Say I want to label the coordinates of a point, but it's coordinates are irrational. Is there a way for me to make the coordinates like this?
[code]
draw((\sqrt{2},\sqrt{3})--(21-3\sqrt{5},\sqrt{67}));
[/code]
Or do I have to just round the value using a calculator?

I think sqrt function is built in Asymptote.

[asy]unitsize(1cm); draw(D(sqrt(2),sqrt(3))--D(23-3*sqrt(5),sqrt(67)));[/asy]

Yes, Asymptote is a full programming language with built-in functions like sqrt(), exp(), log(), sin(), etc. Just write

draw((sqrt(2),sqrt(3))--(21-3*sqrt(5),sqrt(67)));

and, voila:
[asy]draw((sqrt(2),sqrt(3))--(21-3*sqrt(5),sqrt(67)));[/asy]

@Valentin: It is D((2,3)), not D(2,3) (first parentheses for the function, second parentheses for the pair :). I almost thought you discovered a bug in cse5 when I saw your horizontal line)

I suspect you do not know the best news yet: you do not need to compute the coordinates of intersection points yourself. The intersectionpoint() (or IP() in cse5) command does the job for you. I'm writing this because I strongly suspect that your crazy coordinates were obtained by intersecting a circle and a line or something like that :)

No, they were actually made up, but thanks anyways.
Anyways, do you just do something like:
[code]
IP(((0,0)--(1,1)),((1,0)--(0,1)));
[/code]
Or what?

Yes, though you do not need that many parentheses: IP((0,0)--(1,1),(1,0)--(0,1)); is good enough. So, here is an example with a circle and a line:
[code]
path P=D(CR((0,0),3.6)), Q=D((4,0)--(0,5));
D("A",IP(P,Q),W); D("B",IP(P,Q,1));
[/code]
gives
[asy]path P=D(CR((0,0),3.6)), Q=D((4,0)--(0,5));
D("A",IP(P,Q),W); D("B",IP(P,Q,1));[/asy]

Theoretically you can just click on the image and get the code :D

[quote="fedja"][code]
path P=D(CR((0,0),3.6)), Q=D((4,0)--(0,5));
D("A",IP(P,Q),W); D("B",IP(P,Q,1));
[/code][/quote]

How is it determined which intersectionpoint is labeled "A"?

[quote="undefined117"]How is it determined which intersectionpoint is labeled "A"?[/quote]
Notice that there is an extra argument in the second [b]IP()[/b] command. That third argument represents the index of the point that you want. For example: If two paths intersect $ 3$ times, it can be $ 0$, $ 1$, or $ 2$. Also, since $ 0$ is the default, it is not necessary in the first [b]IP()[/b] command.


Hi,

Let P(n) be the number of ways to express a positive integer n in the sum of some positive integers (including itself).
Then
P(1)=1 (1)
P(2)=2 (1+1,2)
P(3)=3 (1+1+1, 1+2, 3)
P(4)=5 (1+1+1+1, 1+1+2, 1+3, 2+2, 4)
P(5)=7 (1+1+1+1+1, 1+1+1+2, 1+2+2, 1+1+3, 1+4, 2+3,  5)
.......

P(n)=?

Is it possible to generalize a formula for P(n) or the relation between P(n+1) and P(n)


search
geometry
incenter
circumcircle
geometric transformation
reflection
homothety

[quote="greentreeroad"]Let AB be a diameter of circle center at O. Pick an arbitrary point P on segment AO and suppose the circle center at P with radius AP intersect AB again at C. Now, let D be one of the two points on the arc of bigger circle(center at O) that DC=DB. Let Q be the center of the circle tangent to DC and the two other circles(i.e. those center at O and P). Prove QC is perpendicular to AB.[/quote]

Posted before, but I am lazy to search for it. I just remember that the proofs were complicated (including mine).

CD cuts the circle (P) with diameter AC again at E. F is the foot of perpendicular from E to AC. Take a circle (Q) centered on the perpendicular to AC at C, tangent to the ray (CD at U and (externally) tangent to (P); there is only one. Parallel to AC through U cuts CQ at V. The right $ \triangle CUQ \sim \triangle AEC$ with their U-, E-altitudes UV, EF are similar with coefficient $ k \equal{} \frac {CQ}{AC} \equal{} \frac {QU}{CE} \equal{} \frac {CV}{AF}.$  Since (P), (Q) with radii PC, QU are externally tangent,

$ (PC \plus{} QU)^2 \equal{} PC^2 \plus{} CQ^2,\ AC \cdot QU \plus{} QU^2 \equal{} CQ^2.$

Using $ CE^2 \equal{} AC \cdot FC$ and substituting for $ CQ^2 \equal{} QU^2 \cdot \frac {AC^2}{CE^2} \equal{} QU^2 \cdot \frac {AC}{FC}$ and calculating radius QU of (Q),

$ AC \cdot QU \plus{} QU^2 \equal{} QU^2 \cdot \frac {AC}{FC},\ QU \equal{} \frac {AC \cdot FC}{AF} \equal{} \frac {CE^2}{AF}.$

It follows that CV = CE. Let the perpendicular lines UV, EF intersect at Q' and let (Q') be a circle with center Q' and radius Q'V, tangent to CQ at V. Since $ \frac {FQ'}{AF} \equal{} \frac {CV}{AF} \equal{} \frac {CQ}{AC} \equal{} k,$ the points Q', Q, A are collinear. From similar right $ \triangle EFC \sim \triangle CUQ,$

$ q \equal{} \frac {Q'V}{QU} \equal{} \frac {FC}{QU} \equal{} \frac {CE}{CQ} \equal{} \frac {CV}{CQ} \equal{} \frac {FQ'}{CQ} \equal{} \frac {AQ'}{AQ},$

so that A is the external similarity center of the circles (Q'), (Q). Inversion in a circle (A) with center A and radius AC takes the circle (P) with diameter AC into the common tangent CQ of (A), (P). The inversion center A is the external similarity center of (Q'), (Q) and (Q') is tangent at V to the inversion image CQ of (P). Consequently, this inversion takes (Q) into (Q'). Let $ W \in (Q')$ be diametrically opposite to V and let tangent to (Q') at W, parallel to CQ, cut the line ECD at K. As EFQ' is midparallel of CQ, KW, it follows that E is the midpoint of CK. As $ AE \perp CK,$ it follows that $ K \in (A).$ Let WK cut the circle (A) again at L. The line KL is inversion image of a circle (O) passing through the inversion center A, centered on the ray (AC, and internally tangent to the circle (Q); there is only one. Let B be the other intersection of (O) with the ray (AC. Since AC = AK = AL, C is incenter of the isosceles $ \triangle BKL$ with circumcircle (O), the line KCD bisects the angle $ \angle LKB$ and DL = DB = DC. QED.

Here is the link which yetti was talking about:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=536144173&t=18266

I did not remember this link, I was talking about a different one: [url]http://www.mathlinks.ro/viewtopic.php?t=38681[/url].

Based on the above solution, i.e., on C being incenter of the isosceles $ \triangle BKL,$ we can generalize:

Let $ \triangle ABC$ be scalene (non-isosceles) with incenter I. The angle bisectors AI, BI, CI cut the triangle circumcircle (O) again at X, Y, Z. (P) is a circle with diameter IX. (Q) is incircle of the curvilinear triangle BIZ, tangent to the rays IB, IZ and internally tangent to (O). (R) is incircle of the curviliear triangle CIY, tangent to the rays IC, IY and internally to (O). Prove that the circles (Q), (R) are both externally tangent to (P).

Here is a very simple solution of the generalization, which can be applied without change to the special Sangaku case of an isosceles $ \triangle ABC$
(see [url]http://www.cut-the-knot.org/Curriculum/Geometry/CirclesAndRegularTriangle.shtml[/url]).

Let $ (O)$ be the circumcircle and $ I$ the incenter of a $ \triangle ABC.$ Perpendicular to $ OI$ at $ I$ cuts $ (O)$ at $ M, N.$ The circumcircle $ (O)$ cuts a circle $ \mathcal J$ with center $ I$ and radius $ IM \equal{} IN$ in diametrically opposite points. Consequently, inversion in $ \mathcal J$ with negative power $ \minus{} IM^2$ (ordinary inversion followed by reflection in the inversion center) takes the circumcircle $ (O)$ to itself and $ X$ to $ A$. The circle $ (P)$ with diameter $ IX$ passing through the inversion center goes to a line perpendicular to $ IX$ at $ A,$ or the external bisector of the angle $ \angle CAB.$ Lines $ BIY, CIZ$ passing through the inversion center also go into themselves. Let $ I_b, I_c$ be the B-, C-excenters of the $ \triangle ABC.$ Inversion image $ (Q')$ of the circle $ (Q)$ is tangent to $ II_b \equiv BI,$ to $ II_c \equiv IZ,$ and externally tangent to the 9-point circle $ (O)$ of the $ \triangle II_bI_c.$ Therefore, $ (Q')$ is its excircle in the angle $ \angle I_bI_cI,$ also tangent to the external bisector $ I_cAI_b$ of the angle $ \angle CAB,$ the inversion image of the circle $ (P).$ Similarly, inversion image $ (R')$ of $ (R)$ is excircle of the $ \triangle II_bI_c$ in the angle $ \angle II_bI_c,$ also tangent to $ I_bI_c.$ As a result, the original circles $ (Q), (R)$ are tangent to the original circle $ (P)$ with diameter $ IX,$ obviously externally.

Great generalization yetti! this is my solution.
Let $ M,N$ be midpoint of $ IB,IC$ respectively. Since $ X$ is circumcenter of triangle $ IBC$, $ \angle IMX \equal{} \angle INX \equal{} 90$ meaning that $ (P)$ pass $ M,N$.So $ (P)$ is circumcircle of triangle $ IMN$. we know that $ (Q)$ tangent to $ IB,IZ$ which are side of triangle $ IMN$. Hence, to prove $ (Q)$ tangent to circumcircle of triangle $ IMN$, we need to show $ (Q)$ is $ I$-exmixtilinear circle of triangle $ IMN$.( I don' know what is exatly name of this).  Let $ (Q)$ touch $ IB,IC$ at $ P_1,P_2$. From well know fact about mixtilinear circle, It is suffice to show $ P_1P_2$ pass $ I_N$ where $ I_N$ is $ N$-excenter of triangle $ IMN$ (In fact, what I mean well known fact about mixtilinear circle is that if A-mixtilinear circle touch $ AB,AC$ at $ P_1,P_2$ respectively, then $ P_1P_2$ pass incenter of $ ABC$. Extroversion must be incenter->excenter) . It is known that $ P_1P_2$ pass incenter of triangle $ BYZ,BCZ$, and we call incenter of traingle $ BCZ$ by $ S$. So consequantly, what we need to show is $ \angle SI_NI \equal{} 90$. In triangle $ BCZ$, since $ ZB \equal{} ZI$, we can show $ S$ lie on circumcircle of $ IBC$.So $ S$ is intersection of circumcircle of triangle $ IBC$ with bisector of $ \angle ICB$. Triangle $ IBC$ and $ IMN$ are homothety with center $ I$ and homothety ratio $ \frac {1}{2}$. Let midpoint of $ IS$ be $ S'$ which is intersection of circumcirlce of triangle $ IMN$ with bisector of $ \angle INM$. From well known fact, $ S'$ is circumcenter of triangle $ II_NM$ and together with fact that $ S'$ is midpoint of $ IS$, $ IS$ is diameter of circumcircle of triangle $ II_NM$. Hecne $ \angle SI_NI \equal{} 90$ and we are done.

we can generalize:

Let quadrilateral $ ABCD$ inscribe in circle $ (O)$. Let diagonal $ AC,BD$ intersect at $ E$. Let incircle of triangle $ ABE$ and $ CDE$ touch $ AC,BD$ at $ X,Y,Z,W$ respectively. $ (P)$ is incircle of the curvilinear triangle $ AEB$, tangent to the rays $ EA$, $ EB$ and internally tangent to $ (O)$.  $ (Q)$ is incircle of the curvilinear triangle $ CED$, tangent to the rays $ EC$, $ ED$ and internally tangent to $ (O)$. Prove that circumcircle of triangle $ EXW$ and $ EYZ$ both externally tangent to $ (P),(Q)$.

Well known lemma: P is arbitrary point on the segment BC of a $ \triangle ABC$ with incircle (I) touchng BC at D. (J), (K) are incircles of $ \triangle ABP, \triangle ACP.$ Their common internal tangent onther than AP goes through D and the angle $ \angle JDK$ is therefore right.

Minor variation: P is arbitrary point on the ray (BC beyond C. (J) is incircle of $ \triangle ABP$ as before, (K) is the A-excircle of $ \triangle ACP.$ The common external tangent of (J), (K) other than AP goes through D and the angle $ \angle JDK$ is therefore right. [hide="Proof is the same."] (J), (K) touch BC at E, F.

$ ED \equal{} PC \plus{} CD \minus{} PE \equal{} PC \plus{} \frac{_1}{^2}(BC \plus{} CA \minus{} AB) \minus{} \frac{_1}{^2}(BC \plus{} CP \plus{} PA \minus{} AB) \equal{}$
$ \equal{} \frac{_1}{^2}(AC \plus{} CP \minus{} PA) \equal{} PF$ $ \Longrightarrow$

$ PD \equal{} PE \plus{} ED \equal{} PE \plus{} PF$ equals to the common external tangent length of (J), (K). The 2 common external tangents of (J), (K) cut segment length on the comon internal tangent BC equal to their common external tangent length PD, therefore their common external tangent other than AP goes through D. [/hide]

Problem: XYZW is isosceles tapezoid with diagonal intersection E $ \Longrightarrow$ $ \odot(EXW) \cong \odot(EYZ)$ are symmetrical WRT the bisector of the angle $ \angle AEB \equal{} \angle DEC,$ same as (P), (Q) $ \Longrightarrow$ it is sufficient to show tangency of (Q) and $ \odot(EYZ).$ $ \triangle ABE \sim \triangle DCE$ $ \Longrightarrow$

$ \frac{EY}{EZ} \equal{} \frac{BE \plus{} EA \minus{} AB}{DE \plus{} EC \minus{} DC} \equal{} \frac{EB}{EC}$ $ \Longrightarrow$ $ YZ \parallel BC.$

Let (I) be the incircle of $ \triangle CDE$ (tangent to CE, DE at Z, W), (J) the incircle of $ \triangle BCD,$ (K) the incircle of $ \triangle BCE,$ and (L) the C-excircle of $ \triangle BCE.$ Let F be foot of perpendicular from J to the angle bisector EL of the angle $ \angle AEB \equal{} \angle DEC.$ KELB is cyclic with a circumcircle (N) because of the right angles at B, E.  JFLB is cyclic with a circumcircle (M) because of right angles at B, F. By the lemma, the angle $ \angle JWL$ is also right and $ W \in (M).$ ZW (parallel to EK) cuts the angle bisector CKL at S. Then

$ \angle WSL \equal{} \angle EKL \equal{} \angle EBL \equal{} \angle WBL$ $ \Longrightarrow$ $ S \in (M).$

$ SW \parallel JF$ $ \Longrightarrow$ JFWS is isosceles trapezoid and $ \angle FWS \equal{} \angle WSJ \equal{} \angle WBJ.$ The $ \triangle FWZ$ is isosceles with FW = FZ,

$ \angle ZFW \equal{} 180^\circ \minus{} 2 \angle FWS \equal{} 180^\circ \minus{} 2 \angle WBJ \equal{} 180^\circ \minus{} \angle WBC \equal{} 180^\circ \minus{} \angle WYZ$ $ \Longrightarrow$

YZFW is cyclic and YF bisects the angle $ \angle EYZ \equal{} \angle WYZ.$ Since EF also bisects the angle $ \angle WEZ,$ F is the Y-excenter of the $ \triangle EYZ.$ Let JF cut CE, DE at U, V, then F is the midpoint of UV. By Thebault theorem, U, V are tangency points with CE, DE of the incircle (Q) of curvilinear $ \triangle DEC$ internally tangent to the circumcircle (O) of the $ \triangle BCD.$ Since the Y-excenter F of the $ \triangle EYZ$ is the midpoint of UV, (Q) is also the mixtilinear Y-excircle of the $ \triangle EYZ,$ externally tangent to its circumcircle. QED.

Dear Yetti, Leonhard Euler and Mathlinkers,
finally, you can see a completely synthetical proof of this nice san Gaku on my site :
http://perso.orange.fr/jl.ayme    vol. 4  La fameuse san Gaku de lma prefecture de Gumma (1803)
Sincerely
Jean-Louis


[i] Let $ a,b,c,d$ be positive real numbers such that :

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

   Prove that :

$ a^3 \plus{} b^3 \plus{} c^3 \plus{}d^3 \plus{} a \plus{}b \plus{} c \plus{} d \leq 8$ [/i]

The inequality doesn't hold for all positive reals.

Isn't that for $ a,\ b,\ c,\ d\geq 2,\ a\plus{}b\plus{}c\plus{}d\equal{}a^2\plus{}b^2\plus{}c^2\plus{}d^2\Longrightarrow a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}a\plus{}b\plus{}c\plus{}d\geq 8$? :wink:

You are right. :blush:  I take a rest, see you.

[quote="nguyenvuthanhha"][i] Let $ a,b,c,d$ be positive real numbers such that :

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

   Prove that :

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} a \plus{} b \plus{} c \plus{} d \leq 8$ [/i][/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58853
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59758

[quote="Dimitris X"][quote="kunny"]The inequality doesn't hold for all positive reals.

Isn't that for $ a,\ b,\ c,\ d\geq 2,\ a \plus{} b \plus{} c \plus{} d \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\Longrightarrow a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} a \plus{} b \plus{} c \plus{} d\geq 8$? :wink:[/quote]

For $ a,b,c,d \ge 2$ the condition cant be true my friend :wink:[/quote]
the numbers $ a,b,c,d$ can not be all greater then $ 2$ at least one of them is greater than $ 1$ and an other one is smaller than $ 1$

I do know it. You should see the former post when you post as rule. :wink:


abstract algebra
linear algebra
matrix
algebra
polynomial
complex numbers
linear algebra unsolved

only scalar multiples of unity commute with every matrix, not all diagonal matrices

yeah, sorry stupid mistake. i've edited it now

[quote="kouboy"]let A be a n x n complex matrix. let K denote the space of all n x n complex matrices and define an operator T on K by T(B)=AB-BA. show that the rank of this operator is atmost n^2-n.

i tried looking at the kernel. the kernel consists of all matrices which commute with A, now we know that scalar multpiles of I commute with any general matrix........am i on the right lines? if not, give me some hints[/quote]
As you noticed, the kernel of this operator consists of all the matrices $B$ which commutes with $A$. Now, there is an interesting result hich you may try to prove(otherwise ask): the dimension of the space of solutions of the equation $AX=XA$ in $M_n(\mathbb{C})$ is $\sum_{\lambda} \sum_{i,j} min(a_i(\lambda),a_j(\lambda))$, where $\lambda$ goes over the eigenvalues of $A$ and $a_i(\lambda)$ is the size of a Jordan block corresponding to $\lambda$.
So, since $\sum_{\lambda} \sum_{i,j} min(a_i(\lambda),a_j(\lambda)) \geq \sum_{\lambda} \sum_i a_i(\lambda)=n$, you get that $dimKer T \geq n$, but $dimKer T +dimIm T =n^2$, so you get the desired result.

well i dont no about jordan forms yet though i dare say i'll be knowing it in another couple of days. is there any other approach ? elementary  i mean. anyway thanks for replying eugene. once i read jordan forms , i'll come back to this prob and try to prove your claim. thanks again

[quote="fredbel6"]Interesting!  But is it necessary that you work over the complex numbers?

I guess it suffices that your characteristic polynomial of A splits in your field?

And what if not?use splitting fields??[/quote]

Actually, the proof i know uses exactly the form of Jordan blocks, so we need to have the characteristic polynomial split over the field. But, who knows, maybe you or someone else will manage to prove this for example using Rational Canonical Form which exists over any field. It would be very nice.


geometry
3D geometry
tetrahedron
incenter
sphere
radical axis
power of a point

For a tetrahedron SABC,the incircle of △ABC has the incenter I and touches BC,CA,AB at points E,F,D respectively.let A',B',C' be points on SA,SB,SC s.t. AA'=AD,BB'=BE,CC'=CF and let S' be the point diametrically opposite to S on the circumsphere of SABC.assume that the line SI is an altitude of SABC.show that S'A'=S'B'=S'C'.
Please :help:  :ewpu:

Let $ \mathcal{T}_a, \mathcal{T}_b, \mathcal{T}_c$ be the spheres with centers $ A,B,C$ passing through $ (D,F,A'),$ $(D,E,B'),$ $(E,FC').$ $I$ is the radical center of the circles centered at $ A,B,C$ passing through $(F,D),$ $(D,E),(E,F).$ Therefore, the perpendicular line  to the face $ ABC$ through $ I$ is the radical axis of $ \mathcal{T}_a, \mathcal{T}_b, \mathcal{T}_c$  $\Longrightarrow$ powers of $ S$ to $ \mathcal{T}_a, \mathcal{T}_b, \mathcal{T}_c$ equal $ k^2.$

$ SB^2 \minus{} (BB')^2 \equal{} SC^2 \minus{} (CC')^2 \equal{} SA^2 \minus{} (AA')^2 \equal{} k^2$

Power of $ B$ with respect to the circumsphere $ (O,R)$ is $ R^2 \minus{} (B'O)^2 \equal{} BB' \cdot B'S$

Note that in the triangle $ \triangle B'SS',$ the segment  $ B'O$ is the median issuing from $B'.$ Thus

$ (B'O)^2 \equal{} \frac {_1}{^2}(SB')^2 \plus{} \frac {_1}{^2}(S'B')^2 \minus{} \frac {_1}{^4}(2R)^2$

Substituting $ (BO')^2$ from the previous expression yields

$ (S'B')^2 \equal{} 4R^2 \minus{} (SB')^2 \minus{} 2BB' \cdot B'S$

$ (S'B')^2 \equal{} 4R^2 \minus{} SB^2 \minus{} (BB')^2 \plus{} 2BS \cdot BB' \minus{} 2BB' \cdot B'S$

$ (S'B')^2 \equal{} 4R^2 \minus{} SB^2 \minus{} (BB')^2 \plus{} 2BB'(BS \minus{} B'S)$

$ (S'B')^2 \equal{} 4R^2 \minus{} SB^2 + (BB')^2$

$ (S'B')^2 \equal{} 4R^2 \minus{} k^2$

Since this latter expression is independent of the chosen vertex, we conclude that

$S'A' \equal{} S'B' \equal{} S'C' \equal{} \sqrt {4R^2 \minus{} k^2}$


complex analysis
integration
logarithms
complex numbers
complex analysis unsolved

Suppose F is holomorphic in the unit disc, and

$ \sup_{0 \le r < 1}$ $ \frac {1}{2\pi}$ $ \int_{ \minus{} \pi}^{\pi}$ $ \log^ \plus{} |F(re^{i\theta})|$ $ d\theta$  <  $ \infty$ ,

where  $ \log^ \plus{} u$=log $ u$  if  $ u \ge 1$, and  $ \log^ \plus{} u \equal{} 0$  if  $ u < 1$.

Then $ lim_{r \to 1} F(re^{i \theta})$  exists for almost every  $ \theta$.

Are you not bracketing the exponent on purpose?

Do you want $ F(re^{i\theta})$ in the two places that it occurs?

Consider the case in which $ F$ has no zeros in the open unit disc. (We can use Blaschke products to arrive at this from the general case.) Then $ F$ has a holomorphic logarithm on the unit disc, and the given condition tells us something about the real part of that logarithm.

Can you proceed from there?

Thanks.
Now, I see that $ F$ has a holomorphic logarithm on the unit disc.
But I don't know what 'the given condition tells us something about the real part of that logarithm' means.


polynomial
algebra
Ring Theory
superior algebra
superior algebra unsolved

in the proof of eisenstein criterion i see that P a prime ideal of R is
considered.
what does it mean that $a_{0}\in P^{2}$ ?
and that $f(x)=x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ is monic and is from R[x] and has all coeffs in P
is reduced modulo P and the resulting polynomial $\widehat{f(x)}=x^{n}$.
where have all the coeffs gone ?
why is it that 1 wich is coeff of $x^{n}$ remains after reduction ?
i know what reduction is but i really dont get whats going on with all
the coeffs disappearing because of it this time.

what i'm actually really intrigued about is the following counter-example
if i have $f(x)=x^{5}+2x^{4}+4x^{3}+1$
now if i reduce it modulo $\mathbb{Z}_{5}$(5 is prime) i get
$\widehat{f(x)}=x^{5}+\hat{2}x^{4}+\hat{4}x^{3}+\hat{1}$ but
from what i know those coefficients don't actually disappear but are
just reduced modulo $\mathbb{Z}_{5}$
moreover if the coefficients are already from $Z_{5}$(supposing),
then after reduction they remain the same(in my oppinion).
where am i wrong ?

in the mean time i understood what an element that is in $P^{2}$
looks like($a \in P^{2}=(p^{2})$ => $p^{2}| a$).
i would like to share that with others as well  :lol:

Sorry, but what book is this?
dummit foote (is that the name? Author?)

this is the book
http://www.emba.uvm.edu/~foote/book_cover.gif

Nevermind. It is here:
http://www.amazon.com/Abstract-Algebra-David-S-Dummit/dp/0471433349

edit: sorry didn't see your post

Is it a good book?

you would have to ask someone with more experience like ZetaX for example.
but in my oppinion its a good book,it keeps stuff going in a practical sense.
it has concrete examples and avoids getting into formalisations and
that kind of things,wich is good,but for my curriculum it isnt sufficient.
i would think that if i would be able to read it all and to read all of serge lang
i would be able to solve harder problems also.
but to answer your question short : for begginers(like me) i think its a very good book

ooooh,now i follow the f i chose as an example earlier does not
respect the hypothesis of eisenstein criterion.
all coeffs have to be divisible by p hence they are all 0 in P
now i get it.


integration
geometry
parameterization
calculus
rotation
real analysis
real analysis theorems

okay, i need some clarification on some formulae.

to compute the volume about the $ x$ axis: $ \pi\int_a^b f^2.$

to compute the volume about the $ y$ axis: $ 2\pi\int_a^b xf(x)\,dx.$

to compute the volume about the $ x$ axis when given $ x \equal{} f(y),$ $ 2\pi\int_a^byf(y)\,dy.$

to compute the volume about the $ y$ axis when given $ x \equal{} f(y),$ $ \pi\int_a^bf(y)^2\,dy.$

to compute the area of surface of revolution about the $ x$ axis $ 2\pi\int_a^bf(x)\sqrt {1 \plus{} (f'(x))^2}\,dx.$

what's the formula to compute the area of surface of revolution about the $ y$ axis?

[quote="Null"]what's the formula to compute the area of surface of revolution about the $ y$ axis?[/quote]
The surface area element is $ dA\equal{}2\pi x\,ds.$

Since $ ds$ can be rendered in various ways depending on whether you're integrating with respect to $ dx,$ $ dy,$ or $ dt$ where $ t$ is a parameter, it's probably best to start out with this simplified form; most of your other formulas should best be written this way as well.

$ ds^2\equal{}dx^2\plus{}dy^2,$ so

$ ds\equal{}\sqrt{1\plus{}\left(\frac{dy}{dx}\right)^2}\,dx$ or $ ds\equal{}\sqrt{\left(\frac{dx}{dy}\right)^2\plus{}1}\,dy$, or $ ds\equal{}\sqrt{\left(\frac{dy}{dt}\right)^2\plus{}\left(\frac{dy}{dt}\right)^2}\,dt.$

i see...

mmm, so i want to compute the area of a surface of revolution given the curve $ y\equal{}x^2$ for $ 0\le x\le3$ about the $ y$ axis, what's the integral?

Probably easiest to do that $ dx,$ so $ ds\equal{}\sqrt{1\plus{}4x^2}\,dx$ and the integral is

$ \int_0^32\pi x\sqrt{1\plus{}4x^2}\,dx$

Which (rarity) can actually be computed in closed form.

That's pretty close to one of the problems on the test I gave this morning; I had it as $ y\equal{}\sqrt{4x},$ $ 0\le x\le 4,$ rotated about the $ x$ axis - but it's the same sort of shape, a parabaloid of revolution.

thanks so much Kent for your time.

last question, are the above formulae i wrote before the correct ones? i mean, i don't want to mess it up by using a wrong formula.

A suggestion: rather than remembering the formulas, it's much more productive to remember the [i]pictures[/i] associated with the formulas.  Understanding how to produce a picture and then use the picture in order to write down the appropriate integral is a skill that might be of some use to you in your life (e.g., if you continue taking mathematics or engineering courses, it helps to have some visual intuition for three-dimensional objects); having the formulas memorized is unlikely to be of much use after you're done with this section of your course.  That said, your formulas look right to me.

[quote="JBL"]A suggestion: rather than remembering the formulas, it's much more productive to remember the [i]pictures[/i] associated with the formulas.[/quote]
That's a very close paraphrase of something I've told my current class maybe a dozen times, since it's worth repeating.


The binomial coefficients $\binom{n}{m}$ can be arranged in rows (with the $n$th row $\binom{n}{0}$, $\binom{n}{1}$, ... $\binom{n}{n}$) to form Pascal's triangle. In which row are there three consecutive entries in the ratio $3 : 4 : 5$?

EDIT: why are problems #2, #3, and #4 on AIME 1992 so easy? #1 is much harder... :P

Unless I missed something, you're right, this is very easy for an AIME problem.

[hide="solution"]Suppose $\binom{n}{m-1},\binom{n}m,\binom{n}{m+1}$ have the ratio $3: 4: 5$. We can easily use the definition $\binom{r}{k}=\frac{r!}{k!(r-k)!}$ to get the equations $\frac{m}{n-m+1}=\frac34$ and $\frac{m+1}{n-m}=\frac45$. These become a pair of linear equations in two variables, which is easy to solve. This gives $m=27$ and $n=62$.[/hide]

#1 isn't really hard, its just worded badly on Kalva.  :P


ceiling function
induction
combinatorics unsolved
combinatorics

Given 2005 distinct numbers $a_1,\,a_2,\dots,a_{2005}$. By one question, we may take three different indices $1\le i<j<k\le 2005$ and find out the set of numbers $\{a_i,\,a_j,\,a_k\}$ (unordered, of course).  Find the minimal number of questions, which are necessary to find out all numbers $a_i$.

I think it's $1003$, actually, and, for general $n$ instead of $2005$, I think it's $\left\lceil\frac n2\right\rceil$.

To show that at least that many are necessary (the part I'm not too sure of), we can do this: We start asking questions so that each one is, if possible, "connected" to the previous ones, in the sense that one of the indices whcih it concerns has already been dealt with in some previous question. If we do this until it's no longer possible, it means that each question we ask adds at most two more indices, so when we have to stop we will have determined all the numbers with the respective indices and we will have used at least $\frac m2$ questions, where $m$ is the number of indices which have been included in our questions so far. After we have to stop, it means that we can start asking questions about the other $n-m$ indices, the ones we haven't touched so far, and the procedure is repeated.

Conversely, in order to show that $\left\lceil\frac n2\right\rceil$ questions suffice, assume first that $n$ is even (but $>4$; for $n=4$ we need $3$ questions, $2$ don't suffice). We ask questions $\{i,i+1,i+2\}$ for odd $i$ between $1$ and $n-3$, and after that we ask the question $\{1,3,n\}$. For odd $n$ we do something very similar.

I do not understand teh first part, grobber. What if we ask for some triple wich intersect with first $m$ numbers after few questions concerning next group? For example: 123, then 234, then 567, then 514?

But the order of the questions does not matter, so we ask them in the order we desire. If after we have asked some questions, there is another question regarding one of the numbers we have asked before, then we ask that one. In this case, we simply reorder the questions so that $514$ is the third one.

It is not clear. aybe, our strategy uses previous answers before asking further questions. Say, we ask 514 only if 2>1>3, else we ask 523.

You ask a question which includes three numbers. After this, if there still are some questions which include on of the numbers used, choose one and ask it. After that, look once more for questions in our list which include one of the numbers already used, and if there are such questions, choose one and ask it. What exactly isn't clear? :? If there are more questions using numbers already used, just choose one of them. We do this until we no longer can, i.e. until there are no questions using the numbers previously used.

Or, I am stupid, I forgot, which problem are you talking about (I thought about 9.4.)

[quote="grobber"]I think it's $1003$, actually, and, for general $n$ instead of $2005$, I think it's $\left\lceil\frac n2\right\rceil$. [/quote]

If we ask instead of triplets for k-tuples, the answer is $\left\lceil\frac {n}{k-1}\right\rceil$ ?

[quote="grobber"]
If we do this until it's no longer possible, it means that each question we ask adds at most two more indices, so when we have to stop we will have determined all the numbers with the respective indices and we will have used at least $\frac{m}{2}$ questions, where $m$ is the number of indices which have been included in our questions so far. After we have to stop, it means that we can start asking questions about the other $n-m$ indices, the ones we haven't touched so far, and the procedure is repeated.
[/quote]

But when we begin each time, we use three indices. So we might be able to determine three indices by that question...

Hmmm I think grobber's solution works (basically induction, right?) and is probably the most intuitive. However, here's another way I will roughly outline. Suppose less than n/2 questions are used. Then more than half the elements have been asked about only once (obviously every element is asked about at least once). Then there exists some question such that two of the elements it concerns were only asked about once. But then we can not distinguish between these two, so we need at least (roughly) n/2 questions.

[quote="dragonfire"][quote="probability1.01"]Then there exists some question such that two of the elements it concerns were only asked about once. [/quote]

Hmm.. how does this come?[/quote]

There are more than n/2 elements with the property of [asked about only once]. Each of these belongs to one of fewer than n/2 questions. Of course, there is some casework for even and odd n, but I think you can figure it out.

Here is an amusing solution that is different from the above.

[hide = Solution]

For showing that $1003$ questions indeed suffice, do the same thing as [b] grobber [/b] did above. Now, we will show that $1003$ questions are indeed necessary. Suppose, for contradiction, that we were somehow able to find out all the numbers with the $1002$ questions $(x_1, y_1, z_1), (x_2, y_2, z_2), \cdots, (x_{1002}, y_{1002}, z_{1002}),$ where the $x_i$'s, $y_i$'s, and $z_i$'s are all in $[1, 2005].$ Call a positive integer $1 \le n \le 2005$ [b] sketchy [/b] if it only appears one time among the $x_i$'s, $y_i$'s, and $z_i$'s. In other words, in the multiset of all $x_i$'s, $y_i$'s, and $z_i$'s (of size $3 \cdot 2005 = 6015$, call this multiset $M$), $n$ only appears once. Else, call $n$ [b] rigorous [/b]. Note that any rigorous $n$ must appear at least twice, since otherwise we would never know what $a_n$ is. Now, let's define the function $f(n)$ for $1 \le n \le 2005$ by letting it equal $\frac1a$, where $a$ is the number of times that $n$ appears in $M$. Observe that the sum $\sum_{k=1}^{1002} (f(x_k) + f(y_k) + f(z_k))$ is exactly equal to $2005$. Now, observe that $f(x_k) + f(y_k) + f(z_k) \le 2$ for all $1 \le k \le 1002.$ Indeed, if it was greater than $2$, then we must have that at least two of $f(x_k), f(y_k), f(z_k)$ are $1$'s, WLOG $f(x_k) = f(y_k) = 1.$ However, this means that the values of $a_{x_k}$ and $a_{y_k}$ are indistinguishable, impossible. Therefore, as $f(x_k) + f(y_k) + f(z_k) \le 2$, we have that $2005 = \sum_{k=1}^{1002} (f(x_k) + f(y_k) + f(z_k)) \le 2 \cdot 1002 = 2004,$ contradiction. Hence, our assumption that we finished in $1002$ questions was incorrect, and the answer is $\boxed{1003}.$

[/hide]


probability
AMC
AIME
AIME II
number theory
relatively prime

Source: AIME II 2001
11. Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 mathches, the probabilities that Club Truncator will win, lose, or tie are each $\frac{1}{3}$. The probability that Club Truncator will finish the season with more wins than losses is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Isn't this an application of the multinomial theorem? Of course, we don't necessarily have to use it here.

Clearly the probability that the team wins more than looses is the same as the probability if the team looses more than wins.  Call that $P$ and add those together.  Then add in the times where the same wins and losses, call this $Q$. This covers all possibilities so: \[2P+Q=1\] We can easily calculate $Q$ because that is just the probability of having let P(win,loss,tie) denote the probability that occurs. Then:

$P(0,0,6)=\frac{6!}{0!0!6!}*\frac{1}{3^{6}}$

$P(1,1,4)=\frac{6!}{1!1!4!}*\frac{1}{3^{6}}$

$P(2,2,2)=\frac{6!}{2!2!2!}*\frac{1}{3^{6}}$

$P(3,3,0)=\frac{6!}{3!3!0!}*\frac{1}{3^{6}}$

Add these to get \[Q=\frac{1}{3^{6}}*(1+30+90+20)=\frac{47}{243}\] Then we can compute $P$: \[P=\frac{1-Q}{2}=\frac{98}{243}\] It follows that $m+n=98+243=\boxed{341}$


let f and g two functions continuous on [a, b] let us suppose that for any x of [a, b] we have f (x) sup g (x).   show that there exists a strictly positive such f (x) sup= g (x) +a

Sorry diane, I don't understand you  :maybe:
What does "f(x) sup g(x)" and "f(x) sup = g(x) + a" mean?
Is "sup" the supremum?

$f,g$ continuous $\Rightarrow$ $f-g$ continuous on $[a,b]$. Hence $f-g$ achieves its minimum on $[a,b]$,  $c$. Because $\forall x \in [a,b]: (f-g)(x) > 0$, we have $c>0$.
Then $f(x) = g(x)+(f-g)(x) \geq g(x)+c > g(x)+\frac{c}{2}( \forall x \in [a,b])$


In the figure, the sides $AB$ and $AC$ of the equilateral triangle $ABC$ have been extended, and the circle tangent to these extended sides and to $BC$, as shown, has radius 4. What is the perimeter of $\triangle ABC$?

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=416[/img]

a) $4 \sqrt 3$
b) $6 \sqrt 3$
c) $8 \sqrt 3$
d) 12
e) 14

[hide]
label the center of the circle $O$ and the perpendicular dropped onto the extension of $AB$, $D$. $OAD$ is a 30-60-90 triangle. Let the point of tangency between $\triangle ABC$ and circle with center $O$ be labeled $P$. $PAB$ is also a 30-60-90 triangle.

so $AO = 8$ and $AP = 4$ thus $AB = \frac{8}{\sqrt{3}}$ and the perimeter = $8\sqrt{3}$ c)[/hide]

[quote="agolsme"][hide]
$OAD$ is a 30-60-90 triangle.[/hide][/quote]
Why is that? :huh:

Because it's a perpendicular so one of its angles is 90 and one of its angles is an angle of the triangle $ABC$ which is equilateral so it's 60. The last one must be 30. :D

[quote="JesusFreak197"]Oh, it's equilateral. I didn't catch that, looking at the drawing. :roll: (Yes, I understand that it's hard to make good drawings. I have trouble with it, too. :P)[/quote]
If you really need help in drawing pictures, you can try a find a Geometer's Sketchpad on the internet.

LOL. My drawing is perfectly fine.  :rotfl:

Plus, if Jesusfreak read the question [b]carefully[/b], it should've been clear.

[quote="JesusFreak197"][quote="shobber"]If you really need help in drawing pictures, you can try a find a Geometer's Sketchpad on the internet.[/quote]

Well, I have never yet drawn a picture on the computer, and I don't think Geometer's Sketchpad helps with normal drawings... does it? Anyway, I generally do pretty well, but a lot of times I'll get proportions way off and stuff. :P[/quote]

Geometer's Sketchpad is really good, but it isn't free :( MetaPost is probably the best though.

[quote="shobber"][hide]Connect the centre of the ex-circle and $A$. Then $h_{BC}=4$, hence: $C=8\sqrt{3}$. [/hide][/quote]

What is $h_{BC}$?


[color=purple] Let an acute triangle $ ABC$ in plane which has three exraii $ r_{a},r_{b},r_{c}$ and three medians $ m_{a},m_{b},m_{c}$. With any $ k\ge0$ prove that
  $ \frac {r_{b}^k \plus{} r_{c}^k}{m_{b}^k \plus{} m_{c}^k} \plus{} \frac {r_{c}^k \plus{} r_{a}^k}{m_{c}^k \plus{} m_{a}^k} \plus{} \frac {r_{a}^k \plus{} r_{b}^k}{m_{a}^k \plus{} m_{b}^k}\ge 3$
  [/color]

[quote="evarist"][color=purple] Let an acute triangle $ ABC$ in plane which has three exraii $ r_{a},r_{b},r_{c}$ and three medians $ m_{a},m_{b},m_{c}$. With any $ k\ge1$ prove that
  $ \frac {r_{b}^k \plus{} r_{c}^k}{m_{b}^k \plus{} m_{c}^k} \plus{} \frac {r_{c}^k \plus{} r_{a}^k}{m_{c}^k \plus{} m_{a}^k} \plus{} \frac {r_{a}^k \plus{} r_{b}^k}{m_{a}^k \plus{} m_{b}^k}\ge 3$
  [/color][/quote]


Da !
Este o ecuatie draguta cu multe solutii


Solve in $\mathbb{R}$:
\[ \begin{displaymath} \left\{ \begin{array}{11} x^2+3y=9 \\ y^4+4(2x-3)y^2-48(x+y)+155=0 \end{array} \right \end{displaymath}  \]


Find the least positive integer $ n$ with the following property: if $ n$ distinct sums of the form $ x_p\plus{}x_q\plus{}x_r,1\le p<q<r\le 5$, equal $ 0$, then $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}0$.

[i]Sava Grozdev, Svetlozar Doychev[/i]

What are $ x_1$, $ x_2$, $ x_3$, $ x_4$, and $ x_5$? Real numbers?


On a standard clock, how many degrees does the minute hand rotate during the time period from 4:12 pm to 4:48 pm of the same day?

The minutes hand travels $ \frac{360}{60}\equal{}6$ degrees every minute. Since $ 36$ minutes elapse during the time period, the minute hand rotates $ 36\cdot6\equal{}\boxed{216}$ degrees during the time period.

Alternately, you could set up an fraction, like $ \frac{36}{60}\cdot360\equal{}\boxed{216}$ degrees in the time period.


F FROM N TO N and F is surjective and bijective mapping  prove that there are a<b<c that F(a)+F(c)=2F(c)

do you mean $ f(a)\plus{}f(c)\equal{}2f(b)$?


algebra
polynomial
Vieta
number theory unsolved
number theory

1.Determine all triples of positive integer (x,y,z) solution to the equation

(x+y)^2+3x+y+1=z^2

2.FInd all integer solution to the equation

(x^2-y^2)^2=1+16y

3.Let a b c d e positive integer such that

     abcde=a+b+c+d+e Find the maximum possible value of max{a,b,c,d,e}

4.Let a and b be positive integer such that ab+1 divides a^2+b^2 Show that a^2+b^2/ab+1 is the square of an integer

thank you

[quote="poonprang"]1.Determine all triples of positive integer (x,y,z) solution to the equation

(x+y)^2+3x+y+1=z^2[/quote]

So $ (x\plus{}y)^2<z^2<(x\plus{}y\plus{}2)^2\equal{}(x\plus{}y)^2\plus{}4x\plus{}4y\plus{}4$ and so $ z\equal{}x\plus{}y\plus{}1$

So $ (x\plus{}y)^2\plus{}3x\plus{}y\plus{}1\equal{}(x\plus{}y\plus{}1)^2$ =$ (x\plus{}y)^2\plus{}2x\plus{}2y\plus{}1$ and so $ x\equal{}y$

Hence the solutions : $ (u,u,2u\plus{}1)$ $ \forall u\in\mathbb N$

[quote="poonprang"]2.FInd all integer solution to the equation

(x^2-y^2)^2=1+16y [/quote]

Obviously $ y\geq 0$ and $ 1\plus{}16y$ is an odd square. So we can write $ 1\plus{}16y\equal{}(2n\plus{}1)^2$ for some $ n\geq 0$

$ \implies$ $ y\equal{}\frac{n^2\plus{}n}{4}$ and $ x^2\minus{}y^2\equal{}\pm(2n\plus{}1)$ $ \implies$ $ x^2\equal{}y^2\pm(2n\plus{}1)$

$ \implies$ $ 2n\plus{}1\geq 2y\minus{}1$ and so $ y\leq n\plus{}1$ and so $ \frac{n^2\plus{}n}{4}\leq n\plus{}1$ and so $ 0\leq n\leq 5$. Then :

$ n\equal{}0$ $ \implies$ $ y\equal{}0$ $ \implies$ $ (x^2)^2\equal{}1$ and two solutions $ (\minus{}1,0)$ and $ (1,0)$

$ n\equal{}1$ $ \implies$ $ y\equal{}\frac 12$ so no solution

$ n\equal{}2$ $ \implies$ $ y\equal{}\frac 32$ so no solution

$ n\equal{}3$ $ \implies$ $ y\equal{}3$ $ \implies$ $ (x^2\minus{}9)^2\equal{}49$ and two solutions $ (\minus{}4,3)$ and $ (4,3)$

$ n\equal{}4$ $ \implies$ $ y\equal{}5$ $ \implies$ $ (x^2\minus{}25)^2\equal{}81$ and two solutions $ (\minus{}4,5)$ and $ (4,5)$

$ n\equal{}5$ $ \implies$ $ y\equal{}\frac {15}2$ so no solution

Hence the solutions : $ (\minus{}1,0),(1,0),(\minus{}4,3),(4,3),(\minus{}4,5)$ and $ (4,5)$

$ 2)(x^{2}\minus{}y^{2})^{2}\equal{}16y\plus{}1$  $ \implies$
$ (x\minus{}y)^{2}.(x\plus{}y)^{2}\equal{}16y\plus{}1$
$ (x\plus{}y)^{2}\leq 16y\plus{}1$  if $ x\geq 5$ $ \implies$
$ y^{2}\plus{}10y\plus{}25\leq 16y\plus{}1$ $ \implies$ $ y^{2}\minus{}6y\plus{}24\leq0$ Contradiction!
$ x\leq 4$  we know if x is a solution -x is  a solution.
$ x\geq 0$
$ i)x\equal{}0$ $ \implies$  $ y^{4}\equal{}16y\plus{}1$ $ \implies$   $ y|1 ,y\equal{}1,\minus{}1$ $ \implies$ we get any solution!
$ ii)x\equal{}1$  $ \implies$ $ y\plus{}1|16y\plus{}1$ $ y\plus{}1|15$
$ y\plus{}1\equal{}1,y\plus{}1\equal{}3,y\plus{}1\equal{}5,y\plus{}1\equal{}15$ $ \implies$
$ y\equal{}0,y\equal{}2,y\equal{}4,y\equal{}14$ $ \implies$ $ (1,0)$ is a solution.
$ iii)x\equal{}2$ $ \implies$  $ y\plus{}4|16y\plus{}1$ $ \implies$ $ y\plus{}4|63$
$ y\plus{}4\equal{}7,y\plus{}4\equal{}9,y\plus{}4\equal{}21,y\plus{}4\equal{}63$  and $ y^{2}\minus{}16y\minus{}5\leq 0$
$ y\leq 16$ $ \implies$ $ y\equal{}3,y\equal{}5$ $ \implies$ we get any solution!
$ iv)x\equal{}3$ $ \implies$ $ y^{2}\plus{}6y\plus{}9\leq 16y\plus{}1$ $ \implies$
$ y^{2}\minus{}10y\plus{}9\leq 0$ $ \implies$  $ 1\leq y \leq 9$
$ y\plus{}9|16y\plus{}1$ $ \implies$ $ y\plus{}9|143$ $ y\plus{}9\equal{}13 ,y\plus{}9\equal{}11$
$ \implies$ $ y\equal{}4,y\equal{}2$ $ \implies$ we get any solution!
$ v) x\equal{}4$ $ \implies$ $ y^{2}\plus{}8y\plus{}16\leq 16y\plus{}1$  $ \implies$
$ y^{2}\minus{}8y\plus{}15\leq0$ $ \implies$ $ (y\minus{}5)(y\minus{}3)\leq0$ $ \implies$ $ y\equal{}3,y\equal{}4,y\equal{}5$
$ \implies$ $ (4,5),(4,3)$ are solution.
all solution $ \{(\minus{}1,0),(1,0),(4,3),(4,5),(\minus{}4,3),(\minus{}4,5)\}$  :wink:

[quote="poonprang"]3.Let a b c d e positive integer such that

     abcde=a+b+c+d+e Find the maximum possible value of max{a,b,c,d,e}
[/quote]

Wlog say $ a\geq b\geq c\geq d\geq e\geq 1$

If $ d\geq 2$, then $ abcde\geq 8a$ while $ a \plus{} b \plus{} c \plus{} d \plus{} e\leq 5a$. So $ d \equal{} e \equal{} 1$ and the equation is $ abc \equal{} a \plus{} b \plus{} c \plus{} 2$
If $ c\geq 3$, $ abc\geq 9a$ while $ a \plus{} b \plus{} c \plus{} 2\leq 3a \plus{} 2\leq 5a$ so either $ c \equal{} 2$, either $ c \equal{} 1$
If $ c \equal{} 2$, the equation is $ 2ab \equal{} a \plus{} b \plus{} 4$ and $ a \equal{} b \equal{} 2$
If $ c \equal{} 1$, the equation is $ ab \equal{} a \plus{} b \plus{} 3$ and so $ a(b \minus{} 1) \equal{} b \plus{} 3$ and so $ b \minus{} 1|b \plus{} 3$ and so $ b\in\{2,3,5\}$

$ b \equal{} 2$ $ \implies$ $ a \equal{} 5$
$ b \equal{} 3$ $ \implies$ $ a \equal{} 3$
$ b \equal{} 5$ $ \implies$ $ a \equal{} 2$, impossible since $ a\geq b$

So $ abcde \equal{} a \plus{} b \plus{} c \plus{} d \plus{} e$ has just 3 ordered solutions in positive integers : $ (2,2,2,1,1),(5,2,1,1,1)$ and $ (3,3,1,1,1)$

Hence the answer : $ 5$

$ 1 \equal{} \frac {1}{bcde} \plus{} \frac {1}{acde} \plus{} \frac {1}{abde} \plus{} \frac {1}{abce} \plus{} \frac {1}{abcd}$
$ a\geq b\geq c \geq d\geq e$ $ \implies$
$ e\geq 2$ $ \implies$ $ \frac {1}{bcde} \plus{} \frac {1}{acde} \plus{} \frac {1}{abde} \plus{} \frac {1}{abce} \plus{} \frac {1}{abcd} < 1$
$ e \equal{} 1$ $ \implies$ $ d\geq 2$ $ \implies$
$ \frac {1}{bcde} \plus{} \frac {1}{acde} \plus{} \frac {1}{abde} \plus{} \frac {1}{abce} \plus{} \frac {1}{abcd} < 1$
$ d \equal{} 1$ $ \implies$
$ abc \equal{} a \plus{} b \plus{} c \plus{} 2$ $ \implies$
$ 1 \equal{} \frac {1}{bc} \plus{} \frac {1}{ac} \plus{} \frac {1}{ab} \plus{} \frac {2}{abc}$ $ \implies$
$ c\geq 3$  $ \frac {1}{bc} \plus{} \frac {1}{ac} \plus{} \frac {1}{ab} \plus{} \frac {2} {abc} < 1$
$ c \equal{} 1$ V $ c \equal{} 2$
$ i)c \equal{} 1,ab \equal{} a \plus{} b \plus{} 3$ $ \implies$ $ a(b \minus{} 1) \equal{} b \plus{} 3$
$ \implies$ $ b \minus{} 1|4$ $ b \minus{} 1 \equal{} 1,b \minus{} 1 \equal{} 2,b \minus{} 1 \equal{} 4$ $ \implies$
$ b \equal{} 2,b \equal{} 3,b \equal{} 5$ $ \implies$ $ a \equal{} 5,a \equal{} 3,a \equal{} 2$
$ a\geq b$ $ \implies$ $ (5,2),(3,3,)$
$ ii)c \equal{} 2,2ab \equal{} a \plus{} b \plus{} 3$ $ \implies$  $ a(2b \minus{} 1) \equal{} b \plus{} 3$
$ 2b \minus{} 1\leq b \plus{} 3$ $ \implies$ $ b\leq 4$
$ (b \equal{} 1,a \equal{} 4),(b \equal{} 4,a \equal{} 1)$ $ \implies$ $ (4,1)$
  :wink:

[quote="cnyd"] $ ii)c \equal{} 2,2ab \equal{} a \plus{} b \plus{} 3$ [/quote]

No, $ 2ab\equal{}a\plus{}b\plus{}4$
 :wink:


Prove that in any triangle $ ABC$ there are the inequalities

$ 1\blacktriangleright\ \ \boxed {\ \frac {1}{2R^2}\ \le\ \frac {1}{b^2 \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} a^2} \plus{} \frac {1}{a^2 \plus{} b^2}\ \le\ \frac {1}{4Rr}\ }$ .

$ 2\blacktriangleright\ \ \frac 12 \plus{} \frac rR\ \le \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\ \le\ 1$ .

[b]Remark.[/b] You can use the first inequality from [url=http://www.mathlinks.ro/viewtopic.php?t=293296][color=red][b]here[/b][/color][/url].

hello, for your first inequality, setting $ R \equal{} \frac {abc}{4A}$ with $ A\equal{}\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$ and $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$ your inequality is equivalent to
$ \frac {8s(s \minus{} a)(s \minus{} b)(s \minus{} c)}{a^2b^2c^2}\le \frac {1}{b^2 \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} a^2} \plus{} \frac {1}{a^2 \plus{} b^2}$ and after some algebraic manipulations we get
$ a^2b^2(a^2 \minus{} b^2)^2(a^2 \plus{} b^2) \plus{} a^2c^2(a^2 \minus{} c^2)^2(a^2 \plus{} c^2) \plus{} b^2c^2(b^2 \minus{} c^2)^2(b^2 \plus{} c^2)\geq0$ which is true.
Sonnhard.

hello, for the second one we have the right-hande side as $ ab \plus{} bc \plus{} ca \le a^2 \plus{} b^2 \plus{} c^2$, this is well-known. The left-hand side is equivalent with
$ \frac {1}{2} \plus{} \frac {4A^2}{s \cdot abc}\le\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$. Let $ c$ the minimum of $ a,b$, then we get after same algebra that the inequation above is equivalent to
$ (a \minus{} b)^2\left(a^2(a \plus{} b \minus{} c) \plus{} ab(b \minus{} c) \plus{} b^2(b \minus{} c)\right) \plus{} c^3(a \minus{} c)(b \minus{} c)\geq0$.
Sonnhard.

hello, to the right-hande side of your first inequality:
we have $ \frac{1}{4Rr}\equal{}\frac{s}{abc}$ with $ R\equal{}\frac{abc}{4A}$ and $ r\equal{}\frac{A}{s}$. So we have to prove
$ \frac{1}{a^2\plus{}b^2}\plus{}\frac{1}{a^2\plus{}c^2}\plus{}\frac{1}{b^2\plus{}c^2}\le \frac{s}{abc}$ with $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$. By the AM-GM inequality we get
$ \frac{1}{b^2\plus{}c^2}\le \frac{1}{2bc}$
$ \frac{1}{a^2\plus{}c^2}\le \frac{1}{2ac}$
$ \frac{1}{b^2\plus{}a^2}\le \frac{1}{2ba}$
adding the right-hande sides above we get $ \frac{1}{2}\left(\frac{a\plus{}b\plus{}c}{abc}\right)$
Sonnhard.


inequalities
function
inequalities proposed

Let $t_{1},t_{2},\cdots,t_{n}$ be positive reals such that $\boxed{t_{1}t_{2}\cdots t_{n}=1}$. Prove that there exists an integer $1\leq k\leq n$ such that $t_{k}(t_{k+1}+1)\geq 2$
 :)

assume the contrary we get that:

$t_{1}(t_{2}+1)<2$
.
.
.
$t_{n-1}(t_{n}+1)<2$
$t_{n}(t_{1}+1)<2$

multiplying the above $n$ inequalities we get that:

$t_{1}t_{2}...t_{n}(t_{1}+1)...(t_{n}+1)<2^{n}$

so $(t_{1}+1)...(t_{n}+1)<2^{n}$

now by $AM-GM$ we have that $t_{i}+1\geq 2\sqrt{t_{i}}$ applying this for the last inequality we get that:

$2^{n}\sqrt{t_{1}...t_{n}}\leq (t_{1}+1)...(t_{n}+1)<2^{n}$

so we get that:

$2^{n}\leq (t_{1}+1)...(t_{n}+1)<2^{n}$

so we get that $2^{n}<2^{n}$ which is a contradiction...

so there exists $k$ s.t $t_{k}(t_{k+1}+1)\geq 2$

Umm, Babak, just a quick note... The contradiction of $\ge$ is actually $<$, not $\le$.

yeah sorry it was a little typo,I edited my post...
tnx  :)

[quote="BaBaK Ghalebi"]
$t_{1}(t_{2}+1)<2$
.
.
.
$t_{n-1}(t_{n}+1)<2$

multiplying the above $n$ inequalities we get that:

$t_{1}t_{2}...t_{n}(t_{1}+1)...(t_{n}+1)<2^{n}$

[/quote]

multiplying the above inequalities you'll get that:

$t_{1}t_{2}...t_{n-1}(t_{2}+1)...(t_{n}+1)<2^{n}$

actually the correct thing is $1\leq k\leq n$ where we'll define $a_{n+1}=a_{1}$ , well i must have said this in the problem...anyway you're solution is correct and we'll have:

$t_{1}(t_{2}+1)<2$
.
.
.
$t_{n-1}(t_{n}+1)<2$
$t_{n}(t_{1}+1)<2$

Now by multiplying these inequalities we'll have:

$t_{1}t_{2}...t_{n}(t_{1}+1)...(t_{n}+1)<2^{n}$

and ...

Then, you can generalize like this:

Let $f$ be a function defined on integers which takes up positive real values, with period $n$ such that $f(2007)f(2008)f(2009)\cdots f(n+2006)=1$. Show that for every integer $l$ there exists infinitely many integers $k$ such that $f(k)(f(k+l)+1)\ge 2$.

The proof is quite similar. Am I right? What do you think about the generalization?

yeah I have edited my post...
and about the generalization,well if for every $l$ there exists such $k$ then by the period of $f$ we`ll have infinitely many such $k$s...
there might be better generalizations which gives another condition on $f$ (different from periodic) and asks for infinetly many $k$s...

Yeah... I agree, there probably is a better generalization, but this is all I can think of right now... It is essentially the same problem as the one sinajackson posted, but it seems kind of more difficult when you see it for the first time because it's more abstract.

yeah at the first look it seems pretty difficult specially the fact that $f$ is a function not a polynomial...


Uh yes. That has happened to me before. What is so significant about it?

What would be interesting if it was a ranking game, and both people got it correct.

Holy.. Yeah that would be interesting. And 1-1!  :lol:

What if that happens on the last question of a countdown?

It probably would result in not a draw, but as if both won at the same time, so both players earn rating points.

Just because they only show 3 digits doesn't mean they entered at the *exact* same time. So probably the one who entered it a nanosecond before.

in a countdown, it would glitch, as always.


trigonometry
algebra
function
domain
rotation
analytic geometry
geometry

[size=134][b]Problem 6[/b] (13 points)[/size]
[size=134][b]Casmir Effect[/b][/size]

[b]Part-A[/b] What is the pressure of ideal gases when the temperature goes to absolute zero? (2 points)

[b]Part-B[/b] Casmir proposed in the 1950’s that vacuum is actually filled with virtual electromagnetic waves, and the energies stored in the vacuum produces an observable force between two parallel metallic plates separated by distance $d$. The electromagnetic energy stored between the plates is given in Quantum Mechanics by $E=\frac12 \sum_{n<p}E_{n}$, where $E_{n}=\eta ck_{n}$, $c$ is the velocity of light and $\eta$ is a fundamental constant called Planck’s constant, $p\, (\gg 1)$ is a large number that depends on the material properties of the plates. $k_{n}\equiv 2\pi/\lambda_{n}$, where $\lambda_{n}$ is the wavelength of a standing electromagnetic wave that fits into the space between the plates.

(a) Assume that the waves take the form of $\sin k_{n}x$, what are the possible values of $k_{n}$ such that at the plate surfaces $x=0$ and $x=d$ the standing waves vanish? (2 points)

(b) Show that the vacuum electromagnetic force is of the form $a/d^{2}$. What is the constant $a$? (5 points)

(c) What is the force due to the vacuum electromagnetic waves outside the two plates? (2 point)

(d) Find the amplitude of the force when $d=1.0\,\text{mm}$. (Hint: If you cannot get $a$ in (b), use dimensional analysis to estimate its order of magnitude.) (2 points)

($\eta=1.05\times10^{-34}\,\text J\cdot\text s$, $p=2000$, $c=3.0\times10^{8}\,\text m/ \text s$)

[size=134][b]Problem 7[/b] (15 points)[/size]
[size=134][b]Electric Current Driven Magnetic Domain Movement[/b][/size]

Electrons and some atoms have permanent angular momentum called [b]spins[/b] ($\vec{\mathbf S}$) that are fixed in amplitude while their directions can be changed by an external torque. Associated with the spins are the magnetic moments that make them as individual magnetic dipoles $\vec{\mathbf m}=\mu_{B}\vec{\mathbf S}$, where $\mu_{B}$ is a constant called magneton. For simplicity, we assume that the electrons and the atoms here have the same spin amplitude and magneton.

(a) Refer to Figure-A below. Dipole-A is at $\vec{\mathbf r}=\vec{\mathbf 0}$ and along $\vec{\mathbf z}_{0}$, and Dipole-B is at $\vec{\mathbf r}=a\vec{\mathbf x}_{0}$ and along $\vec{\mathbf y}_{0}$. Find the torque of Dipole-A on Dipole-B. (3 points)

(b) Find the direction of Dipole-B after a short time $\Delta t$, assuming that the direction of Dipole-A remains unchanged. (5 points)

(c) Figure-B is a schematic of a magnetic domain in $(x_{0},x_{0}+2b)$ that contains many atoms (Dipole-B’s). The directions of the dipoles are pointed in an ordered fashion, as shown. Dipole-B’s are at the fixed sites of equal distance along the x-axis (within the space of $2b$ there are $10^{4}$ dipole-B’s). An electron (Dipole-A) flies by and its action on the Dipole-B [u]is the same as your answer in (b)[/u]. Figure-C shows the initial angular profile of the dipoles. The dipoles at $x_{0}+b$ are along $\vec{\mathbf y}_{0}$, the dipoles at $x\leq x_{0}$ are along $-\vec{\mathbf x}_{0}$, and that at $x\geq x_{0}+2b$ are along $\vec{\mathbf x}_{0}$. Dipoles along $-\vec{\mathbf x}_{0}$ can no longer rotate. Dipoles along $\vec{\mathbf x}_{0}$ cannot rotate either, except for the ones close to the domain edge which can rotate freely as the ones pointing in other directions. Given the electric current carried by the electrons as $I$, find the speed of the movement of the domain center when the current is just turned on. (Take the electron charge $e$ as known) (5 points)

(d) Find the time it takes for the whole domain to pass the $x_{0}+2b$ point. (2 points)

(Hint: Magnetic field $\vec{\mathbf B}(\vec{\mathbf r})$ at position $\vec{\mathbf r}$ due to a magnetic dipole $\vec{\mathbf m}$ at coordinate origin is given by
\[\vec{\mathbf B}(\vec{\mathbf r})=\frac{\mu_{0}}{4\pi r^{3}}\left[\frac{3(\vec{\mathbf m}\cdot\vec{\mathbf r})\vec{\mathbf r}}{r^{2}}-\vec{\mathbf m}\right] \]
Torque $\vec{\mathbf N}$ of magnetic field $\vec{\mathbf B}$ on a dipole $\vec{\mathbf m}$ is $\vec{\mathbf N}=\vec{\mathbf m}\times\vec{\mathbf B}$.)

[size=134][b]Problem 8[/b] (22 points)
[b]Force between a conductor sphere and a conductor plane[/b][/size]

The uniqueness theorem of the theory of electric field ensures that virtual charges could be placed [b][i]outside[/b][/i] the closed space where the electric field is to be determined to mimic the real boundary conditions. For example, in Figure-A below, to find the electric field when a point charge $q$ is placed in front of a large planar conductor at zero potential, a virtual, or 'image' charge $-q$ is placed at the mirror-image position so that the combined contribution of the two charges is to make the potential everywhere on the plane zero. The force on $q$ by the plane is then the same as the force by the image charge.

[b]Part-A[/b]
Consider the force of a conductor sphere of radius $R$ held at zero potential on a point charge $q$ at a distance $d\, (>R)$ from the sphere center. The image charge $q'$ should be inside the sphere (outside the space outside the sphere) and on the line joining the sphere center and $q$, as shown in Figure-B. The combined contribution of $q$ and $q'$ is to make the potential everywhere on the sphere surface zero. Find the position and the value of $q'$. (4 points)

[b]Part-B[/b]
As shown in Figure-C, a problem often encountered in atomic force microscopy is to determine the force of the sample on the probe tip, which is a conductor sphere of radius $R$ held at electric potential $V$ at distance $h_{0}$ from a large conducting sample surface at zero potential. In the following we will apply the 'image charge' method step by step to find the force.

(a) We start by putting a point charge $q_{0}$ inside the sphere such that the sphere surface is an equal-potential one with voltage $V$, while ignoring the effect of the conductor plate. Determine $q_{0}$ in terms of $V$ and $R$. (1 point)

(b) Determine the value and position ($h_{1}$) of the image charge of $q_{0}$ and call it $q_{1}$, such that the combined contribution of them is to make the sample surface a zero-potential plane. (2 points)

(c) The presence of $q_{1}$ now makes the conductor sphere surface no longer equal-potential. Put another point charge $q_{2}$ inside the sphere such that the combined effect of $q_{0}, q_{1}$ and $q_{2}$ is to make the potential on the sphere surface equal to $V$ again. Determine $q_{2}$ and its position $h_{2}$. (2 points)

(d) Repeat (b) to determine the image charge $q_{3}$ of $q_{2}$, and repeat (c) to find the image charge $q_{4}$ of $q_{3}$. Derive the general expression between $h_{2n}$ and $h_{2(n+1)}$, $q_{2n}$ and $q_{2(n+1)}$, and $q_{2n+1}$ and $q_{2(n+1)+1}$, $n=0,1,2,\cdots$ (3 points)

(e) Derive the total force of the sample on the sphere [b][i][u]in the form of a summation over an infinite series[/u][/i][/b]. (2 points)

(f) Suppose the force in (e) is $1.1\times10^{-12}\,\text N$ when $V=V_{0}$, $R=1.0 \times10^{-8}\,\text m$, and $h_{0}=5.0\times10^{-8}\,\text m$, find the force when $V= 2V_{0}, R = 1.0\,\text m$, and $h_{0}= 5.0\,\text m$. (4 points)

(g) Given $R/h_{0}= 1/51$, to determine the force in (e) up to ~1% accuracy, what are the terms that should be kept? (2 points)

(h) Estimate the order of magnitude of the relative error caused by the terms neglected in (g). (2 points)


The areas of two squares differ by 100 square units, and the perimeters of the two squares differ by 10 units. What is the perimeter, in units, of the smaller square?


function
calculus
derivative
parameterization
real analysis
real analysis unsolved

Hey I had a quick question on differentials

When you differentiate

$p_{i}= \sqrt{(x_{i}-x_{u})^{2}+(y_{i}-y_{u})^{2}+(z_{i}-z_{u})^{2}}$

don't you get:

$\Delta p_{i}=-\Big( \frac{ (x_{i}-x_{u}) \Delta x_{u}+(y_{i}-y_{u}) \Delta y_{u}+(z_{i}-z_{u}) \Delta z_{u}}{p_{i}-b_{u}}\Big)+\Delta b_{u}$

and not

$\Delta p_{i}= \frac{ (x_{i}-x_{u}) \Delta x_{u}+(y_{i}-y_{u}) \Delta y_{u}+(z_{i}-z_{u}) \Delta z_{u}}{p_{i}-b_{u}}+\Delta b_{u}$

which is what my book says?

The notation is unclear.  $p$ is a function of up to 6 variables.  Which ones are variables for differentiation and which are treated as fixed parameters?   Also $b_{u}$ is undefined and $\Delta$ usually means something unrelated to this problem.


Prove that $ ab^5 \plus{} bc^5 \plus{} ca^5\ge abc(a^2b \plus{} b^2c \plus{} c^2a)$, for $ a,b,c > 0$.

Hint: Try using AM GM with Power mean inequality.

$ \frac {8ab^5 \plus{} 2bc^5 \plus{} 11ca^5}{21} \ge a^3b^2c$ by AM-GM. Add this cyclically. Can we use Muirhead, I thought it had to be symmetric in all variables?

dgreenb801 i have tried muirhead on this one the case comes of to be somthing like this...

$ a^5c\plus{}a^5b\plus{}b^5a\plus{}b^5c\plus{}c^5a\plus{}c^b\ge\;a^3b^2c\plus{}a^3bc^2\plus{}ab^3c^2\plus{}a^2b^3c\plus{}ab^2c^3\plus{}a^2bc^3$

now we get a sum of two similar inequalities..that we require to prove

i think this can be AM GM (but seriously i don't know how...i'm still trying)

again: see here
http://www.mathlinks.ro/Forum/viewtopic.php?t=103761 :)  :)


If $ a_{0}\equal{}a_{1}\equal{}1$ and $ a_{n\plus{}1}\equal{}2a_{n}\minus{}a_{n\minus{}1}$ then $ a_n$ is a perfect square for all $ n$ and also $ a_{n}\equal{}b_{n}^2$ where $ b_{n}$ is the general term of the sequence given by $ b_{0}\equal{}b_{1}\equal{}1$ and $ b_{n\plus{}1}\equal{}kb_{n}\minus{}b_{n\minus{}1}$ for all $ n\geq1$.

[quote="manjil"]If $ a_{0} \equal{} a_{1} \equal{} 1$ and $ a_{n \plus{} 1} \equal{} 2a_{n} \minus{} a_{n \minus{} 1}$ then $ a_n$ is a perfect square for all $ n$ and also $ a_{n} \equal{} b_{n}^2$ where $ b_{n}$ is the general term of the sequence given by $ b_{0} \equal{} b_{1} \equal{} 1$ and $ b_{n \plus{} 1} \equal{} kb_{n} \minus{} b_{n \minus{} 1}$ for all $ n\geq1$.[/quote]

$ a_n\equal{}1\equal{}1^2$ $ \forall n$
$ b_n\equal{}1$ $ \forall n$ and $ b_{n \plus{} 1} \equal{} 2b_{n} \minus{} b_{n \minus{} 1}$ for all $ n\geq1$


[quote="Iron Fist"]Sherwin buys three cans of beer and two math textbooks. Each math textbook costs four and a half times as much as a bottle of beer. If he spent all of his money on beer instead of beer and math, how many six-packs of beer could he have bought, assuming the price of a can of beer stays the same?[/quote] [hide="Solution"] Let's say the beer costs $x$ dollars, which means the math textbooks each cost $4.5x$. In total, Sherwin spends $3x+2\times4.5x=3x+9x=12x$. This means he can buy 12 beers or $\boxed{2}$ six-packs of beer. :D [/hide]

im truly amazed how did u come up wit that equation... i really feel like everyone here is naturally good at math, im struggling so hard to try to understand

Lol, it comes from practice... and being forced to do math when you're young...

[hide]Let b be the price of beer.
He spends a total of $2(4\frac{1}{2})b+3b=9b+3b=12b$
so he could of bought 2 six packs.[/hide]

[hide]Say the price of a beer is $\text{s}$.

So, the amount of money I was supposed to spend on both mathbooks and beer is $4.5x*2+3x=9x+3x=12x$

So with the amount of money to buy 12 beers, I can get $2$ six packs of beer!  :D

But with my chinese bargaining skills, I could have probably gotten 24 beers for the price of 12.  ;) [/hide]

Yeah, that is my answer.

Sherwin, are you okay?

[quote="lotrgreengrapes7926"][quote="daermon"][hide="answer"]4[/hide][/quote]
That's wrong.  :?[/quote]

Good job lotrgreengrapes7926, you are correct for saying that is incorrect. ;)

The answer is 2, 2 books cost the same amount as 9 beer cans, so there are 12 beer cans in all, so there are 2 packs.

Sherwin, I'm sending you to church every time you get drunk.

oops.  you guys are right.

drunken_math, are you the same Sherwin that came in second at the MathPath ping pong tournament?

[hide]Call the beer price $b$. That means the textbook money is $\frac{9}{2}b$. $3b+2\left(\frac{9}{2}\right)=12b$. $\frac{12}{6}=2$, so he can buy $\boxed{2}$ 6-packs of beer.[/hide]

[hide]he buys beer and math textbooks?!?!?  :| [/hide]
[hide]assume that a beer's price=b and textbook's price is m...We know that m=4,5b...we also know that he has 3b +2m=3b+2 *(4,5b)=3b+9b=12b 12b/6b=[color=red]2[/color][/hide]

[quote="Iron Fist"]Sherwin buys three cans of beer and two math textbooks. Each math textbook costs four and a half times as much as a bottle of beer. If he spent all of his money on beer instead of beer and math, how many six-packs of beer could he have bought, assuming the price of a can of beer stays the same?[/quote]

2 six packs


Let F be an ordered field in which every strictly monotone increasing sequence bounded above converges. Prove that F is complete.

For any Cauchy sequence, consider its limit inferior.


This was the 1st problem in the South African talent search round 5.

Prove that there exist
a)  5 points in the plane so that among all the triangles with vertices among these points, there are 8 right-angled triangles.
b)  64 points in the plane so that there are at least 2005 right-angled triangles with vertices among these points.

Oh, sorry. Well South Africa quite often uses problems from other olympiads, I should have known it would be somewhere

Don't worry about it!  I was just posting the link as a reference.

I think this works for part a: (0,0),(a,0),(0,a),(a,a),(a/2,a/2). By the way how do you get that fancy mth expository font that everyone writes in? Copy it from Word Equation Editor?

You mean $\text {this }$ kind of writing? It's called $\text{\LaTeX} \text { and can be used to typeset mathematics like this: } \sqrt{5+\frac{x}{\sqrt{x+\sqrt{2}}}}.$

Click here to begin how to learn it: http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

:D :D :D


algebra polynomial calculus integration function domain induction

Let $ A\left[x\right]$ be the ring of polynomials with coefficients in a commutative ring $ A$. Let $ f\equal{} a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} .... \plus{} a_nx^n$ be an element in $ A\left[x\right]$.

[b]1.[/b] Show that $ f$ is a unit iff $ a_0$ is a unit in $ A$ and $ a_1$, $ a_2$, ..., $ a_n$ are nilpotent.

[b]2.[/b] Show that $ f$ is a zero divisor iff there exists $ a\neq 0$ in $ A$ such that $ a\cdot f\equal{}0$.

1. Let $g=b_0 + b_1 x + ... + b_m x^m$ be an inverse polynomial of $f=a_0 + a_1 x + \dotsc + a_n x^n$. Then $b_0 a_0 = 1$, so that $a_0$ is a unit in $A$. The trick is now to consider the nilradical of $A$, which consists of the nilpotent elements of $A$, but which is also the intersection of all prime ideals of $A$. So let's take a prime ideal $P$ of $A$. When $\overline{f} , \overline{g}$ are the corresponding polynomials over $A/P$, we have $\overline{f} \overline{g} = 1$, i.e. $\overline{f}$ is a unit in $(A/P)[x]$. But since $A/P$ is an integral domain, this means that $\overline{f}$ is constant (this is well-known, and can be proved by working with the degree of a polynomial). Thus, all coefficients $\overline{a_i}$ with $i>0$ have to be zero in $A/P$, i.e. we have $a_i \in P$. Because $P$ was arbitrary, $a_i$ is nilpotent.

Now let $a_0$ be a unit in $A$ and $a_i$ nilpotent for all $i>0$. We have to find an inverse polynomial $g = b_0 + b_1 x + \dotsc$ of $f$. We'll do that "backwards": The condition $f g = 1$ means $a_0 b_0 = 1$ and $a_0 b_k + \dotsc + a_k b_0 = 0$ for all $k>0$. Well, we find an appropriate $b_0$ because $a_0$ is a unit in $A$. The recursive formula $b_k = -b_0 (a_1 b_{k-1} + \dotsc + a_k b_0)$ determines all $b_i$, s.t. $f$ is a unit in $A[[x]]$. But it is easy to see that the $b_i$ vanish once $i$ is greater than $n=deg(f)$ and the nilpotency indices of the $a_i$. So $f$ is a unit in $A[x]$.

2. If $a*f = 0$ for some $0 \neq a \in A$, then $f$ is of course a zero divisor in $A[x]$. Now let $f$ be a zero divisor in $A[x]$, say $fg = 0$ for some $0 \neq g \in A[x]$. Write $f = a_0 + a_1 x + \dotsc + a_n x^n$. First assume $g a_i=0$ for all $i$. There's a coefficient $a \neq 0$ in $g$. It suffices $a f_i= 0$ for all $i$, hence $af = 0$, and we're done. Else there's a maximal $i \leq n$ s.t. $g a_i \neq 0$. Then $0 = gf = g a_{i+1} x^{i+1} + \dotsc + g a_n x^n$. The potentially hightest monomial in this polynomial is $b a_i x^{m+i}$, when $b x^m$ is the highest monomial in $g$. Thus, $b a_i = 0$, and the hightest monomial in $a_i g$ is smaller than the one in $g$. But every monomial in $a_i g$ occours - apart from the coefficient - in $g$. So we can make an induction and get an $g$ s.t. $g a_i = 0$ for all $i$. But this implies, as we have seen, the existence of some $0 \neq a \in A$ s.t. $a f = 0$.

When you say P von A, do you mean A/P (A mod P)? :lol: thanks for your help

I meant "P of A", see the edit. A/P is the factor ring A modulo P ...

How do we know that the nilradical is the intersection of all prime ideals in A?

Sorry, but why does A/P being a domain imply that f must be constant.   :)

You've posted the first question in another topic. For the second:

When $R$ is a domain, then also $R[x]$, and the degree of polynomials satisfies $\deg(f*g)=\deg(f)+\deg(g)$. So, $f*g=1$ implies $\deg(f) = 0$, s.t. $f$ is constant.

[quote="-oo-"] The recursive formula $ b_k \equal{} \minus{} b_0 (a_1 b_{k \minus{} 1} \plus{} \dotsc \plus{} a_k b_0)$ determines all $ b_i$, s.t. $ f$ is a unit in $ A[[x]]$. But it is easy to see that the $ b_i$ vanish once $ i$ is greater than $ n \equal{} deg(f)$ and the nilpotency indices of the $ a_i$. So $ f$ is a unit in $ A[x]$.
[/quote]

Could u clarify this statement please? why we go to study $ f$ in $ A[[x]]$?how we know $ f$ will be unit of choosing $ b_i$?

Thank you $ \minus{}oo\minus{}$

[color=red][[b]Moderator edit:[/b] The statement by -oo- quoted above is wrong. In fact, the $b_i$ don't necessarily vanish once $i$ is greater than $ n \equal{} deg(f)$ and the nilpotency indices of the $ a_i$. Instead, they necessarily vanish once $i$ is greater than the sum of the nilpotency indices of the $ a_i$. This is not immediately trivial from -oo-'s argument, however. Correct solutions of problem 1 has been given by -oo- in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=89417 . Problem 2 has been solved at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=425496 .][/color]


quadratics
geometry
rectangle
perimeter
algebra
Pythagorean Theorem

Can you help me with these two questions please.

The sides of a right angled triangle are 2x cm, (x+3)cm, and the hypotenuse is (2x + 2.4)cm. Use Pythagorus' theorem to form a quadratic equation in x and then solve it to find x.

The area of a rectangle with length (x + 4.6)cm and width (x - 2.1)cm is 134.63cm squared. a) Form a quadratic equation and solve it to find x to two decimal places. b) What is the rectangle's perimeter to one decimal place?

Charikaar

NO.

these are two questions on AQA past papers that i can't solve.

I know little about pythagraous

a2+b2=c2

but can't use it to solve the above problems so please help.

Charikaar

Okay, so you know the Pythagorean Theorem.  That's a start.  How do you think you should apply it to the first question?

[hide="Hint for problem 1"]Call the expressions for the 2 legs a and b, and call the hypotenuse c.
Through the pythagorean theorem, we know that [tex]a^2+b^2=c^2[/tex] where a and b are the legs and c is the hypotenuse.
We can "plug in" the given lengths of the sides to the pythagorean theorem to get [tex](2x)^2+(x+3)^2=(2x+2.4)^2[/tex][/hide]

P.S: No offense, but I think this problem belongs in the "Getting Started" section of the High School Forum, or mabye even in the upper levels of the Middle School Forum.

thaks for the hint.

i think i can do it now...........let me have a it a go.
i think i will have to use (a+b)^2=a^2+b^2+2ab

charikaar


superior algebra
superior algebra unsolved

Let $ J$ be a finitely generated ideal such that $ J\equal{}J^2$. Prove that $ J$ is principal and generated by an idempotent.

This looks like a straightforward question, but I do not know how to prove this is principal or how this is generated by an idempotent. How should I go about showing this?

http://www.mathlinks.ro/Forum/viewtopic.php?t=181620

  darij


function
vector
algebra
linear equation
calculus
calculus computations

On a recent test, I had a problem of the form
$ ay'' \plus{} by' \plus{} cy^2 \equal{} 0$
The last term was obviously supposed to be $ cy$, but I was just curious whether the problem as printed was actually solvable. I believe it isn't, but I have no idea how to go about proving something like that.

You need to distinguish between the essentially algebraic question "does there exist a technique for writing solutions to this in closed form" and the analytic question "does this differential equation have solutions." I'm usually more interested in the second question than the first.

As for the first question: I don't know of any such techniques, but I won't swear that that's the final answer.

As for the second question: the hypotheses of the existence-uniqueness theorem are met locally at every point. (I am assuming that $ a\ne 0.$) That is, given any point $ (t_0,y_0,y_0')\in\mathbb{R}^3$ there exists a $ \delta>0$ and a unique twice-differentiable function $ \phi$ on $ (t_0\minus{}\delta,t_0\plus{}\delta)$ such that $ a\phi''(t)\plus{}b\phi'(t)\plus{}c(\phi(t))^2\equal{}0,\phi(t_0)\equal{}y_0,$ and $ \phi'(t_0)\equal{}y_0'.$

Of course, the fact that this is a non-linear equation takes away several of our tools. The set of solutions is not a vector space. Independence is not a meaningful word. The Wronskian doesn't do much good. And we can't assume that solutions exist globally. (That is, we might need the $ \delta$ in what I said above.)


I wanted to see if anyone can help me out with this problem:
 \sqrt a 2  + 8abc + \sqrt b 2 +8abc + \sqrt c 2+8abc \leq 3(a+b+c) but you may find this puzzle more interesting:

[quote="Cauchy_Schwartz"]I wanted to see if anyone can help me out with this problem:
 \sqrt a 2  + 8abc + \sqrt b 2 +8abc + \sqrt c 2+8abc \leq 3(a+b+c)[/quote]

Should this be solved using your nickname?  :D
(BTW, maybe it should be \sqrt a 2 + 8bc, not 8abc ? Else, it's not really homogeneous...)

[quote="Cauchy_Schwartz"]but you may find this puzzle more interesting:[/quote]

Well, I feel this should be moved to the "Ultimative Unsolved Challenges" section...

  Darij

Thanks for the observation, you are right it is 8bc, 8ca, 8ab, i did a mistake when i wrote it. I see that my problem wasn't a real mystery, I should try composing something more difficult.

Concerning the inequality, I have posted [url=http://mathlinks.ro/viewtopic.php?p=15777]a proof in the inequalities subforum[/url].

  Darij

yes i found the puzzle very interesting and i bet it is a horse!


For infinitely many points on the plane it is known that distance between any two of them is a whole number. Prove that they all are on the same line.

You should read this book as a reference:
[hide="Challenging Mathematical Problems with Elementary solutions"]http://books.google.com/books?id=nowbdLTlr7YC&dq=problems+with+elementary+solutions+yaglom&printsec=frontcover&source=web&ots=JAgYNY3shM&sig=t5i_8A2yGbSXZDQ6tuwUTgMCHsM#PPA5,M1[/hide]
it is problem 109b there  :wink:


Does there exist a power of 2 such that permuting its digits gives another power of 2?

[i]Iran 1999[/i]

Can this power have 0 in it?
For example is 12 a permutation of 102?
I ask this because the problem would be easier if the 2 powers had the same number of digits.

They have the same number of digits I think :)

I think these two fact can help

1) For every n, n - p(n) divide by 9, where p(n) number, receiving from n by permutating its digits.
2)      1/ 9 < n/p(n) < 9.

[quote="harazi"]The problem is very simple. Suppose that 2^n is formed by a permutation of the digits of 2^m and that m>n. In particular, 2^m and 2^n have the same sum of digits, which means they are congruent modulo 9. From here we find that m-n>=6. Thus, 2^m>=64*2^n and so 2^m has more digits than 2^n, false.[/quote]

Take a look at what Arne said, harazi :
[quote="Arne"]Sorry, zeros can be placed first Mihai :) [/quote].

Now you see that $2^m$ and $2^n$ needn't have the same number of digits.

Yes, I heard that long time before. But I would really like to see a solution of this problem.

it was the first or fourth problem of iran second round. and im sure the number of digits are equale.
and in this case it doesnt hve any solution


function AMC USA(J)MO USAMO calculus derivative Putnam

The number of odd divisors of $n$ equals the number of arithmetic progressions (in steps of one) which sum to $n$.

For example, 15 has 4 odd divisors (1,3,5,15) and 15 is the sum of four arithmetic progressions (15, 1+2+3+4+5, 7+8, 4+5+6).

This can be solved with a simple bijection, but it can also be solved with generating functions.

Let $A(n)$ be the number of odd divisors of $n$.  Let $B(n)$ be the number of ways to express $n$ as an arithmetic progression.  We have,
\[A(n)=[x^n]\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} x^{k(2i-1)}\\
=[x^n] \frac{x}{1-x}+\frac{x^3}{1-x^3}+\frac{x^5}{1-x^5}+\ldots\]
(or you could sum over $i$ then $k$)
\[B(n)=[x^n]\sum_{k=1}^{\infty} \frac{x^{\frac{k(k+1)}{2}}}{1-x^k}\\
=[x^n] \frac{x}{1-x}+\frac{x^3}{1-x^2}+\frac{x^6}{1-x^3}+\ldots\]
Hence we need to show,
\[\frac{x}{1-x} + \frac{x^3}{1-x^3} + \frac{x^5}{1-x^5} + \ldots=\frac{x}{1-x} + \frac{x^3}{1-x^2} + \frac{x^6}{1-x^3} + \ldots\]
I'm not sure how to do this and I can't find this solution anywhere on the internet.  Does anyone have any ideas?

I didn't even understand hoe you het $ A(n) $ and $B(n) $ . Can you plz explain. I'm kinda new in this field!! :(

[quote="Rushil"]I didn't even understand hoe you het $A(n)$ and $B(n)$ . Can you plz explain. I'm kinda new in this field!! :([/quote]

Sure.  $[x^n]$  denotes the coeffecient of $x^n$.  For example, $[x^2]2x^3+5x^2+3x=5$.

To understand the generating function for $A(n)$, note that $(2i-1)$ will hit every odd number.  IF there exsists a $k$ such that $k*(2i-1)=n$, then $(2i-1)$ is an odd divisor of $n$ AND the term $x^{k(2i-1)}$ adds 1 to $[x^n]$.

To understand the generating function for $B(n)$, note that the first arithmetic progression of legnth $k$ is $1+2+\ldots+k=\frac{k(k+1)}{2}$.  Add 1 to each term to get the next arithmetic progression of length $k$. $(1+1)+(2+1)+\ldots+(k+1)=1+2+\ldots+k+k=\frac{k(k+1)}{2}+k$.  It should be clear that if $\frac{k(k+1)}{2}+ik=n$ for some $i$, that $n$ can be expressed as an arithmetic progression of length $k$ AND the term $x^{\frac{k(k+1)}{2}+ik}$ will contribute 1 to $[x^n]$.  Thus,
\[ B(n)=[x^n]\sum_{k=1}^{\infty} \sum_{i=0}^{\infty} x^{\frac{k(k+1)}{2}+ik}=[x^n]\sum_{k=1}^{\infty} \frac{x^{\frac{k(k+1)}{2}}}{1-x^k}  \]

Generating functions are a great tool.  I will use the following examples (which might be hard to follow) of contest problems to make my point.

[b]Problem 5 of the 1986 USAMO:[/b]  [i]If $p$ is a partition, $f(p)=$ number of 1's in $p$, and $g(p)=$ number of distinct integer sizes in $p$.  Show $\sum f(p)=\sum g(p)$, where the sums are taken over all partitions of $n$.

For example, the partitions of 3 are 3, 2+1, 1+1+1.  $f(3)=0$, $f(2+1)=1$, $f(1+1+1)=3$ and $g(3)=1$, $g(2+1)=2$, $g(1+1+1)=1$.

In other words, if you take a random partition of $n$, the expected number of 1's equals the expected number of distinct integer sizes.[/i]

Hopefully it is clear that,

\begin{eqnarray*} \sum f(p)&=& [x^n](x+2x^2+3x^3+\ldots)(1+x^2+x^4+\ldots)(1+x^3+x^6+\ldots)\cdots\\ \ &=&[x^n]\frac{x}{(1-x)^2 (1-x^2)(1-x^3)\cdots} \end{eqnarray*}
Now let,

\begin{eqnarray*} G(x,y)&=&(1+xy+x^2y+\ldots)(1+x^2y+x^4y+\ldots)(1+x^3y+x^6y+\ldots)\cdots\\ \ &=&\left(1+\frac{xy}{1-x}\right)\left(1+\frac{x^2y}{1-x^2}\right)\left(1+\frac{x^3y}{1-x^3}\right)\cdots \end{eqnarray*}

Hopefully it is clear that,

\begin{eqnarray*} \sum g(p)&=&[x^n]\left.\frac{\partial G(x,y)}{\partial y}\right]_{y=1}\\ \ &=& [x^n]\left.G(x,y)\sum_{k=1}^{\infty} \frac{x^k}{1-x^k+x^ky}\right]_{y=1}\\ \ &=&[x^n]\frac{1}{(1-x)(1-x^2)(1-x^3)\cdots} \frac{x}{1-x}\\ \ &=&\sum f(p) \end{eqnarray*}

The derivative was evaluated using the simple formula for differentiation of a product,
\[ \frac{\partial}{\partial x} f_1(x)f_2(x)\cdots f_n(x)=f_1(x)f_2(x)\cdots f_n(x)\sum_{k=1}^{n}\frac{f_k'(x)}{f_k(x)}  \]

[b]Problem B4 of the 1991 Putnam:[/b] [i]$p$ is a prime greater than 2.  Prove[/i]
\[ \sum_{n=0}^{p} {p\choose n} {{p+n}\choose n} \equiv 2^p+1\mod p^2  \]

\begin{eqnarray*} \sum_{n=0}^{p} {p\choose n} {{p+n}\choose n}&=&[x^p]\sum_{n=0}^{p} {{p\choose n}(1+x)^{p+n}}\\ \ &=&[x^p](1+x)^p(1+(1+x))^p\\ \ &=&\sum_{n=0}^{p} {{p\choose n}{p\choose p-n}2^n}\\ \ &\equiv& 2^p+1\mod p^2 \end{eqnarray*}

[b]Problem A1 of the 2003 Putnam:[/b] [i]Let $c(n)$ denote the number of partitions of $n$ such that no pair of integers in a partition differs by more than 1.  For example, the valid partitions of 4 are 4, 2+2, 2+1+1, 1+1+1+1.  So $c(4)=4$.  Compute $c(n)$.[/i]

\begin{eqnarray*} c(n)&=&[x^n]\sum_{k=1}^{\infty} {\frac{x^k}{1-x^k}\frac{1}{1-x^{k+1}}}\\ \ &=&[x^n]\frac{1}{(1-x)^2}\sum_{k=1}^{\infty} \frac{x^k}{(1+x+\ldots+x^{k-1})(1+x+\ldots+x^k)}\\ \ &=&[x^n]\frac{1}{(1-x)^2} \lim_{N\rightarrow\infty} \sum_{k=1}^{N} (\frac{1}{1+x+\ldots+x^{k-1}}-\frac{1}{1+x+\ldots+x^k})\\ \ &=&[x^n]\frac{1}{(1-x)^2} \lim_{N\rightarrow\infty} (1-\frac{1-x}{1-x^{N+1}})\\ \ &=&[x^n]\frac{x}{(1-x)^2}\\ \ &=& n \end{eqnarray*}

[b]Problem A6 of the 1990 Putnam:[/b] [i]Call an ordered pair $(A,B)$ of subsets of $[n]$ "admissible" if $a>|B|$ for each $a\in A$ and $b>|A|$ for each $b\in B$.  How many admissible ordered pairs of subsets of $[10]$ are there?[/i]

Let $c(n)$ denote the number of admissible ordered pairs of subsets of $[n]$.  Hopefully it is clear that,
\[ c(n)=\sum_{a+b\leq n} {{n-b\choose a}{n-a\choose b}}  \]
Thus,

\begin{eqnarray*} c(n)&=&[x^n]\sum_{k=0}^{\infty}x^k\sum_{a=0}^{\infty}{ {a+k\choose a}x^a}\sum_{b=0}^{\infty} {{b+k\choose b}x^b}\\ \ &=&[x^n]\sum_{k=0}^{\infty}\left(\frac{1}{(1-x)^{k+1}}\right)^2\\ \ &=&[x^n]\frac{1}{(1-x)^2}\sum_{k=0}^{\infty}\left(\frac{x}{(1-x)^2}\right)^k\\ \ &=&[x^n]\frac{1}{1-3x+x^2}\\ \ &=&[x^n]\frac{1}{(1-x)^2-x}\\ \ &=&[x^{2n}]\frac{1}{(1-x^2)^2-x^2}\\ \ &=&[x^{2n}]\frac{1}{(1-x-x^2)(1+x-x^2)}\\ \ &=&[x^{2n}]\frac{1}{2}\left(\frac{1+x}{1-x-x^2}+\frac{1+(-x)}{1-(-x)-(-x)^2}\right)\\ \ &=&[x^{2n}]\frac{1+x}{1-x-x^2}\\ \ &=&F_{2n}+F_{2n+1}\\ \ &=& F_{2n+2} \end{eqnarray*}

So $c(10)=F_{22}=17711$.

Note that we used the known generating function for Fibanacci Numbers, $F_n=[x^n]\frac{x}{1-x-x^2}$.

[b]Problem B1 of the 1996 Putnam:[/b] [i]A subset of $[n]$ is selfish if it contains its own cardinality as an element, and the other elements are greater than the cardinality.  Let $b(n)$ denote the number of selfish subsets of $[n]$.  Compute $b(n)$.[/i]

\begin{eqnarray*} b(n)&=&[x^n]\sum_{m=0}^{\infty}x^m\sum_{k=0}^{\infty}{x^k(x+x^2+x^3+\ldots)^{k-1}}\\ \ &=&[x^n]\frac{x}{1-x-x^2}\\ \ &=&F_n \end{eqnarray*}

[b]Problem A5 of the 1956 Putnam:[/b] [i]How many binary strings of length $n$ with $k$ 1's are there such that no 1's are next to each other?[/i]

Let $S$ denote the number of binary strings.

\begin{eqnarray*} S&=&[x^n](x+x^2+x^3+\ldots)(x^2+x^3+x^4+\ldots)^{k-1}(1+x+x^2+\ldots)\\ \ &=&[x^n]\frac{x^{2k-1}}{(1-x)^{k+1}}\\ \ &=&[x^{n-2k+1}]\frac{1}{(1-x)^{k+1}}\\ \ &=&{n-k+1\choose k} \end{eqnarray*}

[b]Unknown Source:[/b] [i]An increasing sequence of nonnegative integers $a_0,a_1,a_2,\ldots$ has the following property:  every nonnegative integer $n$ can be expressed uniquely in the form $n=a_i+2a_j$ where $i$ and $j$ are not necessarily distinct.  Determine $a_{2004}$.[/i]

Let,
\[ F(x)=\sum_{i=0}^{\infty} x^{a_i}  \]
Then the unique representation property gives,
\[ F(x)F(x^2)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty} x^{a_i+2a_j}=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}  \]
Thus,
\[ F(x^2)F(x^4)=\frac{1}{1-x^2}=\frac{F(x)F(x^2)}{1+x}  \]
Thus,
\[ F(x)=(1+x)F(x^4)  \]
By iteration,
\[ F(x)=(1+x)(1+x^4)(1+x^{16})(1+x^{64})\cdots  \]
So, $<a>$ is the sequence of nonnegative integers that are sums of distinct powers of 4.

Hopefully it is clear that the representation of $a_i$ in base 4 [i]looks[/i] like the binary representation of $i$.  Since $2004=11111010100_{(2)}$ it follows that $a_{2004}=11111010100_{(4)}=1396816$.

If you like this stuff, you should look up the Catalan Reccurence which arises in a number of problems.  The idea behind solving the Catalan Reccurence with generating functions can be used in other problems.  For example, the Vandermonde identity,
\[ \sum_{k=0}^{n}{{a\choose k}{b\choose n-k}}=[x^n]\sum_{k=0}^{a} {{a\choose k}x^k}\sum_{m=0}^{b} {{b\choose m}x^m}=[x^n](1+x)^{a+b}={a+b\choose n}  \]
The problem of Derrangements is also something to check out.  The method is nice.  It uses an exponential generating function (to make the generating function converge), and then shows the generating function satisfies a differential equation.

And, the Roots of Unity Filter is a powerful tool.  It allows you create one function from another by extracting every $k$'th power of $x$ for any natural number $k$.  I have seen a couple of contest problems where the Roots of Unity Filter could be used.

I don't understand how Negative_3 got \begin{eqnarray*} \sum f(p)&=& [x^n](x+2x^2+3x^3+\ldots)(1+x^2+x^4+\ldots)(1+x^3+x^6+\ldots)\cdots\\ \ &=&[x^n]\frac{x}{(1-x)^2 (1-x^2)(1-x^3)\cdots} \end{eqnarray*} Why is $(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)\cdots$ in the factorization on the RHS, and what does it mean combinatorially?

P.S.: Correct me if I am wrong but I believe that $(x+2x^2+3x^3+\cdots)$ is on the RHS to signify whether there are $1,2,3,\cdots$ ones in a partition $p$ of $n$. Also, sorry for bumping a really old thread.


Please download the doc file:

[url]http://www.mp95.com/ChinaNationalMiddleSchoolMathTournament2009_session2.doc[/url]

Not everybody has Microsoft Word so here it is as a picture

[quote="HiDN428"]This is a middle school competitions, and has proofs, wow.[/quote]

If without proofs, how can the students be selected? Actually, students focus on proofs
rather than calculations.

mn=3.57
ya need to find  AD,CD,BD.  which are 1.8,15.676,16/5  resp.
then ya calculate the inradius of each triangle. ya add them and get 3.57  ( the two incircles will touch CD at a particular point, through which MN goes.
obviously MN is perp. to CD.


counting
distinguishability
probability
binomial coefficients
AMC

How many non-empty subsets $ S$ of $ \{1, 2, 3, \ldots, 15\}$ have the following two properties?

(1) No two consecutive integers belong to $ S$.
(2) If  $ S$ contains $ k$ elements, then $ S$ contains no number less than $ k$.

$ \textbf{(A) } 277\qquad \textbf{(B) } 311\qquad \textbf{(C) } 376\qquad \textbf{(D) } 377\qquad \textbf{(E) } 405$

i had no other choice but to brute force it out!
wasn't that bad though

If A is the smallest number in the subset...

k = 1:

For all A, there's 1 possibility, 15 possible A, 15 subsets.

k = 2:

A = 2, then we take the sum of A = 4,5,6,...,15 for k = 1.
A = 3, then we take the sum of A = 5,6,7,...,15 for k = 1.

Keep going until you finish k = 5.

Well for $k$ element-subset we much choose from $15-(k-1)$ elements, now the consecutive part is a bit tricky.

We make a 1-1 correspondence. consider the set of sequence ddddddddd, 15-2(k-1) d's, we can choose anyway k d we like, so so far
(15-2(k-1)Ck, now add k-1 "|" each between 2 consecutive d's chosen. Each chosen d represent a number from k to 15, the pth d corresponds to number  k+p-1 (counting the dividing "|").  So yeah $(15-2(k-1))Ck$ is the number of subsets for $k$ and sum that up to 5.

[quote="qweretyq"]For this, it is 15 for k=1 which is trivial then its 1 summation, 2 sums, 3sums, and 4 sums.

[hide]
k=1, 15 (A zero-sum -  $\displaystyle\sum_{x=1}^{15}1$)
k=2, $\displaystyle\sum_{x=1}^{12}x$
k=3 $\displaystyle\sum_{y=1}^{9}{\displaystyle\sum_{x=1}^{y}x}$
k=4$\displaystyle\sum_{z=1}^{6}{\displaystyle\sum_{y=1}^{z}{\displaystyle\sum_{x=1}^{y}x}}$
k=5$\displaystyle\sum_{a=1}^{3}{\displaystyle\sum_{z=1}^{a}{\displaystyle\sum_{y=1}^{z}{\displaystyle\sum_{x=1}^{y}x}}}$

Use TI-89, add em up and you get $\boxed{405}$[/hide]

see the pattern?[/quote]

Solution without TI89:

Can easily calculate table of values:

$1$   $2$   $3$   $4$   $5$   $6$   $7$   $8$   $9$   $10$   $11$   $12$   $13$   $14$   $\boxed{15}$
$1$   $3$   $6$   $10$  $15$  $21$  $28$  $36$  $45$  $55$  $66$  $\boxed{78}$
$1$   $4$   $10$  $20$  $35$  $56$  $84$  $120$  $\boxed{165}$
$1$   $5$   $15$  $35$  $70$  $\boxed{126}$
$1$   $6$   $\boxed{21}$

where each term can be found by adding the term above it, and the term to the left of the one above it.  (Then add, answer E = 405)

[hide="Answer"]Each single-element subset works: $15$.

For two-element subsets, we cannot use $1$, and cannot use consecutive integers, so we essentially have $\binom{13}{2}=78$.

For three-element subsets, we cannot use $1$ or $2$, and we cannot use consecutive integers.

We continue this and notice that we are essentially just removing two numbers each time: the minimum, and one more consecutive integer. We take this to choosing $5$ integers from $7$, so our sum is $\binom{15}{1}+\binom{13}{2}+\binom{11}{3}+\binom{9}{4}+\binom{7}{5}=405\Rightarrow \boxed{E}$. Notice that we cannot do another one, because we would have $\binom{5}{6}$, which is impossible.[/hide]

This problem only took me 4 minutes. :o

[hide="a recursive solution that for some reason produces the wrong answer"]

Let $ A_n$ denote the set of subsets satisfying the two conditions for the set $ \{1, 2, \ldots , n \}$. We partition $ A_n$ into two categories: those containing $ n$ and those not. In the former case, each subset is in $ A_{n\minus{}1}$. In the latter category, we map the subset to a subset of $ A_{n\minus{}3}$ using the following bijection: remove $ n$ and reduce each remaining number by $ 1$.
\[ \{ a_1, a_2, \ldots , a_k, n \} \rightarrow \{ a_1\minus{}1, a_2\minus{}1, \ldots , a_k\minus{}1 \}
\]
Since the least element and the number of elements are both reduced by $ 1$, the conditions are still satisfied. Since $ a_k \leq n\minus{}2$, $ a_k\minus{}1 \leq n\minus{}3$, so we get a subset of $ A_{n\minus{}3}$. We get the following recursive relation: $ a_n \equal{} a_{n\minus{}3} \plus{} a_{n\minus{}1}$. Starting with $ a_1 \equal{}1 , a_2 \equal{} 2, a_3 \equal{} 4$, i get $ a_{15}\equal{}230$, which isn't right.   :(
[/hide]

[quote="Phelpedo"][hide="a recursive solution that for some reason produces the wrong answer"]

Let $ A_n$ denote the set of subsets satisfying the two conditions for the set $ \{1, 2, \ldots , n \}$. We partition $ A_n$ into two categories: those containing $ n$ and those not. In the former case, each subset is in $ A_{n \minus{} 1}$. In the latter category, we map the subset to a subset of $ A_{n \minus{} 3}$ using the following bijection: remove $ n$ and reduce each remaining number by $ 1$.
\[ \{ a_1, a_2, \ldots , a_k, n \} \rightarrow \{ a_1 \minus{} 1, a_2 \minus{} 1, \ldots , a_k \minus{} 1 \}
\]
Since the least element and the number of elements are both reduced by $ 1$, the conditions are still satisfied. Since $ a_k \leq n \minus{} 2$, $ a_k \minus{} 1 \leq n \minus{} 3$, so we get a subset of $ A_{n \minus{} 3}$. We get the following recursive relation: $ a_n \equal{} a_{n \minus{} 3} \plus{} a_{n \minus{} 1}$. Starting with $ a_1 \equal{} 1 , a_2 \equal{} 2, a_3 \equal{} 4$, i get $ a_{15} \equal{} 230$, which isn't right.   :(
[/hide][/quote]

First, that would be $ a_{14} \equal{} 230$.  $ a_{15} \equal{} 387$ is still incorrect though, but it's encouragingly closer.

Watch the element $ \{n\}$ and you'll see what goes wrong.  Adjusting the recurrence accordingly (and $ a_3 \equal{} 3$, yes?) yields the correct answer.

Interesting solution.

please dont post on topics over a year old

Why? It shows another way to solve the problem.

This is the topic linked to in the Contests section of this site. I was always under the impression that such topics are always open for additional discussion, for the purpose of questions/additional solutions.

Revival of [i]any[/i] old topic is not only okay, but positively good, if you wish to contribute something genuinely new and worthwhile, for example, a new solution, a flaw in an old one.

[quote="JesusFreak197"][hide="Answer"] $ \binom{15}{1} \plus{} \binom{13}{2} \plus{} \binom{11}{3} \plus{} \binom{9}{4} \plus{} \binom{7}{5} \equal{} 405\Rightarrow \boxed{E}$. Notice that we cannot do another one, because we would have $ \binom{5}{6}$, which is impossible.[/hide]
[/quote]
I don't understand your solution.  $ \binom{13}{2}$ <- you could be choosing two consecutive integers among the 13 integers.  for example, you could be choosing 7,8.  in other words how can you guareentee you don't have two consecutive numbers in your subset?

I got it right but it took me like 10 minutes to get it.  I just counted the ways. I could've used my time on other easier problems.

[quote="turtlecloud"]how would someone do the method mentioned above by beta involving the dddddd's and a divider with a -1 correspondence? [/quote]

That method is known as stars and bars (or sticks and stones or balls and urns, whatever). It basically gives the number of ways to distribute $k$ indistinguishable objects into $n$ bins and is commonly used in combinatorics. The formula is $\dbinom{n+k-1}{k}$.

Yeah introduction to counting and probability by Dave Patrick is really nice.

Can someone please completely explain the recursive solution?

[quote=MellowMelon]To justify $ {13 \choose 2}$, let's make the selected numbers an ordered pair $ (a,b)$, with both $ a,b \in \{1,2,\ldots,13\}$ and $ a < b$. We will say that this is equivalent to having a 2 element subset containing $ a\plus{}1$ and $ b\plus{}2$. Because $ a \geq 1$, $ b\plus{}2 \geq a\plus{}1 \geq 2$, so condition 2 is satisfied. And because $ b > a$, $ b \geq a \plus{} 1$. This means there is at least one integer between $ b\plus{}2$ and $ a\plus{}1$, so condition 1 is satisfied. Similar reasoning with ordered 3-tuples, 4-tuples, etc. will justify the other binomial coefficients in that sum.[/quote]

Wait does he mean $a-2, b-1$? And also can someone illustrate this with $\binom{11}{3}$?

The largest possible element is $n-k+1$.   Since this process is easily reversible, we have a bijection. You should have: ${15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \choose 8}$

The second condition states that no element in our original configurationcan be less than $k$, which translates to subtracting $k$ from each binomial coefficient. Now we have, after we cancel all the terms ${n \choose k}$ where $n < k$,  ${15 \choose 1} + {13 \choose 2} + {11 \choose 3} + {9 \choose 4} + {7 \choose 5}=  15 + 78 + 165 + 126 + 21 = \boxed{405}. The answer is $E$.

Can you explain the bijection a bit more? Thanks :)

one of the easiest #25s on the 12 I have ever seen.
[hide=Solution]
Obviously, we can use engineer's induction. For the $1$ case, there are $\binom{15}{1}=15$ subsets. For the $2$ case, there are just $\binom{13}{2}$ ways. For the $3$ case, the are $\binom{11}{3}$ ways by the hockey-stick identity. In fact, the hockey-stick identity saves all the computation. Summing over all the cases, ending when $m<n$ for $\binom{m}{n}$, we have the answer is $\boxed{\textbf{(E) } 405}$

Well, I give a interesting way to solve this problem.
First things first, if there is only one element, we have 15=1*15 choices -- ${15 \choose 1}$
If there are two, when a=2 b=4--15 (12 in total) if a=3, b=5--15 (11 in total)....., so all together, we have ${13 \choose 2}$, we just this this situation as the 2-dimensional cases.
If there are three, we have to do the "dimension-decrease" operation. I mean, something like this:
When a=3 b=5, c=7--15 (9 in total) , b=6 , c=8--15 (8 in total) , so all together, we have ${10 \choose 2}$,
When a=4, in the same logic, we get ${9 \choose 2}$,
It is clear that ${10 \choose 2}+{9 \choose 2}+.......{2 \choose 2} = {11 \choose 3}$, so in this case, we "decrease" the "3-dimensional case" to "the sum of 2-dimensional case" and then sum them up!
In the same logic, if there are four, we have ${9 \choose 4}$ in total
If there are five element, we use the meathod above in the same way.
Finally, we get the total answer $\boxed{405}$