factorial
function
number theory
algebra
floor function
prime factorization
number theory open

Hi everybody,

I want to know if there is a general solution to this problem.

22! is divisible by 2^19 and [u]19 is the biggest possible power of 2[/u]
7! is divisible by 2^4 and [u]4 is the biggest possible power of 2.[/u]

Can we have a general solution such as we could know using a formula for any n! the biggest power of 2 dividing n!

We can do it by trial for any n! but I want a general solution

Thank you for any help.

This might be helpful
http://en.wikipedia.org/wiki/De_Polignac's_formula

I don't know if this what you need

Daniel

Thanks a lot.
That's exactly what I need.

[quote="t0rajir0u"]I prefer the following:  let $ s_p(n)$ be the sum of the digits of $ n$ in base $ p$.  Then the number you're looking for is $ \frac {n \minus{} s_p(n)}{p \minus{} 1}$.  It's a good exercise to show that these two are equivalent.[/quote]

Thank you for your elegant formula.

Now I'm going to give you my new idea about how to write some specific factorial n! where n+1 is prime.

Here is my idea. I do not know if it comes with something new. My knowledge in number theory is a tiny one.

Let

S2(n) the sum of the digits of n in base 2.

By using the formula you suggested b = n - S2(n)

If n+1 is prime then we can write

n! = (2^b)*(q*(n+1)-2^(n-b))

where q is an integer maybe with some properties.

Example

22! = 2^19 *(23*93211358756821 - 2^3)

q = 93211358756821 is prime

I do not know if q is always prime or semi-prime because it is just the first part.
We have to expand the sequence of the q numbers to understand if there is some pattern.

The factorial will depend only on q.
If we solve the mystery of q then it is done.
I do not have the tools to expand the sequence of the q's.

The first values of q

1,1,7,1289,35983,37559581,5141708941,93211358756821

Maybe there is a pattern in the sequence.
If it is the case then the factorial could be expressed another useful and workable way

I used the base 2.
If we use another base q will be different and so the sequence.
Using a big base will give a q with few digits.
I do not know how to write a general formula in Latex.
If someone can do it it will be helpful.

The problem is that there's no reason to expect that understanding $ q$ is [b]any[/b] easier than understanding the entire factorial.

Why are you so reluctant to learn number theory?  You seem to like it a lot.  I think you would enjoy your mathematical explorations a lot more if you had the background to carry them out properly.

I think you are right.
I'm going to stop posting and come back to the bases of theory number.
Thank you very much.
I learned more in some posts than I could learn by reading one month.

it is
$ \sum\ \frac {n}{p^k}$  where $ n$ is the component of $ n!$ and $ p$ is the desired prime and $ k$ is natural.

the [u]proof[/u] is as follows:
let $ f(n)$ be the highest power of $ p$ that divides $ n!$.

obviously the factors in $ n!$ divisible by $ p$ are $ p, 2p, 4p.....[\frac {n}{p}]p.$

$ f(n) \equal{} [\frac {n}{p}] \plus{} f([\frac {n}{p}])$

now transforming $ n$ to $ \frac {n}{p}$

$ f([\frac {n}{p}]) \equal{} [\frac {n}{p^2}] \plus{} f([\frac {n}{p^2}])$

by substitution,
$ f(n) \equal{} [\frac {n}{p}] \plus{} [\frac {n}{p^2}] \plus{} f([\frac {n}{p^2}])$

and so it continues hence giving
$ f(n) \equal{} [\frac {n}{p}] \plus{} [\frac {n}{p^2}] \plus{} [\frac {n}{p^3}] \plus{} ....$
and after finite no. of steps the process ends.

in your case, just substitute $ p \equal{} 2$ and you get it.

oops! forgot to  say
in the above proof,
$ [x]$ represents the greatest integer that divides $ x$ ( we call it greatest integer function !)

Let $ \nu_p(n!)$ denote the exponent of the prime $ p$ in the prime factorization of $ n!$.  Another interesting related fact is that
\[ p^{\nu_p(n!)}>q^{\nu_q(n!)}
\]
whenever $ p<q$ are primes and $ \nu_p(n!) \neq \nu_q(n!)$.  Hence, your $ 2^{\nu_2(n!)}$ is the largest prime power.
(This was a monthly problem proposed by Erd\"{o}s)


If you flip a coin 12 times, what is the expected number of 3 consecutives heads?

The probability of getting three consecutive heads is (1/2)^3=1/8. Since there are 12 positions and the desired outcome must happen three in a row, there are 12-3+1 positions, or 10.  We multiply this by the probability of the event occuring once in a set of three tosses.  (1/8)(10)=10/8=

[b]5/4[/b]

[hide]

I'm going to regard each "flip" as a separate coin(so you can follow me grammatically)

Firstly, the total number of combinations of coin flips possible is $2^{12}$, or $4096$

Count the block of 3-heads-in-a-row as equivalent to one coin toss(h).  How many combinations can be made from the tosses of the other 9 coins?  It's $2^9$.  And then there are 10 places where we could stick our thing of heads in there.

But then we have repeats, when there is a "head" coin adjacent to the block of 3.  So we have to subtract $9*2^8$ from that.  And then we have to subtract(from that) when there's a head adjacent to the blocks of four, meaning that we have to add to the cumulative thing $8*2^7$.  And we do that a whole bunch of times,

So I think it comes out to be $10*2^9 -9*2^8+8*2^7-7*2^6 +6*2^5 -5*2^4+4*2^3-3*2^2+2*2^1-1*2^0$

And, if I did not make any mistakes in reasoning(or arithmetic), I believe the answer is $\boxed{\frac{3527}{4096}}$

[/hide]


geometry
trigonometry
trig identities
Law of Cosines

In a triangle ABC , angle ABC=135 degree.Prove that AC^2=AB^2+BC^2+4 area of triangle ABC. :?:

I drew an altitude AD to side BC(extended).
AC^=AD^2+DC^2
=AD^2+(DB+BC)^2
=AB^2-BD^2+BD^2+BC^2+2BD. DC
=AB^2+BC^2+2BD.DC

But RHS has - AB^2+BC^2+4 area of triangle ABC
4area of triangle ABC=4(area of triangle ADC-area of triangle ADB)
=4{1/2AD.DC-1/2 AD.DB}
=4{1/2AD(DC-DB)}
=4{1/2AD(BC)}
=2AD.DC

I am not getting this part-4ar(ABC)
I think there is something wrong in this question. :!:
Just check my steps please.
Thanx :)

Use law of cosines.

[hide]By the law of cosines,
$ AC^2 \equal{} AB^2 \plus{} BC^2 \minus{} 2\cdot AB\cdot BC\cos 135^\circ$
$ AC^2 \equal{} AB^2 \plus{} BC^2 \plus{} \sqrt{2}\cdot AB\cdot BC$

$ [\triangle ABC] \equal{} \frac{1}{2}\cdot AB \cdot BC \cdot \sin 135^\circ$ = $ \frac{\sqrt{2}}{4}AB\cdot BC$.
$ 4 \cdot [\triangle ABC] \equal{} \sqrt{2}\cdot AB\cdot BC$.

So, $ AC^2 \equal{} AB^2 \plus{} BC^2 \plus{} 4\cdot[\triangle ABC]$.[/hide]


difficult problems in geometry is a great one.

[quote="AlphaBetaTheta"]Does anyone know any useful geometry texts to prep for Mathcounts/amc8 and 10
Please pm me[/quote]

The intro to geometry book is great, doing that and the course gives you a good start on geometry, it puts a lot of concepts and strategies to do well at mathcounts/amc 8.
AMC 10 problems have a deep understanding of geo. in particular, do those for fun.

I know AOPs vol.1 has great chapters devoted to geometry.
Two of my favorite are the Angle Chasing and a Potpurri of Geo. Questions! Just keep exploring with texts, that's the most important thing in geometry.

houghton mifflin is okay...
its Algebra txtbook was actually pretty good..
so i guess the geometry is okay[/img]

houghtin mifflin's geometry is just mostly proofs

Also, Houghlon Mifflin is dedicated more to rote memorization and meeting academic cirricula standards since it is a textbook designed for a regular classroom, not problem solving techniques useful for Mathcounts/AMC8/AMC10.

There is also another good web that you can probably purchase a geometry textbook.
http://www.thinkcentral.com/index.htm

This isn't a book, but it might be helpful. It is mostly for high school geometry though:

http://www.gogeometry.com/

"Challenging Problems in Geometry" might help, but it also might be a bit hard for AMC 8/10 levels. It's mostly AMC 12 type problems, but you might want to try it.

Holt Rinestart Winston is good.

Amazon also works.

Here's the Amazon link to "Challenging Problems in Geometry":

http://www.amazon.com/Challenging-Problems-Geometry-Alfred-Posamentier/dp/0486691543/ref=sr_1_1?ie=UTF8&s=books&qid=1250903297&sr=8-1#

It's very cheap too.


analytic geometry
graphing lines
slope
algebra
linear equation

The basic equation is $y=mx+b,$ where $m$ is the slope and $b$ is the y-intercept. Slope -3 and y-int 4 would be $y=-3x+4.$ If you have an x-intercept instead, you have to plug in the $x$ and $y$ values and find $b.$ I hope that's a good explanation.

[quote="Inspired By Nature"]I learned the general form to be y=ax+b, where a is the slope and b is the -int. So our equation is y=-3x+4. This means, that fro every y cordinate value, you multipy it by -3 and add 4 to get the x-cordinate value. Thus you get a graph.

EDIT: i like pie beat me to it. :)[/quote]
The slope is represented by m generally...

[quote="linzk3"]what is the formula for making an equation

I  know the slope is $-3$ and the y-intersept is $4$
does that mean my equation would be $4-3x=y$?[/quote]

Since the format for a linear equation is $y=mx+b$ where $m$ is the slope and $b$ is the y-intercept, the equation would be $y=-3x+4$.

[quote="bpms"]The slope is represented by m generally...[/quote]
Yes, but it doesn't matter what you call the variable. :)

[quote="Inspired By Nature"]I learned the general form to be y=ax+b, where a is the slope and b is the -int. So our equation is y=-3x+4. This means, that fro every y cordinate value, you multipy it by -3 and add 4 to get the x-cordinate value. Thus you get a graph.

EDIT: i like pie beat me to it. :)[/quote]
can i ask which software did you use.

it hard to  draw them on piece of  Paper.

that might be really easy to use a computer software.

I used graphmatica. You can get it from download.com. :)

[quote="linzk3"]what is the formula for making an equation

like if i  know the slope is -3 adn the y-intersept is 4
does that mean my equation would be 4-3x=y
???
pls && thx[/quote]

yes, y=-3x+4


It is given that $ \alpha ,\beta  (\ge \alpha )$are the roots of the equation $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ .
Given that $ af(t)> 0$  , find $ t$  in terms of $ \alpha$ and $ \beta$ .

Hello
any t if satisfies $ af(t)>0$ , it simply means t belongs to $ R\minus{}[\alpha,\beta]$.

[hide="Explanation"]
suppose if $ a$= $ \plus{}ve$ ,the graph is an upwards parabola then f(t)will also be positive .
And if $ a$=$ \minus{}ve$,the parabola would open downwards, then f(t) would be negative.
[/hide]
Thank u.

[quote="kabi"]Hello
any t if satisfies $ af(t) > 0$ , it simply means t belongs to $ R \minus{} [\alpha,\beta]$.

[hide="Explanation"]
suppose if $ a$= $ \plus{} ve$ ,the graph is an upwards parabola then f(t)will also be positive .
And if $ a$=$ \minus{} ve$,the parabola would open downwards, then f(t) would be negative.
[/hide]
Thank u.[/quote]

Thanks for your works. It's just the same as mine. :P

If $ a > 0$ and $ b^2 > 4ac$, then $ t < \alpha$, $ \beta < t$.

If $ a >0$ and $ b^2 \equal{} 4ac$, then $ t\neq \alpha$.

If $ a < 0$ and $ b^2 > 4ac$, then $ \alpha < t < \beta$.

If $ a < 0$ and $ b^2 \equal{} 4ac$, then No solution.

[quote="kunny"]If $ a > 0$ and $ b^2 > 4ac$, then $ t < \alpha$, $ \beta < t$.

If $ a >$ and $ b^2 \equal{} 4ac$, then $ t\neq \alpha$.

If $ a < 0$ and $ b^2 > 4ac$, then $ \alpha < t < \beta$.

If $ a < 0$ and $ b^2 \equal{} 4ac$, then No solution.[/quote]

If $ \boxed{a < 0}$ and $ \boxed{b^2 > 4ac}$, then $ \boxed{\alpha < t < \beta}$.

are you sure?

[quote="kunny"]
If $ a < 0$ and $ b^2 > 4ac$, then $ \alpha < t < \beta$.

.[/quote]
Hello
If $ a<0$  then parabola would open downwards and and any t with $ \alpha < t < \beta$ will give $ af(t)<0$.
Kindly check it.Drawing a graph would be helpful.
Thank u.

[hide="this may be another solution"]

$ af(t)>0 \implies a(at^2\plus{}bt\plus{}c)>0 \implies a^2(t^2\plus{}\frac{b}{a}t\plus{}\frac{c}{a})>0$

$ \implies t^2\plus{}\frac{b}{a}t\plus{}\frac{c}{a}>0$

$ \implies t^2\minus{}(\alpha\plus{}\beta)t\plus{}\alpha\beta>0$
$ \implies (t\minus{}\alpha)(t\minus{}\beta)>0$

If $ \beta > \alpha$
$ \implies t>\beta$ or $ t<\alpha$

If $ \beta \equal{}\alpha$
then $ (t\minus{}\alpha)^2>0$
$ \implies t\in \mathbb{R}\ / \left\{\alpha\right\}$

[/hide]

 :roll:


[quote="Adriana N."]Who think that school is not as interesting as a conquest is?school is boring :lol:  :rotfl:[/quote]
Why did you post such question in "Geometry" topic?
You should post it in "Round table" :)
[color=green]
[[i]Moderator edit[/i]: For a topic as content-free as this one, G&FF][/color]

[color=darkblue][b]Adriana[/b], you wandered here ! Generally, the many things are boring.

In the my opinion, [b]the most important thing is if you like specially something or someone[/b]. [/color]

School sucks. BTW, Erken and koram, like the avatars.

One disadvantage to homeschooling with 2 brothers...

I agree... going to school if you like someone can be funn
I'm homeschooled now... so all I have to look forward to is... studying?

This camp for highly gifted kids in Michigan is the highlight of my year.

for me, it's CTY..
the other ones are all too much geared towards work...
CTY is 60% fun. 40% work....
it's summer.. :D

[quote="7h3.D3m0n.117"][quote="Virgil Nicula"][color=darkblue]In the my opinion, [b]the most important thing is if you like specially something or someone[/b]. [/color][/quote]

Hehe agreed.[/quote]

Exactly  :wink: . Like the Olympiad programmes and...... well let's keep that other part private.

[quote="Adriana N."]Who think that school is not as interesting as a conquest is?school is boring :lol:  :rotfl:[/quote]

conquest of poland is very interesting


what exactly does the chinese remainder theorem state. i looked it up and all i got were these complicated formulas  :mad:

I too am in the same confusion as you, I just cannot seem to find a problem which illustrates CRT.

Like this¿:

Given integers $a_{1}, a_{2}, ..., a_{n}$ and pairwise coprime moduli $m_{1}, m_{2}, ..., m_{n}$, there is an integer $a$ such that $a \equiv a_{i}\mod m_{i}$ for all $i$.
More exactly, the integers $b$ with $b \equiv a_{i}\mod m_{i}$ for all $i$ are exactly those $\equiv a \mod (m_{1}\cdot m_{2}\cdot ... \cdot m_{n})$.


floor function
logarithms
modular arithmetic
number theory

Ten contestants numbered 1-10 stand in order around a circle. Beginning to count at 1, every 5th contestant will be eliminated until only one remains. What is the number of the last remaining contestant?

I think I've made a mistake...I counted wrong...it's actually 4...

As we can see, counting on your fingers, however delightful :P , is rather error-prone and doesn't demonstrate highly developed problem skills. Anyone have a better way?

I think you can use some sort of modular arithmetic algorithm.  I don't know about this.

And also, could you post your solution, because I keep getting three.  :huh:

Each time I try it, I keep getting either 3 or 4. This is very confusing and [i]not[/i] a good method. I was thinking about modular arithmetic, but I'm not sure exactly how to do it. Anyone have ideas?

Doesn't "every fifth contestant" mean that we skip four people every time? So I consistently get $ \boxed9$ just by listing it out and crossing out. I think this is the correct interpretation because "every other person" is equivalent to "every second person," in which we skip one person every time.

Edit: OK, if we let people be numbered $ 1 \minus{} n$, we start at $ 1$, and eliminate every $ k^{th}$ person, then if $ f(n,k)$ denotes the last person in this situation, I'm pretty sure that $ f(n,k) \equal{} f(n \minus{} 1,k) \plus{} k$ if $ f(n \minus{} 1,k) \plus{} k\le n$, and otherwise, $ f(n,k) \equal{} f(n \minus{} 1,k) \plus{} k \minus{} n \plus{} 1$ (I haven't proven this yet). I also know that $ f(n,2) \equal{} n$ for all $ n \equal{} 2^k$ and $ f(n,2) \equal{} 2(n \minus{} 2^{\lfloor{\log_2n\rfloor}})$ for all other $ n$...haven't found a general formula for other $ k$, though.

If it will help anyone, the answer is 3.

it's $ \fbox{3}$. I have totally verified. Just don't know how to do it. I think if you write it out with 2 less each time it might help, like 10, 8, 6, 4, 2, then find some pattern to eliminate.

Oh, sorry, I misunderstood. I thought they started at $ 1$, but they started at $ 5$ (bad wording in the question). Then it is indeed $ \boxed3$--use izzy's method. Another check: I started at $ 1$ and ended at $ 9$, so if I had started at $ 5$, I would've ended at $ 13\equiv\boxed3\pmod{10}$.

But I still don't get it: "beginning to count at 1" definitely has two different meanings.

[quote="isabella2296"]You can use your fingers to simulate this. :)

x  x  x x x x x x x x

The 5th person is eliminated.

x x x x _ x x x x x

The 10th person is eliminated.

x x x x _ x x x x _

The 6th person is eliminated.

x x x x _ _ x x x _

The 2nd person is eliminated.

x _ x x _ _ x x x _

The 9th person is eliminated.

x _ x x _ _ x x _ _

The 8th person is eliminated.

x _ x x _ _ x _ _ _

The 1st person is eliminated.

_ _ x  x _ _ x _ _ _

The 4th person is eliminated

_ _ x _ _ _ x _ _ _

The 7th person is eliminated.

Thus our answer is $ \boxed{3}$.[/quote]

this works.

Yes, i got 3. Is there are faster way , other than listing out the numbers, and canceling them out 1 by 1?

It might be possible with recursion, but I think listing them out is the fastest way


inequalities
geometry
perimeter
conics
ellipse
analytic geometry

Prove that for all triangle with equal perimeter , the largest area has the equilateral triangle .

$ S\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\le \dfrac{(s\plus{}s\minus{}a\plus{}s\minus{}b\plus{}s\minus{}c)^2}{16}$, by AM-GM. Equality is when it's the equilateral triangle, and as RHS is constant, we have came to the conclusion.

I proved it by compare $ S_{1} \equal{} \sqrt {p(p \minus{} n)(p \minus{} n)(p \minus{} n)}$ and $ S_{2} \equal{} \sqrt {p(p \minus{} a)(p \minus{} b)(p \minus{} c)}$

By using AM-GM i proved that $ S_{1} > S_{2}$ where $ p \equal{} \frac {a \plus{} b \plus{} c}{2}$ (a>b>c) and $ p \equal{} \frac {3n}{2}$

Is this right ?

Alternatively, imagine that there is an ellipse with center $ (0,0)$, a major axis formed by the coordinates $ (\pm2x,0)$, a minor axis formed by the coordinates $ (0,\pm x\sqrt3)$, and thus foci at $ (\pm x,0)$. Let the foci be points $ A$ and $ B$ and another point on the ellipse be point $ C$. By the definition of the ellipse, $ CA\plus{}CB\equal{}4x$ is constant, so $ AB\plus{}CA\plus{}CB\equal{}2x\plus{}4x\equal{}6x$, the perimeter of the triangle, is constant as well. Now, it is easy to see that the maximum area is attained when $ C\equal{}(0,\pm x\sqrt3)$ which results in equilateral $ \triangle ABC$.

I think this is a possible sol:
[hide="THEOREM USED"]
The product of $ n$ positive numbers with a given sum is greatest when they are all equal.(TRY TO PROVE IT BEFORE I POST THE PROOF)
In other words, if $ a_i > 0$, $ \sum_{i \equal{} 1}^{n} a_i$ is fixed at some point, say nA. Then we have
$ a_1 a_2 a_3\cdots a_n \leq A^n$
With equality iff all the $ a_i$'s are equal.[/hide]
[hide="SOLUTION"]
I am using some notations:
$ T \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$, $ 2s \equal{}$ perimeter of triangle ABC.
Now, $ 2s \equal{} a \plus{} b \plus{} c$ is fixed.
From the theorem, $ 16T^2$ is greatest when $ (2s \minus{} 2a)(2s \minus{} 2b)(2s \minus{} 2c)$ is greatest,
Which happens iff $ 2s \minus{} 2a \equal{} 2s \minus{} 2b \equal{} 2s \minus{} 2c$ $ \Longrightarrow a \equal{} b \equal{} c$.
Hence proved      :lol:  :)
[/hide]

Your theorem is corollary of AM-GM(post #2)

@Potla: This isn't a different theorem at all but just the equality case of the AM-GM inequality :D .........that's all

[quote="math154"]Alternatively, imagine that there is an ellipse with center $ (0,0)$, a major axis formed by the coordinates $ (\pm2x,0)$, a minor axis formed by the coordinates $ (0,\pm x\sqrt3)$, and thus foci at $ (\pm x,0)$. Let the foci be points $ A$ and $ B$ and another point on the ellipse be point $ C$. By the definition of the ellipse, $ CA \plus{} CB \equal{} 4x$ is constant, so $ AB \plus{} CA \plus{} CB \equal{} 2x \plus{} 4x \equal{} 6x$, the perimeter of the triangle, is constant as well. Now, it is easy to see that the maximum area is attained when $ C \equal{} (0,\pm x\sqrt3)$ which results in equilateral $ \triangle ABC$.[/quote]

You assumed that AB=2x and CA+CB=4x...so you assumed that 2AB=BC+CA which isn't true for a general triangle...

Oops, just realized that. Sorry. :blush:

of course the theorem used in my post #6 follows directly from the AM GM!!! :D
Thanks...


geometry
geometric transformation
dilation

What makes a bond strong or weak?  How would I know that C-Cl is a stronger bond than C-H?

Thanks in advance!

I am not very good at chemistry, but I think the main force involved here is electronegativity, which is basically a measure of how much an atom attracts a proton, and I think that the greater the electronegativity difference between two atoms, the stronger the bond between them. you can probably find a chart of electronegativites for all the elements on the internet, and find that C and Cl differ more than C and H in electronegativity

@ hurdler - Thanks!  :)

I'm going to turn this into a thread of questions that I can't do in preparation for the Local Section of the USNCO.  Some of these questions are on the easy side since I've forgotten much of the stuff I learned in the beginning of the year.  :wink:

What are the elements called for which the 4f subshell is being filled?
(A) transition metals
(B) metalloids
(C) lanthanides
(D) actinides

Which ionic compound has the smallest lattice energy?
(A) NaF
(B) MgO
(C) AlN
(D) MgCl$ _2$

Please include an explanation for each of the questions above.  Thanks!

first one
[hide]lanthanides, it's just something you have to know[/hide]

second
[hide]AlN I'm pretty sure.  They are closest together on the periodic table and have the weakest ionic bond.[/hide]

[quote="kysuke"]second
[hide]AlN I'm pretty sure.  They are closest together on the periodic table and have the weakest ionic bond.[/hide][/quote]

Um, the answer key says that the answer is NaF.  Does anyone know why?

--

Another Question:

During the electrolysis of a dilute solution of sulfuric acid, what substance is produced at the anode?
(A) hydrogen
(B) hydrogen sulfide
(C) oxygen
(D) sulfur dioxide

All of the energy levels listed are allowed EXCEPT
(A) 3f
(B) 4d
(C) 5p
(D) 7s

Well maybe I have the opposite definition of ionic bond strengths.  Questions like those are the ones that kill me.

[hide]If SO4 is present then H2O will oxidize instead so Oxygen gas is created.  I hope this is right.[/hide]
[hide]3f[/hide]

what is an anode?

for the second question, I think 3f does not exist since the smallest for s is 1, for p is 2, so d is 3 and for f is 4

regarding the NaF - My guess is that it's weakest since  all others have charges greater than (less than) +1 (-1) when split into ions i.e. $ F^\minus{}$ is only sharing one electron with Na, while the other molecules have cations that are sharing more than one electron with the anion

then again, i am not good at chemistry....

Suppose we have an ionic compound $ M_nK_p$. The lattice energy of this compound is the energy liberated when n moles of M cations and p moles of K anions in the gas phase are "transformed" into one mole of solid $ M_nK_p$. What factors can affect the value of the lattice energy? Well, the more stable a crystal structure is more energy is released, so we may ask what factors affect the stability of solid ionic compounds. Electrostatic forces are at work here, so Coulomb's law,

$ |F| \equal{} \frac{1}{4 \pi \epsilon} \frac{q_1q_2}{d^2}$,

tells us that the greater the charge of the ions, more strongly they are bounded together. On the other hand

[quote="kysuke"]3f[/quote]

Why?

--

Another Question:

In which species is resonance most useful in explaining the observed bond lengths?
(A) NF$ _3$
(B) NH$ _4^\plus{}$
(C) NO$ _2^\plus{}$
(D) NO$ _2^\minus{}$

Thanks in advance!

[quote="kysuke"]Well a 3f orbital just doesn't exist.  For the resonance one just try to draw out all the diagrams and see which one needs resonance.[/quote]

How can I tell if an orbital exists or not?  Do I have to memorize them?

yes you have to memorize them, but it's really easy. look at my previous post

for the resonance, just draw the lewis diagrams

[hide="what i got"]
i think all but $ NO_2^\plus{}$ have only single bonds[/hide]

[quote="hurdler"]i think all but $ NO_2^ \plus{}$ have only single bonds[/quote]

That's what I got too.  Thanks!

--

What property(ies) of a liquid increases when the temperature is raised?
[list]I. vapor pressure
II. surface tension[/list]
(A) I only
(B) II only
(C) Both I and II
(D) Neither I nor II

Calculate the change in enthalpy, $ \Delta$H, for the combustion of 11.2 L of hydrogen gas, measured at 0 degrees C and 1 atm pressure, to form H2O (g).  Potentially useful information: $ \Delta$H_f of H2O(g) = -241.8 kJ/mol.

Why dont you just think about the problems rather than posting them online. Theyre actually really simple one step problems.

Like if you increase vapor pressure, then you're increasing the avg KE of the particles. And thus ur giving energy for substances to jump out of solution and go to vapor. Thus vapor pressure increases. You can use similar logic for the rest.

The combustion of hydrogen IS the heat of formation of water... so the questions just asking
"If you've got half a mole of hydrogen and you react it with oxygen, how much heat is given off if 241.8kj are given off per mol H2O formed"

[quote="Carcul"][quote="Hurdler"]yes you have to memorize them,[/quote]

No, you don't have to memorize anything. Is that really so hard to understand the caracterization of orbitals through quantum numbers?[/quote]

I guess you are right. What I meant is that you need to know what the 4 letters s,p,d,f represent (like which is which), since you normally cannot fully understand notation (even if they do represent the first letters of some words). The idea is pretty simple though.


(x^2)/(a^2) - (y^2)/(b^2)=1  (x,y are not 0 )

I think there is no integer pair of (x,y) when a , b are integers.

I got a proof , but I am not sure , So I ask for you guys opinion.


function
limit
calculus
derivative
calculus computations

Let $e_{n}$ be the maximum value of the function $(1-x)^{n}(px^{2}+qx+r)\ (n=1,\ 2,\ \cdots)$ at $0\leq x\leq 1$ for the constant numbers $p,\ q,\ r\geq 0.$
Find $\lim_{n\to\infty}e_{n}.$

[hide="Solution"]
${f(x)=(1-x)^{n}\left(p x^{2}+q x+r\right)}$

For the possible maximum we need to test

${x=0\land x=1}$

And Solve

${\frac{\partial f(x)}{\partial x}=0}$

For the first. Its easy to see that

${n\neq 0}$
${f(0)=r}$
${f(1)=0}$

Solving

${\frac{\partial f(x)}{\partial x}=0}$
${(1-x)^{n}(q+2 p x)-n (1-x)^{n-1}\left(p x^{2}+q x+r\right)=0}$
${(1-x)^{n}\left(q+2 p x+\frac{n (r+x (q+p x))}{x-1}\right)=0}$

For the last equation

${\left((1-x)^{n}=0\right)\lor \left(q+2 p x+\frac{n (r+x (q+p x))}{x-1}=0\right)}$

Solving the first

${(1-x)^{n}=0}$
${1-x=0}$
${x=1}$

This solution is not good, since

${f(1)=0}$

And the second equation is undefinied in $x = 1$

Lets solve the next. I did not put the steps of simplification.

${x\neq 1}$
${q+2 p x+\frac{n (r+x (q+p x))}{x-1}=0}$
${(n p+2 p) x^{2}+(-2 p+n q+q) x+(n r-q)=0}$

The solution

${x=\frac{-(-2 p+n q+q)\pm \sqrt{(-2 p+n q+q)^{2}-4 (n p+2 p) (n r-q)}}{2 (n p+2 p)}}$

${x=\frac{2 p-(n+1) q\pm \sqrt{(-2 p+n q+q)^{2}-4 (n+2) p (n r-q)}}{2 (n+2) p}}$

Since we have

${\lim_{n\to \infty }\, f\left(e_{n}\right)=f\left(\lim_{n\to \infty }\, e_{n}\right)}$

If this is not true, the limit does not converge. I think so.... [i]Someone help[/i]

The possible values of x for the maximum is

${x=0}$

${x=\frac{2 p-(n+1) q\pm \sqrt{(-2 p+n q+q)^{2}-4 (n+2) p (n r-q)}}{2 (n+2) p}}$

Lets Calculate

[hide="Tips Used"]
${\lim_{n\to \infty }\, \frac{a n+\left(b\pm \sqrt{\alpha n^{2}+\beta n+\gamma }\right)}{\theta+n \phi }=\frac{a\pm \sqrt{\alpha }}{\phi }}$[/hide]

${\lim_{n\to \infty }\, \frac{2 p-\left((n+1) q+\sqrt{(-2 p+n q+q)^{2}-4 (n+2) p (n r-q)}\right)}{2 (n+2) p}=-\frac{q^{2}+\sqrt{q^{2}-4 p r}q-4 p r}{2 p \sqrt{q^{2}-4 p r}}}$
${\lim_{n\to \infty }\, \frac{2 p-\left((n+1) q-\sqrt{(-2 p+n q+q)^{2}-4 (n+2) p (n r-q)}\right)}{2 (n+2) p}=\frac{q^{2}-\sqrt{q^{2}-4 p r}q-4 p r}{2 p \sqrt{q^{2}-4 p r}}}$

${f\left(-\frac{q^{2}+\sqrt{q^{2}-4 p r}q-4 p r}{2 p \sqrt{q^{2}-4 p r}}\right)=0}$
${f\left(\frac{q^{2}-\sqrt{q^{2}-4 p r}q-4 p r}{2 p \sqrt{q^{2}-4 p r}}\right)=0}$

So, the maximum is in $x=0$

${e_{n}=r}$

${\lim_{n\to \infty }\, f(0)=\lim_{n\to \infty }\, e_{n}=r}$

Analising the plot, this is the right solution.

I think there is a easy solution, just analising the derivative or someting else...

:([/hide]

The simpler solution comes from recognizing that the specific structure of $px^{2}+qx+r$ doesn't matter. Let's phrase it this way:

Suppose $f$ is continuous on $[0,1].$ Let $e_{n}=\max_{x\in[0,1]}(1-x)^{n}f(x).$ Show that $\lim_{n\to\infty}e_{n}=f(0).$


Hint:
[hide]Watch how you get simplify with the parentheses.[/hide]

More specifically:
[hide="something o_O"]
a negative times another negative is a positive
[/hide]

KBabe:

The statement I wrote is false

there is a mistake in one of those steps

That's what we are looking for...

does this explain your[quote]???.....??? [/quote]?

i know what u r looking for..but i don't know what THEY were trying to point out.

[quote]4d + 9d - ( 7 - 6d ) = 4d + 9d - 7 - 6d
[/quote]

look at this statement very carefully.. you are supposed to change the negative sign to the positive

so, the correct answer is supposed to be

4d + 9d -7 + 6d

[quote]KBabe: do y'all learn how to add or subtract a negative number in 6th grade? I don't think so, I don't remember learning it last year... but um, I'll explain it to you sometimes when I see you. I can give you the packet we got in summer math team practice about negative numbers.[/quote]

That's really nice of you to do it for KBabe

[quote="KBabe"]yea. we're bestfriends.  :)  isn't she nice?? ( o wait till u see her ugly side..! it's horrible! heehee i'm j/k)[/quote]

excuse me??? just kidding!!!, Kristy's pretty cool

[quote="shinwoo"]Okay, This is not a place to chat....

so, please try to understand that this is a sacred(?) place[/quote].

understood, so sry sir.

[quote="shinwoo"]Okay, This is not a place to chat....

so, please try to understand that this is a sacred(?) place[/quote]

Lol, that made me laugh :P .

It's OK to chat, just don't get toooo off topic...

[quote]Lol, that made me laugh  .

It's OK to chat, just don't get toooo off topic...[/quote]

Yeah...That's my point..

jli:  Your "correction" is incorrect.  Take another look at it.

whoops :blush: .  it's fixed now...i think

[quote="shinwoo"]4d + 9d - ( 7 - 6d ) = 4d + 9d - 7 - 6d

                             = 4d + 9d - 6d - 7

                             = 7d - 7

there is a mistake in this problem

good mathmaticians always show work . It's very useful..

When you make a mistake, you can go over and see in what step you made the mistake..[/quote]

never actually did the problem, i would just show my step, and then go back and see what's different:

[hide]4d+9d-7+6d
13d-7+6d
19d-7[/hide]

so they accidently forgot to change the negative sign in front of the -6d into a positive sign because there's a negative in front of the parenthesis.... sounds like something i did on my pre-algebra homework  :blush:

[quote="shinwoo"]4d + 9d - ( 7 - 6d ) = 4d + 9d - 7 - 6d

                             = 4d + 9d - 6d - 7

                             = 7d - 7

there is a mistake in this problem

good mathmaticians always show work . It's very useful..

When you make a mistake, you can go over and see in what step you made the mistake..[/quote]

i think that u have to close this top because everybody answered right and other saw it..... :lambda


linear algebra
matrix
ratio
algebra
polynomial
invariant
complex numbers

$A$ and $B$ anticommute if $AB=-BA$.
What is the maximal number of matrices $A_1,..,A_k \in M_n(C)$ such that they pairwise anticommute ?

Alekk, are you sure this is the problem they gave in oral examination? Weren't the matrices $ 2x2$ or something like this? As far as I know, the general problem is extremely difficult.

:blush: . You're right. It is the $2x2$ case.

I don't understand anything anymore. Let us take some arbitrary $n$, some complex numbers $x_1,...,x_n$ different and put the matrix $A_i$ to have $a_{11}=x_i, a_{12}=x_i, a_{21}=-x_i, a_{22}=-x_i$. Either I am completely dumb, or these matrices are different and anticommute since the product of any two of them is 0. Is there some other condition to be fulfilled? Since not we can find even non numerable families of anticummuting matrices. It suffices to take non trivial matrix $A$ such that $A^2=0$ and to consider the matrices $ xA$ with complex x.

sorry again. The matrices are invertible  :)

I thought so . Then it is clear that there are at most 4. Indeed, if there are at least 5, say $A_1,...,A_n$  then we can say that one of them, let's say $A_n$ is a linear combination of the other $n-1$, say $ x_1A_1+...+x_{n-1}A_{n-1}=A_n$. Multiplying first to the left and next to the right by $A_n$ we find $A_n^2=-A_n^2$ and thus $A_n$ is not invertible. I think this is the maximal number, but I'm not sure, I have to find optimal example. See you soon.

It is easy to find 3 anticommuting matrices. Take the matrices $ (0,1,1,0), (1,0,0,-1), (0,-1,1,0)$. I really don't know if there are 4.

Victory!!! If there are 4, we can see from previous argument that they are linearly independent, thus they generate all $M_2(C)$. But then there are $x_1,...,x_4$ such that $x_1A_1+...+x_4A_4=I$. Multiply at left and right by $A_1$ and add the relations. You will find $ x_1 A_1^2=A_1$ and thus $A_1$ is scalar multiple of $I$.THe same works for all, which means they aren't a base, false.

I asked a teacher and he told me that this problem was given in oral examination at ENS Ulm in 2004 in the general case and that the answer is $1+2\cdot v_2(n)$ where $v_2(x)$ is the dyadic valuation of $x$. Yet, he told me that the solution is really monstruous. Did anyone make any progress on this beautiful problem?

I don't know if you are still working on this, but for the benefit of ML users it should be mentioned that this is AMM problem 10456.

If anyone desires, I can even put the proof here, it's not really that monstrous (but very un-obvious, so to speak).

perfect_ratio,
Could you put the proof here, as you suggested?
Thanks

Here is a solution I found some months ago. In my opinion, it's difficult to make simpler. Take $ A_{1},...A_{r}$ an aticommuting family and observe that $ A_{1}$ is similar to its opposite, thus its caracteristic polynomial can be written in the form $ f(x)f(-x)$ for a polynomial $ f$. The anticommutation property ensures that each of $ A_{2},...,A_{r}$ sends $ Ker f(A_{1})$ into $ Ker f(-A_{1})$ and the later into the first. Therefore if we look at $ A_{2}^{-1}A_{j}$ for $ j>2$, these form an anticommuting family sending $ Ker f(A_{1})$ into itself. Since this one has dimension $ n/2$, we can finish the inductive step. The construction of a maximal set is very easy, the difficult part was the majoration.

Here's the follow-up:

Suppose that $ n = 2^{m}$ and $ f_{1}, \ldots, f_{2m}$ anticommute pairwise.
Prove that the products of the form $ f_{i_{1}}f_{i_{2}}\ldots f_{i_{s}}$, where $ 1 \leq i_{1}\leq \ldots \leq i_{s}\leq 2m$, form a basis for $ \mathcal M_{n}\left( \mathbb C \right)$.
Moreover, in this case we have that the square of each $ f_{i}$ is a scalar matrix.

Hi perfect_radio.
I don't understand your post:
1) If $ n=2^{m}$ then you can find $ 1+2m>2m$ anticommuting matrices.
2) The number of products $ f_{i_{1}}{\cdots}f_{i_{s}}$ is $ 2^{2m}-1=n^{2}-1<n^{2}$.

Hello, loup blanc.

[quote="loup blanc"]1) If $ n = 2^{m}$ then you can find $ 1+2m > 2m$ anticommuting matrices.[/quote]
That's right. Harazi proved that we can find $ 2m+1$ anticommuting matrices, but we need only $ 2m$ for the "follow-up".

[quote="loup blanc"]2) The number of products $ f_{i_{1}}\ldots f_{i_{s}}$ is $ 2^{2m}-1 = n^{2}-1 < n^{2}$.[/quote]
I think that we should also consider the "empty product" ($ I_{n}$).

2') The other matrices are of the form $ f_{i_{1}}f_{i_{2}}\ldots f_{i_{s}}$, where $ 1 \leq i_{1}< \ldots < i_{s}\leq 2m$ (notice the $ <$'s instead of $ \leq$'s).

3) The matrices in question are of course invertible.

First, let's show that the square of each $ f_{i}$ is scalar. Let $ \lambda$ be an eigenvalue of $ f_{i}$. Each $ f_{j},\ j\ne i$ acts on the direct sum $ V$ of the $ \lambda$ and $ -\lambda$-eigenspaces of $ f_{i}$, mapping each of those eigenspaces isomorphically onto the other one. Of course, $ V$ is invariant to $ f_{i}$ as well. If $ f_{j},f_{k},\ j\ne k$ were to agree on $ V$, then both $ f_{j}^{-1}f_{k}$ and $ f_{k}f_{j}^{-1}$ would equal $ 1$ on $ V$, contradicting anticommutation. Hence we can restrict our attention to $ V$, assuming all our operators act there. Because of the upper bound previously proved by harazi, we see that $ V$ cannot be proper, and hence it is the whole $ \mathbb C^{n}$. But this means that the square of $ f_{i}$ is $ \lambda^{2}$, and we're done.

Consider now an arbitrary product $ f=f_{i_{1}}\cdot\ldots\cdot f_{i_{s}}$ of $ s$ distinct $ f_{i}$'s (for some $ 1\le s$, i.e. not the emty product). If we show that the trace of $ f$ is zero, we can finish the proof by noticing that the product of any two different products (includding the case when one of the products is empty, i.e. the identity operator) has trace zero, while the trace of the square of any product is non-zero. It follows that those $ 2^{2m}=n^{2}$ products are orthogonal to one another with respect to the bilinear form $ (a,b)\mapsto\mbox{tr}~ab$, so they must be linearly independent and hence a basis for $ \mathcal M_{n}(\mathbb C)$.

It remains to show that $ \mbox{tr}~f=0$ (same notations as in the previous paragraph).

Case 1 - $ s$ is even

Let $ g=f_{i_{2}}\cdot\ldots\cdot f_{i_{s}}\cdot f_{i_{1}}$. On the one hand, we have $ \mbox{tr}~f=\mbox{tr}~g$. On the other hand, we reach $ g$ by starting with $ f$ and commuting $ f_{i_{1}}$ with $ f_{i_{2}},\ldots,f_{i_{s}}$. Each time we perform such a step, the sign changes, and there's an odd number of steps - $ s-1$. Hence $ \mbox{tr}~f=-\mbox{tr}~g$, and we get the desired conclusion that $ \mbox{tr}~f=0$.

Case 2 - $ s$ is odd

Just notice that in this case $ f$ anticommutes with some $ f_{i}$, so $ f$ is similar to $ -f$.


Source: AHSME 1973 #18

If $p \geq 5$ is a prime number, then 24 divides $p^2 - 1$ without remainder
$\text{(A)} \ \mbox{never} \qquad \text{(B)} \ \mbox{sometimes only} \qquad \text{(C)} \ \mbox{always} \qquad \text{(D)} \ \mbox{only if p = 5} \qquad \text{(E)} \ \mbox{none of these}$

[hide="Solution"]
(C)
Factor this
$(p+1)(p-1)$
Each of these is divisible by 2 and one is divisble by 4 so we know that $2^3$ is in the prime factorization.
We also know that $p$ is not divisible by 3 so one of these must be divisible by 3 and, thus, 24 always divides the product.
[/hide]

[hide="solution"]
For every prime $p\ge 5$ exists $k\in\mathbb{N}$ such that $p=6k-1$ or $p=6k+1$.
[hide="explanation"]
Dividing by $6$ can leave reminders $0,1,2,3,4,5$.
Let $p=6q+r$, where $r$ is reminder and $p\ge 5$.

$r=0\implies p=6q\implies 6|p\implies p\text{ is not prime}$
$r=1\implies p=6q+1\implies p\text{ could be prime}$
$r=2\implies p=6q+2\implies 2|p\implies p\text{ is not prime}$
$r=3\implies p=6q+3\implies 3|p\implies p\text{ is not prime}$
$r=4\implies p=6q+4\implies 2|p\implies p\text{ is not prime}$
$r=5\implies p=6q+5=6(q+1)-1\implies p\text{ could be prime}$
[/hide]
If we plug $p=6k\pm 1$ in $p^2-1=(p-1)(p+1)$ we will get $p^2-1=6k(6k\pm 2)=12k(3k\pm 1)$.

$k(3k\pm 1)=k(2k +k\pm 1)=2k^2+k(k\pm 1)$
Observing that $k(k\pm 1)$ is even (because it's product of successive integers) we conclude that $k(3k\pm 1)$ is even too.

Hence, for prime $p\ge 5$, $24|p^2-1$ is always true.
[/hide]


floor function
geometry
induction
combinatorial geometry
combinatorics proposed
combinatorics

Inside a convex $n$-gon we have $2\le k\le\left\lfloor\frac n2\right\rfloor$ points. Prove that we can choose $2k$ vertices of the $n$-gon s.t. the polygon formed by them contains the $k$ points.

P.S.

I consider this combinatorial geometry, and that's why I posted it here.

We use induction on $k$.
For $k=2$, let $A_1,A_2$ be the two interior points. The line $(A_1A_2)$ meets the $n$-gon in two of its sides, say $[XY]$ and $[UV]$. If they are four pairwise distinct points, just consider the convex quadrilateral formed by these four points. Otherwise, wlog, we have $Y=U$ so that the segment $[A_1A_2]$ is contained in the triangle $XYV$ and we may choose any fourth vertex to form a convex quadrilateral.

Now, assume it holds for $k-1\geq 2$, and let's consider a set of $k$ interior points. If their convex hull is a $p$-gon with $p<k$ we are done by the induction hypothesis. Thus, we may assume that they are the vertices of a convex $k$-gon, say $A_1A_2 \cdots A_k$ in that order.
Use the induction hypothesis on the set of all the $A_i$'s except $A_2$ to consider a convex $(2k-2)$-gon which covers these $k-1$ points. If $A_2$ is covered too, we are done. Otherwise, let $M_iM_{i+1}$ be intersected by the segment $[A_1A_2]$ and $[M_jM_{j+1}]$ be intersected by $[A_2A_3]$. If these two sides of the $(2k-2)$-gon are distinct, then one of the vertices of the initial $n$-gon is an interior point of the $k$-gon which is contained in the $n$-gon, and this contradicts the convexity.
Thus the segments $[A_1A_2]$ and $[A_2A_3]$ meets the same side, say $[MP]$ of the $(2k-2)$-gon.
Since $A_2$ is an interior point of the $n$-gon, the set $V$ of vertices of the $n$-gon which lies on the same side of $(MP)$ than $A_2$ is non-empty. If $V$ contains only one point, say $Q$, then adding $Q$ gives a covering $(2k-1)$-gon and we are done. If it contains at least two points, then let $X_1, \cdots, X_r$ be these points, in that order on the $n$-gon. It is easy to see that the quadrilaterals with vertices $M,P,X_i,X_{i+1}$ cover the part of the $n$-gon which is on the same side of $(MP)$ than $A_2$. In particular, we may choose $i$ such that the associated quadrilateral covers $A_2$, and we are done.

Pierre.

Thanks.

I also used induction. The initial step, corresponding to $k=2$ is obvious, and for the induction step I used Caratheodory's Theorem in the stronger form: if a point lies in the convex hull of sveral points in $\mathbb R^n$, then it lies in the convex hull of $n+1$ of them, and one of these $n+1$ points can be chosen arbitrarily. We apply this to our situation to conclude that whenever we add a new point to the $k-1$ we already have, we can choose one of the $3$ vertices of the polygon the convex hull of which is supposed to contain $P$ among the $2(k-1)$ we already have, and we thus add no more than $2$ points to these $2(k-1)$.

I suspect this proof is flawed because it seems different from the other proofs and the official solution, so if someone could check it that would be very helpful.

We use induction on $n$. For $n=2$ it is obviously true.
Now suppose it is true for every convex $(2n-2)$-gon. Consider some convex $2n$-gon and label the vertices $A_1, A_2,..., A_{2n}$.

Lemma: Every group of $k$ vertices for $2\le k\le n-1$ lies inside a convex $(2n-2)$-gon made up of the vertices of the $2n$-gon.
Proof: Suppose there is some group of $k$ not satisfying this condition. Consider the $n$ disjoint triangles $A_1A_2A_3, A_3A_4A_5,...,A_{2n-1}A_{2n}A_1$. If there are two or more triangles that contain no points, so $A_iA_{i+1}A_{i+2}$ and $A_jA_{j+1}A_{j+2}$ with $j>i$. Then we take the convex $(2n-2)$-gon $A_1...A_iA_{i+2}...A_jA_{j+2}...A_{2n}$ which contains all $k$ points. So we must have at most one triangle containing no points. Which is of course only possible if $k=n-1$. Now the same applies if we consider the $n$ disjoint triangles $A_2A_3A_4,...,A_{2n}A_1A_2$. So one triangle from each of these sets must contain no points. If the triangles are disjoint, we can use the same logic above with $A_iA_{i+1}A_{i+2}$ and $A_jA_{j+1}A_{j+2}$. So the triangles have to overlap. WLOG let the empty triangles be $A_1A_2A_3$ and $A_2A_3A_4$. Let $B$ be the intersection of $A_1A_3$ and $A_2A_4$. If triangle $A_1BA_4$ is empty, we use the $(2n-2)$-gon $A_1A_4...A_{2n}$. So it must contain one point $C$. Now note that for all the remaining triangles to contain exactly one point, we must have each point lying in two triangles. However, the only triangles that $C$ can be in are $A_3A_4A_5$ and $A_{2n}A_1A_2$, but these are disjoint, so $C$ lies in at most one triangle, contradiction.

Using the result above, if $2\le k\le n-1$ then the vertices are part of a convex $(2n-2)$-gon and we apply the induction hypothesis. If $k=n$ the result is obviously true. So the induction is complete.

Also I realized after writing this proof that I only proved it for a convex $2n$-gon. But that's because the original ARMO problem asked for a $100$-gon.

Any feedback would be appreciated.

A nice problem :)

[hide="Solution"] Fix $n$ and induct on $k$. The base case $k=2$ is trivial. Now clearly it suffices to consider the case when all $k$ points are the vertices of their convex hull, or else we are done by the inductive hypothesis. Now take some arbitrary $k-1$ of the points. Consider the $2(k-1)$-gon $A_1,A_2,\dots, A_{2k-2}$ that covers these $k-1$ points. If the $k$th point lies inside this, then we are done; if not, then it lies in the region outside the $2(k-1)$-gon and between $A_j$ and $A_{j+1}$ for some $1 \le j<2k-2$. Then consider the vertices of the $n$ gon that lie in between $A_j$ and $A_{j+1}$; call these $B_1,B_2, \dots, B_i$. If the $k$th point lies in one of the triangles $A_jA_{j+1}B_h$ with $1 \le h \le i$ then done; if not then it lies in the region between triangles $A_jA_{j+1}B_h$ and $A_jA_{j+1}B_{h+1}$ so take the new polygon $A_1,A_2, \dots, A_{2k-2}, B_h,B_{h+1}$ which has $2k$ vertices, satisfying the conditions, so we are done. $\Box$

[/hide]


What's the best way to prepare if going for a 5?
Also, where are past exams available?

Thanks,
-diophantient

To be honest, the best way to prepare in my experience was to figure out how to read the quick reference provided. The AP test focuses quite a bit on java's built in classes and such, so being able to quickly look through the API is a useful skill which saves you memorization.


Let p(n) denote the number of partitions of n. Prove that p(n)<1/2(p(n-1)+p(n+1)) and find a relation for P(n) in terms of P(n-1) and p(n+1).

-BOmb

what type of partitions?  where order matters or where it doesn't matter?  like, is 4 = 1+1+2 the same as 4 = 1+2+1?

i suppose that order does not matter.
clearly P(1)=1, P(2)=2, p(3)=3 p(4)=5,...
Now let Q(n) be the number of partitions whose elements are bigger than 1.
so Q(1)=0, Q(2)=1, Q(3)=1, Q(4)=2,...
I shall first prove that {Q(n)} is an increasing sequence.
note that if n=a1+...+ak, with a1<=a2<=...<=ak and where all the terms are considered to be bigger than 1, then the new sequence
(a1,...,ak +1) is a partition for n+1 whose terms are again bigger than 1. so Q(n+1)>=Q(n)
Now obviously p(n)=Q(n)+Q(n-1)+...+Q(2)
Now p(n) - [p(n-1)+p(n+1)]/2 = Q(n) - [Q(n+1)+Q(n)]/2 = [Q(n)-Q(n+1)]/2 <=0 proving the result.
the equality holds only for n=2,4.
-A


Prove, for positive integers $n$: \[\prod_{k=1}^{n}\left( \frac{2k-1}{2k}\right) < \frac{1}{\sqrt{3n}}\]
[hide="Hint"]
Is there a stronger inequality that is more easily proven?
(If you've found this proof, try one without using the stronger inequality.   :P )[/hide]

thats random...

do something like (3n+k)^{-0.5} and choose k so induction works


Suppose that $4^{n}+2^{n}+1$ is prime for some positive integer $n$. Show that $n$ must be a power of $3$.

I provided a generalisation of this in http://www.mathlinks.ro/Forum/viewtopic.php?t=5467 .


Princeton
college
geometry
geometric transformation
reflection
Harvard
MIT

This is some important topic. Because there is no complete definition about the genius, oppinios must be divided.
So what is the genius? Does he (or she - from now on I will only use he, but the [b]she[/b] option will go without saying)  exists?
Is he scared of the big bad world?
Is he destined to live a solitary life because nobody understands him (same as the he case - him will also stand for her) ?
Haw does he see the world?
What does he think about himself?
Are you a genius?
Do you know someone who is? Why do you think that?
Will world ever be ready for the genius? I mean will the world be lead by the highest intellectualy gifted persons? Will genius ever fit into a leading possition?
Any other thoughts are wellcome.

I think that you read too much Eminescu for the Romanian Literature class in high-school :D:D:D

Well I don't care about Eminescu's point of view. I'm interested in your's and mostly I'd like to develop mine.
Why do most of you find it so hard to be serious about my topics in this section.
I just want to have a nice intellectual chat with you people.

I was rather serious. I know that Eminescu wrote a lot of poems about the condition of the genious (see "Scrisoarea I", etc.) and he had about the same questions as you do here ... :)

Nietzsche and Schopenhauer did too. In fact Eminescu's view on the genius was inspired by theirs.
My remark was general. It did not refer only to your post.

did u see "A beautiful mind"... i love that movie...  :D

Yes I did. I liked it too. But John Nash is a special case of genius. His solitude pushed him into schizophrenia, but his beautiful mind (and his wife and child:D ) got him out of it.
Anyway if tha case is real, it was surely embellished in the moovie for commercial purposes.

There is another movie who's name I can't fully remember. It is about a chess genius. There is no happy end though  :( . I think it is called "The Luzhnj defence". I saw it on HBO, so it shouldn't be to hard to find.

Yes, a lot of A Beautiful Mind was embellishment. But it is not at all uncommon for "geniuses" to have psychological disorders. I think for many their disorders are not because they are geniuses, but they are geniuses [i]because of[/i] their disorders. Their mind's attempt to organize the world and overcome their disorder often sets up a situation where i becomes trained to work differently than the average mind.

Psychological disorder is in no way a prerequisite for genius though. Many geniuses are quite functional and well adjusted. So there are probably many paths and mechanisms for genius - not just one.

Do you know more about John Nash? I mean do you know what was embellished in the movie?

There was this tv programme on the Discovery Channel about John Nash, but I only saw parts of it... I don't even know if john Nash still lives! Can ne1 enlight me on that matter?

Yes he still lives and he's doing fine. He's had an honorary position in the Princeton Math Department for years now, though he doesn't teach. I was a student there at the time that he won the Nobel Prize in 94. The discrepencies in the film mostly have to do with his personal life and with what happened in his life leading up to the breakdown. Some of it is probably too speculative or personal to relate here on the internet. You may be able to find some books which recount it though.

Well, there is a biography with the same title ("A Beautiful Mind") also.  I would say it is much "fairer" in terms of portraying Nash as a person, and of course, his illness.  It does get rather disturbing at times, however, but then, that is the true nature of schizophrenia to begin with.

Have you read that? How could one obtain it? Here's something I found on the web:[url]

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Nash.html

[/url]

I heard an anecdote that one of his recommendations for graduate school to Princeton was simply one line: "This man is a genius. Accept him."

Another less known story of a modern mathematician suffering from mental illness is that of Alexander Grothendieck. He was perhaps the greatest mathematician of the 20th century.

Many people prefer David Hilbert as  the greatest mathematician of the 20th century; and he was a very normal man ...

Things like history and geography and so on are compulsory during high school. Fortunately, that ends when you go to university (or so I've heard; I'll see what it's like next year).

I am, of course, referring to universities like Matematica, Politehnica etc. You can study history and become a history teacher, for example, if that's your career choice.

integral0 your system of underrgraduate and graduate education is the best in the world, so be happy :) I will surely be admitted to an US university (doh... :) ) and you won't be facing the problems that the rest of us are facing ...

who knows if I get the right graduate position, I may even become your Algebra teacher :D:D

[quote="Valentin Vornicu"]who knows if I get the right graduate position, I may even become your Algebra teacher :D:D[/quote]

Some changes need to be made in your attitude towards maths to become integral0's college teacher in algebra.
I will only post one: You should start to learn algebra! :D

For now you are too attached to the IMO level maths. You should differentialte serious maths from childish games such as IMO. Remember IMO is just a creche.

[quote="amfulger"]I will only post one: You should start to learn algebra! :D

For now you are too attached to the IMO level maths. You should differentialte serious maths from childish games such as IMO. Remember IMO is just a creche.[/quote]
Indeed I have no clue of what algebra is really about. I guess you do, and all this time I didn't knew ...

And second of all IMO is not a childish game. I belive that most of the people on this forum are here BECAUSE of the IMO. You should not be angry as to IMO just because you never got there ...

You told me last year that the number of visitors of your site exploded when you posted the IMC problems. So perhaps some of the visitors are here because of IMC.
However I think IMC is childish too. Why is this? Because as well as IMO, [b]it only proposes problems that CAN BE SOLVED IN CONTEST TIME[/b].
Because of this, I think IMO and IMC problems are easy. They are not all easy to me, but they are not what you would call spectacular results. Those require months or years of hard work and research and are the real big boy's (or girl's) challange. IMO is just for fun or for CV. You can't compare a IMO participant (even if he has 4 gold medals) to a Fields Winner.

[quote="amfulger"]You told me last year that the number of visitors of your site exploded when you posted the IMC problems. So perhaps some of the visitors are here because of IMC.
[/quote]
you must not be remembering correctly. people didn't take much interest (except maybe moubi :D) in IMC problems. the site exploded when I was the first to put online the [b]IMO[/b] subjects! I know for a fact that the parents of Ana Caraiani found out that she was 5th in the world by accidentally browsing the forum (the IMO official site is really bad).

[quote="amfulger"]Those require months or years of hard work and research and are the real big boy's (or girl's) challange. IMO is just for fun or for CV. You can't compare a IMO participant (even if he has 4 gold medals) to a Fields Winner.[/quote]
IMO requires years of pratice also, just to go there. Thousands of hours spent on math, even if it is merely solving problems. It is much harder to work for 6-7 hrs / day when you are a kid, and everyone else goes and plays soccer, a video game, goes out to pubs, hangs out, you are called a geek, etc. for a few years, then when you are already grownup and that is your JOB. Remember when you are 25 years old you can dedicate to one thing, you don't have other classes to worry about. So from a relatively point a view I belive that the two events are comparable (that is the work that is required for them).

Of course only someone who has done them BOTH can really say for which one did he work, or he felt he worked harder. Of course a Fields Medal is more important than 3 gold medals at the IMO, but don't tell me that IMO is a childish game.

You are not in the position of making such affirmations, or making pseudo-comparisions with things that you haven't (yet :)) experienced.

[quote="Valentin Vornicu"]you must not be remembering correctly. people didn't take much interest (except maybe moubi :D) in IMC problems. the site exploded when I was the first to put online the [b]IMO[/b] subjects! I know for a fact that the parents of Ana Caraiani found out that she was 5th in the world by accidentally browsing the forum (the IMO official site is really bad).[/quote]
I remember correctly. At the IMC award festiviy, you proudly announced that you were the first to post the IMC subjects, and that the number of visitors of the site jumped 4000 (I'm not sure about this number).

[quote="Valentin Vornicu"]IMO requires years of pratice also, just to go there. Thousands of hours spent on math, even if it is merely solving problems. It is much harder to work for 6-7 hrs / day when you are a kid, and everyone else goes and plays soccer, a video game, goes out to pubs, hangs out, you are called a geek, etc. for a few years, then when you are already grownup and that is your JOB. Remember when you are 25 years old you can dedicate to one thing, you don't have other classes to worry about. So from a relatively point a view I belive that the two events are comparable (that is the work that is required for them). [/quote]
I have gone through a similar (to yours) IMO preparation program, so I know what contest preparation is. Only our results differ. I keep my statement that it takes much more to be a successful mathematician that to be an IMO medalist.

[quote="Valentin Vornicu"]Of course only someone who has done them BOTH can really say for which one did he work, or he felt he worked harder. Of course a Fields Medal is more important than 3 gold medals at the IMO, but don't tell me that IMO is a childish game.

You are not in the position of making such affirmations, or making pseudo-comparisions with things that you haven't (yet :)) experienced.[/quote]

I'm sure he would say he worked harder for the Fields medal, but he will include the IMO traninig in that work. Just as creche is part of our education.

Please take a look at the number of people who looked at the posts in the IMO section (tons of guest) and in the IMC section you will see the difference. You misunderstood what I said at IMC, I said that I am again the first to have the questions of an international competition online (even before the official site).

You obivously didn't undertake the same training as I did, as the team of six trains separately for 1 month, and you even were not in the large team of 25 untill our final grade.

What I am saying is that you compare two things (you tend to use the word creche) of which you have no clue of what they really mean by now.

[b]I[/b] cannot compare them (not yet at least) because I know [b]only [/b] one of them well enough to be able to speak about it.

Maybe we could ask Fedor Petrov and / or Myth for their opinion as they have been to IMO and also are graduate students involved in research ...

[quote="Valentin Vornicu"]Please take a look at the number of people who looked at the posts in the IMO section (tons of guest) and in the IMC section you will see the difference. You misunderstood what I said at IMC, I said that I am again the first to have the questions of an international competition online (even before the official site). [/quote]

And to think we speak the same language.

[quote="Valentin Vornicu"]You obivously didn't undertake the same training as I did, as the team of six trains separately for 1 month, and you even were not in the large team of 25 untill our final grade.  [/quote]

I said similar training. What do you think I was doing at that time? Rubbing mint?
You don't always need a teacher to train.
I'm not a natural competitor like you are. This is why I didn't do my best in contests. Maybe I don't do good under pressure or maybe I don't think fast enough for the usual 3-4hrs limit of contests in Romania, but this is not important. What is done is done. I'm past my failures and ready to move on. I hope my mind works fast enough to discover something in what is left of my life.

[quote="Valentin Vornicu"]What I am saying is that you compare two things (you tend to use the word creche) of which you have no clue of what they really mean by now.
[b]I[/b] cannot compare them (not yet at least) because I know [b]only [/b] one of them well enough to be able to speak about it.
[/quote]

When I use creche I use gauss202's expresion. So I have borrowed this word from him. What's wrong in that. I liked the word and I use it. That's it.
Excuse me [b]your royal highness[/b] for dairng to compare things you think I know nothing of.
So I think IMO is a childish game. Think of what is IMO. It is a contest.
That's it.
IMO preparation is the creche (I used [b]the word[/b] again) for research. If you want, I'll upgrade creche to school.
What makes you think you know IMO so well. Maybe some of the gold medalist browsing this site will tell you you know nothing of the IMO and you cannot make comparisons until you win the gold.
Back to research. In research I know (or at least my intuition tells me this is true) it is important to create. And I refer to the real creation. The genius spark, the original kind of creativity. Not the one you get by noticing that a problem you have to solve is similar to another.
IMO just asks you to solve. I am quite aware that there is a great deal of creativity involved in solving IMO problems, but is not the creativity I'm speaking of. This is because the problems are selected so that they can be solved in 4.5hrs.
So you need to be a genius to do something in research and you don't need to be a genius to get a [b]SILVER[/b] medal in IMO. For that you just need to know maths. The Romanian IMO central training program (the 25 team you were talking about) gets enough training so that a student is able to get a SILVER medal at the IMO.

[quote="Mindspa"]Olympiad problem solving is running 100 m, while research is marathon. I think amfulger has a point here, you can't go on with elementary problem solving as your main interest forever if you want to do research.[/quote]

First of all I want to clear the probable confusion created around the last post and claim responsability for it. So I'm the author of the previous post.
There is a problem and I want to advise you to be careful if you don't want to post as guest.
I checked the box on the sign me in automatically each time I visit.
I also checked the box on send me e-mail when a topic gets a reply. When my e-mail service (hotmail) announces me of a reply, I'm lead to the message, but I'm not being logged on, so if I reply I reply as guest. Usualy mathlinks warns me when I'm about to post as guest, but it didn't happen earlier. So be careful to log.

Thanks to Mindspa for agreeing. Pliviu sais this is what A.Wiles said about research. I think you Valentin won't disagree with him on that. I actually think you don't disagree at all, but you are distubed by me posting things about IMO.
I too continue elementary problem solving, but only for training. It is like jogging. So I don't use creche to disconsider IMO training, but to emphasise its importance for the growing up to research process.

Look amfulger you diverged from the subject with futile comparisions between IMO and research. You haven't done either (yet, I say again - you will probably do some research in the future but you are not doing any now) so I don't think you are the most competent person for comparing the two.

[quote]What makes you think you know IMO so well. Maybe some of the gold medalist browsing this site will tell you you know nothing of the IMO and you cannot make comparisons until you win the gold.
[/quote]
 On the other hand I went to two IMOs as a student, and one as Observer to the Leader. I know a lot about IMO, more than the usual participating student does. Also I have participated in 4 large RoMOPs and 2 common trainings one with USA one with Hungary. Again I am a member of the committee that trains the Romanian IMO team, and one of the correctors at the TST. Imagine that I know more about IMO than you.

Furthermore I haven't used a private tutor in my trainings for IMO, but on the other hand you did (as Pliviu told me :D:D) ... so the comment "You don't always need a teacher to train." is again in vain.

Most of Andrew Wiles' research was concentrated on solving a single problem ... so it's basically still problem solving :D:D:D

And I also agree to Mindspa. His metaphore is more adequate than yours.

[quote]Back to research. In research I know (or at least my intuition tells me this is true) it is important to create. And I refer to the real creation. The genius spark, the original kind of creativity. Not the one you get by noticing that a problem you have to solve is similar to another.
[/quote] And if you don't mind me asking how do you know that in reseach it is important to create and to have a genius spark? Other people say that work is much more important in research than the spark. You can be as sparkling as you want, but without tons and tons of work you will not do anything (well anything that hasn't already been discovered :D).

You need to rely your work on ideas. So genius is necessary. Not sufficient - I agree. With so many mathemticians in the world today I think that you can't do without the genius spark. I think that what was easy in ideas and hard in volume of work has already been discovered. New fields are being created so that the hard work can start again.

You don't know what I'm doing. Research or other stuff.

Mindspa used Wiles' words and yes the [b]100m[/b] is a better metaphore then the [b]creche[/b] in terms of the IMO competition, but not in terms of the IMO training.

Wiles' research was concentrated on solving a single problem - true, but in the process he developed some mathematical fields, and this is more important than the proof itself.

You did so much for IMO. But you didn't win the gold, did you? Of course you know more about IMO about me. But you know more about IMO competition spirit, organization ... other unimportant stuff.

I said you don't [b]always[/b] need... This is because you seem to think the the central training is more important than individual training.

[quote="Valentin Vornicu"]Look amfulger you diverged from the subject with futile comparisions between IMO and research.

And if you don't mind me asking how do you know that in reseach it is important to create and to have a genius spark? Other people say that work is much more important in research than the spark. You can be as sparkling as you want, but without tons and tons of work you will not do anything (well anything that hasn't already been discovered :D).[/quote]

Look into your last paragraph to see how divergent I am.

I want to add something more. It is said that a result in 99 percent perspiration and 1 percent inspiration.
I agree.
But there is a tremendous difference between 99% and 100%. The difference between a whole and a part. The 1% is what links the work together in a whole. Without it you only have work. Anybody can do the work.

To give you an exmple. Even though my previous posts wouldn't recommend this as the most fortunate example.
Think of the time you waste in a contest, when trying to solve a problem. You start to rewrite de information in the hypothesis in equivalent forms, trying to link them to the conclusion. You can spend hours doing this without getting the right idea. You must understand how inportant [b]the idea[/b] is. This (contest problem solving) is research at a very small scale.
The [b]you[/b] in my previous paragraph was a general you, like one tries to solve a problem... I don't know what Valentin does while trying to solve a problem.

It is pointless to talk about this as probably we cannot convince each other of the other's opinions. And as a popular Romanian expression goes, the smart guy breaks (or leaves) first, I shall end this topic ... :D


Find the number of all infinite sequence $ a_1,a_2,..$ of positive integer such that :
 $ a_n\plus{}a_{n\plus{}1}\equal{}2a_{n\plus{}2}a_{n\plus{}3}\plus{} 2005$ for all positive integer $ n$.

[quote="math10"]Find the number of all infinite sequence $ a_1,a_2,..$ of positive integer such that :
 $ a_n \plus{} a_{n \plus{} 1} \equal{} 2a_{n \plus{} 2}a_{n \plus{} 3} \plus{} 2005.............(1)$ for all positive integer $ n$.[/quote]

Assume that such infinite sequence $ a_1,a_2,...$ exists.

Then from $ (1)$ we have $ a_{n\plus{}1}\plus{}a_{n\plus{}2}\equal{}2a_{n\plus{}3}a_{n\plus{}4}\plus{}2005............(2)$

$ (2)\minus{}(1)$ we have $ |a_{n\plus{}2}\minus{}a_n|\equal{}2a_{n\plus{}3}|(a_{n\plus{}4}\minus{}a_{n\plus{}2})|>|a_{n\plus{}4}\minus{}a_{n\plus{}2}|$ for $ a_{n\plus{}2}\minus{}a_n \neq 0$.

So $ |a_{2k\plus{}1}\minus{}a_{2k\minus{}1}|$ is strictly decreasing until $ a_{2k\plus{}1}\minus{}a_{2k\minus{}1}\equal{}0$ . But $ |a_{2k\plus{}1}\minus{}a_{2k\minus{}1}|$ cannot decrease infinitely many times as it must always be an integer. So $ a_{2k\plus{}1}\equal{}a_{2k\minus{}1}$ for any large enough $ k$.

Apply this to $ |a_{2k\plus{}2}\minus{}a_{2k}|$ we have $ a_{2k\plus{}2}\equal{}a_{2k}$ for any large enough $ k$ also.

Now for large enough $ n$, we have

$ a_n\plus{}a_{n\plus{}1}\equal{}2a_{n\plus{}2}a_{n\plus{}3}\plus{}2005\equal{}2a_na_{n\plus{}1}\plus{}2005>a_n\plus{}a_{n\plus{}1}$ , contradiction.

So no such sequence exists , i.e. the answer is zero.


Blocks of modeling clay are six inches by one inch. How many whole blocks are needed to mold a cylindrical sculpture seven inches high and four inches in diameter?

The surface area of the sculpture is $ 2\pi r^2\plus{}2\pi rh\equal{}2\pi r(r\plus{}h)\equal{}36\pi\approx113\implies\left\lceil\frac{113}6\right\rceil\equal{}\boxed{19}$.

I think it means the blocks are $6\times1\times1$,  (rectangular prisms)

so it's actually a pretty hard problem. Answer is 8, no idea why.

The blocks are $6 x 2 x 1$ And the volume of the sphere is base times height. And the area of the base is $4pi$ or $12.56$ Multiply it by the height to get $87.92$ and divide $87.92$ by $12$ which is the volume of a block. You get $7.326666666666...$ which rounds up to $8$ blocks because that if you were to get $7$ you'd still need a little more to make up for the $3.26666666666$

I came to this post through FTW, and the problem was copied onto here wrong. The original problem:
A whole block of modeling clay is a right rectangular prism six inches by two inches by one inch. [How many whole blocks need to be unwrapped to mold a cylindrical sculpture seven inches high and four inches in diameter?]

The cylindrical sculpture has radius two inches and uses $\pi(2^2)(7)=28\pi \approx 87.96$ cubic inches of modeling clay. Each block contains $(6)(2)(1)=12$ cubic inches of modeling clay. If we have 7 blocks, then we have 84 cubic inches of clay which is not enough, so we need $\boxed{8}$ blocks of clay, and we see 96 cubic inches are indeed enough.


[quote="DonkeyKong"]Can anyone help me by telling me how Permuation & Combination work?

Also when to use Permutaion & Combination?

Thanks. :lol:[/quote]

permutation is: n nPr k; or it can be written n P k; or (n,k). it works when, say, an order of abc is different than the order cba or bac etc. You do it by $ \frac{n!}{k!}$

combination is: n nCr k, can be writen n C k or $ \binom{n}{k}$ or a big C with n on the top half and k on the bottom half(dont know how to do that). it applies when abc is the same as bac or cab, etc. you do it $ \frac{n!}{(n\minus{}k)!(k!)}$

Hope that helps!!

DonkeyKong, i think this kind of topic is too basic for MATHCOUNTS. I think u should have posted this in Classroom Math

I agree with mihail911, why ignore such a useful resource?

[[Permutation]] and [[Combination]] should give answers.

OH! I didn't know there is a AOPSWiki....

Sorry for bothering you guys... :blush:

[b]combination [/b]is where the order doesnt matter, like comitees (b/c in a comitee, there is no difference between one person and another- so the committee of alex, bob and carol is the same as the commitee of carol, alex and bob)
- "10 ppl choose 3 for a commitee"  is an example of combinations
which would equal 10!/7!3!= 120

[b]permutation[/b] is where order does matter liike elections for
President, vice president, secretary, etc... obviously, secretary is not the same as president!!
-"if you had 10 ppl and you wanted 1 for president, 1 for secretary and 1 for vice president and one person can only hold at the max of 1 office" is an example of permutations.... the answer would be
10(choices for president)*9(choices for VP)*8(choices for secretary) which would equal 720

don't revive a topic. an explanation really isnt necessary anymore. AoPS wiki?


$ a_{n}^{n}\equal{}a_{n}^{n\minus{}1}\plus{}....\plus{}a_{n}\plus{}1$
Find $ lim a_{n}$
The same problem is VN national Olympiad2007.


geometry
3D geometry
octahedron
probability
AMC
AMC 12
AMC 12 B

Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex.
Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each
with equal probability. What is the probability that no two ants arrive at the same vertex?

This question is from AMC12B 2005.  The solution on AoPSWiki:

[url]http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_12B_Problems/Problem_25[/url]


floor function
inequalities
quadratics
geometry
modular arithmetic

A leaf is torn from a paperback novel. The sum of the numbers on the remaining pages is 15000. What are the page numbers on the torn leaf?

This means that $ \frac{n(n\plus{}1)}2\approx15000\implies n(n\plus{}1)\approx30000\implies n\equal{}\lfloor\sqrt{30000}\rfloor\equal{}173$, so there are $ 173$ pages in the book, giving an initial sum of $ \frac{173\cdot174}2\equal{}15051$, so the sum of the pages on the leaf is $ 15051\minus{}15000\equal{}51\equal{}p\plus{}(p\plus{}1)\implies(p,p\plus{}1)\equal{}\boxed{(25,26)}$.

We have the two inequalities $ \frac{n(n\plus{}1)}2 > 15000$ and $ \frac{n(n\plus{}1)}2 \minus{} 2n \plus{} 1 \le 15000$.

This gives us the bounds $ \frac{7\sqrt{2449} \minus{} 1}{2} < n < \frac{7\sqrt{2449} \plus{} 3}{2}$. We know $ n$ is an integer, so $ 173 \le n \le 174$.

$ n \equal{} 174$ gives us a total sum of $ \frac{174 \cdot 175}2 \equal{} 15225$, so the sum of the pages on the leaf is $ 225 \equal{} p \plus{} (p \plus{} 1) \implies (p, p\plus{}1) \equal{} (112, 113)$. However, this is not possible, as the odd number on a page is always lower than the even one.

The other case was already covered by math154.

$ \frac{n(n\plus{}1)}{2}\minus{}2p\minus{}1\equal{}15000\implies \frac{n(n\plus{}1)}{2}>15000$
$ n^2\plus{}n\minus{}30000>0$. Finding the roots of this quadratic, we have $ n\equal{}\frac{\minus{}1\pm\sqrt{120001}}{2}$.

We can find $ \sqrt{120001}$ like this:
$ 120001\equal{}2^2\cdot3\cdot10^4\plus{}1\implies2\cdot10^2\cdot\sqrt{3}<\sqrt{120001}$
$ 1<\sqrt{3}<2$, and it's pretty common knowledge that $ \sqrt{3}\approx 1.73$, so we have $ \sqrt{120001}>200\cdot1.73\equal{}346$.
For our purposes (where the quadratic is greater than zero), we can take $ \sqrt{120001}$ to be $ 347$ (the smallest integer that satisfies the inequality).
Then, $ n\equal{}\frac{346}{2}\equal{}173$.
$ \frac{173(174)}{2}\minus{}2p\minus{}1\equal{}15000\implies \minus{}2p\equal{}15001\minus{}15051\equal{}\minus{}50\implies p\equal{}25$.
The answer is $ \boxed{(25, 26)}$.

As far as I can tell, that's as close to without approximation as possible.

Both you and math154 made an unspoken assumption that the number of pages is the smallest possible. Algebraically though, it is possible to have $ n \equal{} 174$. The reason why it doesn't work is purely contextual (Had the resulting number been 15001, 174 would have been correct). There is no way to really avoid approximating $ \sqrt {120001}$, however.

[quote="mathwizarddude"]Can you method be a little more rigorous (without approximation) please?[/quote]
There's nothing wrong with approximation.  Getting in the ballpark and wiggling your way toward the right answer is perfectly respectable, and in fact in a lot of areas of mathematics where it is not feasible or important to find exact answers our most accurate and most cherished results are just [url=http://en.wikipedia.org/wiki/Prime_number_theorem]really fancy approximations[/url].

[quote="mathwizarddude"]ffao: How did you find that factorization for $ 120 001$?  Is there any divibility rule for 7 (and other numbers as well)?[/quote]
Yes, but it's not really worth using in my opinion.  There is one that tells you what the remainder upon division by $ 7$ is (it's very tedious) and another one that just tells you whether a number is divisible by $ 7$ (which is less tedious, but I always forget it).  For numbers this small it's perfectly fine just to play some arithmetic tricks:

$ 120001 \equal{} 119000 \plus{} 1001 \equal{} 7(17000 \plus{} 11 \cdot 13).$

It helps to have the prime factorizations of two-digit and some three-digit numbers memorized to be able to do this consistently.  I find it just as fast as the divisibility test for $ 7$, and you can do the same thing for larger primes whose divisibility rules are even worse.

There is a trick. Because $ 7,13|1001$, the divisibility rule is to repeatedly subtract the number formed by all but the last three digits and the number formed by the last three digits. In here, this would be $ 120001\equiv001 \minus{} 120\equiv \minus{} 119\equiv0\pmod7$.

It follows directly from the fact that $ 7|1001$ (so this works for all factors of $ 1001$): for $ n\ge3$ (it's useless otherwise) and $ k|1001$,

$ \overline{a_na_{n \minus{} 1}a_{n \minus{} 2}\ldots a_3a_2a_1a_0} \equal{} \overline{a_na_{n \minus{} 1}a_{n \minus{} 2}\ldots a_4a_3}\cdot1000 \plus{} \overline{a_2a_1a_0}$

$ \equiv \minus{} \overline{a_na_{n \minus{} 1}a_{n \minus{} 2}\ldots a_4a_3} \plus{} \overline{a_2a_1a_0}\pmod k$.


I read the solution to the following AMC question, but I didn't quite understand the solution:

25.   For each integer $n\ge4$, let $a_n$ denote the base-n number $0.\overline{133}_n$.  The product $a_4a_5 \ldots a_{99}$ can be expressed as $\frac{m}{n!}$, where $m$ and $n$ are positive integers and $n$ is as small as possible.  What is the value of $m$?
(A) 98       (B) 101       (C) 132       (D) 798       (E) 962

[hide="Solution"]Note that $n^3a_n=133.\overline{133}_n=a_n+n^2+3n+3$, so $a_n=\frac{n^2+3n+3}{n^3-1}=\frac{(n+1)^3-1}{n(n^3-1)}$.
Therefore
$a_4a_5 \ldots a_{99}=\displaystyle \left(\frac{5^3-1}{4(4^3-1)} \right) \displaystyle \left(\frac{6^3-1}{5(5^3-1)}\right) \cdots \displaystyle \left(\frac{100^3-1}{99(99^3-1)}\right)$
$=\displaystyle \left(\frac{3!}{99!}\right) \displaystyle \left(\frac{100^3-1}{4^3-1}\right)=\displaystyle \left(\frac{6}{99!}\right) \displaystyle \left(\frac{99(100^2+100+1)}{63}\right)$
$=\frac{(2)(10,101)}{(21)(98!)}=\frac{962}{98!}$.
So the answer is (E) 962.[/hide]
Specifically, I don't see why:
[hide]$n^3a_n=133.\overline{133}_n=a_n+n^2+3n+3$[/hide]

[hide]

Do you see how they got from $n^3\cdot a_n$ to it equaling $133.\overline{133}_n$?

When you multiple a number in a base $b$ by a power of $b$ you are just changing the decimal place.  For example, in base 10 multiply $23$ by $10^2$ just moves the decimal over $2$ to the right.  Ask if you still don't understand this then I will try to explain it better.

Now let's look at $133.\overline{133}_n$ and call it $X$ for now.  We know that $X=133+.\overline{133}$.  That part on the right is just $a_n$ as defined in the problem.  Now the $133$ part we can convert to base 10 as usual:  $133_n=(n^2+3n+3)_{10}$.

Do you understand that part now?
[/hide]

I understand that part (and the solution) now.  Thanks for your help.


Let $a_{0}= 0$, and $a_{n+1}= \sqrt{3+a_{n}}$

Let $u$ be the least upper bound of this sequence.
Express $u-a_{n+1}$ in terms of $u-a_{n}$


function
algebra
polynomial
Rational Root Theorem

solve
$ (x^2\minus{}3x\plus{}3)^2 \minus{}3(x^2\minus{}3x\plus{}3) \plus{}3\equal{}x$

hello, your equation is equivalent to $ (x\minus{}3)(x\minus{}1)^3\equal{}0$ from here you will get all solutions.
Sonnhard.

$ f(f(x)) \equal{} x$ where $ f(x) \equal{} x^2 \minus{}3x\plus{}3$. So $ f(x) \equal{} x$ gives you two of the roots rightaway.

@hsbhatt

 :o Why would $ f(x) \equal{} x$ give two of the roots?  :blush:

Is it done by observing the properties of the function?
does $ f(x)\equal{}\frac{1}{x}$ also work?

$ f(x)\equal{}x \iff x^2\minus{}3x\plus{}3\equal{}x \iff (x\minus{}3)(x\minus{}1)\equal{}0$.

[quote="chrischris"]@hsbhatt

 :o Why would $ f(x) \equal{} x$ give two of the roots?  :blush:[/quote]
If $ f(x) \equal{} x$ then $ f(f(x)) \equal{} f(x) \equal{} x$.  This is true of any function $ f$.

[quote="Bugi"][quote="aadil"]how did you simplify the expression?[/quote]

Expand, then put divisors of free constant member as the values of x, and use Bezoue to factor the expression.[/quote]
What does "put divisors of free constant member as the values of x" mean? :huh:

[quote="Nora.91"][quote="aadil"]what is bezoue?[/quote]

i think its [url=http://mathworld.wolfram.com/BezoutsTheorem.html]Bezout Theorem[/url][/quote]

Yeah, I meant Bezout (just didn't know spelling), thanks.

@mathwizarddude: again, don't know the right word...

I meant if $ P(x)\equal{}\sum_{i\equal{}0}^n a_ix^i$, then check the values of P(x) for divisors of $ a_0$, and if they are 0, then P(x) is divisible by (x-t), where t is that value.

From $ f(f(x))\equal{}x$, you can get all of the roots of $ f(x)$ by letting $ f(x)$ equal some cyclic function.  If $ f(x)\equal{}1\minus{}x$,for example, we get that $ x^{2}\minus{}2x\plus{}2\equal{}0$.

@Bugi: so is using the rational root theorem basically what you meant (p/q is a root with (p,q)=1 iff $ p|a_0,q|a_n$)?
But after find such rational root $ r$, the standard way would be to use long division to divide the original expression by $ x\minus{}r$, but how does it have to do with Bezout's theorem?

Bezout says that if P(a)=0, then P(x) is divisible by x-a. So before using long division, you can asure that there will be no remainder.

I think in the US that theorem is usually referred to as the factor theorem. The Bezout statements I know, briefly, are that $ ax \plus{} by \equal{} 1$ has a solution for $ (a,b) \equal{} 1$ and that $ f,g$ intersect $ (\deg f)(\deg g)$ times for $ f,g$ polynomials.

EDIT: Yeah, I got some results jumbled up too. :P

[quote="MellowMelon"]$ f \circ g$ has $ (\deg f)(\deg g)$ roots for $ f,g$ polynomials[/quote]
Bezout's theorem for algebraic curves says something quite a bit different, which is about intersection rather than composition.  The first theorem you quote is sort of relevant if you apply it to polynomials instead of integers, but then you're better off calling it the division algorithm.

@MellowMelon: I have no clue what's the name of that theorem in US... We call it Bezout's theorem.

[b]Solution[/b]: no need of [u]Bezout Theorem[/u], factoring it little by little, we have

(x^2 - 3x + 3)^2 - 3(x^2 - 3x + 3) + 3 = x, rearranging

(x^2 - 3x + 3) - 3(x^2 - 3x + 3) - (x - 3) = 0, factoring (x^2 - 3x + 3),

(x^2 - 3x + 3)[(x^2 - 3x + 3 - 3] - (x - 3) = 0, then

(x^2 - 3x + 3)(x^2 - 3x) - (x - 3) = 0, then

[(x^2 - 3x + 3)(x)(x - 3)] - (x - 3) = 0, then

(x - 3)(x^3 - 3x^2 + 3x - 1) = 0, [Note that (x - 1)^3 = (x^3 - 3x^2 + 3x - 1)]

(x - 3)(x - 1)(x - 1)(x - 1) = 0,

Thus, (x - 3)(x - 1)^3 = 0.

Roots are {1, 3).

oh, this is the factorization obtained by [b]Dr Sonnhard Graubner[/b]. A shortcut was obtained by [b]hsbhat[/b]t, that is
(x^2 - 3x + 3) = x, or [b]Brut3Forc3's[/b] factorization (x - 3)(x - 1) = 0 gave the same result. Well, what a cute problem.


trigonometry
geometry
geometric transformation
rotation
calculus
parameterization
parametric equation

Mathnerd314 has a wheel with radius 1 meter. He rolls it on the ground at 1 radian per second. There is a bug sitting on the circumference of the wheel. Find a parametric equation for the bug's position.

[hide]I'm getting
$x=\sin t+2 \pi t$
$y=\cos t$

Is that right?[/hide]

Bonus: How far does the bug travel in one rotation of the wheel? (relative to the ground) This was what I really wanted to find out in this problem. Does it involve calculus?

[quote="Farenhajt"]Do you mean: the distance between the two consecutive points of contact between the bug and the ground, or: the length of the arc traveled by the bug therewith? The first one is easy, just find out the zeros of $y$, and the second one must involve calculus.[/quote]

I mean the length of the arc. Sorry if I misstated the question.

This shape is known as a cycloid, and it has a number of interesting properties which as far as I know all require calculus to prove.


For some reason I can't log in...like if I try logging in, I go straight to the forum afterwards, but the login sign is still there....and for confirmation, I can't post. I found out that the only way I could log in(like now) was to check remember password. Why is this?

As usual please mention your skin, url (aops or ml url), your system and browser. Have you tried erasing the cookies? (a small link in the right of the lower nagivation bar on the index page .. )

Ummm. skin-default?
Url:um...?
Windows
IE 6.0

...?

Try this: [url=http://www.mathlinks.ro/Forum/remove_cookies.php]Delete all cookies set by this board[/url]. You can find this option also on each /Forum/index.php?f=... page you go on the site, on the right-lower corner, on the long horizontal bar which I call navigation bar :D

See if this solves your problem ...


function
inequalities unsolved
inequalities

[quote="chien than"]Let $a_{1}$,$a_{2}$ ,.....,$a_{1945}$ be real numbers satisfying  $\sum_{i=1}^{1945}a_{i}$= 1997$\forall$$a_{i}$ $\in$[1;3]  i=1,2,....,1945
Find max $\sum_{i=1}^{1945}a_{i}^{3}$[/quote]

aaaa? whatever is ai? then take ai=3 and it is not true

Since $a^{3}$ is a convex function in the range $[1,3]$, the sum of the third powers of the $a_{i}$ is maximized by having all but perhaps some of the $a_{i}$ at the endpoints of the prescribed interval.
Suppose $n$ of the $a_{i}$ equal $1$, $1945-n$ equal $3$. From the fact that $\sum_{i=1}^{1945}= 1997$, we have $n+3 \cdot (1945-n) = 1997 \Rightarrow n = 1919$.

Then, $\sum_{i=1}^{1945}a_{i}^{3}\leq 1919+(1945-1919) \cdot 27 = 2621$ as desired. (The max. value is $2621$)


I was playing around with Geometer's Sketchpad today, and this question popped into my head. I don't have any idea how it can be solved, but nonetheless, here goes...

Suppose that there is a circular track. A and B are two runners on the track. They start at one point, heading in the same direction, with B running faster than A. Consider the midpoint P of the line joining A and B. What path would it trace out? (spiral - what kind of spiral?)

Is it a spiral? I think the whole thing depends on the ratio between the speeds of the pts. For example, if speed(B)/speed(A)=2, then the curve is closed and it's tangent to the circle in the START point. On the other hand, if speed(B)/speed(A) is irrational, I don't think the curve is closed, and the pts of tangency with the circle is (I think) dense on the circle.

Can you trace the path of a point with GSP? I use Euklides (because it's free! :D), and I know it can be done, but I don't know how. I'll try to learn that from help.

Yes, the rational and irrational cases are definitely very different, resulting in a closed curve and an unclosed curve with infinitely many tangent points with the circle, which will be dense on the circle as grobber claimed, respectively.  My guess is that it will be something very much like what you get with a spirograph, although I'm not 100% sure.

let the first runner go along the path with parametric coordinates (cos  :theta:  , sin  :theta: ), while the second runner (going k times faster) has coordinates (cos k :theta: , sin k :theta: ). The midpoint is ((cos  :theta:  + cos k :theta: )/2, (sin  :theta:  + sin k :theta: )/2), which doesn't look to inviting. So, seing "spirals" according to Joel, we try changing to polar coordinates: we find that the magnitude of the locus in terms of  :theta:  is

 :sqrt: [(cos^2  :theta:  + 2 cos  :theta:  cos k :theta:  + cos^2 k :theta: )/4 + (sin^2  :theta:  + 2 sin  :theta:   sin k :theta:  + sin^2 k :theta: )/4 ]

= :sqrt: [1/2 (1 + cos k :theta:  cos  :theta:  + sin k :theta: sin  :theta: )]

= :sqrt: [1/2 (1+cos(k :theta: - :theta: ))]

= :sqrt: [cos^2 (1/2 (k-1)  :theta: )]

If we take into account that a polar graph can have negative r, we get the polar equation

r = cos(1/2 (k-1)  :theta: ).

If k is rational (and therefore 1/2 (k-1) is rational as well), then some multiple of 2 :pi:  for  :theta:  will make 1/2 (k-1)  :theta: a multiple of 2  :pi:  thus bringing the curve back around to the starting point (r,  :theta: ) = (1,0). If the k is irrational, then the curve will never get back to the point (1,0), and thus will never close as Joel hypothesized.


It doesn't seem to work for $ x_1\equal{}y_2\equal{}1$, $ x_2\equal{}y_1\equal{}\minus{}1$, $ x_3\equal{}y_3\equal{}0$.

Bu the third condition is not satisfied then.  Anyway, $ \left< x_1, x_2, x_3 \right>$ and $ \left< y_1, y_2, y_3 \right>$ are perpendicular vectors on the plane $ x \plus{} y \plus{} z \equal{} 0$ and can be scaled to have length $ 1$.  The rest appears to be computation.

Ok, I understand up till the x+y+z=0 is a plane, containing two perpendicular vectors as described. How do I continue from there? Thanks!

Suppose $ \{\mathbf{u},\mathbf{v},\mathbf{w}\}$ is an orthonormal set in $ \mathbb{R}^3.$ (That means that each of these vectors has length $ 1$ and their pairwise dot products are zero.) Let $ \mathbf{r}$ be another vector in $ \mathbb{R}^3.$ then:
\[ (\mathbf{r}\cdot\mathbf{u})^2 \plus{} (\mathbf{r}\cdot\mathbf{v})^2 \plus{} (\mathbf{r}\cdot\mathbf{w})^2 \equal{} \|\mathbf{r}\|^2.
\]
Or, doing the same thing without normalizing first: if $ \{\mathbf{u},\mathbf{v},\mathbf{w}\}$ is an orthogonal set, then
\[ \frac {(\mathbf{r}\cdot\mathbf{u})^2}{\|\mathbf{u}\|^2} \plus{} \frac {(\mathbf{r}\cdot\mathbf{v})^2}{\|\mathbf{v}\|^2} \plus{} \frac {(\mathbf{r}\cdot\mathbf{w})^2}{\|\mathbf{w}\|^2} \equal{} \|\mathbf{r}\|^2.
\]
If we believe that, then the result follows from $ \mathbf{u} \equal{} (x_1,x_2,x_3),$ $ \mathbf{v} \equal{} (y_1,y_2,y_3),$ $ \mathbf{w} \equal{} (1,1,1),$ and $ \mathbf{r} \equal{} (1,0,0).$

As for proving what we need, use this:
\[ \mathbf{r} \equal{} \frac {\mathbf{r}\cdot\mathbf{u}}{\|\mathbf{u}\|^2}\,\mathbf{u} \plus{} \frac {\mathbf{r}\cdot\mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v} \plus{} \frac {\mathbf{r}\cdot\mathbf{w}}{\|\mathbf{w}\|^2}\,\mathbf{w}
\]
Just try directly computing the norm of that by taking its dot product with itself, and take advantage of the fact that the terms of the sum are orthogonal.

(Key words: orthogonal basis, orthonormal basis, projection, generalized Fourier coefficients.)

The perpendicular part is because the $ x_1y_1\plus{}x_2y_2\plus{}x_3y_3\equal{}0.$ This just shows the dot product ($ \sum \limits_{i\equal{}1}^n x_iy_i$ for $ x_n,y_n \epsilon R^n)$ is equal to $ 0$. If two vectors are parallel, the dot product is equal to $ 1$.

Thanks everyone! I have managed to solve it! Thanks t0rajir0u for his ideas! I have never thought of vectors.

Let $ \mathbf{x}\equal{}\left<x_{1}, x_{2}, x_{3}\right>$ and $ \mathbf{y}\equal{}\left<y_{1}, y_{2}, y_{3}\right>$ be the vectors. According to the condition, we get $ \mathbf{x}\cdot\mathbf{n}\equal{}\mathbf{0}$ and $ \mathbf{y}\cdot\mathbf{n}\equal{}\mathbf{0}$ and $ \mathbf{x}\cdot\mathbf{y}\equal{}\mathbf{0}$. It remains to be seen that $ \mathbf{x}$ is perpendicular to $ \mathbf{y}$ and the vector n perpendicular to both of them is $ \mathbf{n}\equal{}\left<1,1,1\right>$

Therefore, we can note that $ \left<1,1,1\right>\times\mathbf{x}\equal{}\mathbf{y}$ since all three vectors, $ \mathbf{n}$, $ \mathbf{x}$ and $ \mathbf{y}$ are mutually perpendicular.

Hence $ \mathbf{y}\equal{}\left<x_{3}\minus{}x_{2}, x_{1}\minus{}x_{3}, x_{2}\minus{}x_{1}\right>$ implies that $ y_{1}\equal{}x_{3}\minus{}x_{2}$

Substituting it into the equation

$ LHS\equal{}\frac{x_{1}^2}{|\mathbf{x}|^2}\plus{}\frac{x_{2}^2\plus{}x_{3}^2\minus{}2x_{2}x_{3}}{2|\mathbf{x}|^2\minus{}2\sum x_{1}x_{2}}$

Since $ \left(x_{1}\plus{}x_{2}\plus{}x_{3}\right)^2\equal{}|\mathbf{x}|^2\plus{}2\sum x_{1}x_{2}\equal{}0\Rightarrow |\mathbf{x}|^2\equal{}\minus{}2\sum x_{1}x_{2}$

Also

$ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{}0\Rightarrow x_{1}\equal{}\minus{}x_{2}\minus{}x_{3}\Rightarrow x_{1}^2\equal{}x_{2}^2\plus{}x_{3}^2\plus{}2x_{2}x_{3}\Rightarrow \minus{}2x_{2}x_{3}\equal{}x_{2}^2\plus{}x_{3}^2\minus{}x_{1}^2$

Hence

$ LHS\equal{}\frac{x_{1}^2}{|\mathbf{x}|^2}\plus{}\frac{x_{2}^2\plus{}x_{3}^2\minus{}2x_{2}x_{3}}{2|\mathbf{x}|^2\minus{}2\sum x_{1}x_{2}}\equal{}\frac{x_{1}^2}{|\mathbf{x}|^2}\plus{}\frac{x_{2}^2\plus{}x_{3}^2\plus{}x_{2}^2\plus{}x_{3}^2\minus{}x_{1}^2}{3|\mathbf{x}|^2}\equal{}\frac{2|\mathbf{x}|^2}{3|\mathbf{x}|^2}\equal{}\frac{2}{3}$

Phew, I can finally take a rest. I am having a headache trying to solve this question for one whole day and night!

Thanks everyone!

All right, try this variant:

Suppose $ x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4,z_1,z_2,z_3,z_4$ are real numbers such that $ \sum_{k\equal{}1}^4x_k\equal{}\sum_{k\equal{}1}^4y_k\equal{}\sum_{k\equal{}1}^4z_k\equal{}0$ and $ \sum_{k\equal{}1}^4x_ky_k\equal{}\sum_{k\equal{}1}^4x_kz_k\equal{}\sum_{k\equal{}1}^4y_kz_k\equal{}0$

Prove that

$ \frac{x_1^2}{x_1^2\plus{}x_2^2\plus{}x_3^3\plus{}x_4^2}\plus{} \frac{y_1^2}{y_1^2\plus{}y_2^2\plus{}y_3^3\plus{}y_4^2}\plus{} \frac{z_1^2}{z_1^2\plus{}z_2^2\plus{}z_3^3\plus{}z_4^2}\equal{}\frac34$

The one tool I just took away from enoughsaid05 is the cross product. Can you live without that?

This problem is related to directional cosine.

Let $ X=(x_1,x_2,x_3,x_4)$, etc. Assume that $ X\ne 0$ otherwise the problem explodes. Hence we can immediately normalize all the vectors. Then we must prove

Let $ B=\{b_1=(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}),b_2=X,b_3=Y,b_4=Z\}$. Then the problem conditions stipulate $ B$ is an orthonormal basis for $ \mathbb{R}^4$.

Now let $ u=\left< 1,0,0,0 \right>$.

Lemma: $ u=\sum \left< u,b_i \right>$.
Proof: Since $ B$ is an orthonormal basis for $ \mathbb{R}^4$, $ u$ can be written as a linear combination of $ b_i$. So $ u=\sum k_ib_i$. Taking dot products of both sides with $ b_j$, we get $ k_j$ and the result follows.

Observe that $ \sum\left< u,b_i \right>^2=\sum \left< u,b_i  \left< u,b_i \right>  \right>= \left< u, %Error. "sumb" is a bad command.
_i  \left< u,b_i \right>  \right>= \left< u,u\right>$

Evaluating both sides of the equivalence above gives the result.

[quote="enoughsaid05"]Need help on this question

Given that $ x_{1} \plus{} x_{2} \plus{} x_{3} \equal{} y_{1} \plus{} y_{2} \plus{} y_{3} \equal{} x_{1}y_{1} \plus{} x_{2}y_{2} \plus{} x_{3}y_{3} \equal{} 0$

Prove that

$ \frac {x_{1}^2}{x_{1}^2 \plus{} x_{2}^2 \plus{} x_{3}^2} \plus{} \frac {y_{1}^2}{y_{1}^2 \plus{} y_{2}^2 \plus{} y_{3}^2} \equal{} \frac {2}{3}$[/quote]

Here is the solution for Japanese High School students.

Since $ a \plus{} b \plus{} c \equal{} p \plus{} q \plus{} r \equal{} ap \plus{} bq \plus{} cr \equal{} 0$, we have $ (a \plus{} b \plus{} c)(p \plus{} q \plus{} r) \equal{} 0\Longleftrightarrow (b \plus{} c)p \plus{} (c \plus{} a)q \plus{} (a \plus{} b)r \equal{} 0$.

Thus we can set $ (p,\ q,\ r) \equal{} k(b \minus{} c,\ c \minus{} a,\ a \minus{} b)\ (k\neq 0)$.

$ \frac {p^2}{p^2 \plus{} q^2 \plus{} r^2} \equal{} \frac {(b \minus{} c)^2}{(b \minus{} c)^2 \plus{} (c \minus{} a)^2 \plus{} (a \minus{} b)^2} \equal{} \frac {(b \minus{} c)^2}{2\{a^2 \plus{} b^2 \plus{} c^2 \minus{} (ab \plus{} bc \plus{} ca)\}} \equal{} \frac {(b \minus{} c)^2}{2\{(a \plus{} b \plus{} c)^2 \minus{} 3(ab \plus{} bc \plus{} ca)}$

$ \equal{} \minus{} \frac {(b \minus{} c)^2}{6(ab \plus{} bc \plus{} ca)}$.

$ \therefore \frac {a^2}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {p^2}{p^2 \plus{} q^2 \plus{} r^2}$

$ \equal{} \frac {a^2}{(a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca)} \minus{} \frac {(b \minus{} c)^2}{6(ab \plus{} bc \plus{} ca)}$

$ \equal{} \minus{} \frac {3a^2 \plus{} (b \minus{} c)^2}{6(ab \plus{} bc \plus{} ca)}\equal{} \minus{} \frac {4(b^2 \plus{} c^2 \plus{} bc)}{6(ab \plus{} bc \plus{} ca)}$

$ \equal{} \minus{} \frac {2}{3}\cdot \frac {(b \plus{} c)^2 \minus{} bc}{a(b \plus{} c) \plus{} bc}$

$ \equal{} \frac {2}{3}\cdot \frac {bc \minus{} ( \minus{} a)^2}{a( \minus{} a) \plus{} bc} \equal{} \frac {2}{3}$. $ Q.E.D.$

This is problem GBR-3 (United Kingdom) from the longlist for IMO'1977.
It did not make it to the shortlist.

Note that the sum of the squares of the projections of a
unit vector onto three pairwise orthogonal lines equals 1.

[quote="Altheman"]Assume that $ X\ne 0$ otherwise the problem explodes.[/quote]
Agreed, and that stipulation should have been in the original problem as well. That is, the original problem should have specified $ x_1^2\plus{}x_2^2\plus{}x_3^2>0,\ y_1^2\plus{}y_2^2\plus{}y_3^2>0.$