right)^{\circ}$. Compute the value of $m + n$.

5. Compute, to the nearest integer, the area of the region enclosed by the graph of $13x^2 - 20xy + 52y^2 - 10x + 52y = 563$.

6. Determine the remainder obtained when 1000! is divided by 2003.

7. P is a pyramid consisting of a square base and four slanted triangular faces such that all of its edges are equal in length. C is a cube of edge length 6. Six pyramids similar to P are constructed by taking points $P_i$ (all outside of C) where $i = 1, 2, \dots, 6$ and using the nearest face of C as the base of each pyramid exactly once. The volume of the octahedron formed by the $P_i$ (taking the convex hull) can be expressed as $m + n\sqrt{p}$ for some positive integers $m$, $n$, and $p$, where $p$ is not divisible by the square of any prime. Determine the value of $m + n + p$.

8. A single atom of Uranium-238 rests at the origin. Each second, the particle has a 1/4 chance of moving one unit in the negative $x$ direction and a 1/2 chance of moving one unit in the positive $x$ direction. If the particle reaches $(-3,0)$, it ignites a fission that will consume the earth. If it reaches $(7,0)$, it is harmlessly diffused. The probability that, eventually, the particle is safely contained can be expressed as $\frac{m}{n}$ for some relatively prime positive integers $m$ and $n$. Determine the remainder obtained when $m + n$ is divided by 1000.

9. The value of the sum

$
\sum_{n=1}^{\infty} \frac{(7n + 32)\cdot 3^n}{n\cdot (n+2)\cdot 4^n}
$

can be expressed in the form $\frac{p}{q}$, for some relatively prime positive integers $p$ and $q$. Compute the value of $p + q$.

10. 100 blocks are selected from a crate containing 33 blocks of each of the following dimensions: 13 x 17 x 21, 13 x 17 x 37, 13 x 21 x 37, and 17 x 21 x 37. The chosen blocks are stacked on top of each other (one per cross section) forming a tower of height h. Compute the number of possible values of h.

11. 10 lines and 10 circles divide the plane into at most n disjoint regions. Compute n.

12. Determine the number of permutations of $1, 2, 3, 4, \dots, 32$ such that if $m$ divides $n$, the $m$th number divides the $n$th number.

13. $x$, $y$, and $z$ are distinct non-zero integers such that $-7 \le x, y, z \le 7$. Compute the number of solutions $(x,y,z)$ to the equation

$
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{x+y+z}
$

14. In triangle $ABC$, $BC = 27, CA = 32$, and $AB = 35$. $P$ is the unique point such that the perimeters of triangles $BPC, CPA$, and $APB$ are equal. The value of $AP + BP + CP$ can be expressed as $\frac{p+q\sqrt{r}}{s}$, where $p, q, r$, and $s$ are positive integers such that there is no prime divisor common to $p, q$, and $s$, and $r$ is not divisible by the square of any prime. Determine the value of $p+q+r+s.$

15. $ABCD$ is a convex quadrilateral in which $\overline{AB} \parallel \overline{CD}$. Let $U$ denote the intersection of the extensions of $\overline{AD}$ and $\overline{BC}$. $\Omega_1$ is the circle tangent to line segment $\overline{BC}$ which also passes through $A$ and $D$, and $\Omega_2$ is the circle tangent to $\overline{AD}$ which passes through $B$ and $C$.  Call the points of tangency $M$ and $S$.  Let $O$ and $P$ be the points of intersection between $\Omega_1$ and $\Omega_2$. Finally, $\overline{MS}$ intersects $\overline{OP}$ at $V$. If $AB = 2$, $BC = 2005$, $CD = 4$, and $DA = 2004$, then the value of $UV^2$ is some integer $n$. Determine the remainder obtained when $n$ is divided by 1000.

Wow. A fourth one? Awesome. You should get a prize or something for doing all that work. Same thing with everyone else who made Mock AMCs.

I'll need it early most likely. Is Sunday ok? Happy Holidays.

I've never take mock AIME before but I saw the links to the test and poll were all saying "harder" than real AIME.

Is it okay to take it even you know you are not going to do well on it?

Cuz I would be lucky if I get one right on AIME since I never take AIME before  :(

Silverfalcon, I quote:

[quote="Mildorf"][b]As always, it is open to everyone.[/b][/quote]

ARH I did horrible, I got a 4 :( , 2 stupid mistakes out 6 problems I solved!!! which ruined my vacation because looks like there's no USAMO for me next year.

wow, mock AIME # 4

That's 60 difficult yet great problems that you will have come up with. We should all give him a roundn of applause.

many things:
1.i'll take it...
2 will it be posted in this thread
3. *clapping very loudly*
4. kudos and thanks to you for creating 60 great yet challengin questions

To shake991:

[quote]This is the thread where the 4th Mock AIME contest will appear on Thursday, December 30, from 2-5pm, Eastern U.S. Standard Time (EST=GMT-5hrs)[/quote]

[quote="captcha000"]i'm supposing that for number 8, the other 1/4 chance is that it doesn't move at all?[/quote]

Yes, and that's per second. Each second, there is a 1/4 chance that it stays put, a 1/4 chance it moves one unit in the negative x direction, and 1/2 chance it moves one unit in the positive x direction.

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3742683_2004_mildorf_mock_aime_45]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


invariant
function
limit
probability
advanced fields
advanced fields unsolved

Birkhoff's pointwise ergodic theorem states that if $T$ is a measure-preserving transformation on the measure space $(X, \mathcal{M}, \mu)$, $\mathcal{A}$ is the space of $T$-invariant functions, and $f \in L^{1}(\mu)$, then

$\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n-1}f(T^{k}x) = \text{proj}_\mathcal{A}f \textit{ for a.e. }x.$

However, the proof I have seen for this theorem assumes in a critical step that $\mu$ is a probability measure (i.e. $\mu(X) = 1$). Basically, we show that some dense set of functions has the property, but in passing into the limit, we use the fact that $\mu$ is a probability measure in part of our estimate.

Does the theorem still hold for general $\sigma$-finite measures? Do we care?

Yes and yes. See the statement for infinite measure spaces [url=http://www.math.unc.edu/Faculty/petersen/lecturespdf.pdf]here[/url]; a proof is in "Lectures on ergodic theory" by Halmos.


Prove the following inequality for all positive real numbers $ a,b,c$ and $ k \equal{} 3\sqrt{2}\minus{}3$:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}k\ge\frac{(9\plus{}3k)(a^{2}\plus{}b^{2}\plus{}c^{2})}{(a\plus{}b\plus{}c)^{2}}$

The inequality is  equivalent  to
       $ (3\plus{}3\sqrt2)G5_{3}\plus{}G5_{4}\plus{}P5_{1}\plus{}P5_{2}\geq 0$,

in which
       $ G5_{3}\equal{}\sum abc(a\minus{}b)(a\minus{}c))\geq 0$,
       $ G5_{4}\equal{}\sum bc(b\plus{}c)(a\minus{}b)(a\minus{}c)\geq 0$,
       $ P5_{1}\equal{}\sum b(a^{2}\plus{}bc)(a\minus{}b)(a\minus{}c)\geq 0$,
       $ P5_{2}\equal{}\sum b(c^{2}\plus{}ab)(a\minus{}b)(a\minus{}c)\geq 0$.


Let $ \{a_n\}_{n \equal{} 0}^{\infty}$ as $ a_0 \equal{} a_1 \equal{} 1$, $ a_{n \plus{} 2} \equal{} \frac {1}{a_{n \plus{} 1}} \plus{} \frac {1}{a_n}$ . Find $ \lim_{n \to \infty}a_n$

I don't know how to know that $(a_n)$ have the limit. But if we had that $(a_n)$ have a limit, that: $\lim_{n \rightarrow + \infty} a_n=X$. We have:
$X=\frac{2}{X} \Leftrightarrow X=\sqrt{2}$ or $X=-\sqrt{2}$
Easily, we have $(a_n) > 0$ with all n be the positive integer number. So $X=\sqrt{2}$, or $\lim_{n \rightarrow + \infty} a_n=\sqrt{2}$.


A horizontal line through point $ (\minus{}3,4)$ and a vertical line through point $ (\minus{}1, \minus{}7)$ are drawn in a coordinate plane. What is the sum of the coordinates of the point of intersection of these two lines?

All points on horizontal line have the same y coordinate, and all on vertical have same x coordinate, so the coordinate of the intersection is (-1,4).
Sum: 3


Help me solve this problem.

In Russia school pupils must do some exams before leaving school. Among others, they must do two "selective" exams. This means that school provides a list of available subjects; each pupil selects two different subjects from this list and is going to do this exams. According to rules, pupil isn't allowed to do both exams at the same day, so the school must schedule the exams, i.e. provide some days when pupils will be able to do exams.

One school does not want to warn teachers too much. They want to schedule all the exams into two days in such way that exams on some subjects are on the first day, and exams on all other (and only on them) are on second. You are to write a program, which will determine, if it is possible to schedule exams in this way so that all pupils will be able to do all their selected exams.
There are $n(1 \le n \le 200)$ subjects and $m(1 \le m \le 30000)$

This is just working out whether a given graph (nodes = subjects, edges = pupils) is bipartite, which is very easy (just colour one node white, all adjacent ones black, all adjacent ones white, etc, making sure each node only  gets one colour).


Let $ a, b, c$ be positive integers for which $ ac\equal{}b^2\plus{}b\plus{}1$. Prove that the equation $ ax^2\minus{}(2b\plus{}1)xy\plus{}cy^2\equal{}1$ has an integer solution.


There is this problem given at the romanian informatics olympiad, district round. I'll just write the problem in a mathematical way:

Let $n,k$ be two natural numbers such that $1 \leq n \leq 2.000$ and $2 \leq k \leq 1.000.000.000$. We call a permutation $\sigma \in S_n$ good iff $\sigma^k = \underbrace{\sigma \circ \sigma \circ \ldots \circ \sigma}_{k \, \textrm{times}}$ is the identity permutation.

Find the number of all good permutations $\mod 19997$.

What I did:

1) On the olympiad, I gave little thought on the problem (few minutes) and wrote a backtracking program which generated all permutations. This got me $20$ points. (It's kind of a surprise.... I was thinking of something like $10$ points)

2) At home, I found a second backtracking solution. The idea is the following: we break up $\sigma$ in its cycles. Now, since $\sigma^k$ is the identity, the lengths of all cycles must divide $k$. Put in a vector $d$ the divisors of $k$ smaller than $n$. For obvious reasons, we try to write $n$ as a linear combination of $d_1,d_2,\ldots,d_n$. We do this via a recursive backtracking (the normal version is too mesy for me). We do our best to find a formula for each linear combination: $\prod \frac{n!}{(n - d_i m_i)! k! d_i^{m_i}}$, where $m_i$ are the coefficients corresponding to $d_i$, then sum it up. (actually there's a bit more on the formula... we need to decrease $n$ after each term of the product is computed) And not to forget, $19997$ is a prime. Thus, we precompute the inverses of the numbers, only then we are ready to do the computations, $\mod 19997$ of course. Unbelievablely, this would get you $60$ points.

Now, can you improve this? I think that this idea is good, what do you think?

Try dynamic programming. A[n] = number of such permutations of length n. You build A[n] from a cycle of c (c is from 1 to n and MUST be a divisor of K) and A[n-c]. Official solutions available at http;//olimpiada.info

Thanks... Yesterday I couldn't download the solutions, and I tried to build one by myself.


if $arccosec \frac{1}{x} = k$, and $0<k<\frac{\pi}{2}$ then find the value of $\tan k$ in terms of $x$.

got this one in college today.

[hide]I did it totally wrong to begin with by drawing a right-angled triangle, calling one angle k and the ratio of the opposite sides x, but then had to use pythagoras to find the third side and was left with the answer in terms of some other variables I made up which wouldn't reduce[/hide]

this is getting silly now
note to self: preview post before submit

[quote]If $\text{arccsc}\frac{1}{x} = k$ and $0 < k < \frac{\pi}{2}$,  then find the value of $\tan k$ in terms of $x.$[/quote][hide]
$\text{Since }\csc k = \frac{1}{x}\text{, we have acute angle }k\text{ with: }hyp = 1,\;opp = x.$

$\text{Using Pythagorus, we find that: }adj = \sqrt{1-x^2}.$

$\text{Therefore: }\tan k \;= \;\frac{opp}{adj} \;= \;\frac{x}{\sqrt{1-x^2}} $
[/hide]
A second method --- to be used in moments of sheer desperation . . .
[hide]
$\text{We have: }\;\csc k \;= \;\frac{1}{x}$

$\text{Square: }\csc^2k \;= \;\frac{1}{x^2}$

$\text{Subtract 1: }\csc^2k - 1 \;= \;\frac{1}{x^2} - 1 \;= \;\frac{1 - x^2}{x^2}$

$\text{We have: }\cot^2k \;= \;\frac{1-x^2}{x^2}$

$\text{Then: }\cot k \;= \;\sqrt{\frac{1-x^2}{x^2}} \;= \;\frac{\sqrt{1-x^2}}{x}$

$\text{Therefore: }\tan k \;= \;\frac{x}{\sqrt{1-x^2}} $
[/hide]


there is a much more funny version:
there 10 sets of people; those that unterstand binary system and those that are not.

similar but funnier:
There are 3 kinds of people in this world: those who can count and those who can't  :rotfl:

There are 2 types of people in the world: those who still think the binary joke is funny and those who don't. :rotfl:

Yeah, the binary solution is the one I've known.

Now try to make 1+1= window.

Proof that, in fact, 1+1 does equal 10 [b]in base 10[/b]:

$a=b$
Multiplying by a, $a^{2}=ab$
Subtracting $b^{2}$, $a^{2}-b^{2}=ab-b^{2}$
Factoring, $(a+b)(a-b)=b(a-b)$
Clearing out the common term, $a+b=b$
Since a=b, $b+b=b$
Collecting like terms, $2b=b$
Dividing by a nonzero b, $2=1$

We add 1 to both sides and get the equation $3=2$. Thus $3=1$ by the transitive axiom. Repeatedly doing this, you will eventually work your way up to $10=2$.

:D

haha...that proof is messed up b/c you divide by 0 somewhere :rotfl:
proof:1+1=window
$1+1$
$\downarrow$
$\underline{\overline{1+1}}$
where the overline+underline is the = sign. :rotfl:

[quote="ccy"]My proof goes like this:

So we have 1+1 and we want to get 10.  Since obviously 1=一, we can rewrite the thing 1+1=1+一.  Adding, we get 1+一=十.  Since 十=10, 1+1=10.

QED[/quote]

:rotfl: :rotfl: :rotfl: Wow, I've never seen that one before!!! That's hilarious.

But only because I'm Chinese. :D

[quote="air88"]a friend of mine has a math teacher who asked the class how to make 1+1=10 without changing anything in the problem we racked our brain for many hours and still having come up with nothing we decided to ask people decidedly better at math then us. any suggestions or possible answers to this would be apreciated since this question is driving us crazy.[/quote]
If you want to learn what base 2 (binary) is and how it works, read
[url=http://en.wikipedia.org/wiki/Binary_numeral_system]Binary Numeral System[/url]
starting from the section "Counting in Binary."
If you have any other questions about binary, feel free to ask.

[quote="mathnerd314"]There are 2 types of people in the world: those who still think the binary joke is funny and those who don't. :rotfl:[/quote]
So in actuality there is only 1 type of person in the world.

[quote="mathnerd314"]
[b]
$a=b$[/b]
...
Factoring, $(a+b)(a-b)=b(a-b)$
[b]Clearing out the common term, [/b]$a+b=b$
... [/quote]


[b]Here is the question and my two ways of solution ,,,
Can someone please check if it is right ,,,[/b]

-----------------------------------------------------------------------

Question :
If $ \alpha$,$ \beta$,$ \gamma$ are the roots of the equation
$ 5x^3-8x^2+7x+6=0$

Find the equation whose roots are
$ \alpha^2 + \beta^2 + \beta\alpha, \beta^2 + \gamma^2 + \gamma\beta, \gamma^2 + \alpha^2 + \gamma\alpha$

------------------------------------------------------------------------

Solution 1 :

a cubic equation is in the form of : $ ax^3 + bx^2 +cx +d = 0$
$ \implies 5x^3 - 8x^2 + 7x + 6 = 0$                                           (Given)

$ \therefore \alpha + \beta + \gamma = \frac{-b}{a} = \frac{8}{5}$
$ \alpha\beta + \beta\gamma + \alpha\gamma = \frac{c}{a} = \frac{7}{5}$
$ \alpha\beta\gamma = \frac{-d}{a} = \frac{-6}{5}$

Let $ ax^3 + bx^2 + cx + d = 0$ be the required equation.
We know, $ ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)$, where $ \alpha, \beta, \gamma$ are the roots of cubic equation.

$ \therefore ax^3 + bx^2 + cx + d = a[x - (\alpha^2 + \beta^2 + \alpha\beta)][x - (\beta^2 + \gamma^2 + \beta\gamma)][x - (\gamma^2 + \alpha^2 + \alpha\gamma)]$

$ = a \{x - [(\alpha + \beta)^2 - 2\alpha\beta + \alpha\beta][x - [(\beta + \gamma)^2 - 2\beta\gamma + \beta\gamma][x - [(\alpha + \gamma)^2 - 2\alpha\gamma + \alpha\gamma\}$

$ a[x - (\frac{8}{5} - \gamma)^2 + (\frac{7}{5} - \beta\gamma - \alpha\gamma)][x - (\frac{8}{5} - \alpha)^2 + (\frac{7}{5} - \alpha\beta - \alpha\gamma)][x - (\frac{8}{5} - \beta)^2 + (\frac{7}{5} - \alpha\beta - \beta\gamma)]$

$ =a[x - \frac{64}{25} + \gamma^2 - \frac{16}{5}\gamma) + \frac{7}{5} - \beta\gamma - \alpha\gamma][x - \frac{64}{25} + \alpha^2 - \frac{16}{5}\alpha) + \frac{7}{5} - \alpha\beta - \alpha\gamma][x - \frac{64}{25} + \beta^2 - \frac{16}{5}\beta) + \frac{7}{5} - \alpha\beta - \beta\gamma]$

$ = a[x - \frac{64}{25} - \gamma^2 + \frac{16}{5}\gamma + \frac{7}{5} -  (\alpha + \beta)\gamma][x - \frac{64}{25} - \alpha^2 + \frac{16}{5}\alpha + \frac{7}{5} -  (\beta + \gamma)\alpha][x - \frac{64}{25} - \beta^2 + \frac{16}{5}\beta + \frac{7}{5} -  (\alpha + \gamma)\beta]$

$ = a[x - \frac{29}{25} + \frac{16}{5}\gamma - \gamma^2 - \gamma(\frac{8}{5} - \gamma)][x - \frac{29}{25} + \frac{16}{5}\alpha - \alpha^2 - \alpha(\frac{8}{5} - \alpha)][x - \frac{29}{25} + \frac{16}{5}\beta - \beta^2 - \beta(\frac{8}{5} - \beta)]$

$ = a(x - \frac{29}{25} - \frac{8}{5}\gamma)(x - \frac{29}{25} - \frac{8}{5}\alpha)(x - \frac{29}{25} - \frac{8}{5}\beta)$

$ = a(x^3 - \frac{29}{25}x^2 - \frac{8}{5}\alpha x^2 - \frac{29}{25}x^2 + \frac{841}{625}x + \frac{232}{125}\alpha x - \frac{8}{5}\gamma x^2 + \frac{232}{125}\gamma x + \frac{64}{25}\alpha\gamma x - \frac{29}{25}x^2 + \frac{841}{625}x + \frac{232}{125}\alpha x + \frac{841}{625}x - \frac{24389}{15625} - \frac{6728}{3125}\alpha + \frac{232}{125}\gamma x - \frac{6728}{625}\alpha\gamma - \frac{8}{5}\beta x^2 + \frac{232}{125}\beta x + \frac{64}{25}\alpha\beta x + \frac{232}{125}\beta x - \frac{6728}{3125}\beta - \frac{1856}{625}\alpha\beta + \frac{64}{25}\beta\gamma x - \frac{1856}{625}\beta\gamma - \frac{512}{125}\alpha\beta\gamma)$

$ = a\{x^3 - [\frac{8}{5}(\alpha + \beta + \gamma) + \frac{58}{25}]x^2 + [2(\frac{232}{125}(\alpha + \beta + \gamma) + \frac{64}{25}(\alpha\gamma + \beta\gamma + \alpha\beta) + \frac{2523}{625}]x + \frac{52386}{15625} - \frac{6728}{3125}(\frac{8}{5}) + \frac{64}{25}(\frac{7}{5})\}$

$ = a(x^3 - \frac{122}{25}x^2 + \frac{339}{5}x + \frac{54562}{15625})$

$ = a(5x^3 - \frac{122}{5}x^2 + 339x + \frac{54562}{3125})$

[b]Now how to find the value of this equations 'a' ?? is it supposed to be 5 from the equation $ 5x^3-8x^2+7x+6=0$ ??

My other solution is by using the cardano's method to the equation $ 5x^3-8x^2+7x+6=0$ ,,, and upon getting the three roots ,,, (my results as being $ \alpha = 103.52, \beta = -52.55 + 7.47\iota, \gamma = -52.55 - 7.47\iota$) is substituted in $ a(x - \alpha)(x - \beta)(x - \gamma)$ ,,, but unfortunately my work resulted in more than 10 digits ,,, which i think is either due to wrong calcution or method ,,,

,,, Guys please check it for me and suggest any easier method if any than this lengthy ,,,
,,, I'll post my cardano's way of solution later upon request ,,, as right now it's late night and i'm feeling sleepy ,,,

,,, hope to get the reply soon ,,, ![/b]

[quote="CyanSnow"]Now how to find the value of this equations 'a' ??[/quote]  $ a$ is irrelevant: it's a scaling constant that doesn't change the roots (as long as it's nonzero).  Set it to any convenient value.

What is cardino's method?
Also, is there a better solution b/c the first solution involves tons of computations...  :o

I would use Vieta's...

Let the polynomial be $ ax^3\plus{}bx^2\plus{}cx\plus{}d$.

Then, by vieta's, we would have $ \minus{}\frac{b}{a}\equal{}(\alpha^2\plus{}\beta^2\plus{}\alpha\beta)\plus{}(\beta^2\plus{}\gamma^2\plus{}\beta\gamma)\plus{}(\gamma^2\plus{}\alpha^2\plus{}\gamma\alpha)$

$ \frac{c}{a}\equal{}(\alpha^2\plus{}\beta^2\plus{}\alpha\beta)(\beta^2\plus{}\gamma^2\plus{}\beta\gamma)\plus{}(\beta^2\plus{}\gamma^2\plus{}\beta\gamma)(\gamma^2\plus{}\alpha^2\plus{}\gamma\alpha)\plus{}$
$ (\gamma^2\plus{}\alpha^2\plus{}\gamma\alpha)(\alpha^2\plus{}\beta^2\plus{}\alpha\beta)$

$ \minus{}\frac{d}{a}\equal{}(\alpha^2\plus{}beta^2\plus{}\alpha\beta)(\beta^2\plus{}\gamma^2\plus{}\beta\gamma)(\gamma^2\plus{}\alpha^2\plus{}\gamma\alpha)$

Arrange the terms, and use Vieta's.


For the first one, you may use [hide]Induction[/hide]
For the second one, you may use[hide]a sepcial limits[/hide].
And for the last one, you may use[hide] $(\sqrt{a}-\sqrt{b})^{2}\geq 0$[/hide]

Masoud Zargar

Don't we get #3 automaticaly from the fact that geometric mean is always greater than the harmonic mean??

Yes, we can use GM-AM. And if you don't see that, rewrite it as $a+b\geq 2\sqrt{ab}$ by multiplying both sides with $\frac{a+b}{\sqrt{ab}}$.

[quote="Farenhajt"]It should've read "special", but you don't need that to prove this.

[hide="Proof"]Note that $k^{2}>k(k-1)$, hence $\frac{1}{k^{2}}<\frac{1}{k(k-1)}$ for $k\geqslant 2$. On the other hand, $\frac{1}{k(k-1)}=\frac{k-(k-1)}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}$. Hence

$\frac{1}{2^{2}}+\frac{1}{3^{2}}+\dots+\frac{1}{n^{2}}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\dots+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}$, and the whole sum is therefore less than $2-\frac{1}{n}$ which is less than $2$.[/hide][/quote]
typo  :wink:

[quote="eddie77"] $n \in N^{2}$[/quote]

Not really the correct notation.  If you mean $n$ a square, simply replace every $n$ in your problem with $n^{2}$.  The notation $\mathbb{N}^{2}$ usually denotes the set of ordered pairs / vectors $\left< a, b \right>, a, b \in \mathbb{N}$.

The third inequality is nothing but the GM-HM inequality for two positive reals.

It was intended to be solved without the use of GM-HM.  :wink:

Masoud Zargar


Two circles in a plane intersect at A and B.Starting from A two points move with constant speed,each traveling along its own circle in different sense.The two points return to A simultaneously after one revolution.Prove that  there is a fixed point P in the plane such that the two points are always equidistant from P.

very famous IMO problem
http://www.mathlinks.ro/Forum/viewtopic.php?p=367352#367352


Find out the last two digits of 439^874..

I solved it the following way

We have 439^874 = (400 + 39) ^ 874

= Multiple of 400 + 39^874

So the last two digits will be the same as that of 39^874 = (40 - 1)^874

= Some multiple of 40^2 - 874*40 + 1

= Some multiple of 100 - 874*(100-60) +1

= Some multiple of 100 + 874*60  + 1

so, last two digits are the same as that of 874*60 + 1

It can be observed that the  two digits of 874*60 + 1 are 4 and 1

Is there any better method ????

Hi Santosh

Hope a few  tricks you can enjoy and put it your toolbox.

(The following is not necessary a better or even a faster method but can be helpful in learnig a few tricks and methods which  would be useful in similar problems)

 Note that 100=25*4 ;

 phi(25)=20 and phi(4)=2   and LCM or 20 and 2 is 20
this   means if you take [b]any[/b] number, which is   co-prime to 2 and 5, raised to power 20;  its last 2 digit would be 01.

(Fermat's little theorem is very useful in these cases)

so all you have to do is to divide 874 by 20 and  find the remainder  and thus the last  2 digits digits are same as  39^14..

Another Fun way is to find the values modulo 4 and  25 (Noticing that 100=25*4)

Modulo 4 the answer is 1  (    because  (odd_number)^2 = 1 mod  4)
Now ...
Modulo 25
we want to find 39^14 (modulo 25)   it is easy enough to  calculate it right away but if you want to have fun (or want to do it in your head without paper or pencil) you can notice:

All modulo 25 ..
39= = 14 == 2*7

Now  7^2 == 49 == -1   so 7^14 ==  (- 1)^7 == -1
and 2^7  == 128 == 3  so 2^14  = 9
so 39^14 == -9
You have to find a number which is (-9)   mod 25 and  1 mod 4 , obviously it is 41.

P.S. of course direct calculation is fast enough, but   to know fermat's theorem (so that you know   a^20 == 1 mod 100) and some times little tricks ( breaking down the number into factors) may make  the problem easy . (For example, mod 25 you could have also reduced 39 as  3*13 or 3*(-12) = =-6 ^2 etc..

and  you have  39^14 == 6^28 == 6^8  (remember 6^20 ==1)
== (2^8)*(3^8)  remember 2^7 = = 3; so this becomes 2*(3^9) but remember 3^3 == 27 == 2 so this  2*(2^3) = 16. and the answer again is  41)

Hope the info is useful.

Thanks for all the info Gyan..

Actually I'm not quite conversant with the techniques (esp. the funda of modulus etc) of Number Theory.Have never had enough exposure to number theory except some basic fundamentals..

thanks once again


Hint: consider the effect of $\frac{1}{4}[f(x)+f(ix)+f(-x)+f(-ix)]$ on a polynomial, $f(x)$

I guess that can be used in a solution as follows...

[hide="Try"]
$\frac{1}{4}[(1+1)^{n}+(1+i)^{n}+(1-1)^{n}+(1-i)^{n}]$[/hide]

Expand this on one side, and also use De Moivre's theorem, and the identity follows.


geometry
vector
parallelogram
linear algebra
linear algebra solved

Anyone know to express the area of the triangular region bounded by the three vectors

${\left[\begin{array}{c}x_{1}\\ x_{2}\\ x_{3}\end{array}\right],~\left[\begin{array}{c}y_{1}\\ y_{2}\\ y_{3}\end{array}\right],~ \left[\begin{array}{c}z_{1}\\ z_{2}\\ z_{3}\end{array}\right] }$

in a simple equation? Probably in terms of determinant?

Area of the parallelogram spanned by two vectors is equal to the magnitude of their cross-product. A triangle is one half of a parallelogram.

Volume of the simplex with 4 vertices

${ \left[\begin{array}{c}x_{0,1}\\ x_{0,2}\\ x_{0,3}\end{array}\right],~ \left[\begin{array}{c}x_{1,1}\\ x_{1,2}\\ x_{1,3}\end{array}\right],~ \left[\begin{array}{c}x_{2,1}\\ x_{2,2}\\ x_{2,3}\end{array}\right],~ \left[\begin{array}{c}x_{3,1}\\ x_{3,2}\\ x_{3,3}\end{array}\right]}$

in ${\mathbb R}^{3}$ is given by

${A_{2}=\frac{1}{3!}\det\left[\begin{array}{cccc}x_{0,1}&x_{1,1}&x_{2,1}&x_{3,1}\\ x_{0,2}&x_{1,2}&x_{2,2}&x_{3,2}\\ x_{0,3}&x_{1,3}&x_{2,3}&x_{3,3}\\ 1&1&1&1 \end{array}\right]}$

What I am looking for is the formula of the side area on the simplex bounded by the vertices ${ \left[\begin{array}{c}x_{0,1}\\ x_{0,2}\\ x_{0,3}\end{array}\right],~ \left[\begin{array}{c}x_{1,1}\\ x_{1,2}\\ x_{1,3}\end{array}\right],~ \left[\begin{array}{c}x_{2,1}\\ x_{2,2}\\ x_{2,3}\end{array}\right]}$

So are those column matrices vectors or vertices?

mlok's idea seems simplest. Shift the vertices such that one is on the origin. Take the magnitude of the cross product of the other two, and divide by two.


What state are you from, what was your score at state, and what was your rank?

I'm from Nevada and my score was
[hide] 34 (24/5) and I came in 1st @ state. [/hide]

[quote="nat mc"]I got 40 (28/6) forgot rank and state[/quote]

How can you forget your state!?

26 (20/3), either 39th or 40th in northern CA. :?

[quote="sirorange"]43 (27/8) 2nd in state[/quote]
Are you from MA?

North Carolina. I got 38 (22/8), which I think is really bad, considering I get mid 30s on my nats (I got 38 once)... Anyway, I got second.

Im from TX and my score was...[hide]38(28/5)and i got 2nd place but countdown is official in TX[/hide]

Countdown was official so he got 3rd written and 2nd countdown. 2nd written was a 40. 1st written and countdown was a 46.

How does countdown work in states where it's official? I know that here in CT, where it is unofficial, you just pick letters randomly in an elimination bracket format.

[quote="tarquin"]10th vs 9th, if 10th wins he goes against 8th and so on.[/quote]

So if you get 3rd written, there's no way you can't make nationals?

Anyhow, I got 2nd in CA with a 40 (5 stupid mistakes :)).

The only secure spots for nationals are 1st through 3rd. Any of the other top 10 people may beat 4th place out for a trip to nationals :)

1st in GA
34 (22/6)

1st in CDR
15 points unnoficial

BTW: is kingof21penguins a penguin?

kingof21penguins... don't say my name unless you allow me to say yours...
also, nice grammer :lol:

Darn  :(

At least there is one penguin on this forum.

[hide]Me![/hide]

I am proud to say that our team was first in our state. Our coach has been coaching for 22 years, and now 20 out of 22 years the team she coached has been first.  :first:

:)  :)  :lol:  :P  Tee hee.... tis a secret.....

My team was first in Washington :first: , and this is my school's first year in MATHCOUNTS!  :) I was 2nd with 40 (26/7), and my brother got 5th.

Oh, and by the way, great grammar, all of you, especially kingof21penguins and Xantos C. Guin!

37, which was second in OH. In the unofficial countdown, I got third, having gotten beaten out by the countdown winner who got third written. (!!!)


inequalities
rearrangement inequality
Cauchy Inequality
inequalities unsolved

Prove that if $a,b,c>0$, then
\[ \sqrt{\frac a{a+b}}+\sqrt{\frac b{b+c}}+\sqrt{\frac c{c+a}}\leq2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}\leq\frac3{\sqrt2}.  \]

[quote="hungkhtn"]Do you have solution for it, PVThuan?[/quote]

I created it this afternoon and have not tried to find one.

right side is quite easy ;)

\[ 2\sqrt{1+\frac{abc}{(a+b)(a+c)(b+c)}}\leq\frac{3}{\sqrt{2}}  \]
\[ 4+\frac{4abc}{(a+b)(a+c)(b+c)}\leq\frac{9}{2}  \]
\[ 8+\frac{8abc}{(a+b)(a+c)(b+c)}\leq 9  \]
\[ 8abc\leq (a+b)(a+c)(b+c)  \]
\[ 6abc\leq a^2b+a^2c+b^2a+b^2c+c^2a+c^2b  \]
\[ 6\leq\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}  \]

And as we know well:

\[ 2\leq \frac{x}{y}+\frac{y}{x}  \]

I'm still thinking about the left side :)

Yes, it is really easy. Since we have $(a+b)(b+c)(c+a)\geq 8abc$ which is an application of AM-GM three times.

A related inequality is the following:

Prove that $(1-a)(1-b)(1-c)=abc$ implies
\[ \sqrt{1+abc}\geq\tfrac12(\sqrt a+\sqrt b+\sqrt c). \]

[quote="pvthuan"]Hmm, it's unsolved for months. Here is an almost similar in my book. \[\sqrt{\frac a{b+c}}+\sqrt{\frac b{c+a}}+\sqrt{\frac c{a+b}}\geq2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}.\] [/quote]

See See http://www.mathlinks.ro/Forum/viewtopic.php?t=103080 .

By rearrangement inequality,
$\sum \sqrt{\frac{a}{a+b}}\leq \sum \sqrt{\frac{a}{b+c}}$,
as if wlog $a \geq b \geq c$, then $\sqrt{a}\geq \sqrt{b}\geq \sqrt{c}$
and $\sqrt{b+c}\leq \sqrt{a+c}\leq \sqrt{a+b}$.

So the original inequality can be proved.

Are you sure? Pvthuan's one (I mean the one you quote) has the sign the other way round :)

Ooh I did terrible mistake. sorry :blush:

Humm, it's true if in the final step, we are able to justify the indentity \[(a(b+c)+b(c+a)+c(a+b))(a+b+c)=2(a+b)(b+c)(c+a)+2abc.\] That can be done with a pen in hand  :P
But the following is better. Choose $a+b+c=1$ for sake of simplicity,  the above identity reads \[a+b+c-a^{2}-b^{2}-c^{2}=2(1-a)(1-b)(1-c)+2abc,\] Expanding the first item in the right hand side, we get \[1-(a^{2}+b^{2}+c^{2})=2(1-a-b+ab)(1-c)+2abc\] or \[1-(a^{2}+b^{2}+c^{2})=2(1-a-b-c-abc+ab+bc+ca)+2abc.\] This can be simplified into \[1=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=(a+b+c)^{2}.\] Hmm, it's true as expected.  :P

OK, very good solution, Zhaobin.

[quote="Lovasz"][quote="pvthuan"]Hmm, it's unsolved for months. Here is an almost similar in my book. \[\sqrt{\frac a{b+c}}+\sqrt{\frac b{c+a}}+\sqrt{\frac c{a+b}}\geq2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}.\] [/quote] \[\sqrt{\frac a{b+c}}+\sqrt{\frac b{c+a}}+\sqrt{\frac c{a+b}}\geq{a+b+c \over 3}\cdot\sum_{cyc}{1 \over \sqrt{a(a+b)}}\ge{3(a+b+c) \over \sum_{cyc}\sqrt{a(a+b)}}\ge{3\sqrt{2}(a+b+c) \over 2(a+b+c)}={3 \over \sqrt{2}}\\ \ge RHS.\] [/quote]

First, I don't know what you are doing.
Second, the infimum of L.H.S. is 2, so you must be wrong.

Hmm... first, I used Chebyshev inequality.
Second, you've got a mistake. \[2\sqrt{1+{abc \over (a+b)(b+c)(c+a)}}\le{3 \over \sqrt{2}},\] by AM-GM.

I think you should try a=b=1-, c=0+ on L.H.S., then you will get my meaning.

I think that leads to $2\sqrt{2}\leq 3$ [And that's true  :D ]

[quote="zhaobin"][quote="pvthuan"]Prove that if $a,b,c>0$, then \[\sqrt{\frac a{a+b}}+\sqrt{\frac b{b+c}}+\sqrt{\frac c{c+a}}\leq2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}\] [/quote]
OK,Here is my solution. \[\sqrt{\frac a{a+b}}+\sqrt{\frac b{b+c}}+\sqrt{\frac c{c+a}}=\sqrt{\frac{a(b+c)}{(a+b)(b+c)}}+\sqrt{\frac{b(c+a)}{(b+c)(c+a)}}+\sqrt{\frac{c(a+b)}{(c+a)(a+b}}\]  \[\le \sqrt{\left(a(b+c)+b(c+a)+c(a+b)\right)\left(\frac 1{(b+c)(c+a)}+\frac 1{(a+b)(c+a)}+\frac 1{(a+b)(b+c)\right)}}\]  \[=2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}\] Just use Cauchy-Schwarz's inequality.
I hope it is right :wink:[/quote]

You have nice solutions! :lol:

Never mind. I got another proof.
[quote="pvthuan"] Let $a,b,c$ be positive real numbers. Prove that: \[\sqrt{a \over b+c}+\sqrt{b \over c+a}+\sqrt{c \over a+b}\ge 2\sqrt{1+{abc \over (a+b)(b+c)(c+a)}}.\] [/quote]
[b]Solution:[/b] Let ${\cal N}={abc \over (a+b)(b+c)(c+a)}$ and ${\cal S}=2\sqrt{1+{abc \over (a+b)(b+c)(c+a)}}$. \[\sqrt{a \over b+c}+\sqrt{b \over c+a}+\sqrt{c \over a+b}\ge 2\sqrt{1+{abc \over (a+b)(b+c)(c+a)}}\\ \iff \sum_{cyc}\left({a \over b+c}-{1 \over \sqrt{2}}\right) \ge 2\sqrt{1+{abc \over (a+b)(b+c)(c+a)}}-{3 \over \sqrt{2}}\\ \iff \sum_{cyc}{1 \over \sqrt{2}}\cdot{2a-b-c \over \sqrt{b+c}}\ge{1 \over \sqrt{2}}\left[2\sqrt{2+{2abc \over (a+b)(b+c)(c+a)}}-3\right] \\ \iff \sum_{cyc}(a-b)\left({1 \over \sqrt{b+c}}-{1 \over \sqrt{a+c}}\right) \ge{8{\cal N}-1 \over{\cal S}\sqrt{2}+3}\\ \iff \sum_{cyc}{(a-b)^{2}\over \sqrt{a+c}\sqrt{b+c}\left(\sqrt{a+c}+\sqrt{b+c}\right)}\ge-{\sum_{cyc}c\cdot (a-b)^{2}\over \left({\cal S}\sqrt{2}+3\right)(a+b)(b+c)(c+a)}\\ \iff \sum_{cyc}\left({1 \over \sqrt{a+c}\sqrt{b+c}\left(\sqrt{a+c}+\sqrt{b+c}\right)}+{c \over \left({\cal S}\sqrt{2}+3\right)(a+b)(b+c)(c+a)}\right)(a-b)^{2}\ge 0.\]

Here is another solution by a Vietnamese student. Write it in the form \[\sum_{\rm cyclic}\sqrt{a(a+b)(a+c)}\geq2\sqrt{(a+b+c)(ab+bc+ca)}.\] Squaring both sides gives \[\sum_{\rm cyclic}a(a+b)(a+c)+2\sum_{\rm cyclic}\sqrt{ab(a+c)(b+c)}(a+b)\geq 12abc+4\sum_{\rm cyclic}a^{2}b.\] Multiplying out yields \[a^{3}+b^{3}+c^{3}+2\sum_{\rm cyclic}\sqrt{ab(a+c)(b+c)}(a+b)\geq3\sum_{\rm cyclic}a^{2}b+9abc.\] By Cauchy inequality, \[(a+x)(a+y)\geq (a+\sqrt{xy})^{2}.\] Thus \[2\sum_{\rm cyclic}\sqrt{ab(a+c)(b+c)}(a+b)\geq 2\sum_{\rm cyclic}\sqrt{ab}(a+b)(c+\sqrt{ab})\\ =2\sum_{\rm cyclic}(ab+c\sqrt{ab})(a+b).\] So it's sufficient to show that \[a^{3}+b^{3}+c^{3}+4\sum_{\rm cyclic}\sqrt{ab}c(a+b)\geq\sum_{\rm sym}a^{2}b+9abc.\] Set $x^{2}=a,$.... we have \[[x^{3}+y^{3}+z^{3}+3xyz-\sum_{\rm cyclic}x^{2}(y+z)][x^{3}+y^{3}+z^{3}-xyz+\sum_{\rm cyclic}x^{2}y]\geq0.\]

[quote="pvthuan"]Here is another solution by a Vietnamese student.
...
Set $x^{2}=a,$.... we have \[\left[x^{3}+y^{3}+z^{3}+3xyz-\sum_{\rm cyclic}x^{2}(y+z)\right]\cdot\left[x^{3}+y^{3}+z^{3}-xyz+\sum_{\rm cyclic}x^{2}y\right]\geq0.\] [/quote]
This really is terrible  :) . Who is that boy? He must have been very good at expansion!

[quote="pvthuan"]So it's sufficient to show that \[a^{3}+b^{3}+c^{3}+4\sum_{\rm cyclic}\sqrt{ab}c(a+b)\geq\sum_{\rm sym}a^{2}b+9abc.\] [/quote]
After take a good look to this solution, I think it should be done like this way which is a simpler one.
We are to prove: \[a^{3}+b^{3}+c^{3}+4\sum_{cyc}c(a+b)\sqrt{ab}\ge \sum_{sym}a^{2}b+9abc.\] By AM-GM: \[\sum_{cyc}c(a+b)\sqrt{ab}\ge \sum_{cyc}2abc=6abc.\] It reduce to the famous Schur's: \[a^{3}+b^{3}+c^{3}+3abc \ge \sum_{sym}a^{2}b.\]


Find all functions $ f:\mathbb{R}\to\mathbb{R}$ so that:

$ f(xy)\plus{}f(x\minus{}y)\plus{}f(x\plus{}y\plus{}1)\equal{}xy\plus{}2x\plus{}1$ :)

[quote="Nbach"]Is it IMO shortlisted 2005? Sorry if I forget.[/quote]
$ f(x\plus{}y)\plus{}f(x)f(y) \equal{} f(xy)\plus{}2xy\plus{}1$
this is from shortlist '05 my dear friend  :wink:.
Almost same

another FE from shortlist '05
$ f(x)f(y) \equal{} 2(f(x)\plus{}yf(x))$

This has a very nice and short solution.

$ y \equal{} 0$ gives $ f(0)\plus{}f(x)\plus{}f(x\plus{}1) \equal{} 2x\plus{}1$
$ y \equal{}\minus{}1$ gives $ f(\minus{}x)\plus{}f(x\plus{}1)\plus{}f(x) \equal{}\minus{}x\plus{}2x\plus{}1$
Subtracting, we get $ f(\minus{}x) \equal{}\minus{}x\plus{}f(0)$ or $ f(x) \equal{} x\plus{}f(0)$
So all solutions are of the form $ f(x) \equal{} x\plus{}c$.
Substituting that into the original equation, we get $ c \equal{} 0$.

[quote="nayel"]Sorry if my solution is wrong.

Put $ y \equal{} 0$ in the equation so that $ \boxed{f(0)\plus{}f(x)\plus{}f(x\plus{}1) \equal{} 2x\plus{}1}\quad(*)$. From this it's not difficult to see that $ \boxed{f(x\plus{}2) \equal{} f(x)\plus{}2}\quad(1)$.

Now let $ y \equal{} 1$ in our equation to get $ f(x)\plus{}f(x\minus{}1)\plus{}f(x\plus{}2) \equal{} 3x\plus{}1$. Using $ (1)$, this becomes $ 2f(x)\plus{}f(x\minus{}1) \equal{} 3x\minus{}1\quad(2)$. But we have from $ (*)$, that $ f(x)\plus{}f(x\minus{}1) \equal{} 2x\minus{}1\minus{}f(0)$. Substituting this into $ (2)$, we have $ f(x)\plus{}2x\minus{}1 \equal{} 3x\minus{}1\plus{}f(0)$. And thus $ \boxed{f(x) \equal{} x\plus{}f(0)}\quad(3)$.

Now to determine $ f(0)$, put $ x \equal{} 0$ into $ (*)$ to get $ 2f(0)\plus{}f(1) \equal{} 1$. Also by letting $ x \equal{} 0$ into $ (3)$ we get $ f(1) \equal{} 1\plus{}f(0)$. Solving this system we easily find that $ f(0) \equal{} 0$, implying $ \boxed{f(x) \equal{} x}$.[/quote]

This is my solution  :P


Let triangle regular $ ABC$ with centroid $ G$and $ M$ is any point in
triangle . From $ M$ draw to $ AB ; AC ; BC$at $ D,E,F$such  that$ MD$ and$ AB$
is perpendicular , same at $ ME$ and $ AC; MF$and $ BC$is
perpendicular.Show that:
\[ \overrightarrow{MD}\plus{}\overrightarrow{ME}\plus{}\overrightarrow{MF}\equal{}\frac{3}{2}\overrightarrow{MG}\] :?:

$ \boxed{MD \plus{} ME \plus{} MF \equal{} \frac {3}{2}AG}$, or the altitude of a regular triangle.


induction
number theory proposed
number theory

A sequence of natural numbers $c_1, c_2,\dots$ is called [i]perfect[/i] if every natural
number $m$ with $1\le m \le c_1 +\dots+ c_n$ can be represented as
$m =\frac{c_1}{a_1}+\frac{c_2}{a_2}+\dots+\frac{c_n}{a_n}$
Given $n$, find the maximum possible value of $c_n$ in a perfect sequence $(c_i)$.

let $ C_1 \equal{} 2$ and $ C_i \equal{} 4\times 3^{i \minus{} 2}$ for $ i\geq 2$.we claim that for each $ i$,$ C_i$ is the maximum possible value of $ c_i$.
we first prove by induction on $ i\geq 1$ that $ c_i\leq C_i$.for $ i \equal{} 1$,if $ c_i > 1$,then setting $ (m,n) \equal{} (c_1 \minus{} 1,1)$ shows that $ a_1 \equal{} \frac {c_1}{c_1 \minus{} 1}$ is an integer.this happens exactly when $ c_1 \equal{} 2$.thus $ c_1\leq C_1$.
now suppose that $ c_i\leq C_i$ for $ i \equal{} 1,2,\ldots ,k \minus{} 1$,where $ k\geq 2$.we find the $ a_i$ corresponding to $ (m,n) \equal{} (c_k,k)$.clearly $ a_n\geq 2$.then:

$ c_n \equal{} \sum_{i \equal{} 1}^n\frac {c_i}{a_i}\leq\frac {c_n}2 \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}C_i$

or $ c_n\leq 2\sum_{i \equal{} 1}^{n \minus{} 1}C_i \equal{} C_n$.this completes the inductive step and the proof of the claim.
it remains to be proven that for each $ i$,it is possible to have $ c_i \equal{} C_i$.indeed,we prove by induction on $ n$ that we can have $ c_i \equal{} C_i$ for all $ i$ simultaneously.for $ n \equal{} 1$,if $ m \equal{} 1$ we can set $ a_1 \equal{} 2$;if $ m \equal{} 2$ we can set $ a_1 \equal{} 1$.
assuming that the claim is true for $ n \equal{} 1,2,\ldots ,k \minus{} 1$,where $ k\geq 2$,we prove it for $ n \equal{} k$.if $ m \equal{} 1$,we may set $ a_i \equal{} nC_i$ for $ i \equal{} 1,2,\ldots ,n$.if $ 2\leq m\leq \frac {C_n}2 \plus{} 1$,set $ a_n \equal{} C_n$;if $ \frac {C_n}2 \plus{} 1\leq m\leq C_n$,set $ a_n \equal{} 2$;if

$ C_n \plus{} 1\leq m\leq\frac {3C_n}2 \equal{} \sum_{i \equal{} 1}^n C_i$

set $ a_n \equal{} 1$.in each case,$ 1\leq m \minus{} \frac {C_n}{a_n}\leq\frac {C_n}2$.thus by the induction hypothesis,we may choose positive integers $ a_1,a_2,\ldots ,a_{n \minus{} 1}$ such that

$ m \minus{} \frac {C_n}{a_n} \equal{} \sum_{i \equal{} 1}^{n \minus{} 1}\frac {C_n}{a_n}$

as desired.this completes the proof.

there is latex error in this post can you fix it please


Can anyone help me to start with the Thermodynamic Identity:

(∂s/∂T)_x>0 for all processes where x is an arbitrary intensive or extensive variable of of the system.

I do understand that we have to exploit some property of the properties of the system.Say all of them are exact differential or so...But cannot understand how to do this Identity...
Can anyone please help me to start with it?

Hope this makes sense...Please check it.

First, we recall the definition of "Specific Heat At Constant X", where
"X" is an arbitrary intensive variable (like pressure) or extensive
variable (like volume):

{Specific Heat At Constant X} = c_X = (1/m)*(dQ/dT)X

----> dQ = m*(c_X)*dT ..... for constant X

where "m" is system mass. Thus, from the definition of entropy &
specific entropy (s = S/m), and the Second Law Of Thermodynamics
(for cyclic process), we have:

dS = dQ/T > 0

----> dS = {m*(c_X)*dT}/T > 0 ..... for constant X

----> ds = {(c_X)*dT}/T > 0 (<--- dividing by m, & s = S/m)

----> (∂s/∂T)_X = (c_X)/T > 0 (<--- dividing by dT & at const X)

----> (∂s/∂T)_X > 0


这有北京的么？
刚在北京考了AIME，
考完有什么呀？
还有人考了么？

我在北京, 可压根没听说这事!

[quote="Kalimdor"][quote="nirvanatear"]AIME II?
我只考了AIME I
死了..
对8道[/quote]
neh, I got 10, but still bad. I heard that there's an Ontario kid, John Schneider, got 13, but only in his 8th grade. Man, that's friggin hilarious.
Sorry, can't type chinese under Opera Web Browser right now. IE's cracked.[/quote]
10 is decent.


Can we have three non empty sets $ A_1,A_2,A_3$ such that $ A_i\cap A_j\equal{}\emptyset \quad ,1\le i \le j \le 3$ such that $ \bar{A_1}\equal{}\bar{A_2}\equal{}\bar{A_3}\equal{}\mathbb{R}$ where $ \bar{S}$ indicates closure of set $ S$.

Take $ A_i \equal{} \mathbb{Q} \plus{} y_i$ where $ y_1,y_2,y_3$ are irrational numbers with irrational differences.

@ CommandoGuard: Thanks.

By the above construction, there exist countable number of sets $ A_i$ with the above property. The next natural question to ask will be whether we have the set of $ A_i$'s which is uncountable?

An approach will be looking for a subset $ S$ of $ \mathbb{I}$ (irrationals) such that $ \forall a,b \in S,\quad a \minus{} b\in S$ and cardinality of $ S$ is uncountable.

You can take unions of these cosets of $ \mathbb R / \mathbb Q$. There are $ |\mathbb R|$ many of them, so you can get enough of them. You can even write $ \mathbb R$ as a disjoint union of uncountable many such uncountable subsets.

and just to be a pain in the ass, zetax's approach doesn't need the axiom of choice, while talegari's requires it (at least, to obtain results as strong).


[quote="Treething"]In Fairyland some fairies have three spots while others have five. However, all of them have two wings.

At the Fairyland Convention for Democrats, John counts 138 wings and 255 spots. How many fairies with three spots attended the convention?[/quote]
[hide]
$2x+2y=138$
$3x+5y=255$

Solve to get $x=45$ 3-spotted fairies and $y=24$ 5-spotted fairies.[/hide]

:D [hide=" :D "]t=three spotted fairies, f=five spotted-solve:
$3t+5f=255$ and $2t+2f=138=2(t+f),so, t+f=69$
which gives us $t=45$
[/hide] :D

and splashd, you didn't really answer the question... :P

says who, I said there were 45 3-spotted fairies.

[quote="SplashD"]
[b]Solve to get $x=45$ 3-spotted fairies[/b][/quote]
SplashD did answer the question.

[hide]using elimination, we get that there are 24 5-spotteds and $\boxed{45}$ 3-spotteds[/hide]

[hide]Setting up equation, we can see that the answer is 45.

$45$[/hide]

[hide]Call 3-spotted fairies $x$ and 5-spotted fairies $y$.

We can set up the equations $2(x+y)=138$ and $3x+5y=255$. Solving we find $(x,y)=(45,24)$ so there are 45 3-spotted fairies.[/hide]

[hide]
Let x be the 3-spotted fairies and y be the 5-spotted ones; we have two equations:
$2x+2y=138$
$3x+5y=255$
Solving for x, we get $45$[/hide]

[quote="nebula42"]Many of the posts in this topic were really starting to wander off from the actual subject, so they have been deleted.
Next time, please try to keep the discussion related to the [i]problem[/i].  :wink:
Thanks for your cooperation.  :)[/quote]
I apologize, i kind of got of track.

[hide]138 wings means 69 fairies. If they all have 3 spots, then there would be 207 spots. However, there are 255 spots. Since all 5-spotted faries have two more spots than 3-spotted faries, you would divide 48 by 2 to get 24 5-spotted fairies. 69-24=45

There are [color=green]45 3-spotted fairies[/color].[/hide]


[quote="gemath"]Given triangle $ABC$, $A',B',C'$ are three aribitrary point on segment $BC,CA,AB$ resp, $A_{1},B_{1},C_{1}$ are midpoint of $AA',BB',CC'$,$P$ is an arbitrary point, $A_{2},B_{2},C_{2}$ is midpoint of $PA,PB,PC$ resp. Called $d_{a}$ is line connecting midpoint of $B_{1}C_{2}$ and $B_{2}C_{1}$, similar $d_{b},d_{c}$ prove that $d_{a},d_{b},d_{c}$ are concurrent![/quote]

Dear Gemath,

I think there is something wrong with this problem. See the ilustration below(the red lines are those ones which,according to the problem,should intersect at one point,however,as you can see they do not!).

Hi Tiks, thank for your interest, you are right the post of me is wrong I have fixed it and repost it here:

Given triangle $ABC$, $A',B',C'$ are three aribitrary point on segment $BC,CA,AB$ resp, $A_{1},B_{1},C_{1}$ are midpoint of $AA',BB',CC'$,$P$ is an arbitrary point, $A_{2},B_{2},C_{2}$ is midpoint of $PA',PB',PC'$ resp. Called $d_{a}$ is line connecting midpoint of $B_{1}C_{2}$ and $B_{2}C_{1}$, similar $d_{b},d_{c}$ prove that $d_{a},d_{b},d_{c}$ are concurrent!

Hope there will be a nice solution from anybody!

[b]Lemma:[/b]Given $\Delta ABC$ and three lines $d_{1},d_{2},d_{3}$ are parallel with $BC,CA,AB$, respectively. Let $k_{1}$ is the distance between $d_{1}$ and $BC$, $k_{1}$ is non-negative if $d_{1}$ and $A$ are in the same sides of $BC$ ( my english is not good enough to describe it clearly  :( . Help me, please )and negative if not. Define $k_{2}$ and $k_{3}$ similarly. So $d_{1},d_{2},d_{3}$ are concurent if and only if $\frac{k_{1}}{h_{a}}+\frac{k_{2}}{k_{b}}+\frac{d_{3}}{h_{c}}= 1$
We also define $k_{X/ BC}$ is the distance of $X$ to $BC$, and $k_{X/BC}$ is non-negative if $X$ and $A$ are in the same sides of $BC$, it's negative with not. With this lemma we have $\frac{k_{X/BC}}{h_{a}}+\frac{k_{X/CA}}{h_{b}}+\frac{k_{X/AB}}{h_{c}}=1$
( It's very easy to prove )
[b]Back to the problem:[/b]
Let $M,N$ is the midpoint of $B_{1}C_{2}$ and $B_{2}C_{1}$, we have: $4\overrightarrow{MN}=2(\overrightarrow{B_{1}B_{2}}+\overrightarrow{C_{2}C_{1}})=\overrightarrow{BP}+\overrightarrow{PC}=\overrightarrow{BC}$. It means that $d_{1}\parallel BC$. So $4k_{1}=2(k_{B_{1}/BC}+k_{C_{2}/BC})= k_{B'/BC}+k_{C'/BC}+k_{P/BC}$. Similarly, $4k_{2}= k_{C'/CA}+k_{A'/CA}+k_{P/CA}$ and $4k_{3}=k_{A'/AB}+k_{B'/AB}+k_{P/AB}$
In order to prove that $d_{1},d_{2},d_{3}$ are concurent, we are going to prove that $\frac{4k_{1}}{h_{a}}+\frac{4k_{2}}{h_{b}}+\frac{4k_{3}}{h_{c}}=4$, but it's of course:
$\frac{4k_{1}}{h_{a}}= \frac{k_{B'/BC}}{h_{a}}+\frac{k_{C'/BC}}{h_{a}}+\frac{k_{P/BC}}{h_{a}}\\ \frac{4k_{2}}{h_{b}}=\frac{k_{C'/CA}}{h_{b}}+\frac{k_{A'/CA}}{h_{b}}+\frac{k_{P/CA}}{h_{b}}\\ \frac{4k_{3}}{h_{c}}= \frac{k_{A'/AB}}{h_{c}}+\frac{k_{B'/AB}}{h_{c}}+\frac{k_{P/AB}}{h_{c}}\\ \frac{k_{B'/BC}}{h_{a}}+\frac{k_{B'/AB}}{h_{c}}= \frac{k_{B'/BC}}{h_{a}}+\frac{k_{B'/AB}}{h_{c}}+\frac{k_{B'/CA}}{h_{b}}=1\\ \frac{k_{C'/CA}}{h_{b}}+\frac{k_{C'/BC}}{h_{a}}= \frac{k_{C'/CA}}{h_{b}}+\frac{k_{C'/BC}}{h_{a}}+\frac{k_{C'/AB}}{h_{c}}=1 \\ \frac{k_{A'/AB}}{h_{c}}+\frac{k_{A'/CA}}{h_{b}}= \frac{k_{A'/AB}}{h_{c}}+\frac{k_{A'/CA}}{h_{b}}+\frac{k_{A'/BC}}{h_{a}}=1 \\ \frac{k_{P/BC}}{h_{a}}+\frac{k_{P/CA}}{h_{b}}+\frac{k_{P/AB}}{h_{c}}=1$


Let $ k_y(G)$ and $ k_e(G)$ represent the vertex connectivity and edge connectivity,respectively, of a graph G.
Show $ k_y(G) \leq k_e(G)$ and give an example of  $ k_y(G) < k_e(G)$

thanks~

Let $ K_e$ be a cut edge set with $ k_e$ elements. Starting with the vertex on which highest number of edges of $ K_e$ are incident, we keep removing such vertices iteratively until we delete all the elements of the cut edge set. Since deleting a vertex, automatically deletes at least an edge, the size of the cut vertex set $ k_y$ cannot be larger than $ k_e$.

Here is an example for $ k_y < k_e$:


calculus
derivative
function
calculus computations

Let $ \ f(x)$ be defined for $ \ x>\minus{}1$ and has a continuous derivative. If $ \ f(0)\equal{}1,\ f'(0)\equal{}0$ and $ \ (1\plus{}f(x))f''(x)\equal{}1\plus{}x$, then find $ \ f(x)$.

First make a change of variables:
$ y(x) \equal{} f(x \minus{} 1) \plus{} 1$. Then $ y(x) y''(x) \equal{} x$, or $ y'' \equal{} xy^{ \minus{} 1}$.
Now the bad part of the story: according to [url=http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf]EqWorld[/url], it is not solvable.

Nevertheless, with the given boundary conditions $ y(1) \equal{} y'(1) \equal{} 0$ it might be solvable (though the obvious particular solution $ y(x) \equal{} 2x^{3/2}/\sqrt{3}$ does not work).

[quote="zhoraster"]First make a change of variables:
$ y(x) \equal{} f(x \minus{} 1) \plus{} 1$. Then $ y(x) y''(x) \equal{} x$, or $ y'' \equal{} xy^{ \minus{} 1}$.
Now the bad part of the story: according to [url=http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf]EqWorld[/url], it is not solvable.

Nevertheless, with the given boundary conditions $ y(1) \equal{} y'(1) \equal{} 0$ it might be solvable (though the obvious particular solution $ y(x) \equal{} 2x^{3/2}/\sqrt {3}$ does not work).[/quote]

Can we tell what is the nature of f'(x)

whether it increases or decreases ?

By the way, the conditions are $ y(0) \equal{} 2$, $ y'(0) \equal{} 0$.

Well, the derivative should increase (as long as $ y$ should keep its - positive - sign). So the function should be convex.

hello, try this here
http://www.wolframalpha.com/input/?i=%281%2By%28x%29%29*y%27%27%28x%29%3D1%2Bx
Sonnhard.

Dr Sonnhard Graubner, did you visit the link you gave us?

hello, yes, you must copy the whole line.
Sonnhard.

Lets assume that there exist c that belongs to (-1,+oo) such 1+f(c)=0
by substitution x=c at ODE we taking 0=1+c <-> c = - 1 contradiction. Though f(x)+1 is continuous and cant be equal to zero so is positive or negative at           (-1,+00).
For x=0, f(0)+1=2>0 ,  that means f(x)+1>0 for every x on (-1,+00).
Since f(x)+1 and x+1 are positive that also means f''(x)>0 so f' is strictly increasing

[quote="zhoraster"]Dr Sonnhard Graubner,

Though you kindly gave a broken link, I'm not a complete idiot to be unable to copy and paste.

But your kindly given broken link - even if you copy and paste it - seems to give absolutely nothing for the answer to the question except telling me that it is - holy Moses! - a second order nonlinear ordinary differential equation. What a surprise, perhaps. Oh, it also gives 5 nice pictures!

Thanks, Dr Sonnhard Graubner, you help is highly appreciated.[/quote]

Mhmhmh...well this doesnt help either!

From the fact that f'(0)=0 and that f' is strictly increasing we get f'(x)>0 for x>0 and f'(x)<0 for -1 < x < 0.
That means f is strictly increasing on [0,+00) and strictly decreasing on (-1,0]  and has minimum value f(0)= 1
Dunno if these can help for solution :)


function
Putnam
search
real analysis
real analysis unsolved

Let $ f$ be a real fuction on the real line with continuous third derivative.Prove that there exists a point $ a$ such that
$ f(a)f^{I}(a)f^{II}(a)f^{III}(a)\geq 0$.

kunny you are right, but can you tell me from where can I get solutions of putnam problems!!!


algebra
polynomial
quadratics
geometry
3D geometry

Every monic cubic $ f(h) \equal{} h^3 \plus{} ah^2 \plus{} bh \plus{} c$ factors as $ f(h) \equal{} (h \minus{} r_1)(h \minus{} r_2)(h \minus{} r_3)$ for some complex roots $ r_1, r_2, r_3$.  If you have one root $ r$, divide the original polynomial (in your case $ f(h)\equal{} h^3 \minus{} 12h^2 \minus{} 6$) by $ h \minus{} r$ to be left with a quadratic that you can solve by the quadratic formula.  This will, of course, be horribly messy and disgusting, but it's also guaranteed to work.

Other suggestions:

1. Use the [url=http://en.wikipedia.org/wiki/Cubic_Equation#Root-finding_formula]cubic equation[/url]

2. Use [url=http://en.wikipedia.org/wiki/Newton%27s_Method#Description_of_the_method]Newton's Method[/url]

3. Use [url=http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas]Viete's Formulas[/url] (which is equivalent to JBL's method)

But since you already have one of the roots, the best method is the one outlined by JBL.

Let $ \omega$ be any of the three cube roots of $ 1$.  Multiply your first cube root by $ \omega$ and your second cube root by $ \omega^2$.  That should give you all three roots.

grrr i'm stupid :/

can you do it on my examle step by step i'll be wery grateful

No, but I'll do a simpler example that I've rigged so as to only have integers involved :)

Let's solve the equation $ h^3 \equal{} h^2 \plus{} 32h \plus{} 70$.  I know (by whatever mystical method you know your real root) that this equation has a root $ h \equal{} 7$.  So, I take the associated cubic polynomial $ f(h) \equal{} h^3 \minus{} h^2 \minus{} 32h \minus{} 70$  and divide (long division, synthetic division, whatever) to get $ \frac{f(h)}{h \minus{} 7} \equal{} h^2 \plus{} 6h \plus{} 10$.  The Factor Theorem tells us that the result will be a quadratic polynomial, and the other two roots of the original equation $ f(h) \equal{} 0$ are now the roots of the associated quadratic equation $ h^2 \plus{} 6h \plus{} 10 \equal{} 0$, and this is easy to solve.

Ravi's method works, too, although I don't know how he'd apply it to my "nicer" problem :).

yesss but in my example from division i get [img]http://matematyka.pl/latexrender/pictures/1e52573c9ae6e7b8a2288cb68a4bf49d.gif[/img]

You have to use the exact value, not the approximation.  (That's why I don't want to do your example :P.)

so i should use $ h \approx 12.0413807797668$  :roll:

 :oops: nice

The exact value.  You need to carry out the division exactly (in terms of radicals) to get the correct final answer.  Alternately, Ravi's method lets you circumvent all of that.

ok i solve it and $ \Delta = -1,998$
$ a=1$

now i have

$ h1 = \frac{-b}{2a}+i\frac{\sqrt{\Delta}}{2a}$

$ h2 = \frac{-b}{2a}-i\frac{\sqrt{\Delta}}{2a}$

anybody can help?

please show me step-by-step how to solve those $ h1 \& h2$


show that in any acute triangle holds:
$cos(A-B)*cos(B-C)*cos(C-A)\geq 8cosAcosBcosC$

[hide]
$\sin A=\sin (B+C)$, so we can multiply both sides by the positive $\sin A\sin B\sin C$. The right-hand side becomes $\sin 2A \sin 2B\sin 2C$. For the left-hand side, we can use product-to-sum:
\[\cos (B-C)\sin (B+C)=\frac{\sin 2B+\sin 2C}{2}\]
So our inequality is equivalent to
\[(x+y)(y+z)(z+x)\geq 8xyz\]
where $x=\sin 2A$ etc. Since the triangle is acute, $x,y,z$ are positive, and the inequality follows from AM-GM[/hide]


superior algebra
superior algebra unsolved

How many nonisomorphic groups of order 8 are there?

Five: $ \mathbb{Z}_8, \mathbb{Z}_4\times\mathbb{Z}_2, \mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2, D_4, Q_8$

:ddr: Thanks! That helped me a lot!:ddr:

For a sketch see http://everything2.com/title/groups%2520of%2520order%25208


geometry circumcircle trigonometry ARML AMC

A line drawn from vertex A of equilateral triangle ABC beets BC at D and the circumcircle at P. If PB=5, PC=8, find PD.

lol.. you don't really need to go through your posts and put "any takers" after all of them..  :wink:

well, what i meant to do was to put "any takers" becsaue when the posts go down the list, they are less liable to be answered

Solution 1.

[hide]Quite messy, but a solution nontheless:

First notice that <BPC = 120o, <BPA = <CPA = 60o. Using cosine law on triangle BPC, we obtain

BC2 = BP2 + CP2 - 2 * BP * CP * cos 120o, BC = :sqrt:129.

Let <PCB = x, then by sine law

BC / sin BPC = BP / sin PCB

sin x = (5 * :sqrt:43) / 86;

cos x = :sqrt:(1 - sin2 x) = (7 * :sqrt:129) / 86.

Now, in triangle PCD,

PD / sin x = PC / sin (180 - <CPD - x)

PD * sin (120 - x) = 8 * sin x

PD * (sin 120 * cos x - cos 120 * sin x) = 8 * sin x

Sub in values,

PD * (26 * :sqrt:43 / 172) = 8 * 5 * :sqrt:43 / 86

PD = 40 / 13.[/hide]

Solution 2.

[hide]One just needs to notice that <BPA = <CPA = 60o. K denotes the area.

K (BDP) = BP * DP * sin 60o / 2.

K (CDP) = CP * DP * sin 60o / 2.

K (BCP) = BP * CP * sin 120o / 2.

Since K (BCP) = K (BDP) + K (CDP),

DP * sin 60o * (BP + CP) = BP * CP * sin 120o.

However, sin 60o = sin 120o.

Therefore DP * (BP + CP) = BP * CP.

Therefore DP = BP * CP / (BP + CP) = 5 * 8 / (5 + 8) = 40 / 13.[/hide]

[hide]<BPA and <BCA subtend the same chord BA, hence they're both 60o; similarly, <CPA and <CBA both subtend CA.[/hide]

are these practice meets in preparation for the nyciml ?

my arml coach handed some of those out, and i think they're much easier than these practice meets.

The practice meets are used to determine Stuyvesant's teams to participate in the NYCIML, with the exception of the Senior A team, which was decided at the end of last year.  They are written/compiled by the 5 senior captains (the senior A team), each captain writing problems for his/her particular group of students.  These were used to determine the junior and senior teams, and thus are of higher difficulty.  The groups of problems that I've posted here were the practice meets I used as a senior captain last year.


geometry
geometric transformation
combinatorics unsolved
combinatorics

Real numbers are written on the squares of an infinite grid. Two figures consisting of finitely many squares are given. They may be translated anywhere on the grid as long as their squares coincide with those of the gid. It is known that wherever the first figure is translated, the sum of numbers it covers is positive. Prove that the second figure can be
translated so that the sum of the numbers it covers is also positive.