calculus
integration
number theory
Diophantine equation
algebra solved
algebra

Prove that there exist distinct positive integers a1, a2, ..., am,
with m \geq 2002, such that the number

(a1a2...am)2 - 4(a12 + a22 + ... + am2)

is a perfect square.

[i]Vietnam IMO team selection test, 2002 - Problem 6[/i]

[quote="grobber"]Should we prove this is true for all m>=2002 or for a fixed m>=2002? Maybe it's clear to you guys, but I don't get it... :D[/quote]

Grobber question is interesting to study, but it quite hard. The question is only asked to find some m>= 2002 and distinct number a1, a2, ..., am satisfy the condition. I think this question is related to Markov's equation. I will think about the sln and will come soon.

Namdung

Well, I never said it was easy... :D What's Markov's eqn?

Markov's equation is diophantine equation of the form

x1^2 + x2^2 + ...+xm^2 = kx1x2...xm (1)

It has many interesting properties, but the following property is main:

If (1) has one positive integral sln then it has infinitely many integral sln and (1) has sln with x1 < x2 < ...< xm.

The main idea here is: if (x1, ..., xm) is sln of (1) then (x1, ..., x(m-1), kx1...x(m-1) - xm) is also sln of (1).

Namdung

I have looked it up and I've found many interesting things here:

[url]http://www.qnet.com/~crux/markov.html[/url]

We prove that there exists m>=2002 and distinct a_1,...,a_m such that 1+a_1^2+...+a_m^2=a_1*a_2*...*a_m (**). In this case we will have (prod a_i)^2-4 sum (a_i)^2=(prod a_i-2)^2. Take distinct a,b>2002 (a<b) and put m=4(a^2-1)(b^2-1)-2. Then the equation x^2+y^2+z^2+m-2=xyz has the solution (2a,2b,2ab). We try to create from this particular solution a solution for (**). We won't impose the condition that a_1,...,a_m are distinct yet. So, we have the solution (2a,2b,2ab,1,1,...,1) for (**). Use Viete rules and you will find that (2a,2b,2ab,2a*2b*2ab-1,1,1,...,1) is a solution for (**) ( just consider the equation t^2-t*(2a*2b*2ab)+(2a)^2+(2b)^2+(2ab)^2+m-3=0, which has the solution 1, so it also has the solution 2a*2b*2ab-1). Similarly, (2a,2b,2ab,2a*2b*2ab-1, 2a*2b*2ab*(2a*2b*2ab-1)-1,1,1,...,1) is a solution for (1). It is clear that if we repeat the same argument we will obtain a solution of (1) in which a_1<a_2<...<a_m (since 2a<2b<2ab<2a*2b*2ab-1<...). I hope it is correct.


The radius of circle $O$ is $20$ , so you use 30-60-90 triangles to get $BC$ =  40 :sqrt: 3 and so $AC$ = 40 :sqrt: 3 since traingle is equilateral. Thus the height of triangle ABC would be 60. Triangle BXY (which is the smaller triangle formed by the tangent line parallel to the base of big triangle) would be $ 60-(20+20)$ because you subtract diameter of circle. These triangles are similar, so their inscribed circles radii should also be similar, so $60/20 = 20/x$ then $x=20/3$ $or$ $6$ and $2/3$

hope my answer is right

In any triangle, 2/3 of a median is the distance from the centroid to the vertex the median intersects. In triangle ABC, O is the centroid. Then OB = 40, and the distance from point of tangency between the circles to B = 20. Similarly, you can draw another base for a triangle with centroid P through the point of tangency between the circles. The radius of circle P is 1/3 of the median length of 20, or 20/3 or [b]6 2/3[/b].

radius of O is 20 so, OB is 40
3x=40-20 x=6 2/3

Please hide your answers
[hide]if you take any equilateral triangle and inscribe a circle in it, then the radius of the circle is automatically 1/3 the distance from the center of a side the the opposite vertex, so the distance from the edge of the circle (closest to a vertex) closest to that vertex is the same distance.

Apply that to this prob, we can see that the the smaller circle's raduis is 1/3 the larger circles radius (20)

20*(1/3)=6 2/3

>>6 2/3<<[/hide]

[hide]First, draw all three medians of the triangle (they should all run through the center of the center of the larger inscribed circle and end at the points of tangency of the larger inscribed circle and the triangle). Because the medians intersect each other at the center of the larger circle, you know that the distance from a vertex of the triangle to the center of the circle should be 2/3 of the total median length, and that the part of the median from the center of the circle to the point of tangency is 1/3 of the total median length. Since the latter part of the median is also the radius of the circle, and the radius of that circle is 20 meters (i forgot the units), you know that the whole median length is 60 meters, and the diameter of the circle is 20x2, or 40 meters. Now draw a similar triangle that shares a vertex with the larger triangle and is tangent to the point of tangency of the two circles. Since the remaining median length for this second triangle is 60-40, or 20 meters, and we already know that the radius of a circle inscribed in an equilateral triangle= median length/3, than we calculate the radius of the circle to be 20/3 or 6 and 2/3 meters. [/hide]    :)

Draw a line parallel to the base and tangent to both circles.

[quote="NoSoupForYou"]#3. Circle $O$ of radius $20$ is inscribed in equilateral $\triangle ABC$. Circle $P$ is tangent to circle $O$ and to $AB$ and $BC$. Find the radius of circle $P$. Express your answer as a mixed numeral.

Before 1990, only 4 minutes were given to solve each pair of target problems. I know several ways to solve this problem, but they all require some time. Is there a quick way to solve this?[/quote]

BTW, Alex, this was in the book of the 150 Greatest MATHCOUNTS Problems... So you can find a solution there

[quote="NoSoupForYou"]I found it, but it's a long solution and would take more than 2 min. to do.[/quote]

Oh, I see...


find all real coefficient polynomial $ P(x)$ such that $ P(x)P(x\plus{}1)\equal{}P(x^2\plus{}x\plus{}1)$ for all $ x$

If $ deg P \equal{} 0$, it's easy to see $ P \equiv 0,1$
If $ deg P \equal{} 2n \plus{} 1$, there exists $ x_0$ is a root of $ P(x)$. From the equation $ P(x).P(x \plus{} 1) \equal{} p(x^2 \plus{} x \plus{} 1)$, it follows that $ x_0^2 \plus{} x_0 \plus{} 1$ is a root of $ P(x)$, and so on. It follows that $ P(x)$ has infinitive roots, so $ P \equiv 0$

If $ deg P \equal{} 2n$, it follows that $ a_{2n} \equal{} 1$. Denote $ G(x) \equal{} P(x) \plus{} (x^2 \plus{} 1)^n$, which $ deg G \equal{} k < 2n$, which means

$ P(x)P(x \plus{} 1) \equal{} P(x^2 \plus{} x \plus{} 1)$
$ G(x)G(x \plus{} 1) \plus{} G(x)[(x \plus{} 1)^2 \plus{} 1]^n \plus{} (x^2 \plus{} 1)^nG(x \plus{} 1) \equal{} G(x^2 \plus{} x \plus{} 1)$ (*)

notice that $ deg LHS \equal{} 2n \plus{} k > 2k \equal{} deg RHS$, so (*) holds for all real number $ x$ when $ G(x) \equiv 0$, which follows $ P(x) \equal{} (x^2 \plus{} 1)^n$./.

[quote]If $ deg P \equal{} 2n \plus{} 1$, there exists $ x_0$ is a root of $ P(x)$. From the equation $ P(x).P(x \plus{} 1) \equal{} p(x^2 \plus{} x \plus{} 1)$, it follows that $ x_0^2 \plus{} x_0 \plus{} 1$ is a root of $ P(x)$, and so on. It follows that $ P(x)$ has infinitive roots, so $ P \equiv 0$[/quote]

$ x_n \equal{} x_{n\minus{}1}^2\plus{}x_{n\minus{}1}\plus{}1$ and $ x_0$ is a root of $ P(x)$ so all $ x_n$ are root of $ P(x)$
but, how can prove that sequence $ x_n$ have no cycle? (like $ x_n\equal{}x_m (n<m)$)
 if $ x_n$ have cycle, the root of $ P(x)$ isn't infinite... so $ P \equiv 0$ is not true....

[quote="ll931110"][quote="artofproblemsolving"][quote]If $ deg P \equal{} 2n \plus{} 1$, there exists $ x_0$ is a root of $ P(x)$. From the equation $ P(x).P(x \plus{} 1) \equal{} p(x^2 \plus{} x \plus{} 1)$, it follows that $ x_0^2 \plus{} x_0 \plus{} 1$ is a root of $ P(x)$, and so on. It follows that $ P(x)$ has infinitive roots, so $ P \equiv 0$[/quote]

$ x_n \equal{} x_{n \minus{} 1}^2 \plus{} x_{n \minus{} 1} \plus{} 1$ and $ x_0$ is a root of $ P(x)$ so all $ x_n$ are root of $ P(x)$
but, how can prove that sequence $ x_n$ have no cycle? (like $ x_n \equal{} x_m (n < m)$)
 if $ x_n$ have cycle, the root of $ P(x)$ isn't infinite... so $ P \equiv 0$ is not true....[/quote]

Well, I didn't write it in my solution.
Obviously $ x_{n \plus{} 1} \equal{} x_n^2 \plus{} x_n \plus{} 1 > x_n$  :)
Anyway, thanks to point it out[/quote]

but, root of P(x) can be complex number...
 $ x_n$ is complex number and, $ x_n$ is root of $ p(x)$
then $ x_{n \plus{} 1} \equal{} x_n^2 \plus{} x_n \plus{} 1 > x_n$ is not ture.... ( ex) when $ x_n\equal{}i$ then $ x_{n\plus{}1}\equal{}x_n$)

[quote="artofproblemsolving"][/quote][quote="ll931110"][quote="artofproblemsolving"][quote]If $ deg P = 2n + 1$, there exists $ x_0$ is a root of $ P(x)$. From the equation $ P(x).P(x + 1) = p(x^2 + x + 1)$, it follows that $ x_0^2 + x_0 + 1$ is a root of $ P(x)$, and so on. It follows that $ P(x)$ has infinitive roots, so $ P \equiv 0$[/quote]

$ x_n = x_{n - 1}^2 + x_{n - 1} + 1$ and $ x_0$ is a root of $ P(x)$ so all $ x_n$ are root of $ P(x)$
but, how can prove that sequence $ x_n$ have no cycle? (like $ x_n = x_m (n < m)$)
 if $ x_n$ have cycle, the root of $ P(x)$ isn't infinite... so $ P \equiv 0$ is not true....[/quote]

Well, I didn't write it in my solution.
Obviously $ x_{n + 1} = x_n^2 + x_n + 1 > x_n$  :)
Anyway, thanks to point it out[/quote][quote="artofproblemsolving"]

but, root of P(x) can be complex number...
 $ x_n$ is complex number and, $ x_n$ is root of $ p(x)$
then $ x_{n + 1} = x_n^2 + x_n + 1 > x_n$ is not ture.... ( ex) when $ x_n = i$ then $ x_{n + 1} = x_n$)[/quote]

You're right, but any polynomial whose its degree is odd always has at least one real root./.

[quote="ll931110"][/quote][quote="artofproblemsolving"][/quote][quote="ll931110"][quote="artofproblemsolving"][quote]If $ deg P = 2n + 1$, there exists $ x_0$ is a root of $ P(x)$. From the equation $ P(x).P(x + 1) = p(x^2 + x + 1)$, it follows that $ x_0^2 + x_0 + 1$ is a root of $ P(x)$, and so on. It follows that $ P(x)$ has infinitive roots, so $ P \equiv 0$[/quote]

$ x_n = x_{n - 1}^2 + x_{n - 1} + 1$ and $ x_0$ is a root of $ P(x)$ so all $ x_n$ are root of $ P(x)$
but, how can prove that sequence $ x_n$ have no cycle? (like $ x_n = x_m (n < m)$)
 if $ x_n$ have cycle, the root of $ P(x)$ isn't infinite... so $ P \equiv 0$ is not true....[/quote]

Well, I didn't write it in my solution.
Obviously $ x_{n + 1} = x_n^2 + x_n + 1 > x_n$  :)
Anyway, thanks to point it out[/quote][quote="artofproblemsolving"]

but, root of P(x) can be complex number...
 $ x_n$ is complex number and, $ x_n$ is root of $ p(x)$
then $ x_{n + 1} = x_n^2 + x_n + 1 > x_n$ is not ture.... ( ex) when $ x_n = i$ then $ x_{n + 1} = x_n$)[/quote][quote="ll931110"]

You're right, but any polynomial whose its degree is odd always has at least one real root./.[/quote]

aha I understand about it..
thank you for nice solution :)


When three numbers are added two at a time, the sums are 29, 46, and 53. What is the sum of all three numbers?

[quote="chess64"]just add up the values and divide by 2. i don't have a calc, but you would do $\frac{29+46+53}{2}$[/quote]

that would be 64.. for those too lazy..

[hide]
$x$, $y$, and $z$ are the numbers. We set up the equations:
$x+y=29$
$x+z=46$
$y+z=53$
Now instead of using gaussian elminiation and expending too much effort, we can add these up and divide by 2.
$x+y+z=\frac{29+46+53}{2}$
$x+y+z=64$ which is our answer (the sum of the three numbers)
[/hide]

[quote="mathnerd314"]What's Gaussian elimination? :huh:

By the way, is there any way to determine the individual values?[/quote]

It's a really complicated way to get the answer to problems like these. The method actually gives you the different numbers, answering your 2nd question.  :)

(to learn it click [url=http://astronomy.swin.edu.au/~pbourke/analysis/gausselim/]here[/url])

Once you know the sum of all three numbers is 64, subtract each given sum of two of the numbers to find the third.

64 - 53=11
64 - 46=18
64 - 29=35

It's not complicated at all!

mathmarsha

[quote="nat mc"]In my head I can see that (29+46+53)/2=64[/quote]

That's believable. You transfer a 1 from 29 to 53, so everything's even, and use the split-the-numerator rule:

$\frac{28+1+46+54-1}{2}=14+23+27=14+50=64$. I was just too lazy.  :D  :D


Could anyone provide a solution and explanation for this question?:

1. For how many three-element sets of positive integers {a,b,c} is it true that: (a)(b)(c)=2310?

Welcome to the forum. I see that you posted what appears to be the same problem in another forum. One forum is enough; most of us read most of the forums. Try to get a sense of the level of the problem and put it (usually) in one of the three problem-solving forums: this Getting Started Forum, the Intermediate Forum, or the Advanced Forum. I think the problem you posted is somewhere along the boundary between Getting Started and Intermediate: when in doubt, post HIGHER, as most forums have too many hard problems and too few easy problems, and the difficulty level of Advanced Forum is unbounded above.

[hide]

The prime factorization of 2310 is 2*3*5*7*11.  We can turn this into 3 factors in 3 ways: 2310 *1*1, (some number)(some other number)*1, or the product of three different numbers.  Our first set will obviously be {2310,1,1}.  For the second way, we need to find two numbers that we can multiply to get 2310.  We can choose 3 factors out of 2*3*5*7*11, and that will be our first entry in the set.  The other will be 2310/(first entry).  The last will be 1.  There are 5 choose 3 = 10 possible first entries, and only one corresonding second entry.  For the last way, we need to choose 2 sets of 2 factors from 2*3*5*7*11.  This can be done by (5 choose 2)*(3 choose 2) = 30 ways.  Note that we have 3 choose two after 5 choose 2 since after we choose the first two, there are only three left.  30+10+1 = 41.

[/hide]

Hopefully my writing is understandable.

nr1337, your solution is absolutely correct, except that you should not consider the set {1,1,2310} because it actually belongs to a 2-element[b] set (the question was asking for distinct elements-yes, I know how badly it was worded). Therefore, simply subtract that one set from the 41 and you are left with 40 sets with three distinct factors. [b]

PS-sorry for making a double post tokenadult, I didn't realize that we should just post a question once.  :lol: [/b]

Oh yeah... I forgot that a set can't have two of the same thing.


abstract algebra
vector
superior algebra
superior algebra unsolved

Let b^2<4ac, a,b,c in R (real numbers). How to prove that R[x]/(ax^2+bx+c) is isomorphic to field of complex numbers?

Thanks

$ 4ac>b^2$ gives $ a\not\equal{}0$.
Consider the $ \mathbb{R}$-algebra homo $ \phi: \mathbb{R}[X]\to \mathbb{C}, X\mapsto {\minus{}b\plus{}i\sqrt{4ac\minus{}b^2}\over 2a}$ with kernel $ (aX^2\plus{}bX\plus{}c)$.
The image $ V$ is an $ \mathbb{R}$-vector space with $ \mathbb{R}\subsetneq V\subseteq \mathbb{C}$.
Then $ \dim_\mathbb{R}\mathbb{C}\equal{}2$ gives $ V\equal{}\mathbb{C}$ and $ \phi$ an epi.
This gives the required iso.


calculus
geometry
3D geometry
sphere
geometric transformation
reflection
integration

A building has a floor in the shape of a regular hexagon with each side equal to 60 feet. The roof of this building is a portion of a sphere with center at the exact center of the floor. Only the part of the sphere lying directly above the floor is included in the roof. The roof touches the floor at each vertex of the hexagon and rises as an arch over each side. Find the surface area of the roof.

To simplify, I will first calculate double the area( Original plus reflection in floor) and then halve it.
Thus, we have (Roof+Image) as a sphere out of which 6 circular sectors (each of radius $ \frac {R}{2}$; $ R \equal{} 60\text{ft}$) have been cut out. I calculated the surface area of circular sector of radius $ \frac {R}{2}$ by the integral $ 2\pi R^2 \int_{\frac {\pi}{3}}^{\frac {\pi}{2}} \cos^2\theta d\theta \equal{} \frac {\pi R^2}{2}(\sqrt {3} \minus{} \frac {\pi}{6}) \equal{} A$(say)
We get the answer surface area $ S \equal{} 2\pi R^2 \minus{} 3A$

That's an appropriate overall plan of attack, but the specific calculation of surface area $ A$ is not correct.

I got my mistake. I had taken  sphere to be broken into cylindrical rings whereas it should be conical. Thus the integral is
$ A \equal{} 2\pi R^2 \int_{\frac {\pi}{3}}^{\frac {\pi}{2}}\cos\theta d\theta \equal{} 2\pi R^2 (1 \minus{} \frac {\sqrt {3}}{2})$
Thus we get $ S \equal{} 2\pi R^2(\frac {3\sqrt 3}{2} \minus{} 2)$

Yes, that is correct.

The class is an ordinary "Calculus II" class, and the students are freshmen with AP Calculus AB credit. And they don't have to do the weekly challenges - only a few make the attempt. As for this being #4, as in where are the others?

#1 was $ \int_0^{\frac {\pi}2}\frac {dx}{1 \plus{} (\tan x)^{\sqrt {2}}}$ - very famous, has been posted here several times.

#2 was something I'm going to forget I asked.

#3 was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167489]this.[/url]

As for #5, try searching the forum for the words "cow and silo."


Is there an enviroment which hides all the text inside it, as though it didn't exist? I can't find it from a casual Google, I think because I don't know what I'm looking for.

I'm not sure why you want that since LaTeX is primarily for printing for which hiding text makes no sense. In books answers are given at the back.
The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=cooltooltips]cooltooltips package[/url] allows you to use popup windows in recent versions of Adobe reader.

Well, I'm writing a test and I'd like to put the solutions right after each problem in the tex file. However, I want to be able to change the solution environment to not display its contents before I print the test. Would it be better just to split the file into question and answer documents?

In that case use the comment package \usepackage{comment} which means that everything between \begin{comment}...\end{comment} is treated as a comment so ignored.

If you look at Lars Marsden post at [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/9ae1351905bfe6b9/9fdcedf1be8cbf7e]Sometimes-comment Environment[/url] you'll see a sophisticated system where you can choose whether or not the solutions are printed.


Let $P_n=(1-x)(1-x^2)\cdots(1-x^n)\ (n\geqq 1)$ and $P_0=1$ and let $_nQ_r=\frac{P_n}{P_r\cdot P_{n-r}}\ (0\leqq r\leqq n)$.
 Prove the following equality.

\[_{n+1}Q_r=_nQ_{r-1}+x^r\cdot _nQ_r\ (1\leqq r\leqq n)\]

it's nothing more than trivial computations.


What is the largest n such that 8^n divides evenly in 100! ?
They get 32... I think it is 13. I don't see why I'm wrong.

the largest power of 2 in $ 100!$ is $ \lfloor \frac{100}{2}\rfloor\plus{}\lfloor \frac{100}{4}\rfloor \plus{}\cdots \lfloor \frac{100}{64}\rfloor \equal{} 50\plus{}25\plus{}12\plus{}6\plus{}3\plus{}1\equal{}97$, so the corresct answer is $ \lfloor \frac{97}{3}\rfloor\equal{}32$.

If it's wrong, and you're sure of it, report an error


after she received a 7% raise, leslie's new salary is 22,470 dollars. what was her former salary ?

Let Leslie's old salary be $ x$.

$ x \plus{} .07x \equal{} 22470$

$ 1.07x \equal{} 22470$

$ x \equal{} \boxed{21000}$


x,y,z are positive real numbers  ,prove that
xyz/(1+5x)(4x+3z)(5y+6z)(z+18)<=1/5120

[quote="jacques zhang"]$ x,y,z$ are positive real numbers  ,prove that
$ \frac{xyz}{(1\plus{}5x)(4x\plus{}3z)(5y\plus{}6z)(z\plus{}18)}\leq\frac{1}{5120}$[/quote]
Try $ x \equal{} z \equal{} 1$ and $ y \equal{} 5.$ :wink:

but this Inequality is true  :lol:
Let $ x,y,z > 0$ Prove that
\[ \frac{xyz}{(1\plus{}3x)(x\plus{}8y)(y\plus{}9z)(z\plus{}6)} \leq \frac{1}{7^4}\]
When do we have equality.

My solution:
Applying AM-GM we have:1+x/2+x/2+x/2+...+x/2(6 times for x/2)>=7.seventh  root(x^6/2^6)
                                   x+4y/3+4y/3+4y/3+...+4y/3>=7.seventh root(x.(4y)^6/3^6)
                                   y+3z/2+3z/2+///+3z/2>=7.seventh root[y.(3z)^6/2^6]
                                   z+1+1+...+1>=seventh root(z)
Mutiplying those 3 inequalities above we get:(1+3x)(x+8y)(y+9z)(z+6)>=xyz/7^4
Flipping this inequality over xyz we get the desired result
The equality occures when z=1,y=3/2 and x=2
PS:Nice problem, apollo :)


In a deck of 20 cards, there are 10 hearts, labeled from 1 to 10, and 10 spades, also labeled from 1 to 10. For each $ 1\leq i\leq 10$, the pair of cards (heart and spade) labeled with the number $ i$ is called the $ i^{th}$ pair.

Four cards are randomly drawn from the deck. For each $ 1\leq i\leq 10$, let $ P(A_i)$ denote the probability that the $ i^{th}$ pair are among the four cards that are drawn.

(1) Find the probability $ P(A_1)$.

(2) Find the probability $ P(A_1\cap A_2)$.

(3) Find the probability $ P(A_1\cup A_2\cup A_3)$.

(4) Find the probability that there is at least one $ j^{th}\ (1\leq j\leq 6)$ pair among the four cards that are drawn.

Thank you for helping me to translate the context, nsato.

[hide="number 1"]
[b]Solution A[/b] When we draw the cards, we will first select from $ 20$ cards, then $ 19,18,17$ cards.
Two of them are important and distinct (the 1 heart and 1 spadeand) the other two unimportant and can be treated as the same. Therefore we can draw the cards in $ \frac {4!}{2!} \equal{} 12$ ways...

$ \Rightarrow P(A_1) \equal{} 12\cdot\frac {1}{20}\cdot\frac {1}{19}\cdot\frac {18}{18}\cdot\frac {17}{17} \equal{} \frac {3}{95}$

[b]Solution B[/b]
There are $ \binom{20}{4}$ ways to pick $ 4$ cards. If we have the 1 heart and 1 spade in our four card selection, then there are $ \binom{18}{2}$ ways to pick the remaining two cards

$ \Rightarrow P(A_1) \equal{} \frac {\binom{18}{2}}{\binom{20}{4}} \equal{} \frac {4\cdot 3}{20\cdot 19} \equal{} \frac {3}{95}$
[/hide]

[hide="number 2"]
If $ A_1$ and $ A_2$ are among the four cards, then they make up all four of the cards. Therefore this is one distinct way to select $ 4$ cards out of a possible $ \binom{20}{4}$ ways...

$ \Rightarrow P(A_1 \cap A_2) \equal{} \frac {1}{\binom{20}{4}} \equal{} \frac {1}{4845}$
[/hide]

[hide="number 3"]
$ P(A_1\cup A_2 \cup A_3) \equal{} P(A_1) \plus{} P(A_2) \plus{} P(A_3) \minus{} ( P(A_1\cap A_2) \plus{} P(A_2\cap A_3) \plus{} P(A_3\cap A_1))$
Fortunately we already have these probabilities from question $ 1$ and $ 2$.

$ \Rightarrow P(A_1\cup A_2 \cup A_3) \equal{} 3\cdot \frac {3}{95} \minus{} \frac {3}{4845} \equal{} \frac {8}{85}$
[/hide]

[hide="number 4"]
This is the same as ;
$ P(A_1\cup A_2 \cup A_3\cup A_4\cup A_5 \cup A_6) \equal{} \frac {6\cdot 3}{95} \minus{} \binom{6}{2}\frac {1}{4845} \equal{} \frac {301}{1615}$
[/hide]


inequalities calculus algebra function domain

This problem comes from my A level textbook, meant to be done with calculus, but a non calculus solution is much easier.

Maximise $ S\equal{}\frac{x^2}{2}\plus{}\frac{16000\sqrt{2}}{x}$

[quote="SimonM"]This problem comes from my A level textbook, meant to be done with calculus, but a non calculus solution is much easier.

Maximise $ S \equal{} \frac {x^2}{2} \plus{} \frac {16000\sqrt {2}}{x}$[/quote]
[quote="MellowMelon"][hide="Hint"]$ S \equal{} \frac {x^2}{2} \plus{} \frac {8000\sqrt {2}}{x} \plus{} \frac {8000\sqrt {2}}{x}$[/hide][/quote]
First is, what is the domain of S? If there is no domain specified, I assume is it is for all real numbers x where S is defined and when that happens, there is no point maximizing it, there is no maximum as $ x\rightarrow\infty$.

Second is... how is MellowMelon's hint useful anyways? If we are going to apply AM-GM, we are minimizing... and AM-GM gets tricky especially with functions of reals...

There's no min (x~0) or max (x~ -oo) over the negatives, no max (x~ oo) over the positives. There's only a minimum over the positives, so the problem writer probably intended for the solution to be the minimum value over positive x. In that case, MM's solution works just fine...

Sorry, I meant to say minimize, for positive reals, sorry about the confusion


algebra
polynomial
greatest common divisor
algebra proposed

The sequence of integers $ a_n$ is given by $ a_0 \equal{} 0, a_n \equal{} p(a_{n\minus{}1})$, where $ p(x)$ is a polynomial whose coefficients are all positive integers. Show that for any two positive integers $ m, k$ with greatest common divisor $ d$, the greatest common divisor of $ a_m$ and $ a_k$ is $ a_d$.

For $ m\equal{}k$ everything is obvious. For wlog $ m>k$ we have $ a_m\equal{}\underbrace{p(\ldots(p}_{m\minus{}k}(a_k)\ldots )\equiv\underbrace{p(\ldots(p}_{m\minus{}k}(0)\ldots )\equiv a_{m\minus{}k}\bmod{a_k}$ i.e. we have $ \gcd(a_m,a_k)\equal{}\gcd(a_{m\minus{}k},a_k)$. Using the decomposition $ m\equal{}qk\plus{}r$ with integers $ q,r$ such that $ q>0$ and $ r\equal{}0,1,\ldots,k\minus{}1$ holds, we can generalise this recursion to $ \gcd(a_m,a_k)\equal{}\gcd(a_r,a_k)$. We identify this updating rule "$ (m,k)\mapsto (r,k)$" as the construction principle of the greatest common divisor of $ (m,k)$, i.e. we have $ \gcd(a_m,a_k)\equal{}a_{\gcd(m,k)}$ as claimed.


Putnam calculus derivative integration symmetry function algebra

Suppose that $ f: [0,1]\to\mathbb{R}$ has a continuous derivative and that $ \int_0^1f(x)\,dx\equal{}0.$
Prove that for every $ \alpha\in(0,1),$
\[ \left|\int_0^{\alpha}f(x)\,dx\right|\le\frac18\max_{0\le x\le 1}|f'(x)|\]

Is it just my imagination, or have I seen this problem on AoPS/Mathlinks not too terribly long ago?

I saw this as a Romanian Olympiad Question for Putnam Seniors in 2005. It was given in a book of theirs, which when i get my hands on, will attempt to find the proposer (if stated)

[hide]
Let $ g(\alpha )\equal{}\int_0^\alpha f(x)\ dx$ and $ M\max_{x\in [0,1]} |f^\prime (x)|$. If $ |g(\beta )|\leq \frac{M}{8}$ for all $ \beta \in [0,1]$ such that $ |g(\beta)|$ is maximum, then it holds for all $ \alpha \in [0,1]$.

Consider such a $ \beta$. If $ \beta$ is equal to 0 or 1, or if $ |g(\beta)|\equal{}0$, we're done ($ g$ is identically 0). Otherwise, $ \beta$ is an extreme value of $ g(t)$, and we get $ f(\beta )\equal{}g^\prime (\beta)\equal{}0$.

If $ \beta \leq \frac{1}{2}$, integrate by parts to get
\[ g(\beta )\equal{}\beta f(\beta) \minus{}\int_0^\beta xf^\prime (x)\ dx\equal{}\minus{}\int_0^\beta xf^\prime (x)\ dx\]
As $ |f^\prime (x)|\leq M$ for all $ x\in [0,\beta ]$, we get
\[ \minus{}M\int_0^\beta x\ dx \leq g(\beta) \leq M\int_0^\beta x\ dx\]
But $ \int_0^\beta x\ dx\equal{} \frac{\beta^2}{2}$ and $ 0\leq \frac{\beta ^2}{2}\leq \frac{1}{8}$, so $ |g(\beta )|\leq \frac{M}{8}$.

If $ \beta \geq \frac{1}{2}$, repeat the argument for $ f(1\minus{}x)$.
[/hide]

I set $ F(\alpha) \equal{} \int_0^\alpha f(x)dx$ and did an expansion at $ \alpha$, evaluated at 1. That gives
$ F(1) \equal{} F(\alpha) \plus{} (1\minus{}\alpha)F'(\alpha) \plus{} \frac{(1\minus{}\alpha)^2}{2}F''(\beta)$
with $ \beta \in [\alpha, 1]$.

But we know $ F(1) \equal{} 0$. At the maximum values of $ |F(\alpha)|$ we will have $ F'(\alpha) \equal{} 0$. This leaves
$ F(\alpha) \equal{} \minus{}\frac{(1\minus{}\alpha)^2}{2}F''(\beta)$
We can assume that the maximum occurs at $ \alpha \ge 1/2$ by symmetry, so we have
$ |F(\alpha)| \le \frac{1}{8}|F''(\beta)| \le \frac{1}{8} \max_{0\le x\le 1} |F''(x)|$
which is the desired result.

Are there any equality cases other than linear functions?

[quote="Xevarion"]Are there any equality cases other than linear functions?[/quote]

from the proof you need $ |f'(t)|\equal{}\sup_{0\leq x\leq 1} |f'(x)|$ for all $ a\leq 1/2$ and all $ a\geq 1/2$ (that is, all $ a$); this can only happen if $ f'$ is constant.

[quote="Xevarion"]I set $ F(\alpha) \equal{} \int_0^\alpha f(x)dx$ and did an expansion at $ \alpha$, evaluated at 1. That gives
$ F(1) \equal{} F(\alpha) \plus{} (1 \minus{} \alpha)F'(\alpha) \plus{} \frac {(1 \minus{} \alpha)^2}{2}F''(\beta)$
with $ \beta \in [\alpha, 1]$.

But we know $ F(1) \equal{} 0$. At the maximum values of $ |F(\alpha)|$ we will have $ F'(\alpha) \equal{} 0$. This leaves
$ F(\alpha) \equal{} \minus{} \frac {(1 \minus{} \alpha)^2}{2}F''(\beta)$
We can assume that the maximum occurs at $ \alpha \ge 1/2$ by symmetry, so we have
$ |F(\alpha)| \le \frac {1}{8}|F''(\beta)| \le \frac {1}{8} \max_{0\le x\le 1} |F''(x)|$
which is the desired result.

Are there any equality cases other than linear functions?[/quote]

Yay, that was the same as my solution! :)

Here is another argument.  The idea is to express the integral $ \int_0^\alpha f$ purely in terms of $ f'$.

Integrating by parts, we get the formula
\[ \int_0^\alpha f(x) \, dx = \alpha f(0) - \int_0^\alpha (x - \alpha) f'(x) \, dx.
\]
In particular (when $ \alpha = 1$), we have
\[ \int_0^1 f(x) \, dx = f(0) - \int_0^1 (x - 1) f'(x) \, dx.
\]
Combining these two equations gives
\begin{align*} - \int_0^\alpha f(x) \, dx & = \alpha \int_0^1 f(x) \, dx - \int_0^\alpha f(x) \, dx \\
& = \int_0^\alpha (x - \alpha) f'(x) \, dx - \alpha \int_0^1 (x - 1) f'(x) \, dx \\
& = \alpha \int_\alpha^1 (1 - x) f'(x) \, dx + (1 - \alpha) \int_0^\alpha x f'(x) \, dx. \end{align*}
Taking the absolute value of both sides gives
\begin{align*} \left| \int_0^\alpha f(x) \, dx \right| & \le \alpha \int_\alpha^1 (1 - x) |f'(x)| \, dx + (1 - \alpha) \int_0^\alpha x |f'(x)| \, dx. \end{align*}
Let $ M$ be the maximum value of $ |f'(x)|$.  Then we get
\begin{align*} \left| \int_0^\alpha f(x) \, dx \right| & \le \alpha M \int_\alpha^1 (1 - x) \, dx + (1 - \alpha) M \int_0^\alpha x \, dx \\
& = \frac {1}{2} \alpha (1 - \alpha)^2 M + \frac {1}{2} (1 - \alpha) \alpha^2 M \\
& = \frac {1}{2} \alpha (1 - \alpha) M \\
& \le \frac {1}{8} M\, . \end{align*}

Now, I feel even more depressed. This was exactly my line of approach during the test but I gave up soon after without working on it for a little bit more.

Here's another proof using convexity.  Let $ M$ be the maximum of $ |f'(x)|$.  Define
\[ g(x) \equal{} \int_0^x f(t) \, dt \plus{} \frac{M}{2} x^2.
\]
By taking second derivatives, we see that $ g$ is convex.  Thus
\[ g(\alpha) \le (1 \minus{} \alpha) g(0) \plus{} \alpha g(1).
\]
When we substitute the definition of $ g$, we get
\[ \int_0^\alpha f(t) \, dt \le \frac{M}{2} (\alpha \minus{} \alpha^2) \le \frac{M}{8} \, ,
\]
which is what we want except for an absolute value.  Replacing $ f$ with $ \minus{}f$ gives us the absolute value.

Thanks!

@FieryHydra: Don't feel depressed.  I'm sure you would have figured out a solution without a time constraint.  Math is not a gunfight.

Ravi, or anybody,

Can you explain how you got the first line after you say "combining these two equations" in your proof.

[quote="Ravi B"]
\[ - \int_0^\alpha f(x) \, dx \& = \alpha \int_0^1 f(x) \, dx - \int_0^\alpha f(x) \, dx
\]
[/quote]
Remember that the problem specified that the integral $ \int_0^1 f(x) \, dx$ is zero.  So I added the integral in, which is okay because adding zero doesn't change a number.

Thanks, Ravi B! I find much consolation in your words.

I think when my class meets for the last time this semester to go over a few of the problems, I'll present Ravi's proof. But maybe one more tweak. Again, let $ M\equal{}\max|f'(x)|.$

Define $ g(x)\equal{}\int_0^xf(t)\,dt\plus{}\frac{M}{2}(x^2\minus{}x).$

Now, $ g(x)$ is convex and $ g(0)\equal{}g(1)\equal{}0.$ Hence, $ g(x)\le0$ on $ [0,1].$

This gives us $ \int_0^xf(t)\,dt\le\frac{M}2(x\minus{}x^2)\le\frac{M}{8}.$

And repeat with the sign of the second term reversed.

Nice refinement, Kent!  I like it.  I wonder whether the following tweak is better or worse:
\[ g(x) \equal{} \int_0^x f(t) \, dt \plus{} \frac{M}{8} (2x \minus{} 1)^2.
\]
This time we have $ g(0) \equal{} g(1) \equal{} \frac{M}{8}$, so $ g(x) \le \frac{M}{8}$.  Squares are nonnegative, so we are done.

why is $ x\minus{}x^2 less than 1/4$ in the last line of merry kentfield's proof of ravis

[quote="sexynsmartjenny"]why is $ x \minus{} x^2 less than 1/4$ in the last line of merry kentfield's proof of ravis[/quote]
Because $ x \in [0, 1]$. The maximum of that function will occur at $ x\equal{}1/2$ and the minimum at the endpoints (this is easy to check by writing it as $ 1/4 \minus{} (x \minus{} 1/2)^2$ or taking a derivative).

[quote="sexynsmartjenny"]why is $ x \minus{} x^2 less than 1/4$ in the last line of merry kentfield's proof of ravis[/quote]

This is easy! :lol:

For $ 0\leq x\leq 1,$ we have $ x(1 \minus{} x)\leq \left\{\frac {x \plus{} (1 \minus{} x)}{2}\right\}^2 \equal{} \frac {1}{4}$, of cource, you can get the result just using completing square.

thank you all.  i realized it when i was in the dining hall.  :blush:  :lol:

After sexynsmartjenny asked why $ x \minus{} x^2\le\frac14,$ it took 6 minutes for Xevarion to respond and 10 minutes for kunny; I trust that I didn't claim anything too mysterious there.

---

One final tweak:

For each fixed $ x\in(0,1),$ define $ A(x)$ so that $ \int_0^xf(t)\,dt \equal{} \frac {A(x)}{2}(x \minus{} x^2).$

This is just solving a linear equation; in fact we can write $ A(x) \equal{} \frac {2}{x \minus{} x^2}\int_0^xf(t)\,dt.$

Now define $ g(y) \equal{} \int_0^yf(t)\,dt \minus{} \frac {A(x)}{2}(y \minus{} y^2).$

Note that $ g'(y) \equal{} f(y) \minus{} \frac {A(x)}2 \plus{} A(x)y$ and $ g''(y) \equal{} f'(y) \plus{} A(x).$

$ g(y) \equal{} 0$ when $ y \equal{} 0,y \equal{} x,$ and $ y \equal{} 1.$ Therefore, by Rolle's Theorem, there exist numbers $ a\in(0,x)$ and $ b\in(x,1)$ such that $ g'(a) \equal{} g'(b) \equal{} 0.$ Then a second application of Rolle's Theorem shows that there exists $ \xi\in(a,b)\subset(0,1)$ such that $ g''(\xi) \equal{} 0.$ But that implies that $ f'(\xi) \equal{} \minus{} A(x).$

So we conclude that there exists $ \xi\in(0,1)$ such that

$ \int_0^xf(t)\,dt \equal{} \minus{} \frac {f'(\xi)}2(x \minus{} x^2).$

It follows from this that

$ \left|\int_0^xf(t)\,dt\right|\le\frac12(x \minus{} x^2)\max|f'(t)|\le\frac18\max|f'(t)|.$

----

And with that version, we arrive at what problem this really is, and from what field of mathematics. Yes, it's always been a calculus problem, but it's now revealed as a calculus problem in the service of numerical analysis. In particular, it's an error estimate for an interpolating polynomial. Here's the full theorem:

Suppose $ f$ is $ C^n$ on $ [a,b]$ and $ f^{[n \plus{} 1]}(x)$ exists and is bounded on $ (a,b).$ Suppose we have points $ a \equal{} x_0 < x_1 < x_2 < \cdots < x_{n \minus{} 1} < x_n \equal{} b.$ Then there exists a unique polynomial $ P(x)$ of degree $ \le n$ such that $ P(x_i) \equal{} f(x_i)$ for $ 0\le i\le n.$ Furthermore, for each $ x\in[a,b]$ there exists $ \xi\in(a,b)$ such that

$ f(x) \minus{} P(x) \equal{} \frac {f^{[n \plus{} 1]}(\xi)}{(n \plus{} 1)!}(x \minus{} x_0)(x \minus{} x_1)\cdots(x \minus{} x_n).$

There are really two parts to the proof of this theorem. The first is the existence of the unique interpolating polynomial, but that is purely a matter of linear algebra. (A numerical analysis textbook should discuss efficient algorithms for doing that algebra.) The second part is proving the error estimate. The proof I gave above contains all of the ideas needed for that - just do that same thing on a larger scale.

Ravi, or anyone, can you explain how you got the thrid line in your proof, the one before you take the absolute value of both sides. I understand that you can pulol out the constant, but wouldnt that give us x-ax if we multiplied back through. I'm probably just forgetting certain properties. Any help? :)

well, even if I can understand the proof, then there certainly is no mystery in your proof.

And I don't understand what the person above me is asking.  There are about five proofs of the problem posted in this thread: the first one using Taylor expansion, the second one using integration by parts, the last few using convex functions with several different tricks.  And the last last one is a generalization.

[quote="Ravi B"]
Combining these two equations gives
\begin{align*} - \int_0^\alpha f(x) \, dx & = \alpha \int_0^1 f(x) \, dx - \int_0^\alpha f(x) \, dx \\
& = \int_0^\alpha (x - \alpha) f'(x) \, dx - \alpha \int_0^1 (x - 1) f'(x) \, dx \\
& = \alpha \int_\alpha^1 (1 - x) f'(x) \, dx + (1 - \alpha) \int_0^\alpha x f'(x) \, dx. \end{align*}
[/quote]
kstefani, are you asking how I got the last line of the chain above?  Remember that $ \int_0^1$ is the sum of $ \int_0^\alpha$ and $ \int_\alpha^1$.  Try that replacement and it should work out.  Sorry, I should have explained the proof better before.

Kent, you are right that this stuff is probably old hat for numerical analysts.  I took a numerical analysis course a couple of decades ago in which the professor had a nice method for generating such error estimates with Taylor's theorem.  Thanks for the reminder.

in Xevarion's proof, why does taking F''(B), B in (0,1), imply that we do not have to consider the rest of the taylor expansion, like the 3rd derivative, 4th, etc.?

There may not be a "rest" of a Taylor expansion. We were not told that $ f$ was infinitely differentiable; we were told that it had a continuous derivative (hence, the $ F$ in Xevarion's proof had a continuous second derivative).

Xevarion was not using a Taylor series, and as I said above, you cannot use a Taylor series here, since it might not exist. Xevarion was using a Taylor polynomial, and quoting the Lagrange or mean value form of the remainder in Taylor's theorem. The degree of the Taylor polynomial exactly matched the amount of information given.

Please look up the exact statement of Taylor's theorem.

Let $ M=\max_{x\in [0,1]}|f'(x)|$.  Perform the exchange of variables $ x\rightarrow\frac{t}{\alpha}$ to obtain $ 0=\int_{0}^1f(x)dx=\int_0^{\alpha}f\left(\frac{t}{\alpha}\right)dt$.  By the mean-value theorem, $ \left| f(x)-f\left(\frac{x}{\alpha}\right)\right|\leq\left(\frac{1}{\alpha}-1\right)xM$.  Hence,
\begin{eqnarray*}\left|\int_0^{\alpha}f(x)dx\right| &=& \left|\int_0^{\alpha}\left( f(x)-f\left(\frac{x}{\alpha}\right)\right)dx\right|\\ &\leq& \left(\frac{1}{\alpha}-1\right)M\int_0^{\alpha}xdx\\ &=& \frac{\alpha(1-\alpha)}{2}M\\ &\leq& \frac{M}{8}\end{eqnarray*}
And the conclusion follows. $ \Box$

Hopefully, my solution is correct, for I am very bad at Calculus.

Another proof would be the following:
We want to maximize the LHS and may thus assume that $ f(\alpha) \equal{} 0$ (otherwise the LHS is larger for the next zero to the right or left of $ \alpha$ of for 1 or 0).
Also, by symmetry we may assume that $ \alpha < 1/2$ (otherwise consider $ f(1 \minus{} x)$ instead of $ f(x)$). As the derivative of $ f$ is continuous, the intermediate value theorem tells us immediately that the graph of $ f|_{[0,\alpha]}$ is contained in the union of two right triangles with base $ [0,\alpha]$,
the right angle at 0 and slope $ \pm M$, where $ M : \equal{} \max|f'(x)|$, and thus the integral is bounded by the area of one of these triangles, which is $ 1/2 \times \alpha \times M\alpha \le M/8.$

Actually, there is only very few details left out, and it is easy to fill them in, in order to turn this into a solution that would get points in the competition.  Also, I like the proof a lot, as it requires no
calcultaions whatsoever. :)

I

[quote="Kent Merryfield"]Yes, that's correct. And it seems to be a genuinely different argument than all of the other ones presented in this topic.[/quote]

Suppose that I wrote the exact same thing on the Putnam exam.  Since I have no experience with writing Putnam exams (I'll write my first exam in 2 years :P), how much will it score (from what)?  Do I need to elaborate on certain parts of what I wrote, or can I leave the trivial proofs for claims like $ \frac {a(1 \minus{} a)}{2}\leq \frac {1}{8}$ out?

boxedexe, I do not know what Putnam graders do or will do; I don't think it's safe to speculate. Personally, I would give you full credit, but I'm not one of the graders, so that doesn't mean much.

Sorry to bring up an old thread, but in Xevarion's proof, why can we assume that the maximum is attained at $\alpha \geq 1/2$ due to symmetry? The only thing I could think of use is the fact that we have $\int_0^\alpha f(x)\,dx = \int_{0}^{1-\alpha} f(1-x)\,dx$, but I don't know how this (or something else) would imply the $\alpha \geq 1/2$.

[quote="Morgoth"]Sorry to bring up an old thread, but in Xevarion's proof,....[/quote]

He never used $F(0)=0\ldots$

Note that, $\int_0^x f-x\times\int_0^1 f\le\frac {\left||f'\right||_{\infty}}{8} $ holds true $\forall\,\mathbb {C}^1\mapsto\,[0,1].$
$\text {LHS} $ can be written as $\int_0^1 x\left(f(xy)-f(y)\right)\mathrm{d}y. $
Applying the inequality
$\left|f(xy)-f(y)\right|\le\,(1-x)y||f'||_{\infty}\,\, ,x(1-x)\le\tfrac 14\forall\, x\in [0,1]. $

Here is a bash to do this.

Step 1: Convert everything to $f'(x)$.

Note $f(x)=f(0)+\int_0^x f'(x) \,dx$

Therefore, we write $$ \int_0^1f(x)\,dx= \int_0^1 f(0)+\int_0^x f'(t) \,dt \,dx = f(0) + \int_{0\le t\le x\le 1} f'(t) \,dx \, dt = f(0)+\int_{t=0}^1 (1-t) f'(t) \,dt$$

Therefore, $f(0)=\int_{t=0}^1 (t-1) f'(t) \,dt$

Step 2: Write this as a linear inequality, and finish.

$\int_0^{a} f(x) dx = af(0)+\int_{t=0}^a (a-t) f'(t) \,dt = a\int_{t=0}^1 (t-1) f'(t) \,dt + \int_{t=0}^a (a-t) f'(t) \,dt = \int_{t=0}^a (at-t) f'(t) \,dt + \int_{t=a}^1 (at-a) f'(t) dt$

Note $at-t, at-a<0$. Therefore by triangular inequality this is at most

$\int_{t=0}^a (t-at)M \,dt + \int_{t=a}^1 (a-at)M dt = \frac M2 (a-a^2) \le \frac M8$

Where $M=\max_{0\le x\le 1} |f'(x)|$

[quote=CANBANKAN][quote=Mathological03][quote=CANBANKAN]Here is a bash to do this.

Step 1: Convert everything to $f'(x)$.

Note $f(x)=f(0)+\int_0^x f'(x) \,dx$

Therefore, we write $$ \int_0^1f(x)\,dx= \int_0^1 f(0)+\int_0^x f'(t) \,dt \,dx = f(0) + \int_{0\le t\le x\le 1} f'(t) \,dx \, dt = f(0)+\int_{t=0}^1 (1-t) f'(t) \,dt$$
[/quote]
Why are you allowed to switch variablrs on the integral from dt dx to dx dt[/quote]

I am not an expert, but since $f'(x)$ is bounded and $\int_0^1 f(x) dx=0$ then $|f(x)|$ is also bounded so you can apply Fubini. Please correct me if I'm wrong.[/quote]

Also, the region $0\leq t\leq x\leq 1$ is compact and $f$ is continuous, hence $f$ is bounded. This gives $f'$ integrable.

we replace $\alpha$ to $\alpha x$ so we have $\int_{0}^{\alpha} f(x)dx=\alpha\int_{0}^{1}f(\alpha x)dx$\\

so $\left|\int_{0}^{\alpha}f(x)dx \right|=\left|\alpha \int_{0}^{1}f(\alpha x)dx \right|=\left|\alpha \left(\int_{0}^{1}f(\alpha x)-f(x)\right)dx \right| \leqslant \alpha\int_{0}^{1} \left|f(\alpha x)-f(x)\right|dx$\\

since we have a derivative over $[0,1]$ we apply LMVT to get:\\

$\frac{f(\alpha x)-f(x)}{x(\alpha-1)}=f'(\xi)$\\

so we have $\alpha\int_{0}^{1}\left|f(\alpha x)-f(x)\right|dx \leqslant \alpha\int_{0}^{1} \left|f'(\xi)x(\alpha-1)\right|dx \leqslant \alpha(1-\alpha) \cdot \max_{0 \leqslant x \leqslant 1}|f'(x)|\cdot \frac{1}{2}$\\

also from $\textbf{AM-GM}$ we have $\alpha(1-\alpha) \leqslant \frac{1}{4}$\\

we have $\[ \left|\int_0^{\alpha}f(x)\,dx\right|\le\frac18\max_{0\le x\le 1}|f'(x)|\]$ $\blacksquare$


[color=blue] [size=150]$17{th}$  Iberoamerican Olympiad[/size]
San Salvador, EL SALVADOR. [2002]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 3.[/b] Sea $ P$ un punto interior al triángulo equilátero $ \triangle ABC$ tal que $ \angle APC \equal{} 120^{\circ}.$ Sea $ M$ la intersección de $ CP$ con $ AB$ y $ N$ la intersección de $ AP$ y $ BC.$ Hallar el lugar geométrico del circuncentro de $ \triangle BMN$ al variar $ P.$[/quote]
Si $ O$ es el centro de $ \triangle ABC,$ es claro que $ P$ se mueve en el arco de circunferencia $ AOC$ que mira a $ AC$ bajo $ 120^{\circ}.$ Entonces si $ \odot(BMN)$ corta a $ BO$ en $ O',$ observando el cuadrilátero cíclico $ BNPO',$ se tiene que $ \angle O'PA \equal{} \angle O'BN \equal{} 30^{\circ}.$ Lo cual implica  $ O \equiv O',$ puesto que $ \angle OPA \equal{} \angle OCA \equal{} 30^{\circ}$ $ \Longrightarrow$ La circunferencia $ \odot(BMN)$ pertenece al haz que pasa por los puntos fijos $ B,O.$ El lugar de su centro es pues la mediatriz de $ BO.$

[quote][b]Problema 4.[/b] En el triángulo $ \triangle ABC$ con lados diferentes, sea $ D$  el pie de la bisectriz interior del $ \angle ABC$ y $ E,F$  las proyecciones ortogonales de $ A,C$ sobre la recta $ BD.$ Si $ M$ es la proyección ortogonal de $ D$ en $ BC,$ mostrar que $ \angle EMD \equal{} \angle DMF.$[/quote]
Basta ver que si $ D'$ es el pie de la bisectriz exterior del $ \angle BAC,$ se tiene $ BD' \parallel AE \parallel CF$ y esto significa que la cuaterna $ (B,D,E,F)$ es la proyección de la cuaterna $ (D',D,A,C)$ sobre la bisectriz del $ \angle BAC$ desde el punto del infinito de su gradiente ortogonal. Como $ (D',D,A,C)$ es  armónica, entonces $ (B,D,E,F)$ también lo es. Por propiedad básica de las cuaternas armónicas si $ DM\perp BC,$ entonces se deduce que $ BC$ y $ DM$ son las bisectrices exterior e interior del $ \angle EMF.$


1.Two solid bodies are rotating along two mutually perpendicular stationary axis with constant angular velocities of $4 rad/sec$ and $3 rad/sec$ .Find the relative angular velocity and angular accelaration(both modulus) as seen from one body of the other?

P.S.:This may not be the right forum to ask but still :How do you[i] make[/i] an gif image and [i]attach[/i] that to the question u want to post.?

When you are posting a problem, go a bit lower, and you will see a part "add an attachment" there will be a line were you can write in the filename, or browse from your computer. press browse, select the .gif image you want to attach, then press the "add attachment" button. And you are done. :D  :D

Thanks Thaakisfox but i also asked how i can make a '.gif' image.Do you have any special software to do so.

once you make an image press "save as" and there just under where you can write the name of the file you can choose in what format you want to save it, press .gif and you are done. :D

[quote="Thaakisfox"]once you make an image :D[/quote]
How to do that  :(

Well, use an image program. Which one depends on taste, platform, and so on.

Just sketch something in paint or gimp.

Now this discussion does not belong to the original problem.

thanks everyone now  let's focus on solving the original problem.


Use the Binomial Formula to prove the following:
(1 + 1/n)^n ≥ 2 for all n ∈ N

Use Newton express
$ (x\plus{}y)^{n}\equal{}\sum_{i\equal{}0}^{n}C_{n}^{k}x^{k}.y^{n\minus{}k}$


Hello,

find all primes $ p$ and $ q$ so that $ pq$ divides $ 2^p \plus{} 2^q$.

[hide=" Partial solution"]WLOG, let $ p \geq q$. We first take the case when $ p \equal{} q$, so we have that

$ p^2 | 2^{p \plus{} 1} \implies p \equal{} q \equal{} 2$. So far, we have that $ (2,2)$ works,

Now that we have covered their equality, let $ p > q$. We know that $ pq |2^p \plus{} 2^q \equal{} 2^q(2^{p \minus{} q} \plus{} 1)$.

[b]CASE 1[/b]: $ q \equal{} 2$

$ 2p | 4(2^{p \minus{} 2} \plus{} 1) \implies p |2^{p \minus{} 2} \plus{} 1$

By Fermat's Little Theorem, we have that $ 1 \equiv 2^{p \minus{} 1} \mod p$, so

$ p| 2^{p \minus{} 2} \plus{} 2^{p \minus{} 1} \equal{} 2^{p \minus{} 2} \cdot 3 \implies p \equal{} 3$

So the solution set in this case is $ (3,2)$.

[b]CASE 2[/b]: $ q > 2$

$ pq | 2^{p \minus{} q} \plus{} 1$

By Euler's Theorem, we have that $ 1 \equiv 2^{\varphi(pq)} \equiv 2^{pq \minus{} (p \plus{} q) \plus{} 1} \mod pq$. Hence,

$ pq| 2^{p \minus{} q} \plus{} 2^{pq \minus{} (p \plus{} q) \plus{} 1}$
[/hide]

[quote="RobRoobiks"]The second part can be found by applying mod 3 and seeing that 3 has to be a prime for either p or q, then solve the equation. Sry, can't do it now. On mobile[/quote]

Well that just means that $ 2^{p\minus{}q} \plus{} 1 \equal{} 3k$ for some natural $ k$. So $ q$ doesn't necessarily have to be $ 3$. We could have that $ pq | k$.

Since no one offered a full solution here, maybe this problem is too hard to intermediate topic, so I am asking moderators to move this into pre-oly section, please.


search
inequalities proposed
inequalities

Let $x,y,z>0$ and $xyz=1$.
Prove that $\frac{1}{\sqrt{4+5x}}+\frac{1}{\sqrt{4+5y}}+\frac{1}{\sqrt{4+5z}} \leq 1$.

[i]Edited by Myth[/i]

See http://www.mathlinks.ro/Forum/viewtopic.php?p=126020#p126020
Though no one wrote solution, we have discussed a number of similar problems. They refer to two general theorems: RCF or IPT. Search forum for them.

Myth, but I remember you have posted a solution for the problem given in Mathlinks contest, round 6. This is just a consequence of that general problem. Am i right?.

Really? I did it (?)  :? Probably you are right.
I just want to say that it is the same thing: IPT or RCF.

for this pro, i have two solutions
but can you tell me ho to find solution for general pro in this forum
thanks

what are IPT and RCF?
i remember very long time ago myth you posted sth about it in a thread, but i've lost the link. Can anybody kindly provide me the link?


"let \[a_1,a_2,...,a_n(n>3)\] be real numbers such that
\[ a_1+a_2+...+a_n\geq n\] and \[a_1^2+a_2^2+...+a_n^2\geqq^2\]
prove that \[max{a_1,a_2,...,a_n}\geq 2\]"
Please help me with this

Hmm... I guess you meant this here: http://www.kalva.demon.co.uk/usa/usoln/usol994.html ...

  darij

do someone have another solution?many thanks


all the points on the circumference of the circle are painted with either of the colours red blue or green.
show that there exists an isosceles triangle with all its vertices of same colour.(all the three clours are used in painting the pints on the circumference)

Use Van der Waerden's theorem. Since W(3,3)=27 it's enough to consider a 27-gon with equal sides.


[quote="jmccray"]Three particles are located on the x-axis as follows: (1) 4.00kg at x=0.300 m.  (2) 10.0kg at x=0.700 m and (3)1.50kg at x=0.960m.  (a) Calculate the moment of inertiaof each particle with respect to the y-axis.  (b) Find the total moment of inertia. (c)  Based on the results above, decide whether it is true that the particle with the smallest massnecessarly contributes the smallest amount to the smallest inertia.  Explain.[/quote]

you know the inertia:$I$ can be calculated this way
$I=mr^2$ where $m$ is mass and $r$ is the rdiouse fram the axis $Y$
so we can calcaulate $I$ for each particle
because the axis $Y$ is the same for them all we can add inertia off all in a scaler way $I_{tot}$=$I_1$+$I_2$+$I_3$
and according to $I=mr^2$  it's obviouse that the particle with smallest mass have the smallest contribution in the amount of inertia [/i]

Really?  What if we have a 1 kilogram mass at r=1.0 and a .9 kilogram mass at r=1.0*10^50 ?

[quote="Immanuel Bonfils"]a) 0,36 ; 4,9; 1,3824 $\,kg.m^2$

b)adding, 6,6424 $\,kg.m^2\,$

c)as you can see in a) it isn't true:the distance is "more important" then mass, as it's squared. (I didn't understand the [b]to the smallest inertia.)
[/b]
Sorry but what means  Source: 9.30 ?[/quote]

yeah!I had mistaken!
we should calculate the amount of $I_1$ $I_2$ $I_3$ and the compare them!


inequalities
LaTeX
inequalities proposed

show that for positive reals a,b,c and positive integer n

\sum_{cyclic} {\frac{a^n}{b+c}\ge\frac{ a^n-1*b^n-1*c^n-1}{2}

[quote="geniusbliss"]show that for positive reals a,b,c and positive integer n

$ \sum_{cyclic} {\frac{a^n}{b+c}\ge\frac{ a^n-1*b^n-1*c^n-1}{2}}$[/quote]

show that for positive reals a,b,c and positive integer n

$ \sum_{cyclic} {\frac{a^n}{b+c}\ge\frac{ a^{n-1}*b^{n-1}*c^{n-1}}{2}}$

this is the sum dimitris x

P.S. sry i am kinda new to latex

[quote="geniusbliss"]show that for positive reals a,b,c and positive integer n

$ \sum_{cyclic} {\frac {a^n}{b + c}\ge\frac { a^{n - 1}*b^{n - 1}*c^{n - 1}}{2}}$

this is the sum dimitris x

P.S. sry i am kinda new to latex[/quote]

It is not true for sufficiently big $ a,b,c$
If you wanted $ \sum_{cyc}\frac{a^{n}}{b+c}\geq \frac{1}{2}\sum_{cyc}a^{n-1}$ then it follows from Holder...

I think he mean that:
$ \sum \frac{a^n}{b\plus{}c} \geq \frac{\sum a^{n\minus{}1}}{2}$
if it's true this is my proof:
suppose that $ a\geq b\geq c$
then with Chebychev:
$ S\equal{}\sum \frac{a^n}{b\plus{}c} \geq \frac{1}{3}(a^n\plus{}b^n\plus{}c^n)\left(\sum \frac{1}{a\plus{}b}\right)$ (*)
by Caushy-Shwarz,  $ \sum \frac{1}{a\plus{}b} \geq \frac{9}{2(a\plus{}b\plus{}c)}$
by Chebychev, $ a^n\plus{}b^n\plus{}c^n \geq \frac{1}{3} \left( \sum a\right)\left(\sum a^{n\minus{}1}\right)$
then with replacing in (*) we find that:
$ S\geq\frac{\sum a^{n\minus{}1}}{2}$

ya sorry peine u got the ques right
u could have directly gone for chev with $ \sum_{cyclic}{a^{n - 1}}$

polskimisiek can u please elaborate with holder
i am not so comfortable with holder

Holder's Inequality:

Suppose $ \alpha \plus{} \beta \plus{} \gamma...\omega \equal{} 1$.

$ (a_1 \plus{} a_2... \plus{} a_n)^{\alpha}(b_1 \plus{} b_2... \plus{} b_n)^{\beta}...(z_1 \plus{} z_2... \plus{} z_n)^{\omega} \ge a_1^{\alpha}b_1^{\beta}...z_1^{\omega} \plus{} a_2^{\alpha}b_2^{\beta}...z_2^{\omega}\plus{}... \plus{} a_n^{\alpha}b_n^{\beta}...z_n^{\omega}$  where $ \alpha,.......,\omega$ are postitve reals  and $ a_i,b_i,........z_i$ are  reals.

[quote="peine"]I think he mean that:
$ \sum \frac {a^n}{b + c} \geq \frac {\sum a^{n - 1}}{2}$
if it's true this is my proof:
suppose that $ a\geq b\geq c$
then with Chebychev:
$ S = \sum \frac {a^n}{b + c} \geq \frac {1}{3}(a^n + b^n + c^n)(\sum \frac {1}{a + b})$ (*)
by Caushy-Shwarz,  $ \sum \frac {1}{a + b} \geq \frac {9}{2(a + b + c)}$
by Chebychev, $ a^n + b^n + c^n \geq \frac {1}{3} ( \sum a)(\sum a^{n - 1})$
then with replacing in (*) we find that:
$ S\geq\frac {\sum a^{n - 1}}{2}$[/quote]
My solution is sort of worse than yours.
[ :oops: hide=My solution(complexer than yours)]
Suppose $ a\ge b\ge c$ Then the two sequences $ a^n, b^n, c^n$ and $ \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are monotonically increasing this implies:
$ \sum\frac{a^n}{b+c}\ge\sum\frac{a^n}{a+b}$
$ \sum\frac{a^n}{b+c}\ge \sum \frac{a^n}{c+a}$
_______________________________________________
(add up)
$ \sum\frac{a^n}{b+c}\ge\frac{1}{2} \sum \frac{a^n+b^n}{a+b}$

It is only required to prove now that $ \frac{x^n+y^n}{x+y}\ge \frac{x^{n-1}+y^{n-1}}{2}$, which easily follows from Chebyshev's inequality.

$ \text{QED}$ :oops: [/hide]
I prefer peine's solution over mine....... :rotfl:

no potla, your solution is also nice! very nice!

thanks peine, but I like your solution more than mine! :blush:

c'mon potlla your solution is very nice
i like the approach


[quote="Ignite168"]Baphomet has 45 Yuan (元). If 1 Euro (€) is pegged at 1.25 US Dollars ($\$$) and 1 € is pegged at 10 元, what is the amount of money in 元s Baphomet will have left over once he buys an item costing $\$$3.25?[/quote]

[hide]$\$$3.25 is the same as 26 元. Therefore, Baphomet will have $45-26=\boxed{19}$ 元 left.[/hide]

[hide]I got 19.

Convert 3.25 dollars into euros to get 26

45-26 =19[/hide]

Your solutions are not thorough enough, how did you know how many dollars equaled a Yuan? There's no exchange rate for that in the problem.

[hide="A thorough solution."]
We take fractions.

$\frac{euro}{yuan}=10$

$\frac{euro}{dollar}=1.25$

$\frac{dollar}{yuan}=\frac{euro}{yuan}*\frac{dollar}{euro}=10*\frac{1}{1.25}=\frac{2}{.25}=\frac{2}{\frac{1}{4}}=8$

Therefore, there are 8 yuan in one dollar.
Therefore, there are $8*3.25=24+2=26$? in 3.25.
Therefore, he has $45-26=\boxed{19}$? left.
[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

Hi, i'm studing David Cox book, and stuck with this one.
Let $ f \in \mathbb{K}[x_1,\ldots,x_n]$. If $ f \notin \langle x_1,\ldots,x_n \rangle$ then $ \langle x_1,\ldots,x_n,f \rangle\equal{} \mathbb{K}[x_1,\ldots,x_n]$.
Thanks for any help!

That's a misleading title- this is simply about verifying the fact that $ x_1,x_2,\dots,x_n$ generate a maximal ideal. In particular, it's the ideal of all polynomials with constant term zero, and adding any polynomial with nonzero constant term would immediately get us a nonzero constant, which is invertible.

Thanks jmerry! I think it's clear now, but how it is possible to show that $ \langle x_1,\ldots,x_n \rangle$ generate only polinomials with constant term zero? Sorry for the title.

this is trivial, just review the definition of a generated ideal.


Euler
function
LaTeX
number theory
prime factorization

If mdc (m, o(m)) = 1, where o(m) is the Euler's function, then m = p1p2...pk, where pi's are primes distincts two by two, and for i different from j, then pi doesn't divide pj -1.

Your notation and wording are strangely archaic, I think some people on this forum might have trouble understanding your problem  :huh:  (also, please try to use $LaTeX$ ... ) I'm pretty sure that this is what it is:

Suppose $m$ is an integer such that $gcd(m, \varphi(m)) = 1$.  Show that $m$ is a product of primes $p_1 p_2 ... p_k$ such that for no $i \neq j$ is it true that $p_i | p_j - 1$.

[hide="Solution"]
It is clear that the prime factorization of $m$ cannot have any powers of a prime (or otherwise that prime would divide $m$ and $\varphi(m)$).

For the second part, suppose some $p_i | \varphi(p_j)$ in the prime factorization of $m$.  Then $\varphi(m) = \prod \varphi(p_k)$, and $p_i$ then divides $m, \varphi(m)$.  Contradiction.
[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ K\unlhd G$ has index $ m$, show that $ g^{m}\in K$ for all $ g\in G$.

If $ K\unlhd G$ has index $ m$ and if $ \gcd(m,n)\equal{}1$, show that $ K$ contains every element of $ G$ of order $ n$.

For the first one, I am thinking that since there are $ m$ sets of the form $ gK$, then if $ g^{m}\notin K$, then $ g^{m}K$ is a new coset, which is a contradiction to the fact that there are only $ m$ cosets.  For the second, I've tried to use contradiction but I haven't been able to use the fact that $ m$ and $ n$ are relatively prime.

For the first: $ G/K$ is a group with $ m$ elements. Thus $ (gK)^m\equal{}K$, and $ g^m\in K$.

For the second: find integers $ a,b$ so that $ am\plus{}bn\equal{}1$. If $ x^n\equal{}e$,
$ x\equal{}x^{am\plus{}bn}\equal{}x^{am}x^{bn}\equal{}x^{am}\equal{}(x^m)^a\in K$

If $ G$ is a finite group and $ H$ is such that $ HH \equal{} H$, then $ H$ is a subgroup of $ G$. If $ H \equal{} G$, there is nothing to prove. Otherwise, pick $ g \in G \setminus H.$  In that case we have that $ G \equal{} Hg^{ \minus{} 1} \cup H$. Therefore, $ Hg\equal{}gH$ for every $ g \in G$ and the proof terminates.

[quote="coquitao"]Otherwise, pick $ g \in G \setminus H$. In that case we have that $ G \equal{} Hg^{ \minus{} 1} \cup H$. Therefore, $ Hg\equal{}gH$ for every $ g \in G$ and the proof terminates.[/quote]
Could you elaborate on this?  Thanks.

If $ g\notin H$, then we have the left and right coset decomposition $ H\cup gH\equal{}G\equal{}H\cup Hg$ (both [b]disjoint[/b] unions). Thus $ gH\equal{}Hg$ or $ gHg^{\minus{}1}\equal{}H$.

I discussed this with someone, and I think this requires $ [G: H]\equal{}2$, which is not necessarily true.

[quote="coquitao"]Well, what we've actually proven here is that the condition $ g^{2} \in H$ implies $ [G: H] \equal{} 2$...[/quote]

But that statement is false eg $ G\equal{}\mathbb Z/2\times \mathbb Z/2$, H trivial.  Since all squares are trivial, all squares are in $ H$, but $ [G: H]\equal{}4$. Or $ G\equal{}\{1,\minus{}1,i,\minus{}i,j,\minus{}j,k,\minus{}k\}$ of quaternions and $ H\equal{}\{\pm1\}$, again $ [G: H]\equal{}4$.

I have some doubts you can show the following step without $ [G: H]\equal{}2$:

[quote="coquitao"]Otherwise, pick $ g \in G \setminus H.$  In that case we have that $ G \equal{} Hg^{ \minus{} 1} \cup H$.[/quote]

[quote="JRav"]If $ H$ is a subset of a group $ G$ closed under the group operation, and $ g^{2}\in H$ for all $ g\in G$, why is $ H$ necessarily normal in $ G$?[/quote]

The subgroup N generated by { g^2 : g in G } is normal, and contained in H, so G/N is a group in which (gN)^2 = 1 for all gN in G/N.  Hence G/N is abelian (by some previous, standard exercise), and so H/N is normal in G/N. This means H is normal in G.

I tried proving it more directly, but while I got lots of surprising equalities, none of them was gH=Hg.  I think such a direct proof would be nice to see.

[quote="JackSchmidt"]I think such a direct proof would be nice to see.[/quote]
For any $ g \in G$, $ g^2\equal{}x$ for some $ x \in H$. So $ g \equal{} x g^{\minus{}1}$.  Now for any $ h\in H$, $ g h g ^{\minus{}1}\equal{} x g ^{\minus{}1} h g^{\minus{}1} \equal{} x (g^{\minus{}1}h g^{\minus{}1} h) h^{\minus{}1} \in H$, because the middle term, $ (g^{\minus{}1} h)^2$ is in $ H$ by assumption.  So $ H$ is normal.

This was not easy to find!

Marvelous thanks!  I realized I was trying to prove that if g^2 was in H (for just one g), the gH=Hg, but that is obviously nonsense (g order 2, H order 2 not containing g, G non-abelian order 6).  Using something like (g^-1h)^2 was exactly what I was missing, thanks.

Sorry guys, I was investing the funny index-2-assumption while reading the post I have answered to.  :blush:


induction
number theory
Diophantine equation

Source: 360 PMC
Find all distinct positive integers $x_{1},x_{2},...,x_{n}$ such that
$1+x_{1}+2x_{1}x_{2}+\cdot\cdot\cdot+(n-1)x_{1}x_{2}\cdot\cdot\cdot x_{n-1}=x_{1}x_{2}\cdot\cdot\cdot x_{n}$.

[b]The only solution to the Diophantine equation

$(1) \;\; 1+x_{1}+2x_{1}x_{2}+\ldots+(n-1)x_{1}x_{2}\cdots x_{n-1}= x_{1}x_{2}\cdots x_{n},$

where $x_{1}, \, x_{2}, \ldots ,x_{n}$ are distinct positive integers, is $x_{i}= i$ for all $1 \leq i \leq n.$[/b]

[u][b]Proof:[/b][/u] By (1) $x_{1}|1$ and $x_{2}|1+x_{1}$, hence $x_{1}= 1$ and $x_{2}=2$. Next assume the only distinct solution of (1) when $n=k$ is $(x_{1},x_{2}, \ldots ,x_{k}) = (1,2,\ldots,k) \: (*).$ In other words,  this is the only distinct solution of the equation

$(2) \;\; x_{k}\;=\; \frac{1+x_{1}+2x_{1}x_{2}+\ldots+(k-1)x_{1}x_{2}\cdots x_{k-1}}{x_{1}x_{2}\cdots x_{k-1}}.$

To find the distinct solutions of (1) when $n=k+1$, we transform (1) to

$(3) \;\; x_{k+1}\;=\; k \:+\: \frac{1+x_{1}+2x_{1}x_{2}+\ldots+(k-1)x_{1}x_{2}\cdots x_{k-1}}{x_{1}x_{2}\cdots x_{k-1}}\, \cdot \, \frac{1}{x_{k}}.$

Hence the fraction ${\textstyle \frac{1+x_{1}+2x_{1}x_{2}+\ldots+(k-1)x_{1}x_{2}\cdots x_{k-1}}{x_{1}x_{2}\cdots x_{k-1}}}$ must be an integer. But this fraction is identical to the fraction on the right side of equation (2). According to the induction hypotheses $(*)$ equation (2) has only the distinct solution $(x_{1},x_{2}, \ldots ,x_{k}) = (1,2,\ldots,k).$  Thus by (3)

$x_{k+1}= k+\frac{x_{k}}{x_{k}}= k+1.$

Consequently the induction hypotheses $(*)$ is true for $n=k+1\;\;$. [b]q.e.d.[/b]


given real numbers $x,y,z$ such that
$a+b=6$
$ax+by=10$
$ax^{2}+by^{2}=24$
$ax^{3}+by^{3}=62$
determine $ax^{4}+by^{4}$.
enjoy it  :rotfl:

[quote="Yuriy Solovyov"]Where do you see varible z?  :D[/quote]
oops you are right  :lol:


Να βρεθούν ολοι οι πρώτοι p,q,r που να ικανοποιούν τη σχέση p στην q + q στην p = r

As βαλει καποιος την εκφώνηση σε Latex

Για να τη γράψεις σε latex είναι πολύ εύκολο . Κοίτα

Για την συγκεκριμένη γραψε  p^q+q^p=r   και περιέκλυσέ το σε δολλάρια (Shift 4)

και θα βγει $ p^q\plus{}q^p\equal{}r$  .

$ p^q\plus{}q^p\equal{}r$

Eυχαριστώ Σιλουανέ

Εχουμε : $ p^q \plus{} q^p \equal{} r$
Χωρις βλαβη της γενικοτητας εστω οτι $ q \equal{} 2$ τοτε η εξισωση γινεται:
$ p^2 \plus{} 2^p \equal{} r$
Eπειδη ο $ p$ ειναι πρωτος τοτε θα ισχυει οτι $ p^2 \equal{} 1mod6$.Oμως $ 2^p \equal{} 2mod6$ η $ 2^p \equal{} 4mod6$ και επειδη $ r$-πρωτος τοτε $ 2^p \equal{} 4mod6$ $ \equal{} >$ $ 2^{p \minus{} 2} \equal{} 1mod3$ και επειδη ο $ p$-περιττος πρωτος  τοτε o $ p \minus{} 2$ περιττος  οποτε καταληγουμε οτι δν εχει λυσεις η ισοτιμια γιατι μονο αρτια δυναμη του 2 αφηνει υπολοιπο $ 1mod3$.Μενει να κοιταξουμε τις τετριμμενες λυσεις που παραληψαμε δηλαδη για $ p \equal{} 3$ και $ p \equal{} 5$ .Tελικα μοναδικη λυση ειναι η τριαδα $ (p,q,r) \equal{} (3,2,17)$και $ (2,3,17)$

[quote="Αρχιμήδης 6"]Εχουμε : $ p^q \plus{} q^p \equal{} r$
Χωρις βλαβη της γενικοτητας εστω οτι $ q \equal{} 2$ τοτε η εξισωση γινεται:
$ p^2 \plus{} 2^p \equal{} r$[/quote]
Δεν είναι χωρίς βλάβη της γενικότητας. Πρέπει το ένα από τα p, q να είναι άρτιος, αφού διαφορετικά θα είχαμε πρώτος r =΄περιττός + περιττός = άρτιος, που είναι άτοπο. Άρα το ένα από τα δύο είναι πρώτος και άρτιος, δηλαδή δύο. Λύνουμε για το p = 2 και η δεύτερη τριάδα προκύπτει από q = 2 όμοια με την πρώτη.

Παρεληψα την επεξηγηση γιατι βιαζομουν....Εγω απλα ειπα οτι χωρις βλαβη της γeνικοτητας εστω οτι $ q\equal{}2$....Eχεις καμια αλλη λυση?

Όχι, έκανα ακριβώς την ίδια. (Απλά πιο αναλυτικά  :rotfl:  :rotfl: )

:rotfl: Απλα αργω στο γραψιμο και αφαιρω καμια φορα...
pontios εχεις βρει αλλη λυση?Μπορει καποιος?Για παμε...... :wink:

Ίσως να βγαίνει και με θεώρημα του Fermat (δεν το έχω πολυκοιτάξει), πάντως η λύση θα είναι τεράστια. Προτιμώ τον παραπάνω τρόπο.


[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

Construct a triangle given a side $c$ and the angle bisectors $l_a$ and $l_b$.

[quote="Myth"]Construct a triangle given a side $c$ and the angle bisectors $l_a$ and $l_b$.[/quote]

This problem is problem 98 in

  [url=http://g.boutte.free.fr/articles.htm]В. Б. Фурсенко, Лексикографическое изложение конструктивных задач геометрии треугольника, Математика в школе 5, 6 / 1937[/url].

Generally, this construction problem cannot be solved using compass and ruler.

  Darij


geometry
perpendicular bisector
geometry unsolved

PROBLEM (Lokman GÖKÇE): Point $P$ in the $\triangle ACD$. $m(ACP)=18^{o},m(DCP)=54^{o}$,$m(CDP)=42, m(ADP)=30^{o}$.
Find measure of $m(CAP)$.

Suppose that ∠$PAD$=x.

According to Ceva theorem,

$sinxsin42^{o}sin18^{o}=sin(36^{o}-x)sin54^{o}sin30^{o}$ $\Rightarrow$

$sin(72^{o}-x)=2cos24^{o}sinx$ $\Rightarrow$

$sin(72^{o}-x)sin24^{o}=sinxsin48^{o}$ $\Rightarrow$

$-\frac{1}{2}[cos(96^{o}-x)-cos(48^{o}-x)]=-\frac{1}{2}[cos(x+48^{o})-cos(x-48^{o})]$

$\Rightarrow$ $cos(96^{o}-x)=cos(x+48^{o})$ $\Rightarrow$

$96^{o}-x=x+48^{o}$ $\Rightarrow$ $x=24^{o}$ $\Rightarrow$ ∠$CAP=12^{o}$

look solution, pure geometry  :lol:   [url]http://forogeometras.com/index.php?topic=123.0[/url]

Thx. for intrest.I post my solution:

Since angles of $ \triangle ACD$ are $ 36,72,72$ we can construct circumpentagon of the triangle.$ m(EDP)=30+36=66$.$ m(DCF)=m(BCF)=54$That is, $ CF$ is perpendicular bisector of $ AE$. $ PA=PE$ and $ \triangle APE$ is isosceles.We will prove $ \triangle APE$ equiangular.

Let's draw an equiangular $ \triangle AEP'$.Point $ P'$ must lie on $ CF$.Also, $ EP'=AP'=AE=ED$ and $ \triangle EDP'$ is isosceles.Easly, we see that $ m(P'ED)=48,m(EDP')=m(EP'D)=66$. $ m(EDP)=m(EDP')$ and so, we conclude $ P=P'$.Now, $ m(CAP)=12$

Lokman GÖKÇE


function
real analysis
real analysis unsolved

Consider the differential equation (t^2-3t)y"+ty'-(t+3)y=0 with initial conditions y(1)=2 and y'(1)=1

Find the interval of t where the equation has a unique solution

I am not sure how to tackle this type of problms...Should I develop a power series solution and then demand something?......Or am I wrong!!!

I think that the intent was just to check your understanding of the existence and uniqueness theorem for second order linear ODE. If you divide be the coefficient at $y''$, you'll get an equation $y''+f(t)y'+g(t)y=0$ with certain functions $f$ and $g$ having a couple of discontinuities. Then all you need to do is to find the largest interval containing the initial point $t=1$ that doesn't contain any discontinuities of $f$ or $g$. It is true that sometimes the unique solution may exist beyond this interval but such cases are exceptional and I do not think your teacher or your textbook author was thinking of such possibility here.

The discontinuities are at t=0 and t=(+-)sqrt[3].

So the interval should be
(-infinity,-sqrt[3])U(-sqrt[3],sqrt(3)])U(sqrt[3],+infinity)

Is this what you are trying to say?

Sorry,there was a silly calculation mistake.

(t^2-3t)y"+ty'-(t+3)y=0    <--divide each term by (t2-3t)
y"+(t/(t^2-3t)y'-[(t+3)/(t^2-3t)]y=0

This can be simplified by factoring out the denominators:

y"+(t/t(t-3))y'-[(t+3)/{t(t-3)}]y=0  <----the ts will cancel out in the second term leaving: y"+(1/t-3)y'-[(t+3)/{t(t-3)}]y=0.

Now look at each term and determine which values will make the equation undefined.

t=3 in both the second and third term, and t=0 in the third term.
Therefore t is not equal to 0 or 3.
So, the intervals that t can have are:
(-∞,0), (0,3), and (3,∞).
Noting that the initial conditions require a time of t=1, the first and last intervals are not possible, therefore (0,3) is the correct interval.

[quote]... therefore (0,3) is the correct interval.[/quote]
Yes, that is surely the intended answer.


For all real numbers $ a,b,c,d$, prove that $ \frac{a\plus{}b\plus{}c\plus{}d}{(1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)(1\plus{}d^2)}<1$.

[hide]A quick solution for a weak inequality:
Since $ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)(d^2\plus{}1)>(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}1)>/a/\plus{}/b/\plus{}/c/\plus{}/d/$(e.g:using $ a^2\plus{}1/4 \geq /a/$)(Done since $ a\plus{}b\plus{}c\plus{}d</a/\plus{}/b/\plus{}/c/\plus{}/d/$).
Btw, we can add some terms into the numerator to make it more interesting,such as $ (ab\plus{}bc\plus{}ac)^2/3$ an the inequality is still true.[/hide]


Prove that:
(i) f(x) + g(x) = (f+g)(x)
(ii) f(x) - g(x) = (f-g)(x)
(iii) f(x) * g(x) = (fg)(x)
(iv) f(x) / g(x) = (f/g)(x), if g(x) is not equal to 0.

How else are you defining $ f \plus{} g, f \minus{} g, fg$, and $ f/g$?


Let be $ k\in \mathbb{R}$ constant . Find the values of $ E\equal{}\frac{(a\plus{}b\plus{}c)^3}{abc}$ , where $ a,b,c \in \mathbb{R}$ such that $ \frac{b\plus{}kc}{a}\equal{}\frac{c\plus{}ka}{b}\equal{}\frac{a\plus{}kb}{c}$ .

Case 1: k>1
WLOG assume c is the largest.  Then $ c\ge b\ge a$, or else the 2nd term will be easily larger than the third.  But, if we want the first term not to be larger than the 2nd term, then $ c\ge a\ge b$.  So a=b.  Symmetrically, they all have to be equal.  That means E=27 for this case.

i think this works, because i use it a lot


Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6,
7, 8, 9, or 10, with each number entered on four slips. Four slips are
drawn from the hat at random and without replacement. Let $ p$ be the probability
that all four slips bear the same number. Let $ q$ be the probability that
two of the slips bear a number $ a$ and the other two bear a number $ b\ne a$.
What is the value of $ q/p$?

We can find the ratio of the number of ways instead of the ratio of the probabilities (assume that order does not matter).

The number of ways for them to be the same is $ 10$, and there are $ \dbinom{10}2\dbinom42\dbinom42\equal{}1620$ ways for there to be two distinct pairs of the same number, so our answer is $ \frac{1620}{10}\equal{}\boxed{162}$.

Why do we have to multiply by $\binom{4}{2}$ if the slips of paper for each number are the same thing? Why can't it just be $\frac{\binom{10}{2}}{10}$

[quote=droid347]Why do we multiply by $\dbinom42$ twice?  Doesn't doing is once determine the positions of the $b$ slips?[/quote]

Bump. Same question.

Ohhhh never mind. I got it. The $\dbinom{4}{2}$ is for choosing 2 of the 4 slips from the hat, of the same number.

The order does not matter in this solution.

We can assume that order does matter instead.
To pick 4 of the same number, first pick any slip. Then, the rest of them are determined. So, there are $40*3*2*1$ ways to do this.
To pick 2 of one number and 2 of the other, you can consider the following configurations: $AABB, ABAB, ABBA$
Each of these contributes $40*36*3*3$ possibilities. Thus $q/p=\frac{3*40*36*3*3}{40*3*2*1}=18*3*3=\boxed{162}$


[quote="Shivar"]How would I place large parantheses around the fractions? Sorry my $LaTeX$ is messy; I'm a n00b.

[code]
\\\\
437$\displaystyle\frac{-103 + i\sqrt{8619}}{874}^2
+ 103\displaystyle\frac{-103 + i\sqrt{8619}}{874} + 11 = 0$\\\\
= 437$\displaystyle\frac{10609 + i^2(8619)}{874} + \frac{-10609 + 103i\sqrt{8619}}{874} + 11 = 0$\\\\
= 437$\displaystyle\frac{995}{437} + \frac{-10609 + 103i\sqrt{8619}}{874} = -11$\\\\
[/code]

Thanks very much.  :D[/quote]
If you just used
[code](\frac{1}{2})[/code],
you would have $(\frac12)$.
Instead, try to use
[code]\left( \frac{1}{2} \right)[/code],
which produces $\left( \frac12 \right)$.  :)

By the way, to type the word "LaTeX" in LaTeX, use "\LaTeX", which makes $\LaTeX$. (Wow, that sounded confusing.  :P )

Yay! Thanks.

Edit: This is my other account. Are we allowed to have two? I thought this was a better name. :p


Let $a=\dfrac mn$, $b=\dfrac np$ and $c=\dfrac pm$. Then the inequality becomes
\[ \sum \frac {m^2}{np} + 3 \geq \sum \frac mn + \sum \frac mp \] Multiply both sides by $mnp$ and we get
\[ \sum m^3 + 3mnp \geq \sum mn(m+n) \ \Leftrightarrow \ \sum m(m-n)(m-p) \geq 0 \] which is obviously true :)

[quote="vinoth_90_2004"]When abc=1,
a^2 * b + b^2 * c + c^2 * a + 3 >= a + b + c + ab + bc + ac.[/quote]

You have forgotten the condition that a, b, c are positive (else, the inequality is not true). So the problem should read:

[color=blue][b]Problem.[/b] If a, b, c are three positive reals satisfying abc = 1, then
$a^2b+b^2c+c^2a+3\geq a+b+c+bc+ca+ab$.[/color]

Valentin's solution is cool, but there is a slightly simpler way to state it:

Let $x=\sqrt[3]{a^2b}$, $y=\sqrt[3]{b^2c}$, $z=\sqrt[3]{c^2a}$. Then,

$x^3=a^2b$, $y^3=b^2c$, $z^3=c^2a$;

$x^2y=\left(\sqrt[3]{a^2b}\right)^2\sqrt[3]{b^2c}=\sqrt[3]{a^4b^4c}=ab\sqrt[3]{abc}=ab\sqrt[3]{1}=ab$ and similarly $y^2z=bc$, $z^2x=ca$;

$y^2x=\left(\sqrt[3]{b^2c}\right)^2\sqrt[3]{a^2b}=\sqrt[3]{a^2b^5c^2}=b\sqrt[3]{\left(abc\right)^2}=b\sqrt[3]{1^2}=b$ and similarly $z^2y=c$, $x^2z=a$;

$xyz=\sqrt[3]{a^2b}\cdot\sqrt[3]{b^2c}\cdot\sqrt[3]{c^2a}=\sqrt[3]{a^3b^3c^3}=abc=1$.

But by Schur's inequality,

$x^3+y^3+z^3+3xyz\geq y^2z+z^2y+z^2x+x^2z+x^2y+y^2x$.

Using the above relations, this becomes

$a^2b+b^2c+c^2a+3\cdot 1\geq bc+c+ca+a+ab+b$.

In other words, $a^2b+b^2c+c^2a+3\geq a+b+c+bc+ca+ab$, and the proof is complete.

  darij

hey,

yeah i solved it the same way as valentin, using the a=x/y,etc substitution then it is schur.  :)

but that other substitution was so crazily nice and works so perfectly .. how could you have possibly thought of something like that ?

btw, this problem is equivalant to IMO 01/2 ( the problem with
(a + 1 - 1/b) * (b + 1 - 1/c) * (c + 1 - 1/a) <= 1 when abc = 1. )
it is what you get when you expand it out .

Since they both reduce to Schur they are equivalent.

Bomb

A much simpler solution would be to gradiate it


function
induction
inequalities unsolved
inequalities

If $ a_1,a_2,...,a_k$ are natural numbers,and for $ i\neq j$,$ a_i\neq a_j$,please find  the min of const $ m$ such that
\[ \left(2^{\frac{a_1}{2}}\plus{}2^{\frac{a_2}{2}}\plus{}...\plus{}2^{\frac{a_k}{2}}\right)^2\leq m(2^{a_1}\plus{}2^{a_2}\plus{}...\plus{}2^{a_k})\]
and what about the function $ m(k)$?

[quote="Jianxing113725"]If $ a_1,a_2,...,a_k$ are natural numbers,and for $ i\neq j$,$ a_i\neq a_j$,please find  the min of const $ m$ such that
\[ \left(2^{\frac {a_1}{2}} \plus{} 2^{\frac {a_2}{2}} \plus{} ... \plus{} 2^{\frac {a_k}{2}}\right)^2\leq m(2^{a_1} \plus{} 2^{a_2} \plus{} ... \plus{} 2^{a_k})\]
and what about the function $ m(k)$?[/quote]
I don't check clear but I think that: $ m \equal{} 1 \plus{} \sqrt2$.To prove it we have prove that:
+)$ \left(2^{\frac {a_1}{2}} \plus{} 2^{\frac {a_2}{2}} \plus{} ... \plus{} 2^{\frac {a_k}{2}}\right)^2\leq (1 \plus{} \sqrt2) (2^{a_1} \plus{} 2^{a_2} \plus{} ... \plus{} 2^{a_k})$ (using induction)
+)$ \frac {1 \plus{} \sqrt2 \plus{} \sqrt {2^2} \plus{} ... \plus{} \sqrt {2^k}}{\sqrt{1 \plus{} 2 \plus{} ... \plus{} 2^k}}\rightarrow 1 \plus{} \sqrt2$


calculus
integration
trigonometry
real analysis
real analysis unsolved


Everyone (well not me b4 I read this) knows that both Christmas and New Year's Day always fall on the same day of the week. However, in 1939, the year of the outbreak of World War II, Christmas fell on a Monday, and New Year's fell on a Sunday. Why?

[hide]
Was there some German general named Christmas who was defeated in World War II . . .? :maybe:
 [/hide]

i thnk that is the best answer well perhaps they didn't  have to be german

that's pretty good but isn't he answer I was looking for.

I'll change it.

does it have anything to do with Jewish people?

guys u can do it. I got this in 2 minutes

[hide]New Years Eve was on Sunday?[/hide]

Okay a hint this will probably give away

[hide="hint"]Look at the calender[/hide]

[hide]January 1, 1939 was on a Sunday, and December 25, 1939 was on a Monday.[/hide]

Is is because of the International Date line?

No.

It's a puzzle because it's ambiguous. The first sentence means "New Year's is exactly one week after Christmas". In other words, December 25, 2006 is on Monday, and January 1, 2007 will also be on Monday.

However, the second sentence means that January 1, 1939 and December 25, 1939 were on different days of the week.

Now do you see why? ;)

I hit myself over the head (not really) when I read the answer...


conics
ellipse
trigonometry
analytic geometry
graphing lines
slope
geometry

An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4.$  One vertex of the triangle is $(0,1),$ one altitude is contained in the $y$-axis, and the length of each side is $\sqrt{\frac mn},$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

since the altitude is on the y-axis, and it is equilateral, the triangle is symmetric to the y-axis, and the base of the triangle is parallel to the x-axis, that means the line connecting the top vertex to the left vertex makes a 60 degree angle with the other side, and then the x-axis

so the line $(y-1=(\tan{60})x) \cap (x^2+4y^2=4)=(x_0,y_0)$ is one of the cooridiates of the triangle, and reflected over the y-axis gives us the other, so just find $|2x_0|$ for that side that is parallel to the x-axis

Heh, someone always gets to these problems before me. :D

Anyway, the solution.
[hide="Solution"]Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$

Note that the slope of $\overline{AC}$ is $\tan 60^\circ=\sqrt{3}.$ Hence, the equation of the line containing $\overline{AC}$ is
\[y=x\sqrt{3}+1.\]
This will intersect the ellipse when
\begin{eqnarray*}4=x^{2}+4y^{2}&=& x^{2}+4(x\sqrt{3}+1)^{2}\\ &=& x^{2}+4(3x^{2}+2x\sqrt{3}+1) \implies x=\frac{-8\sqrt{3}}{13}. \end{eqnarray*}
Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\left(\frac{8\sqrt{3}}{13},y_{0}\right)$ and $\left(\frac{-8\sqrt{3}}{13},y_{0}\right),$ respectively, for some value of $y_{0}.$

Since we're going to use the distance formula, the value of $y_{0}$ is irrelevant. Our answer is
\[BC=\sqrt{\frac{m}{n}}=\sqrt{2\left(\frac{8\sqrt{3}}{13}\right)^{2}}=\sqrt{\frac{768}{169}}\implies m+n=937.\]
[/hide]

Let the distance from (0,0) downward to the midpoint of the base of the equilateral triangle be n.
We find the point $ (\frac{n\plus{}1}{\sqrt{3}},\minus{}n)$ is on the ellipse.
Subbing in, $ \frac{n^2\plus{}2n\plus{}1}{12}\plus{}n^2\equal{}1$.
Solving, $ n\equal{}\frac{11}{13}$.
So the side length is is $ \frac{2(n\plus{}1)}{\sqrt{3}}\equal{}\sqrt{\frac{768}{169}}$
And the answer is 937.

why is (n + 1 / $ \sqrt {3}, \minus{} n)$ on the ellipse?
EDIT : NVM i see, it is one of the vertices.

[hide="Cool Solution"]
An orthogonal projection maps the ellipse to a circle of radius $ 1$ (we are essentially squeezing the ellipse into a circle). Note that this transformation preserves area and length ratios. (There's a chapter at the end of AoPS vol 2 that explains this in detail).

Let the side length of the equilateral triangle be $ s$ so that the area is $ A\equal{}\frac{s^2\sqrt 3}{4}$. The area of the ellipse is $ E\equal{}2\pi$.

Now our new figure has a circle with area $ E'\equal{}\pi$. The equilateral triangle maps to an isosceles triangle inscribed in the circle. Let the area of the isosceles triangle be $ A'$.

$ \frac{A'}{A}\equal{}\frac{E'}{E}\equal{}\frac 12\implies A'\equal{}\frac A2\equal{}\frac{s^2\sqrt 3}{8}$.

Let the isosceles triangle have base $ b$. Again, by our ratios, the side length is $ 2b$. Then, dropping a perpendicular and using the Pythagorean theorem, the height is $ b\sqrt 3$. The circumradius is $ 1$, so we can form a right triangle with the circumradius and the base. The height of that triangle is $ b\sqrt 3\minus{}1$ and the hypotenuse is $ 1$.

Using the Pythagorean theorem again, we can solve for $ b$ and find that $ b\equal{}\frac{8\sqrt 3}{13}$.

Then $ s\equal{}2b\equal{}\frac{16\sqrt 3}{13}\equal{}\sqrt{\frac{16^2\cdot 3}{13^2}}\equal{}\sqrt{\frac{768}{169}}$, which makes the answer $ \boxed{937}$.
[/hide]

Nice problem!  If this is number 5 then I should easily qualify for JMO next year (even though I missed AME cutoff by 1.5 points grrrrrr).  I solved it by figuring out the bottom right vertex was at (n, 1-nsqrt3) by 30-60-90 theorem (no trig but essentially the same) and plugged this into the ellipse's equation, yield 13nsquared-8nroot3=0, 13n=8root3, n=8root3/13 = root 768/root 169 and 768+169=937.  Sorry, but I don't know latex.  (I know you're probably thinking just type the radical sign).

Am I the only one that payed attention to the part where it says that "$m$ and $n$ are relatively prime"? If that were true then the form given is incorrect. $\gcd(768, 169) \not= 1$ because they clearly have $3$ in common.

Sorry for the revive, but why is the slope of AC equal to tan(60)?

the angle it makes with the $y$-axis is $60^\circ,$ so the slope is $\tan(60).$


I have the webpage ready with the Olympiad results, but I won't be able to put it up until Monday.  There's a question with a student score (from my school) and the statistician is going to check the original scoresheet at BLS on Monday.  It affects whether the student is a finalist or semifinalist, and we -- the statistician and I -- don't want to release the name either way until we know for sure.

I can, however, tell you the cutoff scores:

The cutoff score for finalists is 102.  Either 99 or 100 students were finalists (depending).
The cutoff score for semifinalists is 90.  Either 148 or 147 students were semifinalists (depending).

The top three scorers in the state were from Lexington.  Noah aced the contest; Sway and Stephen Xu both omitted one.
Finalists:  PA has 21; Lexington, 11; Brookline, 8; Roxbury Latin, 7; four schools with four finalists; three schools with three; and down from there.
Finalists and semifinalists combined:  PA, 26; Lexington, 16; Lincoln-Sudbury, 15; BLS, 12; Brookline, Roxbury Latin, Westford, and Worcester, 10; Weston, 9; Longmeadow, 8; five with seven; two with six; none with five; five with four; and down from there.

There were four Thai scholars with scores at or above 90:  two from Worcester, one from PA, one from Deerfield.
There was one eighth grader with a score of 96, from Newton North.
Only students in grades 9-12 can be finalists or semifinalists, so these five students were not included in the numbers above.

More on Monday.

Rick Yanco
Worcester Academy

The problem was resolved quicker than I thought.  Stats are now online:
http://www.wocomal.org/statistics/olympiad/200708/200708o_level1.html

(Students at WOCOMAL schools are highlighted in yellow.  Students who previously attended WOCOMAL schools are highlighted in light yellow.)

Enjoy.

Rick Yanco
Worcester Academy

I actually got 32nd? Wow, I thought that I did terribly...

same....when i walked out of the testing room, i thought i got both 24 and 25 (i spent 15 min.on this one).. until people at my school pointed out my really stupid mistakes... :oops:

ahh...those 12 points could've really boosted my rank... :bomb:

Yeah, my two dumb mistakes cost me 14th...but still, I guess 4th freshman isn't that bad.

lol...only cost me 13 places...which is still pretty bad

Ahh, one wrong. I must have made an error bubbling, all my answers matched up.

[quote="CatalystOfNostalgia"]Heh, I didn't think my place would be so high (9th), I was expecting more like 20th, but I'm not complaining. Good day for Lexington overall though.[/quote]

hey! i'm 20th!  :mad:

are u mocking me?

meh, i did bad this year....there's always next year  :)

hmm......maybe it's because i'm new here but does lexington and phillip andover always pwn this much?

All I know is that Lexington pwns in middle school, probably in high school as well

[quote="Pineappleperson"]hmm......maybe it's because i'm new here but does lexington and phillip andover always pwn this much?[/quote]

basically.

Man, those Thai PG's have the most complicated names I've ever seen.

[quote="CatalystOfNostalgia"][quote="Pineappleperson"]are u mocking me?[/quote]

No, I just said I would have expected more people to have scored higher than me.[/quote]

lol i can't believe u can't tell i was joking  :rotfl:

We had considerably more qualifiers this year than in past... given my performance within the school, I honestly thought I might be borderline for qualifying, but I guess not.

[quote="e-bert"]How many people took this, also, how does one take it as a PG?[/quote]
Many prep schools offer a postgraduate year, sandwiched between high school and college.  The majority of postgraduates are excellent athletes who either have weak academics (and hence need a year to work on school, so they'll do better in college) or who have had poor exposure (and want to be part of a higher-profile program so D1 coaches will notice them).  e.g., Colt Brennan went to Mater Dei in California for four years, then came to Worcester Academy for a year.  Jarrett Jack spent a PG year at Worcester.  I assume similar athletes go to prep schools too, but I'm only familiar with those who came to WA.

Obviously these types of PG's (excellent athletes) do not typically participate in math competitions.  And students with good academic records (e.g., those who participate in math competitions) aren't going to drop $ \$$20K for a finishing year.  (Athletes can often leverage a PG year into a D1 scholarship, so financially it might make sense.)

The other group of PG's that pass through Worcester Academy are Thai scholars:  international students who have already graduated from high school in Thailand, who spend a year at a US or UK prep school to work on their English while applying for college.  Thai scholars are fully supported by the Thai government.  They are some of the finest students you will ever know (and they are selected for that reason).  If you google the four Thai names on this year's MMO, you'll find that each of them attended an International (something) Olympiad over the past few years.  Choate's two Thai scholars last year were both IMO medal winners.  Worcester had a Thai scholar in 2004-05 who had won a gold at the IMO; he was the top scorer on the MMO level 1 in 10/2004, but was ineligible to advance.  At MIT as a freshman he was in the second group of Putnam scorers, 8-14th place or whatever.

So, answering your question:  to take the MMO as a postgraduate, you need to be spending a PG year at a prep school in Massachusetts that participates in the MMO.  I believe the only three are Andover, Deerfield, and Worcester.

Hope this helps.

Rick Yanco
Worcester Academy

[quote="rjyanco"][quote="e-bert"]How many people took this[/quote]
There were 2,677 participants from 66 schools.  Setting aside the number of schools (which we're working on), the number of students is too low.  Some very good schools offer the exam to a ridiculously small percentage of their students:  ABRHS, just 15 out of 1800+ students in the school (less than 1%).  Mass Academy, 7 out of 91 (7.7%, at a math/science magnet school).  Canton, 11 out of 800+ (1.4%).  Lexington, 33 out of nearly 2000 (1.7%), and PA, 49 out of 1000+ (5% of a hand-picked student body).

So some schools only let their very best students participate, which seems awfully shortsighted to me, since it only costs one dollar per additional student.[/quote]

Hmm, at ABRHS, I believe that anyone can participate. The reason we have only 15 students is because only 15 students signed up. Also, it doesn't cost our school anything, since each student must pay $ \$$1 to take it at ABRHS; I'm guessing most schools don't make their students pay for the exam?

Our school charged for the exam also, we had a lot more people take it, but I think more would if it was better publicised.

[quote="rjyanco"][quote="e-bert"]How many people took this[/quote]
There were 2,677 participants from 66 schools.  Setting aside the number of schools (which we're working on), the number of students is too low.  Some very good schools offer the exam to a ridiculously small percentage of their students:  ABRHS, just 15 out of 1800+ students in the school (less than 1%).  Mass Academy, 7 out of 91 (7.7%, at a math/science magnet school).  Canton, 11 out of 800+ (1.4%).  Lexington, 33 out of nearly 2000 (1.7%), and PA, 49 out of 1000+ (5% of a hand-picked student body).

So some schools only let their very best students participate, which seems awfully shortsighted to me, since it only costs one dollar per additional student.[/quote]

It's difficult to get somebody who isn't serious about math to put aside two hours for a math competition. In our case, this means time that would be spent doing homework to go to bed at a godly hour (we do it at 8pm so no excuse from classes). Even our best students sometimes skip certain contests to study for that upcoming history test or that term paper. This is where Exeter excels - they have a substantial body of students who are willing to devote large amounts of time to practice, at the expense of more immediate demands from courses, sports, etc.


Let $ a,\ b,\ c$ be positive reals such that $ a\plus{}b\plus{}c\equal{}3$.

Prove that $ \frac{b^{2}}{a}\left(1\plus{}\frac{b^{2}}{a}\plus{}\frac{b^{4}}{a^{2}}\right)\plus{}\frac{c^{2}}{b}\left(1\plus{}\frac{c^{2}}{b}\plus{}\frac{c^{4}}{b^{2}}\right)\plus{}\frac{a^{2}}{c}\left(1\plus{}\frac{a^{2}}{c}\plus{}\frac{a^{4}}{c^{2}}\right)\geq 9$

That's right! :)

My solution is using $ x(1\plus{}x\plus{}x^{2})\geq 6x\minus{}3\ (x > 0)$.


geometry
geometric transformation
reflection

[hide]$ \{\pm 4, \pm 5, \pm 7 \}$
$ \{\pm 1, \pm 5, \pm 8 \}$
$ \{0, \pm 3, \pm 9 \}$

Just rearrange the numbers in the sets to reflect that $ a < b < c$.[/hide]

[quote="not_trig"][quote="madness"]how can it be ordered if the triples have a definite constraint?  :D[/quote]

Well, it's ordered for you.

@123456789: an explanation would be nice?[/quote]

Oh, I brute-forced it with a program. Not the way you should do it now that the AMCs don't allow calculators.

[quote="the future"]Find all ordered triples of integers $ (a,b,c)$ such that $ 0 < a < b < c$ and that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 90$[/quote]
[hide="Solution"]
Look at the equation $ \bmod 8$.  Notice that $ a^2 \plus{} b^2 \plus{} c^2\equiv 2\bmod 8$.  Hence, one and only one of $ a,b,c$ is divisble by $ 4$ (this is because any square is congruent to $ 0, 1, 4\bmod 4$ and the only way to partition $ 2\bmod 8$ with ones, zeroes, and fours is $ 0 \plus{} 1 \plus{} 1$.  Hence, one of $ a^2$, $ b^2$, or $ c^2$ is divisible by $ 8$, and thus one of $ a$, $ b$, or $ c$ is divisible by $ 4$).  Now, if $ c \equal{}8$, then $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 90\implies a^2 \plus{} b^2 \plus{} 64 \equal{} 90\implies a^2 \plus{} b^2 \equal{} 26$.  With a quick check, we find that $ (a,b) \equal{} ( 1,5)$ as the only solutions.  Clearly, if $ c \equal{}4$, then $ 90 \equal{} a^2 \plus{} b^2 \plus{} c^2 < 3c^2 \equal{} 48$, which is false.  Hence, let $ b \equal{}4$.  Thus, $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 90\implies a^2 \plus{} b^2 \plus{} 16 \equal{} 90\implies a^2 \plus{} c^2 \equal{} 74$.  A quick check reveals no solutions for this case.  Finally, if $ a \equal{} 4$, then $ b^2 \plus{} c^2 \equal{} 74$.  Another check reveals that $ (b,c) \equal{} ( 5,7)$ are the only solutions to this.  Hence, our solutions for $ (a,b,c)$ are $ \boxed{(4,  5,  7), (1, 5, 8)}$. [/hide]

EDIT: Thanks, [b]diophantient [/b]and [b]firecricket91[/b]!  Solution edited.

[quote="The QuattoMaster 6000"][quote="the future"]Find all ordered triples of integers $ (a,b,c)$ such that $ 0 < a < b < c$ and that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 90$[/quote]
[hide="Solution"]
Look at the equation $ \bmod 4$.  Notice that $ a^2 \plus{} b^2 \plus{} c^2\equiv 2\bmod 4$.  Hence, one and only one of $ a,b,c$ is divisble by $ 4$ (this is because any square is congruent to $ 1\bmod 4$ and the only way to partition $ 2$ with ones and zeroes is $ 0 \plus{} 1 \plus{} 1$).  Now, if $ c \equal{} \pm 8$, then $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 90\implies a^2 \plus{} b^2 \plus{} 64 \equal{} 90\implies a^2 \plus{} b^2 \equal{} 26$.  With a quick check, we find that $ (a,b) \equal{} (\pm 1,\pm 5)$ as the only solutions.  Clearly, if $ c \equal{} \pm 4$, then $ 90 \equal{} a^2 \plus{} b^2 \plus{} c^2 < 3c^2 \equal{} 48$, which is false.  Hence, let $ b \equal{} \pm 4$.  Thus, $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 90\implies a^2 \plus{} b^2 \plus{} 16 \equal{} 90\implies a^2 \plus{} c^2 \equal{} 74$.  A quick check reveals no solutions for this case.  If $ a \equal{} \pm 4$, then $ b^2 \plus{} c^2 \equal{} 74$.  Another check reveals that $ (b,c) \equal{} (\pm 5,\pm 7)$ are the only solutions to this.  Finally, if $ a \equal{} 0$, then $ b^2 \plus{} c^2 \equal{} 90$.  We can easily see that $ (\pm 3, \pm 9) \equal{} (b,c)$.  Hence, our solutions for $ (a,b,c)$ are $ \boxed{(\pm 4, \pm 5, \pm 7), (0, \pm 3, \pm 9), (\pm 1, \pm 5, \pm 8)}$. [/hide][/quote]
$ x^{2} \equiv 0 \mod 4 \Rightarrow x \equiv 0 \mod 2$, not $ x\equiv 0 \mod 4$

[quote="The QuattoMaster 6000"][quote="the future"]Find all ordered triples of integers $ (a,b,c)$ such that $ 0 < a < b < c$ and that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 90$[/quote]
[hide="Solution"]
Look at the equation $ \bmod 4$.  Notice that $ a^2 \plus{} b^2 \plus{} c^2\equiv 2\bmod 4$.  Hence, one and only one of $ a,b,c$ is divisble by $ 4$ (this is because any square is congruent to $ 1\bmod 4$ and the only way to partition $ 2$ with ones and zeroes is $ 0 \plus{} 1 \plus{} 1$).  Now, if $ c \equal{} \pm 8$, then $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 90\implies a^2 \plus{} b^2 \plus{} 64 \equal{} 90\implies a^2 \plus{} b^2 \equal{} 26$.  With a quick check, we find that $ (a,b) \equal{} (\pm 1,\pm 5)$ as the only solutions.  Clearly, if $ c \equal{} \pm 4$, then $ 90 \equal{} a^2 \plus{} b^2 \plus{} c^2 < 3c^2 \equal{} 48$, which is false.  Hence, let $ b \equal{} \pm 4$.  Thus, $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 90\implies a^2 \plus{} b^2 \plus{} 16 \equal{} 90\implies a^2 \plus{} c^2 \equal{} 74$.  A quick check reveals no solutions for this case.  If $ a \equal{} \pm 4$, then $ b^2 \plus{} c^2 \equal{} 74$.  Another check reveals that $ (b,c) \equal{} (\pm 5,\pm 7)$ are the only solutions to this.  Finally, if $ a \equal{} 0$, then $ b^2 \plus{} c^2 \equal{} 90$.  We can easily see that $ (\pm 3, \pm 9) \equal{} (b,c)$.  Hence, our solutions for $ (a,b,c)$ are $ \boxed{(\pm 4, \pm 5, \pm 7), (0, \pm 3, \pm 9), (\pm 1, \pm 5, \pm 8)}$. [/hide][/quote]

I think you have too many solutions since you cannot have negative numbers or 0.

[quote="basted"]By an ordered pair (a,b,c) we mean that (a,b,c) is different from (c,b,a) so YES they're ordered :P
What's the problem? It's an easy task because you don't have many choices (just 1,4,9,16,25,36,49,64,81)[/quote]

Well what i meant was wats the pt of calling it an ordered pair if a<b<c bcos there can only be one such pair  :D

[quote="123456789"][quote="firecricket91"]I think you have too many solutions since you cannot have negative numbers or 0.[/quote]

Why can't they be negative numbers or $ 0$? The set of integers isn't only confined to positive numbers.[/quote]

[quote="the future"]Find all ordered triples of integers $ (a,b,c)$ such that $ 0 < a < b < c$ and that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 90$[/quote]

This kinda states that c>b>a>0, meaning they all must be positive and they must be arranged such that c is the biggest, b second, and so forth.

Oh yes, sorry. I didn't see the $ 0$ in the inequality. I thought it was just $ a<b<c$.