All right. Thanks in advance to anyone who reads this:

My summer is already quite busy, what with me taking a pre-calculus course online, playing golf, and doing a few camps as well. In addition, I was already planning to take an AMC12 problem solving course here at AOPS. However, I just noticed that there is also an AIME course now too.

A brief intro to my mathematical ability:
Well, this year, I didn't even qualify for AIME, but I took both tests for practice (and they were my first ones too). I got a 3 on the first and a 2 on the second, but I feel that if I took another AIME now, I would get at least a 5. But I'm not very confident in AMCing ability. I don't like answering questions fast and my average on the AMC10's have been 123 and AMC12 are 105.

I guess I have a few questions. Just how much of the AIME problem series will help on AMC12 questions, and vice-versa? I'm not sure if I can take both, but if I can, do you think it'll be a good idea? I'm doing a lot of things, but I do love math, and I really want to get better at it (read my sig).

Any help would be appreciated. THANKS!

If you can really score a 5, take the AIME course. It will help you with higher level problems which will help you on the last few AMC 12 problems. Then to get better at speed, take a lot of timed tests. You should be fine.

Also, if I'm correct, what you learn from doing the problems in the AIME class could probably also help on the last few questions on the AMC 10/12. As you practice math problems more, your speed will naturally increase. (At least that's what happened to me.)

That's true because you see [b]how[/b] to arrive at your answer immediately (or at least faster) with more experience.

The AMC class will help you with the easier problems on the AIME, as well.  The AIME class will help you with the harder problems on the AMC class.  There's also the option of taking both...

I have a question about the AIME class. It seems as if each week is assigned to a specific subject. Are there ever practice AIMEs or does the homework focus on just the current week's topic? Additionally, how much time commitment per week is it?

The message board problems focus on that week's topic.

You should expect to spend 3-5 hours/week on the course to get the most out of it.  (The subject classes require a greater time commitment.)

About how much overlap is there in the AIME course and the AMC12 course? I also noticed that there are (or so it appears to be) more topics covered in the AMC12 course. Is there any reason for this and should I take this into acount as well when I decide which course to take?

So, Mr. Rusczyk, if I wanted to take both of these classes concurrently during the summer, I would need to spend about 10 hours a week on it? That doesn't seem too bad, only a bit more than an hour a day, max.

There's not any overlap in actual problems, but there is some overlap in basic concepts used to attack problems.

If you are to get the most out of both classes, then you should expect to be spending 6-10 hours a week on them (collectively).

It's hard to really quantify the concepts covered in the two classes -- I would think of the main difference between the two in terms of difficulty of the problems rather than the subjects covered.


I was in 8th grade when one of my teachers gave me a problem to solve. I as usual was determined to solve the problem, so it took me about 3 years of learning new Math and thinking about the solution to get it right. I have to admit many people helped me on this one despite its very simplisitc look. Enough with time wasting let's see the problem
                 X^2 + Y^2 + Z^2 + W^2 = 2XYZW
1) solve for X,Y,Z given that W=1, and that X,Y,Z must be whole integers.
2) If the equation has an infinite number of integer solutions proof it.(This is the same as proving that it has more than 16 integer solutions). Yet if it does not have an infinite number of integer solutions proof it(this is the same as proving that it has 16 or less integer solutions).

Have fun.

Hints
The first problem can be solved with a parity argument and it is very simple if you follow through with it.
The second problem is harder ok may be harder is an understatement but it is worth trying it. You will need Mazur's theorem and it would be beneficial if you look up a similar problem in the MAA articles namely the problem is a^4+b^4+c^4+d^4=(a+b+c+d)^4 which happens to be the title of the article.


How would I state or prove the following result on an olympiad?

If f is a convex function and a <= b < c <= d, then
[tex]
\frac{f(c)-f(a)}{c-a} \leq \frac{f(d)-f(b)}{d-b}
[/tex]

Hint :

[hide]

use f(t*x + (1-t)y) <= tf(x) + (1-t)f(y)

with x=a, y=c, t=(c-b)/(c-a) to prove that :

 f(b)-f(a)/b-a <= f(c)-f(a)/c-a <= f(c)-f(b)/c-b if a< b < c.[/hide]

Some nice (I like calculus !) results/exercices about convex functions (may require some calculus) :

. f : IR+ -> IR convex and bounded, show that f is decreasing

. f : IR+ -> IR convex and derivable, show that f' is continuous

. show that f is convex iff f((x+y)/2) <= (f(x)+f(y))/2

. f : [a,b]-> IR convex, show that f is continuous on ]a,b[

I may give more if someone interested !

Try handling the following inequality

If f is a convex function and a <= b < c <= d, then
[tex]
\frac{f(c)-f(a)}{c-a} \leq \frac{f(d)-f(a)}{d-a}
\leq \frac{f(d)-f(b)}{d-b}
[/tex]

and breaking it up into two cases.


Billy and Bobbi each selected a positive integer less than 200. Billy's number is a multiple of 18, and Bobbi's number is a multiple of 24. What is the probability that they selected the same number? Express your answer as a common fraction.

[quote="MrRubix"]We need to find the chance that they both picked a common multiple of 18 and 24, less than 200.
[hide]The common multiples are 72 and 144. The chance that Billy picks either of these numbers out of his choices (18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198) is 2/11. The chance that Bobbi picks either of these numbers out of [i]her[/i] choices (24, 48, 72, 96, 120, 144, 168, 192) is 2/8, or 1/4. When we multiply these out we get 2/44, or [b]1/22[/b]
________________
Black holes are where god divided by zero. :rotfl:[/hide][/quote]

doesnt that mean if Billy pics 72, Bobby with the 2/8 choice can pick 144?

But 72 is not 144.  SO i think its 1/44 (2/11*1/8)


Find $x$ such that:$A=\frac{\sqrt{x}}{x\sqrt{x}-3\sqrt{x}+3}\in Z$

\[A=\frac{\sqrt{x}}{(x-3)\sqrt{x}+3}=\frac{x(x-3)-3\sqrt{x}}{x(x-3)^{2}-9}\]
The rest is easy but right away I see $x=1\implies A=1$ works. Do you mean find all?

For the first line of proof you mean $f=0\implies x=0$ instead of other way around, right?

Obviously $x=0$is a solution
Assume $x$ is non zero.Divide both the numerator and denominator by $\sqrt{x}$.

This gives $A = \frac{1}{x-3+(\frac{3}{\sqrt{x}})}$

Thus  $\frac{1}{x-3+(\frac{3}{\sqrt{x}})}$ divides 1.

Hence $\frac{1}{x-3+(\frac{3}{\sqrt{x}})}= 1 or-1$
Case 1: $\frac{1}{x-3+(\frac{3}{\sqrt{x}})}= 1$
Put $\sqrt{x}=t$ we get $t^{3}-4t+3 = 0$
                                   $(t-1)(t^{2}+t-3) =0$
 $t$ has to be positive and hence we get a solution as $x =1$or $x =(\frac{\sqrt{13}-1}{2})^{2}$

Case 2:$\frac{1}{x-3+(\frac{3}{\sqrt{x}})}= (-1)$
The same substitution gives - $t^{3}-2t+3 = 0$

Possibly this might give another solution (I am not able to solve this cubic...I'd be happy if any of u people solve this for me)

I now feel that there can be infinitely many solutions.

 Set $x-3+\frac{3}{\sqrt{x}}= \frac{1}{k}$ where k is any integer.
For every such arbitrary value of k we get a cubic equation which we know has at least one real root.Possibly many such equations even have atleast one positive root and we are done.


Let $a,\ b,\ c,\ d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Find the maximum value of $2a^{2}+2b^{2}+c^{2}+d^{2}+ac+bd$.

[hide]
By Cauchy, we have
\[\left( a(a+c)+b(b+d)\right)^{2}\leq (a^{2}+c^{2})((a+c)^{2}+(b+d)^{2})=(a^{2}+c^{2})(1+2(ac+bd))\]
So if we let $x=a^{2}+c^{2}$ and $y=ac+bd$, we have $(x+y)^{2}\leq x+2xy$, or $x^{2}+y^{2}\leq x$. It follows that
\[x+y\leq x+\sqrt{x-x^{2}}=\frac{1}{2}+\sqrt{x^{2}-x+\frac{1}{4}}+\sqrt{x-x^{2}}\leq \frac{1+\sqrt{2}}{2}\]
where the last part is RMS-AM.

Adding 1 (from $a^{2}+b^{2}+c^{2}+d^{2}$), we get the maximum in the original problem to be $\boxed{\frac{3+\sqrt{2}}{2}}$. There's equality when $a=c=\sqrt{\frac{2+\sqrt{2}}{8}}$ and $b=d=\sqrt{\frac{2-\sqrt{2}}{8}}$.
[/hide]

Your answer is correct and interesting to me. :)


well guys , this problem of AUSTRIA starts to make me angry !

    points D and E are chosen on AB and AC of the triangle ABC in such a way that if F is the intersection of BE and DC then : AE + EF = AD + DF
 prove that AC +CF = AB + BF

Dear Mathlinkers,
this known problem is the URQUHART's theorem.
You can see somme links with Google.
Sincerely
Jean-Louis

We had this at http://www.mathlinks.ro/viewtopic.php?t=6559 , so you don't even have to google...

  darij


[quote="mathVNpro"]Let $ a,b$ be $ 2$ distinct positive interger number such that $ (a^2 \plus{} ab \plus{} b^2)|ab(a \plus{} b)$. Prove that: $ |a \minus{} b| > \sqrt [3] {ab}$.[/quote]

$ (a^2\plus{}ab\plus{}b^2)|ab(a\plus{}b) \Leftrightarrow \begin{cases} (a^2\plus{}ab\plus{}b^2)|a(a^2\plus{}ab\plus{}b^2)\minus{}ab(a\plus{}b) \\ (a^2\plus{}ab\plus{}b^2)|b(a^2\plus{}ab\plus{}b^2)\minus{}ab(a\plus{}b)  \end{cases}$

$ \Leftrightarrow  \begin{cases} (a^2\plus{}ab\plus{}b^2)|a^3 \\ (a^2\plus{}ab\plus{}b^2)|b^3 \end{cases}$

$ \Rightarrow (a^2\plus{}ab\plus{}b^2)| \gcd(a^3,b^3)\equal{}(\gcd(a,b))^3\equal{}(\gcd(a, |a\minus{}b|))^3 \le |a\minus{}b|^3$

So $ |a\minus{}b| \ge \sqrt[3]{a^2\plus{}ab\plus{}b^2}>\sqrt[3]{ab}$

[quote="mathVNpro"]Let $ a,b$ be $ 2$ distinct positive interger number such that $ (a^2 \plus{} ab \plus{} b^2)|ab(a \plus{} b)$. Prove that: $ |a \minus{} b| > \sqrt [3] {ab}$.[/quote]
Let $ a\equal{}px,b\equal{}py$
Then we have that:
$ x^2\plus{}y^2\plus{}xy|pxy(x\plus{}y)$
But $ (x^2\plus{}y^2\plus{}xy,x)\equal{}1,(x^2\plus{}y^2\plus{}xy,y)\equal{}1,(x^2\plus{}y^2\plus{}xy,x\plus{}y)\equal{}1$
So $ x^2\plus{}y^2\plus{}xy|p$
$ p \ge x^2\plus{}y^2\plus{}xy$ $ \boxed{1}$

We have to prove that:
$ p^3|(x\minus{}y)| >p^2xy \Leftrightarrow p|(x\minus{}y)| >xy$

But $ |(x\minus{}y)| >0$,so we finally have to prove that

$ p>xy$,which follows from the $ \boxed{1}$

Another solution:

$a^{2}+ab+b^{2}|ab(a+b) \implies a^{2}+ab+b^{2}\leq ab(a+b)$
We see that equality cannot hold because that implies $a=b$ which contradicts the problem statement.

So we have
\begin{align*}
 a^{2}+ab+b^{2}&<ab(a+b) \\
\frac{a^3-b^3}{a-b}&<ab(a+b) \\
a^3-b^3&<ab(a^2-b^2) \\
a^3-3a^2b+3ab^2-b^3&<ab(a^2-b^2-3a+3b) \\
(a-b)^3&<ab(a^2-b^2-3a+3b) \\
(b-a)^3&>ab(b^2-a^2+3a-3b)  \\
b-a&>\sqrt[3]{ab(b^2-a^2+3a-3b)}
\end{align*}
WLOG let $b>a$, we have $\sqrt[3]{ab(b^2-a^2+3a-3b)}=\sqrt[3]{ab(b(b-3)-a(a-3))}>\sqrt[3]{ab}$ so $b-a=|a-b|>\sqrt[3]{ab}$.

[quote="mathVNpro"]Let $ a,b$ be $ 2$ distinct positive interger number such that $ (a^2\plus{}ab\plus{}b^2)|ab(a\plus{}b)$. Prove that: $ |a\minus{}b|>\sqrt [3] {ab}$.[/quote]

Let $d=(a,b)$ so let $a=dx$ and $b=dy$ where $(x,y)=1$
So now we have
we have
$ \frac{dxy(x+y)}{x^2+y^2+xy} \in Z $
now note that $(x^2+y^2+xy,x) =1 , (x^2+y^2+xy,y)=1$ and similarly since $(x+y,y)=1$ we have
$(x^2+y^2+xy , x+y)=(y^2,x+y) = 1 $ .
since it is a integer , we have $ x^2+xy+y^2 | d  $ so $d \ge x^2+xy+y^2 $
so
$(|a-b|)^3 = (|dx-dy|)^3 = d^3(|x-y|)^3 \ge d^2 \cdot d \ge d^2(x^2+y^2+xy) >  d^2(xy) = (dy)(dy) = ab$
So we are done $\Box$

Let $\gcd(a,b)=g$. Take $a=gc$ and $b=gd$, where $\gcd(c,d)=1$. Then the problem condition gives that $c^2+cd+d^2 \mid g \cdot cd(c+d)$

Now, $$1=\gcd(c,d)=\gcd(c^2,d)=\gcd(c^2+d(c+d),d)=\gcd(c^2+cd+d^2,d)$$
Similarly, we get that $\gcd(c^2+cd+d^2,c)=1$. Also, $$1=\gcd(c,d)=\gcd(c+d,d)=\gcd(c+d,d^2)=\gcd(c+d,d^2+c(c+d))=\gcd(c^2+cd+d^2,c+d)$$
Thus, We have $c^2+cd+d^2 \mid g$ $\Rightarrow g \geq c^2+cd+d^2>cd$. Multiplying both sides by $g^2$, we get that $g^3>g^2cd=ab$. Hence, $$|a-b|^3=g^3|c-d| > g^3 >ab \Rightarrow |a-b|>\sqrt [3] {ab} \text{    } \blacksquare$$

[quote=TheDarkPrince][quote=math_pi_rate]$c^2+cd+d^2 \mid gcd(c+d)$
[/quote]

What do you mean by $\gcd(c+d)$?[/quote]

Perhaps he means $g*c*d*(c+d)$.

[quote=Illuzion]
Perhaps he means $g*c*d*(c+d)$.[/quote]

Oops maybe :D A better choice of variables would have been better ;)

[quote=TheDarkPrince][quote=math_pi_rate]$c^2+cd+d^2 \mid gcd(c+d)$
[/quote]

What do you mean by $\gcd(c+d)$?[/quote]
It meant $g \cdot cd$. Sorry for the mistake. Corrected it.

Took me a solid 30 minutes to realize $ab < a^2 + ab + b^2$.

Set $a \ge b.$ Notice that $a^2 + ab + b^2 \mid a(a^2 + ab + b^2) - ab(a + b) = a^3$, and likewise. $a^2 + ab + b^2 \mid b^3.$ Therefore, we have $a^2 + ab + b^2 \mid \gcd(a, b)^3.$ Now, $$\gcd(a, b)^3 = \gcd(a-b, b)^3 \le (a-b)^3.$$ Moreover, we have that $ab < a^2 + b^2 + ab \le (a-b)^3$ so we conclude.

Let $a=dm$ and $b=dn$, where $d=\gcd(a,b)$. Hence, $\gcd(m,n)=1$. Substituting into the given,
\[d^2m^2+d^2mn+d^2n^2 | d^3mn(dm+dn) \implies m^2+mn+n^2 | dmn(m+n)=.\]

I claim that $\gcd(m^2+mn+n^2, mn(m+n))=1$. To see this, it suffices to prove that $\gcd(m, m^2+mn+n^2)=\gcd(n, m^2+mn+n^2)=\gcd(m+n, m^2+mn+n^2)=1$. The first two are trivial. For the third one, note that $\gcd(m+n, m^2+mn+n^2)=\gcd(m+n, (m+n)^2-mn)=1$ (since $m$ and $n$ are relatively prime).

Hence, we have $m^2+mn+n^2 | d,$ which means that $m^2+mn+n^2 \leq d$. Then,
\[m^2+mn+n^2 \leq d \implies \frac{a^2}{d^2}+\frac{ab}{d^2}+\frac{b^2}{d^2} \leq d \implies a^2+ab+b^2 \leq d^3.\]
However, it is easy to see that $d^3 \leq |a-b|^3,$ so $|a-b|^3 \geq a^2+ab+b^2 > ab.$ Taking the cube root of $|a-b|^3 > ab$, we are done.

WLOG $a \geq b$ \\
$a^2+ab+b^2|ab(a+b)=a^2b+ab^2 \iff a^2+ab+b^2|a^3, b^3 \implies a^2+ab+b^2|d^3$ where $d=(a,b)$. Also let $a=dx, b=dy, (x,y)=1$ of course.
So we have $d  \geq x^2+xy+y^2$. $ |a-b|>\sqrt [3] {ab} \iff d(x-y)^3 > xy$ which follows easily since $d(x-y) \geq x^2+xy+y^2 \geq 3xy > xy$

Let $a=kA,b=kB$ where $k=\gcd(a,b)$, so $\gcd(A,B)=1$. The condition is equivalent to
$$k^2(A^2+AB+B^2) \mid k^3AB(A+B) \implies A^2+AB+B^2 \mid kAB(A+B).$$
Now, $\gcd(A^2+AB+B^2,A)=\gcd(B^2,A)=1$, and likewise $\gcd(A^2+AB+B^2,B)=\gcd(A^2,B)=1$, so we actually have
$$A^2+AB+B^2 \mid k(A+B).$$
This means that $\tfrac{A^2+AB+B^2}{A+B} \leq k$. Now, since $a \neq b$, we have $|a-b|\geq k$, so it suffices to prove that
$$k \geq \frac{A^2+AB+B^2}{A+B} > \sqrt[3]{AB}.$$
I will actually prove this for any positive reals $A,B \geq 1$. Indeed, rewrite the inequality as
$$(A+B)^2>(A+B)\sqrt[3]{AB}+AB.$$
For a fixed value of $A+B$, $AB$ is maximized at $A=B$, which maximizes the RHS but keeps the LHS constant. Hence we only need to consider $A=B=x$. In this case, the inequality becomes
$$4x^2>2x^{5/3}+x^2 \iff x^{1/3}>\frac{2}{3},$$
which is certainly true, implying $|A-B|>\sqrt[3]{ab}$. $\blacksquare$

Let $\gcd(a,b)=d,$ $a=dx,$ and $b=dy.$ Let $z=x^2+xy+y^2$ and we see that $z\mid dxy(x+y).$ But $$\gcd(x,z)=\gcd(y,z)=\gcd(x+y,z)=1$$ so $z\mid d.$ Hence, $a^2+ab+b^2\mid d^3$ so $$ab<a^2+ab+b^2\le d^3\le (a-b)^3$$ and we are done. $\square$

Let gcd(a,b) = d, and a=dx, b=dy.  WLOG, let a>b which implies x>y.  Plugging this in, we get $d^2(x^2+xy+y^2)|d^3xy(x+y)$ simplifying to $x^2+xy+y^2|d(xy)(x+y)$.  We can easily prove that $\gcd(x^2+xy+y^2,xy) = 1$ (by letting p divide both numbers and arriving at a contradiction).  By this fact, we can easily arrive to the fact that $\gcd(x^2+xy+y^2,xy) = 1$ and $\gcd(x^2+xy+y^2,x+y)=1$, so $x^2+xy+y^2|d$.  Therefore, $d \geq x^2+xy+y^2$, meaning $ |a-b|>\sqrt [3] {ab}$, or $d^3(x-y)^3 >xy$ is true as $x-y \geq 1$ and $d^3 $ is clearly greater than $xy$.

Nice problem..
Let $a=gx$ and $b=gy$ where $x$ and $y$ are coprime to each other.
\begin{align*}
a^2+ab+b^2 &\mid  ab(a+b)
\\ g^2 (x^2+xy+y^2) &\mid gx\cdot gy \cdot (gx+gy)
\end{align*}
Now note that as $\text{gcd}(x,y) = 1 \implies \text{gcd}(x,x+y) = 1\implies \text{gcd}(y,x+y) = 1 \implies \text{gcd}(xy,x+y) = 1 \implies \text{gcd}(xy,x^2+xy+y^2) = 1 \implies\text{gcd}(x+y , x^2+xy+y^2) = 1 $ and so $ x^2+xy+y^2 \mid g \implies g \ge x^2+xy+y^2 $
$g^3(x-y)^3 = g^2 \cdot g \cdot (x-y)^3 \ge g^2(x^2+y^2+xy) >g^2xy = ab $                                          $\blacksquare$

Let $a = dx$ and $b = dy$ where $d = \gcd(x, y)$. Then it follows that $$x^2+xy+y^2 \mid dxy(x+y).$$ But as $\gcd(x^2+xy+y^2, xy(x+y)) = 1$, it follows that $x^2+xy+y^2 \mid d$. In particular, $|a-b|^3 \geq d^3 > d^2xy = ab$ as needed.


inequalities
calculus
inequalities proposed

Let $ a,\ b,\ c$ be positive real numbers. Find the minimum value of

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

is the denominator of the question correct ?

ok

use AM>=GM
for each product term

(2a^2 + b^2 + c^2) / 3 >= (2^(1/3))*((abc)^(2/3))

do this for each and multiply all 3

so minimum value should be 2^(3*1/3) = 2

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers. Find the minimum value of

$ P\equal{}\frac {(2a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} 2b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} 2c^2)}{a^2b^2c^2}$[/quote]

WLOG assume that $ abc\equal{}1$. Using AM-GM inequality we have

$ 2a^2 \plus{} b^2 \plus{} c^2\ge 4 a\sqrt{bc}$
$ a^2 \plus{} 2b^2 \plus{} c^2\ge 4b\sqrt{ac}$
$ a^2 \plus{} b^2 \plus{} 2c^2\ge 4c\sqrt{ab}$

=> $ P\ge 4(a\sqrt{bc}\plus{}b\sqrt{ca}\plus{}c\sqrt{ab})\ge4.3\equal{}12$

So minimum value of P is 12.

I haven't received the correct solution yet . :(

[quote="md45"]what's the answer ? isnt it 2[/quote]

If so, then what' are the values of $ a,\ b,\ c$?

[quote="bambaman"]We can put $ x \equal{} a^{2}, y \equal{} b^{2}, z \equal{} c^{2}, a,b,c \in R^{ \plus{} }$:
$ \frac {(2a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} 2b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} 2c^2)}{a^2b^2c^2} \equal{} \prod \frac {x \plus{} y \plus{} 2z}{z} \equal{} \prod (2 \plus{} \frac {x \plus{} y}{z})$
We may normalize the expression x+y+z: $ x \plus{} y \plus{} z \equal{} 1$
Thus we need to find the minimum of $ \prod (2 \plus{} \frac {1 \minus{} z}{z}) \equal{} \prod \frac {1 \plus{} z}{z}, x \plus{} y \plus{} z \equal{} 1$.
$ ln\frac {1 \plus{} x}{x}$ is convex, hence by jensen:  $ \prod \frac {1 \plus{} z}{z} \ge (\frac {1 \plus{} \frac {1}{3}}{\frac {1}{3}})^{3} \equal{} 64$, equality holds when x=y=z.
We can also use holder on $ (2, \frac {x}{z}, \frac {y}{z}), (2, \frac {y}{x}, \frac {z}{x}), (2, \frac {z}{y}, \frac {x}{y})$
Easiest inequality in this forum :}[/quote]

The answer is correct, if you say so, everyone would like to see the solution without using calculus, wouldn't you?  :P

$ \boxed{A.M. \minus{} G.M. Cauchy \minus{} Schwarz, Jensen \leq My\ ability \ < < < Holder, Schur, Muirehead}$

[quote="kunny"]The answer is correct, if you say, everyone would like to see the solution without using calculus, would't you?  :P[/quote]

But Holder inequality can be proved with weighted AM-GM only. Besides, Jensen is fun! Another solution with Holder-

$ \prod (2 \plus{} \frac {x \plus{} y}{z}) \ge ((\prod 2)^{1/3} \plus{} \prod (\frac {x \plus{} y}{z}) ^{1/3})^{3} \ge (2 \plus{} 8^{1/3})^3 \equal{} 64$
Because by AM-GM: $ \prod \frac {x \plus{} y}{z} \ge \prod 2\sqrt {xy}/z \equal{} 8$

EDIT:
by weighted AM-GM: $ \frac{2a^2\plus{}b^2\plus{}c^2}{4} \ge a^{2/4} b^{1/4} c^{1/4}$
If we multiply the 3 permutations of these inequality, we get - $ \frac{(2a^2\plus{}b^2\plus{}c^2)(a^2\plus{}2b^2\plus{}c^2)(a^2\plus{}b^2\plus{}2c^2)}{a^2b^2c^2}\ge 4^{3}$

Sorry, I don't know weighted A.M.-G.M., but it seems that the solution is getting easier than former solution. :lol:

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers. Find the minimum value of

$ \frac {(2a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} 2b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} 2c^2)}{a^2b^2c^2}$[/quote]
  use Cauchy's ineq for 4  positive real numbers :
$ a^2\plus{}a^2\plus{}b^2\plus{}c^2  \geq 4 \sqrt[4]{a^4b^2c^2}$
  thus  we have,  $ LHS \geq \frac{64 \sqrt[4]{a^8b^8c^8}}{a^2b^2c^2} \equal{}64$

[quote="Hong Quy"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers. Find the minimum value of

$ \frac {(2a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} 2b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} 2c^2)}{a^2b^2c^2}$[/quote]
  use Cauchy's ineq for 4  positive real numbers :
$ a^2 \plus{} a^2 \plus{} b^2 \plus{} c^2 \geq 4 \sqrt [4]{a^4b^2c^2}$
  thus  we have,  $ LHS \geq \frac {64 \sqrt [4]{a^8b^8c^8}}{a^2b^2c^2} \equal{} 64$[/quote]

Your solution is shorter than mine.

By the way, Cauchy? :maybe:

Here is my solution:

By $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$,

$ (c \plus{} a)(a \plus{} b) \equal{} a^2 \plus{} ab\plus{}bc \plus{} ca \leq 2a^2 \plus{} b^2 \plus{} c^2$, similarly we have $ (a \plus{} b)(b \plus{} c)\leq a^2 \plus{} 2b^2 \plus{} c^2,\ (b \plus{} c)(c \plus{} a)\leq a^2 \plus{} b^2 \plus{} 2c^2$.

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

$ \geq \{(a \plus{} b)(b \plus{} c)(c \plus{} a)\}^2\geq (8abc)^2\ by\ AM \minus{} GM$.

$ \therefore \frac {(2a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} 2b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} 2c^2)}{a^2b^2c^2}\geq 64$.

The equality holds when $ a \equal{} b \equal{} c > 0$.

[quote="kunny"]$ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$[/quote]

Why? I simply don't understand this passage.

Proof 1: $ a^2\plus{}b^2\geq 2ab,\ b^2\plus{}c^2\geq 2bc,\ c^2\plus{}a^2\geq 2ca$, Adding these inequalities gives $ 2(a^2\plus{}b^2\plus{}c^2)\geq 2(ab\plus{}bc\plus{}ca)$.
$ \therefore a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$.

Proof 2: $ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\equal{}\frac{1}{2}\{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\}\geq 0$.

Proof 3: $ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\equal{}\left(a\minus{}\frac{b\plus{}c}{2}\right)^2\plus{}\frac{3}{4}(b\minus{}c)^2\geq 0$.

See here for (a,b,c)=(2,1,1) in my notations:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=176897

Note that we just apply $ a^2\plus{}a^2\plus{}b^2\plus{}c^2\ge 4\sqrt[4]{a^2bc}$ and analogous inequalities here.

[quote="kunny"]Proof 1: $ a^2 \plus{} b^2\geq 2ab,\ b^2 \plus{} c^2\geq 2bc,\ c^2 \plus{} a^2\geq 2ca$, Adding these inequalities gives $ 2(a^2 \plus{} b^2 \plus{} c^2)\geq 2(ab \plus{} bc \plus{} ca)$.
$ \therefore a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$.[/quote]
Now I've understood. Thanks! :lol:


geometry
analytic geometry
rectangle
invariant
counting
distinguishability

On a $8\times 4$ grid, movements from $A$ to $B$ can only be made to the right or to the above along the edge. How many ways to go from $A$ to $B$ such that the path bisects the area of the grid?

[hide="Suspicion"] I suspect that the sum of the $y$-coordinates of the points on the path is invariant.  If this is true, then a path from $A$ to $B$ can be represented as a sequence of increases of the $y$-coordinates, and the problem is not hard from there. [/hide]

My suspesion

[hide]Just find nonnegtive integer solutions to $a+b+c+d=16$, where $a\geq b\geq c \geq d$.[/hide]

Jelly, wouldn't you need 8 values, one for each column?

Your movements just need to go along the center of the grid as shown in the diagram. So that there're 30 movements satisfying hypothesis.

[quote="Kamil Witek"]Doesn't such a path always have to go through the center of the rectangle?[/quote]

This is a question that is easy enough to answer by experiment.  The answer is "no," with the following example (among others): Move up once, right 5 times, up twice, right once, up once, right twice.

vietkhoa, your comments are woefully under-explained.  If I correctly understand what you're suggesting (counting paths which go through the center of the rectangle), it's also completely wrong: there are $15^{2}= 225$ such paths, not $2 \cdot 15 = 30$, and not all of them bisect the area of the grid, and not all paths which bisect the area of the grid belong to that set.  (If you meant something else, please say what it was.)

[hide="My suspicion"]
hmm well clearly the first row can only have 0-4 white blocks, because anything greater than that would cause an unequal amount of scores. Simplifying the problem to a 4x4 grid, or the first half of the 8x4 grid, the problem now suffices to find the number of paths in the 4x4 grid that corresponds to a solution. i can't find a way to easily calculate this except by brute forcing all $\frac{8!}{4!\cdot 4!}= 70$ possible paths.

[/hide]

[quote="NeverOddOrEven"]hmm well clearly the first row can only have 0-4 white blocks[/quote]

I don't understand what you're trying to say.  The first row in the example diagram given has $5$ white blocks.

In any case, I realized that my suspicion is almost trivially true, since the $y$-coordinate of a point on the path corresponds to the height of the gray squares to the (WLOG) left of that point, and the sum of these heights must be invariant in order for the area of the gray squares to be equal to the area of the right.  The problem is doable from there, but it is late and I can't think of the exact calculation.

[quote="t0rajir0u"][quote="NeverOddOrEven"]hmm well clearly the first row can only have 0-4 white blocks[/quote]

I don't understand what you're trying to say.  The first row in the example diagram given has $5$ white blocks.

[/quote]

sorry for not being clear. if you look at the diagram, i was referring to the very bottom row that has 3 white blocks.

It's cleaner to use Jelly's idea then to use t0r's, I think -- the only problem is that the condition $a \leq 8$ is missing.  Then your counting problem is, "The number of partitions of 16 into 4 parts (some of which may be 0) minus the number of partitions of $n$ into 3 parts (some of which may be 0) for $0 \leq n \leq 7$," which you can write in terms of binomial coefficients.  This might take slightly less time then just listing them a la altheman.


One problem asks:

James and Jasmine decide to play a game of Illusion: The Gathering. James owns five decks, while Jasmine owns four. If they each pick a deck to play with, how many different match-ups are possible?

I don't know this game. Must each player select one of the decks they own (James picks one of five, Jasmine picks one of four)? Or can they pick from any deck (first person picks one of nine, next person picks one of eight)? Do they each use the deck they picked or are both decks combined in some way after they are selected? Depending on how the game is played will determine the answer. For instance, I can see solutions being: 5*4, 9*8, or 9*8/2 (9 choose 2). Anyone know how this game is played?

A quick google search shows that illusion : the gathering isn't a real game.  I would assume it's just for the problem and that they each can only use their own decks.

Illusion: the Gathering is not a real game, however, the name appears to be based upon [url=http://www.magicthegathering.com]Magic: the Gathering[/url].
I would assume that they each pick one of their own decks.


Are there 1000000 positive integers such that the sum of any number of
them (one or more) is never a perfect square?

Yes :) take $2^1$, $2^3$, $2^5$, $2^7$ ... The sum of any number of terms will be $2^a \cdot b$, with $a$ and $b$ odd - hence the sum is never a square.

[quote="Arne"]The sum of any number of terms will be $2^a \cdot b$, with $a$ and $b$ odd - hence the sum is never a square.[/quote]
Do you have a general proof for this you wrote?

It's obvious, but if you insist...

Let $2^{a_1}$, $2^{a_2}$, ..., $2^{a_j}$ be the numbers you choose from the set.

Suppose WLOG $a_1 < a_2 < ... < a_j$ (odd integers).

Then, $2^{a_1} + 2^{a_2} + ... + 2^{a_j} = 2^{a_1} \cdot \left(1 + 2^{a_2 - a_1} + ... + 2^{a_j - a_1}\right)$.

Obviously $a_1$ is odd, and the number inside the brackets is odd.

Hence result.

Obviously you can extended this result from squares to any powers by picking numbers of the form $2^p$, with $p$ prime.


quadratics
algebra
polynomial
linear algebra
matrix
probability
function

1. Consider the set of 2008 equations in 2009 variables: a1+a2= 1, a2+a3=2, a3+a4=3,....., a2007+a2008=2007, a2008+a2009=2008. Find a1+a2009.

2. In a certain town, a cab charges his passenger $ 3 for every 1/6 km traveled when driving faster than x km/hr. It charges$6 per minute when driving slower than x km/hr. When driving at exactly x km/hr, then both methods charge the same amount of money. What is x?

3. Compute the minimum value of a^2+b^2 subject to the constraint 2a-3b=8.

4. If a, b, and c are positive and 9a+3b+c=90, then what is the max value of log a + log b + log c?

5. A patch of grass is n cm tall. Each day, the grass grows 1cm taller. At the end of the week, a gardener comes and cuts the grass down to 1/3 of its height. Assuming the grass continues to grow in this way, and this pattern continues indefinitely for a long time. approximately, how tall will the grass be right after the gardener cuts it?

6. Find the quadratic polynomial p(x) such that p(0)=1, p(x)=p(1-x), and the remainder when p(x) is divided by 1-x is 2.

7. Let x, y, z be positive integers such that x<y<z and let a, b, c, and d be real numbers. If x^a = y^b = z^c = 42^d and 1/a + 1/b + 1/c = 1/d, find x, y, and z.

8. Find all integers A, B, and C such that (x-A)(x-10)+1 can be written as a product (x+B)(x+C) of two integers B and C.

9. A 2x2 matrix is filled with randomly generated, uniformly distributed integers from 1 to 2008 inclusive. What is the probability that the discriminant is even?

10. The function f has the property that for each number x, f(x)+f(1-x)=x^2. If f(19)=94, find f(95).

11. A computer center has 3 printers A, B, and C. Tasks are sent to whichever printer is available. The probability that a task is sent to A, B, and C is 0.6, 0.3, and 0.1. The probability that A, B, and C will fail is 0.01, 0.05, 0.04. If you sent a document and it failed to be printed, what is the chance that B is responsible?

[hide="#1"]
We have $ a_n \plus{} a_{n\plus{}1} \equal{} n$.
Also, $ (a_1 \plus{} a_2) \minus{} (a_2 \plus{} a_3) \plus{} (a_3 \plus{} a_4)\cdots \minus{} (a_{2008} \plus{} a_{2009}) \equal{} a_1 \plus{} a_{2009}$.

So, the answer is $ 1 \minus{} 2 \plus{} 3 \minus{} 4 \plus{} 5 \minus{} 6 \plus{} 7\cdots \plus{} 2007 \minus{} 2008$, or $ ( \minus{} 1)\frac {2008}{2} \equal{} \boxed{ \minus{} 1004}$.
[/hide]

[hide="#2"]
So, driving at exactly $ x$ km per hour, the charge is $ \$ 18$ per kilometer or $ \$ 360$ per hour. So, the answer is $ 360 \div 18$ or $ \boxed{20}$ km per hour.
[/hide]

[hide="#3"]
Substitute $ b \equal{} \frac {2}{3}a \minus{} \frac {8}{3}$ into $ a^2 \plus{} b^2$ to get $ \frac {1}{9}(13a^2 \minus{} 32a \plus{} 64)$. Lowest=$ \frac { \minus{} c_2}{2 c_1} \equal{} \frac {\frac {32}{9}}{\frac {26}{9}} \equal{} \boxed{\frac {16}{13}}$.
[/hide]

[quote="orangefronted"]2. In a certain town, a cab charges his passenger $ \$$3 for every 1/6 km traveled when driving faster than x km/hr. It charges $ \$$6 per minute when driving slower than x km/hr. When driving at exactly x km/hr, then both methods charge the same amount of money. What is x?[/quote]
Since dollar signs make LaTeX, you have to do [code]$ \$ $[/code] to make them appear without messing up the text.

[quote="Textangle"][hide="#1"]
We have $ a_n \plus{} a_n \plus{} 1 \equal{} n$.
Also, $ (a_1 \plus{} a_2) \minus{} (a_2 \plus{} a_3) \plus{} (a_3 \plus{} a_4)\cdots \minus{} (a_{2008} \plus{} a_{2009}) \equal{} a_1 \plus{} a_{2009}$.

So, the answer is $ 1 \minus{} 2 \plus{} 3 \minus{} 4 \plus{} 5 \minus{} 6 \plus{} 7\cdots \plus{} 2007 \minus{} 2008$, or $ ( \minus{} 1)\frac {2008}{2} \equal{} \boxed{ \minus{} 1004}$.
[/hide][/quote]
[hide="Do you mean..."]
$ a_n \plus{} a_{n \plus{} 1} \equal{} n$?[/hide]
[quote="Textangle"][hide="#3"]
Substitute $ b \equal{} \frac {2}{3}a \minus{} \frac {8}{3}$ into $ a^2 \plus{} b^2$ to get $ \frac {1}{9}(13a^2 \minus{} 32a \plus{} 64)$. Lowest=$ \frac { \minus{} b}{2a} \equal{} \frac {\frac {32}{9}}{\frac {26}{9}} \equal{} \boxed{\frac {16}{13}}$.
[/hide][/quote]
[hide="Textangle, again"]You shouldn't use $ a,b$ to mean two different things.  :wink: [/hide]
EDIT: Whoa... I quoted you right when you edited, and added a problem on accident because of that :P
EDIT again...

Blah, you're right, didn't notice that  :D

[quote="orangefronted"]10. The function f has the property that for each number x, f(x)+f(1-x)=x^2. If f(19)=94, find f(95).

11. A computer center has 3 printers A, B, and C. Tasks are sent to whichever printer is available. The probability that a task is sent to A, B, and C is 0.6, 0.3, and 0.1. The probability that A, B, and C will fail is 0.01, 0.05, 0.04. If you sent a document and it failed to be printed, what is the chance that B is responsible?[/quote]
[hide="11"]P(event given condition)=$ \dfrac{P(\text{event and condition})}{P(\text{condition})}$
The condition is that the printer fails and the event is that B is responsible. So our probability is
$ \dfrac{0.05\times0.3}{0.6\times0.01 \plus{} 0.3\times0.05 \plus{} 0.1\times0.04} \equal{} \dfrac{0.015}{0.025} \equal{} \boxed{\dfrac{3}{5}}$
[/hide]
For #10, I have encountered an interesting problem.
f(19)+f(-18)=361
94+f(-18)=361
f(-18)=267
f(-18)+f(19)=324
267+94=324
?[/hide]

For 4 you want to maximize log(abc), so you want to maximize abc, by AM-GM, $ 30 \ge \sqrt [3] {abc}$, so $ abc \le 2700$ and $ log2700 \equal{} 2 \plus{} 3log3$

[hide="numero cuatro"]indeed but your answer is incorrect, we have $ 3a \plus{} b \plus{} \frac {c}{3} \equal{} 30$ by dividing by 3
applying AM-GM to the 3 terms on the left, we see that
$ \sqrt [3]{3a \times b \times \frac {c}{3}} \le \frac {3a \plus{} b \plus{} \frac {c}{3}}{3}$
$ \sqrt [3]{abc} \le 10$
$ abc \le 1000$
and $ log(a) \plus{} log(b) \plus{} log(c) \equal{} log(abc)$
since $ log(abc)$ is maximized when $ abc$ is, and $ abc$ has its maximum at $ 1000$, the maximum of $ log(a) \plus{} log(b) \plus{} log(c)$ is $ log(1000) \equal{} \boxed{3}$[/hide]

[hide="numero cinco"]
we are repeatedly applying the function $ f(x) \equal{} \frac {x \plus{} 7}{3} \equal{} \frac {x \plus{} 3.5 \plus{} 3.5}{3}$
the reason i split up 7 into 2 3.5's is to show that this function makes the distance from 3.5 smaller by a factor of 3
in fact, here is a proof
let $ |x \minus{} 3.5| \equal{} d$
and let $ |\frac {x \plus{} 7}{3} \minus{} 3.5| \equal{} d'$
then $ d' \equal{} |\frac {x \minus{} 3.5}{3}| \equal{} \frac {|x \minus{} 3.5|}{3} \equal{} \frac {d}{3}$
so the height of the grass repeatedly gets closer to 3.5 so the height will be approximately $ \boxed{3.5 cm}$
[/hide]

[hide="numero seis (yes its seis not sies)"]
eheh... if $ p(0)\equal{}1$ and $ p(x)\equal{}p(1\minus{}x)$, then $ p(0)\equal{}p(1\minus{}0) \implies 1\equal{}p(1)$
but since p(x) has a remainder of 2 upon division by 1-x, $ p(x)$ may be rewritten as $ q(x)(1\minus{}x)\plus{}2$ and therefore $ p(x)\equal{}q(x)(1\minus{}x)\plus{}2 \implies p(1)\equal{}q(x)(1\minus{}1)\plus{}2 \implies p(1)\equal{}2$
contradiction
[/hide]

What am I doing wrong with #6?

We are given $ p(0)\equal{}1$.
We can also write $ p(x) \equal{} (1 \minus{} x)q(x) \plus{} 2$, since the remainder when $ p(x)$ is divided by $ (1 \minus{} x)$ is $ 2$. Substituting $ x \equal{} 1$, we get $ p(1) \equal{} 2$. But we're given that $ p(0) \equal{} p(1 \minus{} 0)$. This is not true since $ p(0) \equal{} 1\ne 2 \equal{} p(1)$, contradiction.

EDIT: beaten...

Yes, my bad...The problem in number 6 is incorrect. I'm sorry.

1. There is something wrong with number 1.The answer should be 1004.

2. 20 (ok)

3. 64/13

4. 3

5. 3.5cm

6. Wrong problem

7. 2, 3, 7 (???- Help with this problem)

8. -8, -9, -9 and -12, -11, -11 (???- Help with this problem)

9. 5/8 (???- Help with this problem)

10. 324

11. 0.6

Thanks!

7.

numero siete
assume d=uno, c=dos, b=tres, and a=seis
then
z^2= cuarenta y dos
z=$ \sqrt{42}$
y^3= cuarenta y dos
y=$ \sqrt[3]{42}$
x^6= cuarenta y dos
x=$ \sqrt[6]{42}$

10 has a false condition...

$ f(x)\plus{}f(1\minus{}x)\equal{}x^2$

replace $ x$ with $ 1\minus{}x$...

$ f(1\minus{}x)\plus{}f(x)\equal{}(1\minus{}x)^2$

So $ x^2\equal{}(1\minus{}x)^2$... $ x\equal{}\frac{1}{2}$.

It should probably be $ f(x)\plus{}f(x\minus{}1)\equal{}x^2$

[quote="funtwo"]numero siete
assume d=uno, c=dos, b=tres, and a=seis
then
z^2= cuarenta y dos
z=$ \sqrt {42}$
y^3= cuarenta y dos
y=$ \sqrt [3]{42}$
x^6= cuarenta y dos
x=$ \sqrt [6]{42}$[/quote]

hint: cuarenta y dos = 42
but you cant assume d=1, c=2, b=3, a=6

rewrite x, y, z as $ 42^{x'}, 42^{y'}, 42^{z'}$
respectively, where $ n'\equal{}log_{42}(n) \text{for} n\equal{}x,y,z$
now take log bas 42 of the equation
$ ax'\equal{}by'\equal{}cz'\equal{}d$
and we know $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{d} \implies \frac{d}{a}\plus{}\frac{d}{b}\plus{}\frac{d}{c}\equal{}1$
from $ ax'\equal{}by'\equal{}cz'\equal{}d$, we have $ x'\equal{}\frac{d}{a}, y'\equal{}\frac{d}{b}, z'\equal{}\frac{d}{c}  \implies x'\plus{}y'\plus{}z'\equal{}1$
now raise 42 to the power of both sides
$ 42^{x'\plus{}y'\plus{}z'}\equal{}42 \implies 42^{x'}42^{y'}42^{z'}\equal{}42 \implies xyz\equal{}42$
so x, y, and z are integer factors of 42
notice none of x, y, or z can be 1 since then in the equation $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{d}$, there would be division by zero
so x, y, and z are non-xero divisors of 42 which means they are 2, 3, and 7 (in that order, since x<y<z)


From aops vol 1:

"A chemist has 80 mL of a solution containing a 20% acid. How many mL must be removed and replaced by pure acid in order to obtain a 40% solution?"

The part that really gets me is the "remove". How would you write the equation???

[hide="my equation"]I actually came up with an equation, though I'm pretty sure it's wrong...but it's my best effort.

[0.2(80) + x]/(80+x) = 0.4

16 + x = 32 + 0.4x

x=80/3=25 2/3 mL
[/hide]

[hide]$0.2(80)-0.2x+x=0.4(80)$
We have 80 mL of 20% solution, subtract x mL of 20% solution, add x mL of pure (100%) solution, and we get 80 mL of 40% solution.
The equation itself should be easy to solve.[/hide]


Billy purchased $ 100$ shirts at a cost of $ \$15$ each and planned to resell them.  The first week he sold some of them for $ \$25$ each.  The second week he sold the rest for $ \$20$ each.  If his total profit was $ \$600$, how many shirts did he sell the first week?

[quote="kabi"]hello
$ x\plus{}y\equal{}100$
$ 25x\plus{}20y\equal{}2100$
required answer is 20 shirts.
Thank u[/quote]
Sorry for reviving but I feel I have something to add.
To clarify, kabi means $ 25x+20y=2100 $ because that is total amount of money Billy gets from his customers.  This is $\$2100$ because it is the total profit ($\$600$) plus the amount of money he spent purchasing the $100$ shirts($\$1500$) which totals $\$2100$ as stated by kabi.  Then solving the system of equations we get $\boxed{20 }$ shirts were sold the first week.


1. Find a formula for the sum $ \sum_{k\equal{}1}^n \frac{k}{(k\plus{}1)!}$

2. Find the sum of $ \sum_{k\equal{}2}^n \left(\frac{1}{\log_k(u)}\right)$

3. Find a formula for $ \frac{1}{1 \times 2 \times 3} \plus{} \frac{1}{2 \times 3 \times 4} \plus{} ... \plus{} \frac{1}{n \times (n\plus{}1) \times (n\plus{}2)}$

Can anyone please solve step by step, I just started summations and need to see the procedure of doing these questions

Thanks!

First you are to find $ f(k)$ such that $ a_k \equal{} f(k) \minus{} f(k \plus{} 1)$, then you can calculate the sum $ \boxed{\sum_{k \equal{} 1}^n a_k \equal{} \sum_{k \equal{} 1}^n \{f(k) \minus{} f(k \plus{} 1)\}}$

[quote="kunny"]First you are to find $ f(k)$ such that $ a_k \equal{} f(k) \minus{} f(k \plus{} 1)$, then you can calculate the sum $ \boxed{\sum_{k \equal{} 1}^n a_k \equal{} \sum_{k \equal{} 1}^n (f(k) \minus{} f(k \plus{} 1))}$[/quote]

Thanks for the fast reply kunny, is that for the first question? How would you go about doing it...?

1. Hint

$ \sum_{k \equal{} 1}^n \frac {k}{(k \plus{} 1)!}\equal{}\sum_{k\equal{}1}^n \frac{k\plus{}1\minus{}1}{(k\plus{}1)!}$

[hide="Problem 3"]$ \frac{1}{n(n\plus{}1)(n\plus{}2)}\equiv\frac{A}{n}\plus{}\frac{B}{n\plus{}1}\plus{}\frac{C}{n\plus{}2}$

$ \therefore A(n\plus{}1)(n\plus{}2)\plus{}B(n)(n\plus{}2)\plus{}C(n)(n\plus{}1)\equiv 1$

$ n^2(A\plus{}B\plus{}C)\plus{}n(3A\plus{}2B\plus{}C)\plus{}(2A\plus{}2B\plus{}C)\equiv 1\Rightarrow \begin{cases}A\plus{}B\plus{}C\equal{}0\\ 3A\plus{}2B\plus{}C\equal{}0\\ 2A\plus{}2B\plus{}C\equal{}1\end{cases}$

$ \sim \begin{cases}A\plus{}B\plus{}C\equal{}0 \\ B\plus{}2C\equal{}0\\C\equal{}\minus{}1\end{cases} \;\therefore\; A\equal{}\minus{}1 ,B\equal{}2, C\equal{}\minus{}1$

$ \sum_{k\equal{}1}^{n} \frac{1}{k(k\plus{}1)(k\plus{}2)}\equal{}\sum_{k\equal{}1}^n\minus{}\frac{1}{k}\plus{}\frac{2}{k\plus{}1}\minus{}\frac{1}{k\plus{}2}$[/hide]

As you expand, you will notice the terms will cancel out, leaving you with a generic formula.
This is why finding $ a_k\equal{}f(k)\minus{}f(k\plus{}1)$ solves the problem, because the intermediate terms will cancel each other out.
If you still need the solutions, let me know.


It is not hard to determine the greatest number~$%
f(n)$ such that, for positive~${t_1,\dots,t_n}$, the inequality ${%
\sum_{i=1}^nt_i\sum_{i=1}^n 1/t_i<f(n)}$ implies that each three of the $t_i$%
's are the side lengths of a triangle. The answer is ${f(n)=(n+\sqrt{10}-3)^2%
}$, and the proof is fairly standard.
In our case $n=4$


How do you count the number of multiples of a given number $k$ in the first $n$ rows of pascal's triangle?

This is much easier when $k$ is prime (see: http://mathworld.wolfram.com/LucasCorrespondenceTheorem.html ).  I don't know if there's a simple answer for the general case, and it certainly isn't Intermediate level.

Can someone move this to pre olympiad or something then please?
I'm really interested in the answer.


Propose a synthesis of the compound p-$ (CH_3)_2N\minus{}CH_2CH_2\minus{}C_6H_4\minus{}OH$, which is relevant in the germination process of barley.

phenol+formaldehyde(1 mole)+dimethylamine(1 mole)(mannich reaction).

Wrong: not only there is one carbon missing there, but also both the ortho positions are also active.

i think it should be p-hydroxy toluene. is dere neway to block ortho positions?

Try to propose a different and reasonable synthesis.

What compounds can we start off with?

That $ \minus{}\ce{  N(CH3)2}$ did remind me of the Mannich Reaction but I will have to think about that first.

I am just giving a brief outline of a possible method.

Phenol undergoes Friedel-Craft Acylation (para-position). $ \alpha$ bromination $ \ce{Br2}$/$ \ce{CH3COOH}$. Protect carbonyl group followed by amination of the $ \minus{}\ce{Br}$. Then we can use $ \ce{CH3Br}$ to make the primary amine tertiary. Finally remove the group and reduction of carbonyl to alkyl.

Wow, that's extremely long (and probably not even right, I wrote it on the spur of the moment). Mind giving a hint to a shorter method?

1) Wouldn's also the ring be brominated?
2) Amination with what?
3) You would get a quaternary ammonium salt.

this mite work but its pretty long.
acetaldehyde+ mannich reaction using dimethyl amine to give (Ch3)2NCh2Ch2CHO
+PCC + barbier wieland
+ reduction with DIBAl-H To give aldehyde.
2. phenol->bromination(para product)
Now use si(me)3cl to protect oh group and mg in ether to give organo metallic, which on reaction wid the aldehyde obtained above foll by reaction with socl2 and zn, h+(reduces halide and removes protecting group) gives the product

Here's a possible synthesis from benzene:

[hide="Proposal"]1) Benzene + HNO3/H2SO4 -----> Nitrobenzene

2) Nitrobenzene + ClCH2CN/t-BuOK -----> p-Cyanomethylnitrobenzene

3) p-Cyanomethylnitrobenzene + H2/catalyst -----> p-H2N-CH2CH2-C6H4-NH2

4) p-H2N-CH2CH2-C6H4-NH2 + NaNO2/HCl(pH~1) -----> p-H2N-CH2CH2-C6H4-N2+Cl-

5) p-H2N-CH2CH2-C6H4-N2+Cl- + Cu2O/Cu(NO3)2/H2O -----> p-H2N-CH2CH2-C6H4-OH

6) p-H2N-CH2CH2-C6H4-OH + 2HCHO + 2HCOOH -----> p-(CH3)2N-CH2CH2-C6H4-OH[/hide]

[quote="Carcul"]
4) p-H2N-CH2CH2-C6H4-NH2 + NaNO2/HCl(pH~1) -----> p-H2N-CH2CH2-C6H4-N2+Cl-
[/quote]
Won't the amino group to the left also not get converted into the alcohol(rearranged one)
Though aliphatic diazonium is unstable yet the side products are also expected for your synthesis

Aliphatic amino groups do not undergo diazotization at a pH near 1 (why?). That's why I included that information. With this reaction conditions, aromatic -NH2 can be safely converted to diazonium salt in the presence of aliphatic -NH2 (and this is a procedure described in the literature).

Pratically I feel that it is impossible(Though I dont have enough experience).

[quote="Carcul"]By answering my question above (on pH ~ 1) and remembering the mechanism of diazotization, the answer to your doubt is straightforward.[/quote]
that probably bcoz aliphatic amines being more basic than their aromatic counterparts are completely protonated at ph1.


group theory
abstract algebra
superior algebra
superior algebra unsolved

G is a group without subgroup non trivial, show that G is cyclic with order p, where p is prime

Come on, looking at http://www.mathlinks.ro/Forum/viewtopic.php?t=163138 this shouldn't be impossible for you to do.

[quote="leont"]but I am 40 year old and I am a little, say us, oxidized. always I have liked the mathematics, in special, the algebra. I am read the Herstein's book: abstract Algebra and I need help to begin. I have solved other problems, but this, not if it is easy or difficult, it does not give the imagination me[/quote]
I'm not exactly sure what you are asking? Do you want someone to explain it? If so look below.

[b]Theorem:[/b] Let $ G$ be a group so that if $ H < G$ then $ H \equal{}\{ e\}$. And $ |G| \equal{} p$ for some prime $ p$ or $ p \equal{} 1$. (You made a minor mistake in problem).

[b]Proof:[/b] If $ G \equal{}\{ e\}$ then there is not such subgroup $ H < G$ so this is "vacously true" and so $ |G| \equal{} 1$ which is cyclic. Otherwise if $ G\not \equal{}\{ e\}$ we can chose an element $ a\in G$ so that $ a\not \equal{} e$ and construct the subgroup $ \left < a\right >\leq G$. Now it cant be that $ \left < a\right > \equal{}\{ e\}$, since $ a\not \equal{} e$. So it must mean that $ \left < a\right > \equal{} G$ because otherwise $ \{ e\}\not \equal{}\left < a\right > < G$ which is a contradiction. So $ a$ [i]generates[/i] $ G$ which means $ G$ is cyclic. So $ G\simeq\mathbb{Z}_{n}$, by the isomorphic properties of cyclic groups. But $ \mathbb{Z}_{n}$ has proper non-trivial subgroups unless $ n \equal{} p$ a prime. Q.E.D.


Qs.
In an examination, there are 20 questions with 4 alternatives. There is only 1 correct
alternative. One gets 4 marks for every correct answer & 1 negative mark for every wrong answer.50% are the passing marks. If a student attempts the questions on guesswork, find the probability of his passing.

I need a simplification of the ans. I have arrived at .

Let C be the no. of correct ans. and  W the no. of wrong ans.

Now to pass , the candi. has to obtain a minimum of 50 % of 4*20 i.e. 40 marks.

Now typically, any candidate obtains a score of  4C - W

So , the condition for passing is

  4C - W >= 40

But W = 20 - C

So from above , one obtains  C >= 12.

So the event of the candidate passing could be decomposed into the following set of mutually exclusive and exhaustive sub-events.

 [u]C[/u]          [u] W[/u]

 12                     8

 13                     7

....
.....

  20                    0

So the probability of the candidate passing is given by

( taking into a/c that the P(a qs is marked [b]correctly[/b]) = 1/4 and  P(a qs is marked [b]incorrectly[/b]) = 3/4)

P(E)  = summation ( C(20,r) * (1/4)^r * (3/4)^(20-r) ) where r ranges from 12 to 20.

(Somebody please put the above expression in Latex for better clarity.)

My aim is to simplify the expression  [b]summation ( C(20,r) * (1/4)^r * (3/4)^(20-r) )[/b]

[quote="JBL"]In other words, the question is to find a closed form for $\sum {20 \choose r} \cdot \frac{1}{4^r}\cdot\left(\frac{3}{4}\right)^{20-r}$, for use in the problem shown.[/quote]
[hide="an attempt"](1/4)^r*3/4^20-r*20Cr, and if we just evaluated 1/4^r*3/4^(20-r)=3^(20-r)/4^20 for all values of r. Having this in hand, we now move to the total product. 3^(20-r)(20Cr)/4^20=sum[3^(20-r)(20Cr)]/4^20 using symmetry, we get that that is equal to sum [3^(20-r)(20C(20-r))]. no we invert it, so it is sum[3^r(20Cr)]/4^20, (we inverted the numbers in the sigma and its limits too), at this point, im totally lost. Maybe we can implement Abel Sums?[/hide]

I have a feeling that the only way is to use the binomcdf (binomial cumulative distribution function) on your handy dandy TI-83+.  Unless this specific case has an easy simplification, even the guys at TI can't come up with a closed form for this...


Why doesn't Wisconsin have it's own usergroup? the bay area has one! :weightlift:  :weightlift:  :weightlift:

because wisconsin is public so commoners like you are free to post random stuff here :)

and Bay Area is private because it has enough members

It's because you are so cool that we want you to be able to give us stunning insights.

I know we are (maybe with the exception of [i][b]SOME PEOPLE[/b][/i])

*Cough* [hide]kstan[/hide] *cough* [hide]footballrocks[/hide] *cough*
[hide]Just look at my blog to find out what I mean![/hide]


linear algebra
matrix
real analysis
real analysis unsolved

1. Prove that $ |a| \equal{} | \minus{} a|$.
2. Prove that $ |ab| \equal{} |a\parallel{}b|$
3. Prove that, for any positive real numbers $ c$, there exist natural number $ n$ such that $ n \leq c \leq n \plus{} 1$.
4. Prove that for any real numbers $ b$ and $ c$, there exist rational number $ a$ such that $ b < a < c$.
5. If $ c > 0$, prove that $ |x| < c$ implies $ \minus{} c < x < c$.
6. Prove that $ \sqrt {a^2} \equal{} |a|$.
7. Prove that $ |a \plus{} b| \leq |a| \plus{} |b|$.
8. Prove that for any positive real number $ c$, there exist natural number $ n$ such that $ c < n$.

Is 3. supposed to have a natural $ n$? What if $ c$ is negative? :S

$ \sqrt{x^2}=|x|.$ Now,

$ \left| ab \right|=\sqrt{(ab)^{2}}=\sqrt{a^{2}}\cdot \sqrt{b^{2}}=\left| a \right|\cdot \left| b \right|,$ and it follows that $ \left| a \right|=\left| -(-a) \right|=\left| -1 \right|\left| -a \right|=\left| -a \right|.\quad\blacksquare$

As for 7., we have,

\begin{eqnarray*}
   \left| x+y \right|^{2}&=&(x+y)^{2} \\
 & =&x^{2}+2xy+y^{2} \\
 & \le& \left| x \right|^{2}+2\left| x \right|\left| y \right|+\left| y \right|^{2}=\left( \left| x \right|+\left| y \right| \right)^{2} \\
  \implies\left| x+y \right|&\le& \left| x \right|+\left| y \right|.\quad\blacksquare
\end{eqnarray*}

Healker, for the 5th problem I don't think you can use your definition.

Maybe $ x = {\begin{bmatrix}x \text{ if }x \geq 0 \\ -x\text{ if }x < 0\end{bmatrix}}$


linear algebra
matrix
number theory proposed
number theory

[color=darkred]Let $ m$ and $ n$ be two nonzero natural numbers. In every cell $ 1 \times 1$ of the rectangular table $ 2m \times 2n$ are put signs $ \plus{}$ or $ \minus{}$. We call [i]cross[/i] an union of all cells which are situated in a line and in a column of the table. Cell, which is situated at the intersection of these line and column is called [i]center of the cross[/i]. A transformation is defined in the following way: firstly we mark all points with the sign $ \minus{}$. Then consecutively, for every marked cell we change the signs in the cross, whose center is the choosen cell. We call a table [i]accesible[/i] if it can be obtained from another table after one transformation.
Find the number of all [i]accesible[/i] tables.[/color]

this is in Belarus MO 2007
see http://www.imomath.com/othercomp/Blr/BlrMO07.pdf

the answer is that all tables are accessible. a enable idea is to prove that every table remains itself after two tranformations.

I obviously don't understand this task could someone show me where am I mistaken?
--    ++  ++
--    ++  ++

-+   --   ++
-+   --   ++

They aren't same after 2 transformation on this logic.
The solution to my logic is $ 2^(2m\plus{}2n\minus{}2)$
Can somebody write few transormastions so I understand where is my mistake?

Represent the table as a $ 2n \times 2m$ matrix $ M$ whose elements are set to $ 0$ for plus and $ 1$ for minus and regarded as elements of the ring $ \mathbb{Z}_2$.

Using matrix algebra the transformation is equivalent to $ M \rightarrow M' \equal{} M \plus{} AM \plus{} MB$ where $ A$ is a $ 2n \times 2n$ matrix with all elements set to $ 1$ and $ B$ is a $ 2m \times 2m$ matrix with all elements set to $ 1$

Now apply the transformation twice

$ M' \rightarrow M'' \equal{} M' \plus{} AM' \plus{} M'B \equal{} M \plus{} AM \plus{} MB \plus{} AM \plus{} A^2M \plus{} AMB \plus{} MB \plus{} AMB \plus{} MB^2 \equal{} M$

(We use the fact the $ X \plus{} X \equal{} 0$ over  $ \mathbb{Z}_2$ and $ A^2 \equal{} 0, B^2 \equal{} 0$ because the dimensions of the matrices are even.)

We can conclude that every matrix is accessible because it is the transform of itself transformed.


[quote="manjil"]Hi. I wanted to know the method by which a user is given a rating? Can anyone help?[/quote]

Other people vote you, i.e. your rating is a weighted average of users depending on whether they are regular users, moderators or admins I think. Most people don't vote regular posts and some users like [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=2426]Alex Zhai[/url] posted excellent solutions but have rather a not so good rating. I think the rating does not carry much meaning anyway as it works now.

[quote="manjil"]Thanks orl. Is the rating of others visible to anyone?[/quote]

I can see your rating but this is probably because I am a moderator. But it may not be visible to everybody as you are assumed to rate postings and not preference of users.

Only moderators can see others' ratings, not normal users.

Thanks all for the information. Just a very silly question: How do u become a moderator?

When a moderator is needed, an administrator will choose a responsible person who actively participates in the forum. Basically, if you want to be a moderator, just be helpful and act like one. Chances are, an admin will notice.

spam and spam
last two posts were definetly unnessecary (i know i spelled that wrong...)
remember, if you want to be a mod, you have to act like one, and that means no spamming

and no, i am not a mod of a forum


quadratics
algebra
linear equation
quadratic formula
calculus
calculus computations

Let $v=y'.$ Then you have a first degree differential equation for $v.$ What can you classify it as? It is both a linear equation (which can be solved by the multiplying factor method) and an autonomous, hence separable, equation (which can be solved by separation of variables.) Personally, I prefer to employ the linear equation multiplying factor method in such cases (fewer human errors to be made), but both methods will work.

Then once you have $v=y',$ integrate that to find $y.$

[quote="arjim"]my''(t) = -ky'(t) - mg

1. How can I find the general solution, and 2. if I have the initial conitions y(0)= H, H>0 and y'(0) = 0, how can I find the soluion? How can I sketch the graph of y'(t)?[/quote]

If $m = 0$, then the whole thing is trivial. Suppose $m \neq 0$, then divide through by $m$ and bring everything to one side so that you get $y''+\frac{k}{m}y'+g = 0$.
Now you can treat the equation like a quadratic equation so that you get $a^{2}+\frac{k}{m}a+g = 0$ and by the quadratic formula you get $a = \frac{-\frac{k}{m}\pm \sqrt{(\frac{k}{m})^{2}-4g}}{2}$

Then you have three cases each depends on the sign of the determinant. You can only do this if you find it easy to follow, otherwise I will stick with Kent's suggestion.

akech, that's flat-out wrong. You're describing something that could be appropriate for the homogeneous equation $ay''+by'+cy=0$ but

(1) this equation is not homogeneous, and

(2) there's no $cy$ term.

arjim: I suggest that you completely ignore akech's post. It will only confuse you.

I think that the given differential equation should be $my''(t)=-ky'(t)+mg$, which is related to a falling body.

Whether that's $+mg$ or $-mg$ is entirely a question of whether you have the positive direction for $y$ be up or down. Either one would make sense in the appropriate context. The equation as originally posted has the positive direction as up; kunny's version (with $y$ as "distance fallen") has the positive direction as down. The physics is the same either way, and the solution will be the "same" once you account for the differences in notation.


I was just wondering how you folks in Ivy League school can memorize huge list of dates/words/people in very short amount of time. Any good idea?

I realized that I need to learn this soon because well, I'm not good at it at all.  :blush:

I am not good at memorizing lists. However, I have heard that a popular method is to associate each item on the list with some image. I have a friend who does this and it's pretty effective. Unfortunately, I can't really visualize anything, so this method doesn't work for me.

I memorize things by using them. It's not really fast, but it works.

Silverfalcon, have you actually met people who memorize large amounts of data in short periods of time?  What are you basing those claims on?

After being at MIT for a little over a semester and around some very smart people I don't believe I've met anyone with an insane memory.  Furthermore, there are people who have a vast knowledge of certain topics.  But I don't think this has so much to do with their memory as it does the amount of time they've spent on that subject.  For example, you have a large amount of English words memorized mostly because you use them often every day.  The same would be true if you worked on other topics for large amounts of time.

That said, I think it is pretty apparent that there is a discrepancy between people in terms of memory.  I base this off of people who compete in memory contests.  They obviously have a better memory, or rather, they have practiced memorizing so much that they're very good at it.

Anyway, to answer what you were asking, yes there are ways to more efficiently memorize data.  Xevarion's advice is good.  Also, just repetition with them is a tried and true method, e.g. practice with flashcards, read over the list a bunch, recite it, etc.

[quote="Silverfalcon"]I was just wondering how you folks in Ivy League school can memorize huge list of dates/words/people in very short amount of time.[/quote]

I'm not sure that there really is anyone who can memorize huge amounts of material in a very short time. I don't think most Ivy League students do much memorizing. I think most of them do a lot of reading, rereading, and looking up other sources to make sure they deeply understand a subject. Then the most essential facts are more memorable, because they have a structure of thought to which to connect the facts.

But let's hear from some other students what they think. Do you memorize? Or do you simply remember while you are trying to make sure that you understand?

There are two standard techniques for memorizing lists of things, known as chaining and pegging.

To use chaining, you find some way to memorize the first item on the list, and then for subsequent items you come up with an image associating the previous item with the next. Then you're likely to recall the entire list assuming you don't forget the first one.

To use pegging, you first memorize a list of the same size that is easier to learn than the list you're supposed to memorize. Then figure out how to associate the $i^\text{th}$ item on the simpler list with the $i^\text{th}$ item on the more difficult list.

My most memorable experience with pegging was in preparation for an exam in which I had to be able to recognize the preludes and fugues of the Well-Tempered Clavier and the Mass in B Minor by Bach from a few measures that the professor played on the piano. For the 48 fugues, I first learned the numbers 1 to 48 (which is really not so hard), then a list of words that can be sung with the subjects (see [url]http://www2.rhbnc.ac.uk/~uhwm006/prout.html[/url], but I also changed some of them to make them easier for me), and then listened to the pieces for about 8 hours a day for a week or two. (There aren't too many shortcuts on that part.)

I should say that while I have a rather good memory, I generally hate actually memorizing things, especially if someone asks me to do so. And most of the time when I need to memorize something, I do so through simple repetition.

Interesting responses. Thank you.

I had no basis on stating that Ivy Leaguers have good memorization. However, I assumed that they would be better than some of other students since after all, Ivy Leaguers and students in similiar school levels obviously spend more time on studying, and therefore, maybe memorizing (in addition to reading and researching).

I like the idea of chaining and thinking of images. I actually didn't know about this. Only method I ever used was making acronyms and repeating many times. This method worked effective for short amount of list (say 10-15) but when it reaches higher lists (40-50), I usually can't remember well.  :blush:

[quote="chess64"]When I have to memorize anything, I just repeat it again and again. I memorized 37 chapters of the Qur'an that way, and now I think my memory is pretty good after all that practicing :D[/quote]

yeah, that definetely increases your memory, trust me.

I don't have a very good memory for everyday stuff, and I'm not that good at memorizing lists either. As a musician, I never wanted to be without the sheet music- sometimes I would play something enough times to memorize, but never with that intent.

Even so, I'm very good at retaining knowledge and understanding. If I can fit something into a structure of knowledge, it stays. That doesn't even take much repetition.
Looking at ComplexZeta's example, I wouldn't have much chance at identifying the exact pieces, especially from a list by one composer. I've never tried that kind of memorization, but I don't have perfect recall even among pieces I've played- I make mistakes in identification, and sometimes it takes a while to find a spot I remember. On the other hand, if the assignment was to identify the composers of a variety of short excerpts, I would do very well, even if I had never heard those particular pieces before. Stylistic signatures are memorable to me, since they fit into my understanding of the music and how to play it.

I also use the peg system that ComplexZeta suggested. It really works!

[quote=equinox145111]Does anybody have experience using a memory palace technique?[/quote]

I have a little experience with memory palace technique. I remember I could remember 100 random words in 30mins.

A few good websites for memory tips and practicing memory is [url]https://app.memoryleague.com/?_ga=2.64424428.254029884.1548161591-1812194027.1450831061#!/home[/url]  and [url]https://artofmemory.com/app/[/url].

[quote=equinox145111]Does anybody have experience using a memory palace technique?[/quote]

Yes. The way it works is by visualizing each individual object you want to visualize, and then assigning them to individual places in a familiar place to you. After that everytime you want to recall something, imagine yourself going sequentially through each place, and seeing the objects either have motion or no motion.

Find some relation between the things you want to memorize.  :pilot:


Given a pdf file created with tex, is there any way to get the text file from the pdf?

Thanks,
bpms

The short answer is no. In Adobe Reader you can copy the text or save the whole thing as a text file but you will lose a lot of information (such as equations).

OCR (Optical Character Information) programs may help a little but they aren't very good for mathematics either.

The moral of this is that tex files should always be backed up as you can always re-create pdf files from them.

If you need to choose between keeping a TeX file or a PDF/PS/DVI file, always keep the TeX file. ;)


Hi.i am a student from indonesia and i have a great interest at math olympiad.Can anybody shared me some files about problem solving strategies(or other topics about math olympiad ???)  .I would be thankful for the help. Sory for my bad english. :blush: .


I didn't go to prom this year because I didn't feel like getting invited (I'm only freshman so I have junior and senior year to go to prom) even though I might go next year.

Just wondering, has anyone went to prom this year? It doesn't matter whether you're man or woman. I never been to the prom so I don't know what it's like.

[hide="here"]
Dressed up, picture, dance, food, and afterprom, and etc. That's what I know.[/hide]

I went to my junior prom last year....it was hella fun

I'm going to senior prom this year, May 14th....should be hella fun

Right now, my biggest concern is my gettin-there situation...I want to take a limo but they are gettin hard to come by...so I'm tryin to solidify a travel plan...Cause I don't wanna drive 40 miles to where it's at...Hopefully, I can get a seat for me and my date in one of my friend's limos

[quote="MithsApprentice"]*cough* Rent a ferrari *cough* That's what my friend's doing. Smart kid he is...[/quote]

 :D  He has a solid plan!

I'm a freshman and my boyfriend's a sophomore, thus I'm not going. :)

I'm in eighth grade. Thank God my girlfriend's a senior (just kidding)!

I'm tryin to get my dad to rent me a ferrari or some type of impressive car

It might be expensive, but it'll be worth it...

I can't drive stick though....I never learned how...so maybe they have some nice sports car that has automatic transmission

edit: Ferrari's are pretty expensive to rent....
I see some convertibles that I can get for under $100 though!

[quote="ankur87"]
edit: Ferrari's are pretty expensive to rent....[/quote]

Yeah...if you don't have at least 600 to spare, I wouldn't bother trying. A limo would be okay too...and those would probably only be like 150-200.

Plus, if you get a limo, you can share it with a bunch of other couples, reducing the cost.

A limo would come out to about 60 per person

Around California, they go for about 60-70 an hour

And we need it for 7 hours...Shared with 8-10 people in the limo

But renting a car would be so cool, I been looking into it lately
Hopefully, I can get this mustang convertible I've had my eye on....

60-70 an hour?!?!?!?!?!

gah...you would think renting a car for a day would be cheaper.

I had a lot of fun at our prom last weekend. We kinda ticked off some ppl at this nice restaurant we went to b/c one of the guys in our group kept on eating sugar packets and making his crystal glass sing.  We went bowling after prom and my date destroyed me  :blush: .  It's a good time, but I wouldn't spend too much money, I took my mom's sequoia.


calculus
integration
logarithms
calculus computations

I found this problem written on a board on my dorm floor. I tried to do it but I don't know how....

$\int \frac{2^x 3^x}{4^x + 9^x} dx$

Any help is appreciated, it really has me stumped

I think the first step is to convert $4$ and $9$ to $2^2$ and $3^2$ respectively, but I do not know what I'm doing...

Divide through by $4^x$, then substitute $y=\left(\frac32\right)^x$.

Okay, I'm going to try that

$\int \frac{2^x 3^x}{4^x + 9^x} dx$

dividing by $4^x$

${= \int \frac{\frac{3}{2}^x}{1 + \frac{9}{4}^x}} dx$ (The fractions should be in parentheses but I can't get that to happen in LaTeX....)

${= \frac{1}{2} \int \frac{2u}{1 + u^2}} du$

$= \frac{1}{2} \ln \mid u \mid$

$= \frac{1}{2} \ln \left(\frac32\right)^x$

Is this correct? Thanks for the tip, jmerry
I hope I followed through right

No, not correct yet.

${= \int \frac{\left(\frac{3}{2}\right)^x}{1 + \left(\frac{9}{4}\right)^x}} dx$

[note the use of \left( and \right) ]

Now let $u=\left(\frac32\right)^x$

Then $du=\left(\frac32\right)^x\ln\left(\frac32\right)\,dx$

So this integral becomes

$=\frac1{\ln3-\ln 2}\int\frac{du}{1+u^2}$

... and you're going to wind up with an arctangent in your answer.

Aw, dang

A mistake with the chain rule

Thank, again Kent. You're always very helpful


algebra
polynomial
superior algebra
superior algebra unsolved

show that the polynomial y^(13) + x^(2)y + 5x^(2) + y + 5  in Q[x,y] is irreducible.

I let f(x,y) be the above polynomial.
consider the proper ideal (x), since the image of  f(x,y)/(x) = y^(13) + y + 5  can not be factored into smaller degree, so f(x,y) is irreducible in Q[x,y]

Am I correct???

sorry...I don't get what you mean.... :?:

Irreducibility mod (x) does not imply irreducibility of the original polynomial, as shown by my example. owk

Let $k\ge2$ and $p(x,y)=y^{k}+x^{2}{y}+5x^{2}+y+5$. $p(0,y)=y^{k}+y+5$ is irreducible over $\mathbb{Z}_{2}$ (it's true for $2\leq{k}\leq300$ and I think that it's true for all $k\ge2$) thus $p(0,y)$ is irreducible over $\mathbb{Q}$.
Moreover $y^{k}$ is the term of maximal degree in $y$ and it's independent of $x$.
We may conclude easily that $p$ is irreducible over $\mathbb{Q}$.


Math City has eight streets, all of which are straight. No street is parallel to another street. One police officer is stationed at each intersection. What is the greatest number of police officers needed?

8*7/2=28, because it's the maximum number of intersecting points.


algebra polynomial inequalities calculus integration trigonometry LaTeX

I'm just now posting the problems that we got for our INMOTC Practice test...so they will be available in a few minutes...

Some more problems form our INMOTC
(Click on the nos. to go to the URLs) Some might not work(Because I may be putting their threads later or they are too classical to be made separately..these are boxed..)...I'll be updating the questions after regular intervals...

[url=http://www.mathlinks.ro/viewtopic.php?p=1717903#1717903]1[/url]  $ n$ is of the form $ 3k \plus{} 1$. We start writing nos. $ 1,2,3,\cdots,n$ in any order. At any instant, the sum of the nos. written should not be divisible by 3. In how many ways can these nos. be written?

$ \boxed{2} \frac {a^3}{bc} \plus{} \frac {b^3}{ac} \plus{} \frac {c^3}{ab} \ge a \plus{} b \plus{} c$ If $ a,b,c > 0$

$ \boxed{3} a^3 \plus{} b^3 \plus{} c^3 \ge a^2b \plus{} b^2c \plus{} c^2a$ IF $ a,b,c > 0$

$ \boxed{4} a \plus{} \frac {a}{b(a \minus{} b)} \ge 3$ if $ a \ge b > 0$

$ \boxed{5}$If $ 0 < x < \frac {1}{2}$ Then
\[ x^2(1 \minus{} 2x) \le \frac {1}{27}\]
$ \boxed{6}$ $ \frac {a}{a \plus{} \sqrt {(a \plus{} b)(a \plus{} c)}} \plus{} \frac {b}{b \plus{} \sqrt {(b \plus{} c)(b \plus{} a)}} \plus{} \frac {c}{c \plus{} \sqrt {(c \plus{} a)(c \plus{} b)}} \le 1$ If $ a,b,c > 0$

$ \boxed{7}$ $ \frac {1}{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac {1}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {1}{c^3 \plus{} a^3 \plus{} abc} \le \frac {1}{abc}$ If $ a,b,c > 0$

$ \boxed{8}$ $ \frac {a}{b \plus{} 2c \plus{} 3d} \plus{} \frac {b}{c \plus{} 2d \plus{} 3a} \plus{} \frac {c}{d \plus{} 2a \plus{} 3b} \plus{} \frac {d}{a \plus{} 2b \plus{} 3c} \ge \frac {2}{3}$ If $ a,b,c,d > 0$

$ \boxed{9}$ $ \frac {1}{a^3(b \plus{} c)} \plus{} \frac {1}{b^3(c \plus{} a)} \plus{} \frac {1}{c^3(a \plus{} b)} \ge \frac {3}{2}$ If $ a,b,c > 0$

$ \boxed{10}$ There is an $ 8 \times 8$ chessboard. We have to fill the numbers from $ 1 \minus{} 100$ in them such that each no. is the average of the nos. in its surroundings. Find the no. of such ways. (Nos. can be repeated)

$ \boxed{11}$ The nos. $ 1,2,\cdots, 2n$ are distributed to two sets $ A \equal{} {a_1,a_2,\cdots,a_n} \ , \ B \equal{} {b_1,b_2,\cdots,b_n}$ such that $ a_1 > a_2 > a_3 > \cdots > a_n$ and $ b_1 < b_2 < b_3 < \cdots < b_n$.
Find the value of $ |a_1 \minus{} b_1| \plus{} |a_2 \minus{} b_2| \plus{} \cdots \plus{} |a_n \minus{} b_n|$

$ \boxed{12}$ $ \frac {a^2 \plus{} b^2}{c^2 \plus{} ab} \plus{} \frac {b^2 \plus{} c^2}{a^2 \plus{} bc} \plus{} \frac {c^2 \plus{} a^2}{b^2 \plus{} ca}\ge 3$ If $ a,b,c > 0$

$ \boxed{13}$ $ \frac {a^2}{b(a \plus{} 2b)} \plus{} \frac {b^2}{c(b \plus{} 2c)} \plus{} \frac {c^2}{a(c \plus{} 2a)} \ge 1$ IF $ a,b,c > 0$

$ \boxed{14}$ $ \frac {a}{\sqrt {a^2 \plus{} 8bc}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 8ca}} \plus{} \frac {c}{\sqrt {c^2 \plus{} 8ab}} \ge 1$ If $ a,b,c > 0$

$ \boxed{15}$ No unique Integers $ x,y,z$ exist such that if $ p(x)$ is a polynomial with integer coefficients, $ p(a) \equal{} b,p(b) \equal{} c,p(c) \equal{} a$

$ \boxed{16}$ $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ has $ 3$ real roots, $ a,b,c \in \mathbb{R}$ such that $ \minus{} 2 \le a \plus{} b \plus{} c \le 0$. Show that one of the roots $ \in [0,2]$

$ \boxed{17}$ $ p(x)$ has both integer coefficients as well as integer roots. Show that both $ p(0)$ and $ p(1)$ can't be odd.

Some solutions:
[hide]2. By Am GM, $ \frac {a^3}{bc} \plus{} b \plus{} c \ge 3a$
                     $ \frac {b^3}{ca} \plus{} a \plus{} c \ge 3b$
                     $ \frac {c^3}{ab} \plus{} a \plus{} b \ge 3c$.
Adding, we get the desired inequality.

3.$ \sum a^3 \ge \sum a^2b$ is obvious from rearrangement inequality.

4.Well known. Was discussed in Ineq. marathon.

12. Using AM GM inequality, $ ab \le \frac {a^2 \plus{} b^2}{2}$. So, $ \sum \frac {a^2 \plus{} b^2}{c^2 \plus{} ab} \ge 2\sum \frac {a^2 \plus{} b^2}{(a^2 \plus{} c^2) \plus{} (b^2 \plus{} c^2 )} \ge 3$ by Nesbitt.

5.Rearranging, we need to prove that $ 2x^3 \plus{} \frac {1}{27} \ge x^2$.

Since $ x$ is a positive real, $ x^3 > 0$. The LHS can be written as $ x^3 \plus{} x^3 \plus{} \frac {1}{27} \ge x^2$ by AM GM.  Finish.[/hide]

Some solutions

[hide]$ \boxed{10}$ There are $ 100$ ways of doing this as at a time only 1 number can be used to fill the entire chessboard :D

$ \boxed{15}$ $ p(a) \equal{} b , p(b) \equal{} c , p(c) \equal{} a$ imply that :-
$ (a \minus{} b)|(b \minus{} c)$
$ (b \minus{} c)|(c \minus{} a)$
$ (c \minus{} a)|(a \minus{} b)$

imply that $ a \minus{} b \equal{} b \minus{} c \equal{} c \minus{} a$

implies $ a \equal{} b \equal{} c$
Hence Proved :lol:[/hide]

[hide="@fbaggins"]
There is a small hole in your proof of the 15th.
$ a\minus{}b\equal{}b\minus{}c\equal{}c\minus{}a$ won't always be true.
Their absolute values will be equal.
So you will have to work out those.....
[/hide]

[quote="fbaggins"]@Rijul

but the absolute values will also ensure that a=b=c .

isnt it???[/quote]
Yeah...What I meant to say is that your earlier proof wasn't complete...

Can you provide the answer to number 1 question??? :lol:

My solution to problem 17:- (though not very elegant :P )

[hide]
Since all the coefficients are integral and all roots are also integral,all the coefficients are divisible by the leading coefficient. So we can take the polynomial as
           $ x^{n} \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \cdots \plus{} a_{0}$

Now given $ a_{0}$ and $ 1 \plus{} a_{1} \plus{} \cdots \plus{} a_{n}$ are odd.
So, all the roots must be odd(as product of roots is odd).

Therefore the coefficients are sums of odd nos.
 Now,
     $ a_{i} \equal{}$ sum of $ \dbinom{n}{i}$odd nos.

Since $ \dbinom{n}{i} \equal{} \dbinom{n}{n \minus{} i}$

there are always an even no.of odd coefficients.

implying, the sum of coefficients cannot be odd .

Contradiction.[\Hide][/hide]

@aravind:

Could you explain ineq 4??

 [hide]
Solution to question 11 is $ n^{2}$.
See RMO-2002 question 4.
[/hide]

since pirate asked ..
Problem 4
the inequality is trivial if $ a > 3$,
so let  - $ a\le3$
now we have to prove -
$ a \plus{} \frac {a}{b(a \minus{} b)}\ge 3$
using - $ \frac {a^2}{4} \ge b(a \minus{} b)$ (since $ a \ge b$)
we get,
$ a \plus{} \frac {4}{a} \ge 3$ now add 1 -
$ a \plus{} 1 \plus{} \frac {1}{a} \plus{} \frac {3}{a} \ge 4$,
or,
$ 3 \plus{} \frac {3}{a} \ge 4$,
which is true since $ a\le3$
btw all the other inequalites- some trivial ,some easy and some too well known to ask and i am sure INMO is gonna be a lot tougher..

An additional condition needs to be added in question number 9. abc=1(the question was used in IMO 1995  :D )

[hide="Solution for the last problem posted by aravind srinivas"]

The smallest positive integer $ d$ such that $ a^d \equiv 1 \pmod n$ where $ a$ and $ n$ are co-prime positive integers
is called the [i] order of $ a$ modulo $ n$ [/i] and is represented as $ O_n(a)$ or ord $ _n(a)$.

Also
[b]Claim 1:[/b] if $ a^m \equiv 1 \pmod n$ then $ O_n(a) | m$ (maybe you can try proving this)

Now let us prove the problem

By Euler's theorem we have

$ A^ {\varphi( A^n \minus{} 1) } \equiv 1 \pmod {A^n \minus{} 1}$

Note: We are allowed to use Euler's theorem because $ A$ and $ A^n \minus{} 1$ are co-prime.

If you don't know what $ \varphi (n)$ is please visit [url]http://en.wikipedia.org/wiki/Totient[/url].
Link for Euler's Theorem :[url]http://en.wikipedia.org/wiki/Euler%27s_theorem[/url] .

Also $ O_{A^n \minus{} 1} (A) \equal{} n$ (pretty trivial)

Therefore by claim 1, we get

$ n| \varphi(A^n \minus{} 1)$                                                                                $ \square$

[/hide]

Edit: When I typed this, Rijul saini's proof wasn't still posted. This proof and his proof are not much different (but they aren't the same).

Both Rijul Saini's and Aravind's proof were nice. Both were slightly similar. But I didn't know such things at the time of writing the exam. So I messed up. But now I know. And one more thing, Aravind got the first mark in this exam-25
[quote="Addarsh"][quote]@Addarsh
There are indeed 5 values of k which satisfy the conditions. They are 3, 23, 69, 87 and 2001.
667 does not satisfy.
[/quote]
Harekrishna,ur correct.
I was stupid enough to think that 667 is a prime(sorry i never bothered to check for 23 or 29).hence i missed out 3 cases,yes ur correct.Thanx for pointing out my mistake. :)[/quote]
Even I thought 667 was prime like you.lol

If $ a,b,c$ are positive real numbers such that $ abc\equal{}1$,Prove that:

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

[quote="Addarsh"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$,Prove that:

                            $ a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b} < \equal{} 1$[/quote]

Using the fact that $ abc \equal{}1$ we modify the given equation
to the following form:

$ a^{a}b^{b}c^{c} \geq 1$

so we have to prove( taking log base $ e$ both sides)

$ alna \plus{} blnb \plus{}clnc \geq 0 ......................(*)$

also given condition says

$ ln a \plus{}ln b \plus{}ln c \equal{} 0$

So we get (*) using Chebyshev  over the sequences

$ ( a,b,c)$ and $ (ln a,ln b, lnc)$ which are sorted similarly

 :)  :)

[quote="Addarsh"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$,Prove that:

                            $ a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b} < \equal{} 1$[/quote]

The inequality is equivalent to the following homogenous inequality:

\[ a^a b^b c^c \ge (abc)^{\frac{a\plus{}b\plus{}c}{3}}\]

This follows directly from the convexity of the function $ f: \mathbb{R}^\plus{} \rightarrow \mathbb{R},$ $ f(x) \equal{} x \ln x$ and Jensen's inequality.

You can also use weighted GM-HM inequality followed by AM-GM inequality as follows:

\[ a^a b^b c^c \ge \left( \frac{a \plus{} b \plus{}c}{\frac{a}{a}\plus{}\frac{b}{b}\plus{}\frac{c}{c}} \right )^{a \plus{} b \plus{} c}  \equal{} \left ( \frac{a \plus{} b\plus{} c}{3} \right )^{a \plus{} b \plus{}c} \ge (abc)^ \frac{a\plus{}b\plus{}c}{3}\]

[quote="Addarsh"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$,Prove that:

                            $ a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b} < \equal{} 1$[/quote]

[u]Solution 1[/u]
Write the LHS as - $ (ab)^c (bc)^a (ca)^b$.So we have to prove that -
\[ (ab)^c (bc)^a (ca)^b \le 1\]
By Weighted AM-GM inequality we have -
\[ \left[(ab)^c (bc)^a (ca)^b\right]^\frac{1}{a\plus{}b\plus{}c} \le \frac{1}{a\plus{}b\plus{}c}\left(a.bc \plus{}b.ca\plus{}c.ab\right) \equal{} \frac{3}{a\plus{}b\plus{}c} \le 1\]
(By Myself)

[u]Solution 2[/u]
WLOG:$ c\ge b\ge a$
$ abc\equal{}1$, so $ c\ge 1$,so $ ab \le 1$  and  $ a \le 1$
Now:
\[ a^{b\plus{}c}\cdot b^{c\plus{}a} \cdot c^{a\plus{}b}\]
\[ \equal{}(abc)^a\plus{}b a^{c\minus{}a} b^{c\minus{}b}\]
\[ \equal{}a^{c\minus{}a}b^{c\minus{}b}\]
\[ \equal{}(ab)^{c\minus{}b} a^{b\minus{}a}\]
\[ \le 1^{c\minus{}b}1^{b\minus{}a}\]
\[ \equal{}1\]
Therefore $ a^{b\plus{}c}\cdot b^{c\plus{}a} \cdot c^{a\plus{}b} \le 1$

(by mathlinks user: rem)

[u]Solution 3[/u]
Note that -
\[ a^{b\plus{}c}\cdot b^{c\plus{}a} \cdot c^{a\plus{}b} \equal{} \frac{1}{a^ab^bc^c} \le 1\]
the last inequality is true because for $ a,b,c$ individually greater or lesser than 1 we have -
\[ a^a \ge a , b^b \ge b, c^c\ge c\]

(by mathlinks user:Maharjun)

How about this , -
Let $ a,b,c \in \mathbb{R}_0^\plus{}$, Prove that -
\[ 4(a\plus{}b\plus{}c) \ge 3(a\plus{}(ab)^\frac{1}{2} \plus{} (abc)^\frac{1}{3})\]

[quote="geniusbliss"]
How about this , -
Let $ a,b,c \in \mathbb{R}_0^ \plus{}$, Prove that -
\[ 4(a \plus{} b \plus{} c) \ge 3(a \plus{} (ab)^\frac {1}{2} \plus{} (abc)^\frac {1}{3})\]
[/quote]

[hide="Solution"]
By AM-GM inequality, we have

\[ \frac{1}{4} ( a \plus{} 4b ) \ge \frac{1}{4} \left ( 2 \sqrt{a.4b} \right ) \equal{} \sqrt{ab}\]

Similarly

\[ \frac{1}{12} ( a \plus{} 4b \plus{} 16c ) \ge \frac{1}{12} \left ( 3 \sqrt[3]{a.4b.16c} \right ) \equal{} \sqrt[3]{abc}\]

Adding the two inequalities above, we get

\[ \frac{a\plus{}4b\plus{}4c}{3} \ge \sqrt{ab} \plus{} \sqrt[3]{abc}\]
\[ \Leftrightarrow 4(a\plus{}b\plus{}c) \ge 3 \left ( a \plus{} \sqrt{ab} \plus{} \sqrt[3]{abc} \right )\]

$ \mathbb{Q.E.D.}$[/hide]

Here's another nice inequality:

Given $ a, \, b, \, c \in \mathbb{R}^\plus{}$, such that $ a^2\plus{}b^2\plus{}c^2 \equal{} 3abc$, prove that \[ \frac{a}{b^2 c^2 } \plus{} \frac{b}{c^2 a^2 } \plus{} \frac{c}{a^2 b^2 } \ge \frac{9}{a\plus{}b\plus{}c} .\]

duh

trivial from cauchy schwarz

$ {(a\plus{}b\plus{}c)(a/b^2c^2\plus{}b/a^2c^2\plus{}c/a^2b^2)>\equal{}(a/bc\plus{}b/ac\plus{}c/ab)^2\equal{}9}$

qed

In $ \triangle ABC$ $ BD$ and $ CE$ are bisectors of $ \angle ABC$ and $ \angle ACB$ respectively and $ I$ is the incenter of the triangle. Line $ L$ is perpendicular to segment $ DE$ and passes through point $ I$ . Point $ M$ is the intersection of line $ L$ and segment $ DE$ , Point $ N$ is the intersection of line $ L$ and segment $ BC$. Prove that \[ \angle BAC \equal{}60^{\circ} \iff IN\equal{}2 \cdot IM\]

Hint - Line L is the angle bisector of Angle BAC.  :)

[quote="Rijul saini"]In $ \triangle ABC$ $ BD$ and $ CE$ are bisectors of $ \angle ABC$ and $ \angle ACB$ respectively and $ I$ is the incenter of the triangle. Line $ L$ is perpendicular to segment $ DE$ and passes through point $ I$ . Point $ M$ is the intersection of line $ L$ and segment $ DE$ , Point $ N$ is the intersection of line $ L$ and segment $ BC$. Prove that
\[ \angle BAC \equal{} 60^{\circ} \iff IN \equal{} 2 \cdot IM\]
[/quote]

As sourabh pointed out , with $ I$ as incenter a circle can be drawn thru $ D$ and $ E$ . Thus the line perpendicular to $ DE$ and thru $ I$ is the perpendicular bisector line of triangle $ AED$ again forcing it to be the angle bisector of $ \angle A$.
I'll post the complete sol. when i have time.

[quote="sahilsharma94"]
As sourabh pointed out , with $ I$ as incenter a circle can be drawn thru $ D$ and $ E$ . Thus the line perpendicular to $ DE$ and thru $ I$ is the perpendicular bisector line of triangle $ AED$ again forcing it to be the angle bisector of $ \angle A$.
I'll post the complete sol. when i have time.[/quote]

[hide="@both Sahil and Sourbh - "]Oh really?
If that is the case would you show me the proof of this proposition?[/hide]

Note that $ AED$ is an equilateral triangle. So perpendicular bisector of $ DE$ will also be the angle bisector of angle $ BAC$.  :)

I am really very sorry. I made some wrong assumptions.
Sorry again.
Here is my solution.
By angle-chasing we get $ \angle{EID} \equal{} 120$.
observe  that Quadrilateral $ AEID$ is cyclic.
Hence, we get $ EID$ to be isosceles.
Now , Triangle $ BEI$ and triangle $ BNI$ are congruent thus, $ IN \equal{} IE$
since, $ \sin{30} \equal{} \frac {1}{2}$, we get the required result.


geometry
algorithm
complex analysis
Functional Analysis

I am curious...do employers of math majors pay a lot of attention to whether a student just out of college has an undergraduate pure math degree or whether he/she has an applied math degree? I am asking because one university that I may attend next year offers a regular (pure) mathematics degree, but not an applied one (they only offer the applied math degree to graduates). However, undergraduate students can "concentrate" on areas of applied mathematics in the regular mathematics degree by taking various related electives.

I hear that financial firms don't really care (from another thread on this forum), but what about most technical firms? Any input would be appreciated. Thanks.

The [i]name[/i] of the degree isn't as important as the skill set that you graduate with. If you get a degree in "pure math", but take lots of courses in cognate subjects -- its not much different than getting a degree in "applied math".

It is important, however, to back up any degree in pure math with a cognate in some other subject area/s (for example: computer science, finance, business, physics, biology... whatever interests you). Unless you go on to grad school and then go into academia, its almost certain that you will end up using your math degree in some kind of an applied setting. (And even if you do go into academia, a knowledge of other subject areas is still extremely useful).

That's not to say that a pure math degree is useless - it just means that it's more valuable when coupled with some other subject area/s. But the actual name of the math degree isn't all that important.

gauss202,

Thanks for your response. I am rather naive about the process of writing a resume and job hunting, but how would employers distinguish among a candidate with a math degree who focused on pure mathematics and a candidate with a math degree who focused on applied mathematics by solely examining their resumes, for both candidates have the same degree name?

I find this discussion somewhat relevant.  I am currently pursuing a major in (pure) mathematics (with no minors; I have no desire to take the lower-level courses required to get a major/minor in, say, economics)

However, by the time I graduate, I'll have taken courses in applied complex analysis, applied functional analysis, PDEs (I skipped the ODEs course), dynamical systems, optimization (two courses), numerical analysis, and mathematical finance.  Is there anything that I can do to ensure that, say, in the job market, I don't get classified as having only 'pure' math skills?

One common trick is to include a list of "relevent courses taken" on your resume. This alerts whoever is reading your resume that you diversified your studies within your pure math major.

After that, its up to you to "talk yourself up" during your interview about your knowledge and experience (without being obnoxious). I majored in pure mathematics, but mentioning that I'd done my senior thesis on wavelets seemed to get me a fair amount of attention when applying for jobs. Unfortunately, the next question that I usually got was -- had I taken many courses in programming or did I know much about programming; to which my answer was no, not much.

I took an introductory course in programming, but I didn't get any experience programming within my major or during my summer work experience. I don't think I ever even used as much as a calculator in my college math classes.

Some employeres didn't see this as a problem, and some did. One employer said that they would only hire me if I agreed to spend a year learning programming in grad school first.

So my advice is to definitely get some experience with programming (preferably in algorithms and data structures) - and to do it early on in your college career. Some pure math majors have a loathing for programming -- but really its much more intellectually stimulating than you might think.

Maybe not as fun as proving theorems, but still fairly interesting. And it doesn't mean that you have to become a programmer when you get out of school (unless you want to). Its just another tool to have in your toolbag if you need one.

Would you say that pure programming skills (e.g, C++, Java, some other high level language) are regarded more highly than proficiency with mathematical software such as Maple or Matlab?

[quote="gauss202"]Yes.[/quote]

Emphatically, I agree.

How much programming do you need? Is Computer Science AP sufficeint, or is the focus more on understanding how programs work?

If all you have is computer science AP and haven't done any coding between that and graduating from college, you'll have a tough time convincing a potential employer that you can program.  If you continue coding, though, you should be ok.

mathematical science is applied, right?  :blush:
that's what it looks like i'll be majoring since i'm more than likely going to clemson and they don't offer a general mathematics degree.

Mathematical Sciences is Clemson's name for their entire math department. Within the department they offer both a Bachelor of Sciences degree and a Bachelor of Arts degree. The actual difference between the two degrees is pretty minor, but they may be viewed differently by employers (with the BS degree possibly considered more rigorous as it has more requirements).

Clemson tends to be a fairly applications-oriented kind of school, so it wouldn't surprise me if most of their math classes were also applications-oriented. You can find out more about their math department and the degrees that they offer from here http://www.math.clemson.edu/ug_prog/ug_major.html


geometry
incenter
circumcircle
geometry proposed

[quote="grobber"]Let I be the incenter of ABC. AI, BI, CI intersect the circumcircle of ABC again at A', B', C' respectively. Show that the area of A'B'C' >= the area of ABC.[/quote]

Yes, that's quite known. Solution at http://www.kalva.demon.co.uk/short/soln/sh883.html

[quote="grobber"]I remember Valentin posting a question (I can't remember the author right now) about the locus of the pts X s.t. if AX, BX, CX intersect the circumcircle of ABC in A'B'C' then S[A'B'C']>=S[ABC].[/quote]

This problem was shortly discussed on the Hyacinthos newsgroup:

[quote="Hyacinthos message #6178"][b]From:[/b] gil1993 (Lascu Mircea)
[b]Date:[/b] Wed Dec 18, 2002 0:20am
[b]Subject:[/b] Problem.

Let M be a point in the interior of the triangle ABC. The lines
AM,BM and CM intersect the triangle's circumcircle in the points
A',B',C', respectively. Find the locus of M for which [A'B'C']>=
[ABC]. [/quote]

[quote="Hyacinthos message #6179"][b]From:[/b] Jean-Pierre Ehrmann
[b]Date:[/b] Wed Dec 18, 2002 8:56am
[b]Subject:[/b] Re: Problem.

Dear Hyacinthists
--- In Hyacinthos@yahoogroups.com, "gil1993 <gil1993@y...>"
<gil1993@y...> wrote:
> Let M be a point in the interior of the triangle ABC. The lines
> AM,BM and CM intersect the triangle's circumcircle in the points
> A',B',C', respectively. Find the locus of M for which [A'B'C']>=
> [ABC].

I presume - may be, I'm wrong - that [ABC] means area(ABC).
In this case, as the triangles have the same circumcircle, the ratio
S'/S of their areas is (A'B'/AB)(B'C'/BC)(C'A'/CA).
If o(M)=power of M wrt the circumcircle, the inversion (M,o(M))maps
A->A',..., hence
S'/S = -o(M)^3/(AM.BM.CM)^2
Your condition is equivalent to
o(M)^3 + AM^2 BM^2 CM^2 <= 0.
o(M)^3 + AM^2 BM^2 CM^2 = 0 is a tricircular sextic entirely inside
the circumcircle (apart the isolated points A,B,C) and going through
O.
I don't see how it is possible to construct this sextic.
Do you have some idea
Best regards from France. Jean-Pierre[/quote]

[quote="grobber"]I remember Valentin posting a question (I can't remember the author right now) about the locus of the pts X s.t. if AX, BX, CX intersect the circumcircle of ABC in A'B'C' then S[A'B'C']>=S[ABC]. This problem might be a good start for that more general problem (or maybe not.. :)).[/quote] yes I remeber it was the second question on the forum :D quite an antique one :D


linear algebra
matrix
linear algebra unsolved

On an arbitrary field, prove that any frobenius matrix is product of two symetric matrices.
Conclude that any matrix is a product of two symetric matrices.

Suppose now that $K$ is a field of characteristic different of $2$ and such that any element
of $K$ is a square. Then prove that any matrix over this field is similar to a symetric matrix.

sorry .. it is another name used to call companion matrix (http://mathworld.wolfram.com/CompanionMatrix.html) .. or maybe it is just in the book I was reading :) Anyway, companion matrix seems to be the usual name.


What are some good tips to prepare for olympiad exams, other than doing the problems?

Any good books or websites(in particular) you can recommend me?

The website that turned me into an IMO participant from an amateur math enthusiast is this: http://www.mathlinks.ro/  :wink:

I would also recommend the books "Art and Craft of Problem Solving" by Paul Zeitz, for those learning from the basics; and "Problem Solving Strategies" by Arthur Engel, for more advanced learners. But I think the most perfect way of preparing for Olympiads is simply by doing more and more problems!

I find [url=http://www.mathlinks.ro/viewtopic.php?p=1199472#1199472]this post[/url] very inspiring, take a look at it if you're interested. :)

solve problems  :D

seriously, sign up for an olympiad preparation course, if any. and solve more problems...

actually, it mostly depends on your current level. More "advanced guys" are inclined just simply to do problems...

and the last suggestion : try to do more combinatorics problems...

Combinatorial problems allow you to think about them from different points of view, as there is no general way of solving such a problem. This helps a lot in the long run.

[quote="leoxnlin"][quote="Erken"]
and the last suggestion : try to do more combinatorics problems...[/quote]

Why combinatorics especially?[/quote]

Well, generally there is no evidence which would confirm the correctness and validity of my suggestion. Though it's a bit awkward to explain -  combinatorics is a foundation of the the whole olympiad math. Try to think of a math problem (not necessary combinatorial) like of a mountain peak, that is hidden by clouds or darkness, and you need to get to the top of this mountain. Consequently you need to find the right way and on the way to the top you should outdo several hurdles. Perhaps, this is not a really good comparison, but enough to get the main point -
so by improving your combinatorial skills, you gradually improve your "skill of finding a way to the top". Combinatorial problems. much more than other problems, "persistently demand" not knowledge, but the original way of thinking and analyzing, i.e ability to find an approach and to overcome all drawbacks.


Is it possible to sum (not integrate but add up)

 10-k/10 from k=1..10-x

You can evaluate any finite sum by just adding the terms. In this case, there aren't any shortcuts.

The sum $ \sum_{x\in I} f(x)$ on an interval $ I\equal{}[a,b]$ in general does not converge.

Is this sum $ \sum_{k \equal{} 1}^{10\minus{}x} 10 \minus{} \frac{k}{10}$ or $ \sum_{k \equal{} 1}^{10\minus{}x} \frac{10\minus{}k}{10}$?
Either way, the sum can be evaluated using the formulas $ \sum_{k \equal{} 1}^n {c} \equal{} cn$, with c a constant, and $ \sum_{k \equal{} 1}^n {k} \equal{} \frac{n(n\plus{}1)}{2}$

In the first instance the sum is $ 10(10\minus{}x) \minus{} \frac{(10\minus{}x)(11\minus{}x)}{20}$, and in the second instance the sum is $ (10\minus{}x) \minus{} \frac{(10\minus{}x)(11\minus{}x)}{20}$


[quote="Lovasz"]Let:
\[ 2x+y-2 \ge 0  \]
\[ 2x-y-2 \le 0  \]
\[ 2y-x-4 \le 0  \]
Then:
\[ x^2+y^2 \ge \frac45  \][/quote]

Are we supposed to prove that $x^2+y^2 \geq \frac{4}{5}$ ???

[quote="Arne"]Seems like it :)[/quote]

ok

There is, in fact, a very simple graphical approach to this question. Just think about it in two dimensions :)

The first three inequalities define the region enclosed by a triangle.
Let A, B, and C be the three vertices of the triangle.
AB: 2x + y - 2 = 0
BC: 2x - y -2 =0
CA: 2y -x -4 = 0
After calculation, A = (0,2), B = (1, 0) and C =($\frac83$,$\frac{10}3$).
Note that the origin and the whole triangle ABC lies on oppsite side of the line AB.
( Maybe some can help to provide a figure.)
Let P(x,y) be any point inside the triangle.
Then OP is [b]minimum[/b] when OP is perpendicular to OA and P lies on OA.
In this case, OP = $\mid\frac{2(0)+(0)-2}{\sqrt{2^2+1^2}}\mid$=$\frac{2}{\sqrt{5}}$.
Finally, $x^2+y^2\geq {OP}^2=\frac{4}{5}$ as required.

Actually, just the first condition $2x+y\ge2$ is sufficient: \[ (2x+y)^2+(2y-x)^2\ge 4+0 \Leftrightarrow x^2+y^2 \ge \frac45. \]