$\{a,b,c\}\subset (0,\infty )\Longrightarrow$ $\sum_{\mathrm{cyc}}\frac{1}{a(a+b)}\ge\frac{2(a+b+c)+3\sqrt [3]{abc}}{a^{2}b+b^{2}c+c^{2}a+3abc}\ .$

The given inequality is the same as $\sum_{cyc}\left(b\cdot\frac{a+c}{a+b}+c\frac{c+b}{a+b}+\frac{bc}{a}\right)\geq2\sum{a}+3\sqrt[3]{abc}$ $\Leftrightarrow$ $\sum_{cyc}a\left(\frac{a+c}{b+c}+\frac{c+b}{a+c}\right)+\sum_{cyc}\frac{bc}{a}\geq\left(\sum_{cyc}a\cdot2\right)+3\sqrt[3]{abc}$ from where it can be seen why it holds.

Congratulations on your 2500th post! Thats Birch & Swinnerton Dyer!

As to your inequality, I am wondering you came up with it...

[hide="failed attemptl; overestimation"]\[\frac{9}{2}\geq \frac{(a+b+c)^{3}}{a^{2}b+b^{2}c+c^{2}a+3abc}\\ \Longleftrightarrow 9\sum \left(a^{2}b\right)+27abc\geq 2(a+b+c)^{3}=2\left[\sum \left(a^{3}\right)+3\sum \left(a^{2}b\right)+3\sum \left(a^{2}c\right)+6abc\right] \\ \Longleftrightarrow 3\sum \left(a^{2}b\right)+15abc\geq 2\sum\left(a^{3}\right)+6\sum \left(a^{2}c\right)\]doesn't hold for $1,1,2$ :( [/hide]


[color=blue]Two points, $A$ and $B$, are locate on the ground a certain distance $d$ apart. Two rocks are launched  simultaneously from point $A$ and $B$ with equal speeds but at different angles. Each rock lands at the launch point of the other. Knowing that one of the rocks is launched at an angle $\theta >45^{\small 0}$ with the horizontal.
What is the minimun distance between the rocks during the flight?[/color]

is it
[hide]
$d \sqrt{ \frac{1-sin(2 \alpha) }{2} }$ , where $\alpha$ is one of the angles?
I'll put up a solution tomorrow...
[/hide]

[color=blue]Your answer is correct. Let post the solution[/color]

[hide="all right..."]

Firstly, you know that they land at each other's spots.  Meaning that they have the same ranges.
Range is given by the equation $\frac{v^2sin(2 \theta) }{g}$
In this case, if we denote the angles of the velocities with $\alpha$ and $\beta$, it can be seen that(because of how the sine function looks), $\alpha + \beta=90$

Then, we can write the equations concerning the motion of the two projectiles.
Denoting the initial velocities of the projectiles as $v$, and the time spent in air as $t$,

$\triangle y = vt(sin(\alpha) - sin(\beta) )$
and $\triangle x = d- vt(cos(\alpha) + cos(\beta) )$

where $\triangle x$ and $\triangle y$ are the displacements in the x and y directions between the two balls

But because the angles are complements, you can write
$\triangle y = vt(sin(\alpha) - cos(\alpha) )$
and $\triangle x = d- vt(cos(\alpha) + sin(\alpha) )$

So we want the distance between the rocks- square both those expressions:
$(\triangle y)^2 = (vt)^2(1-sin(2 \alpha) )$
$(\triangle x )^2= d^2+ (vt)^2(1+sin(2 \alpha) ) - 2vtd(cos(\alpha) + sin(\alpha) )$

Adding the two together, defining $r$ as the distance...
$r^2= d^2+2v^2t^2 - 2vtd(cos(\alpha) + sin(\alpha) )$

Minimizing $r^2$ would still give you the minimum $r$ in this case.. so you have a quadratic, and we find that the vertex lies at

$t= \frac{d(cos(\alpha) + sin(\alpha) ) }{2v}$
and... plugged back in to $r^2$ and taking the square root(to find the distance $r$...
I get the result posted above

ugh-ness
This is an ugly way to do it, though.  Anyone have a cooler way?

[/hide]


A circular table has exactly sixty chairs around it. There are N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest possible value of N?

[quote="AstroPhys"][hide]
Every-other chair filled, $ N=30$.
[/hide][/quote]

Incorrect.

[hide="Hint"]We can do even better. note that If X are the people and _ are the chairs,
_ X _ _ X _

forces the new person to sit next to some one.
In other words, every third chair is seated.[/hide]

Its n/3 then ceiling that so its 60/3=20, celing still 20.


Given the plane, $ 2x \minus{} 3y \minus{}5z \equal{} 100$, give four points which lie on the plane.

[quote="guile"]Given the plane, $ 2x \minus{} 3y \minus{} 5z \equal{} 100$, give four points which lie on the plane.[/quote]

The obvious ones are where two variables are zero.

Hence $ (50,0,0), (0,\minus{}\frac{100}{3},0), (0,0,\minus{}20)$

Then I'd let two be 1, giving $ (54,1,1)$


Hi,

I'm developing a program and I need to approximate the number of the integer factors of a number(I will allocate that much memory for the array to hold the factors.)

How can i do this?

Thanks

I think the integer part of the square root of the number should suffice.  e.g
$\sqrt{100}= 10$ and the factors are $1,2,4,5,10,20,25,50,100 = 9$.
Or you may choose not to count 1 and the number itself. In which case it has 7 factors.


LaTeX
number theory unsolved
number theory

i believe this question is asked by others before, but i can't find the post... anyway here's the problem:

Find all solutions to the equation x^p+y^p=p^z
where p is a prime and x,y,z are positive integers.

thanks very much.

Take $p$ to be odd. Clearly, we may assume that $p$ doesn't divide $x$, otherwise we would have $p|y\Rightarrow p^p|x^p+y^p$, so we will be able to divide bot $x,y$ by $p$ and get smaller values, doing this until $p$ doesn't divide $x$ anymore.

Since $p$ is odd, we have $x+y|x^p+y^p\Rightarrow x+y=p^t$. We thus have $x^p+(p^t-x)^p=p^z$. When we expand $(p^t-x)^p$, $x^p$ will be reduced and we will eventually have $M_{p^{2t}}+x^{p-1}p^t=p^z$. This clearly isn't divisible by $p^{t+1}$ because $p$ doesn't divide $x$, so $p^z<p^{t+1}\Rightarrow z=t$, which is obviously absurd because it means $x+y=x^p+y^p$.
This means that for odd $p$ there are no solutions.

There's the case $p=2$. We have $x^2+y^2=2^z$. We may again assume that $x,y$ are odd by dividing the whole eqn by $4$ until we get odd $x,y$. At that point, we will have a sum of two oidd squares, which is $M_4+2$, equal to a power of $2$, so the power of $2$ should be $1$. The solutions are $(x,y,z)=(2^{2k},2^{2k},2k+1),\ k\ge 0$.

yes,$(p, x, y, z)=(3,1,2,2)$ gives $1^3+2^3=3^2$, which is obviously true. it is also a counterexample to a part of grobber's solution : he said "there is no solution for p odd".
finally, this problem doesn't look like a famous theorem.

Here's what I did wrong (luckily, it can be fixed, but it's really boring): when I expanded the binomial $(p^t-x)^p$ I forgot the binomial coefficient. it's $M_{p^{2t}}+x^{p-1}p^{t+1}$ (notice the difference?; in the first post it was blabla $+x^{p-1}p^t$). If $t>1$ then $z=t+1\Rightarrow \frac {x^p+y^p}{x+y}=p\le \sqrt{x+y}$. We can dismiss this case, eventually after considering some small cases, like $x=y=1$ etc. We're left with $t=1$, getting $x^p+(p-x)^p=p^z$ (because $x+y=p^t=p$). We get $M_{p^3}+p^2x^{p-1}=p^z\Rightarrow z=2$. This implies $x=1$ because otherwise the left part would be too large. The eqn becomes $1+(p-1)^p=p^2$. There clearly is a very small upper bound for $p$: $p< 4\Rightarrow p=3$. We check this, see it fits, and we're done, we obtain the solutions you guys noticed.

These aren't all the solutions, as we assumed that $p$ didn't divide $x$. The solutions for the case $p$ odd are $(p,x,y,z)=(3,3^t,2\cdot 3^t,2+3t)$.

[edit: a $\$ was missing, and that caused a mess!:)]

some problems with LaTeX, l see ! tough in my mind the solution is right


salam be hameye barobachehaye olampiadd !bacheha farda bad az olampiad biyayn net va tooye hamin topic soalha va javab haro bezanim ta bebinim ke che kar kardimm emtehan(hamoon kari ke paral ham ba bachehaye mathlinks kardim)ishala ke hamamoon khoob midim!

are,nazare kheili khoobie,hodoode sa@e 3 fek konam monaseb bashe.

mishe, yecki javab ha ro bezane!
[ye bakhshi, az soorat soal ham benevise ke code haye mokhtalef, betoonan check konan]

chera emsal asoon bood?

ye makhroote naghes ba shoahaye ghaedeye3 va 5 va yale 10 rooye ye dayere mighalte va be jaye khodesh barmigarde shoaye dayere cheghadre?
lotfan age in soala hal kardin hallesha bezarin.
doroste emsal soala asoontar az ghabl bood :wink:

javabe soalet mishe 15 [man dar avordam va fek mikonam doroste]
valli neveshtane, rah halesh toolaniye!

valli idash ine ke in makhroote naghes mesle, makhroote kamel
micharkhe va markaze oon dayere ,rase makhroote kamele.

man 24 ta zadam [nemidoonam chand tash doroste!]
lotfan, ye adame khair khahi javab ha ro bezane :?:

bache ha harkas biad bege chand ta zade va javabasham bezane (code bargasham bege).in fek konam sari tarin rah baraye peyda kardane javabaye dorost bashe.man 24 ta zadam

avallan in ke sevomam, chi kare esmam dari!? :?:
dovom, in ke khodet chi kar kardi?

javab hayi ke man fek mikonam doroste :

A + B + C : 16
majmooe mohit va masahat : 100
loole va lamp : 4
adade seraghami ba raghame motevali motemayez : 648
parekhat haye ghate y = 0 : 24
mostatile haye moteshabeh : 2
emtehane testi : 4
gharbale eratesten : 4141
x + y + z = 0 : 1 / 4
2^2^5 - 1 : 4294967295
majmooeha AUB = BUC : A=BUD

baghiye ro felan hal nadaram!

dorood in ham pasokhe man be moghaddamati ke fek mikonam doroste :
(gharbal)->4141
(Harkat Az A be B) -> 8
(Adade Bozorgtar az 10 ) -> 3
(Kokab khanoom) -> 3
(Majmooe haye A,B,C,D)-> A = B $\cup$ D
(Khotoote  mostatile 2*5) -> 4 (man be eshtebah zadam 3 :(( )
(Davandegan ba sorat haye 7 va 5 ) -> 13
(Davarane karte morbbaii) -> "Z" barax
(BAA + BA = CDAD) -> 15
(Mohit + Masahat = 140 ) -> 100 (shak daram)
(Haft Tas) -> 105
(adade 4 raghami $abab$) -> 8
(otaghe 5*5 va kashi 1*3) -> 8
(hasel jame 7 adad tabii kamtar az 10 bar 3 bakhshopazire) -> 12 ta mishe ( vali man ino ehtebah zadam )
(mostatile abcd va paye ertefa do mosallas m,n masahate AMCN) -> $\frac{{12}}{5}$
(tope footbale charmi)-> 20
(tedade arghame $3^{40}$) -> 20
(faseleye noghte rooye dayerye mohati)-> 10
(3mohre sefid va 3 mohre siah dar morbbae 3*3) -> 12
(11 khane siah dar jadvake 2*12) -> 44
(cheraghe tooye ostovane) -> 4 (shak daram)
(hajme majmooe) -> nazadam
(morbbaha va mosalllas haye ghaemozaviye) -> $8\sqrt 2$
(morabba ABCD va noghteye M) -> $\frac{{13}}{32}$
(parekhathaye ke mehvar x ro ghat mikonanad) -> 24
(x+y+z = 0) -> $\frac{{1}}{4}$
(mosallase 3,4,5) -> $\frac{{12}}{5}$
(nomreye 89 emtehane testi) -> 4
(makhroote naghes) -> 15
(rang amizi vojooh mkaab) -> nazadam

ba in hesab man 28 ta zadam( ke motmannem 2tash ghalate )
(man onlinam ki hal dare ba man to Yahoo chat kone idim ham ke in zire )

khob dobare hal peyda kardam!

noghtaye daroon mosalas, ba azlae 5, 4 ,3 : 12 / 5
masahate chahar zelee : 13 / 32
makhrote naghes : 15
otaghe 5 * 5 : 7
dayere mohati moraba : 10
x^2 + x + 8 = y^2 : 1
n, 2n, 3n, ..., 1000n : 2607
ta kardan az ghotr : 12 / 5
x^2 = [x^2] : 2
pahlevanha : 7

khob [u]nazaretoon[/u] ro begid!

kollan be nazaretoon melat chi kar kardan!? :wink:

salaam  man riaziioo   << moghaddamati na >>
27 ta zadaaaam fek konaam 3taash ghalaate ta fardaa pasokhnamaro kamel mizaraaaaaam    :wink:  :D

man fek mikonam [be ehtemale ziyad] javabe in soal 7e! :?:

avallan, ke khob tashaboh benevis ( x / x + 10 = 3 / 5 ==> x = 15)
dovoman mage be nazare to kaf chand!? :huh: [/i]

ok,khob baba fahmidam.
badesham goftam ke kaf mishe 10,11
dar zemn behtare too in topic  inghadr cherto pert nanevisim.

javabe soala  http://www.matholamp.blogsky.com

[quote="splash"]javaba too http://imo.blogfa.com hast

rah e esbat e matris chie?! chera bishtar az 25 nemi she?!

[u][i][b]mane kheng :D [/b][/i][/u] hamoon aval vase 3 dar 3 va do dar do ro ke didam hamin tori zadam 9 :D !

ByE[/quote]
daghighan mane kheng ham hamin karo kardam!!!!hamine migam soalle asoon khoob nist chon joob khoresh khodastt soalle asoon eyne emrooz!!!!!!!!bacheha khabari az kaf gereftid begid hatmannn

[quote="amirhtlusa"]soal a ro as koja mishe did?  :maybe:[/quote]
be ehtemale ziad ta 1 hafte dge tooye site bashgah http://www.ysc.ac.ir bezanan!!

be nazare shoma momkene kaf balaye 15 bashe!?

[quote="parinaz.sa"]man ke migam nemisheee hameye moallema migann az 11 nemitoone balatar biyadd aghaye eslahpazir ke khodesh tooye bashgah hast dge migoft 8 goft  age kheili khodemoono bekoshim mishee 9!!!!!!!!!!!!!!![/quote]

are eslas gofte hol o hosh e hast! kas e Dge i nabood hame etoon az eslas porCDn?! :D

P.N : eslas too komite hast?! :ninja:  fekr nemi konam faghat oon ja dars mi de!

[quote="parinaz.sa"]man ke migam nemisheee hameye moallema migann az 11 nemitoone balatar biyadd aghaye eslahpazir ke khodesh tooye bashgah hast dge migoft 8 goft  age kheili khodemoono bekoshim mishee 9!!!!!!!!!!!!!!![/quote]

bebakhshid emkan dare ie bare dige begin eteghade aghaie behman eslahpazir darbareie nomreie kaf chi boode? :blush:
ba tavajo be inke sherkat konandegan nesbat be parsal 5 barabar shodan :huh:

ki mige teadade sherkat konandeha 5 barabar shode!? :?:
sale pish 80 hezar o khoordeyi, emsal ham 100hezar nafar bood!  :mad:

bad mage be nazare to   kafe chande?

aghaye eslahpazir 2shanbe azash porsidam goft 10 albate barkhi az moallema ke ba bacheha bishtar dar ertebat hastand va midoonan bacheha chi kar kardan migan 11 ,12 !!!

too madreseh ma e'elam kardan balaye 40% hatman ghabool hast,amma kamtar az oon momkene ghabool nashe. :)

hala az hame ina begzarim khob maloome balaye 40 ghaboole vali be nazaretoon computer balaye chand ghaboole ba tavajoh beinke sakhtie soalaa binazir boood !!! intori ke amaro argham neshoon mide balaye 25 hatmaan ghaboole !! valla nemidooim :-??


[quote="Genecks"]I'm primarily looking for a way to do this problem. I suppose I have to use interval notation and then insert a number between the values of each interval to check for validity. But I'm not sure exactly the correct way to go about this:

$\frac{x^{2}-5x+4}{x-2}> 0$

I figure I have to list each interval, such as the following:

$(-\infty, 4)$
(1, 4)
(2, 4)
$(4, \infty)$

What else do I do?

I'm not necessarily asking for the picture, just what to do with the interval notation.[/quote]

don't let the fraction throw you off. since you are just looking for the sign of that function (since the other side of the inequality is 0) then act as if they were multiplied. factor the numerator, and go from there.

[hide]
$\frac{(x-4)(x-1)}{x-2}>0$

x cannot be 4, 1, or 2, obviously.
When x>4, the fraction is positive, check.
When 2<x<4, the fraction is negative, nope.
When 1<x<2, the fraction is positive, check.
When x<1, the fraction negative, nope.

So you're answers are (1, 2) U (4, inf)
[/hide]


USAMO geometry trapezoid inequalities trigonometry

A circle $\omega$ is inscribed in a quadrilateral $ABCD$. Let $I$ be the center of $\omega$. Suppose that \[
(AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2.
\] Prove that $ABCD$ is an isosceles trapezoid.

(AI + DI)^2 + (CI + BI)^2 <= ( (AI^2 + DI^2)^1/2 + (BI^2 + CI^2)^1/2 )^2 by the Cauchy Schwarz Inequality (easily seen by expanding).
However, AI*DI sin<AID = rAD, BI*CI sin(pi-<AID)=rBC where r is w's inradius.  And by Cosine Law on AID, (AI^2 + DI)^1/2 = (AD^2 + 2AI*ID cos<AID)^1/2 = (AD^2 + 2rAD cos<AID/sin<AID)^1/2 <= AD + r cos<AID/sin<AID with equality iff cos <AID=0, i.e. <AID=pi/2.
Similarly, we find that (BI^2 + CI^2)^1/2 <= BC + r cos(pi-<AID)/sin(pi-<AID) = BC - r cos <AID/sin<AID with equality iff <AID = pi/2, as before.
Therefore, (AI + DI)^2 + (CI + DI)^2 <= ( (AI^2 + DI^2)^1/2 + (BI^2 + CI^2)^1/2 )^2 <= (AD + BC)^2 = (AB+CD)^2.  Equality holds iff <AID = pi/2 and AI/BI=CI/DI (the Cauchy equality condition).  i.e. Equality holds iff AID ~ BIC.  Therefore, <DAB=<CBA, <ADC=<BCD from which it follows immediately that ABCD is an isosceles trapezoid.

There is a minor error above.  The first inequality should be (AI + DI)^2 + (CI + BI)^2  ( (AI^2 + BI^2)^1/2 + (DI^2 + CI^2)^1/2 )^2.
Then, instead of working with triangles AID and BIC, follow the exact same procedure with triangles AIB and CID.  The same procedure and same conclusion will follow.

Alternately we can do a bash (though it is better than it looks)... let the length of the tangents to the incircle from $ A,B,C,D$ be $ a,b,c,d$ respectively. First we prove a lemma.

[b]Lemma:[/b] If $ r$ is the inradius of a tangential quadrilateral $ ABCD$ then
\[ r^2 = \frac {abc + bcd + cda + dab}{a + b + c + d} = \frac {ab(c + d) + cd(a + b)}{a + b + c + d}.\]
[i]Proof:[/i] Let $ \alpha = \tan{A/2},\beta = \tan{B/2},\gamma = \tan{C/2},\delta = \tan{D/2}$. Letting $ s_1 = \sum\alpha$, $ s_2 = \sum\alpha\beta$, $ s_3 = \sum\alpha\beta\gamma$, and $ s_4 = \alpha\beta\gamma\delta$, we can prove by induction that
\[ 0 = \tan\pi = \tan((A/2) + (B/2) + (C/2) + (D/2)) = \frac {s_1 - s_3}{1 - s_2 + s_4},\]
so
\[ s_1 = s_3,\]
and the lemma follows.$ \blacksquare$

By Cauchy,
\begin{align*} (AI + DI)^2 + (BI + CI)^2 & = AI^2 + BI^2 + CI^2 + DI^2 + 2(AI\cdot DI + BI\cdot CI) \\
& \le AI^2 + BI^2 + CI^2 + DI^2 + 2\sqrt {(AI^2 + BI^2)(DI^2 + CI^2)} \\
& = a^2 + b^2 + c^2 + d^2 + 4r^2 + 2\sqrt {(a^2 + b^2 + 2r^2)(c^2 + d^2 + 2r^2)}, \end{align*}
with equality only when $ AI/DI = BI/CI$.* We claim that this is at most
\[ (AB + CD)^2 = (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd),\]
so we need to prove that
\[ ab + ac + ad + bc + bd + cd\ge2r^2 + \sqrt {(a^2 + b^2 + 2r^2)(c^2 + d^2 + 2r^2)}.\]
Using our lemma and the substitutions $ u = a + b$, $ v = ab$, $ w = c + d$, and $ x = cd$ for convenience (so $ r^2 = (ux + vw)/(u + w)$), this is equivalent to
\[ (u + w)(v + x + uw) - 2(ux + vw) \\
\ge\sqrt {uw(u(u + w) - 2(v - x))(w(u + w) + 2(v - x))}\]
or
\begin{align*} (uw(u + w) + (u - w)(v - x))^2 & \ge uw(u(u + w) - 2(v - x))(w(u + w) + 2(v - x)) \\
u^2w^2(u + w)^2 + 2uw(u^2 - w^2)(v - x) + (u - w)^2(v - x)^2 & \ge u^2w^2(u + w)^2 + 2uw(u^2 - w^2)(v - x) - 4uw(v - x)^2 \\
(u + w)^2(v - x)^2 & \ge0, \end{align*}
as desired. Equality occurs when $ ab = cd$, so
\[ r^2 = \frac {ab(c + d) + cd(a + b)}{a + b + c + d} = \frac {ab(a + b + c + d)}{a + b + c + d} = ab = cd.\]
But this can only be true if (note similar triangles with $ r/a = b/r$ and so on)
\[ \angle AIB = \angle CID = 90^\circ.\]
Now from *, we have $ AI/DI = BI/CI$, so $ \triangle AIB$ and $ \triangle CID$ are similar right triangles with the same length $ r$ of altitude to the hypotenuse, so they are congruent, and $ a = d,b = c$ so $ a + b = c + d$ and we conclude that $ ABCD$ is an isosceles trapezoid.

Here's a more trigonometric solution...

Without loss of generality, let $\omega$ have radius 1. Let $\omega$ be tangent to $AB$, $BC$, $CD$, and $DA$ at $W$, $X$, $Y$, and $Z$, respectively. It is clear that $AI$ bisects $\angle ZIW$, $BI$ bisects $\angle WIX$, etc. Let $m \angle AIW = 2a$, $m \angle WIX = 2b$, $m \angle XIY = 2c$, $m \angle YIZ = 2d$. By some simple trigonometry, $AI = \sec a$, $BI = \sec b$, $CI = \sec c$, $DI = \sec d$, and $AB + CD = \tan a + \tan b + \tan c + \tan d$. Furthermore, $a,b,c,d < \frac{\pi}{2}$ and $a+b+c+d = \pi$. We wish to show that $a=b$ and $c=d$.

The given equation can be rewritten as $(\sec a + \sec d)^2 + (\sec b + \sec c)^2 = (\tan a + \tan b + \tan c + \tan d)^2$. Expanding,
$\sec^2 a + \sec^2 b + \sec^2 c + \sec^2 d + 2 \sec a \sec d + 2 \sec b \sec c = (\sec^2 a - 1) + (\sec^2 b - 1) + (\sec^2 c - 1) + (\sec^2 d - 1) + 2 (\tan a \tan b + \tan a \tan c + \tan a \tan d + \tan b \tan c + \tan b \tan d + \tan c \tan d)$.
Rearrange this to get $\sec a \sec d + \sec b \sec c + 2 = \tan a \tan d + \tan b \tan c + (\tan a + \tan d)(\tan b + \tan c)$,
or $(\sec a \sec d - \tan a \tan d + 1) + (\sec b \sec c - \tan b \tan c + 1) = (\tan a + \tan d)(\tan b + \tan c)$.
This rearranges to
${\frac{1 - \sin a \sin d + \cos a \cos d}{\cos a \cos d} + \frac{1 - \sin b \sin c + \cos b \cos c}{\cos b \cos c} = (\frac{\sin a \cos d + \sin d \cos a}{\cos a \cos d}}) ( \frac{\sin b \cos c + \sin c \cos b}{\cos b \cos c} ),$
which rearranges again to
$\frac{1 + \cos(a+d)}{\cos a \cos d} + \frac{1 + \cos(b+c)}{\cos b \cos c} = \frac{\sin(a+d) \sin(b+c)}{\cos a \cos b \cos c \cos d}.$

Clearing denominators, we get that

$(\cos b \cos c)(1 + \cos(a+d)) + (\cos a \cos d)(1 + \cos(b+c)) = \sin(a+d) \sin(b+c).$

Note that $f(x) = \log \cos x$ over $0 < x < \frac{\pi}{2}$ has $f''(x) = - \sec^2 x < 0$, so $f$ is concave. By Jensen's inequality, it follows that $\log \cos \frac{x+y}{2} \geq \frac{\log \cos x + \log \cos y}{2}$, that is, $\cos x \cos y \leq \cos^2 \frac{x+y}{2}$ for any $x,y \in (0, \frac{\pi}{2})$, with equality holding iff $x=y$. Since $1 + \cos(a+d), 1 + \cos(b+c) \geq 0$, it follows that

$\sin(a+d) \sin(b+c) = (\cos b \cos c)(1 + \cos(a+d)) + (\cos a \cos d)(1 + \cos(b+c)) \leq \cos^2 \frac{b+c}{2} (1 + \cos(a+d)) + \cos^2 \frac{a+d}{2} (1 + \cos(b+c))$.

Let $\alpha = a+d$ and let $\beta =b+c$; then $\alpha+\beta = \pi$. We have that $\sin \alpha \sin \beta \leq \cos^2 \frac{\beta}{2} (1 + \cos \alpha) + \cos^2 \frac{\alpha}{2} (1 + \cos \beta)$ with equality holding iff $a=d$ and $b=c$.

On the other hand,
$\begin{align*}\cos^2 \frac{\beta}{2} (1 + \cos \alpha) + \cos^2 \frac{\alpha}{2} (1 + \cos \beta)
&= \frac{1 + \cos \beta}{2} (1 + \cos \alpha) + \frac{1 + \cos \alpha}{2} (1 + \cos \beta) \\
&= (1 + \cos \beta)(1 + \cos \alpha) \\
&= (1 + \cos \beta)(1 + \cos(\pi - \beta)) \\
&= 1 - \cos^2 \beta \\
&= \sin^2 \beta \\
&= \sin \alpha \sin \beta,\end{align*},$
so equality always holds. Hence, we must have that $a=d$ and $b=c$, which completes our proof.

Here's a completely algebraic solution.  WLOG $\omega$ has radius $1$, and let $a$, $b$, $c$, $d$ be the lengths of the tangents from $A$, $B$, $C$, $D$ to $\omega$.  It follows that $a+b+c+d = abc+bcd+cda+dab$ ($\star$) as many people noted.  Then, the content of the problem is to show that \[ (\sqrt{a^2+1}+\sqrt{d^2+1})^2 + (\sqrt{b^2+1}+\sqrt{c^2+1})^2 \le (a+b+c+d)^2 \] subject to ($\star$), with equality only when $a=d=\tfrac1b=\tfrac1c$.

Let $S = ab+bc+cd+da+ac+bd$. Then the inequality is \[ \sqrt{(a^2+1)(d^2+1)} + \sqrt{(b^2+1)(c^2+1)} \le S - 2. \] Now, by [b]USAMO 2014 Problem 1[/b] and the condition ($\star$), we have that $(a^2+1)(b^2+1)(c^2+1)(d^2+1) = (S - abcd - 1)^2$.  So squaring both sides, the inequality becomes \[ (ad)^2+(bc)^2 + a^2+b^2+c^2+d^2 \le S^2 - 6S + 2abcd + 4. \] To simplify this, we use the identities \begin{align*} S^2 &= 6abcd + \sum_{\text{sym}} a^2bc + \frac14\sum_{\text{sym}} a^2b^2 \\ (a+b+c+d)^2 &= (abc+bcd+cda+dab)(a+b+c+d) \\ &= 4abcd + \frac12\sum_{\text{sym}} a^2bc \end{align*} So $S^2+2abcd = \tfrac14\textstyle\sum_{\text{sym}} a^2b^2 + 2(a^2+b^2+c^2+d^2) + 4S$ and the identity we want to prove reduces to \[ 2S \le (ab)^2+(ac)^2+(bd)^2+(cd)^2 + 4 + a^2 + b^2 + c^2 + d^2. \] This is immediate by repeated applications of AM-GM, and the equality case is when $ab=ac=bd=cd=1$, $a=d$, $b=c$.

Construct a point $J$ such that $JBC\sim IAD$ and $JBCI$ is cyclic. Let the ratio of similarity $\frac{AD}{BC}$ be $r$, and let $JB=a,JC=b, IB=c,IC=d$. Note that $AI=ra$ and $BI=rb$. Additionally,
\[
r = \frac{AD}{BC} = \frac{[IAD]}{[IBC]} = r^2\frac{[JBC]}{[IBC]}= r^2 \frac{JB\cdot JC \sin \theta}{IB\cdot IC \sin \theta} = r^2 \frac{ab}{cd}
\]
so $r = \frac{cd}{ab}$. Finally, it's well known that $AB+CD=AD+BC$. So the condition reduces to
\[
(ra+rb)^2 + (c+d)^2 = (AD+BC)^2 = (1+r)^2 BC^2
\]
or equivalently
\[
c^2d^2(a+b)^2+a^2b^2(c+d)^2=(ab+cd)^2 BC^2
\]
Now a few more computations yield $BC^2 = \frac{(ac+bd)(ad+bc)}{ab+cd}$ so the condition is
\begin{align*}
c^2d^2(a+b)^2+a^2b^2(c+d)^2&=(ab+cd)(ac+bd)(ad+bc)\\
\Leftrightarrow abcd(a^2+b^2-2ab+c^2+d^2-2ac)&=0\\
\Leftrightarrow a=b, c=d.
\end{align*}

[quote=math154]
[b]Lemma:[/b] If $ r$ is the inradius of a tangential quadrilateral $ ABCD$ then
\[ r^2 = \frac {abc + bcd + cda + dab}{a + b + c + d} = \frac {ab(c + d) + cd(a + b)}{a + b + c + d}.\]
[i]Proof:[/i] Let $ \alpha = \tan{A/2},\beta = \tan{B/2},\gamma = \tan{C/2},\delta = \tan{D/2}$. Letting $ s_1 = \sum\alpha$, $ s_2 = \sum\alpha\beta$, $ s_3 = \sum\alpha\beta\gamma$, and $ s_4 = \alpha\beta\gamma\delta$, we can prove by induction that
\[ 0 = \tan\pi = \tan((A/2) + (B/2) + (C/2) + (D/2)) = \frac {s_1 - s_3}{1 - s_2 + s_4},\]
so
\[ s_1 = s_3,\]
and the lemma follows.$ \blacksquare$
[/quote]
Why does this prove the lemma?

[quote=pi37]Construct a point $J$ such that $JBC\sim IAD$ and $JBCI$ is cyclic
\end{align*}[/quote]
How can we be sure such a point even exists?

Sorry for coming to a old problem but is it possible to solve with complex numbers in a elegant way? As the solutions are mainly a bunch of calculations I asking myself if using complex is a good idea.

Let $a,b,c,d$ denote the length of the tangents from $A,B,C,D$, respectively, and let $\angle AID=\theta$ As others have noted above, we have $$r^2=\frac{abc+bcd+cda +dab}{a+b+c+d}$$
Note that we easily get $\angle BIC=180^{\circ}-\theta$ from angle chasing. The problem falls to the following:
[b]Claim:[/b] $AD\cdot BC=4r^2$
[i]Proof:[/i] We apply Law of Cosines on $\triangle AID$ and $BIC$. Plugging this into the condition we get $$(AI+DI)^2+(BI+CI)^2=AD^2+BC^2 +2AI\cdot DI(\cos\theta +1)+2BI\cdot CI(1-\cos\theta)=(AB+CD)^2=(AD+BC)^2 $$ where the last equality comes from Pitot’s theorem. Rearranging, applying reverse power reduction on cosine, and using trig in right triangles gives $$AI\cdot DI(\cos\theta+1)+BI\cdot CI(1-\cos\theta)=AD\cdot BC$$
$$\implies 2AI*DI\cdot \cos^2\frac{\theta}{2}+  2BI*CI\cdot \sin^2\frac{\theta}{2}=AD\cdot BC$$
$$\implies 2r^2+2r^2=4r^2=AD\cdot BC$$ hence the claim is proven.
We can use the radius formula to get $$4\cdot \frac{abc+bcd+cda +dab}{a+b+c+d}=(a+d)(b+c)$$ Rearranging and completing squares gives $$(b+c)(a-d)^2+(a+d)(b-c)^2=0\implies a=d, b=c$$ It is easy to see that this implies $ABCD$ is an isosceles trapezoid, as desired.

Let $\omega$ have radius $1$. Let $a=\sqrt{AI^2-1},b=\sqrt{BI^2-1},c=\sqrt{CI^2-1},d=\sqrt{DI^2-1}$. Then we have
\[a^2+2\sqrt{a^2+1}\sqrt{d^2+1}+d^2+b^2+2\sqrt{b^2+1}\sqrt{c^2+1}+4=(a+b+c+d)^2.\]
By cancelling, we have
\[\sqrt{a^2+1}\sqrt{d^2+1}+\sqrt{b^2+1}\sqrt{c^2+1}=ab+bc+cd+da+ac+bd-2.\]
At this point, note that by the tangent addition formula we have
\[\frac{a+b}{1-ab}=-\frac{c+d}{1-cd}\implies a+b-acd-bcd=-c-d+cab+dab\implies a+b+c+d=abc+bcd+cda+dab.\]
Let $ab+bc+cd+da+ac+bd=S$. Then, the key identity from USAMO 2014/1 is that
\[(a^2+1)(b^2+1)(c^2+1)(d^2+1)=(S-abcd-1)^2.\]
Now, square to yield
\[a^2d^2+b^2c^2+a^2+b^2+c^2+d^2+2+2[S-abcd-1]=S^2-4S+4\implies\]
\[a^2d^2+b^2c^2+a^2+b^2+c^2+d^2=S^2-6S+4+2abcd.\]
Let $T=a+b+c+d$ and rearrange this as
\[a^2d^2+b^2c^2-2abcd=(S-2)^2-T^2 = \]
\[\frac 12\sum_{\mbox{sym}}a^2b^2 + \sum_{\mbox{sym}}a^2bc+6abcd-2\sum_{\mbox{sym}}ab+4-\sum_{\mbox{cyc}}a^2-(a+b+c+d)(abc+bcd+cda+dab)\iff\]
\[(a^2+d^2)(b^2+c^2)+\frac 12\sum_{\mbox{sym}}a^2bc+4abcd-2\sum_{\mbox{sym}}ab+4=0.\]
Rewrite this last equality as
\[0=(ac-1)^2+(ab-1)^2+(cd-1)^2+(bd-1)^2+\frac 12 \sum_{\mbox{sym}}a^2bc + 4abcd-4ad-4bc-2(a+d)(b+c).\]
So it is sufficient to check
\[\frac 12 \sum_{\mbox{sym}}a^2bc + 4abcd\ge 4ad+4bc+2(a+d)(b+c)\]
and find the equality case(s).
Multiplying one side by $a+b+c+d$ and the other by $abc+bcd+cda+dab$ yields that we want to check
\[7\sum_{\mbox{cyc}}a^2bcd+\frac 12\sum_{\mbox{sym}}a^3bc+\sum_{\mbox{sym}}a^2b^2c\ge 2[a^2bcd+abcd^2+ab^2cd+abc^2d+cd^2a^2+d^2a^2b+ab^2c^2+db^2c^2]+\]
\[\sum_{\mbox{sym}}a^2b^2c+6\sum_{\mbox{cyc}}a^2bcd.\]
Cancelling, we need to check
\[\frac 12\sum_{\mbox{sym}}a^3bc\ge (a+b+c+d)abcd+2[cd^2a^2+d^2a^2b+ab^2c^2+db^2c^2].\]
By the inequalities $abd(a-d)^2\ge 0,acd(a-d)^2\ge 0,abc(b-c)^2\ge 0,bcd(b-c)^2\ge 0$, it suffices to show
\[a^3bc+b^3ad+c^3ad+d^3bc\ge (a+b+c+d)abcd.\]
Finally, use cyclic versions of
\[\frac{2a^3bc}{3}+\frac{d^3bc}{3}\ge a^2bcd\]
to finish. Then, we extract that the equality case satisfies $a=d,b=c,ac=1$. Hence the tuple is $(a,b,c,d)=(a,\tfrac 1a,\tfrac 1a,a)$. This implies the desired.

Here is a somewhat different algebraic solution: WLOG the inradius is $1$. We are given
$$(\sqrt{w^2 + 1} + \sqrt{z^2 + 1})^2 + (\sqrt{x^2 + 1} + \sqrt{y^2 + 1})^2 = (w+x+y+z)^2,$$
where $w,x,y,z$ are positive reals and we have to prove $w = z, x = y$ and $wx = 1$. Expanding, we get $$4 + 2\sqrt{w^2 + 1}\sqrt{z^2 + 1} + 2\sqrt{x^2 + 1}\sqrt{y^2 + 1} = (w+x+y+z)^2 - (w^2 + x^2 + y^2 + z^2).$$ Using the fact that

$\sqrt{w^2 + 1}\sqrt{z^2 + 1} = \sqrt{(wz - 1)^2 + (w+z)^2}$ we can rewrite the new equation as:
$$4 + 2\sqrt{(wz - 1)^2 + (w+z)^2} + 2\sqrt{(xy - 1)^2 + (x+y)^2} = (w+x+y+z)^2 - (w^2 + x^2 + y^2 + z^2).$$
Also, the condition is:
$$\frac{wz - 1}{w+z} = \frac{1 - xy}{x + y} = K$$ where WLOG $K \ge 0$ (otherwise, we can switch the pair (w,z) and (x,y)).
Now, let $u_1 = w + z, u_2 = wz, v_1 = x + y, v_2 = xy.$ By the condition, we get that $u_2 = 1 + Ku_1$ and $v_2 = 1 - Kv_1$. Substituting into the equation, we get: $$(w+x+y+z)^2 = (u_1 + v_1)^2,$$ $$2\sqrt{(wz - 1)^2 + (w+z)^2} = 2u_1\sqrt{K^2 + 1},$$ $$2\sqrt{(xy - 1)^2 + (x+y)^2} = 2v_1\sqrt{K^2 + 1},$$ and $$w^2 + x^2 + y^2 + z^2 = u_1^2 - 2u_2 + v_1^2 - 2v_2 = u_1^2 + v_1^2 - 4 - 2K(u_1 - v_1).$$
The equation can now be rewritten as:
$$4 + 2u_1\sqrt{K^2 + 1} + 2v_1\sqrt{K^2 + 1} = 2u_1v_1 + 4 + 2K(u_1 - v_1),$$ which can be simplified further:
$$u_1v_1 + K(u_1 - v_1) = (u_1 + v_1)\sqrt{K^2 + 1}.$$
We use SFFT here to write the equation as:
$$(u_1 - (\sqrt{K^2 + 1} + K))(v_1 - (\sqrt{K^2 + 1} - K)) = 1.$$

Now observe that $u_1^2 = (w+z)^2 \ge 4wz = 4u_2 = 4 + 4Ku_1$. Similarly, $v_1^2 \ge 4 - 4Kv_1$. So, $$(u_1 - 2K)^2 \ge 4K^2 + 4 \implies |u_1 - 2K| \ge 2\sqrt{K^2 + 1}$$ and similarly $$|v_1 + 2K| \ge 2\sqrt{K^2 + 1}.$$ Since $u_1$ and $v_1$ are both positive, the absolute value signs cannot change any values, so $u_1 \ge 2(\sqrt{K^2 + 1} + K)$ and $v_1 \ge 2(\sqrt{K^2 + 1} - K)$. Using our SFFT equation, we check that equality actually holds everywhere, so $u_1v_1 = 4, u_1^2 = 4u_2,$ and $v_1^2 = 4v_2$. Thus $w = z, x = y, $ and $wx = 1$ as desired.

WLOG inradius $r = 1$.Let $AC \cap BD = X$ (WLOG $X$ lies on $\overrightarrow{AB}$), $AB,BC,CD,DA$ tangent to $\omega$ at $P,Q,R,S$ and the length of tangent line from $A,B,C,D,X$ to $\omega$ are $a,b,c,d,l$ respectively. Let $p,q$ be the length of $AD,BC$ respectively ($p = a+d,q=b+c$).
By well-known $[\triangle] = rs$, $r^2 = 1 = \frac{lad}{l+a+d} = \frac{lbc}{l-b-c}$. So, $ad = \frac{l+p}{l} = 1+\frac{p}{l}$. Similarly, $bc = 1-\frac{q}{l}$. By the problem equation,
\begin{align*}
a^2+b^2+c^2+d^2 +4 + 2(\sqrt{(a^2+1)(d^2+1)} + \sqrt{(b^2+1)(c^2+1)}) &= (p+q)^2 \\
\iff  p^2-2(1+\frac{p}{l}) + q^2 - 2(1-\frac{q}{l}) +4 + 2(\sqrt{(1+\frac{p}{l})^2 + p^2 - 2(1+\frac{p}{l}) +1} + \sqrt{(1-\frac{q}{l})^2 + q^2 - 2(1-\frac{q}{l}) +1})  &=p^2+2pq+q^2\\
\iff p = q \left( \frac{1+\sqrt{1+l^2}}{1-\sqrt{1+l^2}+ql}\right)
\end{align*}
Note that $p,q$ obtain the minimum value when $X,I,Q$ and $X,I,S$ are collinear. So $p \geq \frac{2(1+\sqrt{l^2+1})}{l}$ and $q \geq \frac{2(-1+\sqrt{l^2+1})}{l}$[hide=Proof]$\angle AIS + \angle DIS$ is constant and $AD = \tan \angle AIS + \tan \angle DIS$. Since $\tan$ is convex function, the minimum holds when $\angle AIS = \angle DIS$[/hide]. Consider function $f(x) = \frac{x}{1-\sqrt{1+l^2}+xl}$. For $q_1>q_2 \geq \frac{2(-1+\sqrt{l^2-1})}{l}$, we get $f(q_1)-f(q_2) = \frac{(q_1-q_2)(1-\sqrt{1+l^2})}{(1-\sqrt{1+l^2}+q_1l)(1-\sqrt{1+l^2}+q_2l)}<0$. Therefore, $f$ is deceasing function. Hence, $$ \frac{2(1+\sqrt{l^2+1})}{l} \leq p = f(q) \leq f\left(\frac{2(-1+\sqrt{l^2+1})}{l} \right) =  \frac{2(1+\sqrt{l^2+1})}{l}.$$
The equality holds when $X,I,Q$ and $X,I,P$ are collinear. Therefore, $ABCD$ is an isosceles trapezoid.


My school isn't offering the B tests, but a school near me is.  I got permission to go to that school to take the AMC 10 B, but now the department math head over there is saying that an experienced teacher said my test couldn't be sent off to be graded.  Thus, the department head said the AMC 10 B would essentially be a practice test for me.

I understand that the school can't send my scores, but what if I score above a 120 after reading the solutions manual and checking my own paper?  Can I just tell my math teacher to order an AIME for me?  Or do I just not get to take the AIME?

Thanks.

It is actually I think possible for you to take the 10B officially.  I just don't know the specifics, but I believe that math head person is mistaken.  :blush:

Does anybody know for certain if the school can send my test to the AMC department?
I really need to know tonight...

I believe you can officially take it (I've heard of a lot of people taking it at other schools). Yes, that school can mail it, but you might have to take the AIME there as well if you make it.

[quote="NightFlarer"]
I understand that the school can't send my scores, but what if I score above a 120 after reading the solutions manual and checking my own paper?  Can I just tell my math teacher to order an AIME for me?  Or do I just not get to take the AIME?

Thanks.[/quote]

Umm... you cannot take the AIME unless you officially take the AMC. There's no way they would let people just check their own solutions and then order an AIME. Mailing of the AIME is based on AMC scores that they receive; I'm pretty sure they only mail the AIME to schools where someone scored high enough to qualify (officially, of course).


[hide="woot!!!"]

[hide="woot!!!"]
[hide="click please"]

$16=.64x$
$x=25$

[/hide]
[/hide]
[/hide]

[quote="joejia"]Sixteen is 64% of what number?[/quote]

[hide]In school, my math teacher set up proportions:

$\frac {16} {x} = \frac {64} {100}$

In other words:

$\frac {part} {whole} = \frac {percent} {100}$

Then, using cross multiplication, you could solve this:

$ x \times 64 = 16 \times 100$

$64x = 1600$

$x = 25$[/hide]

Please solve it and provide us an answer.
[hide].64x=16
x=25[/hide]

in the future you should use the hide button ;)


Let $ a$ be a natural number shich is greater than $ 3$ and consider the sequence $ (u_n)$ $ (n\equal{}1,2,...)$ defined inductuvely by
$ u_1\equal{}a$
$ u_{n\plus{}1}\equal{}u_n\minus{}[\frac{u_n}{2}]\plus{}1$ for all $ n\equal{}1,2,...$
Prove that there exists $ k\in N^*$ such that $ u_n\equal{}u_k$ for all $ n\geq k$


A white cylindrical silo has a diameter of 30 feet and a height of 80 feet.  A red stripe with a horizontal width of 3 feet is painted on the silo, as show, making two complete revolutions around it.  What is the area of the stripe in square feet?
(sorry about picture)

[hide]When we unravel the stripe from the silo.  It becomes a parallelogram which can be rearranged to be a rectangle with dimensions of 3x80.  Making the area = 240.[/hide]

[hide]But how do you know it will be 80 when you unravel it?[/hide]


Friends! please, try to solve in 3 min.

2. Let T be the set of ordered triples (x,y,z) where x,y and  z are integers with 0<=x,y,x<=9. Players A and B play the following guessing the game. Player A chooses a triple (x,y,z) in T, and player B has to discover A's triple in as few moves as possible. A move consists of "B gives A a triple (a,b,c) in T, and A replies by giving B the number |x+y-a-b|+|y+z-b-c|+|z+x-c-a|.Find the minimum number of moves that B needs to be sure of determining A's triple"

If we say the points (0,0,0), (9,9,9), and some other point like that ( :? ) we will get three simultanious equations, and we know if the absolute value signs are positive or negative. Its an idea, can anyone finish it?

BTW, do the numbers b pick have to be integers? If not, he can get it in one go!


Can the class of 3SUM hard problems be solved in subquadratic time? These problems can be reduced from the problem of determining whether, given three sets of integers, A, B, and C with total size n, there are elements a element of A, b  B, and c element of C such that a + b = c.
 ;)

What notion of reducibility do you mean, for $O(n^2)$ problems?


given a circle K ,a line d is drawn not intersecting K. M,N  are two points varing on line d  such that the circle with diameter MN  is externaly tangent to K . prove that there exists a point P in the plane such that for any such  segment  MN  , the angle MPN is constant
PLEASE HELP :D  :)

as the pic says.
Consider AD*AB-CA^2 ,and you can find out
thats a constant k^2.Then we know there
exist F on OC,such that  CF=k
Then AD*AB=AF^2 holds for every R.
by now we have ADF AFB are similar.
so angle AFB=angle ADF,which is a constant.


inequalities
induction
inequalities unsolved

Let $a_1,a_2,...$ be nonegative integers satisfying $ a_{m+n}\leq a_m+a_n\ \ (m,n\in N)$ Prove that;

$ a_n\leq ma_1+(\frac{n}{m}-1)a_m$

If $n\leq m$, the inequality is $\Leftrightarrow a_n+a_m(1-\frac nm)\leq na_1$, and we use the fact that $a_x\leq xa_1$. If $n>m$, let $n=mk+r$, $r<m$. Now applying the hypothesis $k$ times, we get $\ds a_n\leq ka_m+a_r\leq ka_m+ra_1=\frac {mk+r-m}ma_m+\frac {m-r}ma_m+ra_1\leq (\frac nm -1)a_m+a_1(r+\frac {m-r}m\cdot m)$
, and this the end.

by induction we have [tex]b_n=na_1-a_n[/tex] is a sequence of positive integer !
after that we have to substitute and we got the condition :
[tex]b_n>=(n/m-1)b_m[/tex] for all m and n
it's evident if [tex]n<=m[/tex]
well if [tex]n>m[/tex]
so
by taking n=qm+r with 0<=r<m
[tex]b_n>=qb_m+b_r>=(n/m-1)b_m[/tex]
cause q is=[n/m] and b_r>=0
bye

Just a concretisation:
It's China 1995 problem 6!


[quote="kunny"]E.1 Factor the following expresisons.

(1) $ (x \plus{} y)^2 \minus{} (y \plus{} z)^2.$

(2) $ x^2 \plus{} 2xy \plus{} y^2 \plus{} 4x \plus{} 4y \plus{} 3.$[/quote]

$ (x\plus{}y \plus{} y \plus{} z)(x \plus{} y \minus{} (y \plus{} z))$ the difference of 2 squares

$ \equal{}(2y \plus{} x \plus{} z)(x \minus{} z)$

(2) $ (x\plus{}y)^2 \plus{} 4(x\plus{}y) \plus{} 3$

let u = (x+y), then

u^2 + 4u + 3

(u+1)(u+3)

$ \therefore (x\plus{}y\plus{}1)(x\plus{}y\plus{}3)$

Or for the second:
It is equivalent to:
$ x^2\plus{}2(y\plus{}2)x\plus{}y^2\plus{}4y\plus{}3\equal{}0$
The discriminant is $ 4$.
So $ x_1\equal{}\frac{\minus{}2y\minus{}4\plus{}2}{2} \Leftrightarrow x_1\equal{}\minus{}y\minus{}1$
and $ x_2\equal{}\frac{\minus{}2y\minus{}4\minus{}2}{2} \Leftrightarrow x_2\equal{}\minus{}y\minus{}3$
So it is $ (x\plus{}y\plus{}1)(x\plus{}y\plus{}3)$

Right. :)

Alternative Solution:

$ x^2 \plus{} 2xy \plus{} y^2 \plus{} 4x \plus{} 4y \plus{} 3 \equal{} x^2 \plus{} (2y\plus{}4)x \plus{} (y \plus{} 1)(y \plus{} 3)$

$ \equal{} (x \plus{} y \plus{} 1)(x \plus{} y \plus{} 3)$

[quote="kunny"]Factor the following expresisons.

(1) $ (x \plus{} y)^2 \minus{} (y \plus{} z)^2.$

(2) $ x^2 \plus{} 2xy \plus{} y^2 \plus{} 4x \plus{} 4y \plus{} 3.$[/quote]

(1)  $ x^2 \plus{} 2xy \plus{} y^2 \minus{} y^2 \minus{} 2yz \minus{} z^2 \equal{} (x \minus{} z) (x \plus{} z) \plus{} 2y (x \minus{} z) \equal{} (x \minus{} z) (x \plus{} 2y \plus{} z)$
(2) $ (x \plus{} y)^2 \plus{} 4 (x \plus{} y) \plus{} 3 \equal{} (x \plus{} y \plus{} 1) (x \plus{} y \plus{} 3)$


number theory
greatest common divisor
linear algebra
matrix
algebra
polynomial
number theory unsolved

The problem goes as follows --

We are given a GCD Table , which may be taken as a matrix such that its entries are gcd(i,j) [i and j being the index].
So for example ( 3 X 4 ) GCD Table will look like --

                              1 1 1 1
                              1 2 1 2
                              1 1 3 1

Given x , y which are the dimensions of the Matrix, we need to find the number of primes in the GCD Table with dimensions x and y.

for example for

x = 100 , y = 65 , Sol = 1806.
x = 10   , y = 10 , Sol = 30.

Can anyone suggest a fast way to compute this ?

Thanks.

You can determine the gcd of two numbers in a polynomial time via Euclid's algorithm. Assume that the hardest gcd computation takes time $ t$. Then you can find all entries of the table in time $ xyt$. Furthermore, testing primality of a given integer can be done in a polynomial time by AKS primality test. Now you need to choose at most $ xy$ primes, and this can also be done in a polynomial time with respect to $ x$ and $ y$.


a) Find the product $ \prod\limits_{i \equal{} 0}^\infty  {\left( {1 \plus{} x^{2n \minus{} 1} } \right)}$

b) i) if a series $ \sum\limits_{n \equal{} 1}^\infty  {a_n }$with positive terms converges, then the series $ \sum\limits_{n \equal{} 1}^\infty  {A_n }$where $ A_n  \equal{} \sqrt {\sum\limits_{k \equal{} n}^\infty  {a_k } }  \minus{} \sqrt {\sum\limits_{k \equal{} n \plus{} 1}^\infty  {a_k } }$also diverges, and $ a_n  \equal{} o(A_n )$

ii) if a series $ \sum\limits_{n \equal{} 1}^\infty  {a_n }$with positive terms diverges, then the series $ \sum\limits_{n \equal{} 2}^\infty  {A_n }$, where $ A_n  \equal{} \sqrt {\sum\limits_{k \equal{} 1}^n {a_k } }  \minus{} \sqrt {\sum\limits_{k \equal{} 1}^{n \minus{} 1} {a_k } }$also diverges, and $ A_n  \equal{} o(a_n )$.

I understand i) and ii)  but I do not understand the author's conclusion that it follows from them that no convergent/divergent series can serve as a universal standard of comparison to establish the convergence/divergence of other series.

Also how can we prove that if $ b_n  \equal{} o\left( {\frac{1}{n}} \right)$ then one can always construct a convergent series $ \sum\limits_{n \equal{} 1}^\infty  {a_n }$ such that $ b_n  \equal{} o\left( {a_n } \right)$?

I understand part b now. That is a rather surprising theorem.


A circle of radius 2 is centered at $ O$. Square $ OABC$ has side length 1. Sides $ \overline{AB}$ and $ \overline{CB}$ are extended past $ b$ to meet the circle at $ D$ and $ E$, respectively. What is the area of the shaded region in the figure, which is bounded by $ \overline{BD}$, $ \overline{BE}$, and the minor arc connecting $ D$ and $ E$?

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=610[/img]

$ \textbf{(A) } \frac {\pi}3 \plus{} 1 \minus{} \sqrt {3} \qquad \textbf{(B) } \frac {\pi}2\left( 2 \minus{} \sqrt {3}\right) \qquad \textbf{(C) } \pi\left(2 \minus{} \sqrt {3}\right) \qquad \textbf{(D) } \frac {\pi}{6} \plus{} \frac {\sqrt {3} \minus{} 1}{2} \\
\qquad \indent \textbf{(E) } \frac {\pi}{3} \minus{} 1 \plus{} \sqrt {3}$

Connect $OD$ and $OE$. Now $OD = OE = 2$ and $OA = OC = 1$, so by Pythagoras's theorem, it is easy to see that $CE = DA = \sqrt{3}$. Furthermore, $BE = DB = \sqrt{3} - 1$.

Now connecting $OB$ we have divided the sector into 3 parts, namely triangles $ODB$ and $OBE$ (which are congruent) and the area we wish to find. The area of triangle $ODB =$ area of triangle $OBE = \frac{1}{2} \cdot 1 \cdot (\sqrt{3} - 1) = \frac{\sqrt{3} - 1}{2}$.

So the area of both the triangles is $\sqrt{3} - 1$ square units.

It is also relatively easy to see that $mDOE = \frac{\pi}{6}$ radians. Thus the area of sector $ODE$ is $\frac{1}{2} \cdot 2^2 \cdot \frac{\pi}{6} = \frac{\pi}{3}$ square units.

And finally, the desired area is $\frac{\pi}{3} - (\sqrt{3} + 1) = \frac{\pi}{3} + 1 - \sqrt{3}$,

hence $(A)$.

Typo! You meant $\sqrt3-1$ in that last expression, not $\sqrt3+1$. ;)

I don't get this:
You can connect $OE$ and that equals the radius which is $2$. Since $OC$ is $1$, you can find that $CE$ is $\sqrt3$. $BE$=$\sqrt3-1$. Isn't the shaded region also $1/4$ of a circle with center $B$? So my answer was $(\sqrt3-1)^2/4$. So I got $C$.
What did I do wrong?


I finished reading freakonomics today and I want to say that it is an interesting and unique book that I enjoyed even tohugh I want to say that if I were to become an economist, I would not write about the things Levitt and Dubner wrote because they lack a`unifying theme and r too practical.
Your opinions?

Why do they need a unifying theme?

Where's the brightline for being "practicle" and being "too practical" ?

I am just saying I like more theoretical rather then practical. And also, the practical results of Levitt's work doesnt seem very generalized, so there are like thousands of such situations that have different asnwers with a minor change.

There's a few papers here which the book is based on-
the theoretical stuff which was stripped from the book.
I particularly like the one about soccer penalty takers and goalkeepers

http://pricetheory.uchicago.edu/levitt/LevittCV.html

Yes, I know about that site. Thanks though, I started reading econ blogs, a lot of the info I dont understand.

the mathsy bits?

my favourites are:
http://www.knowledgeproblem.com
http://www.marginalrevolution.com
http://www.domoecon.com
http://www.divisionoflabour.com
angry-economist.russnelson.com

I read marginalrevolution too, I think its the most popular. Domo econ is yours right?

No, I think its canadian. I've neither the time nor ability to
host a worthwhile site, not on my own anyway...

I read Freakonomics,I liked the part on parenting: "Swimming pools vs. pistols", as it treads on one of my other favourite topics also ,Psychology.


Is there a difference between multiplying matrices and cross product?  Thanks!

There's a big difference.  There are minus signs in the definition of the cross product, for example.  Who told you they were the same?

(In general, anyway.  There's a way to "embed" the cross product so that it corresponds to multiplication of certain $ 3 \times 3$ matrices, but it's not perfect; see the [url=http://en.wikipedia.org/wiki/Cross_product#Conversion_to_matrix_multiplication]Wikipedia article[/url].)


geometry
rhombus
number theory theorems
number theory

Let p=1+4k a prime
S={(a,b,c)\in NxZ, 4ab+c=p}
1) Show that S=/= \emptyset , is finite set
S is included in (N\{0})xZ\{0}
Show that S_1={(a,b,c) \in S, a>b+c} and
S_2={(a,b,c)\in S , a<b+c} are a partition of S
(this means S=S_1 \cup S_2, and S_1 \cap S_2= \emptyset )
Find an explicit bijection from S_1 to S_2
2) Prove that f:S_1---> S_1, f(a,b,c)=(a-b-c,b,-2b-c) is an involution
( means f(f(a,b,c))= (a,b,c) )
Find the fixed point of f (means (a,b,c) such that f(a,b,c)=(a,b,c))
Deduce that 4|(card(S) - 2)
3)Let S_3={(a,b,c) \in S, a=/=b}, prove that 4|card(S_3)
Finally deduce that p is the sum of two squares
(there exist x,y integer p=x+y)

This is Don Zagier proof, it was published in AMM american mathematical
monthly i don't remember the issue
oubi

hi
about that proof what do you mean by card (S)?
and how do you finally deduce that P is the sum of 2 squares?

$card P=\mid P \mid$ the number of elements that $P$ contain

[quote="Moubinool"]Let p=1+4k a prime
S={(a,b,c)\in N�xZ, 4ab+c�=p}
1) Show that S=/= \emptyset , is finite set
S is included in (N\{0})�xZ\{0}
Show that S_1={(a,b,c) \in S, a>b+c} and
S_2={(a,b,c)\in S , a<b+c} are a partition of S
(this means S=S_1 \cup S_2, and S_1 \cap S_2= \emptyset )
Find an explicit bijection from S_1 to S_2
2) Prove that f:S_1---> S_1, f(a,b,c)=(a-b-c,b,-2b-c) is an involution
( means f(f(a,b,c))= (a,b,c) )
Find the fixed point of f (means (a,b,c) such that f(a,b,c)=(a,b,c))
Deduce that 4|(card(S) - 2)
3)Let S_3={(a,b,c) \in S, a=/=b}, prove that 4|card(S_3)
Finally deduce that p is the sum of two squares
(there exist x,y integer p=x�+y�)

This is Don Zagier proof, it was published in AMM american mathematical
monthly i don't remember the issue
�oubi[/quote]

Moubinool, what equation do $a,b,c$ fulfill? i'm getting many question marks inside a black rhombus in your proof  :(

Ive the same problem, but changing the table of signs away from Unicode worked.

i don't understand. how did you do that?

OK, the correct word is: chartable (not table of signs).
It's some setting of your Browser, so it depends on our computer.

OK, copying worked, so here is the original text:
 [quote="Moubinool"]Let p=1+4k a prime
S={(a,b,c)\in N²xZ, 4ab+c²=p}
1) Show that S=/= \emptyset , is finite set
S is included in (N\{0})²xZ\{0}
Show that S_1={(a,b,c) \in S, a>b+c} and
S_2={(a,b,c)\in S , a<b+c} are a partition of S
(this means S=S_1 \cup S_2, and S_1 \cap S_2= \emptyset )
Find an explicit bijection from S_1 to S_2
2) Prove that f:S_1---> S_1, f(a,b,c)=(a-b-c,b,-2b-c) is an involution
( means f(f(a,b,c))= (a,b,c) )
Find the fixed point of f (means (a,b,c) such that f(a,b,c)=(a,b,c))
Deduce that 4|(card(S) - 2)
3)Let S_3={(a,b,c) \in S, a=/=b}, prove that 4|card(S_3)
Finally deduce that p is the sum of two squares
(there exist x,y integer p=x²+y²)

This is Don Zagier proof, it was published in AMM american mathematical
monthly i don't remember the issue
µoubi[/quote]

This proof can also be found in the famous "Proofs from the book" (Springer)

How does one construct an explicit bijection from $S_1$ to $S_2$?

try $f(a,b,c)=f(b,a,-c)$ and show that there's no point such that $b+c=a$ (if I remember correctly  :D )


$ pK_{a} (\ce{H3BO3}) \equal{}9.24$
$ pK_{a} (\ce{HCOOH}) \equal{}3.74$

A $ 20.00mL$ aliquot of an equal mixture of the two monoprotic acids eahc of $ 0.100M$ is titrated with $ 0.100M$ $ \ce{NaOH}$. Calculate the $ pH$:

1. Initially

2. Half way to the first equilibrium

3. At the first equivalence point

4. Half way between the first and second equilibrium point

5. At the second equilibrium point

It will probably be better if you guide me through rather than just giving me the answer/solution.

Start by writing the equilibrium reaction for each acid.

$ \ce{H3BO3} \rightleftharpoons \ce{H\plus{}} \plus{} \ce{H2BO3\minus{}}$

$ \ce{HCOOH} \rightleftharpoons \ce{H\plus{}} \plus{} \ce{HCO2\minus{}}$

Oh yeah to show that I have tried:

[hide="Thoughts"]

I think the treatment of this problem will to some extent resemble the titration of a polyprotic weak acid.

- Initially Methanoic acid is dominant also halfway to the equivalence point

-First equivalence point, boric acid is dominant but would we have to take methanoate ions into account?

- Hafway between first/second equivalence point boric acid is dominant

- At the second equivalence point use the $ pK_b$ of boric acid.[/hide]

i think ur equation is wrong....
cos H3BO3 behaves as a weak acid in solution beacuse it accepts OH- from water and there by releases a proton....i.e.H+

Yes i think the eqn should be as:

$ \ce{H3BO3} + \ce{H2O} \rightleftharpoons \ce{B(OH)4^{ - }} + \ce{H^{ + }}$

That is why $ \ce{H3BO3}$ is better written as $ \ce{B(OH)3}$

 :)

Yes it is a non-protic acid and releases the H+ from H2O

Let's start by 1). BanishedTraitor, how can we calculate the pH?

The $ pK_a$ for methanoic acid is about 6 orders of magnitude greater  :rotfl:  (sorry it was a joke I might tell you guys about some day).

So I am assuming that the contribution from Boric acid is minimal at this stage (this can be confirmed).

Say $ [\ce{HCO2\minus{}}] \equal{}x$ then $ K_{a} \equal{} \frac{x^{2}}{0.1\minus{}x}$ which gives a $ pH$ of around $ 2.37$.

That approximation looks ok, and the deominator 0.1 -x should not in this case be reduced to 0.1.

Side note: how about if we had an equimolar mixture of formic and acetic acids? (pKa for acetic acid is 4.75)

Now going to 2), how to calculate the pH here?

Henderson-Hasselbalch? It looks like methanoic acid seems act as a buffer at this point. I am confused at when I start to take boric acid into account.

That's right. And because we are halfway the calculations are very simply (why?). We can start worring about boric acid when formic acid titration is over.

Just a question. Why do we assume that all the $ \ce{NaOH}$ react with methanoic acid?

cos NaOH in solution completely dissociates into Na+ and OH- ions...........

Because formic acid is quite a stronger acid than boric acid, as shown by their pK values.

Carcul do you have hard equilibrium problems? I am not very good at equilibrium (but I am not terrible either).

What kind of equilibrium problems are you referring to?

Just equilibriums in general. Compleximetric, Titration, Electrochemical. Anything goes really. I need improve a lot. Can you also teach me some useful methods of solving problems?

Anyway, let's go back to the original problem.

[quote="Carcul"]And because we are halfway the calculations are very simply (why?).[/quote]

I am still waiting this question to be answered.

The contribution from boric acid is negligible. The calculation is simple because $ pH \equal{} pK_a$ of formic acid because we are in a buffering zone (and the ratio of conjugate base to the acid is $ 1$).

That's right. How about the pH at the first equivalence point?

Yeah that's where the problem is. If I treat this like a diprotic acid titrated with a base then I had to consider an equilibrium established between boric acid and methanoate ion. I really don't know how that works. Or am I on th wrong track?

Well it's the first equivalence point so I am not sure anymore.........

Sorry I can't really see why only considering formic acid is valid  :( .

Again? Isn't formic acid a much stronger acid than boric acid?


What is the nearest integer to $ (2\plus{}\sqrt3)^4$?

[quote="GameBot"]What is the nearest integer to $ (2 \plus{} \sqrt3)^4$?[/quote]

$ (2^1\plus{}3^{\frac12})^4 \equal{} 2^4 \plus{} 3^2 \equal{} \boxed{25}$

$ (a\plus{}b)^n\ne a^n\plus{}b^n$ :wink:

Instead, consider $ (2\plus{}\sqrt3)^4\plus{}(2\minus{}\sqrt3)^4\equal{}194$ from the Binomial Theorem. Since $ (2\minus{}\sqrt3)^4<0.5$, the closest integer is $ \boxed{194}$.

Argh...I thought that was true.  /me is too dumb to do math.

Some people memorized that $ \sqrt{3} \approx 1.732$ and taking to the fourth power gives us $ 193.9842\ldots \approx \boxed{194}$, but your way is much better.


vector
algebra
binomial theorem
algebra unsolved

Prove that if A has n elements, then A has $2^{n}$ subsets :D

Classic.
Let $S\equiv\{s_{1},s_{2},\dots, s_{n}\}$.
We can describe without ambiguity every subset $A\subset S$ with a vector $(v_{1}, v_{2}, \dots, v_{n})$ where
$v_{i}=1$ if $s_{i}\in A$,
$v_{i}=0$ if $s_{i}\not\in A$.
And, of course there are $2^{n}$ such vectors (choice of $0$ or $1$ for every element).

This solution is good, but it is not suitable for junior section.
We have $\frac{n}{1!}$ subsets that contains 1 element
              $\frac{n(n-1)}{2!}subsets that contains 2 elements$
              ........
If that, we have to prove the relation:
$\frac{n}{1!}+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+....+\frac{n(n-1)(n-2)...3.2}{n-1!}+\frac{n(n-1)(n-2)...3.2.1}{n!}$
$=2^{n}$

Come on! The vector thing was only to make it short.
Question number $1$: Is element $1$ in the subset? Yes/No.
Question number $2$: Is element $2$ in the subset? Yes/No.
...
Question number $n$: Is element $n$ in the subset? Yes/No.
Since there are $n$ possible independent answers "Yes/No", then there are $2^{n}$ different subsets.
But I agree this is combinatorics and not algebra.

By the way, your relation is Newton's Binomial Theorem with $(1+1)^{n}$. But, in fact, that theorem is proved from combinatorics, and not the other way. So it would be almost circular logic.


integration
logarithms
calculus
calculus computations


He knew that if x=-y, that was a solution so then he knew that x+y was a factor then, and then he did long division on that assuming that x was a variable and y was a constant.

Actually I just used the identity

If n is odd,
$x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-\ldots+y^{n-1})$

But mathgeniuse^ln(x) explains it well too.

Also, for completion:

For any positive integer n
$x^{n}-y^{n}=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+\ldots+y^{n-1})$


Let a and r be real numbers with r 6= 1. Prove by induction, that
a + ar + ar2 + ar3 + · · · + arn−1 =
a (rn − 1)
r − 1
for n = 1, 2, 3 . . .

I can't understand your syntax.  Please try to clarify it.

$ a \plus{} ar \plus{} ar^2 \plus{} \ldots ar^{n \minus{} 1} \equal{} \frac {a(r^n \minus{} 1)}{r \minus{} 1}$, most likely. Try subtracting similar version of the left side, for a noninductive solution. The induction is completely straightforward.

Yeah sorry about that forgot to format properly. I actually got this the second i put it up! Frustrating. It wouldn't let me post to say that straight away. Thanks anyway.

If you want the non-inductive solution,
$ s_{n}\equal{}a\plus{}ar\plus{}\cdots\plus{}ar^{n}$ and $ rs_{n}\equal{}ar\plus{}ar^{2}\plus{}\cdots\plus{}ar^{n\plus{}1}$

Then just subtract the second from the first.

$ a^n\minus{}b^n\equal{}(a\minus{}b)[(a^n\minus{}1)\plus{}(b^1)(a^n\minus{}2)\plus{}(b^2)(a^n\minus{}3).........(a^1)(b^n\minus{}2)\plus{}(b^n\minus{}1)]$
applying here,$ \frac {a(r^n \minus{} 1)}{r \minus{} 1}$ $ \equal{} \frac{a(r\minus{}1)((r^n\minus{}1)\plus{}(r^n\minus{}2)\plus{}....1)}{r\minus{}1}$

here im having problems with latex, such as  in a^(n-1), -1 is not in power. i hope u all can understand.

Do

[code]
$a^{n-1}$
[/code]

for $ a^{n\minus{}1}$.


A polynomial $P(x)$ leaves a remainder of $2x$ when divided by $(x-1)^{2}$, and a remainder of $3x$ when divided by $(x-2)^{3}$. Find the sum of the $x^{3}$ coefficient and the $x^{2}$ coefficient of $P(x)$.

here is the not rigirous way to do it:

let $p(x) =(ax+b)(x-2)^{3}+3x$, divide by $(x-1)^{2}$, and the remainder is $(2a+3b+3)x-3a-4b=2x$...so we get a system...$a=4$. $b=-3$

so we have found a suitable polynomial, that being $f(x)=(4x-3)(x-2)^{3}+3x=4x^{4}-27x^{3}+66x^{2}-65x+24$

hence our sum is $-27+66=\boxed{39}$

i don't see how this problem is uniquely defined... $p(x)=f(x)+q(x) *( x-2)^{3}(x-1)^{2}$ is also a solution for $q(x)$ a polynomial, taking $q(x)=x^{3}$ obviously changes the answer

It is troublesome.  $P(x)$ should only be uniquely defined $\bmod (x-1)^{2}(x-2)^{3}$, and the $x^{3}, x^{2}$ coefficients of the above polynomial don't sum to $0$...


calculus
integration
trigonometry
calculus computations

Evaluate the iterated integral
\[ \int_0^1\int^1_x\sin{y^2}\, dy\, dx\]

Change the order of integration, it will become much easier.

Let me do the solution for you.

Change the order of the integral from $ \int \int _R \sin{y^2} dydx$ to $ \int \int _R \sin{y^2} dxdy$, where $ R$ is the region bounded by $ x \equal{} 0$, $ y \equal{} 1$, and $ y \equal{} x.$

Now, $ \int \int _R \sin{y^2} dxdy \equal{} \int_{0}^{1}\int_{0}^{y}\sin{y^2}dxdy \equal{} \int_{0}^{1}y\sin{y^2}dy$

$ \equal{} \minus{}\frac{1}{2}\left(\cos{1^2}\minus{} \cos{0^2}\right) \equal{} \dfrac{1 \minus{} \cos{1}}{2}$.


probability
calculus
derivative
function
integration
absolute value

A particle moves along the x-axis, starting from 0 at t=0, such that, at any point in time, its speed is either 10 or -10. If it reaches 20 before it reaches -10, then Bob wins. If it reaches -10 before it reaches 20, then Alice wins. Find, with justification, the probability that Alice wins.

[hide="Solution"]Let $ P_A$ represent the probability that Alice wins and $ P_B$ represent the probability that Bob wins.

Now firstly, the probability that the particle reaches -10 before it reaches 10 is exactly $ 1/2$.  Now, consider the remaining $ 1/2$ chance that the particle reaches 10 first.  Then, consider whether the particle will reach -10 or 20 first... this setup is symmetric to the original though the point 5 on the x-axis.  So, the probability that the particle reaches -10 first is $ P_B$, and overall, we have:

$ P_A\equal{}\frac{1}{2}(1\plus{}P_B)$

$ P_A\plus{}P_B\equal{}1$

Solving, we obtain $ P_A\equal{}2/3$[/hide]

I like how the particle moves continuously instead of stopping in between to change direction. :D

[quote="The Zuton Force"]A particle moves along the x-axis, starting from 0 at t=0, such that, at any point in time, its speed is either 10 or -10.[/quote]
I don't think this situation is possible.  Let $ f(t)$ be the position of the particle at time $ t$.  Then your condition appears to be that $ f'(t) \equal{} \pm 10$.  But if the derivative is $ \minus{} 10$ at one time and $ 10$ at another time, then the derivative has to have all in-between values (such as 7) too, by the Intermediate Value Theorem for Derivatives.  As the previous poster indicated, there is no way to switch instantaneously between a speed of -10 and 10, not even theoretically.

Sorry to use calculus notation in the Intermediate Forum, but the problem seems to require it.

So you're saying that since the position is not twice differentiable, it cannot be described as the effect of some force, and is thus physically impossible?

If that's the case, let me clarify the point of my problem. The particle serves as a sort of marker of a position on the x axis. The fact that the speed can either be +/- 10 at any point in time gives some relationship between the position at some time and the possible positions at some other time, as well as the probability distribution of these possible locations.

I'm not talking about twice-differentiable or about physical possibility.  Even the single-differentiable problem is mathematically impossible.  There is no differentiable function that has derivatives $ \minus{}10$ and $ 10$ and nothing else.  And if there is no single such function, then we can't impose a probability distribution on them.

The usual way to generalize discrete random walks to continuous ones is to use Brownian motion.  But that's quite a different condition from your $ \pm 10$ derivative condition.

It sounds like TZF doesn't want a function at all, but a distribution.

I guess it was poorly formulated then. The idea was to make it sort of like absolute value in terms of how it can increase or decrease.

What if I add the two conditions:

1. The speed is +10 or -10, unless it is changing direction, at which point the speed is undefined
2. In any finite-length time interval, the particle changes direction a finite number of times, if at all

Also, [b]Ravi B[/b], I'm a bit confused about what you said. Let $ S \in \mathbb{R}^{\geq 0}$ be a random subset of the non-negatives. I can define the function $ f(x) \equal{} 20 \left( \mathbf{1}_S (x) \minus{} \frac {1}{2} \right)$. And I can integrate this function. And I have an initial condition on the function, x(0)=0. Why, then, can I not just call the integral the position function? And couldn't the problem equivalently be formulated along the lines of:

[i]Find the probability that there is a solution to the equation $ |I(t) \minus{} 5| \equal{} 15$, and that this solution is equal to the minimum solution to $ I(t) \equal{} 20$, where $ I$ is the integral of $ f$ from $ 0$ to $ t$, and $ f$ is defined based on a randomly chosen subset $ S$ of the nonnegative numbers.[/i]

Your problem is in how you plan on constructing $ S$.  What probability distribution on subsets of the non-negatives are you assuming?  Is it well-defined?  Are you allowing non-measurable sets?  (If you want an "even" distribution, almost all of the sets you pick will be non-measurable.)

[quote="The Zuton Force"]Also, [b]Ravi B[/b], I'm a bit confused about what you said. Let $ S \in \mathbb{R}^{\geq 0}$ be a random subset of the non-negatives. I can define the function $ f(x) \equal{} 20 \left( \mathbf{1}_S (x) \minus{} \frac {1}{2} \right)$. And I can integrate this function. And I have an initial condition on the function, x(0)=0. Why, then, can I not just call the integral the position function?[/quote]
First, as t0r pointed out, almost surely $ S$ will not be measurable, which means you can't integrate.  Even if $ S$ is measurable, you don't know that the derivative of the integral of $ f$ gives you back $ f$; the Fundamental Theorem of Calculus has a continuity condition.

You still seem to be trying to come up with a differentiable function whose derivative has range $ \{ \minus{} 10, 10\}$, but that's impossible.  Google for Darboux's theorem, aka the intermediate value theorem for derivatives.  For example, if $ f'(1) \equal{} 10$ and $ f'(2) \equal{} \minus{} 10$, then somewhere in the middle the derivative will be zero.  (Consider the maximum point.)

I still think that Brownian motion is a good way to properly formulate your original problem.  That's the usual way to model a continuous random walk.

Edit: Changed "minimum point" to "maximum point".

Ah, I see, thanks. How problematic. Unfortunately, I don't know how to formulate it with Brownian motion.

So then, what happens when I add the restriction I mentioned in the previous post? The reformulation (slightly adjusted) was that the particle moves according to the following rules:

- The time derivative of the position is either 10 or -10 [i]when it exists[/i]
- The derivative does not exist iff the particle is changing directions, which happens instantaneously
- If the particle changes direction at t=q, then there exists numbers p and r such that p<q<r and such that there is only one direction change in the open time-interval (p,r)

It seems like the picture you have in your head is of a continuous piecewise linear function with an infinite number of discontinuities.  As far as I know, however, the number of discontinuities needs to be countable, and then the answer depends on how you pick the location of the discontinuities.  And if you want to pick them randomly, I still don't know if there's a sensible way to pick a random countable subset of the reals (or of an interval).


topology
real analysis
real analysis unsolved

Prove that sequence $ \frac{f_{n\plus{}1}}{f_n}$ converges, but in solution use Banach fixed point theorem.

$ (f_n)$ is Finobacci sequence.

[hide="hint"]Let $ a_{n} \equal{} \frac{f_{n\plus{}1}}{f_{n}}$. We are searching for $ X \subset \mathbb{R}$ and $ f: X \to X$ s.t.
1. $ (X, |\cdot|)$ is a complete metric space,
2. $ f$ is contraction mapping,
3. $ a_{n\plus{}1} \equal{} f(a_{n})$.[/hide]
[hide="solution"]It is easy to see that $ X \equal{} \left[\tfrac{3}{2}, 2\right]$ and $ f(x) \equal{} 1 \plus{} \frac{1}{x}$ satisfy 1.-3. above, if we consider $ a_{n}$ for $ n\geqslant 2$.
Now convergence of $ a_{n}$ follows straightforward from Banach Fixed Point Theorem.[/hide]


The numbers $a, b, c, d$ (where $a<b<c<d$) can be paired up in 6 different ways. The four lowest sums of those pairs are $1, 2, 3, 4$. What are all the possible values of $d$?

[hide="Hint"]These numbers are not neccesarily integers, and not neccesarily positive.[/hide]

[hide]The two smallest sums are $a+b < a+c$. Now for the third and fourth, we can have $b+c$  and $a+d$ in any order. If we have $b+c < a+d$, then

$a+b=1\\ a+c=2\\ b+c=3\\ a+d=4$

From these we get $d-b=3$, $b-a=1$. Thus $b=1$, and $d=3$.

If we have $a+d < b+c$, then

$a+b=1\\ a+c=2\\ a+d=3\\ b+c=4$

From these we get $d-b=2$, $b-a=2$. Thus $b=\frac{3}{2}$, and $d=\frac{7}{2}$.

Thus $d=4, \frac{7}{2}$.[/hide]

I took this from illinois (ILML) so they must use some or all of the same questions in other places.  CNML, hmm, can't figure it out.

Woah, in Illinois, really? I thought CNML (Canadian National Math League) was Canada-only. Well, perhaps they just share questions. Interesting questions sometimes, though :)

The Mathleagues all have the same questions? I had this one in the California ML today.  :)

I got 3.5 and 4 as well.

I also got 3.5 and 4, but I didn't get the mark because I forgot that 3.5 should be written as $\frac72$  :(

What??? That's stupid grading. I put 3.5 and got it right!


calculus
trigonometry
inequalities
derivative
real analysis
real analysis unsolved

Solve the differential equation

$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+5y=3\cos 2x-\sin 2x$ given that $y=0$ and $\frac{dy}{dx}=0$ when $x=0$.
Then , show that the particular solution for this differential equation satisfy the inequality

$|y|<e^{2\pi}-1$ for the range $0\leq x\leq 2\pi$

I only manage to solve the  first part which is to find the particular solution . But I dont know how to prove the inequality part . Can someone help me ?

[quote="hpe"][quote="shyong"]I only manage to solve the  first part which is to find the particular solution . But I dont know how to prove the inequality part . Can someone help me ?[/quote]Well what is the solution that you found? Can you find its maxima and minima on $[0,2\pi]$?[/quote]

i found $y=e^x(\frac{1}{17}\cos 2x+\frac{25}{34}\sin 2x)-\frac{1}{17}\cos 2x-\frac{13}{17}\sin 2x$

but I dont know how to use this to prove the inequality part since the derivative of $y$ is quite messy and hard to find the turning point thing ....


[quote="helijet"]What is the pattern to this:
5,,6,5,9,6,15,10,26,19,44,35,71,60,109,?[/quote]
It goes like this:
5+1^2=6
6-1=5
5+2^2=9
9-3=6
6+3^2=15
15-5=10
It alternates adding perfect squares and subtracting odd numbers in order.


trigonometry
algebra
polynomial
algebra proposed

Let $ a\equal{}2\sin\frac{\pi}{19},$ $ b\equal{}2\sin\frac{7\pi}{19},$ $ c\equal{}2\sin\frac{11\pi}{19}$ and prove

$ a\plus{}b\plus{}c\equal{}\frac1a\plus{}\frac1b\plus{}\frac1c$ :D

I think Arqady may like that one. I dedicate it to him. ;)

hi... this is nice  :D  let $ \frac{\pi}{19}\equal{}\omega$ and note that the identity is equivalent to

$ \dfrac{\cos 2\omega}{\sin \omega}\plus{}\dfrac{\cos 14\omega}{\sin 7\omega}\plus{}\dfrac{\cos 22\omega}{\sin 11\omega}\equal{}\dfrac{1}{2\sin \omega}\plus{}\dfrac{1}{2\sin 7\omega}\plus{}\dfrac{1}{2\sin 11\omega}$

which is the same as

$ 4\cos 2\omega \sin 7\omega \sin 11\omega\plus{}4\cos 14\omega\sin \omega \sin 11\omega\plus{}4\cos 22\omega\sin \omega\sin 7\omega\equal{}2\sin 7\omega \sin 11\omega\plus{}2\sin \omega \sin 11\omega\plus{}2\sin \omega\sin 7\omega$

the lhs equals
$ \cos 6\omega\plus{}\cos 2\omega\minus{}\cos 20\omega\minus{}\cos 16\omega\plus{}\cos 24\omega\plus{}\cos 4\omega\minus{}\cos 26\omega\minus{}\cos 2\omega\plus{}\cos 28\omega\plus{}\cos 10\omega\minus{}\cos 30\omega\minus{}\cos 14\omega$

using that $ \cos (19\omega\pm k)\equal{}\minus{}\cos k$ it follows that this equals $ \cos \omega\plus{}\cos 4\omega\plus{}\cos 6\omega\plus{}\cos 7\omega\minus{}\cos 8\omega\minus{}\cos 9\omega$.

the rhs equals $ \cos 10\omega\minus{}\cos 12\omega\plus{}\cos 6\omega\minus{}\cos 8\omega\plus{}\cos 4\omega\minus{}\cos 18\omega$

which is the same as

$ \cos \omega\plus{}\cos 4\omega\plus{}\cos 6\omega \plus{}\cos 7\omega\minus{}\cos 8\omega\minus{}\cos 9\omega$

and we're done  :D

There you go!
My proof actually shows a bit more: it proves (in a similar way) the identity
$ (\csc x\plus{}\csc 7x\plus{}\csc 11x)\minus{}4(\sin x\plus{}\sin 7x\plus{}\sin 11x)$ $ \equal{}\frac{\sin 19x }{\sin 7x \sin 11x } (1\plus{}2\cos6x\plus{}2\cos10x)$.

Another funny identity, much harder though:
With the same $ \omega\equal{}\frac{\pi}{19}$, prove
$ \frac{7}{\cos2\omega\plus{} \cos14\omega\plus{}\cos22\omega}\plus{} \frac{\sqrt{19}}{\sin2\omega\plus{} \sin 14\omega\plus{}\sin 22\omega}\equal{}\minus{}4$ :D :D

can i ask how you went through this identity that you claim? i mean, maybe proving it may be easy, but how to obtain it...

good question.  :lol:
Well, I stumbled upon the fact that the minimal polynomials of  $ \frac{1}{\cos2\omega\plus{} \cos14\omega\plus{}\cos22\omega}$ and $ \frac{\sqrt{19}}{\sin2\omega\plus{} \sin 14\omega\plus{}\sin 22\omega}$ (they must be cubic for algebraic reasons) have the same discriminant.

By the way, the original identity of this thread is linked to the fact that both $ \frac{\sqrt{19}}{2(\sin\omega\plus{} \sin 7\omega\plus{}\sin 11\omega)}$ and $ \frac{2\sqrt{19}}{\csc\omega\plus{} \csc 7\omega\plus{}\csc 11\omega}$ have the same minimal polynomial. And you know what? It is $ x^3\minus{}19x\plus{}19$ :) :)


$ a,b,c \in R^\plus{}$
abc=1
prove that
$ \dfrac{1}{\sqrt{b\plus{}\frac{1}{a}\plus{}\frac{1}{2}}}\plus{}\dfrac{1}{\sqrt{c\plus{}\frac{1}{b}\plus{}\frac{1}{2}}}\plus{}\dfrac{1}{\sqrt{a\plus{}\frac{1}{c}\plus{}\frac{1}{2}}} \ge \sqrt{2}$

[quote="Matematika"]$ a,b,c \in R^ \plus{}$
abc=1
prove that
$ \dfrac{1}{\sqrt {b \plus{} \frac {1}{a} \plus{} \frac {1}{2}}} \plus{} \dfrac{1}{\sqrt {c \plus{} \frac {1}{b} \plus{} \frac {1}{2}}} \plus{} \dfrac{1}{\sqrt {a \plus{} \frac {1}{c} \plus{} \frac {1}{2}}} \ge \sqrt {2}$[/quote]
Let $ a\equal{}\frac{x}{z} , b\equal{}\frac{y}{x} , c\equal{}\frac{z}{y}$ ; $ (x,y,z >0)$

So $ LHS \equal{} \frac{1}{\sqrt{\frac{y\plus{}z}{x}\plus{}\frac{1}{2}}} \plus{} \frac{1}{\sqrt{\frac{z\plus{}x}{y}\plus{}\frac{1}{2}}}\plus{}\frac{1}{\sqrt{\frac{x\plus{}y}{z}\plus{}\frac{1}{2}}}$ , let $ x\plus{}y\plus{}z \equal{} 1$

$ \boxed{\frac{1}{\sqrt{\frac{1}{x}\minus{}\frac{1}{2}}}\ge \sqrt{2} x}$ . So $ LHS > \sqrt{2}$ ( try : $ a\equal{}n , b \equal{} \frac{1}{n} , c \equal{} 1 (n \rightarrow \infty))$ :)

I don't get it maybe I'm stupid but what are you doing after supstitution?
mathlifecro I don't understand what you mean.
 :blush:

I get it now thank you very much; nice solution :)

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141304
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175467


function
logarithms
algebra unsolved
algebra

Find all continuous function on $[0;1]$ such that \[f(xy)=f(x)+f(y)\;\forall x,y,x+y\in [0;1].\]

[quote="p.v.ninh"]Find all continuous function on $[0;1]$ such that
\[f(xy)=f(x)+f(y)\;\forall x,y,x+y\in [0;1]. \]
[/quote]

Note that
\[f(0 \cdot y) = f(0)+f(y).\]
Therefore, $f(y) = f(0)-f(0 \cdot y) = 0$. The answer is $f(y) \equiv 0$.

(BTW, if $f$ were defined on $(0,1]$ then the solution would be $f(x) = c \log x$.)


Suppose that the points $ z_1, z_2,...,z_n$ all lies on one side of a line passing through the origin. Prove that it's true for the points $ \frac{1}{z_n}, \frac{1}{z_2}, ..,\frac{1}{z_n}.$

The conjugates of the $ z_k$ lie on the same side of the symmetric of the line (with respect to the axis $ Ox$) that has all $ z_k$ on one side.
But a semi-plane is invariant to homotheties of positive factor, and $ \frac {1} {z_k} \equal{} \frac {\overline{z_k}} {\parallel{}z_k\parallel{}^2}$.


[quote="PenguinIntegral"][quote="Silverfalcon"]Find the value of

\[\sqrt[3] {5 + 2 \sqrt{13}} + \sqrt[3] {5 - 2 \sqrt{13}}\]

Solve without calculator.[/quote]
[hide="Hint"]
13 is prime.
(x+y(rootz))^3=(x+y(rootz))*(x+y(rootz))^2=(x+y(rootz))*(x^2+zy^2+2xy*rootz)
[/hide][/quote]
Ive seen this somewhere ...
[hide="Sol"]
use $(a+b)^3$,
$10+3ab(x)=x^3$
$10-9x=x^3$
Solve for $x$
[/hide]

[quote="Gyan"]Also, if you are lucky enough to hang around  with radicals (pun intended :)) like:
 $1+\sqrt{13}$         then you can  see that:
[hide] $(1+\sqrt{13})^3   =1+3\cdot \sqrt{13}+39 + 13\sqrt {13} = 40+16 \sqrt {13}= 8(5+2 \sqrt  {13})$[/hide]

 So ...
[hide] $\sqrt[3] {5 + 2 \sqrt{13}} = \frac 1 2  +  \frac {\sqrt {13}}  2 $  and

 $ \sqrt[3] {5 - 2 \sqrt{13}} =  \frac 1 2  -  \frac {\sqrt {13}}  2 $ [/hide]

 So the sum is obviously[hide] $1$ [/hide][/quote]
Nice method....


integration
calculus
real analysis
real analysis unsolved

Let $f \in C([0,1])$. Prove that there exists $c \in (0,1)$ such that \[\int_{0}^{1}f(x) \, dx = (1-c)f(c)\]

This has been posted and solved somewhere in either this forum or "Calculus Computations and Tutorials." The moderators might know where the original post is.


for the linear map T: $ (E,\parallel  \parallel ) \to (F, \parallel \parallel ^\prime )$ these statements are equivalent:
1)T is continuos
2)T is continuos in 0
3)the set {$ { \parallel T(x) \parallel ^ \prime \mid \parallel x \parallel < 1}$} is bounded.

All comes from LINEAR property. Try to use definition and you can see it.


Write an equation for the locus of pts that include (2,0) and are equidistant from the line y = 3x-1.

[hide]The locus is a line parallel to the given one. As we know the slope is $3$, we have
$0 = 2(3) + b \implies b = -6$
Thus, the equation is $\boxed{y = 3x - 6}$
This seems far too easy for intermediate....[/hide]

[quote="Andreas"][hide]It is the line parallel to $y = 3x - 1$ which contains $(2, 0)$, so $y = 3x + 2$.[/hide][/quote]
$(2,0)$ is on the x-axis. ;) :P


[img]http://i822.photobucket.com/albums/zz142/xiaosanfelix/ChessProblem1.jpg[/img]
Find the best move for black (no check or checkmate). Worth [b]2[/b] Points

Chess Code:
Here is the code for chess, that is used here, and used everywhere.

For a pawn to go to g2 say g2
For a rook to go to f1 say Rf1
For a knight to go to a3 say Na3
For a bishop to go to b8 say Bb8
For a queen to go to d5 say Qd5
For a king to go to e3 say Ke3
If a rank let say rook to defeat the queen on c3 you say Rxc3(x means kill)
To castle king side say 0-0
To castle queen side say 0-0-0
If a rank let say bishop moves to c3 to checks the king say Bc3+(+ means check)

Example: If a pawn defeats a bishop on c3 and checking the king. xc3+

Post the answer on here and hide it.

This is taken from shredderchess, but the requirement
[quote](no check or checkmate)[/quote]
is wrong. How can another move but Rd4+, winning the white knight, be better? (moreover, this is the correct one admitted on the site).

What I actually meant was like no checks or checkmate in puzzle.

[img]http://i822.photobucket.com/albums/zz142/xiaosanfelix/ChessProblem2.jpg[/img]

Best move for black.

Yes it is from shredder chess

[hide]1... Qa3 with the menace 2... Qb2++ mate. If 2.xc4, then ... bxc4, opening the file for Black's rook at a8, with quick mate; if 2.Qxc4, then ...bxc4 or dxc4 and White loses the Queen.[/hide]

[img]http://i822.photobucket.com/albums/zz142/xiaosanfelix/ChessProblem3.jpg[/img]

Best move for white.

[img]http://i822.photobucket.com/albums/zz142/xiaosanfelix/ChessProblem4.jpg[/img]

Best move for black.

[img]http://i822.photobucket.com/albums/zz142/xiaosanfelix/ChessProblem5.jpg[/img]

Best move for white

[img]http://i822.photobucket.com/albums/zz142/xiaosanfelix/ChessProblem6.jpg[/img]

Best move for white

I guess this is over then right?
Cause people haven't posted on this for like 10 days.
So I guess it should be locked.

Sudden drop :ninja:

[hide]If you go R*g4 you have a chance for promotion.[/hide]
I mean for the first problem.

first of all yes thats true but IMO the black rook can quickly counter that.

Also, this thread was last posted on in "October 26,[b]2009[/b]" Please dont revive threads that are almost a year old  :wink:


An 'unfair' coin has a 2/3 probability of turning up heads. If this coin is tossed 50 times, what is the probability that the total number of heads is even?

[hide="What I got"]$\frac{1}{3}$[/hide]

I posted this before. I forgot where though.

For some reason I can't seem to find it in the search... :?

[hide]
Well, first, the answer should be very close to $\frac{1}{2}.$
I get $\frac{3^{50}+1}{2*3^{50}}$
Let us call the probability that we are looking for $N$.
$N=\binom{50}{0}\left ( \frac{2}{3} \right) ^0 \left( \frac{1}{3} \right) ^{50}+\binom{50}{2}\left ( \frac{2}{3} \right) ^2 \left( \frac{1}{3} \right) ^{48}+...+\binom{50}{50}\left ( \frac{2}{3} \right) ^{50} \left( \frac{1}{3} \right) ^0$
Hmmm... This looks a lot like a binomial expansion :D
$1=\left( \frac{1}{3} +\frac{2}{3} \right)^{50}=\binom{50}{0}\left ( \frac{2}{3} \right) ^0 \left( \frac{1}{3} \right) ^{50}+\binom{50}{1}\left ( \frac{2}{3} \right) ^1 \left( \frac{1}{3} \right) ^{49}+...+\binom{50}{50}\left ( \frac{2}{3} \right) ^{50} \left( \frac{1}{3} \right) ^0$
Also,
$\frac{1}{3^{50}}=\left( \frac{1}{3}-\frac{2}{3} \right) ^{50}=\binom{50}{0}\left ( \frac{2}{3} \right) ^0 \left( \frac{1}{3} \right) ^{50} - \binom{50}{1}\left ( \frac{2}{3} \right) ^1 \left( \frac{1}{3} \right) ^{49} \pm ...+\binom{50}{50}\left ( \frac{2}{3} \right) ^{50} \left( \frac{1}{3} \right) ^0$
Adding the two expansions together, we get $2N$. Thus $N=\frac{3^{50}+1}{2*3^{50}}$
[/hide]

Hey chess, how'd you get $\frac{1}{3}$?  Common sense tells you its going to be very close to $\frac{1}{2}$...

because there is am unfair coin, it would not necessarily be 1/2, sorry if i am missing something

[quote="Altheman"]because there is am unfair coin, it would not be 1/2[/quote]but this deals with odd/even not heads/tails.  The prob. of getting 35 heads is close to 36 heads for example.

[quote="chess64"]$(1/3)^{50}+(1/3)^{48}(2/3)^2+\cdots+(2/3)^{50}= \text { what?}$
I don't know what it is, I thought $1/3$ but I guess I made a big mistake somewhere so i don't know.[/quote]
Thie LHS is not even right.  You have to account for the fact that there are multiple ways to get, say, 2 heads and 48 tails, but there are not neccesarily the same number of ways to get some other combination say, 0 heads and 50 tails.  You can look at the beginning of my post to see the right summation.

$\sum_{k=0}^{25}\binom{50}{2k}\left(\frac{2}{3}\right)^{2k}*\left(\frac{1}{3}\right)^{50-2k}$
but i don't know how to simplify more...it is close to the binomial theorem at least

EDIT: wait, that is what somebody posted earlier...i see how to do it now

Also there is an intutive (recursive) way to think about this problem, Suppose you knew the answer for 49 flips, Can you calculate it for 50?

In general:

[hide]Let  $p_n$ is the prob. for $n$  , then for $n+1$ throws,   You get  $p_{n+1} = (1/3) p_n +(2/3) (1-p_n)$

We know $p$ is about (1/2) so we might as well have $p_n =  (1/2)+x_n$ so $(1-p_n) = (1/2)-x_n$

we get (after putting these values in equation above  $x_{n+1} = -  (1/3)x_n$
[hide](Which translates into $x_n =  k (-1/3)^n$   where $k$ is a constant,.
$k$ can be calculated by checking   the case for $n=1$
[/hide]
And we will get $p_n$ as something like $(1/2) + (1/2) (-1/3)^n$
[/hide]

I think I did this one too intuitively... I knew it was about half, so I thought about it for a little and got the right answer. :P

Here's another slick way to solve such problems.

[hide="Solution"]
Let $x_i$ be ${-1}$ if the $i$th toss is heads, and ${+1}$ otherwise.  Then $E(x_i)$, the average value of $x_i$, is ${-\frac{1}{3}}$.  By independence, the average value of the product of the $x_i$ is
\[
  E(x_1 x_2 \cdots x_n) = E(x_1) E(x_2) \cdots E(x_n) = \left( -\frac{1}{3} \right)^n \, .
\]
Let $p$ be the (asked-for) probability that the number of heads is even.  By splitting into odd and even cases, we also have
\[
  E(x_1 x_2 \cdots x_n) = p \cdot (+1) + (1-p) \cdot (-1) = 2p - 1.
\]
Setting the two equations equal to each other, we get
\[
  2p - 1 = \left( -\frac{1}{3} \right)^n \, .
\]
Solving for $p$, we get
\[
  p = \frac{1}{2} \left[1 + \left( -\frac{1}{3} \right)^n \right] \, .
\]
[/hide]


linear algebra
matrix
induction
vector
linear algebra unsolved

Prove  that a matrix that is both unitary and upper triangular must be a diagonal matrix.

Here's that induction laid out.

Suppose the theorem has already been proved for $ n\times n$ matrices. (It's trivial for $ 1\times 1,$ of course.)

Let $ A$ be an upper triangular unitary $ (n \plus{} 1)\times (n \plus{} 1)$ matrix. Write
\[ A \equal{} \begin{pmatrix}B & v \\
0 & \lambda\end{pmatrix}
\]
Where $ B$ is upper triangular, $ v$ is an $ n$-vector, and $ \lambda\in \mathbb{C}.$

Since $ A$ is unitary, we have $ A^*A \equal{} I.$ In detail,
\[ \begin{pmatrix}B^* & 0 \\
v^* & \overline{\lambda}\end{pmatrix}\begin{pmatrix}B & v \\
0 & \lambda\end{pmatrix} \equal{} \begin{pmatrix}B & v \\
0 & \lambda\end{pmatrix} \equal{} \begin{pmatrix}B^*B & B^*v \\
v^*B & v^*v \plus{} |\lambda|^2\end{pmatrix} \equal{} \begin{pmatrix}I & 0 \\
0 & 1\end{pmatrix}.
\]
Hence, $ B^*B \equal{} I.$ Hence $ B$ is unitary as well as upper triangular, hence by the inductive hypothesis, $ B$ is diagonal.

Then we have that $ B^*v \equal{} 0.$ Since $ B^*,$ being unitary, is invertible, that implies that $ v \equal{} 0.$

Examining the consequences of what we have proved, we see that $ A$ is diagonal, proving the theorem.


AMC
USA(J)MO
USAMO
inequalities
inequalities unsolved

1. [Canada 1997] Prove that \[\frac{1}{1999}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6} ... \frac{1997}{1998}<\frac{1}{44}\]

2. [Poland] Let $a,b,c$ be positive numbers. Prover that \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \le \frac{a^8+b^8+c^8}{a^3b^3c^3}\]

3. [USAMO 1999 packet} Let $a,b,c,d$ be positive such that $abc=1$. Prove that \[\frac{a}{1+b+c+d} + \frac{b}{1+a+c+d}+\frac{c}{1+d+a+b} + \frac{d}{1+a+b+c}\ge 1\]

2 is Muirhead or AM-GM, 3 is Jensen and then AM-GM, and 1 follows by letting the big product thingy be A, letting B be 2/3 * 4/5 * ... * 1996/1997, getting some trivial bounds between A and B, and noting AB = 1/1998

About the first one, here is the general one:

       $\frac{1}{2}\cdot\frac{3}{5}\cdot ... \cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$ ;)

 This can be proved very easy by induction. ;)

The third problem follows from the well-established idea of making numerators equal and using AM-HM. All that remains to show is $a+b+c+d\ge 4$. which is AM-GM.


trigonometry
geometry
3D geometry
sphere

When two bubbles cling together in midair, their common surface is part of a sphere whose center D lies on the line passing through the centers of the bubbles. Also, the angles ACB and ACD both have a measure of 60 degrees.

[img]http://i641.photobucket.com/albums/uu136/snowman4160/2.jpg[/img]

A) Show that the radius r of the common face is given by r=(ab)/(a-b)
B)What shape does the common face take if the two bubbles have a radii of the same length?

[hide="(1)"]$ [BCD] \equal{} [BCA] \plus{} [ACD]$
$ \therefore \frac {1}{2}ar\sin{120^\circ} \equal{} \frac {1}{2}ab\sin{60^\circ} \plus{} \frac {1}{2}br\sin{60^\circ}$
Rearranging gives $ r \equal{} \frac {ab}{a \minus{} b}$[/hide]
[hide="(2)"]I would guess that the common face would be flat if $ a\equal{}b$.  Intuitively, a flat face corresponds to part of a surface of a sphere with 'infinite radius', which is kind of what you get if you substitute $ a\equal{}b$ into the formula for $ r$ in part (1).[/hide]

What formula did you use to come up with the step beginning with the 1/2arsin120=?

He used the trig area formula:

$ A \equal{} \frac {ab\sin{C}}{2}.$

In the steps before that, he broke up the total area into two areas and used the trig area formula on each area and equated.

For part 1, how did you rearrange the trig expressions to get the answer?

Ok, I'll work it out. I'll assume that you know how to evaluate $ \sin{\theta}$ for certain special values of $ \theta$, like the ones used in the problem. We'll divide by $ \frac {1}{2}\sin{60^\circ}$ and then simplify from there. Note that $ \sin{60^\circ} \equal{} \sin{(180^\circ \minus{} 60^\circ)} \equal{} \sin{120^\circ}$.

$ ar\sin{120^\circ} \equal{} ab\sin{60^\circ} \plus{} br\sin{60^\circ} \\
ar \equal{} ab \plus{} br \\
(a \minus{} b)r \equal{} ab \\
r \equal{} \frac {ab}{a \minus{} b}.$

Oops sorry I didn't realise that people had asked questions.  Thank you sunehra! :)


A fair, 8-sided die has faces numbered 1,2,3,4,5,6,7, and 8. If the die is rolled twice, what is the probability that the sum of the two numbers will be 9? Express your answer as a common fraction.

There are $ 8 \cdot 8$ total possible rolls.

To get 9, we need one of the following:
(1,8)(2,7)(3,6)(4,5)(5,4)(6,3)(7,2)(8,1)

There are $ 8$ of these, so the probability is $ \frac{8}{8 \cdot 8} \equal{} \frac{1}{8}$


Let $ a_1, a_2, b_1, b_2 > 0$, if $ \frac {1}{a_1} \plus{} \frac {1}{a_2} > \frac {1}{b_1}\plus{}\frac {1}{b_2}$. Is it true that $ \frac {1}{a_1 \plus{} \lambda} \plus{} \frac {1}{a_2 \plus{} \lambda} > \frac {1}{b_1 \plus{} \lambda}\plus{}\frac {1}{b_2 \plus{} \lambda}$ for any $ \lambda > 0$?  :maybe:

You can multiply $ a_{1}, a_{2}, b_{1}, b_{2}$ and $ \lambda$ by any positive number. So it is false for all $ \lambda>0$.


inequalities
function
inequalities proposed

$ a,b,c \geq 0 ; ab\plus{}bc\plus{}ca\equal{}3$
Prove that :
$ a(a\plus{}b^3)\plus{}b(b\plus{}c^3)\plus{}c(c\plus{}a^3)\plus{}2abc(a\plus{}b\plus{}c) \geq 12$

[quote="mai quoc thang"]$ a,b,c \geq 0 ; ab \plus{} bc \plus{} ca \equal{} 3$
Prove that :
$ a(a \plus{} b^3) \plus{} b(b \plus{} c^3) \plus{} c(c \plus{} a^3) \plus{} 2abc(a \plus{} b \plus{} c) \geq 12$[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ t\geq1$ and your inequality is equivalent to
$ \sum_{cyc}(a^3b \plus{} 4a^3c \minus{} 4a^2b^2 \minus{} a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(5a^3b \minus{} 4a^2b^2 \minus{} a^2bc)\geq3(a \plus{} b \plus{} c)(a \minus{} b)(a \minus{} c)(b \minus{} c)\Leftrightarrow$
$ \Leftrightarrow15u^2v^2 \minus{} 18v^4 \plus{} 3uw^3\geq(a \minus{} b)(a \minus{} c)(b \minus{} c)u.$
If $ (a \minus{} b)(a \minus{} c)(b \minus{} c)\leq0$ your inequality is true.
But when $ (a \minus{} b)(a \minus{} c)(b \minus{} c)\geq0$ it remains to prove that
$ 9(5u^2v^2 \minus{} 6v^4 \plus{} uw^3)^2\geq27(3u^2v^4 \minus{} 4v^6 \plus{} 6uv^2w^3 \minus{} 4u^3w^3 \minus{} w^6)u^2\Leftrightarrow$
$ \Leftrightarrow u^2w^6 \plus{} (3u^4 \minus{} 2u^2v^2 \minus{} 3v^4)uw^3 \plus{} (2u^2 \minus{} 3v^2)^2v^4\geq0.$
If $ 3u^4 \minus{} 2u^2v^2 \minus{} 3v^4\geq0$ the inequality is proven.
Hence, we can assume  $ 3u^4 \minus{} 2u^2v^2 \minus{} 3v^4 < 0$ id est, $ 1\leq t < \frac {1 \plus{} \sqrt {10}}{3}.$
Thus, it remains to prove that
$ (3u^4 \minus{} 2u^2v^2 \minus{} 3v^4)^2u^2 \minus{} 4(2u^2 \minus{} 3v^2)^2u^2v^4\leq0,$
which is equivalent to $ t(t \minus{} 1)^2(3t^2 \plus{} 2t \minus{} 9)\leq0,$
which is true for $ 1\leq t < \frac {1 \plus{} \sqrt {10}}{3}.$

[quote="mai quoc thang"]$ a,b,c \geq 0 ; ab \plus{} bc \plus{} ca \equal{} 3$
Prove that :
$ a(a \plus{} b^3) \plus{} b(b \plus{} c^3) \plus{} c(c \plus{} a^3) \plus{} 2abc(a \plus{} b \plus{} c) \geq 12$[/quote]
Homogenizing:
$ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca) \plus{} 3 (ab^3 \plus{} bc^3 \plus{} ca^3) \plus{} 6abc(a \plus{} b \plus{} c) \ge 4(ab \plus{} bc \plus{} ca)^2 \iff$
$ 4\sum_{cyc} ab^3 \plus{} \sum_{cyc} a^3b \ge 4 \sum_{cyc} a^2b^2 \plus{} \sum_{cyc} a^2bc$
And consider the function $ f(a,b,c) \equal{} 4\sum_{cyc} ab^3 \plus{} \sum_{cyc} a^3b \minus{} 4 \sum_{cyc} a^2b^2 \minus{} \sum_{cyc} a^2bc$
If we differentiate it with the respect to the sum of $ a$, $ b$ and $ c$ we get:
$ f'_{\sum} (a,b,c) \equal{} \sum_{cyc} 5a^3 \plus{} 3ab^2 \minus{} 6ab^3 \minus{} 2abc \equal{} \sum_{cyc} 3a(a \minus{} b)^2 \plus{} (a^3 \minus{} abc)$
So $ f'_{\sum} \ge 0$ for all $ a,b,c \ge 0$. Assume wlog $ c \le a,b$, then we have:
$ f(a,b,c) \ge f(a \minus{} c,b \minus{} c,0)$
And the inequality is strict when $ c \neq 0$ and $ a > c$ or $ b > c$.
Then $ f(x,y,0) \equal{} 4x^3y \plus{} y^3x \minus{} 4x^2y^2 \equal{} xy(2x \minus{} y)^2$.
So there are equality when $ a \equal{} b \equal{} c$ or $ c \equal{} 0$ and $ 2a \equal{} b$ or cyclic permutations. Since $ ab \plus{} bc \plus{} ca \equal{} 3$ we have equality when:

$ (a,b,c) \in \{(1,1,1); (\frac {\sqrt {6}}{2}, \sqrt {6}, 0), (\sqrt {6}, 0, \frac {\sqrt {6}}{2}), (0, \frac {\sqrt {6}}{2}, \sqrt {6}) \}$

[quote="mai quoc thang"]$ a,b,c \geq 0 ; ab \plus{} bc \plus{} ca \equal{} 3$
Prove that :
$ a(a \plus{} b^3) \plus{} b(b \plus{} c^3) \plus{} c(c \plus{} a^3) \plus{} 2abc(a \plus{} b \plus{} c) \geq 12$[/quote]
The inequality is equivalent to the homogeneous inequality:
$ (a^3b\plus{}b^3c\plus{}c^3a)\plus{}4(ab^3\plus{}bc^3\plus{}ca^3) \ge 4(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\plus{}abc(a\plus{}b\plus{}c),$
or
$ ab(a\minus{}2b)^2\plus{}bc(b\minus{}2c)^2\plus{}ca(c\minus{}2a)^2 \ge abc(a\plus{}b\plus{}c),$
that is,
$ [ab(a\minus{}2b)^2\plus{}bc(b\minus{}2c)^2\plus{}ca(c\minus{}2a)^2](c\plus{}a\plus{}b) \ge abc[(a\minus{}2b)\plus{}(b\minus{}2c)\plus{}(c\minus{}2a)]^2.$
which follows from Cauchy Schwarz Inequality. :)

[quote="nguoivn"]It's only: $ \sum\ ab(a \plus{} c \minus{} 2b)^2 \geq\ 0$
Because equality occurs when $ a \equal{} b \equal{} c$ and $ a \equal{} 2b$; $ c \equal{} 0$
So, we'll think immediataly about that hint  :P[/quote]
yes ..... you are right :D


Let be given a triangle $ABC.$ Prove that: $\sum\cos^{2}\frac{B-C}{2}\ge 24\prod\sin\frac{A}{2}$

use this:$\cos{\frac{B-C}{2}}=\frac{b+c}{a}\cdot \sin{\frac{A}{2}}$

then our ineq is $\Longleftrightarrow \sum{(\frac{a+b}{c})^{2}\cdot \sin^{2}{\frac{A}{2}}\geq 24\prod{\sin{\frac{A}{2}}}}$

then AM-GM :$\sum{(\frac{a+b}{c})^{2}\cdot \sin^{2}{\frac{A}{2}}\geq 3\sqrt[3]{(\frac{\prod{(a+b)}}{abc})^{2}\cdot (\prod{\sin{\frac{A}{2}}})^{2}}}$.

so it remains us to prove that : $(\frac{\prod{(a+b)}}{abc})^{2}\geq 8^{3}\prod{\sin{\frac{A}{2}}}$

but : $(\frac{\prod{(a+b)}}{abc})^{2}\geq 64$ (AM-GM) so it remains : $\prod{\sin{\frac{A}{2}}}\leq \frac{1}{8}$ , which is true and well known.

Thank you for your solution.
And now, I'll present my solution:
We have:
\begin{eqnarray*}\cos^{2}\frac{A-B}{2}+\cos^{2}\frac{B-C}{2}+\cos^{2}\frac{C-A}{2}&=&1+\frac{1}{2}[\cos{(A-B)}+\cos{(B-C)}]+\cos^{2}\frac{C-A}{2}\\ &=&1+\cos\frac{C-A}{2}\cdot\cos\frac{A+C-2B}{2}+\cos^{2}\frac{C-A}{2}\\ &=&1+\cos\frac{C-A}{2}\cdot\left(\cos\frac{A+C-2B}{2}+\cos\frac{C-A}{2}\right)\\ &=&1+2\cos\frac{A-B}{2}\cdot\cos\frac{B-C}{2}\cdot\cos\frac{C-A}{2}\end{eqnarray*}
Hence: $\sum\cos^{2}\frac{B-C}{2}\ge 24\prod\sin\frac{A}{2}\Longleftrightarrow 1+2\prod\cos\frac{B-C}{2}\ge 24\prod\sin\frac{A}{2}$, what is truly because in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=124908]Trig Inequality3[/url] we have: $\begin{cases}\prod\sin\frac{A}{2}\le\frac{1}{8}\\ \prod\sin\frac{A}{2}\le\frac{1}{8}\prod\cos\frac{B-C}{2}\end{cases}$
[b]Remark[/b] $\boxed{\sum\cos^{2}x+\cos^{2}(\sum x)=2+2\prod\cos(y+z)\quad\forall x,\,y,\,z}$


arithmetic sequence
combinatorics unsolved
combinatorics

The vertices of a regular $7$-gon are colored white or black. Prove that there are vertices of the same color, which form an isosceles triangle.

Easy, assume it is not true.
Label the vertices with numbers 1,2...7. WLOG there are (at least) 4 white vertices and vertex 4 is white. Now since triangles 147, 246 and 345 are isosceles exactly one vertex in each pair  (3,5) (2,6) and (1,7) is white. Again WLOG 3 is white but now 347 is iscosceles so 7 is black and so 1 is white now 123 and 164 are iscoseles which means no of 2 and 6 is white. Contradiction.

For what regular $n$-gons is the assertion valid?

For $n\leq 6$ it is not true. For example for n=6 1,2,4 white and 3,5,6- black and there are no equilateral triangle. For $n\geq 7$ it is true.

Do you have a proof, Baronov?  Because it is true for $n = 5$, since any three vertices of a regular pentagon form an isosceles triangle, and false for $n = 8$ (take 1, 2, 5, 6 black and 3, 4, 7, 8 white).

[quote="Baronov"]you are right JBL i've made mistake. But now i think that for $n\geq 9$ it is true because if you color numbers $1,2...9$black and white then there is an arithmetic progression with length 3 and so there is an isosceles triangle. The proof includes checking many cases, so i won't write it here.[/quote]
Can you send my mail?
Adress: huyvan_vmo@yahoo.com

That sounds better :).  That result (about an arithmetic sequence) is the fact that the van der Waerden number $n(2, 3)$ is equal to 9; and indeed, the result does follow for all $n \geq 9$.


[quote="lambern314"]Hello there, I did the following question:

What is the sum of:

5+55+555+5555......

and so on, for any coeffecient, (in this case 5), however my solution was too long, and took me a little while to get, I didn't like it, so I was wondering if there was an easier and more elagant way of solving this problem.

Thanks[/quote]
[hide]
if you are only adding 5 numbers, it wouldnt be that hard just to add them

another way is by factoring out the fives to get:

$5(1+11+111+1111+11111\cdots)$

in the case of 5 numbers, it becomes $5\cdot 12345=61725$[/hide]

[quote="lambern314"]Sorry, I meant if the series  is infinite[/quote]
Then its infinity..

I apologize again, my wording is off, but I want to find an easy way of deriving the formula to find the partial sum of this series with any coeffecient.
For example if I want to sum
6+66+666+6666....
to the first 50 terms, then there is an equation that will describe this partial sum,

and more generally, describe the partial sum of

n(1+11+111+1111...) to any number of terms, to any value of n.

dear Astro,

I dont think your solution is correct, because the sum of
5+55=60

while your formula says
30
for the first 2 terms
(sorry this is for the first formula that you got)

[quote="AstroPhys"]Corrected it now, put it wrong into the $\LaTeX$ and did not bother to check if it was right before simplifying.  :oops:
It is not simplified though.[/quote]

Sorry to be hypercritical, but its still not correct, you neglected the 5's in the middle

:oops: Wow, by this time the solutions to this that I posted should be ignored.
It is not hypercritical, if they are wrong, they are wrong.

yeah.. i am pretty sure that my solution is correct.

Thank you everyone, my solution was not very pretty...
Im wondering if you guys have any suggestions for any interesting geometric series that I would have fun generallizing and summing. (not too hard please...), i need to practice these kinds of questions.

Thanks


This is almost basic problem but I think my book has some extra steps.

The question is:

Given: ABCD is a rhombus; DE = BF
Prove: AECF is a rhombus.

My answer

1. DO = BO; CO = AO 1. Definition of Parallelogram
2. EO = FO 2. Subtraction
3. DB :perp: AC 3. Defintion of Parallelogram
4. EF :perp: AC 3. Since EF is a part of DB, I can't think of a theorem or corollary or postulate.
5. AECF is a rhombus. 5. Def. of Rhombus

My book answer

Key step Proof

1. DO = BO; AO = CO (Diagonals of Paralleogram bisect each other.)
2. EO = FO (Subst. Prop = )
3. AECF is a parallelogram.
4. BD :perp: AC (Diagonals of rhombus are :perp:)
5. :tri: COE :cong: :tri: COF (SAS)
6. CE :cong: CF (CPCTC)
7. AECF is a rhombus.

My question is, when I need to prove the above problem, do I HAVE to say that those consecutive sides of a quadrilateral is congruent? I instead proved that their diagonals perpendicular so it's rhombus.

Maybe it's not sufficient but I'm not so confident.

It depends a little bit on what order you've taken theorems in.  For example, if your class hasn't shown the converse of the theorem, "A rhombus is a figure whose diagonals bisect each other and are perpendicular," then you haven't got enough basis to use your proof.  Yours proof is valid, though, since the converse of that theorem is true, but you may not have done enough in class to prove it that way.

You do have a couple of places in your proof where you wrote parallelogram instead of rhombus, I think.

Hmm..

I see..

I think I must have slipped off.. More proof practice required before test on Monday then!


calculus
integration
calculus computations

if $ \varphi: [0,\infty)\to \mathbb{R}$ is convex and $ \varphi(0)\equal{}0$, $ \minus{}1<\alpha<\infty$, A is a definite real number.
Prove $ \int_0^A y^\alpha \exp \left( \minus{} \frac{\varphi(y)}{y} \right) dy < \infty$
how to prove this?
thanks

I guess $ \varphi$ should be non-negative and this is missing from the formulation. Otherwise with $ \varphi(x) \equal{} \minus{}\sqrt{x}$ the integral diverges.


Let $0<a<1,0<b<1,0<c<1$.Prove the following inequality.

\[a+b+c-abc<2\]

You are right, mozilla. :)  This is the entrance exam for university.

Here are another solution. Maybe someone will post.

kunny

I think it's very difficult for an entrance exam.


Prove that for each $ n$ natural , $ \frac {10^{n} \plus{} 5}{15}$ is natural.

[hide]So we need to prove both 3 and 5 go into the numerator. Clearly, 5 divides the expression, since both numbers are divisible by 5. $ 10^n\equiv{1^n}\pmod{3}$, so the expression is congurent to 6 (mod 3), so it is divisible by 3. Now, since n is natural we are done.[/hide]

My solution:

$ 10^{n}$ is a number with 1 and $ n$(0) $ + 5$ . It i divisible by 5 . $ \frac {10^{n} + 5}{5} = 20....0_{n} + 1$
and $ 20....0_{n} + 1$ has a sum of digit equal to 3 hence this number is divisible by 3.
From that $ {10^{n} + 5}$ is divisible by $ 3*5 = 15$