probability
greatest common divisor
number theory
relatively prime

When rolling a certain unfair six-sided die with faces numbered 1,
2, 3, 4, 5, and 6, the probability of obtaining face $ F$ is greater
than $ 1/6$, the probability of obtaining the face opposite face $ F$
is less than $ 1/6$, the probability of obtaining each of the other
faces is $ 1/6$, and the sum of the numbers on each pair of opposite
faces is 7. When two such dice are rolled, the probability of
obtaining a sum of 7 is 47/288. Given that the probability of
obtaining face $ F$ is $ m/n$, where $ m$ and $ n$ are relatively prime
positive integers, find $ m\plus{}n$.

I don't understand what this problem  means when it talks about being relatively prime.

In other words, it wants your final answer in simplest form.

[quote="ernie"]If you have relatively prime numbers $ a$ and $ b$, then $ \text{gcf}(a,b) \equal{} 1$.[/quote]
so whats the answer

I don't really understand how they said $47/288$ was equal to that stuff in the 4th line

[hide] If you are given a value, just substitute into the problem. So $p(F)=\frac{m}{n}$ and so $p[op(F)]=\frac{1}{3}-\frac{m}{n}=\frac{n-3m}{3n}.$ So the probability of getting a sum of 7 is if we get F and op(F) or a pair of numbers not including F and op(F) which sum to 7. So the first case is easily $\frac{n-3m}{3n} \cdot \frac{m}{n}.$ The second case is $\frac{4}{6} \cdot \frac{1}{6}$ (note every one of the first 4 numbers forms a bijection with the second). So we have $$\frac{n-3m}{3n} \cdot \frac{m}{n}+\frac{4}{6} \cdot \frac{1}{6}=\frac{47}{288}.$$ By easy calculations we find our solutions are $(m, n)=(5, 24), (1, 8).$ $(5, 24)$ is the only one that works, so $5+24=29$ is our answer. [/hide]

But that nasty algebra should have reminded me not to substitute so wildly. Instead, just let the probability of obtaining face $F$ be $(1/6)+x$. This is easily seen since the problem says F is greater than $\frac{1}{6}$ so it's pretty motivated.

This is nasty!! Im on the easy level and i got this problem aargh :furious:

Yeah. I'm on normal and I got like 5 level 25 problems.

[hide=my sol]
First, denote the probability of getting face $n$ as $P(n)$. (in my original solution I denoted it as $f_n$, but that's irrelevant). Which face $F$ is is irrelevant because the number on the face does not have any effect on the probability. Let $F$ be the face and $G$ be the opposite face. WLOG we can assume that F is face 1 and that G is face 6. (original solution used 2 and 5 but like I said that has no effect on the final answer). Then, $P(\text{faces sum to 7})=2P(1)P(6)+2P(2)P(5)+2P(3)P(4)=\frac{47}{288}$. Then, we know that $P(2)=P(3)=P(4)=P(5)=\frac{1}{6}$. From this we get that $2FG=\frac{5}{96}$ and that $FG=\frac{5}{192}$. (remember that $FG=P(1)P(6)$). We also know that the probabilities must sum to $1$. This gives us that $F+G=\frac{1}{3}$ (add the probabilities for each face together and subtract the constant ones from both sides to get this expression). Substituting in the system of equations, we get $F(\frac{1}{3}-F)=\frac{5}{192}$. Expanding and solving with the quadratic formula, we get the solutions $\frac{1}{6}\pm\frac{1}{24}=\{\frac{3}{24},\frac{5}{24}\}$. We know that it must be $\frac{5}{24}$ because 1. $F>\frac{1}{6}$ and 2. it is the only solution that works when plugged back in to the system. Because $\gcd(5,24)=1$, the integers are already coprime (i.e. the fraction is simplified already). Adding, we get that $5+24=\boxed{29}$.
[/hide]
actually surprised i got a aime problem correct, first try
this was also my last problem before counting victory :)


limit
calculus
derivative
function
calculus computations

$ f'(c)$ exists if and only if $ f'_{\plus{}}(c)$ and $ f'_{\minus{}}(c)$exist and are equal?

how would i show this first for $ f'_{\plus{}}(c) \equal{} \lim_{x\rightarrow c\plus{}}\frac{f(x)\minus{}f(c\plus{})}{x\minus{}c}$ and similarly for $ f'_{\minus{}}(c)$?
(would i have to expand on the standard definition of the derivative?)

Also, does $ f'(c)$ have to exist for the left hand and right hand limits to be equal?

The derivative is really just a special kind of limit. So your question can be restated in terms of limits of the function $ g(x) \equal{} \frac {f(x) \minus{} f(c)}{x \minus{} c}$.
Yes, $ \lim_{x\to c} g(x)$ exists if and only if both one-sided limits exist and are equal. Use the standard definition of the limit to prove this.

(Edited to correct a typo)

[quote="mstrip"][quote="mlok"]The derivative is really just a special kind of limit. So your question can be restated in terms of limits of the function $ g(x) \equal{} \frac {f(x) \minus{} f(x)}{x \minus{} c}$.[/quote]

Could you explain this part a little bit more? I don't see how the terms can be restated.[/quote]
He want to write :
$ g(x) \equal{} \frac {f(x) \minus{} f(c)}{x \minus{} c}$
You can definion as
$ f'(x) \equal{} \lim_{h\to 0}\frac {f(x \plus{} h) \minus{} f(x)}{h}$

for the "$ \rightarrow$"

def. of derivative for $ f'(c) \equal{} \lim_{x\rightarrow c}\frac{f(x)\minus{}f(c)}{x\minus{}c}$

since $ f(c)$ continuous, then $ f(c)\equal{}f(c\plus{})\equal{}f(c\minus{})$.

then $ f'(c) \equal{} \lim_{x\rightarrow c}\frac{f(x)\minus{}f(c\plus{})}{x\minus{}c}$ which is the definition for the right hand limit, similarly with the left.


A truck travels 40 miles from point A to point B in exactly 1 hour. When the truck is halfway between point A and point B, a car starts from point A and travels at 50 miles
per hour.  How many miles has the car traveled when the truck reaches point B?

(1) 25       (3) 50
(2) 40       (4) 60

[hide] The truck is half an hour away at the halfway point, so in half an hour, the car travels 25 miles.  You sure this shouldn't be in getting started?[/hide]

Isn't this pretty obvious? The truck had 1/2 hour left and the car would've went 25 miles... obviously (1).

[quote="symmetry"]What is the equation?[/quote]

uhh.. d=rt
d=0.5(50)
d=25.

0.5 is half an hour (since the truck was halfway to an hour trip).

Here's a hint:

Getting started is for high-school level problems. Maybe this should be in something like The Basics or Classroom Math?

c'mon..this is way to easy for getting started


n vectors are given in the plane so that any one of them is collinear to the
sum of the others. Prove that they are all collinear.

What if the sum of the vectors equal to 0?

I will assume that the sum of the vectors is nonzero, because otherwise the problem does not hold. Let S be the sum of the vectors, and let a be any vector in the bunch. We then have s-a is parallel to a, which implies s is parallel to a. (To be more precise, since s-a and a are parallel, let s-a = ka for a real number k, which then implies s=(k+1)a and so s and a are parallel). So, all the vectors in the bunch are parallel to s, so they are all parallel to each other.


Yesterday night, after I finished my last Alcumus problem of the day, my percentage was about 72%. I just checked again, and it became 65%. Was the algorithm for determining percentage changed or something?

The percentage, as well as total number of correct, incorrect, and given up were being computed incorrectly for a brief period of time due to bugs in other parts of the system. They should now be correct. The percentage algorithm is simply total number correct divided by total number of problems.


advanced fields
advanced fields unsolved

Hello, let X,Y topological spaces and $ \phi: X \rightarrow Y$ an omeomorphism.

The equivalence relation $ \sim$ is defined over Y, and we define the relation $ \bowtie$ on X like this:

$ x_1 \bowtie x_2$ iff $ \phi(x_1) \sim \phi(x_2)$

Can we say there is something between $ X/ \bowtie$ and $ Y / \sim$ given with the quotient topologies??

Probably there is a lack of givens in this problem, I will expose the particular case in which I need it:

Let P = [0,1]x[0,1] in R^2 and we define the usual equivalence relation whose quotient is the torus T.

Let Q the square cut along the median of one side and let Q/~ the quotient space of the relation ~ which identificates a cut (the usual cut and paste relation).

What is an explicit omeomorphism between T and a quotient of Q/~ which identificates the sides?

[quote="dphexen"]Can we say there is something between $ X/ \bowtie$ and $ Y / \sim$ given with the quotient topologies??[/quote]
They are homeomorphic. Indeed, $ X$ and $ Y$ are essentially the same space, to which you apply the same operation (taking a quotient).


pigeonhole principle
combinatorics unsolved
combinatorics

In a multiple choice test there are 4 problems and three possible answers to each problem. A groups of students took the test. It turns out that for any three students, there was a problem that the three students answered differently. What is the maximal possible number of students?

[quote="Fabian May"][quote="sunchips"] It turns out that for any three students, there was a problem that the three students answered differently. [/quote]
do u mean that there's one problem where not all the students had the same answer OR there's a problem with 3 different answers from 3 different students??[/quote]

For any three students, there exists a problem that the three students had three different answers to (e.g. student 1 may have answered A, student 2 may have answered B, student 3 may have answered C).

Perhaps Pigeonhole, counting in two ways to get bound? some good construction of large set of students?

some help and ideas please, PLEEEAAASSSEEE

sunchips :help:

[quote="Deedlit"]For each choice of pairs of answers to each problem, there are at most two people who selected from those pairs of answers.  There are 81 choices of pairs of answers, so summing the number of students over all of them gives at most 162 students.  Each student gets counted 16 times, so there are at most 10 students.
An example for 9 is

1111
1222
1333
2123
2231
2312
3132
3213
3321[/quote]

Sorry, I'm not getting your solution. What do you mean by [i]"For each choice of pairs of answers to each problem, there are at most two people who selected from those pairs of answers"? [/i]There are three possible answers for each problem. And what do you mean by [i]"Each student gets counted 16 times"[/i]?

Thanks for you help and solution!

Sunchips


geometric series
real analysis
real analysis unsolved

Let $ z \in \mathbb{C}$. Show that the series
\[ \frac{z}{1 \minus{} z^2} \plus{} \frac{z^2}{1 \minus{} z^4} \plus{} \frac{z^4}{1 \minus{} z^8} \plus{} \frac{z^8}{1 \minus{} z^{16}} \plus{} \cdots\]converges to $ \frac{z}{1\minus{}z}$ if $ |z| < 1$, converges to $ \frac{1}{1 \minus{} z}$ if $ |z| > 1$, and diverges if $ |z| \equal{} 1$.

$ \frac {z}{1 \minus{} z^2} \equal{} z(1\plus{}z^2\plus{}z^4\plus{}...)$, similarly for your other terms. Now $ \frac {z}{1 \minus{} z} \equal{} z\plus{}z^2\plus{}z^3\plus{}...$, so your goal should be to see that you get $ z^n$ for each positive integer $ n$ one and only one time.

WWW is talking about $ |z|<1.$ You need a slightly different argument, albeit also based on geometric series ideas, for $ |z|>1.$

In particular, write:

${ \frac1{1-z}=\frac{-z^{-1}}{1-z^{-1}}=-(z^{-1}+z^{-2}+z^{-3}}+\cdots)$

${ \frac{z}{1-z^2}=\frac{-z^{-1}}{1-z^{-2}}=-(z^{-1}+z^{-3}+z^{-5}}+\cdots)$

${ \frac{z^2}{1-z^4}=\frac{-z^{-2}}{1-z^{-4}}=-(z^{-2}+z^{-6}+z^{-10}}+\cdots)$

As for $ |z|=1:$ $ \left|\frac{z^{2^k}}{1-z^{2^{k+1}}}\right|\ge\frac{1}{2}$ for all $ |z|=1.$ The series diverges because the terms don't go to zero, and that includes infinitely many points at which the infinitely many of the terms are undefined.

That's a nice trick for $ |z| > 1$.

Thanks!

[quote="Kent Merryfield"]WWW is talking about $ |z| < 1.$ You need a slightly different argument, albeit also based on geometric series ideas, for $ |z| > 1.$

In particular, write:

${ \frac1{1 - z} = \frac { - z^{ - 1}}{1 - z^{ - 1}} = - (z^{ - 1} + z^{ - 2} + z^{ - 3}} + \cdots)$

${ \frac {z}{1 - z^2} = \frac { - z^{ - 1}}{1 - z^{ - 2}} = - (z^{ - 1} + z^{ - 3} + z^{ - 5}} + \cdots)$

${ \frac {z^2}{1 - z^4} = \frac { - z^{ - 2}}{1 - z^{ - 4}} = - (z^{ - 2} + z^{ - 6} + z^{ - 10}} + \cdots)$

As for $ |z| = 1:$ $ \left|\frac {z^{2^k}}{1 - z^{2^{k + 1}}}\right|\ge\frac {1}{2}$ for all $ |z| = 1.$ The series diverges because the terms don't go to zero, and that includes infinitely many points at which the infinitely many of the terms are undefined.[/quote]

I'm sure i'm missing something obvious here but could you help me understand why this trick works for |z| > 1 but not |z| < 1?

What's $ 1 \plus{} 2 \plus{} 4 \plus{} 8 \plus{} \ldots$?   :wink:

Ah right I see!  How emarassing, I'm too tired to be doing this sheet  :blush:


What is the remainder when you divide $6^{83}+8^{83}$ by $49$?

[hide="Hint to method 1"] Write as $(7-1)^{83}+(7+1)^{83}$. [/hide] [hide="Hint to method 2"] $\varphi(49) = 42$ and $48 = 6 \cdot 8 \equiv-1 \bmod 49$. [/hide]

Look in the resources section. It's from AIME 1983.


Mr. Byrd can buy a 757-page Algebra book for $ \$14.99$, or he can make copies of it for 5 cents a page. In either case, a $ 9\%$ sales tax applies. What is the number of dollars in the positive difference between the costs for buying and copying the book? Express your answer to the nearest cent.

if he copies the book, then it would cost $ 757 \times 0.05 \times 1.09 \equal{} 41.2565$
this is because there are 757 pages and each page costs 5 cents
then we multiply by 1.09 to add on the tax

if he buys the book, it would cost $ 14.99 \times 1.09 \equal{} 16.3391$
this is because the book costs 14.99 and we have to add the tax

therefore, the difference is $ 41.2565\minus{}16.3391 \equal{} 24.9174$
it said round to the nearest cent, so our answer is 24.92


Need to solve for X:
Would appreciate any help. Thanks

Csc(2X) = Sec(X+30)

I'm guessing by the title, this is a trig problem.  Trigonometry does not belong in this forum.  Please correct me if necessary

I would take reciprocals of both sides and use the identity $\cos \theta =\sin (90^\circ-\theta )$. Now the equation looks something like $\sin A=\sin B$, which can be shown to be equivalent to $\sin \frac{A-B}{2}\cos \frac{A+B}{2}=0$. (To prove this, use trig addition and the fact that $\frac{A+B}{2}$ and $\frac{A-B}{2}$ have sum $A$ and difference $B$.)


Let the edges of the graph be coloured blue and red ($b$ and $r$). We will call a path $v_1v_2v_3$ an $xy$ path if $v_1v_2$ is of the colour $x$ and $v_2v_3$ is of the colour $y$. Fix two vertices $v_1$ and $v_2$ and consider all paths $v_1v_sv_2$. We have $12$ such paths. If there are two $rr$ or two $bb$ paths among them we get the desired quadrilateral. Otherwise there exists at least $10\ \ rb$ and $br$ paths. Without loss of generality we may assume that there are at least $5\ \ rb$ paths $v_1v_iv_2$. Let the set of such $v_i$'s be $X$. Fix any $v_3v_4$ with $v_3, v_4\in X$. We may assume it is red. Consider all the neighbours of $v_4$ in $X$. If there is such a neighbour $v_5\neq v_3$ that $v_4v_5$ is red then $v_1v_3v_4v_5$ is a desired quadrilateral. Otherwise take any two neighbours $v_6$ and $v_7$ and note that $v_2v_6v_4v_7$ is a desired quadrilateral. Q.E.D.

[quote="silouan"]If a $G_{14}$  is colored with two colors,  there will  be a unicolored  quadrangle .[/quote]

I wonder if 14 aren't too much vertices. What's wrong with for instance, the following argument.
Take any vertex $V_1$. Then it is connected with at least another seven vertices with say red color. Let $X$ be the set of such vertices, and $V_2$ a vertex in $X$. Then $V_2$ is at most connected with red color to one other vertex in $X$. Thus there are at most 3 red edges in the $G_7$ formed by points in $X$ and there is one such point which is not connected with a red edge to any point in $X$.  Let $V_7$ be such point, which is connected by blue color to every other point in $X$. Then one can clearly see that it is enough for the statement to be true that there be four points in $X$ besides $V_7$ and not six which is the number with which you end up if you start with a $G_{14}$.
By the way, you can probably attack the $G_{80}$ problem the same way

Daniel


It would be nicer if you wrote it in latex... :wink:

Pierre.

$ f$ is the crossfoot of $ n$ in binary representation.
$ 2^{10} < 2008 < 2^{11} \minus{} 1$.
So $ M \equal{} 10$.
If $ n<\equal{}2008$ with $ f(n)\equal{}10$,
then $ n$ in its binary representation has at most $ 11$ digits, $ 10$ of which are $ 1$.
So $ n\equal{}\overline{1\cdots101\cdots1}\equal{}2^{11}\minus{}1\minus{}2^k$ for some $ 0\le k\le 10$.
Then $ n\le 2008$ gives $ 2^k\equal{}2^{11}\minus{}1\minus{}n\ge 40$ and $ k\ge 6$.
So there are $ 5$ possibilities $ n\in\{1023,1535,1791,1919,1983\}$.


Any positive integer $m$ can be written uniquely in base $3$ form as a string of $0$'s, $1$'s and $2$'s (not beginning with a zero). For example,
$98=(1\times81)+(0 \times 27)+(1 \times 9)+(2 \times 3)+(2 \times 1) = (10122)_{3}$.

Let $c(m)$ denote the sum of the cubes of the digits of the base $3$ form of $m$; thus, for instance
$c(98)=1^{3}+0^{3}+1^{3}+2^{3}+2^{3}=18$

Let $n$ be any fixed positive integer. Define the sequence $(u_{r})$ by
$u_{1}=n$ and $u_{r}=c(u_{r-1})$ for $r \geq 2$.

Show that there is a positive integer $r$ which which $u_{r}=1,2$ or $17$.

[hide="Solution"] Let $d(m)$ denote the number of digits of $m$ in base $3$.  then

$c(m) \le 8 d(m)$

Let $N$ be the smallest integer such that $8 d(N) \le N$.  Experimenting, we find that $d(N) = 3$ and therefore that $N = 24_{10}= 220_{3}$.

Therefore, for $u_{i}> N$ we will have $u_{i+1}< u_{i}$.  In other words, there exists $I$ such that $\forall i > I: u_{I}< N$.  Hence WLOG $n < N$.

The maximal value of $u_{2}$ occurs when $u_{1}= 212_{3}$, which gives $u_{2}= 17_{10}$.

For $u_{1}\neq 212_{3}$ we have $u_{2}\le 16_{10}= 121_{3}$ and therefore $u_{3}\le 10_{10}$ unless $u_{2}= 22_{3}$, in which case $u_{3}= 16_{10}, u_{4}= 10_{10}$ anyway.

If some $u_{i}= 10_{10}= 101_{3}$ then $u_{i+1}= 2$.  Otherwise, some $u_{i}< 10_{10}$ and therefore $u_{i+1}\le 9_{10}, u_{i+2}\le 2_{10}$ (since we eliminated the case for $22_{3}$.)

Very messy casework, but the point is that everything becomes bounded and then decreases quickly. [/hide]


probability
geometric transformation
rotation
complementary counting

Twenty five of King Arthur's knights are seated at their customary round table.  Three of them are chosen - all choices of three being equally likely - and are sent off to slay a troublesome dragon.  Let $P$ be the probability that at least two of the three had been sitting next to each other.  If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

mathkid wrote:25 knights are seated at a round table and 3 are chosen at random. Find the probability that at least two of the chosen 3 are sitting next to each other.

[hide]11/46.

I broke it up into 2 cases.  The one where 2 were adjacent, and the one where all 3 were adjacent.  Obviously the denominator will be 25C3.  For 2 adjacent pieces, you have to think as if the 2 pieces joined together to make one piece.  Thus, there would only be 24 pieces.  Since it is a circle, the third piece cannot be on either side of the "piece".  So there are 25-4=21 places to put the third piece.  Hence, there is 24x21 ways for two adjacent pieces.  For all three of them to be adjacent, there are 25 ways.

So, the answer would be (24x21+25)/25C3=11/46.[/hide]

how do we know that there are 25 for three adjacent?

I don't think there are 22C3 ways to place non-adjacent pices...

[edit]-yeah, the rotations are goofy as JBL said

Okay, I'm not 100% sure why your method doesn't work, but I think it may have to do with rotations.  However, I recommend for problems of this type that you try to use the principle of inclusion-exclusion, which is what Syntax was doing.  The reason is that it can't fail, basically, where as other methods lead you into these complicated questions of whether you are counting all of the cases right and whether there are distinctness issues and so on.

One more question just struck me:

There are 25 different ways to arrange 3 adjacent knights, which I can see clearly.

But why aren't there 25 ways to arrange two adjacent knights? I understand you made them into 1 piece, making 24.. but why can't you do it the same way as with 3 adjacent ones?

[quote="mathkid"]One more question just struck me:

But why aren't there 25 ways to arrange two adjacent knights? I understand you made them into 1 piece, making 24.. but why can't you do it the same way as with 3 adjacent ones?[/quote]

It's a bit different when all 3 are one piece because you don't have any other pieces to put anywhere.

[quote="churchilljrhigh"]there are 24 pieces but there are only 23 pieces for the third and can't be the two by the combined piece so there is only 21.

and i don't see clearly why there is 25 for three adjacent?[/quote]

I don't understand your first question.

draw it out if you really have to for your second question to understand the concept.  Note that it's a circle and not a line.

I don't see how come its always 25 ways for one knight, pair of knights, trio of knights.

Just use your way to do it, mathkid. The first knight you add in doesn't matter. The second knight you add in, the probability is 21/23 because now you have 23 knights and 23 spaces. the third knight you add in, the probability is 20/24. 21/23*20/24=35/46. 1-35/46=11/46. I think you forgot that if you add one knight in, you also have one more space.

Hmm... I am kind of confused.

I decided to use PIE....

[hide]

There are 25 pairs of people in the round table, and there are 23 other ways to select the third person.  Thus there are 25*23 ways to select a position where two people are adjacent.  Then there is 25 different ways we can choose three people all adjacent to each other.

The probability would then be (25*23-25)/C(25,3)=11/92.  However, I saw that my answer is 1/2 of the real answer.  It dawned on me that I have to multiply something by 2.  What is it?[/hide][/hide]

[quote="mathkid"]25 knights are seated at a round table and 3 are chosen at random. Find the probability that at least two of the chosen 3 are sitting next to each other.

For this, I found the number of ways three could be chosen with no adjacent.
I imagined 22 knights seated. I would have to place each of the three nights in between two that were already seated. There are 22C3 ways to do this.
So I would think the final probability would be 1- (22C3)/(25C3)

What am I doing wrong?[/quote]

The fatal flaw in your assumption is that you failed to account for rotations. We can demonstrate this case with only 6 knights. By your assumption, there is only one way to place such that the three knights are not adjacent. However, there are in fact two ways (choose the three knights that are already seated).

Therefore, what you found was the number of ARRANGEMENTS such that all three knights are not adjacent, but you did not find all possible CHOICES.

Imagine for a second that you have 25 points around a circle.

[hide]It is obvious that there are 25 ways to select 3 adjacent points, simply by the fact that there are 25 possibilities for our leftmost position of the trio. Similarly, there are 25 ways to select 2 adjacent points, and 21 possibilities for the last point such that only two are adjacent.

Thus our probability is ((25+25*21)/25C3), which is 11/92[/hide]

If you have the two knights in a pair, why can't the rightmost knight be switched with the leftmost knight to yield 25*2 possibilities for seating a pair of knights? Similarly, for seating 3 together, why don't you have to account for the number of arrangements of the three knights themselves?

What's wrong with this solution:

[hide="bogus"]We will use complementary counting: finding the probability of no knight being next to any other one.

First, notice that there are $\dbinom{25}{3}$ ways to choose 3 knights from 25.

Now we must count the number of ways to place knights such that no knight is next to another. We place three knights, and ignore the rest. Note that there must be at least one empty chair between any two knights; thus, removing three chairs reduces the problem to choosing 3 spots for the knights. There are $22$ chairs left, and $\dbinom{22}{3}$ ways to put knights in the chairs.

Thus the probability is $1-\dfrac{\dbinom{22}{3}}{\dbinom{25}{3}}=\boxed{\dfrac{38}{115}}$[/hide]

[quote="bobthesmartypants"]What's wrong with this solution[/quote]
It omits the case in which one of the three knights is at the head of the table. To see this, consider the case where there are $6$ knights instead of $25$.

Thus $\tbinom{22}3$ should be replaced by $\tbinom{22}3+\tbinom{21}2$.

No. of arrangements in which
(a) the three knights sit together ,
 $21! \times 22 \times 3!$, as we may first let the other 22 knights be seated then there are 22 slots to allowed to the group of the three knights , and there are 3! ways to permute after the group of 3 knights is inserted.
(b) two knights together and the other knights separated,
$21! \times 22 \times 21 \times C_2^3 \times 2$ as we may first let the other 22 knights be seated,then divide the 3 knights into 2 groups of sizes 2 and 1, there are 22 slots for the group of size 2 to be inserted, after that , (22-1) = 21 slots remains, and there are two ways for the group of size 2 to permute.
Hence, the required probability $=\frac {21! \times 22 \times 3!+21! \times 22 \times 21 \times C_2^3 \times 2}{24!}$
$=\frac {22 \times 6+22 \times 21 \times 6}{24 \times 23 \times 22 } = \frac {11}{46}$.
The final answer is $11+46=57$.

[hide = Solution]
For the case with $n$ people, we can constructively count the no. of desired arrangements. Denote $n$ points on a circle.
[u]Case 1: Exactly 2 people are adjacent. [/u]
There are $n$ sets of adjacent points. Let us choose $A,B$ as our adjacent points. The third point must be non-adjacent to the given 2 points. Thus we have eliminated 4 choices for the final point ($A,B$ and the other two points adjacent to A&B). Running through all the pairs gives $n(n-4)$.
[u]Case 2: All three people are adjacent. [/u]
This is just $n$ ways.
So our final answer is $$P = \frac{n(n-4)+n}{\binom{n}{3}} = \frac{n(n-3)}{\binom{n}{3}}.$$
Letting $n=25$ gives $$P = \frac {11}{46}.$$

[hide="All probability, very little counting"]WLOG, seat the first person in seat $3$.

In $\frac{2}{24} = \boxed{\frac{1}{12}}$ of the cases, when we seat the second person, we're done - seats $2$ or $4$.

In $\frac{2}{24} = \frac{1}{12}$ of the cases, when we seat the second person in seat $1$ or $5$, we still have to seat the third person.  These cases are special, because there are only $3$ places the last person can go - one spot between, and one spot on either side.  That's $\frac{3}{23}$ of the time.  Overall, this gives $\frac{1}{12} \cdot \frac{3}{23} = \boxed{\frac{3}{276}}$ of the time.

In $\frac{20}{24} = \frac{5}{6}$ of the cases, the first two people are far enough from each other that there are now $4$ good seats for the third knight.   This means the third knight has a $\frac{4}{23}$ chance of being next to one of the original knights.  Overall, this gives a probability of $\frac{5}{6} \cdot \frac{4}{23} = \boxed{\frac{10}{69}}$.

Adding the three boxed probabilities gives $\frac{11}{46}$, for a bubble in answer of $\boxed{57}$.[/hide]

[url=https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_7]Here[/url] are more solutions.

A nice solution by 'Syntax Error' however (24x21+25)/25C3=23/100 and not 11/46.
Here is an improvement,
Again break into 2 cases. One where 2 knights are adjacent and the one where 3 knights are adjacent. The denominator must be 25C3. There are a total 25 ways to select adjacent pairs. For any adjacent pair the knight on either the left or the right must not be picked (to avoid picking 3 knights that are adjacent). So the 3rd knight can be picked from the remaining 25-4=21 pool of knights. Hence there are a total of 25x21 ways to pick knights such that exactly two are adjacent. To pick exactly 3 adjacent knights, there are 25 ways.
So, the answer must be (25x21+25)/25C3=11/46.
And 11+46=57

Very simple solution:
Break this up into 2 cases

Case 1: only 2 adjacent
There are 25 ways to choose the 2 adjacent people and then 21 ways to seat the third knight as he cannot sit in the place of either knight, or the remaining adjacent seat for each knight.

Case 2; all 3 adjacent
25 ways trivially

Add up for 25*22/25*4*23 for 11/46 giving 57.

Say order doesn't matter. If exactly two are next to each other, then there are $25 \cdot 21$ ways. If all three are next to each other, then there are $25$ ways. Then the answer is $\frac{25 \cdot 22}{\binom{25}{3}} = \frac{25 \cdot 22}{25 \cdot 4 \cdot 23} = \frac{11}{46} \implies \boxed{57}$.

Actually , the whole thing got messed up cuz at the very beginning, #Syntax wrote (24*21+25)/25C3 as 11/46 , but it's not practically.

[quote=club52]What's wrong with my solution?
[hide=Bogus]The answer is 1 - the number of ways to arrange 3 knights without them touching. So we seat one knight at the table. There is then a 22/24 chance that the next knight will sit apart from the first, but first we break it into cases. There's a 20/24 chance that there are 4 seats that no knight could ever occupy, leaving a (22/24)*(19/23) chance for that case. The other case is (2/24)*(20/23). Adding them together and subtracting from 1, we have 41/138, and since they are coprime, the answer is 179.[/hide][/quote]

Overall, you were pretty spot on. Your second case is correct, since there are only 2 seats where only 3 seats are unsittable. But in your first case, you multiply by 22/24. This is weird considering before you already said that there's a 20/24 chance that there are 4 seats that no knight could occupy, so you should be multiplying 19/23 by 20/24 instead

it doesn't add up to 1, since there are already 2 seats that just don't work. This way, it adds up to 22/24, which is correct since the 2 remaining seats are the ones previously mentioned.

[hide=7]We use complementary counting (even though casework is easier), so we want to count the number of ways to pick three knights such that they have at least one knight in between them. WLOG, let the three knights be $K_1$, $K_2$, and $K_3$, and the number of knights in between $K_1$ and $K_2$ $a$, the number of knights in between $K_2$ and $K_3$ $b$, and the number of knights in between $K_3$ and $K_1$ $c$. We know $a+b+c=22$ and $a, b, c \ge 0$ and by stars and bars, there are $\binom{21}{2}$ valid triplets of $(a, b, c)$. Then, there are $25$ possible options for $K_1$, after which $K_2$ and $K_3$ are fixed after picking $a, b, c$. However, note that this overcounts by a factor of $3$ because for each triplet of knights, we count it three times using our counting method designating each of the knights as $K_1$. Therefore, the probability that each pair of knights have at least one knight in between them is $\frac{\frac{25\binom{21}{2}}{3}}{\binom{25}{3}}=\frac{35}{46}$, so the probability that at least two of the three knights are sitting next to each other is $\frac{11}{46}$, so the answer is $57$.[/hide]


Find the area of region bounded by $ y\equal{}x^2$ and $ y\equal{}9$
Then find a line that makes the area $ \frac{1}{3}$ of that.

Calculus does not belong in the high school forums.  Post problems like this [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]here[/url].

The problem was too easy to post in an another forum.


calculus
integration
trigonometry
derivative
real analysis
real analysis unsolved

Problem: Evaluate \[ \int^{\pi/2}_0 \dfrac{1}{1\plus{}\tan^{\sqrt{2}}{x}} dx\]

I know of one solution using [hide="hint"] differentiation under the integral sign [/hide], but my friend tells me there is another. In any case, any solutions are welcome.

Let $ x \equal{} \arctan m$. Then $ dx \equal{} \frac {dm}{1 \plus{} m^2}$ and our integral will be equal to

$ I \equal{} \int_{0}^{\infty}{\frac {1}{(1 \plus{} m^2)(1 \plus{} m^{\sqrt {2}})}\;dm}$

Now put here $ m \equal{} \frac {1}{t}$. Then $ dm \equal{} \minus{} \frac {dt}{t^2}$ and

$ I \equal{} \int_{0}^{\infty}{\frac {t^{\sqrt {2}}}{(1 \plus{} t^2)(1 \plus{} t^{\sqrt {2}})}\;dt}$

Sum this two integrals we get:

$ 2I \equal{} \int_{0}^{\infty}{\frac {1 \plus{} t^{\sqrt {2}}}{(1 \plus{} t^2)(1 \plus{} t^{\sqrt {2}})}\;dt} \equal{} \int_{0}^{\infty}{\frac {1}{1 \plus{} t^2}\;dt} \equal{} \frac {\pi}{2}$

So, $ I \equal{} \frac {\pi}{4}$.

Symmetry. $ \tan(\frac{\pi}{2}\minus{}x)\equal{}\frac1{\tan x}$, and that's all you need. The other arguments are just complicated ways of getting at this.

[quote="jmerry"]Symmetry. $ \tan(\frac {\pi}{2} \minus{} x) \equal{} \frac1{\tan x}$, and that's all you need. The other arguments are just complicated ways of getting at this.[/quote]
yes,  :oops:

Notice also that the value of this integral is independent on the exponent of tan x (this has been shown several times already on this forum).


function limit integration trigonometry Euler logarithms calculus

is $\sum_1^n \frac{|\sin(k)|}{k} - \frac{2\ln(n)}{\pi}$ bounded ?

I also struggled a lot woth this question, but I have no answer. Do you have any?

Like harazi, I don't have the answer either.

If $f$ is a continuous function periodic of period $P$ and $\frac{\alpha}P$ is irrational, then

$\lim_{n\to\infty}\frac1n\sum_{k=1}^nf(k\alpha)=\frac1P\int_0^Pf(x)\,dx.$

Thus, if $G(n)=\sum_{k=1}^n\left(|\sin k|-\frac2{\pi}\right),$ then $\lim_{n\to\infty}\frac{G(n)}n=0.$

That's the good news. The bad news is that we don't have any good estimate at all about the rate of that convergence.

Now, consider the sum we are interested in and use summation by parts:

$\sum_{k=1}^n\frac1k\left(|\sin k|-\frac2{\pi}\right)=\frac{G(n)}n +\sum_{k=1}^{n-1}\frac{G(k)}{k(k+1)}.$

The boundary term goes to zero, so we are left looking at the last sum. The problem is that all we can really say about $G(k)$ is that $G(k)=o(k)$ and that's not good enough to guarantee the convergence of that sum.

Here's an attempt to prove that the sequence is bounded.  The proof uses Fourier series.  I will pretend to be Euler and not worry about convergence issues or switching limits.  Hopefully in a later post, I or someone else (Kent?) can justify the steps taken.

I will attempt to prove that the sequence is not just bounded, but converges to a finite limit.  By well-known results about the harmonic series, that convergence is equivalent to the convergence of the infinite series \[\sum_{k = 1}^\infty \frac{2 - \pi |\sin k|}{k}.\]

The function $|\sin x|$ is a continuous, periodic function.  As such, it has a Fourier series.  I could calculate the Fourier coefficients, but I was lazy and looked up the result in a handbook:\[2 - \pi |\sin x| = \sum_{j = 1}^\infty \frac{4}{4j^2 - 1} \cos(2jx).\]
Dividing by x and summing over all integers gives the identity \[\sum_{k = 1}^\infty \frac{2 - \pi |\sin k|}{k} = \sum_{k = 1}^\infty \sum_{j = 1}^\infty \frac{4}{4j^2 - 1} \frac{\cos(2jk)}{k}.\]Next, is it legal to switch the order of summation?  Don't know, but I'm going to anyway:
\[\sum_{k=1}^\infty \frac{2- \pi |\sin k|}{k} = \sum_{j=1}^\infty \frac{4}{4j^2 - 1} \sum_{k=1}^\infty \frac{\cos(2jk)}{k}.\]

To evaluate the final sum, I will use the following identity, which I also saw in a chapter on Fourier series:
\[\sum_{k = 1}^\infty \frac{\cos(2kx)}{k} = - \ln(2 |\sin x|),\]
provided that x is not a multiple of $\pi$.  We then get
\[\sum_{k=1}^\infty \frac{2 - \pi |\sin k|}{k} = \sum_{j = 1}^\infty \frac{-4 \ln(2 |\sin j|)}{4 j^2 - 1}.\]

We need to analyze the convergence of the sum on the right.  The only trouble can come if j is too close to a multiple of $2 \pi$.  In that case the term $-\ln(2 |\sin j|)$ could become huge.  Fortunately, an integer cannot be too close to a multiple of $2 \pi$.  We can use  [url=http://mathworld.wolfram.com/IrrationalityMeasure.html]Hata's theorem[/url], which I used in a post on a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]"tough infinite series"[/url].  Using that theorem, we get that $1/|\sin j| = O(j^8)$.  Thus $|\ln(2|\sin j|)| = O(\ln j)$.  The absolute version of the sum we are interested in is bounded by $\sum_j \frac{\ln j}{4j^2 - 1}$.  Because that latter sum converges, our sum converges absolutely, and in particular converges.  So we are done.

How does it look?

[quote="Ravi B"]Next, is it legal to switch the order of summation?[/quote]
Unfortunately, it isn't: the sum is not absolutely convergent.

I'm still thinking about how parts of this might be salvaged; it is a beautiful idea.

Yes, you are right.  In retrospect, my post was laughably naive.  Nevertheless, it did contain a good idea.  In this post I will make the proof rigorous.  The main change is to make the sums finite, and explicitly handle the error term.

Recall the Fourier series
\[2 - \pi |\sin x| = \sum_{j = 1}^\infty \frac{4}{4j^2 - 1} \cos(2jx).\]
(In my initial post, I assumed that pointwise convergence follows from continuity, but is that true?  My book seems to require that the function be differentiable from the left and right.  Fortunately $|\sin x|$ is differentiable from the left and right.)  All the terms of the sum beyond the $m$th term add up in absolute value to $O(1/m)$.  Thus we can replace the infinite sum with a sum of $m$ terms, and be off only by $O(1/m)$.

Recall also Kent's function $G$.  I will show that $|G(n)| = O(n^{7/8})$.  That will be good enough for Kent's summation by parts to converge.  By using the Fourier series above, we have
[tex]\begin{eqnarray*}
\pi |G(n)|
  & = & \sum_{k=1}^n \left( \sum_{j=1}^m \frac{4}{4j^2-1}
               \cos(2jk) + O(\frac{1}{m}) \right) \\
  & = & O(\frac{n}{m}) + \sum_{k=1}^n \sum_{j=1}^m
              \frac{4}{4j^2 - 1} \cos(2jk) \\
  & = & O(\frac{n}{m}) + \sum_{j=1}^m \frac{4}{4j^2-1}
                                           \sum_{k=1}^n \cos(2jk).
\end{eqnarray*}
[/tex]
At this point, I will use the handy trig identity
\[\sum_{k=1}^n \cos(kx) = \frac{\sin((n+1/2)x)}{2\sin(x/2)} - \frac{1}{2}.\]
We get
[tex]\begin{eqnarray*}
\pi |G(n)|
  & = & O(\frac{n}{m}) + \sum_{j=1}^m \frac{4}{4j^2-1}
                                          \frac{\sin((2n+1)j)}{2 \sin j} \\
  & \leq & O(\frac{n}{m}) +
          \sum_{j=1}^m \frac{4}{4j^2-1} \frac{1}{|2 \sin j|} \\
\end{eqnarray*}
[/tex]
Now recall that Hata's theorem shows that $1/|\sin j|$ is $O(j^8)$.  Thus each term of the sum is $O(j^6)$.  Thus the sum itself is $O(m^7)$.  Thus we have shown that
$|G(n)| = O(\frac{n}{m})  + O(m^7)$.  Choosing $m = n^{1/8}$ shows that $|G(n)| = O(n^{7/8})$.  So we're done.

Does the proof look right this time?

I think I also managed to prove the boundness of the sequence. My idea was to prove that in fact there is a constant $C$ which does not depend nor on x between 0 and $2\cdot\pi$ nor on n such that ${ |\sum_{k=1}^{n}{\frac{coskx}{k}}}|\leq C -2min(lnx, ln(2\cdot\pi-x))$ (*). For this it suffices to see that the derivative of the function $ f(x)=\sum_{k=1}^{n}{\frac{cos kx}{k}}$ verifies $|f'(x)|\leq\frac{1}{sin\frac{x}{2}}$ and also the sequence of derivatives of such functions calculated in $\pi$ for n varying is bounded. Then we can write that there exists a constant K that does not depend on x between 0 and $\pi$ and does not depend on n such that ${ |\sum_{k=1}^{n}{\frac{coskx}{k}}}|\leq K+\int_{x}^{\pi}{\frac{1}{sin\frac{t}{2}}dt}$. A simple estimation on the later integral gives the existence of a constant $G$ such that for all n and all x between 0 and $\pi$ we have ${ |\sum_{k=1}^{n}{\frac{coskx}{k}}}|\leq G-2lnx$ .  It is easy to see that for this constant C we actually have (*) true. Thus if $c_n(f)$ are the Fourier coefficients of $ f(x)=|sinx|$ then we have $f(x)=\sum_{k\geq-1}{c_k(f) cos kx} $ and  $\sum_{k=1}^{n}\frac{f(k)}{k}=\sum_{j>-1}c_j(f)\sum_{k=1}^{n}{\frac{coskj}{k}}$. Next, use that ${ |\sum_{k=1}^{n}{\frac{coskx}{k}}}|\leq C -2min(lnx, ln(2\cdot\pi-x))$ (*) and a well known estimation of Mahler that there exists a constant $r$ such that for all sufficiently large $p,q$ we have ${|\frac{p}{q}-\pi}|\geq\frac{1}{q^r}$ and we will deduce that $|\sum_{k=1}^{n}\frac{f(k)}{k}-\frac{2}{\pi}lnn|\leq D\sum_{j>0}{c_j(f) lnj}$ ( with D an absolute constant depending only on G and r ) and the last series converges. Hope I did not again an idiot mistake.

Upps, sorry Ravi B, I did not see your posts. I hope I did not rewritte your ideas. Anyway, if it is not your idea, it surely has a bug somewhere. As usual...

Harazi, your proof looks good.  Our proofs have some similar parts, and some differences.  Vive la difference!

[quote="Ravi B"](In my initial post, I assumed that pointwise convergence follows from continuity, but is that true? My book seems to require that the function be differentiable from the left and right.[/quote]
The Fourier series of a continuous function is Cesaro (or Abel) summable to itself, with uniform convergence. If in fact the partial sums of the Fourier series converge, they can only converge to the same number (Abel's Theorem or its Cesaro analogue.) Note that your Fourier series is in fact absolutely and uniformly convergent. (It was predictable that the Fourier coefficients for a continuous function whose derivative has a jump discontinuity but is otherwise well-behaved would be $O(j^{-2}).$) Hence you have pointwise convergence of your series. Your argument is sound on that point, at least.


Yes; the announcer clearly stated that the competitor has 3 seconds to [b]begin[/b] his/her answer.

[quote="kobebryant24"]IN the mathcounts national cd round, a competitor has 3 seconds to begin their answer. so couldn't they just begin their answer within 3 seconds and then for instance, if the answer was a fraction, they could pause in the middle? thats what david yang did[/quote]

You can. But the judges know whether you are stalling or not.

Like if you say: fifty.....thousand......three-hundred thirty-six?

They know you are stalling. The judges have a choice whether accepting it or not.

At state countdown, the question was "What power of 27 is 9?" and I buzzed in while he was reading the question. I thought for about 2.9 seconds and then started and said "One" but then stopped. I thought for about .5 more seconds and finally said "2 over 3" and the mod accepted it and I got it right, sending me to the next round. I don't know why it took me that long and why he still accepted it, but I guess as long as you don't pause for more than three seconds, it is accepted. For example, you can say "2" after 2.9 seconds and then say "thirds" after another 2.9 seconds.

If you go a little over once, I think it will be ok. I remember one guy would buzz in and be like "The... answer... is........." on every question and the judges told him after the second time to stop or the answer would not count.

[quote="alex960531"]North Carolina's State MATHCOUNTS countdown reader was hilarious, he let people buzz in before he finished reading the question and allowed them to say the answers after he finished reading[/quote]

This same thing happened my 8th grade year, and I was very disappointed to someone take advantage of it by immediately buzzing in then working during the problem reading, he went up 2-0 on me in the semis like this, I squeaked out a win 3-2, I had beaten the same person in the Chapter cd final, so I think he really wanted to win.  Several coaches raised concerns, and I think they fixed it this year, it left an opportunity for someone to take advantage of the system.

[quote="alex960531"]North Carolina's State MATHCOUNTS countdown reader was hilarious, he let people buzz in before he finished reading the question and allowed them to say the answers after he finished reading[/quote]

Huh? So you have to buzz in after the speaker is done reading the question? That would totally stink! OK will let you buzz while the speaker is still reading.

But I think I'm not getting your point.

He's saying if you buzzed in while the moderator was reading the question, you didn't have to answer until the afore-mentioned moderator finished, like:

Moderator: Blah blah blah -
Contestant: *buzz*
Moderator: Blah blah blah blah blah blah blah blah blah [...]
Contestant: <answer>

nope, you're understanding correctly. the moderator let you buzz in whenever, but then finished the question before allowing you to answer. basically, if you see a really long or really easy question, you buzz in as quick as possible, and then take allll the time you want to answer. wonderful.

Wait; so can you do that in the Nats, too?

no. in nats, the moderator stops reading, a three second timer begins, and you have three seconds to begin your answer.

Wow...I've never thought of that....
Interesting...
I don't think the Mathcounts ppl know what to do if this happens...
You should tell them about this..lol

[quote="PowerOfPi"]Does anybody know the rule if someone buzzes in at 42 seconds, doesn't start the answer for more than three seconds while the other person buzzes in at 43 seconds? Someone could take advantage of this if he/she does not know the answer, counts seven seconds after the 10-second warning, buzzes in, and stalls for three seconds while the other person is fantically trying to get the buzzer to sound because he knows the answer. The person frantically trying to make the buzzer work (me) advances if he/she gets it right and doesn't advance if he/she doesn't answer.[/quote]

But the problem is: Can you count 7-seconds?

[quote="PowerOfPi"]Does anybody know the rule if someone buzzes in at 42 seconds, doesn't start the answer for more than three seconds while the other person buzzes in at 43 seconds? Someone could take advantage of this if he/she does not know the answer, counts seven seconds after the 10-second warning, buzzes in, and stalls for three seconds while the other person is fantically trying to get the buzzer to sound because he knows the answer. The person frantically trying to make the buzzer work (me) advances if he/she gets it right and doesn't advance if he/she doesn't answer.[/quote]

In our chapter that happened..

We were using a buzzer system that kept who buzzed in,
So after the other kid buzzed in and stalled,
It automatically was my turn to answer even though the time limit was over.


A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled?

$ \textbf{(A)}\ 1404 \qquad \textbf{(B)}\ 1462 \qquad \textbf{(C)}\ 1604 \qquad \textbf{(D)}\ 1605 \qquad \textbf{(E)}\ 1804$

[hide="Answer"]This is essentially just in base 9, so we have $2\cdot 9^3+5=\boxed{1463}$.[/hide]

That's not one of the choices. You probably made a tiny mistake somewhere. Though that was an extremely elegant approach (in my opinion; I didn't see it) :)

(it's off by 1)

How can it be off by 1? That doesn't make any sense...

Whoops, heheh... I was getting rid of the 9 rather than the 4, and it's already past the four... so it should be

[hide="Answer"]$\boxed{1462}$[/hide]

It's an old topic but I have a question about this problem. Is the answer just 2005-the number of numbers with at least one '4'?

[hide]Put in base 9, $2005\implies 2\times9^{3}+4=1462, \boxed{B}$.[/hide]

[quote="andy_tok"]Thanks, but I'm just wondering if what i said is right.[/quote]

Yes, but approaching the problem that way is so much messier.

why is the answer 2004 baser 9 insted of 2005 base 9 can some1 explain that a little better?

0 1 2 3 5 6 7 8 9 <-- odometer digit system
0 1 2 3 4 5 6 7 8 <-- base nine digit system

We can make a one-to-one correspondence between the odometer number system and the base-nine number system.

As you can see by how the numbers line up, 0's and 2's are the same in both systems, but a 5 in the odometer digit system is equivalent to a 4 in the base nine number system.

Thus, 2005 in the odometer system can be treated as 2004 in base 9.

oh thanks it seems i dont fully understand base numbers yet...i wonder if there are some base number questions in mc. :maybe:

Mathcounts has them, I remember probably a few every year. If you haven't learned them, I suggest you do. It's not that hard conceptually.

Why do we make 2005 into 2004?

12 year bump lol

Because 5 is never shown by the odometer, so we must make into 2004.

Wait, isn't 4 never shown? I thought 5 was shown.

Yes doesn't get shown. Basically base 9 kills this. Just remember that 002004 doesn't exist lol


Let $a,b,c$ be positive real numbers that satisfy following inequalities
\[4(ab+bc+ac)-1\geq a^2+b^2+c^2\geq 3(a^3+b^3+c^3)\]
Find all such $a,b,c$.

we have to find all positive real numbers $ a,b,c$ such that:

$ 4(ab\plus{}bc\plus{}ca)\minus{}1\geq a^2\plus{}b^2\plus{}c^2\geq 3(a^3\plus{}b^3\plus{}c^3)$    (*)

note that $ a,b,c$ are positive so according to Cauchy-Schwartz inequality we have:

$ a^3\plus{}b^3\plus{}c^3\equal{}\frac{a^4}a\plus{}\frac{b^4}b\plus{}\frac{c^4}c\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{a\plus{}b\plus{}c}$

hence according to (*) we will have:

$ a^2\plus{}b^2\plus{}c^2\geq 3\frac{(a^2\plus{}b^2\plus{}c^2)^2}{a\plus{}b\plus{}c}$

$ \Rightarrow a\plus{}b\plus{}c\geq 3(a^2\plus{}b^2\plus{}c^2)$  (1)

again using Cauchy we get that:

$ a^2\plus{}b^2\plus{}c^2\geq\frac{(a\plus{}b\plus{}c)^2}3$  (2)

now using the last two inequalities we get that:

$ a\plus{}b\plus{}c\geq (a\plus{}b\plus{}c)^2$

$ \Rightarrow\boxed{a\plus{}b\plus{}c\leq 1} (I)$

now we know that $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ so according to (*) we must have:

$ 4(ab\plus{}bc\plus{}ca)\minus{}1\geq ab\plus{}bc\plus{}ca\Rightarrow ab\plus{}bc\plus{}ca\geq\frac 13$

now we have:

$ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca)\geq 3(ab\plus{}bc\plus{}ca)\geq 1$

$ \Rightarrow\boxed{a\plus{}b\plus{}c\geq 1} (II)$

now according to (I),(II) we get that:

$ a\plus{}b\plus{}c\equal{}1$

now on the other hand according to (1) and (2) we have:

$ 1\equal{}a\plus{}b\plus{}c\geq 3(a^2\plus{}b^2\plus{}c^2)\geq (a\plus{}b\plus{}c)^2\equal{}1$

hence we must have:

$ \Rightarrow a\plus{}b\plus{}c\equal{}3(a^2\plus{}b^2\plus{}c^2)\equal{}(a\plus{}b\plus{}c)^2$

thus:

$ 3(a^2\plus{}b^2\plus{}c^2)\equal{}(a\plus{}b\plus{}c)^2$

$ \Rightarrow 2(a^2\plus{}b^2\plus{}c^2)\equal{}2(ab\plus{}bc\plus{}ca)$

$ \Rightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\equal{}0$

$ \Rightarrow a\equal{}b\equal{}c$

and also we had $ a\plus{}b\plus{}c\equal{}1$ thus:

$ a\equal{}b\equal{}c\equal{}\frac 13$

is the only solution.

Cauchy-Schwartz inequality hmm - that was all i need
tnx

By $ \frac {4(a \plus{} b \plus{} c)^2}{3} \minus{} 1 \geq 4(ab \plus{} bc \plus{} ca) \minus{} 1 \geq a^2 \plus{} b^2 \plus{} c^2 \geq \frac {(a \plus{} b \plus{} c)^2}{3}$, we have $ a \plus{} b \plus{} c \geq 1$, with equality iff a=b=c.

Now, $ a^2 \plus{} b^2 \plus{} c^2 \geq 3(a^3 \plus{} b^3 \plus{} c^3 ) \geq (a^2 \plus{} b^2 \plus{} c^2 )(a \plus{} b \plus{} c) \geq a^2 \plus{} b^2 \plus{} c^2$.

Since we must have equailty, we must have a=b=c, a+b+c=1. This implies $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$

btw, this should belong to the inequalities section  :P


trigonometry
quadratics
email
geometry
complex numbers

I'm looking for a good book for my son to learn about complex numbers.
He has had pre-algebra and a little algebra (most of the Key to Algebra workbooks in school).

Any suggestions for a good intro to complex numbers at about an Algebra I level?  When I explain it to him, I quickly get into trigonometry---I can't seem to stay away from e^(i omega)!

Applications would be good, so an intro to electronics that includes an intro to complex numbers might be good.

Reading level doesn't matter much, but a narrative or visual format would work better than a formula-heavy approach.  (The "Cartoon Guide to ..." series is about the right level for him, but he also likes "Physics of Superheros".)

Our own Introduction to Algebra book should be available in late March, and will include a couple of chapters on complex algebra: no trig or Euler's formula, just enough to be able to find non-real solutions to quadratic equations, and some simple complex-number arithmetic (up to the level of dividing one complex number by another).

The book will be in the same style as [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]our other Introductory-level textbooks[/url], and will include lots and lots of sample problems.

Thanks.

Could you post a reply on this message or otherwise inform me when the new book becomes available?  It sounds like just about the right level for my son.

[quote="engineering_prof"]Could you post a reply on this message or otherwise inform me when the new book becomes available?  It sounds like just about the right level for my son.[/quote]If you click on the "Stay Informed" link on the left side of the page, you can join our mailing list, and we'll send an email message when the book is available.

[quote="Phelpedo"]Titu's [i]Complex Numbers from A ... to Z[/i] is pretty good.[/quote]Maybe so, but it's definitely not at an "Algebra I" level.

Yes, A..to Z is a brilliant book but not really the required level.
I trust the AOPS book'll be good, and if you can find it my
preferred text is Hobson's Plane and Advanced Trigonometry, under supervision the more advanced material can be overlooked and it is sure to be rewarding.

We have the AoPS Algebra 1 text now, and it does look to be about the book I was looking for.  For various external reasons, we've been using the Intro to Geometry book first, but we'll get back to Algebra over the summer.


I found two very complicated composite function questions that I have never seen in class before. However, I am interested in learning how to solve each question below.

Find functions f and g so that f(g(x)) = H.

1) H(x) = sqrt{x^2 + x + 1}

2) H(x) = (1 - 1/(x^2))^2

1) $f(x)=\sqrt{x}$, $g(x)=x^2+x+1 \Rightarrow f(g(x))=\sqrt{x^2+x+1}$.

2) $f(x)=x^2$, $g(x)=1-\frac{1}{x^2}\Rightarrow f(g(x))=\left(1-\frac{1}{x^2}\right)^2$.

Am I missing something here?  :?


Condsider two triangle $AHM$ and $MDC$, prove or disprove. That there exist a circle with $A,H, D$ and $C$ on it.

[quote="nutz_for2.718281828"]There is no proof, or disproof(prove me right or wrong!!) because quad AHDC can either be cyclic or non-cyclic.[/quote]
Yep. :huh:


you are given any 2 points, could u [color=red][b]USE COMPASS ONLY[/b][/color] to get the midpoint???
remember...COMPASS ONLY!!
This question drives me crazy for a few days already...i hope someone can help me find out the solution.. :(

Draw a line between the two points.
Then place each point of the compass on each point and draw two circles, with one of the points being the center of each.
Now draw another line through the points where the two circles intersect.
The intersection of this line and the one you drew initially is the midpoint of the segment made from the two points.

But you are using only a compass, so you do not have a straightedge.

I think this is it: [url]http://www.cut-the-knot.org/do_you_know/compass14.shtml[/url]

:huh: No straightedge  :( .
That site has an elegant solution  :) .

Thanks,kohjhsd. I think that is the only solution for it...
Without straightedge, it is pretty hard to think about bisect the line (for me :D )
Anyway... :idea: I hope my insomnia disappears because of this question.... :rotfl:

just a little thing...

The ________ Theorem states that anything that can be constructed with compass and straightedge can be constructed with compass alone, if what you want to construct is a line, one can construct two points on that line.

The ________ Theorem states that anything that can be constructed with compass and straightedge can be constructed with straightedge and a compass that is "rusty," that is, you can't change the radius.

Even better than the rusty circle thing:
you can still construct the same if you are given a straightedge and one single drawn circle only.

Rather surprisingely, you can do even more (trisect angles for example) if you are given a marked ruler.

What about this?
Can you only use compass to construct the circumcenter of a triangle?

Umm....I got another question ...also it uses compass only
Question: Given a circle....can u find out its center of the circle by using compass only??? :)
Thx guys... :D

the Mohr-Mascheroni Theorem states the compass-alone constructibility; the Poncelet-Steiner Theorem gives the minimum condition for only a straightedge.

constructions using a marked ruler rotatable around any point are called 'neusis' constructions and were frowned upon even by classical Greek mathematicians.  using such tools also allows one to construct a regular heptagon.


[quote="swapnilshwetank"][b]if a+b+c=1, show that:
a/(1+bc) + b/(1+ac) +c/(1+ab)>=9/10[/b][/quote]
since cauchy shawrz ineq
$ \sum \frac {a}{1 \plus{} bc} \ge \frac {(a \plus{} b \plus{} c)^2}{a \plus{} b \plus{} c \plus{} 3abc}$
and $ 3abc\le \frac {1}{9}$ ($ AMGM$)we have done


probability
function
binomial coefficients
expected value
probability and stats

I have an exam later, and while practicing, found these 4 problems I can't complete

In a class of 15 people, exactly 3 got an A. If 2 people are randomly chosen from this class, what is the probability that at least one of these 2 people got an A?

A box contains 3 defective lights and 5 non-defective lights. The lights are tested one at a time without replacement. What is the probability that the 3 defective bulbs will be found in the first 3 tests?

A medical test detects H.I.V. Among those who ahve H.I.V, the test will detect the disease with probability 0.95; among those who do NOT have HIV, the test will falsely claim that HIV is present with probability 0.0125. Among those who take this test, 4% have HIV. The test is given to Lucille Jones, and indicates taht she has HIV. What is the probability that Lucille actually has HIV?

The probability that both Barb and Bob are both on the same committee of 4 which is chosen from their class of 10 people is:

Any help is appreciated. It'd be nice if you could give the steps you took, though just an answer would also give me something to work off of

Words that apply to these four problems: hypergeometric, hypergeometric, Bayes, hypergeometric.

The general setup for a hypergeometric probability distribution: We have a pool of $ N$ objects, of which $ r$ are of a certain type. (Say, we have a bucket with $ N$ colored balls, of which $ r$ are red balls.) We choose, without replacement, a sample of $ n$ of these objects. Our random variable, $ Y,$ is the number that are of the certain type. (Such as letting $ Y$ be the number of red balls in our chosen sample.)

The probability function: $ P(Y \equal{} y) \equal{} \frac {\binom{r}{y}\binom{N \minus{} r}{n \minus{} y}}{\binom{N}{n}}$

For this to be anything other than zero, all of the the binomial coefficients must make sense - so we must have $ n\le N, 0\le y\le r,$ and $ 0\le n \minus{} y\le N \minus{} r.$

The expected value of such a distribution is its common sense value: $ E(Y) \equal{} \frac {nr}N.$

The variance is $ V(Y) \equal{} n\left(\frac {r}{N}\right)\left(\frac {N \minus{} r}N\right) \left(\frac {N \minus{} n}{N \minus{} 1}\right).$

If we hold $ n$ (the sample size) fixed but let $ N$ (the pool size) tend to $ \infty,$ while keeping $ \frac {r}{N}$ (the proportion of the pool with the desired property) constant and equal to $ p,$ then the distribution tends in the limit towards a binomial distribution. This, too, is common sense.

By the way, in kenn4000's solution to #1, the denominators should be $ \binom{15}{2},$ not $ \binom{14}{2}$ as he wrote. I assume that's just a typo. That particular problem falls into what I wrote above with $ N \equal{} 15, n \equal{} 2,$ and $ r \equal{} 3.$

yeah typo,

professor kent, does this relate to the mathematica solutions i always see on this website that say
hypergeom(a,b,c)? or just coincidentally the same name?


There is a circular table with 8 identical chairs. In how many ways can 5 people seat in the table?

[hide="Solution"]Fix one dude in some chair. There are 7 chairs left, and $ \frac{7!}{(7\minus{}4)!}\equal{}\boxed{840}$ ways to place people.[/hide]

How come it's $ 7!/(7\minus{}4)!$ instead of $ 7!/4!3!$?

Well, assumably the people are distinct.


What is the largest circle within which the Maclaurin series for the funtion [b]tanhz[/b] converges to [b]tanhz[/b]?

$\tanh z$ is a meromorphic function - analytic except at isolated poles. The poles are the zeros of the denominator, which is $\cosh z=\cos iz.$ The closest poles to the origin are at $\pm\frac{i\pi}2,$ so the radius of convergence of the Maclaurin series is $\frac{\pi}2.$


Please refer to attached figure....

The numbers $1$ to $6$ appear once in each row and each column. The clues refer to the sum of the numbers in the designated squares. So, for example, $BCD5 = 10$ would mean that the numbers in squares $B5$, $C5$ and $D5$ add up to $10$.

$ABC5 = 10$	$BCD3 = 7$	$DEF1 = 8$	$E123 = 15$
$A23 = 8$		$C12 = 7$		$D456 = 15$	$E56 = 4$
$A56 = 10$	$C56 = 4$		$EF2 = 11$	$F34 = 7$

Find all the numbers?

Solve this problem the way one would solve a Sudoku except that instead of having a 3x3 box dictate where the numbers go, use the hints given in the sums.  The solution is:
[hide]163425
456231
312564
524613
641352
235146[/hide]


modular arithmetic
number theory proposed
number theory

Determine all integers $n$ satisfying:

1. $n = p^2 + (p+5)^2$, where $p$ is a prime number which is the sum of two squares of prime numbers.
2. $n \not\equiv 3 \pmod{10}$.

$n = 2p^2 + 10p + 25$. So $2p^2 + 10p + 25 \not\equiv 3 \bmod 10$. Hence $p^2 \not\equiv 9 \bmod 10$
So we must have $p \equiv 1 \bmod 10$ or $p \equiv 5 \bmod 10$.
If $p$ is the sum of two squares of primes, then we must have $p \equiv 1 \bmod 4$.
$p = q^2 + r^2$. Then $r = 2$ or $q = 2$.
Thus we have $p \equiv q^2 + 4 \bmod 10$. $q^2 \equiv 1, 5, 9$. Hence $p = 5$ is the only solution.


trigonometry
algebra
system of equations
algebra proposed

it's more like a computational problem: substitute $a=2 \cos{x}$ etc. and it turns out that the solutions are $(a,b,c,d,e)=(2, \frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2})$


[hide]Draw the diagram. Since the tangent line has slope -1 and y-intercept 2k, the radius that's perpendicular to the tangent line must have slope 1. You have a right isosceles triangle formed. Applying Pythagora's theorem, you find

k + k = (2k)2

k = 1/2.[/hide]

Tom, I think that you may have mistaken the OP's state of the problem with x^2+y^2=k with x^2+y^2=Sqrt(k), because x^2+y^2=k yields a radius of Sqrt(k), not k; thus, your right isosceles triangle is not right nor isosceles.[/quote]

Max:

You worked out that x=y.. now just remember that if its tangent then [hide]it must intersect the circle.. ie you got two equations in terms of x and k..[/hide]

http://osiris.dyndns.info/temp/analytic.gif

[hide]It's a right isosceles triangle with hypotenuse 2k and sides :sqrt:k. Is that not right?[/hide]

Well, maybe Im being dumb for some reason, but by Pythagorean, Sqrt(sqrt(k)^2 + sqrt(k)^2) = Sqrt(2*sqrt(k)^2) = Sqrt(2k), right?

x+y=2k => y = -x + 2k, so the y-intercept is 2k. The length of the segment from the origin to the point of intersection of the line with the y-axis has length 2k. So, (:sqrt:k)2 + (:sqrt:k)2 = (2k)2 by Pythagora's theorem.

Ah, my bad. For some reason, I misunderstood your original solution the first time (not your fault, I musta been tired or something).

Or, if you don't want a diagram, you could do what I said before.. ie:

[hide]x=y so x+x = 2k so x=k, and x:^2:+y:^2: = k so 2k:^2: = k. k=1/2. [/hide]


I need to know how to do basic single replacement, decomposition, double replacement, combination, and combustion reactions.  My teacher gave us a short handout but it's not clear at all.

Combination: A + B -> AB
Sounds simple enough but Ca + C -> CaC2
Do I use oxidization numbers for the product?

Decomposition: AB -> A + B
Simply the opposite but the problems are never as simple as that.  I just have no idea how to determine what the products are.

Single Replacement: A + BC -> AC + B
I have to make sure A is more reactive than B with the activity series and then I replace it with B and balance with oxidization numbers.  Probably the one I understand the most (I hope). Also if A is a nonmetal do the same thing and it'd replace the C in BC.

Combustion: These ones I think are easy.  It's always something reacting with O2 and each element in the reacting compound make an oxide.

Double Reaction: AB + CD -> AD + BC
I'm not sure on the special rules for this one.

So that's pretty much it.  Maybe we can get a solid thread going on how to do reactions in general since there seems to be a lack of resources on it.

[quote="Carcul"][quote="Kysuke"]and each element in the reacting compound make an oxide.[/quote]

Wrong. When the compound contains nitrogen, it appears in the products as $ \ce{N2}$, not as a nitrogen oxide.[/quote]

Really? my teacher said with limited oxygen it makes NO and with excess oxygen it makes NO2.

When I wrote the above, what I had in mind was the combustion of nitrogen containing organic compounds, where molecular nitrogen is *really* the major nitrogen product.


limit
function
integration
real analysis
real analysis unsolved

The limit depends on both $f$ and the exact nature of the $a_{i}$. It doesn't have to exist.
For example, suppose $f(\frac14)=-1$ and $f$ is zero otherwise. If $a_{n}$ is the $n$th odd number, that minimum is equal to $-1$ half of the time and zero the other half.

@jmerry...I forgot some part of the hypothesis :blush:  I modified a little the statement (f is defined on [0,1] and is positive-I think this was the key)
@all: So c'mon :)

By "positive", you mean nonnegative; all strictly positive Riemann integrable functions have positive integral.

Let $x^{n}_{i}=\frac{a_{1}+a_{2}+\cdots+a_{i}}{a_{1}+a_{2}+\cdots+a_{n}}$ for $0\le i\le n$ ($x^{n}_{0}=0$).  We have $x^{n}_{i}-x^{n}_{i-1}\le\frac{a_{n}}{\sum_{j=1}^{n}a_{j}}\to 0$ as $n\to\infty$, so the Riemann sums $\sum_{i=1}^{n}(x^{n}_{i}-x^{n}_{i-1})f(x^{n}_{i})$ converge to $\int_{0}^{1}f(x)\,dx=0$. The minimum $f(x^{n}_{i})$ must be at most the weighted average, and thus $\limsup_{n}\min_{i}f(x^{n}_{i})\le 0$. The terms of this sequence are nonnegative, so the limit exists and is zero.

As for why $\frac{a_{n}}{\sum_{j=1}^{n}a_{j}}$ tends to zero-
if $a_{n+1}<2^{1/m}a_{n}$ for $n\ge N$, the next $m$ terms after $a_{N}$ are less than $2a_{N}$, and $a_{N+m}\le 2A_{N}\le \frac{2}{m+1}\sum_{j=N}^{N+m}a_{j}\le \frac{2}{m+1}\sum_{j=1}^{N+m}a_{j}$.
Such an $N$ exists for each $m$.

Nice jmerry
I have a question...How do you got the idea to use powers of 2? :maybe:
I solved that part using Cesaro-Stolz and  I think is much easier that way :P  However, I find your idea very interesting :)


MATHCOUNTS
Ring Theory
AMC
AIME
geometry
search
Alcumus

[b]Do lots of problems.[/b] If the problem is money and/or time, the internet is also a great resource. See some links [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286559]here[/url].

For AMCs, try [url=http://www.math.ksu.edu/main/handbook/ProblemSets]here[/url] and for MathCounts try [url=http://www.mnspe.org/mathcounts/coach.htm]here[/url] and [url=http://mathcounts.org/Page.aspx?pid=1524]here[/url].

Try For the Win, and Alcumus(when it's running again), they have MATHCOUNTS and MATHCOUNTS level problems. Does your school have a math club if it doesn't see if you can start one. If your homeschooled I sure you can find other students you can work with on Art of Problem Solving.  Also you could look at previous problem to see if you know how to do them. And if you want to you could always take an AoPS introductory or mahcounts problem series class.

why do i think this should go into the MATHCOUNTS forum?

You could just use screencapture to get the images that you don't print out.

I would try doing some aops questions and see the type of problems they use. Then I would go online and search for some problems... I would also try to find previous years of mathcounts problems online.

Hope this helps!

Try doing the online Alcumus, go on mathcounts.org for problems of the week and old sprint rounds, or go online and search for target rounds and old sprint rounds for practice.

[quote="batteredbutnotdefeated"][quote="topofmath"][quote="shoobish"]Thanks Power of Pi,
Do you knwo where I can find printable past AMC exams and Past state mathcounts tests. I am aware that there are some at The art of problem solving but i rather able to find PDF files of these tests.
I am going to try and do as many math problems as i can.

Thanks for your help.[/quote]

State Mathcounts I don't believe you can get them without the books, except for 2009 and 2007. AMC 8 and 10 are on [url]http://mathjunk.googlepages.com/[/url]

All of the AMC 8,10,12, AIME are on AoPS Wiki.

Hope that helps.:)[/quote]

yes, but you can't print out the geometry problems. They will appear as asymtote.[/quote]

Not if you print them directly from AoPS wiki. Although I do agree that the PDF makes it easier. You can try the [url=http://www.mathlinks.ro/Forum/resources.php]Olympiad Resources[/url] page.

[quote="shoobish"]Hello World,
does anybody know the best way to train for mathcounts without any type of
formal classes or textbooks?

Please Advise.
Please.[/quote]
you should try memorizing the squares, cubes, prime numbers, and your multipication facts oh yeah and the different formulaes for certain problems

Also, you should make sure you understand the answers, not just memorize them so you can reuse it in similar problems.

[quote="AwesomeToad"][quote="shoobish"]Thanks Power of Pi,
Do you knwo where I can find printable past AMC exams and Past state mathcounts tests. I am aware that there are some at The art of problem solving but i rather able to find PDF files of these tests.
I am going to try and do as many math problems as i can.

Thanks for your help.[/quote]

State Mathcounts I don't believe you can get them without the books, except for 2009 and 2007. AMC 8 and 10 are on [url]http://mathjunk.googlepages.com/[/url]

All of the AMC 8,10,12, AIME are on AoPS Wiki.

Hope that helps.:)[/quote]

Wow, that person HAD to password protect the mathcounts section.

and mathacle.com is now out of business.  :sigh:  I didn't finish with it yet :(

and the people who have everything possible aren't exactly willing to give stuff away..........so I have very little.

1) LOL I have every single test there is. I'll post 3 on this forum and password protect them (can you password protect on AoPS)? I'll give the passwords to some people if I feel like it and they ask nicely (mayback for example)

2) I have something better than solving problems. MODIFYING PROBLEMS. Take a problem and use letters for all the variables. Construct a new problem with different numbers that can be solved by the same process. If you can do that, it means you have understood every nook and cranny of the problem.


calculus
derivative
integration
real analysis
real analysis unsolved

I can't solve,,,help!

Let $ I(a,b)\equal{}\int_0^1 |x^2\plus{}ax\plus{}b|dx$ where $ a,b$ are reals.
Determine the minimum value of $ I$ and $ (a,b)$ gives the minimum.

Equivalently, find the minimum of $ I \equal{} \int_0^1|(x \minus{} r_1)(x \minus{} r_2)|dx$ where $ r_1,r_2\in \mathbb{R}$ (why can we disregard the case when $ r_1,r_2$ are complex?).  Assume without loss of generality that $ r_1\leq r_2$.  There are 6 cases: (1) $ r_1\leq r_2\leq 0$, (2) $ r_1\leq 0\leq r_2 \leq 1$, (3) $ r_1\leq 0 < 1 \leq r_2$, (4) $ 0\leq r_1 \leq r_2 \leq 1$, (5) $ 0\leq r_1 \leq 1 \leq r_2$, (6) $ 1\leq r_1 \leq r_2$.  I'll only do case 4 (in some sense the hardest case, and the one that happens give the minimum):
In case 4 we have
\[ I \equal{} \int_0^{r_1}(x \minus{} r_1)(x \minus{} r_2)dx \minus{} \int_{r_1}^{r_2}(x \minus{} r_1)(x \minus{} r_2)dx \plus{} \int_{r_2}^1(x \minus{} r_1)(x \minus{} r_2)dx \\
\equal{} \frac {1}{3}(1\plus{}r_2^3 \minus{} r_1^3) \minus{} \frac {1}{2}(r_1 \plus{} r_2)  \plus{} r_1r_2(1 \plus{} r_1 \minus{} r_2)
\]
By the usual methods (partial differentiation), we find the minimum is $ \frac {1}{16}$ achieved at $ r_1 \equal{} \frac {1}{4},r_2 \equal{} \frac {3}{4}$ or equivalently $ \boxed{a \equal{} \minus{} 1,b \equal{} \frac {3}{16}}$.

It remains, of course, to check that the other cases do not yield a minimum.  You can do this by directly computing each minimum in a similar manner as above.
[hide="remark"]  Actually for cases 1, 3, and 6, you don't need to compute anything; they cannot yield a minimum for essentially the same reason we ignored the possibility of complex roots.  Cases 2 and 5 are less clear, but you can argue them away as well, though it's a bit more subtle.  But direct computation works just fine, too.[/hide]

JoeBlow,Thank you very much.I'm really poor at Analysis with many variables...confusing...
but I try it!


binomial coefficients
combinatorics proposed
combinatorics

How do I simplify this

$\sum_$ $\left( \begin{array}{c}k \\ i \end{array}\right)$ $p^{i}\cdot (1-p)^{(}k-i)$

into a form of something like this

$(1+(1-2p)^{k+1}-2 \cdot (1-p)^{k+1}) / 2$

I know that $\left( \begin{array}{c}k \\ i \end{array}\right)$
$= k! / (i! \cdot (k-i)!)$

thanks!

Hrm?  You should be using { } in place of ( ) if you want to raise more than one character into an exponent.

Also, use _{i = }^{} to put bounds on the summation.

Also, you can use \binom{}{} instead of an array to get binomial coefficients, and \cdot instead of * to get a dot for multiplication.

any help on how to solve it to make it look like that form?
i fixed the post a little bit, thanks!

$\sum_{i = 0}^{k}\binom{k}{i}p^{i}(1-p)^{k-i}= 1$, so I think it may be some difficulty to put it into the form you desire.

$i$ is from 2 to k but all the i's are even: 2,4,6...k

$ \sum_{i=0}^{[\frac{k}{2}]}{ k \choose 2i}p^{2i}(1-p)^{k-2i}=\frac{\sum_{i=0}^{k}{k \choose i}p^{i}(1-p)^{k-1}+\sum_{i=0}^{k}{k \choose i}(-1)^{i}p^{i}(1-p)^{k-i}}{2}$
Is that what you had in mind? (if you dont need $i=0$ just kick it out of the sum)


If a,b,c are different numbers and if $a^3+3a+14=0,b^3+3b+14=0,\text{and } c^3+3c+14=0$ find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

[hide="answer"]Notice that $a=b=c$, so simply we need to transform the polynomial, let us give it the variable $x$, which is $x^3+3x+14=0$ into the polynomial $14x^3+3x^2+1=0$ to find out the sum of the reciprocals of the roots, which is simply $\frac{-3}{14}$[/hide]

[quote="José"][quote="Iversonfan2005"]Notice that $a=b=c$[/quote]

In the problem say that $a$, $b$ and $c$ are different.[/quote]

oh well, my problem still works out, because I let $a,b,c$ be the roots themselves :)

Factoring  x ^ 3 +3x +14, get:
(x + 2)(x ^ 2 - 2x + 7). Now this is 0 for x = a, b, c.
Wolog, a= -2, and b and c are different roots of
x ^ 2 - 2x + 7 = 0.(Because a, b, c are different).
Then:
1/a +1/b + 1/c = 1/(-2) + (b+c)/bc.          (1)
In x ^2 - 2x + 7 = 0, product of the roots, or bc, is 7, and sum of the roots, or b + c, is - ( - 2) = 2.
Get:
(b + c) / bc = 2/7. Substitute in (1):
1/a + 1/b + 1/c = -1/2 + 2/7 =  - 3/14

now you see how efficient my way is by simply transforming the polynomial :)

Notice that a,b,c are roots of x^3 + 3x +14 = 0. We then know that
abc = -14 and ab + bc + ac = 3. Also,
1/a + 1/b + 1/c = (ab + bc + ac)/(abc) = -3/14.

That's 4 lines. Even more efficient! :lol:


Find the largest positive integer n with the property that n+6(p^3 +1) is prime whenever p is a prime number such that 2 ≤ p < n. Justify your answer.

[hide="The rest is computation"] It is easy to calculate that $ n < 20$ because when $ 2\le p < 20$ the expression $ 6(p^{3}\plus{}1)$ covers all residues $ \bmod 5$.  We also know that $ gcd(n, 6) \equal{} 1$; it remains to test $ n \equal{} 19, 17, 13, 11, 7, 5$. [/hide]


Can someone post random counting distinguishability ball and box problems. Thanks

Try searching, [url=http://www.mathlinks.ro/viewtopic.php?p=1584859#1584859]here[/url].

Also, read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308193]this[/url].


algebra
polynomial
group theory
superior algebra
superior algebra unsolved

[b]Problem [/b]Let F be a finite field of characteristic p. Show that every element of F is algebraic over the prime field $ Z_p \leq F$. (Hint: Let $ F^*$ be the set of nonzero elements of F. Apply group theory to the group $ < F^*,\cdot >$ to show that $ \alpha \in F^*$ is a zero of some polynomial in $ Z_p[x]$ of the form $ x^n \minus{} 1$.

My first question is about the second sentence. Is it true that every field of characteristic p has Z_p contained in it or is that an assumption we make just for this problem?

It's not an assumption, but a fact. If $ F$ is of characteristic $ p$, then $ \mathbb Z_p$ can be identified with $ \{ 0,1,2,...,p \minus{} 1 \} \subset F$.


Let $ x,y \geq 0$. Prove that $ 3(x- \sqrt[3]{xy}+y) \geq 4 \sqrt{xy}-1  $  and $3(x- \sqrt[3]{xy}+1) \geq 4 \sqrt{x}-y$. When does eqaulity occur?

3(x-(xy)^(1/3)+y) \geq 6 \sqrt (xy) -3(xy)^(1/3)
witre xy as z^6 we have LHS \geq 6z 3 -3z 2 and RHS=4z 3  +1.
Thus we only need to prove 2z 3 -3z 2 +1>=0, which is immidiate since 2z 3 -3z 2 +1=(z-1) 2 (2z+1).
Equality occurs iff x=y and z=1, which is equivlent to x=y=1.

By AM-GM we have y+2 \sqrt x \geq 3(xy) 1/3 .
so we have LHS-RHS=3x-4 \sqrt x+y-3(xy) 1/3 +3 \geq 3x-6 \sqrt x+3=3( \sqrt x-1) 2  \geq 0.
Equality occurs iff x=y=1.


linear algebra
matrix
complex numbers
complex analysis
linear algebra unsolved

Let $ A\in \mathcal{M}_{n}(\mathbb C)$ , $ n\in \mathbb N^*$ , with $ \det(I_{n}\plus{}A^k)\equal{}1 ,$ $ \forall k\in \mathbb N^*$ .Prove that $ A^n\equal{}O_{n}$

By Schur's Lemma, WLOG we may assume that $ A$ is upper triangular with diagonal elements $ \lambda_1\,\dots,\lambda_n.$ The theorem comes down to this:

If we assume that $ \prod_{j\equal{}1}^n(1\plus{}\lambda_j^k)\equal{}1$ for all $ k\in\mathbb{N},$ can we prove that $ \lambda_1\equal{}\lambda_2\equal{}\cdots\equal{}\lambda_n\equal{}0?$

look here
http://www.mathlinks.ro/viewtopic.php?t=180240
if all eigenvalues are complex nobody give a proof :( (if all is real is trivial )

Hmmm, I think we discussed this really many times on this forum. It reduces to proving that if $ z_1,z_2,...,z_n$ are complex numbers and all sums of $ k$-powers of them are $ 0$, then all numbers are $ 0$. Does Newton's formulae ring a bell? If not, we have actually seen (I think it was used by Fedja in one of my problems) that the superiour limit of the sequence $ |z_1^k\plus{}...\plus{}z_n^k|^{1/k}$ is $ \max |z_i|$.This can be obtained either by complex analysis or by diophantine aproximation. All this is discussed, as I said, in several topics.

[quote="harazi"]Hmmm, I think we discussed this really many times on this forum. It reduces to proving that if $ z_1,z_2,...,z_n$ are complex numbers and all sums of $ k$-powers of them are $ 0$, then all numbers are $ 0$. Does Newton's formulae ring a bell? If not, we have actually seen (I think it was used by Fedja in one of my problems) that the superiour limit of the sequence $ |z_1^k \plus{} ... \plus{} z_n^k|^{1/k}$ is $ \max |z_i|$.This can be obtained either by complex analysis or by diophantine aproximation. All this is discussed, as I said, in several topics.[/quote]
Hello,Harazi,
How can I see can this problem reduce to:
"proving that if $ z_1,z_2,...,z_n$ are complex numbers and all sums of $ k$-powers of them are $ 0$, then all numbers are $ 0$"?

Well, just note $ x_i$ the eigenvalues of the matrix and use the fact that $ (1\plus{}x_1^k)(1\plus{}x_2^k)...(1\plus{}x_n^k)\minus{}1$ is the sum of $ z^k$ over all $ z$ equal to products of some $ x_i$'s. These will be the $ z_i$'s in my statement.

Thank you,Harazi.To remind others,that in your post the "n" is not the "n" in posabogdan's post.

Another question,Is the problem still be true for just we have the condition $ k\equal{}1,\cdots,n$ instead of $ k \in \mathbb{N}$?


Let M be the midpoint of BC in triangle ABC. Extend AM to H so that AHC is a right triangle. It is known that AB=2MH, find the ratio of <BAM to <MAC

[img]http://i778.photobucket.com/albums/yy69/Thamus/tri.jpg[/img]
here is a picture I made on paint.


Hello
$ (ln(x))^2\minus{}2ln(x)\equal{}0$
$ x\equal{}1,e^2$
Thank u


[quote="ely_en"]Let $ a, b, c > 0$ Show that:
$ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\ge \frac {a(b^2 \plus{} c^2)}{a^2 \plus{} bc} \plus{} \frac {b(c^2 \plus{} a^2)}{b^2 \plus{} ca} \plus{} \frac {c(a^2 \plus{} b^2)}{c^2 \plus{} ab}$[/quote]
[hide]
After expanding we obtain that your inequality is equivalent to
$ \sum_{sym}(a^8bc\minus{}2a^6b^3c\plus{}2a^6b^2c^2\plus{}a^5b^5\plus{}4a^5b^4c\minus{}2a^5b^3c^2\minus{}2a^4b^4c^2\minus{}2a^4b^3c^3)\geq0,$
which is obvious. [/hide]

[quote="micf"]arquady, can you show me, why this inequality is obvios? Unfortunately it isn't obvious for me yet... :D[/quote]
For example,
$ \frac{1}{2}\sum_{sym}(a^8bc\minus{}2a^6b^3c\plus{}a^6b^2c^2)\equal{}\sum_{cyc}(a^8bc \minus{} a^6b^3c \minus{} a^6c^3b \plus{} a^6b^2c^2) \equal{}$
$ \equal{} abc\sum_{cyc}(a^7 \minus{} a^6b \minus{} a^6c \plus{} a^5bc) \plus{} abc\sum_{cyc}(a^6b \plus{} a^6c \minus{} a^5b^2 \minus{} a^5c^2)\geq0,$
 which is true by Schur and Muirhead. :wink:


search
LaTeX
number theory unsolved
number theory

Show that for any integer n > 10000, there are integers a, b such that$n < a^2 + b^2 < n + 3\sqrt[4]{n}$.

Weird...

Indeed, I could find it... :
http://www.mathlinks.ro/Forum/viewtopic.php?t=62830
http://www.mathlinks.ro/Forum/viewtopic.php?t=292

:)

[quote="Mathx"][quote="ZetaX"]Posted a lot of times before![/quote]

I couldn't find it using search function.[/quote]
How couldn't you find it¿¿¿
I simply entered your $\LaTeX$ code (n < a^2 + b^2 < n + 3\sqrt[4]{n}) and got a link!!!

Yes, the same goes for me [b]ZetaX[/b], but, I think his problem (and it appears he isn't the only one) is that he might have searched for "Russia 2002", or something like that.


I submitted my solutions on October 18. Do you accept solutions sent this early? If so, when will you start logging the solutions?

We do accept them early, but we won't start logging them until after the submission deadline.

[quote="SplashD"]Around when would you expect the second round solutions to be all graded?[/quote]
The NSA grading session is scheduled for December 18. I don't know whether any other groups have grading sessions after that. So the grading and the recording of the grades will be done no earlier than December 23.

Erin Schram
USAMTS grader

Expect the scores to be released something during the week of December 25-29.  (This is not a guarantee.)

And we can't discuss the problems until after the math jam?

Do not start discussing the problems until after the Math Jam.  We expect to put up Round 3 problems next week.


calculus
inequalities unsolved
inequalities

find the maximum of
$ \frac{x}{x^2\plus{}1}\plus{}\frac{y}{y^2\plus{}1}\plus{}\frac{z}{z^2\plus{}1}$  such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$

[quote="Dr Sonnhard Graubner"]hello, are $ x,y,z$ real numbers?
Sonnhard.[/quote]

$ x,y,z \in \mathbb{R}^{\plus{}}$

hello, with the help of calculus i have got
$ \frac{x}{x^2\plus{}1}\plus{}\frac{y}{y^2\plus{}1}\plus{}\frac{z}{z^2\plus{}1}\le \frac{3}{2}$ the equal sign holds if $ x\equal{}y\equal{}z\equal{}1$.
Sonnhard.

$ (x\minus{}1)^2\ge0\Rightarrow x^2\plus{}1\ge2x\Rightarrow \frac{1}{2}\ge\frac{x}{x^2\plus{}1}$ with equality only for $ x\equal{}1$.
$ \Rightarrow \frac{x}{x^2\plus{}1}\plus{}\frac{y}{y^2\plus{}1}\plus{}\frac{z}{z^2\plus{}1}\le\frac{3}{2}$ with equlity only for $ x\equal{}y\equal{}z\equal{}1$.Hence these values verify the condition that $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$ the maximum is equal to $ \frac{3}{2}$


Princeton
college
USAMTS
geometry
geometric transformation
reflection
calculus

Some conversations I've had in real life prompt me to ask this rather open-ended question. Suppose you knew a high school student who likes math, wants to study math in a good college math program, and is willing to study hard to make that happen. What would you recommend that he do to increase his chances of getting into Princeton? Any and all advice will be much appreciated.

My advice to such a student would be to do what they're probably already doing. That is, to seek out as many opportunities and resources as possible, demonstrate their interest in mathematics, and improve their understanding of mathematics. This includes taking challenging courses, checking out books from the library, participating in summer science/math programs, and participating in the USAMTS and other science/research competitions.

And of course it goes without saying, they should keep their grades high in all of their courses, especially science and mathematics.

Basically, though, there's no magic recipe for getting into a good undergraduate school. Just follow your passion for what you love to do and it will reflect in your application.

Thanks, Gauss202. I'm not sure if the friend who raised this question is even aware of some of those math opportunities, and the issue of other courses besides math and grades in those is something I don't know much about in connection with that student. This gives me something to talk about the next time our families meet.

Following up on the first helpful reply here, I'm wondering what kind of previous (math) course work would be good preparation for Princeton's math sequence.

There are different entry points into the math sequence at Princeton (and probably at any University).

There were many math majors at Princeton who had completed the standard high school sequence up to Calculus BC, and no other courses. I'd say about half of the math majors were in this category. The other half had taken at least a college multivariable calculus course and possibly more before matriculating. Almost all of the math majors in my class (10 out of 14) started in MAT 217, Honors Linear Algebra. 1 started out in MAT 204, Multivariable Calculus, and maybe 3 or 4 started in something higher.

I'd say you need at least a year long exposure to calculus, and if possible, some exposure to how to read, construct, and write a mathematical argument. But if you haven't, that's okay.

I had taken multivariable calculus (equivalent to MAT 204), and an independent study course that introduced me to mathematical proofs via set theory, elementary point set topology, and abstract algebra (just up to group theory). Both of these courses were extremely helpful when I went to college. I started in MAT 217, Honors Linear Algebra, and concurrently took MAT 319 (now MAT 331), Complex Analysis.

Princeton now has a course similar to the independent study course that I did in high school called MAT 215, Analysis in a Single Variable. It's a course meant to transition students from high school mathematics to college mathematics via advanced calculus (i.e. introduction to analysis). I suspect that most prospective math majors start out in this course. So a solid grounding in high schoool calculus and the ability to deal with some mathematical abstraction should be the only prerequisite to entering the math department.


[quote="nguoigo"][quote="N.T.TUAN"]See here  http://www.mathlinks.ro/LaTeX/AoPS_L_About.php or http://www.mathlinks.ro/index.php?f=123
$ \frac{(b+c)^{2}}{8a^{2}+(b+c)^{2}}+\frac{(c+a)^{2}}{8b^{2}+(c+a)^{2}}+\frac{(a+b)^{2}}{8c^{2}+(a+b)^{2}}\geqq1$.[/quote][/quote]
It's right for non-negative number but not for real number
Ex:a=1,b=-1,c=0

[quote="Cyberjunkie"]
 $ \Leftrightarrow$:$ 7(a^{2}+b^{2}+c{2})\leq 12(ab+bc+ca)$
which appears to be valid. :maybe:[/quote]
It's not.
Just take $ b=c=1$ and you'll have the quadric $ 7a^{2}-24a+2\leq 0$ wich is false $ \forall a\geq 0$

[quote="nguoigo"]Let $ a$, $ b$, $ c$ be three real numbers. Prove that
$ \frac{\left(b+c\right)^{2}}{8a^{2}+\left(b+c\right)^{2}}+\frac{\left(c+a\right)^{2}}{8b^{2}+\left(c+a\right)^{2}}+\frac{\left(a+b\right)^{2}}{8c^{2}+\left(a+b\right)^{2}}\geq\frac{7}{12}$.[/quote]

Wrong, e. g. for a = 100, b = - 100, c = 1.

Your left hand side tends to $ \frac29$ for a = 1, b = -1, $ c\to 0$. Don't ask me (for now) whether $ \frac29$ is its infimum...

@Svejk: You forgot that a, b, c are arbitrary reals, not necessarily positive.

  Darij

[quote="darij grinberg"]

@Svejk: You forgot that a, b, c are arbitrary reals, not necessarily positive.

  [/quote]

[quote="rangdong27"][quote="nguoigo"][quote="N.T.TUAN"]
$ \frac{(b+c)^{2}}{8a^{2}+(b+c)^{2}}+\frac{(c+a)^{2}}{8b^{2}+(c+a)^{2}}+\frac{(a+b)^{2}}{8c^{2}+(a+b)^{2}}\geqq1$.[/quote][/quote]
It's right for non-negative number but not for real number
Ex:a=1,b=-1,c=0[/quote]

The original problem asked about arbitrary reals. But your solution to the second one is nice, I just thought it referred to the original problem. As we have 2 problems now, maybe split the topic?

  darij


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let $ R$ be a finite ring. Prove that :

a) $ R$ has not zero divisor if and only if $ \forall a,b \not\equal{}0$ the equation $ ax\equal{}b$ has answer.

b) $ R$ is noncommutative and integral domain if and only if $ R$ is skew field.

a) Suppose $ ax \equal{} b$ always has a solution. Suppose we have zero divisors, so $ ab \equal{} 0$ for some $ a,b \neq 0$. But $ bx \equal{} 1$ has a solution, say $ c$, so $ bc \equal{} 1$. Multiplying $ ab \equal{} 0$ with $ c$ gives $ abc \equal{} a(bc) \equal{} a \equal{} 0$, which is a contradiction since we assumed $ a \neq 0$.

Now suppose $ R$ has no zero divisors and that $ R$ has $ n$ elements. Consider the product of an element $ a \neq 0$ with all elements in $ R$. There is $ n$ such products. All must be different, since $ ax \equal{} ay$ implies that $ a(x\minus{}y) \equal{} 0$ which is impossible since $ x \neq y$ and $ a \neq 0$ and we have no zero divisors. So we have $ n$ different elements, and since $ R$ has $ n$ elements, one of them must be $ b$.

b) Well I think $ R$ can be a skew field and a commutative integral domain (all fields are skew fields, even the commutative)

Suppose $ R$ is a skew field, i.e. all elements has a multiplicative inverse. If $ ab \equal{} 0$, multiplication with $ a^{\minus{}1}$ gives that $ b \equal{} 0$, so we have no zero divisors (i.e. $ R$ is an integral domain). Now suppose $ R$ is an integral domain. As before, look at an element $ a \neq 0$ and all possible left products. They must all be different (as above). Thus, one must be 1, so $ ab \equal{} 1$ for some $ b$. Thus, all elements have a multiplicative inverse.


Two different cuboids are placed together, face-to-face to form a large cuboid.The surface area of the large cuboid is 3/4 of the total surface area of the original two cuboids.
Prove that the lengths of the edges of the large cuboid may be labelled x, y and z where 2/z = 1/x + 1/y.

[hide="My Solution"]
Because the two cuboids can be placed face-to-face to form another cuboid, they must have a face in common which have dimensions $ a$ and $ b$. Let the differing dimensions be $ c$ and $ d$. We can form the following equation based on the problem statement:

$ \frac{3}{4}(2(2ab \plus{} ac \plus{} bc \plus{} ad \plus{} db)) \equal{} 2((d \plus{} c)(a) \plus{} (d \plus{} c)(b) \plus{} ab)\\
3(2ab \plus{} ac \plus{} bc \plus{} ad \plus{} db)) \equal{} 4((d \plus{} c)(a) \plus{} (d \plus{} c)(b) \plus{} ab)\\
6ab \plus{} 3ac \plus{} 3bc \plus{} 3ad \plus{} 3db \equal{} 4ad \plus{} 4ac \plus{} 4bd \plus{} 4bc \plus{} 4ab\\
2ab \equal{} ac \plus{} bc \plus{} ad \plus{} db$

Now we rearrange the equation given to the form we got above.

$ \frac{2}{z} \equal{} \frac{1}{x} \plus{} \frac{1}{y}\\
2xy \equal{} yz \plus{} xz$

Now suppose we let $ x \equal{} a$, $ y \equal{} b$, $ z \equal{} c \plus{} d$.

$ 2(a)(b) \equal{} (b)(c \plus{} d) \plus{} (a)(c \plus{} d)\\
2ab \equal{} ac \plus{} bc \plus{} ad \plus{} db$

This equation is exactly the same as the equation we obtained above, thus, the edges of the larger cuboid can be labeled $ x$, $ y$, and $ z$ such that $ \frac{2}{z} \equal{} \frac{1}{x} \plus{} \frac{1}{y}$. $ \Box$
[/hide]


inequalities
triangle inequality
algebra unsolved
algebra

Let $P$ be a polynom of degree $n \geq 5$ with integer coefficients given by $P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_0 \quad$ with $a_i \in \mathbb{Z}$, $a_n \neq 0$.

Suppose that $P$ has $n$ different integer roots (elements of $\mathbb{Z}$) : $0,\alpha_2,\ldots,\alpha_n$. Find all integers $k \in \mathbb{Z}$ such that $P(P(k))=0$.

Nice problem! Let me post my solution of this:

Let $ P(x) \equal{} a_nx(x \minus{} \alpha_2)\cdots (x \minus{} \alpha_n)$. We shall prove that the equation $ P(x) \equal{} \alpha_2$ has no integer solution, so WLOG, none of the equations $ P(x) \equal{} \alpha_k$ has integer solutions. Thus all the integer solutions of $ P(P(x)) \equal{} 0$ are exactly the roots of $ P(x)$.

Since all the $ \alpha_k$-s are distinct, and $ n > 4$ it follows that if $ |P(k)| \equal{} |a_nk(k \minus{} \alpha_2)\cdots (k \minus{} \alpha_n)| \geq 4 \equal{} 1 \cdot 1 \cdot 2 \cdot | \minus{} 2| \cdot | \minus{} 1|$ for all integers $ 0\neq k \neq \alpha_i$, $ i \equal{} \overline{2,n}$. Therefore if $ k$ is an integer solution of $ P(x) \equal{} \alpha_2$ then $ |\alpha_2|\geq 4$.

Now we have $ (|k|\cdot |k \minus{} \alpha_2|) \mid \alpha_2$, and as such $ |k| \neq |\alpha_2|$ and also $ |k \minus{} \alpha_2| \neq |\alpha_2|$ (for obvious reasons), therefore we have $ 2 |k| \leq |\alpha_2|$ and also $ 2 |k \minus{} \alpha_2 | \leq |\alpha_2|$. Adding these last two relations up side by side we obtain
\[ |\alpha_2| \leq |k| \plus{} |k \minus{} \alpha_2| \leq |\alpha_2|
\]
by the triangle inequality, so we must have $ |\alpha_2 | \equal{} 2|k|$ and also $ |\alpha_2| \equal{} 2|k \minus{} \alpha_2|$ but then $ |k|\cdot |k \minus{} \alpha_2| \equal{} \frac {|\alpha_2|^2}4 \geq |\alpha_2|$ which means that $ 2\leq |a_n (k \minus{} \alpha_3)\cdots (k \minus{} \alpha_n)| \leq 1$ which is a contradiction. The problem is thus finished :P

[quote="pbornsztein"]Try the others Valentin, maybe you will qualify for french team  :D

Pierre.[/quote]The ones I solved were already posted :) I was too late in posting this year :P :P

Bad luck..., maybe will you qualify next year.  :P

Pierre.

hi
i have the same probleme but generally
find the solutions of$P(P(x))=0$
if $deg(p(x))=n$and$p(x)$has n root

[quote="Valentin Vornicu"]
\[ |\alpha_2| \leq |k| + |k-\alpha_2| \leq |\alpha_2| \] by the triangle inequality, so we must have $|\alpha_2 |=2|k|$ and also $|\alpha_2|=2|k-\alpha_2|$ [/quote]

Why is that? I don't understand.

[quote="Valentin Vornicu"]which means that $2\leq |a_{n}(k-\alpha_{3})\cdots (k-\alpha_{n})| = 1$ [/quote].

No, It is $2\leq |a_{n}(k-\alpha_{3})\cdots (k-\alpha_{n})| \leq 1$ .
@indybar: If $A\leq B,C\leq D$ and $A+C=B+D$ then $A=B$ and $C=D$ :)

[hide="My solution"]

since $ 0$ is a root of $ P(x)$ $ \implies a_0 = 0$

let $ P(x_i) = \alpha_i$

$ \& {a_n}{x_i}^n + {a_{n - 1}}{x_i}^{n - 1} + {a_{n - 2}}{x_i}^{n - 2} + \ldots \ldots + {a_2}{x_i}^2 + {a_1}{x_i} = {\alpha _i} \\
 \\
\& \implies {x_i}|{\alpha _i} \implies {x_i}|{\alpha _i} - {x_i}$

$ \& P\left( {{\alpha _i}} \right) = 0 \\
\& P\left( {{x_i}} \right) = {\alpha _i} \\
\& \implies {\alpha _i} - {x_i}|P\left( {{\alpha _i}} \right) - P\left( {{x_i}} \right) \implies {\alpha _i} - {x_i}| - {\alpha _i} \implies {\alpha _i} - {x_i}|{x_i}$

$ \implies {\alpha _i} - {x_i} = \pm {x_i} \implies {\alpha _i} = 0{\text{ or }}{x_i} = \frac {{{\alpha _i}}} {2}$

the case of $ \alpha_i = 0$ is trivial. let us consider the case of $ {\alpha_i \over 2} = x_i$

let $ P'\left( x \right) = {a_n}{x^{n - 1}} + {a_{n - 1}}{x^{n - 2}} + {a_{n - 2}}{x^{n - 3}} + \ldots \ldots + {a_2}x + {a_1}$

$ \& P'\left( {{\alpha _i}} \right) = 0 \\
\& P'\left( {{x_i}} \right) = \frac {{{\alpha _i}}} {{{x_i}}} = 2 \\
 \\
\& P'\left( x \right) = {a_n}\left( {x - {\alpha _1}} \right)\left( {x - {\alpha _2}} \right)\left( {x - {\alpha _3}} \right) \ldots \ldots \left( {x - {\alpha _n}} \right),{\text{ }}n \ge 4 \\
 \\
\& \implies \left| {P'\left( x \right)} \right| = \left| {{a_n}\left( {x - {\alpha _1}} \right)\left( {x - {\alpha _2}} \right)\left( {x - {\alpha _3}} \right) \ldots \ldots \left( {x - {\alpha _n}} \right)} \right| \\
\ge \left| {{a_n}} \right|\left| { - 1} \right|\left| 1 \right|\left| 2 \right|\left| { - 2} \right| \ge 4{\text{ }}\left[ {If{\text{ }}x \in {\Bbb Z}} \right] \\
 \\
\& \implies 2 \ge 4$

Contradiction

Hence for no $ x \in \Bbb Z$ we have $ P(x) = \alpha_i$
hence the only values of $ k$ are the roots of $ P(x) = 0$
[/hide]

My solution:
$P(x)=0$ only at those points where $x$  $\epsilon$  $\{0,\alpha_2,\alpha_3...\alpha_n\}$
Hence $P(P(k))=0$ iff   $P(P(k))$ $\epsilon$ $\{0,\alpha_2,\alpha_3...\alpha_n\}$
$P(k)=0$   implies that      $k$  $\epsilon$  $\{0,\alpha_2,\alpha_3...\alpha_n\}$. $P(k)=\alpha_i$  for some $i$  $\epsilon$  $\{0,\alpha_2,\alpha_3...\alpha_n\}$   implies that  $a_{n}k(k-\alpha_{2})\cdots (k-\alpha_i) \cdots (k-\alpha_{n})=\alpha_i$  with  $a_n,\alpha_i,k \ne 0$. Hence we have $k|\alpha_i$ implies that $k|k-\alpha_i$.  Whereas we have $k-\alpha_i|\alpha_i$ implies that  $k-\alpha_i|k$. Hence $|k-\alpha_i|=|k|$ meaning that $2k=\alpha_i$ as $\alpha_i$ is non-zero . Plugging it back into the equation , we have $a_{n}(k-\alpha_{2})\cdots (k-\alpha_i) \cdots (k-\alpha_{n})=2$ . As all the $\alpha_i$'s are distinct this means that all the $k-\alpha_i$'s are distinct , but they are at least four in number and must belong to the set $\{-2,-1,1,2\}$ with both $2$ and $-2$ not occuring together , which is impossible. Hence as a conclusion , $k$ can take only the values of roots of $P(x)$


A spherical volume of radius $a$ is filled with charge of uniform density $\rho$.  We want to know the potential energy $U$ of the sphere of charge, that is, the work done in assembling it.  Calculate it by building the sphere up layer by layer, making use of the fact that the field outside a spherical distribution of charge is the same as if all the charge were at the center.  Express the result in terms of the total charge $Q$ in the sphere.

Let's see... The work needed to bring a charge $\text dq$ from infinity to a distance $r$ from a spherical charge $q$ is $\text dW = kq\text dq/r$, where $k = 1/4\pi\varepsilon_0$. The charge $q$ and the distance $r$ are connected by $q = \rho V = \rho \frac 4 3 r^3\pi$ (and $\text dq = \rho \text dV = 4 \rho r^2 \pi \text d r$), so the work needed for building up a spherical volume of radius $a$ will be \[ W = \int \text dW = \int_0^a \frac{16}{3} k \rho^2 \pi^2 r^4 \text dr = \frac{16}{15} k \rho^2 \pi^2 a^5 = \frac{3kQ^2}{5a}. \] This is the energy of the total charge.