How many whole numbers between $ 100$ and $ 400$ contain the digit $ 2$?

\[ \textbf{(A)}\ 100 \qquad
\textbf{(B)}\ 120 \qquad
\textbf{(C)}\ 138 \qquad
\textbf{(D)}\ 140 \qquad
\textbf{(E)}\ 148
\]

There are 19 between 100 and 199 (102,112,120,121,...,132,142,...), 100 between 200 and 299 (all of them), and an additional 19 between 300 and 400 (same as between 100-199).  So that is a total of 138.

There are a total of 400-100+1=301 numbers between 100 and 400, inclusive. The total number of numbers that don't have any digits equal to 2 is 2(hundreds digit can be either 1 or 3)*9(tens digit can have any number except 2)*9(the ones digit has the same number of choices as tens)+1(don't forget to add the single number 400, since it wasn't counted before) =163. So 301-163=138<--Answer

I think complementary counting is quicker here, so the second solution. :)


calculus
integration
trigonometry
calculus computations

Compute
$ \int \frac{dx}{\cos^{3}x\minus{}\sin^{3}x}$

First notice that

$ D \equal{} \cos^{3}x \minus{} \sin^{3}x \equal{} (\cos x \minus{} \sin x)(1 \plus{} \cos x \sin x)$

Set $ x \plus{} \frac {\pi}{4} \equal{} t$ and this gives us
$ D \equal{} \sqrt {2} \cos t (1 \minus{} \frac {\cos 2t}{2}) \equal{} \frac {1}{\sqrt {2}} \cos t (3 \minus{} 2\cos^{2}t)$

Thus
$ \frac {1}{D} \equal{} \frac {\sqrt 2}{3}(\frac {1}{\cos t} \plus{} \frac {2\cos t}{3 \minus{} 2\cos^{2}t})$

Thus the integral can be split into two obvious integrals, one of $ \sec t$ and the second  can be gotten by the subsitution $ \sin t \equal{} z$.

(I hope the working is right).


[hide="Solution"]We have obvious lower bounds on $ x$ and $ y$; $ x \ge 4$ and $ y \ge 3$. $ \sqrt[3]{1331_x} \equal{} 11_x$. Similarly, $ \sqrt{121_y} \equal{} 11_y$. We therefore have that $ x\plus{}y\plus{}2\equal{}9$. Now, $ \boxed{x\equal{}4}$ and $ \boxed{y\equal{}3}$ are the only valid solutions due to the established lower bounds..[/hide]


I think about 121:

In the first line you find 1 possible point (0;10), in the second 2, because the point (0;9) can't be the final one (you can only after an odd number of steps).
So the total is 1+2+3+4+5+6+7+8+9+10+11+10+9+8+7+6+5+4+3+2+1=121

[quote="mathnerd314"]Mathnerd314 is trapped in the coordinate plane!

He starts at the origin. At each step, he can move one unit north, one unit south, one unit east, or one unit west. He will be airlifted out after ten steps.

How many different points are possible as the final point?[/quote]

[hide]If the final coordinates are $(x,y)$ then $-10\le x+y \le 10$ and $x+y$ is even. All points that satisfy these conditions can be reached...so then there are $11$ possible $x$ and $11$ possible $y$, and $11\cdot 11=\boxed{121}$.[/hide]


Let $ p$ be a positive prime integer and $ a,b,c$ be natural numbers,find all $ (p,a,b,c)$ such that
$ \frac{1}{p}\equal{}\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}$

$ \frac {1}{p} \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$
$ 1 \equal{} \frac {p}{a^2} \plus{} \frac {p}{b^2} \plus{} \frac {p}{c^2}$
$ (abc)^2 \equal{} p(bc)^2 \plus{} p(ac)^2 \plus{} p(ab)^2 *$
$ p|(abc)^2$
$ p^2|(abc)^2$

At least one of a, b, or c is divisible by p.
Now, divide the right-hand side of * by p.
Since p divided the entire expression at least twice...

$ p|((bc)^2 \plus{} (ac)^2 \plus{} (ab)^2)$

At least two of these three terms are divisible by p.
Because the entire expression is divisible by p, the third term must be as well.
Then at least two of a, b, or c are divisible by p.
Let these be a and b WLOG.

$ 1/a^2 < \equal{} 1/p^2$
$ 1/b^2 < \equal{} 1/p^2$
$ 1/c^2 > \equal{} 1/p \minus{} 2/p^2$
$ c^2 < \equal{} \frac {p^2}{p \minus{} 2}$

If a, b, and c are all odd, then p is of the form 4n+3.
At least two of a, b, or c are divisible by 3.

Not sure what else...


Let x, y, z>0. Find min:
\[ \frac{4x}{x\plus{}y\plus{}2z} \plus{}\frac{y\plus{}3z}{2x\plus{}y\plus{}z} \minus{}\frac{8x}{x\plus{}y\plus{}3z}\]

It's really difficult to solve this ex. I see it is strange to us.

I'm not positive but it seems like if y and z are very close to 0 and x is really large you will get close to -4 and i don't see how it could get any smaller than that because if you have a large enough y and z (compared to x) it won't be negative because of the middle term.

Oh, thank you but i really need a fully done solution. It's a nice and interesting math problem. But i can't figure any solution for it.

[hide="Solution"]
\[ \frac {4x}{x \plus{} y \plus{} 2z} \plus{} \frac {y \plus{} 3z}{2x \plus{} y \plus{} z} \minus{} \frac {8x}{x \plus{} y \plus{} 3z} \\
 \\
 \\
> \frac {4x}{x \plus{} y \plus{} 2z} \minus{} \frac {8x}{x \plus{} y \plus{} 3z} \\
 \\
 \\
\equal{} 4\left(\frac {x^2 \plus{} xy \plus{} 3xz \minus{} 2x^2 \minus{} 2xy \minus{} 4xz}{(x \plus{} y \plus{} 2z)(x \plus{} y \plus{} 3z)}\right) \\
 \\
 \\
\equal{} \minus{} 4\left(\frac {x(x \plus{} y \plus{} z)}{(x \plus{} y \plus{} 2z)(x \plus{} y \plus{} 3z)}\right) \\
 \\
 \\
> \minus{} 4\left(\frac {x(x \plus{} y \plus{} z)}{x(x \plus{} y \plus{} z)}\right) \\
 \\
 \\
\equal{} \boxed{ \minus{} 4}.\]
But
\[ \lim_{y,z\to0}\frac {4x}{x \plus{} y \plus{} 2z} \plus{} \frac {y \plus{} 3z}{2x \plus{} y \plus{} z} \minus{} \frac {8x}{x \plus{} y \plus{} 3z}\equal{}\minus{}4\]
as well, so we're done.[/hide]


Triangle ABC is acute-angled.
Perpendiculars are drawn from the vertices to the opposite sides and extended to meet the circumcircle.
If the extended parts of the perpenculars from A, B and C  have lengths  $\alpha$, $\beta$ ,$\gamma$ respectively, find the least value of :

$\frac{a}{\alpha}$ +$\frac{b}{\beta}$ +$\frac{c}{\gamma}$

[hide]Answer $\sqrt 108$[/hide]

Uhm...3 by AM-GM and equality when the triangle is equilateral? Or did I misunderstand the question?


What is the last digit of $ (7^{5})^{3}$?

$ (7^5)^3 \equal{} 7^{15}$

We look for a pattern for the last digits of the powers of 7.

$ 7^1: 7$
$ 7^2: 9$
$ 7^3: 3$
$ 7^4: 1$
$ 7^5: 7$
$ 7^6: 9$

And now we see that the pattern is $ 7, 9, 3, 1, 7, 9 \cdots$
So the 15th term of this pattern is $ 7,9,3,1,7,9,3,1,7,9,3,1,7,9, \boxed {3}$
We could have just done $ \dfrac {15}{4}$ and found the remainder is 3, then found the 3rd term of the pattern, which is $ 7,9, \boxed {3}$

[hide="Click here"]

The pattern for 7 is 7,9,3,1. Thus 7^5 last digit is 7, and  the pattern repeats. Since the number is cubed, the last digit is the third number in our pattern, which is [b]3[/b].[/hide]


algebra
polynomial
quadratics
function
analytic geometry
graphing lines
slope

Find the quadratic polynomial that best approximates a function f near x0 = 1 if the tangent line at x0 = 1 has equation y = 4x - 1 and if f''(1) = 6

[hide] from Taylor's theorem, f(x) = P(x) + R(x) and f(x) = f(1) + f'(1)(x-1) + 3(x-1)^2 + R(x)

Thus, as approximation,
f(x) = f(1) + f'(1)(x-1) + 3(x-1)^2
f'(x) = f'(1) + 6(x-1)
f''(x) = 6

Am I in the good track? [/hide]

You're part way there. Now, what are $ f(1)$ and $ f'(1)?$ (There is sufficient information given to determine that.)

[hide]
f''(x) = f''(1) = 6
f'(1) = 4, from the slope of the tangent line at x = 1
at x = 1, y = 4(1) - 1 = 3, so f(1) = 3

So, the quadratic polynomial is f(x) = 3 + 4(x-1) + 3(x-1)^2

[/hide]

pareto wrote :

"Thus, as approximation,
f(x) = f(1) + f'(1)(x-1) + 3(x-1)^2
f'(x) = f'(1) + 6(x-1)
f''(x) = 6

Am I in the good track?"

R and its derivatives can’t be ignored
f(x) = f(1) + f'(1)(x-1) + 3(x-1)^2 + R(x)
f'(x) = f'(1) + 6(x-1) + R’(x)
f''(x) = 6+ R”(x)


[quote="vamathletes"]a problem: kind of easy, kind of hard[/quote]

This has been once posted some while ago.

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=133862[/url]


math was easy except for 5-6 questions
physics was fun :D
chem was stupid(the last 5ques in D..all inorganic..i had no idea)

Yeah..
Guessed the inorganic questions myself..
Anyone knows on what site the solutions will be available

i don kno..but why was physics soo easy?
i think i may have made too many mistakes :(

I m pretty much in same situation...

[url]http://www.prernaclasses.com/AIEEESolutions2009.htm[/url]
not much authentic though!!

maths..well i resorted to my guessing :rotfl:

yes i too wanted to kno the same....(i m not interested though :rotfl: )

betw,wat will b the answer of that charge density question???..... :o

that P(r) question u mean.

it was opt 4 of paper code c

really i need to kno defn of gud scor

[hide]
me gettin 340+ acc prerna, career pt
[/hide]

me getting ~~325 according to prerana :(

i m also getting above 330

but how wud i kno wat's the  4th option of code c :rotfl:
do u mean prerna's ans is correct??

a good rank in AIEEE actually does no good :rotfl:

Q r1^2 / 4 pi epsilon R^4

[quote]
a good rank in AIEEE actually does no good
[/quote]

but itz a gud morale booster for bitsat. esp after jee

lol
:D
in karnatake..a score of 300 might get ur a state rank of 30 or so so lol
no competition

not fr me..i m getting more in chem in Phy...which i dont like:p

lol
happened with jee
as well
chem 92 phy 88
O_O
and math..i had planned for 70% in phy 60% in chem and 50% in math
but math i got 101
wow
lol
and...
hmm...
lol..chem was way too easy...i missed one dumb question..shortest bond lengt.i marked longest
:(

yaar, the question in chemistry about physisorption; the  enthalpy  of  adsorption is low and positive.isn't it correct.it's written in ncert that delta H=20 - 40kJ and obviously it will low.  :huh:

so all the options are  correct

hmm..
its a mistake..but in our 1st pu tb.its given as -ve..adsorption is negative..anyways..going by logic..
more stable intermediate is formed.so energy shud be released
my total =~~330+/- 5(331 to be exact-)

It was an easy paper. And what is the slowest step in Cannizaro? Isnt it the base attack?

donno.all other options seemed to be veru wrong. so i marked tht one :D

I am getting ~380, with few silly mistakes only in physics. electron charge forgot to take -ve. And that resistance temp dependence.

[quote="rituraj007"]
a good rank in AIEEE actually does no good :rotfl:
[/quote]
IIIT Hyderabad is one of the best institutions in the country for computer science. With a closing rank of ~2800 for any course, I think it is the institution most sought by those who've cleared AIEEE.

Ofcourse, for someone like rituraj007 who is only interested in biology, AIEEE is of no good.

sorry to change the topic..but..u guys got ur top 1% certificate for nsec?

me gettin 348 to be precise (acc to narayana)
as usual screwed maths and phy,
maths 106
phy 110
chem 132 (wow for once i feel like top 1%   :winner_first:  )

http://www.ankur-gupta.com/blog/aieee/aieee-2009-cutoff

it seems more than 100 aopsians will figure in aieee top 10 according to above statistics  :D

[hide]
ankur gupta shud work wit aaj tak for predicting election results  :rotfl:
[/hide]

[quote]
IIIT Hyderabad is one of the best institutions in the country for computer science.
[/quote]
better than cse at pilani ????

:maybe:
i donno..why i do so bad.335 acording to narayna..screwed physics and mth
:(
math 90
a mere 90 :|

hmm..u guys checked the resonance solns?me getin 340 acc to dem
:O ..paper was too easy this time i suppose...a 400+scorer if u cud avoid mistakes

[quote="sumanth0811"]paper was too easy this time i suppose[/quote]

lol toi has an article on how the sudden change in pattern made aieee 2009 difficult  :rotfl:

i don't understand, what was so difficult.
paper was straightforward compared to jee 09

acc to resonance
maths 110
phy 118
chem 132
total 360  :D

according to ankur gupta i should be TOP 10!!!!
me
144 -chem
129 in physics
114 in maths

385 in all.

i m getting 330  ..i shud forget that ,just 70 in maths :rotfl:

yeah he says 375 + top 10

and already 3 over here are that much :rotfl:

btw my marks have improved

maths 119   ( it seems fiitjee has gvn wrong ans ( many more ...aakash is most accurate)for bijection Q  correctans : Both stat 1 ,stat2 are true  as for f-1(x) to exist it has to be bijection ?? )
chem  124
phy    144

[quote="_MM_"]Is it true that the AIEEE results are coming out today??? :maybe:[/quote]

No.

I think June 5th or 6th.

so what about this:::

http://results.gov.in/result2announced.aspx

My friends told me that it was mentioned in some newspaper as well :huh:

EDIT : Its now official  http://cbseresults.nic.in/

Results already out!!!

http://www.cbseresults.nic.in

Nora...now u can post ur score :P :rotfl:

[quote="rituraj007"]:ninja:  :mad:[/quote]
 :huh: .

[quote]  :ninja:  :mad: [/quote]

:wacko:    :what?:    :huuh:

[quote="_MM_"]

:wacko:    :what?:    :huuh:[/quote]
?


LaTeX
trigonometry
geometry
geometric transformation
reflection
function
limit

Let X  is a point inside the triangle ABC (AC≠BC) ,
    ∠A=α,   ∠B=β,    ∠ACX=ϕ,   and ∠BCX=ψ.
    Prove that    ((sinαsinβ)/(sin(α-β)))=((sinϕsinψ)/(sin(ϕ-ψ)))   if and only if  X is laying on the median from C.

If you just type with LATEX code sandwiched between dollar signs,

[code]Let $X$ be a point inside the $\triangle ABC$ ($AC \neq BC$), $\angle A = \alpha$, $\angle B = \beta$, $\angle ACX = \phi$, and $\angle BCX = \psi$.
Prove that $\frac{\sin \alpha \sin \beta}{\sin (\alpha - \beta)} = \frac{\sin \phi \sin \psi}{\sin (\phi - \psi)}$, if and only if $X$ is laying on the median from $C$.[/code]
it will show as:

Let $ X$ be a point inside the $ \triangle ABC$ ($ AC \neq BC$), $ \angle A \equal{} \alpha$, $ \angle B \equal{} \beta$, $ \angle ACX \equal{} \phi$, and $ \angle BCX \equal{} \psi$.
Prove that $ \frac{\sin \alpha \sin \beta}{\sin (\alpha \minus{} \beta)} \equal{} \frac{\sin \phi \sin \psi}{\sin (\phi \minus{} \psi)}$, if and only if $ X$ is laying on the median from $ C$.

$ \frac {\sin \alpha \sin \beta}{\sin (\alpha \minus{} \beta)} \equal{} \frac {\sin \phi \sin \psi}{\sin (\phi \minus{} \psi)}\ \Longleftrightarrow\ \frac {\sin (\alpha \minus{} \beta)}{\sin \alpha \sin \beta} \equal{} \frac {\sin (\phi \minus{} \psi)}{\sin \phi \sin \psi}\ \Longleftrightarrow$

$ \cot \beta \minus{} \cot \alpha \equal{} \cot \psi \minus{} \cot \phi$.

WLOG, assume $ \alpha > \beta$. Let $ M$ be the midpoint of $ (AB)$ and $ F$ the foot of the altitude from $ C$.
Denote $ \angle ACM \equal{} \phi_0$, $ \angle BCM \equal{} \psi_0$ and $ \angle AMC \equal{} \mu_0$, $ \angle BMC \equal{} \nu_0$, where $ \mu_0 \plus{} \nu_0 \equal{} 180^\circ$.
Reflect $ A$ in $ CF$ into $ A'$. Then $ (A'B) \equal{} 2 (FM)$.

$ \cot \beta \minus{} \cot \alpha \equal{} \cot \widehat{FBC} \minus{} \cot \widehat{FA'C} \equal{} \frac {(FB) \minus{} (FA')}{(CF)} \equal{} \frac {(A'B)}{(CF)} \equal{}$

$ \equal{} \frac {2 (FM)}{(CF)} \equal{} 2 \cot \widehat{FMC} \equal{} 2 \cot \mu_0 \equal{} \cot \mu_0 \minus{} \cot \nu_0\ \Longrightarrow$

$ \cot \alpha \plus{} \cot \mu_0 \equal{} \cot \beta \plus{} \cot \nu_0$

$ \cot \psi_0 \minus{} \cot \phi_0 \equal{} \cot (\alpha \plus{} \mu_0) \minus{} \cot (\beta \plus{} \nu_0) \equal{} \frac {\cot \alpha \cot \mu_0 \minus{} 1}{\cot \alpha \plus{} \cot \mu_0} \minus{} \frac {\cot \beta \cot \nu_0 \minus{} 1}{\cot \beta \plus{} \cot \nu_0} \equal{}$

$ \equal{} \frac {\cot \alpha \cot \mu_0 \minus{} \cot \beta \cot \nu_0}{\cot \alpha \plus{} \cot \mu_0} \equal{} \frac {\cot \alpha \cot \mu_0 \plus{} (\cot \alpha \plus{} 2 \cot \mu_0) \cot \mu_0}{\cot \alpha \plus{} \cot \mu_0} \equal{}$

$ \equal{} 2 \cot \mu_0 \equal{} \cot \mu_0 \minus{} \cot \nu_0 \equal{} \cot \beta \minus{} \cot \alpha$

Let $ X'$ be the trace of $ X$ in $ (AB)$. $ f(X') \equal{} \cot \psi \minus{} \cot \phi$ is a strictly increasing function of $ X'$, when $ X'$ moves from $ A$ to $ B$, with $ \lim_{X' \rightarrow A} f(X') \equal{} \minus{} \infty$, $ \lim_{X' \rightarrow B} f(X') \equal{} \plus{} \infty$. Therefore, $ f(X') \equal{} \cot \beta \minus{} \cot \alpha$ for a unique $ X' \in (AB)$. But $ f(M) \equal{} \cot \psi_0 \minus{} \cot \phi_0 \equal{} \cot \beta \minus{} \cot \alpha$.

[size=134][color=darkblue][b]I modified a bit the enunciation for to be it more simply and more clearly ![/b][/color][/size]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle , $ A\ne B$ and let $ P\in (AB)$ be a point for which denote $ m(\widehat {ACP}) = x$ and $ m(\widehat {BCP}) = y$ .

Prove that $ \frac {\sin A\sin B}{\sin (A - B)} = \frac {\sin x \sin y}{\sin (x - y)}$ if and only if $ PA = PB$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g $ A > B$ . Hence and $ x > y$ . Denote the circumcircle $ w$ of the $ \triangle ABC$ and the second point

$ M\in CP\cap w$ . Construct the points $ X\in w$ , $ Y\in w$ for which $ MX\parallel AB\parallel CY$ . Observe that $ m(\widehat {CMY}) = A - B$ ,

$ m(\widehat {XYM}) = x - y$  and $ \boxed {\ \frac {PC}{PM} = \frac {CB\cdot CA}{MA\cdot MB}\ }$ $ \Longleftrightarrow$ $ \frac {PC}{PM} = \frac {\sin A\sin B}{\sin x\sin y}$ . Observe that $ \frac {CY}{MX} = \frac {\sin (A - B)}{\sin (x - y)}$ .

In conclusion, $ \frac {\sin A\sin B}{\sin (A - B)} = \frac {\sin x \sin y}{\sin (x - y)}$ $ \Longleftrightarrow$ $ \frac {PC}{PM} = \frac {CY}{MX}$ $ \Longleftrightarrow$ $ P\in XY$ $ \Longleftrightarrow$ $ PA = PB$ ([u]by simmetry[/u] !).

[b][u]Remark.[/u][/b] The boxed relation results immediately from the well-known property (prove easily !) :[/color]

[quote="Virgil Nicula"][color=darkred]In a cyclical convex quadrilateral $ ABCD$ denote $ \{\begin{array}{c} AB = a\ ,\ BC = b \\
 \\
CD = c\ ,\ DA = d \\
 \\
AC = e\ ,\ BD = f\end{array}$ and $ P\in AC\cap BD$ .

Then $ \frac {PA}{da} = \frac {PB}{ab} = \frac {PC}{bc} = \frac {PD}{cd} = \frac {e}{ad + bc} = \frac {f}{ab + cd} = \sqrt {\frac { - p_w(P)}{abcd}}$ ,

where $ p_w(P) = \overline {PA}\cdot\overline {PC} = \overline {PB}\cdot\overline {PD}$ is  the power of the point $ P$ w.r.t. the circle $ w$ .[/color][/quote]


hmm.. im missing something here

the probability that a randomly chosen positive divisor of 10^99 is divisible by 10^88 is m/n lowest terms. find m+n

[hide="Answer"]$10^{99}=2^{99}\cdot 5^{99}=10^{88}\cdot 2^{11}\cdot 5^{11}$
Therefore, $12\cdot 12=144$ of these factors are divisible by $10^{88}$, and there are a total of $10,000$ factors, so we have $\frac{144}{10000}=\frac{9}{625}\Rightarrow 9+625=\boxed{634}$.[/hide]

that's what i don't get
wouldn't it be 144 factors that are NOT divisible by $10^{88}$
since if the exponent of the 2 or 5 went above 88 then it would NOT be divisble by $10^{88}$

No, because for example, if it was $2\cdot10^{88}$, then it divided by $10^{88}$ is $2$.  This is the same for any combination of the $2^{11}\cdot5^{11}$ group, which is why $144$ of them ARE divisible by $10^{88}$.

Isnt it?
$\frac{89\cdot89}{100\cdot100}$
since all the factors of $10^{88}$ are factors of $10^{99}$.
something must be wrong since this is a AIME question

No; $89\cdot 89$ is the number of factors that aren't divisible by $10^{88}$, which has nothing to do with the original equation. You need to look at only factors that contain $10^{88}$.

ah see i had the same problem p.c. had
now i get it
"divisible by 10^88"
not
"the number divides 10^88"
ok thanks


algebra
polynomial
search
calculus
calculus computations

hi all could someone post a proof that evry complex polynome has a root.

Do you mean that?
http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra


Are the 2005 AMC/AIME problems out for sale yet?

Yeah, they were out for sale a long time ago.  Check the UNL/AMC website.

you can find them on the resources page also.


This is a pre-calc problem im stuck on.
Click on the image to see it large and read it.
Any help would be greatly appreciated.
[url=http://img63.exs.cx/my.php?loc=img63&image=mathproblem5oe.jpg][img]http://img63.exs.cx/img63/4647/mathproblem5oe.th.jpg[/img][/url]

Oh and incase there is a problem with this post, it is not a calculus problem and is a 10th grade high school subj. which I am in. Thx.

You're going to have to hope some very charitable person comes across your thread if you expect help on a problem as dry as this..

:rotfl: Very dry problem indeed :rotfl: Thx AntonioMainenti :rotfl:

Are you looking for help on all 10, or a specific number.

Perhaps it would be easier and less time-consuming for you to tell us where it is you get stuck, and what concepts need a clearer explanation.

[quote="cr00k in a n00k"]Thanks for the help u guys.  :no:
Anyway do you know of anyone that could help me out?[/quote]

Maybe your teacher :?:

If you show some work or ask about a specific part then you will have a higher chance of being answered.

well, can you tell me the name of the book? or some links where i can read the book online?


This is pretty much the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28142]kunny's problem[/url], only easier...

$A=\int_{0}^{\pi}xf(\sin x)dx$
use
$y=\pi-x$ to relate to $\int_{0}^{\pi}f(\sin x)dx$

$\int\limits_0^\pi  {\frac{{x\sin x}}{{1 + \cos ^2 x}}dx =  - \int\limits_0^\pi  {\frac{{x\left( { - \sin x} \right)}}{{1 + \cos ^2 x}}dx = } } $

$ =  - \int\limits_0^\pi  {x\left[ {\arctan \left( {\cos x} \right)} \right]^\prime  dx = } $

$ =  - \left[ {x\arctan \left( {\cos x} \right)\left| \begin{array}{l}
 \pi  \\
 0 \\
 \end{array} \right. - \int\limits_0^\pi  {\arctan \left( {\cos x} \right)dx} } \right] = $

$\mathop  = \limits_{\mathop {\scriptstyle \varphi \left( x \right) = \arctan \left( {\cos x} \right) \Rightarrow  \hfill \atop
  {\scriptstyle  \Rightarrow \varphi \left( {\frac{\pi }{2} - h} \right) =- \varphi \left( {\frac{\pi }{2} + h} \right),\forall h \in \left( {0,\frac{\pi }{2}} \right) \Rightarrow  \hfill \atop
  \scriptstyle  \Rightarrow \int\limits_0^\pi  {\arctan \left( {\cos x} \right)dx = 0}  \hfill}}\limits^ \uparrow  }  - \left[ {x\arctan \left( {\cos x} \right)} \right]_0^\pi   = \frac{{\pi ^2 }}{4}$


Find all non empty subset $ S$ of $ \mathbb{N}: \equal{} \{0,1,2,\ldots\}$ such that $ 0 \in S$ and exist two function $ h(\cdot): S \times S \to S$ and $ k(\cdot): S \to S$ which respect the following rules:
i) $ k(x) \equal{} h(0,x)$ for all $ x \in S$
ii) $ k(0) \equal{} 0$
iii) $ h(k(x_1),x_2) \equal{} x_1$ for all $ x_1,x_2 \in S$.
[i](Pierfrancesco Carlucci)[/i]

[hide]Substituting $ x_1 \equal{} 0$ into (iii) yields $ h(k(0), x_2) \equal{} 0$ for all $ x_2 \in S$. But (ii) tells us that $ k(0) \equal{} 0$, so $ h(0, x_2) \equal{} 0$ for all $ x_2 \in S$. But (i) tells us that $ k(x_2) \equal{} h(0, x_2)$ for all $ x_2 \in S$, implying that $ k(x) \equal{} 0$ for all $ x \in S$. But then, $ h(0, x_2) \equal{} 0 \equal{} h(k(x_1), x_2) \equal{} x_1$ for all $ x_1 \in S$, implying that $ x_1 \equal{} 0$ for all $ x_1 \in S$. In other words, the only possible set $ S$ can be is $ \{0\}$. In addition, the set $ \{0\}$ trivially satisfies the conditions of our problem (by setting $ k(0) \equal{} 0$ and $ h(0,0) \equal{} 0$), so we are done. [/hide]

k(.) is constant and injective since k(x)=h(0,x)=h(k(0),x)=x and x=h(k(x),0)=h(0,0)=k(0)=0 for all x in S,so S={0}.


algebra
polynomial
Vieta
Vieta Jumping
IMO Shortlist
IMO
IMO 2007

Let $a$ and $b$ be positive integers. Show that if $4ab - 1$ divides $(4a^{2} - 1)^{2}$, then $a = b$.

[i]Author: Kevin Buzzard and Edward Crane, United Kingdom [/i]

Standard problem using Vieta Jumping (finally I know how this method is called - big thx to the german team :D).

Standard manipulations of $ 4ab-1\mid (4a^{2}-1)^{2}$ shows that this implies $ 4ab-1\mid (a-b)^{2}$.
Let's assume now that there exist some postive integers $ a\neq b$ so that $ 4ab-1\mid (a-b)^{2}$. Let $ k=(a-b)^{2}/(4ab-1)$.

Let $ S=\{ (a,b): a,b\in \mathbb{Z}^{+}, (a-b)^{2}/(4ab-1)=k\}$. Since $ \mid S\mid \geq 1$, there is a pair $ (A,B)\in S$ which minimizes $ a+b$ over all $ (a,b)\in S$. wlog assume that $ A > B$. Consider now the quadratic equation
\[ \frac{(x-B)^{2}}{4xB-1}=k \ \ \ \Leftrightarrow x^{2}-x(2B+4kB)+B^{2}+k=0 \]
which has roots $ x_{1}=A$ (obviously) and $ x_{2}=2B+4kB-A=(B^{2}+k)/A$ (from Vieta). This implies that $ x_{2}\in \mathbb{Z}^{+}$, so $ (x_{2}, B)\in S$. By the minimality of $ A+B$, we however have $ x_{2}\geq A$, ie $ (B^{2}+k)/A\geq A$ and therefore $ k\geq A^{2}-B^{2}$.

Thus, $ (A-B)^{2}/(4AB-1)=k\geq A^{2}-B^{2}$ and hence $ A-B\geq (A+B)(4AB-1)$, clearly impossible for $ A,B\in\mathbb{Z}^{+}$. $ \Box$

Let $ a-b=k\Longleftrightarrow b=a-k>0$ $ \therefore k<a$.
$ 4ab-1=4a(a-k)-1=4a^{2}-4ka-1$, Dividing $ (4a^{2}-1)^{2}=16a^{4}-8a^{2}+1$ by $ 4ab-1=4a^{2}-4ka-1$ gives the quotient of $ 4a^{2}+4ka+4k^{2}-1$ and residue of $ 4k^{2}(4ka+1)$. $ \therefore 4ab-1|(4a^{2}-1)^{2}\Longleftrightarrow k=0$ or $ 4ka=-1$. Since $ k,\ a$ are integers, yielding $ k=0\Longleftrightarrow a=b$. Q.E.D.

If $ 4ab-1|4a^{2}-1$ it follows that $ 4a^{2}-1=4abk-k$ and $ kb=\frac{4a^{2}+k-1}{4a}.$ If $ k=1$ we get $ a=b.$ If $ k>1$ we have that $ 4a|k-1$ and $ k-1>0$ therefore $ k\geq 4a+1.$ But now we have $ 4a^{2}-1=k(4ab-1)\geq 16a^{2}b+4ab-4a-1\geq 16a^{2}-1$ which is a contradiction.
Let's suppose that there is some $ p-$ prime divisor of $ 4ab-1$ such that if $ p^{\alpha}\parallel 4a^{2}-1$ and $ p^{\beta}\parallel 4ab-1$ then $ \beta>\alpha.$ Since $ p^{\beta}|(4a^{2}-1)^{2}$ it follows that $ b\leq 2\alpha.$ Since $ GCD(2a-1,2a+1)=1$ we have that either $ p^{\alpha}\parallel 2a-1$ or $ p^{\alpha}\parallel 2a+1.$ In both cases using that $ p^{\alpha}|4ab-1$ we get that either $ p^{\alpha}|2b-1$ or $ p^{\alpha}|2b+1.$ Since $ \beta\leq 2\alpha$ and the above we conclude that $ 4ab-1|(4a^{2}-1)(4b^{2}-1)=16a^{2}b^{2}-1-4a^{2}-4b^{2}+2$ and thus \[ 4ab-1|2a^{2}+2b^{2}-1.\] Therefore  there exists some $ m\in\mathbb{N}$ such that \[ (*)\ \ \ \ 2b^{2}-4abm+2a^{2}+m-1=0.\] Suppose that there is at least one couple $ (a,b)$ with $ a<b.$ We choose such couple for which $ a-min$ and $ b$ is the maximal for this $ a.$ Using Viete we get that there is some $ b'\in\mathbb{N}$ which is solution with $ a.$ If $ b'<a$ we get a contradiction for $ a.$ If $ b'<b$ we get a contradiction for $ b.$ thus $ b'=a$ or $ b'=b.$ If $ b'=a$ we get $ m=1$ and $ b=a.$ If $ b'=b$ it follows that the descriminant of (*) should be 0, or $ 4a^{2}m^{2}-4a^{2}-2m+2=0$ or $ 4a^{2}=\frac{2m-2}{m^{2}}$ and therefore $ m=1$ and again $ a=b.$

Hope it's correct.

Nice problem. This one will go into PEN soon. :) A little easy for a number 5 though - although not as easy as Kunny pretends it to be.

$ 4ab-1|b^{2}(4a^{2}-1)^{2}$ implies $ 4ab-1|(a-b)^{2}$.
$ (a-b)^{2}=(4ab-1)k$ so that $ a^{2}-(2b+4kb)a+b^{2}+k=0$.
assume there is $ (a,b)$ satisfying this condition and $ a \neq b$. wlog, let $ a>b$. then by Vieta's theorem, $ (b, \frac{b^{2}+k}{a})$ is a solution too, since $ \frac{b^{2}+k}{a}=2b+4kb$ is a positive integer. we can check that $ \frac{b^{2}+k}{a}<b$ bu the direct substitution of $ k$. therefore, if there is a solution $ (a,b)$ with minimum $ b$, we can get a solution with smaller minimum, contradiction.

[quote="iwan_no2000"][quote="gabrielsebastian"][b]LEMMA 1[/b]
Call $ (a,b)$ a "good" pair if it satisfies $ 4ab-1|(4a^{2}-1)^{2}$. If $ (a,b)$ is good, then so does $ (b,a)$

[b]LEMMA 2[/b]
Call an integer $ m$ "great" if it satisfies
i) $ m|(4a^{2}-1)^{2}$
ii) $ m \equiv-1 (mod 4a)$
If $ m$ is great, then so does $ \frac{(4a^{2}-1)^{2}}{m}$

[Karena bukti kedua lemma tidak sulit, maka bukti diserahkan pada pembaca]

We'll use induction on $ a$ so that if $ (a,b)$ is good, then $ a = b$.
If $ a = 1$, then $ 4b-1|9$ and it's clear that $ 4b-1 = 3 \Leftrightarrow b = 1 = a$.
Assume it's true for any $ a < n$.
Consider a good pair $ (n,b)$ (This pair exists since setting $ n = b$ works well).
If $ n < b$, then let $ R = 4nb-1$. By Lemma 2, we have $ S = \frac{(4n^{2}-1)^{2}}{R}$ satisfies $ S \equiv-1 (mod 4a)$. Set $ b' = \frac{S+1}{4a}$. Since $ b > n$, then $ R > 4n^{2}-1$ and thus $ S < 4n^{2}-1 \Leftrightarrow b' < n$. Hence, $ (n,b')$ is good, too.
In particular, any good pair $ (n,b)$ with $ n < b$ correspondences to a good pair $ (n,b')$ with $ n > b'$.
Now, consider the case if $ n > b$. By Lemma 1, we have $ (b,n)$ is a good pair, too. But, by induction hypothesis, all good pair $ (x,y)$ with $ x < a$ satisfies $ x = y$. Since $ b < a$, then $ b = n$, contradicting the condition $ b < n$.
Now, we are left for the case $ n = b$. Indeed, this is true since $ 4nb-1 = 4n^{2}-1$ divides $ (4n^{2}-1)^{2}$.
This completes our induction and our proof.[/quote]

What does "Karena bukti kedua lemma tidak sulit, maka bukti diserahkan pada pembaca" mean?[/quote]

I'm sorry, he just copied the solution from Indonesian forum. It means that because the proof for both lemmas aren't too hard, so it's given to the reader..

[quote="kunny"][quote="milo"]kunny, how can you be sure that $ k=0$ or $ 4ka =-1$?

As far as I can see, you only get that $ 4ab-1$ divides $ 4k^{2}(4ka+1)$, not $ 4k^{2}(4ka+1) = 0$...[/quote]

[quote="kunny"]Let $ a-b=k\Longleftrightarrow b=a-k>0$ $ \therefore k<a$.
$ 4ab-1=4a(a-k)-1=4a^{2}-4ka-1$, Dividing $ (4a^{2}-1)^{2}=16a^{4}-8a^{2}+1$ by $ 4ab-1=4a^{2}-4ka-1$ gives the quotient of $ 4a^{2}+4ka+4k^{2}-1$ and residue of $ 4k^{2}(4ka+1)$.[/quote]

$ \therefore (4a^{2}-1)^{2}=(4ab-1)\{4a^{2}+4(a-b)a+4(a-b)^{2}-1\}+4(a-b)^{2}\{4(a-b)+1\}$
$ =(4ab-1)(8a^{2}-4ab-1)+16a^{2}(a-b)^{2}.$
Since we have $ 8a^{2}-4ab=4a(2a-b)\neq 1$ for $ a,\ b\in \mathbb{N}$,
$ \therefore 4ab-1|(4a^{2}-1)^{2}\Longleftrightarrow 16a^{2}(a-b)^{2}=0$, yielding $ a=b$. Q.E.D.

How about this time?

kunny[/quote]

the same mistake as last one,how can you sure that $ 16a^{2}(a-b)^{2}=0$?

[quote="kunny"]If $ 16a^{2}(a-b)^{2}\neq 0$, then $ 4ab-1$ is not a divisor of $ (4a^{2}-1)^{2}$, so $ 16a^{2}(a-b)^{2}=0$, yielding $ a=b$.
Is this a right reasoning, or wrong the whole of my solution basically?[/quote]

You have only shown that $ 4ab-1$ divides $ 16a^{2}(a-b)^{2}$.
Since $ 4ab-1$ and $ 16a^{2}$ are relative prime, this implies that $ 4ab-1$ divides $ (a-b)^{2}$.

The main part of the argument is still missing.

Fairly standard vieta jumping.

First, note that
$$4ab-1 \mid (4a^2-1)^2 \implies 4ab-1 \mid (4a^2-4ab)^2 \implies 4ab-1 \mid (a-b)^2,$$
since $\gcd(4a,4ab-1)=1$. Hence we only have to show that if $4ab-1 \mid (a-b)^2$, then $a=b$. Suppose otherwise, namely that there exists some pair of distinct positive integers $(a,b)$ which satisfy the divisibility condition. Call such pairs "good". Among all good pairs, choose the pair $(A,B)$ such that $A>B$ and $A+B$ is minimal. Now let $0<k=\tfrac{(a-b)^2}{4ab-1}$. Consider the quadratic
$$\frac{(x-B)^2}{4xB-1}=k \iff x^2-B(4k+2)x+(B^2+k)=0.$$
It is clear that one root $x_1$ of this quadratic is $A$. Let the other root be $x_2$. From Vieta's, we have $A+x_2=B(4k+2)$ and $Ax_2=B^2+k$. The first equation implies $x_2$ is an integer, and the second implies that $x_2$ is positive, so $(x_2,B)$ is also good. By the minimality of $(A,B)$, we must have $x_2 \geq A$. This means that
$$B^2+k=Ax_2 \geq A^2 \implies k \geq A^2-B^2.$$
On the other hand, we have
$$\frac{(A-B)^2}{4AB-1}=k \geq A^2-B^2 \implies (A-B)^2\geq (A^2-B^2)(4AB-1) \implies A-B \geq (A+B)(4AB-1),$$
which is impossible for $A,B \in \mathbb{Z^+}$ (we divide by the positive $A-B$ to reach the third inequality). Hence no good pairs can exist, as desired. $\blacksquare$

Another solution by Vieta Jumping.
Thank's the hint by #2

$4ab-1\mid(4a^2-1)^2b^2=(4a^2b-b)^2\Rightarrow\dfrac{(a-b)^2}{4ab-1}\in\mathbb{Z}$
Let $S=\{(a,b)\in\mathbb{N}^2:a<b,\dfrac{(a-b)^2}{4ab-1}\in\mathbb{N}\}$
Let $(x,y)\in S$ s.t. $\forall\,(x',y')\in S,x<x'\lor(x=x'\land y\leq y')$
Consider the equation $f(b)=\dfrac{(x-b)^2}{4xb-1}=k$
That is $b^2-2(2k+1)xb+(x^2+k)=(b-y)(b-y')=0$ ,where $b=y,y'$ with $x<y$
$\Rightarrow\begin{cases}y+y'=2(2k+1)x\Rightarrow y'\in\mathbb{Z}\\yy'=x^2+k\Rightarrow y'>0\end{cases}\therefore y'\in\mathbb{N}$
By assuming: $y\leq y'$
$\Rightarrow 0<(x-y)(x-y')=x^2-2(2k+1)x^2+(x^2+k)=k(1-4x^2),\to\leftarrow$

Nothing new but here is my solution.

Notice that we have $(4ab-1, 4ab) = 1$, so $$4ab-1\mid (4a^2-1)^2 \implies 4ab-1 \mid (4a^2-4ab)^2 = (4a)^2(a-b)^2 \implies 4ab-1 \mid (a-b)^2.$$ Assume FTSOC that $(A, B)$ is an integer solution to $k = \frac{(a-b)^2}{4ab-1}$ with $a \ne b$, such that $A>B$ and $A$ is as small as possible. We consider the quadratic $x^2-x(2B+4Bk)+(B^2+k)$ which has $A$ as a root by our hypothesis. Notice that the other root $r = 2B+4Bk-A = \frac{B^2+k}{A}$, so $(r, B)$ is also a solution which means $A < r$. However, $$k = \frac{(A-B)^2}{4AB-1} < (A-B)^2 \le A^2-AB \le A^2-B^2$$ yet $r > A \implies \frac{B^2+k}{A} > A \implies k > A^2-B^2$, a contradiction.

Therefore, this shows that $a = b$ as desired. $\blacksquare$

Note \[4ab-1|(4a^2-1)^2 \implies 4ab-1|(4a^2b-b)^2 \implies 4ab-1|(a-b)^2\] where we can perform the second step as $\gcd(4ab-1,b^2)=1$. Now we Vieta jump. Consider a pair $(a,b)$ that works and FTSOC assume $a\neq b$ and WLOG $a>b$. Let $k=\frac{(a-b)^2}{4ab-1}$. Then we have \[a^2-a(2b+4kb)+b^2k=0\], so if $(a,b)$ is a solution, then $(\frac{b^2+k}{a},b)$ is also a solution. Now consider the pair $(a,b)$ with $a>b$ WLOG and let it be the pair with the minimum maximum value of the pair (i.e. $a$, or the max value of the pair, is the minimum across all pairs). Then we also need to have $\frac{b^2+k}{a}\geq a$. We show this is impossible.

Now first note that $a  \geq k$. To prove this, note we need to show \[a(4ab-1) \geq (a-b)^2 \implies 4a^2b \geq a^2+b^2 \implies 2a^2+2a^2 \geq a^2+b^2\] which is obvious. Thus, we need to show that $\frac{b^2+a}{a} \geq a$ or $b^2+a \geq a^2$ is impossible. We do this by maximizing $b$ i.e. set it to $a-1$. This turns the inequality to $(a-1)^2+a \geq a^2 \implies a^2+1 \geq a^2+a$ which is a contradiction as $a>1$, so we are done.

[hide=Solution]Let $(a,b)$ be the smallest solution to our equation.

[b][color=red]Claim:[/color][/b] $4ab-1\mid(a-b)^2.$
[i]Proof:[/i] We know $4ab\equiv 1\pmod{4ab-1}$ and $$(4a^2-1)^2\equiv 0\pmod{4ab-1}\implies ((4ab)a-b)^2\equiv 0\pmod{4ab-1}.$$ Thus, $(a-b)^2\equiv 0\pmod{4ab-1},$ as desired. $\blacksquare$

FTSOC assume that $a\neq b,$ and assume WLOG $a>b.$ We know $(a-b)^2=c(4ab-1),$ or $a^2-2b(1+2c)+(b^2+c)=0.$ Let $a'$ be the second solution for $a$ to this equation. Notice that $(a',b)$ is also a solution, so $a\le a'.$ Hence, $$a\le \frac{b^2+c}{a}\implies a^2-b^2\le c\le(a-b)^2=a^2-2ab+b^2.$$ Therefore, $2ab\le 2b^2\implies a\le b,$ a contradiction. $\square$[/hide]

Consider a solution $a,b$ with $a+b$ minimal.
$4ab-1\mid(4a^2-1)^2$ implies $4ab-1\mid(a-b)^2$, so let $\frac{(a-b)^2}{4ab-1}=k$. Assume FTSOC that $a\ne b$, WLOG $a>b$. We have:
$$a^2+(-4kb-2b)a+b^2+k=0.$$
There is another solution $a_0$ with:
$$a_0=\underbrace{4kb+2b-a}_{\in\mathbb Z}=\underbrace{\frac{b^2+k}a}_{>0}.$$
Thus $a_0\in\mathbb N$. We have $(2ab-1)(b-a)<0$ and hence the inequalities:
$$2ab^2+a-b<2a^2b$$
$$2b^3+a-b<2a^2b$$
$$2b(a-b)^2+4ab^2-b<4a^2b-a$$
$$\frac{2b(a-b)^2}{4ab-1}+b<a$$
$$4kb+2b-a<a$$
$$a_0<a,$$
contradiction.

[hide=Solution]
Since $4ab-1\mid16a^4-8a^2+1\mid256a^4b^4-128a^2b^4+16$, this means that $0\equiv256a^4b^4-128a^2b^4+16\equiv1-8b^2+16b^4\equiv(4b^2-1)^2\pmod{4a}$, so $4ab-1\mid(4a^2-1)^2$ if and only if $4ab-1\mid(4b^2-1)^2$. Suppose that there exists a solution with $a\neq b$ such that $a+b$ is minimal, and assume without loss of generality $a<b$. Then, we claim that $b'=\frac{\frac{(4a^2-1)^2}{4ab-1}+1}{4a}$ satisfies $4ab'-1\mid(4a^2-1)^2$. We have $4ab'-1=\frac{(4a^2-1)^2}{4ab-1}$, so it divides $(4a^2-1)^2$. Since the numerator of the fraction is an integer, it suffices to show that the numerator is divisible by $4a$, which is true since $\frac{(4a^2-1)^2}{4ab-1}\equiv\frac1{-1}\equiv-1\pmod{4a}$. Therefore, since $(4ab'-1)(4ab-1)=(4a^2-1)^2$ and $b>a$, this means that $b'<a$, so $a+b'<a+b$, which is a contradiction.

Therefore, we must have $a=b$.

See that:
$$4ab-1\mid(4a^2-1)^2\implies4ab-1\mid(4a^2-1)^2-(4ab-1)(4a^3+a^2-2ab)=(a-b)^2.$$
Suppose, FTSOC, that there exists a solutions $(a_0,b_0)$, with $a_0+b_0$ minimal and, WLOG, $a_0>b_0$. We have:
$$\frac{(a_0-b_0)^2}{4a_0b_0-1}=k\implies a_0^2-a_0(2b_0+4b_0k)+b_0^2+k=0,$$
and if $x$ is the other root of this quadratic, we have $x+a_0=2b_0+4b_0k$, which implies $x\in\mathbb{Z}$, and $x=\frac{b^2+k}{a_0}$. But since $a_0,b_0>0$, we have $k<(a_0-b_0)^2=(a^2-a_0b_0)+b_0(b_0-a_0)<a_0^2-b_0^2\implies x=\frac{b^2+k}{a_0}<\frac{a_0^2}{a_0}=a_0$, which is a contradiction if $x\neq b_0$. But if $x=b_0$, we have $a_0=b_0(4k+1)$ and $4b_0^2(4k+1)-1\mid16b_0^2k^2\implies4b_0^2(4k+1)-1\mid4b_0^2k-k$, which is false by its size. $\blacksquare$

Observe that $$4ab - 1\mid (4a^2-1)^2 \iff 4ab-1 \mid (4a^2b-b)^2 \iff 4ab-1 \mid (a-b)^2.$$ Next let $a+b$ be minimal and suppose $b \neq a$. Then there exists a positive integer $c$ such that $$c(4ab-1) = (a-b)^2 \iff a^2-(2b+4bc)a + b^2+c=0.$$ Thus, the positive integer solution $\left(\frac{b^2+c}a, b\right)$ works as well, and by minimality we must have $\frac{b^2+c}a \geq a$. So $$b^2+(a-b)^2 \geq b^2+c \geq a^2 \iff a^2-b^2 \leq a-b,$$ which is impossible; thus we have a contradiction.

Note that since $\gcd(b,4ab-1)=1$, we have $4ab-1\mid b^2(4a^2-1)^2$. Expanding, we see that $b^2(4a^2-1)^2\equiv (a-b)^2 \pmod{4ab-1}$, so we get $4ab-1\mid (a-b)^2$. Now assume for contradiction that $a\neq b$, and WLOG, $a>b$. Fix $k$,and consider all positive integer values of $x$ such that $k=\tfrac{(x-b)^2}{4bx-1}$. This is $$x^2-2bx-4bkx+b^2+k=0.$$ We know that $a$ is a solution, and there is another solution $x_1$. Since by Vieta's, we have $x_1=2b+4bk-a$, so $x_1$ is an integer, since $k,a,b$ are integers. Again by Vieta's, we have $x_1=\tfrac{b^2+k}{a}$, which is positive, so $x_1$ is a positive integer.

    We now prove that $x_1<a$. We want to show that $x_1=\tfrac{b^2+k}{a}<a$, which is equivalent to showing that $k=\tfrac{(a-b)^2}{4ab-1}<a^2-b^2$. Suppose for the sake of contradiction that $k>a^2-b^2$, then we get $\tfrac{a-b}{4ab-1}>a+b$, since $a-b>0$ by assumption. Multiplying by $4ab-1$ (which is positive since $a,b$ are positive integers), expanding, and simplifying, we get $a>2ab+4b^2$. This is clearly false, since $a,b$ are positive integers and $2ab>a$, so $k<a^2-b^2$, i.e. $x_1<a$.

    Thus if an ordered pair $(a,b)$ of distinct positive integers satisfies $k=\tfrac{(a-b)^2}{4ab-1}$, where $k$ is a positive integer, then a smaller ordered pair of positive integers also satisfies this relation, which is a contradiction due to infinite descent. Therefore $a\neq b$ is impossible, i.e. we must have $a=b$.

We have $$(4ab-1)|(16a^4-8a^2+1)$$ $$(4ab-1)|(16a^4b-8a^2b+b)$$ $$(4ab-1)|(4a^3-8a^2b+b)$$ $$(4ab-1)|(4a^3-2a+b)$$ $$(4ab-1)|(4a^3b-2ab+b^2)$$ $$(4ab-1)|(a^2-2ab+b^2)$$ $$(4ab-1)|(a-b)^2.$$\\

Note that by symmetry, if $(a,b)$ is a solution, then $(b,a)$ is also a solution. We will then use Vieta jumping.\\

Claim: If $n=\frac{(a-b)^2}{4ab-1}$, then $\frac{b^2+n}{a}<a.$ Expanding this out, we get $n<a^2-b^2$, or $(a-b)^2<(a+b)(a-b)(4ab-1)$ so $(a-b)<(a+b)(4ab-1)$, which is obviously true since $a+b>a-b$ and $4ab-1>1.$\\

Letting $n=\frac{(a-b)^2}{4ab-1},$ we have that $$a^2+a(-2b-4bn)+b^2+n.$$ Suppose FTOSC that there exists a solution with $a>b$ (symmetry can be used for $a<b$), and further suppose that $(a_1,b_1)$ is the solution with the smallest value of $a_1$ that has $a_1>b_1$. If $(a_1,b_1)$ is a solution, by Vieta jumping, setting $a$ equal to $2b_1+4b_1n-a_1=\frac{b_1^2+n}{a_1}$ is a solution, so $(\frac{b_1^2+n}{a_1},b_1)$ is also a solution. However, we have that $\frac{b_1^2+n}{a_1}<a_1$ by our claim, contradicting the minimality of $a$.

Firstly observe that $(a-b)^2\equiv (4a^2b-b)^2\equiv (4a^2-1)^2 \pmod {4ab-1}.$ Assume $a>b$ and denote by $(A,B)$ the solution of $\frac{(a-b)^2}{4ab-1}=k\in \mathbb{Z}$ which minimizes $a+b$. We have a Vieta jump $$(A,B)\rightarrow (A_0,B)=(4Bk+2B-A,B)=\left( \frac{B^2+k}{A},B\right),$$ with $A_0\in \mathbb{Z}^+.$ But clearly $2Bk<A-B\implies A_0+B<A+B,$ the contradiction completes our proof.

If $4ab - 1$ divides $(4a^{2} - 1)^{2}$, then we have:

$$\frac{(4a^2 - 1)^2}{4ab - 1} = k$$

where $k$ is a positive integer.

Expanding the left side:

$$\frac{16a^4 - 8a^2 + 1}{4ab - 1} = k$$

Multiplying both sides by $4ab - 1$:

$$16a^4 - 8a^2 + 1 = k(4ab - 1)$$

Expanding the right side:

$$16a^4 - 8a^2 + 1 = 4kab - k$$

Isolating $16a^4$:

$$16a^4 = 4kab + 8a^2 + k - 1$$

Since $a$ and $b$ are positive integers, both sides must be positive, which implies that $k > 0$.

Thus, $4kab + 8a^2 + k - 1 > 0$. Factoring the left side:

$$(2a\sqrt{k})^2 + (2\sqrt{k})^2 + k - 1 > 0$$

Since $k > 0$, both $(2a\sqrt{k})^2$ and $(2\sqrt{k})^2$ are positive, so the left side is always positive. However, $k - 1 < 0$ for all $k > 0$, so we have a contradiction. Hence, it must be the case that $a = b$.

A neat Vieta Jumping Exercise, assume $a \neq b$ and $a > b$, now we take the minimal $a+b$ which satisfies this equation. Note that $4ab - 1 \mid (a-b)^{2}$.
$$\frac{(a-b)^{2}}{4ab-1} = c \implies (a-b)^{2} = c(4ac-1)$$
Just take $a' = 2b + 4bc - a = \frac{b^{2}+c}{a}$, which satisfies the above quadratic equation and contradicts the minimality of $a+b$ since $a' < a$, so we're done.

AFTSOC WLOG $a > b$ with minimal $a+b$ which satisfies this equation. Notice $$(4ab-1)\mid(4a^2b-b-a(4ab-1))^2=(a-b)^2\iff\frac{(a-b)^2}{4ab-1} = c \implies (a-b)^{2} = c(4ab-1)\implies a^2-(4bc+2b)a+(b^2+c)=0.$$ But notice taking $a_0=\frac{b^2+c}{a}$ gives a smaller solution, which is easily seen because substituting and multiplying by $\frac{a^2}{b^2+c}$ returns to the original equation equaling 0; in particular, this is a contradiction because we assumed minimum, yet it's easy to check that $\frac{b^2+c}{a}<a$ making a smaller solution. As a result, a must equal b.


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations
number theory unsolved

If the difference of cubes of two consecutive integers can be represented as a square of an integer , then this integer is the sum of squares of two consecutive integers.


Consider the equation x^2+y^2=z^2+18, where x,y,z are positive integers.

Prove there are an infinite number of solutions where x=y.

spoiler:

[hide]

let the equation be 2y^2-z^2=18...

it can be rewritten as y^2/9-z^2/18=1 which is a hyperbola and therefore proof by definition...

but let's say we want to prove it ourselves...

let n be an arb. natural number ....

let y = n

n^2/9 - z^2/18 = 1

or

z_1 =  :sqrt: (2n^2-18)

now let y = n + 1

(n+1)^2/9 - z^2/18 = 1

or

z_2 = :sqrt: (2(n+1)^2-18)

(1) If z_2 > z_1 then infinite n's exist.

:sqrt:(2(n^2+2n+1)-18) > :sqrt:(2n^2-18)

n^2+2n+1 >  n^2 which is proof by (1)[/hide]

Hmm, is that rigorous enough?. Can you just let y=n, and then y=n+1????

I reckon I'm just an old country hick, but I can't figure for the life of me how infinite integer solutions follow from the definition of a hyperbola, 'cause I reckon a^2/pi + b^2/e = 1 is just the same old kind of hyperbola and I'll be darned if it's got any rational solutions whatsoever.

well, i've gotta say, thank you very much for all these problems. could you do something like this every week? (because i'm on all day saturday, and half the day sunday)

Besides, spoon, even if it is a hyperbola, you still need to prove there are infinitely many lattice points on that hyperbola.

So, i Guess the question remains unanswered!

PS. *there's a reason y i listed it as "Very Difficult"

Spoiler:

[hide]I worked this out while in Physics AP class today (I have it independent study, so I could do whatever I want), and I think I got the answer...

Let M_n be the set of all natural y values that will yield a natural z in the equation:

z=:sqrt:(2y^2-18)

y = 3 is the first value that will work, at point (3,0) which is also the vertex.

so M_1 = 3.

y = 9 is the second value, which hits (9,12)

so M_2 = 9

For n > 2

Making M_n recursive:

Let M_n+1 = 6M_n - M_n-1

M_3 = 6M_2 - M_1 = 6*9-3 = 51, whose z value = 72

M_4 = 6M_3 - M_2 = 6*51-9 = 297, whose z value = 2448...

Assuming M_n is finite, we take the max of M_n and apply the recursive forumla, thus producing M_n+1 which is greater than M_n, and therefore M_n must have infinite terms. Proof by contradiction.

What do you think, does this satisfy the proof?

[/hide]

Hope that's right... g2g to class, won't be home till after 10:00 PM... peace.

[color=cyan]A good approach.   However, one thing is missing.  While you demonstrated you could find a few more members of the sequence, you didn't prove that, given any 2 members, your method will get you another one.  Which it should, I believe.  You need to do a general proof for this, something inductive.  Better yet, find the general form of numbers generated by your formula and prove that the general form satisfies the conditions, which will mean that all of the individual members will, which is good.[/color]

Hmm, i believe the formal proof of this question is done with..............--> [hide]Pell-Like equations[/hide]

How do you know that any term of that sequence will satisfy the relation at all?

Yes I realized all of this but I thought I should give it a shot anyway.... does anyone have any suggestions as to how to go about finding a closed form expression for the recursive definition...

Ok, as for the closed form expression:

Let M_n=a^n

Thus, a^(n+1)=6*a^n-a^(n-1)
a^(n-1)*(a^2-6a+1)=0.
This means either a^(n-1) or (a^2-6a+1) is equal to zero, but the former can't be equal to zero, so the latter must be.
Solving for a,

a=3 :pm: 2*sqrt[2]

Now, we plug this back into the original equation:

M_(n+1)=C_1*(3+2*sqrt[2])^n+C_2*(3-2*sqrt[2])^n.
where C_1, C_2, are constants.

Now, let n=0,1. FRom there, we can solve for the constants, and hence get the closed form expression.

I can plug this into mathematica and solve, but as you've all noticed, im pretty lazy, so i'll leave the rest to u guys...

enjoy

yeah..i got bored in calc class today and found that if 2a^2 = b^2 + 18 , then (the smallest solution is (3,0))  if (c,d) is a solution, the next largest solution will be (3c + 2d, 4c+3d).


geometry
geometric transformation
rotation
linear algebra
matrix
trigonometry
rotation matrix

I'm trying to understand Itten's description of his Fire Tower. The math for the construction can be seen in this paper http://www.emis.de/journals/NNJ/MarRos-en.html

Can anybody explain the "analysis in plan part"? especially how the transformation matrices are found(the rotation and scale matrices angle and factor are found).

[asy]pair A1=(2,-2), B1=(2,2), C1=(-2,2), D1=(-2,-2); pair A2=(2,-1), B2=(1,2), C2=(-2,1), D2=(-1,-2); draw(A1--B1--C1--D1--cycle); draw(A2--B2--C2--D2--cycle); draw(A2--(0,0)--A1); label("$O$",(0,0),NW); label("$A_1\left(\frac{1}{2},-\frac{1}{2}\right)$",A1,SE); label("$B_1$",B1,NE); label("$C_1$",C1,NW); label("$D_1$",D1,SW); label("$A_2\left(\frac{1}{2},-\frac{1}{4}\right)$",A2,E); label("$B_2$",B2,N); label("$C_2$",C2,W); label("$D_2$",D2,S);[/asy]The rotation matrix is found by calculating $\angle A_{1}OA_{2}=\angle B_{2}A_{2}B_{1}=\tan^{-1}{\frac{1}{3}}$.
The general form for the rotation matrix is $\left| \begin{array}{cc}\cos{\theta}&-\sin{\theta}\\ \sin{\theta}& \cos{\theta}\end{array}\right|$. \begin{eqnarray*}\left| \begin{array}{cc}\cos{\left(\tan^{-1}{\frac{1}{3}}\right)}&-\sin{\left(\tan^{-1}{\frac{1}{3}}\right)}\\ \sin{\left(\tan^{-1}{\frac{1}{3}}\right)}& \cos{\left(\tan^{-1}{\frac{1}{3}}\right)}\end{array}\right|=\left| \begin{array}{cc}\frac{3}{\sqrt{10}}&-\frac{1}{\sqrt{10}}\\ \frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\end{array}\right| \end{eqnarray*}

The general form for the scaling matrix is $\left| \begin{array}{cc}k & 0\\ 0 & k\end{array}\right|=k\left| \begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right|$ where $k$ is the scaling factor. Notice that it is just $k$ times the identity matrix. $k=\frac{OA_{2}}{OA_{1}}=\frac{\sqrt{\left(\frac{1}{2}\right)^{2}+\left(-\frac{1}{4}\right)^{2}}}{\sqrt{\left(\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)^{2}}}=\frac{\sqrt{10}}{4}$


probability
videos
conditional probability
Support

Is there a video for conditional probability?

I've gotten a couple problems but no videos that deal with conditional probability...Has anyone unlocked a conditional probability video?

Well, of the around 70 videos i have "unlocked", i did not get a conditional probability video.


integration
calculus
calculus computations

Help me to check the solution, thanks.

$ *\ \ {{\left( \arcsin x \right)}^{'}}\equal{}\frac{1}{\sqrt{1\minus{}{{x}^{2}}}}\equal{}{{\left( 1\minus{}{{x}^{2}} \right)}^{\minus{}1/2}}\equal{}\sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i}}}{i!{{.2}^{i}}}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\plus{}1\text{,   where }\left| x \right|<1$

$ \Rightarrow \arcsin x\equal{}\int{\left( \sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i}}}{i!{{.2}^{i}}}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\plus{}1 \right).dx}\equal{}\sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i\plus{}1}}}{i!{{.2}^{i}}\left( 2i\plus{}1 \right)}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\plus{}x$

$ *\ \ {{\left( \arccos x \right)}^{'}}\equal{}\frac{\minus{}1}{\sqrt{1\minus{}{{x}^{2}}}}\equal{}\minus{}{{\left( 1\minus{}{{x}^{2}} \right)}^{\minus{}1/2}}\equal{}\minus{}\sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i}}}{i!{{.2}^{i}}}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\minus{}1\text{,   where }\left| x \right|<1$

$ \Rightarrow \arccos x\equal{}\minus{}\int{\left( \sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i}}}{i!{{.2}^{i}}}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\plus{}1 \right).dx}\equal{}\minus{}\sum\limits_{i\equal{}1}^{n}{\frac{{{x}^{2i\plus{}1}}}{i!{{.2}^{i}}\left( 2i\plus{}1 \right)}.\prod\limits_{k\equal{}1}^{i}{\left( 2k\minus{}1 \right)}}\minus{}x$

$ \Rightarrow \arccos x\plus{}\arcsin x\equal{}0\text{  ?}$

You forgot about the constant when you integrated.

Any indefinite integral includes an arbitrary constant- which means that your equation is actually $ \arccos x\plus{}\arcsin x\equal{}C$ for some constant $ C$ on $ [0,1]$. This is true, and easy to see from the right triangle definition.


Welcome to my chess club! the only requirements to join is to know how to play the game! because there is no place to play on forum or here any games you wish to play should be played here [url]http://www.freechess.org[/url] now i know how to play and am fairly decent but i have no idea of the clubs going on (im only in 6th grade :blush: ) so i will need a staff of president and vice prez now requirements for those will be given out as soon as this club has enough members :D JOIN NOW!!! :idea:

[hide="if you join"]to play the game there is two ways:
1. if you know how to notate make a private chatroom and play there. (by notating moves)
2. is to come to  the freechess website. to play here follow the instructions:
1. go to [url]http://www.freechess.org[/url]
2.go to login now.
3. login as a guest. for playing someone in particular i will post instructions when we hold our first tourney
4. go to find opponent and put your chess settings for the games when you click send seek a player will see it and hopefully join if you dont want to make a game instructions on how to play differently will be posted in one of my next posts (too lazy right now)
5. PLAY CHESS![/hide]

oooh i join
i wanna be prez ;)

and i'll lose every game but what hte heck
 :D

yay mathblitz!

I wanna be prez too!!!! (srry angie)

You made a copy of the topic, so should we just post in this one?

yes... he said to delete the other.

Also, i dont see login... where is it?

my bad the the website is [url]http://www.freechess.org[/url]

ill edit the other one

yeah... i hate those ppl who go around making parody websites (like .com instead of .org, and common spelling mistakes) to make life harder for everbody.

I would prefer yahoo. Otherwise, I'll join.

Sounds like fun. I'll do it even though I'm not very good at chess. :)

as long as you can play the game oh and nerd you can use anything its just that i prefer fics but it doesnt matter just as long as it works

ok ok well hold elections and then my first chess tourney with instructions

ok sign ups for prez elections are here sign up now then we will have vice prez elections

[hide="sign up now!"]prez list:

1. me![/hide]

[quote="evilhamster"]hello, may i join? I know I'm jumping in a little late but I haven't been on AoPS for some time.
Also, could you please attach ratings to the descriptions of playing level?
Currently, I'm probably above average (1700) because USCF median rating is about 1500[/quote]

I think you should be decent or higher... heh heh...

Or I'll just drop down one level... =\

[quote="splatyango"]yes i no hes really good
hes my brother btw
and once in some random tourney he beat a master! ummm... so if hes joining the tourney i feel sorry for who has to play him
if its me, i might as well forfeit
or maybe just play it live for the heck of it[/quote]
Hey I didn't beat a master...
I beat a guy who beat a master( actually, several, and I beat him twice) and I would have beaten him three times BUT FOR THAT FREAKIN QUEEN TO F2 MOVE WHICH I COMPLETELY MISSED!!! :bomb:  :cursing:  :furious:  ARRRGGHHHH I HATE U DAN MATTHEWS (actually i don't because it was only my fault that I didn't do enough tactics :oops: )
I have outplayed a master, and I have beaten 2.5 1900's in a row... :jump:

um sorry. although then again, i can just tell u BECAUSE UR IN THE FREAKING ROOM NEXT TO ME!!!!!!!!!!!!!!!!! oh well. i like online

Hmm, looks like a good place to ask.

I want to learn how to play chess.  I know the basic rules, and played maybe one game ever, but I want to get better.  What's the best way to do that?  Is it just to play and lose at a ton of games?

There are several good chess books out there. If you are willing to spend some money, you can buy several and get really good by learning techniques and playing games. Also, I don't know of any free computer chess engines out there--although you can also pay to get one--but if you can get one, that would be great. The reason I don't recommend live chess until you get better is because if you try to play as guest, then you could be playing some really good person who played a prank on the little buds that play as guests, and if you get an account, others who just start usually don't make accounts (although that may change!) and you'll be facing really good people anyway. But computer engines don't have feelings, and they won't get really angry at you if you win against them, or brag if you lose.  :D Also, they come in every skill and style.

Was that too long? If it was, please ask m to clarify. I don't know if that really made sense.  :lol:

I'm not sure if this is something I'm willing to spend money on right now.  Though maybe I'll check out the library.

I want to get better because I have a lot of friends who play, and it seems like a fun game.  The problem is if I'm losing to them all the time, I don't think I'll get better, it's just discouraging losing all the time and I don't feel like I'm learning anything.  Looking for a free chess engine is a good idea...I'll google it when I'm off work!

Anyone wanna play on yahoo sometime....Idc when.

icc? icc? icc anyone?

if no one has an icc account, then this can't be a chess club

http://www.chessclub.com/

um come on, AoPS is [i]math[/i], not many people here have such a desire for chess/enough skill to apply for icc. and besides, daddy wont let me apply. =]

Do you play Chinese Chess? :) I like playing Chinese Chess.

[quote="crazychinesemaniac"]did this fail?[/quote]

I don't think that!  :?:


calculus
integration
calculus computations

Let $ J \equal{} \int_0^1 x^2(x \minus{} 1)^2e^{2x}\ dx$

$ J \equal{} \int_0^1 x^2 \cdot (x \minus{} 1)^2 \cdot \sum_{n \equal{} 0}^{\infty} \dfrac{\left(2x\right)^n}{n!}\ dx \equal{} \sum_{n \equal{} 0}^{\infty} \dfrac{2^n}{n!} \left[ \dfrac{1}{n \plus{} 5} \minus{} \dfrac{2}{n \plus{} 4} \plus{} \dfrac{1}{n \plus{} 3} \right]$

Let:
$ f(x, y) \equal{} \sum_{n \equal{} 0}^{\infty} \dfrac{x^n}{n! (n \plus{} y)} \implies x^y f(x, y) \equal{} \sum_{n \equal{} 0}^{\infty} \dfrac{x^{n \plus{} y}}{n! (n \plus{} y)}$

$ \dfrac{d}{dx} \left(x^y f(x, y) \right) \equal{} \sum_{n \equal{} 0}^{\infty} \dfrac{x^{n \plus{} y \minus{} 1}}{n!} \equal{} e^x x^{y \minus{} 1}$

Now consider the case y = 5:

$ x^5 f(x, 5) \equal{} (x^4 \minus{} 4x^3 \plus{} 12x^2 \minus{} 24x \plus{} 24)e^x \plus{} C \implies C \equal{} \minus{} 24$
$ f(2, 5) \equal{} \dfrac{1}{4} \left(e^2 \minus{} 3\right)$

The case y = 4:

$ x^4 f(x, 4) \equal{} (x^3 \minus{} 3x^2 \plus{} 6x \minus{} 6)e^x \plus{} C\implies C \equal{} 6$
$ 2 \cdot f(2, 4) \equal{} \dfrac{1}{4} \left( e^2 \plus{} 3 \right)$

The case y = 3:
$ x^3 f(x, 3) \equal{} (x^2 \minus{} 2x \plus{} 2)e^x \plus{} C \implies C \equal{} \minus{} 2$
$ f(2, 3) \equal{} \dfrac{1}{4} \left( e^2 \minus{} 1\right)$

combining these we get $ J \equal{} \dfrac{1}{4} \left(e^2 \minus{} 7\right)$

I think you introduce complexity in this solution.
Try to remove brackets and split to few integrals and integrate them by parts.

To dcc.

That's correct answer. Maybe no one will solve as yours, but your solution is very interesting to me, thank you.

:) you are right.
I'm trying to find universal and poverful methods of solutions which are easy for understanding.
My idea was to find general formula for $ x^{n}e^{x}$ to remove the brackets and then just to substitute and make remaining calculations. This approach make the problem routine excercise. You are right it is not so beautiful but is easier for understanding
for not so experienced people.

hello, you can make the ansatz
$ \int x^2(x\minus{}1)^2e^{2x}dx\equal{}(ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}f)e^{2x}$ with reals
$ a,b,c,d,f$.
Sonnhard.

hello, so your result is $ \frac{1}{4}e^2\minus{}\frac{7}{4}$.
Sonnhard.

That's right, but the answer have been posted in #4.


Probability - Think About It Part 1

We tackle a very complicated-looking probability problem by realizing that it is much simpler than it appears.

Where are the problems?
Or is this a vid title?


Prove that $ cos 40^{o}(1\minus{}2sin 10^{o})\equal{}\frac{1}{2}$

Oh,that solution is nice. Nowaday,I will send you some problems and I will receive a nice solution from you.


geometry
rectangle
incenter
exterior angle

Prove that in any triangle, the projection of a vertex onto the interior and exterior bisectors of the other two angles are collinear.

[hide="Proof"]
Let us call the traingle we are to deal with $ABC$.  Let the projections of $A$ onto the interior angles of $B$ and $C$ be $P$ and $Q$ respectively, and let the projections of $A$ onto the exterior angles of $B$ and $C$ be $R$ and $S$ respectively.  The sum of an interior angle and an exterior angle is $90^{\circ}$ because if we have an angle $\alpha$, $\text{(interior angle)+(exterior angle)=\frac{\alpha+180^{\circ}-\alpha}{2}=90^{circ}$; exterior and interior angle bisectors of a traingle are said to be orthogonal.  As a result, quadrilateral $APBS$ is a rectangle-->cyclic.

$\angle ABS=\angle APS=\frac{\pi-\angle B}{2}=\frac{\pi-\angle B-\angle C}{2}+\frac{\angle C}{2}=\angle A/2+\angle C/2$

Let the incentre be $O$, thus quadrlateral $APOQ$ is cyclic since it has two opposite 90 degree angles.
$\angle APQ=\angle AOQ=\angle A/2+\angle C/2$.  This implies that $\angle AOQ\angle =\angle APS$, thus $P$, $Q$, and $S$ are collinear.  In order to prove that all four points, $P$, $Q$, $R$, and $S$ are collinear, we could do the same thing with the other side of of the traingle.

And the proof is complete.
[/hide]
I would like to see your proofs.

Masoud Zargar

Mmmm!
 Let  $D , E$ the projections of $A$ on the bisectors of angle $B$ and it's exterior. Produce $AD , AE$ to meet line $BC$ at $K, L$ respectively. Then triangles  $BAK , BAL$are isosceles , so $D , E$are midpoints of  $AK , AL$ . But then $DE // KL$and hench  line $DE$  passes through the midpoints of $AB$ and $AC$ . But this line is constant. To the same line belong the other two points, too.

 [u]Babis[/u]

Dear Mathlinkers,
see
http://jl.ayme.pagesperso-orange.fr/  vol. 4  An unlikely concurrence p. 4

Sincerely
Jean-Louis

Notation: Let $A_1, A_2, A_3, A_4$ be the reflections of $A$ across the external bisector of angle $B$, the internal bisector of angle $B$, the internal bisector of angle $C$, and the external bisector of angle $C$, respectively. Also, let $P_1, P_2, P_3, P_4$ be the intersections of $AA_1$ with the external bisector of angle $B$, $AA_2$ with the internal bisector of angle $B$, $AA_3$ with the internal bisector of angle $C$, and $AA_4$ with the external bisector of angle $C$, respectively.

Approach: By a property of reflections, $AA_1 \perp BP_1$, $AA_2 \perp CP_2$, $AA_3 \perp BP_3$, and $AA_4 \perp CP_4$. As all of our reflecting lines are angle bisectors, $A_i$ lies on $BC$, with $i=1,2,3,4$. Note that, by definition of reflection, $P_i$ bisects $AA_i$. So $P_1P_2 \| A_1A_2$, $P_2P_3 \| A_2A_3$, and $P_3P_4 \| A_3A_4$. Therefore, as $A_i$ are collinear, $P_i$ are collinear as well.


Let $ c_{n, m}$ be the number of permutations of $ \{1, \ldots, n\}$ which can be written as the product of $ m$ transpositions of the form $ (i, i+1)$ for some $ i=1, \ldots, n-1$ but not of $ m-1$ suct transpositions. Prove that for all $ n \in \mathbb{N}$,
\[ \sum_{m=0}^\infty c_{n, m}t^{m}= \prod_{i=1}^{n}(1+t+\cdots+t^{i-1}). \]

[quote="Kouichi Nakagawa"]Let $ c_{n, m}$ be the number of permutations of $ \{1,\ldots, n\}$ which can be written as the product of $ m$ transpositions of the form $ (i, i\plus{}1)$ for some $ i \equal{} 1,\ldots, n\minus{}1$ but not of $ m\minus{}1$ suct transpositions. Prove that for all $ n\in\mathbb{N}$,
\[ \sum_{m\equal{}0}^{\infty}c_{n, m}t^{m}\equal{}\prod_{i\equal{}1}^{n}(1\plus{}t\plus{}\cdots\plus{}t^{i\minus{}1}).\]
[/quote]
By induction.
For $ n \equal{} 2$ is trivial, now supose that $ \sum_{m \equal{} 0}^{\infty}c_{n, m}t^{m}\equal{}\prod_{i \equal{} 1}^{n}(1\plus{}t\plus{}\cdots\plus{}t^{i\minus{}1})$ for some $ n$, and consider the set $ \{1,\ldots, n\plus{}1\}$. Let $ \pi(i)$ the permutation consisting in to transpose the elements in the places $ (i,i\plus{}1)$. Now if we have a permutation $ \pi(a_{1})\pi(a_{2})\ldots\pi(a_{m})$ of this set and some $ a_{i}\equal{} n$, we can assume that $ i \equal{} m$ becouse, as is wellknown, the permutations are asociative and conmutative. So we can first move the first $ n$ numbers and later $ n\plus{}1$. Consider the permutations of $ \{1,\ldots, n\plus{}1\}$which can be written as the product of $ m$ transpositions of the form $ (i, i\plus{}1)$ is easy to see that there are $ c_{n, m}$  with $ n\plus{}1$ in the $ n\plus{}1$th place, $ c_{n, m\minus{}1}$ with $ n\plus{}1$ in the $ n$th place .... and $ c_{n, m\minus{}n}$ with $ n\plus{}1$ in the first place, that is $ c_{n\plus{}1,m}\equal{} c_{n,m}\plus{}c_{n,m\minus{}1}\plus{}...\plus{}c_{n,m\minus{}n}$, or $ c_{n\plus{}1,m}\equal{} c_{n,m}\plus{}c_{n,m\minus{}1}\plus{}...\plus{}c_{n,0}$ if $ m < n$
 therefore
$ \sum_{m \equal{} 0}^{\infty}c_{n\plus{}1, m}t^{m}\equal{}\sum_{m \equal{} 0}^{\infty}c_{n,m}t^{m}\plus{}\sum_{m \equal{} 1}^{\infty}c_{n,m\minus{}1}t^{m}\plus{}\ldots\plus{}\sum_{m \equal{} n}^{\infty}c_{n,m\minus{}n}t^{m}$
$ \equal{}\sum_{m \equal{} 0}^{\infty}c_{n, m}t^{m}\plus{}t\sum_{m \equal{} 0}^{\infty}c_{n, m}t^{m}\plus{}...\plus{}t^{n}\sum_{m \equal{} 0}^{\infty}c_{n, m}t^{m}\equal{}$
$ \equal{} (1\plus{}t\plus{}\ldots\plus{}t^{n})\sum_{m \equal{} 0}^{\infty}c_{n, m}t^{m}\equal{} (1\plus{}t\plus{}\ldots\plus{}t^{n})\prod_{i \equal{} 1}^{n}(1\plus{}t\plus{}\cdots\plus{}t^{i\minus{}1}) \equal{}$ $ \prod_{i \equal{} 1}^{n\plus{}1}(1\plus{}t\plus{}\cdots\plus{}t^{i\minus{}1})$

Please excuse my english but i dont have another, in the next i'll try to solve problems without text. :blush:
I edited it to corret a mistake.


inequalities
Euler
complex numbers
geometry unsolved
geometry

If $ z_1,z_2$ are two complex numbers, prove that $ 2(|z_1|^n\plus{}|z_2|^n)\leq |z_1\minus{}z_2|^n\plus{}|z_1\plus{}z_2|^n$, for every $ n\geq\ 2$.

Presume that inequality is true for $ n\in\mathbb{R}$, and  prove that is true for n+2.

  So \[ |z_1\minus{}z_2|^{n\plus{}2}\plus{}|z_1\plus{}z_2|^{n\plus{}2}\ge \frac{|z_1\minus{}z_2|^2\plus{}|z_1\plus{}z_2|^2}{2}(|z_1\minus{}z_1|^n\plus{}|z_1\plus{}z_2|^n)\]and then\[ |z_1\minus{}z_2|^{n\plus{}2}\plus{}|z_1\plus{}z_2|^{n\plus{}2}\ge (|z_1|^2\plus{}|z_2|^2)2(|z_1|^n\plus{}|z_2|^n)\ge 2(|z_1|^{n\plus{}2}\plus{}|z_2|^{n\plus{}2})\]

  Thus it suffices to prove inequality for n=2 and n=3.

  For n=2 we have Euler Theorem $ 2(|z_1|^2\plus{}|z_2|^2)\equal{}|z_1\minus{}z_2|^2\plus{}|z_1\plus{}z_2|^2$

 For n=3 see [url=http://www.mateforum.ro/viewtopic.php?t=2030]here[/url]


One of the sides of a triangle is divided into segments of $ 6$ and $ 8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $ 4$, then the length of the shortest side of the triangle is:
$ \textbf{(A)}\ 12\text{ units} \qquad\textbf{(B)}\ 13\text{ units} \qquad\textbf{(C)}\ 14\text{ units} \qquad\textbf{(D)}\ 15\text{ units} \qquad\textbf{(E)}\ 16\text{ units}$

[hide]
Assume the side lengths are $ 14$, $ a\plus{}6$, and $ a\plus{}8$.

Using heron's and $ A\equal{}sr$, and setting them equal, we get:

$ \sqrt{(14\plus{}a)(a)(8)(6)}\equal{}4(14\plus{}a)$

$ 42a\plus{}3a^2\equal{}196\plus{}28a\plus{}a^2$

$ 2a^2\plus{}42a\minus{}196\equal{}0\implies a\equal{}7$

The shortest side length is $ 7\plus{}6\equal{}13\implies\boxed{\text{B}}$[/hide]

Let $x$ be the unmarked length. Then the triangle has side lengths $14, x+8, x+6$, and hence semiperimeter $x+14$. Heron and $a=rs$ give $4=\frac{(x+14)(x)(6)(8)}{x+14} \implies x=7 \implies \boxed{13}$.


x,y,z    are positive real numbers such that  $ xyz\equal{}1$
prove that

$ \frac{x\minus{}1}{y\plus{}1}\plus{}\frac{y\minus{}1}{z\plus{}1}\plus{}\frac{z\minus{}1}{x\plus{}1} \geq 0$

it is equal to:
$ \sum(x^2z \minus{} z \plus{} x^2 \minus{} 4)\geq 0$
then,we should show botht:
    $ \sum x^2\geq \sum x$
    use $ xyz \equal{} 1$
    we have $ \sum\frac {x}{yz}\geq x \plus{} y \plus{} z$
    use Cauthy ineq,we have the answer.
And to prove,$ \sum x^2z\geq3$
it is easy by using AM-GM ineq

See hear:[url]http://www.mathlinks.ro/viewtopic.php?t=190663[/url]   :wink:
Why you double post?


Hope you don't mind if I use calculus....
The error in this "proof" is very subtle, making for a cool party trick:

We all know our old friend integration by parts:

$\int u\ dv = uv - \int v\ du$

Now apply it to the integral $\int  \frac{1}{x\ln x} dx$, setting $u = \frac{1}{\ln x}$ and $dv = \frac{1}{x}$:

$\int  \frac{1}{x\ln x} dx = \frac{\ln x}{\ln x} - \int \frac{-1}{(\ln x)^2} \cdot \frac{1}{x} \cdot \ln x\ dx$

$= \frac{\ln x}{\ln x} + \int \frac{1}{x\ln x}\ dx$. Therefore,

$0 = \frac{\ln x}{\ln x} = 1$

 :ninja:

Nice. :)
[hide]Don't forget to include the constant when evaluating an indefinite integral![/hide]

ok, nice question and solution

 :rotfl:  :D

i didnt fnd the mistake on my own. man my calc teacher would kill me. nice trick. ill have to show it to her and see if she can find the mistake.


A train leaves Boston at 8:00 a.m. and travels the 216 miles to New York at a speed of 90 mph. At what time in the morning does the train arrive in New York?

It will take the train 2.4 hours, since 216/90 is 2.4.
2/5 of an hour is 24 mins, so the train will take 2 hrs 24 mins, and therefore will arrive at 10:24.


I found this in Hardy and Wright's [i]Introduction to the Theory of Numbers[/i] that I got from the library. I've heard about the inclusion-exclusion principle but never knew what it was. When I read this it seemed like something that could have that name. Is this right?

Anyway, show that if there are $N$ objects and some properties $\alpha,\beta,\gamma,\ldots$, and $N_\alpha$ of them have property $\alpha$, $N_\beta$ have property $\beta$, and so forth, $N_{\alpha\beta}$ have both $\alpha$ and $\beta$, $N_{\alpha\beta\gamma}$ have $\alpha$ and $\beta$ and $\gamma$, and so on, the number of these objects having none of the properties is\[N-N_\alpha-N_\beta-N_\gamma-\ldots+N_{\alpha\beta}+\ldots-N_{\alpha\beta\gamma}-\ldots\]I hope it is clear exactly what the terms in the sum are. This is how it was written in the text.

Why, yes, it is inclusion-exclusion.

A simple proof might be to find how many times an object with $k$ properties is counted.

It's a been a little while. If no one can prove it or wants to post the solution I will..

As I said, consider an object that has $k$ properties. See how many times it's counted in the formula. That's a simple proof

The easiest way to see it is to just draw 3 circles that overlap each other.

Yes, thats an easy way for three variables, but it is impossible to create a diagram with four circles that accuarately represents a data set, and therefore you cant extend that approach to N without using non-circular figures

[quote="Altheman"]this is kind of an intuitive idea, where you overcount the number of something twice, so you want to subtract that double count to get the total[/quote]That's true, but the proof is pretty nice too. I'll just go ahead and post it.

[hide="proof of the inclusion-exclusion principle"]Let $O$ be one of the $N$ objects, and suppose that it has $k$ of the properties $\alpha,\beta,\gamma,\ldots$. Consider how many times it is counted in the sum above, which can be written as \[N-\sum_{\text{cyc}}{N}_\alpha+\sum_{\text{cyc}}{N}_{\alpha\beta}-\sum_{\text{cyc}}{N}_{\alpha\beta\gamma+\ldots}\]

If $k=0$, then it is counted only once, in the $N$ term. It thus adds $1$ to the sum, which is what we want.

If $k>0$, then $O$ is counted once in $N$, $k$ times in $\sum_{\text{cyc}}{N}_\alpha$, $\frac{k(k-1)}2$ times in $\sum_{\text{cyc}}{N}_{\alpha\beta}$, and so on.

Thus in the sum it is counted $1-k+\frac{k(k-1)}2-\frac{k(k-1)(k-3)}6+\ldots$ times, but this is the binomial expansion of $(1-1)^k=0$.[/hide]

Edit: bob123, as Scrambled pointed out, that only works when there are three properties $\alpha,\beta,\gamma$. But yes, that is a good graphical explanation.


inequalities
real analysis
real analysis unsolved

Let $ \zeta \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {1}{2^{3^n}}}$. Prove, that inequality $ \left|\zeta \minus{} \frac {p}{q}\right| < \frac {c}{q^3}$ has infinitely many positive integer solutions in numbers $ p,\; q$ if $ c > 1$ and finit number of solutions if $ c \equal{} 1$.

case when $ c \equal{} 1$ is easy :
suppose for $ c \equal{} 1$ we have infinity many sequance$ \frac {p_{n}}{q_{n}}$ and $ q_{n}\to \infty$ say
$ 2^{3^{m \minus{} 1}} < q_{m} < 2^{3^{m}}$
then you will have: (rewrite as ):

$ |\sum_{j \equal{} 0}^{m}\frac {q^{3}_{m}}{2^{3^{j}}} \plus{} \sum_{k \equal{} m \plus{} 1}^{\infty}\frac {q^{3}_{m}}{2^{3^{k}}} \minus{} p_{m}q^{2}_{m}| < 1$
try to show that for big $ m$:
$ \sum_{j \equal{} 0}^{m}\frac {q^{3}_{m}}{2^{3^{j}}} \minus{} p_{m}q^{2}_{m} > 1$


Find all f(x), decreasing in $ [0,\plus{}\infty ]$ and satisfy $ f(x\plus{}f(y))\equal{}\frac{y}{xy\plus{}1}$, $ \forall x,y>0$

$ f(x+f(y)=\frac{1}{x+\frac}1}{y}}.$
Let $ z=x+f(y)$, then $ f(z)=\frac{1}{z-f(y)+\frac{1}{y}}\to f(y)=\frac{1}{y}+const,$ Obviosly $ const =0$.

let $ P(x,y)$ the assertion $ f(x\plus{}f(y))\equal{}\frac{y}{xy\plus{}1}$

$ P(1,x) \implies f(1\plus{}f(x))\equal{}\frac{x}{x\plus{}1}$

$ P(\frac{1}{x\plus{}1},1\plus{}f(x))\implies f(\frac{1}{x\plus{}1}\plus{}\frac{x}{x\plus{}1})\equal{}\frac{(x\plus{}1)(f(x)\plus{}1)}{(x\plus{}1)\plus{}(f(x)\plus{}1)}$

so $ f(x)\equal{}\frac{(x\plus{}1)(1\minus{}a)\plus{}1}{a(x\plus{}1)\minus{}1}$ with $ a\equal{}f(1)\neq 0$ (else we got $ f(1)\equal{}\frac{1}{2}$ wich is absurd )

reciprocally we find $ a\equal{}1$ so $ f(x)\equal{}\frac{1}{x}$ $ \forall x \neq 0$


Support
integration
real analysis
real analysis unsolved

How to prove that if $f,g: \mathbb{R}^{n}\to \mathbb{R}$ has compact supports then $\text{spt }f*g$ is compact?

Suppose $\text{supp}(f)=A$ and $\text{supp}(g)=B.$

Now suppose that $f*g(x)\ne0.$ Since $f*g(x)=\int f(x-y)g(y)\,dy,$ then there must be some $y$ (actually, a set of positive measure of such $y$) such that $x-y\in A$ and $y\in B.$

That means that $x\in A+B$ where $A+B=\{a+b: a\in A,b\in B\}.$

And that means that the support of $f*g$ is contained in the closure of $A+B$. To finish what you want, you need this:

[b]Lemma:[/b] If $A$ and $B$ are compact, then $A+B$ is compact.


geometry
combinatorics proposed
combinatorics

A banana plantation is located next to a desert. The plantation owner has 3000 bananas that he wants to transport to the market by camel, across a 1000 miles stretch of desert. The owner has only one camel, which carries a maximum of 1000 bananas at any moment in time, and eats one banana everymile it travels.

What is the largest number of bananas that can be delivered at the market?  :wink:

(Note: A banana is eaten uniformly within each mile, so the answer can involve a fraction of banana)

[hide]

Let $B(x)$ be the total amount of the bananas at the position $x\in[0,1000]$. Of course it is $B(0)=3000$

We will also denote as $x_{n}$, the position in which we have $n$ bananas. So $x_{0}=3000$

Suppose that we transport all the bananas from the point $0$ to the point $x$.
We transport $1000$ bananas and we go back to $0$
Once again we transport $1000$ bananas and we go back to $0$
The last time we just transport $1000$ bananas.
The camel travelled a total distance $5x$. So the cost for this trip is $5x$ bananas.

So, at the position $x$ we have $B(x) = B(0)-5x$ bananas.
The position $x$ will be an intermediate station.
After $x$, we will make [u]only two[/u] courses. So, it is necessary that $B(x)\leq 2000 \Leftrightarrow$
$3000-5x\leq 2000 \Leftrightarrow x\geq 200$

So $x_{2000}=200$, that is $B(200)=2000$

After this point, we make two courses:
The camel transports $1000$ bananas, goes back to the point $200$ and transports again $1000$ bananas to the point $200+y$
The camel travels a distance $3y$, so the cost is $3y$ bananas
(notice that, if we would made 3 courses then we would pay 5y bananas)

So, at the position $200+y$ we have $B(200+y)=B(200)-3y = 2000-3y$ bananas

$B(200+y)\leq 1000 \Leftrightarrow$
$2000-3y\leq 1000 \Leftrightarrow$

$y\geq 333.\overline 3$

So, $x_{1000}=533.\overline 3$

That is, at the position $533.\overline 3$ we have $1000$ bananas.

The final move is a single transportation of $1000$ bananas for the last $466,\overline 6$ miles

This means that, at the position $1000$ we will have $533.\overline 3$ bananas[/hide]

3 yrs ago this was a challenge trick problem in my geometry class, we had to write up proof and everything. I got it wrong :mad: but I think I could easily have done it now. Nice answer!


[hide]

4.

1/2+1/3+1/4 is greater than 1

[/hide]

Yup, yup. Kind of boringly easy, but this is just a middle school regional (all the middle schools in my city plus one other school) math competition...it's even easier than the MATHCOUNTS school competition...it even features [b]arithmetic[/b] as a problem category, can you believe that!? :lol:

[quote="ComplexZeta"]I presume you want the smallest positive integral value of n?[/quote]
Yup, don't think too hard about it :)

Here's a challenge for you: If (1/1)+(1/2)+(1/3)+...+(1/n-1)+(1/n)=2, what is n?

[hide]1...1+1=1 (when I last checked   )[/hide]

If n=1 all you have will be 1/1, which is 1 the last time I checked, but hey, someone proved 1=2 before so I guess you're right :lol: </jk>
Anyways, If n=1, 1/1, if n=2, (1/1)+(1/2), and so on until n, which can be any rational number; all other number must be whole. Example: If n=4.3, (1/1)+(1/2)+(1/3)+(1/4)+(1/4.3):approx:2.316

1 doesn't equal two, unless someone changed the proof since we went over in in Algebra  :lol:

We disproved it here before, remember? Since A=B, you can't divide by (A-B) since that's just 0.

...it was a joke, fyi.

I know, I'm the one that posted the proof to see who could figure out the error in it. I think you missed the laughing smiley.  :lol:

Oh whoops :lol:

...anyways, any go on the second one?


geometry
rhombus
parallelogram
geometry open

The     number       of     rhombus     that     can   be   inscribed    in    a   parallelogram
such      that     each     side       of      the    parallelogram   contains        precisely
one  vertex   of   the  rhombus  is(please explain with mathematical argument)

A)0      B)1        C)0  or 1       D)0 or 2
E)infinitely many

.. we call the parallelogram ABCD and the rombus EFGH such that the point E is in the AB side, F in BC, G in CD and H in DA. For the Varignon's theorem we know that, in any quadrilateral , can be inscribed a parallelogram joining the middle points of every side. Seen that a parallelogram is a quadrilateral, the theorem is evidently applicable. At this moment i think it's useless u to say that the problem can be solved only demostrating the Varignon's theorem. However i think the right answer is B.


linear algebra
matrix
superior algebra
superior algebra unsolved

I have the non-commutative ring R which is
\[ \begin{matrix} \mathbb Z & \mathbb Q \\ 0 & \mathbb Q \end{matrix}\]

How can I show that every right ideal in the ring R is projective and finitely generated. And how can I use this to show that the right global dimension of R is 1.

I don't understand what you mean. A ring is a set with two laws of compositions. You gave two sets and an integer 0. What do you mean?

I assume he wants that ring to consist of those matrices that have that entries at that positions.


Good morning people this is the PERUVIAN TEAM RESULTS: [official]

PER1 [Ricardo Ramos]       7 7 3 6 0 0 = 23 (Silver)
PER2 [Cesar Cuenta]         7 4 0 6 7 1 = 25 (Silver)
PER3 [Fernando Manrique] 7 7 7 6 7 1 = 35 (Gold, the First Gold for Peru ;))
PER4 [Amilcar Velez]         7 4 0 7 0 0 = 18 (Bronze)
PER5 [Tomas Angles]         7 7 0 7 0 1 = 22 (Silver)
PER6 [Ivan Munoz]             7 7 0 4 0 0 = 18 (Bronze)
------------------------------------------------------------------
                                            TOTAL    = 141ptos.

Good job for each peruvian contest :)

Carlosbr

[Moderator says: Do not write titles in capitals!]

[quote="Jutaro"]Impressive work!! Congratulations to Peruvians :lol: By the way, do you know which country had Latin America's top result??[/quote]

Yes.. BRAZIL ...still ;)

Renan Henrique      7 7 2 7 7 0 =30  Medalla de plata
Henrique Pondé      7 7 0 7 7 1 =29  Medalla de plata
Rafael Tupynambá  7 5 0 7 7 1 =27 Medalla de plata
Davi Lopes Alves   7 5 0 7 7 0 =26  Medalla de plata
Régis Prado          7 4 0 7 7 0 =25   Medalla de plata
Marcelo Matheus   7 1 0 7 0 0 =15  Medalla de bronce

The ranking of ibero countries is:{only the first 10}
              Team Score   Total Raking
1. Brazil          152       -16-
2. Peru            141       -17-
3. Mexico         087       -37-
4. Argentina     085       -39-
5. Spain           082       -43-
6. Colombia     077       -45-
7.Costa Rica     065      -55-
8. Portugal       055       -67-
9. Chile            049      -69-
10. El Salvador   031      -76-

Carlosbr

Congratulation for all peruvian team

I am Israel Diaz Acha an Math Olympiad Peruvian Trainer and I have to say that I feel very excited with the results .

The first gold medal for Peru since our first partipation in 1988 . ... Congratulations in special to Fernando Manrique because He have given a good example to all math peruvian comunity and to whole country .
I hope that Peruvian GAYS continue studiying and training a lot in order to get better results .

Bye Bye

As the Brazilian Deputy Leader, I want to congratulate the Peruvian team for its impressive performance. Special kudos for Fernando!

And Brazil also had its best result ever. Had we repeated our usual scores, Peru would beat us.

But I can assure you that Brazilian students will keep on studying harder and harder to get even better results in the next years!


In how many ways can 8 people sit around a round table if Pierre and Thomas want to sit together, but Rosa doesn't want to sit next to either of them?

We treat Pierre and Thomas as one person (with two permutations) so that there are $ 2\cdot\frac {7!}7 \equal{} 1440$ ways to seat the people with only the restriction that Pierre and Thomas are next to each other. Then, given this, there is a $ \frac46 \equal{} \frac23$ probability that Rosa doesn't sit next to either of them. Thus, our final answer is $ \frac23\cdot1440\equal{}\boxed{960}$.

Another solution: The number of ways in which P and T sits together is $ 6!\times 2\equal{}1440$. The number of ways in which P and T are together and R sits next to one of them is $ 5!\times 2\times 2\equal{}480$. So the number required is $ 1440\minus{}480\equal{}\boxed{960}$.

The answer is [hide]960[/hide].
The way I got the answers are [hide]Solution 1: We choose any seat for Pierre, and then seat everyone else relative to Pierre. There are 2 choices for Thomas; to the right or left of Pierre. Then, there are 4 possible seats for Rosa that aren't adjacent to Pierre or Thomas. The five remaining people can be arranged in any of $5!$ ways, so there are a total of $2\cdot 4\cdot 5!=960$ valid ways to arrange the people around the table.

Solution 2: The total number of ways in which Pierre and Thomas sit together is 6! times 2 = 1440. The number of ways in which Pierre and Thomas sit together and Rosa sits next to one of them is 5! times 2 times 2 = 480. So the answer is the difference $1440 - 480 = \boxed{960}$. [/hide]

[font=times new roman][hide=Simple Steps][b]Step 1 -  Simplify to Basic Structure[/b]
There are 8 people, which implies that there are $8\times7\times6\times5\times4\times3\times2\times1=40320$ ways to seat them.

[b]Step 2 -  Account for Symmetry[/b]
The table can be rotated into 8 positions so we divide by 8: $\tfrac{8\times7\times6\times5\times4\times3\times2\times1}{8}=5040$

[b]Step 3 - Consider Neighbor Restrictions[/b]
Two of the people actually must sit together, so we can temporarily combine them into one person (Remove one seat), and then multiply by 2 to account for the ways to order them. We thus have $\tfrac{7\times6\times5\times4\times3\times2\times1}{7}\times2=1440.$

[b]Step 4 - Consider Separation Restrictions[/b]
One of our people doesn't want to sit next the two neighboring people, so there are two less seats that they can sit in (the ones surrounding the neighbors): $\tfrac{7\times4\times5\times4\times3\times2\times1}{7}\times2=960.$

As we derive through these steps, $\boxed{960}$ is our final answer. $$\definecolor{color}{rgb}{0.0, 0.0, 0.0} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

[hide=another way to get the answer]
We can treat Pierre and Thomas as one person, PierreThomas. Since there were $8$ people before, there are now $7$ people, considering our new super-person, PierreThomas.

There are $7$ ways to choose a spot for PierreThomas, and $4$ ways to choose a spot for Rosa ($6$ spaces are left, and $2$ of them are next to PierreThomas). Then, there are $5!$ choices for the next five people. So, there are $\frac{7\cdot4\cdot5\cdot4\cdot3\cdot2}{7}$ combinations. We need to divide by $7$ because rotations aren't the same. So, we get $4\cdot5\cdot4\cdot3\cdot2$ choices.

But, we need to multiply by $2$ again, since for every seating, PierreThomas can also be ThomasPierre. So, the answer is $4\cdot5\cdot4\cdot3\cdot2\cdot2=\boxed{960}$.[/hide]

Alternatively, instead of placing our restrictions first (Pierre, Thomas and Rosa), we can first place the other five people in a circle in $5!/5 = 4!$ ways (we fix one of the five people to account for rotational symmetry). Then we have $5$ spots between each of the five people where we can insert Pierre and Thomas, and $4$ remaining spots where we can insert Rosa (as she cannot be next to them). Remembering to permute Pierre and Thomas in $2!$ ways, we have $4! \cdot 5 \cdot 4 \cdot 2! = \boxed{960}$ ways.

we dont count for first seating as rotation
2 spaces so x2
then 5 people exlcuding rosa, two times so 5x4
next we have the rest including rosa
4x3x2x1
= 960


vector
linear algebra
linear algebra unsolved

Let $ X,Y\leq V$, $ V$ vector space and $ A$ base for $ X\cap Y$ and $ B$ base for $ X$ such that $ A\subseteq B$, and $ C$ base for $ Y$ such that $ A\subseteq C$. Prove that $ B\cup C$ is base for $ X\plus{}Y$.

It's easy to see that $ B\cup C$ generate $ X\plus{}Y$, but I have somewhat of a hard time proving that vectors from $ B\cup C$ are linearly independent. We don't assume that  $ X,Y,V$  have finite dimension.

Let $ W$ denote the complement sub-space of $ X\cap Y$ in $ Y$ with associated basis $ B\setminus (A\cap B)\equal{}B\setminus A$.

Now let $ v_1,\ldots,v_n$ be any finite selection of vectors out of the (possibly infinite) set $ A\cup B$ with $ 0\equal{}\sum_{i\equal{}1}^na_iv_i$ and $ a_i\in K$, the field over which the vector space $ V$ is defined. If the vectors are all contained in $ B$ (resp. $ A$) we conclude $ a_i\equal{}0$, since they must be independent as a basis for $ Y$ (resp. $ X$). Wlog we can assume $ v_1,\ldots,v_m\in A\,,\,v_{m\plus{}1},\ldots,v_n\in B\setminus A$, then $ w\equal{}\sum_{i\equal{}m\plus{}1}^na_iv_i\in W\subseteq Y$ and on the other hand $ w\equal{}\minus{}\sum_{i\equal{}1}^ma_iv_i\in X$. Now $ X\cap W\equal{}X\cap (Y\cap W)\equal{}(X\cap Y)\cap W\equal{}\{0\}$. This implies $ w\equal{}0$ and therefore $ \forall i\equal{}1,\ldots,m: a_i\equal{}0$, since $ A$ is a base and $ \forall i\equal{}m\plus{}1,\ldots,n: a_i\equal{}0$, since $ B\setminus A$ is a base for $ W$.


Heres a problem that I don't know the solution of / any 'easy' way to do it. It was from the Australian Maths Competition which nobody here would have heard of, just a 30 multichoice question test. Real fun. Heres the number 30 in the senior exam from 2002:

How many 1s are there in the result of
1 + 11 + 111 + ... + 11..111 (which is 2002 ones)

Any ideas?

I get 224 for this. Hopefully that's one of the answer choices. I'll wait and let some other people work on it before I post my idea though.

Very nice. Yes this was one of the solutions.

I did it essentially the same way, just using a slightly different way to sum the series.

yeah i was thinking like zeta. but i didnt know how to carry it out. and im lazy


Is it possible to simplify this more in a meaningful manner?

$\pi r^2 \frac{sin^{-1}(\frac{c}{2r})}{180} - \frac{c}{2} \sqrt{r^2 - \frac{c^2}{4}}$

If it is, please post your method of simplification.

isn't this the equation for the segment in the circle??
where c, is the cord, r is the radius of circle?

Yeah, it is. People stopped replying to it. This formula works, but is very "clumsy".

[quote="4everwise"]I suppose that this cannot be simplified then.[/quote]
yep i looked at it and there are no signs of additional simplifying


function
combinatorics unsolved
combinatorics

Two sequences $ P(m,n)$ and $ Q(m,n)$ are defined as follows ($ m,n$ are integers). $ P(m,0)\equal{}1$ for $ m\geq 0$,$ P(0,n)\equal{}0$ for $ n\geq 1$. $ P(m,n)\equal{}0$ for $ m,n<0$. $ P(m,n)\equal{}\sum_{j\equal{}0}^{n}P(m\minus{}1,j)$ for $ m\geq 1$. $ Q(m,n)\equal{}P(m\minus{}1,n)\plus{}P(m\minus{}1,n\minus{}1)\plus{}P(m\minus{}1,n\minus{}2)$ for $ m\geq 1$. Express $ Q(m,n)$ in terms of $ m$ and $ n$ for $ m\geq 1$.

yes, work with generating functions, e.g.,

$ F_m(z)=\sum_{n=0}^{\infty}P(m,n)z^n$, then $ F_0(z)=1$ and for arbitrary $ m$ we have:

$ F_m(z)\cdot\sum_{n=0}^{\infty}z^n=\sum_{k=0}^{\infty}z^k\sum_{v=0}^kP(m,v)=F_{m+1}(z)$ from which we conclude: $ F_m(z)=(1-z)^{-m}$, i.e. for $ m>0$ we have $ F_m(z)=\frac{1}{(m-1)!}\frac{\partial^{m-1}(1-z)^{-1}}{\partial z^{m-1}}=\sum_{n=0}^{\infty}\left(n+m-1\atop m-1\right)z^n$, i.e. $ P(m,n)=\left(n+m-1\atop m-1\right)$.

This gives for $ m>0,n>2$ the identies $ Q(m,0)=1, Q(m,1)=m,Q(1,2)=1,Q(1,n)=0$ and for $ m,n>1$:

$ \begin{array}{lcl}Q(m,n) & = &\left(n+m-2\atop m-2\right)+\left(n+m-3\atop m-2\right)+\left(n+m-4\atop m-2\right)\\[0.25cm]
 & = & \frac{(n+m-4)!}{(m-2)!n!}\cdot (2n^2-9n+6+(n-m)^2))\end{array}$

(modulo typos  :blush: )

and there were typos in the last step, which can be more elegantly (?) written as:

$ \forall n,m>1: Q(m,n)\equal{}\left(n\plus{}m\minus{}1\atop m\minus{}1\right)\minus{}\left(n\plus{}m\minus{}4\atop m\minus{}1\right)$


Please suggest ideas for the following problem:

Given the set of consecutive positive integers {1, 2, 3, ..., x, ...}.  The numbers in this set are arranged in ascending order.  A purple mark is placed under each number of the set.  Of those with one purple mark, beginning with the second (that is #2), a second purple mark is placed under every seventh element in the ascending order.  Of those with two purple marks, beginning with the second (that is #9), a third purple mark is placed under every seventh element in the ascending order.  Of those with three purple marks, beginning with the second (that is #58), a fourth purple mark is placed under every seventh element in the ascending order.  This pattern of marking continues.  If x is the smallest number with 8 purple marks under it, find the value of x.

Interesting problem...but you can use a simple summation to find x.

In order to find x, you would use the following equation:  1+(7^0)+(7^1)+(7^2)+(7^3)+(7^4)+(7^5)+(7^6)=x
The reason for this is that the interval between two numbers with one purple mark is 1, for two purple marks is 7, for three is 49, and, for any general n, is 7^(n-1).

You could add an extra layer to it to that the next time, you use the 4th, then the 8th, and you could make the intervals change according to some interesting pattern as well.  You could also have an alternation between adding and removing marks, and you could add or remove different numbers of marks each time.  That would make the problem thoroughly confusing confusing at first glance  :lol:


Doug and Dave shared a pizza with $ 8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $ \$8$, and there was an additional cost of $ \$2$ for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?

$ \textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

[hide]Cost of plain pizza=$\$8$
Cost of pizza+anchovies on half=$\$8$+$\$2$=$\$10$
Cost of plain pizza slice=$\$1$
Number of anchovie slices=$4$
therefore, cost of one anchovie slice=$\$1$+($\$2$/$4$)=$\$1.50$
Dave eats 4 anchovie slices and one plain slice.
Therefore Dave pays $4$x$\$1.50$+$\$1$=$\$7$
Doug pays $\$10$-$\$7$=$\$3$
Dave's payment-Doug's payment= $\$7$-$\$3$=$\$4$

$D$[/hide]

because there is no need to bump these ancient threads :D

I mean it's technically not a bad thing in small amounts, but too much of anything is bad.

RFT if you have a different and unique solution, bumping is fine. But essentially repeating a solution or writing a trivial solution is considered useless bumping. :)

[quote=RedFireTruck][quote=AOPqghj]looks like we'll be seeing a lot of 2006 AMC solutions now[/quote]

[quote=Alex-131]I agree there are a lot of posts about them since RFT decided to solve each and every one of them[/quote]

why is everybody alwats so mad at me :([/quote]

Also you can post your solutions on AoPS wiki, I think that would be better if you're going to post in mass amounts.

but bumping is so much fun! (especially the part where the number under my name goes up)

[quote=samrocksnature]Feeling trolly?[/quote]

yes lol


geometry
3D geometry
analytic geometry
Putnam

How many vertices does a cube in 4 dimensions have? How any vertices does a cube in 137 dimensions have?

[hide="Solution"]
Any hypercube in $ n$ dimensions can be defined as a transformation on the unit hypercube in that dimension, which has every vertex whose coordinates are chosen from $ {0,1}$. There are $ 2^n$ of them.[/hide]

on expansion about hypercubes:

what is the radius of the largest circle that can be entirely contained in a 4-d hypercube?

this was putnam 2008 problem b2 or b3 i think.


[quote="SUPERMAN2"]Let a,b,c be the positive numbers satisfying $ (ab)^3\plus{}(bc)^3\plus{}(ca)^3\equal{}1$.Find the minimum value of A=$ ab(a\plus{}b\minus{}c)\plus{}bc(b\plus{}c\minus{}a)\plus{}ca(c\plus{}a\minus{}b)$[/quote]
The minimal value is $\sqrt3.$ :wink:


Wilma and Betty ran a 100-meter race at top speed, and Wilma finished when Betty had 10 meters to go. They decided to run again, but this time Wilma gave Betty an advantage. Wilma's starting point was 10 meters behind the original starting point. Given that Wilma and Betty run at the same speeds as the previous race, how many meters will Betty be from the finish line when Wilma crosses it?

Since Wilma runs $ 100$ meters in the same time Betty takes to run $ 90$ meters, we know that Wilma's speed is $ \frac{10}{9}$ times Betty's. So we let $ B \equal{} \text{Betty's speed(it doesn't matter what it is or the label)}$. Suppose for the sake of ease that Betty's speed is $ \frac{1\text{meter}}{\text{min}}$
Next, when they run again, while Wilma runs $ 110$ meters, Wilma covers $ \frac{110}{\frac{10}{9}} \equal{} 99$ in the same time that Betty covers $ 100$ (because she runs at the same speed and the same distance). That means they were $ \boxed{1\text{meter}}$ away when Wilma crossed.


Does anyone have any college interview tips or know where I can find some tips for college interviews (preferably free and preferably for college, not job)?

[url=http://talk.collegeconfidential.com/showthread.php?s=40135703eeda47870d4f986037426857&t=6372]College Interview Questions that You Should Prepare For![/url]

[url=http://talk.collegeconfidential.com/showthread.php?s=40135703eeda47870d4f986037426857&t=6963]Interview Etiquette[/url]

[url=http://talk.collegeconfidential.com/showthread.php?s=40135703eeda47870d4f986037426857&t=6371&highlight=interview+tips]Interview Questions You Should Know About[/url]

[url=http://talk.collegeconfidential.com/showthread.php?s=40135703eeda47870d4f986037426857&t=684]MIT Alumni Interview[/url]

[url=http://talk.collegeconfidential.com/showthread.php?s=40135703eeda47870d4f986037426857&t=7993]Interview Tips[/url]

All of these threads are from College Confidential, an obsessive, occasionally badly misleading, but usually useful online discussion forum. Spend more time preparing your application and doing your schoolwork than chatting online there.

Hope this helps! Good luck.

[quote="rrusczyk"]Over the last 5 years I've done a fair amount of interviewing for Princeton.  In early December, I'll do a Math Jam to discuss my opinions on the matter if there's sufficient interest in my doing so.[/quote]

I'm sure there are quite a few AoPSers who would like to hear your point of view on college interviews. When my son and I met the Princeton representatives (one admissions office officer, and one alumni representative) here in Minnesota this year, we mentioned that there are some Princeton grads in California who have put together a great online site for math discussion and who are raising the visibility of Princeton among math-liking young people.

[quote="rrusczyk"]Over the last 5 years I've done a fair amount of interviewing for Princeton.  In early December, I'll do a Math Jam to discuss my opinions on the matter if there's sufficient interest in my doing so.[/quote]

There is...rest assured.

Just make sure I can make it :-D

Hi, JS1527,

Good to see you posting again. How are you liking college? Are you using the Apostol calculus textbook in your class now? I'm curious about how your studies are going.

Hey tokenadult,

Yea, I'm very very busy...  I'm actually not taking the same classes I planned to.  I'm taking Calculus in 3D (multivariable) and I use Stewart.  It's pretty difficult as it is...  I'm also taking an intro to discrete math /proof writing course.  Next semester...differential equations and linear algebra.  Glad to hear from you too, and good to see you're still around to give advice.  As for how much I like college, all I can say is that it's different, and harder than good old public school, but you get used to it and I'm doing well.  Now all I'm left with is deciding on my major...which is tough.  Anyway, I'll try to check up on AoPS as I can...

[quote="JS1527"]Mathnovice, the best thing I can tell you is to BE YOURSELF.  I know you might be nervous, but you can necessarily research how to "be yourself" by looking at articles and such.  Don't talk about grades and GPA or w/e, because they already have that on file, tell them something new...  Try to bond w/your interviewer.  Just chat and don't bore them.  It's not wrong to talk about your favorite sports team (go red sox!) or how you love swimming or how your favorite band is Something Corporate.  Anyway, just be original....they don't care too much about school /academics, they want to know about YOU, so be yourself.  Don't sweat it!

- Julie[/quote]

My MIT interviewer never really gave me a chance to have a conversation with him. He only asked question after question : (

:(

Well, as long as you didn't really screw it up you should be fine.  I'll keep my fingers crossed for you.

[quote="mathnovice"]My MIT interviewer never really gave me a chance to have a conversation with him. He only asked question after question : ([/quote]

That's not necessarily a bad thing. Perhaps some interviewers give high scores/reviews/etc. for just a straight-forward interview...


calculus
derivative
function
induction
logarithms
algebra
binomial theorem

How do you prove that: $(a^n)'=\frac1n \cdot a^{n-1}$

This might not be the nicest proof, and I'm not the greatest at calculus (I've just learn a little on my own time), but I do have a suggestion (anyone feel free to correct me on incorrect statements).

Consider the function $f(x) = x^n$. Then, by the definition of the derivative, we know $f'(x)$ is equal to the limit, as h approaches 0, of $\frac{f(x+h)-f(x)}{h}$, which in our case is $\frac{(x+h)^n-x^n}{h}$. Now, it is clear that the $x^n$ terms will cancel. By binomial theorem, what we have left is $\frac{\sum_{k=1}^n\binom{n}{k}x^{n-k}h^k}{h}$, or $\sum_{k=1}^n\binom{n}{k}x^{n-k}h^{k-1}$. It is clear that the only term that is not multiplied by $h$ is the first term, when $k=1$. Since all other terms will zero out (because $h$ tends to zero), the derivative will be equal to $\binom{n}{1}x^{n-1}$, which is equivalent to $nx^{n-1}$.

I hope that helps!

[quote="Kent Merryfield"]I know three proofs:

1. Use the binomial theorem to expand $(x+h)^n$. This is what nolachrymose did.

2. Use the geometric series factorization:

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1}).$

3. Use the product rule in the inductive step of a proof by induction on $n.$[/quote]

$y=x^n\Rightarrow \ln y=n\ln x\Rightarrow y'/y=n/x\Rightarrow y'=ny/x\Rightarrow y'=nx^{n-1}$

It holds for all $n$, which is a nice benefit.

man, everyone beats me to these easy ones. lol


If $m|ab^n+cn+d, n\geq 0,n\in\mathbb{Z}$,prove that $m|c^2$

This isn't true (for example $m=2, a=d=1, n=0, c=3$ would imply that $2|9$). Was there a typo?

Well,this is the question,lets say $m\neq a \neq b\neq c\neq d$

Still doesn't work. Try m=13, a=1, b=2, n=3, c=4, d=6.
You can see quite quickly that c and m must be independent. Fixing c, you can get any m to satisfy the first condition with an appropriate d.