Let $ A,B$ are 2 points in the plane. $ C$ is not a fix point. Construct two squares: $ ACED, CBFG$ . Prove that: $ DG$  pass to a fix point

try co ordinate geometry  along with the proof of pythagoras
that involves construction of a square inside another square[/img]

the problem is most probably wrong or else there is a typing mistake as it can be disproved by just taking 3 co ordinate values for point $ C$

The problem is true only if you replace $ DG$ by $ DF$.

Let me first assume that the squares $ ACED$ and $ CBFG$ are oriented in the same sense. Then, the midpoint of the segment $ DF$ coincides with the point $ T$ such that triangle $ BTA$ is an isosceles right triangle (with $ AT \equal{} TB$ and oriented in the same way as the squares $ ACED$ and $ CBFG$), and thus is fixed. See http://www.mathlinks.ro/Forum/viewtopic.php?t=16186 for proofs (should I explain the equivalence?).

The problem also holds if the squares $ ACED$ and $ CBFG$ are oriented in different senses, and in this case the fixed point is an infinite one - this is pretty easy (show that the points $ D$ and $ F$ are the images of the points $ A$ and $ B$ under the spiral similitude with center $ C$, angle 45° and factor $ \sqrt2$).

  darij

thanks for correcting the question darij :)  :)


geometry vector circumcircle parallelogram

let $ABC$ be a triangle $H$ is the orthocenter  and $AB = CH$.
Find $\angle C$

remark: this is Spanish olympiad Q

To solve, let a,b,c be vectors from the circumcenter to the 3 vertices.

Lemma:  a+b+c is the orthocenter.  Proof: only one point H satisfies $(a-c )\bullet( H - b) = 0$, and we see H = a+b+c satisfies.

Now condition is $|a-b| = |a+b|$.  Interpreted geometrically with a parallelogram of sides a and b, it implies a and b are at right angles.

Now AOB = 90deg so that ACB = 45 deg.

[quote="Singular"]

$(a-c )\bullet( H - b) = 0$,

.[/quote]
i don't understand what's that$(a-c )\bullet( H - b) = 0$ :?:


Suppose $ f(x) \equal{} (1 \plus{} x)(1 \plus{} x^2) \cdots (1 \plus{} x^n) \equal{} \sum_{k \equal{} 0}^{m}a_k x^k$, prove that $ \{a_n\}_{n \equal{} 1}^m$ is unimodal.

aha, it is the famous grid shading problem


The incircle (I) of triangle $ ABC(AB>AC)$ tangents to BC at E. AE meets (I) at E and D. Let $ F\in AC$ suct that $ CF\equal{}CE$. Let G be the intersection of CF and BD. Prove that $ CF\equal{}FG$.

I don't think that those problems are true  :huh:

http://www.mathlinks.ro/Forum/viewtopic.php?t=204652
http://www.mathlinks.ro/Forum/viewtopic.php?t=204655

I made a solution for solving the problem, i am sure that it is true and very nice.

Should be: Let $ F \in AE$ be such that $ CE \equal{} CF.$
_________________________________________

$ U, V$ are tangency point of the incircle $ (I)$ with $ AB, AC.$ $ UV$, tangent to $ (I)$ at $ D,$ and tangent $ BC$ to $ (I)$ at $ E$ are polars with respect to $ (I)$ of the collinear points $ A, D, E,$ they are concurrent at $ K.$ The cross ratio $ \frac {\overline{KB}}{\overline{KC}} \cdot \frac {\overline{EC}}{\overline{EB}} \equal{} \minus{} 1$ is harmonic (by Menelaus for the transversal $ UVK$), hence $ \overline{EK} \equal{} \overline{EC} \plus{} \overline{CK} \equal{} \overline{EC} \plus{} \frac {\overline{EC} \cdot \overline{BC}}{\overline{BE} \minus{} \overline{EC}} \equal{} \frac {2\ \overline {BE} \cdot \overline{EC}}{\overline{BE} \minus{} \overline{EC}}.$ The isosceles $ \triangle KDE \sim \triangle CFE$ are centrally similar with center $ E$, $ CF \parallel KD.$ Let a parallel to $ BD$ through $ F$ cut $ BC$ at $ M.$ The $ \triangle MCF \sim \triangle BKD$ are centrally similar with center $ E$ and coefficient $ k \equal{} \frac {\overline{EC}}{\overline{EK}} \equal{} \frac {\overline{BE} \minus{} \overline{EC}}{2\ \overline{BE}},$ hence $ \overline{EM} \equal{} k \cdot \overline{EB} \equal{} \minus{} \frac {\overline{BE} \minus{} \overline{EC}}{2}$ and $ \overline{BM} \equal{} \overline{BE} \plus{} \overline{EM} \equal{} \frac {\overline{BE} \plus{} \overline{EC}}{2} \equal{} \frac {\overline{BC}}{2}.$ Thus $ M$ is the midpoint of $ \overline{BC}.$ Since $ BG \parallel MF,$ $ MF$ is a midline of the $ \triangle BCG$ and $ F$ is the midpoint of $ \overline{CG}.$

[quote="mr.danh"] [color=darkred]The incircle $ w \equal{} C(I)$ of triangle $ ABC\ \ (c > b)$ touches it in the points $ D\in BC$ , $ E\in CA$ , $ F\in AB$ .

Denote $ \{D,T\} \equal{} AD\cap w$ and let $ K\in AD$ be a point suct that $ CK \equal{} CD$ . Denote  $ L\in CK\cap BT$ . Prove that $ KC \equal{} KL$.[/color] [/quote]

[color=darkblue][b][u]Proof (similarly with Yetti's).[/u][/b] Denote the midpoint $ M$ of the side $ [BC]$ and $ N\in EF\cap BC$ . Observe easily that $ \boxed {\ MD \equal{} \frac {c \minus{} b}{2}\ }$ .

From the well-known property $ IN\perp \overline {ATD}$ obtain $ N\in TT$ . Thus, $ CK \equal{} CD$ $ \implies$ $ \widehat {DKC}\equiv\widehat {CDK}\equiv\widehat {DTN}$ $ \implies$

$ \widehat {DKC}\equiv\widehat {DTN}$ $ \implies$ $ KC\parallel TN$ . Denote the point $ M_1\in BC$ , $ KM_1\parallel BT$ . Therefore, $ \triangle M_1KC\sim\triangle BTN$ $ \implies$ $ \frac {DM_1}{DC} \equal{} \frac {DB}{DN}$ .

Since $ \frac {NB}{NC}\equal{}\frac {DB}{DC}\equal{}\frac {p\minus{}b}{p\minus{}c}$ ascertain easily that $ DN \equal{} \frac {2(p \minus{} b)(p \minus{} c)}{c \minus{} b}$ . Thus, $ M_1D \equal{} \frac {DB\cdot DC}{DN} \equal{} (p \minus{} b)(p \minus{} c)\cdot\frac {c \minus{} b}{2(p \minus{} b)(p \minus{} c)}$ ,

i.e. $ \boxed {\ M_1D \equal{} \frac {c \minus{} b}{2}\ }$ . In conclusion, $ M_1D \equal{} MD$ , i.e. $ M_1\equiv M$ $ \implies$ $ KM\parallel BT$ $ \implies$ $ KC \equal{} KL$ .[/color]

We restate the notations of the problem, like to the [b][size=100]Virgil Nicula’s[/size][/b] ones.

We denote as $ D,\ E,\ F,$ the tangency points of the incircle $ (I)$ of $ \bigtriangleup ABC,$ with its side-segments $ BC,\ CA,\ AB$ respectively and let be the points $ T\equiv (I)\cap AD$ and $ K$ on $ AD,$ such that $ CD = CK.$

Then, we must prove that $ CK = KL,$ where $ L\equiv CK\cap TB.$

$ \bullet$ Let $ S$ be, the intersection point of the sideline $ BC,$ from the tangent line of the incircle $ (I),$ at point $ T.$

The pencil $ T.SBDC,$ is in harmonic conjugation ( easy to prove ) as well and so, it is enough to prove that $ CK\parallel TS.$

But it is true, because of $ \angle CKT = \angle CDK = \angle SDT = \angle STK.$

So, we conclude that $ CK = KL$ and the proof is completed.

Kostas Vittas.


geometry
number theory
greatest common divisor
invariant

The vertices of $\triangle ABC$ are lattice points on the xy-plane.
If both $AB$ and $AC$ contain odd numbers of lattice points (excluding the vertices themselves), prove that the number of lattice points on $BC$ (excluding the vertices) is also an odd number.

this might have to do with the formula for area that gives the area of a triangle with lattice point verticies

I would guess that Pick's theorem may be useful.

Position the axes so that A is the origin.  Suppose $B = (bp,bq)$ and $C = (cr,cs)$, where p,q and r,s are coprime.

We have the equation of the line AB to be $y = bqx/bp$.  We are concerned with integer solns to $bpy = bqx$.  Obviously (0,0) and (bp,bq) work.  So clearly the equation reduces to $py = qx$ and we must have $x$ a multiple of $p$ etc. (since p,q coprime) so that we have solutions $kp, kq$ for k = 0...b.  So an equivalent condition for AB to have odd # of lattice points is that the gcd of the ordinate and abscissa is even. (*)

Now position the axes so B is the origin by a translation.  Then the gcd of the ordinate and abscissa of C is even (since we have translated by an even number to both the ordinate and abscissa of C, keeping invariant their parity).  But similar to (*) we deduce that BC has an odd # of lattice points.


geometry
rectangle
geometric transformation
reflection
homothety
trigonometry
IMO Shortlist

Let $ ABC$ be a triangle and take three similar rectangles inscribed $ ABC$ (a rectangle inscribed $ ABC$ has two vertices on side a and  two remaining vertices on two other sides of $ ABC$). Let $ A',B',C'$ be center of these rectangles (cyclic with $ A,B,C$), respectively.

a) Prove that $ AA',BB',CC'$ are concurrent.

b) Let $ A'',B'',C''$ are reflections of $ A',B',C'$ through $ BC,CA,AB$, resp. Prove that $ AA'',BB'',CC''$ are concurrent.

Note that, part a) is generalization of the problem on the post [url=http://www.mathlinks.ro/viewtopic.php?t=17463]IMO ShortList 2001, geometry problem 1[/url].

$ \bullet$ Let $ PQRS$ be the rectangle relative to $ A$ such that $ P \in AB$ and $ Q \in AC.$ Construct outwardly on $ BC$ the rectangle $ BCC_aB_a$ similar to $ PQRS.$ Homothety with center $ A$ and coefficient $ \frac {_{AQ}}{^{AC}}$ takes $ PQRS$ into $ BCC_aB_a$ $\Longrightarrow$ their centers $ A'$ and $ A_1$ are collinear with the homothetic center $ A$ $ \Longrightarrow$ $ A_1 \in AA'.$ Likewise, $ B_1,C_1$ lie on $ BB',CC'.$

$ \triangle BCA_1,\triangle CAB_1, \triangle ABC_1$ are 3 similar isosceles triangles erected outwardly on the sides of $ \triangle ABC.$ By Kiepert theorem, $ AA',BB',CC'$ concur at the outer Kiepert perspector relative to $ \angle CBA_1 \equal{} \angle ACB_1 \equal{} \angle BAC_1 \equal{} \phi.$

$ \bullet$ Homothety with center $ A,$ taking $ PQRS$ into $ BCC_aB_a,$ takes the reflection $ A''$ of $ A'$ about $ BC$ into the reflection $ A_2$ of $ A_1$ about $ B_aC_a.$ Let $ M_a$ denote the midpoint of $ BC.$

$ \tan \widehat{M_aCA_2} \equal{} \frac {M_aA_2}{M_aC} \equal{} \frac {3A_1M_a}{M_aC} \equal{} 3\tan\phi$

Therefore, $ \triangle A_2BC,\triangle B_2CA$ and $ \triangle C_2AB$ are three similar isosceles triangles erected outwardly on the sides of $ \triangle ABC$ $ \Longrightarrow$ $ AA'',BB'',CC''$ concur at the outer Kiepert perspector relative to $ \tan^{ \minus{} 1}(3 \tan \phi).$


calculus
integration
logarithms
function
calculus computations

a. Evaluate $ I_{1} \equal{} \int_{0}^{\infty} \frac{te^{\minus{}t}}{1 \minus{} e^{\minus{}2t}} dt$

b. Evaluate $ I_{2} \equal{} \int_{0}^{1} \frac{t^{2} \log t }{t^2 \minus{} 1} dt$

c. Find $ I_{3} \equal{} \int_{y\equal{}0}^{\infty} \int_{x \equal{}y}^{\infty}  \frac{(x \minus{} y)^{2}\log \left(\frac{x \plus{}y}{x \minus{} y}\right)}{xy \sinh (x \plus{}y)} dx dy$

Here's $ I_1:$

Since $ \sum_{k\equal{}0}^{\infty}e^{\minus{}2kt}\equal{}\frac1{1\minus{}e^{\minus{}2t}},$ we have

$ I_1\equal{}\int_0^{\infty}\sum_{k\equal{}0}^{\infty}te^{\minus{}(2k\plus{}1)t}\,dt\equal{}\sum_{k\equal{}0}^{\infty}\int_0^{\infty}te^{\minus{}(2k\plus{}1)t}\,dt$

$ \equal{}\sum_{k\equal{}0}^{\infty}\frac1{(2k\plus{}1)^2}\int_0^{\infty}ue^{\minus{}u}\,du\equal{}\sum_{k\equal{}0}^{\infty}\frac1{(2k\plus{}1)^2}$

This is a well-known sum closely related to $ \zeta(2).$ Its value is $ \frac34\zeta(2)\equal{}\frac{\pi^2}{8}.$

$ I_{2}\equal{}\minus{}\sum_{k\equal{}0}^{\infty }\int_{0}^{1}t^{2k\plus{}2}\ln tdt\equal{}\sum_{k\equal{}0}^{\infty }\frac{1}{\left( 2k\plus{}3\right) ^{2}}\equal{}\allowbreak \frac{\pi ^{2}}{8}\minus{}1$

Thank you all for your solutions.
By setting $ x \equal{} \frac {u \plus{} v}{2}$ and $ y \equal{} \frac{u \minus{} v}{2}$ and then again setting $ v \equal{} uw$ in the resultant integral reduce $ I_{3}$ to solving for $ I_{1}$ and $ I_{2}$.

here i have a doubt

In the second problem it can also be reduced to $ I_{2}\equal{}\sum_{k\equal{}0}^{\infty }\int_{0}^{1}t^{\minus{}2k}\ln tdt\equal{}\sum_{k\equal{}0}^{\infty }\frac{\minus{}1}{\left( \minus{}2k\plus{}1\right)^{2}}\equal{}\allowbreak\frac{\minus{} \pi^{2}}{8}\minus{}1$

How does this happen or am i wrong ?? please help me .  :(

The key in both of the solutions supplied by Kent and Indochina is to use the geometric series. So if your answer differs from the correct answers already supplied, then just show your work and someone will point out an incorrect step.

vijaymenon: note that negative powers of $ t$ are greater than $ 1$ on the interval $ (0,1).$ Hence $ \sum_{k\equal{}0}^{\infty}t^{\minus{}2k}\ln t$ is a divergent series on the interval in question. Furthermore, for $ k\ge 1,$ $ \int_0^1t^{\minus{}2k}\ln t\,dt$ is a divergent integral.

One more note: on the interval in question, $ \ln t<0$ and $ t^2\minus{}1<0.$ Hence, $ \frac{t^2\ln t}{t^2\minus{}1}>0$ and the result of the integral should be positive. (I would also add that the integrand tends to finite limits at both ends of the interval, and can be considered as if it were continuous on $ [0,1]$ - hence we can be sure of convergence of the integral.)

IndoChina's computation is correct, and I don't see any plausible alternatives to his methods.

[quote="Kent Merryfield"]Here's $ I_1:$

Since $ \sum_{k \equal{} 0}^{\infty}e^{ \minus{} 2kt} \equal{} \frac1{1 \minus{} e^{ \minus{} 2t}},$ we have

$ I_1 \equal{} \int_0^{\infty}\sum_{k \equal{} 0}^{\infty}te^{ \minus{} (2k \plus{} 1)t}\,dt \equal{} \sum_{k \equal{} 0}^{\infty}\int_0^{\infty}te^{ \minus{} (2k \plus{} 1)t}\,dt$

$ \equal{} \sum_{k \equal{} 0}^{\infty}\frac1{(2k \plus{} 1)^2}\int_0^{\infty}ue^{ \minus{} u}\,du \equal{} \sum_{k \equal{} 0}^{\infty}\frac1{(2k \plus{} 1)^2}$

This is a well-known sum closely related to $ \zeta(2).$ Its value is $ \frac34\zeta(2) \equal{} \frac {\pi^2}{8}.$[/quote]
Sorry, but I dont understand this:

$ \sum\limits_{k\equal{}0}^{\infty }{\int\limits_{0}^{\plus{}\infty }{t.{{e}^{\minus{}(2k\plus{}1)t}}\,dt}}\equal{}\sum\limits_{k\equal{}0}^{\infty }{\frac{1}{{{(2k\plus{}1)}^{2}}}}.\int\limits_{0}^{\plus{}\infty }{u.{{e}^{\minus{}u}}\,du}$
And I have not learned this follow formula yet. Please tell me about this function:

$ \sum\limits_{k\equal{}0}^{\infty }{\frac{1}{{{(2k\plus{}1)}^{2}}}}\equal{}\frac{3}{4}\zeta (2)\equal{}\frac{{{\pi }^{2}}}{8}$

The first line you quote is simply the ordinary substitution $ u\equal{}(2k\plus{}1)t$ or $ t\equal{}\frac{u}{2k\plus{}1}.$

As for the computation of $ \zeta(2)\equal{}\sum_{n\equal{}1}^{\infty}\frac1{n^2}\equal{}\frac{\pi^2}6:$ there are many different computations, and there are links to places where these computations have been collected together. It's quite a famous problem, and much about it has appeared on this site over the years. (Someone else can supply the links.)

chuong01,
If you are familiar with double integrals, then doing the following exercises may help in the evaluation of the series in question.

Evaluate: $ J_{1} \equal{} \int_{\minus{}1}^{1}\int_{\minus{}1}^{1} \frac{1}{1 \plus{} 2xy \plus{} y^2}dxdy$

Evaluate: $ J_{2} \equal{} \int_{\minus{}1}^{1}\int_{\minus{}1}^{1} \frac{1}{1 \plus{} 2xy \plus{} y^2}dydx$


Is it just me, or is no one going to the mathcounts forum anymore. If so, I think everyone is flocking to this site (like me), cause it's so good. This site is cool 8-) !

Mathcounts.org forum?  What's the address?

just go to http://www.mathcounts.org and then meeting place. but this site and forum is much better to me. after you go to meeting place there are hmmmmmmm i think 4 different forums you can go to. but one of them you have to be a coordinator of somat like that. if you post something it takes a few days because of screening.

Yah, it like took 4 days once. Well, this site is WAY better because of instant posting, and smileys and all sorta stuff.

Only to Syntax : Your LoL is usually "lool". You always do that, but maybe your forgot it this time and made a Synatx Error. (Don't kill me Neal)

Just forget about the last post. And it's so amusing there aren't any posts on that MathCounts.org forum, and there's like 50 new posts here.  :lol:

humph chinaboi. one more acronym "lau" i made it up. it stands for laughing at u. lool!

What is "lool"?

What is "lolol"?

I'll stick to the unabbreviated words. :lol:

It's supposed to stand for laughing out loud, but with these guys, you never know...

Maybe "laughing out of leprosy"? sort of like, laughing out of happiness?

Hey, it's Syntax. You don't really know what he can think of. Right Neal?

yup yup yup. dipdidpdip, hoot hoot hoot. lool. lau. make sure to spread lau to whoever you see. lool. maybe i should do something productive. ha. lool. namaste jall

whoa. you have a lot of freetime on ure hands. how long did it take for you to think that up. well namaste in hindi means like hello goodbye greeting type of thing. and jall is just like yall which is just like you. so i basically said bye everyone. nuff said

Now you see what I mean. Anyways, I'll make sure not to spread the "lau", Neal. See Ya later.

Yep, Syntax is indeed Neal, and Neal, splash1u27 is Amy.

representin from the hood jall. anyways. im the normal one. its martin, you dont have to worry about him. hes just oddly. anyways. spread the acronym "lau" it rocks the hizhouse. yep yep. martin or chinaboi is the one who has a bedtime of like 8:30 pm. shame. some people just have no life. and hes a teenager. tisk tisk. well if you havent figured out by now im sorta mad at him because of his stupidity. anyways. nuff said

:-x Hey! I went to bed late enough um... this morning... hee. I don't wanna go to bed that early. It's parents. Oh yah, I was at Neal's house at 7:30, and all the computer's were used up, so couldn't get on. Was Josh also on at 7:30??? Anyways, gtg. Hey Neal, look at the time I posted this. Way past "my bedtime". anyways, bye.

wooooooooot. i only went to sleep about 6 hours later. so shup

[quote="Ex274513"]"Namaste jall"? Does that mean

[color=#CC3399]Not As Much As Some Thought[/color] [Syntax] [color=#CC3399]Error[/color]

[color=#3399CC]Just A Little Left[/color]

:?: :mrgreen: :lol:[/quote]

Thats no abbreviation! That means "Hello, a deception!"....doesn't really make much sense. :?

of course it doesnt make much sense. just like i end every post with nuff said. to stop my ranting. anywho. nuff said

lau is outdated lolt kicks ***(*)

5 posts to 1000 NEAL GO FOR IT WHURM!!

[quote="sugarcookie34"]Chinaboy, what's the whole point in this lol thing?  Didn't I tell you lol or LOL, not LoL?  Are you trying to get yourself in the crowd?  I'll try lOl! :shock:  I think this site is way better than MATHCOUNTS site!  By the way, is Syntax Error, Neal?[/quote]
I'm going with AOL :lol:

Um, what's the plural of a*s anywho?


For a curve $x=y=z^{2}$, the differential length vector $d\bold{l}$ at a point $(x,y,z)$ is $dx\bold{i_{x}}+dy\bold{i_{y}}+2zdz\bold{i_{z}}$.

What is the physical meaning of $d\bold{l}$?

$2 z dz\bold{i_{x}}+2 z dz\bold{i_{y}}+dz\bold{i_{z}}$.

Are you sure you don't mean that?  If not, I don't think I can help?

oh yes... you are absolutely right! Sorry my bad.  :oops:

Well, do you understand it, or do you still need an explanation.
I'm sorry but I rather refrain from doing work for nothing, it is so so SO hot here right now  :roll:  :wink:

[quote="JBL"]Please don't double post -- you posted this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=578216]here[/url] as well.[/quote]
I am sorry, but I posted this problem 4-5 days ago in the Calculus/Real Analysis forum. After not getting any response for so many days, I started thinking that maybe topic would be more appropriate for Matrix/Linear Algebra forum. It wasn't until today that anybody took the time to answer my question in the Calculus/Real Analysis forum.  :(

Sorry for the trouble.  :oops:


geometry
circumcircle
perpendicular bisector
USAMO
power of a point

Let ${\cal C}_1$ and ${\cal C}_2$ be  concentric circles, with ${\cal C}_2$ in the interior of  ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular  bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof,  the ratio $AM/MC$.

Let $O$ be the common center of the concentric circles $c_1, c_2$. The tangency point B is the midpoint of the chord AC, because AC is perpendicular to the radius OB of the circle $c_2$, and O is also the center of the circle $c_1$. The power of the point A to the circle $c_2$ is $AE \cdot AF = AB^2$. But since B is the midpoint of AC and D the midpont of AB, $AD \cdot AC = \frac{AB}{2} \cdot 2 AB = AB^2$ as well. Hence, the quadrilateral CDEF is cyclic. The intersection M of the perpendicular bisectors of its diagonals CE, DF is its circumcenter. If this circumcenter is to be on one of its side CD, it must be the midpoint of this side, $DM = MC = \frac{DC}{2}$. Since $DC = \frac 3 2 AB$, $DM = MC = \frac 3 4 AB$ and $AM = AD + DM = \frac{AB}{2} + \frac 3 4 AB = \frac 5 4 AB$ and $\frac{AM}{MC} = \frac 5 3$.

[quote]But since M is the midpoint of AC[/quote] But since B is the midpoint of AC.  :)

[b]Problem:[/b] Two circles are concentric. A chord $AC$ of the outer circle touches the inner circle at $Q$. $P$ is the midpoint of $AQ$. A line through $A$ intersects the inner circle at $R$ and $S$. The perpendicular bisectors of $PR$ and $CS$ meet at $T$ on the line $AC$. What is the ratio $AT/TC$?

By Power of a Point we have $(AR)(AS)=AQ^2 = \frac{1}{2}2(AQ)^2 = (AP)(AC)$, so $\triangle ARP\sim\triangle ACS$. Hence we have $\angle ACS=\angle ARP\implies PRSC$ is cyclic. Hence $T$ must be the center of its circumcircle and must also lie on the perpendicular bisector of $CP$. Hence, $T$ is the midpoint of $CP$. This yields $CT=\frac{3}{8}(CA)\implies\frac{AT}{TC}=\frac{5}{3}$.

Why, maplestory?  Is there something wrong with the other solutions?

yes, they assumed that M is not on AB but rather reflected across line OB (O is the center of the concentric circles) so when they found AM/MC they actually found MC/AM, now they have to take the reciprocal to get 3/5.

can any one please give a bash type solution as well ;)  I tried both complex and bary bash but the calculations are getting huge(alas!)

Between I agree with [quote=maplestory]You guys are wrong its actually 3/5...[/quote] The answer is coming 3/5
See p8 in the attachment

Yufei probably just made a typo.  Note that from $\tfrac{MC}{AC}=\tfrac38$ we actually get \[\dfrac{AM}{MC}=\dfrac{AC-MC}{MC}=\dfrac{AC}{MC}-1=\dfrac83-1=\dfrac53,\] not $\tfrac35$ as he suggests.

[quote=Phie11]Wait it says that M is on AB, so I took it to be M on segment AB. However, according to the solution, M is on segment BC. Did they mess up the problem or did they mean line AB?[/quote]

I believe they are referring to line $AB$, not segment $AB$ because that is how they defined point $C$ as well.

Some wishful thinking tells us that M should be the midpoint of $\overline{DC}$. Since $M$ is the intersection of the perpendicular bisectors of $\overline{DE}$ and $\overline{CF}$, it suffices to show that quadrilateral $DEFC$ is cyclic. However notice that $\overline{AC}$ is the radical axis of $C_1$ and $C_2$, hence by PoP, we have that $\overline{AE}\cdot\overline{AF}=\overline{AD}\cdot\overline{AC}$, implying that $DEFC$ is cyclic. Through direct computation, we see that $\frac{\overline{AM}}{\overline{MC}}=\frac{5}{3}$. $\square$

By power of a point, $$AE \cdot AF = AB^2 = AD \cdot AC,$$ so $CDEF$ is cyclic. But by definition, $M$ must be the cyclic center of $(CDEF)$, and thus $MC = MD$. Then $$\frac{MA}{MC} = \frac{\frac 52 AD}{\frac 32 AD} = \boxed{\frac 53}.$$

Note that $$AD\cdot AC=(1/2 AB)(2AB)=AB^2=AE\cdot EF,$$ so $CDEF$ is cyclic. Then, $M$ is the circumcenter of $CDEF$, and since $M$ lies on $DC$, $M$ is the midpoint of $DC$. Hence, $M$ is 5/8 of the way from A to C, so the answer is $5/3.$

We have AF*AE=1/2AB*2AB=AD*AC, so CDEF is cyclic. Then M is the circumcenter and because it lies on CD consequently the midpoint. Calculating simply (find BM=1/8AC) we get an answer of 5/3.

By Power of a Point, $AE \cdot AF = AB^2$ and by the midpoint condition we see that $AD \cdot AC = \tfrac12 AB \cdot 2AB = AB^2$, so $AE \cdot AF = AD \cdot AC$. By Converse of Power of a Point we find that $CDEF$ is cyclic.

[b]Claim:[/b] $M$ is the midpoint of $CD$.
[i]Proof.[/i] Since $M$ is the intersection of the perpendicular bisectors of $DE$ and $CF$, we see that it is the circumcenter of $(CDEF)$. From here, it immediately follows that $M$ is the midpoint of $CD$, as desired. $\square$

Finally, note that $MC=\tfrac38 AC$, so $AM=\tfrac58 AC$, which gives $\tfrac{AM}{MC}=\tfrac53$. $\blacksquare$

Headsolving without coherency [verbatim]:D[/verbatim]

Write $AC\cdot AD=AB^2=AE\cdot AF$ so that $CDEF$ is cyclic. It follows that $M$ is the center with $MC=MD$, whence it's clear that $\tfrac{AM}{MC}=\tfrac{5}{3}$. $\square$

[asy]
            size(250,250);
            draw(circle((0,0),5)); draw((-4,3)--(4,3)); draw(circle((0,0),3));

            label("$A$", (-4,3), NW); filldraw(circle((-4,3),0.05),black);
            label("$B$", (0,3), N); filldraw(circle((0,3),0.05),black);
            label("$C$", (4,3), NE); filldraw(circle((4,3),0.05),black);
            label("$D$", (-2,3), NW); filldraw(circle((-2,3),0.05),black);
            label("$M$", (1,3), N); filldraw(circle((1,3),0.05),black);

            label("$E$", (-1.918677,2.306225775), W); filldraw(circle((-1.918677,2.306225775),0.05),black);
            label("$F$", (2.918677,0.6937742252), SE); filldraw(circle((2.918677,0.6937742252),0.05),black);

            draw(circle((1,3),3)); draw((-4,3)--(2.918677,0.6937742252));
[/asy]

I claim that the ratio $\frac{AM}{MC}$ is always $\frac{5}{3}$.

[color=#00f][b]1. $B$ is the midpoint of $AC$.[/b][/color]
Note that since $AC$ is tangent to $\mathcal C_2$, we have that $\angle OBA=90$. However, since $AC$ is also a chord of $\mathcal C_1$, which has center $O$, this also means that $B$ is the midpoint of $AC$.

[color=#00f][b]2. (Power of a Point Converse) $CDEF$ is cyclic.[/b][/color]
Note that the power of $A$ with respect to $\mathcal C_2$ is
\[AB^2=BC^2=AD(AC),\]
since by \textbf{(1)}, $B$ is the midpoint of $AC$, and by problem conditions, $D$ is the midpoint of $AB$. However, note by Power of a Point on $A$ with respect to $\mathcal C_2$, we have
\[AB^2=AE(AF) \iff AE(AF)=AD(AC).\]
Finally, using the converse of Power of a Point, we have that $CDEF$ must be cyclic.

[color=#00f][b]2. (Similar Triangles) $CDEF$ is cyclic.[/b][/color]
Similarly, as in the first proof of \textbf{(2)}, we find that
\[AE(AF)=AC(AD),\]
which actually gives us that
\[\frac{AD}{AF}=\frac{AE}{AC},\]
meaning that by SAS similarity, $\triangle ADE \sim \triangle AFC$. Through angle chasing, we then have that
\[\angle AFC + \angle EDC = \angle AFC + (180-\angle ADE) = \angle AFC + (180-\angle AFC) = 180,\]
meaning that $CDEF$ is cyclic.

[color=#00f][b]3. (Finishing) $M$ is the midpoint of $CD$.[/b][/color]
Note that since $CDEF$ is cyclic, and the perpendicular bisectors of $DE$ and $CF$ intersect at $M$, $M$ must be the circumcenter of $(CDEF)$. However, since $M$ is on $CD$, this means that $M$ must be the midpoint of $CD$. Therefore, since $D$ is the midpoint of $AB$, and $B$ is the midpoint of $AC$, we find that $\frac{AM}{MC}=\frac{5}{3}$, finishing the problem.

Let the common centre be $O$ and the radius of $C_1$ be $R_1$ and of $C_2$ be $R_2$.
Now from PoP we have: $AB^2 = AO^2-R_2^2 = R_1^2-R_2^2 = CO^2-R_2^2 = CB^2$.
So: $AB=BC$.
Again from a PoP: $AB.AC=2AB.\frac{AC}{2}=AB^2=AE.AF$. $ \Leftrightarrow DEFC$ is cyclic.
Now we have that $M$ is the midpoint of $DC=3AD$, so $AM=\frac{5}{2}AD$ and because of $AC=4AD$, $MC=\frac{3}{2}AD$.
So: $\frac{AM}{MC}=\frac{5}{3}$

We first claim that $CFED$ is cyclic. This is because

$AB^2 = AE \cdot AF \Rightarrow AD \cdot AC = AE \cdot AF \Rightarrow \triangle ACF \sim \triangle AED$
This proves our claim. From the given perpendicularity constraints we can see that $M$ is the circumcenter of $CFED$. This gives our answer of $\boxed{\frac{5}{3}}$ . $\blacksquare$

Notice that

\[AE \cdot AF = AB^2=AB \cdot BC = AD \cdot AC\]

implying that $CDEF$ is cyclic. Hence, it is clear that $M$ must be the midpoint of $CD$ at which point we easily find the ratio to be $\boxed{\frac{5}{3}}$.

[Refer to #48 for a diagram, I was too lazy to make one]

From Power of a point, $AD \cdot AC = AB^2 = AE \cdot AF$, so quadrilateral $CDEF$ is cyclic. Point $M$ must be the circumcenter of $CDEF$, so $M$ is the midpoint of $CD$, and the ratio $AM/MC = \frac{5}{3}$.

[b][color=#00f]Claim:[/color][/b] $DEFC$ is cyclic

[i]Proof:[/i]
By PoP $$AE \cdot AF=AB^2=2AB \cdot \tfrac{AB}{2}=AD \cdot AC$$
hence  $DEFC$ is cyclic. $\square$
Since $M$ is the intersection of the perpendicular bisectors of $DE$, and $CF$, $M$ is the center of $(DEFC)$.
So:
$$MC=\frac{AC-AD}{2}=\frac{AC-\tfrac{1}{2}AB}{2}=\frac{AC-\tfrac{3}{4}AC}{2}=\frac{3}{8}AC.$$
$$\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\tfrac{3}{8}AC}{\tfrac{3}{8}AC}=\frac{\tfrac{5}{8}AC}{\tfrac{3}{8}AC}=\frac{5}{3}.$$

Note $B$ is the midpoint of $\overline{AC}$ so \[AF\cdot AE=AB^2=(\tfrac{1}{2}AB)(2AB)=AD\cdot AC\] and $CDFE$ is cyclic. Hence, $M$ is the center of this circle and thus the midpoint of $\overline{CD}$. WLOG if $AD=2$, then $DB=2$ so $BC=4$ so $CM=3$ so $AM/CM=5/3$. $\square$

By power of a point $AE*AF=AB^2=AD*AC$, so $DEFC$ is cyclic.
Because $M$ is the intersection of the perpendicular bisectors of $DE$ and $FC$, $M$ is the center of the circumcircle, so $MD=MC$
Let $AD=a, BM=x$
$MC=MD=a+x$, so $MC+MB=a+2x=2a$, so $x=\frac{a}{2}$. Then $\frac{AM}{MC}=\frac{2.5a}{1.5a}=\boxed{\frac{5}{3}}$

By Power of a Point with respect to $\mathcal{C}_2,$ we get $$AB^2 = AE \cdot AF.$$

Then, since $D$ is the midpoint of $\overline{AB},$ it follows that $$AD \cdot AC = AB^2 \implies AD \cdot AC = AE \cdot AF.$$

Hence, by the converse of Power of Point, quadrilateral $CDEF$ is cyclic.

Since the perpendicular bisectors of $\overline{DE}$ and $\overline{CF}$ intersect at $M,$ this tells us that the circumcenter of $CDEF$ is $M,$ so $M$ is on the midpoint of $\overline{CD}.$

Therefore, if we let $AD=x,$ we then get
\begin{align*}
CD=3x &\implies DM=MC=1.5x \\
&\implies AM=2.5x \\
&\implies \frac{AM}{MC} = \frac{2.5x}{1.5x} = \boxed{\frac{5}{3}}.
\end{align*}
$\square$

[asy]
draw(circle((2,1),1));
draw(circle((2,1), sqrt(5)));
pair O=(2,1), A=(0,0), B=(2,0), C=(4,0), D=(1,0), M=(5/2,0);
draw(A--C);
dot(D);
dot(B);
dot(M);
dot(O);
dot(A);
dot(C);
label("$A$", A, SW);
label("$B$", B, S);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, S);
label("$O$", O, S);
pair E=(2-2sqrt(5)/5, 1-sqrt(5)/5), F=(2+2sqrt(5)/5, 1+sqrt(5)/5);
dot(E);
dot(F);
label("$E$", E, NW);
label("$F$", F, NE);
draw(A--F);
pair G=D/2+E/2, H=F/2+C/2;
draw(G--M--H);
draw(D--E);
draw(F--C);

[/asy]
We claim the answer is $\boxed{\dfrac{5}{3}}.$

Let $O$ be the center of both $C_2$ and $C_1.$ Since $AB$ is tangent to $C_2,$ $\angle OBA=90^{\circ},$ so $AC$ is bisected by $OB$ and $AB=BC.$ By power of a point, $AE\cdot AF=AB^2=\dfrac{AB}{2}\cdot 2AB=AD\cdot AC,$ so $DEFC$ is cyclic. Note that the perpendicular bisectors of chords $DE$ and $FC$ in the circumcircle of $DEC$ intersect at the circumcenter of $DEC,$ so $M$ is the circumcenter of $DEC$ and $DM=CM.$ As such, $\dfrac{AM}{MC}=\dfrac{AD+DM}{MC}=\dfrac{\left(AD+\dfrac{DC}{2}\right)}{\left(\dfrac{DC}{2}\right)}=\dfrac{\left(AD+\dfrac{3AD}{2}\right)}{\left(\dfrac{3AD}{2}\right)}=\dfrac{5}{3},$ as desired. $\Box$


let $ b_{n}\equal{}a_{n}\minus{}1$, so $ b_{n\plus{}1}\equal{}\frac{b_{n\minus{}1}}{2n}$ so $ b_{n}\rightarrow 0$ so $ a_{n}\rightarrow 1$.


what is the different betwin molal and molar?

[quote="g1sk8terz"]did u mean what is the diff between molal and molar?

well molal is mols of solute over kg of solvent, while molar is mols of solute over liters of solution...[/quote]

thanks,,,

the mistake edited. ;)


We have from the given that:
P(Q(Q(x)))=Q(Q(Q(x))) , so polYnomials P and Q take for a sufficiet big nr of values(as many as Q(Q(x))) take the same value, so they must be identical!
I hope I've explained it well enough :) ,at least the idee should be clear!

I think a problem from a recent Romanian TST stated that $P=Q$ if $P\circ P=Q\circ Q$, but I'm not sure right now. I'd have to check.

[quote="Xixas"]Myth, your polynomials do not satisfy the second condition.[/quote]
It was counterexample for grobber's post.

[quote="Myth"][quote="grobber"]$P$ and $Q$ were monic complex polynomials in that problem.[/quote]
It is obvious statement for real polynomials, but complex... I need additional reasoning.[/quote]

Yes, for real polynomials you don't need any messy coefficient identification: you can do it by some analysis, studying what happens when $x\to\infty$. I think that's what I did when I first saw the problem, because I didn't read the text properly and missed "complex polynomials". On the other hand, I think we have to resort to identifying the coefficients if $P,Q$ are complex, and I don't like such problems, so I won't try to solve it :).

Actually, at this time I am writing solution, which uses behavour when $x\to\infty$.

[quote="Xixas"]P(x) and Q(x) are polynomials of positive degree such that for all x P(P(x))=Q(Q(x)) and P(P(P(x)))=Q(Q(Q(x))). Does this necessarily mean that P(x)=Q(x)?[/quote]
From $P(P(P(x)))=Q(Q(Q(x)))$ we immediately obtain $P$ and $Q$ have the same leading coefficient.
Suppose $P(x)\neq Q(x)$, then $P(x)=a_nx^n+...+a_kx^k+p_{k-1}x^{k-1}+...+p_0$ and $P(x)=a_nx^n+...+a_kx^k+q_{k-1}x^{k-1}+...+q_0$, where $p_{k-1}>q_{k-1}$ (for example).
0) $n=1$. Any questions?
1) $a>0$. Then there is $M$ s.t. $P(x)$ and $Q(x)$ increase on $[M,+\infty]$, $P(M)>M$, $Q(M)>M$, $P(x)>Q(x)$ on $[M,+\infty]$. We see that $P(P(x))>Q(P(x))>Q(Q(x))$ on $[M,+\infty]$.
2) $a<0$. Almost the same argumentation, but a bit more boring. :?

[quote="Xixas"]P(x) and Q(x) are polynomials of positive degree such that for all x P(P(x))=Q(Q(x)) and P(P(P(x)))=Q(Q(Q(x))). Does this necessarily mean that P(x)=Q(x)?[/quote]

Let $S$ be the image of $P(P(x))$. Since $\deg P > 0$, $S$ has infinitely many elements.

Since $P(P(P(x)))=Q(Q(Q(x)))$, we see that if $y \in S$, then $P(y) = Q(y)$. Two polynomials agree on infinitely many values imply that they're identical. Hence $P = Q$.

It is strange. What if we remove condition $P(P(x))=Q(Q(x))$.

[quote="grobber"]Yes, sorry about that, $P$ and $Q$ were monic complex polynomials in that problem. It's from a TST in $2000$.[/quote]Yes I remember that problem. Nice one, solved it quickly, with equalizing coefficients ;)


How do we solve this with no use of calculators or logs
4^n < 2^100 + 3^100 < 4^(n+1)

The trick is clearly the 3^100, right?  So,[hide] you're going to want to try and estimate something like (3/4)^n and see if you can use that.  Because what's going to happen is that the 2^100/4^n is probably going to be very, very, very small, so you may not even have to think about it.[/hide]

Maybe you can use the fact that [hide]44 is just a little larger than 35[/hide].

[hide]An idea perhaps :

2^100 + 3^300 = 4^50 * (1 + 4^25 * (1 + 1/8)^50))

So we can can now say that n >= 75.

(1+x)^50 > f(x)=1 + 50x + 1225x (binomial)

f(1/8) is still tractable without a computer (<cheat>But of course I used one since I'm at home and noone is looking at me  </cheat>). f(1/8) = 1689/64 ~26 > 4^2.

So n >= 77.

Since C(50,n) (1/8)^n quickly becomes tiny, 77 is not to far from the actual value.[/hide]

[hide] We must have 4^n-2^100<3^100 and n is as big as possible. factor 2^100 out, we have (4^(n-50))-1<(3/2)^100. Note that (3/2)^4 is very close to 5. sub that in, (4^(n-50))-1<5^25+S, S=(3/2)^100-5^25. Obviously, n>75. If n=80, then 4^30>5^25 because 4^6>5^5. 80 is very close to what the answer will be because S is not going to be a small number. I couldn't get the exact answer although looks like n=79 or 78.  [/hide]

 I edited my solution, I just figured out how to get the exact answer. [hide]4^6=5^5+971. (3125+997)^5-1>3125+S, because 3125+S=(5+1/16)^25, or (3125+625(5)/16+125(10)/256+25(10)/4096+..)^5 <(but very close) to (3125+201)^5. Let's say n=79, then ((3125+997)^5)/4 should be<(3125+201)^5, or (4096/3326)^5<4.(4096/326)^5<(1+1/4)^5 and (1+1/4)^5<4, which is true. so 79 works[/hide]

Is it possible to get an exact answer without using a calculator? No one seemed have gotten an exact answer without using calculator or logs.I think the answer is [hide]79      [/hide]    by using logs and calculator.

I gave you folks a hint, but you didn't listen.    So I'll finish it off.

I will use the inequality (1 + x)n :ge: 1 + nx for x :ge: -1 and n a nonnegative integer.  The proof of that inequality is by induction on n.

The rest is in spoiler: [hide]

We have 480 / 3100 = (256/243)20 = (1 + 13/243)20 :ge: 1 + (13)(20)/243 = 1 + 260/243 > 2.  Thus 480 > 2 3100 > 2100 + 3100.  That's one direction.

We have 325 / 420 = (243/256)5 = (1 - 13/256)5 :ge: 1 - (13)(5)/256 = 1 - 65/256 = 191/256.  Thus 350 / 440 :ge: (191/256)2 = 36841/65536 > 1/2.  Thus 3100 / 480 > (1/2)2 = 1/4.  Hence 479 < 3100 < 2100 + 3100.  That's the second direction.

Thus, assuming that n must be an integer, we have n = 79.

[/hide]


Both $ \bigtriangledown$ and $ \oslash$ are operations in the set $ \{ \plus{} , \minus{} , \times , \div \}$, and $ \frac{12 \bigtriangledown 2}{9 \oslash 3} \equal{} 8$. What is the value of $ \frac{4 \bigtriangledown 3}{2 \oslash 5}$?

We find out that the triangle means multiplication and the circle means division. Therefore, answer:
4*3/(2/5)=4*3*5/2=[b]30[/b]


Does it bug anyone else how there are the same number of natural numbers as there are whole numbers (and rational numbers and integers and blah blah). I mean, I know the reasons and everything about one to one mappings but it just doesn't seem [i]right[/i] to me.. All these numbers, add on one more number and you get the same. Argh.

It used to bother me, but I have now accepted that arithmetic for infinities doesn't work quite the same way that it does for finite quantities. What really bothered me is that between any two rationals is an irrational, and between any two irrationals is a rational, but even though any transfinite quantity +/-1 is itself, there are more irrationals than rationals. But now I accept these things, since I do know their proofs, and you can't argue with a proper proof.

Question: Are there just as many real numbers as there are irrational numbers?

I would suppose so. Seeing as the number of real numbers is a "bigger" infinity than the number of rational numbers, subtracting a "smaller" infinity from the "bigger" infinity should leave the "bigger" infinity. Just like subtracting one from infinity leaves infinity.

Speaking of how [color=#0099FF]infinity - 1 = infinity[/color], then

Shouldn't [color=#0099FF]1/infinity[/color] be 0?

Shouldn't [color=#0099FF]infinity/infinity[/color] be 1?

And what do you do with [color=#0099FF]infinity/0[/color]?

Somehow infinite numbers seem a little more real than they used to. :mrgreen:

Shouldn't infinity/infinity be 1?

No, consider 2x/x. For any x except zero, that will equal two, since it's a horizontal line. But, if you plug in x = infinity, you get inf./inf. = 2. Inf./inf. can equal anything, depending on how you got the infinities.

infinity/0
Either +inf. or -inf., depending on how you got it.

There's some junk with limits that will help with your questions regarding (infinity/infinity) or 0/0. As for the "equal number", the accepted method that helped me "comprehend" the situation is by establishing a bijection. For example, with whole numbers and natural numbers, if you list them "all" out, you have:

Wholes:  0, 1, 2, 3, 4, ...
Naturals: 1, 2, 3, 4, 5, ...

Notice how 0 and 1 pair up, 1 and 2, 2, and 3, and so on. This is a bijection (one to one correspondence), and this can continue forever, so there are an equal number of wholes and naturals. The same "logic" can be applied to natural numbers and multiples of some integer as well.

1, 2, 3, 4, 5, ...
2, 4, 6, 8, 10, ...

[quote="Bookworm"]How do you figure out what infinity/0 is?  I thought you couldn't divide by zero.[/quote]

Limits...take the case y = 1/x. As you get closer to 0 for x, y approaches +/- infinity. So when you divide by 0, you get +/- infinity in any case. I think. Any ideas?

Basically (read: very informally) infinity/0 = infinity * (1/0) = infinity * infinity = infinity. The +/- thing is a bit weird, but we can reasonably say that infinity=-infinity. In complex analysis, we usually let all infinities be the same point (for example on a Riemann sphere, it's the north pole).

Yeah. Well basically if we have a fraction x/y and increase x the fraction increases, and decrease y it increase. Either way as x->oo or y->0 it will go to infinity.

When we get to things like 0/0 and infinity/infinity, you can use L'Hopitals rule, the limit equals the limit of the thing where you differentiate top and bottom separately. Eg lim x->0 of (sin x)/x = lim x->0 (cos x)/1 = 1.
(Just learnt that this year so felt like telling you all :))

L'hopital's rule also works for 0/0, and other forms which are equivalent to one of those two.

BTW, here's an example of a really weird problem. Let's say you choose one integer out of the set of all integers. What is the probability that it's odd? Common sense tells you that it is 1/2. But remember, probability can be defined as <sample space>/<total space>, or the number of desired possibilities over the total number of possibilities. In this case, that would be <# of odd integers>/<total # of integers>, and since those two sets are the same size, the answer should be 1! (read as 1 in a very surprised way, not as 1 factorial)

[quote="MysticTerminator"]BTW, here's an example of a really weird problem. Let's say you choose one integer out of the set of all integers. What is the probability that it's odd? Common sense tells you that it is 1/2. But remember, probability can be defined as <sample space>/<total space>, or the number of desired possibilities over the total number of possibilities. In this case, that would be <# of odd integers>/<total # of integers>, and since those two sets are the same size, the answer should be 1! (read as 1 in a very surprised way, not as 1 factorial)[/quote]

I don't think that's right. Just because there's a bijection between the two doesn't mean that the quotient is the same. It's still lim_{x\rightarrow \infty} {x/(2x)}=1/2.

[quote="Bookworm"]Thanks for answering my question.
Sorry if I ask a lot of questions, but what do you mean by "the limit equals the limit of the thing where you differentiate top and bottom separately"?[/quote]

Sorry I was trying to say it as informally as possible, probably got too informal. Basically lim (x->a) f(x)/g(x)  =  lim (x->a) f'(x)/g'(x).

As for the odd number probability thing, I think it could be equal to 1 I'm not sure. I know there are examples where even though its obviously not 1, it does equal 1. For example, whats the probability a random positive integer has the digit 3 in it? That turns out to be 1, I think.

Whoa... don't quite follow you there about every number having a three in it. Does that use the same sort of logic that I used in my "ever number is an odd number" thing?

BTW, about L'Hopital's rule, does anyone have a good way to visualize it, or "intuitively" understand it. I mean, it doesn't make any sense that if the two functions are both 0 or infinity at the point where you are taking the limit, their ratio will be equal to their slopes at that point. Should I just accept it and move on, or does anyone have an intuitive explanation of it?

Oh yeah,
As an example of L'Hopital's rule, in case it isn't clear, let's say we take the limit as x goes to 3 of (x^2-9)/(x^2-2x-3). If you just plug in x=3, that will be 0/0. Now, we can do that the nice little way of canceling out a factor of (x-3) on both the numerator and denominator, leaving (x+3)/(x+1). Pluggin in x=3 now gives 1.5.
-OR-
We could differentiate the top and bottom to give 2x/(2x-2)=x/(x-1), and the plug in x=3 to get the same answer. Of course, you don't usually use L'hopital's rule on things you can do other ways, I just wanted to show it worked.

[quote="TripleM"]For example, whats the probability a random positive integer has the digit 3 in it? That turns out to be 1, I think.[/quote]

Yes, that's definitely 1, since if you pick a random positive integer, chances are it's going to be really large and have many digits. Basically, the probability that an n-digit number has a 3 in it is 1-(9/10)^n, and the average number of digits in a random positive integer is infinitity, so when n goes to infinity we get 1. That's really handwavy, but that's the general idea behind problems like that.

really? so it's not [u]exactly[/u] one? b/c if u let handwavy things start to get by you, you could say 0 is about 1, and then you could get all sorts of results, like your school is actually a giant carrot.

In this case, ``handwavy'' means ``correct, but probably not good enough to go in a proof for, say, the USAMO.''

It's .999bar. Call that n.
10n = 9.9999bar
-  n =   .9999bar
 9n =  9
   n = 1

So it's 1, not just "about 1."

Yes, I agree. But in the case of the problem with the 3s, it is slightly more difficult since you have to do work with the expected value of the number of digits. This is infinity with uniform distribution, but I think you have to use umbral calculus to prove the result rigorously. I don't think it's really that easy, but it's not too hard to give an intuitive ``proof'' of the desired result.

Yeah I think it comes down to how you can pick a random integer. Which is basically impossible..

It's impossible with uniform distribution. It's not impossible with, say, the Poisson Distribution.

There are two general themes here. One, is not to confuse ordinal numbers with cardinal numbers. For example, the number 4 is different from a *set* with four members. In one instance 4 is in itself an algebraic object, and in the other it is just describing the cardinality of the set. This was one of the pitfalls of early set theory when Whitehead tried to define natural numbers by the equivalence class of all sets having that cardinality. There are some serious problems with doing that. In the 70s however, John Conway came up with a system of incorporating transfinite numbers into the system of natural numbers with what he called surreal numbers. There's a very good exposition of this in his book [u]On Numbers and Games[/u]. I highly recommend it.

The other theme I see here is to not confuse a sequence of numbers with it's limit. This is best illustrated by Zeno's Paradox, but in the opposite sort of way than some of you are doing it. For example with the 3s digit problem, you are applying properties of the finite sample spaces (i.e. numbers having n digits), to the limit space (numbers having a countably infinite number of digits). But there's no justification for being able to do that. There are many examples in mathematics where something is true for each finite case in a sequence, but not true for the limit case. The best you can say about that 3s digit problem is that the probability can be made as arbitrarily close to 1 as you want by increasing the number of digits. But once you talk about an infinite number of digits you have to redefine the probability distribution on your sample space.[/u]

Precisely. See, so even though if you let n=infinity and the probability is one, that won't work for like a two digit number (which is possible) (even though choosing a two digit number technically has probability 90/infinity=0). I suppose some guy who (again) had two much spare time showed that there were an equal number of the numbers that had threes in them as there were numbers, or something like that.

Another cute example of the pitfals of infinity is something I found on the math.sci newsgroup. Suppose you have the following infinite sequence 3 31 314 3141 31415 314159... where the nth term in the suquence is the truncation of the first n digits of pi. Then is pi contained in this sequence?

[quote="Ex274513"]Wow...[color=#3399CC]infinity[/color] is a hard idea to comprehend.

It is an idea, right? :mrgreen:[/quote]

The reason why people have so much trouble with the idea of infinity is because we don't deal with it on an everyday basis. In our everyday lives we only deal with the finite; so our intuitions are bad when it comes to infinity. This is why Cantor, who first developed much of our modern understanding of infinity, was derided by even well established mathematicians of his day.

But this is where the strength of mathematical thinking comes through. It forces you to put aside your everyday intuitions, which can be misleading, and to use the mathematical tools which you know to look at things in a different light. Then once you have done that enoough you build a new intuition, much the same was as we have done with complex numbers.

In response to gauss's example, I say... no. The nth term is just a very good approximation to Pi* 10^(n-1). The [u]limit[/u] of the sequence will have Pi in it (actually it won't b/c it goes to infinity, but if you created a new sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, etc., the limit of that would have Pi in it.) The terms themselves will just be approximations.

Sort of thinking in calculus terms here: the [u]limit[/u] of (x^2-9)/(x-3) at 3 will be 6, meaning that as you get closer to 3, you can make the value get as close to 6 as you want, but the value itself will never be 6.


Let $a$ and $c$ be fixed positive numbers. For each real number $t$ let $(x_t, y_t)$ be the vertex of the parabola $y = ax^2+bx+c$. If the set of vertices $(x_t, y_t)$ for all real values of $t$ is graphed in the plane, the graph is

A. a straight line
B. a parabola
C. part, but not all, of a parabola
D. one branch of a hyperbola
E. None of these

[hide]we know that the vertex of this parabola is at the point $(x_t,y_t)=\left(\frac{-b}{2a},y_t\right)$.
we then plug this into the equation and find that the set of verticies is the equation
$\displaystyle y_t-c=\left(\frac{b^2}{4a}-\frac{b^2}{2a}\right)= \frac{-b^2}{4a}$
since the values of $a$ and $c$ are fixed we can disregard them. this basically leaves the equation $y_t=\left(\frac{-1}{4}\right)(b^2)$ whcih is a parabola.
since the equation of the verticies is $\displaystyle\left(\frac{-b}{2},\frac{-b^2}{4}\right)$

did i do this correctly of have i made a mistake somewhere?  :oops:[/hide]

You did it right, I also got a [hide="answer"]parabola[/hide].


algebra
polynomial
Vieta
inequalities
algorithm
number theory proposed
number theory

find all positive integers such that $ \frac{a^2\plus{}b^2\plus{}a\plus{}b\plus{}1}{ab}$ is an integer

[hide]there is trivial solution (1,1)
let $ k \equal{} \frac {a^2 \plus{} b^2 \plus{} a \plus{} b \plus{} 1}{ab}$  fix k
\[ a^2 \minus{} a(kb \minus{} 1) \plus{} b^2 \plus{} b \plus{} 1 \equal{} 0\]
if we have a $ (x,y)$ couple satisfying that condition we can suppose that $ x > y$ (because x cannot be equal to y unless they are not equal to 1).
so if $ (x,y)$ is a solution $ (y,ky \minus{} 1 \minus{} x)$ satisfies too. We have $ ky \minus{} 1 \minus{} x > 0$ because $ (ky \minus{} 1 \minus{} x).x \equal{} y^2 \plus{} y \plus{} 1$
so we have another couple $ (y,ky \minus{} x \minus{} 1)$   we have $ x > y$ there our aim is to prove $ y > ky \minus{} x \minus{} 1$
\[ ky \minus{} x \minus{} 1 < y < \equal{} > kxy \minus{} x < yx \plus{} x^2\]
$ kxy \equal{} x^2 \plus{} y^2 \plus{} x \plus{} y \plus{} 1$ so this inequality is equal with $ yx > y^2 \plus{} y \plus{} 1$
we have two case here if $ x \geq y \plus{} 2$ in every step of algorithm  we can always reach a smaller solution and it should stop at (a,1)
where a=3 or a=1.

case 2 : if x=y+1 somewhere, it is clear by parity there will be no solution.
all of the cases we observe that k=5 and with this algorithm we can reach infinite number of solutions[/hide]
not a solution

I think there are infinitely many solutions. I've got $ (1,3),(3,13),(13,61),\ldots$. From one solution, we can generate another.. I don't know how to write the general form, but we certainly cannot use Vieta jumping.

edit: we use vieta jumping for proving if this number is an integer it is 5. it might help to find recursive formula

with $ x_1 \equal{} 1$ and $ x_2 \equal{} 1$ all solutions are $ (x_n,x_{n\plus{}1})$
$ x_{n \plus{} 2} \equal{} 5.x_{n \plus{} 1} \minus{} x_n \minus{} 1$

[quote="stvs_f"]sorry,Where can i find some information about Vieta jumping? :!:[/quote]

Maybe http://www.google.fr/search?q=vieta+jumping

 :P

[quote="stvs_f"]sorry,Where can i find some information about Vieta jumping? :!:[/quote]

Yimin Ge from austria wrote a great paper about it. [url]http://www.georgmohr.dk/tr/tr09taltvieta.pdf[/url]


How do i obtain the logarithm of an expression like

x = (a^m - b^m) / (a^n - b^n) also m > n (don't think this is significant)

[hide]

I tried

mlog(a/b) - nlog(a/b) .. know this is wrong cant just figure out why :(

[/hide]

Let me re-type this:

$ x\equal{}\frac{a^{m}\minus{}b^{m}}{a^{n}\minus{}b^{n}}$

When you say "obtain the logarithm", what do you mean?  Now to take the log of the expression, or express $ x$ using logs?

[quote="JRav"]Let me re-type this:

$ x \equal{} \frac {a^{m} \minus{} b^{m}}{a^{n} \minus{} b^{n}}$

When you say "obtain the logarithm", what do you mean?  Now to take the log of the expression, or express $ x$ using logs?[/quote]

Okay to be more clear. I need to obtain log x

hello, it is clear that
$ \log(x)\equal{}\log\left(\frac{a^m\minus{}b^m}{a^n\minus{}b^n}\right)\equal{}\log(a^m\minus{}b^m)\minus{}\log(a^n\minus{}b^n)$ with $ a^m\minus{}b^m>0$ and $ a^n\minus{}b^n>0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, it is clear that
$ \log(x) \equal{} \log\left(\frac {a^m \minus{} b^m}{a^n \minus{} b^n}\right) \equal{} \log(a^m \minus{} b^m) \minus{} \log(a^n \minus{} b^n)$ with $ a^m \minus{} b^m > 0$ and $ a^n \minus{} b^n > 0$.
Sonnhard.[/quote]

I found out my mistake!!

Btw, Is there any way to solve/Simplify further $ \log\left(a^m \minus{} b^m\right)$ in general?

[quote="solveitall"]
Btw, Is there any way to solve/Simplify further $ \log\left(a^m \minus{} b^m\right)$ in general?[/quote]

$ \log\left(a^m \minus{} b^m\right)$

=$ \frac{\log a^m}{\log b^m}$

=$ \frac{m\log a}{m\log b}$

Any mistakes?

Yes, there is a fundamental mistake. You're making your own rule for logarithms: difference of logarithms is a quotient of arguments, but the difference of arguments ISN'T quotient of logarithms.

$ \log_2 (32 \minus{} 16) \equal{} \frac {\log_2 32}{\log_2 16} \equal{} \frac {5}{4}$

but, $ \log_2 (32 \minus{} 16) \equal{} 4$

[quote="Euclid_great"][quote="solveitall"]
Btw, Is there any way to solve/Simplify further $ \log\left(a^m \minus{} b^m\right)$ in general?[/quote]

$ \log\left(a^m \minus{} b^m\right)$

=$ \frac {\log a^m}{\log b^m}$

=$ \frac {m\log a}{m\log b}$

Any mistakes?[/quote]

Yes there is. This is the same trap which i too fell into before i realised. Difference inside cant be expressed in the form of $ a/b$

Right now the only way out to simplify this i can think of this expressing

$ \left(a^m \minus{} b^m\right) as \left(a \minus{} b\right) \left(a^{n\minus{}1}...\right)$ :(

Ah yes, a very fundamental mistake I might want to highlight in a problem I will make in the TPP (see link below).

It's not worth it to try to simplify some binomial expansion.  Just leave it as it is.


H,A,T,B,O, and X are digits from 0 to 9. Repeats are allowed.HATBOX and BOXHAT are six digit numbers.
$ 9\times HATBOX \equal{} 4\times BOXHAT$
Find a HATBOX and BOXHAT for which this is true.

[quote="Icy"]I think I might be on to something here, but I can't take it to the end...

Assign to $ x$ the 3-digit number $ HAT$, and $ BOX$ to $ y$.

So...

$ 9(1000x\plus{}y) \equal{} 4(1000y\plus{}x)$

$ \frac{9}{4}\equal{}\frac{1000y\plus{}x}{1000x\plus{}y}$...so...taking the fraction into two equations...with k being some number

$ 9k \equal{} 1000y\plus{}x$ and $ 4k \equal{} 1000x\plus{}y$, and adding them together...

$ 13k \equal{} 1001y\plus{}1001x$, so $ x\plus{}y \equal{}\frac{13k}{1001}$

This seems promising to me, except for the fact that I don't have an answer...and I don't know how to proceed from here. And I guess all the digits could just be 0, but I think there are more solutions.

EDIT: Well...what if $ k$ was just assigned some multiple of 1001 such as...100100...then the numerator would be 1300, which could be the sum of many different 3 digit numbers... I extremely doubt it's right though, as it doesn't make too much sense on inspection...[/quote]

 13 is a factor of 1001. :wink:

So...$ \frac{1001}{13}\equal{}77$...but I don't know where to go from there....
$ k\equal{}77(x\plus{}y)$, but I think I'm still missing something obvious that'll give an answer.

[quote="Icy"]I think I might be on to something here, but I can't take it to the end...

Assign to $ x$ the 3-digit number $ HAT$, and $ BOX$ to $ y$.

So...

$ 9(1000x\plus{}y) \equal{} 4(1000y\plus{}x)$

$ \frac{9}{4}\equal{}\frac{1000y\plus{}x}{1000x\plus{}y}$...so...taking the fraction into two equations...with k being some number

$ 9k \equal{} 1000y\plus{}x$ and $ 4k \equal{} 1000x\plus{}y$, and adding them together...

$ 13k \equal{} 1001y\plus{}1001x$, so $ x\plus{}y \equal{}\frac{13k}{1001}$

This seems promising to me, except for the fact that I don't have an answer...and I don't know how to proceed from here. And I guess all the digits could just be 0, but I think there are more solutions.

EDIT: Well...what if $ k$ was just assigned some multiple of 1001 such as...100100...then the numerator would be 1300, which could be the sum of many different 3 digit numbers... I extremely doubt it's right though, as it doesn't make too much sense on inspection...[/quote]

$ \frac{9}{4}\equal{}\frac{1000y\plus{}x}{1000x\plus{}y}$
$ 9(1000x\plus{}y)\equal{}4(1000y\plus{}x)$
$ 9000x\plus{}9y\equal{}4000y\plus{}4x$
$ 8996x\equal{}3991y$
$ 692x\equal{}307y$
so then if $ x\equal{}307$ and $ y\equal{}692$
$ 2769228\equal{}2769228$

If 692 and 307 are relatively prime (haven't checked, but I think so) then there are no more solutions.

Well 307 is prime and 692 is not a multiple of 307 so I am pretty sure 307 and 692 are relatively prime :P .


compute the limit:

lim(x-->0) x^4 (1+2+..+[1/x^2])
where [a]=a-{a}

thank you very much! :)

Let S(x) = 1+2+...+[1/x 2 ].

Assume that x2 < 1.

Let k be the integer such that 1/(k+1) < 1/x 2   \leq  1/k.
Then k/(2(k+1))  \leq  x4 S(x)  \leq (k+1)/(2k).

Since k tends to +oo as x tends to 0, we deduce that lim x 4 S(x) = 1/2.

Pierre.

1+2+..+[1/x^2]
what are the terms after 2, the general term in the previous sum?

but
[quote]Let k be the integer such that 1/(k+1) < 1/x  1/k. [/quote]
since x tends to 0=>1/x tends to oo results that it can't be smaller than 1/k with k integer
i think u meant k<=1/x<k+1 with k integer
 :D
ain't that right?


algebra
polynomial
superior algebra
superior algebra unsolved

How do you show that $ \mathbb{F}_{p}\left[x\right]$ contains an irreducible polynomial of degree $ 2$ for each prime $ p$?

I know that there is a theorem generalizing this to polynomials of all degrees, but is there a simple proof for this specific case that doesn't require invoking the general theorem?

Let $ p$ be odd. The map $ x \mapsto x^2$ is not injective, thus not surjective, thus there is an $ a$ that isn't a square. Take $ x^2\minus{}a$.


integration
calculus
function
complex numbers
real analysis
real analysis unsolved

I hope this topic is not posted here yet.

Identify $ x!$ as $ \Gamma(x\plus{}1)$. Then, does the equality

$ \int_{\minus{}\infty}^{\infty} \binom{n}{x} \, dx \ \equal{} \ 2^n$

hold for a nonnegative real number $ n$? (where $ \binom{n}{x} \equal{} \frac{n!}{x!(n\minus{}x)!}$ is the binomial coefficient)

I proved that it holds for $ n$ nonnegative integer, but the methid I used seems unable to be extended to the general case.

According to numerical tests the formula seems correct for $ n\in\mathbb{R^\plus{}}$.
A partial result: if $ \phi(n)\equal{}\int_{\minus{}\infty}^{\infty}\binom{n}{x}dx$ for $ n>0$, then $ \phi(n\plus{}1)\equal{}2\phi(n)$.
Proof: $ \forall{y}\in\mathbb{R},\phi(n)\equal{}\int_{\minus{}\infty}^{\infty}\binom{n}{x\plus{}y}dx$; moreover $ \forall{x,y}\in\mathbb{R},\binom{n}{x}\plus{}\binom{n}{x\plus{}1}\equal{}\binom{n\plus{}1}{x\plus{}1}$. Thus $ \phi(n)\plus{}\phi(n)\equal{}\phi(n\plus{}1)$.    QED

It is actually much better than "a partial result". Note that the identity is true for complex $ n$ with positive real part as well, so, when we multiply the integral by $ 2^{ \minus{} z}$, we get an analytic function that is $ 1$-periodic in the right half-plane and, therefore, can be extended to the entire plane. Now all we need to do is to show that this function grows not too fast when we go up the imaginary axis (any estimate short of $ e^{\pi|z|}$ will suffice) to conclude that it is actually constant. This reduces the problem to the [b]estimation[/b] of the integral instead of its [b]exact computation[/b]. I have no time to do it now but I hope it is not too difficult :).


Which word from the bottom  group belongs in the top group?

blast,paper,box,bank

juice,bag,cradle,carpet

why?

[quote="Ihatepie"][quote="mustafa"][hide]bag, because the first letter is either a b or an upside-down b[/hide][/quote]

ur answer is right but the reason isn't wat I was thinking of[/quote]What's your reason?

[quote="Ihatepie"]try to figure it out

[hide="the reason"]u can put the word sand before each word and get another word.[/hide][/quote]

that is creative..did u just think of that...or was that liek the actual answer? :D

i've heard problems like that b4 so I made up a new one.


topology
real analysis
real analysis unsolved

Hi guys,

I have been working on this problem and made some progress but now I am stuck.

Suppose A is an arbitrary indexing set, and C={B_a : a in A} is a collection of closed balls in R^n. Let E be the union of the elements of C and let D be the boundary of E. Prove D has measure zero.

Ok well the problem suggested to first assume that all the balls have radius at least one. I did that and managed to solve the problem, but now I am stuck because I am not sure how to use my case to solve the more general case. Any advice? P.S. This is a homework assignment so plz don't finish it for me, just give me a hint on how I can use my case to solve the more general one.

Thanks.

You've got a wrong problem here: you can easily construct a closed set of positive measure with empty interior. Its complement can be represented as a union of closed balls of positive radii. I'm afraid that your instructor merely forgot that the boundary of the union may be larger than the union of the boundaries :)


Can any one post problem ARML 2000-2008?

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201069[/url] for problems after 2003.


let a,b,c >0
a+b-c>0
b+c-a>0
c+a-b>0
Find min $ \sum \frac {a}{\sqrt{a^2 \plus{} 3bc}}$

[quote="fesha3t"]let a,b,c >0
a+b-c>0
b+c-a>0
c+a-b>0
Find min $ \sum \frac {a}{\sqrt {a^2 \plus{} 3bc}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163238


:?: ABCD is a convex quadrilateral. Diagonals AC and BD are joined. Given tht area of pentagon ABOCD = Ar(BOC). Now two points are located on BC ( P and Q) such tht OP||AB and OQ||DC.
Prove that : 2 Ar(POQ) = Ar(AOB) + Ar(COD).


A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards.  Two of the suits ($ \heartsuit$ and $ \diamondsuit$, called `hearts' and `diamonds') are red, the other two ($ \spadesuit$ and $ \clubsuit$, called `spades' and `clubs') are black.  The cards in the deck are placed in random order (usually by a process called `shuffling'). In how many different ways can we pick two red cards?  (Note: Order matters in the sense that drawing an ace of hearts followed by jack of diamonds is different than drawing a jack of diamonds followed by ace of hearts.)

Can anyone explain why we don't divide by 2 in this case

@above You only divide by 2 if order doesn't matter. However, the problem says that order matters, so you don't divide by 2.

[hide=sol]The first position has $26$ options, the second has $25$, so the answer is $26\cdot25=\boxed{650}$[/hide]
[hide=Why my answer is good]It is nice and quick. [/hide]


geometry
rectangle
search
combinatorics proposed
combinatorics

It is problem numbers six from one of the TST's of IRAN.
We have a polygon which can be coverd by $100$ rectangles but can not be covered by $99$ rectangles .
Prove that it can not be covered by $100$ triangles.

By covering I mean that all area of the shape must be covered and non of the rectangles shall cover each other.

I really felt the life with solving this:
Here are some things i have done ,but it is the most usuall approch :

1.Each side of this polygon is prependicular to the two other sides near it.
2.our dear polygon has even number of sides.
3.any polygon with $2n$ sides can be covered with n-1 triangle(here with $\frac{2*100}{2}-1$)

I remember seeing this problem before around here. I think Myth solved it, and it was posted by a Romanian member of the forum, but I forgot which one. Anyway, you have enough key words to search for it (Myth, polygons, triangles, etc.) :).

The last move for the bane of the polygon:

This polygon can not be covered by $99$ rectangles so $n-1 > 99$ .
So our dear polygon has at least $202$ sides.
So it has ,at least, $101\pi$ degree in its internal angles.
But i assure you $99$ trinagle can not provide that much angles to cover the polygon.

Well i didnt manage to find the one that grobber mentioned but i found a nice
topic somewhere from Myth to his very few friends:

$\text{Myth has a heart too}$

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=covering&t=1896[/url]


If  $x$  is  a positive integer  find te solutions of

$2^x\equiv 2x  (mod 3)$

$2^6 = 1\ mod \ 3$ so $2^x - 2 \cdot x$ is periodic with period at most $6$. Now it just the matter of 6 easy computations.

$x \in 1+6 \cdot \mathbb{N}$ or $x \in 2+6 \cdot \mathbb{N}$

It is a godd solution ,but not elementary at all. does ayone has an elementary solution?

Well, $2^x \equiv 2x$ is modulo $3$ equivalent with $(-1)^x \equiv -x$. Now consider two cases:

$x \equiv 0 \mod 2$, then $1\equiv -x \mod 3$, i.e. $x \equiv 2 \mod 3$. This has exactly one solution mod $6$ (Chinese remainder theorem); $x \equiv 2 \mod 6$.

$x \equiv 1 \mod 2$, then $-1 \equiv -x \mod 3$, i.e. $x \equiv 1 \mod 3$. And this would result in $x \equiv 1 \mod 6$.

So, the solution set is $S=\{x\ |\ x \equiv 1 \mod 6 \ \ \wedge \ \ x \equiv 2 \mod 6\}$


induction
inequalities
algebra unsolved
algebra

[quote="reason"]Let $ (w_{n})$ $ n\in{\mathbb{N}}$
$ w_{n} \equal{} \prod_{k \equal{} 1}^{n}(1 \plus{} \frac {1}{k^{3}})$
prove that:
$ (\forall{n}\in{\mathbb{N} \minus{} {0}})$  $ w_{n} < 3$[/quote]
Prove by induction that the inequality $ \prod_{k \equal{} 1}^{n}(1 \plus{} \frac {1}{k^{3}}) < 3 \minus{} \frac{1}{n}$ holds for every $ n > 1$.

Now try to prove the much stronger inequality :

 $ \prod_{k \equal{} 1}^{n}(1 \plus{} \frac {1}{k^{3}}) < \frac{49}{20}$

BTW,take a look here for some more interesting facts :

http://www.mathlinks.ro/viewtopic.php?t=276003


search
abstract algebra
superior algebra
superior algebra theorems

Does anybody know why Lagrange's Theorem is attributed to him?? After all Guass never used the concept of a group in his famous Disquisitiones Arithmeticae(1801), so Lagrange could have hardly formulated such a result himself. Please have sources to back up your information. Also, does anyone know how Cauchy himself( if he did  ), proved Cauchy's Theorem: that for any prime, p, dividing a finite group, the group has an element of order p. Did Cauchy really know about the Class Equation of a group??? I will be researching these questions myself, so I will let others know. If I should not be proposing such questions here, let me know also.

Mozart

Lagrange essentially proved his theorem for Galois groups. He did not call the set of permutations a group though. Here is a reference (in Russian  :oops: )
S. S. Gindikin, "Rasskazy o fizikah i matematikah", 3rd edition, Moscow, MCCME, IMU, 2001, page 281.
http://www.mccme.ru/free-books/gindikin/lagrange.zip

I asked a colleague of mine who knows more about history than I do, and this is his reply:

[quote]Here is what I know:

The word group is due to Galois who also proved Lagrange's periodicty of continued fractions theorem.

Lagrange did discuss the set of permutations that fixed polynomials of many variables--he probably observed easily that the set was closed under multiplication and also that the size had to be a divisor of n!

Gauss used cosets without naming them so--and he certainly used the fact that they partiition the group (again without using the word)

I would consult Kline (the big book)--I do remember reading a long article on Lagrange in Historia Mathematica (I wonder if that is on line). Another book would be Van Der Waerden History of Algebra (but a quick scan did not provide any obvious references). It did say that absatract groups do not occur until 1870 by Jordan and Kronecker.
[/quote]

Do you remember the volume # in Historia Mathematica. It appears to be online,but there is no way to search the document. Thanks.


Prove:There exists no positive integer $ n$ such that for every rational number $ r$,
$ \exists b,a_1,a_2,...,a_n \in Z$ so that $ r\equal{}b\plus{}\frac{1}{a_1}\plus{}...\plus{}\frac{1}{a_n}$

Next time when someone links to an old thread I suggest continuing there  :wink:


function
induction
algebra unsolved
algebra

Find all functions $ f: \mathbb Z \rightarrow \mathbb Z$ satisfying:
(i) $ f(0)\equal{}1$,
(ii) $ f(f(n))\equal{}n$, and
(iii) $ f(f(n\plus{}2)\plus{}2)\equal{}n$ for all integers $\mathbb Z$.

[quote="amparvardi"]find all functions $ f: Z \rightarrow Z$ satisfying:
$ ( \imath)$ $ f(0) \equal{} 1$
$ ( \imath \imath)$ $ f(f(n)) \equal{} n$
$ (\imath \imath \imath)$ $ f(f(n \plus{} 2) \plus{} 2) \equal{} n$   .  $ \forall n \in Z$[/quote]

$ f(f(n))\equal{}n$ shows that $ f(n)$ is injective.

Then $ f(f(n\plus{}2)\plus{}2)\equal{}n\equal{}f(f(n))$ $ \implies$ $ f(n\plus{}2)\equal{}f(n)\minus{}2$

Then $ f(0)\equal{}1$ and $ f(n\plus{}2)\equal{}f(n)\minus{}2$ $ \implies$ $ f(2n)\equal{}1\minus{}2n$

Then $ f(0)\equal{}1$ $ \implies$ $ f(1)\equal{}f(f(0))\equal{}0$ and $ f(n\plus{}2)\equal{}f(n)\minus{}2$ $ \implies$ $ f(2n\plus{}1)\equal{}\minus{}2n\equal{}1\minus{}(2n\plus{}1)$

And it is easy to check back that this indeed is a solution.

Hence the solution: $ \boxed{f(x)\equal{}1\minus{}x}$

[quote="amparvardi"][quote="pco"]

Then $ f(0) \equal{} 1$ and $ f(n \plus{} 2) \equal{} f(n) \minus{} 2$ $ \implies$ $ f(2n) \equal{} 1 \minus{} 2n$

[/quote]

How do you can say since ''$ f(0) \equal{} 1$ and $ f(n \plus{} 2) \equal{} f(n) \minus{} 2$ $ \implies$ $ f(2n) \equal{} 1 \minus{} 2n$''?[/quote]

In order to see it :
$ f(0)\equal{}1$ and then $ f(2)\equal{}f(0\plus{}2)\equal{}f(0)\minus{}2\equal{}\minus{}1$ and then $ f(4)\equal{}f(2\plus{}2)\equal{}f(2)\minus{}2\equal{}\minus{}3...$
$ f(0)\equal{}1$ and then $ f(\minus{}2)\equal{}f(0\minus{}2)\equal{}f(0)\plus{}2\equal{}3$ and then $ f(\minus{}4)\equal{}f(\minus{}2\minus{}2)\equal{}f(\minus{}2)\plus{}2\equal{}5...$

In order to prove it : just use induction


logarithms
trigonometry
limit
algebra
polynomial
function
real analysis

Well, IMHO this is a Real analysis problem...

note that $ \ln (\sin x)^{\frac {1}{x}} \equal{} \frac {\ln \sin x}{x}$.

$ \lim_{x\to 0}\frac {\ln \sin x}{x} \equal{} \minus{} \infty$ (prove it using L'Hôpital rule or knowing the growth rate of log and polynomial function) so $ \lim_{x\to 0} (\sin x)^{\frac {1}{x}} \equal{} e^{ \minus{} \infty} \equal{} 0$.

On the other hand, $ \ln \left(\frac {1}{x}\right)^{\sin x} \equal{} \minus{} \sin x \ln x$

$ \lim_{x\to 0} \minus{} \sin x \ln x \equal{} 0$ which you can also prove knowing that sine is asymptotically equivalent to $ x$ in the limit process towards zero, or by L'Hôpital's rule, taking the sine down in the denominator.

Hence, $ \lim_{x\to 0}\left(\frac {1}{x}\right)^{\sin x} \equal{} e^{0} \equal{} 1$

and as both limits exist, your desired limit is their sum: $ 1$.


Solve for the system of linear equations

[img]http://webwork2.math.utah.edu/webwork_system_html/tmp/equations/f0/693379284ca3188721313e621d10211.png[/img]

Thanks

[hide]
Input the 4x5 matrix on your trusty TI-84 and use rref

$(w,x,y,z)=(2,4,-1,3)$.[/hide]

ok, thanks for that answer. i already knew how to do that. the problem is when i go to solve it manually i dont get that answer. Anyone what to take the time and post the work on this. I know theres going to be alot

well, im sure playing around with substitution and subtracting/adding the equations will also work... eventually. Right?

hello, you can solve this system by the so called Gaussian elimination, see here
http://mathworld.wolfram.com/GaussianElimination.html
Sonnhard.

Add 4 times the 3rd to the 1st to obtain $ 13w\minus{}2x\minus{}5y\equal{}23$. Add 2 times the 3rd to the 2nd to obtain $ 8w\minus{}3x\minus{}3y\equal{}7$. Subtract 3 times the 3rd from the 4th to obtain $ \minus{}5w\minus{}x\plus{}4y\equal{}\minus{}18$. Now we wish to solve the systems:
$ 13w\minus{}2x\minus{}5y\equal{}23$ $ (1)$
$ 8w\minus{}3x\minus{}3y\equal{}7$   $ (2)$
$ \minus{}5w\minus{}x\plus{}4y\equal{}\minus{}18$  $ (3)$

Subtract 2 times the 3rd from the 1st to get $ 23w\minus{}13y\equal{}59$. Subtract 3 times the 3rd from the 2nd to get $ 23w\minus{}15y\equal{}61$. Now simply subtract these two equations to get $ 2y\equal{}\minus{}2$ so $ y\equal{}\minus{}1$ and from this $ w\equal{}2$. Then everything else follows: $ x\equal{}4$ and finally $ z\equal{}3$.


calculus
integration
trigonometry
function
calculus computations

[quote="kg2100"]Can we use $\int_{0}^{1}x^{x}\ dx=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{n}}$ to solve this $\int_{0}^{\pi}(\sin{x})^{\sin{x}}\ dx$ or $\int_{0}^{1}(\sin{x})^{\sin{x}}\ dx$ ?[/quote]

Let me try, instead,
$\int_{0}^{\pi/2}\sin x^{\sin x}dx$
Write this as,
$\int_{0}^{\pi /2}\sin x^{\sin x}\cdot \frac{\cos x}{\cos x}dx = \int_{0}^{\pi /2}\sin x^{\sin x}\cdot \frac{\cos x}{\sqrt{1-\sin^{2}x}}dx$

Let $t=\sin x$ then,

$\int_{0}^{1}t^{t}\cdot \frac{1}{\sqrt{1-t^{2}}}dt$

But I dont think this can be simplified, to $\int_{0}^{1}t^{t}dt$.  :huh:

[quote="yagaron"]no. we cant. because x is a variable, not a function, so we cannot make a substituin of x for a function[/quote]
What do you mean?

Suppose we have ${x^{x}dx}$

Let's make a substituition ${x=\sin (y)}$

${x^{x}dx=\cos (y) \sin^{\sin (y)}(y) dy}$

we will get the same solution as your formula if and only if ${\cos (y)=1}$ for all $y$, but this is impossible.

More general case

${f(x) dx}$
${x=G(y)}$
${f(x) dx= f(G(y)) G'(y) dy}$

Will have the same formula if and only if ${G'(y)=1\land G(y)=y}$

so we cannot make this substituition


Three circles, $ C_{1}, C_{2}$ and $ C_{3}$ intersect. $ C_{1}$ intersects $ C_{2}$ at $ X$ and $ W$. $ C_{1}$ intersects $ C_{3}$ at $ W$ and $ Z$, $ C_{2}$ intersects $ C_{3}$ at $ W$ and $ Y$.

From the point $ A$ on $ C_{1}$ a line is drawn through $ X$ to meet $ C_{2}$ at $ B$. From $ B$ a line is drawn through $ Y$ to meet $ C_{3}$ at $ C$. From $ C$ a line is drawn through $ Z$ to meet $ C_{1}$ at $ A$. So we have a triangle $ ABC$. Is it necessarily the case that we get a triangle?

Beginning with another point $ A'$ on $ C_{1}$, the process is repeated to get another triangle $ A'B'C'$ Are the triangles similar? Are they congruent?

Sorry I can't post a diagram, but I hope this is sufficient to draw a picture.

[hide=" One approach"]

Join $ X$, $ Y$ and $ Z$ to $ O$, the point where all three circles meet.

Then angle chasing shows that $ OYCZ$ is a cylic quadrilateral since $ AXOZ$ and $ XBYO$ are cyclic quadrilaterals.

For the second part, it seems that the trianlges are similar but not necessaily congruent.

[/hide]


LaTeX
quadratics
algebra
quadratic formula

hello, do you mean
$ 5x\left(1\plus{}\frac{1}{x^2\plus{}y^2}\right)\equal{}12$
$ 5y\left(1\minus{}\frac{1}{x^2\plus{}y^2}\right)\equal{}4$
Sonnhard.

hello, writing your both equations in the form
$ 1\plus{}\frac{1}{x^2\plus{}y^2}\equal{}\frac{12}{5x}$
$ 1\minus{}\frac{1}{x^2\plus{}y^2}\equal{}\frac{4}{5y}$
and adding them we get
$ \frac{6}{5x}\plus{}\frac{2}{5y}\equal{}1$ and solving for $ x$ we get
$ x\equal{}\frac{6y}{5y\minus{}2}$
setting this in your equation (1) and simplifying and factoring we get
$ (y\minus{}1)(5y\plus{}1)(5y^2\minus{}4y\plus{}4)\equal{}0$, from here you will get all solutions of your system.
Sonnhard.

Take (1)+ i (2) to get

$ x\plus{}yi\plus{}\frac{x\minus{}yi}{x^2\plus{}y^2}\equal{}\frac{12}{5}\plus{}i\cdot \frac{4}{5}$

Letting $ z\equal{}x\plus{}yi$, we get

$ z\plus{}\frac{1}{z}\equal{}\frac{12}{5}\plus{}i\cdot \frac{4}{5}$

we can use the quadratic formula to finish from here...


Hi, I'm having a hard time figuring out how to do this;

If $ a,b,c,d$ are in continued proportion, meaning $ \frac{a}{b}\equal{}\frac{b}{c}\equal{}\frac{c}{d}$, prove that

$ \frac{a}{b\plus{}d} \equal{} \frac{c^3}{c^2d \plus{}d^3}$

Is there a way to do this by means of invertendo, alternando, componendo and dividendo?

[hide="Hint"]Let $ r\equal{}\frac{a}{b}$.  Now express everything in terms of $ d$ and $ r$.[/hide]


The lower and upper deciles of any distribution are the points that mark off the lowest 10% and the highest 10%. What are the lower and upper deciles of the standard normal distribution?

I doubt that there is a closed form formula for that one: take your computer and compute a good approximation ( the erf function should be helpful)


A number $a$ has  $27$ digits and all of them are equal to $1$.Which is the remainder in the division $a: 27$ ?

[u] Babis[/u]

wow i was kind of lost on that expl. could you explain more...?(it looks sort of like base conversion) :lol:

[hide="You might want to think about it like this instead..."] We can write this number as

$111111111 \cdot 1000000001000000001 = 111 \cdot 1001001 \cdot 1000000001000000001$

Now,

$111 = 99+9+3$
$1001001 = 999999+999+3$
$1000000001000000001 = 999999999999999999+999999999+3$

Which are all divisible by $3$, so the whole thing is divisible by $27$.

You could do this in a somewhat more advanced way if you remember that a number is divisible by $3, 9$ if and only if the sum of its digits is divisible by $3, 9$.  This is basically because $10^{k}-1 = 999...$ which is divisible by $3, 9$. [/hide]

[hide]just take \frac{111111111111111111111111111}{27} = 4115226337448559670781893 since this is a whole number then there is no remainder so the remainder = 0[/hide]

Care to repeat that operation (without mechanical help) for something larger?  Perhaps if I asked you to divide $\underbrace{111...111}_{1997 times}$ by $1997$?   :o


[size=200]Teh MAZE[/size]

I will allow up to 30 people to play. The object of the game is to go from the Start to the Finish.
You have no map, but it gradually grows as you travel with trial and error.

[u]Rules[/u]

1. I was evil in creating this maze. Acknowledge that.
2. There are only 4 directions you can go: Up, Down, Right, Left (Do not post North, South, East, West. Just don't)
3. To move, put it in the form: (your username) goes (up, down, right or left). For example, I might do: andrewjjiang97 goes right. (This is not implying anything) If you do not use this format, I will ignore your move.
4. Do not post if you are not in this game.
5. You do not have to put /join or something like that. Just post the direction you want to go in. (see format above)
6. I will not draw any maps except for the one below.
7. Have fun!!

S

S=start

[i]bluecarneal goes down. S/he finds a note. There seems to be a word etched into paper.It has the word "in" on it.
[/i]

To see what you have put /view inventory.

me is he.

[b]Bluecarneal moves left[/b]

[i]bluecarneal hits a brick wall. Ouch!! If you hit 2 more walls, you die.[/i]

[b]tennis123 moves down[/b]
If we die, are we simply out of the game? Do we only have 1 life?

[b]math_explorer moves right[/b]
So can we make our own maps? (And post them here?)
Are our mazes all the same? (Or we start at different places in the same maze? Or something else?)
Can we interact with each other?
Can we make random guesses on the forum about the design or anything?
Sorry for all these questions, but, well, I don't want to be modkilled. BTW, I'm a he as well :)

[i]tennis123 goes down. S/he finds a note. There seems to be a word etched into paper. It has the word "in" on it.[/i]

@tennis123 you have unlimited lives

[i]math_explorer moves right. Nothing happens... [/i]

@math_explorer Yes you can make your own maps and post then here. However you want to be the first to the finish. All the mazes are the same, but as you just saw, there are unlimited "notes". You can interact with each other.

[b]math_explorer moves up[/b]
When we die, do we go back to the origin?
Do we really need the notes to, say, unlock something, or is it merely information? Or should you tell us that?

[i]math_explorer moves up. When he reaches there, there seems to be a black metallic door. When you test the handle, it clicks open. Now, choose a direction to venture forth.[/i]

Yes, when you die you go back to the origin. I will not answer your last question.

I know. You're suppose to create it by trial and error

[i]bluecarneal goes up. Hey, isn't this were I started? he wonders...[/i]

Just FYI, you can post multiple moves at a time.

btw I'm to the left of where I started.

bluecarneal moves up, left, up, right, as far as walls allow.

pacman moves right, up, up, right, right, down.

Previous moves: down, left
Hits wall once
[b]FlyAgaric moves right[/b]

Me moves like this: DLRULDRUURDLRURLDURDLRUDRULLRD
I bet I won't die.

he's not on vacation, he's playing my game.

BTW, my game is Legend of the Black Dragon.
There's still one spot!  Join!

Please stop advertising your RPG.

[b]randomguy64 goes right and then down.[/b]

Before this post, I successfully did RUURL. I omitted wall-crashes.

[i]pirgraph goes left, then up... Nothing happens...

math_explorer goes down, ,finds note with "in", then goes down into a wall, then  he goes right down left right and as he goes right again, a mystic fog covers his body and kills him.

pacman2812 goes right, up, up through the door, right, and right into a wall, then down.

FlyAgaric moves right...

aback goes down, finds note with "in" then left into a wall, right, up into a wall, left, and finally down into another wall dying...

randomguy64 goes right, then down...[/i]

piegraph, please STOP advertising otherwise I will have you banned...

lol

[quote="andrewjjiang97"]piegraph, please advertising otherwise I will have you banned...[/quote]
Whaaaaaaaaaaat?  :|

[b](Respawned again, :( ) math_explorer moves down, right, right, right, left, down, left, down, left, down, right.[/b]

pacman goes up, left, up, up, right, right, down.
*fails*

down up down right left right down left up right left down left

[b]randomguy goes up twice[/b]

(I did RUURLRD successfully)


UHH....No! I've never heard of that definition of scene be4.

[quote="h_s_potter2002"]The topic basically says it all, are you scene?[/quote]

Shravas...whats the point of this? Were you trying to find out how many people knew what scene meant or something?

I think I speak for many people when say: What the heck!?

[quote="bleumoose"]*bursts into laughter* i think AoPS and the scenesters are like, polar opposites by definition.[/quote]

That's why I started it.

for the most part scene/emo people annoy me.

haha, the moment I saw the topic I knew everyone would be like "WTH??"

[quote="WarpedKlown1335"]Yeah scene kids are pretty gay by definition, lots of them at my HS.[/quote]

I'm going to "call you out" on this. It may not seem like a big deal to you, or maybe not to anyone here, but I take very personal offense when someone uses "gay" as a derogatory term.  I have some extremely intelligent, beautiful, and talented homosexual friends, and I love them dearly. So please, think before you say something like that.

yeah...well about i would say 70% of ppl at my high school are like that..

i don't like stoners, and there is a whole bunch at my school

[quote="bleumoose"]I have... homosexual friends.... [b]I love them dearly[/b].[/quote]
Was this intentional?It seems like a funny blunder, but yes, gay people are fine.
Execpt for the annoying ones who think they should get special treatment because they're gay...but that is another story.

Musta missed something. I know 'about' these kids.. never met one. No self identity.

no. :huh:
never met any of these kinda people anyway...


Hey MA people,
MAML I is in about a week, how are YOU prepping for it?

Wow its already in about a week? I need to sign up asap then..

When is the exact date?


A set $ S$ of points in the $ xy$-plane is symmetric about the origin, both coordinate axes, and the line $ y \equal{} x$. If $ (2, 3)$ is in $ S$, what is the smallest number of points in $ S$?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 4 \qquad
\textbf{(D)}\ 8 \qquad
\textbf{(E)}\ 16$

[hide="Solution"]
Because the point (2,3) is not on the origin, an axis, or the line y=x, we know that there will be no weird situations. If (2,3) is reflected across the origin, we have 2 points. When those points are reflected across the axes we have 4 points. When [i]those[/i] points are reflected across the line y=x, we have o total of [u][b]8[/b][/u] points in S. [b]D[/b][/hide]


Saw this on a part of the site (About Us\Contact Us):

[quote]If you would like to report an incident of copyright infringement by a member of our Community on our Forum, please contact Richard Rusczyk via one of the methods above.[/quote]

Just wondering, but what does this mean?  And what does this mean for us?

Does "copyright infringement" mean posting questions from copyrighted sources, like the AMC 10/12 exams?  Or does it have to do with posting large quantities of questions?  Because in that case, the "Resources\Websites" section might be dangerous, both for AoPS Incorporated and for me, mathfanatic.

Any input from anyone would be appreciated.

Why would the resources/websites list be dangerous for you?  If anything on those sites violates copyright, then that's the owner of that site's problem, not yours (or ours at AoPS).

What it means is a legal grey area in some regards, but I think pretty clear-cut in others.  Throwing out problems from various sources for conversational purposes on the board is fine.  If we (meaning AoPS) gathered up, say, a bunch of Olympiad problems and started selling them on the site, that wouldn't be fine.  I'm not comfortable reproducing them en masse, either (i.e. I'm happy linking to the Scholes site, but I wouldn't host such a site without a clear ok from either a lawyer or the folks whose tests I was reproducing).

Furthermore, for educational purposes, the copyright laws are decidedly less strict than they are for commercial purposes.  I can use USAMO problems in the weekend classes we hold here in San Diego (and do quite often) to my heart's content.  I can't box 'em up and sell 'em to the students as USAMO collections.

Also, I wouldn't worry too much about the AMC coming after you for citing a problem here or there (particularly if you note where it came from).  I would want you to discuss my problems if I produced them.  I wouldn't want you releasing every single one to everyone in an easy to get at format if I were selling them, but I'd be very happy to have you throwing out problems occasionally (moreover, unless you were throwing them out and claiming you wrote them, I think I'd be powerless to stop you unless you did it on a scale that was damaging to me).

As for why that information about copyright infringement is on the site at all, it's a result of the Digital Millenium Act, which requires that information be present.  There's an ongoing war over intellectual property (see the recent Disney laws for one of the more absurd recent government decisions).  The internet intensifies it, as it's easier than ever to reproduce and distribute information.

Hope that answers your questions.

Okay, thanks for clearing that up.  I just didn't like the sound of the sentence:
[quote]...an incident of copyright infringement by a member of our Community on our Forum...[/quote]

Which now makes perfect sense legally: it identifies that if there is a copyright infringement made [i]by a member of the community[/i] (for instance - although I don't think so - me), it is made by a member of the community, and not AoPS Inc.  That is, if I started posting advertisements that I was selling collections of problems, then it ought to be me that gets in trouble, not you.

Not that I think anyone would try to do that because 1) there are enough problems out there for free that 2) it would be too much trouble for 3) not enough profit.  And 4) because we are "mathletes" with morals and selling someone else's things on someone else's site would be completely wrong.

So in summary, no problems.  That disclaimer is, as I understand, just there for legal purposes?

'k.
_____________________

"Boy howdy, all this legal stuff is making me hungry...for math!"

=> mathFANATIC


It's christmas, so as a gift, I will attempt to write my first problem. :)

$4xy({(x+y)}^2-2xy)=a$

$6\left({\frac{a}{4(x^2+y^2)}}\right)^2=b$

$(x^2+y^2)(x^2+y^2)-\frac{2b}{6}=c$

Find $a+b+c$ given that $x=\frac{3}{17}$ and $y=\frac{31}{17}$

What would $a+b+c$ be in terms of x and y?

Don't try to brute force this, the solution is simpler than it looks.

This is a very simple question.

All you have to do is plugging in the values.

This is quite nice for Christmas!  :D

Actually, I didn't intend for you to plug in x and y from the beginning, I changed their values to make it harder to do that. The answer remains the same though.

we can imply the result might be (x+y) or something related, because the sum of the given numbers is 3/17+31/17=2


[hide]Ok, so they are effectively rushing towards each other at 200 miles per hour.  Thus, 1 minute before impact, they will be $ \frac{200}{60}=3 \frac{1/3}$ miles apart. [/hide]

[quote="k00lperson"]This is the problem:

Two trains are headed directly towards each other at rates of 8o miles per hour and 120 miles per hours. How many miles apart are they 1 minute before impact? Express your answer as a mixed number.[/quote]

Ok....so they are heading towards each other at the rate of 80 + 120 = 200 mph. At 60 minutes they are 200 miles, so at 1 minute they will be 200/60= 10/3= 3 1/3 miles apart. Hope that helps!!


probability
trigonometry
floor function
geometry
search
greatest common divisor

Please read each part of the hidden text.

[hide="Round 1"]
Welcome to the COP the 4th Round 1. I welcome you and here are few things you have to know.

[hide="rules"]
Please note that:

1. You must send your answers (not your solutions. If you want to, you can, however)
2. You have a week to do these 5 problems.
3. To go to Round 2, you must get all five problems right.
4. Check your answer carefully so you get 4 and miss one.[/hide]

[hide="Problems"]
1. Laurence is choosing one card from a standard deck of 52 playing cards.  Let probability of choosing face or spade card is $\frac{n}{m}$ with $(m,n) = 1$. Compute $mn$.

2. Let $y_1 = \frac{1}{1 + \cos x}$ and $y_2$ is another way to write $y_1$ without using cosine. Find $y_2$.

3. Let $f(x) = \lfloor x \rfloor$. Find $S$ if:

$S = f(\frac{1}{3}) + f(\frac{2}{3}) + f(\frac{4}{3}) + f(\frac{7}{3}) + f(\frac{11}{3}) + f(\frac{16}{3}) +... f(\frac{211}{3})$

4. Phil walks from the center of the circle $(x-5)^2+(y+5)^2 = 100$. From that point, he walks seven units up and nine units east. Call this point $z_2$ and the original point he started $z_1$. The midpoint of segment connecting $z_1$ and $z_2$ is $z_3$. Write the equation of the circle with radius 17 that has $z_3$ as its center. Express the equation with the fractions.

5. Given that two points, $c_1$ and $c_2$, are on the circumeference of the circle $O$ that arc ${c_1 O c_2} = 60^\circ$. If $\overline{c_1 O} = \overline {c_2 O} = \overline {c_1 c_2} = 9$, find the difference between the area of the sector $c_1 O c_2$ and the area of $\Delta {c_1 O c_2}$. Express your answer in fraction with $\pi$.[/hide]

PDF version of this test is also available at this post.[/hide]

We send them to you by PM, right?  And on the fourth rule, I assume you mean so you [b]don't[/b] get 4 and miss one.  :P

PM is correct.

Thanks for pointing that out.

for #2, there could be multiple ways too answer.... and do you want it simplified or not?

True.. That is only problem.

So, for #2, send me the steps you did to get the answer.

Then I'll later post all the answers of #2. :)

In what form do you want #4 expressed?
With the specifications given, it could be anything with fractions.

For question 1, are face cards just Jacks, Queens or Kings, or do they also include Aces?

#1: It's a regular 52 playing cards.

#4:

In form of $(x - h)^2 + (y - k)^2 = r^2$

Hmm.. my question better phrased:

Are aces included as face cards?

Do you mean angle instead of arc on problem 5?

[quote="TheBB"]I'm not used to math in english, so excuse me, but I would like a few things clarified:

#1: (m,n)=1 means that m and n are mutually prime, right. Or generally, (a,b)=c means that the greatest common divisor of a and b is c?

#2: I guess sine is not allowed either? :)

#5: "Arc $c_1Oc_2$" means the curve you get if you trace along the circle from $c_1$ to $c_2$, correct? And what is with the lines? Does it mean the distance between these two points? And in that case, which point is O? Actually, come to think of it, can't you just generally clarify this one? :)[/quote]

Number one: Yes
Number two: There is no restriction on using sine. :)
Number five: Curve, yes.  The distance between the two points (straight line). Which point is O? Circle P has center P. :)

Can we enter this if we haven't participated in the previous rounds?

This is the first round so you may enter.

All you have to do is solve these five problems and send the answers by PM.

Sorry if this is really obvious to you guys, but...

On #3, what are the absolute-value (| |) like symbols around $x$?

Greatest integer function.

$\lfloor \frac{3}{4} \rfloor$ is equal to 0 while $\lfloor \frac{5}{4} \rfloor$ is equal to 1.

This basically takes the greatest integer of $x$.

For more info, look here:

 [url=http://mathworld.wolfram.com/FloorFunction.html]From Mathworld[/url]

My english is not so good and I have an "unknown-word" question. At #1 I understood tha face cards=Queens, Kings & Jacks, but Spade cards means?! ... (in romanian  :D )

From what I know, any expression other than cosine is allowed in your answer.

Do you know which ones clubs are? Spades are the other black cards. The have the point at the top, and if you put your thumb and index finger of each hand together (index finger touching index finger, thumb touching thumb), it looks kind of like a spade... I hope that last part didn't confuse you.

Go to this site (from Google search engine)

http://images.google.com/images?q=Spade&hl=en

[quote="tarquin"]Spades = Small shovels.[/quote]

What? Spades = big shovels

IF that's supposed to be fraction then no it's not equal ti 8.

$\left\lfloor \frac 89 \right\rfloor = 0$

Hm, he might have meant the comma to be a decimal point, which in that case it would be $8$.

Yes. The comma was a decimal point. Thanx :P

if x is an integer and
if x  <=  n < x+1
then [n]=x

What is a sector? Is the area between a arc and a chord?

Answers must be turned by this Friday. 5/6/05.

Please send them in! :)

Yes, sin can be used in #2. A sector is the area inside a circle bounded by two radii.

Where can we discuss the solutions and when?

Solution in PDF Format will be posted in tomorrow afternoon 1:00 EST.

Please do not talk about the solutions until I have posted PDF solution document.

Thanks and Regards,

Silverfalcon

All right. And we discuss the problems in this thread?

I'll post the separate thread for solutions.

It's easier to have two different threads for problems and solutions and that's how I did on the past ones.

Thats great. Thank Silverfalcon for all the efforts made to this competition.  :)


Euler
limit
logarithms
inequalities proposed
inequalities

Let $\gamma=$ Euler's constant,i.e.
$ \gamma:=\lim\limits_{n\to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} - \ln{n}\right) .$
[b] (Without using a PC),  prove or disprove that [/b]  $ \pi^2 < 8e^{\gamma}  . $

I wonder if we can prove this using the fact that :
$\gamma =\lim\limits_{n\to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} - \ln{n}\right) .$
and
$\frac{\pi^2}6 = \lim_{n \to \infty} \left(1 + \frac 1{2^2} + \frac 1{3^2} + \ldots + \frac 1{n^2}\right) .$

 :?:


Let $ \operatorname{frac}(x): \equal{}x\minus{}\operatorname{floor}(x)$ where $ \operatorname{floor}(x)$ is the largest integer $ \leq x$. Prove that $ \operatorname{frac}((3/2)^n)<1\minus{}(3/4)^n$ as $ n>1$.


Im down in san diego and one of my teachers up north wants problems to present for a putnam masterclass. I would like to present the first paul erdos problem in this book, so if anyone could tell me what it was id be so grateful. Thanks a bunch,
J.D.

The first problem from Paul Erdos is as follows:

Prove that every infinite sequence of distinct positive integers contains either
(a)an infinite subsequence such that for every pair of terms, neither term ever divides the other,or

(b)an infinite subsequence such that in every pair of terms one term always divides the other.


function
calculus
integration
complex analysis
complex analysis solved

let f be an analytic function in the once_punctured disc G={ z: 0<|z-zo|<R }
show that if

double integral over G of |f(x+iy)|^2 dxdy < infinity

then f is removable singularity at zo. if the power 2 in the integral is replaced by p>0
then what can u say about the nature of the singularity at zo?

appreciate ur help


The points $ (x, y)$ represented in this table lie on a straight line.  When the equation of this line is written in the form $ y \equal{} Ax \plus{} B$, what is the value of $ A \plus{} B$?\\

\begin{tabular}{c|c}
 $ x$ & $ y$ \\ \hline
 $ 2$ & $ 7$ \\
 $ t\minus{}2$ & $ v$ \\
 $ t$ & $ v\plus{}6$
\end{tabular}

The points (x,y) represented in this table lie on a straight line. When the equation of this line is written in the form y=Ax+B, what is the value of A+B?

$ \\
\begin{tabular}{c|c}
x & y \\ \hline
2 & 7 \\
t\minus{}2 & v \\
t & v\plus{}6
\end{tabular}$

...I LaTeXified it.

slope : $ \frac{v\plus{}6\minus{}v}{t\minus{}(t\minus{}2)}\equal{}3$

(x,y) = (2,7).

so y=3x+1.

A=3, B=1.

A+B=4.

answer : 4

Instead of using the slope formula, we see that from $t-6$ to $t$ (an increase in $x$ by 2), we have $v \Rightarrow v+6$ (an increase in $y$ by 6). Slope, essentially, is the change in $y$ over the change in $x$, so we have the slope as $\dfrac{6}{2}=3$.


[quote="integer"](x+y) = a (mod c)
(x*y) = b (mod c)

specifically:

(x+y) = 17 (mod 35)
(x*y) = 2  (mod 35)[/quote]

yes.  Chinese remainder theorem says that you can assemble the solutions (if any exist) from solutions modulo 5 and 7.  There can be up to 2 solutions in each case, so one expects (at most) 2x2=4 solutions in total.