calculus
integration
ratio
trigonometry
number theory
prime factorization
AMC

1. Given two arithmetic sequences: 1, 8, 15, 22, . . .   and 7, 18, 29, 40, . . .   The smallest number they have in common is 29.  Compute the third-smallest number they have in common.

2. Compute the geometric mean of the positive integral factors of 969.  (The geometric mean of a set of n numbers is the nth root of their product.)

3. a and b are two digits, and 7ab/ab4 = 7/4.  (That is a ratio of two three-digit numbers.)  Find ab (a two-digit number).

4. AB is the diameter of a circle of radius 2 and a side of equilateral triangle ABC.  The circle intersects BC at D.  Compute the length of AD.

For number 3, you get 2/7 for a correct answer.  For number 4, you get 0/7.

ahhh

[hide]root 3, 2*root 3, its all good[/hide]

but then again, stupid mistakes cost me big on Mandlebrot

i hate that sh*t

[quote="JBL"]1. Given two arithmetic sequences: 1, 8, 15, 22, . . .   and 7, 18, 29, 40, . . .   The smallest number they have in common is 29.  Compute the third-smallest number they have in common.
[/quote]

The common difference of the first one is 7.  The common difference of the second one is 11.  So every 77 numbers they should coincide.  The first number they coincide at is 29.  The third should be 29 + 2(77) = 183

In the spirit of D[hide]airy[/hide] Q[hide]ueen[/hide] - I mean darkquantum - I'm going to answer each question separately.

Quote:3. a and b are two digits, and 7ab/ab4 = 7/4. (That is a ratio of two three-digit numbers.) Find ab (a two-digit number).

[hide]7ab/ab4 = 7/4

(700 + 10a + b)/(100a + 10b + 4) = 7/4

4(700 + 10a + b) = 7(100a + 10b + 4)

2800 + 40a + 4b = 700a + 70b + 28

(700-40)a + (70-4)b = 2800 - 28

(70-4)10a + (70-4)b = 28*99

10a + b = 28*99/66 = 28*3/2 = 42

That was showing a lot of work.  Oof.[/hide]

Both look good to me.  Perhaps I should argue for a justification on number 1, but I'll take it.  7/7, 7/7.  You can write a little bit less for number 3 if you let 10a + b = x, since you never need to deal with the two parts seperately.

[quote]2. Compute the geometric mean of the positive integral factors of 969. (The geometric mean of a set of n numbers is the nth root of their product.)

3. a and b are two digits, and 7ab/ab4 = 7/4. (That is a ratio of two three-digit numbers.) Find ab (a two-digit number). [/quote]

I'll do 3 first.

a and b are integers.        1 :le: a :le: 9
                                          0 :le: b :le: 9

(700 + 10a + b)/(100a + 10b + 4) = 7/4

2800 + 40a + 4b = 700a + 70b + 28

2772 = 660a + 66b

42 = 10a + b

(a,b) = (4,2)*

Therefore, 42 is the number we're looking for.

* I didn't need to do that... *

Now to 2.

969 has a prime factorization of (3)(17)(19).  Therefore, there are 8 positive integral factors of 969.  We're looking for the geometric mean of all the factors of 969 though.  Among all the factors, 3, 17, and 19 each appear 4 times, meaning that the product of all the factors of 969 is ((3)(17)(19))^4.  The geometric mean is just
:sqrt:969.

in fact, for #2, [hide]969 can be replaced by any number x, and the answer will still be :sqrt:x. This is because the product of all the factors of x will be x^(d(x)/2), where d(x) is the number of factors of x. This is because, for every factor, there is a corresponding factor such that those two factors multiplied will give x. So we have x multiplied by itself for exactly half the number of factors that it has. Now, if we want to take the GEOMETRIC MEAN !!! of all the factors, we take the product of the factors to the (1/d(x)) power. This gives us x^(1/2), or :sqrt:x.[/hide]

7/7 for both, warpedklown.  MysticTerminator's solution is where the question originated from.  I thought that was a rather pretty identity.

I've suspected that, but I didn't know how to prove that for all integers.  Cool!

Since nobody seems to have given a correct solution to number 4, [hide] We can form a triangle ABD. The length of AB is 4, and the measure of angle B (ABC is equilateral) is 60.

Pause to call upon my memory of trig ratios....

OK. tan60=x/4

1.732 :approx: x/4

6.928 :approx: x[/hide]

Actually, darkquantum later amended his answer to the correct value.  You are off by a little bit.
There has yet to be a solution to the question, however.
Also, why do you use decimal approximations?  Why not just use the exact value :rt3:?

Wow!  That was a rather simple way of doing #1... I did Diophantine Equations  :oops:  Same result nevertheless...

JBL, darkquatom said 2*sqrt3, but topper showed that it is 4*sqrt3. which is right?

Sorry, I didn't make that clear -- 2:rt3: is the correct answer.  Topper must have made a slight error soemwhere, but as s/he didn't post a solution, I can't say where.

:edit:  Actually, looking again, I can say where -- topper used the wrong trig function.


[quote="inom"]$sin3z-cos3z=\sqrt\frac{3}{2}$
Solve for z?[/quote]
[hide]
Squar both sides and simple algebraic computation yields:
$-\frac{1}{2}=2\cos(3z)\sin(3z)$.
This is equal to $-\frac{1}{2}=\sin(6z)$..
This is pretty much it... :D [/hide]

[hide]Let $3z = x$
After squaring both sides we get $1 - 2\sin x\cos x = \frac{3}{2} \Rightarrow 2\sin x\cos x = -\frac{1}{2} \Rightarrow \sin 2x = -\frac{1}{2}$.
Hence $z = \frac{11}{36}\pi + k\frac{\pi}{3}$. [/hide]

[hide]
square both sides to get $1-\sin{6z} = \frac{3}{2}$.  then we get $6z = \frac{(7\pm12k)\pi}{6}, \frac{(11 \pm 12k)\pi}{6}$, divide by 6 to get $z = \frac{(7\pm12k)\pi}{36}, \frac{(11 \pm 12k)\pi}{36}$
[/hide]


The incircle of the triangle ABC touches the sides BC, AC, and AB at points A', B', and C', respectively. It is known that AA'=BB'=CC'. Does the triangle ABC have to be equilateral?

(I am very interested in ingenious solution of this problem, because I found an ugly one using Stewart's theorem and tons of algebra during the contest).

I remember this problem being discussed on the Hyacinthos newsgroup, but I don't succeed to find the proper messages anymore. But why should I if there is a very simple solution?

Indeed, from AA' = BB' = CC', it follows that triangle ABC is equilateral.

[i]Proof.[/i] We first show:

[b]Lemma 1.[/b] If a > b, then AA' < BB'.

[i]Proof.[/i] Since the two tangential segments from a point to a circle are equal, we have CA' = CB'. Thus, the triangle A'CB' is isosceles, and we get

$\measuredangle A^{\prime}B^{\prime}C = \frac{180^{\circ}-\measuredangle A^{\prime}CB^{\prime}}{2} = \frac{180^{\circ}-C}{2} = 90^{\circ}-\frac{C}{2}$.

But < AB'A' = 180 - < A'B'C, and thus

$\cos \measuredangle AB^{\prime}A^{\prime} = -\cos \measuredangle A^{\prime}B^{\prime}C = -\cos \left(90^{\circ}-\frac{C}{2}\right) = -\sin\frac{C}{2}$.

Also, it is well-known that AB' = s - a and BA' = s - b, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, with a > b, we have s - a < s - b, so that AB' < BA'.

Now, after the Cosine Law in triangle AA'B', we have

$AA^{\prime}\ ^2 = AB^{\prime}\ ^2 + A^{\prime}B^{\prime}\ ^2 - 2\cdot AB^{\prime}\cdot A^{\prime}B^{\prime}\cdot \cos \measuredangle AB^{\prime}A^{\prime} = AB^{\prime} ^2 + A^{\prime}B^{\prime}\ ^2 + 2\cdot AB^{\prime}\cdot A^{\prime}B^{\prime}\cdot \sin\frac{C}{2}$

(in the last step, we used $\cos \measuredangle AB^{\prime}A^{\prime} = -\sin\frac{C}{2}$). Similarly,

$BB^{\prime}\ ^2 = BA^{\prime} ^2 + A^{\prime}B^{\prime}\ ^2 + 2\cdot BA^{\prime}\cdot A^{\prime}B^{\prime}\cdot \sin\frac{C}{2}$.

Now, comparing these two formulas, we directly see that AA' < BB' because of AB' < BA' (and because of $\sin\frac{C}{2}>0$).

Lemma 1 is thus proven.

Now, for our problem, if the triangle ABC were not equilateral, we would have one side which greater than the other; WLOG we could assume a > b, and Lemma 1 would yield AA' < BB' in contradiction to our hypothesis AA' = BB' = CC'. So the problem is solved.

  Darij

Hmm... Maybe some shortl solution?
Oh.. I again realized that Darij post is too long ;)
We see that $CB'=CA'$, so from cosine theorem
\[AA'^2-BB'^2=(AC^2+CA'^2-2AC\cdot CA'\cos C)-(BC^2+CB'^2-2BC\cdot CB'\cos C)=(AC-BC)(AC+BC-2CB'\cos C),\]
where $BC,AC>CB'=CA'$, $\cos C\leq 1$. Thus if $AC>BC$ then $AA'^2>BB'^2$.


A positive integer has all single-digit prime factors. None of its prime factors occurs more than twice in the prime factorization, but the number is not a perfect square. What is the greatest possible number with these properties?


Good evening.

Due to requests for an easier MATHCOUNTS practice round, Nosepuppy and I have compiled a Mock MATHCOUNTS Individual Round, complete with Sprint Round and Target Round. The difficulty of the rounds are intended to be between Chapter Competion difficulty and State Competition difficulty.

If you choose to take this test, please send either of us your answers to both rounds all at once (please do not send every pair of target round questions as you finish them). Also, because the tests will be up for awhile, please take the correct amount of time for each one:

Sprint Round: 40 minutes
Target Round: 6 minutes per pair

To make the test more worthwhile, the top three scorers will recieve [b]a special treat[/b]. The test should be posted at some time tonight, and will be up until next Sunday. If you have any questions, please feel free to send Nosepuppy or me a private message. Good luck!

here's the stuff, kids!

SPRINT:
[url]http://www.boomspeed.com/drakonar/SPRINT.pdf[/url]

TARGET:
[url]http://www.boomspeed.com/drakonar/TARGET.pdf[/url]

DIAGRAM(you'll need it for target #8):
[url]http://www.boomspeed.com/drakonar/diagram.GIF[/url]

[size=200]ATTENTION:  THE CONTEST WILL END [b]FRIDAY SEPTEMBER 10TH,[/b] NOT SUNDAY SEPTEMBER 12TH AS PREVIOUSLY STATED.[/size]

the reason for this is: we are trying to get the grading finished by sunday night so we don't have to worry about this after AHSIMC starts.

we will have the results back to you by Sunday the 12th.  At night most likely.

thanks!

I'm taking it right now.

EDIT:
Ok, done!

Target round was funny.

an hour or two from now you'll have your scores.

[quote="JGeneson"]We have not got many submissions...

Remember the deadline is tomorrow (or maybe Saturday).

Do not miss out on [b]a special treat[/b].[/quote]

Don't be surprised at a small turnout. I think there isn't as much enthusiasm for mathcounts this year on the forums. I guess we'll see as Chapters get closer.

I'm a mathcounts coach and am planning to use this for my students. This is my second year coaching and last year we were crushed. I'm still trying to learn all of the tricks to pass on to my students. Could I possibly get the key to this sprint round.

a total of appx 3 people took the contest.

prize for everyone who took it:

free gmail invitation from me!

i know, i know, everyone and their brothers already have 6 gmail accs, but just in case.

due to ahsimc and other arbitrary activities JGen and I have been partaking in, the answer key is still not entirely finished.  DNAS did pretty well though.  sadly, not perfect, but still well.

so if you took part in the contest, PM me your email address (yes, pm them to me so i can verify you took the contest) or post it here, and i'll email you an invite.

I don't want a gmail invite, but I would like to see the answers, to see if I got them right,

Same here. (already got one.)

Or just the score would suffice.

[quote=kootrapali]Is there an answer key?[/quote]

you're killing me. whats with the bump


Let $a,b,c$ be non-negative real numbers. Prove that
\[\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ca}}+\frac{1}{\sqrt{4c^{2}+ab}}\ge \frac{4}{a+b+c}. \]
This problem is in chapter II of my 2006 inequalities book (seemingly unsolved in Mathlinks). I have a very nice solution to this one.

[quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers. Prove that
\[\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ca}}+\frac{1}{\sqrt{4c^{2}+ab}}\ge \frac{6}{a+b+c}. \]
[/quote]
Something is not right, hungkhtn. :wink:

Well it should have at least one of a,b,c positive, not non negative

I am sorry. I did edit it. Everything is Ok now.

It is great to hear that this hard inequality can be solved just by classical way (in Vietnamese book, only mixing variables way is presented. So I am waiting for a clever Mathlinker who can solve me this solution.

[quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers. Prove that
\[\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ca}}+\frac{1}{\sqrt{4c^{2}+ab}}\ge \frac{4}{a+b+c}. \]
[/quote]
Let $a+b+c=1.$
Hence, $\sum_{cyc}\frac{1}{\sqrt{4a^{2}+bc}}=\sum_{cyc}\left(\frac{b+c}{2}\cdot\frac{1}{\sqrt{\frac{(b+c)^{2}(4a^{2}+bc)}{4}}}\right)\geq\frac{1}{\sqrt{\sum\frac{(b+c)^{3}(4a^{2}+bc)}{8}}}.$
Thus, remain to prove that $(a+b+c)^{5}\geq2\cdot\sum_{cyc}(b+c)^{3}(4a^{2}+bc)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^{5}-a^{4}b-a^{4}c+a^{3}bc)+4\cdot\sum_{sym}(a^{4}b-a^{3}b^{2})+\sum_{cyc}(19a^{3}bc-18a^{2}b^{2}c)\geq0.$:)
Jensen forever! ( Myth )
What is your proof, hungkhtn? Thank you.

Very nice, Arqady. I am using Holder for the first step, and the next is similar to you.

I use

\[\left(\sum \frac{1}{\sqrt{4a^{2}+bc}}\right)^{2}\left(\sum (a+b)^{3}(4a^{2}+bc)\right)\ge 8(a+b+c)^{3}.\]

The rest is similar to you. Congratulation for your very nice solution  :lol:


Ignoring the blob in the middle, there are $ 16$ 1-by-1 squares, $ 9$ 2-by-2 squares, $ 4$ 3-by-3 squares, and $ 1$ 4-by-4 square. Adding these up, $ 1\plus{}4\plus{}9\plus{}16\equal{}30$ squares.

Now, returning to the additional "decoration" in the middle: there are $ 4$ tiny squares, plus $ 1$ big square made from the union of these $ 4$ squares, making $ 5$ squares from this.

The total is $ 30\plus{}5\equal{}\fbox{35}$ squares.

If we're talking about squares with sides parallel to the grid, then it's $ (1^2 \plus{} 2^2 \plus{} 3^2 \plus{} 4^2) \plus{} (1^2 \plus{} 2^2) \equal{} \boxed{35}$; otherwise, it's

$ \frac {5^2(5^2 \minus{} 1)}{12} \plus{} \frac {3^2(3^2 \minus{} 1)}{12} \equal{} \boxed{56}$.

What formula are you using for the second part of your solution?

Ah...the number of squares in an $ n\times n$ grid of dots is $ \frac {n^2(n^2 \minus{} 1)}{12}$...major time saver that I have no idea how to derive (seems elegant, but rather complicated to prove)...

Edit: I got it originally from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192810&search_id=644715908&start=20]here[/url].


linear algebra
matrix
function
superior algebra
superior algebra solved

X belongs to M_n(Z)(with n>=2). If (n-1) elements of the first(principal) diagonal are even and the rest of the elements of the matrix (n^2-n+1) are odd, then please prove that det(X)<>0.

thank you very much! :)

consider the determinant with elements 0 and 1 from the field Z/2Z. It has all elements aij=1 for all i and j, but only when i=j<n aii=0.

the general formula for the determinant is \sum s(f(n))a1f(1)a2f(2)...anf(n) where the sum is taken over all permutations f(x) with n elements of the set {1,2,...,n} and s(f(n)) represents (-1) at the number of inversions in f (the sign of the permutation).

It is easy to see that this sum has n! terms, and only some of them are 0-s. In fact for each permutation f that has a fixed point less than n, the corresponding term in the sum above is 0. So the number of terms equal with 1 is in fact the number of permutations (bijective functions) of the set {1,2,...,n} for which f(i)  is different from i, for all i<n.

Let us denote by an the number of permutations which don't have a fixed point at all (not even n). Then the number we are seeking above is obviously an-1 + an.

There are two well known ways to determine an. I shall only present one of them: it is easy to see that an fulfills the recurrence relationship:  an+1 = n(an + an-1.

(The other method to prove the latter is to use the Inclusion - Exclusion principle and we obtain the relationship  an = n! ( 1- 1/1! + 1/2! - 1/3! + ... + (-1)n 1/n!  from which we can check the recurrence above).

Because a1=0, a2=1, a3=2, etc. we prove that a2k+1 is even and a2k is odd, for all k-s. Easy thing to do using the recurrence.

But that means that the sum of two consecutive terms is odd, thus the number of 1-s in the sum above is odd, so the determinant is equal with 1 in Z/2Z. It obviously cannot be 0 in Z.

in Z/2Z the determinant Valentin obtained could have been computed easier.
Add the last line all other lines, then add those new lines to the last and obtain I_n, the det of which is 1.

I thought about that too, but I found more quickly the solution above (although it obviously is much longer and harder to write :D).


geometry
inequalities
inequalities proposed

I that problem,a,b,c are three sides of the striangle ABC,and S is the value of its area

Let $a=x+y, b=y+z, c=z+x$, we have,

$(x+y)^2(x+y+4z)^2\geq 48(x+y+z)xyz$

it is equivalent to this Inequalities,

$(x+y)^2(x+y-2z)^2+12z^2(x-y)^2+12z(x-y)^2(x+y)\geq 0$

which is obvious!

[quote="quangphuc789"]It's so familiar for us to solve the ineq :
    $a^2+b^2+c^2\geq4\sqrt3S+(a-b)^2+(b-c)^2+(c-a)^2$
But the ineq I've just tried is more interesting:
    $2ab+2ac-a^2\geq4\sqrt3S$
Can you solve by the shortest way???[/quote]

  It isn't hard. Because the inequality $a^2+b^2+c^2\geq4\sqrt3S+(a-b)^2+(b-c)^2+(c-a)^2$ is more strict than your inequality. Apply this inequality and from $b^2+c^2 \geq 2bc$ the result is proven immediately.


group theory
abstract algebra
topology
advanced fields
advanced fields unsolved

Compute the fundamental group of the space obtained from two tori $ S^1 \times S^1$ by identifying a circle $ S^1 \times \{x_0\}$ in one torus with the corresponding circle $ S^1 \times \{x_0\}$ in the other torus. (Hatcher 1.2.8)

Attempted Solution:

I think we want to use van Kampen's theorem but I don't really know what cover to use.

Cover it by the two tori. To get an open cover, you can add some open neighbourhood of that common circle to each component.

So, let $ A_i \equal{} T_i \cap W$ (i=1,2) where W is some neighborhood of the identified circle (so $ \pi_1(W) \equal{} \mathbb{Z})$. Then van Kampen tells us that $ \pi_1(X) \equal{} \left( \mathbb{Z} \times \mathbb{Z}\right) * \left( \mathbb{Z} \times \mathbb{Z}\right)/N$ where N is the normal subgroup of  $ \left( \mathbb{Z} \times \mathbb{Z}\right) * \left( \mathbb{Z} \times \mathbb{Z}\right)$ generated by (n,0)(-n,0) where (n,0) is meant as an element of the first $ \left( \mathbb{Z} \times \mathbb{Z}\right)$ factor and (-n,0) is meant as an element of the second $ \left( \mathbb{Z} \times \mathbb{Z}\right)$ factor. Hmm, I'll need to do some more work with free groups before I can simplify that.


geometry
geometric transformation
reflection

Q. A train leaves from Newcastle for Sydney at 7:45 am while another train leaves from Sydney for Newcastle at 8:15 am. The first train arrives in Sydney 40 minutes after the trains pass while the train to Newcastle arrives 1 hour 40 minutes after they pass. Assuming that the trains travel at constant speeds (not necessarily the same), at what time did they pass?

Answer

I will be referring to the trains from Sydney and Newcastle as $ S$ and $ N$ respectively.

Let the distance between Sydney and Newcastle be $ d$

Let the point at which the trains pass be $ x$ from Sydney, let S reach $ x$ at $ t$ minutes after departing.

Let the speed of $ S = v_{1}$

Let the speed of $ N = v_{2}$

$ S$ travels a distance of $ d-x$ in $ 40$ minutes:

$ \Rightarrow v_{1}= \frac{d-x}{40}$

But $ S$ travels a distance of $ x$ in $ t$ minutes:

$ \Rightarrow \frac{d-x}{40}= \frac{x}{t}$

$ \Rightarrow d-x = \frac{40x}{t}$

$ N$ travels a distance of $ x$ in $ 100$ minutes:

$ \Rightarrow v_{2}= \frac{x}{100}$

But $ N$ travels a distance of $ d-x$ in $ t-30$ minutes (since $ N$ departs $ 30$ minutes after $ S$):

$ \Rightarrow \frac{x}{100}= \frac{d-x}{t-30}$

$ \Rightarrow d-x = \frac{x}{100}(t-30)$

Equating both expressions for $ d-x$

$ \frac{40x}{t}= \frac{x}{100}(t-30)$

$ \Rightarrow \frac{40}{t}= \frac{t-30}{100}$

$ \Rightarrow t^{2}-30t-4000 = 0$

$ \Rightarrow (t-80)(t+50) = 0$

$ \Rightarrow t = 80$ minutes

They pass $ 80$ minutes after 7:45 hence they pass at 9:05

Wow,,~ thank you!
I was having a problem because there seemed to be too many variables.
It works out quite nicely.

Thanks heaps!!

[b]sludgethrower[/b],

great. Thanks, man. I used to make similar ones in school but the solution was not as elegant. Cheers

_________________
Open your mind. Let the harmonious reflection of nature moves our hearts.

[url=http://www.printsplusposters.com]Art Prints & Art Posters[/url]


In a equilateral triangle, each side of is stretched from to vertex to a point on the opposite side so that the segment divides the side in the ratio 1:2. What is the ratio of the two triangles? Is this the same for non-equilateral triangles? How about when the ratio of the sides is 1:n ?

It's hard to understand your question, but I believe you mean that a segment is drawn from the vertex to the opposite side, splitting it into a $ 1: 2$ ratio.  In this case, the areas are also in a $ \boxed{1: 2}$ ratio.  This is also true for non-equilateral triangles.  And if the segment divides the side into a $ 1: n$ ratio, then the ratio of the areas is similarly $ \boxed{1: n}$.

This really shouldn't be in intermediate, but in general, if two triangles have the same altitude, then the ratio of their bases is the ratio of their areas.  This is extremely trivial to prove, but a very important fact in plane geometry.

[quote="smellypoopy and Arrange your tan"]In an equilateral triangle, each side is extended from a vertex to a point on the opposite side so that the original side divides the extended side in the ratio 1:2. What is the ratio of the area of the original triangle to the area of the triangle formed with the extended sides? Is this the same for non-equilateral triangles? How about when the ratio of the sides is 1:n ?[/quote]

. . . . . A *

. B * . . . . . C *

Extend the line segment from point B to point A to
point A' (not shown) so that the length of BA' = 2
times the length of BA.

Extend the line segment from point B to point C to
point C' (not shown) so that the length of BC' = 2
times the length of BC.

Connect point A' to point C' with a line segment.

The ratio of the area of the original triangle to the
area of this "extended" triangle is 1:4.

The same would be true for triangles which are
not equilateral.

If the ratio of the sides is $ 1: n$, then the ratio of the
areas of the triangles is $ 1: n^2$.


function
calculus
derivative
linear algebra
matrix
calculus computations

looking at

d2y/dx2)-3(dy/dx)+2y=e^5x

Look for particular of form,
y=Ce^{5x}

y' = 5Ce^{5x}

and y''= 25Ce^{5x}

then 25Ce^{5x} - 3(5Ce^{5x}) + 2(Ce^{5x}) = e^{5x}

giving 12c = 1
and c=1/12

so the general solution is y= Ae^x + Be^{2x} + 1/12 e^{5x}

is this right now ?

[quote]is this right now ?[/quote]
You have the power to answer that question for yourself. You claim that this function is the solution to a particular differential equation. OK, then, compute its derivative and its second derivative, then plug all of that back into the original differential equation. Does it satisfy that differential equation? (In this case, yes it does.)

There's one more test, one which arises from the existence-uniqueness theorem. You didn't give initial conditions, and you're claiming this is the general solution. Furthermore, your solution contains two unknown constants, $ A$ and $ B.$ Now, suppose you imposed completely arbitrary initial conditions at some point. Would you always be able to solve the resulting equations for $ A$ and $ B,$ and uniquely so? (In this case, yes: you get two linear equations in the two unknowns, and the determinant of the coefficient matrix is nonzero.)


vector
linear algebra
linear algebra solved

Hi everyone, sorry that my first post is a proof, needless to say I've been a lurker here for quite some time, wish I could help answer some of these tough problems posted in this section, but to put it simply, I'm just not smart enough. (hehe...yet!!)

Let $v$ be in $R^n$ and $W$ be a subspace of $R^n$.  Suppose that $w_1$ and $w_2$ are vectors in $W$ such that both $v - w_1$ and $v - w_2$ are orthogonal to every vector in $W$.  Prove that $w_1$ = $w_2$.
I am supposed to consider the vector $w_1-w_2$ in $W$.

While I cannot use the Theorem stating that there exists a unique vector $w$ in $W$ and a unique vector $z$ in $W^\bot$ such that $v = w + z$.

Here is what I have started with,

My thoughts are that I have to use the Gram-Schmidt process, however I don't think I'm on the right track.

$w_1 \bullet (v - w_1)$ = $w_1 \bullet v - w_1 \bullet w_1$ = $w_1 \bullet v - ||w_1||^2$

Where would I go from here? I appreciate any and all help!! Thanks guys!

[b]Theorem[/b] In an inner product space $V$, if $ \langle y,x\rangle = \langle z,x\rangle$ for all $x\in V$ then $y=z$.

So you are saying that if I take, $<w_1 , v - w_1>$ = $<w_2 , v - w_2>$ then I will get $w_1 =  w_2$....

I'm a bit confused... :?

If anyone can help, I would greatly appreciate it, I have a big test tomorrow and this proof is supposed to be on it.

So I hear anyway,

 Thanks guys!!


Here are a couple of questions on my birthday(Today!!) :D

1)solve-
$z^6+z^4+z^3+z^2+1=0$

2)Find all possible solutions for x,y,z-
$x^2+y^2+z^2=2xyz$

Solution method for first one:
Set $w = z + \frac{1}{z}$, then multiply the whole equation with $z^{-3}$ and the equation will transform into $w^3 + 2w + 1 = 0$.

For the second one I'm guessing x + y + z = 0, based on some inequality that I can't quite recall.

You can factor the first one into $(z^2-z+1)(z^4+z^3+z^2+z+1)=0$, whose roots are fairly simple to find if you know DeMoivre's Theorem (Take roots of $z^3=-1$ except for $-1$ and take fifth roots of unity except for $1.$

I THINK, but am not sure, that you use AM-GM for two variables on the second one.  Because $a+b\geq 2\sqrt{ab}$, set $a=x^2$ and $b=y^2+z^2$ and you get that $x^2+y^2+z^2 \geq 2xyz$ with equality holding when $x^2=y^2+z^2.$  When you then set $a$ to $y^2$ and later $z^2$, you get three equations:

$x^2=y^2+z^2$

$y^2=x^2+z^2$

$z^2=x^2+y^2$

When you add these, you get $x^2+y^2+z^2=2(x^2+y^2+z^2)$, so therefore, $x^2+y^2+z^2 =0.$  This can only occur when $x=y=z=0.$  Someone please critique this, as I'm not too sharp when it come to inequalities.

And a very [size=150][color=red]Happy Birthday [/color][/size] to you!

for the second one, you can show that x=y=z by descent, showing that they are divisible by 2's indefinately, and the only number that satisfies that is 0

[quote="Kalle"][quote="pkothari13"]You can factor the first one into $(z^2-z+1)(z^4+z^3+z^2+z+1)=0$, whose roots are fairly simple to find if you know DeMoivre's Theorem (Take roots of $z^3=-1$ except for $-1$ and take fifth roots of unity except for $1.$[/quote]
Very nice solution, but a couple of questions: 1) How did you find the factorization? 2) I understand why fifth roots of unity will be factors of the right factor, but why will roots of $z^3 = -1$ solve $z^2 - z + 1 = 0$? Please explain. Seems like a useful technique.

edit: is it because if $s = 1 - z + z^2$, then $s(-z - 1) = (-z + z^2 - z^3) - (1 - z + z^2) = -z^3 - 1)$. And so $s = \frac{z^3 + 1}{z + 1}$ and this is only 0 when z^3 = -1 (but not z = -1). Yeah, that's smart. Thanks![/quote]

well whenever you see $z^n + z^{n-1} + ....1 = 0$, then by the geometric series formula, you know that sum equals $\displaystyle\frac{z^{n+1}-1}{z-1}$, and $z^k = 1$ hints at finding the roots of unity.  However, since you divide by $z-1$  you know you have to exclude 1 as a root.

I knew the other factorization by recognizing $a^3+b^3 = (a+b)(a^2-ab+b^2)$.

Does my solution for the second question work?  And altheman, can you show me your method?

[quote="Arne"]Do you want to find all INTEGER solutions for the second equation, shadysaysurspammed?

By the way, happy birthday :)[/quote]

Thanks :D
And i think you have to find all the solutions,including integral.
Guys,i really can't understand the second one,coulkd someone please explain it all at one time?

[quote="Arne"][quote="pkothari13"]I THINK, but am not sure, that you use AM-GM for two variables on the second one.  Because $a+b\geq 2\sqrt{ab}$, set $a=x^2$ and $b=y^2+z^2$ and you get that $x^2+y^2+z^2 \geq 2xyz$ with equality holding when $x^2=y^2+z^2.$  When you then set $a$ to $y^2$ and later $z^2$, you get three equations:

$x^2=y^2+z^2$

$y^2=x^2+z^2$

$z^2=x^2+y^2$

When you add these, you get $x^2+y^2+z^2=2(x^2+y^2+z^2)$, so therefore, $x^2+y^2+z^2 =0.$  This can only occur when $x=y=z=0.$  Someone please critique this, as I'm not too sharp when it come to inequalities.[/quote]

No, this is definitely wrong.

Shadysaysurspammed, I am pretty sure you only need to find all integer solutions. It's pretty much impossible to find all real solutions the the equation (three variables and just one equation...)[/quote]

O.k then do so because i dont have the answer to it.

[quote="Arne"]
No, this is definitely wrong.

Shadysaysurspammed, I am pretty sure you only need to find all integer solutions. It's pretty much impossible to find all real solutions the the equation (three variables and just one equation...)[/quote]

meh, it was worth a try.  why doesn't it work? is it because am-gm only works for positive integers?

[quote="Altheman"]
$ 4n^2+y^2+z^2=4nyz $ since the right side is divisible by 4, the other side is also,  it is also divisible by 2,...... therefore we substitute $ y = 2m $ and $ z = 2p $, giving us
$ 4n^2+4m^2+4p^2=8mnp $
$ n^2+m^2+p^2=2mnp $
the variables are essentially the same as before , so the process can be repeated indefinately, and that means that x,y, and z are integers that are divisible by $ 2^{\infty} $ and the only integer with that property is 0, therefore
$ x=y=z=0 $[/quote]

Nice aregument. I believe there is a typo, the equation should come out to be:
$ 4n^2+4m^2+4p^2=16mnp $ (istead of $ 4n^2+4m^2+4p^2=8mnp $)
But  the argument given above still holds as RHS  of  $ n^2+m^2+p^2=4mnp $ is still even.

yes, i believe you are correct, i guess that i missed that

[quote="Altheman"][quote="shadysaysurspammed"]
2)Find all possible solutions for x,y,z-
$ x^2+y^2+z^2=2xyz $[/quote]
since the right side is even, the left side is even, then the squares can be even, odd, odd or even, even, even, so one of them must be even so we substitute $ x = 2n $, giving us
$ 4n^2+y^2+z^2=4nyz $ since the right side is divisible by 4, the other side is also, and the residues for perfect squares in mod 4 are 0 and 1, and since it has to be divisible by 4, it has to have a residue of 0 in mod 4, since it is divisible by 4, it is also divisible by 2, therefore we substitute $ y = 2m $ and $ z = 2p $, giving us
$ 4n^2+4m^2+4p^2=8mnp $
$ n^2+m^2+p^2=2mnp $
the variables are essentially the same as before , so the process can be repeated indefinately, and that means that x,y, and z are integers that are divisible by $ 2^{\infty} $ and the only integer with that property is 0, therefore
$ x=y=z=0 $[/quote]

Cool solution.niceone.
thanks guys. :lol:


[quote="Scrambled"][quote="AntonioMainenti"]Write a polynomial $p(x)$ of any degree $n$ with nonnegative integer coefficients. Tell me $p(1)$, then choose an integer $k>p(1)$ and tell me $p(k)$, and I will tell you what $p(x)$ is (as long as you don't make some polynomial with rediculously large degree or coefficients.. keep them under 10 to be safe).

For those who know why this works and want to post it, please hide your solutions!

[hide="no peeking until you try it.."]Let the polynomial be \[p(x)=a_0x^n+\ldots+a_{n-1}x+a_n.\] Then $p(1)=m,$ where \[m=\sum_{i=0}^na_i.\] Since the coefficients are nonnegative, we have $m\ge{a_i}$ for $0\le{i}\le{n}.$ Pick an integer $k>m,$ so that $k>a_i$. Now \[p(k)=a_0k^n+\ldots+a_{n-1}k+a_n=(a_0\cdots{a}_n)_k,\] the last part being the [i]unique[/i] base $k$ representation of the number since $k>a_i.$

Thus when we have $p(k)$ written in base $k,$ the polynomial's coefficients are written out in order for us, so writing out the polynomial is a piece of cake![/hide][/quote]
er are you sure about this method? what if i give you x^2+3 and plug in 1, and 2, and give you 3x+1, whcih gives us the same thing[/quote]If $p(x)=x^2+3$, then $p(1)=4$, but $2\not>4$. The second value for $x$ needs to be greater than $p(1)$. Also, where did you get $3x+1$?

I'm pretty sure this is more interesting than how much attention you guys are giving it..  :?

[quote="white_horse_king88"]Cool. I'm interested.

Here. P(1) = 2.

P(3) = 12.

I hope I followed directions right.[/quote]$P(x)=x^2+x$.

[quote="WindSlicer"]Here I go

$P(1) = 7$
$P(8) = 1631$[/quote]Are you sure about $P(1)$? I think you might have given $P(0)$. My answer is $P(x)=3x^3+x^2+3x+7$.

[quote="WindSlicer"]Ah, blah, I missed the 'nonnegative' coefficients.. bleh :P But nice job at guessing that one.. lemme make a new one (sorry for the mistake)

$P(1) = 20$
$P(21) = 65820$[/quote]That's better  :D  $P(x)=7x^3+2x^2+5x+6$.

Oh, I had negative coefficients too. Good thing it worked out.  :D

The idea can be used to solve my coin problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=coin&t=13133

[quote="amirhtlusa"]yes, what you are saying makes sense. because when you find the base $k$ of $p(k)$ you do the opposite of exactly what the reader does to obtain $p(k)$...
now do you know why $k$ has to be bigger than $p(1)$?[/quote]

its because otherwise the base will be smaller than the numbers, which does not work since the numbers in base n have to be less than n, right?

so you could also ask for f(n) where n greater than all the coefficients

[quote="amirhtlusa"][quote]its because otherwise the base will be smaller than the numbers, which does not work since the numbers in base n have to be less than n, right? [/quote]
that's correct.

[quote="nat mc"]Can someone explain in a different way?  I don't seem to understand it.[/quote]
so basically you want others to make their own polynomial then they have to choos a number such as $k$ that $p(1)<k$.
they tell you their $p(k)$. then all you need to do is to convert the number that they have given you to the base $k$ (for example if they say $p(3)=12$ the base $3$number would be $(110)_3$. means that the polynomial is $1x^2+1x+0=x^2+x$[/quote]
Then why do you need $p(1)$

[quote="dangvh"]haha i hope i didnt over do it  :D

f(1)=11

f(12)=3484261

I hope i did this right... ;)[/quote]

$x^6+2x^5+4x^2+3x+1$

[quote="dangvh"]haha i hope i didnt over do it  :D

f(1)=11

f(12)=3484261

I hope i did this right... ;)[/quote]
$p(x)=x^6+2x^5+4x^2+3x+1$

[quote="amirhtlusa"]you do not need the $P(1)$, it is unusfull to you. all you need to know is that the number $k>p(1)$.[/quote]
Ok I get it now. Thanks

People will be even more impressed if you don't ask for f(1). Just say: [i]think of a polynomial with natural coefficients and compute f(1). Now take any integer n>f(1) and tell me n and f(n). I will tell you your polynomial.[/i]


You can find the pas problems of IMC International Mathematics
Competition at

http://www.imc-math.org/

Can anybody tell the criteria and how to participate in a IMC

it is simple. you must be an undergraduate student.
you then have to announce the organizers by e-mail of your participation and have ready a few hundred dollars  (or maybe the university will pay for a larger delegation - including a teacher for grading the papers).

[quote="Peter VDD"]It would be nice if you could create forums for the other past IMC's :) [so we can discuss alternate solutions and so][/quote]You can discuss all the previous problems in the main IMC forum ;)


geometry
3D geometry
sphere
vector
induction
geometry unsolved

There are $ n$ points ($ n \geq 4$) on a sphere with radius $ R$, and not all of them lie on the same semi-sphere. Prove that among all the angles formed by any two of the $ n$ points and the sphere centre $ O$ ($ O$ is the vertex of the angle), there is at least one that is not less than $ \displaystyle 2 \arcsin{\frac{\sqrt{6}}{3}}$.

Can anyone help?

My approach was to transform this to vector language for unit vectors on the sphere
and apply induction but I failed. :)

First, note that $\cos (2\arcsin (\frac{\sqrt{6}}{3})) =\frac{-1}{3}$, so we need to show that the cosine of at least one such angle is smaller than $\frac{-1}{3}$.
Suppose to the contrary that the cosines of all such angles are greater than $\frac{-1}{3}$.
Consider any three points from those $n$ points.
If there's a semi-sphere that contain all three points, there's a point from other $n-3$ point that lie on the opponent semi-sphere.
If there's no semi-sphere that contain all three points, choose any other points from other $n-3$ points.
The vectors from the origin to four points we chose will partition the whole sphere into $4$ disjoint parts.
There's a part with volume at least $\frac{1}{4}$ of the sphere.
Let the three vectors from the origin that form this part pointed at three points $V_1,V_2,V_3$ on the sphere.
Let three spherical angle of spherical triangle $V_1V_2V_3$ are equal to $A,B,C$ degree.
It's well known that the volume of the part is equal to $\frac{A+B+C-180}{360}$ times the total volume of the sphere.
Hence, $A+B+C\geq 360$. And note that $A,B,C\in (0,180]$.
Also, by spherical trigonometry, $\cos (V_iOV_j)$ where $i\neq j\in \{ 1,2,3\}$ is equal to $\frac{\cos (A)+\cos (B) \cos (C)}{\sin (B)\sin (C)}$ or two other cyclic permutation of $A,B,C$.

If there exists, WLOG it's $A$, angle $A$ that $\cos (A)>0$, which means $A<90$.
We get that $\cos (B)\leq 0$, otherwise $B<90$ will leads to $C>180$, impossible. Similarly, $\cos (C)\leq 0$.
In summary, $A\in (0,90)$ and $B,C\in [90,180]$.
We know that $\frac{\cos (B)+\cos (A) \cos (C)}{\sin (A)\sin (C)} > \frac{-1}{3}$, which gives $3\cos (B)+2\cos (A) \cos (C)\geq -\cos (C-A)$.
So, $-\cos (C-A)\leq 0$, this gives $C-A\leq 90$. Similarly, $C-B\leq 90$.
Hence, $360\leq A+B+C\leq 3A+180\Rightarrow A\geq 60$.
So, $-cos (C-A)\leq 3\cos (B)+2\cos (60)\cos (C)\leq 0+\cos (C)=\cos (C)$.
We get $\cos (C) +\cos (C-A)=2\cos (C-\frac{A}{2})\cos (\frac{A}{2}) \geq 0\Rightarrow C-\frac{A}{2}\leq 90$. Similarly, $B-\frac{A}{2}\leq 90$.
Hence, $360\leq A+B+C\leq 2A+180\Rightarrow A\geq 90$, contradiction.

Now, we can conclude that $\cos (A),\cos (B),\cos (C)\leq 0$.
In other words, $A,B,C\in [90,180]$.
Let $p,q,r\in [0,90]$ that $p=180-A,q=180-B,r=180-C$.
We've $p+q+r=540-(A+B+C)\leq 180$. And WLOG $p\geq q\geq r$.
Transform our inequalities to $\cos (p)\cos (q) -\cos (r) > \frac{-1}{3}\sin (p) \sin (q)$ and two other cyclic permutations.
Hence, $\cos (p)\cos (q) +\frac{1}{3}\sin (p) \sin (q)\geq \cos (r)$.
By C-S, we get $\cos^2 (r)\leq \Big( \cos^2 (p)+\frac{1}{3}\sin^2 (p)\Big) \Big( \cos^2 (q)+\frac{1}{3}\sin^2 (q)\Big) = \Big( 1-\frac{2}{3}\sin^2 (p)\Big) \Big( 1-\frac{2}{3}\sin^2 (q)\Big)$.
Since $p\geq q\geq r$, we easily get $p\geq 60\Rightarrow \sin (p)\geq \frac{\sqrt{3}}{2}$ and $\cos (r)\geq \cos (q)$.
Hence, $\cos^2 (q)\leq \frac{1}{2}\times \Big( 1-\frac{2}{3}\sin^2 (q)\Big) \Rightarrow 1\geq 2\cos^2 (q)+\frac{2}{3}\sin^2 (q)\Rightarrow \cos (q)\leq \frac{1}{2}$.
This gives $q\geq 60$. So, $\sin (q)\geq \frac{\sqrt{3}}{2}$ too.
So, $\cos^2 (r)\leq \frac{1}{4}\Rightarrow \cos (r)\leq \frac{1}{2}$.
This gives $r\geq 60$. But we've $p+q+r\leq 180$, so $p=q=r=60$.
But that case gives us $\cos (p)\cos (q) -\cos (r) = \frac{-1}{3}\sin (p) \sin (q)$, contradiction since the sign must be strictly greater.
This contradiction completes the proof.


vector
calculus
integration
percent
calculus computations

Evaluate the following integral in two ways: 1) $ \int_{C}\textbf{F}\cdot\mbox{d\textbf{s}}$ and 2) by finding its scalar potential.  I'm 99% percent sure this is a conservative field.
$ \int_{C}(2y\minus{}3z)dx\plus{}(2x\plus{}z)dy\plus{}(y\minus{}3x)dz$;
$ C$ is the line segment from the point $ (0,0,0)$ to $ (0,1,1)$ followed by the line segment from $ (0,1,1)$ to $ (1,2,3)$.

I keep getting different answers for the line integral and the scalar potential so please, do both methods.  Thank you.

The field must be conservative, or the problem would not be phrased that way. The potential is $ f(x,y,z)\equal{}2xy\minus{}3xz\plus{}yz$, so the answer should be $ f(1,2,3)\minus{}f(0,0,0)\equal{}4\minus{}9\plus{}6\equal{}1$.

Are we allowed to just skip the intermediate point?  If so, is it because it is a common point of the segments?  That is the same computation I got.  And by means of direct evaluation of the line integral, how would we evaluate the two separate components since there are two distinct segments, $ (0,t,t)$ and $ (t,t\plus{}1,2t\plus{}1)$?


calculus
integration
calculus computations

:mad:

[integrate]xe^(-x^2) dx
  from 1 to infinity

\[ \int_1^\infty x e^{\minus{}x^2} dx \equal{} \frac{1}{2e}.
\]

Let $ u \equal{} \minus{}x^2$ so that $ du \equal{} \minus{}2x \, dx$.


limit
integration
calculus
calculus computations

$ \lim_{n\to \plus{}\infty} \frac{\frac{1} {\sqrt 1}\plus{}\frac{1}{\sqrt 2}\plus{}\cdots\plus{}\frac{1}{\sqrt n}}{\sqrt n}$

 :D

[quote="vinskman"]$ \lim_{n\to \plus{} \infty} \frac {\frac {1} {\sqrt 1} \plus{} \frac {1}{\sqrt 2} \plus{} \cdots \plus{} \frac {1}{\sqrt n}}{\sqrt n}$

 :D[/quote]

Riemman Sum:

$ \lim_{n\to \plus{} \infty} \frac {\frac {1} {\sqrt 1} \plus{} \frac {1}{\sqrt 2} \plus{} \cdots \plus{} \frac {1}{\sqrt n}}{\sqrt n}$

$ \equal{}\lim_{n\to\infty} \frac{1}{n}\sum_{k\equal{}1}^n \frac{1}{\sqrt{\frac{k}{n}}}\equal{}\int_0^1 \frac{1}{\sqrt{x}}\ dx\equal{}2$.


Prove that, for any positive real numbers $a,b,c$ and $k\ge 2$,
\[ \frac{a^k+b^k}{a+b}+\frac{b^k+c^k}{b+c}+\frac{c^k+a^k}{c+a}\ge\sqrt{ \frac{3(a^k+b^k+c^k)(a^{k-1}+b^{k-1}+c^{k-1})}{a+b+c} } \]

Seem to be easy.
Chebyshev's and Jensen's might work.

hmm... unsuccesful, this generalization from Cezar's problem may cost a lot of calculation for its solution.
Indeed, SOS again.

This is so nice  :) , but so difficult  :blush: .

Has anyone tried?  :wink:

Thank you very much.


Take two glasses, A and B. A is initially full of water and B wine. Remove a teaspoon of wine from A and put it in B, then a teaspoon of the mixture in B and put it in A.

In what proportions are the water and wine now mixed in A and B?

What are the proportions of water and wine in each glass after n iterations?

It depends on the volume of the glasses and the spoon.

It is quite interesting to see a brainteaser in Number Theory Open Questions :rotfl:

You're right. I got distracted, somehow.

[hide] this is actually an interesting riddle but actually the ratio is 1:1[/hide]

I don't see how.

Let's assume the glass is 1 cup, and 1 cup= 48 teaspoons

the mixture is 1:48...

I'm still working on this problem. After n iterations n -> oo, we should get a geometric progression.


calculus
integration
function
real analysis
real analysis theorems

Given $ \triangle \equal{} [a,b] \times [c,d] \subset R^2$ and $ f: \triangle \rightarrow R$ a function defined in $ \triangle$ .
Note $ \alpha, \beta : [c,d] \rightarrow R$
Also given:
    1> f is continuous in $ \triangle$
    2> $ \forall x \in [c,d] , \alpha(x) \in [a,b], \beta(x) \in [a,b]$
    3> $ \alpha , \beta$ are derivable in $ [c,d]$
    4> $ \frac{\partial f}{\partial x}$ is continuous in $ \triangle$

Then, the function $ g : [c,d] \rightarrow R, x \rightarrow g(x) \equal{} \int^{\beta(x)}_{\alpha(x)} f(t,x)dt$ is derivable in $ [c,d]$ and for all $ x \in [c,d]$, we have :
$ g'(x) \equal{}  \int^{\beta(x)}_{\alpha(x)}\frac{\partial f}{\partial x}(t,x)dt \plus{} \beta '(x) f(\beta(x),x)\minus{}\alpha '(x) f(\alpha(x),x)$

I found this theorem in a book but the proof in this book is wrong and I really want to see how to prove it. Cound anyone give me a beautiful proof ?

Could anyone help me, please ! I found it useful because we can use this theorem to calculate the derivation of an integral which depends on a variable. I think proving it is also interesting, please help !

PS: I managed to correct the proof in the book. There was just a typo that lead to the wrong solution. Thanks.


inequalities
calculus
derivative
trigonometry
inequalities solved

This is easy:

Since -1<x,y,z<1, we have 0<1+x, 1-x  and (1+x)(1-x)=1-x^2<=1, and similarly. So

1/(1+x)(1+y)(1+z) >= (1-x)(1-y)(1-z):=a>0.

Thus we want to show that

a+1/a>=2,

which is well known to be true with "=" if and only if a=1. Thus, the proposed inequality is true and we have "=" if and only if x=y=z=0.

[quote="hxtung"]Let real numbers $ x,y,z\in ( \minus{} 1,1)$. Prove that

$ \frac {1}{(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)} \plus{} \frac {1}{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)}\geq 2$.[/quote][b]Proof[/b] $ M_{ \minus{} 3}\left[G(1 \minus{} x,1\minus{}y,1\minus{}z),G(1 \plus{} x,1\plus{}y,1\plus{}z)\right]$

$ \leq A\left[A(1 \minus{} x,1\minus{}y,1\minus{}z),A(1 \plus{} x,1\plus{}y,1\plus{}z)\right] \equal{} 1$.

Actually, the following stronger inequalities holds :

$ \frac {1}{3(1 \minus{} x)^3 \plus{} 3(1 \minus{} y)^3 \plus{} 3(1 \minus{} z)^3 \plus{} 23(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}$

$ \plus{} \frac {1}{3(1 \plus{} x)^3 \plus{} 3(1 \plus{} y)^3 \plus{} 3(1 \plus{} z)^3 \plus{} 23(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)}\geq \frac {1}{16},$

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

have f(x,y,z) = LHS
Show that it is conves with respect to each variable
Since lim x to -1 makes 1/(1+x) go to infinity
and lim x to 1 makes 1/(1-x) go to infinity
take the derivative set equal to zero for minimum
minimum = 2

Since $ \minus{}1<x,y,z<1$ we get  $ 1\plus{}x, 1\plus{}y, 1\plus{}z, 1\minus{}x, 1\minus{}y, 1\minus{}z >0$
By AM-GM  we have
$ \frac 1 {(1\minus{}x)(1\minus{}y)(1\minus{}z)} \plus{} \frac 1 {(1\plus{}x)(1\plus{}y)(1\plus{}z)} \geq 2\sqrt {\frac 1 {(1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2)}}$
But $ (1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2) \leq 1$

Hence,$ \frac 1 {(1\minus{}x)(1\minus{}y)(1\minus{}z)} \plus{} \frac 1 {(1\plus{}x)(1\plus{}y)(1\plus{}z)} \geq 2$
Equality holds iff $ x\equal{}y\equal{}z\equal{}0$

[quote="hxtung"]Let $ x$, $ y$, $ z$ be real numbers in the interval $ \left]\minus{}1, 1\right[$. Prove that
$ \frac{1}{\left(1\minus{}x\right)\left(1\minus{}y\right)\left(1\minus{}z\right)} \plus{} \frac{1}{\left(1\plus{}x\right)\left(1\plus{}y\right)\left(1\plus{}z\right)}\geq 2$.[/quote]
by AM-GM we have
$LHS\ge 2\sqrt{\frac{1}{(1-x^2)(1-y^2)(1-z^2)}}$
Now we want to prove $(1-x^2)(1-y^2)(1-z^2)\le 1$
Have $x^2\ge 0\Rightarrow 1-x^2\le 1\Rightarrow (1-x^2)(1-y^2)(1-z^2)\le 1$

[quote=hxtung]Let $ x$, $ y$, $ z$ be real numbers in the interval $ \left]\minus{}1, 1\right[$. Prove that
$ \frac{1}{\left(1\minus{}x\right)\left(1\minus{}y\right)\left(1\minus{}z\right)} \plus{} \frac{1}{\left(1\plus{}x\right)\left(1\plus{}y\right)\left(1\plus{}z\right)}\geq 2$.[/quote]
[url]https://artofproblemsolving.com/community/c6h299899p1778577[/url]
[b]Mathematics Magazine[/b]


If $ \frac{a\plus{}b}{b\plus{}c}\equal{}\frac{c\plus{}d}{d\plus{}a},$ then

a) $ a\equal{}c$

b) $ a\plus{}b\plus{}c\plus{}d\equal{}0$

c) $ a\equal{}c$ or $ a\plus{}b\plus{}c\plus{}d\equal{}0$

d) $ a\plus{}b\plus{}c\plus{}d\equal{}0$ if $ a\equal{}c$

e) $ a(b\plus{}c\plus{}d)\equal{}c(a\plus{}b\plus{}d)$

Note: Justify your answer  :wink:

[hide="Answer"]
$ \frac{a\plus{}b}{b\plus{}c}\equal{}\frac{c\plus{}d}{d\plus{}a}\\
\Rightarrow bc\plus{}bd\plus{}c^2\plus{}cd\equal{}ad\plus{}a^2\plus{}bd\plus{}ab \\
\Leftrightarrow b(c\minus{}a)\plus{}d(c\minus{}a)\plus{}(c\minus{}a)(c\plus{}a)\equal{}0 \\
\Leftrightarrow (c\minus{}a)(a\plus{}b\plus{}c\plus{}d)\equal{}0$
Which means $ \boxed{c\equal{}a}$ or $ \boxed{ a\plus{}b\plus{}c\plus{}d\equal{}0}$
[/hide]

[hide="Another way"]
Take the cross products:

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

$ a^2\plus{}ab\plus{}ad\plus{}bd\equal{}c^2\plus{}cb\plus{}cd\plus{}bd$

$ a^2\plus{}ab\plus{}ad\equal{}c^2\plus{}cb\plus{}cd\plus{}bd$

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

If $ a\equal{}c$, then the two sides are equal.

If $ a\plus{}b\plus{}c\plus{}d\equal{}0$, the two sides are also equal:

We rewrite $ (a\plus{}b\plus{}d)$ as $ 0\minus{}c$ and $ (c\plus{}b\plus{}d)$ as $ 0\minus{}a$:

$ a(0\minus{}c)\equal{}c(0\minus{}a)$

$ \minus{}ac\equal{}\minus{}ca$, which is true by commutativity.

Therefore, either $ a\equal{}c$ or $ a\plus{}b\plus{}c\plus{}D\equal{}0$, so $ \boxed{C}$.

We can produce a simple counterexample to disprove answer $ D$:

Let $ a\equal{}c\equal{}1$ and $ b\equal{}d\equal{}2$.

$ \frac{1\plus{}2}{2\plus{}1}\equal{}\frac{1\plus{}2}{2\plus{}1}$, where $ a\plus{}b\plus{}c\plus{}d$ does not equal $ 0$.
[/hide]

[quote="Discrete_Math"]
$ a^2 \plus{} ab \plus{} ad \equal{} c^2 \plus{} cb \plus{} cd \plus{} bd$
[/quote]
I think there's not $ bd$ at $ RHS$


Hey guys,

Anyone know of some prestigious/competitive piano competitions? I personally don't play piano anymore, but a friend was looking for some. Please also offer your opinions on these competitions if you've participated.

Thanks in advance.

Where do your friend live? I don't want your friend to go cross country to compete  :lol:

Greater Houston. But I was kinda thinking of national competitions. Lol, especially those where you submit tapes or something to qualify or w/e.

This isn't exactly a competition, but your friend can go try out for the International Guild of Musicians or something like that...you have to memorize 16 songs and perform them (at least for piano)


iso for the #6 on the AIME1 (2008) i don't get the solution at all, so do u guys think i should take an intermediate number theory course at AoPS?  Or can u guys help me to refer to some textbooks that teaches it?
http://www.artofproblemsolving.com/Wiki/index.php/2008_AIME_I_Problems/Problem_6

lol I can get the problem but I still dont get azjps' solution...

That would be my fault - sorry. My solution was worded weirdly and over-formalized. I was going through some of the other solutions and fixing them, but I forgot this one; is it clearer now?

thanks, its quite clear now, but i still wondering how did u get the pattern in the beginning of the problem


In a triangle ABC, let D be a midpoint on BC. If  angle ADB= 45degrees and angle ACD=30 degrees, determine the angle BAD.

Construct $BE \perp AC$ with $E$ on $AC$ and join $ED$.
Clearly $\angle{DAE}=15^{\circ}$
Moreover, $BEC$ is a right angled triangle and $D$ is the midpoint of the hypotenuse $\Longrightarrow BD=DC=DE$.
$\Longrightarrow \angle{DEC}=\angle{DCE}=30^{\circ}$
$\Longrightarrow \angle{BED}=\angle{BDE}=60^{\circ} \Longrightarrow$ $BED$ is an equilateral triangle.
So $AE=DE=BD=DC=BE$

Now, let $\angle DAB=x$
Since $AE=BE$, it follows that $\angle{ABE}=\angle{BAE}=x+15^{\circ}$
Also $\angle{CBA}+\angle{BAE}+\angle{ACB}=180^{\circ}$
$\Longrightarrow (60^{\circ}+x+15^{\circ})+(x+15^{\circ})+30^{\circ}=180^{\circ}\Longrightarrow x=30^{\circ}$

Hence $\angle DAB=30^{\circ}$.

[quote="prime04"]_______________________________
Construct BE perpendicular to AC with E on AC
Join ED
clearly ang(DAE)=15'

also,BEC is a right-triangle
hence D being mid-pt of hypotenuse,BD=DC=DE
thus ang(DEC)=ang(DCE)=30'
implies ang(BED)=ang(BDE)=60' gives ang(ADE)=15'
since ang(ADE)=ang(DAE)=15' we have AE=DE=BD=DC

we get that BED is an equilateral triangle
hence:
AE=DE=BD=DC=BE
let ang(BAD)=x
since AE=BE we have ang(ABE)=ang(BAE)=x+15

by triangle sum property:
ang(CBA)+ang(BAE)+ang(ACB)=180'

-> (60+x+15)+(x+15)+30=180'
->x=30'

hence ang(BAD)=30'[/quote]

thanks prime04

Sorry to interupt but I think it's an RMO question....although easier than the rest...


If $ x$ satisfies the equation $ \arccos x\plus{}\arccos2x\plus{}\arccos 3x\equal{}\pi$, for what ordered triple of integers $ (a,b,c)$ will $ x$ also satisfy $ ax^3\plus{}bx^2\plus{}cx\minus{}1\equal{}0$?

$ \arccos x\plus{}\arccos 3x\equal{}\pi\minus{}\arccos 2x$

Apply cos to both sides:

$ 3x^2\minus{}\sqrt{1\minus{}x^2}\cdot\sqrt{1\minus{}9x^2}\equal{}\minus{}2x$

$ 3x^2\plus{}2x\equal{}\sqrt{9x^4\minus{}10x^2\plus{}1}$

Sqare both sides:

$ 12x^3\minus{}14x^2\minus{}1\equal{}0\Rightarrow a\minus{}12, \ b\equal{}\minus{}14, \ c\equal{}0$


algebra
polynomial
function
calculus
derivative
topology
complex analysis

Maybe this problem is way too hard, so please don't flame me I just think that this special case of the general case is much simpler (the general case was only proved in 1984).
Let $ D$ be the unit disk and $ f$ be a one-to-one polynomial with $ f(0) \equal{} 0$ and $ f'(0) \equal{} 1$ on $ D$.
If $ \deg f \equal{} n \geq 1$ and $ a_0,...,a_n$ are its coefficients prove that $ |a_k|\leq k$ for $ 0\leq k\leq n$.

Alas, the general case can be reduced to this special case very easily: if $ f(z)$ is a one-to-one analytic function in the disk such that $ f(0)\equal{}0$, $ f'(0)\equal{}1$, then $ r^{\minus{}1}f_n(rz)$ is also one-to-one for fixed $ r<1$ and sufficiently large $ n$ (here $ f_n$ is the $ n$-th Taylor polynomial of $ f$). So, if you can do this "special case", you can do the deBranges theorem as well at no extra cost.

[quote="fedja"]Alas, the general case can be reduced to this special case very easily: if $ f(z)$ is a one-to-one analytic function in the disk such that $ f(0) \equal{} 0$, $ f'(0) \equal{} 1$, then $ r^{ \minus{} 1}f_n(rz)$ is also one-to-one for fixed $ r < 1$ and sufficiently large $ n$ (here $ f_n$ is the $ n$-th Taylor polynomial of $ f$). So, if you can do this "special case", you can do the deBranges theorem as well at no extra cost.[/quote]
What is $ r$, the radius of the disk? It seems you are expanding (or shrinking) a general disk onto the unit disk. So that reduces the general case into a more special case on unit disks. But what I do not understand is how its Taylor polynomials would imply the general case.  :oops: , thank you for your answer.

I have a different question on this problem. What is so special about shlifit functions? I am guessing if $ f(z) \equal{} a_0\plus{}a_1z\plus{}...$ is analytic then it can be transformed into a shlifit function by $ a_1^{\minus{}1}(f(z)\minus{}a_0)$ (if $ a_1\not \equal{} 0$), and so almost any analytic function can be thought of these special terms. However, what about the special case $ a_1\equal{}0$? Is it uninteresting in some way?

[quote="kouboy"]what is a shlifit function? i only heard of schlicht[/quote]
That's what I meant.  :oops: . I function is a schlicht function on an open set $ \Omega$ when $ f$ is analytic and $ f(0) \equal{} 0$ and $ f'(0) \equal{} 1$ and $ f$ is one-to-one.

[quote="jmerry"]In that special case, it's not even locally one-to-one.[/quote]
What do you mean?

The equation $ f(x)\equal{}a_0$ has a double root at zero, so $ f$ is two-to-one on neighborhoods of the origin. If we're looking for injective complex functions, zeros of the derivative are the first thing we can rule out.

[quote="jmerry"]The equation $ f(x) \equal{} a_0$ has a double root at zero, so $ f$ is two-to-one on neighborhoods of the origin. If we're looking for injective complex functions, zeros of the derivative are the first thing we can rule out.[/quote]
Oh, you are talking about the case when $ a_1 \equal{} 0$ in the infinite series expansion. When you said "special case" I was think polynomials in general.


I don't know where to ask this but oh well.

Given the coordinates of 2 points on 2 rectangles (4 points total) the lower left vertex and the upper right vertex of each rectangle, is there a general term/forumla to determine a) if they overlap and b) (if a is true) the area of the overlapping region?

Thanks

Yes.  Call the coordinates of the lower left point of rectangle 1: $(x_1, y_1)$ and the upper left of rect 1 $(x_2, y_2)$.  Then the lower left of rect 2 call $(x_3, y_3)$ and the upper right call $(x_4, y_4)$.  Then if $y_4>y_1$, the overlapping rectangle's top right coordinate's y value is: $max(y_2, y_3)$, otherwise it is $max(y_1, y_4)$.
And the x value of that coordinate is, if $x_4>x_1$, $max(x_2, x_3)$, otherwise it is $max(x_1, x_4)$.

Do something similar for the lower left coordinate of the new rectangle.  These two coordinates will be on a horizontal or vertical line if the rectangles have no intersecting area, otherwise you can find the area by finding the distance between x and y coordinates and multiplying them together.

It's probably better to just draw a picture than to use a formula on this.  That's how I got the formula.

thanks, i figured it out late last night, and got pretty much the same thing you did :D.


function
real analysis
real analysis unsolved

A function $ f\colon \mathbb R\to\mathbb R$ has the [b]Intermediate Value Property[/b] (IVP) if the following holds: for any $ a,b,c\in\mathbb R$ such that $ f(a)<c<f(b)$ there is $ x$ between $ a$ and $ b$ such that $ f(x)\equal{}c$.

The Intermediate Value Theorem says that any continuous function $ f\colon \mathbb R\to\mathbb R$ has the IVP. The converse is false.

As we all know, the sum of two continuous functions is continuous. And it's not hard to show that the sum of two functions with the IVP does not necessarily have the IVP. Suppose $ f$ is continuous and $ g$ has the IVP; does $ f\plus{}g$ have the IVP?

We have the Lebesgue theorem that say there is at least a function which has in all points discountinuity of type 2 with IVP(Darboux Property).So let $ f: \mathbb{R}\rightarrow\mathbb{R}$ a function with this property,and $ g = f|_{[0,1]}$,a surjective function.(you can find one on wikipedia,i don't remeber now the link).
Let $ h: \mathbb{R}\rightarrow[0,1];h(x) = \frac {1}{2} + \frac {1}{\pi}\cdot arctgx$ ,and $ k: [0,1]\rightarrow\mathbb{R} ;k(x) = 0 if x\in\{0,1\} \& k(x) = \tg\(\pi\cdot x - \frac {\pi}{2}\) ,x\in(0,1)$.Let $ w: \mathbb{R}\rightarrow\mathbb{R},w(x) = k(g(h(x)))$
Now let $ M = \{x\in\mathbb{R}|w(x) = x\} \& u: \mathbb{R}\rightarrow\mathbb{R},u(x) = w(x),x\in\mathbb{R} - M,u(x) = 1 + x,x\in M$.
You can verify that u has IVP and $ p = u - Id_{\mathbb{R}}$doesn't have IVP(there isn't $ x\in\mathbb{R}$ with $ p(x) = 0$,and there are x,y with p(x)>0 and p(y)<0)


What is the code to have a background color of like sky blue?

[color=#87CEEB]SkyBlue 	87 CE EB 	135 206 235[/color]
[color=#87CEFA]LightSkyBlue 	87 CE FA 	135 206 250[/color]
[color=#00BFFF]DeepSkyBlue 	00 BF FF 	  0 191 255[/color]

From: [url]http://en.wikipedia.org/wiki/Web_colors[/url]. Others may look better for your purposes.


[hide]There are 113 ones digits, for each of the 113 numbers.  9 numbers have only one digit, so 102 numbers have a tens digit.  And 14 numbers have hundreds digits.  So there are 113+102+14 = 229 digits. [/hide]

[hide]Memorize that there are 189 digits from 1-99 (10-99 is 180 digits and 1-9 is 9 digits). Then from 100-113 there are 14 numbers and three digits per number so the answer is 189 + 42 = 231[/hide]

Hm? But you don't count floor 1 as it's labeled "lobby"...

:blush:

It seems I misread the problem to read: The lobby is labled as the first floor. Apologies.

Yeah, now that I reread it, it is pretty oddly worded.

Heh, yeah. It's pretty oddly worded. By the way, nobody has gotten the right answer yet. Tarquin's method was correct, but he forgot something that happens in most hotel elevators...

(Hopefully most people will be familiar with the trick)

[quote="WindSlicer"][hide]Elevators don't have a thirteenth floor.. the thirteenth floor deal works like this. you subtract 2 from the number of digits and add 3(since you will now have floor 114), so, in essence, you are adding 1.

and I'm still not sure whether you are saying "1" is on the elevator or not.

If it is, then 232
If not, then 231

Is that the trick?
[/hide][/quote]

 :D

[quote="yif man12"]A hotel has an elevator. The hotel contains 113 total floors, if one doesn't count the basements. If the lobby is listed in the elevator as the first floor, how many total digits does the elevator have on the buttons combined?[/quote]

[hide="Solution"]
1 to 9 = 9
10 to 99 = 180
100 to 113 = 42

[b]231[/b]

If you want to be tricky, [b]229[/b].
[/hide]


The prime numbers are added in order starting with $ 2$: $ 2$, $ 2 \plus{} 3$, $ 2 \plus{} 3 \plus{} 5$, and so on. How many of the first 12 such sums are also prime?

[quote=Kestzh]Is there a non-bashing way which probably doesn't involve casework, because I missed teh 45 second mark with casework by a little.[/quote]

I did also. Does anyone else have a solution?


logarithms trigonometry algebra function domain

I know how to calculate the $\ln$ of a complex number $z$, take the $\ln$ of the magnitude of $z$ then add to it the amplitude of $z$ in the complex plane times $i$.  For example $\ln{(3+4i)} = \ln{\left|3+4i\right|} + i\tan^{-1}{\left(\frac{4}{3}\right)}$, but I can't prove this.  I think it may involve domain stretching.  Can someone help me?

Maybe this will help
$x=e^{\ln{x}}\Rightarrow i=e^{\ln{i}}$
The $\ln$ of $i$ using the above formula is $\frac{\pi}{2}i$.
Thus $i=e^{\frac{\pi}{2}i}$.  This is well known.
From this we can calculate $i^i$ which equals $e^{\frac{-\pi}{2}}$ (interestingly this is a real number).

[quote="Jimmy"]
From this we can calculate $i^i$ which equals $e^{\frac{-\pi}{2}}$ (interestingly this is a real number).[/quote]

Not quite true - it is a value of $i^i$, but $i^i$ actually has infinitely many values :)

[quote]$i^i$ actually has infinitely many values :)[/quote]

Can you show me how this is true?

You should look up the definition of branch cut; that should clarify certain things here.

[quote="Jimmy"]Is it because the amplitude of $i$ can be written as $2k\pi+\frac{\pi}{2}$?[/quote]

Yes.  By the way, one way to see that you have a correct method to calculate the complex log is to look at the expression $e^{x+iy}$ where $x, y \in \mathbb{R}$ and then try to reverse-engineer the result.


hello

please help me to understand this question...
thank u

a) Find a recurrence relation for the number of n-digit sequences over the alphabet {0, 1, 2} containing two consecutive 1’s. For example, the strings 0011112021 and 11000000 contain two consecutive 1’s but 1010121212 does not

b) What are the initial solutions?

c) How many 6-digit sequences are there containing two consecutive 1’s

a) When they ask you to find a recurrence relation for the number of n-digit sequences, they're basically asking you to find the number of n-digit sequences in terms of the number of n-1 digit sequences.  Or more generally, it could be in terms of how many n-2 digit sequences there are, etc.

b) Here they want the number of 1-digit and 2-digit sequences.  The initial solutions of a recurrence relation are the solutions yuo need to "get the chain started".  Like suppose you have

$a_{n}$ = $a_{n-2}$ + 2

Then in order to find $a_{n}$ for any n you'd need to find two solutions, $a_{1}$ to find all the odd-indexed solutions (like $a_{3}$ and $a_{5}$, etc.), and $a_{2}$ to find all the even-indexed solutions ($a_{4}$, $a_{6}$).

c) This should be trivial once you get (a) and (b).

Maybe someone can explain this better than I can.  :maybe:


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Which of the following numbers is greater, $ \int_{0}^{1}\sqrt[7]{1-x^{4}}dx$ or $ \int_{0}^{1}\sqrt[4]{1-x^{7}}dx$?

I think the easy way is just to use the Beta function if you know it.

Let $ u=x^{4}$ in the first integral.  It becomes $ \int_{0}^{1}\frac{1}{4}(1-u)^{\frac{1}{7}}u^{-\frac{3}{4}}du =\frac{1}{4}\frac{\Gamma(\frac{1}{4}) \Gamma(\frac{8}{7})}{ \Gamma(\frac{39}{28})}= \frac{1}{28}\frac{\Gamma(\frac{1}{4}) \Gamma(\frac{1}{7})}{ \Gamma(\frac{39}{28})}$

Similarly let $ u=x^{7}$ in the second integral.  It should equal:

$ \frac{1}{28}\frac{\Gamma(\frac{1}{4}) \Gamma(\frac{1}{7})}{ \Gamma(\frac{39}{28})}$

(Assuming no mistakes)

Which means the integrals are equal.

(Edited moments later, since I goofed on the Gamma evaluations)

@wushu_1984: Actually, even if you don't know Beta Integrals, there's still hope. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=127429]this topic[/url], mainly fedja's and Kent Merryfield's posts. I find edoo's approach more natural (although requiring some "advanced" math), it doesn't seem at all obvious that the two functions are inverses of each other :blush:

Thanks edoo. I do actually know the Beta function (I'm a Statistics student). I should have seen that, come to think of it.


It's not it  :maybe:

but it contains nice problems anyway

[quote]Too easy for IMO Wink[/quote]

Maybe too easy for Boreico, Grinberg, ... but for IMO :noo: Chinese problems are often difficult. :wink:


Following facts are given:
$ 1$. Two numbers, each comprising of two digits.
$ 2$. Sum of the two digits of the greater number is more than that of the smaller number.
$ 3$. Difference of the two numbers is four times the value of one.
$ 4$. $ 2$ raised to the power of one number is twice of that of the other.
$ 5$. Square of one is less than $ 25$ times of that of the other

what is the value of the smaller number ?

[quote="ElChapin"][quote="xxxyyyy"]

$ 4$. $ 2$ raised to the power of one number is twice of that of the other.

[/quote]

I don't understand your fourth condition...
 :?:

Let $ m$ and $ n$ be the numbers, $ 2^m \equal{} 2*2^n$??? don't think so, but that is what I understand

Could you please explain it more?

 :)[/quote]

Yes, it is exactly so.

[hide="How is this possible"]
Let the smaller be $ x$ and the larger be $ y$.  By condition 5 we have $ 2^y\equal{}2 \cdot 2^x$, so $ 2^{y\minus{}x}\equal{}2^1$.  Thus, $ y\minus{}x\equal{}1$. From this we see that $ x$ and $ y$ must be consecutive integers.

But then condition 3 states that "the difference of the two numbers is four times the value of one".  Since the numbers are consecutive, the difference of the two numbers is $ \pm 1$.  Then the value of one must be $ \pm 1/4$ which is clearly not a two-digit number.[/hide]

[quote="Dushi"]Hey guys..
Went thru the statements.. couldn't come up with the formula.. but trial and error gives me the answer as 10.
Took me around 3 and a half minutes to come up with this

Cheerz[/quote]

can you give some explanation as to how u came to such answer?


I've noticed that when colonialism became unpopular, the usa and its allies were in power.  The soviet union, which still practiced colonialism, was at a disadvantage because anticolonialism feelings spread.  So US had many allies (supporters of democracy), and russia didnt have many allies because there werent that many communist states, especially because many communist countries were under Soviet rule.

China was still in power because it only invaded tibet and the united states doesnt give a **** about tibet.

Please correct me if i stated any baseless ideas or fallacies
Feel free to rant about the two governments.

um...so what's the conclusion you're making based on that?...or is it just a list of facts and opinions?

That's a bit oversimplified - on a number of grounds. First of all, it depends on what you mean by "colonialism". Some people would say that the US still engages in a form of socio-political colonialism with the endorsement of most of it's people. Not to mention maintaining certain outright territories still at this point.

Secondly, as to when colonialism ever became unpopular in the West, or even if it ever did, is questionable . Most western powers engaged in outright colonialism well into the 1960s and 70s - and most installed puppet governments in places that they controlled after they left.

But when it comes down to it, people's moral values tend to lag behind utilitarian values. If colonialism provides benefits and advantages to them, they will endorse it as long as it does. So in that sense it is not a coincidence.


Two questions:

(a)Will it always be free?

(b)Will we be able to use it after we complete the class?

a) i'm not exactly sure
it was posted in the sticky, but nobody ever replied to it

b) i'm not sure what class you're talking about, but if you mean the C&P AoPS class, then yes
in fact, you can take all the classes AoPS gives and still have access to Alcumus
if you take the C&P class, all that's going to be affected in alcumus is probably your score, since you will probably do better

(a) We haven't made a firm decision on this, but expect it to be free for quite a while.

(b) Yes, you'll have access to it after your C&P class ends.  (I like the way vallon thinks in stating that the only thing that will be affected is your score!)

[quote="cmpgm88@yahoo.com"]Speaking of points, do they have any purpose, yet?

If not, perhaps you might consider a "prize" system, in which you might win a gift certificate, t-shirt, or the like.[/quote]

Please consider that this is a FREE system.

Think of your "passed" and "mastered" badges as "items" earned from your points to be your prize.

There is one prize actually...

The more badges and mastered things you have, you get more confidence.

Very elegantly and deeply said, 007math...

Plus you get the prize that you do better at school and contests. That's a huge [color=green][size=200]+[/size][/color].

[quote="Joe10112"]woah! how do u do the big green +???[/quote]

Above the part where you type your post, there are two drop-down menus, one for "Font colour" and one for "Font size". For that plus sign, I used [color=green]green[/color] and [size=200]huge[/size].

hmmmmmmmm......

[color=green][size=200]+[/color][/size]

yay!

Please test stuff like that in the Test forum, not here.

As far as the future of Alcumus goes, of course we would all love to have it be free  :D , and in order for it to be free, we should cherish the fact that the guys at AoPS are so nice to give us such a helpful probability course.

EDIT: Oh wow, I just noticed an interesting trend in our "locations"...

sorry

where is the test forum?

(and who is rating our posts?)

The test forum is located at the bottom of the Forums page.  A example picture:

[img]19475[/img]

Posts are rated from other AoPS users with $ >25$ posts.

That's true, as well as the greatest prize of all,

[size=200]KNOWLEDGE[/size]

knowlegde is good

you get to use it in the future.

and also pwn ppl at math

lol

When I first read the topic title, I thought it was going to be about a topic about subjects. So my question- How soon from now will other subjects be added to Alcumus? Would we be able to select which topic we wanted? Also, are you permitted to give more details on the time, if it happens, when the program will not be free?

When we have further information, it will be posted in the stickies here on the forum.


calculus
derivative
calculus computations

When is the magnitude of $ \frac{d}{dt}40\plus{}30e^{\minus{}2t}$ largest for $ t\geq0$?

Are you asking when $ 40\plus{}30e^{\minus{}2t}$ is maximized, or when its derivative is maximized?  Either way, you're going to have to take a derivative (once or twice) and set it to 0.

I am looking for when the derivative is maximized. So are you saying I should take the double derivative, set it equal to $ 0$, and solve for $ t$?

Yes, it's the same procedure-you're just taking an extra derivative.  A real-life example would be if you're trying to find out when velocity is maximized.

So it appears that when

[hide]$ t\equal{}0$ the magnitude of $ \frac {d}{dt}40 \plus{} 30e^{ \minus{} 2t}$ is largest at $ 120$.[/hide]


Let $\Delta ABC$ be an isoceles triangle with $AB=AC$. Suppose the angle bisector of $\angle B$ meets $AC$ at $D$ and that $BC=BD+AD$.
Determine $\angle A$.

[hide="Confusion"]I'm getting $\angle A=180^\circ$  :?:  :!: [/hide]

[quote="tim1234133"][hide="I reckon that"]
$\angle A=50^\circ$   :maybe:  [/hide][/quote]

Then BC would be too short... By logic, A must be obtuse.

Provided, of course, the statement is correctly written.

:wallbash: I found $\frac{A}{2}$ and forgot to double my answer- I mean $\angle A = 100^{\circ}$

[hide="For the record, here is my solution"]
Let $\frac{A}{2}=\theta$ and $\frac{B}{2}=\phi$.

Since $\angle A$ is obtuse and $CD>DA$, there exist two points $T$ on $BC$ such that $DT=AD$.

Let $T$ be the one of these points that is closer to $C$.
By the Sine Rule:

$\frac{DB}{\sin 2\theta}=\frac{DA}{\sin \phi}=\frac{DT}{\sin \phi}=\frac{DB}{\sin \angle DTB}$

Hence $\sin DAB=\sin DTB$ and since we have $T$ nearer $C$ (than the other possible point $T$) $\angle DAB+\angle DTB=180^{\circ}$

Hence $\angle DTB=2\theta$ and so triangle $CTD$ is similar to $CAB$, and so $CT=DT=DA$. Then $BT=BD$. Hence $\angle DTB=\angle TDB=180^{\circ}-2\theta$.

Considering triangle $DTB$, $360^{\circ}-4\theta+\phi=180^{\circ}\Rightarrow 4\theta-\phi=180^{\circ}$.
Considering triangle $ABC$, $2\theta+4\phi=180$

Equating these two gives $2\theta=5\phi$ and $\theta+2\phi=90^{\circ}$
giving $\theta=50^{\circ}$ and $\angle A= 100^{\circ}$
 [/hide]

Never mind, I've found the mistake in my reasoning. Good job, tim  :)


Hello everyone  :lol:

Find the maximum value that can take $ x\plus{}y\plus{}z$, knowing that $ x,\;y,\;z$ are integers and $ x^2\plus{}y^2\plus{}z^2<xy\plus{}3y\plus{}2z$.

Greetings  :D

[quote="TALENTO"]Hello everyone  :lol:

Find the maximum value that can take $ x \plus{} y \plus{} z$, knowing that $ x,\;y,\;z$ are integers and $ x^2 \plus{} y^2 \plus{} z^2 < xy \plus{} 3y \plus{} 2z$.

Greetings  :D[/quote]

First, if $ (x,y,z)$ verifies $ x^2 \plus{} y^2 \plus{} z^2 < xy \plus{} 3y \plus{} 2z$, so does $ (|x|,|y|,|z|)$ and, since $ |x| \plus{} |y| \plus{} |z|\geq x \plus{} y \plus{} z$, we can consider only non negative numbers.

Now $ x^2 \plus{} y^2 \plus{} z^2 < xy \plus{} 3y \plus{} 2z$ may be written $ (2x \minus{} y)^2 \plus{} 3(y \minus{} 2)^2 \plus{} 4(z \minus{} 1)^2 < 16$

So $ z\in\{0,1,2\}$ and :
$ z \equal{} 2$ $ \implies$ $ (2x \minus{} y)^2 \plus{} 3(y \minus{} 2)^2 < 12$ $ \implies$ $ y\in\{1,2,3\}$
$ z \equal{} 2$ and $ y \equal{} 3$ $ \implies$ $ (2x \minus{} 3)^2 < 9$ and $ x\leq 2$ giving for $ (2,3,2)$ : $ x \plus{} y \plus{} z \equal{} 7$
$ z \equal{} 2$ and $ y \equal{} 2$ $ \implies$ $ (2x \minus{} 2)^2 < 12$ and $ x\leq 2$ giving lower $ x \plus{} y \plus{} z$
$ z \equal{} 2$ and $ y \equal{} 1$ $ \implies$ $ (2x \minus{} 1)^2 < 9$ and $ x\leq 1$ giving lower $ x \plus{} y \plus{} z$

$ z \equal{} 0$ implies the same equation than $ z \equal{} 2$ with lower $ x \plus{} y \plus{} z$, so must not be considered.

$ z \equal{} 1$ $ \implies$ $ (2x \minus{} y)^2 \plus{} 3(y \minus{} 2)^2 < 16$ $ \implies$ $ y\in\{0,1,2,3,4\}$
$ z \equal{} 1$ and $ y \equal{} 4$ $ \implies$ $ (2x \minus{} 4)^2 < 4$ and $ x\leq 2$ giving for $ (2,4,1)$ : $ x \plus{} y \plus{} z \equal{} 7$
$ z \equal{} 1$ and $ y \equal{} 3$ $ \implies$ $ (2x \minus{} 3)^2 < 13$ and $ x\leq 3$ giving for $ (3,3,1)$ : $ x \plus{} y \plus{} z \equal{} 7$
$ z \equal{} 1$ and $ y \equal{} 2$ $ \implies$ $ (2x \minus{} 2)^2 < 16$ and $ x\leq 2$ giving lower $ x \plus{} y \plus{} z$
$ z \equal{} 1$ and $ y \equal{} 1$ $ \implies$ $ (2x \minus{} 1)^2 < 13$ and $ x\leq 2$ giving lower $ x \plus{} y \plus{} z$
$ z \equal{} 1$ and $ y \equal{} 0$ $ \implies$ $ (2x)^2 < 4$ and $ x \equal{} 0$ giving lower $ x \plus{} y \plus{} z$

Hence the result : $ 7$

Thank you very much [b]pco[/b]

Greetings  :D


geometry
geometric transformation
reflection
incenter
trapezoid
rhombus
circumcircle

Let $ ABC$ be a triangle and $ (O)$ be its circumcircle. Let $ d$ be an arbitrary line and $ d_1$, $ d_2$, $ d_3$ the reflections of $ d$ into the sides $ BC$, $ CA$, $ AB$, respectively. Call $ A_1\equal{}d_2\cap d_3$, $ B_1\equal{}d_3\cap d_1$, $ C_1\equal{}d_1\cap d_2$, then prove that the incenter of triangle $ A_1B_1C_1$ lies on the circle $ (O)$.

Let $ p \parallel d$ pass through the triangle orthocenter $ H.$ Reflections $ p_1, p_2, p_3$ of $ p$ in $ BC, CA, AB$ concur at the pole $ P \in (O)$ of the Simsom line $ s \parallel p \parallel d$: The A-altitude $ AH$ meets $ (O)$ again at the reflection $ H_1$ of $ H$ in $ BC,$ the line $ p$ meets $ BC$ at $ P_1$ and $ p_1 \equiv H_1P_1.$ Parallel to $ p$ through $ A$ meet $ (O)$ again at $ A'$ and perpendicular from $ A'$ to $ BC$ meets $ (O)$ again at $ P.$ Since $ AH_1 \parallel A'P$ (both $ \perp BC$), $ AH_1A'P$ is a cyclic trapezoid, isosceles, $ \angle HH_1P \equiv \angle AH_1P \equal{} \angle H_1AA' \equal{} \angle H_1HP_1 \equal{} \angle HH_1P_1$ $ \Longrightarrow$ $ P \in p_1.$ Similarly, $ P \in p_2, P \in p_3.$

Let $ \varrho$ be the distance of $ (d, p).$ Lines $ p, p_1, d, d_1$ form a rhombus, the distance of $ (d_1, p_1)$ and of $ (d_1, P \in p_1)$ is also $ \varrho.$ Similarly, the distance of $ (d_2, P \in p_2)$ and the distance $ (d_3, P \in p_3)$ is $ \varrho$ $ \Longrightarrow$ $ P \in (O)$ is incenter of the $ \triangle A_1B_1C_1.$

Lemma: let a triangle $ \triangle ABC$, I is the incenter and $ I_a$ the A-excenter. Then $ AI_a$ pass through the circumcenter of $ \triangle BCI_a$ an of $ CBI$.
Proof: The perpendicolar line from C to $ CI_a$ meet the perpendicolar line from B to $ BI_a$ on the incenter so $ II_a$ is a diameter of the circumcircle of $ \triangle BCI_a$.

The line d meet BC,CA,AB on X,Y,Z. Call K the incenter of $ \triangle A_1B_1C_1$. For the lemma we have that the angle bisector of $ \angle B_1A_1C_1$ pass through the circumcenter of $ \triangle AYZ$ a.s.o. Call $ d_1'$ the isognonal line w.r.t. $ \triangle ABC$ of AK a.s.o. So $ d_1'$ is the A-altitude of $ \triangle AYZ$ and then $ d_1' \perp d$. Equally we have $ d_2' \perp d$ and $ d_3' \perp d$, then $ d_1' \parallel d_2' \parallel d_3'$.


Do you see a relationship in today's date (meaning month and
day of the month) and the letters in "PIE" or "pie?"

A hint
[hide]the way the letters look[/hide]

Additional hint
[hide]the way the letters read backwards[/hide]

The letters mirrored look similar to 314(3/14)


FTW Asymptote LaTeX MATHCOUNTS Alcumus Mafia ARML

What is the main reason you are on AoPS? (I didn't do a poll because mine always screw up)

- Math classes
- Chatting on FTW / GFF
- Asymptote / LaTeX / Geogebra
- Improving mathcounts score on FTW
- Alcumus
- AoPS is addicting once I tried it
- I love math more than my parents (jk)
- Other (state reason)

- Chatting on FTW / GFF
- AoPS is addicting once I tried it
-i heart math

-Mafia
-Games+FF
-I love math
-Virus-free :lol:

This doesn't look like a game :noo:.

Anyway, for FTW, of course.

there isnt a poll section anymore so i think its fine here.

really, math is probably what attracts most people.

Actually, this probably should go in FF.

Anyway, I randomly found a link to AoPS on another site, clicked on it, looked around a bit, registered, and the rest is history!    :lol:

well, I registered for ARML because someone on my school's math team told me to, then at ARML, the coordinator for our team recommended this, so I signed up

and that's it...

I was one of two on my MC team without an aops account in seventh grade in 06, and then I remembered a few months later.

For math, and it's good practice for contests.

1. Ask any kind of question to AoPSers who are 10 times, 100 times more accomplished than me.
2. Discuss random stuff. Participate in debates (?)
3. Search for friends I know outside AoPS
4. Use AoPS wiki

math classes:20%

CHATTING AND CDS ON FTW:80%

I got this thing in the mail about AoPS FTW... so i registered, and that's about it. :) Yay AoPS FTW!

Hm I do not think this adequately represented. +1 and -2. Lets go. H&H #1000000

Friend-21
junk mail-18
Teacher-20
Searching math games on google-20

I'm going to ignore the above Hurt or heal and say most people here were probably attracted by math and socializing with other mathimatically gifted people.

Friend-22
junk mail-16
Teacher-20
Searching math games on google-20

Hmm...i think my signature says blue mop...

- Math classes
- Chatting on FTW / GFF
- AoPS is addicting once I tried it
- Well paying job involving mathematics :yup:

Math, FTW, Alcumus, and classsssssssssssssssssses.


Consider a circle $ \Gamma(O,R)$ and a point $ P$ found in the interior of this circle. Consider a chord $ AB$ of $ \Gamma$ that passes through $ P$. Suppose that the tangents to $ \Gamma$ at the points $ A$ and $ B$ intersect at $ Q$. Let $ M\in QA$ and $ N\in QB$ s.t. $ PM\perp QA$ and $ PN\perp QB$. Prove that the value of $ \frac {1}{PN} \plus{} \frac {1}{PM}$ doesn't depend of choosing the chord $ AB$.

$ PM \equal{} AP \cdot sin BAQ$ and $ PN \equal{} PB \cdot sin BAQ$ (since BAQ=ABQ)
So we need to show $ \frac {1}{AP \cdot sin BAQ} \plus{} \frac {1}{BP \cdot sin BAQ}$ is constant.
Or $ (\frac {1}{sin BAQ})(\frac {AP \plus{} BP}{AP \cdot BP})$ is constant.
$ AP \cdot BP$ is constant, so we need to show $ \frac {AB}{sinBAQ} \equal{} \frac {AB}{cosOAB} \equal{} \frac {AB}{(AB/2)/R)} \equal{} \frac {R}{2}$ is constant, clearly true.


integration
real analysis
real analysis unsolved

for any positive integer n, define M(n)={Sum from i=1 to n}Mu(i),
where Mu(i) is the Moebius Mu function.
(1)Prove M(n)=O(n) by the Prime number Theorem.
(2)Show that if we can prove M(n)=O(n 1/2 +epsilon )then we have the Riemann Hypothesis.
(3)Prove that M(n)=O(n a ) (1/2<a<1) is equivlent to zeta(s+it)have no zero where s \geq a.

There's something I don't get about $(1)$: Doesn't $M(n)=O(n)$ mean that there is a positive constant $A$ s.t. $|M(n)|<An$ for all positive integers $n$? But this is obvious because $-n\leq M(n)\leq n$ for all positive integers $n$. Maybe you meant $o(n)$ instead? (small letter instead of capital letter).

It's all based upon the fact that $\displaystyle \frac 1{\zeta(z)}=z\int_1^{\infty}\frac {M(x)}{x^{z+1}}dx$, where $\displaystyle M(x)=\sum_{1\leq n\leq x}\mu(n)$.

I don't know how to derive this. It was "stolen" (:D) from here :

[url]http://en.wikipedia.org/wiki/Mertens_conjecture[/url]


1. what is the sunm of the next three terms in the geometric sequence of 1, -2, 4, ...?

2. give that 5^(x+1) = 30, what is the value of 5^(3x+1)?

3. find the numeral respresenting the last year of this century (20th) which is divisible by 2, 3, 4, and 37.

4. what is the probability taht lea and dan select diagonals of the same length from a regular hexagon if they select different diagonals? express your answer as a common fraction.

5. the dial on the lock of a school locker has 40 numbers. if no combinations uses the same number all three times, how many three number combinations are possible on the lock?

6. what is the probability that when even selects a positive even integer less than twenty and theo picks a positive multiple of 3 less than thirty, they pick the same number? express your answer as a common fraction.

1.[hide]-8+16-32=-24[/hide]

20th century means the 1900's, right?

4.[hide]there are 6 diagonals with one length, 3 diagonals of another length, so

(2/3)*(5/8)+(1/3)*(2/8)=(5/12)+(1/12)=(6/12)=(1/2)[/hide]

Question on #5, can two numbers be used in a row?  All the problems involving lockers normally include that requirement.

6.[hide]So even can pick 2,4,6,8,10,12,14,16,18 and theo must pick a 3,6,9,12,15,18,21,24,27...so the chance that evne picks 6,12,or 18 is 1/3 and the chance that theo picks the same number even does is 1/9, so 1/3*1/9 is 1/27.[/hide]

5.[hide]wait, it wouldn't make a difference if you couldn't use two of the same numbers in a row...Well, it would be 40*40*39 i think, which is 1600*40-1600=64000-1600=62400[/hide]

2.[hide]

For number 2 if you raise both sides of 5^(x+1)=30 you get 5^(3x+3)=30^3 but your looking for 5^(3x+1) so you divide it by 5^2(because when you divide like bases you subtract exponents)  so the answer is 30X30X30/5X5 or just simply 1080[/hide]

3. 3*4*37=444  there are none because no multiple of 444 is between 1900 and 2000.

[quote="eryaman"]20th century means the 1900's, right?[/quote]
Close, but not exactly.  The 20th century means the years 1901 to 2000.  The 1900's means the years 1900 to 1999.


probability
function
vector
real analysis
Functional Analysis
probability and stats

On a complete probability space $ (\Omega, \mathcal{F}, \mathbf{P})$ consider integrable random variables $ X_1, \cdots, X_n$ and $ X_{n\plus{}1} \equal{} X_1$, as well as sub-$ \sigma$-algebras $ \mathcal{F}_1, \cdots, \mathcal{F}_n$ of $ \mathcal{F}$ for which
\[ \mathbf{E}(X_{i\plus{}1} | \mathcal{F}_i) \equal{} X_i \] a.e. holds for every $ i \equal{}  1, \cdots,n$. Show that $ X_1 \equal{} \cdots \equal{} X_n$ a.e.

[quote="hjbrasch"]any condition on the sigma-algebras such as ascending/descending ?[/quote]

No condition.

If $ \psi$ is any smooth function with $ \psi''(x)>0$ for all $ x$ (say, $ \sqrt {1\plus{}x^2}$), then $ E\psi(X)$ never increases after taking conditional expectation and decreases unless the conditional expectation coincides with $ X$ itself. This is enough to solve the problem.

Beautiful solution, Fedja. But it is really hard to find this approach :)
I'm also interested in this idea:
Suppose  $ X_i\in \ L_2(\Omega)$, then conditional expectation $ E(\cdot|F)$ is just an orthoprojection on the subspace of $ F$-measurable functions. So we have to deal with some identities in the Hilbert space, and i'm sure there must be a way to give the solution using this. Any comments are appreciated.

Well, it is pretty much the same story: the norm of the projection is not greater than the norm of the vector, so the only chance to get the vector back after several projections is not to change it during each step :).

Oh yes i should repeat some basic functional analysis  :blush:

Another approach for $ L^2$ case: $ \sum E[(X_i\minus{}X_{i\plus{}1})^2] \equal{} 2 \sum E(X_i^2) \minus{} 2 \sum E(X_i X_{i\plus{}1})$ is zero because $ E(X_i X_{i\plus{}1} | \mathcal{F}_i) \equal{} X_i E(X_{i\plus{}1} | \mathcal{F}_i) \equal{} X_i^2$.

This exercise has a hint in the book : "Assume first that the $ X_i$'s are square-integrable; then truncate"
But I can't see how to generalize to $ L^1$ using truncation. :huh:


trigonometry
quadratics
ARML
linear algebra
matrix
algebra
quadratic formula

Let $ x$ be in radians with $ 0<x<\frac{\pi}{2}$.  If $ \sin(2\sin x)\equal{}\cos (2\cos x)$, then $ \tan x \plus{} \cot x$ can be written as $ \frac{a}{b\minus{}\pi^c}$ where $ a$, $ b$, and $ c$ are positive integers.  Compute the sum of $ a\plus{}b\plus{}c$.

----

For $ k>1$, let $ S_k$ denote the sum of the $ k$ consecutive integers starting with $ k$.  Compute the smallest value for $ k$ such that $ S_k$ is a perfect square.

#2 - $ S_2 \equal{} 2 \plus{} 3 \equal{} 5$

$ S_3 \equal{} 3 \plus{} 4 \plus{} 5 \equal{} 12$

$ S_4 \equal{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \equal{} 22$

$ S_5 \equal{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9 \equal{} 35$

We can derive a little formula:

$ S_k \equal{} \frac {k(k \plus{} 2k \minus{} 1)}{2} \equal{} \frac {k(3k \minus{} 1)}{2}$

If that is a perfect square, then $ 2(3k \minus{} 1) \equal{} k$ so $ k \equal{} \frac {2}{5}$ or $ 3k \minus{} 1 \equal{} 2k$ so $ k \equal{} 1$ but it's given that $ k > 1$ and is an integer so there is no solution??

#1)
[hide]
Begin by noting that $ \cot{x} \plus{} \tan{x} \equal{} \frac{\sin^2{x} \plus{} \cos^2{x}}{\sin{x}\cos{x}} \equal{} \frac{1}{\sin{x}\cos{x}}$

Since our angles are restricted to the 1st quadrant, $ \cos{u} \equal{} \cos{v}$ implies $ u \equal{} v$

$ \sin(2\sin x)\equal{}\cos (2\cos x)$
$ \cos(2\sin x \minus{} \frac{\pi}{2})\equal{}\cos (2\cos x)$
$ 2\sin x \minus{} \frac{\pi}{2} \equal{} 2\cos x$

Rearranging and squaring yields:

$ \sin^2x \minus{} 2\cos{x}\sin{x} \plus{} \cos^2x \equal{} \frac{\pi^2}{16}$

Substituting for $ \sin^2x \plus{} cos^2x$ and rearranging yields:

$ \cos x\sin x \equal{} \frac{16 \minus{} \pi^2}{32}$

Substituting this into our expression derived earlier yields:

$ \tan x \plus{} \cot x \equal{} \frac{32}{16 \minus{} \pi^2}$

Thus, $ (a,b,c) \equal{} (32,16,2)$ , and $ a \plus{} b \plus{} c \equal{} \boxed{050}$[/hide]

[quote="314.math"]

For $ k > 1$, let $ S_k$ denote the sum of the $ k$ consecutive integers starting with $ k$.
Compute the smallest value for $ k$ such that $ S_k$ is a perfect square.[/quote]

k = 81

Work:

k(3k - 1)/2 = x^2, for x the square of an integer

3k^2 - k = 2x^2

3k^2 - k - 2x^2 = 0

Solve for k using the quadratic formula:

k = [1 +/- sqrt(1 + 24x^2)]/6

When x = 99, k = 81

(This was found by setting y = [1 + sqrt(1 + 24x^2)]/6 in a TI-83
calculator, and then scrolling ahead in the table of values.)

This is from last ARML, I think, and very few people got this within the time limit.

[b]Arrange your tan[/b] had a good idea back there. He solved the quadratic equation for $ k$, giving:

$ k \equal{} \frac {1 \plus{} \sqrt {1 \plus{} 24 n^2}}{6}$

There must be an integer $ m$ such that $ \sqrt {1 \plus{} 24 n^2 } \equal{} m \iff \boxed{m^2 \minus{} 24n^2 \equal{} 1}$.

This is an instance of [url=http://en.wikipedia.org/wiki/Pell%27s_equation]Pell's equation[/url], and we can solve it by examining the continued fraction representation of $ \sqrt {24}$. We can determine that the coefficients are 4, 1, 8, 1, 8, 1, 8, etc...

How could I not notice the much nicer Pell equation formed by the discriminant? :o And Pell equations are on ARML?

I have no idea what the intended solution was. Maybe someone will remember from last year.

In any case, the Pell's equation solution isn't bad, so I'll post it. Continued from my previous post:

[hide="Solution"]Going through the possible continued fraction representations, we get the following $ (n,m)$ values:
$ \left \{ 4,1 \right \} \equal{} 5 \implies (m,n) \equal{} (5,1)$
$ \left \{ 4,1,8 \right \} \equal{} 44/9 \implies (m,n) \equal{} (44,9)$
$ \left \{ 4,1,8,1 \right \} \equal{} 49/10 \implies (m,n) \equal{} (49,10)$
$ \left \{ 4,1,8,1,8 \right \} \equal{} 436/89 \implies (m,n) \equal{} (436,89)$
$ \left \{ 4,1,8,1,8,1 \right \} \equal{} 485/99 \implies (m,n) \equal{} (485,99)$

We find $ n \equal{} 99$ gives $ k \equal{} 81$.[/hide]

Alternatively, if you like guess-and-check, you can look for a perfect square value of k that works.

Begin by strategically taking the equation mod 3 to reduce the exponent on k. We find k=0 or 1 mod 3. Note that k=0 mod 3 means k=0 mod 9 as well.

Let's start first with perfect square multiples of 9. 9 and 36 don't work, but 81 does! Now, we write down 81 on our answer sheet and see if we can find a smaller solution to replace it with, which we know must be 1 mod 3.

Begin by looking at the other perfect squares, all of which are 1 mod 3. None of them work, and we feel a bit better. Finally look at the rest of the numbers that are 1 mod 3, and they don't work. Or you ran out of time before you got to check all of them, but it doesn't matter because you got the right answer.

Can anyone explain what Pell equations are? I looked it up on wiki but I don't really understand. All I know is that it's a way to solve Diophantine equations.

The other question deserves attention, too! It's a decent question. If you don't know where to start:
[hide="Hint"]Find the solution set of $ \cos A = \sin B$, and then substitute.[/hide]

[b]@ modularmarc101:[/b] Pell's equations are a particular class of Diophantine equations of the form $ x^2 - Ny^2 = 1$, where $ N$ is a given integer. There happens to be a general technique for finding the solution set of $ (x,y) \in \mathbb{N}^2$. That's really all. It happens that Pell's Equations (or equations that can be reduced to a Pell's equation by some elementary substitutions) show up all over the place, so it is a good idea to know how to solve them.

In this particular case, a Pell's equation came from setting the square-root term in the quadratic formula to an integer. Can you see how the quadratic equation can give rise to Pell's Equations in general (though possibly with a few substitutions)?

Also, in this case, it may have been easier to use the recursive method of generating solutions (under solution techniques in the Wiki) rather than cranking through the convergents:

$ \binom{x_{n + 1}}{y_{n + 1}} = \left( \begin{array}{cc} x_1 & N y_1 & y_1 & x_1 \end{array} \right) \binom{x_n}{y_n}$

In this case, it was very easy to see that $ (m,n) = 5,1$ was the fundamental solution.

Thanks and whats  $ \left(\begin{array}{cc}x_{1}& N y_{1}& y_{1}& x_{1}\end{array}\right)$ ?

I expressed the recursive definition in terms of matrix multiplication, because I find it easier to remember that way.

It is equivalently expressed as:
$ x_{n\plus{}1} \equal{} x_1 x_n \plus{} N y_1 y_n$
$ y_{n\plus{}1} \equal{} y_1 x_n \plus{} x_1 y_n$