What is the tens digits of 5^2005. I have lots of trouble with problems like these. 2^255 divided by 5 and find the remainder is an another example.

This all can easily be done by modulo. (If you know it)

sorry, i do not know what modulo is. Could you please explain it?

$ n\equiv m\mod k$
That mean if we divide a number $ n$ by any number $ k$ we have the remainder $ m$.(Nowhere near the Classroom Math)

Congruency modulo. Sorry wrote only modulo. :blush:

[hide]

Notice that for $ n\ge2$, $ 5^n\equiv 25\mod 100$. Thus, the ten's digit is 2.

Look for a pattern

$ 2^1\equiv 2 \mod5$

$ 2^2\equiv4\mod5$

$ 2^3\equiv3\mod5$

$ 2^4\equiv1\mod5$

$ 2^5\equiv2\mod5$

...

This repeats in groups of 4s. So, we need to find the modulo 4 residue of 225. We have $ 225\equiv1\mod4$. So, $ 2^{225}\equiv2^1\equiv2\mod5$[/hide]

These types of problems usually require you to find a pattern. For example, you could notice that 5 raised to any integer greater than 1 ends in "25".

For the second one, note the final digits in the powers of two:

2^1 => 2

2^2 => 4

2^3 => 8

2^4 => 6 (just the last digits)

2^5 => 2

2^6 => 4

And the pattern repeats.

255 is congruent to 3 mod 4, so the final digit would be the same as the one in 2^3.
In other words, 8.
The remainder is therefore 3.

[quote="lifeisacircle"][hide]

Notice that for $ n\ge2$, $ 5^n\equiv 25\mod 100$. Thus, the ten's digit is 2.

Look for a pattern

$ 2^1\equiv 2 \mod5$

$ 2^2\equiv4\mod5$

$ 2^3\equiv3\mod5$

$ 2^4\equiv1\mod5$

$ 2^5\equiv2\mod5$

...

This repeats in groups of 4s. So, we need to find the modulo 4 residue of 225. We have $ 255\equiv3\mod4$. So, $ 2^{225}\equiv2^3\equiv3\mod5$[/hide][/quote]

Edited a mistake. I accidentally thought it asked for the remainder when 2^[i]225[/i] is divided by 5.


integration
trigonometry
calculus
calculus computations

\[ Evaluate :  \int \frac{\sqrt{x \plus{} \sqrt{25 \plus{} x^2}}}{x} dx\]

Let $ u\equal{}\sqrt{\sqrt{25\plus{}x^{2}}\plus{}x}$

$ \int \frac{\sqrt{\sqrt{25\plus{}x^{2}}\plus{}x}}{x}dx$

$ \equal{}\allowbreak \int \allowbreak \left( 2\plus{}\frac{10}{u^{2}\minus{}5}\minus{}\frac{10}{u^{2}\plus{}5}\right) du\allowbreak$

$ \equal{}\allowbreak 2u\minus{}2\sqrt{5}\tanh ^{\minus{}1}\frac{u}{\sqrt{5}}\minus{}2\sqrt{5}\tan ^{\minus{}1}\frac{u}{\sqrt{5}}\plus{}C$

$ \equal{}\allowbreak 2\sqrt{\sqrt{25\plus{}x^{2}}\plus{}x}\minus{}2\sqrt{5}\tanh ^{\minus{}1}\frac{\sqrt{\sqrt{25\plus{}x^{2}}\plus{}x}}{\sqrt{5}}\minus{}2\sqrt{5}\tan ^{\minus{}1}\frac{\sqrt{\sqrt{25\plus{}x^{2}}\plus{}x}}{\sqrt{5}}\plus{}C$

:roll:

[quote="IndoChina"]Let $ u \equal{} \sqrt {\sqrt {25 \plus{} x^{2}} \plus{} x}$

$ \int \frac {\sqrt {\sqrt {25 \plus{} x^{2}} \plus{} x}}{x}dx$

$ \equal{} \allowbreak \int \allowbreak \left( 2 \plus{} \frac {10}{u^{2} \minus{} 5} \minus{} \frac {10}{u^{2} \plus{} 5}\right) du\allowbreak$

$ \equal{} \allowbreak 2u \minus{} 2\sqrt {5}\tanh ^{ \minus{} 1}\frac {u}{\sqrt {5}} \minus{} 2\sqrt {5}\tan ^{ \minus{} 1}\frac {u}{\sqrt {5}} \plus{} C$

$ \equal{} \allowbreak 2\sqrt {\sqrt {25 \plus{} x^{2}} \plus{} x} \minus{} 2\sqrt {5}\tanh ^{ \minus{} 1}\frac {\sqrt {\sqrt {25 \plus{} x^{2}} \plus{} x}}{\sqrt {5}} \minus{} 2\sqrt {5}\tan ^{ \minus{} 1}\frac {\sqrt {\sqrt {25 \plus{} x^{2}} \plus{} x}}{\sqrt {5}} \plus{} C$

:roll:[/quote]

your solution very nice , IndoChina

thanks

IndoChina

small request  , please

may be explain for me steps the substitute till this step : $ \equal{} \allowbreak \int \allowbreak \left( 2 \plus{} \frac {10}{u^{2} \minus{} 5} \minus{} \frac {10}{u^{2} \plus{} 5}\right) du\allowbreak$

Sure...

Let $ u \equal{} \sqrt {\sqrt {x^{2} \plus{} 25} \plus{} x}\Rightarrow \sqrt {x^{2} \plus{} 25} \plus{} x \equal{} u^{2}..(1)$

Thus $ \frac {1}{u^{2}} \equal{} \frac {1}{\sqrt {x^{2} \plus{} 25} \plus{} x}\frac {\sqrt {x^{2} \plus{} 25} \minus{} x}{\sqrt {x^{2} \plus{} 25} \minus{} x} \equal{} \frac {\sqrt {x^{2} \plus{} 25} \minus{} x}{25},$

Hence $ \sqrt {x^{2} \plus{} 25} \minus{} x \equal{} \frac {25}{u^{2}}..(2)$

1)$ \minus{}$(2) gives $ x \equal{} \frac {25}{u^{2}} \minus{} u^{2}$

Evaluate $ dx$ and the substitution is complete.. Hope this helps... :)

IndoChina,is there anything that you don't know?You are really good! :lol:

hi IndoChina there is a mistake from (1) -- (2) [code]x[/code] should be equal to [code](u^2)/2 - 25/2(u^2)[/code]


Find the function $f: \mathbb{R}\rightarrow\mathbb{R}$ such that  for every $x,y\in\mathbb{R}$ we have $f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y)).$

Put $y=0$ and denote $b=f(0)$:

$f(x^3)-b=x^2(f(x)-b)\implies f(x^3)=x^2(f(x)-b)+b$

Then the initial equation becomes

$x^2(f(x)-b)-y^2(f(y)-b)=(x^2+xy+y^2)(f(x)-f(y))$

Put $y=1$ and denote $c=f(1)$:

\begin{eqnarray*}x^2(f(x)-b)-c+b=(x^2+x+1)(f(x)-c) &\iff& x^2f(x)-bx^2-c+b=x^2f(x)+xf(x)+f(x)-cx^2-cx-c\\ &\iff& (x+1)f(x)=cx^2+cx-bx^2+b\end{eqnarray*}

This gives $f(x)=\frac{cx(x+1)-b(x+1)(x-1)}{x+1}=(c-b)x+b$. This formula satisfies $f(0)=b, f(1)=c.$

Denote $a=c-b$ which gives the general solution $f(x)=ax+b$. Checking proves that this is a solution for any real $a,b.$

[quote="silouan"]$f(x)=x$[/quote]

Any proof silouan?????.


[quote="ln(dx/dy)"]140 tickets were sold, some at whole price and some at half price, neither a fractional amount. The total profit was 2001 dollars.  How much money was raised from the whole tickets?

${2^{n^{2}}+1}$=a prime number (this has nothing to do with the problem, it's just something I found recently)[/quote]

I believe that when you have ${2^{n^{2}}}$ that means you square the variable first.  Therefore as an example, ${2^{3^{2}}+1={2^{9}}+1=512+1=513}$ which is divisible by 3 among other things.  Do you mean ${(2^n){^{2}}+1}$ ??

If so that is equivalent to ${2^{2n}}+1$ which again when n=3 yields ${2^{6}}+1=64+1=65$ which is divisible by my lucky 13!

After further consideration, I think the rule is ${2^{2^{n}}+1}$ is prime.

Does that make sense?  Maybe just a typo :)

Josh Frost

Actually, no. That was something Fermat or Pascal or Eular found(can't remember which). But it was proven untrue later. When n=5, it's a composite number.

THanks Filletwho,

So does anyone know if there is any such rule at all???

I think it was Mersenne. Although $2^{2^n}+1$ is not always prime, it does give us the largest known prime, according to a math poster I once saw, but I forgot what $n$ they used.

And for your problem, I think we need the price; otherwise there are many solutions.

Mersenne primes are primes of the form $2^p - 1$, where $p$ is a prime number.  (A challenge: show that if $2^n - 1$ is a prime number, then $n$ is a prime number.)  Not all such numbers are prime, but they do appear reasonably frequently.  The largest known primes are currently Mersenne Primes.  In fact, I am currently running a program on my computer as part of the [url=http://www.mersenne.org/prime.htm]Great Internet Mersenne Primes Search (GIMPS)[/url] which checks Mersenne numbers to see if they are prime.  (I'm currently working on $2^{24102049}-1$ -- 58% finished.)

Fermat Primes are numbers of the form $2^{2^n} + 1$.  To quote [url=http://mathworld.wolfram.com/FermatPrime.html]Mathworld[/url] on the subject: [quote]Fermat  conjectured in 1650 that every Fermat number is prime and Eisenstein proposed as a problem the proof that there are an infinite number of Fermat primes. At present, however, the only Fermat numbers $2^{2^n}+1$ for $n \geq 5$ for which primality or compositeness has been established are all composite, and it seems unlikely that any more will be found using current computational methods and hardware..[/quote] It is perhaps surprising that Fermat did not discover that $2^{2^5} + 1 = 2^{32} + 1 = 4294967297$ is divisible by 641.  On the other hand, I would have been too lazy to do that out by hand as well.

Going further and further on the tangent just     to give a, little less known  trivia if you want to show-off :) ...
[hide][quote]Mersenne primes are primes of the form $2^p - 1$ , where p is a prime number. (A challenge: show that if 2^n - 1 is a prime number, then n is a prime number.) [/quote]

$p = 2^n - 1 $ is prime if and only if

$p$ divides  $Cosh (2^{n-2} \ln (2+ \sqrt {3})) $

As useless result  as one can think  of but  may come in handy to impress  :D
(When n >2, of course. BTW this is the Famous Lucas  (+Lehmer)  test , I think this  (or some thing like this) is the form    how Lucas first reported his result ..:)[/hide]


A juice company makes two kinds of juice: A and B. A gallon of a juice A is made by mixing 3 quarts of pumpkin juice and 2 quarts of carrot juice, while one gallon of juice B is made by mixing 5 quarts of pumpkin juice and 3 quarts of carrot juice. A profit of 30 dollars is made on every gallon of juice A sold and 20 dollars is made on every gallon of juice B sold. If the company has only 100 gallons of pumpkin juice and 120 gallons of carrot juice, and assuming they sell ALL of their juice, how many gallons of each type of juice should be made in order to maximize its profit?

I want to graph this on the x-y coordinate plane and draw some lines as inequalities to find the feasible region.  How can I tell which things to use as the x and y coordinates? How do I get the inequalities from that information? What is the equation for profit in terms of x and y?

call one x, one y (for x/y axis), then find intercepts by seeing when it maxes out and just meets a limit (so like 100/3 for a and 20 for b) so if we do a as y and b as x then

graph y<100/3-5/3x i think it is and then you should be able to do the rest :)


parameterization
real analysis
real analysis unsolved

Can anyone explain to me why in a logistic model when the parameter exceeds the bifurcation value, the population tends to zero unless we set the parameter to zero.

I don't think the notation for this is universal enough for it to be clear what you mean when you refer to "the parameter." Can you provide more details, including the formulas for what you are talking about?


the function f is defined by f(n) = f(n-1) + f( n-2) it is also true that f(1) = 3 and f(3) =10, what is the value of  f(5) =?

hello, $ f(3)\equal{}f(2)\plus{}f(1)$ from here we get $ f(2)\equal{}7$, further $ f(4)\equal{}f(3)\plus{}f(2)$, $ f(4)\equal{}10\plus{}7\equal{}17$ and $ f(5)\equal{}f(4)\plus{}f(3)\equal{}17\plus{}10\equal{}27$.
Sonnhard.


I am having problems applying this problem with the given information?
If the bob on the bottom of a 22 cm pendulum traces a 9.5 cm arc on each swing, through what angle in degrees does the pendulum rotate on each swing.

would i use s=theta x r to solve?

With theta in radians and r being the length of the pendulum, it would work. If they did not give you the length of the pendulum string, you do not have enough information.
If later they ask you for the amplitude of the "simple harmonic motion," remember that it is only half of the total swing.

[quote="madiemae"]so i would solve for theta , making , theta=s/r

theta=9.5cm/22cm

the answer would be in cm and i need degrees for theta.  so , do i need to covert cm to radians and then to degrees?

this is the part i get confused at.[/quote]Both quantities in your division are in cm. The result of that division is just how many times 9.5 cm can fit into 22 cm. It is just a number, with no units, but represents radians. You could convert radians to degrees, or solve the problem a different way.

If radians confuse you, you could look at the problem another way. The angle/arc the pendulum moves through is part of a circle, whose entire circunference is $ 2\pi R$ where $ R$ is the radius, which is the length of the pendulum (22 cm). $ 2\pi(22 cm) \equal{} 138.23 cm$. The pendulum moved through just a fraction of the whole circle. That fraction is the fraction of 360 degrees that you are looking for.
$ \frac{9.5 cm}{138.23 cm}\equal{}0.0687...$  and  $ (0.0687...)(360 degrees) \equal{} 24.7...degrees$

To convert radians to degrees or the other way around take into account that a whole circle is 360 degrees or $ 2\pi$ radians, so $ \pi$ radians is the same as 180 degrees (half a circle). You could divide the $ \frac{9.5}{22}$ radians you had by $ \pi$ to find how many half circles, and then multiply by 180 degrees to find the angle in degrees.


algebra
polynomial
modular arithmetic
superior algebra
superior algebra unsolved

Let $ x^r\minus{}1 \in \mathbb F_p[x]$ . We know that its only root is $ 1$ in $ \mathbb F_p[x]$. How do I prove that $ x\plus{}j$, with $ j \in \{0,\ldots,r\minus{}1\}$ are units in $ \mathbb F_p[x]/ \frac{x^r\minus{}1}{x\minus{}1}$ ?

the units are exactly those polynomials (mod $ (x^r\minus{}1)/(x\minus{}1)$), which are coprime to $ (x^r\minus{}1)/(x\minus{}1)$. the rest is clear.

And in the case of $ r$ prime but not $ p \equiv 1 \pmod r$?

In that case, $ \frac{x^r\minus{}1}{x\minus{}1}$ is irreducible, it generates a maximal ideal, and the quotient is a field.

Why $ \frac {x^r \minus{} 1}{x \minus{} 1}$ is irreducible? Just because $ r$ is prime? Irreducible in $ \mathbb F_p[x]$, I mean.


if sin t=4/5

what is sin(pi - t)

and what is sin(pi + t)

Actually, the sine function has a period of $ 2\pi$, i.e. $ \sin(2\pi\plus{}t)\equal{}\sin t$

With the angle sum and difference identities, you can get:

$ \sin(\pi \plus{} t) \equal{} \sin \pi \cos t \plus{} \cos \pi\ sin t \equal{} 0 (\cos t) \plus{} (\minus{}1)(\sin t) \equal{} \minus{}\sin t$

$ \sin(\pi \minus{} t) \equal{} \sin \pi \cos t \minus{} \cos \pi \sin t \equal{} 0 (\cos t) \minus{} (\minus{}1)(\sin t) \equal{} \sin t$

Since you know the value of $ \sin t$, you can solve the problem from here.

[quote="FantasyLover"]well since sin function repeats with cycle of $ \pi$

$ \sin (\pi \plus{} t) \equal{} \frac {4}{5}$

and $ \sin(\pi \minus{} t) \equal{} \sin( \minus{} t) \equal{} \minus{} \sin(t) \equal{} \minus{} \frac {4}{5}$

EDIT:

oh, oops.

sry :oops:[/quote]

Isn't $ sin(\pi \minus{}t)\equal{}sin(t)$?

[quote="Bugi"][quote="FantasyLover"]well since sin function repeats with cycle of $ \pi$

$ \sin (\pi \plus{} t) \equal{} \frac {4}{5}$

and $ \sin(\pi \minus{} t) \equal{} \sin( \minus{} t) \equal{} \minus{} \sin(t) \equal{} \minus{} \frac {4}{5}$

EDIT:

oh, oops.

sry :oops:[/quote]

Isn't $ sin(\pi \minus{} t) \equal{} sin(t)$?[/quote]

nope.

$ \sin(\minus{}\theta)\equal{}\minus{}\sin\theta$

[quote="FantasyLover"][quote="Bugi"][quote="FantasyLover"]well since sin function repeats with cycle of $ \pi$

$ \sin (\pi \plus{} t) \equal{} \frac {4}{5}$

and $ \sin(\pi \minus{} t) \equal{} \sin( \minus{} t) \equal{} \minus{} \sin(t) \equal{} \minus{} \frac {4}{5}$

EDIT:

oh, oops.

sry :oops:[/quote]

Isn't $ sin(\pi \minus{} t) \equal{} sin(t)$?[/quote]

nope.

$ \sin( \minus{} \theta) \equal{} \minus{} \sin\theta$[/quote]

Yeah,but I am sure that $ sin(\pi\minus{}\alpha)\equal{}sin\alpha$

say $ \alpha \equal{} 0$

$ \sin \pi \equal{} \sin0$? nope.

blah

$ \alpha\equal{}30$ (degrees)

$ \sin 30 \equal{} \sin 60$?

.......

Ouch, I'm very sorry!

I thought all time of $ 2\pi$ and I was writing $ \pi$!!!

Again I'm sorry!

lol, no need to be sorry.

meh i thought sin graph rotates with cycle $ \pi$ too before(post #1)... blah


[quote="towersfreak2006"]Source: M&IQ 2003

Prove that:
$\displaystyle\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-c)(b-a)}+\frac{1}{c(c-a)(c-b)}=\frac{1}{abc}$

for all sets of distinct nonzero numbers $a,b,c$.

This is actually a problem from the Equations and Expressions chapter of AoPS volume 2, but I'm not convinced by the explanation, so if somebody could clarify, it would be greatly appreciated.  :)[/quote]

[hide]Multiply each side by (abc)(a-b)(a-c)(b-c)

bc(b-c) - ac(a-c) + ab(a-b) = (a-b)(b-c)(a-c)
cb 2 - bc 2 - ca 2 +ac 2 + ba 2 - ab 2 =
cb 2 - bc 2 - ca 2 +ac 2 + ba 2 - ab 2

QED[/hide]

[hide="Click for solution"]
Adding the first two terms gives $\frac{(a-b)(a^2-ac-b^2+bc)}{ab(c-b)(a-c)(a-b)^2}=\frac{((a+b)(a-b)-c(a-b))}{ab(c-b)(a-c)(a-b)}=\frac{(a+b-c)}{ab(c-b)(a-c)}$
Now add this to the las term: $\frac{ab-c(a+b-c)}{abc(c-b)(c-a)}=\frac{(a-c)(b-c)}{abc(c-b)(c-a)}=\frac{1}{abc}$
 :D[/hide]

OKay, well I was actually hoping somebody would explain it the way the book did it, that is:

$\displaystyle\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-c)(b-a)}+\frac{1}{c(c-a)(c-b)}=\frac{1}{abc}$

Show $\displaystyle\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-c)(b-a)}+\frac{1}{c(c-a)(c-b)}-\frac{1}{abc}=0$ by first writing the left side with common denominator $abc(a-b)(a-c)(b-c)$ to get $\displaystyle\frac{bc(b-c)-ac(a-c)+ab(a-b)-(a-b)(a-c)(b-c)}{abc(a-b)(a-c)(b-c)}$ and taking the numerator as a polynomial in $c$.
So, we have $f(c)=bc(b-c)-ac(a-c)+ab(a-b)-(a-b)(a-c)(b-c)$. Three roots of this function are $c=a$, $c=b$, $c=0$, and since $f(c)$ is a second degree polynomial, it must be zero everywhere.

So let me get this straight...it doesn't matter that in the actual problem $a,b,c$ are nonzero since the polynomial has three roots and thus it must be congruent to the zero function everywhere? Also, what exactly does setting the numerator as a function of $c$ do? Is it like saying "okay, we don't really care about $a,b$ since no matter what they are, the function will be zero anyways"?

It maxes perfect sense.  They prove the graph of the function is a horizontal line through zero no matter what a,b, and c are.  This is basically saying the identity is true always.  It is the same thing to say if y=x then y-x is zero no matter what y and x are.

When you set the numerator as a function of c, you think of a and b as constants and vary c. You look for values of c that make the function 0. Since you can find three different values, the function must be always 0, since it's second degree.

Well, you know any polynomial of degree n can have at most n zeroes.  That's the fundamental theorem of algebra.  Since f(c) is of degree 2 at most, it can have more than 2 roots only if it is not a polynomial, or the 0 function.

Note to towers: I thought it was fishy the first time I saw this argument back in high school, too.


analytic geometry
vector
geometry unsolved
geometry

$\Delta ABC$ is right angled in $A$
$D$ is outside the plane $(ABC)$
$BD \perp (ABC)$

$M=p_{A}BC$ (i.e. $AM \perp BC , M \in [BC]$ )
$N=p_{A}DC$ (i.e. $AN \perp DC , N \in [DC]$ )
$P=p_{B}AD$ (i.e. $BP \perp AD , P \in [AD]$ )
$Q=p_{B}DC$ (i.e. $BQ \perp DC , Q \in [DC]$ )

proove that $(AMN) \parallel (BQP)$


Let A and B be 2 square matrces from M n (R)

and A^m * B =A^m + B        ,m from N

Proove that A*B=B*A

...hope i got it well


A 2-inch elastic band is fastened to the wall at one end, and there's a bug at the other end. Every minute (beginning at time 0), the band is instantaneously and uniformly stretched by 1 inch, and then the bug walks 1 inch toward the fastened end. Will the bug ever reach the wall?

It seems to me that it the bug shouldn't, but I've having seen problems like these appear in several places, it's probably not as simple as I think it is  :maybe:

If wall to wall is 3 inches, then the bug will reach the wall for a brief second before having to walk back, according to the wording of the question.  However, that's probably not what the writer intended.

P.S. - This seems like more of a riddle to me?  Maybe?


Let $ a,b,c > 0$Prove
\[ \sum\sqrt {\frac {a^3}{a^2 \plus{} ab \plus{} b^2}}\ge\frac {\sum\sqrt {a}}{\sqrt {3}}
\]

[quote="secrets"]Yes,this problem is old.(posted by Phan Thanh Nam)[/quote]
Could you give me a link?
Thank you very much.

Sorry,I can't find the link.I know this problem through secrets in inequalities(Vietnamese verison) :)

[quote="secrets"]Yes,this problem is old.(posted by Phan Thanh Nam)[/quote]
not true , this problem is posted by Le Trung Kien ! :P  :P

[quote="Cai Ni"]Let $ a,b,c > 0,$ prove

$ \sum_{cyc}\sqrt {\frac {3a^3}{a^2 \plus{} ab \plus{} b^2}}\ge \sqrt {a}\plus{}\sqrt {b}\plus{}\sqrt {c}.$[/quote][quote="iqtimo"]$ a \equal{} x^{2} , b \equal{} y^{2}, c \equal{} z^{2}$ and[/quote]$ \sqrt{\frac{x^2\plus{}xy\plus{}y^2}{3(x^2\minus{}xy\plus{}y^2)}}\geq\frac{x^2\plus{}2xy\minus{}y^2}{2x^2}$

$ \Longrightarrow\sum_{cyc}\frac{x^3}{\sqrt{3(x^4\plus{}x^2y^2\plus{}y^4)}}\geq\sum_{cyc}\frac{x(x^2\plus{}2xy\minus{}y^2)}{2(x^2\plus{}xy\plus{}y^2)}\geq\frac{x\plus{}y\plus{}z}{3}.$

[quote="Ji Chen"]$ \sqrt {\frac {x^2 \plus{} xy \plus{} y^2}{3(x^2 \minus{} xy \plus{} y^2)}}\geq\frac {x^2 \plus{} 2xy \minus{} y^2}{2x^2}$[/quote]$ \frac {4x^4(x^2 \plus{} xy \plus{} y^2) \minus{} 3(x^2 \minus{} xy \plus{} y^2)(x^2 \plus{} 2xy \minus{} y^2)^2}{(x \minus{} y)^2}$

$ \equal{} x^4 \minus{} 3x^3y \plus{} 9xy^3 \minus{} 3y^4$

$ \equal{} \frac {(x^2 \plus{} 2xy \minus{} y^2)(x^2 \minus{} 5xy \plus{} 7y^2) \plus{} x^4 \minus{} 3x^3y \plus{} 4x^2y^2 \minus{} xy^3 \plus{} y^4}{2}.$

[quote="Cai Ni"]Let $ a,b,c > 0,$ prove

$ \sum_{cyc}\sqrt {\frac {3a^3}{a^2 \plus{} ab \plus{} b^2}}\ge \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}.$[/quote]
Let $ a\equal{}x^{2},b\equal{}y^{2},c\equal{}z^{2}$
$ \sum_{cyc}\sqrt {\frac {3a^3}{a^2 \plus{} ab \plus{} b^2}} \geq \sum_{cyc}\frac{3x^3}{2\left(x^{2}\plus{}y^{2}\right)\minus{}xy} \geq \sum_{cyc}x$


An aqueous solution of acetone, Ch3COCH3, is 10% acetone by weight. What is the mole percentage of acetone in this solution?

How would I approach this? Any help would be greatly appreciated.

[hide="I am not so good at chemistry"]
Pretend that there are 100g of solution: 90g water (18g/mol) and 10g acetone (about 58g/mol).

Thus there are 5.0 moles of water and 0.17 moles of acetone.
The mole percentage of acetone is

$\frac{0.17}{5+0.17}=3.3\%$[/hide]


topology
advanced fields
advanced fields unsolved

I have a question that:

A complex algebraic curve in $ \mathbb{C}^2$ is never compact,

where a complex algebraic curve is $ C\equal{}\lbrace{ \left( x,y \right) \in \mathbb{C}^2 : P(x,y)\equal{}0 \rbrace}$,

and $ P(x,y)$ is a nonconstant polynimial with complex coefficients and no repeated factors.

It's not bounded (I assume you want standard topology, not Zariski).


geometry
geometric transformation
reflection
geometry proposed

In triangle $\triangle ABC$, $P$ is an interior point such that

$m(\widehat{PBC}) = c, m(\widehat{PCB}) = 90+b$,

$m(\widehat{BAP})=a, m(\widehat{PAC}) = b$.

If $D$ is a point on the ray $[BC$  ($C$ is between $B$ and $D$) satisfying

$m(\widehat{CAD})=a+b$,

then find $m(\widehat{CDP})$

A' is the reflection of A wrt BC.
As AC is the bisector of <BAD, we have AB/AD = A'B/A'D = CB/CD;
hence the circle AA'C is the locus of M such as MB/MD = CB/CD
But, as <CPA = 90+<ACB = 180-<CA'A, it follows that AA'CP are concyclic and PB/PD = CB/CD
Hence PC is the bisector of <BPD and <CPD = <BPC = 90-b-c.
So <CDP = <PCB-<CPD = (b+90)-(90-b-c) = 2b+c


[b]Problem.[/b]  [i]Let $\triangle ABC$ be a triangle with its incircle touching the sides $BC, AC, AB$ at $D,E,F$, respectively.  Construct circles with diameters $AD, BE, CF$.  Prove that their radical center is the triangle's orthocenter $H$.[/i]

Enjoy! :)

[hide]Call the circles with diameters $AD$, $BE$, $CF$ $C_{1}, C_{2}, C_{3}$, respectively.  Let the altitudes from $A$, $B$, and $C$ hit the sides of the triangle at $X$, $Y$, and $Z$, respectively.  Because $m\angle{AXD}=90^{\circ}$, we must have that $X$ lies on $C_{1}$.  Similarly, $Y$ and $Z$ lie on $C_{2}, C_{3}$, respectively.  Then the power of $H$ wrt $C_{1}$ is equal to $HA*HD$.  But this is equal to $HE*HB$, because $\triangle{AHE}\sim \triangle{BHD}$.  This value is the power of $H$ wrt $C_{2}$.  Similarly, $H$ has the same power wrt $C_{3}$, and so $H$ is the radical center of $C_{1}, C_{2}, C_{3}$.[/hide]


hello, your inequality is aequivalent with
$ x^2\plus{}y^2\plus{}z^2\minus{}(x\plus{}y\plus{}z)\plus{}x^2z\plus{}xy^2\plus{}yz^2\minus{}3 \ge 0$.
This is true, because
$ \frac{x^2z\plus{}xy^2\plus{}yz^2}{3}\ge \sqrt[3]{x^3y^3z^3}\equal{}xyz\equal{}1$
and
$ x^2\plus{}y^2\plus{}z^2 \ge \frac{(x\plus{}y\plus{}z)^2}{3} \ge x\plus{}y\plus{}z$
because $ x\plus{}y\plus{}z \ge 3$.
Sonnhard.

[quote="FOURRIER"]$ x,y,z > 0$ , $ xyz \equal{} 1$

Then prove: $ \frac {x \minus{} 1}{y \plus{} 1} \plus{} \frac {y \minus{} 1}{z \plus{} 1} \plus{} \frac {z \minus{} 1}{x \plus{} 1} \ge 0$

Morroccan olympiad  :blush:[/quote]
hi
what do you mean by blush do you want to say something by it ? do you think that moroccan olympiad is bad ? if you want to say something say it fourrier (mathsmaroc i know you)


[quote="tnk"]a,b,c >0. Prove that:

$\frac{a^2}{b^3}  + \frac{b^2}{c^3}+ \frac{c^2}{a^3}$ >= $ \frac{a}{b^2}+ \frac{b}{c^2} + \frac{c}{a^2}$[/quote]

Using AM-GM,
$14\frac{a^2}{b^3}  + 2\frac{b^2}{c^3}+ 3\frac{c^2}{a^3}$ >= $19\frac{a}{b^2}$

Adding this with two other similar inequalities and dividing by 19 we get the results

[quote="E. Leung"]Using AM-GM,
$14\frac{a^2}{b^3}  + 2\frac{b^2}{c^3}+ 3\frac{c^2}{a^3}$ >= $19\frac{a}{b^2}$
[/quote]

I don't see how you got this. Could you please explain?

I have a solution for this problem :
 $ \frac{a^2}{b^3} + \frac{1}{b} \geq 2.\frac{a}{b^2} $
 It's enough to prove :
 $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{a}{b^2} + \frac{b}{c^2} +\frac{c}{a^2} $
 [b]Case 1:[/b] $ a \geq b \geq c $
 $ \to \frac{a}{b^2} + \frac{b}{c^2} +\frac{c}{a^2} \geq min (\frac{1}{c} + \frac{b}{c^2} + \frac{a}{b^2} , \frac{1}{c} + \frac{1}{a} +\frac{1}{b} ) $
  It's easy to show
 $ \frac{b}{c^2} + \frac{a}{b^2} \geq \frac{1}{a} + \frac{1}{b} $
 => Q.E.D
 [b]Case 2 [/b]: the same case 1


:help:
use fermat's last theorem to deduce the solutions of a^(n+p) + b^(n+q)=c^(n+r); in non-zero integers, where p<q<r are consecutive positive primes and n is a positive integer.

Who told you this was a simple problem or that FLT said anything about it?  This really isn't any easier than the hardest cases of [url=http://en.wikipedia.org/wiki/Beal%27s_conjecture]Beal's conjecture[/url].

well,i didn't write the problem is simple............... :o  :maybe: ....   and the problem was "use FLT to deduce.....". So you have to use FLT, isn't it?

[quote="udyam.bnghs"]well,i didn't write the problem is simple............... :o[/quote]
It's right there in the subtitle.

[quote="udyam.bnghs"]and the problem was "use FLT to deduce.....". So you have to use FLT, isn't it?[/quote]
I doubt it.  What's the source of the problem?

a mathematician gave me............. :love:

but now i think you're right. cuz beal's conjecture is a generalised result of FLT....


One day the main page informed me that aops was down, and also presented a problem. It went something like this:

---
Suppose the number n is the listing of the sequence of positive integers:

n = 1234567891011121314151617181920212223...

Is there any point where n is a palindrome? (Same forwards and backwards)
---

Did anyone figure this out?

I thought about it a bit, but not enough to answer. I suspect that it's never a palindrome, but I have no clue how to go about proving it.

having two topics for the same problem is usually not the best thing to do, but we'll just skip this one ;)

if anyone's interested the other topic in which this problem has been discussed is
http://www.mathlinks.ro/Forum/viewtopic.php?t=19872

[quote="TripleM"]This was posted (and solved) not too long before it went down, in one of the advanced sections I think. But its not too hard for intermediate topics people to think about - if you want a hint:
[hide="Hint"]Think about places where you have lots of 0s in a row.[/hide][/quote]

Or may be a liitle more clear (Sorry if it is already posted)

 [hide="Hint"]  if n is the  largerst possible number such that 10^n - 1 is less than the last number (maximum  digit number of the type "99999..."). This sequence can appear only once (So it must occur in the middle .. but left of this is 8 and either it has no digit to the right or has 1 - Contadiction.....  [/hide]i

[quote="h_s_potter2002"]Why wouldn't n=1 work?  Isn't 1 a palindrome?[/quote]

From the Kalva Website:

[quote]17.  The numbers from 1 to [b]n > 1[/b] are written down without a break. Can the resulting number be a palindrome (the same read left to right and right to left)? For example, if n was 4, the result would be 1234, which is not a palindrome.[/quote]

just for the difficulty level, i think this was on the Olympiad problem solving class

Immediately a way to prove this came to me, but like all of my proofs, it is very strange. Please give me some comments - i.e. "you can't do that" or "that's completely wrong" - and I would be most grateful.

Answer:

No. The first instance of n consecutive zeros that appears if you list the natural numbers in order starting from 1 is 10^n. What this means is that assuming there is a point when this number becomes palindromic, then there cannot be a power of 10 to the left of the halfway line. This is because if there is, then that means that there are n consecutive zeros to the right of the halfway line, but at most n-1 consecutive zeros to the left, and this is a contradiction because the number should be a palindrome.

Note: The halfway line would be the imaginary vertical line drawn that separates the number into two equal sections. It can be drawn through a number (for example the halfway line for 12321 would be a vertical line through the 3).

So now we have established that a power of ten cannot be to the right of the halfway line. Ok, so the natural argument that follows would be that there can be a palindrome without any power of 10 to the right of the halfway line. But even so, there cannot be a palindrome, for this reason:

Since the numbers 1-10 are placed on the left of the halfway point (we know this because 1234567891011121314151617181920 is not palindromic, and neither are any of the "previous" numbers), and the numbers get longer as you go on (soon you will be adding numbers like 100, and 1000, and so on to the end), what it means is that the whole long 1234567891011121314... thing is in a sense shifted to the left. The point of this is that if the last power of 10 placed in the number is say 10^n, then 10^(n-1), 2*(10^(n-1)), 3*(10^(n-1)), 4*(10^(n-1)), and 5*(10^(n-1)) will all be to the LEFT of the halfway line. This is another contradiction because there are more consecutive n-1 zeros on the left than on the right.

Therefore the answer is no.

If you could comment, I would appreciate it, especially one of the moderators. What would I receive point wise if this was an official solution to an offical USAMTS problem? What could I do to make this more clear?

[quote="hatsup123"]Does it mean that somewhere in n ther's a palindrome? Because if it was, then 1001 would be a palindrome[/quote]

No


Has anything been confirmed about the Georgia site yet? That is, ARML will have a separate site in Georgia next year?

Why would they need another site? Don't be a wimp, 12 hour bus rides build character!

Where did you get such an idea?  I [i]live[/i] in Georgia, and I haven't heard anything about this.

[quote="Xantos C. Guin"]I also live in GA, and I have not heard anything about this. What a coincidence.

Personally, I dislike the idea. Even though the trip would be cheaper, it would be a whole lot shorter and less fun.[/quote]
But the trip for the people in Florida, Alabama, Louisiana, and some other states would be an enormous amount cheaper. For Florida people we'd end up saving $100+

You can read an older post on the same subject here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=526379794&t=63847

I think ARML has been considering for several years whether to open a site in the southeast.  It certainly would be convenient for the teams coming from that area and would probably allow for the creation of more teams from the region.  I imagine, however, that it's a fairly major undertaking to get it started.

[quote="h_s_potter2002"][quote="pianoforte"]I was under the impression that Alabama goes to Las Vegas, despite the other two sites being much closer.[/quote]

That's because plane tickets to Las Vegas are cheaper than Penn State and Iowa.  But I'm sure if there was a southeastern site, then Alabama would go there.[/quote]

Plane tickets?? Who sends math teams to ARML on a plane? :ninja:

[quote="roadnottaken"]Plane tickets?? Who sends math teams to ARML on a plane? :ninja:[/quote]
The western site has gone from Las Vegas to San Jose back to Las Vegas. When my son competed in 2000 for the nearly-single-school team known as Peninsula South Bay, they flew, LAX to Las Vegas. In my first year of coaching, 2002, our team (Southern California) flew from Orange County to San Jose, and our two neighboring teams (Peninsula South Bay and North Hollywood) also flew. We first used a bus in 2003. A bus changes the dynamics: once you've decided to utilize a bus, you might as well bring two or even three teams.

Last year at Las Vegas, the following teams arrived by air: Taiwan, Philippines, Washington, Alabama. I'm not sure what arrangements Northern California made. I think New Mexico, Colorado, Arizona, and Utah all used some form of ground transportation, and SFBA, Southern California, and San Diego all used buses.

A year or two ago, the talk about a southeastern site was focusing on finding a place in Florida. That doesn't seem to have happened. Both Atlanta and Orlando would have the same advantages that Las Vegas has for air travel; Atlanta would be closer for ground travel for most teams other than Florida. But either site begs the question: what university campus (semester calendar) has the both the facilities and the willingness to host the event?

[quote="JBL"]People for whom the other options involve [url=http://maps.google.com/maps?f=d&hl=en&geocode=&saddr=Alabama&daddr=University+of+Iowa&sll=37.13804,-89.104905&sspn=12.476246,17.95166&ie=UTF8&z=6&om=1]14 hour[/url] or [url=http://maps.google.com/maps?f=d&hl=en&geocode=&saddr=Alabama&daddr=Penn+State+University&sll=36.72412,-82.335145&sspn=12.544898,17.95166&ie=UTF8&z=6&om=1]15 hour[/url] drives?[/quote]

It takes about 15 hours to get from Dallas to Iowa City, and another 5 more for the kids who get on the bus in Houston.

Hell, I'd love to have an ARML site in Florida. Especially if it were at FSU (though I'm aware that the flight situation would be terrrrible). I could trip and fall into the ARML contest. Orlando only has one university that could possibly handle it.. while Atlanta has a couple (Georgia Tech, Emory) and the advantage of being a Delta HUB/the largest airport... ever... or something like that. It'd all be good stuff. It really does make a ton of sense to have a southestern site, if some university would just take responsibility. And wow, some of you guys are really hardcore.. taking 15 hour bus rides and what not. It'd boost participation from Florida teams. Alot of good teams are based in the SE anyway.

[quote="Kent Merryfield"]I think New Mexico, Colorado, Arizona, and Utah all used some form of ground transportation, and SFBA, Southern California, and San Diego all used buses.
[/quote]

For the past two years, the Colorado team has flown to Las Vegas on Southwest Airlines, $ \$$118 rt.  It's cheaper than taking a bus.

Once you've been on an ARML bus, you understand a bit more the appeal of 15-hour bus rides.  Mine was about five hours, and I would have traded it for a 15-hour ride.

[quote="joml88"]I agree that a long bus ride for ARML is really fun.  I've gone 4 times now (I think the ride takes at least 12 hrs(?)).  I enjoyed each of the bus rides.  To be sure, I've taken the chinatown/greyhound bus from Boston to New York multiple times and haven't had a good time on any of those (they're only 4.5 hrs).  So I'm not some kind of bus enthusiast in particular ;)

I was at the coaches meeting this past year at Penn State and one of the topics discussed was opening another site in the south.  Those in charge of ARML seem to be interested in opening a new site but, iirc, need to find the right place for it.[/quote]

If I were you I wouldn't trust those Chinatown buses...
But with a shorter bus ride, there would be more time to have fun when you actually get to the site. I mean, there's alot more to do when you can stand up and run around instead of sitting in a moving vehicle. Besides, Penn State is so crowded, it would be beneficial to direct the southern teams going there (NC, SC, Florida, Georgia, etc.), so Penn State isn't as crowded.

With a shorter bus ride, I'm pretty sure teams leave later, arriving around the same time as most people
With a long bus ride, you get almost the same 'fun time at the site' as everyone else, but also fun time on the bus

Alright let me clear some things up.

1) If we were to have a bus for the FL team, it would be around a 20 hr bus ride.
2) I had received an email from Mr. Sullivan about how there might be a SE site in Athens (University of Georgia) because Penn State was too crowded.
3) Unfortunately, I haven't heard anything more about this after that email in late June.

So no, just because you live in Georgia clearly doesn't mean that you know  everything that goes on there.  :)

Ok I heard today that there will be a Georgia ARML site, from teh people who will be organizing it. This makes me sad, but you Floridians were right.

Aye, agreed with Billy. The southern site will be held in the University of Georgia. However, I do regret that we won't go to Penn State anymore considering that the bus ride was probably one of the more entertaining parts about ARML. The ride for us now shortened from about 13 hours or so to about 45 (minutes). The positive aspect is that we don't have to pay for anything.

I've been in the grading room in Iowa for the past 6 years. Every year the topic of a new site comes up, and every year it's still a couple of years away.

I understand that a lot of team organizers / parents / coaches would like shorter bus rides and possibly cheaper travel, but I also imagine that it takes a certain mass of students to make it affordable to rent out the needed amount of space at whatever university ends up hosting.

That being said, I'm sure there'll be new site(s) in the next few years; ARML is on the verge of a major expansion. It's already practically a worldwide competition as is.

From what I heard at UGA though, it will be up from this May. That being said, I'm not sure if there will be an ARML meet this year or not at UGA.

I just received this email message:

[quote="Steve Adrian"]To All ARML Coaches,

We are pleased to announce that for the 2008 ARML Competition we will add the University of Georgia in Athens, Georgia to our present three sites of Penn State, Iowa and UNLV. For those in the South this should greatly reduce your travel costs and time. For any team flying to the meet, Atlanta is a huge hub and there is direct transportation to the University in Athens for a very moderate fee. We will send out more information on this option in the near future. Having been to the University I can tell you that the food is great, the entire campus is air conditioned,  and Southern Hospitality is at its finest. If you haven't already decided to attend ARML at the University of Georgia and it is a practical alternative for you, I strongly suggest that you give it a try. Your students will experience a fantastic time.

Steven L. Adrian
ARML Executive Director [/quote]

Ah, so it's finally happening for real!  Thanks for the info.

Out of curiosity, does anyone from the Texas team know which site you're likely to go to this year?  It seems like there will be a lot less competition at Iowa if Texas leaves.

[quote="worthawholebean"]There's only one Texas team? That seems rather... silly.[/quote]

there are 3 texas teams

But only one Texas organization. And there are some very large distances that may make meeting as a team difficult.

I think in most cases, when people count "teams," they really mean organizations. The average number of teams per organization is around two, and three-team or greater organizations are common. In the case of California, there are now four organizations. Last year, those four organizations fielded a total of nine teams.

I could see a plausible geographic split in which Texas had three organizations: an East/Southeast Texas team, meeting somewhere in the Houston metro area; a Central/South/West Texas team, probably meeting in Austin, and a North Texas team meeting somewhere near Dallas. Or combine any two of those together, split off the other, and have two organizations. And someone from Lubbock is still nowhere near anyone else, but that can't be helped.

Wow. The year after I organize an AMP reunion at Iowa City, I find out I myself may not be going there. I really hope Texas keeps going to Iowa. Otherwise, it could be just Chicago from Iowa that places top 10 nationally.

Well, I can't be so sure TX would go to GA. It's 805 miles from Dallas to U of Iowa and 861 miles to UGA. We really could go either way.

[quote="Karth"][quote="worthawholebean"]I'm glad there's a southeastern site, though. Penn State was crowded last year - maybe this year we won't have to take the tests sitting in the stadium with boards on our laps.[/quote]

Actually, I thought that wasn't that bad... Penn State isn't that crowded (population density wise) because of the large stadium. In addition, the stadium made it a bit more exciting.[/quote]

I couldn't stand the stadium... plus the food lines were awful.

In my efforts to make sure NC continues to go to Penn State, I decided to make a map of closest sites. The only states for which a nontrivial portion is closer to Georgia than the alternate site appear to be North Carolina, Virginia, Arkansas, and Texas.

good times... good times... good old voronoi diagrams.

that picture makes it look like the new site location is some sort of joke on whoever happens to live near laredo, texas.

i wish i could go to iowa next year and see how things have changed, but i think 2008 will find me at vegas.

[quote="solafidefarms"][quote="worthawholebean"]I'm glad there's a southeastern site, though. Penn State was crowded last year - maybe this year we won't have to take the tests sitting in the stadium with boards on our laps.[/quote]

Isn't it an ARML tradition to accidentally forget to turn in your lapboard and return it the next year? ;)[/quote]

ROFL I did that twice... whoops did I say that out loud?  :D And Carol still has hers from two years ago... that's what we used to play bridge this year and during the trip up to Duke. Good times...  :)


Somehow I'm not able to edit my posts immediately after I posted it, I always get

[quote]
 Consecutive posts not allowed for less than 72 hours
[/quote]

I've noticed it in the inequalities-forum, but I also tried it in the Test-Forum.


calculus
integration
function
algebra unsolved
algebra

Does there exist a function $ f$ for integers such that $ f(f(n) \minus{} 2n) \equal{} 2f(n) \plus{} n$ is true for any integer $ n$?

1) if you want say $ f: \mathbb{N}\to \mathbb{N}$ where $ \mathbb{N}\equal{}\{1,2,.....\}$ ?? :?:
________________________________________
we take $ g(n)\equal{}f(n)\minus{}2n$ (we must have $ f(n)\ge 2n\plus{}1$ )
$ g(n)\equal{}g(m)$ gives $ f(n)\minus{}f(m)\equal{}2(n\minus{}m)$ and  $ 2f(n)\plus{}n\equal{}f(g(n))\equal{}f(g(m))\equal{}2f(m)\plus{}n \Rightarrow 5(m\minus{}n)\equal{}0$
then $ g$ is injective.
and we have $ g(g(n))\plus{}2g(n)\equal{}2(g(n)\plus{}2n)\plus{}n\Rightarrow g(g(n))\equal{}5n$
Now it's easy to prove that $ \exists A,B$ partition of $ \mathbb{N}\minus{}5\mathbb{N}$
such that $ A\equal{}\{a_1,a_2,....\},B\equal{}\{b_1,b_2,...\}$ and $ g(a_i)\equal{}b_i,g(b_i)\equal{}5a_i$
and for $ k\in \mathbb{N}\minus{}5\mathbb{N}$ $ g(5^ak)\equal{}5^ag(k)$

there are infinity of solutions,

for example if we take $ a_{2i}\equal{}5i\plus{}1,a_{2i\plus{}1}\equal{}5i\plus{}2$ and $ b_{2i}\equal{}5i\plus{}3,b_{2i\plus{}1}\equal{}5i\plus{}4$
we have $ f(5^{a}(4h\plus{}1))\equal{}g(5^{a}(4h\plus{}1))\plus{}2.5^a(4h\plus{}1)\equal{}5^{a}(5h\plus{}3)\plus{}2.5^a(4h\plus{}1)$
$ f(5^{a}(4h\plus{}2))\equal{}...$
..
..


April fools day is a long long long way off.
I though of making this topic last year. Did somebody do it finally?

[quote="jackdillon"][quote="ashpotter"]April fools day is comming soon. Wondering if you guys have any pranks?

Here's my favorite prank: Supergule a penny with the heads side up on a side walk where kids cross alot.[/quote]

I do this with a dollar, cause then adults try to pick it up to :P[/quote]

you'd waste a dollar on that?

[quote="vicseb"]A related one is to put a dollar on the ground in a quiet place where an adult would pass by. When they bend over to pick it up rip a piece of paper or something related. If you want take a picture of their shock and amazement[/quote]

or go a step further and very sneakily undo their belt

bonus points if they are of the opposite gender

Tape down the button on your kitchen sink sprayer. Don't forget-- [b]use clear tape[/b]. :D

[quote="ashpotter"]
Here's my favorite prank: Supergule a penny with the heads side up on a side walk where kids cross alot.[/quote]

I saw that on a lunch table in my school once... no matter how hard we tried it wouldn't come off :(

Put sticky notes on people's back. This works especially well here because my there is one freshman school for both high schools in my district, so I can put a sticky note cheering for my school onto the back of someone from the opposing school. Unfortunately April 1st is a Sunday this year.

Set up a password for people in your family to access the computer.

Hm, that's all I got. I'm not very creative.

Put Vaseline on the teacher's doorknob. Both sides.

If you have a sister, then take all her dolls and some yarn. Tie the yarn aroung the neck, leg, arm, etc.. Hang them all over the place. I did it on my sister once when she got back from school  :lol:

With cooperation with your math teacher:

Convert your standard math class into an AP App Calc class for about 10 minutes (just enough for them to try and do the drill) and then switch back.  :huh:


algebra
polynomial
number theory solved
number theory

A splendid problem from a russian olympiad:
   Let P a polynomial with integer coefficients. Then in the sequnce P(n) (n>=1) there are infinitely many terms with the same sum of digits.
   I would like a complete solution to this problem.

So let's give a complete solution....

Let S(x) denotes the sum of the digits of x.

If P is constant the result is trivial, so we may assume that
P(x) = a_p*x p  + ... + a_1*x + a_0, where p > 0.

First case : If all the a_i's are non-negative.
Let k be an integer such that 10 k > max{a_i}, and n_j = 10 k+j . Thus, for each j  \geq 0, and for each in {0,1,...,p},
since 10 (k+j)(i+1) = 10 i(k+j) 10 k 10 j > a_i*10 i(k+j) , we deduce that
P(n_j) = (a_p)0...0(a_(p-1))0...0(a_0) where (a_i) denotes the decimal expansion of a_i.
Thus S(P(n_j)) = sum{ 0 \leq i \leq p} a_i = constant.

Second case : If at least one of the a_i's is negative.
As we may prove the result with -P), we may assume that a_p > 0.
Let k be the greatest subscript such that a_k < 0.
Note that k
 0 and P(x + (-a_k)) = sum {0 \leq i \leq p} a_i(x-a_k) 2  i .

From the binomial expansion formula and since -a_k > 0, we deduce that Q(x) = sum{k+1 \leq i \leq p} a_i*(x-a_k) i  is a polynomial with nonegative integer coefficients. Moreover the coefficient of x k  in Q(x) is q_k \geq a_p*(-a_k) p-k \geq -a_k since a_p \geq 1 and p-k \geq 1.

Let R(x) = sum{0 \leq i \leq k} a_i*(x-a_k) i. Then  is a polynomial with integer coefficients and degree k. Moreover the coefficient of x k  in R(x) is r_k = a_k.

It follows that P(x-a_k) = Q(x) + R(x) has degree p and integer coefficients b_j and b_j  \geq 0 for j = k,...,p.
Let P_1(x) = P(x - a_k). If P_1 has nonegative coefficients, we use the same reasoning on P_1, and so on. Since the greatest subscript k' for which the coefficient of x k'  in P_1 is k' < k, we deduce that with a finite number of steps, we will obtain of polynomial P_q with nonegative integer coefficients. The procedure ensures that P_q(x) = P(x + a) where a is a nonegative integer, that is P(x) = P_q(x-a).

We then use the first case on P_q, and only keeping the n such that n > a, we deduce that there exists an integer s and an infinite number of positive integers n such that S(P(n)) = s, and we are done.

Pierre.

Thank you very much. It seems natural for me not to have solved this one after I had seen this solution.


ceiling function
pigeonhole principle
modular arithmetic

A prime number $ p$ divides $ a^2\plus{}2$ for a natural number $ a$. Prove that $ p$ or $ 2p$ is of the form $ x^2\plus{}2y^2$ for some natural numbers $ x,y$.

[quote="nguyentrang"]A prime number $ p$ divides $ a^2 \plus{} 2$ for a natural number $ a$. Prove that $ p$ or $ 2p$ is of the form $ x^2 \plus{} 2y^2$ for some natural numbers $ x,y$.[/quote]

[hide]
ok, since p is prime, $ a^2\plus{}2\equal{}p$
then 2p is of the form $ 4\plus{}2a^2$
now let's set x as 2 and y as a

too simple?
[/hide]

[quote="CircleSquared"][quote="nguyentrang"]A prime number $ p$ divides $ a^2 \plus{} 2$ for a natural number $ a$. Prove that $ p$ or $ 2p$ is of the form $ x^2 \plus{} 2y^2$ for some natural numbers $ x,y$.[/quote]

ok, since p is prime, $ a^2 \plus{} 2 \equal{} p$
[/quote]

I dont understand why that shd be true. Maybe i am missing sth?  :maybe:

$ p$ divides $ a^2\plus{}2$ means that $ p$ is a factor of $ a^2\plus{}2$, not that $ a^2\plus{}2$ is a factor of $ p$, which CircleSquared thought.

The problem might as well just ask to show $ p$ is of the form $ x^2\plus{}2y^2$, for if $ 2p\equal{}x^2\plus{}2y^2$, then $ x$ is even, so...

[hide]
Let $ S$ be the set of ordered pairs of integers $ (u,v)$ such that $ 0\leq u<\sqrt{p}\sqrt[4]{2},0\leq v<\frac{\sqrt{p}}{\sqrt[4]{2}}$. The number of elements in $ S$ is $ \lceil \sqrt{p}\sqrt[4]{2}\rceil \lceil \frac{\sqrt{p}}{\sqrt[4]{2}}\rceil >p$, so by Pigeonhole, there exist $ u,u^\prime ,v,v^\prime$ such that $ (u,v),(u^\prime ,v^\prime)$ distinct elements in $ S$ and
\[ u\minus{}av\equiv u^\prime \minus{}av^\prime \pmod{p}\Rightarrow x\equiv ay\pmod{p}\]
where $ x\equal{}u\minus{}u^\prime ,y\equal{}v\minus{}v^\prime$. Then $ x,y$ not both 0, and
\[ |x|<\sqrt{p}\sqrt[4]{2},|y|<\frac{\sqrt{p}}{\sqrt[4]{2}}\Rightarrow 0<x^2\plus{}2y^2<2\sqrt{2}p\]
But $ x\equiv ay\pmod{p}$ implies $ x^2\plus{}2y^2\equal{}(x\plus{}ay)(x\minus{}ay)\plus{}(a^2\plus{}2)y^2$ is divisible by $ p$, so either $ p$ or $ 2p$ is of the form $ x^2\plus{}2y^2$. Of course, as mentioned above, this means $ p$ must be of that form.
[/hide]


Is anybody aware of an answer to the following question?

"1" has the shortest "egyptian" representation 1 = 1/2 + 1/3 + 1/6 with all denominators different and greater than 1.
We denote the length of  "1" by L(1) = 3.

Similarly we look at entire numbers n greater than 1.
What about L(n) for small values of n such as 2, 3, 4, 5?
What about L(n), as n tends to infinity? Are there any asymptotic results?

Thanks for your help!

Just one example with the following denominators: 2,3,4,5,6,8,9,10,12,15,30,36,360
and notice that $ 10 \le L(2)$


suppose that i don't know chinese (true anyway). given any chinese character on a text.
how can i type exactly that same character on here?

i guess there's a formula to type chinese characters on here. can someone teach me how to type?

thanks

i am confused too...if you don't know Chinese, what do you need it??

[quote="Connie"]Actually there IS a kind of type that can separate a character into several parts and type it--but it's not popular and it's hard to remember![/quote]
But it requires one to know how to WRITE a character to an extent.

[quote="Connie"]Actually there IS a kind of type that can separate a character into several parts and type it--but it's not popular and it's hard to remember![/quote]
do you mean i need to recognize "part" of the character?

e.g character, 望
part, 月

there are over 200 such parts right?
you mean i need to know all these parts and be able to recognize them in a character?

[quote="Connie"][quote]suppose that i don't know chinese (true anyway). given any chinese character on a text.
how can i type exactly that same character on here?

i guess there's a formula to type chinese characters on here. can someone teach me how to type? [/quote]
So sorry...
Chinese is different from Korean that if you don't know the character, you cannot type it (why don't you just copy it?)
Actually there IS a kind of type that can separate a character into several parts and type it--but it's not popular and it's hard to remember![/quote]

that's very popular in Hong Kong. Usually typist use this method--Changjei. I use this method to type too.

[quote="liyi"]Sometimes you need to know how we write a characters, i mean, the order of strokes.
for example, 王
you need to know we write two horizontal strokes, then the horizontal one, finally a horizontal stroke.[/quote]
two horizontal strokes, then the horizontal one, finally a vertical stroke?

you mean stroke order of calligraphy?
i have a formula for that,

top-bottom e.g. 五
left-right e.g. 大 (exception right-left 力 ...)
horizon-vertical e.g. 十
middle-sides (if symmetric) e.g. 水, 山, 小
outside-inside 月
(exception closed rectangle e.g. mouth, the third (bottom) stroke last; 巡 inside-outside)

can i type using this formula?

[quote="siuhochung"]that's very popular in Hong Kong. Usually typist use this method--Changjei. I use this method to type too.[/quote]
may be you can show me this method right?

[quote="analysis"]
[quote="siuhochung"]that's very popular in Hong Kong. Usually typist use this method--Changjei. I use this method to type too.[/quote]
may be you can show me this method right?[/quote]
well, you may search for 倉頡輸入法. It has many rules so it is really hard to show you here.

[quote="analysis"][quote="siuhochung"]well, you may search for 倉頡輸入法. It has many rules so it is really hard to show you here.[/quote]
我不知道漢語

[url]http://www.fed.cuhk.edu.hk/readwrite/typing/[/url][/quote]

as others said before, if you don't know chinese, it is really hard to type chinese....
at least you have to know the order of strokes, or its pronunciation to type chinese.

[quote="siuhochung"]as others said before, if you don't know chinese, it is really hard to type chinese....
at least you have to know the order of strokes, or its pronunciation to type chinese.[/quote]
guess i can't type chinese, unless i spend a few years to study 漢語


trigonometry
symmetry
geometry
Pythagorean Theorem
perpendicular bisector

Cyclic pentagon $ ABCDE$ has a right angle $ \angle{ABC} \equal{} 90^{\circ}$ and side lengths $ AB \equal{} 15$ and $ BC \equal{} 20$. Supposing that $ AB \equal{} DE \equal{} EA$, find $ CD$.

[hide] CA is a diameter of length 25. Now since CA is a diameter then <CEA must be right, so CE equals 20. And similarly, if we let CD = x we get that AD is sqrt(625-x^2).   Now just do Ptolemy on CDEA and solve for x.  I get x=7 [/hide]

[hide="Trig solution"]
Start out like nateharman to find the radius is 25/2. Call the center of the circle O. We have angle AOE+angleEOD is 4arcsin(3/5) using trig. But angle DOC is 2arcsin(x/25), where x is CD. So adding these up, setting it equal to 180, and solving gives x=7.  [/hide]

[hide="Another way"]By the Pythagorean Theorem we get $ AC\equal{}25$.  Now let's find $ BE$.  Since $ AC$ is a diameter, by symmetry $ AC$ is the perpendicular bisector of $ BE$.  Since the area of triangle $ ABC$ is 150, the altitude to $ AC$ (which is $ BE/2$) has length 12.  Therefore, $ BE\equal{}24$, so $ AD\equal{}24$ also.  Then right triangle $ ACD$ is a 7-24-25 triangle, so $ CD\equal{}7$.[/hide]


inequalities
LaTeX
inequalities unsolved

[quote="TV-TV-07"]Let a, b, c be real nonnegative numbers. Prove that:
                              1/(a^2-bc+1)  +  1/(b^2-ca+1)   +   1/(c^2-ab+1)   <= 3 :ninja:[/quote]
Try $ a^2\minus{}bc\plus{}1\rightarrow0^\plus{}.$ :wink:

Yah more precisely take $ a\equal{}4,b\equal{}\frac{27}{10}\equal{}2.7$ and $ c\equal{}2$
I am afraid the inequality isn't true. Probably u've made typing mistakes or something else. And try  using latex while witing mathematical symbols plz...

Oni.


The sum of the areas of circle O and circle P, which are tangent to each other, is $ 28\pi$ and the distance between the centers is 6. Find the length of the external tangent.

[hide]
say o is the bigger one
call the radii of o and p,  a and b respectively
draw the perpendiculars from o and p to the external tangent, call the intersections m and n respectively
now drop the perpendicular from p to om, it intersects at d

it's clear now that pd^2+od^2=op^2
but pd=mn
so mn^2=(a+b)^2-(a-b)^2
or mn^2=4ab

a^2+b^2=28
(a+b)^2=36
so 2ab=8, 4ab=16
so mn=4

weird solution, probably wrong

[/hide]

[hide]
We start given

$ r_1 \plus{} r_2 \equal{} 6$
and
$ r_1^2 \plus{} r_2^2 \equal{} 28$  (from the area information)

now we square the first one

$ 36 \equal{} (r_1 \plus{} r_2)^2 \equal{} r_1^2 \plus{} r_2^2 \plus{} 2r_1r_2 \equal{} 28 \plus{} 2r_1r_2$
$ r_1r_2 \equal{} 4$

To get the length of the external tangent, we look at the right triangle whose hypotenuse is the two radii together. I will try and edit in an asymptote picture.

Ah, here we go. The triangle I'm speaking of is $ \triangle ABC$. Sorry about the lack of labeling. I hacked this together.
[asy]pair A=(0,1), B=(2.8,2);
draw(Circle(A,1));
draw(Circle(B,2));
draw((0,0)--A--B--(2.8,0)--(0,0));
draw(A--(2.8,1));

dot(A);dot(B);
label("A", A, NW);
label("B", B, NE);
label("C", (2.8,1), E);[/asy]

$ (AC)^2 \plus{} (BC)^2 \equal{} (AB)^2$

$ l^2 \plus{} (r_2 \minus{} r_1)^2 \equal{} (r_1 \plus{} r_2)^2$

cranking that out

$ l^2 \equal{} 4r_1r_2 \equal{} 16$

$ l \equal{} 4$
[/hide]


How many distinct positive integer divisors does $ 2\times 3\times 3\times 3\times 5\times 5$ have?

The formula for the number of factors is:

Given a number n, where n's prime factorization is $ \displaystyle n \equal{} {p_1}^{e_1} \cdot {p_2}^{e_2}\cdot{p_3}^{e_3}\cdots{p_k}^{e_k}$ the number of factors is $ (e_1\plus{}1)\cdot(e_2\plus{}1)\cdot(e_3\plus{}1)\cdots(e_k\plus{}1)$.

So in this case, it would be $ 2*4*3\equal{}24$.

Now, let's see how the formula works:

The idea is that if you factor N into its prime factorization, then you know that its factors' prime factorizations are hiding in there as well.
F.E.
$ 24\equal{}2^3 3^2$.  Anything that divides 24 has to be built out of some number of 2's and some number of 3's.  A factor could have any power of two from 0 to 3, and any power of three from 0 to 2.  So, you chose a power of two from that set of 4 powers, a power of three from that set of 3 powers, and in total that's $ 3\times 4\equal{}12$ possible factors.

So now you can see how the formula works... for each prime power $ p^{e_i}$, you can chose $ 0, 1, 2, ... , e_i$, that is, $ e_i\plus{}1$ choices.  Do that for each prime and its power and then multiply the $ e_i\plus{}1$ all together. Viola

$ 2\times 3\times 3\times 3\times 5\times 5$ $=$ 2^1*3^3*5^2

Then we take each exponent and add 1 to it and multiply them to get the positive integers of  $ 2\times 3\times 3\times 3\times 5\times 5$ or 1350

 $(1+1)\times (3+1)\times (2+1) = 2\times 4\times 3 = 24$

Ignore me... I can't Latext


How does one spell this guy's name? I've seen:

Viete
Vite
Vieta
Vita

Which one?

Vite is the French spelling. Vieta is the Italian spelling. Since he was French, I use the former.

Oh, so its with the  rather than the ? Ok, thx.


Ok, I don't know if this is appropriate for this forum, but I found this problem particularly easy:

If $a+b+c=0$, then prove that $a^{3}+b^{3}+c^{3}=3abc$.

Thanks for the fast replies. Can this be generalized to more variables? Like for example,
$a^{2}+b^{2}=2ab$ $\textrm{if}$ $a+b=0$
$a^{3}+b^{3}+c^{3}=3abc$ $\textrm{if}$ $a+b+c=0$
Is it true that
$a^{4}+b^{4}+c^{4}+d^{4}=4abcd$ $\textrm{if}$ $a+b+c+d=0$
$a^{5}+b^{5}+c^{5}+d^{5}+e^{5}=5abcde$ $\textrm{if}$ $a+b+c+d+e=0$
etc...

I'm just curious.

[quote="vishalarul"]$a^{2}+b^{2}=2ab$ $\textrm{if}$ $a+b=0$[/quote]

This is not true.  $a+b = 0 \implies a^{2}+b^{2}=-2ab$.

Edit:  The others are also untrue.  Simply let $a = 0$ and let $b, c, d, ... \neq 0$.

Yes, your solution is standard. I have hinted this factorization, .... :(

No problem. Here is another solution.
We can use the equality
$(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a).$


Let $ x,y,z \geq 0$ satisfy x+y+z=1 .Prove that:
$ \frac{xy}{z\plus{}1}\plus{} \frac{yz}{x\plus{}1}\plus{} \frac{zx}{y\plus{}1}\leq \frac{1}{4}$


function
real analysis
real analysis theorems

Prove the following two theorems:

1. Sums, differences, and products of continous functions are themselves continuous functions.  The quotient of a pair of continuous functions is continuous except where the denominator equals zero.

2. A continuous functions of a continuous function is continuous function.

If you can do sums, then you have differences (constant multiple of a continuous function is continuous and $ f\minus{}g\equal{}f\plus{}(\minus{}1)g$).

Suppose $ f$ and $ g$ are continuous.

Let $ \varepsilon>0$ and $ c\in\mathbb{R}$.

Choose $ \delta_1>0$ and $ \delta_2>0$ such that $ |x\minus{}c|<\delta_1\Rightarrow |f(x)\minus{}f(c)|<\frac{\varepsilon}{2}$ and $ |x\minus{}c|<\delta_2\Rightarrow |g(x)\minus{}g(c)|<\frac{\varepsilon}{2}$.

Now let $ \delta\equal{}\min(\delta_1, \delta_2)$.

Examine $ |(f\plus{}g)(x)\minus{}(f\plus{}g)(c)|$ when $ |x\minus{}c|<\delta$ and use triangle inequality.


probability
logarithms
expected value
combinatorics unsolved
combinatorics

We have n strings and begin tying their ends together until only loops remain. What is the expected number of loops?

[hide="big hint"]
Let the answer be $ L_n$. On the first step, with probability $ 1/(2n\minus{}1)$ you create a loop from a single string, so the expected number of loops is $ 1\plus{}L_{n\minus{}1}$. With the complementary probability you just merge two strings into a longer one, so the expected number is $ L_{n\minus{}1}$.
[/hide]

Thanks. In a rush. At first glance seems to be 1 + 1/3 + 1/5 + 1/7 + ..., which is?

Yes, that's right. The expected number of loops with $ n$ strings is $ L_n \equal{} \sum_{k\equal{}1}^n \frac{1}{2k\minus{}1}$. This doesn't have any "nice" formula, just as the similar $ H_n \equal{} \sum \frac{1}{k}$ doesn't. You got sort of an answer in your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1264420#1264420]other post[/url] on this topic, but I don't know that it's more illuminating than the sum itself. Asymptotically, this is about $ \frac{1}{2}\ln n\plus{}$some constant, I suppose.


There are 2001 points in the plane. It is known that the area of any triangle with vertices at the given points is less or equal than 1 cm^2. Prove that there exists a triangle with area no more than 4 cm^2, which contains all 2001 points.

Can someone show me how to do this problem?

This looks really like Korean math olympiad(1995): Consider finitely many points in the plane such that, if we choose any three points A,B,C among them, the area of triangle $ABC$ is always less than 1. Show that all of these points lie within the interior or on the boundary of a triangle with area less than 4.

[hide]
First construct $MLN$. Now constructs midpoints triangle $ABC$(A on the side $ML$, B on the side $MN$,etc.). Let the triangle $ABC$ has the largest area. (BTW this is the extreme principle)
Say the area of $ABC$ is less than or equal to 1, then the area of $MLN$ is less than or equal to 4 ( by the similarity).
Now the set of points should lie in or on the $MLN$. Suppose there's a point P outside $MLN$, and draw triangle $ACP$. The area of $ACP$ is clearly larger than that of $ABC$. this contradicts  that $ABC$ has the maximal area.
[/hide]


Prove that for every convex quadrilaterals $ABCD$, we have:
$\frac{\sqrt{AC^2+BD^2}}{2}\ge min${$AB,BC,CD,DA$} where $min${$AB,BC,CD,DA$}
is the minimum of $AB,BC,CD,DA$

Nobody is interested in this problem? :(

A hint:
Denote $k=\frac{AC^2+BD^2}{2}$.Construct two circles $(C,k),(A,k)$, they intersect each other at $B'$ and $D'$($B'$and$B$ are in the same half-side to $AC$).
It's easy to show that $BD=B'D'$
We have to prove that at least $B$ or $D$ is inside one of that two circles.
Let's try now, you will find the happiness :lol:


Find the largest integer $ n$ for which $ 12^{n}$ evenly divides 20!.

[hide="Solution"]$ 20! \equal{} 2^{18} \times 3^{8} \times 5^4 \times 7^2 \times 11 \times 13 \times 17 \times 19$.

$ 12^8 \equal{} 2^{16} \times 3^8$, so $ 12^8$ divides $ 20!$, yet $ 12^9$ does not because the power of $ 3$ exceeds $ 3^8$..

$ \boxed{n\equal{}8}$.

[/hide]

There's a neat little formula. The largest $ n$ such that $ p^n|m!$ ($ p$ being prime) is given by
\begin{eqnarray}
n&=&\sum_{k=1}^{\infty}\left\lfloor\frac{m}{p^k}\right\rfloor
\end{eqnarray}

This leads us to an [hide="alternative solution"]
Applying $ (1)$, the largest power of $ 3$ that divides $ 20!$ is $ 6+2=8$.

The largest power of $ 2$ is $ 10+5+2=17$, which means there's enough $ 4$s to make $ 12$s from all the $ 3$s, which makes the answer $ \boxed{8}$.
[/hide]


Hi. Now i'm giving a problem, below,see.

Let $ p$ be a prime number. Prove that there exists a prime number $ q$ such that for every integer $ n$, the number $ n^{p} \minus{} p$ is not divisible by $ q$.

As you remember it's from IMO 2003/6, the hardest problem of that year's contest. And all of we know several solutions to it. But the thing that i say is, can one find please any solution to it, be the easiest to notice and seem good a bit, and be much shorter than the others.

and please note, i wanna it for i'm mad on this not for the other reasons


trigonometry
geometry
function
search
power of a point
geometry unsolved

Line $l$ tangents unit circle $S$ in point $P$. Point $A$ and circle $S$ are on the same side of $l$, and the distance from $A$ to $l$ is $h$ ($h > 2$). Two tangents of circle $S$ are drawn from $A$, and intersect line $l$ at points $B$ and $C$ respectively.

Find the value of $PB \cdot PC$.

Consider the circle $(S)$ of radius r = 1 and the line $l$ directed. Let $k$ be a directed line parallel to the line $l$ at a distance $h$ and A a point on this line outside of the circle $(S)$. Let A be a point on the line $k$ outside of the circle $(S)$ and H the foot of a normal from the point A to the line $l$. Let $b, c$ be the 2 directed tangents from the point A to the circle $(S)$ intersecting the line $l$ at points B, C and let

$\beta = \widehat{(b, k)} = \widehat{(b, l)} = \measuredangle ABH$

$\gamma = \widehat{(c, k)} = \widehat{(c, l)} = 180^\circ - \measuredangle ACH$

be the directed angles formed by the directed tangents $b, c$ with the directed lines $k, l$. (A directed tangent must have the same direction as the directed circle at their tangency point. As opposed to 2 undirected lines, which form 2 different complementary angles, 2 directed lines form a unique angle.) Then

$\frac{r}{PB} = \frac{PS}{PB} = \tan \widehat{SBP} = \tan \left(90^\circ - \frac{\widehat{ABH}}{2}\right) = \tan \left(90^\circ - \frac{\beta}{2}\right) = \frac{1}{\tan \frac \beta 2}$

${\frac{r}{PC} = \frac{PS}{PC} = \tan \widehat{SCP}} = \tan \frac{\widehat{ACH}}{2} = \tan \left(90^\circ - \frac{\gamma}{2}\right) = \frac{1}{\tan \frac \gamma 2}$

The power of the directed line $k$ to the directed circle $(S)$ is equal to

$\tan \frac{\widehat{(b, k)}}{2} \cdot \tan \frac{\widehat{(c, k)}}{2} = \tan \frac{\beta}{2} \cdot \tan \frac{ \gamma}{2} = \frac{PB \cdot PC}{r^2}$

and it does not depend on the position of the point A on the directed line $k$. Obviously, this is a dual concept of the power of a point to an undirected circle, which does not depend on the angle of the circle secant from this point. I guess it is not considered a pure geometry, because it contains trigonometric functions :P. To calculate the power of the directed line $k$ to the directed circle $(S)$, we can move the point A into the foot $A_0$ of a normal from the tangency point P to the line $k$, because then $\gamma_0 = \beta_0$ and

$\tan \frac{\beta}{2} \cdot \tan \frac{ \gamma}{2} = \tan \frac{\beta_0}{2} \cdot \tan \frac{ \gamma_0}{2} = \tan^2 \frac{\beta_0}{2}$

Let $U_0, V_0$ be the tangency points of the directed tangents from the point $A_0$ to the directed circle $(S)$ intersecting the directed line $l$ at points $B_0, C_0$.

$\frac{r}{h - r} = \frac{U_0S}{A_0S} = \cos \widehat{A_0SU_0} = \cos \widehat{A_0BP} = \cos (180^\circ - \beta_0) = -\cos \beta_0$

$\cos \beta_0 = \cos^2 \frac{\beta_0}{2} - \sin^2  \frac{\beta_0}{2} = \cos^2  \frac{\beta_0}{2} \left(1 - \tan^2 \frac{\beta_0}{2}\right) = \frac{1 - \tan^2 \frac{\beta_0}{2}}{1 + \tan^2 \frac{\beta_0}{2}} = \frac{1 - x}{1 + x}$

Solving for x, we get

$x = \tan^2 \frac{\beta_0}{2} = \frac{1 - \cos \beta_0}{1 + \cos \beta_0} = \frac{1 + \frac{r}{h - r}}{1 - \frac{r}{h - r}} = \frac{h - r + r}{h - r - r} = \frac{h}{h - 2r}$

$PB \cdot PC = r^2 \tan^2 \frac{\beta_0}{2} = \frac{r^2h}{h - 2r}$

and for a unit circle r = 1,

$PB \cdot PC = \frac{h}{h - 2}$

HI
yetti , your solution is nice but it is very long. ;)
Let [b]AB[/b]=c and [b]AC[/b]=b and [b]BC[/b]=a  and (a+b+c)/2  =k
In [b]ABC[/b] we know that [b]S[/b](the area of triangle) = k*r
r=1  ===> [b]S[/b]=k=a*h/2 ==>k-a = a*(h-2)/2
[b]PC[/b]=k-c and [b]PB[/b]=k-b
we khow [b]S[/b] = [(k-a)*(k-b)*(k-c)*k]^(1/2) ===>[b]PC*PB[/b] = k /k-a
===>[b]PB*PC[/b]=(a*h/2)/(a*(h-2)/2)

so  [b]PB*PC[/b] = [b]h/h-2[/b] ;)  :lol:

Length of a solution depends mainly on how many details and comments is the solver willing to put down and perhaps how general the solution is. Still, if a problem was phrased "Find the shortest possible solution of ...", I would not touch it, because I do not search for brevity, but for illumination.

Yetti

Naser
yetties solutions are not just for finishing the probleme but to find most possible facts about the argumentsd generlising it as more as possible .our dear [b]yetti[/b] is one of the best guys in calculating and his solutions are just elegant in forms of calculating . if you want to see some nice peices of [b]art of problem solving[/b] then read the solution of yetti for the most hardest problems with elgant calculatings.

I agree with you, Lupin! But such Yeti often solved a lot of the difficult problems.
In the triangle ABC there is the relation: $h_a=\frac{2r(s-b)(s-c)}{|r^2-(s-b)(s-c)|}$
 In the forum http://www.mathlinks.ro/Forum/viewtopic.php?t=42518 I proposed the following problem:
"the ecuation $h_a=\frac{2(s-b)(s-c)x}{|x^2-(s-b)(s-c)|}$ has the solutions r,$r_a$ and the ecuation ... has the solutions $r_b,r_c.$"  This problem [b]even Yeti[/b] solved on one-two pages. From r=1, $h_a=h\ge 2.$ results $PB.PC=(s-b)(s-c)=\frac{h}{h-2}.$ because $r^2\le (s-b)(s-c)$.


geometry
incenter
circumcircle
inequalities
geometry proposed

$ABC$ is an triangle.$I$ incenter of $ABC$.$AI,BI,$ and $CI$ meet the circumcicle of $ABC$ at $A',B'$ and $C'$
  recpectively.$S_I$ it's area of interseqtion of the interior triangle $ABC$ and the interior of triangle $A'B'C'$
      Prove  That    $S_I\geq (2/3)S_{ABC}$ ;)

Notice that $B_1C_1$ is the midperpendicular of $AI$. From this you can calculate the area $S_I$ which is then easily proved not less then $\frac{2}{3}S$

Mabe you rigte,becous this is not hard problem,if you can
solve this problem[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58189]www.mathlinks.ro/Forum/viewtopic.php?t=58189[/url] ;)  :D

[i]Dear Tiks[/i], I saw later this problem. Here is my solution for it.
I note the area $[XYZ]$ of the triangle $[XYZ]$. Prove easily that $[AA_1A_2]=\frac{Rr^2}{a}$ a.s.o. So that,
$\sum [AA_1A_2]=\frac{Rr^2(ab+bc+ca)}{abc}=\frac{r}{4s}\cdot (ab+bc+ca)\le$$\frac{r}{4s}\cdot \frac 13 (a+b+c)^2\le \frac{sr}{3}=\frac 13 S.$
Therefore, $\sum [AA_1A_2]\le \frac 13 S$, i.e. $[A_1A_2B_1B_2C_1C_2]\ge \frac 23 S.$

[u]Remark.[/u] I hope solve a generalization of the your problem for the any point $P$ from the interior of the triangle $ABC$ (at this time it is true for $P: =I$ and $P: =G$). I suppose that and you try the same thing !. If this problem will be true generally, then it becomes a splendid problem. I wish you (and me) every success !

[quote="Tiks"]Dear Virgil I saw your solution It is similiar to my sloution,but this problem is easy :(  :D .
I think this problem is not true for all interior point of $ABC$,I remember that I prove
that for ortosenter$H$,or for circumsentre$O$ we have $S_H\leq 2/3S$or$S_O\leq 2/3S$
for any acute triangle$ABC$,but my solution was be wery long,so I can't remember it. . :([/quote]
O.K. I am sorry that the points $O$ and $H$ don't respect the above inequality !. I note the set $\Omega$ of the points $P$ which respect the mentioned inequality. Then come on  prove that $\Omega$ is a convex set, i.e. for any points $X\in \Omega$, $Y\in \Omega$ and from the interior of the triangle $ABC$ we have $(XY)\subset \Omega$ or what is the contour  line between the sets $\Omega$ and $\neg \Omega$ with the computer (who knows use it !).


probability
MATHCOUNTS
FTW
expected value

On the average, how many times must a die be thrown until one gets a 6?

i think it should be 6 times as each number has equal prob. of coming on each throw. so, on average one can get 6 on 6 throws

If the first term is term 1, then the chance of rolling a six on term n is $ n *\frac { (5)^{n \minus{} 1}} {6^n}$

we multiply by n because there are n ways to choose which turn up to n we rolled a six in.
Summing as above gives the result.

[hide="Or..."]
We could approach this recursively. Let $ E$ be the expected number of throws until we get a 6. Clearly, there is a $ \frac 16$ probability that we could get a 6 on the first throw.
\[ E \equal{} \frac 16(1) \plus{} [\dots]
\]
Also, there is a $ \frac 56$ probability that we [b]don't[/b] get a 6 on the first throw, so we have wasted one roll and we're back to where we started (except that we used up one roll already), so the expected number of flips is now $ 1 \plus{} E$.
\[ E \equal{} \frac 16(1) \plus{} \frac 56(1 \plus{} E) \\
E \equal{} \frac 16 \plus{} \frac 56 \plus{} \frac 56E \\
E \equal{} 1 \plus{} \frac 56E \\
\frac 16E \equal{} 1 \\
E \equal{} 6,
\]
as desired.
[/hide]

i havnt done much study in prob. as i dont find it interesting. and thats why, i am not able to understand solution. yongyi! please explain your solution further.

Basically $ E$ is just the expected number of throws. We're using it at the same time as we're trying to solve it.

Now, if we tried to solve it manually, the cases would go on to infinity and result in an infinite series, much like 1=2 did :P

So I choose to solve it recursively. (This is a very common method of finding the expected number of trials until one reaches a desired event.)

First, we know that 1/6 of the time, the desired number (in this case, 6) will land up on the very first throw, and we're done, so the expected number of throws to get a 6 is 1, 1/6 of the time.

That's the first part of the equation.
\[ E \equal{} \frac 16(1) \plus{} [\dots]
\]
Now, the cool part is that the throws are independent, so once we've "wasted a throw", so to say, by not throwing a 6 on the first throw, we're back we're we started and we can use $ E$ again. But the crucial point here is that [b]we already wasted a throw, so the expected number of throws until we get a 6, [i]given that we did not throw a 6 on the first try[/i], is now incremented by 1.[/b]

So that accounts for the second part of the equation. So we now have
\[ E \equal{} \frac 16(1) \plus{} \frac 56(1 \plus{} E)
\]
And we can proceed to solve it like any other equation.

Great solution yongyi, It is not easy to find many creative approaches to this straightforward problem. Are you the same Yongyi from mathcounts countdown?

Massachusetts? Yes :D I was 3rd.

I assume you are from Massachusetts as well? What place did you get?

oh im in HS too old for MC  :D
What books have you done?

Only Intro to Counting and Probability :blush:

Because my coach had it and let me borrow it.

I'm in 9th grade now, so I'm too old for Mathcounts as well :lol:

But my brother isn't. He's in 6th grade.

I was tutored in 5th and 6th grade, and that taught me to think critically about puzzles, etc. But not MATHCOUNTS stuff or AIME/USAMO.

Hm, so maybe I don't know so much :P

Lol... This is the first time I've ever had to think about where I learn all this stuff. I think most of it's from the AoPS website... I browse the Olympiad section and look at the proofs there/attempt to prove them myself. Then I look at the past solutions to AIME/USAMO/AMC in the AoPSWiki.

Then the rest (like this one), I explored myself.

[quote="mathemonster"]How do you know so much if you havent done aops v1 or v2? tutoring?[/quote]

The technique Yongyi used on this problem has been used on a few other problems on this forum. I believe it also appears in the Intro Counting and Probability book.

I know the member named Ravi once made a post with a lot of probability problems that could be solved using this "expected value/recursion" technique.

im not refering to this problem- that was easy. Im saying to have placed 3rd at Nats, and his other posts.

Yongyi didn't place third at Nats- he was talking about Massachusetts state.

oh LOL....
I c... There was a kid at CD who was also named Yongyi i saw on tv.
Btw how do i get my FTW rating to not show???


In the row of Pascal's triangle that starts with 1 and then 12, what is the fourth number?

The first number is $ \binom{12}{0}$, the second is $ \binom{12}{1}$, the fourth is $ \binom{12}{3}\equal{}\boxed{220}$.

[quote="abhinand"]is there another way or pattern do solve this problem?[/quote]
If the second number is 12, then it is row 12. Then, the forth term is 12 choose 3, or 220.


I dont actually get how the pattern generates ...  :huh:

Well, I'm not a maths genius, but from what I can see with that, you already know what comes after 1+,2+, and 3+. so, you can now complete the equation.

And don't ask me to do it, by brain switches off after you get a fraction on top of a fraction [i]on top of another fraction[/i]!!!

Hope this helps!

Starting with the first number, you go through each column, from top to bottom, one after another, and you'll see the numbers go 132132132132...
But we can see that because each column has a number of numbers $1\mod{3}$ or $2\mod{3}$, a power of 2, we have three fixed fractions, we know that whenever you have a $1$, you'll have $\frac{3...}{2...}$, added to it, , for 2, $\frac{2}{1}$, and for 3, $\frac{1}{3}$.  Hence, we can let these three continuted fractions be $a$, $b$ and $c$, respectively.  Then we have
$1+a=1+\frac{3+c}{2+b}=1+\frac{3+\frac{1+a}{3+c}}{2+\frac{2+b}{1+a}}$, etc.  I'm sure you can substitute in and solve...
The rows alternate between beginning with 1 and 3, so i think you can use that...

[quote="prototypex24"]Saw this at the $F^3$ of the Iowa City West High. Couldn't find the solution so I decided to post it as a challenge:

Evaluate

$1+\frac{3+\frac{1+\frac{3+...}{2+...}}{3+\frac{1+...}{3+...}}}{2+\frac{2+\frac{2+...}{1+...}}{1+\frac{3+...}{2+...}}}$[/quote]

Let $x=1+\frac{3+\cdots}{2+\cdots}$, $y=3+\frac{1+\cdots}{3+\cdots}$, $z=2+\frac{2+\cdots}{1+\cdots}$.

Then, $x=1+\frac{y}{z}$, $y=3+\frac{x}{y}$, $z=2+\frac{z}{x}$

or $xz=z+y$, $y^2=3y+x$, $xz=2x+z$

So $xz=z+y=2x+z\Rightarrow y=2x$ and put into second equation

$(2x)^2=3(2x)+x\Rightarrow 4x^2=7x$. But $x\ne 0$, so $x=\frac74$

PS: $y=2x=\frac72$ and $z=\frac{2x}{x-1}=\frac{14}3$

good job 1000th user  :)  Nice solution!


Let $ r$ be a real number such that the sequence $ (a_{n})_{n\geq 1}$ of positive real numbers satisfies the equation $ a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{m \plus{} 1} \leq r a_{m}$ for each positive integer $ m$. Prove that $ r \geq 4$.

Let $ S_n \equal{} \sum_{i \equal{} 1}^n a_i \forall n \ge 0$. We have

$ r(S_n \minus{} S_{n \minus{} 1}) \ge S_{n \plus{} 1}$.

([hide="Motivation to the following"] Moreover, $ S_n$ is (strictly) increasing. Then $ r(S_n \minus{} S_{n \minus{} 1}) \ge S_{n+1} > S_n \Rightarrow S_n(r \minus{} 1) > rS_{n \minus{} 1} \Rightarrow S_n > \frac {r}{r \minus{} 1}S_{n \minus{} 1}$. In view of the fact that $ S_{n \minus{} 1} > 0$, we get $ r > 1$, which allows the last implication. Now we can substitute this in the formula we have to obtain

$ r(S_{n \plus{} 1} \minus{} S_n) \ge \frac {r}{r \minus{} 1}S_n$

This gives another bound of $ S_{n \plus{} 1}$ in terms of $ S_n$. Well, there is no meaning in making the counts here, but three iterations of this process give us $ r > 3$ (we just need to look the denominator $ r^2 \minus{} 4r \plus{} 3$). This is sufficient to conjecture that the answer is going to come from here, if we make "infinite iterations").[/hide]

Let us suppose $ S_{n \plus{} 1} \ge kS_n \forall n \in \mathbb{Z}_ \plus{}$ and some real $ k$. We have

$ r(S_n \minus{} S_{n \minus{} 1}) \ge S_{n+1} \ge kS_{n} \Rightarrow S_n(r \minus{} k) \ge rS_{n \minus{} 1}$

$ S_{n \minus{} 1} > 0$ gives $ r \minus{} k > 0$ and $ S_n \ge \frac {r}{r \minus{} k}S_{n \minus{} 1}$. In other words, $ S_{n \plus{} 1} \ge \frac {r}{r \minus{} k}S_n \forall n \in \mathbb{Z}_ \plus{}$.

Now, as we know that $ S_{n \plus{} 1} > S_n$, if we define $ a_0 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} \frac {r}{r \minus{} a_n}$, we see that this sequence is well defined and all its terms are $ > 0$. Suppose $ r < 4$. Now, we can prove that this sequence is increasing. In fact,

$ \frac {r}{r \minus{} a_n} > a_n \Longleftrightarrow r(a_n \minus{} 1) < a_n^2$

and $ r(x \minus{} 1) < 4(x \minus{} 1) \le x^2 \Longleftrightarrow (x \minus{} 2)^2 \ge 0$. But $ a_n \le \frac {S_2}{S_1}$, therefore $ a_n$ is bounded. Since this sequence is increasing, it is convergent to some limit $ L$. Now it is easy!

$ L \equal{} \frac {r}{r \minus{} L} \Rightarrow L^2 \minus{} Lr \plus{} r \equal{} 0$

We must have $ 0 \le \Delta \equal{} r^2 \minus{} 4r$.

why you take $ a_{n + 1} = \frac {r}{r - a_n}$  and please explain this $ a_n \le \frac {S_2}{S_1}$

[quote="feliz"]... if we define $ a_0 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} \frac {r}{r \minus{} a_n}$, we see that this sequence is well defined and all its terms are $ > 0$. ...[/quote]
This is false in general. If we consider the function $f(x) = \dfrac {r} {r-x}$ and we consider its iterates at $x=1$ we may run into trouble, especially when $r<4$; just see that for $r=2$ or $r=3$ we reach a value equal to $r$, and so we cannot continue. The claim $a_n \leq \dfrac {S_2} {S_1}$ is also unjustified (and other numerous typos and errors I have corrected).

However, the idea is good. As there, define $S_n = \sum_{i=1}^n a_i$, so the sequence $(S_n)_{n\geq 1}$ is increasing, forcing $r>1$. Now take $K = \{k \in \mathbb{R}_+^* \mid S_{n+1} \geq kS_{n} \textrm{ for all } n \in \mathbb{Z}_+^*\}$, and $\ell = \sup K$. This supremum exists, since $1\in K$ and $K$ is upper bounded by $\dfrac {S_2} {S_1}$; moreover $\ell\geq 1$, and $\ell \in K$.  But now $\ell S_{n+1} \leq S_{n+2} \leq r(S_{n+1} - S{n})$ is equivalent to $(r-\ell)S_{n+1} \geq rS_n$, forcing $\ell < r$. Therefore $S_{n+1} \geq \dfrac {r} {r-\ell} S_n$, and this for all $n$, so $\dfrac {r} {r-\ell} \in K$, therefore $\dfrac {r} {r-\ell} \leq \ell$. This writes as $\ell^2 - r\ell + r \leq 0$, so we must have for its discriminant $\Delta = r^2 - 4r \geq 0$, thus $r\geq 4$.

Let us finally notice that $r=4$ is indeed reachable, by taking, for any positive real number $a$, the values $a_1=a$, and $a_n=2^{n-2}a$ for $n\geq 2$; then $S_{n+1} = 2S_n$, so $S_{n} = 2^{n-1}a$, $\ell = 2$, $r=4$ and equalities allover.

Can someone post full and correct solution???

Can someone explain this part in [i][b]mavropnevma[/b]'s[/i] solution:
[quote=mavropnevma]
Now take $K = \{k \in \mathbb{R}_+^* \mid S_{n+1} \geq kS_{n} \textrm{ for all } n \in \mathbb{Z}_+^*\}$, and $\ell = \sup K$. This supremum exists, since $1\in K$ and $K$ is upper bounded by $\dfrac {S_2} {S_1}$; moreover $\ell\geq 1$, and $\ell \in K$.  [/quote]

Here is what I believe to be a very natural and straightforward solution.
Clearly $\frac{a_{n+1}}{a_n} \le r$ is bounded. So let $L=\limsup \frac{a_{n+1}}{a_n}$ (this sounds fancy but is really just the smallest essential upper bound on the ratios, i.e. upon ignoring the first finitely many values).
Now for any fixed $L_1<L<L_2$, we have $\frac{a_{n+1}}{a_n}<L_2$ for all sufficiently large $n$ and hence for any fixed $k$, we have for all large $n$, that
\[a_n+a_{n-1}+a_{n-2}+\dots+a_{n-k} \ge \left(1+\frac{1}{L_2}+\frac{1}{L_2^2}+\dots+\frac{1}{L_2^k}\right) a_n.\]
Moreover, for infinitely many $n$ we have $a_{n+1}>L_1a_n$ and hence
\[r \ge L_1+1+\frac{1}{L_2}+\frac{1}{L_2^2}+\dots+\frac{1}{L_2^k}.\]
Now the rest is straightforward: Since $L_1<L<L_2$ were arbitrary, we get that
\[r \ge L+1+\frac{1}{L}+\frac{1}{L^2}+\dots+\frac{1}{L^k}.\]
Moreover, as $k$ was arbitrary, we get
\[r \ge L+1+\frac{1}{L}+\frac{1}{L^2}+\dots=L+1+\frac{1}{L-1}=2+(L-1)+\frac{1}{L-1} \ge 4\]
as desired (note that $L>1$ is already implied from the first inequality by the fact that $r$ is finite).

Let $S_n=\sum_{i=1}^{n}a_i$ then $S_{n+1}>S_n.$
We have: $S_{n+1} \leq r\left ( S_n-S_{n-1} \right ) \, \rightarrow \frac{S_{n-1}}{S_n}+\frac{1}{r.\frac{S_n}{S_{n+1}}} \leq 1$
Define $x_n=\frac{S_{n-1}}{S_n}$ then $1 \geq x_n+\frac{1}{r.x_{n+1}}, \, (1) $
Assume $r<4$, then: $1\geq x_n + \frac{1}{4.x_{n+1}} \geq 2\sqrt{\frac{x_n}{4.x_{n+1}}} \rightarrow x_n \leq x_{n+1} \leq ... \leq 1$
So $\textrm{lim}x_n=L \leq 1.$
From $(1)$, we know: $1 \geq L+\frac{1}{r.L} \rightarrow r \geq \frac{1}{L.(1-L)}\geq 4,$ adsurb.

We can prove this by contradiction. Suppose that r < 4. We will show that this leads to a contradiction.

Since r < 4, we have a_1 + a_2 + … + a_{m+1} <= (r)a_m < 4a_m for each positive integer m.

If we let S_m = a_1 + a_2 + … + a_m, we can write:

S_{m+1} = S_m + a_{m+1} < 4a_m + a_{m+1} = 4(a_1 + a_2 + … + a_m) + a_{m+1} = 4S_m + a_{m+1}

Subtracting 4S_m from both sides, we get:

S_{m+1} - 4S_m < a_{m+1}

This simplifies to:

(1-4)S_m < a_{m+1}

Since S_m is the sum of the first m terms of the sequence, it is a positive real number. Therefore, the left-hand side of the inequality is negative, which means that the inequality has no solutions. This is a contradiction, so our assumption that r < 4 must be false.

Therefore, we must have r >= 4.

I hope this helps! Let me know if you have any questions.


percent
conics
parabola
probability
parameterization
percent increase
number theory

Sorry for the tardiness but here it is!:
1'.) The retail price of an item underwent the following changes from year to year over a four year period: a 25% increase, a 25% decrease, a 20% increase, and finally a 20% decrese. What percent increase or decrese was the final retail price over the original retail price of the item?
2'.) Kaitlin is now eight years younger than half her mother's age. In k years (k is a positive integer), Kaitlin will be one-third her mother's age then. What is the oldest possible age now for Kaitlin's mother?
3'.) Given the following base 5 addition: abcd (5)+bdc (5)=aadb (5) where a and b is not 0 and a, b, c, and d are different digits, find the four digit number abcd (5).
4'.) Find the length of the line segment joining the center of the circle C, which is {(x,y)l 42x :^2: +42y :^2: -84x+252y+84=0} to the vertex of the parabola P, which is {(x,y)l y :^2: =8x+4y+12}.
5'.) Thirty children were polled about their likes or dislikes of [u]Friends[/u] or [u]Will and Grace[/u]. The number of children who like [u]Friends[/u] is twice the number who like [u]Will and Grace[/u]. The number who dislike [u]Friends[/u] equal the number who like [u]Will and Grace[/u]. There are four children who like neither. Find the number of children who like [u]Friends[/u] but don't like [u]Will and Grace[/u].
6'.) Write the following expression in simplest radical form with the smallest possible index for the radical: ((12^(1/3))*(72^(1/6)))/(((108^(1/3))^(1/4))*((3^(1/3))^(1/2)))
7'.) Urn A contains 6 red and 4 green marbles. Urn B contains 7 red and 3 green marbles. If 2 marbles are drawn at random from each urn, what is the probability that from these drawn, there will be 2 red and 2 green marbles? Write the answer in the form a/b where a and b are relatively prime whole numbers. (2 points)

Kouunn wo innoru!

P.S. If anyone speak any foreign language other than Japanese, English :), Spanish, and French please PM me.

1) [hide]10.  you do this(1.2*1.25)(.8*.75) is .9 of the original, which is a ten percent decrease [/hide]

1)[hide]10% decrease[/hide]

3)[hide]3142[/hide]

[hide]1)1.25*.75)*1.2)*.8=.9 worth or 90%

2) solving equations and setting k=1 we get 44.

7) combine them and them look at working/toatl= 91/323[/hide]

Scrambled: +1 point
theone853: 2 points
ml2: 1 point (1 is so close...just read the question again, I know I should've copied the bolds and the underlines...sorry 'bout that, same with 7, it's both 2 not 2 and 3)

6 I'm still checking...I might've screwed up some of the parenthesis...anyways work on the others while I "check please" :)

Scores are up!

2)[hide]44

it took me so long cuz i didn't realize a negative wasn't possible.[/hide]

2 you got right, 7 you're close...but not quite. It's a much simpler fraction than that...1 point awarded, scores are updated.

1.[hide]10% decrease[/hide]

7. [hide]1/3[/hide]

They're both right, good jobe! :lol: 3 points awarded, scores updated.

P.S. I also updated what problems were left unsolved.

P.P.S. I believe now all problems are solvable...I hope :)

hmm for 3 [hide]did you want me to just say 3124 or say 3124 in base 10 (in this case 164.) otherwise the addition 3124+142=3321.

[/hide]

5. [hide]16[/hide]

6. [hide](32/3) all to 12th root[/hide]

No it's not that, it's just that the number's off by a tiny bit.

You got 5 right, not 6 though...scores are updated

Sorry for the tardiness, but here I am...

[hide]

1. 10% decrease

2. 48

3. 3142 (base 5)

4.  :sqrt: 34

5. 16

6. 2* 4th root of 3

7. 1/3[/hide]

Oooh, so close to a topper sweep....you got everything tight except for 2, which is oh so close...think about the parameters of each variable, what are the restrictions? (Hint: Use common sense on this one :lol:)

Anyways 7 points awarded, updated, yadayadaya :lol:

Oh wait, I only updated topper's...sorry about that, wow another clean sweep by ml2 with a total of 19 points! topper's right behind with a 15! *whistle* tough competition we got here :)

P.S. Did anyone notice for like a couple of hours ml2 had a 165 points 'cause of my typo? No? :lol:

i did! and i thought you meant 16.5 but that;s impossible, so i don't konw what you meant.

Still no deadline, so here goes...

[hide]

2)

use a coor. graph, set mom's age as y, daughter's age at x.

y=2x+16 and

y=3x+2k are given.  Fiddling gives 16-2K=X. X must be positive, so K has to be 0 to 7. As K decreases, X increases, and as X increases, Y increases, so as K decreases, Y increases, and the smallest legal possible value of K is 1, so we have

y=2x+16 and

y=3x+2, whose solution is (16,44). 44 is the answer.[/hide]


find all positive integers

\[n^{3}-5n+10 = 2^{k}\]

LHS is congruent to $0,1,3,5$ or $6$ modulo $7$, and RHS is congrunet to $1,2$ or $4$ modulo $7$, so both must be congruent to $1$ modulo $7$. Therefore $3\mid k$, and $n^{3}-5n+10$ must be perfect cube. But for $n>2$ we have $(n-1)^{3}<n^{3}-5n+10<n^{3}$, and so $n\leq 2$. Only solution is $n=2$ and $k=3$.


$ P$ is a point on the exterior of a circle centered at $ O$.  The tangents to the circle from $ P$ touch the circle at $ A$ and $ B$.  Let $ Q$ be the point of intersection of $ PO$ and $ AB$.  Let $ CD$ be any chord of the circle passing through $ Q$.  Prove that $ \triangle PAB$ and $ \triangle PCD$ have the same incentre.

We have $ QP.QO \equal{} QA.QB \equal{} QC.QD$ so $ PCOD$ is inscribed so $ \angle CPO \equal{} \angle OPD$
Let $ PO$ intersect $ (O)$ at $ I,K$
$ (P,Q,I,K) \equal{} \minus{} 1$ and $ \angle IDK \equal{} 90^o$ so $ DO$ bisector $ \angle PDC$ so $ I$ is incenter of $ PCD$
Easy to prove $ I$ is incenter of $ PAB$
So $ PAB$ and $ PCD$ have the same incenter

Let PO intersect circle O at X and E and let D be an arbitrary point on circle O. Let line DQ be extended past Q to intersect the circle again at a point C.
By construction, (E, X; Q, P) is a harmonic bundle.
Also, angle EDX is a right angle.
Hence, DX bisects angle QDP and circle O is the apollonian circle of segment EX with respect to Q.
Hence, BX bisects angle ABP, AX bisects QAP, and CX bisects angle QCP.
Hence the incenter of triangle PAB is point X, which also happens to be the incenter of triangle PCD.

Let segment $OP $ intersect the circle at $M $.
Observe that $OCPD $ is cyclic. Since, $OC = OD$, so, $OP $ bisects $\angle CPD $.
Also, $\angle PDC = \angle MOC = 2\angle MDC$ (since $O $ is the center). Thus, $MD $ bisects $\angle PDC $. Hence, $M $ is the incenter of both $\Delta PAB $ and $\Delta PCD $. This proves the result.