source AMC!2

how many empty subsets s of ${1,2,3,.....15}$ have the following:

1st: no two consecutive integers belong to S

2nd/ if s contains k elements than s contains no number less the K

ryan22: Did you check what you just typed before clicking the Submit button? 'Empty sets' is is the title and the words 'empty subsets' is also on the problem... but did you know that there is only one type of empty set? Typoes do occur but this is definitely not a typo. Can you correctly state the problem please?

this is AMC 12A 2006 #25, look for it in the amc section (even if it is locked, you can still look at topics)

and he obviously means nonempty subsets


geometry parallelogram rectangle rotation conics hyperbola symmetry

The points of intersection of $ xy \equal{} 12$ and $ x^2 \plus{} y^2 \equal{} 25$ are joined in succession. The resulting figure is:

$ \textbf{(A)}\ \text{a straight line} \qquad\textbf{(B)}\ \text{an equilateral triangle} \qquad\textbf{(C)}\ \text{a parallelogram}$
$ \textbf{(D)}\ \text{a rectangle} \qquad\textbf{(E)}\ \text{a square}$

[quote="grn_trtle"][hide="Tentative"]Plotting the first gives a rotated hyperbola and a circle on the origin. Given the symmetry of both, and the fact that the circle hits the hyperbola at two places (working out the system of equations shows this), it should be $ \boxed{\textbf{(D)}\ \text{a rectangle}}$.[/hide][/quote]
[hide="I agree"]The hyperbola is in the first and third quadrant and both figures (the hyperbola and the circle) are symmetrical with respect to the diagonals $ y\equal{}x$ and $ y\equal{}\minus{}x$, so we would have two (or none, if the numbers were different) intersections in the first quadrant, and two in the third quadrant, symmetrical with respect to the diagonals. That makes a rectangle with the long sides parallel to $ y\equal{}x$.[/hide]
[hide="and if we wanted the coordinates of those intersections..."]The points are at distance 5 from the origin, and the $ x$ and $ y$ corrdinates multiply to give $ 12$. Does that remind anyone of a famous right triangle with leg lengths $ x$ and $ y$, and hypotenuse length $ 5$?[/hide]


calculus
integration
real analysis
real analysis theorems

My lecturer asked us to figure out how to determine the direction of parametric curves $(x,y)=(x(t),y(t))\in\mathbb{R}^{2}$. I came up with something (2 pages) , if you have time, give a comment about it.

Your write-up looks very nice. There is an interesting question that you have not discussed: what if the quantity $\Delta$ is positive for some values of $t$ and negative for others? Assuming that the curve is closed and does not cross itself, it should still be possible to decide whether it's oriented clockwise or counterclockwise. However, integration with respect to $t$ may be required.

thanks mlok  :) I have been thinking about simple closed curves. One idea:
find one $t=t_{0}$ that produces maksimum of $y(t)$ and then run test $\Delta$ for $(x_{0},y_{0})=(x(t_{0}),y(t_{0})-1)$ and that particular $t=t_{0}$. That simplifies $\Delta=-x'(t_{0})$. Sometimes this can be inconclusive.
Although, integration also gives the answer.


complex analysis
complex analysis unsolved

Determine explicitly the largest disc about the origin whose image under the mapping w=z^2 + z is one to one.

Appreciate ur help

Radius $0$¿  :?
When the real $r>0$ lies in the disk, then also $r^2+r > r$ should do. Now take $s$ as supremum of all such $r$ to get $s^2=0$.


Teams A, B, C and D competed against each other once.
A won 3 matches, drew 0, lost 0, scored 5 goals and let in 1 goal.
B won 1, drew 1, lost 1, scored 2 goals and let in 2.
C won 0, drew 2, lost 1, scored 5 goals and let in 6.
D won 0, drew 1, lost 2, scored 3 goals and let in 6.
(a) Find (with proof) which team won in each of the six matches.
(b) Find (with proof) the scores in each of the six matches.

[hide]
(a) Because each team competed against each other once, we have the following match ups:
1. $ A$ vs. $ B$
2. $ A$ vs. $ C$
3. $ A$ vs. $ D$
4. $ B$ vs. $ C$
5. $ B$ vs. $ D$
6. $ C$ vs. $ D$

Because team $ A$ won $ 3$ matches and it only played $ 3$ matches, team $ A$ won the first three games.

Because team $ B$ won a game and team $ D$ lost in game other than with $ A$, team $ B$ won match $ 5$.

Because no one else one and there are only draws left, matches $ 4$ and $ 6$ resulted in draws.

(b) I'll let someone else do it, I'm too lazy right now to work through the possibilities.
[/hide]

Is this difficult? Or too easy? Sunehra?

I make the scores for matches 1 to 6:

A v B: 1-0
A v C: 2-1
A v D: 2-0
B v C: 1-1
B v D: 1-0
C v D: 3-3

the first one is easy. A won all three matches against B, C, D. B and C lost 1 match. so it must be with A. D lost 2 matches. one must be with A. the only other team which won a match was B. so the other one was with B.

finding the scores of the individual matches is very complicated. I'm getting 12 variables and 8 equations connecting 3 of them each. :(
some clever guessing might be needed  :wink:

I found my answer as follows (as far as I can remember):

I built up the scores in the 6 matches starting from

1-0
1-0
1-0
0-0
1-0
0-0

Since B scored 1 more goal and conceded 2, there are 2 possibilities:

2-1
1-0
1-0
0-0
1-0
3-3

or

1-0
1-0
1-0
1-1
1-0
3-3

For the first situation, we need 2 more goals against D, implying A scored >=6 goals. Therefore we are left with the second situation.

We now need 1 more goal against C and 2 more against D, so the scores must be:

AvB: 1-0
AvC: 2-1
AvD; 2-0
BvC: 1-1
BvD: 1-0
CvD: 3-3.

(Any ideas on the problem posted as "circular arcs and a right-angled-triangle"?)


I didn't know where to put this topic

But can anyone help me to find something about steiner trees

Sorry everyone for reviving old thread, but the question is unanswered.
Look here:
http://en.wikipedia.org/wiki/Steiner_tree


real analysis
function
real analysis unsolved

Hi! I'm having difficulties proving the following problem:

Let $ f,g\colon \textbf{R}\to\textbf{R}$ be continuous functions. Prove that if $ f\equal{}g\quad(\lambda \minus{} a.e.) \Rightarrow f\equal{}g$, where $ \lambda$ denotes the Lebesgue measure, and $ a.e.$ stands for almost everywhere.

 :blush:

Show that if $ f \neq g$ at some point, then there is a small neighborhood of this point in which $ f \neq g$.  (It may simplify your life to subtract $ g$ from both functions and show that if $ f \minus{} g \neq 0$ at some point then $ f \minus{} g \neq 0$ in a neighborhood of this point.)

let say $ f(x_0)\minus{}g(x_0)\neq 0$ then there is a neighborhood , say $ V\equal{}(\minus{}\delta\plus{}x_0, x_0\plus{}\delta)$ such that $ f\neq g$  and the measure of $ V>\delta$


geometry
projective geometry
geometry solved

Let ABCD be the equilateral triangle and Ao,Bo,Co be the midpoints
of segments BC,CA,AB .The lines d1,d2,d3 each of which parallels two
others,through Ao,Bo,Co respectively , cut BoCo,CoAo,AoBo respectively
at A1,B1,C1 . Prove that AA1,BB1,CC1 are concurrent

There is a very powerful generalization (actually a classical theorem):

[b]Cevian Nests Theorem.[/b] Let ABC be an arbitrary triangle, and A', B', C' three points on the sides BC, CA, AB such that the lines AA', BB', CC' concur. Let also A", B", C" be three points on the sides B'C', C'A', A'B' of triangle A'B'C' such that the lines A'A", B'B", C'C" concur. Then, the lines AA", BB", CC" concur, too.

In your case, A1lqdSchool, A', B', C' are the midpoints Ao, Bo, Co of BC, CA, AB (the lines AA', BB', CC' concur at the centroid of triangle ABC), and A", B", C" are the points A1, B1, C1 (the lines A'A", B'B", C'C" are parallel - i. e., we can say they concur at an infinite point). Hence your problem is a special case of the Cevian Nests Theorem.

Here is my favorite proof of the Cevian Nests Theorem:

We will use a lemma:

[b]Menelaos Theorem for quadrilaterals.[/b] If X, Y, Z, W are points on the sidelines AB, BC, CD, DA of a quadrilateral ABCD, and these points X, Y, Z, W are collinear, then

  AX / XB * BY / YC * CZ / ZD * DW / WA = 1

with directed segments.

This is very easy to prove. (Incidentally, the converse is not true, unlike the converse of the Menelaos theorem for triangles.)

Now, let the lines AA", BB", CC" meet the sides BC, CA, AB of the triangle ABC at the points X, Y, Z, respectively. In order to prove the Cevian Nests Theorem, we have to show that the lines AA", BB", CC" concur. This will be clear by Ceva once we succeed to show

  BX / XC * CY / YA * AZ / ZB = 1.

Well, apply the Menelaos Theorem for quadrilaterals to the quadrilateral BCB'C' with the collinear points X, A, A", A on its sides (you are reading right - the point A occurs twice, it is actually the point of intersection of two sides of the quadrilateral!). Then, you obtain

  BX / XC * CA / AB' * B'A" / A"C' * C'A / AB = 1,

so that

  BX / XC = AB / C'A * A"C' / B'A" * AB' / CA
          = (AB' / C'A) * (AB / CA) * (A"C' / B'A").

Similarly,

  CY / YA = (BC' / A'B) * (BC / AB) * (B"A' / C'B"),     and
  AZ / ZB = (CA' / B'C) * (CA / BC) * (C"B' / A'C").

Hence,

  BX / XC * CY / YA * AZ / ZB
= (AB' / C'A) * (AB / CA) * (A"C' / B'A")
* (BC' / A'B) * (BC / AB) * (B"A' / C'B")
* (CA' / B'C) * (CA / BC) * (C"B' / A'C")
= (AB' / C'A * BC' / A'B * CA' / B'C)
* (AB / CA * BC / AB * CA / BC)
* (A"C' / B'A" * B"A' / C'B" * C"B' / A'C").

Now, all three bracketed products are = 1, because after Ceva, the concurrent lines AA', BB', CC' yield

  AB' / C'A * BC' / A'B * CA' / B'C = 1,

and the concurrent lines A'A", B'B", C'C" yield

  A"C' / B'A" * B"A' / C'B" * C"B' / A'C" = 1,

and finally, AB / CA * BC / AB * CA / BC = 1 is obvious. Hence, you have

  BX / XC * CY / YA * AZ / ZB = 1,

and after Ceva, the lines AX, BY, CZ concur, i. e. the lines AA", BB", CC" concur. Qed..

See http://www.mathlinks.ro/Forum/viewtopic.php?t=6579 for an extension of the Cevian Nests Theorem with a proof (which is actually the above proof after a few modifications).

  Darij

This would be my favorite proof for the cevian nests theorem.
It's a projective geometry proof.
Take the line joining the intersection points of the 2 sets of concurent cevinas to infinity to get the picture bellow.

In the picture, AA',BB',CC' are parallel. So are A'A",B'B" and C'C".
If you can prove that AA", BB" and CC" are concurent, you solve the problem.


Solve in R the ecuation x^n=[x]^2+{x}^2 where n is natural, n>=3 and x={x}+[x].

***

Write x=k+p, with k integer and 0<=p<1. We must have (k+p)^n=k^2+p^2. Suppose that |k|>3. Thus, k^2+1>(k+p)^n=|k+p|^n>=(|k|-|p|)^n>(|k|-1)^n. It follows that k^2+1>(|k|-1)^3, from where 4k^2>|k|^3, false.
   So k is -3,-2,-1,0,1,2 or 3. If k=0 we find that p=0 and so x=0. If k>0, then k^2+p>(k+p)^3, which cannot hold (if it is so, then k^3<k^2+1, so k<=1, so k=1 and 1+p>(1+p)^3, false). If k=-3, then p^2+9=|3-p|^n>2^n and so n<4. Thus, n=3 and (p-3)^3=p^2+9, false, since p-3<0. If k=-1 we must have |p-1|>1, false. So the only case is k=-2 and (p-2)^n=p^2+4. Of course, n=2q, with q>=2. But, unfortunately, this equation has a solution between 0 and 1 for any q. So, what can I say? Can anyone see if I was wrong?

After killing my brain with this problem I have managed to find a solution:

here it goes! hope it is good! :D

Lets 1st consider the case when x>=0.
It is obious that 0 and 1 are solutions!
Lets take x in ]0,1[ and the ecuation is equivalent to [x]^2+{x}^2=x^3 and this is equivalent to the equation x^2=x^3, which has no solution in ]0,1[.
Lets take x in ]1,2[  and the equation is equivalent to 1^2+(x-1)^2=x^n which is equivalent to x^n-x^2+2x-2=0.
We consider the function:
f:[1,2]-->R, f(x)=x^n-x^2+2x-2 which is strictly increasing (just consider it's derivate) and f(1)=0. So it has solutions in ]1,2[.
Lets take x>=2 so we get now that:

x^n>=[x]^n=[x]^2*[x]^(n-2)>=[x]^2*2^(n-2)>=2[x]^2>=[x]^2+1>[x]^2+{x}, so no solutions here!

Lets take x<0 the equation has :
1) for n=2k the solution x=-1
2) no solutions for n=2k+1; just follow the procedure above!
hope it's okay!

cheers! :D :D

I just thought about it and it seems pretty straight forward.

Denote by k=[x] and r={x}. If k = 0 => r=0 => x=0. Ok, if k>1 then it's easy to see that (k+r)^n \geq k^n \geq  k^2+1 > k^2+r^2 because n >2. So in fact if k>0 then k=1 and we have (r+1)^n = 1+r^2. But (r+1)^n > (r+1)^2 = r^2+2r+1 \geq r^2+1 so again we have no solution.

Okay so k<0. Then it's obvious that n=2m, with m\geq 2, because otherwise, as x<0, x^n < 0  contradiction. So we have ((k+r)^2)^m= k^2+r^2.

But (k+r)^{2m} = (-k-r)^{2m} \geq (-k-1)^{2m} \geq (-k-1)^4 which is greater than k^2 +1 > k^2 + r^2.

Indeed (k+1)^4 \geq k^2 +1 <=> (k+1)^4 - (k+1)^2 \geq -2k <=> (k+1)^2k(k-2) \geq -2k <=> (k+1)^2 (2-k) \geq 2 which is true (remember that k \leq -1). Well it's almost done, because the only flaw is when k=-1. So the equation becomes (r-1)^{2m} = r^2 +1. It's easy to check that f(r)=(r-1)^{2m}-r^2-1 has as root r=0 and (r-1)^{2m} \leq (r-1)^2 = r^2+1 - 2r < r^2 +1 for 0<r<1.

Summing up all solutions are x \in { 0, -1 }.


In general, for each of the following stress applications, would the equilibrium shift to the right or to the left?

[list](a) Concentration decreases.
(b) Concentration increases.
(c) Volume decreases (pressure increases).
(d) Volume increases (pressure decreases).
(e) Temperature decreases.
(f) Temperature increases.
(g) An inert gas is added.
(h) A catalyst is added.
[/list]

Thanks!

Two side notes:

1) The effect of concentration may depend on the concentration of reagent/product.

2) Adding or removing solids does not interfere with the equilibrium, since the concentration of a component in the solid phase is constant.

[quote="jdlkem"]General concepts to note:
1. The principle (LeChatelier's) is a guideline to equilibrium. One could picture it as a two-pan balance. At equilibrium, the pans are of equal height; this configuration is preferred. However, if there is more mass (higher concentration) on one side, then there is a difference between the two heights. To alleviate this "stress" and reach equilibrium, one may subtract more to the "heavier" side or add more to the "lighter" side. Also keep in mind that everything is related to the direction of the reaction as written.
2. PV=nRT. Concentration is really n/V. So, concentration = P/RT. One usually assumes T = constant in this case (also either pressure or volume is constant).
3. It depends on whether the written reaction is exo- or endothermic (exo = "heat" on the products side, endo = "heat" on the reactants side).
4. do inert gases add to the "concentration" of a gas?
5. what do catalysts do? (hint: alternate pathway) do they affect concentration?

other possible stress applications could include:
adding more solid reactant (if present)
adding more solvent (in a solution)
a reactant or product undergoes a side reaction[/quote]

So if there was a two pan balance, and the left pan was more concentrated than the right pan, would the equilibrium shift left or right?

in the case of concentrations, equilibrium is unbalanced. The actual equilibrium constant or position of equilibrium is not altered in any way; only temperature can affect it afaik.

$ \Delta G = - RT\lnK$

The position of the system with respect to equilibrium is altered, and thus the system spontaneously alters itself to equilibrium.

probably.

It is not the concentrations by themselves but their combination in the [i]reaction quotient[/i] that will tell us in what direction will the equilibrium shift. Here's an example:

- In one of his experiences on the ammonia synthesis $ \ce{N2} \plus{} 3\ce{H2} \rightleftharpoons 2\ce{NH3}$ at 500 K, Haber verified that, in a given instant, the concentrations of the components were the following: $ [\ce{H2}] \equal{} 3.0 \times 10^{\minus{}3}\,\,M$, $ [\ce{NH3}] \equal{} 2.0 \times 10^{\minus{}3}\,\,M$, and $ [\ce{N2}] \equal{} 1.0 \times 10^{\minus{}3}\,\,M$. At 500 K, the equilibrium constant has the value K = 61. In what direction will the reaction go?

[quote="Carcul"]It is not the concentrations by themselves but their combination in the [i]reaction quotient[/i] that will tell us in what direction will the equilibrium shift. Here's an example:

- In one of his experiences on the ammonia synthesis $ \ce{N2} \plus{} 3\ce{H2} \rightleftharpoons 2\ce{NH3}$ at 500 K, Haber verified that, in a given instant, the concentrations of the components were the following: $ [\ce{H2}] \equal{} 3.0 \times 10^{ \minus{} 3}\,\,M$, $ [\ce{NH3}] \equal{} 2.0 \times 10^{ \minus{} 3}\,\,M$, and $ [\ce{N2}] \equal{} 1.0 \times 10^{ \minus{} 3}\,\,M$. At 500 K, the equilibrium constant has the value K = 61. In what direction will the reaction go?[/quote]

To the right?

[quote="i_thirst_for_knowledge"]I think the equilibrium would shift to the left.
use the concentration to calculate the reaction quotient Q.
Q would be equal something like 148148, much larger than K. it means the product is more concentrated than the reactions. thus, the reaction should shift to the left.[/quote]

Assuming this is correct (which it probably is) we see the point that Carcul was trying to illustrate. The exact shift in equilibrium needs to be calculated quantitatively (to determine tha vale of $ Q$). I recall posting a similar problem before (and Carcul posted one before that).

Should I post more problems related with equilibrium, or all the doubts are gone?


Square all two digits number
How many quadratic numbers are there which is palindrome??

Hello
Search  for quadratic number field  Jrav
And IW@IT It will be a fun for u if u prepare a computer program on this.
Thank u.

[quote="JRav"]What is a quadratic number?[/quote]

Oh, I mean the number in the form $ x^2$

sorry for my wrong wording

Uh, wouldn't there be an infinite amount? For example, consider $ 101^2, 10101^2, 101010101^2$, etc.

Oops, I forget again :blush:
The problem should be like this

Square all two digits number
How many quadratic numbers are there which is palindrome?
So the quadratic numbers come from all two digits number
I've edited my problem... s
sorry for my mistake :blush:

I think JRav used a program... it doesn't seem like there's a good way to solve it without doing that... am I correct?

hmm... well, you can rule out most values of $ x$ (where $ x^2$ is palindrome) since if $ x^2$ has an even number of digits $ x$ must be divisible by $ 11$.

No, I didn't use any programs.  First note that the only possible roots are 10-31, since $ 10^{2}\equal{}100$ and $ 31^{2}\equal{}961<1000<32^{2}$.  Now obviously we can't have any number that ends in 0, so that knocks out every multiple of 10.  We can't have any number that ends in 2, 3, 7, or 8 since no number squared yields a ones' digit of 2, 3, 7, or 8.  Hence we lose all numbers in the following sets: 200-399, 700-899.  We've managed to knock out 400 of 900 choices right there.  Now we're not left with too many choices: $ 1n1$, $ 4n4$, $ 5n5$, $ 6n6$, $ 9n9$.  Now it's easy to check that $ 121$, $ 484$, and $ 676$ are the only ones that are actually squares.

[quote="JRav"]No, I didn't use any programs.  First note that the only possible roots are 10-31, since $ 10^{2} \equal{} 100$ and $ 31^{2} \equal{} 961 < 1000 < 32^{2}$....[/quote]

Is this a mistake?  :huh:

[quote="Bugi"]Palindroms-perfect squares-between 100 and 10000- that's the problem.[/quote]

[quote="IW@IT"]Square all two digits number
How many quadratic numbers are there which is palindrome? [/quote]

[quote="leoxnlin"]
Is this a mistake?  :huh:
[/quote]

JRav didn't wrote full solution.

[quote="leoxnlin"]hmm... well, you can rule out most values of $ x$ (where $ x^2$ is palindrome) since if $ x^2$ has an even number of digits $ x$ must be divisible by $ 11$.[/quote]

That means, that we have to check only these situations between 1000 and 10000:

$ 33^2,44^2,55^2,66^2,77^2,88^2,99^2$

No one of them is palindrome.

Bachukas is correct.  I posted a solution for the first few cases so you could see how it's done.  My solution holds only for three digit squares so yes, it is incomplete.

thx for answering my problem :)

Another problem

$ A$ and $ B$ are four digits number. $ C$ is five digits number. If $ A,B,$ and $ C$ are palindrome numbers and satisfy $ A\plus{}B\equal{}C$, find all possible values of $ C$..

the first digit of C must be 1 so the last digit must be 1. so the last digits of A and B must sum to 1 or 11. this means the first digits of A and B also sum to 1 or 11. but we know that C is 5 digits so they must sum to 11. we can just assume two values that sum to 11:
.9xx9
.2xx2
1xxx1
now in the middle we have the sum of two two digit palindromes with 1 added to the hundreds and ones digits. Using the method we used at the beginning, we find that the two digit palindromes must sum to 121. Adding 1 to the first and last digits we get $ \boxed{C\equal{}12221}$


abstract algebra
superior algebra
superior algebra solved

Hi all.

If R is a noetherian ring with 1 and $f: R\to R$ an epimorphism, then show that $f$ is injective.

Hanno

One way to characterize Noetherian rings is to say that there's no infinite strictly increasing sequence of ideals $I_1\subset I_2\subset\ldots$.

Assume now that $f$ is not injective. Then $I_1=\ker f$ is non-trivial, and $R/I_1\cong R$. If $\varphi_1$ is an isomorphism from $R/I_1$ to $R$, then let $I_2$ be an ideal of $R$ satisfying $I_2\supset I_1,\ I_2/I_1=\varphi^{-1}(I_1)$. We then have $R/I_2\cong (R/I_1)/(I_2/I_1)\cong R/I_1\cong R$. We continue this process, finding $I_3\supset I_2\supset I_1$, and so on, constructing an infinite strictly increasing sequence of ideals of $R$ and contradicting the fact that $R$ is Noetherian.

Hi.

Wow, that's a very nice solution, grobber! Quite similar is the one using the sequence $Ker(f)\subsetneqq Ker(f^2)\subsetneqq Ker(f^3),...$.

Hanno


When gold sold for $ \$16$ an ounce, Johnny found $ \$6$ worth of gold at his claim. Gold presently sells for $ \$328$ an ounce. How many dollars would Johnny's amount of gold be worth today?


Given a number n, let S(n) be the sum of it's digits.
Example S(38)=11;S(121)=4.
Find a solution of the equation or prove that there isn't a solution:
a)$ S(n) \plus{} S(n^2) \equal{} 2008$
b)$ S(n) \plus{} S(n^2) \equal{} 2009$

It is very known that $ n\equiv S(n) \mod 3$, then $ S(n)\plus{}S(n^2)\equiv n\plus{}n^2 \mod 3$

 so $ S(n)\plus{}S(n^2)$ is $ 0$ or $ 2$ mod $ 3$.

Since $ 2008\equiv 1\mod 3$ there are no solutions for a)

 :)


IS that the number$100 000 000 000 000 000 001$ is praime :rotfl:  :?:


If $ x,y,z\in \mathbb{R}_\plus{}$ such that $ xyz\equal{}x\plus{}y\plus{}z\plus{}2$ show that :

$ 2(\sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx})\le x\plus{}y\plus{}z\plus{}6$

[quote="alex2008"]If $ x,y,z\in \mathbb{R}_ \plus{}$ such that $ xyz \equal{} x \plus{} y \plus{} z \plus{} 2$ show that :

$ 2(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})\le x \plus{} y \plus{} z \plus{} 6$[/quote]
Here is my solution:This inequality is equivalent to:
 $ \sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z} \le \sqrt{2(x\plus{}y\plus{}z\plus{}3)}$
Since $ xyz\equal{}x\plus{}y\plus{}z\plus{}2$,Put $ x\equal{}\frac{b\plus{}c}{a},y\equal{}\frac{c\plus{}a}{b},z\equal{}\frac{a\plus{}b}{c}$.
This  inequality becoms:
$ \sum \ \sqrt{\frac{b\plus{}c}{a}} \le \sqrt{2(\sum \ a)(\sum \ \frac{1}{a})}$
Which is obvious true by Cauchy-Schwarz inequality


quadratics
superior algebra
superior algebra unsolved

Suppose Ok is the ring of integers in some quadratic field K=(sqrt(d)) and Ok is a UFD.
Trying to show that if k is prime in Ok, then k|p for some unique prime number in Z.

Also, if q is prime in Z, is it prime in Ok?

If $ k$ is prime in $ \mathcal{O}_K$, it divides its norm $ N(k) \equal{} k \cdot \bar{k}$.  This automatically gives the answer to the second question as "no."


Show that $2^{n}-1$ is not divisible by $n$ for integer $n \geq 2$.

Let $p$ be the smallest prime such that $p|n$.
We have that:
$p|n|2^{n}-1$ wich is the same as:
$2^{n}\equiv1(\mod p)$
Let $d=\text{ORD}_{p}2$ we have that:
$d<p$ and $d|n$
$\implies d=1$
$\implies 2\equiv 1 (\mod p)$
$\implies p|1$ contradiction.
QED

Why is $d=1$? All I can see is that $d | p-1$ from Fermat's Theorem.


calculus integration logarithms function algebra domain trigonometry

Hello...  :maybe:

Does anyone here know how to solve these 2 integrals?
Please help!

(1) $\int_{0}^{\infty}\frac{\ln(1+x+x^{2})}{1+x^{2}}dx$

(2) $\int_{0}^{\infty}\frac{\ln^{3}(1+x^{2})}{1+x^{2}}dx$

Thanks!
-Nasim

For 1), I would consider the branch $f(z) = \frac{log(z^{2}+z+1)}{z^{2}+1}$ $(|z| > 0,-\frac{\pi}{3}< argz < \frac{7\pi}{3})$ of the multiple-valued function $\frac{log(z^{2}+z+1)}{z^{2}+1}$. This branch is analytic everywhere in the indicated domain except at the point $z = \pm i$. It will not be a short argument. I have to go now and will come back later

Thanks, akech! I will wait for your reply...  :(

Earlier, I had tried the $f(z)$ you specified over the following domain:
$ \bigodot = \{ \begin{array}{lll}1 \leq |z| < \infty \\ \frac{2\pi}{3}\;\;\;\leq\;\;\; \angle z \;\;\;<\;\;\; \frac{8\pi}{3}\end{array} $
since
$\;\;\;\;\; z^{2}+z+1=0 \;\;\;\;\;\Rightarrow\;\;\;\;\; z=-\frac{1}{2}\pm \imath \frac{\sqrt{3}}{2}$

[ I took the +ve root, for which the branch point becomes
$\;\;\;\;\; z_{bp}= \pi-tan^{-1}{\sqrt{3}}= \frac{2\pi}{3}\;\;\;\;\;\;\;\;\;\; |z| =+\sqrt{(\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}= 1$
Hence, my "non-primary" branch $\bigodot$ defined as above. ]

But it didn't give me what I wanted
as my C-shaped contour then gives me $\int_{1}^{\infty}$ instead of  $\int_{0}^{\infty}$

[quote="didilica"]and for 2) you can use the substitution $x=\tan \theta$.[/quote]

hello.  :|
i tried making the substitution $x=\tan\theta$ for (2), but it gets me into a spaghetti!
am i doing it wrong?

$\int_{0}^{\infty}\frac{\ln^{3}{(1+x^{2})}}{1+x^{2}}\; dx \\ = \int_{0}^{\frac{\pi}{2}}\ln^{3}{(1+\tan^{2}{\theta})}\; d\theta \\ = ( \ln^{3}{(1+\tan^{2}{\theta})}\cdot \theta ) \; |_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}3\; \ln^{2}{(1+\tan^{2}{\theta})}\frac{2\tan\theta \cdot \sec^{2}{\theta}}{1+\tan^{2}{\theta}}\cdot \theta \; d\theta \\ = 4\pi \lim_{\wp \rightarrow \infty}\ln^{3}{\wp}\;\;\;-\;\;\; 6 \int_{0}^{\frac{\pi}{2}}\theta \cdot \tan\theta \cdot \ln^{2}{(1+\tan^{2}{\theta})}\; d\theta \\ = 4\pi \lim_{\wp \rightarrow \infty}\ln^{3}{\wp}\;\;\;-\;\;\; ( 6\; \ln^{2}{(1+\tan^{2}{\theta})}\int \theta \cdot \tan\theta \; d\theta \; ) \; |_{0}^{\frac{\pi}{2}}\\ \hspace* \qquad+\;\;\; 6 \; \int_{0}^{\frac{\pi}{2}}{\Large (}2\; \ln{(1+\tan^{2}{\theta})}\frac{2\tan\theta \cdot \sec^{2}{\theta}}{1+\tan^{2}{\theta}}\int \theta \cdot \tan\theta \; d\theta \; ) \; d\theta \\ \\ \\ \hspace* \qquad \ldots \;\;\;\;\; \ldots \;\;\;\;\; \ldots$

hello!  :maybe:

i tried to get a [i]closed [/i]expression for the sum of your inf. series but failed!

let me ask, would you also try to get an inf. series for-

$\int_{0}^{\infty}\;\frac{\ln^{2}{(1+x^{2})}}{1+x^{2}}\;dx$

and will you be able to find a [i]closed [/i]expression for it's sum ?

because i was able to get a [i]closed [/i]expression for the sum regarding  $\ln^{2}{(1+x^{2})}$ ; i used method of substitution to calculate the integral. for the sum regarding $\ln^{3}{(1+x^{2})}$ , i tried something similar, but it proved to be painstaking and didn't take me far...

You can use this method:

Let $I=\int_{0}^{\infty}\frac{\ln^{3}(1+x^{2})}{1+x^{2}}dx$.

The substitution $x=\tan\theta$ shows that

$I=-8\int_{0}^{\frac{\pi}{2}}\ln^{3}(\cos \theta)d\theta$.

On the other hand we have that:

$\frac{\sqrt{\pi}}{2}\frac{\Gamma(\alpha)}{\Gamma(1/2+\alpha)}=\int_{0}^{\pi/2}\cos^{2\alpha-1}\theta d\theta$.

It follows that

$I=-\frac{\sqrt{\pi}}{2}\frac{d^{3}}{d\alpha^{3}}\left(\frac{\Gamma(\alpha)}{\Gamma(1/2+\alpha)}\right)\biggm|_{\alpha=1/2}$

As for the other integral, namely $J=\int_{0}^{\infty}\frac{\ln^{2}(1+x^{2})}{1+x^{2}}dx$ use the same technique and you should get, provided that my computations are correct, that

$J=\frac{\pi}{2}\left(\frac{\pi^{2}}{3}+4\ln^{2}2\right)$.


hi!!
solve the following equation:

$ \sqrt{2x\plus{}11}\minus{}\sqrt{2x\minus{}1}\minus{}2\equal{}0$

thanx.

$ \left\{ \begin{array}{ll} \sqrt {2x \plus{} 11} \minus{} \sqrt {2x \minus{} 1} \equal{} 2 & \quad \\
\sqrt {2x \plus{} 11} \plus{} \sqrt {2x \minus{} 1} \equal{} 6 & \quad \end{array} \right.\Longleftrightarrow \left\{ \begin{array}{ll} \sqrt {2x \plus{} 11} \equal{} 4 & \quad \\
\sqrt {2x \minus{} 1} \equal{} 2 & \quad \end{array} \right.\Longleftrightarrow \boxed{x \equal{} \frac 52}$

thanx for your solution,  but $ x\equal{}\frac{5}{2}$ :)


Let $X=[1,16]\cap\mathbb{N}$ . Please divide $X$ into 2 parts $A, B$ such that $|A|=|B|$ and $\sum_{i\in A}i^2=\sum_{j\in B}j^2$.


ratio
geometric sequence
number theory proposed
number theory

Prove that the set $\{[n\sqrt(2003)]\} (n \geq 1)$ contains arbitrarily long geometric progressions with arbitrarily large common ratio :)

We will find a geometric progression with n terms and ratio r, among terms of the given sequence. According to Kronecker's theorem we can choose k such that $\ k \sqrt 2003-[k \sqrt 2003]< \frac {1}{r^n} $. Now the sequence $\ [k \sqrt2003], [rk \sqrt 2003]...[r^{n-1}k \sqrt 2003]$ is a geometric progression with n terms and ratio r. Since we have chosen r,n arbitrarly the conclusion is true.


Does Simply Connected imply connected in algebraic topology?


Express the value of $ \frac{2 \minus{} \frac13}{2 \plus{} \frac13}$ as a common fraction.

I see nothing better than doing it out.
$ \frac{\frac{5}{3}}{\frac{7}{3}}\equal{}\frac{\frac{5}{3}}{\frac{3}{7}}\equal{}\frac{5}{7}$


I just got a very dumb idea. We should make a list of dumb polls to ask. For example,

DO YOU THINK THIS POLL IS DUMB?
Possible choices: YES, NO

And by the way, I don't know how to make a poll. Can someone teach me? Thanks.

Don't let this get out of hand.  :wink:

There is already one in the Indiana forum, and we should keep it that way.

[quote="i_like_pie"][quote="Kevin K."]Oh sorry, but I didn't copy that one...  :blush:

Is there a way to find out whether somebody has posted something similar without having to check everywhere?[/quote]
[url=http://www.artofproblemsolving.com/Forum/search.php]SEARCH[/url][/quote]

AMAZING
I never knew we could search.
I guess that's just how noobish i am of an aops poster.

A+ for sarcasm, F for self-esteem. :rotfl:


Solve the equation
$ \sqrt{x\plus{}1}\plus{}\sqrt{x\plus{}33}\equal{}\sqrt{x\plus{}6}\plus{}\sqrt{x\plus{}22}$

Squaring both sides we get

$ 2x\plus{}34\plus{}2\sqrt{(x\plus{}1)(x\plus{}33)} \equal{} 2x\plus{}28\plus{}2\sqrt{(x\plus{}6)(x\plus{}22)}$, substracting $ 2x\plus{}28$ in both sides we get

$ 2\sqrt{(x\plus{}1)(x\plus{}33)} \equal{} 6\plus{}2\sqrt{(x\plus{}6)(x\plus{}22)}$, dividing by 2

$ \sqrt{(x\plus{}1)(x\plus{}33)} \equal{} 3\plus{}\sqrt{(x\plus{}6)(x\plus{}22)}$, squaring both sides again

$ (x\plus{}1)(x\plus{}33) \equal{} 9\plus{}6\sqrt{(x\plus{}6)(x\plus{}22)}\plus{}(x\plus{}6)(x\plus{}22)$

$ x^2\plus{}34x\plus{}33 \equal{} 9\plus{}6\sqrt{(x\plus{}6)(x\plus{}22)}\plus{}(x^2\plus{}28x\plus{}132)$

$ 6x\minus{}108 \equal{} 6\sqrt{(x\plus{}6)(x\plus{}22)}$, dividing by 6

$ x\minus{}18 \equal{} \sqrt{(x\plus{}6)(x\plus{}22)}$, squaring both sides once again

$ x^2\minus{}36x\plus{}324 \equal{} (x\plus{}6)(x\plus{}22) \equal{} (x^2\plus{}28x\plus{}132)$ then

$ \minus{}36x\plus{}324 \equal{} 28x\plus{}132$

$ \minus{}64x \equal{} 132 \minus{} 324 \equal{} \minus{}192$

$ x\equal{}3$.

yes,and the work can be simplified by using a substitution
i was wondering if there is a simpler way to solve it

[color=darkblue]This problem can done by this way:

+) Let $ a\equal{}\sqrt{x\plus{}1}, b\equal{}\sqrt{x\plus{}33}, c\equal{}\sqrt{x\plus{}6}, d\equal{}\sqrt{x\plus{}22} \Longrightarrow a\plus{}b\equal{}c\plus{}d$, (1).

+) We have $ a^2\minus{}b^2\equal{}\minus{}32, c^2\minus{}d^2\equal{}\minus{}16 \Longrightarrow \frac{a^2\minus{}b^2}{c^2\minus{}d^2}\equal{}2\Longrightarrow a\minus{}b\equal{}2(c\minus{}d)$, (2).

+) We have $ a^2\minus{}c^2\equal{}\minus{}5, d^2\minus{}b^2\equal{}\minus{}11 \Longrightarrow \frac{a^2\minus{}c^2}{d^2\minus{}b^2}\equal{}\frac{5}{11} \Longrightarrow a\plus{}c\equal{}\frac{5}{11}(d\plus{}b)$, (3).

+) We have $ a^2\minus{}d^2\equal{}\minus{}21, c^2\minus{}b^2\equal{}\minus{}27 \Longrightarrow \frac{a^2\minus{}d^2}{c^2\minus{}b^2}\equal{}\frac{7}{9} \Longrightarrow a\plus{}d\equal{}\frac{7}{9}(c\plus{}b)$, (4).

From (1), (2), (3), (4) we get a system of $ a, b, c, d$. Easy we get $ (a;b;c;d)\equal{}(2;6;3;5) \Longrightarrow x\equal{}3$.[/color]

[b][size=150][color=darkred]Thanhnam2902[/color][/size][/b] :P


For what value of x does the expression $ \frac{x^{10}}{25^x}$ yield a value of $ 25$ ?

Hi;

[hide]

By inspection the only possibility I can get is when x = -1 because

1 / ( 1 / 25 ) = 25. I cannot find any others.[/hide]

25^x*25=25^x+1 which equals 5^(2x+2). Since this only has values of 5 in its in prime decomposition x cannot me greater than 1. By inspection we can tell that x is -1.

I tested everything else, except -1.   :blush:

Thanks guys


Let an equitaletar triangle $ABC$(all the angles are $60$). and a point M such that
angle$AMB=40$ degrees and angle$AMC=30$ degrees. If angle$ABM>90$ degrees find the angle $CBM$. :)

Rotate $\triangle ACM$ about A through an angle of $60^\circ$.


modular arithmetic
number theory unsolved
number theory

Let $ p,q$ be distinct primes. Prove that there are positive integers $ a,b$ such that the arithmetic mean of all divisors of the number $ n \equal{} p^a\cdot q^b$ is also an integer.

We have the AM$ \equal{} \frac {(1 \plus{} p \plus{} p^{2} \plus{} p^{3}........p^{a})(1 \plus{} q \plus{} q^{2} \plus{} ....... \plus{} q^{b})}{(a \plus{} 1)(b \plus{} 1)}$
We just got to prove $ a ,b$ exists for the AM to be an integer .
So I am setting a value for $ a$
Let $ a$ be an integer such that $ p \equiv 1\bmod{a \plus{} 1}$
So we have $ p^{2} \equiv 1\bmod{a \plus{} 1}$
                .
                .
                .
$ p^{a} \equiv 1\bmod{a \plus{} 1}$
So $ p \plus{} p^{2} ................ \plus{} p^{a}\equiv 1 \plus{} 1 \plus{} 1 \plus{} ........1 \equiv 1* a\equiv a \bmod{a \plus{} 1}$
So $ 1 \plus{} p \plus{} p^{2} ................p^{a} \equiv a \plus{} 1\equiv 0\bmod{a \plus{} 1}$
Since the values of $ a$ and $ b$ are independent of each other ,I mean the value of $ a$ is not going to affect the value of $ b$ so in a similar way we can set $ q\equiv 1\bmod{b \plus{} 1}$ and finish the problem .

@srinath.r: your solution works for p,q odd. It can be simplified by putting a=b=1. You didn't solve the case $ p\equal{}2$.

sorry , even if  $ p \equal{} 2$ ,I will set $ a \equal{} 1$. So ,I can chose 'a' accordingly such that $ p \equiv 1 \bmod{a \plus{} 1}$ ,so I dont think I need to solve $ p \equal{} 2$ ,separately ,do I ?

Suppose p=2,q=5. According to your proof one could choose b=1 or b=3. If you put a=1 (which doesn't satisfy $ 2\equiv 1\pmod{a\plus{}1}$, by the way), the mean is 3*6/4 or 3*156/8, neither of which is integer.

Your solution works unless $ p\equal{}2$ and $ q\equiv 1\pmod{4}$ and unfortunately I don't know how to improve it.

Yes ,point noted :) ,forgot $ a$ is a positive integer .
If $ p \equal{} 2$ and $ q$ is odd ,chose $ b \equal{} q$ and let $ a \plus{} 1 \equal{} 1 \plus{} q^{1} \plus{} .... \plus{} q^{q \minus{} 1}$,there is no problem with $ q \equiv 1 \bmod{4}$ when you consider it separately ,I mean in the case when $ q \equal{} 5$ ,set $ b \equal{} 3$ ,you get $ 4 \mid 156$

I've failed to spot an easy way how to state the whole proof (idea remains): one of p,q is allways odd, say q. Then we let b=1. We put a=(q+1)/2-1 and the mean is $ \frac {(1 \plus{} \cdots \plus{} p^a)(1 \plus{} q)}{\frac {1 \plus{} q}2\cdot 2} \equal{} (1 \plus{} \cdots \plus{} p^a)$, which is integer.


inequalities
algebra
polynomial
inequalities proposed

Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (a) \ \ \ \ \sum x^2(x \minus{} y)(x\minus{}z) \ge \frac {3(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy\plus{}yz\plus{}zx}$;

$ (b) \ \ \ \ \sum (x\minus{}y)^2(x\minus{}z)^2 \ge \frac {4(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy\plus{}yz\plus{}zx}$;

$ (c) \ \ \ \ \sum (x^2 \minus{} y^2)(x^2 \minus{} z^2) \ge \frac {6(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy\plus{}yz\plus{}zx}$.

Nice Vasc! Do you have some Identities in these inequalities? Like this one,
\[ \sum(b \plus{} c)(a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \equal{} \sum c(a \minus{} b)^2(a \plus{} b \minus{} (k \plus{} 1)c)^2\]
and
\[ \sum (a \minus{} b)(a \minus{} c)(a \minus{} kb)(a \minus{} kc) \equal{} \frac 12 \sum (a \minus{} b)^2(a \plus{} b \minus{} (k \plus{} 1)c)^2\]
.
If yes can u share us?

[quote="Vasc"]Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (a) \ \ \ \ \sum x^2(x \minus{} y)(x \minus{} z) \ge \frac {3(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;
[/quote]
Let $ x\plus{}y\plus{}z\equal{}3u,$ $ xy\plus{}xz\plus{}yz\equal{}3v^2$ and $ xyz\equal{}w^3.$
Hence, we need to prove that $ f(w^3)\geq0,$ where
$ f(w^3)\equal{}w^6\plus{}(4u^3\plus{}12uv^2)w^3\plus{}81u^4v^2\minus{}138u^2v^4\plus{}40v^6.$
But $ f$ is increasing function.
Thus, $ f$ gets the minimal value when $ w^3$ gets an extremal value. Id est,for the proof it remains to check two cases:
1) $ b\equal{}1$ and $ c\equal{}0$;
2) $ b\equal{}c\equal{}1.$

[quote="Vasc"]Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (a) \ \ \ \ \sum x^2(x \minus{} y)(x \minus{} z) \ge \frac {3(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;

$ (b) \ \ \ \ \sum (x \minus{} y)^2(x \minus{} z)^2 \ge \frac {4(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;

$ (c) \ \ \ \ \sum (x^2 \minus{} y^2)(x^2 \minus{} z^2) \ge \frac {6(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$.[/quote]
The following is my proof:
(a) Without loss of generality, we may assume that $ z\equal{}\min\{x,y,z\}.$ Then, we write our inequality as
$ (x\minus{}y)^2(x^2\plus{}xy\plus{}y^2\minus{}xz\minus{}yz)\plus{}z^2(x\minus{}z)(y\minus{}z) \ge \frac{3(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2}{xy\plus{}yz\plus{}zx}.$
If $ x\equal{}z$ or $ y\equal{}z,$ we see that the inequality is trivial. Otherwise, let $ a\equal{}x\minus{}z,b\equal{}y\minus{}z,$ then $ a,b >0.$ We have $ (x\minus{}y)^2\equal{}(a\minus{}b)^2,$ $ z^2(x\minus{}z)(y\minus{}z) \ge 0,$ $ x^2\plus{}xy\plus{}y^2\minus{}xz\minus{}yz\equal{}x(x\minus{}z)\plus{}y(y\minus{}z) \plus{}xy \ge a^2\plus{}b^2\plus{}ab,$ and $ xy\plus{}yz\plus{}zx \ge ab.$ Thus, it suffices to prove that
$ (a\minus{}b)^2(a^2\plus{}ab\plus{}b^2) \ge \frac{3(a\minus{}b)^2a^2b^2}{ab},$ or $ (a\minus{}b)^4 \ge 0.$
This is true and it completes our proof.

(b) If $ x\equal{}y,$ or $ y\equal{}z,$ or $ z\equal{}x,$ we have equality. Alternatively, without loss of generality, assume that $ x>y>z,$ we rewrite our inequality as
$ \frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{(y\minus{}z)^2}\plus{}\frac{1}{(z\minus{}x)^2} \ge \frac{4}{xy\plus{}yz\plus{}zx}.$
Using the identity $ \sum \frac{1}{(x\minus{}y)(y\minus{}z)} \equal{}0,$ we see that the left hand side is equal to $ \left( \sum \frac{1}{x\minus{}y}\right)^2,$ and thus we are left to prove
$ \left( \frac{1}{x\minus{}y}\plus{}\frac{1}{y\minus{}z}\plus{}\frac{1}{z\minus{}x} \right)^2 \ge \frac{4}{xy\plus{}yz\plus{}zx}.$
By the AM-GM Inequality, we have
$ \left( \frac{1}{x\minus{}y}\plus{}\frac{1}{y\minus{}z}\plus{}\frac{1}{z\minus{}x} \right)^2  \ge 4\cdot \frac{1}{x\minus{}y} \left( \frac{1}{y\minus{}z}\plus{}\frac{1}{z\minus{}x}\right) \equal{}\frac{4}{(x\minus{}z)(y\minus{}z)}.$
On the other hand, it is clear that $ (x\minus{}z)(y\minus{}z) \le xy\plus{}yz\plus{}zx.$ Therefore, from the above inequality, we can easily deduce the result.

(c) Take $ a\equal{}x^2,b\equal{}y^2,c\equal{}z^2$ and assume that $ a \ge b \ge c.$ Then, our inequality is equivalent to
$ (a\minus{}b)(a\minus{}c)\plus{}(b\minus{}c)(b\minus{}a)\plus{}(c\minus{}a)(c\minus{}b) \ge \frac{6\left( \sqrt{a} \minus{}\sqrt{b} \right)^2\left( \sqrt{b} \minus{}\sqrt{c}\right)^2\left( \sqrt{a}\minus{} \sqrt{c}\right)^2}{\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}},$
or equivalently,
$ (a\minus{}b)^2\plus{}(a\minus{}c)(b\minus{}c) \ge \frac{6(a\minus{}b)^2(b\minus{}c)^2(a\minus{}c)^2}{ \left( \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\right) \left( \sqrt{a}\plus{}\sqrt{b}\right)^2\left( \sqrt{a}\plus{}\sqrt{c}\right)^2\left( \sqrt{b}\plus{}\sqrt{c}\right)^2}.$
Now, let $ m \equal{}a\minus{}c,n\equal{}b\minus{}c,$ ($ m \ge n \ge 0$). If $ n\equal{}0,$ the inequality is trivial so we may assume that $ m \ge n>0,$ then we have
$ \sqrt{a}\plus{}\sqrt{b} \ge \sqrt{m}\plus{}\sqrt{n}, \sqrt{a}\plus{}\sqrt{c} \ge \sqrt{m}, \sqrt{b}\plus{}\sqrt{c} \ge \sqrt{n}, \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca} \ge \sqrt{mn},$
and $ (a\minus{}b)^2\equal{}(m\minus{}n)^2,(a\minus{}c)^2\equal{}m^2,(b\minus{}c)^2\equal{}n^2.$ Thus, it suffices to prove that
$ (m\minus{}n)^2\plus{}mn \ge \frac{6m^2n^2(m\minus{}n)^2}{mn\sqrt{mn}\left( \sqrt{m}\plus{}\sqrt{n}\right)^2},$
or equivalently,
$ (m\minus{}n)^2 \plus{}mn \ge 6\sqrt{mn} \left( \sqrt{m} \minus{}\sqrt{n}\right)^2.$
This can be written as
$ m^2\minus{}6m\sqrt{mn}\plus{} 11mn \minus{}6n\sqrt{mn} \plus{}n^2 \ge 0,$
or
$ \left( {m} \minus{}3\sqrt{mn}\plus{}{n}\right)^2 \ge 0.$
The last one is obviously true, thus our proof is completed.

Your solutions are not nice, its [b]VERY NICE[/b]. Congratulations Can.

Nice solutions, Can_hang2007.  :lol:

Can you apply the same ideas to prove that

$ (d) \ \ \ \ \sum (x \minus{} y)(x \minus{} z)(x\minus{}2y)(x\minus{}2z) \ge \frac {5(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;

$ (e) \ \ \ \ \sum (x \minus{} y)(x \minus{} z)(x\plus{}2y)(x\plus{}2z) \ge \frac {9(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}\ ?$

[quote="Vasc"]Nice solutions, Can_hang2007.  :lol:

Can you apply the same ideas to prove that

$ (d) \ \ \ \ \sum (x \minus{} y)(x \minus{} z)(x \minus{} 2y)(x \minus{} 2z) \ge \frac {5(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;

$ (e) \ \ \ \ \sum (x \minus{} y)(x \minus{} z)(x \plus{} 2y)(x \plus{} 2z) \ge \frac {9(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}\ ?$[/quote]
The below attachment is my solution for (e). I also found that (e) is a corollary of (d), but for (d), the inequality is so strong. I have only an ugly solution by pqr, and I dont want to write it out. I am still trying to find a nice solution for it. What is your solution, dear Vasc?

[quote="Vasc"]Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (a) \ \ \ \ \sum x^2(x \minus{} y)(x \minus{} z) \ge \frac {3(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;

$ (b) \ \ \ \ \sum (x \minus{} y)^2(x \minus{} z)^2 \ge \frac {4(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;

$ (c) \ \ \ \ \sum (x^2 \minus{} y^2)(x^2 \minus{} z^2) \ge \frac {6(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$.[/quote]

Let : $ 0 \le x \le y \le z$

1.  ($ x \equal{} y$)  -  trivial , equality holdes for : ($ x \equal{} y \equal{} z$)

2. $ x < > y$ then :

$ x \equal{} x$ , $ y \equal{} x \plus{} a$ , $ z \equal{} x \plus{} b$

a) in (a) , (b) , (c)  $ \rightarrow \boxed{min(LHS *(xy \plus{} yz \plus{} zx ))}$ holdes for $ x \equal{} 0$

b) we can assume : $ y \equal{} 1$ ;

So only need to prove this inequalities for $ x \equal{} 0$ , $ y \equal{} 1$ and $ z \ge 1$

(a) $ z^4 \minus{} 4z^3 \plus{} 6z^2 \minus{} 4z \plus{} 1 \equal{} (z \minus{} 1)^4 \ge 0$

(b) $ z^4 \minus{} 6z^3 \plus{} 11z^2 \minus{} 6z \plus{} 1 \ge 0 \Leftrightarrow (z^2 \minus{} 3z \plus{} 1)^2\ge 0$

(c) $ z^4 \minus{} 6z^3 \plus{} 11z^2 \minus{} 6z \plus{} 1 \ge 0 \Leftrightarrow (z^2 \minus{} 3z \plus{} 1)^2\ge 0$

[quote="Vasc"]Nice solutions, Can_hang2007.  :lol:

Can you apply the same ideas to prove that

$ (d) \ \ \ \ \sum (x \minus{} y)(x \minus{} z)(x \minus{} 2y)(x \minus{} 2z) \ge \frac {5(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$;

$ (e) \ \ \ \ \sum (x \minus{} y)(x \minus{} z)(x \plus{} 2y)(x \plus{} 2z) \ge \frac {9(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}\ ?$[/quote]

Let : $ 0 \le x \le y \le z$

1.  ($ x \equal{} y$)  -  trivial , equality holdes for : ($ x \equal{} y \equal{} z$)

2. $ x < > y$ then :

$ x \equal{} x$ , $ y \equal{} x \plus{} a$ , $ z \equal{} x \plus{} b$

a) in (a) , (b) , (c)  $ \rightarrow \boxed{min(LHS *(xy \plus{} yz \plus{} zx ))}$ holdes for $ x \equal{} 0$

b) we can assume : $ y \equal{} 1$ ;

So only need to prove this inequalities for $ x \equal{} 0$ , $ y \equal{} 1$ and $ z \ge 1$

(a) $ z^4 \minus{} 8z^3 \plus{} 18z^2 \minus{} 8z \plus{} 1 \ge 0 \Leftrightarrow (z^2 \minus{} 4z \plus{} 1)^2\ge 0$

(b) $ z^4 \minus{} 8z^3 \plus{} 18z^2 \minus{} 8z \plus{} 1 \ge 0 \Leftrightarrow (z^2 \minus{} 4z \plus{} 1)^2\ge 0$

[quote="can_hang2007"] For (d), the inequality is so strong. I have only an ugly solution by pqr, and I dont want to write it out. I am still trying to find a nice solution for it. What is your solution, dear Vasc?[/quote]
Actually, the following generalization holds for $ k\ge 0$:

$ \sum (x \minus{} y)(x \minus{} z)(x \minus{} ky)(x \minus{} kz) \ge \frac {(k\plus{}3)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$.

After multiplying by $ xy\plus{}yz\plus{}zx$, we get a symmetric homogeneous polynomial inequality of degree six, with equality for
$ x\equal{}y\equal{}z, \  \frac x{k}\equal{}y\equal{}z,\   y\equal{}z\equal{}0 \  and \  \frac y{z}\plus{}\frac z{y}\equal{}k\plus{}2$.
For $ k\ge 0$, my solution is not nice. For $ k\equal{}2$, I have 2 nonclassic solutions based on new theoretical results.

[quote="can_hang2007"]  I also found that (e) is a corollary of (d) [/quote]
Indeed, this follows from

$ \sum x(y\plus{}z)(x \minus{} y)(x \minus{} z) \ge \frac {(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xy \plus{} yz \plus{} zx}$,

with equality for $ x\equal{}0$ and for $ x\equal{}y\equal{}z$.

A similar nice inequality is

$ \sum yz(x \minus{} y)(x \minus{} z) \ge \frac {(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{x^2\plus{}y^2\plus{}z^2}$,

with equality for $ x\equal{}y\equal{}z$ and for $ y\equal{}z\equal{}0$.

[quote="Sergic Primazon"]
1.  ($ x \equal{} y$)  -  trivial , equality holdes for : ($ x \equal{} y \equal{} z$)
2. $ x < > y$ then :
$ x \equal{} x$ , $ y \equal{} x \plus{} a$ , $ z \equal{} x \plus{} b$    [/quote]

I think this method solves (a), (b) and (c)


if u>0 is any real number and x<y, show that there exists a rational number r such that, x<ru<y

$x/u < r < y/u$. Do you believe that the rationals are dense in R?


geometry
geometric transformation
geometry theorems

i have a question??? consider a square ABCD take a point P  on BC and construct the square APRS.  are ABCD  and APRS  homothetic??? if so may i conclude that S, D,C collinear???

no, they wouldn't be homothetic, because corresponding sides would have to be parallel, which they aren't; instead they're related by a spiral similarity centered at A.  you can, however, say S, D, and C are collinear, *if* the square APRS is drawn with R and S on the same side of AP as C and D.  (because then $\triangle ABP \equiv \triangle ADS$, which means that $\angle ADS$ is a right angle, and hence that $SDC$ is a straight line.)


inequalities
Cauchy Inequality
inequalities unsolved

Maybe I have a solution.
let [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/18170bfde2c211f70add51d7ecbb71e8.gif[/img]
then [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/aef3feb153d42b72362a3a19ae780e68.gif[/img]
use the AM-GM,we will get that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a33b2bfb7ec172fc8ea852000fd0282f.gif[/img] :D

I've made a terrible mistake, of course is ${P(\frac{1}{2},\frac{1}{2},\frac{1}{2})}=125$

[quote="ehku"]Thanks [b]suweijie[/b]. It's my solution. But I don't know what the AM_GM is. Is it Cauchy inequality : $a_1+a_2+..+a_n \ge n\sqrt[n]{a_1a_2...a_n} $ ???[/quote]
yes it is.
as far as i know only vietnamese calls amgm as cauchy.... so are you from vietnam? :)


algebra
polynomial
symmetry
linear algebra
Galois Theory
superior algebra
superior algebra unsolved

Just out of curiosity suppose $ a,b$ are algebraic numbers prove that $ a\plus{}b,a\cdot b$ are algebraic numbers. [b]Without[/b] using any results about field theory just by first principles (otherwise it is trivial).

it's worthless to think about this. the standard proof of existence of an algebraic equation for $ a\plus{}b$ just uses linear algebra and is not constructive.

[quote="-oo-"]it's worthless to think about this.[/quote]

Wrong.

[b]EDIT:[/b] -oo-'s claim that "the standard proof of existence of an algebraic equation for $ a\plus{}b$ just uses linear algebra and is not constructive" is also wrong. The linear-algebraic proof [b]is[/b] constructive.

See also Theorem 5 in [url=http://www.cip.ifi.lmu.de/~grinberg/#integrality]my notes [i]A few facts on integrality[/i][/url] for a different proof (similar to the linear-algebraic one).

  darij

we'll do it for $ a \plus{} b$; the proof for $ ab$ isn't too far off. let $ p(x)$ and $ q(x)$ be monic polynomials over the field $ F$ with $ p(a) \equal{} 0$, $ q(b) \equal{} 0$. let $ a \equal{} a_1, a_2, ..., a_m$ be the roots of $ p(x)$, $ b \equal{} b_1, ...,b_n$ the roots of $ q(x)$. now observe (like a holiday!) that the polynomial $ P(x) \equal{} \prod_{j \equal{} 1}^n p(x \minus{} b_j) \equal{} \prod_{j \equal{} 1}^n \prod_{i \equal{} 1}^m (x \minus{} a_i \minus{} b_j)$ has $ a \plus{} b$ as a root. it only remains to check that the coefficients of $ P(x)$ belong to $ F$. in fact, the coefficients of $ P$ are symmetric polynomial expressions in the $ b_j$, with the coefficients of these symmetric polynomial expressions being products of coefficients of $ p$ (which are of course in $ F$). now, the well-known theorem on symmetric polynomials tells us that each of these symmetric expressions in the $ b_j$'s can be expressed as polynomials, with coefficients in $ F$, in the [i]elementary symmetric polynomials [/i]of the $ b_j$. but the elementary symmetric polynomials in the $ b_j$ belong to $ F$ because they're the coefficients of the polynomial $ q$. so the coefficients of $ P$ belong to $ F$. done.

We can perhaps motivate the above remarks by noting that the symmetry of the polynomials guarantees that they will be fixed under any element of the Galois group (since it will just permute the roots and the expression is symmetric with respect to all the roots).

ok I've changed my point of view  :whistling:

thx pleurestique

edit: erm, does your proof work in the non-separable case, too?


John divided his souvenir hat pins into two piles. The two piles had
an equal number of pins. He gave his brother one-half of one-third of
one pile. John had 66 pins left. How many pins did John originally
have?

lets say that two piles each had 6x pins.

if john gives his 1/6 of one pile to his brother, it becomes 5x and 6x.

11x=66, and x=6.

whole pile is 12x, so its 72.

answer : 72

We know that he gave away 1/12 so he still has 11/12 so that means that 1/12=6 so he had 72.


Hwo many degrees clockwise does the hourhand of a 12-hour clock rotate in a 90 minute period? 180 minute period?

[hide]
360 degrees/ 60 minutes=6 degrees per minute.

x90=540

x180=1080

540/12 hours=45degrees

1080/12=90 degrees[/hide]


function
algebra
polynomial
induction
number theory
relatively prime
number theory unsolved

The function $ f : \mathbb{N} \to \mathbb{Z}$ is defined by $ f(0) \equal{} 2$, $ f(1) \equal{} 503$ and $ f(n \plus{} 2) \equal{} 503f(n \plus{} 1) \minus{} 1996f(n)$ for all $ n \in\mathbb{N}$. Let $ s_1$, $ s_2$, $ \ldots$, $ s_k$ be arbitrary integers not smaller than $ k$, and let $ p(s_i)$ be an arbitrary prime divisor of $ f\left(2^{s_i}\right)$, ($ i \equal{} 1, 2, \ldots, k$). Prove that, for any positive integer $ t$ ($ t\le k$), we have $ 2^t \Big | \sum_{i \equal{} 1}^kp(s_i)$ if and only if $ 2^t | k$.

Lemma. If $ \gcd(a,b)\equal{}1$ then any prime dividing $ a^{2^n}\plus{}b^{2^n}$ is $ 1\mod 2^{n\plus{}1}$.

Proof. $ a^{2^n}\plus{}b^{2^n}\equiv 0\mod p$, so $ p\not|a,b$. Hence, $ (ab^{\minus{}1})^{2^n}\plus{}1\equiv 0\mod p$, but $ x^{2^n}\plus{}1$ is the $ 2^{n\plus{}1}$-cyclotomic polynomial, so $ 2^{n\plus{}1}|p\minus{}1$.

Or simply,

$ (ab^{\minus{}1})^{2^n}\equiv \minus{}1\mod p$
$ (ab^{\minus{}1})^{2^{n\plus{}1}}\equiv 1\mod p$

so $ 2^{n\plus{}1}|ord_p(ab^{\minus{}1})|p\minus{}1$. Either way, we have proved the lemma.

From basic facts about linear recurrences, we find that $ f(n)\equal{}499^n\plus{}4^n$, so we have $ p(s_i)|f(2^{s_i})\equal{}499^{2^{s_i}}\plus{}4^{2^{s_i}}$. Therefore by the lemma, $ p(s_i)\equiv 1\mod 2^{s_i\plus{}1}\implies p(s_i)\equiv 1\mod 2^{k\plus{}1}$ (since $ s_i\ge k$ as given). Then

$ \sum_{i\equal{}1}^k p(s_i)\equiv k\mod 2^{k\plus{}1}$. Since $ 2^{k\plus{}1}$ is larger than $ k$, the greater power of $ 2$ to divide $ \sum_{i\equal{}1}^k p(s_i)$ is the greatest power of $ 2$ to divide $ k$, so we are done.

[b]Solution[/b]:
Instead of considering a function we will talk about sequence of the integer numbers,i.e let $ a_n$ be a sequence of integer numbers,satisfying to the following relations:
\[ a_{n \plus{} 2} \equal{} 503\cdot a_{n \plus{} 1} \minus{} 1996\cdot a_n
\]
Let's try to find an explicit formula to our sequence.We shall prove that for every $ n\in\mathbb{N}$,we have that $ a_n \equal{} 4^{n} \plus{} 499^{n}$.

-We can proceed by induction to prove the formula,indeed:
It is not hard to check that for $ n \equal{} 0$ and $ n \equal{} 1$ our formula works.
If it's true for all numbers $ \leq n$,we shall show that it's still true for $ n \plus{} 1$:
By the recurrent relation it follows that:
$ a_{n \plus{} 1} \equal{} 503\cdot\left(4^{n} \plus{} 499^{n}\right) \minus{} 1996\cdot\left(4^{n \minus{} 1} \plus{} 499^{n \minus{} 1}\right) \equal{} 4^{n \plus{} 1} \plus{} 499^{n \plus{} 1}$.
$ \square$

Now let's return to our initial problem.As we know $ p_i$ is some prime divisors of $ a_{2^{i}}$,respectively.

-We shall show that if $ p$ divides $ 4^{2^{n}} \plus{} 499^{2^{n}}$ then $ p\equiv 1$ mod $ 2^{n \plus{} 1}$.
It is well-known fact,but anyway I'll show its proof.
-As we know $ 4^{2^{n}}\equiv \minus{} 499^{2^{n}}$ mod $ p$.Obviously $ p$ is relatively prime with $ 4$ and $ 499$.Therefore,there exist some number $ x$,such that $ 499\cdot x\equiv 1$ mod $ p$.
Let's multiply our relation to $ x^{2^{n}}$:
$ \left(4\cdot x\right)^{2^{n}}\equiv \minus{} 1$ mod $ p$.
Or $ \left(4\cdot x\right)^{2^{n \plus{} 1}}\equiv 1$ mod $ p$.
Hence if $ d$ is order of $ 4x$ modulo $ p$ then $ d$ is the power of two,but simultaneously,as we know $ \left(4x\right)^{2^{n}}\equiv \minus{} 1$ mod $ p$,therefore $ d \equal{} 2^{n \plus{} 1}$.Hence $ p\equiv 1$ mod $ 2^{n \plus{} 1}$,here we've just used Fermat's theorem.
$ \square$

-Suppose that $ s \equal{} min\{s_1,s_2,\dots,s_k\}$,hence $ p_i\equiv 1$ mod $ 2^{s \plus{} 1}$,therefore,$ \sum_{i \equal{} 1}^k p_i\equiv k$ mod $ 2^{s \plus{} 1}$.
As we know $ t\leq k\leq s < s \plus{} 1$,therefore,
$ \sum_{i \equal{} 1}^k p_i\equiv 0$ mod $ 2^{t}\Longleftrightarrow k\equiv 0$ mod $ 2^{t}$.

$ \blacksquare$.


Alright, next year as a sophomore, I am planning on applying to several math/science programs since I am absolutely determined to do something useful next summer. But then I have a question. Can I turn down acceptances, just like I am to colleges? Are summer programs very much like colleges?

How about any others? Cost is certainly a major concern (I want programs that are generous in offering scholarships and that do not require application fees).. I've also started late on problem-solving and have had a lackadaisical freshman year So I want programs that aren't too competitive (i.e, ones with at least 50% acceptance rate).

Any other suggestions as well?

We have a list of links to several excellent summer programs.

Click on "Resources" on the left side of your screen, then click the "Programs" tab.

If you want to stay close to home, also consider SIMUW:
[url]http://www.math.washington.edu/~simuw/[/url]
It's free(!), and they only take students from the Pacific NW, BC, and Alaska.  I know most of the instructors there (I used to work at UW) and they are excellent.

In general, applying to a competitive summer program does NOT obligate you to attend should you be accepted (as many people apply to more than one), but make sure you read each application carefully for exceptions.

THanks for the wonderful information! :)

The SIMUW program is totally awesome!!!  I attended for the first two years of the program, and it has been the best experience in my life.   You get exposed to lots of awesome math topics, the UW environment is wonderful, and the people I met there were really cool and smart.  This program made me want to be a mathematician!

  I would definitely recommend applying to SIMUW.

Thanks theunknot - I was just accepted and it's very likely that I'll attend SIMUW. :)


obviously x=3...i think we can play around with that as a function of x to show that it is increasing after a while...so there are limited solutions...ther might be a second one though...

Here's an easier-to-read version of the problem:
[quote="kcbu"]Solve $\left( \sqrt[3]{7+\sqrt{15}}\right)^{x}+\left( \sqrt[3]{7-\sqrt{15}}\right)^{x}=14$[/quote]

Let $f: \mathbb R\to  \mathbb R$ $f(x)=\left( \sqrt[3]{7+\sqrt{15}}\right)^{x}+\left( \sqrt[3]{7-\sqrt{15}}\right)^{x}-14$
Because f in sum of two increasing functions $\rightarrow$ $f$ is increasing $\rightarrow$ $f$ is injective so it has unique solution $f(3)=0$

[quote="diophantient"]
If $a+b+c=0$ then $a^{3}+b^{3}+c^{3}=3abc$
$(7+\sqrt{15})^{x}+(7-\sqrt{15})^{x}-2744=42(\sqrt[3]{34})^{x}$
Can this be used?[/quote]
Maybe in other equations some identies like this one above could be very usefull ,but in this case there's no point in using it because it only gets complicated.

[quote="bilot4"][quote="diophantient"][quote="kcbu"]solve
$\large(\sqrt[3]{7+\sqrt{15}})^{x}+(\sqrt[3]{7-\sqrt{15}})^{x}=14$[/quote]
[hide]If $a+b+c=0$ then $a^{3}+b^{3}+c^{3}=3abc$
$(7+\sqrt{15})^{x}+(7-\sqrt{15})^{x}-2744=42(\sqrt[3]{34})^{x}$[/hide]
Can this be used?[/quote]

Could u explain how did u find the number $-2744$ :blush:
How is this related to what is given :?:[/quote]$2744=14^{3}$.
diophantient used that $\left(\left(\sqrt[3]{7-\sqrt{15}}\right)^{x}\right)^{3}+\left(\left(\sqrt[3]{7+\sqrt{15}}\right)^{x}\right)^{3}+\left(-14\right)^{3}=3\left(\sqrt[3]{7-\sqrt{15}}\right)^{x}\left(\sqrt[3]{7+\sqrt{15}}\right)^{x}(-14)$

It follows the same idea .It has unique root $x=1$


geometry
circumcircle
incenter
ratio
trigonometry
power of a point
radical axis

part a
Four points are concyclic
iff angle TCI=angle TYI
iff angle TK1S=angle TK2X
which is true.

part b
i only managed to prove the external bisector of angle C passes through I... can anyone guide me what to do next?

2*angle ACB = angle AK1B.
2*angle BCI = angle SK1B.
Extend SK1 to meet AB at Z. Since SK1//K2X and K2X is perpendicular to AB, K1Z is perpendicular to AB.
Since AK1B is isoceles, angle AK1Z=angle BK1Z.
So angle AK1S=angle BK1S.
So 2*Angle AK1S + Angle AK1B = 360
2*Angle BCI + Angle BCA = 180
So CI is the external bisector of angle C.

I did same as you siuhochung b) but to complete solution
I think the following idea can be used:

Let $YC$ and $YT$ intersect $k_1$ in $D$ and $Z$.
$ZB$ intersects $XY$ in $J$.
Using first idea $BJ$ is external bisector of $DBC$.
From a) $BTJX$ are concyclic.

<$XJB$ = < $BTX$ = <$SAB$ = <$BCI$
Hence $BCIJ$ are concyclic.

And we need to prove < $ABD$ = <$IBJ$
sorry I didn't finish I found mistake.

First, clearly if the following statement(converse of the problem) is true, then the original conclusion is als otrue.
[i] $k_1$ be the circumcircle of a $\triangle ABC$. The external bisector of C meets $k_1$ at S. T is a point chosen on arc BC. Let $ST\cap AB=X$ and I be the excenter of ABC wrt BC. The circumcircle of TIC intersects XI again at Y. Then the circumcircle of XYT,$k_2$ is tangent to $k_1,BX,CY$.[/i]

Proving BX and $k_1$ are tangent to $k_2$ is easy, so I will only prove for CY. Note that it is equivalent to $\angle TIS=\angle TXI\Leftrightarrow SI^2=ST\cdot SX$. By similar triangle, we also have $SB^2=ST\cdot SX$. Thus we only have to show that SI=SB, but this is trivial by angle chasing using IB is the external bisector of B.

I know I have missed lots of details, but all of them can be obtained by angle chasing.
Anyway, this problem really surprises me because it cannot be solved by the "ordinary" method.

b)Let CY meet AB in $T_1$ and AC meet XY in R.
From [b]Menelaos[/b] for $\triangle{AT_1C}$:

$\frac{CY}{AX}=\frac{CR}{AR}$.(1)
From [b]Caysey[/b] applied to A,B,k2,C:

$a=\frac{CY}{AX}\cdot{c}+\frac{BX}{AX}\cdot{b}$.(2)
Using (1) and (2) and the converse of [b]transversal theorem [/b]we obtain the conclusion.

Let me give a remark on where the problem (apparently) comes from.

In the paper

[url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, [i]Sawayama and Thébault's Theorem[/i], Forum Geometricorum 3 (2003) pp. 225-229[/url]

(I have cited this paper in numerous threads, e. g. in http://www.mathlinks.ro/Forum/viewtopic.php?t=35337 ), a nice result - which is Lemma 1 in the paper - is shown:

[b](1)[/b] Let ABC be a triangle, and D a point on its side BC. Let $\mathcal{C}_1$ be a circle touching the segments DC and DA at two points E and F and touching the circumcircle $\mathcal{C}_2$ of triangle ABC at a point K. Then, the line EF passes through the incenter I of triangle ABC.

During the proof of Lemma 1, the following two additional facts are established:

[b](2)[/b] If the line KE meets the circle $\mathcal{C}_2$ at a point M (apart from K), then the line AM passes through the point I.
[b](3)[/b] The points A, K, F, I are concyclic.

Now, let's "extravert" - i. e. state the analogous result for external points:

[b](1')[/b] Let ABC be a triangle, and D a point on the extension of its side BC beyound C. Let $\mathcal{C}_1$ be a circle touching the segments DC and DA at two points E and F and touching the circumcircle $\mathcal{C}_2$ of triangle ABC at a point K. Then, the line EF passes through the B-excenter $I_b$ of triangle ABC.
[b](2')[/b] If the line KE meets the circle $\mathcal{C}_2$ at a point M (apart from K), then the line AM passes through the point $I_b$.
[b](3')[/b] The points A, K, F, $I_b$ are concyclic.

Now, we change some notations: $A,B,C\to C,A,B$, $E\to X$, $F\to Y$, $K\to T$, $M\to S$, $\mathcal{C}_1\to k_2$, $\mathcal{C}_2\to k_1$, $I_b\to I_a$. These statements then become:

[b](1'')[/b] Let ABC be a triangle, and D a point on the extension of its side AB beyound B. Let $k_2$ be a circle touching the segments DB and DC at two points X and Y and touching the circumcircle $k_1$ of triangle ABC at a point T. Then, the line XY passes through the A-excenter $I_a$ of triangle ABC.
[b](2'')[/b] If the line TX meets the circle $k_1$ at a point S (apart from T), then the line CS passes through the point $I_a$.
[b](3'')[/b] The points C, T, Y, $I_a$ are concyclic.

But this is exactly the configuration of our problem, just seen from a slightly different viewpoint. Now, from [b](1'')[/b] and [b](2'')[/b], it follows that the point $I_a$ lies on the lines XY and CS, so that $I_a=XY\cap CS=XY\cap SC$. Comparing this with the problem which states $I=XY\cap SC$, we see that $I_a=I$. Thus, the point I coincides with the A-excenter $I_a$ of triangle ABC. This solves problem [b](b)[/b]. Now, problem [b](a)[/b] follows from [b](3'')[/b] using $I_a=I$.

  Darij

in the pictures an "ordinary" solution.

pic.1
------
<TCI = <TYI ==> TCYI CIclico ==> <TXY = <TYC = < TIC ==> SI^2 = ST*SX

<SAT = <XYT = <AXT ==> SA^2 = ST*SX  ==>SA=SI

<SAT = <AXT ==> <SBA = <ATX = <SAB ==>SA=SB

pic.2
------
SC _|_ CF ==> SC bisettrice esterna di <C

<ASK = < AEC = 2<AIK ==> EC =EI

<BEA = <BSA = 2<BIA ==> BE =EI

CE = BE ==> <BAI = <CAI

[quote="darij grinberg"]

(I have cited this paper in numerous threads, e. g. in http://www.mathlinks.ro/Forum/viewtopic.php?t=35337 ), a nice result - which is Lemma 1 in the paper - is shown:

[b](1)[/b] Let ABC be a triangle, and D a point on its side BC. Let $\mathcal{C}_1$ be a circle touching the segments DC and DA at two points E and F and touching the circumcircle $\mathcal{C}_2$ of triangle ABC at a point K. Then, the line EF passes through the incenter I of triangle ABC.

Darij[/quote]

The following argument is also applicable to give a proof of this Lemma

<TCI = <TYI ==> TCYI ciclico ==> <TXY = <TYC = < TIC ==> SI^2 = ST*SX

<SAT = <XYT = <AXT ==> SA^2 = ST*SX ==>SA=SI

<SAT = <AXT ==> <SBA = <ATX = <SAB ==>SA=SB

SC _|_ CF ==> SC bisettrice esterna di <C

<ASK = < AEC = 2<AIK ==> EC =EI

<BEA = <BSA = 2<BIA ==> BE =EI

CE = BE ==> <BAI = <CAI

To complete the proof of the Thebault theorem, instead of that used in the paper, one can use the following argument.

[I refer to the picture in the  paper  Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003).]

If I = GF/\EH let U and V be the intersection of c(PEI) with GF and GE respectively. Since GF _|_ EH, then UP// GE therefore VE=UP. Now, as GF//DP, UP=GD so we have that VE=GD.
Since QGD ~ GIE, is easy to deduce that IGQ ~ IEV, then IQGV is cyclic which means that <QIV = 90° and then I lies on PQ.

Somewhere here in Bulgaria-2 also 'hiding' a nice  proof that the radical axis of

   circle ABC and it's  A-excircle is parallel to the line through traces of

   B- and C-bisectors on sides  AC and AB.

   T.Y.

   M.T.

[quote="armpist"]Somewhere here in Bulgaria-2 also 'hiding' a nice  proof that the radical axis of

   circle ABC and it's  A-excircle is parallel to the line through traces of

   B- and C-bisectors on sides  AC and AB.

[/quote]

Is perhaps this?

if Ia is the excenter of ABC and K and L the ends of a chord of circumcircle of ABC through the feet of B- and C-bisector on AC and AB respectively, then c(IKL), where I is the incenter of ABC, is homotetic with c(ABC) with center Ia and ratio 2 (*). Therefore, if P=IaK/\c(ABC), FP//KI so <IKIa = < IaAK then IaK is tangent at K to the circumcircle c(AIK) from which  IaK^2 = IaI*IaA. Similarly we have that IaL^2 = IaI*IaA, then IaK = IaL. This implies that IaO is othogonal to LK which is equivalent to the claim.

(*) this is a lemmaRL22022006.

Hello RL !

   Initially I made a naming mistake in my drawing, but after overdue correction

   I think that this very well may be it.  Nice idea! Hope Darij likes it,

   otherwise back to the drawing board.

   T.Y.

   M.T.

Ciao MT.

I'm not sure to understand exactly what you are referring to.

What was the naming mistake?  Why you are asking if this like to Darij?

[quote="N.T.TUAN"][quote="Sailor"] converse of [b]transversal theorem [/b][/quote]
What is it?[/quote]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=20289[/url]

Very nice! Thank you very much, tdl and darji.

I think I've got a slightly different proof for (b), proving that I is the A-excenter of ABC.
From a quick angle chase, we can see S is the midpoint of the arc AB, in the same segment as B; and IC is the external angle bisector of <ACB.
Thus it suffices to prove that, the E, the A-excenter of ABC lies on XY.
Let R be the midpoint of arc AC in circle k1, that lies in the same segment as B.
It's easy to see RB and SC intersect at E. Applying Pascal's Theorem to A,R,B,S,C,T - {RB,SC},{AB,ST},{RT,AC} are collinear - i.e. E lies on XZ where Z is where RT intersects intersects AC. In order to prove E lies on XY, it is now sufficient to prove X,Y,Z are collinear.
Let the tangents to k2 at X,Y intersect at P. Using Menelaus in triangle ACP, and noting that XP=YP, now reduces the problem to proving AZ/CZ=AX/CY. But noticing that RT is the external bisector of <ATC, we have AZ/CZ=AT/CT. We now have to prove AT/CT = AX/CY.

Unfortunately, for this seemingly simple problem I don't have a Euclidean proof, but a trig-approach is simple enough: forgetting about all the unnecessary points, angle chase the angles in quadrilateral ACXY and triangles ACT,ATX,TXY,TCY. Using the quadrilateral extension of trig-ceva (for lines AT,XT,YT,CT) gives an equation, which after some manipulation, reduces to the sine-rule equivalent of AT/CT=AX/CY.

Dear Mathlinkers,
for a) we can use a converse of the pivot theorem applied to the triangle IST with C on IS, T on SX, Y on IX, pivot being T.
Sincerely
Jean-Louis

Dear Mathlinkers,
for b) we can use the result of Protassov in an extraversion, I being the A-excenter of ABC.
http://perso.orange.fr/jl.ayme  vol. 2  Protassov
Sincerely
Jean-Louis

Here's my solution:

Let $CT\cap k_2=D$, since $T$ is the center of homothetic of the two circles, hence $CS\parallel DX\Rightarrow \angle XYT=180-\angle XDT=180-\angle SCT$ and a) follows.

For b) let $O_1,O_2$ denote the circumcenters of the two circles. Since $(SO_1\parallel XO_2)\perp AB$, hence $SB=SA\Rightarrow CI$ externally bisects $\angle C$. Let $(IYCT)\cap CB=E$, Since $\angle BTS=\angle BCS=180-\angle XYE$, therefore $BT\cap YE=F\in k_2$ $\Rightarrow \angle BFE=\angle CYT=\angle TEB\Rightarrow BE^2=BT*BF=BX^2$ $\Rightarrow BE=BX$, since $\angle XTC=\angle IEB=\angle IXB$, we can deduce that $BI$ externally bisects $\angle B$ and we done.

[quote=Bulgaria MO 2005/P2]Consider two circles $k_{1},k_{2}$ touching externally at point $T$. a line touches $k_{2}$ at point $X$ and intersects $k_{1}$ at points $A$ and $B$. Let $S$ be the second intersection point of $k_{1}$ with the line $XT$ . On the arc $\widehat{TS}$ not containing $A$ and $B$ is chosen a point $C$ . Let $\ CY$ be the tangent line to $k_{2}$ with $Y\in k_{2}$ , such that the segment $CY$ does not intersect the segment $ST$ . If $I=XY\cap SC$ . Prove that :

(a) the points $C,T,Y,I$ are concyclic.

(b) $I$ is the excenter of triangle $ABC$ with respect to the side $BC$.[/quote]

Here's an elementary and most important , complete solution

$\textbf{(a)}$
Consider the homothety $\mathcal{H}$ taking $k_2$ to $k_1$ . Then $\mathcal{H} : X\leftrightarrow{S}$ .
So $\mathcal{H} : \overarc{XT}\leftrightarrow{\overarc{TS}}$ and $\angle TAS=\angle TXA=\angle TCI=\angle TYI$ , as desired .

$\textbf{(b)}$
It's easy to see $SA$ is tangent to $(ATX)\implies{SA^{2}=ST.SX}$ , but also $SI$ is tangent to $(ITX)\implies{SI^2=ST.SX}$ , similarly $SB$ is tangent to $(TXB)\implies{SB^2=ST.SX}$ , hence $SA=SI=SB$ .
\[\angle BCI=180-\angle BCS=90-\frac{\angle BCA}{2}\] , so $CI$ is the external bisector of $\angle ACB$ . Analogously , since $\angle SBI=\angle SBI=90-\frac{\angle BAC}{2}$ , we get
\[\angle CBI=90-\frac{\angle ABC}{2}\] . Thus $I$ is the $A$-excenter of $\triangle ABC$ . $\blacksquare$

This problem is same as [i]Curvilinear Incircle[/i] configuration but here circles are externally tangent. The proof is same as the proof of curvilinear incircle.

By homothety, $S$ is the midpoint of arc $AB$ containing $T$ and $\widehat{ST}=\widehat{XT}$. So \[ \angle TCI=\frac{1}{2}\widehat{ST}=\frac{1}{2}\widehat{XT}=\angle TYI \] So $CTIY$ is cyclic.

Notice that $\triangle SCT\sim\triangle SIY$. So $SA^2=SB^2=SI^2=ST\cdot SX$ implies $I$ is the $A-$excenter of $\triangle ACB$ because $CI$ is the angle bisector of exterior angle $\angle ACB$.


Find the smallest positive or zero integer  $z$  that makes the quantity
$\frac{(z+ 3)^{16}}{289}$ a positive integer.

[hide]For the number to be an integer, the top number has to have the factors of 289, which prime factorization is $17^2$.
Therefore, the top part must have 17 as a factor and the $^{16}$ on the top takes care of the $^2$ on the bottom.
So the minimum of $z+3$ is 17 and solving gives $z=\boxed{14}$[/hide]

[hide]289 is a perfect square (17^2) so the least positive value of z+3 must be 17 since 17 is prime. z=14.[/hide]


Prove that an integer $n$ is a sum of two squares if and only if $2n$ is also a sum of two squares.

This is what the book had:
[hide="solution"]
$n=a^2 + b^2$, so $2n = 2a^2 + 2b^2 = a^2 - 2ab + b^2 + a^2 + 2ab + b^2 = (a-b)^2 + (a+b)^2$.
[/hide]

but i thought that was not strong enough of a proof, because it only proves that having 2n as sum of two squares is just an option. It does not prove WHY it has to be. Am I assuming it wrong, or am I correct?

If my assumption is correct, what should I do to solidify the proof? Maybe show that the converse works?

Yeah, I think you're right. I always learned that if and only if means you have to go both ways:

[hide]
Prove: If an integer $n$ is a sum of two square, then $2n$ is also a sum of two squares.
Let $x,y$ be integers that satisfy $n=x^2+y^2$. Then $2n=2x^2+2y^2=(x^2+2xy+y^2)+(x^2-2xy+y^2)=(x+y)^2+(x-y)^2$.
Since $x,y$ are integers, then $x+y$ and $x-y$ are also integers, and $2n$ is a sum of two squares.

Prove: If $n$ is an integer such that $2n$ is a sum of two squares, then $n$ is a sum of two squares.
Let $r,s$ be integers so that $2n=r^2+s^2$. Then $\displaystyle n=\frac{r^2}{2}+\frac{s^2}{2}=\left(\frac{r^2}{4}+\frac{rs}{2}+\frac{s^2}{4}\right)+\left(\frac{r^2}{4}-\frac{rs}{2}+\frac{s^2}{4}\right)=\left(\frac{r}{2}+\frac{s}{2}\right)^2+\left(\frac{r}{2}-\frac{s}{2}\right)^2$.
Furthermore, $2n$ is even, so $2n\equiv0\pmod4$ since it is a perfect square and $r^2+s^2\equiv0\pmod4$. For any integer $m$, $m^2\equiv0\pmod4$ or $m^2\equiv1\pmod4$, so $ r^2\equiv s^2\equiv0\pmod4$ implying $4|r^2$ and $4|s^2$.
Then $2|r$ and $2|s$, so $\frac{r}{2}$ is an integer, as is $\frac{s}{2}$. Then both $\left(\frac{r}{2}+\frac{s}{2}\right)$ and $\left(\frac{r}{2}-\frac{s}{2}\right)$ are integers, so $n$ is also a sum of two squares.
[/hide]


geometry
circumcircle
trigonometry
trig identities
Law of Sines
geometry unsolved

$ABC$ is a triangle. $M$ is the midpoint of $BC$. $\angle MAB = \angle C$, and $\angle MAC = 15^{\circ}$. Show that $\angle AMC$ is obtuse. If $O$ is the circumcenter of $ADC$, show that $AOD$ is equilateral.

Hey! He has apoint you know.(pun intended!)What is $D$?

VISIT THIS LINK FOR THE CORRECT SUM.

http://www.mathlinks.ro/viewtopic.php?p=1420925#1420925

The OP has made a typo, $ M\equal{}D$.

Let $ \angle BCA \equal{} \gamma$. Then we have that $ \angle ADC \equal{} \gamma$, $ \angle CDA \equal{} \frac {\pi}{12}$, $ \angle ABC \equal{} \pi \minus{} (2\gamma \plus{} \frac {\pi}{12})$. By the law of sines we have:

$ \frac {CD}{AD} \equal{} \frac {\sin(\frac {\pi}{12})}{\sin(\gamma)} \equal{} \frac {BD}{AD} \equal{} \frac {\sin(\gamma)}{\sin(\frac {\pi}{12} \plus{} 2\gamma)}\Rightarrow \sin^2(\gamma) \equal{} \sin(\frac {\pi}{12})\sin(\frac {\pi}{12} \plus{} 2\gamma)\Leftrightarrow$

$ \frac {1 \minus{} \cos(2\gamma)}{2} \equal{} \frac {\cos(2\gamma) \minus{} \cos(\frac {\pi}{6} \plus{} 2\gamma)}{2}$

Let $ x \equal{} 2\gamma$, and note that $ 2\cos(x) \minus{} 1 \minus{} \cos(x \plus{} \frac {\pi}{6}) \equal{} 0\Rightarrow 0<x\leq \frac{\pi}{2}$ (or else the LHS would be negative). We have that:

$ 2\cos(x) \minus{} 1 \minus{} \cos(x \plus{} \frac {\pi}{6}) \equal{} 0\Leftrightarrow$
$ \cos(x) \minus{} (\cos(\frac {\pi}{6})\cos(x) \minus{} \sin(\frac {\pi}{6})\sin(x)) \equal{} 1 \minus{} \cos(x)\Rightarrow$
$ \frac {2}{\cos(x)} \equal{} 4 \minus{} \sqrt {3} \plus{} \tan(x)\Rightarrow (4 \minus{} \sqrt {3} \plus{} \tan(x))^2 \equal{} 4(1 \plus{} \tan^2(x))\Rightarrow$
$ \tan(x) \equal{} \sqrt {3}\vee \tan(x) \equal{} \frac {8 \minus{} 5\sqrt {3}}{3} < 0$
We conclude that $ \tan(x) \equal{} \sqrt {3}\Rightarrow x \equal{} \frac {\pi}{3}\Rightarrow \gamma \equal{} \frac {\pi}{6}$

[b]i)[/b] $ \angle ADC\equal{} \pi\minus{}(\frac{\pi}{12}\minus{}\gamma)\equal{}\frac{3\pi}{4}>\frac{\pi}{2}$
[b]ii)[/b] $ \angle DOA\equal{}2\angle DCA\equal{}2\gamma\equal{}\frac{\pi}{3}$, but since $ OA\equal{}OD$, $ OAD$ is equilateral, as desired.

Angle BAD=Angle ACD=$X$.
In Triangle ABD,

                       BC/sin$X$=AB/sin$X$+15
In Triangle ABC,

                        2BC/sin$X$+15=AB/sin$X$.
Then we get $X$ =30 degree.
O is the circumcentre of a circle which is passes through the point A.D and C.
So,angle AOD =2*30 degree=60 degree.
AO=DO.
So,angle OAD=angle ODA=60 degree.
From here we know that OAD is an equilateral triangle.

[hide=Solution]
Note that \(AB\) is tangent to \((AMC)\), so considering the power of \(B\) with respect to \((AMC)\), we get
\[BA^2 = BM\cdot BC \qquad \implies \qquad \sqrt{2}=\frac{a}{c}=\frac{\sin(\alpha+15^{\circ})}{\sin(\alpha)}\]
and from inspection we obtain \(\alpha=30^{\circ}\).

Therefore \(\angle AMC = 135^{\circ} > 90^{\circ}\), and \(\angle AOM = 60^{\circ}\), so we are done.
[/hide]


Let s(n) be the sum of the divisors of n. An integer n is solitary if s(n)/n doesn't equal s(m)/m for all integers m not equal to n. Prove that a number is solitary if (n,s(n))=1.

gcf(n, s(n)) = 1.

Therefore, s(m)/m = s(n)/n iff s(m) = x*s(n) and m = x*n.  Assume m = x*n.  It follows that

[tex]\displaystyle s(x*n) = x*s(n)[/tex]

[tex]\displaystyle s(n) = (p_1^{k_1}+p_1^{k_1-1}+...+1)(p_2^{k_2}+p_1^{k_2-1}+...+1)\cdots[/tex]

[tex]\displaystyle \mbox{Let }x = p_a^{q_a}p_b^{q_b}\cdots[/tex]

[tex]\displaystyle s(x*n) [/tex]
[tex]\displaystyle = \cdots(p_a^{k_a+q_a}+p_a^{k_a+q_a-1}+...+1)\cdots(p_b^{k_b+q_b}+p_b^{k_b+q_b-1}+...+1)[/tex]

[tex]\displaystyle x*s(n) [/tex]
[tex]\displaystyle = \cdots(p_a^{k_a+q_a}+p_a^{k_a+q_a-1}+...+p_a^{q_a})\cdots(p_b^{k_b+q_b}+p_b^{k_b+q_b-1}+...+p_b^{q_b})[/tex]

Clearly, s(x*n) > x*s(n)

QED

Good. Can someone extend this? Why does the argument break down if
gcf(n,s(n)) isn't 1.

[quote="probability1.01"]s(x*n) > x*s(n)[/quote]

Here is another way to prove [b]probability1.01[/b]'s inequality. For each divisor d of n, x*d is a divisor of x*n. So each term in the sum of x*s(n) is a term in the sum of s(x*n). But 1 appears in s(x*n) and not in x*s(n) if x>1.


calculus
integration
function
limit
derivative
trigonometry
complex analysis

Yesterday I attended a talk by Prof. Chris Lennard of University of Pittsburgh.
One of the slide is this,

Integrals and Concavity

Consider the following "integral representation of the power function": For all $p \in \mathbb{R}$ with $0 < p < 1$, for all $t \in \mathbb{R}$ with $t \ge 0$,
\[
t^p  = \frac{1}
{{k_p }}\int_{\lambda  = 0}^\infty  {\frac{{\lambda t}}
{{1 + \lambda t}}\frac{{d\lambda }}
{{\lambda ^{p + 1} }}}
\]
here $k_p$ is a constant, depending only on p, but independent of t, i.e.,
\[
t^p  = \mathop {\lim }\limits_{A \to 0^ +  } \mathop {\lim }\limits_{B \to \infty } \frac{1}
{{k_p }}\int_{\lambda  = A}^{\lambda  = B} {\frac{{\lambda t}}
{{1 + \lambda t}}\frac{{d\lambda }}
{{\lambda ^{p + 1} }}}
\]

My questions:
1. Why and how is this related to power functions?
2. How to evaluate this integral?
3. How is it related to the concavity or convexity of a function? (Professor mentioned that this integral provides another test for convexity besides the wellknown second derivative test)

Thank you!

[quote="Buffalo"]1. replace $\lambda t$ by $\mu$.
2. you need some complex analysis, contour integral. $k_p=\frac{\pi} {\sin p \pi}$.
3. you could go to Prof. Chris Lennard.[/quote]

Can you elaboraqte more?
What exactly is a power function?
Thanks.


Let $\{a_n\}\ (n=0,1,2,\cdots)$ be the sequence such that $a_{n+3}=10a_{n+2}-31a_{n+1}+30a_n\ (n=0,1,2,\cdots).$

(1) The cubed equation $x^3=10x^2-31x+30$ has distinct three real roots $\alpha,\ \beta,\ \gamma\ (\alpha<\beta<\gamma).$　Find the values of $\alpha,\ \beta,\ \gamma.$
(2) Let $\alpha,\ \beta$ be the values in given $(1).$ Let $b_n=a_{n+1}-\alpha a_n (n=0,1,2,\cdots),$ express $b_{n+2}$ in terms of  $b_{n+1},\ b_n,$ and let $c_n=b_{n+1}-\beta b_n (n=0,1,2,\cdots),$ express $c_{n+1}$ in terms of $c_n.$
(3) If $a_0=1,\ a_1=2,\ a_2=5,$ then find $c_n,\ b_n,\ a_n.$

c) easy? Maybe you will use characterestic equation,but in this case Japanese highschool students are not allowed to use this directly. :lol:


probability
integration
expected value
calculus
calculus open

Given two independent random variables $X_1$, $X_2$ with standard normal distribution, expected value 0, variance 1:
\[ P(X\leq x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{\frac{-t^2}{2}}dt \]
the cumulative probability distribution of $Y=\max(X_1,X_2)$ is
\[ P(Y\leq y)=\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^ye^{\frac{-t^2}{2}}dt\right)^2. \]

Is it possible to express neatly the expected value and variance of $Y$?

By the way, i think we can use following lemma to copmute $D\xi$.
${\bf Lemma}$ If $\xi_1,\xi_2 \simeq N(0,1)$ then $\eta_1=\xi_1+\xi_2$ and $\eta_2=\xi_1-\xi_2$ are independent.
${\bf Proof}$ First of all $\eta_1\simeq N(0+0,1+1)$ and $\eta_2 \simeq N(0+0,1+1)$. So
$f_{\eta_1}(t) = f_{\eta_2}(t) = \frac{1}{\sqrt{4\pi}} e^{-\frac{t^2}{4}}.$
Now let $\eta=(\xi_1,\xi_2)$, then we have
$F_{\eta}(t_1,t_2) = \int\limits_{\begin{array}{cc} x-y\le t_1 && x+y\le t_2 \end{array}} f_{\eta}(x,y) dxdy =$
$=\left| \begin{array}{cc} x-y = a && x+y = b \end{array}\right| =\frac{1}{2} \int\limits_{\begin{array}{cc} a\le t_1 && b\le t_2 \end{array}} f_{\eta}(\frac{a+b}{2},\frac{b-a}{2}) dadb =$
$=\frac{1}{2} \int\limits_{\begin{array}{cc} a\le t_1 && b\le t_2 \end{array}} f_{\eta_1}(\frac{a+b}{2}) f_{\eta_2}(\frac{b-a}{2}) dadb = \frac{1}{2} \int\limits_{\begin{array}{cc} a\le t_1 && b\le t_2 \end{array}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(a+b)^2}{8}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(a-b)^2}{8}} dadb =$
$= \frac{1}{2} \int\limits_{\begin{array}{cc} a\le t_1 && b\le t_2 \end{array}} \frac{1}{\sqrt{2\pi}} e^{-\frac{a^2+b^2}{4}} \frac{1}{\sqrt{2\pi}} dadb = \frac{1}{2} \int_{-\infty}^{t_1} \frac{1}{\sqrt{2\pi}} e^{-\frac{a^2}{4}} da \int_{-\infty}^{t_2} \frac{1}{\sqrt{2\pi}} e^{-\frac{b^2}{4}} \;db =$
$= \int_{-\infty}^{t_1} f_{\eta_1}(a) da \int_{-\infty}^{t_2} f_{\eta_1}(b) db = F_{\eta_1}(t_1)F_{\eta_2}(t_2)$.
So $\eta_1$ and $\eta_2$ is independent.

Then $X_1-X_2$ and $X_1+X_2$ independent it follows that $|X_1-X_2|$ and $X_1+X_2$ also independent.
Now we can write $D\xi = D \frac{X_1+X_2 - |X_1-X_2|}{2} = \frac{1}{4} \left(D(X_1+X_2) + D|X_1-X_2| \right)=$
$= \frac{1}{4} \left( EX^2_1 + EX^2_2 + EX_1^2 + EX_2^2 +2EX_1X_2 - (E|X_1-X_2|)^2 \right) =$
$= \frac{1}{4} \left( 4 - \frac{1}{\pi}\right) = 1- \frac{1}{4\pi}.$


geometry algebra function domain search \/closed

Surely you meant 12 KB?

Using a randomly generated 80 by 80 picture...

GIF 128 no dither setting on Photoshop can only yield about 5.7 KB.

JPEG compression grows exponentially less reliable as quality is increased. 80% quality yields a very clear picture, and it's only 8.9 KB.

Of course, JPEG with no quality loss is larger than 14 KB. But by then it behaves much like BMP.

All in all, 12 KB is enough. ^_^

2?? 20?

I think 12 and 20 isn't that different...I agree with Osiris, plus I have all kinds of compression tips and tricks. Basically if you have Adobe you're all set to go, and you could even use Paint to convert a picture into GIF, which will significantly drop the quality but it should be fine if seen from far away (keep in note our avatars' kind of small...)

I think the main reason there's a limit to avatar size (both the area and the memory) is that it takes up too much space in the server. 550 users all having 12KB avatar would give you 6.6 MB, and the members are increasing...If we increase to 20KB, That'll take up 11MB (I'm significantly simplifiing this since it's not all that simple, but give or take it's around there), so that's 4.4MB increase...That can add up.

This is a math site after all, so don't expect much...Do the math :wink:  :lol:

Tare, even 11MB isnt too much. Even at 1000 members, we'd have somewhere around (a bit less...) 20MB which isnt too much.

Seeing as how some computers out there now have a TB of hard drive...

Mmm, I guess it could be a possibility after we move to that server.

Btw, when are we moving? I heard its sometime in the winter but no one's said anything other than that.

We still expect to move sometime in the winter.

To a server dedicated only to AoPS so we can have more bandwidth, stable connection and/or more server space.

Doubtful. DNS servers take care of the IP resolving, so unless you're accessing the site through its IP, you should be good.

I don't think so. What has happened at another website that I went to, all that happened was for a while you actually had to type in the new IP (xxx.xx.xx.xx type thing) until the actual name took over.. so we may have to do that for a while..

Doubtful, I believe Mr. Rusczyk planed ahead of time since I believe AoPS is getting a dedicated server purchased by AoPS Inc. so they (Mr. Rusczyk and Mr. Crawford) had to configurate everything and get internet connection and such so they would probably move after they ensure the domain name and change the info on the DNS server but it's all up to them...

Btw is there a reason why you can't access AoPS through IP address? Just wondering...

No reason that you can't, other than it's a bunch of numbers to remember, and it isn't guaranteed to stay constant. Oh, and you'd have to know what you're doing well enough to get the IP... not that it's particularly hard.

Um you need abolutely no knowledge in these over-helpful days, just go to a tools website or just use Command Prompt, they even made a help file for it, unbelievable: "Command Prompt for Dummies" :lol: </jk>

Here's what I came up with when I tried to go to AoPS through IP:
"The website for 66.118.156.82 is
Coming Soon!
For more information, please click here to contact us.
Search The Web:
Powered By WEBMASTERS.COM"

I wonder why...maybe the IP address already handed over...but then how is AoPS still usable? Hmm...

a.) I used ping
b.) I used tracert
c.) I used pathping
d.) I used Samspade
e.) I'm a psychic
f.) I'm secretly a co-founder of AoPS :lol:
g.) lucky guess
h.) trial and error :lol:
i.) The Fibbonacci rabbits told me :lol:
j.) all of the above

We'll be at the same http://www.artofproblemsolving.com address, but on a new server.  For many of you, you'll just show up one day and the site will look different, the message board will be faster, and so on.

I'll publicize more details when the switch comes...

mmmkay.  One thing: why will the site look different?

The board will be reorganized again like the Etc. and Books Music... combine and spliting and stuff.

We also picked out a message board style that we believe will be an upgrade in readability (in terms of being easy on the eyes).

I also like the general layout Richard put together.  I think after people get used to the change most will prefer the new style.


geometry
3D geometry
icosahedron
sphere
trigonometry
geometry solved

what is the acute angle between the main diagnoles of an icosahedron?

[color=red][Moderator edit: Cross-posting with http://www.mathlinks.ro/Forum/viewtopic.php?t=58942 .][/color]

Select 3 vertices $A_1, A_2, A_3$ forming an equilateral triangle $\triangle A_1A_2A_3$, Let $A_4$ be the 4th vertex forming a congruent equilateral triangle $\triangle A_1A_2A_4 \cong \triangle A_1A_2A_3$ with the vertices $A_1, A_2$. Let $A_4B_4$ be one of the 6 main icosahedron diagonals, i.e., $B_4$ is a diametrally opposite vertex to $A_4$ on the circumscribed sphere. The arc $A_4A_3B_4$ is the sphere main semicircle. Let $M_{12}$ be the midpoint of $A_1A_2$, lying in the plane of this main semicircle. Denote $\theta = \angle B_4OA_3$ (the angle subtended by the equilateral triangle edge $a = B_4A_3$) and $\phi = \angle A_3OM_{12} = \angle M_{12}OA_4$. Obviously,

$\angle B_4OA_3 + \angle A_3OM_{12} + \angle M_{12}OA_4 = \theta + 2 \phi = \pi$

$\phi = \frac \pi 2 - \frac \theta 2,\ \ \ \cos \phi = \sin \frac \theta 2$

$a = B_4A_3 = 2OB_4 \cos \frac{\widehat{B_4OA_4}}{2} = 2r \cos \frac \theta 2$

where $r = OB_4 = OA_3 = OA_4 = OA_1 = OA_2$ is the circumscribed sphere radius. Now, $A_3M_{12} = A_4M_{12} = \frac{a \sqrt 3}{2}$ are the altitudes of the equilateral triangles $\triangle A_1A_2A_3, \triangle A_1A_2A_4$ and

$OM_{12} = OA_1 \cos \widehat{A_1OM_{12}} = OA_1 \cos \frac{\widehat{A_1OA_2}}{2} = r \cos \frac \theta 2$

Using the cosine theorem for the triangle $\triangle A_4OM_{12}$ with the sides $OA_4 = r$, $OM_{12} = r \cos \frac \theta 2$, $A_4M_{12} = \frac{a \sqrt 3}{2}$ and the angle $\angle A_3OM_{12} = \phi$, we have

$OA_4^2 + OM_{12}^2 - 2\ OA_4 \cdot OM_{12} \cos \widehat{A_4OM_{12}} = A_4M_{12}^2$

$r^2 + r^2 \cos^2 \frac \theta 2 - 2r^2 \cos \frac \theta \cos \phi = \frac{3a^2}{4} = 3r^2 \sin^2 \frac \theta 2$

Reducing by $r^2$, substituting $\cos \phi = \sin \frac \theta 2$ and then using $1 - \sin ^2 \frac \theta 2 = \cos ^2 \frac \theta 2$, we get

$1 + \cos^2 \frac \theta 2 - 3 \sin ^2 \frac \theta 2 - 2 \cos \frac \theta 2 \sin \frac \theta 2 = 0$

$2 \left(\cos^2 \frac \theta 2 - \sin^2 \frac \theta 2 \right) - 2 \cos \frac \theta 2 \sin \frac \theta 2 = 0$

$2 \cos \theta - \sin \theta = 0,\ \ \ \tan \theta = 2,\ \ \ \cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}} = \frac{1}{\sqrt 5},\ \ \ \theta \doteq 63.435^\circ$

Compare with [url=http://mathworld.wolfram.com/PlatonicSolid.html]Platonic Solid[/url] from Mathworld (the last table giving the angles $\beta$ subtended by the edges of the Platonic solids).


A track meet awards 6,5,4,3,2, and 1 for 1st, 2nd,3rd,4th 5th , and 6th places, respectively. Teams A's only two runners originally earned 6 points in the first race with both runners finishing in the top6, but then the winner of the first race was disqualified. How many points did Team A  have after the ranks were adjusted?

[hide]Obviously, no one came first, since the score would be atleast 7. So it goes up 2 higher, so 8.[/hide]

EDITED. Why am I such a bad reader?

Uh, I had no clue, so I just said 6. Probably, so I'll switch.

[quote="Math Geek"][hide]We know that the top person wasn't from team A cuz the total for team A was only 6.  Thus, if that person got kicked out, that means everyone else moves up a rank.  Thus, the total points for team A increases by 1 point per person, so the answer should be $ \boxed {8}$.[/hide]

Hmm, if he gets disqualified, does that mean that the people move up, because if it doesn't, then that explains the difference between xpmath's and my solution.[/quote]

yeah, I think it does mean they move up.  which does make the answer $ \boxed{8}$.


geometry
geometric transformation
reflection
logarithms
symmetry

Consider an analog clock with an hour and a minute hand but no numerals.  At time $ T$ A.B., a mirror image of the clock shows a time $ X$ A.M. that is 5 hours and 28 minutes later than $ T$.  Compute $ T$ in terms of hours and minutes.

Thanks in advance!

A mirror image which way? Left and right, or up and down?

This seems wrong...

[hide="Solution"]
Let the affix of the hour hand be $ h\equal{}e^{i\alpha}$ and the affix of the minute hand be $ m\equal{}e^{i\beta}$, where $ \alpha$ and $ \beta$ are degree values (such that $ 0$ minutes corresponds to $ 0^\circ$, $ 60$ minutes to $ 360^\circ$, and so on).

After 5 hours and 28 minutes, the hours hand should have traveled $ \frac{360^\circ}{12\text{ hours}}\cdot5\text{ hours}\plus{}\frac{360^\circ}{12\text{ hours}}\cdot 28\text{ minutes}\cdot\frac{1\text{ hour}}{60\text{ minutes}}\equal{}164^\circ$.

After 28 minutes, the minutes hand should have traveled $ 28\text{ minutes}\cdot\frac{360^\circ}{60\text{ minutes}}\equal{}168^\circ$.

A reflection (assuming it's up and down) is equivalent to taking the conjugate, so we have $ \overline{h}\equal{}h\cdot e^{i164^\circ}$. From this we get $ e^{\minus{}i\alpha}\equal{}e^{i(\alpha\plus{}164^\circ)}$. Taking the complex logarithm of both sides, $ \minus{}\alpha\equal{}\alpha\plus{}164^\circ\implies \alpha\equal{}\minus{}82^\circ\equiv 278^\circ$.

Similarly, $ e^{\minus{}i\beta}\equal{}e^{i(\beta\plus{}168^\circ)}\implies \minus{}\beta\equal{}\beta\plus{}168^\circ\implies \beta\equiv 276^\circ$.

Now converting backwards, $ 278^\circ\cdot\frac{60\text{ minutes}}{360^\circ}\equal{}\frac{139}{3}\text{ minutes}$.

$ 276^\circ\equal{}46\text{ minutes}$.

The time is, then $ \frac{277}{3}\text{ minutes}\equal{}\boxed{\text{1 hour, } 32\frac{1}{3}\text{ minutes}}$.
[/hide]

[quote="Octahedr0n"][quote="rd5493"]Let me take a crack at this one.

[hide]
Since $ 3: 00$ AM and $ 9: 00$ AM are $ 6$ hours apart, and $ 4: 00$ AM and $ 8: 00$ AM are $ 4$ hours apart, the desired time is about a quarter of the time from $ 3: 00$ to $ 4: 00$ AM.  More precisely, since $ 5$ hours, $ 28$ minutes equals $ 328$ minutes, the hands at $ T$ AM are $ 164$ minutes behind $ 6$ AM, so we subtract $ 2$ hours and $ 44$ minutes from $ 6: 00$ AM to get [size=150]$ 3: 16$[/size][/hide][/quote]

I don't quite understand what you did.  Could you please explain a bit more?  Thanks![/quote]

Since the time we are trying to find in 5 hours and 28 minutes, we see that this is around 6 and 4 hours, but it is closer to 6 hours than 4.  Because of this, then we see that the time is between 3 and 4 AM.  Since 5 hours 28 minutes equals 328 minutes, we can divide by 2 because of the symmetry at 6:00.  This means that that the time will be 164 minutes from 6:00, and since 164 minutes equals 2:44,
$ 6: 00\minus{}2: 44\equal{}3: 16$

I hope this clears things up...
If anyone can explain it better, please do so...

[quote="rd5493"][quote="Octahedr0n"][quote="rd5493"]Let me take a crack at this one.

[hide]
Since $ 3: 00$ AM and $ 9: 00$ AM are $ 6$ hours apart, and $ 4: 00$ AM and $ 8: 00$ AM are $ 4$ hours apart, the desired time is about a quarter of the time from $ 3: 00$ to $ 4: 00$ AM.  More precisely, since $ 5$ hours, $ 28$ minutes equals $ 328$ minutes, the hands at $ T$ AM are $ 164$ minutes behind $ 6$ AM, so we subtract $ 2$ hours and $ 44$ minutes from $ 6: 00$ AM to get [size=150]$ 3: 16$[/size][/hide][/quote]

I don't quite understand what you did.  Could you please explain a bit more?  Thanks![/quote]

Since the time we are trying to find in 5 hours and 28 minutes, we see that this is around 6 and 4 hours, but it is closer to 6 hours than 4.  Because of this, then we see that the time is between 3 and 4 AM.  Since 5 hours 28 minutes equals 328 minutes, we can divide by 2 because of the symmetry at 6:00.  This means that that the time will be 164 minutes from 6:00, and since 164 minutes equals 2:44,
$ 6: 00 \minus{} 2: 44 \equal{} 3: 16$

I hope this clears things up...
If anyone can explain it better, please do so...[/quote]

I still don't get your solution.  For example, when you say that 5 hours and 28 minutes is around 6 and 4 hours, why don't you use 6 and 5 hours instead?  After all, 5 hours and 28 minutes is closer to 5 hours than 6 hours.

He's just figuring out the range in which the time has to be in.  Since the mirror image of $ 4: 00$ is $ 8: 00$, making their difference $ 4$ hours, and that of $ 3: 00$ and $ 9: 00$ is six hours, you know that the desired time is somewhere in between this.  Moreso, because you know that $ 3: 30$ will give you a difference of five hours, and $ 28$ minutes is nearly half of an hour, you would expect the desired time to be somewhere around $ 3: 15$.

As for grn_trtl's response, I don't see how it can possibly be right.  However, the method is very interesting ;]

lol yeah it's wrong, but I think you can do this with complex numbers if you don't mess it up.


Let  $ a,b,c$  be nonnegative real numbers. Prove that

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}1\ge \frac{1}{2}\sqrt{\frac{11(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\plus{}14}$

[quote="tardi"]Let  $ a,b,c$  be nonnegative real numbers. Prove that

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} 1\ge \frac {1}{2}\sqrt {\frac {11(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 14}$[/quote]
Your inequality is equivalent to the following:
$ \sum_{sym}(4a^7b\plus{}5a^6b^2\minus{}4a^5b^3\minus{}5a^4b^4\plus{}7a^6bc\plus{}12a^5b^2c\minus{}14a^4b^3c\minus{}5a^3b^3c^2)\geq0,$
which obviously true.

[quote="tardi"]From this follows Iran's Inequality

$ \sum \frac 1{(a\plus{}b)^2}\ge \frac {9}{4(ab\plus{}bc\plus{}ca)}$[/quote]
______________
want to show it? Thanks!

[quote="sandumarin"]If $ a,b,c>0 $   prove that:
$ 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+1\ge\sqrt{19\frac{a^2+b^2+c^2}{ab+bc+ca}+30} $[/quote]
The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[ 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+1\ge\sqrt{\frac{70(a^2+b^2+c^2)}{ab+bc+ca}-21}\]

Dear xzlbq! Check please the following inequality.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge\sqrt{\frac{28(a^2+b^2+c^2)}{ab+bc+ca}+8}\]
Thank you!


geometry
3D geometry
perimeter
Pythagorean Theorem

a paper cone, when cut along its slant height and opened out, forms a semicircle of radius 10. What is the altitude of the original cone?

how would you do this?

are there any important geometric properties involved in solving this problem? if so, please tell me them.  :)

[hide="Solution"]The semicircular arc is equal to the perimeter of the cone base, hence $10\pi=2r\pi\implies r=5$.

The radius of the semicircle is equal to the slant height of the cone, hence $s=10$.

By Pythagorean Theorem, $H=\sqrt{s^{2}-r^{2}}=\sqrt{75}=5\sqrt{3}$[/hide]


prove that $\sum_{i=1}^n \binom{i+1}{2} = \binom{n+2}{3}$...a combinatorial or block walking solution prefered

$\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\ldots+\binom{n+1}{2}=\binom{n+2}{3}$.

$\binom{2}{2}=1=\binom{3}{3}$.

The rest of the sum telescopes from the simple identity $\binom{n-1}{k}+\binom{n-1}{k+1}=\binom{n}{k+1}$.

A block-walking argument would simply be showing again that $\binom{2}{2}=\binom{3}{3}$ and then going along a visual representation of the above identity (which along Pascal's basically says a term is the sum of the two terms above it).

[hide="counting argument"]$\binom{n+2}{3}$ is the number of ways to choose 3 elements from a set of $n+2$.

Suppose the $k+2$nd element is the last one chosen. Then there are 2 elements left to choose out of the previous $k+1$.

$k$ can range from 1 (pick the first three elements) to $n$ (the last element is one of the ones picked).

Therefore $\sum_{k=1}^n \binom{k+1}{2}=\binom{n+2}{3}$. Done.[/hide]