We have triangle ABC and his area S. Point M is the middle of interval BC. There is point K on median AM (AK:AM=1:3). Through the point K and vertex B we have straight line crossing side AC in point L. Find area of AKL triangle. Could you help me ?:)

[hide]
Basically, you need to use the concept of area ratios.

Consider the area of $\triangle AMB$ and $\triangle AMC$.
If we regard $BM$ as the base of $\triangle AMB$ and $MC$ as the base of $\triangle AMC$, the height of the two triangles are equal.

Recall the formula $\text{Area of }\triangle=\frac12bh$.
Since the two triangles have the same $h$, the area ratio of the two triangles is the same as the ratio of the length of the two bases, which is $1: 1$ in this case.

Then use the same idea for $\triangle AKB$ and $\triangle BKM$.
You should be able to finish the rest :) [/hide]

EDIT- Actually, the problem was not that simple. (I misread the problem.) See my other post for more hints.

I'm counting and I have Area of triangle AMC and AMB = 1/2 * S

area of AKB=3/8S and KMB=1/8S and equation Area (ALK) + Area (LKMC)=1/2S

Area (LKMC)= Area (LCB) - Area (KMB)

to full hapiness i need Area (LCB) but from where ?

thanks for everything

krzysiek wrote: (...) Point M is the middle of interval BC (...)

I`m sorry, but I`m not an english speaker; here in my country we speak spanish... Well, that what krzysiek wrote means that M is the midpoint of BC??????.

Yes, I meant that. AM is median <=> M is midpoint of interval BC

Oh wait, I thought the problem asks to find the area of $\triangle AKM$... My bad, I was in a hurry, so I didn't fully read the problem.

Anyway, you need to use some other techniques as well.

In order to find the area of $\triangle AKL$, you need to find $BK: KL$. You can try a method called "mass points". This should be very helpful. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56325]this thread[/url] for some info on mass points.


geometry
perpendicular bisector
power of a point
radical axis
geometry unsolved

A circle is tangential to sides $AB$ and $AD$ of convex quadrilateral $ABCD$ at $G$ and $H$ respectively, and cuts diagonal $AC$ at $E$ and $F$.  What are the necessary and sufficient conditions such that there exists another circle which passes through $E$ and $F$, and is tangential to $DA$ and $DC$ extended?

[quote="orl"]A circle is tangential to sides $AB$ and $AC$ of convex quadrilateral $ABCD$ at $G$ and $H$ respectively, and cuts diagonal $AC$ at $E$ and $F$.  What are the necessary and sufficient conditions such that there exists another circle which passes through $E$ and $F$, and is tangential to $DA$ and $DC$ extended?[/quote]

Assume that when correcting the typo, the problem would begin "A circle is tangential to sides $AB$ and $BC$..."

The 1st circle $(P)$ tangent to the sides $AB$ and $AC$ is centered on the internal bisector of the angle $\angle ABC$. The points $E, F$ lie both inside of the segment $AC$. The 2nd circle $(Q)$ tangent to the lines $DA, DC$ has to be centered the internal bisector of the angle $\angle CDA$. In order to pass through the points $E, F$, it also has to be centered on the perpendicular bisector of the segment $EF$. Let $Q$ be the intersection of the the internal bisector of the angle $\angle ABC$ with the perpendicular bisector of the segment $Q$. Let $(Q)$ be a circle centered at this point and passing through the points $E, F$. The quadrilateral diagonal $AC \equiv EF$ is the radical axis of the circles $(P), (Q)$. Iff the circle $(Q)$ is tangent to the lines $DA, DC$ at points $K, L$, the tangent lengths $AG = AK$ to the circles $(P), (Q)$ from the vertex $A$ are equal and the tangent lengths $CH = CL$ to the circles $(P), (Q)$ from the vertex $C$ are also equal, because these 2 vertices are on the radical axis of the circles $(P), (Q)$. The tangent lengths $BG = BH$ to the circle $(P)$ from the vertex $B$ are equal the tangent lengths $DK = DL$ to the circle $(Q)$ from the vertex $D$ are also equal. Consequently,

$AB + CD = (AG + BG) + (CL + DL) = AK + BH + CH + DK =$

$= (BH + CH) + (DK + AK) = BC + DA$

which is equivalent to the quadrilateral $ABCD$ being tangential. Since the circles $(P), (Q)$ are supposed to be different, the given circle $(P)$ must not be identical with the quadrilateral incircle $(I)$.


+
++
+++
++++
+++++

The grid of pegs is shown above.

There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color? (Any two pegs of the same color are indistinguishable.)

[/img][/b]

I think there is only one way...
y
ry
gry
bgry
obrgy


hey i'm interested. could i join you guys?

i'm sorry i can't go.. stupid science olympiads regionals are the same day.. (i would not go to science olympiad if i could, but my teams kinda relying on me.. so i guess i have to go to that..)
well, good luck forming a team

oh bad luck blueboat.  :(  raleigh's is feb. 9, so i actually can come.  yay.  i guess you'll have to hope it snows  :D

hrmm...i will ask vivek ok?  and maybe i will not rely on AIM and just email jeremy.  OK.

i think that registration has closed...are we going?

guys. listen. if you are registered. you should like join one of our teams.

Hmm this is a possibility. Assuming one of your teams only has three people. Or something like that. I think we have five... but we might get more in the near future.

OK this is getting extremely annoying because I can't get a team together. Plane tickets are also annoying. jb05 is also annoyed and is sending me angry emails. Therefore, I think we just won't have a team this year.  :( Sorry.

The Wootlings team currently has 5 people and apparently no prospects of getting more.  If 3 of you want to join with us, that would be great . . .

Or maybe you should join with us, in the interest of not being called wootlings. Granted, we all are taking / have taken woot... Hmm.

Oh, and I am open to combining the teams (provided I don't do geometry), but I want to make sure the others agree first.

Currently, I am at a "no" status, but it could change.  I'm not really sure what my parents want to do, and they have not given an entirely definitive answer.

As to merging teams, that would probably be required in order to obtain a complete team, in any case.

Have you made a decision about merging teams?  I need to know soon.

Oh sorry we've dealt with it. No merging necessary.

[quote="not_trig"]Hmm... we have me, Berman, Bryce, and Pom. Vitek/Jane/Ray probably cannot go, and Jeremy and Vivek (because I can't seem to contact them, the former because he has not been on AIM, the latter because... I'm too lazy to look it up... wait maybe I should) have not responded.

So, we may need to merge with another team.

Also... plane tickets, etc., will incur a large cost. Berman dislikes SkyBus and wants to leave from Wilmington. He also would like everyone to land in the -boston- airport at around the same time so we can meet up.

Of course, going on SkyBus early and then bus ride to Boston airport... hmm... tricky.

And we need to sign up for subjects, email me what you want, for now... until we get something figured out.[/quote]

Are there any openings on any teams currently?

My team didn't get accepted...

2009 presents many possibilites (not really.) But why don't we try anyway?

people who need to go:
berman (duh)
damien (also duh)
calvin (yes, i'm serious. he's insanely smart (he got all 10 bus questions, remember?))
bryce
gray?
pom?
and 2 other random people


Sometimes, we have to analize critically not only the reasoning provided by a book, but also an answer proposed by a teacher. Yesterday, a chemical engineering student of organic chemistry showed me the following problem (from Morrison),

"Propose the several steps of a possible laboratory synthesis of the following compounds, using benzene, toluene, and alcohols with four carbons or less as the only organic starting materials, and any inorganic reagent necessary:

(a) Triphenylchlorometane.  (b) 1,2-Dichloro-1-phenylethane. (c) Phenylcyclopropane."

together with an answer for (c) provided by a teacher:

"(c) Benzene + cyclopropanol + H+ -----> Phenylcyclopropane".

Any comments?

For 1 i would suggest Friedel Craft's  Reaction of Excess Benzene with $ \ce{CHCl3 / AlCl3}$

Any comments are to be made regarding the given answer for (c).

I think 1 should have excess benzene with $ \ce{CCl4}$

For 2 it would be
1.Gatterman Aldehyde synthesis From Benzene
2.Perkin Condensation of Benzaldehyde using Acteic Anhydride and Sodium Acetate to get Cinnamic Acid
3.Decarboxylation
4.add chlorine to double bond
To get Acetic Acid oxidise Ethanol.
To get Acetic anydride React Ketene with Acetic Acid

Yes that's right madness excess of $ \ce{CCl4}$
good u have learnt Mchem

for (c) you can react benzene with ethanol in presence of a good dehydrating agent( preferably P2O5 INSTEAD OF H2SO4 TO PREVENT SIDE REACTIONS) to get ethyl benzene foll by free radical chlorination and elimination to get styrene which on reacting wid carbene should give c.

actually 1 isn't that easy $ \ce{CCl4}$ is non polar hence doesn't show Friedel Craft's reaction  :rotfl:  :rotfl:

I will say this one more time: the purpose of my post above is to make people think [i]about the given answer for (c)[/i], not to provide synthesis for (a), (b) and (c). Anyway,

[quote="Pardesi"]For 2 it would be
1.Gatterman Aldehyde synthesis From Benzene
2.Perkin Condensation of Benzaldehyde using Acteic Anhydride and Sodium Acetate to get Cinnamic Acid
3.Decarboxylation
4.add chlorine to double bond
To get Acetic Acid oxidise Ethanol.
To get Acetic anydride React Ketene with Acetic Acid[/quote]

That looks like a complicated synthesis to me.

[hide="My answer for 2"]1) Toluene + CrO2Cl2 (Étard reaction) -----> Benzaldehyde

2) Benzaldehyde + CH2=PPh3 -----> styrene

3) styrene + Cl2 -----> 1,2-Dichloro-1-phenylethane.[/hide]

that's complicated because we have limitations on the choice of the Reagents we can use for our reactions now since u have used Wittig u have to be able to synthesise the Ylide provided u don't assume that as an inorganic Reagent.

[hide="Ylide preparation"]$ CH_3OH \plus{} SOCl_2 \longrightarrow CH_3Cl$

$ CH_3Cl \plus{} PPh_3 \longrightarrow CH_3PPh_3^\plus{}Cl^\minus{}$

$ CH_3PPh_3^\plus{}Cl^\minus{} \plus{} PhLi \longrightarrow CH_2\equal{}PPh_3$

Phenyl lithium preparation:

$ C_6H_6 \plus{} Fe/Br_2 \longrightarrow C_6H_5Br$

$ C_6H_5Br \plus{} 2Li \longrightarrow C_6H_5Li \plus{} LiBr$

Triphenylphosphine preparation:

$ 3C_6H_5Li \plus{} PCl_3 \longrightarrow (C_6H_5)_3P \plus{} 3LiCl$.[/hide]


Find all f: R --> R satistifying

f(x+y) + f(xy) = f(x) + f(y) + f(x).f(y) for all x, y from R.

We make x=y=0 and we get f(0)=0. Then we make y=1 and we get f(x+1)=f(1)[f(x)+1] (*). We have 2 cases: if f(1)=0, then f(x+1)=0 for all x, so f(x)=0 for all x. If f(1)=/=0 we can continue like this: we make x=-1 in (*) and we get f(-1)=-1. We make y=-1 in our first relation and we get f(x-1)+f(-x)=-1. But we know, from (*), that f(x)=f(1)[f(x-1)+1]=f(1)*(-f(-x)), so f(x)=-f(1)*f(-x) (**).

We make x=-1 in (**) and we get (f(1))^2=1, so f(1) is either 1 or -1. If f(1)=-1, then f(x)=f(-x) and, by making y->-y in our first relation we get f(x+y)=f(x-y) for all real x and y, so f(x)=f(1)=-1 for all real x, but this fction doesn't verify the relation we were given. This means that f(1)=1, so f(-x)=-f(x). We make y->-y in the first relation and we get f(x-y)-f(xy)=f(x)-f(y)-f(x)f(y). We add this with the first relaton to get f(x+y)+f(x-y)=2f(x), so f(a)+f(b)=2f((a+b)/2) (***).

(***) => f(x+y)+f(xy)=2f((x+y+xy)/2); f(x)+f(y)=2f((x+y)/2) => f(x)f(y) = 2(f((x+y+xy)/2)+f(-(x+y)/2)) = 4f(xy/4), so f(1)*f(xy)=4f(xy/4)=f(x)f(y), so

(1) f(xy)=f(x)f(y)
(2) from (1) and the first relation we get f(x+y)=f(x)+f(y)

I didn't continue, but I don't think it's difficult from here on. f(x) is either 0 for all x or x for all x.

indeed it is not that difficult.
f(x2)=f2(x) \geq 0 thus f is non-decreasing (because f verifies the Cauchy equation), and that means that for each increasing sequence xn -> r and each decreasing sequence (both containing only rationals) yn -> r we have

f(xn) \leq f(r) \leq f(yn) that is
xn \leq f(r) \leq yn and taking n-> \infty we get f(r) =r  for all irrationals r.

Thus f(x)=x for all reals x (or f(x)=0 for all reals x :D).

Indeed. This last part of the problem is quite well-known. I think I saw it in a book containing old Gheorghe Titeica problems (for those who don't know what this is, it's a popular Romanian maths contest (below college level).

I think there is an easier way of solving this problem. Of course, we have f(0)=0. Take y=1 and find that f(x+1)=f(1)(f(x)+1). Replace y by y+1 and use this result.You will find that f(1)(f(x+y)-f(x)-f(y)-f(x)f(y))=f(x)-f(x+xy).Hence -f(1)f(xy)+f(x+xy)=f(x) and this means that f(1)f(a)+f(b)=f(a+b) for all a and b with b<>0. If f(1)<>1 we obtain (interchange a and b) that f is zero. If f(1)=1, then f(x+y)=f(x)+f(y) and f(xy)=f(x)f(y) for all x and y and now the problem is trivial.


geometry
conics
ellipse
area of a triangle
geometry proposed

Space geometry 1
 I've found the sol for it but my friend told me that we can use space geometry to solve it.I'll thank u a lot if u can show me this sol.
        Let (O,R) be a circle with radius R. Three points A,B,C on (O,R). Find max of area(ABC)

Using this problem, try to show the following: given an ellipse, find tha maximal area of a triangle inscribed in it. It has a cute idea.

I don't know how one could use space geometry to solve the problem you've posted, and I have to admit that I don't know why one would try.. It's pretty easy to see that the maximum oft he area is reached for the equilateral triangle.


modular arithmetic
calculus
integration
number theory

consider the set of natural nos(S:1,2,3,4,.......)

two nos are randomly removed from a subset of this set and the average of the remaining nos of the subset is equal to  105/4........

then  find the subset of S wich contains the nos  (1,2,3,....n) i.e. find n and the two nos that have been removed from this set..

So the equations is $ 105/4 \equal{} (n(n\plus{}1)/2\minus{}x\minus{}y)/(n\minus{}2)$, where $ x<y \le n$ are the numbers we removed.  From this, we can immediately see $ 4|n\minus{}2$, so $ n \equiv 2 \pmod 4$.

Let $ n\equal{}4k\plus{}2$.  Our equation reduces to $ 105k \equal{} (2k\plus{}1)(4k\plus{}3)\minus{}x\minus{}y$.  Hence we have $ x\plus{}y \equal{} 8k^2\minus{}115k\plus{}3$.  Taking our discriminant, $ 115^2\minus{}4*8(3\minus{}x\minus{}y) \ge 0$ ie $ 13129\plus{}32x\plus{}32y \ge 0$ or $ x\plus{}y > \minus{}410.28125$ of course useless...

Anyway, the problem is equivalent to solving $ 105k \equal{} (2k\plus{}1)(4k\plus{}3)\minus{}x\minus{}y$ for positive integers $ x<y \le 4k\plus{}2$...

perhaps: $ 105/4 < n(n\plus{}1)/(2(n\minus{}2))$ so $ 105 < 2n(n\plus{}1)/(n\minus{}2)$ ie $ 105n\minus{}230 < 2n^2\plus{}2n$ so $ 0 < 2n^2\minus{}103n\plus{}230$ so $ n > 51.2$.  Similarly we should be able to find an upper bound for $ n$.  I'll work more on this later.

ok so our other bound can be found by maximizing $ x$ and $ y$.  We then have

$ 105/4(n\minus{}2) \ge n(n\plus{}1)/2\minus{}n\minus{}(n\minus{}1) \equal{} 1/2n^2 \minus{}3/2n\plus{}1$
$ 0 \ge 1/2n^2 \minus{} 27.75n \plus{} 53.5$
$ 0 \ge 2n^2 \minus{} 111n\plus{}214$
$ 2 \le n \le 53.5$

Putting our bounds together, $ n \equal{} 52, 53$ but neither of these meet the $ n \equiv 2 \pmod{4}$ requirement.  Thus there are no solutions.

If i say u that the integers removed are consecutive , the problem becomes a little bit easy ,
so try now..!!! :lol:

Maybe facis ruled out smaller numbers because $ 2 \cdot 105 \neq 230$?

Yes facis that was ur mistake 2*105=210, so now we get the values 49,50,51,52,53..(as n is integral)
out of which only 50 is of the form 4k+2..
so 50 is the answer no we can get x and y because y=x+1 as they were consecutive...

thanks for ur help..

oops... that was stupid.  sorry about that.  I thought it was strange I was getting no solution but I double checked my work...  I guess I didn't check that with the calculator.  But I think you guys get the idea anyway: look for bounds, a little number theory if possible, then check to see if you can find x and y.

btw, if $ (x,y) \equal{} 7,8$ works, then so does $ (6,9)$, $ (5,10)$, $ (4,11)$, $ (3,12)$, $ (2,13)$, and $ (1,14)$, and of course we could reverse the x and y values here to get another $ 7$ solutions, if we drop the restriction that $ x < y$.


Find all answers to this equation $x,y \in N$
$x^y=y^x$

[hide]

It is true for $x=y$

WLOG let $x>y$ $x=y^k$ , we have $y^{ky}=y^{y^k}$ $\implies ky=y^k$

again let $k=y^p$ then obtain $y^p=p+1$ . When $p=0$,$k=1$ and $x=y$
When $p=1$ , $y=2$ , $k=2$ ,$x=4$ .

When $y>2$ we have $p+1>2^p$ which is not true for positive integer $p$ . Hence the answer is $(x,y)=(k,k) , (2,4) , (4,2)$ for natural $k$
[/hide]


hello everybody!
please help me to solve this problem.

triangle ABC is given. AD, BE and CF are altitudes of ABC. AD meet BE at H.
P, Q and K are points on BH, DH and CH respectively.
if S(APC) + S(BQC) + S(AKB) = S(ABC) then prove that  HPQK is cyclic.(S(ABC) = the area of the triangle ABC.)

as mentioned we have:
$ KF.AB\plus{}PE.AC\plus{}QD.BC\equal{}2.S_{\Delta ABC}\equal{}HF.AB\plus{}HE.AC\plus{}HD.BC \implies KH.AB\plus{}PH.AC\equal{}QH.BC$
let's suppose that the circumcircle of $ \Delta PKH$ intersect $ DH$ at $ H,Q'$.we want to prove that $ Q'\equal{}Q$.
or in other words $ QH\equal{}Q'H$.we have
$ \Delta PKH \sim \Delta ABC \implies \frac{PK}{BC}\equal{}\frac{PQ'}{AB}\equal{}\frac{KQ'}{AC}\equal{}s$ and also we have $ KH.PQ'\plus{}PH.KQ'\equal{}Q'H.PK \implies s.(KH.PQ'\plus{}PH.KQ')\equal{}(Q'H.PK).s \implies  KH.AB\plus{}PH.AC\equal{}Q'H.BC$
but we had $ KH.AB\plus{}PH.AC\equal{}QH.BC$ so we can conclude $ Q'H.BC\equal{}QH.BC \implies Q'H\equal{}QH$
DONE!

i've made two typo:
in the 6-th line we have $ \Delta PKQ' \sim \Delta ABC$
and in the 7-th line the expression must be multiplied by $ \frac{1}{s}$
(sorry)


abstract algebra
vector
advanced fields
advanced fields unsolved

Let $ M_i\subset B(H_i)$, $ i\equal{}1,2$ be maximal abelian von Neumann algebras such that they are $ *$-isomorphic. Prove that they are spatial isomorphic, i.e. $ \exists u: H_1\to H_2$ unitary such that the map $ x\mapsto uxu^*$ from $ M_1$ to $ M_2$ is an isomorphism.

[quote="Kuba"]Let $ M_i\subset B(H_i)$, $ i \equal{} 1,2$ be maximal abelian von Neumann algebras such that they are $ *$-isomorphic. Prove that they are spatial isomorphic, i.e. $ \exists u: H_1\to H_2$ unitary such that the map $ x\mapsto uxu^*$ from $ M_1$ to $ M_2$ is an isomorphism.[/quote]
This is Theorem 9.3.1 in Kadison and Ringrose, [i]Fundamentals of the theory of operator algebras, Vol. II[/i]. The essence of the proof goes like this. If $ M_2$ is countably decomposable then it has a separating cyclic vector, which gives rise to a vector state. The transpose of the *-isomorphism, applied to this state, gives a state of $ M_1$ which is also a vector state. From there, it is easy to build up a unitary map from $ H_1$ to $ H_2$ that implements the *-isomorphism.

For the general case, you need to consider a maximal family of countably-decomposable projections in $ M_1$, and apply the provious result to the restrictions of $ M_1$ to the ranges of these projections.


To be able to walk to the center $ C$ of a circular fountain, a repair
crew places a 16-foot plank from $ A$ to $ B$ and then a 10-foot plank
from $ D$ to $ C$, where $ D$ is the
midpoint of $ \overline{AB}$ . What is the
area of the circular base of the
fountain? Express your answer
in terms of $ \pi$.

The circular region of $ AB$ forms an arc.  And then you have the bisector of $ AB$, $ DC$.

[asy]draw((5,0)--(5,16.4));
draw((-3,10)--(13,10));
dot((5,0));
dot((5,16.4));
dot((-3,10));
dot((13,10));
dot((5,10));
label("$A$",(-3,10),W);
label("$B$",(13,10),E);
label("$D$",(5,10),NE);
label("$C$",(5,0),S);
label("$E$",(5,16.4),N);
label("$8$",(1,10),N);
label("$8$",(9,10),N);
label("$10$",(5,5),E);[/asy]

$ AB$ is a secant that forms an arc (which I do not know how to draw :P).  $ CDE$ is the radius.

A well known theorem is that:

$ AD \cdot DB \equal{} CD \cdot DE$

$ 8 \cdot 8 \equal{} 10 \cdot DE$

$ DE \equal{} 6.4$

Therefore, the radius is $ 8.2$, and the area would be $ 67.24\pi$.

That is probably wrong. xD

Actually...

[hide="Solution"]

Since $AD=8$ and $DC=10$, $AC$ must equal $\sqrt{164}$.

Since $AC$ is the radius, we can easily calculate the area of the circle which comes out to be $\boxed{164\pi}$
[/hide]


suppose u have $ 4-chloro but-1-ene$  and $ 1-chloro butane$ then which will react faster via $ E_{2}$

[hide="Answer"]4-Chlorobut-1-ene reacts faster, because the double bond stabilize the incipient double bond that is forming in the transition state.[/hide]


Hi, I need some advice :maybe: . Ive studied math olympiad for about a year. I could do about 7 in the NZIMO problems for camp (see attachment), my goal if to represent NZ next year at IMO.

What material, books, should I use. How long should I study a day inorder to achieve my goal? :ninja:  :maybe:

i don't know what exactly you want to accomplish at IMO, but the time spent problem solving is pretty much as much as you can to get to that level,

and for problem 12, what is P? is it the point of tangency between circumcircle CA_1B_1 and $\Omega$?


[img]http://upload.wikimedia.org/wikipedia/commons/thumb/0/0e/Rubik%27s_cube_variations.jpg/250px-Rubik%27s_cube_variations.jpg[/img]

For 2*2*2 Pocket Cube , the all possible permutations is $\frac{8!\cdot3^{3}}{24}=3674160$

$\frac{8!\cdot3^{8}\cdot12!\cdot2^{12}}{12}\approx4.3\cdot10^{19}$ which for 3*3*3 Rubik's Cube

$\frac{8!\cdot3^{7}\cdot(24!)^{2}}{24\cdot(4!)^{6}}\approx7.4\cdot10^{45}$ which for 4*4*4 Rubik's Revenge

$\frac{8!\cdot3^{7}\cdot12!\cdot2^{10}\cdot(24!)^{3}}{(4!)^{12}}\approx2.8\cdot10^{74}$ which for 5*5*5 Professor's Cube

So can we find a formula of all possible permutations for the general n*n*n Cube ?


Let $S_n$ denote the sum of the elements in the $n$th set of the sequence of sets of squares $\{1\}$, $\{4,9\}$, $\{16,25,36\}$,$\ldots$. Find a formula for $S_n$.

[hide="interesting"]
after some trial and error we see that the last term of the $n$th set is the square of $\frac{n^2+n}{2}$, and the first term of the $n$th set is the square of $\frac{n^2-n+2}{2}$ (Theorem of finite difference helps greatly in finding these formulas).  Then let $f(n) = \frac{n(n+1)(2n+1)}{6}$, a.k.a. the sum of squares from 1 to $n$.  then $S_n = f(\frac{n^2+n}{2}) - f(\frac{n^2-n+2}{2}-1)$, or $f(\frac{n^2+n}{2}) - f(\frac{n^2-n}{2})$.  heavy simplifying yields $\frac{n(3n^2+1)(n^2+2)}{12}$. Now that's right :lol:
[/hide]
edit: fixed algebraic error.

Try again  :lol: . You can plug in $n=2$ to see that the formula isn't right.
You started off correctly. Must have been an algebra error

the answer can be expressed in the product of the two numbers because the sum of second brackett is a prime number. But the method is true so you have done some mistakes in simplifying.

[quote="drunner2007"]Looks good. Good job  :)

Edit: nvm, there's another error somewhere[/quote]
wait...I plug in 7 values and they all work, and it must be polynomial function.  Since it's degree 5, I think the rest should work...

by theorem of finite difference, I got

$1 + 12\binom{n-1}{1} + 52\binom{n-1}{2} + 101\binom{n-1}{3} + 90\binom{n-1}{4} + 30\binom{n-1}{5}$

Is this the right one??? or else I got my data wrong...

Not sure if I explained enough.
[hide]$\{1\},\{2,3\},\{4,5,6\},\{7,8,9,10\},\{11,12,13,14,15\},\cdots$
$\{1\},\{4,9\},\{16,25,36\},\{49,64,81,100\},\{121,144,169,196,225\},\cdots$
My approach is to sum up all of the terms before the terms up to the nth set, sum up all of the terms up to the (n-1)th set and find the difference.
The series involved is $1^{2}+2^{2}+3^{2}+\cdots+x^{2}=\frac{1}{6}x(x+1)(2x+1)$ and we need to find the upper bounds of both sums.
The last term in the nth set is $1+2+3+\cdots+n=\frac{1}{2}n(n+1)$.
The last term in the (n-1)th set is $1+2+3+\cdots+(n-1)=\frac{1}{2}n(n-1)$
Now plug these in to the $\sum_{1}^{n}{x^{2}}$ formula and subtract them:
\begin{eqnarray*}\frac{1}{6}\left[\frac{1}{2}n(n+1)\right]\left[\frac{1}{2}n(n+1)+1\right]\left[n(n+1)+1\right] = \frac{1}{24}n(n+1)(n^{2}+n+2)(n^{2}+n+1) \\ \end{eqnarray*}
\begin{eqnarray*}\frac{1}{6}\left[\frac{1}{2}n(n-1)\right]\left[\frac{1}{2}n(n-1)+1\right]\left[n(n-1)+1\right] = \frac{1}{24}n(n-1)(n^{2}-n+2)(n^{2}-n+1) \\ \end{eqnarray*}
Subtracting these two (it was a lot of expanding) gives:
$\boxed{S_{n}=\frac{n}{12}(n^{2}+2)(3n^{2}+1)}$[/hide]


How many pairs of integers (x, y) are there for which x^2 − y^2 = 81?

[hide]Factoring and finding factors of 81, we have
$ (\pm41,\pm40),(\pm15,\pm12),$ and $ (\pm9,0)$ for a total of $ 10$ pairs of integers.[/hide]

[hide]
$ (x+y)(x-y) = 81$

since x and y are integers, we know it will fall into one of the following factorizations
$ \pm1 * \pm81$
$ \pm3 * \pm27$
$ \pm9 * \pm9$
$ \pm27 * \pm3$
$ \pm81 * \pm1$

equating those back to the original x+y and x-y factors, we have (x,y) pairs of
$ (\pm41, \mp40)$
$ (\pm15, \mp12)$
$ (\pm9, 0)$
$ (\pm15, \pm12)$
$ (\pm41, \pm40)$

or 10 ordered pairs
[/hide]


induction
geometry
3D geometry
complex numbers
calculus
calculus open

Let a be a complex number
Let $(x_n)_{n \in \mathbb{N}}$ be a sequenece defined by
$x_0 \in \mathbb{C}$ and $x_{n+1}=\frac{x_n+\frac {a}{x_n}}2$.

It is known that if xn converges, then it converges to one of the roots of the equation x2=a.
When does (xn) converge? And to which of the roots does it converge to?
--------------------------------------------------------------------------------
I'm not sure that is an open question, but it is open to me.
Not that I've squeezed my brains out on it.
--------------------------------------------------------------------------------
There is a general construction for xn .

Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be a differentiable function.
Let $x_0 \in \mathbb{C}$.
Define $x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}$.

Same as before, if (xn) converges, then it converges to one of the roots of the equation f(x)=0.

Can you give conditions for x0 so that xn converges? Other than x0 is a root of f(x)=0 and not a root of f'(x)=0 :D?

I think I have some hints, which are not far from the solution:

    Take r such that r^2=a. Then put x_n=r*(a_n). We have a_(n+1)=((a_n)^2+1)/2a_n. Thus, (a_(n+1)-1)/(a_(n+1)+1)=((a_n)-1/((a_n)+1))^2. By an easy induction we have [1+a_n]/[a_n-1]=[ (a_0+1)/(a_0-1)]^2^n. Take x=(a_0+1)/(a_0-1). Then a_n= 1+2/(x^2^n-1). Thus, we need to find x complex such that x^2^n converges to a number different from 1. But if x^2^n converges, so does |x|^2^n (I haven't studied convergence for complex sequences, but it seems natural) . Thus, |x|<=1. If |x|<1, x^2^n converges to 0 and thus a_n converges to -1 and so x_n converges to -r. Suppose |x|=1. Take x=cos t+i sin t. Then sin (t*2^n) converges and I think this problem was already discussed on this forum. As far as I remember t must be of the form k*pi/2^p for k and p integers.

The original problem, the square root, has a simple and easy characterization, which is not at all typical of the general type of problem. If the starting point lies in the right hand half-plane (real part > 0), then all iterates lie in the right hand half plane, and the sequence converges to the positive square root. If the starting point is in the left hand half plane, then all iterates lie to the left and the sequence converges to the negative square root. If the starting point lies on the imaginary axis, then all iterates are pure imaginary and the sequence cannot converge. (There are countably many pure imaginary starting points for which the sequence reaches exactly zero in some finite number of steps and cannot be further defined; for all other pure imaginary starts, it just wanders around.)

That's deceptively simple. If you were trying to find the cube root of $a$ using Newton's method, by letting $\displaystyle z_{n+1} = \frac {2z_n + \frac a{{z_n}^2}}3$, and then trying to partition the complex plane into the three sets for which the sequence starting within that set will tend to a specific one of the three cube roots of $a$, along with a "borderline" set for which the sequence fails to converge, then these sets intertwine in a complex and beautiful fractal pattern. The same can be said for higher roots. These fractal patterns are well known and wouldn't be too hard to reproduce for someone who knows a little programming.


Ok, here is the problem... also... if there is some sort of formula or generalization that I can use to solve it, would you please tell me... thank you!! :D
Problem:

John won $\$750000$ in a lottery. He invests it at 4% per annum compounded semi-annually. How much could he withdraw at the end of every 6 months for the next ten years?

Let's work out a similar problem.  Suppose an amount L is deposited at the beginning of the year into a fund bearing compound interest at an effective annual rate of i.  What is the level amount K that can be withdrawn at the end of each year, such that at the end of n years, the fund is depleted?

The naive solution is to consider what happens to the balance at the end of each year.  In the first year, the fund has grown to (1+i)L, just before the first withdrawal.  The balance at the beginning of year 2, just after the withdrawal, is then (1+i)L - K.  This again grows to (1+i)((1+i)L - K) at the end of year 2, and after the second withdrawal, is now at (1+i)((1+i)L - K) - K.  It is not hard to see that at the end of n years, the balance is

$(1+i)^{n}L-\left((1+i)^{n-1}+(1+i)^{n-2}+\cdots+1\right)K = (1+i)^{n}L-Ks_{\overline{n|}i}= 0.$

Therefore, $L = K a_{\overline{n|}i}$, or $K = \frac{Li}{1-(1+i)^{-n}}$.

The solution of your problem is similar, with $n = 20$ payments, and $i = i^{(2)}/2 = 0.02$, since the fund is compounded semiannually.

The non-naive approach is to see that the accumulated value of the initial deposit of L must equal the accumulated value of the withdrawals, since the fund is depleted.  This leads us directly the the above relationship.


modular arithmetic
algebra proposed
algebra

Find all positive integers $ n$ such that there exists sequence consisting of $ 1$ and $ - 1: a_{1},a_{2},\cdots,a_{n}$ satisfying $ a_{1}\cdot1^2 + a_{2}\cdot2^2 + \cdots + a_{n}\cdot n^2 = 0.$

I think you have a small error in typo. :wink:
Although,it is a well-known problem of Erdos and Suranyi.You can see more in here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=30998[/url]

I'm sorry, it should be $ a_{1}\cdot1^2 \plus{} a_{2}\cdot2^2 \plus{} \cdots \plus{} a_{n}\cdot n^2 \equal{} 0.$

[quote="No Reason"]I think you have a small error in typo. :wink:

Although,it is a well-known problem of Erdos and Suranyi.You can see more in here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30998[/url][/quote]
I don't think it's a same problem.

Here is my solution:
Obviously $n \equiv 3, 0 \pmod4 $. It cannot happen $n=3$ or $n=4$. Let $n \ge 7$.
Now:
$n=7$ works: $1^2+2^2+4^2+7^2=3^2+5^2+6^2$
$n=8$ works: $1^2+4^2+6^2+7^2=2^2+3^2+5^2+8^2$
$n=11$ works: $11^2+9^2+7^2=rest$
$n=12$ works: $12^2+10^2+9^2=rest$
Now if $n$ works then also $n+8$ works because: $(n+1)^2+(n+4)^2+(n+6)^2+(n+7)^2=(n+2)^2+(n+3)^2+(n+5)^2+(n+8)^2$. End of story.


Do you support my idea for a group that decides when to do tournaments, as described in ernie's thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213520&sid=b481b095d4cb8c8571293af7af8c09d1
Please describe why/why not.

Copy/paste power! Here is the idea:

I say we should have a group of people who run the tournaments, and we should have a monthly tournament. In this monthly tournament we should have a bunch of divisions, like the sub-1300 division, the sub-1400, etcetera. We should also have team tournaments. If we have one group organizing tournaments, no chaos will ensue. Maybe the people of AOPS could vote who will run tournaments? Please comment on my idea.

I support it entirely.  It is a way that people who want to host tourneys can host tourneys, without all the mania.

I think this group idea is good. This way, as ernie said, the people who really like leading ppl by making tourneys can do so. Also, by making tourneys a group effort everyone there will probably be many different viewpoints, and by collaborating them together, tourneys will be happier for everyone!!

(lol nice avatar ernie)

...its the internet..you cant tell other ppl who to do...

Yes, we can.  Trust me, mewto has a lot of ways to go from here...

I'm not saying tell you what to do, but if everyone decides to stop doing every single tournament, people will stop making them. And will the people who voted no say [b]why[/b] they dont support it?

Eat that 25% of the people who voted!  EAT THAT!  :D

I want to :D I like tourneys and i think its a good idea to have ppl help. Its a good idea because then tournaments could be arranged easier


calculus algebra polynomial function geometry AMC AIME

Hi folks,

For those of you that are in some type of pre-Calc course (or have already taken it), what topics out of the norm would you like to see in that course?

Fulton

I don't know what's in a typical Pre-Cal class, but [i]I'd[/i] enjoy combinatorics and Stirling numbers being discussed. Somehow I doubt that Stirling numbers would make  many people happy.

dude

we should discuss lagrange interpolations

we know that it pwnz all polynomials and rational functions

i mean really

can you think of anything that beats lagrange

well karamata does

but lagrange wins over all

because karamata is a turnip and is thus very advanced

but apart from that i'd like a ton of geometry

not the weird kind where they make up their own theorem names

like the house builder theorem

i mean who came up with that

its such a random name

but i'd like cevalaus = ceva + menelaus, and other such stuff because it'd be useful

I'm interesting in all topics concerning infinity and wouldn't have minded learning more infinite stuff in pre-cal. I'll learn more about infinity in AP calculus next year though, won't I? Some more geometry (like cyclic quadrilaterals) may have helped me on the AIME (but I have AoPS for that).

What is the house builder theorem? I did a google search and only found one result.

You will learn limits next year in Calc (or more about them). And there are certainly limits as x approaches infinity

$\lim_{x\to\infty}f\left(x\right)$

I don't know that qualifies as learning about infinity. Are you talking about infinitesimals? Or perhaps the sizes of infinities?

I haven't them all yet, but there is a really good MathJam series here on AOPS about that. It's called "Cardinalities and Infinity". Check it out.

There is also a book called "The Art of the Infinite" by Kaplan that I believe is pretty good.

Fulton

Yeah, I like limits and also sequences are fun (such as $1/2+2/4+3/8+4/16+5/32+...$)

So does it converge? How can you prove it? Can you show what it converges to? (Now you see the teacher in me coming out! :-))

I just figured it out a couple weeks ago. I was so happy.
Let $A = 1/2+2/4+3/8+4/16+5/32+...$
$2A = 1/1+2/2+3/4+4/8+5/16+6/32+...$
Subtracting the first equation from the second yields
$A = 1+1/2+1/4+1/8+1/16+1/32+...$
Which is an infinite geometric sequence with a first term of 1 and a common ration of $1/2$
$F/(1-R)=1/(1-1/2)=2$
Thus the original expression is equivalent to $2$.

Lemma: $\sum_{i=1}^\infty nx^{n}= \frac{x}{(1-x)^{2}}$

[hide="Proof of Lemma"]This is true if one takes the derivative and multiplies by $x$ the commonly known:

$\sum_{i=0}^\infty x^{n}= \frac{1}{1-x}$[/hide]

Proceeding, the sequence is equivalent to:

$\sum_{i=1}^\infty \frac{n}{2^{n}}.$

By our lemma, substituting $x=1/2$ yields: $\frac{1/2}{(1-1/2)^{2}}= \frac{1/2}{1/4}= 2.$

@Twiz: Whoa, that's a pretty sweet solution. Could you describe the motivation behind it?

Okay, time for my 50th post!
@Karth:
I'm not quite sure what you mean by motivation; if you mean how did I figure it out, I've just been playing around with sequences a lot lately. I've tried everything with a lot of sequence but have found that many sequences that look nice do not have nice solutions or require calculus (which I'm taking next year). For example I played around with:
$\sqrt{2+\sqrt[3]{2+\sqrt[4]{2+\sqrt[5]{2+...}}}}$
and posted it somewhere on AoPS, but someone told me that it requires calculus. If anyone can give me a non-calculus solution, I'd really appreciate it.

[quote="gtf"]Karth, I'm missing something. Shouldn't the geometric series in the proof of your lemma start with $i=0$?[/quote]

Heheh, you're right... I'll fix it.

So far, I've been writing random stuff on the board in precalc (like Ceva, DeMoivre, Menelaus, a tiny bit of Euler). I don't think anyone in my precalc class is a user on AoPS, however.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170097

I have posted the same question some time before (it was from a UGA problem, in fact), and the answers given in the two replies are not pre-calculus at all...my method was similar to Twiz's (doing differences), and the higher the power is, the more "chaotic" the formula appears to be...


Muirhead για $ \left(6,\minus{}\frac{1}{2},\minus{}\frac{1}{2}\right) \succ (5,0,0)$ και είσαι έτοιμος!

Αλέξανδρος

[quote="cretanman"]Muirhead για $ \left(6, \minus{} \frac {1}{2}, \minus{} \frac {1}{2}\right) \succ (5,0,0)$ και είσαι έτοιμος!

Αλέξανδρος[/quote]

Πολύ γρήγορος!!!!! Δεν την έβαλες σε hint τουλάχιστον για να την προσπαθήσουν και οι άλλοι. Δεν πειράζει όμως θα βάλω ακόμη μία. :roll:

Παιδιά μπορεί κανείς να μου πει ποια είναι η ανισότητα Muirhead?  :huh:

Ανισότητα Muirhead

Αν για τους πραγματικούς αριθμούς $ a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}$ ισχύει ότι $ (a_{1},a_{2},...,a_{n})\succ (b_{1},b_{2},...,b_{n})$, τότε για κάθε $ x_{i}>0$ όπου $ i\equal{}1,2,3,...,n$ ισχύει ότι $ \sum_{sym}\prod^{n}_{i\equal{}1}x^{a_{i}}_{i}\geq \sum_{sym}\prod^{n}_{i\equal{}1}x^{b_{i}}_{i}$

[quote="Γιώργος"]Ανισότητα Muirhead
b]$ (a_{1},a_{2},...,a_{n})\succ (b_{1},b_{2},...,b_{n})$[/b][/quote]
Αυτό τι υποτίθεται ότι παριστάνει;  :huh:

το θεμα λυνεται και με την ανισοτητα των βαρων.
θετοντας $ x\equal{}a^2 ,y\equal{}b^2, z\equal{}c^2$
exeis na apodeikseis oti $ a^{13} \plus{}b^{13} \plus{} c^{13}>\equal{}a^{11}bc \plus{}ab^{11}c \plus{} abc^{11}$

omws opws eipa apo tn anisotita twn barwn pairnume oti:

$ 11/13a^{13} \plus{} 1/13b^{13} \plus{} 1/13c^{13}>\equal{}a^{11}bc$

$ 1/13a^{13} \plus{}11/13b^{13} \plus{} 1/13c^{13}>\equal{}ab^{11}c$

$ 1/13a^{13} \plus{}1/13b^{13} \plus{} 11/13c^{13}>\equal{}abc^{11}$

prosthetontas kata meli tis parapanw anisotites pairnume tn zitumeni. :wink:

[quote="Γιώργος"]Έστω $ n\geq 2$ θετικός ακέραιος και $ a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}$ πραγματικοί αριθμοί. Λέμε ότι η $ n$-άδα $ (a_{1},a_{2},...,a_{n})$ μεγιστοποιεί τη $ n$-άδα $ (b_{1},b_{2},...,b_{n})$

c)$ \sum^{k}_{i \equal{} 1}a_{i} \equal{} \sum^{k}_{i \equal{} 1}b_{i}$

Παραδείγματα:

$ (4,2)\succ (3,3)$ γιατί $ 4\geq 3$ και $ 4 \plus{} 2 \equal{} 3 \plus{} 3$

$ (4,1,1,0)\succ (3,2,1,0)$ γιατί $ 4\geq 3$          , $ 4 \plus{} 1\geq 3 \plus{} 2$,         $ 4 \plus{} 1 \plus{} 1\geq 3 \plus{} 2 \plus{} 1$,      $ 4 \plus{} 1 \plus{} 1 \plus{} 0 \equal{} 3 \plus{} 2 \plus{} 1 \plus{} 0$[/quote]

Μήπως στη συνθήκη c εννοείς $ \sum^{n}_{i \equal{} 1}a_{i} \equal{} \sum^{n}_{i \equal{} 1}b_{i}$? Ή μήπως εννοείς ότι στην b πρέπει η ισότητα να ισχύει για κάποιο k? :roll:
Please διευκρίνησέ το λίγο.Ευχαριστώ

Διόρθωσα το λάθος μου. :blush:  Check it again...

[quote="Athinaios"]το θεμα λυνεται και με την ανισοτητα των βαρων.
θετοντας $ x \equal{} a^2 ,y \equal{} b^2, z \equal{} c^2$
exeis na apodeikseis oti $ a^{13} \plus{} b^{13} \plus{} c^{13} > \equal{} a^{11}bc \plus{} ab^{11}c \plus{} abc^{11}$
[/quote]

αυτό αποδεικνύεται πολύ εύκολα με αναδιάταξη


calculus
integration
trigonometry
function
complex analysis

[quote="ksun"]Find

Integral from 0 to infinity of (sinx+1)/((x-2)(x-1)x(x+1)(x+2)).[/quote]
$\int_{0}^{+\infty}\frac{(sinx+1)dx}{(x-2)(x-1)x(x+1)(x+2)}$

[quote="jmerry"]That's obviously divergent, since it has poles on the positive real axis.[/quote]

I don't understand your argument. E.g. one can prove $\int_{0}^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$

That's irrelevant. The function $f(z)=\frac{\sin z}z$ has no poles anywhere. It has a removable singularity at zero.

There would be some hope for an integral like

$\int_{-\infty}^{\infty}\frac{\sin(\pi x)}{(x+2)(x+1)x(x-1)(x-2)}\,dx$

But that wasn't what what asked.

OK, admittedly my example was completely irrelevant. But I'm still not seeing the obviousness of the argument. Consider for instance the integral $\int_{0}^{+\infty}\frac{dx}{(x-1)^{2n+1}}$ has -if I did not make any errors- (Cauchy principal) value $\frac{1}{2n}$ whenever $n\in\mathbb{N}_{0}$. However the integrand has a pole 1 of order $2n+1$.

Sure, it has a principal value- but that's not the default meaning of an integral.


calculus
derivative
analytic geometry
graphing lines
slope

[quote="phongthong"]from x 2  = x + x + ... + x (x times)
if we differentiate on LHS. we get : 2x
but on RHS. we get 1 + 1 + .. + 1 = x

why this is not true.? Can anyone explane me[/quote]

why should it be true?

yes you have made proper steps but just because 2 things are equal does not mean their derivatives have to be equal

ex.
x^6 = x^2 * x^2 * x^2
does not mean that
6x^5 = 2x * 2x * 2x

derivative is slope at a point and has nothing to do with your basic equation, for you cannot relate the slope of 2 different elements in such a way.

I think it has something to do with you saying x+x+x+...+x (x times) where x is the variable.  Because x is the variable, both the number of x's and the value of them change.  Also, the x+x+x+...+x argument doesn't work if x is not an integer.  How many x's would you have if x=3.14?  However, I cannot pin down exactly where the fault is.

Edit (to respond to jason)
" ex.
 x^6 = x^2 * x^2 * x^2
 does not mean that
 6x^5 = 2x * 2x * 2x"

That isn't true...  $x^6 = x^2 * x^2 * x^2$ then differentiating RHS would use the product rule, so you'd get $6x^5= 2x*x^2*x^2+2x*x^2*x^2+2x*x^2*x^2=6x^5$
I think the mistake has more to do with the discreteness of the x+x+x+...+x argument.

well yeah, it doesnt work except for integers, but besides that, consider the fact that x must be a constant if you are using the "sum" rule for derivatives. if x is a constant, then both derivatives are 0.

by the way - this example is simply false - forgot to use product rule

[quote]
ex.
x^6 = x^2 * x^2 * x^2
does not mean that
6x^5 = 2x * 2x * 2x [/quote]

[quote="jasonpingpong"]yes you have made proper steps but just because 2 things are equal does not mean their derivatives have to be equal

ex.
x^6 = x^2 * x^2 * x^2
does not mean that
6x^5 = 2x * 2x * 2x

derivative is slope at a point and has nothing to do with your basic equation, for you cannot relate the slope of 2 different elements in such a way.[/quote]

Actually, your example is not true because of the way the product rule for derivatives is.  The product rule states that
\[\frac{d}{dx} f(x) \cdot g(x) = ( \frac{d}{dx} f(x) ) g(x) + ( \frac{d}{dx} g(x) ) f(x)\]

Thus $\frac{d}{dx} ( x^2 \cdot x^2 \cdot x^2 ) = \frac{d}{dx} ( (x^2 \cdot x^2) \cdot x^2 )$
\[= \frac{d}{dx} (x^2 \cdot x^2) \cdot x^2 + ( (x^2 \cdot x^2) \frac{d}{dx} x^2\]
\[= [ ( \frac{d}{dx} x^2 ) x^2 + x^2 ( \frac{d}{dx} x^2 ) ] x^2 + x^4 \cdot 2x\]
\[= [ 2x \cdot x^2 + x^2 \cdot 2x ] x^2 + 2x^5\]
\[= [ 4x^3 ] x^2 + 2x^5\]
\[= 6x^5\]

Furthermore, if two things are equal, their derivatives are equal.

And you [i]can[/i] take the derivatives of separate terms (separated by addiiton) - that is, the following example would be correct:

\[3x = x + x + x\]
\[\frac{d}{dx} 3x = \frac{d}{dx} x + \frac{d}{dx} x + \frac{d}{dx} x\]
\[3 = 1 + 1 + 1\]

So the reason why phongthong's proposition is false is because you're taking the derivative of a variable number of terms, not for any of the reasons you cited.


The curve y = 1/(1 + x^2) is called a witch of Maria Agnesi.  Find the equation of the tangent line to this curve at the given point
( -1, 1/2).


group theory
abstract algebra
real analysis
superior algebra
superior algebra solved

Prove that if a group $ G$ satisfies $ \left|G\right|=2k$, with $ k$ odd, then $ G$ has a subgroup of index 2.

(please try to avoid, if possible, Sylow's theorems; because I have not seen those up till now)

Thanks in advance

Hi.

Here is something with sylow. Perhaps one can change this a little bit:

Consider the subgroup $G'\leq S_{G}$ that is given by Cayley's theorem. Then consider the normal subgroup $G'\cap A_{G}$. Since every element of odd order lies in $A_{G}$, $A_{G}$ contains all p-Sylowgroups for $p>2$. Therefore we have $|G': G'\cap A_{G}|\leq 2$.
There must be an element of order two which is mapped to a product of m disjoint transpositions, so its sign is -1 and hence $|G': G'\cap A_{G}|=2$ so we have the desired subgroup in G'.

Maybe it is possible to do this without knowledge about the existence of p-Sylowgroups, but I don't see how...

[quote="Halmos"][quote="Gockel"]Here's another try:

We look again at the Group $G'\leq S_{G}$ which is given by Cayley's theorem. G contains an element of order 2 (which can be proven even without using Cauchy's theorem). The corresponding element of $G'$ must be a product of k disjoint transpositions and has therefore sign -1.
So we get $G'A_{G}=S_{G}$ and hence $G'/G'\cap A_{n}\cong G'A_{G}/A_{G}=S_{G}/A_{G}\Rightarrow |G': G'\cap A_{G}|=2$.
Again $G'\cap A_{G}$ is the desired subgroup.[/quote]

Beautiful! A pitty I have to study differential equations and real analysis too. I could do Algebra 24/7 :D[/quote]

Ok. Exams are coming up and I'm reviewing my problems. I couldn't follow Gockel's answer too well. I'm also aware of the fact there have been several threads about this problem. But most of them are -at this moment- too advanced for me (when using Sylow or tricks with $\mathcal{A}_{k}$). I tried to find a proof now myself. Would someone be so kind to correct or confirm me.

First of all I proved some preliminary results

i) [b]If $G$ is some finite group and $\pi: G \to S_{G}$ is the left regular representation. Then we know that if $|x|$ is even and $|G|=mn$ for some $m$, then $\pi(x)$ is a product of $m$ $n$-cycles.[/b]

We let $G$ act on itself. Since $|x|=n$ we consider the subgroup $H=<x>$. Then we consider the cosets of $H$. Since this partitions $G$, we can aswell let $x$ act on those cosets. If we let $x$ act on one of those cosets it is clear that this gives us an $n$-cycle in $S_{G}$. Since we have $m$ such cosets, we see that $\pi(x)$ is a product of $m$ $n$-cycles. If $|x|$ is even and $|G|/|x|$ is odd, then $\pi(x)$ will be an odd permutation.

The actual problem:

ii) [b]Prove that if |G|=2k, with k odd, then G has a group of index 2. [/b]

Now if we consider more specifically, a group $G$ with $|G|=2k$ whereas $k$ is odd. Then we can produce with Cauchy an element $x$ of order $2$. Because of the facts mentioned above, $\pi(x)$ will be be an odd permutation. Now if we observe in our group $G$ the subgroup $A=\{g | \pi(g) \ is \ even \}$, we must have that  $A$ is an index 2 subgroup or all of $G$. Because $\pi(x)$ is odd, left multiplication by $x$ on $A$ is a bijection between $A$ and $B=\{g | \pi(g) \ is \ odd \}$. Hence $A$ is a subgroup of index 2 in $G$, and we're done.

Can somebody confirm this?

Thanks in advance

I don't understand why for $x\in G$ of order 2, the image of $x$ must be odd (an element of order 2, can be even or odd).

I think $|G|=2k$ where $k>1$ is odd. But then $2k\geq 6$. So $A_{2k}$ is simple. So $\pi(G)=G'$ can't be a subgroup of $A_{2k}$ (since $G$ has a normal subgroup of order 2). Hence $G'A_{2k}=S_{G}$ and we can follow the argument of Gockel.

[quote="barra"]I don't understand why for $x\in G$ of order 2, the image of $x$ must be odd (an element of order 2, can be even or odd).
[/quote]

Because of the arguments in part i). if $|x|=2$ then $\pi(x)$ is the product of $k$ transpositions ($2$-cycles). Or did I go wrong somewhere? If you see a flaw in the proof of i) please let me know.

thanks in advance


Starting for (0,0) on a coordinate plane, I can move up,down,left or right one unit at a time.  How many different points could I end up on after 16 steps?

[hide="hint"]try for 1 step, then for 2..you'll find a easy recursive pattern[/hide]

Let (a,b) be the ending point. Note that we can end at (a,b) iff 2|(a+b). The rest is simple counting.


probability
analytic geometry
rotation
trigonometry
conics
parabola
geometry

no solutions since Yufei said don't include them.

* = "short answer": only answer was required for the contest.
& = "long answer": a written justification/proof was required for the contest.

Questions are not exact wording, but all essentials are there, and it is very very similar to exact wording in most cases.

1a*.  (a,a) is on the line 3x - y = 10.  Find a.
1b*. A = (0, -1), B = (2,0), C on AB such that BC = 2AB with C in the first quadrant.  Find C.
1c&. O = (0,0), B = (40,0), D = (88,0), A is on the y axis s.t. AB = 50, C is s.t. BC = 50 and CD || AO.  Find the midpoint of A and C.

2a*.  If y = 2x + 3 and 4y = 5x + 6, find x.
2b*.  If a,b,c are numbers such that -3b + 7c = -10, b-2c = 3, a+2b-5c = 13, find a.
2c&.  A and B write a contest: 2 times A's score is 60 more than B's score, and 2 times B's score is 90 more than A's score.  Find the average
score of A and B.

3a*.  If $2^x = 2(16^12) + 2(8^16)$, find x.
3b&.  If f(x) = 2x-1, find all real x such that $(f(x))^2 - 3(f(x)) + 2 = 0$.

4a*.  Cards 1 to 6 are placed in a box, and 2 randomly picked without replacement.  Find probability that the smaller of the two numbers is less than or equal to 4.
4b&.  B and H are on the y axis, with $BH = 400$, and B closer to the origin; $\angle HLP = 60^{\circ}$, $\angle BLP = 30^{\circ}$.  Find LP.

5a*.  A goat starts at (0,0) and moves 1 unit up on the first move, 2 units right on the 2nd move, 3 units down on the 3rd move, 4 units left on the 4th move, and continues in this fashion in a spiral.  Find the coordinates of the goat if it moves 55 units total.
5b&.  Determine all possible values of r such that 4, 4r, 4r^2 is an arithmetic sequence.

6a*.  An equilateral triangle of sidelength 3 is inside a circle so that two of the vertices, B and C, are on the circle, and the third vertex, A, is the center of the circle.  The triangle is rotated about C so that A is now on the circle, then about A so that B is on the circle, until the triangle returns to its original position, where it stops.  Determine the length travelled by B during this time.
6b&.  ABD is a triangle, and C is a point under side BD.  AB, BC, CA, CB = 5,6,7,4 respectively.  Find CD.

7a*.  If $f(x) = \sin^2 x - 2\sin x +2$, find the min/max of f(x).
7b&.  The parabola $y = -\frac{1}{4}(x-r)(x-s)$ intersects the axes at the points $(-k,0), (3k,0), (0,3k)$.  Find k, and the coordinates of the vertex of the parabola.

8a&.  Given $f(x) = \{ -4$ if $x<-5$, $-x$ if $-4\leq x \leq 5$, -5 if $x>5 \}$, graph $g(x) = \sqrt{25 - (f(x))^2}$, and state the shape of each portion of the graph.
8b&.  Two circles are tangent at B, A is on one circle, C on the other, such that ABC is collinear.  Show that the tangent to one circle at A is parallel to the tangent at C of the other circle.

9%.  The circles $(x-p)^2 + y^2 = r^2$ and $x^2 + (y-p)^2 = r^2$ have centers C and D.  The circles intersect at distinct points A and B, with x coords. a and b respectively.

a)  prove a+b=p, $a^2 + b^2 = r^2$.
b)  If r is fixed, and p is found to maximize area of CADB, prove either A or B is the origin.
c)  If p, r are integers, determine the minimum possible distance between A and B, and show an example of p, r that give this distance.

10%.  f(n) is defined as follows: for every positive integer n, consider n lockers in a row.  Starting with 1, we close every 2nd open locker.  When we reach the end, we turn around and close every 2nd open locker.  This repeats until 1 locker is left open, and this is the value of f(n).  [In example, when n=15, we first close 2,4,6,8,10,12,14; then 13,9,5,1; then 7,15; then 3, leaving f(15) = 11.]

a) Find f(50)
b) Prove there is no positive integer n so that f(n) = 2005.
c) Prove that there are infinitely many positive integers n such that f(n) = f(2005).

---

(I believe that #1c ihas been written wrong.) 2  The D coordinates are actually (88,0), not (98,0).  I read over the other questions and it looks like there are no further errors though.  So you might want to fix that...  :lambda

2 and half hours and yes it's like a geometry test

[quote="shobber"]when i first see its name, i think it is an All Geometry Contest...

How long are you given to solve all these problems?[/quote]

Its a variety of problems.

If you want to see the past tests, go to http://www.cemc.uwaterloo.ca/english/contests/euclid.shtml


Let x_i (i=1,..,n) be n real positive numbers such that the sum of n-1 of them is greater than remaining number. Prove that

(x_1)^2/(-x_1+x_2+..+x_n) + (x_2)^2/(x_1-x_2+x_3+..+x_n)+..+(x_n)^2/(x_1+..-x_n) \geq (x_1+..+x_n)/(n-2)

Sorry but I don't remember if this very famous and well known inequality was already posted in the forum.


rotation
geometry
geometric transformation
trigonometry
trig identities
Law of Cosines

Prove that through every point inside a circle (exept of center) there can be drawn a chord of rational length.

[hide="Hint"]Well, you may make the chord varie, while still passing through the point. Prove that as it varies, it's length varies, and since it's length variation is continuous, by the intermediate value theorem, somewhere in "between" it has to have a rational length. :P [/hide]

Didn't I say that I need a high school solution, terms "continuity" and "intermediate value theorem" are very sofisticated for a high school student.

Essentially, he's saying that if you have 2 different lengths of chords, then every length inbetween those 2 lengths must be attainable, since otherwise, you would have "skipped over" a length when you were rotating the chord. Therefore, it suffices to show that there exists a rational number in every interval [a,b] $ a\ne b$, where a s the smaller of the 2 chords, and b is the larger of the 2 chords.

To do that, write a,b in decimal form. Then, since they are different, at some point along the decimal line, starting from left and going to the right, the 2 decimal digits will differ. Cut off the remaining stuff to the right from b, and you have your rational number.

e.g. a=312.443567893... and b=312.443574836... they are first different when a=312.44356, and b=312.44357.

Hence, 312.44357 is your rational number.

Fill in all the detals you need, the gist of a high school proof is here.

I think rofler's proof is about as high-school as you're going to find.

Just to clear up one more thing, you can find two chords with different length by taking a diameter and a non-diameter (as long as the point isn't the center).

Well, ya, I tried to simplify "intermediate value theorem" and "continuity" into the simplest terms I could find :P

This also (kind of) proves that there exist infinitely many rational numbers in any subset of real numbers, no matter how small. (Just truncate the digits :D)

[quote="yedi"]Yes guys, that everything rests on intuition. The existance of rational number between two reals can be shown to a high school child, but no one strictly mathematicaly proved that every lenth between a and b is attainable, which is needed.[/quote]take two chords, one of them is the diameter, the other is non diameter. rotate the non diameter chord until it coincides the diameter. as we rotate the chord, the length increases very smoothly. so it is clear that every length between the first (non diameter) and the second (diameter) chord  is attainable.

[quote="mathking123"]is that considered a strict, mathmatical proof?[/quote]are you referring to my proof? I don't know exactly what a "strict" proof is, but I think my proof is mathematical and valid.

Instead of using continuity to use the Intermediate value theorem, we can solve this by giving a formula for some chords and show it takes on all values in some interval by exhibiting an inverse function.
[hide]

Say our circle is of radius $ r$ centered at the origin. WLOG (by rotation) we can say our point is $ P = (0,y)$ for some $ y < r$,$ y\neq0$ Then the longest chord is the diameter through $ P$, which has length $ a = 2r$. Consider the chord perpendicular to the radius through $ p$, which has length $ b = 2\sqrt {r^2 - y^2}\neq r$ by Pythagoras. We have also $ b < a$, and $ b$ is the shortest length of a chord.

Now consider the chord which makes an angle of $ \theta$ with the chord of length $ b$ (There are 2 but they have the same length.) By law of cosines and solving, you can find that it has length $ 2\sqrt {y^2 \sin^2\theta + r^2 - y^2}$. Let $ f(\theta) = 2\sqrt {y^2 \sin^2\theta + r^2 - y^2}$

Between the lengths of the two chords there is a rational number. Call it $ c$. Then there exists a value of $ \theta$ such that $ f(\theta) = c$, name $ \theta = \pm \arcsin\left(\frac1{2y}\sqrt {c^2 - 4r^2 + 4y^2\right)}$. Since $ c^2 < 4r^2$ and $ c^2 > 4r^2 - 4y^2$ (because those are the shortest and longest chords), this is always defined.

[/hide]

Does that work?


Calculati: $\mathop {\lim }\limits_{n \to \infty } (\sin \frac{a}{1} + \sin \frac{a}{2} + ... + \sin \frac{a}{n}),a \in (0,\pi )$

Use  $x-\frac{x^3}{6}\leq \sin x \leq x$

the limit is $+\infty$


complex analysis
advanced fields
advanced fields unsolved

Let $f$ be a holomorphic map of the unit disc |z|<1 into itself, which is not the identity map . Show that $f$ can have, at most, one fixed point.

In fact, it either has 1 fixed point in the unit disc, or 2 on the boundary of the unit disc if it's conformal.

It suffices to show that a map holomorphic on the unit disc with two distinct fixed points is the identity.  With f such a map, take [i]a[/i] and [i]b[/i] distinct and in the unit disc such that f(a) = a and f(b) = b.  For z in the disc, let g(z) = (a - z)/(1 - <a>z), where <> denotes conjugate.  It is well known that g is an analytic automorphism on the unit disc and is its own inverse.   For z in the disc, let h(z) = g(f(g(z))).  We have h mapping from the unit disc to itself, and h(0) = 0; furthermore, h(g(b)) = g(f(g(g(b)))) = g(b) != 0, so g(b) is a fixed point of h.  Thus the hypotheses of the Schwarz Lemma apply to h, and we have h(z) = z for all z in the disc.  But then, for all z in the disc, z = g(f(g(z))).  Applying g to each side yields g(z) = f(g(z)) for all z in the disc, and since g is an automorphism on the unit disc, we are done; f is the identity.


Let be given a triangle $ABC$. Prove that: $\frac{\sum\cos{(B-C)}}{2}\le\sum\cos A\le\frac{3}{2}$
[b]Part 2[/b] was posted here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=120170

$\sum(cos(B-C))$ $\leq$ 2$\sum(cosA)$

$\Leftrightarrow$  $\sum(cos(B-C))$ $\leq$ $\sum(cosA)$ - $\sum(cos(B+C))$

$\Leftrightarrow$  $\sum{cos(B-C)+cos(B+C)}$ $\leq$ $\sum(cosA)$

$\Leftrightarrow$  $\sum(2cosB.cosC)$ $\leq$ $\sum(cosA)$

$\Leftrightarrow$ 2$\sum2RcosBcosC$ $\leq$ $\sum2RcosA$

If the perpendiculars from the vertices A,B and C meet the opposite sides in L,M and N and P is the orthocentre it is known that PL =2RcosBcosC..... and PA =2RcosA........

 This information and the Erdos Mordell inequality( 2$\sum(PL)$ $\leq$ $\sum(PA)$ ) allow us to see that the final inequality above is true.

We have: $\frac{\sum\cos{(B-C)}}{2}\le\sum\cos{A}\Longleftrightarrow\sum\cos{(B-C)}+\cos{(B+C)}\le\sum\cos{A}\Longleftrightarrow\sum 2\cos{B}.\cos{C}\le\sum\cos{A}$
On the otherhand: $\sum\cos{A}=\frac{2}{3}.\frac{3}{2}\sum\cos{A}\ge\frac{2}{3}\left(\sum\cos{A}\right)^{2}\ge\sum 2\cos{B}\cos{C}$


algebra
polynomial
integration
complex analysis
function
advanced fields
advanced fields unsolved

$a_n$ sequence in Z such that the power serie $\sum_{n\geq0}a_n.z^n$ has a radius
of convergence $R\geq 1$, let $f(z) = \sum_{n\geq0}a_n.z^n $ for
$ |z| <1$. Suppose f is bounded by M on the open disk
 $D=(z \in C; |z|<1)$
Prove that f is a polynomial

I'll go with a method motivated by Fourier series.

[b]Lemma:[/b] If $\displaystyle f(z) = \sum_{n=0}^\infty a_nz^n$ has radius of convergence greater than $r$, then $\displaystyle \frac1{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta = \sum_{n=0}^\infty r^{2n}|a_n|^2$.

For now, I'll leave the proof of this lemma to the reader. If requested, I could supply the proof.

If $|f(z)| \le M$ on the open unit disc, then for $\displaystyle r<1, \frac1{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta \le M^2$. Hence by the lemma,
$\displaystyle \sum_{n=0}^\infty r^{2n}|a_n|^2 \le M^2$. Take the limit as $r\to1$ (It's increasing in $r$, so it must have a limit, and the interchange of sum and limit is justified by Abel's Theorem) to get that $\displaystyle \sum_{n=0}^\infty |a_n|^2 \le M^2$.

But the coefficients are integers. If $k$ of the coefficients are different than zero, then the sum above must be at least $k$, hence $k\le M^2$. Since only finitely many of the coefficients are nonzero, the power series is a polynomial.

The set of bounded holomorphic functions on the unit disc is also known as the Hardy space $H^\infty$. The problem asked to prove this: if $f \in H^\infty$ and the coefficients of the power series for $f$ are integers, then $f$ is a polynomial.

I actually proved this in two steps, involving the Hardy space $H^2$.
First, $H^\infty \subset H^2$.
Second, the theorem is true for $f \in H^2$.

Interestingly, the bound for $f$ constrains the number of nonzero terms in the polynomial but does not constrain the degree of the polynomial.


function
induction
inequalities
algebra
partial fractions
geometric series

Define the Fibonacci numbers $F_n$ by $F_0 = 0, F_1 = 1$, and $F_{n} = F_{n-1} + F_{n-2}, n > 1$.

1)  Evaluate $\sum_{k=1}^{\infty} \frac{1}{F_k}$.

2)  Find the function given by $\sum_{k=0}^{\infty} \frac{F_k x^k}{k!}$.  Can you generalize for $a_n$ any second-order linear recurrence with initial conditions $a_0, a_1$?

the first part is irrational
i believe Erdos proposed it first and it was proven by Jeannin.
its something around $3.35....$ if i recall correctly

edit: proving the convergence is easy though ;)

[quote="JBL"]For the second part: Let $f(x) = \sum_{k=0}^{\infty} \frac{F_k x^k}{k!}$.  Then $f'(x) = \sum_{k=0}^{\infty} \frac{F_{k+1} x^k}{k!}$ and $f''(x) = \sum_{k=0}^{\infty} \frac{F_{k+2} x^k}{k!}$ and so $f''(x) = f'(x) + f(x)$.  The generalization to any linear recurrence (second order or even higher) is equally straight forward.  Then the question is reduced to solving a linear differential equation with constant coefficients, which has a well-known solution set.[/quote]

What is this solution set?  I tried $f(x) = ab^x$ for suitable $a, b$ but $f(0) = 0 \implies a = 0$.  :(

[hide="the second one"]
Write $F_n$ in closed form:

Let $g(t)=\sum_{n=0}^\infty F_nt^n$. Then $g(t)-tg(t)-t^2g(t)=t$, since $F_0=0$, $F_1-F_0=1$, and the recursion clears higher power terms.

This means $g(t)=\frac{t}{1-t-t^2}$. Let $\phi=\frac{1+\sqrt{5}}{2}$ and $\bar{\phi}=\frac{1-\sqrt{5}}{2}$. Then using partial fractions, $g(t)=\frac{A}{1-\phi t}+\frac{B}{1-\bar{\phi} t}$ for some constants $A$ and $B$. Putting this over a common denominator, $B(1-\phi t)+A(1-\bar{\phi} t)=t$.

Substitute $\frac{1}{\phi}=\phi-1$ for $t$ to get $A(2+\bar{\phi})=\phi-1$. Writing this in radical form, $A=\frac{1}{\sqrt{5}}$. Similarly, substituting in $\frac{1}{\bar{\phi}}$ for $t$ gives $B(2+\phi)=\bar{\phi}-1\Rightarrow B=-\frac{1}{\sqrt{5}}$.

Expanding the fractions as geometric series and setting coefficients equal, $F_n=\frac{\phi^n-\bar{\phi}^n}{\sqrt{5}}$. Summing the power series, $\sum_{n=0}^\infty \frac{F_n x^n}{n!}=\boxed{\frac{e^{\phi x}-e^{\bar{\phi} x}}{\sqrt{5}}}$.
[/hide]

All second-order differential equations with constant coefficients have solutions of the form $y = a\cdot e^{\alpha x} + b\cdot e^{\beta x}$ where $a, b$ are arbitrary parameters.

[quote="maokid7"]so does anyone have a soltion for

$\displaystyle\sum_{k=0}^\infty\frac{1}{F_k}$?
and how would you go about showing that it converges?  :([/quote]
[b][size=109][u]Proof of convergence [/u][/size][/b]

[b]Claim[/b]: $F_k \geq k^2$ for all natural $k\geq12.$

We use induction on $k$ to prove the claim is true.
[u]Base case[/u]:
$F_{12}=144$ (we just compute it by hand) and $12^2=144.$ So the inequality is true for $k=12$.

[u]Induction case[/u]:
Let $F_m \geq m^2$ for some $m > 12.$
Then, $F_{m+1} = F_m + F_{m-1} \geq m^2 + {(m-1)}^2$
$\therefore F_{m+1} \geq {(m+1)}^2$
$\Longleftrightarrow m^2 + {(m-1)}^2 \geq {(m+1)}^2$
$\Longleftrightarrow m(m-4) \geq 0$
$\Longleftrightarrow m \leq 0$ or $m \geq 4$.
Since, $m > 12$, the [b]claim[/b] is true.

Now, for all natural $k\geq12$,
$F_k \geq k^2$
$\Longrightarrow \frac{1}{k^2} \geq \frac{1}{F_k}$
$\Longrightarrow \displaystyle\sum_{k=12}^\infty\frac{1}{F_k} \leq \displaystyle\sum_{k=12}^\infty\frac{1}{k^2} < \displaystyle\sum_{k=1}^\infty\frac{1}{k^2} = \frac{\pi^2}{6}$
$\Longrightarrow\displaystyle\sum_{k=1}^\infty\frac{1}{F_k}$ is convergent.

(I am sorry I don't know how to compute the sum of the series.)

thanks. i was having trouble finding an upper bound.
i had no problem finding a lower, namely $\displaystyle\sum_{n=0}^\infty\frac{1}{2^n}$.


function
algebra
functional equation
number theory proposed
number theory

If a function $f : \mathbb{R} ^2 \longrightarrow \mathbb{R} $ satisfies the functional equation $ f(x_1, y_1) f (x_2 , y_2) = f(x_1 x_2 + y_1 y_2, x_1 y_2 - y_1 x_2 ) $ $\forall x_1, x_2 , y_1 , y_2 \in \mathbb{R} $, prove that $ \exists $ a multiplicative function $ m : \mathbb{R} \longrightarrow \mathbb{R} $ and a signum function $ \sigma : R \longrightarrow  \pm 1 $ s.t.
$ f(x , y) = \sigma (x , y) m ( \sqrt{x^2 + y^2 } ) $

[hide]this problem has a nice number thoery connection!!![/hide]

Let $z_k = x_k + i y_k$.   Then  $f(z_1) f(z_2) = f( \bar{z_1} z_2 )$.
Let $m$ be $f$ restricted to the real line; it is multiplicative.
Let $|z_2| =1$ and proceed from there.


[quote="Ditdo"]Some three non-negative real numbers $p,q,r$ satisfy $p+q+r=1$. Prove that:
\[ 7(pq+qr+rp)\le\ {2+9pqr}.  \][/quote]
Very old inequality, just homogenize then use Schur's inequality.

yec , it's very old
 Prove that :
$7(pq+pr+rp)(p+q+r) \leq (p+q+r)^3 +9pqr$
   hence , with ineq's Schur ,  we done complete

[quote="Ditdo"]Some three non-negative real numbers $p,q,r$ satisfy
                                                  $p+q+r=1$.
Pove that
                                  $7(pq+qr+rp)\le\ {2+9pqr}$.[/quote]
Use $(1-2p)(1-2q)(1-2r)\le pqr$ and $8pqr\le (1-p)(1-q)(1-r)$
 ;)

thanks boys..your hints have gived new ideas to me ;)


inequalities
geometry
inequalities proposed

Hello guys could you help me with this problem  :oops: :

In an acute triangle with altitudes $ h_a,h_b,h_c$ show the inequality:

$ (a/h_a)^2\plus{}(b/h_b)^2\plus{}(c/h_c) \ge 4$

[quote="Nebraska boy"]Hello guys could you help me with this problem  :oops: :

In an acute triangle with altitudes $ h_a,h_b,h_c$ show the inequality:

$ (a/h_a)^2 \plus{} (b/h_b)^2 \plus{} (c/h_c) \ge 4$[/quote]

$ LHS \equal{} \frac{1}{4S^2}(a^4\plus{}b^4\plus{}c^4) \ge \frac{4s^4}{27S^2}\equal{} \frac{4s^2}{27r^2} \ge 4$

well 1/2*a*h_a = area K
and the same applies for b,c

so (a/h_a)^2 = a^2/4K^2
and similiarly applies for b,c

therefore your sum i s equal to (a^2 + b^2 + c^2)/4K^2
Then you can use Weitzentock's inequality

hello, we have $ A\equal{}\frac{1}{2}ah_a$ and from here we get $ \frac{a^2}{2A}\equal{}\frac{a}{h_a}$ and analogous we get
$ \left(\frac{a^2}{2A}\right)^2\plus{}\left(\frac{b^2}{2A}\right)^2\plus{}\left(\frac{a^2}{2A}\right)^2\geq4$
with $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ we get
$ a^4\plus{}b^4\plus{}c^4\geq16A^2$
or equivalent to
$ (a^2\minus{}b^2)^2\plus{}(b^2\minus{}c^2)^2\plus{}(a^2\minus{}c^2)^2\geq0$
which is always true.
Sonnhard.

[quote="Nebraska boy"]Thanks a lot guys for your very nice solutions , may u help me with this related inequality i think that the solution is similar:

in the acute angled triangle $ r_a,r_b,r_c$ are exinradios show the ineq:

$ (a/r_a)^2 \plus{} (b/r_b)^2 \plus{} (c/r_c)^2 \ge 4$

Thanks in advance[/quote]

By AM-GM we have : $ r_{a}^{2}\equal{}\frac{p(p\minus{}b)(p\minus{}c)}{p\minus{}a}\leq \frac{a^2p}{4(p\minus{}a)}$

$ \Leftrightarrow \frac{a^2}{r_{a}^{2}} \geq \frac{4(p\minus{}a)}{p}$

We have : $ \sum \frac{a^2}{r_{a}^{2}} \geq  \sum \frac{4(p\minus{}a)}{p}\equal{}4$

Done ! :)


For all real numbers $x,y$ define $x\star y = \frac{ x+y}{ 1+xy}$. Evaluate the expression \[ ( \cdots (((2 \star 3) \star 4) \star 5) \star \cdots ) \star 1995. \]

[i]Macedonia[/i]

[hide]another way to solve this problem

i'll use * instead star  :D

clearly: \[(a * b) * c = a * (b * c)\]

Now:Notice that : \[a*b*c = \frac{{a + b + c + abc}}{{1 + ab + ac + bc}}\]

\[a*b*c*d = \frac{{\sum\limits_{cyc} a  + \sum\limits_{cyc} {abc} }}{{1 + \sum\limits_{cyc} {ab}  + abcd}}\]

by induction we can prove - If n is even - : \[{a_1}*{a_2}*....*{a_n} = \frac{{\sum\limits_{i = 1}^n {{a_i} + \sum\limits_{1 \leqslant k < l < j \leqslant n} {{a_k}{a_l}{a_j} + .... + \sum\limits_{i = 1}^n {\frac{{\prod\limits_{i = 1}^n {{a_i}} }}{{{a_i}}}} } } }}{1+{\sum\limits_{1 \leqslant s < t \leqslant n} {{a_s}{a_t} + ...... + \sum\limits_{1 \leqslant s < t \leqslant n} {\frac{{\prod\limits_{i = 1}^n {{a_i}} }}{{{a_s}{a_t}}} + \prod\limits_{i = 1}^n {{a_i}} } } }}\]

and then it's easy

define:\[P(x) = \prod\limits_{i = 2}^n {\left( {x - i} \right)} \]

\[\therefore 2*3*....*1995 = \frac{{\frac{{P( - 1) - P(1)}}{2}}}{{\frac{{P(1) + P( - 1)}}{2}}} = \frac{{P( - 1) - P(1)}}{{P(1) + P( - 1)}}\]

:) :) :)[/hide]

I'm not sure if this one is simpler: in $x*y$ we recognise the addition formula for $\tanh$ . It is defined as $\tanh x:={{\sin hx}\over {\cosh x}}={{e^x-e^{-x}}\over {e^x+e^{-x}}}={{e^{2x}-1}\over {e^{2x}+1}}$. Then indeed we have $\tanh(a+b)={{\tanh a+\tanh b}\over {1+\tanh a\cdot \tanh b}}$. Hence * describes adding the arguments within the $\tanh$ function. This shows that * is associative and commutative.
Now we want to calculate $2*\cdots *n$ for $n\ge 3$. For this we need the inverse of $\tanh$: unfortunately the range is $(-1,1)$, but no problem since $a*b=(1/a)*(1/b)$. Hence we can also calulate $(1/2)*\cdots *(1/n)$.
Explicit calculation gives $\tanh x=a\iff x={{\ln(1+a)-\ln(1-a)}\over 2}$ for $|a|<1$.

Hence ${1\over 2}=\tanh {{\ln (3/2)-\ln(1/2)}\over 2}$ up to ${1\over n}=\tanh{{\ln ({{n+1}\over n})-\ln ({{n-1}\over n})}\over 2}$.
Adding all $\ln$-terms gives $2*\cdots *n=\tanh ({1\over 2}\ln {{n^2+n}\over 2})={{n^2+n-2}\over {n^2+n+2}}$.

Of course, once you know this formula, you can also give a proof using induction since ${{n^2+n-2}\over {n^2+n+2}}* (n+1)$ gives the formula for $n+1$.

Nice idea! But I don’t really get how this is P1 and still looking for something simple


The number of teeth in three meshed circular gears $A, B, C$ are $x, y, z$, respectively. (The teeth on all gears are the same size and regularly spaced.) The angular speeds, in revolutions per mintue, of $A, B, C$ are in the proportion:

$(A) x: y: z$
$(B) z: y: x$
$(C) y: z: x$
$(D) yz: xz: xy$
$(E) xz: yx: zy$

[hide]
The time between each tooth for all three gears is constant, since the gears are meshed. We have $\omega=\frac{\theta}{t}$. Since $\theta=\frac{2 \pi}{n}$, where $n$ is the number of teeth, we have that $\omega \sim \frac{1}{n}\implies \omega_{a}: \omega_{b}: \omega_{c}= \frac{1}{x}: \frac{1}{y}: \frac{1}{z}\implies \boxed{D}$.[/hide]

[quote="cincodemayo5590"][hide]
The time between each tooth for all three gears is constant, since the gears are meshed. So, we have $\omega=\frac{\theta}{t}$ Since $\theta=\frac{2 \pi}{n}$, where $n$ is the number of teeth, we have that $\omega \sim \frac{1}{n}\implies \omega_{a}: \omega_{b}: \omega_{c}= \frac{1}{x}: \frac{1}{y}: \frac{1}{z}\implies \boxed{D}$.[/hide][/quote]
But $\frac{1}{x}: \frac{1}{y}: \frac{1}{z}$ isn't one of the choices...

[hide]
Multiply everything by $xyz$ to get $yz : xz : xy$.[/hide]


Prove that if you choose $(n+1)$ elements from the set $\{1,2,3,\dots,2n\}$ prove that there are atleast numbers that are relatively prime.

Did you solve this problem by yourself the first time you solved it? I remember although it seems very easy in retrospect, a group and I struggled with this one...

[quote="Farenhajt"]If $1$ is chosen into the subset, then it's trivial since $1$ is coprime with all the remaining numbers. Hence assume $1$ is not there. Take the smallest available prime, $2$, and pick all the numbers which AREN'T coprime with it. In the set there are $n$ even numbers available, therefore $(n+1)$-st must necessarily be odd and thus coprime with $2$. If we skip $2$ and take any higher prime, we'll have even less of its multiples available, hence we'll get to the desired situation even faster.[/quote]

[quote="t0rajir0u"]Or:

We always have two consecutive elements.

 :rotfl:[/quote]

[b]Farenhajt Consecutiveness Conjecture[/b]: Someone will always destroy your nice and wordy and lengthy proof in the following post.

 :rotfl:

[quote="Phelpedo"]A variation: prove that there are always 2 number that aren't relatively prime (given a selection of n+1 of the 2n terms).[/quote]

Consider the set of largest odd divisors of the $n+1$ elements; since there are only $n$ odds in $\{1,2,\ldots,2n\}$, two share the same largest odd divisor. And one of them has a larger power of $2$ so it divides the other, which is a stronger statement than they aren't relatively prime.

Oh yeah...that's actually what I meant to say.

And that was the problem i was referring to also...


Two integes are randomly selected from the set of integers greater than or equal to -5 and less than or equal to 5. The two numbers need not be different. What is the probability that the sum of the two integers is less than their product?

[hide]There are three cases, negative negative, negative zero, and positive positive.
For both negatives, you have 5*5=25 ways.
For negative zero, you have 1*5+5*1=10 ways.
For both positive, the only ones that don't work are 11, 12, 13, 14, 15, 22, and their reverses. So out of 5*5=25, 10 don't work. That's a total of 25-10=15 for both positive.
25+10+15=50. The total number of ways is 11*11=121. So the probability is $\frac{50}{121}$ [/hide]

[hide]$-5\ge x\ge 5$
$-5\ge y\ge 5$

$x\le 0$ works for all. That is $35$ ways. Then, you have the positives. For this one, $12$ DO NOT work. So, for the positives, you have $15$.

Then, you add those up $\implies 50$

The total is $121$, because you have $11$ values for $x$ and $11$ values for $y$. So, the answer is $\boxed{\frac{50}{121}}$.[/hide]

h34ni,, ur solution is wrong, even if u got the right answer

[hide]use simon's favorite factoring trick.

x+y<xy

xy-x-y+1>1

(x-1)(y-1)>1

0,0 is one turning point and 2,2 is the other.

so, the total number of successful ways is 36+16-2=50

answer is 50/121

I think this is a more systematic approach.[/hide]

Sorry to revive the topic, but I thought that order doesn't matter. (3,1 is the same as 1,3) How do you know that in this case, order matters?

Order doesn't matter because the outcome doesn't change. I'm pretty sure you are correct.

Yeah, but if you calculate the probability when order doesn't matter, you get a denominator of 66, but the answer has a denominator of 121.

Vinayak, order [b]doesn't[/b] matter.

But why do the answer looks like the order matters? Here is a simple problem to explain:

[quote="Sample Problem"]I have a coin, a regular dice and 4 cards (ace, two, three and four).

I flip the coin first.

[list]
[*]If it is heads, I will throw the dice.
[*]If it is tails, I will take a card from the four cards.[/list]

What is the probability of getting number 3? (either from the card or the dice)[/quote]

If you answer $\frac{2}{10}=\frac{1}{5}$, it is wrong.
Why? Because probability of getting a value from the dice is different from getting a value from the dice.

You can't simply count the wanted outcomes, but you have to pay attention to the probability of the outcomes.

It's similar to the original problem that started this thread. The probability of getting $3$ and $1$ is twice larger than $3$ and $3$. So, it looks like we count $(3,1)$ and $(1,3)$ different, although it is not.

Hopefully this helps.

Note: Well, the answer to the sample answer is $\frac{5}{24}$.


inequalities induction function calculus

If $a>1$ and $b>2$ are positive integers, show that $a^{b}+1 \geq b(a+1)$, and determine when equality holds.

Easy induction on a.

We have that $(a+1)^{b}= \sum_{i=0}^{b}\binom{b}{i}a^{i}> a^{b}+b$

Thus if $a^{b}+1 \ge b(a+1)$ then $(a+1)^{b}+1 > b(a+2)$

Since equality occurs for $a=2, b=3$ the inequality is proven and we have shown that the equality occurs only if $a=2, b=3$

[hide="I was stumped for a while until I read the words integers :wink: "]Let $f(a)=a^{b}+1-b(a+1)$ and $f'(a)=ba^{b-1}-b=b(a^{b-1}-1)>0$ since $a>1, b-1>1$.
The function is continuous and we want to find $a$ that minimizes $f(a)$ so this must be the smallest possible value of $a$ in the interval, since in the interval the function is increasing. It follows that for all $b$, it suffices to test $a=2$ as the minimum value of the difference of the two sides of the inequality. Plugging in, we need only prove
\[2^{b}+1\geq 3b \]
yet clearly the difference is again an increasing function so equality if at all must hold at $b=3$, where it does (since this is minimum $b$ on interval).

IF this contest doesn't allow calculus all the derivative parts have easy workarounds with the same principle. I know its a lame proof :D
[/hide]

Students participating in this Olympiad must be less than 15. So they are supposed not to know a thing about calculus! :D

Hey! Why can't I edit my post? It says I need to talk to a moderator to do that.

I wanted to add the $2^{b}+1 > 3b$ can easily be shown for $b > 3$ (The proof I gave above was incomplete)

You can easily proceed by induction on $b$.
Equality iff $a$ = $2$ and $b$ = $3$.


calculus
integration
trigonometry
calculus computations

Calculate: $ \int \frac{sinx}{(2\minus{}cosx\plus{}sinx)^{2}} dx$

Standart substitution $ u \equal{} \tan\left(\frac {x}{2}\right)$ will work.

Yes I know it is an easy integral but I found it is in some way "beautiful".


While I was doing mathcounts warm up 18 (Jan 06) I encountered this question:

The local sandwhich shop has 3 cash registers. In how many diffrent ways can 4 customers line up at the registers?

The answer is 360 but I don't know how you get it. :)

[hide]First, you look at the different ways the people could line up.
(After each combination, I have the number of ways they could line up that way)

4 0 0 - $4!$, which is 24
3 1 0 - 4p3, which is $4!$ = 24
3 0 1 - same thing, 24
2 1 1 - With each combo of 2 people, there are 2 ways for the other 2 people to line up at the other 2 registers. So 4p2 * 2, which is 24
2 0 2 - the same as that, you can probly see why, 24
2 2 0 - 24
1 1 2 - up there, similar, 24
1 2 1 - 24
1 3 0 - 24
1 0 3 - 24
0 1 3 - 24
0 2 2 - 24
0 3 1 - 24
0 4 0 - 24

Hmm... looks like I'm missing one... but hopefully you understand what I meant, someone smarter can point out whatever I missed :lol: [/hide]

You missed 0 0 4 :)

[hide="I did it this way..."]
There are 3 places for the first person to go.
Then there are 4 places for the second person to go (including in front of the 1st person).
Now there are 5 places for the third person to go, then 6 places for the fourth person to go.
3*4*5*6 = [color=green]360[/color][/hide]

LadyKn1ght's solution is so... pwnzor!

I probably would've done something similar to Klebian's way, but I would've just put certain groups together. For example, instead of counting 004 040 and 400 seperately, I would've just counted it once and multiplied the result by 3...


Given $\{1,2,...,30\}$ how many subsets have the property that the sum of the elements exceeds $232$?

If possible provide the motivation as well as a solution if you know what I mean.
Because I have a rather nasty way of doing this by considering many cases, there got to be a better way.

Posted many times.

[hide="hint"] $1+2+\cdots+30 = 465$. [/hide]


I have a few questions about pascals triangle:

1: Who Is Pascal?
2:What is Pascal's Traingle?
3:What is it used for
4: What are the multiples of 2 in Pascals triangle??

Any answers would be greatly apreciated :)

[hide="3"] Lots of things. It has countless applications, including combinations, Binomial Theorem, and triangular numbers, square numbers, pentagonal numbers, etc.[/hide]

[hide="4"] Umm...All mutiples of 2, or even numbers, are in Pascal's triangle. [/hide]

1: [url]http://en.wikipedia.org/wiki/Blaise_Pascal[/url]
2: [url]http://en.wikipedia.org/wiki/Pascal%27s_triangle[/url]

Every positive integer appears in Pascal's Traingle at least once. All positive integers except 2 appear at least twice.

If you are talking about modulo 2, then there are nice traingle patterns
      1
      11
     101
     1111
    10001
    110011
   1010101
   11111111
  100000001
  1100000011
 10100000101
 111100001111
1000100010001


ratio
geometry
circumcircle
rectangle
power of a point
radical axis
geometry proposed

I managed to solve IMO 1996/2 using inversion today :)
I know it is possible to solve IMO 1999/5 and IMO 1993/2 using inversion.
Now my question is: how do you do that?

Btw,

[u]IMO 1993/2[/u]

D is a point inside (acute) triangle ABC such that ADB = ACB + 90° and AC·BD = AD·BC.
(a)  Calculate the ratio AB·CD/(AC·BD).
(b)  Prove that the tangents at C to the circumcircles of ACD and BCD are perpendicular.

[u]IMO 1999/5[/u]
The circles C1 and C2 lie inside the circle C, and are tangent to it at M and N, respectively. C1 passes through the center of C2. The common chord of C1 and C2, when extended, meets C at A and B. The lines MA and MB meet C1 again at E and F. Prove that the line EF is tangent to C2.

So, I would really like to see inversion solutions.
Grobber (inversion-freak), maybe this is something for you? :)

I did try to solve 1999/5 with inversion (I haven't tried anything for the other one yet). However, i don't know if inversion is the best way to approach it. Here's an outline (I have the proofs but no time):

1) Let C be a circle, and let C1 and C2 be circles internally tangent to C in M and N s.t. they have a common chord AB (AB is a chord in C, but it contains the common chord of C1 and C2). Let MA and MB intersect C1 at E1 and F1 respectively and NA and NB intersect C2 at E2, F2 respectively. Prove that E1E2 and F1 F2 are, in fact, the common tangents of C1 and C2 (maybe you could use inversion somewhere in the proof for this).

2) Let C1 and C2 be circles s.t. C1 passes through the center of C2. Let the common tangents of C1 and C2 touch C1 at E and F. Prove that EF is tangent to C2. We really don't need any inversion for this part.

I think it's obvious what you get by combining (1) and (2).

Well, what you said is the classical solution.
But I saw our deputy invert this situation once ... but I don't remember how...
Actually, that worked out nicely :) and it is different from yours, Grobber.

Well, I even have a hard time finding a good inversion pole (one for which the resulting configuration is nicer than the given configuration). I'll give it more thought, but I only apply inversion when it's obvious what inversion to apply! :D  :D  :D

And for IMO 1993/2, our deputy did something with inversion with respect to a circle of Apollonius I believe ... But again, I don't know anymore what he exactly did. Btw, can someone explain me the use of circles of Apollonius? :? Where are they good for?

Hey, I solved IMO 1993/2 :)

Just invert with respect to a circle centered at D.
The images A', B', C' of A, B, C form an isosceles, rectangled triangle (easy to prove).
The conclusions follow quite easily.

I thought I had a solution different from yours, but it's basically the same. I inverted everything with respect to B. The fact that ADB=ACB+90 translates to the circumcircles of ADB and ACB being orthogonal, and since inversion conserves the angle of 2 curves, the lines that these 2 circles are turned into are orthogonal. Then, by using that other hypotesis: AC*BD=AD*BC, we get A'D'C' to be an isosceles and right triangle (just what you got, but you used another inversion). We can derive that ratio pretty easily: 2^1/2.

As for part b, I'm sure it can also be derived from here, but a very simple angle-chase suffices (we just have to use the hypothesis abt the angles).

Hi. I see this topic only today. This circle of problems (like 1993.2) is well known. Some basic fackts about Apollonius circles of triangle. All they very simple but interesting.

1) Apollonius circle of triangle is the set of points X such that
AX/BX=AC/BC. (and so defines other two Apollonius circles)

2) The tangent to circumcircle af ABC at point C pass through the center of Apollonius circle assosiated with side AB

3) Apollonius circles are ortogonal to circumcircle of ABC

4) All Apollonius circles pass through the point, that is isogonal to Fermat (Torrichelli) point R. The foots of perpendiculars from this point to sides are the vertexes of equilateral.

5) Common chord of Apollonius circle and circumcircle is a simedian of triangle.

6) So the Lemuan point L have equal degree for all three Apollonius circles

7) So Apollonius circle have the common radical axis.

8) This axis pass through O,R,L and perpendicular to its centerline.

9) The second common point R_1 of Apollonius circles is the invertion of R for circumcircle. The foots of perpendiculars from R_1 to sides of triangle are vertexes of equilateral too.

and so on....

That inversion with respect to Apollonius' circle might be a good idea, but I don't know exactly how to use it :D Such an inversion with respect to the Apollonius' circle corresponding to side AB would leave C and D unchanged and turn A into B (and vice-versa).

I think that inversion with Respect to circumcirle is better, because it's ortogonal to all Apollonius circles

[quote="Fedor Bakharev"]8) This axis pass through O,R,L and perpendicular to its centerline.[/quote]

Sorry, maybe this is a stupid question, but what do you mean by the term "centerline"?

I think he's referring to the centers of the 3 Apollonius circles. What he means is, I think, this: the common radical axis of the 3 Apollonius circles passes through O, R and L and is perpendicular to the centerline of the 3 Apollonius circles.

The centerline is, actually, Lemoine's line: the line which contains the intersection point between BC and the tangent in A to the circumcircle and the analogous points for B and C.


Fie sirul $a_n=\int_0^n\ln(1+e^{-x})dx$
Demonstrati ca e convergent si aflati limita sa.

Since $\ln(1+e^{-x})<e^{-x}$, therefore the result immeduatelly follows.
I hope, I correctly understood this problem  :blush:

hmm... da Myth... ai inteles-o bine... dar problema este la Liceu - Bacalaureat.. adik... nu e grea :)
imi place ca raspunzi la mai toate posturile mele :D .. dar.. sper sa nu fiu rau.. pb asta nu itzi era destinata tie chiar delok!! :D

hmm... Yes Myth... you understood it correctly... but the problem is in the High-School, Baccalaureat section... so it is not so demanding :)
I like that you answer to almost all of my posts :D .. but... I hope I am not mean if I say that this problem was not for you... :D


Cái này em thấy trong sách của bác Thuận(mà không có đáp án), ai có lời giải NGẮN GỌN thì cho em nha:
Cho a,b,c>0.Chứng minh rằng
(a/(a+b))^2+(b/(b+c))^2+(c/(c+a))^2>=3/4


$ f(x) \equal{} x^3 \minus{} x \minus{} 1$
Prove that $ f$ has [i]only[/i] one real root $ r \in \left(1, \frac 32 \right)$.
Prove that if $ a$ is a root and $ a \notin \mathbb{R}$ then $ |a| \in \left( \sqrt {\left(\frac 23\right)}, 1\right)$.

$ f(1) \equal{} \minus{}1$ and $ f(\frac{3}{2}) \equal{} 0.875$. Since there is a transition of $ y$ from being negative to positive in the interval for $ x$, the line must've crossed the x-axis, in other words a root.

For the second part:
If $ a\rightarrow(0.544331,1)$, then plugging in the two limits for the interval, gives us $ f(a)\rightarrow(0.161283,\minus{}1)$. That means that for some value of $ a$ inside the given interval, gives us a root.

For the first part I think you only proved the existence but not the uniqueness of the root...
For the second part you didn't show the root must be non-real...

For the first one, could you just graph it?
For the second one, i thought the R meant rational...
Oh well, lets wait for someone else to solve..

modular marc: graphs don't really prove.  I mean, they do provide insight but for a real proof, you need to examine the properties of the function, which typically explain why a graph looks a certain way.  And R means real.

first, note $ f'(x) \equal{} 3x^2 \minus{} 1$ so the function has a max at $ x \equal{} \minus{} \frac {1}{\sqrt {3}}$ and a min at $ x \equal{} \frac {1}{\sqrt {3}}$, is increasing on $ ( \minus{} \infty, \minus{} 1/\sqrt {3}) \cup (1/\sqrt {3}, \infty)$ and decreasing on $ ( \minus{} 1/\sqrt {3}, 1/\sqrt {3})$.  Now $ f( \minus{} 1/\sqrt {3}) \approx \minus{} .615 < 0$, , and $ 0 > f( \minus{} 1/\sqrt {3}) > f(x) > f(1/\sqrt {3})$ for all $ x \in ( \minus{} 1/\sqrt {3},1/\sqrt {3})$ since the function decreases over the interval.  Hence there are no solutions in the interval.  Similarly, $ f(x) < f( \minus{} 1\sqrt {3}) < 0$ for all $ x < \minus{} 1/\sqrt {3}$ since the function is increasing over the interval.  Thus there can only be zeros in the interval $ (1/\sqrt {3},\infty)$ and since the function is strictly increasing over the interval, there can only be one zero (because if $ x < y$ then $ f(x) < f(y)$.  If either one is the zero, then we know that the other is not a zero.  So $ f(x) \equal{} f(y) \equal{} 0$ can only occur if $ x \equal{} y$).

Now note that $ f(1) \equal{} \minus{} 1$ and $ f(3/2) \equal{} .875$.  By intermediate value theorem, there must exist $ r \in (1,3/2)$ such that $ f(r) \equal{} 0$.

for the second part, let $ a \equal{} re^{i\theta}$, with $ \theta \in [0,2\pi)$.  since $ a$ is complex, $ \theta \neq 0, \pi$.  plugging in, $ r^3 e^{3i\theta} \minus{} re^{i\theta} \minus{} 1 \equal{} 0$ so $ r^3e^{3i\theta} \minus{} re^{i\theta} \equal{} 1$.  Clearly $ r \equal{} 0$ is not a solution.  Equating real and imaginary parts, we have $ r^3 \cos 3\theta \minus{} r \cos \theta \equal{} 1$ and $ r^3 \sin 3\theta \minus{} r \sin \theta \equal{} 0$.  From the second equation, we have $ r^2 \equal{} \sin \theta / \sin (3\theta)$ so $ r(\sin \theta \cos 3\theta/\sin 3\theta \minus{} \cos \theta) \equal{} 1$ ... not sure this is useful...

Oh..I don't even think this problem's at my level, I'm only in 8th grade..

yeah, probably not.  what i've done is calculus (I can't think of a way to do it without.  But perhaps doable with inequalities...).  $ f(x)$ approaches $ \minus{}\infty$ as $ x$ approachess $ \minus{}\infty$ since the degree is odd and the leading coefficient positive, and $ f(x)$ approaches $ \plus{}\infty$ as $ x$ approaches $ \plus{}\infty$.  This quickly gives the increasing/decreasing once you find critical points.  For the second part, I used the formula $ e^{ix} \equal{} \cos x \plus{} i\sin x$

anyway, to continue: I ended up with $ r\equal{}\pm\sqrt{sin \theta/\sin 3\theta} \equal{} \frac{1}{\sin \theta \cot 3\theta \minus{} \cos \theta}$ which has solutions $ (\theta,r) \approx (.70385772, \minus{}.868837); (2.4377349,.86883696); (3.8454504,.86883696); (5.5793276, \minus{}.868837)$.  Note all the solutions fall nicely within the desired bound; in fact there appears to be only 1 value for $ |r|$ for all of them. (I don't think estimating is the best way to do this.  In any case, the trig is sure to be not pretty.)

Does anyone have a different approach? esp. for part 2.

Actually, part 1 can be much easier.

$ f(\sqrt {\frac {4}{3}}) < 0$ and $ f(\frac {3}{2}) \equal{} 0.875$, so there must be a real root $ r \in (\sqrt {\frac {4}{3}}, \frac {3}{2})$
Then we factorize it: $ x^3 \minus{} x \minus{} 1 \equal{} (x \minus{} r)(x^2 \plus{} rx \plus{} (r^2 \minus{} 1))$
$ Discriminant < 0 \Leftrightarrow 4 \minus{} 3r^2 < 0$, which is obviously true . This proves the uniqueness

For part 2, similarly we have $ r \in (\sqrt{\frac{5}{3}}, 1.4)$

So $ \frac{2}{3} <r^2\minus{}1<1$ $ \Leftrightarrow |a| \in \left( \sqrt {\left(\frac 23\right)}, 1\right)$ because $ |a|^2\equal{}r^2\minus{}1$ (it's easy to see that the two other roots are complex conjugates.)

Once you prove that there is only one real root, call it $ r$. If the other roots are complex, then they are conjugates (!). So the roots are $ r$, $ z$, $ \overline{z}$.

Now the product of the roots is $ |z|^2r\equal{}1$ by vieta. Since $ r$ is in (1,3/2), |z| is in root (2/3), 1 by that relation.


Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?

$ \textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$

[hide]The smaller circle has area pi.  The large circle has area 9pi.  The blue area has area 9pi-pi=8pi.  The ratio is 8pi/pi=8. D.[/hide]


has anyone seen numbers on CBS?

the season is over now, but I thought it did a pretty good job and it showed the public of "who needs math"ians that everyone needs math...

does this go in this forum??
Is this the first topic about this??

If the answer to one of those is no, then sorry

Yay! Awesome show! There was something about Math & Media and NUMB3RS was mentioned, but I don't think any topic was dedicated to it. This is probably the best forum to put it in as well.

I liked most of them a lot, but like the last one about the manhunt didn't have enough math :D but I guess that's expected of a TV show. Everybody here should watch it; I find it very interesting and a good way to show people what math is like, despite maybe not the perfect representation of it.

I saw in the Mathcounts newsletter that apparently one of the characters (no doubt the mathy one) was made Mathcounts National Champion! :D  I thought it was pretty funny/cool.

ah, but the funniest part, no doubt is that the actor who plays Charlie (the math whiz) has trouble... with algebra.  :rotfl: (read this in a newspaper column... didn't clip it, so can't cite it, sorry).

I love NUMB3RS!! Great show.. Yeah, I saw the guy that plays Charlie on the news one time after a show.. and he was at Caltech trying to do some math, and he couldn't do anything!! He was just writing down meaningless signs on the board.. and really slow too. Hahaha.

great show!!! but.....

it's got a thread devoted to it here:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=numb3rs&t=24524]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=numb3rs&t=24524[/url]

and has also been discussed here:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=numb3rs&t=33179]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=numb3rs&t=33179[/url]

try to go there rather than discuss it here...

Heard of the show but I never saw it (I don't watch that much TV...only Seinfeld)

I'll check it out

[quote="Mariag"]the show was kinda cool - though i only saw the first and last episodes :(
do you guys think they would have a rerun or just have it out on DVD?[/quote]

Reruns are going on right now I think. I just watched an episode today... I'm pretty sure it was the first one.

the first one was the one with the water sprinkler... charlie keeps looking at it and going "hmmmm..." was that this one?

i missed the first one...

most of the shows were about computers and binary code though, though.

Next season everyone should watch it cuz it gets a  :10:

[quote="paladin8"]
Reruns are going on right now I think. I just watched an episode today... I'm pretty sure it was the first one.[/quote]
what channel, CBS again? Same time?

Friday nights at 10 (Eastern and central (don't know about others)) on CBS

i'm going to rent the season from clockbuster.  The show looks awesome.

[quote="stupidkid"]Uhh...i was reading the MathCounts newsletter and it said that the guy that played Charlie was a former MathCounts national champion.  Is that true?[/quote]

No, I think the character he plays is a former MathCounts champion...


Five points $ O,A,B,C,D$ are taken in order on a straight line with distances $ OA\equal{}a$, $ OB\equal{}b$, $ OC\equal{}c$, and $ OD\equal{}d$.  $ P$ is a point on the line between $ B$ and $ C$ and such that $ AP: PD\equal{}BP: PC$.  Then $ OP$ equals:

$ \textbf{(A)}\ \frac{b^2\minus{}bc}{a\minus{}b\plus{}c\minus{}d} \qquad
\textbf{(B)}\ \frac{ac\minus{}bd}{a\minus{}b\plus{}c\minus{}d} \qquad
\textbf{(C)}\ \minus{}\frac{bd\plus{}ac}{a\minus{}b\plus{}c\minus{}d} \\
\textbf{(D)}\ \frac{bc\plus{}ad}{a\plus{}b\plus{}c\plus{}d} \qquad
\textbf{(E)}\ \frac{ac\minus{}bd}{a\plus{}b\plus{}c\plus{}d}$

Call BP x. therefore PC=c-b-x

$ \frac {b \minus{} a \plus{} x}{d \minus{} c \plus{} (c \minus{} b \minus{} x)} \equal{} \frac {x}{c \minus{} b \minus{} x}$

crossmultiplying and mass-simplifying gives us

$ x \equal{} \frac {(c \minus{} b)(b \minus{} a)}{d \minus{} c \plus{} b \minus{} a}$

and then $ OP \equal{} b \plus{} \frac {(c \minus{} b)(b \minus{} a)}{d \minus{} c \plus{} b \minus{} a} \equal{} \frac {bd \minus{} ac}{ \minus{} a \plus{} b \minus{} c \plus{} d} \equal{} \frac {ac \minus{} bd}{a \minus{} b \plus{} c \minus{} d} \equal{} \boxed{B}$

[hide="Solution"]
For slightly simpler computation, just let $ x \equal{} OP$ so that all lengths, from left to right, are $ a,b \minus{} a,x \minus{} b,c \minus{} x,d \minus{} c$, and we have

$ \frac {x \minus{} a}{d \minus{} x} \equal{} \frac {x \minus{} b}{c \minus{} x}\implies \minus{} x^2 \plus{} ax \plus{} cx \minus{} ac \equal{} \minus{} x^2 \plus{} bx \plus{} dc \minus{} bd$

$ \implies(a \minus{} b \plus{} c \minus{} d)x \equal{} ac \minus{} bd\implies x \equal{} \boxed{\textbf{(B)}\ \frac {ac \minus{} bd}{a \minus{} b \plus{} c \minus{} d}}$.
[/hide]


Prove that no seven positive integers, not exceeding 24, can have sums of all subsets different.

:D  Do you hear the sound of pigeons flying?

Pierre.

We assume a_1<a_2<...<a_7 (if two numbers are equal then we are ready).
Then all the sums of the subsets except a_1 lie in the interval [a_2,a_1+a_2+...a_7]. Alltogether there are a_3+a_4+...a_7+2 possibilities for the sums: so atmost 24+23+...+20+2=112. But we have 2^7-1=127 sums. Now the box principle applies the result.