Let $ x,y,z\geq0$.Prove that:
$ 23(x^3\plus{}y^3\plus{}z^3)\plus{}17(x^2y\plus{}y^2z\plus{}z^2x)\geq37(xy^2\plus{}yz^2\plus{}zx^2)\plus{}9xyz$

[quote="Inequalities Master"]Let $ x,y,z\geq0$.Prove that:
$ 23(x^3 \plus{} y^3 \plus{} z^3) \plus{} 17(x^2y \plus{} y^2z \plus{} z^2x)\geq37(xy^2 \plus{} yz^2 \plus{} zx^2) \plus{} 9xyz$[/quote]
$ \sum_{cyc}(23x^3\plus{}17x^2y\minus{}37xy^2\minus{}3xyz)\geq\sum_{cyc}(23x^3\plus{}14x^2y\minus{}37xy^2)\equal{}$
$ \equal{}\sum_{cyc}x(x\minus{}y)(23x\plus{}37y)\equal{}\sum_{cyc}\left(x(x\minus{}y)(23x\plus{}37y)\minus{}20(x^3\minus{}y^3)\right)\equal{}$
$ \equal{}\sum_{cyc}(x\minus{}y)^2(3x\plus{}20y)\geq0.$

Very nice your proof,arqady.I am very impressed. :first:


geometry
3D geometry
percent
ratio
percent increase

If the length of each edge in a cube is increased by the same percent and the result was a cube that had 25% more volume than it originally had, by what percentage were the edges increased?

I don't know if this is right or not...but if the volume increased by 25%, than the ratio of volumes is $ 1: 1.25$. So the cube root is the ratio of edges, or $ 1: (1.25)^{\frac{1}{3}}$. That makes the percent increase $ (1.25)^{\frac{1}{3}}\%$? This seems too...straightforward to be right.

Because a $ 25\%$ increase means an increase by a factor of $ 1.25$.  This is the relevant scale factor to cube root to find the scale factor of the sides.  The percent increase is $ (100\%)\times (\sqrt[3]{1.25}\minus{}1)$.

That's the definition of percent increase.  The side lengths increase by a factor of $ \sqrt[3]{1.25}$.

Oh duh...sorry.Forgot to think before speaking.


Show that every triangle can be dissected into nine convex nondegenrate pentagons.

Nearly in this way (see picture).
It's clear that if the statement is true for one triangle, it holds for any other one, as we can transform it affinely.

Is $9$ the minimal number of pentagons that are necessary to fill the triangle ?


Demonstrate that there exists only one triangle such that the three sides are 3 consecutive numbers (a, a+1, a+2), and one of the angles is the double of other inside the triangle.

Please explain it step-by-step :lol:  Thanks!

[quote="AndrewTom"]There is an infinite number of solutions. See here: http://math.exeter.edu/RPARRIS/peanut/commtri.pdf

Hope this is helpful.[/quote]Citing the article is not enough. The article solves for 2-commensurable triangles but it does not prove that there are an infinite number of triangles of the form $ (a,a\plus{}1,a\plus{}2)$ with such property. Your job here is to prove that there is only one such triangle (or an infinite of them if that's your claim).


algebra
polynomial
number theory solved
number theory

let f(x) be a polinomial of degree d in Z[x] and p a prime .we know that
a-f(0) = 0
b-f(1) = 1
c-for all natural number n we have f(n) = 0 or 1 ( mod p)

prove that d>= p-1.

Suppose we have $d<p-1$
$\sum_{i=0}^{p-1}f(p)$ is not divisible by $p$
So for 1<i<p we have $\sum_{k=1}^{p-1}k^i $ is not divisible by p .contradiction
Also it is from India 1998

[quote="Omid Hatami"]Suppose we have $ d < p \minus{} 1$
$ \sum_{i \equal{} 0}^{p \minus{} 1}f(p)$ is not divisible by $ p$
[/quote]
would it be $ \sum_{i \equal{} 0}^{p \minus{} 1}f(i)$? :maybe:


29. Two circles are externally tangent, and have a common external tangent line, with points of tangency P and Q. , and the radii of the circles are and . The center of the smaller circle (the one which contains point Q) is S, and a line through S is drawn tangent to the larger circle. The point of tangency of this line is T, as shown. If ST= cm, find the length of the larger radius .

"the radii of the circles are __ and __". Please fill in the blanks so that this problem is solvable.  :P

radii of the circles are R cm and R minus 1 cm

PQ segment equals 10 cm

also, a diagram and the length of ST are required.

[quote="now a ranger"]it doesn't give any lengths. and i described the picture i think

well its on http://www.mualphatheta.org/National_Convention/Past_Tests.html

2005 Theta Geometry[/quote]
Why don't you just look at the solutions?

they don't like give a clear view on the ones i posted. I can't really comprehend them without people explaining  :( I was like where in the world is so and so from when i just see they worked it out .


function
symmetry
algebra unsolved
algebra

find all $ f: R$ to $ R$ such that $ f(1)\equal{}1$ and
$ f(f(x)y\plus{}\frac{x}{y})\equal{}xyf(x^2\plus{}y^2)$
for all real numbers x and y with y isn't 0
I need your help  :blush:

Note that the right side is symmetric in x and y, so
$ f(f(x)y \plus{} \frac {x}{y}) \equal{} f(f(y)x \plus{} \frac {y}{x})$
Subbing in x=1,
$ f(y \plus{} \frac {1}{y}) \equal{} f(f(y) \plus{} y)$
So we see that $ f(y) \equal{} \frac {1}{y}$ is a possibility, and checking it, we find it works.
But I still have to prove there are no other solutions. However, any other function would break the symmetry of the left side.
In fact, can we just start out by saying that since the right side is symmetric in x and y, the left side must be too?

[quote="dgreenb801"]Note that the right side is symmetric in x and y, so
$ f(f(x)y \plus{} \frac {x}{y}) \equal{} f(f(y)x \plus{} \frac {y}{x})$
Subbing in x=1,
$ f(y \plus{} \frac {1}{y}) \equal{} f(f(y) \plus{} y)$
So we see that $ f(y) \equal{} \frac {1}{y}$ is a possibility, and checking it, we find it works.
But I still have to prove there are no other solutions. However, any other function would break the symmetry of the left side.[/quote]
We didn't know that the function is injective so how can you know that $ f(y) \equal{} \frac {1}{y}$ is a possibility... :maybe:

[quote="MellowMelon"]That's probably not intended to be a complete solution. In any case, $ f(0) \equal{} a, f(x) \equal{} 0 \forall x \neq 0$ for any real $ a$ is another set of solutions, so there is more work to be done.[/quote]
f(0) should be 0
if $ f(0)$ isnt 0 substitute $ x\equal{}0,y\equal{}\frac{y}{f(0)}$
and we will get that
$ f(y)\equal{}0$ which is contradiction because $ f(1)\equal{}1$
so $ f(0)\equal{}0$


Find the largest possible value of k for which 3^11 can be expressed as the sum of k consecutive integers.

2*3^11. This is true in general. (Consider the numbers from -3^11+1 to 3^11.)

Hmm... that's what I got too, but this is an AIME problem and apparently the answer is [hide]486, which equals 2*3^5[/hide].  I haven't the faintest idea how this answer was come up with.

But here's the exact wording of the problem:

Find the largest possible value of k for which 3^11 is expressible as the sum of k consecutive integers.

hmm, what year AIME was this on? i think i have seen the problem (and solution) before, but I don't quite remember...

Fierytycoon

This was 1987 AIME #11.

But I believe there is an error in my copy of the problems (unofficial).

Perhaps they meant [i]positive[/i] integers.

[quote="mathfanatic"]This was 1987 AIME #11.

Perhaps they meant [i]positive[/i] integers.[/quote]

That must be true, because otherwise Zeta's answer is correct.

The sum of a, a+1, a+2, up to a+(k-1) is the sum of k consecutive positive integers (a>0). These have sum ka + (k-1)k/2 = k(2a+k-1)/2.

So k(2a+k-1) = 2*3^11. But 2a>1, so k is the smaller of these two factors, and so it definitely is less than 3^6.. and the next smallest possibility is 2*3^5, which works with 2a+2*3^5-1 = 3^6, ie 2a = 3^5 + 1, a is half this.


Find all square numbers $ S_1$ and $ S_2$ such that $ S_1 \minus{} S_2 \equal{} 1989.$

[hide="solution"]Let $ S_1\equal{}x^2$ and $ S_2\equal{}y^2$. Thus $ x^2\minus{}y^2\equal{}(x\minus{}y)(x\plus{}y)\equal{}1989\equal{}3^2*13*17$. WLOG, let $ x$ and $ y$ be positive. Thus $ x\minus{}y<x\plus{}y$. Now only half of the factors of 1989 will be suitable to be $ x\minus{}y$, since $ 44^2\equal{}1936<1989<2025\equal{}45^2$, and thus 1989 isn't a square. Now there are $ (2\plus{}1)(1\plus{}1)(1\plus{}1)\equal{}12$ factors of 1989, so only 6 of them will work for $ x\minus{}y$. They are not hard to find; since we have the restriction that they must be less than 45: 1, 3, 9, 13, 17, 39. Then $ x\plus{}y$ takes on these values: 1989, 663, 221, 153, 117, 51. Solving for $ x$ and $ y$ gives

$ (x,y)\in \{ (995,994), (333,330), (115,106), (83, 70), (67,50), (45, 6)\}$.

Squaring, we get that

$ (S_1,S_2)\in \{ (995^2,994^2), (333^2,330^2), (115^2,106^2), (83^2, 70^2), (67^2,50^2), (45^2, 6^2)\}$.[/hide]

you bounded 1989 to prove its not a square however, you can just not this by looking at the prime factorization. Also we know that only 6 work for integer values of x,y because the sum of the two factors must be even. How do we know that there are the same number of factor pairs who sum to be even, as who sum to be odd?

His argument is not that half of them sum to an even number and half sum to an odd number. In fact, none of the factor pairs can sum up to an odd number, because all the factors are odd, and two odd numbers add to an even number.

Here's what he's saying there:
- 1989 is not a perfect square, therefore two numbers that multiply to 1989 cannot be equal. One must be greater
- Since $ y>0$, we must have that $ x\plus{}y > x\minus{}y$.
- If there are 12 factors of 1989, there are 6 factor-pairs.
- Each factor pair corresponds to one value of $ x\minus{}y$, since $ x\minus{}y$ is the lower of the two numbers in a factor pair

[quote="Wichking"]and someone acctuly thought this is an IMO problem ????? :stink:[/quote]
A good amount of Longlist problems is ridiculous. I don't get how their authors could think they are interesting, difficult or at least nice.

Um... Recall that this is 1989. Not 30 years later was a high-school basics Number Theory problem.

That has nothing to do with the year. IMO problems got more difficult since then, but not _that_ much. Additionally, as said, this problem is just plain stupid for much more reasons...


Source: Polya Competition, 1994

2.8.  A truck develops engine trouble one mile down a straight highway from a service station.  The truck continues to the service station at a constant speed, emitting smoke all the while.  The smoke is carried to the service station by the wind at a steady rate of 20 miles per hour.  The smoke envelops the service station for exactly two minutes.  Find the difference between the fastest and slowest possible speeds (in miles per hour) of the truck.

3.2.  Acacio's teacher asked him to find the sum of an arithmetic series.  The teacher gave him the first term (-1994), the common difference, and the number of terms.  When Acacio went to solve the problem, he confused the common difference and number of terms (the two numbers were not the same) - yet he got the right answer anyway!  What was that answer?

3.5. Compute log2[sin(6)*sin(42)*sin(45)*sin(66)*sin(78)].  (All angle measures are in degrees.)

wyra is [hide]correct[/hide]   .  There are still several problems left, though.  Someone do them!

mathfanatic wrote:Source: Polya Competition, 1994

3.2.  Acacio's teacher asked him to find the sum of an arithmetic series.  The teacher gave him the first term (-1994), the common difference, and the number of terms.  When Acacio went to solve the problem, he confused the common difference and number of terms (the two numbers were not the same) - yet he got the right answer anyway!  What was that answer?

[hide]-1994. an easy way to do it would be if there were only 1 term of -1994 and it would be the sum no matter what the common difference was. another way would be by manipulating the sum forumula thingie. but yeah, its all goode.[/hide]

mathfanatic wrote:Source: Polya Competition, 1994

2.8.  A truck develops engine trouble one mile down a straight highway from a service station.  The truck continues to the service station at a constant speed, emitting smoke all the while.  The smoke is carried to the service station by the wind at a steady rate of 20 miles per hour.  The smoke envelops the service station for exactly two minutes.  Find the difference between the fastest and slowest possible speeds (in miles per hour) of the truck.

[hide]48. at 20 mph it would take 3 minutes for the smoke to reach the station.  that means the truck can take 1 or 5 minutes to reach the station, or 60 mph and 12 mph respectively. thus:48[/hide]

Hmmm ... would you mind doing it out with the formula anyhow?  (I'm not sure 1 term counts as a sequence ...)

mathfanatic wrote:Source: Polya Competition, 1994

3.2.  Acacio's teacher asked him to find the sum of an arithmetic series.  The teacher gave him the first term (-1994), the common difference, and the number of terms.  When Acacio went to solve the problem, he confused the common difference and number of terms (the two numbers were not the same) - yet he got the right answer anyway!  What was that answer?

fine fine, i thought it was a slick way though. but the algebraic approach:

[hide]call the difference d and number of term things n so then n(2x(-1994)+(n-1)d)/2 is the sum, or so according to aops v1.  then use a grouping identity thing switching around d and n. multiply by 2 yada yada. divide around. end up with 3988=nd or something like that. so then that factors to 2x1994, 1x3988, and 4x997 not including the sign thingie. since the first term is -1994 just plug in and stuffs and you get -1994.[/hide]

I agree with you that it's slick -- in fact, I've seen a lot of competition problems that say things like, "Bob takes a number x > 1 and blah blah blah," and the easiest solution is just to realize that there isn't any particular reason that Bob couldn't actually have taken 1, except that then the problem would be too easy :).

heh, well for no point the sum from -1994 to 1993 with common difference 1 is -1994


Has anyone competed in this contest and won? Or knows someone who did? I competed for the Fountainhead last year... and I want to do for Atlas Shrugged, but I'm curious if anyone participated in it from this forum.  :)

I did the Fountainhead contest this year, not sure if results have been announced, but I noticed two horrible typos the morning after submitting so I doubt I won.


function
logarithms
conics
ellipse
calculus
derivative
calculus computations

Let $ c(t)$ and $ m(t)$ represent the number of cats and mice, respectively, at time $ t$.

We have $ \left \{ \begin{array}{ccc} c' & = & c (m-8) & m' & =  & m(7-c) & c(0) & = & 5 & m(0) & = & 10 \end{array}$

Prove that $ c$ and $ m$ are periodic in $ t$.

Consider the function $ g(c,m) \equal{} m \minus{} 8\ln m \plus{} c \minus{} 7\ln c.$ On the quadrant $ c > 0,m > 0,$ this function has a unique minimum value at $ c \equal{} 7,m \equal{} 8$ (the equilibrium point for the system) and tends to $ \infty$ as either $ c$ or $ m$ tends to either $ 0$ or $ \infty.$ Hence the level curves $ g(c,m) \equal{} K$ are simple closed loops containing $ (7,8)$ in the interior. (Near that equilibrium point, these curves are nearly ellipses; further away, they take on more distorted shapes.)

Now we compute the time derivative:

$ \frac {d}{dt}g(c,m) \equal{} \frac {\partial g}{\partial c}\cdot \frac {dc}{dt} \plus{} \frac {\partial g}{\partial m}\cdot\frac {dm}{dt}$

$ \equal{} \left(1 \minus{} \frac7c\right)c(m \minus{} 8) \plus{} \left(1 \minus{} \frac8m\right)m(7 \minus{} c) \equal{} (c \minus{} 7)(m \minus{} 8) \minus{} (m \minus{} 8)(c \minus{} 7) \equal{} 0.$

Hence $ g(c,m)$ remains constant over time and the trajectory of the system is confined to one particular level curve of $ g.$ (The one that goes through the initial condition.)

We need to show a few more things. (I won't, but they aren't too difficult - just detail-intensive). We need to show that the system always travels in the same direction along that curve, and with at least a certain minimal speed. Hence in some finite time, it will arrive back at the initial point. When that happens, since this is an autonomous system, repetition and periodicity will occur.


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

Prove that, for every $ n\in\mathbb{N}$, exists lattice point $ X(k,l,m)$ (with $ \gcd(k,l,m) > 1$), such that for every lattice point $ Y(p,q,r)$ ($ XY\leq n$), $ \gcd(p,q,r) > 1$

You mean in space?

Let $ n>0$ be a given integer.
Consider a set $ \{p_{i,j,k}/\minus{}n \leq i,j,k \leq n \}$ of pairwise distinct primes.
From the chinese remainders theorem, there exist integers $ x,y,z$ such that for each $ i,j,k$, we have
$ x\equal{}\minus{}i \mod[p_{i,j,k}],y\equal{}\minus{}j \mod[p_{i,j,k}],z\equal{}\minus{}k \mod[p_{i,j,k}]$.
Let $ X(x,y,z)$.
For each $ i,j,k \in \{\minus{}n, \cdots, n\}$, the point $ M(x\plus{}i,y\plus{}j,z\plus{}k)$ is such that $ p_{i,j,k}$ divides $ \gcd (x_M,y_M,z_M)$ so that $ \gcd (x_M,y_M,z_M)>1$.

Now, each point $ M$ such that $ XM \leq n$ is one of the form above (the square with center $ X$ and side $ 2n$ contains the sphere with center $ X$ and radius $ n$), then we are done.

This can be generalized to any finite dimension.

Pierre.


Βρείτε τις τιμές του $ a$ για τις οποίες η εξίσωση $ {(x\minus{}1)}^2$=|$ x\minus{}a$|έχει ακριβώς 3 λύσεις.

Προφανως εννοεις χωρις την πολλαπλοτητα
Στα γρηγορα (αν δεν εκανα λαθος) ειναι τα $ 3/4,1,5/4$ ; ( Δεν βαζω αναλυτικη λυση για να μην την χαλασω στους προσπαθουντες)


[quote="iamagenius"]for those of you who have the program mathematica how did you get it? surely you didnt actually pay the \$1000...  :ninja:[/quote]

some people can get pirated versions hehe i haven' heard of mathematica but it is possible

I am reasonably sure that you can get a nice discount if you are a student; that's how I got my copy.

but if your school has it or something can i just download their copy on to my computer?

But even if you can't buy a copy, there's a [url=http://www.quickmath.com]nice website[/url] that allows you access to some of the functions of Mathematica on a server.

[quote="b-flat"]Almost certainly not.  I think Mathematica only allows itself to be on one computer at a time, per copy.

What do you mean by "download their copy?"  Like from their website or something?[/quote]
i mean like zip the file and email it to myself.

I am guessing that that way would be illegal.  Besides, the attachment would be huge.

i've heard about stuff like crack or keygen. i tried downloading a crack for mathematica but it didn't work. does any one know why?

Just as PDF files of AMC's are taken down for copyright reasons, Mathematica is a copyrighted software. Please no posting of links to cracks/keygens websites or asking for pirated stuff. *ehem* [size=59]use PM[/size] *ehem*

This topic should go to the forum: [color=blue]High School>>>Other Problem Solving Topics.[/color]

Isn't pirating like illegal or something?

Don't you read copyright law? I don't think the $\$250,000$ is worth $\$1000$?

[quote="#H34N1"]Isn't pirating like illegal or something?

Don't you read copyright law?[/quote]
I THINK you should be asking this to yourself....but.....

[quote="iamagenius"]i mean like zip the file and email it to myself.[/quote]

This wouldn't work because Mathematica is not "one file."  Also, as someone else said, it would be huge.

Many schools have licencing agreements with Mathematica that allows their students to use the program for the duration of their enrollment.  So, for example, I need to get a new password from our system administrator every September to use Mathematica for the year.

Also, the student version is substantially cheaper than the full deal, though not objectively speaking "cheap."

[quote="AMC37.5"]Some universities have site licenses for Mathematica or similar programs. Also, they may offer discounts for students (?)[/quote]

If you buy it online (or win it at ARML!), I think you can download the installer (it's about 450 MB) from Wolfram. You probably have to register it when it installs.

wow.. about half of our crap on our computer is pirated stuff...  :D

Go to HCSSiM. It's a 6 week summer program and they give you a student version for free. There's a forum for alumni somewhere on AoPS. If you want to email me or ask me any questions about it, please feel free. I cannot speak highly enough of that program. I encourage everyone to apply. Check out the website: http://www.hcssim.org


PM me some random words [b]in the right category[/b] to me and I will post the ridiculous stories!!!
It will be up to you to vote for whose is the best.

[color=red][size=200]DUE NOV 23[/size]
[/color]
[b]Sign ups:[/b]

Format:

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Same animal as above::
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[u][i][b]I have recieved from:[/b][/i][/u]
evilhampster
aleph0

wow talk about comic relief :rotfl:

[u][i][b]I have recieved from:[/b][/i][/u]
evilhampster
aleph0
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You must send in your entry by 7:00 PST or it will not be accepted.

I would like you to post the results here in this thread instead of starting a new topic each.

Okay.

Will do in future.

Is it possible to add a poll afterwards?

Another question.

Should I not make a new thread for each Mad Libs #1, #2?

Probably :wink:

Just make a votecount like in mafia.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

$ A \equal{} \begin{pmatrix}2 & 4 & 4 \\
8 & 6 & 8 \\
0 & 4 & 6\end{pmatrix}$  prove that $ det(I \minus{} A^{2006})$ is not equal to 0

[hide]
We will prove that none of the eigenvalues of $ {A}^{2006}$ is $ 1$, (Then none of the eigenvalues of $ I\minus{}{A}^{2006}$ is $ 0$ so the multiplication of the eigenvalues of $ I\minus{}{A}^{2006}$ (the determinant of the matrix) is not $ 0$.
Note that the characteristics polynomial of $ A$ is the value of $ x$  that statisfies :
$ (2\minus{}x)(6\minus{}x)(6\minus{}x)\plus{}32.4\minus{}32(2\minus{}x)\minus{}32(6\minus{}x)\equal{}0$
$ (2\minus{}x)(6\minus{}x)(6\minus{}x)\plus{}32(2x\minus{}4)\equal{}0$
$ (2\minus{}x)(6\minus{}x)(6\minus{}x)\minus{}64(2\minus{}x)\equal{}0$
$ (2\minus{}x)(x^2\minus{}12x\minus{}28)\equal{}0$
$ (2\minus{}x)(x\minus{}14)(x\plus{}2)\equal{}0$
$ x\equal{}2$ or $ x\equal{}14$ or $ x\equal{}\minus{}2$
Then the eigenvalues of $ {A}^{2006}$ is $ x\equal{}{2}^{2006}>1$, or $ x\equal{}{14}^{2006}>1$ or $ x\equal{}{\minus{}2}^{2006}>1$. Then all eigenvales of $ {A}^{2006}$ is greater than $ 1$..
[/hide]

Alternately, $ I\minus{}A^{n}\equiv I\mod 2$ for all $ n\ge 1$, so its determinant is odd. No detailed calculation required.


Mods can move this one around as they will so long as i know its location.
dunno if its in the right forum, so the first responses please tell me.

I don't think it is, but AoPs is CRRRRAZY as folk.

[hide=" :wallbash: "]ok basically, I saw a cool degenerating sin wave.
simply understood to be the product of the sin wave and $e^{-mx}$

I thought it would be much MUCH cooler if, starting at (0,0) a system of circles could be drawn that touches 4 things:

$\pm e^{-mx}$ 1 solution for each
$h\pm r$ 1 solution for each tangent circle; a continuance in this system.

I can't even get 1 circle to appear at x-coordinate h that matches the first characteristic (but i can get close!).

[/hide]

savior gets this cup :  :first:

two questions

1) What do you mean by system of circles?

2) What do you mean by "touches". Do you mean "is tangent to?"

by system i am referring to the fact that there is a circle tangent to said circle, that meet the lister requirements.

tangent i do believe is the correct term for this.

one more thing, for those who attempt to solve, it is false to assume the radius is e^{-mx} for point h.

I uploaded a very rough approximation...just plugging in until it looked ok.
[hide="Here it is"] [url=http://imageshack.us][img]http://img477.imageshack.us/img477/8537/aopsmf6.png[/img][/url]
[/hide]
EDIT: notice I use $m = \frac{1}{5}$

P.S.
             I got sick of the third circle. You get the idea, I assume.


logarithms
number theory
inequalities open
inequalities

Using log 2 and log 3

Prove that 2+6+18+54+...... upto 30 terms is greater than $2*(10^{14})$

[quote="Minh Can"]2+6+18+..+2.3^(29) = 3^(30) - 1 = 2.0589.10^(14) - 1 > 2.10^(14). only need a ti-83. suggestion: may be useful to look into mathematica or maple V because it can help us handling large numbers and solve a larger class of problem ( number theory, algebra, etc..). :lol:[/quote]

how did you equal $3^{30}$ to 2.0589*$10^{14}$
can you please explain that step


[hide]I don't know if this has been asked before although I doubt it!![/hide]

I am an international student in India interested in applying to the US. I am planning to take the SAT II Math Level 2, Physics and Chemistry exams. I already have the Barron's and Princeton review prep books. They are good , but have too little practice for the actual exam and the matter covered is quite simple(we in India go in a much greater detail for IIT). The official book(with just one test) is not available in India and I don't like the Kaplan books at all - too elementary -just telling about the millions they have spent.

Anyways, I would like your suggestions on Online notes and test papers which I can use for practice for the above said SAT II tests. Cost is not a factor. Please help!!!!! :help:

[url=http://www.sparknotes.com/testprep/sat2/]Sparknotes[/url] has some good SATII prep stuff. I'm planning on taking Math IIC and Physics in October, too.

Yes I saw sparknotes!! I want to know your opinion (or if you've heard from someone) on the practice tests available at Sparknotes. DO they exactly conform to the actual exam?? Any other suggestions???

I heard from someone here on AoPS that used ONLY Sparknotes prep for SAT II Physics and got an 800. No doubt he is very intelligent, but I would assume that the materials available there should offer some good preparation.


If $ \dfrac{\overline{AE}}{\overline{EB}}\equal{}\dfrac{1}{3}$ and $ \dfrac{\overline{CD}}{\overline{DB}}\equal{}\dfrac{1}{2}$, find $ \dfrac{\overline{EF}}{\overline{FC}}\plus{}\dfrac{\overline{AF}}{\overline{FD}}$.

[hide]Draw BF and extend it to meet CA at M.
By Ceva's Theorem, we have
$ \frac{\overline{AE}}{\overline{EB}}.\frac{\overline{BD}}{\overline{DC}}.\frac{\overline{CM}}{\overline{MA}}\equal{}1\Rightarrow\frac{\overline{CM}}{\overline{MA}}\equal{}\frac{3}{2}$
Applying Menelaus' Theorem to $ TriangleAEC$, with $ MFB$ as the transversal, we have
$ \frac{\overline{EF}}{\overline{FC}}.\frac{\overline{CM}}{\overline{MA}}.\frac{\overline{AB}}{\overline{BE}}\equal{}1\Rightarrow\frac{\overline{EF}}{\overline{FC}}\equal{}\frac{1}{2}$
Similarly, the theorem on $ TriangleCAD$, with $ BFM$ as transversal gives
$ \frac{\overline{DF}}{\overline{FA}}.\frac{\overline{AM}}{\overline{MC}}.\frac{\overline{CB}}{\overline{BD}}\equal{}1\Rightarrow\frac{\overline{AF}}{\overline{FD}}\equal{}1$
Therefore we have
$ \frac{\overline{EF}}{\overline{FC}}\plus{}\frac{\overline{AF}}{\overline{FD}}\equal{}\frac{3}{2}$[/hide]

[hide="Alternate Solution"]Mass points.
Assign weight $ 1$ to $ B$ and $ 3$ to $ A$.  Thus $ E$ has a weight of $ 4$.  Also, $ C$ has a weight of $ 2$, so $ D$ has a weight of $ 3$.  Thus $ AF\equal{}FD$ and $ FC\equal{}2EF$, so $ \frac{EF}{FC}\plus{}\frac{AF}{FD}\equal{}\boxed{\frac{3}{2}}$.[/hide]


Subtracting LHS from $1990$ and each term of the RHS from $2$, and then dividing both sides by $1991$, it is equivalent to prove that
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{1990} = \frac{1}{996}+\frac{1}{997}+\frac{1}{998}+....+\frac{1}{1990}$

Generally, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{2n}$ is seen to be true for positive integer $n$ by induction

[quote="lajanugen"]Subtracting LHS from $1990$ and each term of the RHS from $2$, and then dividing both sides by $1991$, it is equivalent to prove that
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{1990} = \frac{1}{996}+\frac{1}{997}+\frac{1}{998}+....+\frac{1}{1990}$

Generally, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{2n}$ is seen to be true for positive integer $n$ by induction[/quote]
No need to use induction, Let plus two hands by $\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$.
note that  $\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}=2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})$

[quote="lajanugen"]Subtracting LHS from $1990$, and then dividing by $1991$, then we have
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{1990}$[/quote]

I do not know how you got here.


Find the sum of all even numbers up to 512.... :D.

Only positive numbers, thanks lotrgreengrapes7926....

[quote="ajai"]Find the sum of all even numbers up to 512.... :D.[/quote] Inclusive? You should state that it's only positive integers.  :wink: [hide] $2+4+...+512=2(1+2+...+256)=2(\frac{256\cdot257}{2})=256\cdot257=\boxed{65792}$ [/hide]

[hide]256x514/2, so [color=red]65792[/color][/hide]There is another way I tried that seems like it should work, but isn't working  :( since the sum of the first n odd #'s is n^2, shouldn't it just be 256^2 + the 256th triangular which is 98432?  :maybe:


trigonometry
real analysis
real analysis unsolved

Source: Concurs by ,, Ecole Polytechnique "-1857. (A=3142, B=157, Answer: 1999)
[b]
If $A,B$ are positive integers , how many (real) roots has the equation $\; \; x=A\sin{x}+B \; \; \; ,\; \; x\in {\mathbb R}\; ,\;$   ?
[/b]

Let us denote $g(x) = sin x - mx - n$, where $m = \frac{1}{A}, n = \frac{-B}{A}, m\in(0,1], n < 0$
$g'(x) = cos x - m$ has roots $x_k = arccos (m) + 2k\pi, y_k = -arccos (m) + 2k\pi, k \in Z$
We should count k such that $g(x_k) \cdot g(x_{k+1}) \leq 0$ and the same for $y_k$
If $p = arccos (m), C = g(x_k)$ because $x_{k+1} = x_k + 2\pi$ the above condition means $C \cdot (C - 2m\pi) \leq 0, C \in [0,2m\pi]$
This is equivalent, after computations with $t - 1 \leq k \leq t$ where $t = \frac{-n + sin p - mp}{2m\pi}$, so
$k = [t - 1]$. The number of solution then is $|k|$ because k just founded may be negative.
The same works for $y_k$ and $t = \frac{-n - sin p + mp}{2m\pi}$.
Finally the number of solutions is:
$|[\frac{-n + sin (arccos (m)) - m(arccos (m))}{2m\pi} - 1]| + |[\frac{-n - sin (arccos (m)) + m(arccos (m))}{2m\pi} - 1]|$. :)

[the particular case m = 1/3142, n = -157/3142, gives indeed 999, so i might be right in my proof  :blush: ]


algebra
polynomial
LaTeX
integration
calculus
calculus computations

y^{"} + y' = x

characteristic eqn:
$ r^2\plus{}r\equal{}0$
r=0 or -1

so, $ y\equal{}C_1\plus{}C_2e^{\minus{}x}$

How do I find the particular solution to this question?

$ Y_p(x)\equal{}Ax\plus{}B$
$ Y_p'\equal{}A$
$ Y_p"\equal{}0$

But this yields the wrong answer

The solution has $ Y_p(x) as Ax^2\plus{}Bx$
Why is it like this ??

hello, for a particular solution you must make a ansatz like this:
$ y_p\equal{}Ax^2\plus{}Bx\plus{}C$.
Sonnhard.

Here's a general answer to your question.

We'll observe a differential equation $ y^{(n)} + p_1y^{(n - 1)} + ... + p_ny = f(x)$ where $ p_i$ are real constants, and $ f(x)$ an $ m$-th order polynomial.

1. If $ p_n\neq 0$ then there is an $ m$-th order polynomial which is a particular solution of our equation. (this is not your case!)

2. If $ p_n = p_{n - 1} = ... = p_{n - \alpha + 1}$ (or equivalently, if $ 0$ is $ \alpha$-ordered root of the ODE's characteristic equation), there is an $ m + \alpha$-th order polynomial which is a particular solution of our equation (this is our case- $ \alpha = 1$).

[color=green](edited a $ \text{\LaTeX}$ typo)[/color]

Your differential has the form ay'' +by' +cy =g(x), where a=b=1 ,c=0, g(x)=x (polynomial with degree 1)
To be sure that $ Y''_p(x) \plus{}Y'_p(x)$ is a polynomial with degree n, you must choose $ Y_p(x)$ as polynomial with degree n+1 ...
Sometimes  may $ Y_p(x)$ have to be polynomial with degree n+2

$ y'' \plus{} y' \equal{} x\Longrightarrow e^xy'' \plus{} e^xy' \equal{} xe^x\Longrightarrow e^xy' \equal{} \int xe^x\ dx$.

$ \therefore e^xy \equal{} (x \minus{} 1)e^x \plus{} A\Longrightarrow y' \equal{} (x \minus{} 1) \plus{} Ae^{ \minus{} x}$.

$ \therefore y \equal{} \frac {1}{2}x^2 \minus{} x \plus{} C_1x \plus{} C_2$.

The particular solution is $ y \equal{} \frac {1}{2}x^2 \minus{} x$.

ok thanks for everyones replies I got it now!


the question:
Elisabeth invested 500 pounds for 8 years. At what interest rate, compounded semi annually, would she need to earn for the value of her investment to double?

Equation: A= P(1+r/n)^nt

okay i know how to put all the information into the equation but i dont know how to get rid of the exponent to open up the brakets, if someone could explain or show me how to do it that would be nice
thank you for any help
Magic Man

Use:
$ \log (a^nb^m) \equal{} n\log a \plus{} m\log b$
The application of this is fairly obvios, I think....

\begin{align*}
A&=P\left(1+\frac rn\right)^{nt}\\
2 = \frac AP = \left(1+\frac rn\right)^{nt}
\end{align*}
Set n=2, t=8. You don't even need logs.


This was a spiffy kewl problem -

Find all rational points on the unit circle, implicitly. For those who have already seen this please don't show off meaninglessly and post a solution.

Enjoy!
IW.OAV

For a hint read the crappy hint below...

[hide]It has something to do with number theory... But if you send me SASE with money I can send you the solution.[/hide]

There , its now a spoiler

[hide]Ohh, someone asked me the other week to parametrize the solutions to x^2 + y^2 = 1, I suppose this is what they meant. Now I know what the problem is, I can try to solve it  I think they said it was a Lewis Carroll problem or something?[/hide]

I guess that's what the problem is saying... [I don't mean to over react but please keep those hints in spoiler form... I don't wanna be a dick head or something like that tho...]

Oh sorry, didn't realise that was a hint. But thats basically what implicitly means isn't it?

I will give up the solution once I receive a huge lump sum of cash or once someone thinks they have it solved. Or no one solves it.. [which is impossible come on we have New Zealand Olypmiaders here...]

Key word being New Zealand :P no wait, maybe the key word was olypmiaders.. but really, the annoying thing about IMO and things like that is that if you take all the problems you've seen before you can easily take 6 that you can't solve, put them in the IMO and you get 0. Well no full solutions anyway. It almost comes down to some luck on the day.

Yes that is what it means (refferring to an earlier post)

Honestly, it's all luck... Who can flip the coin the most, or who can close their eyes and pick a multiple choice option the best. I mean it's all luck. It all depends if you can pick the right option and follow it for an hour, or do you pick the wrong option and get stuck on following some BS path for an hour and a half..

My two rupees.
IW.OAV

I would replace the word luck with intuition.

Well, I think I solved it, came to me when I was just thinking about it in my head, strangely enough..

[hide]

Let x^2 + y^2 = 1, with x=a/b, y=c/d.

We have a^2/b^2 + c^2/d^2 = 1 for integers a,b,c,d. Now that means c^2/d^2 = (b^2-a^2)/b^2, and so we can scale c and d so that we can assume d=b. Now we get c^2 = b^2 - a^2, which is a nice easy pythagorean triple, and we end up with x=(m^2-n^2)/(m^2+n^2) and y = (2mn)/(m^2+n^2) for integers m,n.

My exact thinking went something along the lines of:

Hmm this is a Lewis Carroll problem (as I heard). Wonder if its a trick question, maybe x=1 and y=0 is the only solution or something silly like that. No wait, 36/100 + 64/100 = 1. Oh yeah, any pythagorean triple would work, thats when those denominators are equal. So is it possible to have them not equal? No of course they're equal, when you subtract one to the '1' side the denominators equal there..

in case you were wondering

[/hide].

A try :

(a/b) + (p/q) = 1 is the same as (aq) + (bp) = (bq)
and a + b = c is the same as (a/c) + (b/c) = 1

Now the > 0 solution of a + b = c (with (a,b,c)=1) are well known :

a = 2uv, b = u - v, c = u + v

The solutions are then :
(+-2uv/(u + v), +-(u - v)/(u + v))
and
(+-(u - v)/(u + v), +-2uv/(u + v))

Here's a hint for a more geometric way to think about this problem (this method is actually ueeful for understanding the arithmatic of rational points on elliptic curves as well).

[hide]Say you have a rational point (p,q) on the unit circle. Consider a line passing through (1,0) and (p,q). Then the equation for this line will have rational coefficients right?

Well what about the converse of this? If you take a line through (1,0) with rational coefficents, does it necessarily intersect the unit circle in a rational point?

So what can you say from this? How can you use this as a method for generating all of the rational points on the circle?[/hide]

Yes you can think of it like that as a way to start the problem, and Yes pythagorean triplets do have something to do with the problem. I found it pretty entertaining myself...

Koblitz' Introduction to Elliptic Curves and Modular Forms has an excellent and readable discussion of these and related points in the first chapter.


I took the Introduction to Number Theory class this summer. Now I want to continue studying it. When will the Intermediate class likely be offered?

Also, does anyone know of a book that would be helpful? I've looked on the AoPS wiki but the next book they have listed (Number Theory: A Problem Solving Approach by Titu Andreescu and Dorin Andrica) is listed at the Olympiad level and I'm definitely not ready for that. Any suggestions?

Thanks a bunch!

We will likely offer the Intermediate class this coming summer.

The only things I can think of are Number Theory by Andrews and Introduction to the Theory of Numbers by Hardy and Wright, but those too are pretty difficult.

Like the above poster, I think a classic NT textbook would not be too bad. NT is one of those topics that even an intro course provides quite though provoking material, usually it is in college. I have the book by Niven. The stuff it dwelves into is probably at the olympiad level, but I think if you're ready to jump into theory, any good textbook would work. (Hardy, Stark, Niven)

is AoPS going to come out with a interm number theory book??

Eventually, but it won't be in the next 12-18 months.

Dover books has a number of number theory books you can search for on Amazon.  I have these:

Excursions in Number Theory (Dover Books Explaining Science) by C. Stanley Ogilvy &
An Adventurer's Guide to Number Theory by Richard Friedberg

I had started reading the 2nd book (1st one newly purchased), but our AoPS Intro. to Number Theory book just came, so I think we'll be busy with that for a while.

Snee's Mom

Ive got a couple of number theory e-books if your interested. PM me if you want it


trigonometry
inequalities
integration
logarithms
real analysis
real analysis unsolved

Here is my solution:

I will use the following formula:

$ \frac{\sin x}{1-2t\cos x+t^{2}}=\sum_{m=1}^{\infty}t^{m-1}\sin mt$, $ |t|<1$.

$ \sum_{m=1}^{\infty}\frac{\sin mx}{m}=\frac{\pi-x}{2}$.

We have, using the first formula above, that

$ F(x)=\frac{1}{\sin x}\left[\sin x \frac{\pi-x}{2}-\sin(n+1)x \sum_{m=1}^{\infty}\frac{\sin mx}{m+n}+\sin nx \sum_{m=1}^{\infty}\frac{\sin mx}{m+n+1}\right]$.

Now,

$ \sum_{m=1}^{\infty}\frac{\sin mx}{n+m}=\cos(nx)\left(\frac{\pi-x}{2}-\sum_{k=1}^{n}\frac{\sin kx}{k}\right)-\sin nx \left(A(x)-\sum_{k=1}^{n}\frac{\cos kx}{k}\right)$

and

$ \sum_{m=1}^{\infty}\frac{\sin mx}{n+m+1}=\cos(n+1)x\left(\frac{\pi-x}{2}-\sum_{k=1}^{n+1}\frac{\sin kx}{k}\right)-\sin(n+1)x\left(A(x)-\sum_{k=1}^{n+1}\frac{\cos kx}{k}\right)$,

where

$ A(x)=\sum_{m=1}^{\infty}\frac{\cos kx}{k}=\frac{1}{2}\ln \frac{1}{2(1-\cos x)}$, $ 0<x<2\pi$, however the value of $ A$ is not important for our calculations.

Putting all these things together we get that, assuming that I made no computational mistakes,

$ F(x)=\sum_{k=1}^{n}\frac{\sin kx}{k}>0$

which is positive for $ x\in(0,\pi)$.

The preceding inequality is called if I am not making a mistake Jackson's inequality.

PS. Alex do you happen to know what is the original solution that Fejer proposed for this problem?


geometry
cyclic quadrilateral
power of a point
radical axis
geometry proposed

let ABCD a cyclic quadrilateral and  point P such that <PDC+<BAP=<BPC .the perpendiculars through P meet  AB,AD,DC at E,F,G repectively .prove that BPC~EFG.
*sorry for last mistakes.

did any one solve this problem?
i have a solution for this problem but that is long  and i need a simple way for it.

I dunno if this is "long".......forgive me if it is.......

Mark point T such that <CPT=<CDP and <TPB=<PAB (hence <CPB =<CDP+<PAB). Then PT is tangent to both circles PAB and CPD, which must be extenally tangent to each other. So PT is the radical axis of the two circles.

But AB and CD are the radical axes of circles ABCD  and PAB, and circles ABCD and PDC respectively. Thus AB, CD and PT are concurrent, say at K. Assume K is nearer to AD then BC (the other case is similar). Let PK cut AD at X.

<GEF=<PEF - <GEP=<PAD - <GKP = (<AXK-<APK)-<GKP = (<KDA+<GKP)-<PBA-<GKP=<KDA-<PBA=<ABC-<PAB=<CBP. Similarly, <EGF=<BCP. The result follows.