inequalities
function
vector
graph theory
combinatorics unsolved
combinatorics

$ 1650$ students are arranged in $ 22$ rows and $ 75$ columns.  It is known that in any two columns, the number of pairs of students in the same row and of the same sex is not greater than $ 11$.  Prove that the number of boys is not greater than $ 928$.

I like this kind of problems. :)

Let $ A_1,...,A_{75}$ be subsets of $ B\equal{}\{1,2,...,22\}$ definded like it folows:

Let $ j\in B$,  then $ j\in A_i \Longleftrightarrow$ student at $ (i,j)$ is boy.

We are interested at $ \sum_{i\equal{}1}^{75}|A_i| \equal{}: a$.

Define bipartite graph like this:
 For $ b\in B$ and pair $ \{A_i,A_j\}$:
  \[ b\sim \{A_i,A_j\} \Longleftrightarrow x\in A_i\cap A_j \vee x\in A'_i\cap A'_j\]

Let $ b\in B$ be in $ x$ set among $ A_1,A_2...$. Then degree of $ b\in B$ is
 \[ d(b) \equal{}  {x\choose 2} \plus{}{75\minus{}x \choose 2} \equal{} x^2\minus{}75x \plus{}75\cdot 37\equal{}: f(x)\]

By asumption each pair $ \{A_i,A_j\}$ is connected with at most 11 elements from $ B$,
so $ a_{i,j}\leq 11$, where $ a_{i,j}$ is degree of pair $ \{A_i,A_j\}$.

Now we have (double counting):
\[ \sum_{b\in B}d(b) \equal{} \sum_{i<j}a_{i,j}\leq {75\choose 2}\cdot 11\]

On the other hand left side is by Jensen inequality (function $ f$ is convex):

\[ 22f({a\over 22})\leq \sum_{b\in B}d(b)\]

From here we can easly get $ a\leq 520$.

Probably the mistake in the above proof is just in the last computation - clearly $a$, the number of boys, cannot be at most $520$, as the problem being symmetrical in boys and girls, it will mean $g$, the number of girls, is also at most $520$, but the total number of children is $22\cdot 75 = 1650$.

Alternative Solution. Consider a $2m \times n$ array, where on any pair of columns the number of gender differences of children on a same row is at least $m$. Denote by $b_i$ the number of boys on row $i$, by $g_i = n - b_i$ the number of girls, by $B = \sum_{i=1}^{2m} b_i$ the total number of boys, by $N$ the total number of such gender differences in the array.

Then $N = \sum_{i=1}^{2m} b_i(n-b_i) \leq 2m\left(\dfrac {\sum_{i=1}^{2m} b_i} {2m}\right)\left(n - \dfrac {\sum_{i=1}^{2m} b_i} {2m}\right) = \dfrac {B(2mn - B)} {2m}$ by Jensen's inequality, since the function $\varphi(x) = x(n-x)$ is concave.
On the other hand (by double counting over all pairs of columns) $N \geq \binom {n} {2} m = \dfrac {mn(n-1)} {2}$.
It follows $B^2 - 2mnB + m^2n(n-1) \leq 0$, so $m(n-\sqrt{n}) \leq B \leq m(n+\sqrt{n})$. Under the conditions of the problem, with $m=11$ and $n=75$, it follows that $730 \leq B \leq 920$.

[b]The punch line however comes now![/b] Replace the boys by $+1$ and the girls by $-1$; the statement of the problem now reads that the column vectors, seen as real vectors in $\mathbb{R}^{2m}$, have pairwise non-positive dot product. It can be proven though that the maximum number of such vectors in $\mathbb{R}^{d}$ is $2d$, with equality only if $d$ of them are pairwise orthogonal, while the other $d$ are negative multiples of the first $d$ (thus also pairwise orthogonal, with the pairs of collinear vectors having as dot product the product of their norms with a minus sign), which of course can be achieved (take the $\pm$ versors of the axes).
However, our vectors need have entries equal to $\pm 1$; or a complete set of such pairwise orthogonal vectors can exist only if $d$ is a multiple of $4$ - check Hadamard matrices topic (the conjecture is that they always exist if $4 \mid d$).
What this all means in our case is that the [b]object of the problem is empty[/b]; $n=75$ is too large as compared to $2d=4m=44$, while even that cannot be achieved, since $d=2m=22$ is not a multiple of $4$, so not even $44$ columns may be possible in such an array!

Just fall back on the general proof for $m$ and $n$, keeping in mind the limitations described in the above.

Nice solution with double counting (maybe similar to above not sure)Number of boys is $X$.We double count number of pairs boy,boy or girl,girl that are in same row(this number is $T$ ).Number rows from $1$ to $22$ (starting from top) and let $d_i$ be number of boys in row $i$.Now $T=\sum_{i=1}^{22}\binom{d_i}{2}\binom{75-d_i}{2}$ and from quadratic arithmetic inequality we get $T\ge\frac{1}{2}(\frac{X^2}{22}-X+\frac{(1650-X)^2}{22}-(1650-X))$ Now we see that this identity grows as $X$ grows and it starts when $X\ge824.5$ so $\frac{1}{2}(\frac{X^2}{22}+\frac{(1650-X)^2}{22}-1650)\ge\frac{1}{2}(\frac{929^2}{22}+\frac{(1650-929)^2}{22}-1650)=30604,136\cdots$ .On the other hand in any pair of columns there is less than or equal to $11$ pairs so $T\le11\binom{75}{2}$ from which follows $T\le30525$ .Collecting $30525\ge T \ge30604,136\cdots$ contradiction and we are done .


Prove that

$ {n\choose 0}^{2}\plus{}{n\choose 1}^{2}\plus{}{n\choose 2}^{2}\plus{}\cdots{n\choose n}^{2}\equal{}{2n\choose n}$

Now for some reason I'm given an outline for a proof, which states:

a) Verify that $ (1\plus{}x)^{n}\left( 1\plus{}\frac{1}{x}\right)^{n}\equal{}\frac{(1\plus{}x)^{2n}}{x^{n}}$

b) Show that the term containing no x in the expression on the right of a) is $ {2n\choose n}$

c) Expand $ (1\plus{}x)^{n}$ using the binomial theorem, then show that the coefficient of the term containing no $ x$ on the left of a) is $ {n\choose 0}^{2}\plus{}{n\choose 1}^{2}\plus{}{n\choose 2}^{2}\plus{}\cdots\plus{}{n\choose n}^{2}$

I don't understand what  "the term containing no x in the expression on the right of a)" is referring to. Can anyone help/give a hint or something?

For example, when you expand $ (x\plus{}1/x)^{n}$, there will be some terms where the $ x$'s cancel out.  So, basically they're asking for the terms in the expansion in which the $ x$'s cancel out.  Expand it and it should become apparent, I think.

[hide="a"]
$ (1\plus{}x)^{n}(1\plus{}1/x)^{n}\equal{} ((1\plus{}x)(1\plus{}1/x))^{n}$
$ \equal{} (x\plus{}2\plus{}1/x)^{n}$
$ \equal{}\left(\frac{x^{2}\plus{}2x\plus{}1}{x}\right)^{n}$
$ \equal{}\frac{(x\plus{}1)^{2n}}{x^{n}}$[/hide]

[hide="b"]
Expand the top part of the fraction.
$ \binom{2n}{0}x^{2n}\plus{}\binom{2n}{1}x^{2n\minus{}1}\plus{}\dots\plus{}\binom{2n}{n}x^{n}\plus{}\dots$
As you can see, the term with exactly $ x^{n}$ will cancel out with the $ x^{n}$ in the denominator.[/hide]

Ok, but how does that help in reaching the final result? You're only taking one part the whole expansion  :maybe:

Edit: I think I figured it out. All you need to do is expand everything you can with the binomial theorem, then simplify

There is actually a very good proof using sum and difference sequences, but that would require a lot of lemmas and a bit of prerequisite understanding...

You're one is a bit simpler :-)

How about this?

Consider trying to pick $ n$ objects from $ 2n$ possible objects.
 which can be done in $ \binom{2n}{n}$ ways of course. Now we'll count the same thing a little differently.

Split the $ 2n$ objects into two groups with $ n$ objects each. Now how many ways can we pick $ n$ objects from these?

well we can pick 0 object from the first group, and $ n$ objects from the second group.
or we could pick 1 objects from the first group and $ n\minus{}1$ objects from the second group.
$ \vdots$
or we could pick $ n$ objects from the first group and 0 object from the last group.

so for the first option there we can do it in $ \binom{n}{0}\binom{n}{n}$ choices.
for the second can do it in $ \binom{n}{1}\binom{n}{n\minus{}1}$
$ \vdots$
and finally the last way to do it is $ \binom{n}{n}\binom{n}{0}$

Remembering that $ \binom{n}{r}\equal{}\binom{n}{n\minus{}r}$ we get (when we add them all up)

$ \binom{n}{0}^{2}\plus{}\binom{n}{1}^{2}\plus{}\cdots\plus{}\binom{n}{n}^{2}\equal{}\binom{2n}{n}$

Sorry to be a hassle, but Archimedes c) makes beautiful sense but for b), how are we getting rid of all the terms that are not $ {2n\choose n}$? Or is it somehow sufficient to show that when no x terms are present, the two formulas match? This is uber confusing.

Nobody's a hassle.

Okay, basically, we're trying to prove that $ a \equal{} b$.
To do that, we prove that $ x \equal{} y$, and then prove that $ a$ is the same part of $ x$ that $ b$ is of $ y$, or in this case, $ a$ is the constant term (the term where the variables cancel out) of $ x$ and $ b$ is the constant term of $ y$.
...if that doesn't make it more confusing...

[quote="archimedes1"][hide="c"]
Expand $ (1\plus{}x)^{n}$:
$ \binom{n}{0}\plus{}\binom{n}{1}x\plus{}\binom{n}{2}x^{2}\plus{}\dots\plus{}\binom{n}{n}x^{n}$
Now expand $ (x\plus{}1/x)^{n}$:
$ \binom{n}{0}\plus{}\binom{n}{1}x^{\minus{}1}\plus{}\binom{n}{2}x^{\minus{}2}\plus{}\dots\plus{}\binom{n}{n}x^{\minus{}n}$
Clearly, when we multiply these two expressions to get the value on the LHS of a), the term without an $ x$ in them will be
$ \binom{n}{0}^{2}\plus{}\binom{n}{1}^{2}\plus{}\dots\plus{}\binom{n}{n}^{2}$.[/hide]
[quote="Anaxerzia"]There is actually a very good proof using sum and difference sequences...[/quote]
I've heard of another proof using paths on Pascal's Triangle.  I think it's pretty cool how there are so many different ways of proving the same thing.[/quote]
They are actually the same proof xD.

The pathwalking proof, the proof by generating functions, and the proof by counting are all essentially the same proof (that is, they all represent the same kind of counting), and each approach generalizes readily in exactly the same way to [b]Vandermonde's identity[/b].  In fact, most of the difficulty in this exercise reduces to recognizing that it is a special case of Vandermonde's identity.


A square peg just fits inside a round hole. What fraction of the hoel is occupied by the peg?

Whoops Typo there you know i meant peg not hole! I'll fix it!


linear algebra
matrix
algebra
polynomial

Hi everyone,
Why the determinant of a matrix is equal to the product of its eigenvalues and why the trace is equal to the sum of its eigenvalues?

Two arguments:

The first one involves merely looking at the characteristic polynomial. Assume $A$ is an $n\times n$ matrix.

$P_{A}(\lambda)=\det(A-\lambda I)= (-1)^{n}\lambda^{n}+(-1)^{n-1}(\text{trace}A)\lambda^{n-1}+\cdots+\det A.$

Assuming that this polynomial splits and can be factored as $(-1)^{n}(\lambda-\lambda_{1})\cdots(\lambda-\lambda_{n})$ then we can apply Viete's relations between the roots and coefficients of a polynomial. That is, $\text{trace}A=\sum_{i}\lambda_{i}$ and $\det A=\prod_{i}\lambda_{i}.$

The second proof involves a version of Schur's lemma. The matrix $A$ can be shown to be similar to an upper triangular matric $T.$ From the fact that $\det (AB)=(\det A)(\det B),$ we can show that $\det T=\det(P^{-1}AP)=\det A$ and from the fact that $\text{trace}(AB)=\text{trace}(BA),$ we can show that $\text{trace}T=\text{trace}(P^{-1}AP)=\text{trace}A.$

But the trace and determinant of the upper triangular matrix $T$ are determined solely by the elements on the diagonal. Those elements are the eigenvalues of $T$ and can in turn be shown to be the eigenvalues of $A.$

Both proofs require only that the characteristic polynomial splits - in other words that there exist (counting algebraic multiplicities) $n$ eigenvalues.


geometry
3D geometry
Pythagorean Theorem

The slant height of a cone is 13 cm, and the height from the vertex to the center of the base is 12 cm. What is the number of cubic centimeters in the volume of the cone? Express your answer in terms of $ \pi$.

Since the slant height, the vertical height, and the radius of the base form a triangle, we can use the Pythagorean Theorem to find that the radius of the base is $ 5$. The area of the base is $ 5^2\pi\equal{}25\pi$. The area of the cone is $ \frac13 25\pi \cdot12\equal{}100\pi$

$ Answer\equal{}100\pi$


There is a triangle AOB with angle O = 30 degrees. The length of the side AB is 1 unit. What is the maximum length of side AO?

$ AO \equal{} \frac{1}{\sin 30} \sin x \equal{} 2 \sin x \le 2$ when $ \angle OBA \equal{} 90^\circ$.


geometry
3D geometry
sphere
topology
set theory
geometry unsolved

Can you proove/disproove that the set R^3 can be represented as an union of disjoint circles, all of them having radius 1 ? I repeat, circles, not spheres or balls!

it's an interesting question.  i think it's for algebraic topology to answer...

actually, some set theory, but yet it's quite interesting indeed :)

so you know the answer and you're saying the proof is elementary??

yes, i know the answer and find the problem quite challenging.
although i don't know what it means an 'elementary' proof to you,
i intend no to post the solution for now in order for people to enjoy the problem. What i can say is that it uses set theory and nothing more - do you consider these tools elementary? If so, then is IS elementary.


geometry
geometric transformation
reflection
number theory
divisibility tests

What is the units digit of the product:

$ [\left(14^5 \mod 5\right)(2738756 \mod 63)] \mod 9$?

Option 1:  Long division.  This is straightforward.

Option 2:  Chinese Remainder Theorem.  $ 2738756 \equiv 2 \plus{} 7 \plus{} 3 \plus{} 8 \plus{} 7 \plus{} 5 \plus{} 6 \equiv 2 \bmod 9$ and $ 2738756 \equiv 2 \minus{} 2 \cdot 7 \minus{} 3 \cdot 3 \minus{} 8 \plus{} 2 \cdot 7 \plus{} 3 \cdot 5 \plus{} 6 \equiv 6 \bmod 7$, hence by inspection $ 2738756 \equiv 20 \bmod 63$.

There is a test some people will teach you for divisibility by $ 7$, but it doesn't tell you the remainder $ \bmod 7$.  The test I used is that

$ d_0 \plus{} 10 d_1 \plus{} 10^2 d_2 \plus{} 10^3 d_3 \plus{} 10^4 d_4 \plus{} 10^5 d_5 \plus{} 10^6 d_6 \plus{} ... \equiv d_0 \plus{} 3d_1 \plus{} 2d_2 \minus{} d_3 \minus{} 3d_4 \minus{} 2d_5 \plus{} d_6 \plus{} 3d_7 \plus{} ...$

The coefficients $ 1, 3, 2, \minus{}1, \minus{}3, \minus{}2$ repeat with period $ 6$ and reflect the fact that

$ 10^0 \equiv 1 \bmod 7$
$ 10^1 \equiv 3 \bmod 7$
$ 10^2 \equiv 2 \bmod 7$
$ 10^3 \equiv 6 \equiv \minus{}1 \bmod 7$
$ 10^4 \equiv 4 \equiv \minus{}3 \bmod 7$
$ 10^5 \equiv 5 \equiv \minus{}2 \bmod 7$
$ 10^6 \equiv 1 \bmod 7$

We can construct similar divisibility tests for larger primes like $ 13$, but they are unwieldy (even the one for $ 7$ is troublesome).  This is a generalization of the divisibility tests for $ 9$ and $ 11$, which are simple in base $ 10$.

[quote="1=2"][quote="ProtestanT"][quote="i_like_pie"][hide]$ [\left(14^5 \mod 5\right)(2738756 \mod 63)] \mod 9$

$ [\left(( \minus{} 1)^5 \mod 5\right)(2738756 \mod 63)] \mod 9$

$ 4\cdot20\equiv\boxed{8}\mod 9$[/hide][/quote]
Is there a better way of getting the $ 20$ from $ 2738756 \mod 63$ part without using a calculator?[/quote]

mod 9 instead of mod 63, since we are taking the whole expression mod 9, and 63 is divisible by 9.[/quote]
How does this work?  Could you elaborate?

[quote="1=2"][quote="xpmath"]t0rajir0u already said it.[/quote]

where?[/quote]

Please read my post more carefully.

[quote="t0rajir0u"]Option 2:  [b]Chinese Remainder Theorem.[/b]  $ 2738756 \equiv 2 \plus{} 7 \plus{} 3 \plus{} 8 \plus{} 7 \plus{} 5 \plus{} 6 \equiv 2 \bmod 9$ and $ 2738756 \equiv 2 \minus{} 2 \cdot 7 \minus{} 3 \cdot 3 \minus{} 8 \plus{} 2 \cdot 7 \plus{} 3 \cdot 5 \plus{} 6 \equiv 6 \bmod 7$, hence by inspection $ 2738756 \equiv 20 \bmod 63$.[/quote]

archimedes1:  The remainder of a number $ \bmod 63$ is uniquely determined by its remainder $ \bmod 7$ and its remainder $ \bmod 9$.  This is because $ 7$ and $ 9$ are relatively prime.  Google "Chinese Remainder Theorem" for more information.


t(5) means t subscipt 5

suppose that it is claimed that t(5)=47 is the fifth term of a sequence which obeys the recurrence relation:

t(n+1) = 3t(n)-2t(n-1)

by finding possible values for t(0) and t(1) verify that the claim could be correct...

Thought about this for a while, then felt pretty stupid when I saw the obvious answer: $T_0=47$, $T_1=47$, and all the subsequent $T_k$ are equal to 47.

It turns out that this is not the only possible sequence.

There is a way to, given a recursion like the one you gave, determine a closed form.  First, we suppose that $T_k=n\times x^k$ and plug that in to the given recursion; we get $nx^{k+1}=3nx^{k}-2nx^{k-1}$.  We can simplify the above by dividing both sides of the equation by $nx^{k-1}$.  We then solve: $x^2=3x+2=0$, $(x-2)(x-1)=0$, and we have two solutions for $x$; 2 and 1.  This tells us that the recursion is of the form $T_k=a\times2^k+b\times1^k$.  We can actually choose $a$ and $b$ arbitrarily, as long as they satisfy $a\times2^5+b=47$.  Once we choose such values for $a$ and $b$, we can just set $T_0=a+b$ and $T_1=2a+b$, and we have a pair of terms that can begin such a sequence.  To get the relatively trivial sequence described above, for instance, we would set $a=0$ and $b=47$, but we can also choose $a=1$ and $b=15$ to get a non-trivial sequence that satisfies $T_5=47$ (it begins with the terms $T_0=16$ and $T_1=17$).

random_mathematician, you have already posted this, and the other two problems you recently posted, in the intermediate forum. Please do not post the same thing in multiple forums..


Let $ a,b,c$ be positive real numbers and let $ p$ be a real number such that $ 0 < p < 1$. Prove that $ \frac{a}{\left(b\plus{}c\right)^p}\plus{}\frac{b}{\left(c\plus{}a\right)^p}\plus{}\frac{c}{\left(a\plus{}b\right)^p}\geq \frac{1}{2^{p}}\left(a^{1\minus{}p}\plus{}b^{1\minus{}p}\plus{}c^{1\minus{}p}\right)$

$ \frac{a}{(b+c)^p}+\frac{b}{(c+a)^p}+\frac{c}{(a+b)^p}\geq \frac{1}{2^{p}}(a^{1-p}+b^{1-p}+c^{1-p})$ can be re-written to;

$ \frac{a}{(b+c)^p}+\frac{b}{(c+a)^p}+\frac{c}{(a+b)^p}\geq \frac{a}{(2a)^{p}} + \frac{b}{(2b)^{p}}+ \frac{c}{(2c)^{p}}$ $ \Leftrightarrow$

$ \sum_{cyc} \frac{a}{(b+c)^{p}} - \frac{a}{(2a)^p}\geq0$  $ \Leftrightarrow$

$ \sum_{cyc} \frac{ a[(2a)^{p} - (b+c)^{p}]}{[2a(b+c)]^{p}}\geq0$  but we can assume $ a\geq b \geq c$ $ \Rightarrow$

$ 2a\geq b + c$ $ \Rightarrow$ $ (2a)^{p}\geq (b+c)^{p}$

[quote="Papalakis"]
$ \sum_{cyc} \frac { a[(2a)^{p} - (b + c)^{p}]}{[2a(b + c)]^{p}}\geq0$  but we can assume $ a\geq b \geq c$ $ \Rightarrow$

$ 2a\geq b + c$ $ \Rightarrow$ $ (2a)^{p}\geq (b + c)^{p}$[/quote]

But the second term of the sum is $ \frac { b[(2b)^{p} - (c + a)^{p}]}{[2b(c + a)]^{p}}$ .
Do you know if it is positive or negative ?


Hey can anyone give me a crash course on graph theory? I need to use it for the Mandelbrot team competition.

Discrete mathmatics has always been one of my weak points...

Actually, unless SamV has changed the contest a lot since I was working with him on it, the subjects are given out ahead of time so students will seek out a crash course in the subject.  However, if the round has already started, it would be inappropriate for someone to answer this question if they've seen the test (obviously).

I just need to review the general concepts of graph theory. I'll check out Paul Zeitz's book.

Yes, the Mandelbrot Team Topics are given out 2 days prior to the examination.


well, it would be 85 total, but since we stop at the 30 degree mark, we must do 85-30=55 degrees

A basic postulate, called the Angle Addition Postulate.  (Hmmm, I wonder why its named that...)


algebra
function
domain
algebra unsolved

Solve the equation $\sqrt{5-\sqrt{5-x}}=x$ for $x>0$

[quote="pontios"]Let $f(x)=\sqrt{5-x}$

Domain of $f: \ (0,5]$

The given equation becomes $\sqrt{5-f(x)}=x$, or simply $f(f(x))=x$

One obvious solution is the root of the equation $f(x)=x$

$\sqrt{5-x}= x \Rightarrow$

$x^{2}+x-5 = 0 \Rightarrow$

$x = \frac{\sqrt{21}-1}{2}$[/quote]
its missing one solution man... bye!

I think pontios just wanted to point out 1 solution.

After I analysed, there is not any other solution.


Evaluate limit
\[ \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{1}{\binom{n}{k}}  \]

Ok i'm asking for confirmation now. its
[hide]  $\frac{8}{3}$ or $2$  [/hide]

Denote An the sequence.

For n big enough:
Cn,3 > n^2
Cn,4 > n^2
...
Cn,n-3 > n^2

Then :
Cn,1 = Cn,n-1 = 1/n
Cn,0 = Cn,n = 1

So 2 < An < 2 + 2/n + (n-3)/n^2 -> 2, An -> 2.

mareleG, for completeness you need to account for ${n\choose 2}$ and ${n\choose n-2}$ in your proof. But you're basically correct.

Thanks.
Something that made me unsure with this is the relation \[ \sum_{k=0}^{n-1} \frac{1}{\binom{n-1}{k}} = \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k}  \]

$\lim_{n \rightarrow \infty} \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k}$  Looks like $\frac{8}{3}$.

But My friend has a proof that the limit is 2. :D


When I try to use LaTeX it keeps taking me out of math mode.  I think it might be because my / are backwards.  (Aren't they supposed to be facing the other way?)

For example when I type $ /sqrt{4}$  that is what I get....

:| Have you tried the other way?  Or, say, looking at an introductory resource like the [[LaTeX]] section of the wiki?

What other way?  I've read the introduction of LaTeX and I think I get it.  However, as soon as I type in something after the "dollar sign" (in TeXnic Center) it turns green.  The same thing happens when I try to put it directly on here.

Yes, they are backwards. You're supposed to use the backslash, above the Enter (or Return) key: \
What you've been using is the slash, which is incorrect: /

Green text in Texnic Center indicates math mode, the mode that the dollar signs initiate and end. What's that "same thing" you're talking about, and why is it problematic?

You wrote, "Aren't they supposed to be facing the other way?"  I asked, "Have you tried the other way?"  You asked, "What other way?"   :|   (The answer to your question is, "Yes, they're suppose to be facing the other way."  I do admit that I'm curious why you didn't simply [i]try[/i] using a \ slash and see how it worked.)

The green is TeXnic Center's way of telling you that you're in math mode.  By "the same thing" I assume you mean the failure to produce well-formatted mathematics, not turning green  :wink:  :roll:

Edit: Bictor posted while I was choosing my phrasing somewhat delicately.  Bictor, are you sure you haven't mixed up the names back slash and forward slash?

$ \backslash$ is \backslash, so that's what I'm going by. :)

I feel really stupid now!
I never knew there was another black slash facing the other way  :oops:


$\sqrt[n]{\prod_{k}\left(1+\frac{1}{x_{k}}\right)}\geq 1+\frac{1}{\sqrt[n]{\prod_{k}x_{k}}}$

By AM-GM we have

$\frac{\sum_{k}x_{k}}{n}=\frac{1}{n}\geq\sqrt[n]{\prod_{k}x_{k}}\implies \frac{1}{\sqrt[n]{\prod_{k}x_{k}}}\geq n$

Therefore $\sqrt[n]{\prod_{k}\left(1+\frac{1}{x_{k}}\right)}\geq 1+n$, which is the result.

[quote="amcavoy"]$\sqrt[n]{\prod_{k}\left(1+\frac{1}{x_{k}}\right)}\geq 1+\frac{1}{\sqrt[n]{\prod_{k}x_{k}}}$

[/quote]

How did you get that?

[hide]
$\prod_{k=1}^n ( 1+\frac{1}{x_k} )=\prod_{k=1}^n ( 1+[ \underbrace{\frac{1}{nx_k}+\frac{1}{nx_k}+\cdots +\frac{1}{nx_k}}_n] )$. By AM-GM, each term being multiplied is greater than or equal to $\frac{n+1}{(nx_k)^\frac{n}{n+1}}$, so the product is greater than or equal to $\frac{(n+1)^n}{n^\frac{n^2}{n+1}( \prod_{k=1}^n x_k} )^\frac{n}{n+1}$.

Using AM-GM to put a bound on the product, $\prod_{k=1}^n x_k\leq \frac{1}{n^n}\Rightarrow \frac{1}{\prod_{k=1}^n x_k}\geq n^n$. Using this in the above inequality, the product is greater than or equal to $(n+1)^n$.
[/hide]

May I ask which book this came from?  Just the other day I found one of these in a book, but it was just a specific case where $n=3$.

[quote="amcavoy"]I don't know the name of the inequality, but in two variables it is

\[ \sqrt{\left(a_{1}+a_{2}\right)\left(b_{1}+b_{2}\right)}\geq\sqrt{a_{1}b_{1}}+\sqrt{a_{2}b_{2}}.  \]
I expanded that to $n$ instead of $2$ and my first equation followed.  (I used $a_{1}=b_{1}=\cdots =1$ and $a_{2}=\frac{1}{x_{1}}$, $b_{2}=\frac{1}{x_{2}}$, etc.)

While we're at it, does anyone know what that inequality is called (the exponents on each term can be different; in this case they were all the same so it looks like Cauchy)?[/quote]

This is a variation of an AM-GM.

thanks for solutions!

the polish title is "Kółko matematyczne dla olimpijczyków" (~"Olimpic math classes")  author Henryk Pawłowski

hi,

here is my solution :

$S = (x_1+1)(x_2+1)\dots(x_n+1) \times\frac{1}{x_1x_2\dots x_n}$

$S = (x_1+x_1+x_2+\dots+x_n)\dots(x_n+x_1+\dots+x_n)\dots$

$S \geq \bigg((n+1)\sqrt[n+1]{x_1^2x_2\dots x_n}\bigg)\dots\bigg((n+1)\sqrt[n+1]{x_1\dots x_n^2}\bigg)\times\frac{1}{x_1x_2\dots x_n}$

$S \geq (n+1)^n$
 :D .

[quote="durt"]By AM-GM:
$\frac{\left(1+\frac{1}{x_1}\right)+\left(1+\frac{1}{x_2}\right)+...+\left(1+\frac{1}{x_n}\right)}{n} \geq \sqrt[n]{\left(1+\frac{1}{x_1}\right) \left(1+\frac{1}{x_2}\right) ... \left(1+\frac{1}{x_n}\right)}$
So
$\frac{n + \left(\frac{1}{x_1} + \frac{1}{x_2} + ... +\frac{1}{x_n}\right)}{n} \geq \sqrt[n]{\left(1+\frac{1}{x_1}\right) \left(1+\frac{1}{x_2}\right) ... \left(1+\frac{1}{x_n}\right)}$
And
$1+\frac{1}{n}\left(\frac{1}{x_1} + \frac{1}{x_2} + ... +\frac{1}{x_n}\right) \geq \sqrt[n]{\left(1+\frac{1}{x_1}\right) \left(1+\frac{1}{x_2}\right) ... \left(1+\frac{1}{x_n}\right)}$
Here I had to make a little conjecture. If someone knows how to prove this I'd be glad to know.  ;)  So I guessed this:
If $x_1 + x_2 + ... + x_n = a$ then $\frac{1}{x_1} + \frac{1}{x_2} +... +\frac{1}{x_n} \geq \frac{n^2}{a}$
So
$1+\frac{1}{n} \frac{n^2}{a} = 1+n$
Therefore
$\left(1+\frac{1}{x_1}\right)\left(1+\frac{1}{x_2}\right)...\left(1+\frac{1}{x_n}\right)\geq\left(n+1\right)^n$[/quote]

That conjecture can be easily proven from Cauchy (provided the $x_k$ are positive), because
\[ \left( x_1+x_2+\cdots +x_n \right) \left( \frac{1}{x_1}+\frac{1}{x_2}+\cdots +\frac{1}{x_n} \right) \geq (1+1+\cdots +1)^2 \]
But it doesn't prove the given inequality because it goes like
\[ \left( 1+\frac{1}{x_1} \right) \left( 1+\frac{1}{x_2} \right) \cdots \left( 1+\frac{1}{x_n} \right) \leq \text{something} \geq (n+1)^n \]
which really says nothing about how the left side compares to $(n+1)^n$.


LaTeX
modular arithmetic
counting
distinguishability
symmetry

[quote="cyprusad"]Find the number of ways to select [i][b]n[/b][/i] objects from [i][b]3n [/b][/i]objects of which [i][b]n[/b][/i] are identical and rest are different.[/quote]
[hide="I believe it is"]$\frac{\binom{3n}{n}}{n!}= \frac{3n!}{(n!)^{2}(2n)!}$

We divide by $n!$ because $n$ objects are identical.[/hide]

No it's not that.

I'm not too good with LaTeX, but the answer is supposed to be:

2^(2n-1) + (1/2)2n[b]C[/b]n

But I'm not quite getting it.

Suppose you want to choose $n-k$ of the $n$ objects that are indistinguishable. $k$ can range from $0$ to $n$. This leaves $k$ of the distinguishable objects, and there are $\binom{2n}{k}$ ways to choose these.

We then need to calculate $\sum_{k=0}^{n}\binom{2n}{k}=\frac{1}{2}\left( \sum_{k=0}^{2n}\binom{2n}{k}+\binom{2n}{n}\right)=\frac{1}{2}2^{2n}+\frac{1}{2}\binom{2n}{n}=2^{2n-1}+\frac{1}{2}\binom{2n}{n}$

[quote="vishalarul"][quote="The Zuton Force"]$\sum_{k=0}^{n}\binom{2n}{k}=\frac{1}{2}\left( \sum_{k=0}^{2n}\binom{2n}{k}+\binom{2n}{n}\right)$ [/quote]
How did you get that?[/quote]

Symmetry of $\dbinom{2n}{k}$ about $k=n$.


What is Jones Reagent? Give the mechanism of its reaction with an alcohol???

jone's reagent is $ CrO_{3}\minus{}H_{2}SO_{4}$
it oxidises prmary alcohols to acids

That's right, but the reagent also includes acetone as a cosolvent. It's also used in the oxidation of secondary alcohols to ketones, and in the oxidation of alcohols containing double or triple bonds (which remain untouched).


topology
advanced fields
advanced fields unsolved

Let $f: X\to Y$ continuous function.

Show that the mapping $g: X\to X \times Y$ defined as $g(x)=(x,f(x))$ is an imbedding (embedding?) of $X$ to $X \times Y$ .

Thanks!

Here an embedding means a map that is a homeomorphism onto its image. We must prove that $g$ is continuous on $X$ and $g^{-1}$ is continuous on $g(X)$. The first part is straightforward (use the definition of the product topology). For the second part, observe that $g^{-1}$ is a restriction of the projection $X\times Y\to X$.


Let $ a, b, c$ be positive reals with sum $ 1$. Set \[ x \equal{} a\minus{}a^2, \quad y \equal{} b\minus{}b^2, \quad z \equal{} c\minus{}c^2.\] Prove that \[ 6\plus{}44(xy\plus{}yz\plus{}zx)\geq 17(x\plus{}y\plus{}z)\plus{}108xyz.\]

[quote="djuro"]Let $ a, b, c$ be positive reals with sum $ 1$. Set
\[ x \equal{} a \minus{} a^2, \quad y \equal{} b \minus{} b^2, \quad z \equal{} c \minus{} c^2.
\]
Prove that
\[ 6 \plus{} 44(xy \plus{} yz \plus{} zx)\geq 17(x \plus{} y \plus{} z) \plus{} 108xyz.
\]
[/quote]

Denote $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$ then we have

\[ x \plus{} y \plus{} z \equal{} 2q
\]
\[ xy \plus{} yz \plus{} zx \equal{} q^2  \plus{} r
\]
\[ xyz \equal{} r\left( {q \minus{} r} \right)
\]

Then the inequality becomes
\[ 6 \plus{} 44\left( {q^2  \plus{} r} \right) \ge 34q \plus{} 108r\left( {q \minus{} r} \right)
\]

\[ \Leftrightarrow 108r^2  \plus{} \left( {44 \minus{} 108q} \right)r \plus{} 44q^2  \minus{} 34q \plus{} 6 \ge 0
\]

Denote $ f\left( r \right) \equal{} 108r^2  \plus{} \left( {44 \minus{} 108q} \right)r \plus{} 44q^2  \minus{} 34q \plus{} 6$
\[ \Rightarrow f'\left( r \right) \equal{} 216r \plus{} 44 \minus{} 108q \ge 216r \plus{} 8 > 0
\]
Thus $ f$ is increasing and then it follows Schur. :)

[quote="djuro"][quote="Honey_S"]
Thus $ f$ is increasing and then it follows Schur. :)[/quote]

Can you explain this better, please?[/quote]

If $ q \le \frac{1}{4}$ then with $ r \ge 0$ we have

\[ f\left( r \right) \ge f\left( 0 \right) \equal{} \left( {22q \minus{} 6} \right)\left( {2q \minus{} 1} \right) \ge 0
\]

If $ q \ge \frac{1}{4}$ then with $ r \ge \frac{{4q \minus{} 1}}{9}$ we have
\[ f\left( r \right) \ge f\left( {\frac{{4q \minus{} 1}}{9}} \right)
\]
Easy to check that $ f\left( {\frac{{4q \minus{} 1}}{9}} \right) \ge 0$ too. :)


What is the least possible positive difference between the sum of five different positive integers and the greatest of them?

Let the integers be $ a,b,c,d,e$ such that $ a<b<c<d<e$. It follows that $ a\leq e\minus{}4$, $ b\leq e\minus{}3$, $ c\leq e\minus{}2$, and $ d\leq e\minus{}1$. Adding all of these up, we have $ a\plus{}b\plus{}c\plus{}d\leq4e\minus{}10$, so $ \minus{}(a\plus{}b\plus{}c\plus{}d)\geq10\minus{}4e$, and $ e\minus{}(a\plus{}b\plus{}c\plus{}d)\geq10\minus{}3e$. Since $ e\geq5$, and $ |10\minus{}3e|\leq|10\minus{}3(e\plus{}1)|$ for all $ e$ satisfying this, our answer is $ |10\minus{}3\cdot5|\equal{}\boxed5$. Of course, intuition tells you that the minimum occurs when $ (a,b,c,d,e)\equal{}(1,2,3,4,5)$, and since this is MC... :wink:

Obviously this is smallest when the integers are 1,2,3,4,x=x+10. x+10-x=10


John has a 1/3 chance of getting an A and independently, Anne has a 3/5 chance of getting an A. I look at one of their papers on the teacher's desk and see an A. I can't see whose paper it is; what is the probability that it is Anne's?

The probability that both are $ A$'s is $ \frac13\cdot\frac35\equal{}\frac15$, in which case the final probability is $ \frac15\cdot\frac12\equal{}\frac1{10}$.

The probability that John didn't get an $ A$ and Anne did get an $ A$ is $ \frac23\cdot\frac35\equal{}\frac25$, in which case the final probability is $ \frac25\cdot1\equal{}\frac25$ (you must realize that somebody must have gotten an $ A$, so it must have been Anne's in this case).

We don't consider the possibility that John gets an $ A$ and Anne doesn't, as then it is impossible for Anne to have gotten the $ A$.

Adding, we get $ \frac15\plus{}\frac25\equal{}\boxed{\frac35}$.


[hide]Square the first equation.
$(x\sqrt{x}-5\sqrt{x})^2=x^3-10x^2+25x=18$
I see that 2 solves the equation for 8-40+50=18.  Factor x-2 out of the equation $x^3-10x^2+25x-18$.  We have $x^2-8x+9$.  The solutions to this is $4\pm\sqrt{7}$.  We test these three equations for x since our original method involved squaring.  We notice that plugging in two yields $-3\sqrt{2}$.  Let's factor the $\sqrt{x}$.  We have $(\sqrt{7}-1)(\sqrt{4+\sqrt{7}})$.  After squaring the term not in the square root, we have $(8-2\sqrt{7})(4+\sqrt{7})$.  Notice the square root terms cancel and we get $\sqrt{18}$ which is $3\sqrt{2}$.  Also, notice that this process on $4-\sqrt{7}$ is the same because we move the negative sign from one place to the other.  This is the same as the plus case.
The answers are $4-2\sqrt{2}-\sqrt{7}$ and $4-2\sqrt{2}+\sqrt{7}$[/hide]

well, asking for $x-2\sqrt{2}$ doesn't make very much sense...

[hide]Square both sides to get $x^3-10x^2+25x=18\Rightarrow x^3-10x^2+25x-18=(x-2)(x^2-8x+9)=0$.

But looking at the original equation, $x$ must be greater than 5 or the left side would be negative. Clearly 2 is extraneous.

Then $x^2-8x+9=0$. This has solutions $4\pm \sqrt{7}$; only $4+\sqrt{7}$ is greater than 5.

Noting that $\sqrt{4+\sqrt{7}}=\frac{\sqrt{2}}{2}\left(1+\sqrt{7}\right)$, the left side is $(x-5)\sqrt{x}=\left(-1+\sqrt{7}\right)\left(1+\sqrt{7}\right)\frac{\sqrt{2}}{2}=3\sqrt{2}$ which checks.

So $x$ must be $4+\sqrt{7}\Rightarrow \boxed{4-2\sqrt{2}+\sqrt{7}}$.[/hide]


I'm taking the ACT for the second time saturday.  this time I might be able to get straight 36's, if i can pay attention.  i can do all the math, and the rest is just interpreting.  any tips?

English:
Say each of the sentences in your mind, and then the answer should immediately come to you.

Math:
I think you can do math by yourself.

Reading:
Read the stories, look at the questions but not the answers, and come up with answer yourself.

Science:
Same tactics as reading, but with graphs, tables, etc.

English is all luck about which sentences you get. The "read them out loud and you should see the answer" trick never worked for me because every single way sounded grammatically correct to me.

I forgot to put this on too.

For all the parts of the test, if you don't know it, guess it. Because of the the way the ACT is scored, if you miss it it won't go against you, and if you get it right, it will help you.

I'm fine on everything but english, really. i already have a perfect in science.
can your scores carry over? Like, if i get a 36 on everything but science  (probably not), will i be able to carry over the 36 on science so i have straight 36's?

they just average the scores ex. 36+36+36+35 will give you a composite score of 36.

Colleges usually do NOT take the best-score-from-each-section, from whatever sitting, approach that they take with SAT I scores when they look at ACT scores in evaluating admission files. But if you've had strong section scores before, you are well on the way to scoring well next time. My son calls the ACT "the American Reading Test," because it's really all about reading. Read carefully, don't skip any opportunities to mark answers, and you should do fine.


Hello!

Does anybody know any book where I can find detailed explanations about "Binomial Interpolation"? :?:

I was searching the Internet but could not find any good source. I have acquired 5 books of Numerical Analysis but none of them has a topic specific to this subject. I found a lot of information about Lagrange, Newton's Divided Differences and splines.  :maybe:

Thanks!  :lol:

This topic belongs either in the "Books and Journals" section, or in one of the college sections (Advanced Topics?) where someone who knows something about it will find it.


I got it, but it took me 11 steps...  I'll see if I can improve on this during my classes this morning (I'm late already or I'd post what I have).

TripleM wrote:If * is an associative operation on a set S, and A, B, C are three elements of S such that
A*B = B*A
A*C = C*A
A*B*A = A
B*A*B = B

Prove B*C = C*B.

(associative means X*(Y*Z) = (X*Y)*Z for all X,Y,Z in S).

Here it goes (expanded in an attempt at clarity):

[hide]

B*C  = B*C

        = B*A*B*C

        = B*(B*A)*C

        = B*B*(C*A)

        = B*B*C*A*B*A

        = B*B*(A*C)*B*A

        = B*(A*B)*C*B*A

        = B*A*B*C*(A*B)

        = B*A*B*(A*C)*B

        = B*A*C*B

        = B*(C*A)*B

        = B*C*A*B*A*B

        = B*(A*C)*B*A*B

        = (A*B)*C*B*A*B

        = A*B*C*(A*B)*B

        = A*B*(A*C)*B*B

        = A*C*B*B

        = (C*A)*B*B

        = C*(B*A)*B

        = C*B

[/hide]

QED


function
logarithms
algebra
functional equation
algebra unsolved

Find the  function $f$

$f(x)-\frac{1}{2} f(\frac{x}{2})=x^2$
 I have corrected the problem previous was wrong

Just subtitute $x$ by $\frac{1}{x}$  :!:

$f(x)=\frac12f\left(\frac{x}{2}\right)+x^2$
$f(x)=\frac14f\left(\frac{x}{4}\right)+\frac18x^2+x^2$
$f(x)=\frac18f\left(\frac{x}{8}\right)+\frac{1}{64}x^2+\frac18x^2+x^2$
...
$f(x)=\frac{1}{2^n}f\left(\frac{x}{2^n}\right)+\frac{1-(1/8)^n}{1-(1/8)}x^2$
Let $n=\log_2x$.  Then,
$f(x)=\frac{1}{x}f(1)+\frac{1-(1/x)^3}{7/8}x^2$
$f(x)=\frac{C}{x}+\frac87x^2-\frac{8}{7}\frac1x=\frac{C}{x}+\frac87x^2$

$f(x)$ satisifies for all complex C and all complex x (except for $f(0)$ which is defined as 0, based on the original equation).

[quote="draperp"]
Let $n=\log_2x$.  Then,
$f(x)=\frac{1}{x}f(1)+\frac{1-(1/x)^3}{7/8}x^2$
.[/quote]
I am wondering if you can do that as $n$ is integer and $x$ is real :?:

My partial solution is very similar to draperp's, and I get stuck at the same place.

a) $x=0$ $\to$ $f(0)=0$

b) $x\neq 0$
$f(x)=x^2+\frac{1}{2}f(\frac{x}{2})=x^2+\frac{1}{2}(\frac{x}{2})^2+\frac{1}{4}f(\frac{x}{4})=\dots$
$f(x)=\sum_{n=0}^{+\infty}{2^{-3n}x^2}+L(x)$, $L(x)=lim_{n\to +\infty}2^{-n}f(2^{-n}x)$.
$f(x)=\frac{7x^2}{8}+L(x)$, $L(x)=lim_{k\to 0}kf(kx)$.

We can see that $L(x)=0$ if $f(kx)$ is finite when $k\to 0$. Therefore, the only solution that is continuous in $x=0$ is $f(x)=\frac{7x^2}{8}$.

Substituting the general solution $f(x)=\frac{7x^2}{8}+L(x)$ in the functional equation, we get another functional equation $L(x)=\frac{1}{2}L(\frac{x}{2})$.
One solution is $L(x)=\frac{C}{x}$, but I should find a general solution.

[quote="AYMANE"]
I am wondering if you can do that as $n$ is integer and $x$ is real :?:[/quote]

You're right that one can't do that rigorously, but it is so easy just to extend a rule from integers to all reals.

One things that you [u]can[/u] do rigorously is use a similar procedure to rewrite $f(x)$ in terms of $f\left(2^nx\right)$.
$f(x)=2^nf\left(2^nx\right)-\frac{8n-1}{7/8}$
Whereas orginially, I could only rigorously define $f(x)$ in terms of $f(a)$ when $x=\frac{2^na}$, for whole n, now one could extend it to all integer n.

Also, I think that it must be given that $f$ is continuous.
$...f(1/4),f(1/2),f(1),f(2),f(4)...$ seems to be independent of $...f(1/6),f(1/3),f(2/3),f(4/3),f(8/3)...$ unless $f$ is continuous.

In that case, I think that you could use all real n.


geometry
perpendicular bisector
geometry proposed

Show how to construct triangle $ \triangle ABC$ if followings are given: $ \frac{a}{h_{a}}\equal{}\frac{m}{n}$, value of angle $ \angle BAC$ and length of median $ m_{b}$.

Let $ B'C' \equal{} m$ and a line $ \varepsilon \parallel B'C'$ such that $ d(B'C',\varepsilon) \equal{} n$
Let $ O$ be a point on the perpendicular bisector of $ B'C'$ such that $ \angle B'OC' \equal{} 2\angle BAC$ (the point $ O$ and the line $ \varepsilon$ should be on the same half-plane of $ B'C'$)
Draw the circle $ (O)$ passing through $ B'$
Let $ A$ be one of the points where this circle intersects the line $ \varepsilon$ (if it does)

The triangle $ AB'C'$ has the property $ \frac {a}{h_{a}} \equal{} \frac {m}{n}$ and $ \angle B'AC' \equal{} \angle BAC$

Let $ B'M'$ be the median of this triangle

The triangle $ ABC$ we are searching for has:
$ B\in (AB'$
$ C\in (AC'$
and $ BC\parallel B'C'$
Then its median $ BM$ is parallel to $ B'M'$

So it's enough to find the point $ B\in (AB'$ such that $ \frac {AB}{AB'} \equal{} \frac {m_b}{B'M'}\Leftrightarrow AB \equal{} AB'\frac {m_b}{B'M'}\ ,$ constructible


abstract algebra
category theory
linear algebra
matrix
Ring Theory
superior algebra
superior algebra solved

assume that $R$ is noetherian ring and $M$ is finitely generated  $R$ module if $p\in supp_{R}(M)$ then prove that $Hom_{R}(M,R/p)\neq 0$

Since localisation is an exact functor, $Hom_{R}(M,R/p)$ is in bijection to $Hom_{R_{p}}(M_{p},R_{p}/pR_{p})$. $p\in Supp M\Leftrightarrow M_{p}\neq 0$.

It is therefore sufficient to assume that $R$ is a local noetherian ring with maximal ideal $m$ and residual field $k=R/m$. We want to prove that if $M$ is finitely generated and nonzero, then $Hom_{R}(M,k)\neq 0$.
Let $R^{m}\stackrel{\varphi}{\to}R^{n}\to M\to 0$ be a finite presentation of $M$. This exists because $M$ is finitely generated over $R$ which is noetherian. We have $M\simeq coker\varphi$. Try not to confuse the maximal ideal $m$ and the exponent in $R^{m}$.
I will prove that we can choose $m,n,\varphi$ such that $Im\varphi\subset mR^{n}$.
Let $f_{i}$ and $e_{j}$ be bases in $R^{m}$ and $R^{n}$ respectively.
Let $\varphi(f_{i})=\sum_{j=1}^{n}a_{ij}e_{j}$ with $a_{ij}\in R$.
Assume one of the $a_{ij}$'s is not in $m$. Then it is invertible in $R$ because $R$ is local. WLOG assume $a_{11}$ is invertible. It is easy to see that we can chase the bases for $R^{n}$ and $R^{m}$ such that the matrix of$\varphi$with respect to these bases has only$0$s on the first row and the first column, except for the entry$1\times 1$which is$1$. This means that$\varphi$splits as$\varphi 1\oplus\varphi|_{R^{m-1}}: R\oplus R^{m-1}\to R\oplus R^{n-1}$. By considering$\varphi'=\varphi|_{R^{m-1}}: R^{m-1}\to R^{n-1}$, we have an$R$-modules application such that$M\simeq coker\varphi'$. Continue this inductive process untill there is no invertible entry in the matrix of$\varphi$. Now consider the exact sequence$0\to Hom(M,k)\to Hom(R^{n},k)\to Hom(R^{m}\to k)$. Obtained by applying the left-exact functor$Hom(\cdot,k)$to the exact sequence$R^{m}\stackrel{\varphi}{\to}R^{n}\to M\to 0$. The application$Hom(R^{n},k)\to Hom(R^{m}\to k)$is given by$f\to f\circ\varphi$. Since$f$has values in$k=R/m$and the matrix of$\varphi$has entries in$m$, we get the the application between the Homs above is actually $0$. In view of the exactness of the sequence above, $Hom(M,k)\simeq Hom(R^{n},k)$. $Hom(R^{n},k)$ is obviously nonzero. In fact it is isomorphic to $k^{n}$. ($n\neq 0$ because $M\neq 0$)

So $Hom(M,k)\neq 0$.
------------------------------------------------------------------------------------
There is a more general proposition:
Let $R$ be a ring (commutative with $1$), and let $M,N$ be $R$-modules. Set $I=Ann N$.
a) If $I$ contains an $M$-regular element, then $Hom_{R}(N,M)=0$. (for $x\in R$ to be $M$-regular means that the multiplication by $x$ on $M$ is injective)
b) Conversely, if $R$ is noetherian and $M,N$ are finitely generated, $Hom(N,M)=0$ implies that $I$ contains an $M$-regular element.
----------------------------------------------------------------------------------
Point (b) applied to our problem says that if $Hom(M,R/p)=0$, then $I=Ann M$ contains an $R/p$ regular element which means an element $x\in R\setminus p$. $(R\setminus p)\cap Ann M\neq\emptyset$ means that $M_{p}=0$ which contradicts $p\in Supp M$.


In how many ways can 36 be written as the product $ a \times b \times c \times d$, where $ a, b, c$ and $ d$ are positive integers such that $ a \le b \le c \le d$?

A common mistake.  You forgot the negative numbers.  I'm too lazy to state all the combinations, so I won't.  :D

9 ways to write them as positive factors, and I believe 9 ways to write them including negative factors, so $ \boxed{18}$ ways in total. Am I correct or incorrect?

I believe you are incorrect, because for example:

you could have -1,-1,3,12; -1,-1,-3,-12; 1,-1,-3,12, and etc.

Oh, then I'm right!
Is there any other way to do this because if I play a 20 second game, I can't get the answer unless I memmed it.

[quote="mewto55555"][quote="GameBot"]In how many ways can 36 be written as the product $ a \times b \times c \times d$, where $ a, b, c$ and $ d$ are [b][size=200]positive integers[/size][/b] such that $ a \le b \le c \le d$?[/quote]

 :rotfl:  :D[/quote]

THATS HOW I GOT THAT WRONG AT STATES LAST YEAR!  :D

Ah. Who reads the questions anyway?  :roll:

So answer = $ \boxed{9}$.

36 [b]does[/b] have 9 factors.  Doesn't mean you aren't crazy though.  :D


HMMT
trigonometry
MATHCOUNTS
Support
geometry
area of a triangle
Heron\u0027s formula

I think it obviously could have been run better (like our team round being quite messed up), but overall it was pretty fun. Even without the very end... :wink: Also we got very lost twice...Though that was our fault.

i cant believe that BLS actually got top ten in team round: i didn't hear our school name called
other than that, i could have done better
and if your smart, you get to be a proctor?

For no preparation, my team (Lunenburg) did pretty decent.  I enjoyed myself; some great problems.

Oh btw we did not receive solutions for Theme (our room was the one that screwed up the team questions too...) Can someone explain #8? Is it just a ton of case work? I spent like 20 minutes on the test on that one, looking for a good way then screwing up the case work.

hey perfect, were you in hall b?
we were and we didn't receive solutions to themes either, and those problems were hard...
esp. the chessboard

Ugh we were in Jefferson 250. Also I didn't think the Theme was all that hard...Like I don't see how my 40 won it. I expected at least 45s, because I still don't think there exists a reasonable solution to #8.

[quote="CatalystOfNostalgia"]So clearly it has been clearly established who the superior school in the state is. Our unofficial Guts Round team was pretty beast. We got the first 32 questions right and Sergei decided not to try to prove our bound on 33 and spend 25 minutes on the guessing questions at the end. It turned out we got 33 wrong but we did get 8 points out of our (Sergei's) intense trig bashing and Taylor Series approximations on #34. Too bad it didn't count. Also we had a slight time advantage because we didn't have to run (they just gave us all the problems).

But being a proctor (and getting a free T-shirt!) was fun.[/quote]

They should've made Lexington's fifth team official...what a dumb rule :P

Anyways...I'm still waiting for HMMT to post the questions so I can attempt them myself. (Being sick makes the wait even worse.)

I don't think anyone got Theme solutions, I was told at the proctor meeting that they probably forgot to put them in when stuffing envelopes. And also we had two college kids on our "team," which probably makes it unofficial.

I'm beginning to realize how amazingly great the problems were. Toning down problem difficulty can often result in a bunch of silly MathCounts and/or pure computational problems, but this definitely was not so; there were many nice and easy problems that you had to think at least a little bit to solve, as well as some towards the end of the Guts round that challenged our unofficial team of five MOPpers and Julia (two blue), two of which now go to MIT. Also, they were very well suited to the competition, seeing that the top scores were very much in line with the usual top scores at the big HMMT (with the exception of General 1, of course, but that still wasn't too bad), or a little bit higher. Of course, this was after barring the top-scoring students from this year...but I'm not bitter, I had just as much fun not competing.

And seeing that there was so much screwing up going on, on the organizers' part (according to Sway), it was hardly recognizable (probably more however for the teams with the messed up team rounds), and it was very well run, seeing that it was only announced like two minutes ago. I was surprised by the turnout, I guess it's a job well done on the part of the publicists (half of which come from Lexington; that's how we do things).

[quote="Smartguy"]your Sam T?![/quote]

Actually he is SamL

Actually I am Sam T

Also what exactly was your "support" of which you spoke?

The problems were pretty good, I agree.  Shame on whichever team looked up the answer to #34 on their iPhones or whatever it must've been, though (I was grading Guts at the leftmost table in the front!) - kind of missed the point there :|

[quote="CatalystOfNostalgia"]uh, me yelling "yeah sam" extremely loudly right when they called your name?[/quote]

I did not hear that.

Also, we jokingly said, "Hey let's look it up on your iPhone" when we saw #34...But didn't because a) why cheat, and b)we were right near the front :P

[quote="CatalystOfNostalgia"]Seeing that we had 25 minutes, our idea was to go outside and ask Sway to log in for us on one of the computers in the Science Center lobby. But we still got 17 with our analytical guessing.[/quote]

does anyone know the username and password for logging in?

when are the results going to be posted on the hmmt website?

They actually had been posted on the website, but I guess they took it down for the moment.

can someone post the answers to the theme round?

[hide] [geogebra]04fc1b2be564f63b96a4cdb1be4320ea15a64c03[/geogebra]
Note: 21/8 in the pic is because tangents to a point are equal.

Solve for x with Pythagoras and get x=1/2. so the area is (1/2)(3)(7/8)=21/16[/hide]

Can someone post solutions to #5 triangle and #'s 3-5 chessboard?

[hide="Triangles #5"][code]A triangle has altitudes of length 15, 21, and 35. Find its area.[/code]

Start with triangle $ ABC$. Let us denote the area of triangle $ ABC$ by $ [ABC]$, the sidelengths as $ a$, $ b$, and $ c$, and the corresponding altitudes by $ h_a \equal{} 15$, $ h_b \equal{} 21$, and $ h_c \equal{} 35$, respectively.

We'll start with the fact that $ [ABC] \equal{} \frac {1}{2}15 a \equal{} \frac {1}{2}21 b \equal{} \frac {1}{2}35c$. We can rewrite this as: $ \frac {a}{2} \equal{} \frac {[ABC]}{15}$, $ \frac {b}{2} \equal{} \frac {[ABC]}{21}$, $ \frac {c}{2} \equal{} \frac {[ABC]}{35}$.

Take Heron's formula in the following form:

$ [ABC] \equal{} \sqrt {\left(\frac {a}{2} \plus{} \frac {b}{2} \plus{} \frac {c}{2}\right)\left(\frac {a}{2} \plus{} \frac {b}{2} \minus{} \frac {c}{2}\right)\left(\frac {b}{2} \plus{} \frac {c}{2} \minus{} \frac {a}{2}\right)\left(\frac {c}{2} \plus{} \frac {a}{2} \minus{} \frac {b}{2}\right)}$.

Then substitute for each of $ \frac {a}{2}$, $ \frac {b}{2}$, $ \frac {c}{2}$, then factor out $ [ABC]$ and simplify:

$ [ABC] \equal{} [ABC]^2\sqrt {\left(\frac {1}{15} \plus{} \frac {1}{21} \plus{} \frac {1}{35}\right)\left(\frac {1}{15} \plus{} \frac {1}{21} \minus{} \frac {1}{35}\right)\left(\frac {1}{21} \plus{} \frac {1}{35} \minus{} \frac {1}{15}\right)\left(\frac {1}{35} \plus{} \frac {1}{15} \minus{} \frac {1}{21}\right)}$

$ [ABC] \equal{} \frac {1}{\sqrt {\left(\frac {1}{15} \plus{} \frac {1}{21} \plus{} \frac {1}{35}\right)\left(\frac {1}{15} \plus{} \frac {1}{21} \minus{} \frac {1}{35}\right)\left(\frac {1}{21} \plus{} \frac {1}{35} \minus{} \frac {1}{15}\right)\left(\frac {1}{35} \plus{} \frac {1}{15} \minus{} \frac {1}{21}\right)}}$

$ [ABC] \equal{} \boxed{245\sqrt {3}}$
[/hide]

I think for numbers 3 and 4 on Chessboard it's all a bunch of casework, so I won't post a solution (that, and I'm not too sure how the casework works myself).

[hide="Chessboard #5:"][code]Suppose that at some point Joe B. has placed 2 black knights on the original board, but gets bored of chess. He now decides to cover the 34 remaining squares with 17 dominos so that no two overlap and the dominos cover the entire rest of the board. For how many initial arrangements of the two pieces is this possible?[/code]

Look at the fact that a domino would cover a black and a white square. Since we have 18 black squares and 18 white squares, and the knights eliminate two of these squares, we can conclude that the knights must cover exactly one white and one black square. (A proof that all such cases work is beyond the scope of this solution.) Thus the answer is simply $ 18*18 \equal{} \boxed{324}$.[/hide]

Ugh, I was so close to solving Triangle Problem #5.....I gave up while playing with Heron's formula.

Chessboard Q #3 is supposed to be done on a 4x4 chessboard according to the official solution but the part about breaking the 6x6 chessboard into a 4x4 chessboard comes on #4 so...


the product of 2 quadratic non-residues is a residue (in an odd prime modulus)

how do you prove this?

My proof seems too easy :maybe:

Since Legendre's symbol is multiplicative (it follows directly from Euler criterion), $ \left(\frac {ab}{p}\right) \equal{} \left(\frac {a}{p}\right)\left(\frac {b}{p}\right)$

As $ \left(\frac {a}{p}\right) \equal{} \left(\frac {b}{p}\right) \equal{} \minus{} 1$ by the problem statement, obviously $ \left(\frac {ab}{p}\right) \equal{} 1$ qed.

[hide]First show there are $ (p \minus{} 1)/2$ quadratic residues (QR), thus $ (p \minus{} 1)/2$ quadratic nonresidues (QNR). Let $ a$ be a QNR. If $ b_1, b_2, \ldots b_{(p \minus{} 1)/2}$ are the QRs, then $ ab_1, ab_2, \ldots ab_{(p \minus{} 1)/2}$ are all of the QNRs. Let $ c$ be another arbitrary QNR (possibly equal to $ a$). $ ac$ can't be any of $ ab_1, ab_2, \ldots ab_{(p \minus{} 1)/2}$, so the only thing it can be is a QR, as desired.[/hide]

[hide="My personal preference..."] is, as hsiljak noted, to use Euler's criterion:

[b]Lemma:[/b]  $ \left( \frac {a}{p} \right) \equiv a^{\frac {p \minus{} 1}{2} } \bmod p$. [/hide]


integration
calculus
function
trigonometry
logarithms
calculus computations

I have a problem, it's may be easy  :blush: :
      $\int\frac{1}{x^4+ax^2+1} ?$

My hint is $\frac{1}{x^4+ax^2+1}=\frac{A}{x^2-a_1}+\frac{B}{x^2-a_2}$, where $a_1$ e $a_2$ are the roots of $x^2+ax+1=0$. I found $A=-B=\frac{1}{\sqrt\ a^2-4}$. And now it´s just evaluate the integral by arctan function, write?

On top of mffmartinelli's step, you can continue partial fraction decomposition on his two fractions, this will then trivial the problem.

$I=\int \frac{1}{x^4+ax^2+1} dx,\ J=\int \frac{x^2}{x^4+ax^2+1} dx.$

$I+J=\int \frac{\left( x-\frac 1x\right) '}{\left( x-\frac 1x\right)^2+2+a} dx,\ J-I=\int \frac{\left( x+\frac 1x\right) '}{\left( x+\frac 1x\right)^2+a-2} dx$, a.s.o.

[u]Remark.[/u] There are five cases: $a<-2$; $a=-2$; $-2<a<2$; $a=2$; $a>2$.

$\int \frac1{x^4+ax^2+1}\,dx =$
$\int \frac{\frac{-1}{\sqrt{a^2-4}}}{x^2+\frac{a+\sqrt{a^2-4}}{2}}+\frac{\frac1{\sqrt{a^2-4}}}{x^2+\frac{a-\sqrt{a^2-4}}{2}}\,dx =$
$\frac{-1}{\sqrt{a^2-4}} \frac1{\sqrt{\frac{a+\sqrt{a^2-4}}{2}}} \tan^{-1} (\frac{x}{\sqrt{\frac{a+\sqrt{a^2-4}}{2}}}) +\frac1{\sqrt{a^2-4}} \frac1{\sqrt{\frac{a-\sqrt{a^2-4}}{2}}} \tan^{-1}(\frac{x}{\sqrt{\frac{a-\sqrt{a^2-4}}{2}}})$

I won't bother simplifying, but anything of the form $\sqrt{x \pm \sqrt{y}}$ can be written in the form $a \pm \sqrt{b}$ and then when you have that in the denominator you can simplify by multiplying by $1$ (it's conjugate over it's conjugate).

[quote="kunny"]This is not easy, I think. We are to classify some cases. ;)[/quote]
Yes  :)  for example if $|a|<2$.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $(G,\cdot)$ be a group with $p^3$ elements where $p$ is a prime and for $x\in G$ we denote with $C(x)=\{y\in G|xy=yx\}$.
Find the cardinal of $\{C(x)\mid x\in G\}$.

no... I mean what I wrote
and
$\bigcup_x C(x)=G$ because if $x\in Z(G)$ then $C(x)=G$

Plainly, $\#\{C(x) \mid x \in G\} \in \{1,p,p^2\}$: it is $1$ if $G$ is abelian; $p$ if it is the (unique) non-abelian group of order $p^3$ and exponent $p^2$; and $p^2$ if it is the unique non-abelian group of order $p^3$ and exponent $p$.

I think some minor corrections need to be made. For example, if we look at the multiplicative group $\{\left(\begin{array}{ccc}1&a&b\\0&1&c\\0&0&1\end{array}\right)|a,b,c\in\mathbb Z_p\}$, then I think there are $p+2$ elements in that set (including $G$ itself, which is a commutator of the elements in the center).

P.S.: that group has exponent $p$, by the way.

Actually, if the group is non-abelian and has exponent $p$, it's easy to show hat that is indeed the answer. If we disregard the entire group $G$, we should show that there are $p+1$ other commutators.

First we can show that each commutator has cardinal $p^2$. Conversely, it's easy to show that each subgroup of order $p^2$ is a commutator for one of its non-central elements. We thus need to count the subgroups of order $p^2$. Since each group of order $p$ is contained in a group of order $p^2$, we do this by counting the subgroups of order $p$ of $G$:

There are $\frac{p^3-1}{p-1}=p^2+p+1$ subgroups of order $p$ in $G$. One of these is $Z(G)$, which is also common to all the subgroups of order $p^2$. If we let $n$ be the number of subgroups of order $p^2$, since each such subgroup has $\frac{p^2-1}{p-1}=p+1$ subgroups of order $p$, one of which is $Z(G)$, and the intersection of each two is precisely $Z(G)$, it means that the total number of subgroups of order $p$ must be $np+1$. We know that this is equal to $p^2+p+1$, so $n=p+1$.

Together with the entire group $G$, we get $p+2$ commutators whn $G$ is non-abelian and of exponent $p$.


fin all n such
$3n^{4}+3n^{2}+1$ is a perfect square

I believe only 0 is solution but I don't know...

Doesn't that show 3n4+3n2+1 is not a perfect cube, but the original question was to show it's not a perfect square?

[quote=maxal]Let $ 3n^4\plus{}3n^2\plus{}1\equal{}m^2$ then
\[ y^2 \equal{} x^3\minus{}15x\minus{}22\]
where $ x\equal{}\frac{2\plus{}n^2\minus{}2m}{n^2}$ and $ y\equal{}\frac{2(2\plus{}3n^2\minus{}2m)}{n^3}$.

The above elliptic curve has rank $ 0$ and the only rational points it contains are $ (x,y)\equal{}(\minus{}2, 0)$ and the infinity.
Therefore, there is the only solution $ (n,m)\equal{}(0,1)$ to the original equation.[/quote]

What do you mean by rank?

[quote=dgreenb801]If $ 3n^4\plus{}3n^2\plus{}1\equal{}m^3$, then $ (n^2)^3\plus{}m^3\equal{}(n^2\plus{}1)^3$, contradiction by Fermat's Last Theorem unless $ (n,m)\equal{}(0,1)$!
Still wonder if there is a more elementary way...[/quote]

This subcase of fermat's last theorem is simpler then the general theorem.
If a.b.c coptimes in pairs and a³+b³=c³ then (a+b)(a²-ab+b²)=c³ note that gcd(a+b,a²-ab+b²) is 1 or 3 then we have quite easy case checking.


相邻的国家不能涂同种颜色,因此只要证明地图中不可能存在五个两两相邻的国家,不就可以了吗.
证明:在地图中,一定存在某个国家与周围至少三个国家两两相邻,形成至少四个国家两两相邻(这显然存在,否则不需要四种颜色就可以涂满).

任选一个这样的国家记为A涂上红色,任选三个与它两两相邻的国家记为B,C,D,分别涂上颜色黄,蓝,绿, 如图,此时地图被分为1,2,3,4, 四个区域,

假设存在第五个国家E与它们两两相邻,若在区域1中显然不能与国家C相邻,故E不可能在区域1中.同理,E也不可能在区域2,3,4中.

故不存在第五个国家与A,B,C,D两两相邻,由B,C,D的任意性,A不能与相邻国家形成五个两两相邻的国家.

由国家A的任意性,地图中不可能存在五个两两相邻的国家,故不需要涂第五种颜色,所以四种颜色就可以将地图涂满.
[/b]

地图中即使不存在五个两两相邻的国家,四色猜想也未必成立唷
考慮 9 个国家分為三組: A1,A2,A3,A4,B1,B2,B3,B4,C
A之中兩兩相邻, B1與A2,A3,A4相邻, B2與A1,A3,A4相邻,
 B3與A1,A2,A4相邻, B4與A1,A2,A3相邻, C與B1,B2,B3,B4相邻
那麼地图中就不可能存在五个两两相邻的国家(B內的国家一定不在五个国家內,
C只與B內的国家相邻,所以也不在五个国家內)
可是,B1的顏色一定與A1相同,B2的顏色一定與A2相同.....
B1,B2,B3,B4已經有四種顏色,C不可能被填色
所以,如果這種情況存在,四色猜想就不成立了,
可是,你的證明沒有排除這種情況呢

總之四色定理早就在機器上得到證明, 而那麼多前輩花了那麼多精神和時間也未能找出理論依據明, 相信這並不是一個用初等方法就能解決的問題. 有時用初等的角度去分析一些高深的問題, 也許會有一點點的收穫, 但光用初等知識去解決那些問題似乎是天方夜譚...


Let $ a=\sin^{2}\frac{\pi}{5},\ b=\sin^{2}\frac{2\pi}{5}$. Prove the following statements.

(1) $ a+b$and $ ab$are rational numbers.

(2) $ (a^{-n}+b^{-n})(a+b)^{n}$ is integer for any positive integers $ n$.


What is the greatest integer less than $ \frac{\minus{}21}4$?

[quote="AIME15"]-21/4=-4.25, so the answer would be -5[/quote]

I'm getting -21/4 = -5.25.

Yes, it is $ \minus{}5.25$, so the answer is actually $ \minus{}6$.


Problem 1:
[hide]
A certain clock loses 15 minutes every hour.
Suppose the clock is set to the correct time at 9 AM.
What will the correct time be when the clock first shows 10 AM?
[/hide]

Problem 2:
[hide]
In the addition problem below, each letter represents a digit and different letters represent different digits
What four digit number does DEER represent?

IN + RID = DEER
[/hide]

Problem 3:
[hide]
I have a drawer which contains 40 socks in the following numbers and colors: 12 tan, 9 brown, 11 gray, and 8 blue. Suppose I am blindfolded. What is the fewest number of socks I must pick from the drawer to be absolutely certain that I have 2 socks of the same color among those I have picked?[/hide]

[hide="#2"]Since DEER is a 4-digit number produced by adding a three-digit and 2-digit number, the three-digit number must have a hundreds digit of 9 in order to yield such a number. DEER must have a thousands digit of 1. So D=1 and R=9. Also, any two single digit number summed together can only yield a maximum sum of 18, so E=0 since 9+1=10. I+I=10, so I=5. R=9, so 1+N=9 and N=8. Thus DEER is the number $ \boxed{1009}$.[/hide]

[hide="#3"]There are 4 colors of socks in the drawer. Using the pigeonhole principle, it can take up to 5 draws before finding a matching pair, since 4+1=5. Therefore you need to draw at least $ \boxed{5}$ socks in order to guarantee yourself a matching pair.[/hide]

[hide="1"]

First, we have that the clock loses 15 minutes every hour. So, it take 45 minutes for the clock to show the time 10 AM, therefore the correct time is 9:45 AM.

[/hide]

[hide="#1"]technically for #1, the clock must go backwards 23 hours in wrong time, so in real life 23*60/15=92 hours have elapsed. the time now is 17:00=5:00PM[/hide]

i think?

[hide="i disagree with you both?"] okay, so in an hour, this clock ACTUALLY goes 45 minutes, or 3/4 of how far it needs to. so, after one hour, its 9:45. after 20 more (real) minutes, it'll show 10:00. i think the answer is 10:20.

i think qwerty interpreted the question wrong, and i, quite frankly, have no idea what mz did.

[/hide]

fail.

Nikeballa was correct.

Answer:
[hide]
A short way: how long is one minute on the slow clock?
Ans: 4/3 of a minute.
So we multiply 60 minutes by 4/3 and get 80 minutes.
The elapsed time in one hour is 45 minutes.
So the elapsed time is 60 minutes for the slow clock and 80 minutes for the real clock.
So the real clock went 80 minutes from 9:00 AM,
Therefore, the time is 10:20.
[/hide]

[quote="mz94"]oops. hehe i misinterpreted it in this way:

assume we ahve 2 clocks. one is the messed up one, the other is the correct one. they both are set to 9AM. when the correct one goes to 10AM the incorrect one goes to 8:45AM. WOW I FAIL[/quote]

I think you misinterpreted this way: The clock does not travel but it loses 15 minutes every hour, so it travels backwards.

Actually, the problem meant that a certain clock travels 15 minutes less than a real clock in 1 hour.
From this, it shows this problem is actually like a rate problem:
example: fill the swimming pool with the drain open. etc.

So actually, it is about a clock that travels 3/4 of the time as any other real clock.

what a coincidence, it was 10:20 some time ago!  :D

[quote="cgyao15"]21 minutes, actually (since you posted!!)

 :starwars:  :bruce:[/quote]

Technically, you are not supposed to put useless posts like that.
Useless posts =  :spam:
I heard that if you want to be funny, go to the AOPS fun factory.


There are k books of mine among n books. We put them in a shelf ramdomly. Which is the possibility that there are p books of my who are placed continuously?

If I haven't made a stupid mistake somewhere, the problem is essentially equivalent to finding the Taylor expansion
\[ \frac {1}{1 \minus{} w \minus{} z \plus{} w^pz} \equal{} \sum_{\ell,m\ge 0}F_{\ell,m}w^\ell z^m
\]
More precisely, the probability in question is just
\[ 1 \minus{} \frac {k!(n \minus{} k)!}{n!}(F_{k,n \minus{} k \plus{} 1} \minus{} F_{k \minus{} 1,n \minus{} k \plus{} 1})
\]
The coefficients $ F_{\ell,m}$ can be found quite fast from the recurrence relations $ F_{0,0} \equal{} 1$,
\[ F_{\ell,m} \equal{} F_{\ell \minus{} 1,m} \plus{} F_{\ell,m \minus{} 1} \minus{} F_{\ell \minus{} p,m \minus{} 1}
\]
where, as usual, we put $ F_{\ell,m} \equal{} 0$ if $ \ell$ or $ m$ is negative but I'm afraid there is no nice closed form formula for them. If such an answer is satisfactory for you, I'll post the details.


Let a,b,c be three real numbers. Prove that: $ \sum_{cyc}^{}\frac{(b\plus{}c)^{2}}{(b\plus{}c)^{2}\plus{}4a^{2}}\geq \frac{3}{2}$

10 or more topics for 1 problem??? MODS lock all of them (except one :) )

[quote="duythuc_lqd"]Let a,b,c be three real numbers. Prove that: $ \sum_{cyc}^{}\frac {(b \plus{} c)^{2}}{(b \plus{} c)^{2} \plus{} 4a^{2}}\geq \frac {3}{2}$[/quote]
This problem is wrong!!!
Try $ a\equal{}b\equal{}1,c\equal{}0$ :wink:

oh I'm sorry. a,b,c be three positive real numbers


search
number theory unsolved
number theory

1/ An integer n> 1 and  a prime number p are such that n divides p - 1 , and p divides n^3 - 1. Show that 4p -3 is the square of an integer.

2/ Let n be a positive integer such that A=2 + 2*sqrt(28*n^2+1) is an integer. Show that A is the square of an integer

2/ Let n and q be integers with n>= 5 , 2 =<q  =< n . Prove that  q- 1 divides [ (n-1)! / q ]
( [x] is the greast integer not exceeding x)

It seems that #1 doesn't hold: For $n=2$ and $p=7$ we have $2|7-1$ and $7|2^3-1$, but $4\cdot 7+3=31$ is not a square of an integer.

Likewise for $n=3$ and $p=13.$

Maybe it should be $4p-3$?

sorry , I have a mistake in typing.
for problem#1
" Show that 4*p-3 is the square of an integer. "

Posted before.
And I will not give any link because it's trivial to find and some people need to learn how to search (and I hope noone else posts a link).


calculus integration parameterization function LaTeX

Sorry if the question is a bit hard to read, since I'm still not very familiar with Latex.

A continuous random variable X has density of the form
fX (x) =1/b [e((x−a)/b)]/[1 + e((x−a)/b)]^2, for − ∞ < x < ∞,
where the parameters a and b satisfy −∞ < a < ∞ and b > 0.

Show that the density function integrates to 1.

Thanks for the help.

By the way, this site has a link for learning Latex.  It doesn't take very long to learn, and is very useful; also, you can hover your mouse over (or use it to click on) a Latex image to see the code, and you can learn it that way.

If $fX(x)=\frac{e((x-a)/b)}{b(1+e((x-a)/b))^2}$, as you wrote, then the integral is not always $1$ (try $a=b=1$).  However, if $fX(x)=\frac{e^{\frac{x-a}{b}}}{b(1+e^{\frac{x-a}{b}})^2}$, then it is:
[hide]$\int_{-\infty}^\infty \frac{e^{\frac{x-a}{b}}}{b(1+e^{\frac{x-a}{b}})^2}dx=\int_{-\infty}^\infty \frac{du}{(1+u)^2}$, where $u=e^{\frac{x-a}{b}}$, so the value of the integral is $-\frac{1}{e^{\frac{x-a}{b}}+1}$ evaluated at infinity (which is $0$) minus $-\frac{1}{e^{\frac{x-a}{b}}+1}$ evaluated at negative infinity (which is $-1$), so the integral is $1$.[/hide]

Thank you so much. I can't believe I forgot about substitution! Totally slipped my mind.

I can't use Latex because I'm not using my own computer, therefore I can't install MiKTeX and the like.

Actually, there is a second part to this question, which I thought I could solve after knowing the answer to the first part. Here's my attempt at write it in Latex:

\displaystyle\int^infty_-infty e^xt \displaystyle\frac{e^\frac{x-a}{b}}{b(1 + e^\frac{x-a}{b}}^2} dx

But since I can't read it in the proper form I don't know if it makes sense or not.

If it makes no sense, the second part basically requires me to integrate the original equation (the second one that kokosai-san suggested, and solved) but this time with e^xt multiplied to the function within the integral. The new equation is supposed to be called a "moment generating function". I tried integration by parts, and then tried to integrate the second part of the answer, but everything cancelled out...sigh. Help me out, guys!

Perhaps you should read the guidelines of each forum [u]before[/u] posting because there is another forum for calculus questions.

Here is the link to the appropriate forum [url=http://www.mathlinks.ro/Forum/index.php?f=296]CALCULUS FORUM[/url]

[EDIT]You don't need to install $\LaTeX$ in order to put formulas in here. This forum has a compiler, all you need to do is put dollar signs and remember to enable BBCode


Hi.

Does anyone have a proof of the following:
letting f_n denote the nth Fermat number (of the form 2^(2^n)+1), prove all factors of f_n take the form 2^(n+2)*k+1, where k is an interger.

vinoth

Assume $p$ is a prime which divides $F_n=2^{2^n}+1$. Take $k$ to be the order of $2$ in $Z_p$. $2^{2^{n+1}}=1 (mod\ p)\Rightarrow k|2^{n+1}$, so $k$ is a power of $2$. However, $2^{2^n}=-1(mod\ p)$, so $k\geq 2^{n+1}\Rightarrow k=2^{n+1}$.

At the same time $2^{p-1}=1(mod\ p)\Rightarrow 2^{n+1}|p-1$. We check the fact that $2^{n+2}|p-1$ for $n\leq 2$, and for $n\geq 3$ we have $p=8t+1$, so $(\frac 2p)=1$, so there exists $a$ s.t. $a^2=2(mod\ p)$, so the order of $a$ must be $2\cdot 2^{n+1}=2^{n+2}\Rightarrow 2^{n+1}|p-1$, and we're done.


integration
function
calculus
logarithms
trigonometry
real analysis
real analysis unsolved

if you know something about Digamma function http://en.wikipedia.org/wiki/Digamma_function then this problem is really easy.
expand your integral as the serie and  you will get :
$ \int_{0}^{1}{\frac {x^{n}}{x^{m} \plus{} 1}} \equal{} \frac {1}{m} \sum_{i \equal{} 0}^{\infty} \frac {( \minus{} 1)^{j}}{j \plus{} n/m} \equal{} calculation \equal{} \frac {1}{2m}\sum_{j \equal{} 0}^{\infty}\left(\frac {1}{j \plus{} n/2m} \minus{} \frac {1}{1 \plus{} j} \plus{} \frac {1}{1 \plus{} j} \minus{} \frac {1}{j \plus{} (m \plus{} n)/2m }\right) \equal{} \frac {1}{2m} \left( \psi \left( \frac {n \plus{} m}{2m}\right) \minus{} \psi \left( \frac {n}{2m} \right) \right)$
where
\[ \psi\left(\frac {p}{q}\right) \equal{} \minus{} \gamma \minus{} \frac {\pi}{2}\cot\frac {\pi p}{q} \minus{} \ln q \plus{} \sum_{n \equal{} 1}^{q \minus{} 1}\cos\frac {2\pi n p}{q}\ln\left(2\sin\frac {\pi n}{q}\right)
\]

Or another way: it's true by the integral representation of Digamma function. you just need to write difference of Digamma functions..


inequalities
search
inequalities unsolved

From the Bulkan inequality we have that
$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq\frac{3}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$ (1)

Put in this where a the b and where b the a and we have
$\frac{1}{b(1+a)}+\frac{1}{a(1+c)}+\frac{1}{c(1+b)}\geq \frac{3}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$

Now by the rearrengement
$\frac{1}{a(1+a)}+\frac{1}{b(1+b)}+\frac{1}{c(1+c)}\geq\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \frac{3}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$

Summing these three we have
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c})\geq \frac{9}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$

which is stronger than yours...... :wink:

what is Bulkan Inequality anyway? I've search thru the forum and just found this topic itself.

This is a well know inequality! But I don't sure that is Balkan ineq!

Well I also wonder what is bulkan inequality

I realized similar one which is in balkan island 2006 Q1.

But how can you prove the stronger one,
which is

1/a(1+b) + 1/b(1+c) +1/c(1+a) >= 3/   (abc)^1/3 (  1+ (abc)^1/3 )

The Balkan inequality is \[\sum\frac{1}{a(b+1)}\geq\frac{3}{1+abc}.\]

[quote="Suh Joon-Choi."]:D silouan,  please prove the stronger one you mentioned.

 :lol:[/quote]

Let $abc=t^{3}, a=t\frac{q}{p}, b=t\frac{r}{q}, c=t\frac{p}{r}$. Then the original is equivalent to $\sum_{cyc}\frac{p}{t(q+tr)}=\frac{1}{t}\frac{p}{q+tr}$.

By CS, $\sum\frac{p}{q+tr}\sum p(q+tr) \geq (p+q+r)^{2}$
$\iff \sum\frac{p}{q+r}\geq\frac{(p+q+r)^{2}}{(t+1)(pq+qr+rp)}$. So $\sum_{cyc}\frac{p}{t(q+tr)}=\frac{1}{t}\frac{p}{q+tr}\geq\frac{(p+q+r)^{2}}{t(1+t)(pq+qr+rp)}\geq\frac{1}{t(1+t)}$, and the proof ends up.

[quote="dyd"]Wew, first time I saw your "stronger" inequality, I was thinking you had proved it using $f(x,y)=\frac{1}{e^{x}(1+e^{y})}$ then Jensen or blah3 :blush:[/quote]

Unfortunately, it isn't convex.


calculus
integration
real analysis
real analysis unsolved

Use Green's theorem to prove the change of variables formula for a double integral where f(x,y)=1

double integral of 1 over S dxdy = double integral of |d(x,y)/d(u,v)| over T dudv.
Where S is a regular region in the xy plane with piecewise smooth boundary dS that corresponds to the region T in the uv-plane under the one to one, C2 transformation given by x=x(u,v), y=y(u,v).

See [url=http://books.google.com/books?id=Emra0A_QDWoC&pg=PA474&lpg=PA474&dq=green's+theorem+change+of+variables&source=bl&ots=GNxYI-ohAa&sig=Fwzxoj_iqxAmiwH3awI3GJnuyAQ&hl=en&ei=DEa6SbKoLMH7tgfs2ZnEDQ&sa=X&oi=book_result&resnum=6&ct=result#PPA475,M1]here[/url], starting on the bottom of page 474.  There are some pages missing but it seems pretty good.


Given a square with side length 2. Let $ M$ be the mid point of $ AB$. For a point $ P$ on $ CM$, except, end points, find the minimum area of the sum of two triangles formed by broken line $ APMDA$.

Let $ d$ will be the distance from $ P$ to $ AB$. Then $ S(d)\equal{}S_{ADP}\plus{}S_{APM}\equal{}2\cdot\frac{1}{2}\cdot(1\plus{}\frac{d}{2})\plus{}1\cdot\frac{1}{2}\cdot d \equal{} 1\plus{}d$. Obviously minimum area will be when $ d\equal{}0$, but then $ P\equal{}M$.

[quote="Yuriy Solovyov"]Let $ d$ will be the distance from $ P$ to $ AB$. Then $ S(d) \equal{} S_{ADP} \plus{} S_{APM}$.[/quote]

That's not what I meant.

Hi Kunny,
Do you mean if we let I be the intersection of AP and DM then we have to minimize $ S_{AID}\plus{}S_{MIP}$


modular arithmetic
number theory
greatest common divisor

1. What is the average of 1/8 and 1/10?
2. Suppose today is Tuesday. What day of the week will it be 100 days from now?
3. A motorist made a 60-mile trip averaging 20 miles per hour. On the return trip, he averaged 30 miles per hour. What was the motorist’s average speed for the entire trip?
4. The four digit numeral 3AA1 is divisible by 9. What digit does A represent?
5. A set of marbles can be divided in equal shares among 2, 3, 4, 5, or 6 children with no marbles left over. What is the least number of marbles that the set could have?

[hide="Number 1"]$ \frac{\frac18 \plus{} \frac1{10}}{2} \equal{} \boxed{\frac9{80}}$[/hide]

[hide="Number 2"]$ 100 \equiv 2\pmod{7} \implies \boxed{\text{Thursday}}$[/hide]

[hide="Number 3"]$ \frac{2}{\frac1{a} \plus{} \frac1{b}} \implies \frac{2ab}{a\plus{}b} \implies \frac{2 \times 20 \times 30}{50} \implies \boxed{24}$[/hide]

[hide="Number 4"]We have either:

$ 3\plus{}A\plus{}A\plus{}1 \equal{} 9$

or

$ 3\plus{}A\plus{}A\plus{}1\equal{}18$

The first equation leaves a decimal, but the second leaves $ \boxed{7}$[/hide]

[hide="Number 5"]$ \text{gcd}(2,3,4,5,6) \equal{} \boxed{60}$[/hide]

It's supposed to be 2 over all the stuff, not 1.


Let $M$ be the midpoint of side $BC$ of a triangle $ABC$. Let the median $AM$ intersect the incircle of $ABC$ at $K$ and $L,K$ being nearer to $A$
than $L$. If $AK = KL = LM$, prove that the sides of triangle $ABC$ are in the ratio $5 : 10 : 13$ in some order.

Hint:Notice that AI=IM and BI bisects angle B. It's easy to prove that ABI and BIM are congruent.
Then, suppose AB=x+y, BC=2x+2y, AC=3x+y, AK=$\frac{x}{\sqrt{2}}$
By using $AB^2+AC^2=\frac{BC^2}{2}+2AM^2$
We get the proof completed.□

[quote="Rushil"]I didn't quite understand your solution. Can you explain a bit!!![/quote]
Construct $IQ \perp AM$, then $KQ=LQ$. So $AQ=AK+KQ=LQ+LM=QM$. Hence $\triangle{QAM}$ is isoscele. Therefore, $\triangle{BIA} \cong \triangle{BIM}$. Is that clear now?

Apologoies for being a bit slow, (and for bringing up an old topic), but I don't understand this step:

[quote]Then, suppose AB=x+y, BC=2x+2y, AC=3x+y, AK=$\frac{x}{\sqrt{2}}$ [/quote]

Why do we know that there are $x$ and $y$ that fit these equations?

[color=darkblue] [b]Remark.[/b] The proposed problem at INMO - 2005 appeared in G.M.B. (Romania) under the [b]Marcel Tena[/b]'s signature.

[b]Proof.[/b] Suppose w.l.o.g. that $b>c\ .$ The incircle $w$ of the triangle $ABC$ touches the sides $AB$, $BC$ in the points $F$, $D$.
Denote $AK=KL=LM=x$ and the power $p_{w}(X)$ of the point $X$ w.r.t. the circle $w\ .$ We know that $AF=p-a$ and $2\cdot MD=b-c$. Therefore,
$p_{w}(A)=x\cdot 2x=AK\cdot AL=AF^{2}$ and $p_{w}(M)=x\cdot 2x=ML\cdot MK=MD^{2}$ $\Longrightarrow$ $2x\sqrt 2=2(p-a)=b-c$ $\Longrightarrow$ $a=2c\ \ (1)\ .$
$m_{a}=3x$ $\Longrightarrow$ $4m^{2}_{a}=36x^{2}$ $\Longrightarrow$ $2(b^{2}+c^{2})-(2c)^{2}=36\cdot \frac{(b-c)^{2}}{8}$ $\Longrightarrow$ $4(b+c)=9(b-c)$ $\Longrightarrow$ $13c=5b\ \ (2)\ .$
The relations $(1)$ and $(2)$ $\Longrightarrow$ $\boxed{\ \frac{a}{10}=\frac{b}{13}=\frac{c}{5}\ }\ .$

[b]An extension.[/b] Let $M$ be the point of the side $BC$  of a triangle $ABC$ so that the incircle cuts the segment $[AM]$ in three equal segments. Then,

$\boxed{\ \frac{MB}{MC}=m\ne\frac{c}{b}\Longrightarrow \{\begin{array}{ccc}m<\frac{c}{b}& \Longrightarrow & \frac{a}{(m+1)(9m+1)}=\frac{b}{9m+1}=\frac{c}{m(9m+17)}\\\ m>\frac{c}{b}& \Longrightarrow & \frac{a}{(m+1)(m+9)}=\frac{b}{17m+9}=\frac{c}{m(m+9)}\end{array}\ }\ .$[/color]

This problem has been discussed before on another link.

[quote=Moonmathpi496][quote="Rushil"]Let $ M$ be the midpoint of side $ BC$ of a triangle $ ABC$. Let the median $ AM$ intersect the incircle of $ ABC$ at $ K$ and $ L,K$ being nearer to $ A$
than $ L$. If $ AK \equal{} KL \equal{} LM$, prove that the sides of triangle $ ABC$ are in the ratio $ 5 : 10 : 13$ in some order.[/quote]
We shall prove that $ c: a: b\equal{}5: 10: 13$.
Here $ IK\equal{}IL\equal{}r$, $ \angle IKA\equal{}\angle ILM$, and $ AK\equal{}LM$. So $ \triangle IKA \cong \triangle ILM$.
So $ IA\equal{}IM$. So, $ \triangle IEA \cong \triangle IDM$.
These yields $ AB\equal{}BM\equal{}\frac{BC}{2}$.
We have $ KL^2\equal{}\frac{2b^2\plus{}2c^2\minus{}a^2}{9\cdot 4}\equal{}\frac{b^2\minus{}c^2}{9\cdot 2}$
Now we calculate the power of $ M$ wrt the incircle.
We get\[ AK \cdot AL\equal{}AE^2 \iff 2KL^2\equal{}(s\minus{}a)^2 \iff \frac{b^2\minus{}c^2}{9}\equal{}\frac{(b\minus{}c)^2}{4} \iff 5b\equal{}13c\]
And we are done!  :)[/quote]

Use power of point to get $MD=AE$,we have $BE=BF$ this implies $AB=BM or 2AB=BC$, for simplicity take $AB=1$,use $ME=CE-CM=[(2)^{1/2}/3]×[1/2]×(2b^2+2c^2-a^2)^{1/2}$  where b and c are for their usual notations simplify to get $b=AC=13/5$ now put them in ratio to get$5:10:13$ and hence done :cool:

[quote=OlympusHero]Hello, can I please have a hint here? Thanks![/quote]

[hide=hint]Prove first that $AB=BM$,then use pop to find AK and lastly cosine law on B w.r.t two triangles which you figure out.

[hide=power of point praktise]denote $X,Y,Z$ as where the incircle touches the sides $BC=b, AB=a, AC=c$ respectively.\\
we have $AY=(s-a)$ and $XM=\frac{b-c}{2}$ where $s=a+b+c$\\
from pop we have $2AK^2=(s-a)^2 \implies AK=\frac{s-a}{\sqrt{2}}$\\
also from pop $2ML^2=\frac{(b-c)^2}{4}$ which gives $s-a=\frac{b-c}{2}\implies 2c=a$\\
now since $AM$ is median we have $AM^2=\frac{2(b^2+c^2)-a^2}{4}$ but from above relation we have $AM^2=\frac{9(s-a)^2}{2}$\\
equating we get $\frac{9}{4}\cdot (b-c)=b+c$\\
so we get $5b=13c\implies$ $\boxed{\ \frac{a}{10}=\frac{b}{13}=\frac{c}{5}\ }\ .$ $\blacksquare$[/hide]


trigonometry
exterior angle
geometry proposed
geometry

Two circles C and C' have centers O and O' and meet at M and N. The common tangent closer to M touches C at A and C' at B. The line through B perpendicular to AM meets the line OO' at D. BO'B' is a diameter of C'. Show that M, D and B' are collinear.

 :oops:  :oops:

[quote="minhtuan_lequydon_dn"][quote="kaka__2004"]how to do it?
anybody can?[/quote]
The problem is very nice but my solution so long and bad  :(  :([/quote]
well, could you post it?

The common external tangent $ AB$ cuts the center line $ OO'$ at the external similarity center $ E$ of the circles $ \mathcal C, \mathcal C'.$ $ BD, EM$ cut the circle $ \mathcal C'$ again at $ U, V,$ respectively. $ BV \parallel AM$ by similarity of $ \mathcal C' \sim \mathcal C$ $ \Longrightarrow$ $ (BD \equiv BU) \perp BV$ $ \Longrightarrow$ $ UV$ is a diameter of $ \mathcal C'.$ By Pascal theorem for degenerated cyclic hexagon $ BBMUB'V,$ the intersections $ O' \equiv BB' \cap UV,$ $ E \equiv BB \cap MV,$ $ D \equiv BU \cap MB'$ are collinear $ \Longrightarrow$ $ M, D, B'$ are collinear.

$ MB'\cap OO'\equal{}D'$. We will prove that $ AM \perp BD'$

Let $ m(BAM)\equal{}\alpha , m(ABM)\equal{} \beta , m(D'BB')\equal{} \theta , m(O'D'B')\equal{} \gamma$. We will prove that $ \alpha\equal{}\theta$

By sines theorem; $ \frac{sin\beta}{sin\theta}\equal{}\frac{sin\gamma}{sin(\beta\plus{}\theta\plus{}\gamma)} \Longrightarrow$

$ cot\gamma\equal{}\frac{sin\theta\minus{}sin\beta.cos(\beta\plus{}\theta)}{sin\beta.sin(\beta\plus{}\theta)}$

Let $ |OM|\equal{}x$ and $ |O'M|\equal{}y$ so $ |AM|\equal{}2ysin\alpha$ and $ |BM|\equal{}2xsin\beta$.

$ \frac{|AM|}{|BM|}\equal{}\frac{sin\beta}{sin\alpha} \Longrightarrow$

$ \frac{x}{y}\equal{}\frac{sin^2{\alpha}}{sin^2{\beta}}$. In the other hand $ \frac{x}{y}\equal{}\frac{sin(2\alpha\plus{}\beta\plus{}\gamma)}{sin(\gamma\minus{}\beta)} \Longrightarrow$

$ \frac{sin(2\alpha\plus{}\beta).cot\gamma\plus{}cos(2\alpha\plus{}\beta)}{cos\beta\minus{}sin\beta.cot\gamma}\equal{}\frac{sin^2{\alpha}}{sin^2{\beta}} \Longrightarrow$

$ cot\gamma\equal{}\frac{sin^2{\alpha}.cos\beta\minus{}sin^2{\beta}.cos(2\alpha\plus{}\beta)}{sin\beta.[sin(2\alpha\plus{}\beta)sin\beta\plus{}sin^2{\alpha}]}\equal{}\frac{sin\theta\minus{}sin\beta.cos(\beta\plus{}\theta)}{sin\beta.sin(\beta\plus{}\theta)}$

Because of $ \frac{sin^2{\alpha}.cos\beta\minus{}sin^2{\beta}.cos(2\alpha\plus{}\beta)}{sin\beta.[sin(2\alpha\plus{}\beta)sin\beta\plus{}sin^2{\alpha}]}\equal{}\frac{sin\alpha\minus{}sin\beta.cos(\beta\plus{}\alpha)}{sin\beta.sin(\beta\plus{}\alpha)}$

$ \frac{sin\alpha\minus{}sin\beta.cos(\beta\plus{}\alpha)}{sin\beta.sin(\beta\plus{}\alpha)}\equal{}\frac{sin\theta\minus{}sin\beta.cos(\beta\plus{}\theta)}{sin\beta.sin(\beta\plus{}\theta)} \Longrightarrow$

$ sin\alpha.sin(\beta\plus{}\theta)\minus{}sin\theta.sin(\alpha\plus{}\beta)\equal{}sin\beta.sin(\theta\minus{}\alpha)$

$ sin\beta.sin(\theta\minus{}\alpha)\equal{}sin\alpha.sin(\beta\plus{}\theta)\minus{}sin\theta.sin(\alpha\plus{}\beta)\equal{}sin\beta.sin(\alpha\minus{}\theta) \Longrightarrow$

$ sin(\alpha\minus{}\theta)\equal{}0 \Longrightarrow$

$ \alpha\equal{}\theta$ and solution is done  :wink:

[quote="crazyfehmy"]$ MB'\cap OO' \equal{} D'$. We will prove that $ AM \perp BD'$

Let $ m(BAM) \equal{} \alpha , m(ABM) \equal{} \beta , m(D'BB') \equal{} \theta , m(O'D'B') \equal{} \gamma$. We will prove that $ \alpha \equal{} \theta$

By sines theorem; $ \frac {sin\beta}{sin\theta} \equal{} \frac {sin\gamma}{sin(\beta \plus{} \theta \plus{} \gamma)} \Longrightarrow$

$ cot\gamma \equal{} \frac {sin\theta \minus{} sin\beta.cos(\beta \plus{} \theta)}{sin\beta.sin(\beta \plus{} \theta)}$

Let $ |OM| \equal{} x$ and $ |O'M| \equal{} y$ so $ |AM| \equal{} 2ysin\alpha$ and $ |BM| \equal{} 2xsin\beta$.

$ \frac {|AM|}{|BM|} \equal{} \frac {sin\beta}{sin\alpha} \Longrightarrow$

$ \frac {x}{y} \equal{} \frac {sin^2{\alpha}}{sin^2{\beta}}$. In the other hand $ \frac {x}{y} \equal{} \frac {sin(2\alpha \plus{} \beta \plus{} \gamma)}{sin(\gamma \minus{} \beta)} \Longrightarrow$

$ \frac {sin(2\alpha \plus{} \beta).cot\gamma \plus{} cos(2\alpha \plus{} \beta)}{cos\beta \minus{} sin\beta.cot\gamma} \equal{} \frac {sin^2{\alpha}}{sin^2{\beta}} \Longrightarrow$

$ cot\gamma \equal{} \frac {sin^2{\alpha}.cos\beta \minus{} sin^2{\beta}.cos(2\alpha \plus{} \beta)}{sin\beta.[sin(2\alpha \plus{} \beta)sin\beta \plus{} sin^2{\alpha}]} \equal{} \frac {sin\theta \minus{} sin\beta.cos(\beta \plus{} \theta)}{sin\beta.sin(\beta \plus{} \theta)}$

Because of $ \frac {sin^2{\alpha}.cos\beta \minus{} sin^2{\beta}.cos(2\alpha \plus{} \beta)}{sin\beta.[sin(2\alpha \plus{} \beta)sin\beta \plus{} sin^2{\alpha}]} \equal{} \frac {sin\alpha \minus{} sin\beta.cos(\beta \plus{} \alpha)}{sin\beta.sin(\beta \plus{} \alpha)}$

$ \frac {sin\alpha \minus{} sin\beta.cos(\beta \plus{} \alpha)}{sin\beta.sin(\beta \plus{} \alpha)} \equal{} \frac {sin\theta \minus{} sin\beta.cos(\beta \plus{} \theta)}{sin\beta.sin(\beta \plus{} \theta)} \Longrightarrow$

$ sin\alpha.sin(\beta \plus{} \theta) \minus{} sin\theta.sin(\alpha \plus{} \beta) \equal{} sin\beta.sin(\theta \minus{} \alpha)$

$ sin\beta.sin(\theta \minus{} \alpha) \equal{} sin\alpha.sin(\beta \plus{} \theta) \minus{} sin\theta.sin(\alpha \plus{} \beta) \equal{} sin\beta.sin(\alpha \minus{} \theta) \Longrightarrow$

$ sin(\alpha \minus{} \theta) \equal{} 0 \Longrightarrow$

$ \alpha \equal{} \theta$ and solution is done  :wink:[/quote] this very nice but so long

[quote="PeterScales"]angleDBB' = angle MAB
angleDB'B = angleMBA
angle BDM = angleMAB + angleMBA = exterior angle of triangleBDB'
.: B', D and M are collinear.

Regards, Peter Scales.[/quote]
Your solution don't right  :maybe:

First of all it's sufficient to prove that $ \triangle MAB \sim \triangle DBB'$.
---We can evaluate(easy) that $ \angle MAB = \angle DBB'$.
---If $ P = MN\cap AB$ then $ PA^2 = PM.PN = PB^2$
--- $ \angle APM = 180^\circ - \angle AO_1O_2$ ($ APO_1Q$ is cyclic where $ Q = O_1O_2\cap MN$ )= $ \angle BO_2D$ ($ AO_1\parallel{}BO_2$).
---$ \triangle MAP \sim \triangle DBO_2$
--- $ PA = PB$, $ BO_2 = B'O_2$ and $ \triangle MAP \sim \triangle DBO_2$ $ \Longrightarrow$ $ \triangle MAB \sim \triangle DBB'$
So the problem is easy!

[quote="Flakky"]First of all it's sufficient to prove that $ \triangle MAB \sim \triangle DBB'$.
---We can evaluate(easy) that $ \angle MAB = \angle DBB'$.
---If $ P = MN\cap AB$ then $ PA^2 = PM.PN = PB^2$
--- $ \angle APM = 180^\circ - \angle AO_1O_2$ ($ APO_1Q$ is cyclic where $ Q = O_1O_2\cap MN$ )= $ \angle BO_2D$ ($ AO_1\parallel{}BO_2$).
---$ \triangle MAP \sim \triangle DBO_2$
--- $ PA = PB$, $ BO_2 = B'O_2$ and $ \triangle MAP \sim \triangle DBO_2$ $ \Longrightarrow$ $ \triangle MAB \sim \triangle DBB'$
So the problem it's easy![/quote]
Yeah  :P  :P


function
inequalities proposed
inequalities

Set $ p\equal{}x\plus{}y\plus{}z$, $ q\equal{}xy\plus{}yz\plus{}zx$, $ r\equal{}xyz$, $ a\equal{}\frac{x^3(y\plus{}z)(x(y\plus{}z)\plus{}4yz)}{q^2}$ and $ b,c$ similarly. We prove this instead, which is a bit more general: if $ f: \mathbb R^\plus{}\rightarrow\mathbb R$ is a concave and increasing function then

$ \sum f(a)\leq \sum f\left(x^2\plus{}\frac{r}{p}\right).$

Assume that $ x\ge y\ge z$. Then $ a\ge b\ge c$, because for example $ q(a\minus{}b)\equal{}(x\minus{}y)(xy(x\plus{}y)\plus{}z(x^2\plus{}y^2)\plus{}4xyz)\ge0$. And, of course, $ x^2\plus{}\frac{r}{p} \ge y^2\plus{}\frac{r}{p} \ge z^2\plus{}\frac{r}{p}$. So we can apply Karamata-Abel theorem and the only thing that we have to prove is

$ a\geq x^2\plus{}\frac{r}{p}$, which, after expanding, is $ 2xy(x^2\minus{}z^2)\plus{}2xz(x^2\minus{}y^2)\plus{}xy^2(x\minus{}z)\plus{}xz^2(x\minus{}y)\plus{}yz(x^2\minus{}yz)\ge0$;

$ a\plus{}b\geq x^2\plus{}y^2\plus{}\frac{2r}p$, which, after expanding, is $ x^2y(x\minus{}z)\plus{}xy^2(y\minus{}z)\plus{}z(x^3\plus{}y^3\plus{}\minus{}2xyz)\plus{}xy(xy\minus{}z^2)\ge0$, and

$ a\plus{}b\plus{}c\leq x^2\plus{}y^2\plus{}z^2\plus{}\frac{3r}p$, which, after expanding, is $ \sum_{cyc}x(x\plus{}2y\plus{}2z)(y\minus{}z)^2\ge0$.

That's it.


search
ARML
AMC
AMC 12
AMC 12 B
conics
function

I just got Geometer Sketchpad 3.0 from my school today, but I'm having a hard time using it.  I didn't get an instruction manual, and the help feature isn't being very helpful.  Can anyone give me some pointers?

For example, Is there a way to select points and have a circle touch all the points?  I know you can do an arc with 3, but an arc isn't enough!

No, there is no way to do that.  You could do two different arcs through the points, but otherwise you have to construct the center (using perpendicular bisectors) and draw a circle like that.

[quote="Ragingg"]For example, Is there a way to select points and have a circle touch all the points?  I know you can do an arc with 3, but an arc isn't enough![/quote]

This isn't a Round Table question; it really belongs on the Other Problem Solving Topics Forum or somewhere like that.

My son A+MATH saw your question, opened up Geometer's Sketchpad in another window, and found out how to draw a circle including any arbitrary three points. Select all three points. From the construct pull-down menu select Arc through three points. Then unselect the constructed arc, select the three points again, use construct arc through three points AGAIN, and you will have your circle. Unselect all the arcs, and then you drag on any of the points and the constructed circle will still intersect all three of them.

Hope this helps!

Please think ahead about what forum to post to before posting a question.

Regarding tokenadult's post on the incorrect forum, I still stand by my decision to put this in the Round Table.  It says "for serious discussion," and we are discussing Geometer's Sketchpad.  This really has nothing to do with problem solving, but how to use a mathematical program.

Hmm... The method isn't working out for me.  It gets hazy around select the 3 points again. Can you please reclarify?

Select them in a different order.  If you selected ABC before, select BAC now.  Then make the arc again.

[quote="Ragingg"]Is there any way I can draw an angle bisector, but get it to be a segment instead of a ray?[/quote]

If you want a segment instead of a ray you can draw the ray, create the point on the ray where you want the segment to end, hide the ray, then draw the segment from the angle to that point.

One last question... Is there any way I can draw in nice right angle signs?  You know... the box.

i don't get it how can you install sschool software on your own computer?

Not legally.

Anyhow, the best way to do right angle signs is messy -- create an angle bisector, drop perpendiculars, and then hide everything and put in segments.  On version 4, everything gets nicer because you can save tools to remember how to do annoying tasks like that, but I don't know if you can do that on version 3.  I don't think you can.

I bought it through the school.. They said they could get a discount.

Oh, maybe Key Curriculum Press cuts deals with schools and groups of students, I don't know.  Could be.

The old version (3.0) may be available to schools from students after the schools upgrade to the current version (4.0), depending on how the license terms define rights for disposing of old copies.

By far the best way to learn Geometer's Sketchpad is to simply play around with it. Another place to ask questions is the

[url=news:geometry.software.dynamic]geometry.software.dynamic[/url]

newsgroup on Usenet. All the developers of those kinds of software hang out on the newsgroup.

I have version 3 -- they do have macros for that sort of thing.  I've never really used them though because they didn't seem worth the hassle of setting up.

[quote="JBL"]On version 4, everything gets nicer because you can save tools to remember how to do annoying tasks like that, but I don't know if you can do that on version 3.  I don't think you can.[/quote]

macros are awesome they make typing alot of stuff in Visual Basic shorter.

I used to like vB, but after taking a semester-long programming class that turned out to be pure ****, I changed my mind  ":pi: radians".  (Ok not that funny though.)

I don't like VB. At all. Seriously.

C++ / (eh... maybe Java too) all the way!

problem with Java: NO POINTERS TO POINTERS ! :(

pointer = a variable that stores the memory address of another variable.

pointer to pointer = a variable that stores the memory address of a pointer.

C++ has them. Java doesn't.

I have used them often so don't say they sound silly. Most people think pointers sound silly in the first place, until they learn data structures. Try implementing a binary search tree without pointers or references :).

There was a person on my ARML team in 2003 who made a calculator program to solve #4 on the ARML power round. This person also said that he got 150 on the AMC12B in 2003 just because of his calculator.

My friend actually used his calculator to calculate humongous coin flippings to get an answer :)

For anyone who knows VB: is there anyway to make connect two remote computers that are not in a LAN using winsock or other components? :?

Hi, i have question about the sketchpad:
How can i draw the Conic sections and inverse function??

What I suggest is that you use a tool in GSP, or it would be rather difficult.
[url=http://64.233.187.104/search?q=cache:t7LThWJJMdAJ:www.mste.uiuc.edu/courses/ci336sp04/folders/jlin/+ellipse.gsp&hl=en]here[/url]
This is a great site for tools
 :D

[i]Note: Edited by moderator to make it fit on the screen.  Please use the url tags to make long web addresses not look long.  Or, use [url=http://tinyurl.com]the tinyurl webtool[/url].[/i]

I believe this is a good site for GSP tools, too, as long as you understand Chinese
[url=http://64.233.187.104/search?q=cache:wO7_aXJMbxMJ:train.ilc.edu.tw/infointea/%E6%95%B8%E5%AD%B8/ljchen/mytools/gsp3.1/%E7%AF%84%E4%BE%8B/+ellipse.gsp&hl=en]here[/url]

[size=125]
By the way, if you are still having trouble for elipse here's a solution
1.) Construct a circle
2.) Place a point inside the circle
3.) Create a point on the circle
4.) Create a segment between the point in the circle and the point on the circle
5.) Create the midpoint of that segment
6.) Use the midpoint and line and create a perpendicular
7.) Animate the point on the circle (not the one that controls the size of the circle[/size] ;)

[quote="WarpedKlown1335"]I wonder if anyone ever makes a calculator program to solve a math problem on an AMC for example.  Because I did that for #25 on 12B :-D.[/quote]

I do that all the time for problems I am too lazy to solve by hand.  (e.g. 'Find all x such that [ . . .] is true, if x<100')

Is there a way to limit the amount of lines in a locus?  :?

You're doing the locus of a curve?  I'm not sure.  You could create a series of points on the curve and construct their individual loci.

[quote="JBL"]You can use the "locus" command to create conic sections.  For inverse functions, can you reflect the graph of a function?[/quote]

easy, construct a point on the function, its reflection across a line constructed on the function f(x)=x (it needs to be constructed because the program doesn't recognize the function itself as a line), and the locus of the the point on the function and its reflection

Can anyone please show me how to sketch the picture of the problem at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=48943]here[/url]?

That is, how to draw a big circle tangant to some small circles?

shobber, did you buy your sketchpad?

i have a demo version and i can't use the locus properly; for example, how can i draw the locus of a midpoint when another point moves on a given line or circle?

another question: you have a circle + a point; how do you draw the tangent to the circle from the point?

please help me....

[quote="shobber"]Construct a segment with center and point, then contruct perpendicular line from the point to the radius. then you get the tangent.[/quote]

but we can't take any radius, can we?

please explain some more. i don't understand  :(

[edit:]

[quote]nevermind. i finally got it :)

you construct the circle with diameter OA (center of circle - the point given)

this may be exactly what shobber said, but didn't understand his formulation.[/quote]

can anyone explain me how to use the locus ?
 :?:  :?  :blush:  :?:


Solve the Inequality |x-2| <= 2(x)^2 – 9x + 9
a.	X> (4 – (2)^(1/2))/2
b.	X< (5 + (3)^(1/2))/2
c.	Both a & b
d.	None of these

I got the solution as a.... the given solution is c....

now here is how i tested for b....

(5 + (3)^(1/2))/2 = 5+ 1.7 / 2 = 3.35

so testing for 2.... | 0 | < = 8 - 18 + 9

am i missing something???? and is | 0 | 0 or 1????

More help.... inequalities seems to be one area of trouble for me... any suggestions how i can strengthen my skills in this area.. i'm just scared of them....(now i have nearly 2 pages of inequalities with sqr roots and stuff!)....i badly want to master this area...seems challenging.... but not sure how to...so if anybody had felt like me and did master it...how did u do it???

The question is very poorly stated.  Instead of doing the question as written, solve the following question:

Find all values of $x$ such that $|x-2| \leq 2x^{2}-9x+9$.

There are two tricky parts to this question: the absolute value and the quadratic.  First, you should deal with the absolute value.  Then, you should deal with the quadratic.

[quote="fizsle"]Solve the Inequality $|x-2| \le 2x^{2}-9x+9$
a.	$x> \frac{4-\sqrt2}{2}$
b.	$x< \frac{5-\sqrt2}{2}$
c.	Both a & b
d.	None of these

I got the solution as a.... the given solution is c....

now here is how i tested for b....

(5 + (3)^(1/2))/2 = 5+ 1.7 / 2 = 3.35

so testing for 2.... | 0 | < = 8 - 18 + 9

am i missing something???? and is | 0 | 0 or 1????

More help.... inequalities seems to be one area of trouble for me... any suggestions how i can strengthen my skills in this area.. i'm just scared of them....(now i have nearly 2 pages of inequalities with sqr roots and stuff!)....i badly want to master this area...seems challenging.... but not sure how to...so if anybody had felt like me and did master it...how did u do it???[/quote]

How did you get $|0|$?

When you have something like $|x|\le y$, it can be written as $-y\le x\le y$.

So in this case, $|x-2| \le 2x^{2}-9x+9$ becomes $-(2x^{2}-9x+9)\le x-2\le 2x^{2}-9x+9$ then solve both inequalities separately. :)

[quote="lotrgreengrapes7926"]When you have something like $|x|\le y$, it can be written as $-y\le x\le y$.

So in this case, $|x-2| \le 2x^{2}-9x+9$ becomes $-(2x^{2}-9x+9)\le x-2\le 2x^{2}-9x+9$ then solve both inequalities separately. :)[/quote]

You've got to make sure that you'll never have $2x^{2}-9x+9 < 0$, i.e. $x\notin\left]\frac{3}{2},2\right[$.
Anyway, nice method.

hey lotrgreengrapes,

Irrespective of the method used to solve...shouldn't the values satisfy the given inequality??? :maybe:

You aren't thinking hard enough about what the question is asking.  An inequality is a restriction.  If I say $x > a$ [i]and[/i] $x < b$, that means that $x$ must satisfy [i]both restrictions at the same time[/i].  Your choice to plug in (2) satisfies the second but not the first.

If you really want to do this question without actually doing it (i.e. by trying to out-smart the multiple-choice) you would first have to check how those numbers (in answers (a) and (b)) compare to each other.  After determining that the number in answer (a) was smaller, you'd have to check 1 number less than the number in answer a, one number between the two numbers, and one number larger than the number in answer b.

It's not clear to me what you think you can accomplish by this method when one of the answer choices is "none of the above," though.  You should just do the problem honestly rather than trying to game it.

As a side note: why on earth would you suppose that $|0| = 1$?

use interval testing...that makes all of these problems trivial