inequalities
function
calculus
derivative
logarithms
rearrangement inequality
inequalities proposed

[color=darkred]Let $ m,n\in \mathbb{N}$, $ n\ge 2$ and numbers $ a_i > 0$, $ i \equal{} \overline{1,n}$, such that $ \sum a_i \equal{} 1$. Prove that

$ \small{\dfrac{a_1^{2 \minus{} m} \plus{} a_2 \plus{} ... \plus{} a_{n \minus{} 1}}{1 \minus{} a_1} \plus{} \dfrac{a_2^{2 \minus{} m} \plus{} a_3 \plus{} ... \plus{} a_n}{1 \minus{} a_1} \plus{} ... \plus{} \dfrac{a_n^{2 \minus{} m} \plus{} a_1 \plus{} ... \plus{} a_{n \minus{} 2}}{1 \minus{} a_1}\ge n \plus{} \dfrac{n^m \minus{} n}{n \minus{} 1}}$[/color]

Oops... I did a mistake.... it should be

$ \small{\dfrac{a_1^{2 \minus{} m} \plus{} a_2 \plus{} ... \plus{} a_{n \minus{} 1}}{1 \minus{} a_1} \plus{} \dfrac{a_2^{2 \minus{} m} \plus{} a_3 \plus{} ... \plus{} a_n}{1 \minus{} a_2} \plus{} ... \plus{} \dfrac{a_n^{2 \minus{} m} \plus{} a_1 \plus{} ... \plus{} a_{n \minus{} 2}}{1 \minus{} a_n}\ge n \plus{} \dfrac{n^m \minus{} n}{n \minus{} 1}}$

I solved the problem looking the function $ F(m) \equal{} \sum_{i \equal{} 1}^{n}\frac {a_i^{2 \minus{} m}}{1 \minus{} a_i} \minus{} \frac {n^m}{n \minus{} 1}$ with first derivative - $ F'(m) \equal{} \minus{} \sum_{i \equal{} 1}^{n}\frac {a_i^{2 \minus{} m}}{1 \minus{} a_i}\ln a_i \plus{} \frac {n^m}{n \minus{} 1}\ln\frac {1}{n}\ge 0$. :roll:

Hence $ LHS \minus{} \frac {n^m}{n \minus{} 1}\ge \sum_{i \equal{} 1}^n\frac {1 \minus{} a_i}{1 \minus{} a_{i \plus{} 1}} \minus{} \frac {n}{n \minus{} 1}\ge n\minus{}\frac{n}{n\minus{}1}$ and therefore $ LHS\ge RHS$  :wink:
However, it's obvious but very ugly and because of that I didn't write anything :oops:

my solution:

First step is the same as the official solution
our inequality is equivalent to $ \sum_{i\equal{}1}^{n}\frac{a_{i}^{2\minus{}m}\minus{}a_{i\minus{}1}}{1\minus{}a_{i}}\ge\frac{n^{m}\minus{}n}{n\minus{}1}$

then using the rearrangement inequality we have

$ \sum_{i\equal{}1}^{n}\frac {a_{i\minus{}1}}{1\minus{}a_i} \le \sum _{i\equal{}1}^{n}\frac {a_i}{1\minus{}a_i}$

so it suffices to prove

$ \sum_{i\equal{}1}^{n}\frac {a_{i}^{2\minus{}m}\minus{}a_i}{1\minus{}a_i} \ge \frac{n^{m}\minus{}n}{n\minus{}1}$

the case of $ m\equal{}0,1,2$ is obvious because $ LHS$ is just equal to $ RHS$

for $ m \ge 3$

$ \frac {a_{i}^{2\minus{}m}\minus{}a_i}{1\minus{}a_i}\equal{}\sum_{k\equal{}0}^{m\minus{}2} \frac {1}{a_{i}^k}$

so it remains to prove

$ \sum_{k\equal{}0}^{m\minus{}2} \sum_{i\equal{}1}^{n} \frac {1}{a_{i}^k} \ge \frac{n^{m}\minus{}n}{n\minus{}1}$

now applying the Holder inequality we have

$ \sum_{i\equal{}1}^{n} \frac {1}{a_{i}^k}\equal{}(\sum_{i\equal{}1}^{n} \frac {1}{a_{i}^k}) ( \sum_{i\equal{}1}^{n} a_i )^k \ge n^{k\plus{}1}$, for $ k\equal{}0,1,...,m\minus{}2$

so

$ \sum_{k\equal{}0}^{m\minus{}2} \sum_{i\equal{}1}^{n} \frac {1}{a_{i}^k} \ge \sum_{k\equal{}0}^{m\minus{}2} n^{k\plus{}1}\equal{}\frac{n^{m}\minus{}n}{n\minus{}1}$

Q.E.D.

[quote="linboll"]my solution:

First step is the same as the official solution
our inequality is equivalent to $ \sum_{i \equal{} 1}^{n}\frac {a_{i}^{2 \minus{} m} \minus{} a_{i \minus{} 1}}{1 \minus{} a_{i}}\ge\frac {n^{m} \minus{} n}{n \minus{} 1}$

then using the rearrangement inequality we have

$ \sum_{i \equal{} 1}^{n}\frac {a_{i \minus{} 1}}{1 \minus{} a_i} \le \sum _{i \equal{} 1}^{n}\frac {a_i}{1 \minus{} a_i}$

so it suffices to prove

$ \sum_{i \equal{} 1}^{n}\frac {a_{i}^{2 \minus{} m} \minus{} a_i}{1 \minus{} a_i} \ge \frac {n^{m} \minus{} n}{n \minus{} 1}$

the case of $ m \equal{} 0,1,2$ is obvious because $ LHS$ is just equal to $ RHS$

for $ m \ge 3$

$ \frac {a_{i}^{2 \minus{} m} \minus{} a_i}{1 \minus{} a_i} \equal{} \sum_{k \equal{} 0}^{m \minus{} 2} \frac {1}{a_{i}^k}$

so it remains to prove

$ \sum_{k \equal{} 0}^{m \minus{} 2} \sum_{i \equal{} 1}^{n} \frac {1}{a_{i}^k} \ge \frac {n^{m} \minus{} n}{n \minus{} 1}$

now applying the Holder inequality we have

$ \sum_{i \equal{} 1}^{n} \frac {1}{a_{i}^k} \equal{} (\sum_{i \equal{} 1}^{n} \frac {1}{a_{i}^k}) ( \sum_{i \equal{} 1}^{n} a_i )^k \ge n^{k \plus{} 1}$, for $ k \equal{} 0,1,...,m \minus{} 2$

so

$ \sum_{k \equal{} 0}^{m \minus{} 2} \sum_{i \equal{} 1}^{n} \frac {1}{a_{i}^k} \ge \sum_{k \equal{} 0}^{m \minus{} 2} n^{k \plus{} 1} \equal{} \frac {n^{m} \minus{} n}{n \minus{} 1}$

Q.E.D.[/quote]

see my solution here [url]http://ddbdt.co.cc/forum/showthread.php?p=969#post969[/url][/url]


If you have a repelling force in which the strength varies with the inverse square of the distance (i.e. Columb's Law for like charges), imagine having one object fixed. Then, if you have the other start some distance away from this fixed object, obviously the strength of the force diminishes as this object gets further and further away. Say you let the forces act until they get d (distance) away from each other. Is there a way to calculate the speed of the non-fixed object as we let d go to infinity?

Here is the exact wording of a problem I got:
An electron starts from rest 72.5 cm from a fixed point charge with Q=-0.125 microC (1 microC = 1 x 10^-6 C). How fast will the electron be moving when it is very far away?

The constants used are the charge on an electron is 1.60 x 10^-19 C, and its mass is 9.11 x 10^-31 kg.

Thanks in advance!

applying the conservation of energy:
$\frac{1}{4\pi \varepsilon_{0}}\frac{eQ}{d}=\frac12 mv^{2}$
from here the velocity:
$v=\sqrt{\frac{eQ}{2\pi \varepsilon_{0}m d}}$
($e$ is the charge of the electron, $\varepsilon_{0}$ is the dielectrical constant, $d$ is the distance between the charge and the electron at the beginning.)

Using calculus:
the equation of motion:
$\frac{1}{4\pi \varepsilon_{0}}\frac{eQ}{s^{2}}=m \frac{dv}{dt}$
using:$v=\frac{ds}{dt}$
we get:
$\frac{1}{4\pi \varepsilon_{0}}\frac{eQ}{m}\frac{ds}{s^{2}}=vdv$
if at beginning the electron was at $d$ distance from the charge and it goes very very far:
$\frac{eQ}{4\pi \varepsilon_{0}m}\int_{d}^{\infty}\frac{ds}{s^{2}}=\int_{0}^{v}vdv$
from  here:
$v=\sqrt{\frac{eQ}{2\pi \varepsilon_{0}m d}}$
And this is just the same as before.

Thanks. That really helped me figure it out. :)


AMC AIME geometry USA(J)MO USAMO \/closed

Hard question to answer.

Depends on your comfort with counting.  If you don't find the post-test for the counting class pretty easy, I'd recommend the counting class (followed up by the intermediate counting class, which we'll offer again sometime in 2004).  Otherwise, I'd recommend the AMC/AIME class.

The counting class should give you virtually all the tools to tackle counting problems on MATHCOUNTS/AMC (and a little on the AIME) - it's a pretty thorough exploration of fundamental counting problems.  If you generally tend to make mistakes on those problems now, this class will help with that.  If you already cover those problems very easily, then go for the AMC/AIME class.

[quote="Rep123max"]If I had to choose between the AMC/AIME problem series class and the MATHCOUNTS/AMC Counting class which would you recommend?[/quote]

What are your goals?

The AMC/AIME class will examine the last ten problems from AMC/AHSME exams and many AIME problems of all levels.  It is a class built for the kinds of students who can already score 2-10 problems on the AIME.  Dozens of topics and problem ideas will be discussed.

Our subject classes are designed to give a more thorough view of each individual subject -- something good for any student who wants to build a comprehensive base in each of the subject areas that are covered on the USAMO, though the subject classes are certainly also very helpful with AMC and AIME problems.


attatchment contains all information.............

would u all mind shifting it to an appropriate  forum. i am curious to know the answer


On my Get Problem page, the info isnt showing up, I have tried clicking refresh, but it didnt work. please reply someone.

Have you possibly ran out of [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints]karma[/url]?

no i havent, i fixed it, sorry for your time. :blush:


find the least value and the greatest value of the expression
 $P= \frac{x+y}{1+z}+ \frac{z+y}{1+x}+ \frac{x+z}{1+y}$ where x,y,z are real numbers belonging to the segment $[1/2;1]$

The given expression is convex in each of the variables, thus it reaches its maximum at the endpoint of the given interval...

Pierre.


function
calculus
integration
real analysis
real analysis unsolved

let I=[a,b] and f be riemann integrable function from I to R satisfying
  /f(x)/<=integral /f/ on [a,x] for all x in I.show f(x)=0 for all x in I.
  P.S. /X/=mod x. :huh:

Let $g: [a,b]\to R$$g(x)=|f(x)|$. We have that g is positive and $g(x) \leq \int_{a}^{x}g(t)dt$ $\forall$ $x\in [a,b]$. We need to prove that g is constant 0.

Let $M\geq 0$ such that $g(x)\leq M \forall x\in [a,b]$. Than $g(x)\leq M(x-a)$. We substitute x with t and reintegrating we get that $g(x)\leq M\frac{(x-a)^{2}}{2}$. After we reiterate this n times we get:
$0\leqq(x)\leq M\frac{(x-a)^{n}}{n!}$ $\forall x\in [a,b]$. Taking the limit we get that g(x)=0 $\forall x\in [a,b]$.

This problem also appeared in a strange form with a different solution $\int_{a}^{b}f^{n}(t)dt\leq (\int_{a}^{b}f(t)dt)^{n}$ but the conclusion is obvious once you know the above problem :)

BTW arline, I like your problems :)  and I met most of them in the last weeks. I wonder if we were working the same book (mine is an olympiad romanian book for analysis and superior algebra...) What's your source?

I've also seen this in a book, but with a slightly different solution:
Fix $x \in [a,b]$. Let $M = \sup_{t \in [a,x]}|f(t)|$. We have that $|f(t)| \leq \int_{a}^{x}|f(u)| \, du \leq (x-a) M$ for all $t \in [a,x]$. Taking sups, we get that $M \leq (x-a)M$. If $|x-a| < 1$, then this implies $M = 0$, so $f \equiv 0$ on $\left[ a, a+1 \right)$. We can easily cover the whole interval.

[quote]This problem also appeared in a strange form with a different solution $\int_{a}^{b}f^{n}(t) \, dt \leq \left( \int_{a}^{b}f(t) \, dt \right)^{n}$ but the conclusion is obvious once you know the above problem :)[/quote]
What do you mean?

Maybe not strange was the right word to use :oops: ...'trap problem' is more expressive :roll:


inequalities
function
logarithms
inequalities proposed

[quote="nguoivn"]Given $ a, b, c$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ \frac {1}{2 \plus{} a^3b^3} \plus{} \frac {1}{2 \plus{} b^3c^3} \plus{} \frac {1}{2 \plus{} c^3a^3} \geq\ 1$

With this ineq, we can solve it easily by Am-Gm. But the following (of him) is harder and I still have no solution:  :oops:
$ \frac {1}{2 \plus{} a^5b^5} \plus{} \frac {1}{2 \plus{} b^5c^5} \plus{} \frac {1}{2 \plus{} c^5a^5} \geq\ 1$
(with the same condition)[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3 \equal{} x.$
Hence, $ \sum_{cyc}\frac {1}{2 \plus{} a^5b^5}\leq1\Leftrightarrow(abc)^{10} \plus{} (a^5 \plus{} b^5 \plus{} c^5)(abc)^5\leq4\Leftrightarrow$
$ \Leftrightarrow w^{30} \plus{} (243u^5 \minus{} 405u^3v^2 \plus{} 135uv^4 \plus{} 15(3u^2 \minus{} v^2)w^3)u^{10}w^{15} \minus{} 4u^{30}\leq0$
$ \Leftrightarrow w^{30} \plus{} 15(3u^2 \minus{} v^2)u^{10}w^{18} \plus{} 27(9u^4 \minus{} 15u^2v^2 \plus{} 5v^4)u^{11}w^{15} \minus{} 4u^{30}\leq0$
Let $ f(x) \equal{} x^{10} \plus{} 15(3u^2 \minus{} v^2)u^{10}x^6 \plus{} 27(9u^4 \minus{} 15u^2v^2 \plus{} 5v^4)u^{11}x^5 \minus{} 4u^{30}.$
We obtain: $ f''(x) \equal{} 90x^3(x^5 \plus{} 5(3u^2 \minus{} v^2)u^{10}x \plus{} 6(9u^4 \minus{} 15u^2v^2 \plus{} 5v^4)u^{11})\geq0$ because
$ g(x) \equal{} x^5 \plus{} 5(3u^2 \minus{} v^2)u^{10}x \plus{} 6(9u^4 \minus{} 15u^2v^2 \plus{} 5v^4)u^{11}$
 is an increasing function of $ x$ and we need to check only
$ 9u^4 \minus{} 15u^2v^2 \plus{} 5v^4\geq0$ for $ w^3 \equal{} 0,$ which is obviously true.
Thus, $ f$ is a convex function.
Hence, $ f$ gets the maximal value on bounds of $ w^3.$
Id est, it remains to check the original inequality for $ b \equal{} c$ and for $ c \equal{} 0,$ which are true.

The case $ b\equal{}c$ is not easy to check  :maybe: . Hope that we'll find the nicer proofs for it  :)

I am sure that no one has enough free time to check these stuffs  :D  properly. If this problem is proposed in IMO, it would be a nice surprise. :P

By Holder ,the inequality is equivalent with :

$ a^3 (2\plus{}a^3 b^3 ) \plus{} b^3 (2\plus{}b^3 c^3 ) \plus{} c^3(2 \plus{}c^3 a^3) \geq 9$

Is this true ?  :blush:

[quote="enndb0x"]By Holder ,the inequality is equivalent with :

$ a^3 (2 \plus{} a^3 b^3 ) \plus{} b^3 (2 \plus{} b^3 c^3 ) \plus{} c^3(2 \plus{} c^3 a^3) \geq 9$

Is this true ?  :blush:[/quote]

You shouldn't say "is equivalent with" but rather "would follow from",  that's different  :!:

I think the stronger also trues: $ a^5b^5c^5(a^5\plus{}b^5\plus{}c^5) \leq\ 3$
With $ a, b, c > 0$ and $ a\plus{}b\plus{}c\equal{}3$.
The general problem: Find all $ k$ for this ineq is trues (with the same condition):
$ a^kb^kc^k(a^k\plus{}b^k\plus{}c^k) \leq\ 3$

[quote="nguoivn"]I think the stronger also trues: $ a^5b^5c^5(a^5 \plus{} b^5 \plus{} c^5) \leq\ 3$
With $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
[/quote]

You can try $ a\equal{}1 \ ; \ b\equal{}0,9 \ ; \ c\equal{}1,1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

Mạc Thị Thanh Hoa

Thank you, Blueflower. Maybe the following is trues:
$ a^4b^4c^4(a^4\plus{}b^4\plus{}c^4) \leq\ 3$  :maybe:
With $ a^3b^3c^3(a^3\plus{}b^3\plus{}c^3) \leq\ 3$, I had had solution by Am-Gm.

[quote="nguoivn"]Thank you, Blueflower. Maybe the following is trues:
$ a^4b^4c^4(a^4 \plus{} b^4 \plus{} c^4) \leq\ 3$  :maybe:
[/quote]

Try $ a\equal{}2 \ ; \ b\equal{}c\equal{}1$

P/S : [hide]nguoivn : anh có biết anh Zi ( Materazzi ) nhà em đi đâu rồi ko ạ :(([/hide]

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Mạc Thị Thanh Hoa

With the problem

$ (abc)^3(a^3\plus{}b^3\plus{}c^3) \le 3$ then we can actually prove by Mixing Varable although it seems to us a little bit messy

[quote="Blueflower"][quote="nguoivn"]Thank you, Blueflower. Maybe the following is trues:
$ a^4b^4c^4(a^4 \plus{} b^4 \plus{} c^4) \leq\ 3$  :maybe:
[/quote]

Try $ a \equal{} 2 \ ; \ b \equal{} c \equal{} 1$

P/S : [hide]nguoivn : anh có biết anh Zi ( Materazzi ) nhà em đi đâu rồi ko ạ :(([/hide]

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Mạc Thị Thanh Hoa[/quote]
Dear my friend,but $ a\plus{}b\plus{}c\equal{}3$
:)Chị là em gái hay bạn gái anh materazzi :)

[quote="Blueflower"][quote="nguoivn"]Thank you, Blueflower. Maybe the following is trues:
$ a^4b^4c^4(a^4 \plus{} b^4 \plus{} c^4) \leq\ 3$  :maybe:
[/quote]

Try $ a \equal{} 2 \ ; \ b \equal{} c \equal{} 1$

P/S : [hide]nguoivn : anh có biết anh Zi ( Materazzi ) nhà em đi đâu rồi ko ạ :(([/hide]

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Mạc Thị Thanh Hoa[/quote]

But $ a\plus{}b\plus{}c\equal{}3$, Blueflower. I think this ineq is trues.
From it, we have a nice result:
Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that: $ \frac{a\plus{}b\plus{}c}{3} \geq\ \sqrt[16]{\frac{a^4\plus{}b^4\plus{}c^4}{3}}$
A similar result and well-known (with the same condition): $ \frac{a\plus{}b\plus{}c}{3} \geq\ \sqrt[10]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}$
[hide]@Blueflower: Ec, sao anh bit dc. Ku Thang' len day ma` con` bi nguoi` yeu duoi?. Po tay ong em nay` lun:| [/hide]

[quote="nguoivn"]
From it, we have a nice result:
Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that: $ \frac {a \plus{} b \plus{} c}{3} \geq\ \sqrt [16]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}$[/quote]
which is wrong. Try $ a\equal{}1.44$ and $ b\equal{}c\equal{}\frac{5}{6}.$

[quote="nguoivn"]
@Blueflower: Ec, sao anh bit dc. Ku Thang' len day ma` con` bi nguoi` yeu duoi?. Po tay ong em nay` lun:| [/quote]

[hide]Ai mà thèm yêu anh ấy chứ  :blush:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

Mạc Thị Thanh Hoa[/hide]

By the way, the following inequality is true.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3} \geq\ \sqrt [17]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}\]

[quote="arqady"]By the way, the following inequality is true.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3} \geq\ \sqrt [17]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}\]
[/quote]
Did you have solution for it, arqady?
I think with $ abc\equal{}1$, this ineq is trues: $ \sqrt[10]{\frac{a^3\plus{}b^3\plus{}c^3}{3}} \geq\ \sqrt[17]{\frac{a^4\plus{}b^4\plus{}c^4}{3}}$  :maybe:

[quote="nguoivn"][quote="arqady"]By the way, the following inequality is true.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that
\[ \frac {a \plus{} b \plus{} c}{3} \geq\ \sqrt [17]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}\]
[/quote]
Did you have solution for it, arqady? [/quote]
Here is my proof.
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ \frac {a \plus{} b \plus{} c}{3} \geq\ \sqrt [17]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}\Leftrightarrow$
$ \Leftrightarrow u^{17}\geq(27u^4 \minus{} 36u^2v^2 \plus{} 6v^4 \plus{} 4uw^3)w^{13}\Leftrightarrow f(x)\leq0,$ where $ x \equal{} w^3$ and
$ f(x) \equal{} 4ux^{\frac {16}{3}} \plus{} 3(9u^4 \minus{} 12u^2v^2 \plus{} 2v^4)x^{\frac {13}{3}} \minus{} u^{17}.$
Hence, $ f''(x) \equal{} 4\cdot\frac {16}{3}\cdot\frac {13}{3}ux^{\frac {10}{3}} \plus{} 3\cdot\frac {13}{3}\cdot\frac {10}{3}(9u^4 \minus{} 12u^2v^2 \plus{} 2v^4)x^{\frac {7}{3}} \equal{}$
$ \equal{} \frac {26w^7}{9}\left(32uw^3 \plus{} 15(9u^4 \minus{} 12u^2v^2 \plus{} 2v^4)\right) \equal{}$
$ \equal{} \frac {26w^7}{9}\left(32uw^3 \plus{} 15\cdot\frac {a^4 \plus{} b^4 \plus{} c^4 \minus{} 12uw^3}{9}\right) \equal{}$
$ \equal{} \frac {26w^7}{27}(36uw^3 \plus{} 5(a^4 \plus{} b^4 \plus{} c^4))\geq0.$
Thus, $ f$ is a convex function.
Id est, $ f$ gets a maximal value on bounders of $ w^3,$ which holds when
1) $ w^3 \equal{} 0$ ( for homogeneous inequality ), which is true and
2) when  $ b \equal{} c.$
Let $ b \equal{} c \equal{} 1$ in homogeneous inequality.
Hence, we need to check here $ \left(\frac {a \plus{} 2}{3}\right)^{17}\geq\frac {a^4 \plus{} 2}{3}\cdot a^{\frac {13}{3}},$
which is equivalent to $ g(a)\geq0,$ where $ g(a) \equal{} 17\ln(a \plus{} 2) \minus{} \ln(a^4 \plus{} 2) \minus{} \frac {13}{3}\ln a \minus{} 16\ln3.$
But $ g'(a) \equal{} \frac {17}{a \plus{} 2} \minus{} \frac {4a^3}{a^4 \plus{} 2} \minus{} \frac {13}{3a} \equal{} \frac {2(a \minus{} 1)(13a^4 \minus{} 12a^3 \minus{} 12a^2 \minus{} 12a \plus{} 26)}{3a(a \plus{} 2)(a^4 \plus{} 2)}.$
Easy to show that $ 13a^4 \minus{} 12a^3 \minus{} 12a^2 \minus{} 12a \plus{} 26 > 0.$ Hence, $ a_{min} \equal{} 1.$
Hence, $ g(a)\geq g(1) \equal{} 0.$  Done! :)
[quote="nguoivn"]
I think with $ abc \equal{} 1$, this ineq is trues: $ \sqrt [10]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \geq\ \sqrt [17]{\frac {a^4 \plus{} b^4 \plus{} c^4}{3}}$  :maybe:[/quote]
It's not true. Try $ a \equal{} 1.21$ and $ b \equal{} c \equal{} \frac {10}{11}.$ :wink:

[quote="arqady"]Hence, we need to check here $ \left(\frac {a \plus{} 2}{3}\right)^{17}\geq\frac {a^4 \plus{} 2}{3}\cdot a^{\frac {13}{3}},$
which is equivalent to $ g(a)\geq0,$ where $ g(a) \equal{} 17\ln(a \plus{} 2) \minus{} \ln(a^4 \plus{} 2) \minus{} \frac {13}{3}\ln a \minus{} 16\ln3.$[/quote]
I must say that I was suprised by this one! Very nice! :)

Thank you, Mathias_DK!
[quote="nguoivn"]The case $ b \equal{} c$ is not easy to check  :maybe: [/quote]
We need to check only that
$ (3\minus{}2b)^5\plus{}2b^5\plus{}2b^{10}(3\minus{}2b)^5\minus{}b^4(3\minus{}2b)\minus{}b^3(3\minus{}2b)^2\minus{}b^2(3\minus{}2b)^3\minus{}b(3\minus{}2b)^4$
 is positive on $ [0,1.5]$ or maybe a root of him is small than $ 1$ or something similar.

Thank you, arqady with your proof  :lol:
I'm trying to use Am-Gm to solve it  :)

[quote="nguoivn"]
With $ a^3b^3c^3(a^3 \plus{} b^3 \plus{} c^3) \leq\ 3$, I had had solution by Am-Gm.[/quote]

Dear nguoivn, can you please explain better how to solve $ \sum \frac{1}{1\plus{}a^3b^3}\geq 1$ by AM-GM?

Thank you very much. :)

[quote="nguoivn"]
[hide]@Blueflower: Ec, sao anh bit dc. Ku Thang' len day ma` con` bi nguoi` yeu duoi?. Po tay ong em nay` lun:| [/hide][/quote]

[hide]@nguoivn : đàn em vô tội  :lol: [/hide]
[hide]@Blueflower: anh đây anh đây :( .... chạy theo anh lên tới đây luôn mới ghê chứ :([/hide]


The measure of the supplement of an angle is three times the measure of its complement. How many degrees are in the measure of the angle?

The supplement of an angle is always 90 degrees more than the complement. So, let x be the complement, and
x+90=3x
x=45

Since the complement is 45 degrees, the angle is 90-45= $ \boxed{45}$

Let the angle be $ x.$ The complement of angle $ x$ is $ 90\minus{}x$ and the supplement is $ 180\minus{}x.$ We then have the equation $ 180\minus{}x\equal{}3(90\minus{}x).$Solving, we get that $ \angle x\equal{}45^\circ.$


LaTeX
real analysis
real analysis unsolved

1.Let $f:R\to R$ derivable such that: for any $a,b$ real numbers with $a<b$ exists and it's unique a point
 $c$ in $(a,b)$ such that $f^'(c)=f(b)-f(a)/b-a$.Prove that $f$ has no inflexion points.

 2. Let $f:R\to R$ derivable such that $f^{''}(x)>=0$ and $f(x)<=c$ for every $x$ real number.
     Prove that $f$ is constant.

I hope you do understand what I've written.

cezar learn latex see top of mathlinks Latex help

[quote="cezar lupu"]1.Let $f:R\to R$ derivable such that: for any $a,b$ real numbers with $a<b$ exists and it's unique a point
 $c$ in $(a,b)$ such that $f^{'}(c)=\frac{f(b)-f(a)}{b-a}$.Prove that $f$ has no inflexion points.

 2. Let $f:R\to R$ derivable such that $f^{''}(x)\geq 0$ and $f(x)\leq c$ for every $x$ real number.
     Prove that $f$ is constant.[/quote]

I don't understand what you're saying in 1)

For 2), it is clear that $f'$ is everywhere increasing.  Say that we have $f'(x)>0$ for some $x$.  Then there exists $a>x$ and $c\in(x,a)$ such that $f'(c)=\frac {f(a)-f(x)}{x-a}\leq \frac {c-f(x)}{a-x}<\epsilon$ for every $\epsilon>0$.  In particular, we can find a $c$ such that $f'(x)>f'(c)$, a contradiction.

Otherwise, $f'(x)<0$ for all $x$.  Fix a point $x$.  Then there exists a $c\in(a,x)$ such that $f'(c)=\frac {f(x)-f(a)}{x-a}\geq \frac {f(x)-c}{x-a}$.  As $a\to -\infty$, we can find arbitrarily small $f'(c)$, which contradicts that $f'$ is everywhere increasing.

To understand better, we call a inflexion point a number $x$ a root for the equation $f''(x)=0$.All $x$ which verify
 that equality are called inflexion points. ;)


[quote="007math"][quote="tcjy"]The age of a man is the same as his wife's age with the digits reversed. The sum of their ages is 99 and the man is 9 years older than his wife. How old is the man?[/quote]

Let $ 10a \plus{} b$ be the age of the man.  Therefore, his wife's age is $ 10b \plus{} a$.  We get:
\[ 11a \plus{} 11b \equal{} 99
\]

\[ 9a \minus{} 9b \equal{} 9
\]
Solving this system, we get $ (a,b) \equal{} (5,4)$, so the man is $ \boxed{54 \text{ years}}$[/quote]

this is the correct answer.

my way is to guess and check.

umm..
u just revived a month old post, if u haven't noticed... :o

dude this is such an old contest problem! :roll:

54
why are all of the problems that are like this have the same answer?

plz in the future refrain from answering the same question unless
1) u have a different solution
2) u think the other person is wrong
thx :lol:  :wink:

[quote="Twilight5331"]If pi is 3.14159265358979323846, what is the square root of pi?[/quote]
I see you are new, but if you want to post a new problem, you should post a new topic unless it is related.
BTW, the answer can be found using your calculator, which is $ \approx1.772453851$.

I hate to be one of those people who are always criticizing other peoples' posts...

[quote="isabella2296"]4. Keep in mind that each thread is only, and I repeat, only, for purposes of solving the problem shown.

-- Be careful not to stray off topic. [/quote]

Moderators, please lock.

What is sum of number from 11 to 55 with reversed digits ?
 :roll:

uhh dude
[quote]I see you are new, but if you want to post a new problem, you should post a new topic unless it is related.
BTW, the answer can be found using your calculator, which is .
[/quote]
soo don't need to post a np... :wink:

What is the expanded value of x(x+5)(x+5+3a)?  Good luck.

No offense but [b] DUDE R U BLIND! CAN YOU READ WAT I POSTED BEFORE U![/b]

[quote="tcjy"]The age of a man is the same as his wife's age with the digits reversed. The sum of their ages is 99 and the man is 9 years older than his wife. How old is the man?[/quote]
A number reversed so that it is 9 less than the original number has to have a ones digit 1 less than the 10s so you find two reasonable tens digits like 10 20 30 that when doubled is around the 90s or 100s which it should be 5 then subtract 1 and use that as your ones digit so 54 is the mans age and revearsed is 45 which 45+54=99.  Or you could use algebra which is slower

FOR THE LAST TIME STOP REVIVING.
there my angers out. rip now.

Why do you have a problem with people doing what they want with their posting privileges? If you're going to post, at least contribute to the discussion, not be a "moderator" Also there was never a rule against reviving.

[quote="cgyao15"]No offense but [b]DUDE R U BLIND! CAN YOU READ WAT I POSTED BEFORE U![/b][/quote]
Not the best thing to say to get on admins' good side ;)

Again, moderators/admins, please lock this thread.

[i]Contents originally posted by my sister while I was logged on under the Automatic Login option, now deleted by me.[/i]

the wife's age= (99-9)/2=45
the man is 54


the sequence $u_n$ has the following properties:

$\forall n \exists i>n$ s.t. $u_i>0$
$\forall n \exists j<n$ s.t. $u_j<0$
$\forall i\in\mathbb{Z^+}: i-j|u_i-u_j$

prove every positive integer appears exactly once in $\{u_n\}$

There is a missing hypothesis, because if $\{U_n\}$ has the desired properties then so has $\{kU_n \}$ for each non-zero integer $k$, but in this last case you will only have the multiples of $k$...

Pierre.


limit integration function calculus algebra polynomial Gauss

Show that [tex]\displaystyle\lim_{n\rightarrow\infty}n\int_0^1 x^nf(x)\,dx = f(1)[/tex], where the function [tex]f[/tex] is continuous at 1 and Riemann integrable on the interval [tex][0,\,1][/tex].

(Edited) Thanks to JBL for pointing out an error in the statement.

This was one of my favorite exercises using [tex]\epsilon-\delta, \epsilon-\-N[/tex] language for limit problems when I was a freshman.

[quote="JBL"]I assume that you also want the function to be integrable.  In that case, our function f must be a uniform limit of step functions [tex]f_n:[0, 1]\to\mathbb{R}[/tex], defined by points [tex]\{a_0, a_1, ..., a_m\}\subset[0,1], a_i<a_{i+1}, a_0 = 0, a_m = 1[/tex], f constant on [tex](a_i,a_{i+1}][/tex].  Since the convergence is uniform, we may switch our limit outside the integral.  It is easy to verify the given statement in the case of step functions and the desired result follows for all integrable functions immediately.[/quote]

Do you need the condition that [tex]f[/tex] is continuous at 1?

No, that's fine. I was just wondering because the last few problems you've posted have been rather advanced for a secondary school student. They're good questions though.

Hmm, that's a good question.  It is definitely easy to verify for step functions ... I think the nature of the convergence and how I constructed the functions (so that each of the step functions are continuous at 1) implies that f is continuous at 1.  At least, I hope it does, because otherwise there is obviously something wrong with my proof :)

Perhaps I should actually try to do it with epsilons :)

I think we need the condition that f is continuous at 1.  Otherwise we could change f(1) to anything we want, and falsify the equation.

Right, we need it.  I purported to have solved the question.  Thus, if I am correct, I must have used that property somewhere.

JBL,

Assume that [tex]f(x)=1[/tex] for all [tex]x\in[0,\,1][/tex], then [tex]\displaystyle\lim_{n\rightarrow\infty}n\int_0^1 x^nf(x)\,dx = \lim_{n\rightarrow\infty}\frac{n}{n+1}=1[/tex]. On the other hand, if we switch the limit and the integral, we have [tex]\displaystyle\int_0^1\lim_{n\rightarrow\infty}\left(n x^nf(x)\right)\,dx = 0[/tex]. Therefore, your switching method doesn't work here.

It's time to try to do it with epsilons. :-)

JBL wrote:f must be a uniform limit of step functions.

I don't think this is true for *all* Riemann integrable functions, only a strict sub-set (so-called "ruled functions"). For instance f(x) = sin(1/x), f(0)=0 is Riemann integrable but not ruled.

I think this may work :

[hide]Let eps > 0, there exist a > 0 such as |1-x|<=a => |f(x)-f(1)|< eps, or f(x) = f(1) + g(x) with |g(x)|< eps.

Let M >= 0 a bound of |f|

I(n) = int (0..1)(x^n f(x)) = int (0..a)(x^n f(x)) + int (a..1)(x^n f(x))

I1(n) = | int (0..a)(x^n f(x)) | <= a^n / (n+1) M

int (a..1)(x^n f(x)) = int (a..1)(x^n f(1)) + int (a..1)(x^n g(x))

I2(n) = int (a..1)(x^n f(1)) = f(1) (1-a^n)/(n+1)

I3(n) = | int (a..1)(x^n g(x))| <= eps (1-a^n)/(n+1) <= eps

I1(n) -> 0, so for n >= N0, |I1(n)| < eps

I2(n) -> f(1), so for n >= N1, |I2(n) - f(1)| < eps

So for n > max(N1,N2), | I(n) - f(1) | < 3*eps. QED.[/hide]

belenos, excellent job! (The main idea is right, but you missed n in the front of the integral).

The original version of this problem assume that [tex]f[/tex] is continuous on the whole interval [tex][0,\,1][/tex]. Then I realized that it can be approximated uniformly by a sequence of polynomials (the therorem itself is more advanced than this problem!). Using the approximation (I am not sure whether it works or not) might defeat the original purpose (an exercise for epsilons :-)).

I am going to ask you the same question that gauss asked me a while ago :-) in this thread. [quote]what grade are you in?[/quote]
But I am putting up the reason first so that I don't scare you off.  I am just wondering because you solved many "Advanced " problems posted here. By the way, you don't have to answer my question if you don't feel comfortable.

[quote="belenos"][quote="fuzzylogic"] what grade are you in?[/quote]
No problem, see private message  :)[/quote]

You are certainly smarter than me! :-)

[quote="fuzzylogic"][quote="gauss202"]Fuzzy what grade are you in?[/quote]

Retired stock trader. math is my hobby. A lot of stuff is coming from memory. Sometimes, statements in my post may not be entirely accurate. Sorry for the inconvenience if that happens.[/quote]

whew. all this time I thought you were a brilliant 8th grader for some reason.

I actually wasn't trying to switch that limit -- if I was correct and we can write every Riemann-Integrable function as a uniform limit of step functions, then we write f as a limit of functions of x and switch that limit outside the integral.  Because that limit would be uniform, we should be able to do the switching.

I had thought that Riemann-integrable functions were equivalent to ruled functions, but I could be wrong -- I don't even know explicitly what Riemann-integrable means.  Are you sure Sin(1/x) is riemann-integrable?

Riemann integrable :

it's explained (briefly!) there : http://planetmath.org/encyclopedia/RiemannIntegral.html

sin(1/x) :

a very strong (and beautiful) result from Lebesgue is the following :

A bounded function is Riemann integrable iff the set of points at which f is discontinuous has Lebesgue measure zero.

sin(1/x) is discontinuous only at 0.


This is a type of magic square. The eight sums from the numbers in each of the 3 rows, in each of the 3 columns, and in each of the 2 main diagonals are equal. What is the value of $ n$?

[asy]for(int i = 0; i<4; ++i)
{
	draw((0,i)--(3,i),linewidth(1));
}

for(int j = 0; j<4; ++j)
{
	draw((j,0)--(j,3),linewidth(1));
}

label("$n-3$",(.5,.5));
label("3",(.5,1.5));
label("$n+1$",(.5,2.5));

label("$n+2$",(1.5,.5));
label("$2n-9$",(1.5,1.5));
label("$1$",(1.5,2.5));

label("$2$",(2.5,.5));
label("$n$",(2.5,1.5));
label("$n-1$",(2.5,2.5));[/asy]

We can just use two of the rows, columns, or diagonals to write a linear equation, and solve.  Using the first column and the second column yields:

\begin{align*}
(n+1)+(3)+(n-3)&=(1)+(2n-9)+(n+2)
\\2n+1&=3n-6
\\n&=\boxed{7}
\end{align*}


group theory
abstract algebra
superior algebra
superior algebra theorems

Hi,

Is it true that if $G_1$ is isomorph to $G_2$ and the normal subgroups $H_1$ of $G_1$ and $H_2$ of $G_2$ are isomorph, the quotient groups $G_1/H_1$ and $G_2/H_2$ are isomorph ? It's true if it's the same morphism restricted but is it true in general ?

Thank you

Consider $C_4 \times C_2 = \langle g \rangle \times \langle h \rangle$.  The subgroups $\langle g^2 \rangle$ and $\langle h \rangle$ are isomorphic, being $C_2$, but the quotients are $C_2 \times C_2$ and $C_4$.


1. Bachukas draw a triangle on a plane, what's the probability that a triangle is acute?
(This problem is equivalent to: 'a+b+c=180; a,b,c - positive real numbers. What's the probability that a<90,b<90 and c<90?')

2. Bachukas draw a hexagon on a plane, what's the probability that a hexagon is convex?

I guess that answer to my first problem is $ 1/2$ and the answer to my second problem is $ 1/2^3\equal{}1/8$, am I right?

For your first one (of course, I'm not sure), I'm leaning towards $ \frac12$ as well.

For your second one, I think it may be $ \left(\frac12\right)^5$ or $ \frac1{32}$.  But I think it may be smaller.

Well, the question more complex than you could imagine :)

http://www.jstor.org/pss/3616588
http://jd2718.wordpress.com/2007/09/01/prepuzzle-puzzle-what-is-a-random-triangle/
http://www.ratio.huji.ac.il/dp_files/dp235.doc

For the first one, there are $ 4$ conditions:
a) $ a < 90$
b) $ b < 90$
c) $ 180 \minus{} a \minus{} b < 90\implies a \plus{} b > 90$
d) Also, we must have $ a\plus{}b<180$, as this is a triangle.

Now, on a graph with $ a$ as the x-axis and $ b$ as the y-axis, this looks like
[geogebra]d122927f186efb65c75eeb127df10815db9cda85[/geogebra]

You can tell that the probability is $ \frac14$.

The hexagon is much more complicated. :wink:

EDIT: Diagram fixed. Bachukas, that's only considering integer values, which isn't "rigorous".

Thanks hsiljak, answer to my first problem is $ \frac{1}{4}$, not $ \frac{1}{2}$  :o
I found fast way of solving that problem:
What would be the answer, if degrees could be only positive integers?

It's $ \frac{\binom{92}{2}}{\binom{182}{2}}\equal{}\frac{4186}{16471}\approx\frac{1}{4}$

Solution to my second problem:

The probability that a random triangle is convex - 1.

The probability that a random quadriteral is convex - 1/2 :
We choose a point in one of the triangle's side, we can move that point either inside, or outside of the triangle.

The probability that a random pentagon is convex - 1/2*1/2:
We choose a point in one of the convex quadriteral's side, we can move that point either inside, or outside of the convex quadriteral.

The probability that a random hexagon is convex - 1/2*1/2*1/2:
We choose a point in one of the convex pentagon's side, we can move that point either inside, or outside of the convex pentagon.

What are you thinking about this solution?  :maybe:


[quote="williams"]:D solve for x and y if x+y=5,and x^y+y^x=17[/quote]
[hide]$ (2,3)$
[/hide]

[hide]and $ (3,2)$. You now need to show that these are the only solutions, i.e. that the curves $ x^{5 \minus{} x}$ and $ 17 \minus{} (5 \minus{} x)^x$ intersect in exactly two points-it isn't hard, with or without calculus.[/hide]


I was just curious if there was any way you could have a more structured Location thing in the Profile box... Would it be possible to have a list of countries, and if they were from the U.S.  a list of states to choose from as well?

Also, is there any way you can allow students to put in their grade?  I think that would be nifty!

Lemme know.  :D

Won't happen soon, but we'll put it on the todo list.

[quote="Ragingg"]Also, is there any way you can allow students to put in their grade?  I think that would be nifty![/quote]

Would be quite hard I think... since there are many students who are partially in more than 1 grade :)

[quote="Ragingg"]Would it be possible to have a list of countries[/quote]

Isn't that there? :?
when selecting flag I could not type my country, could only select it in a list...


function
floor function
induction
algebra

Calculate the sum :
\[ \left[\frac {n \plus{} 1}{2}\right] \plus{} \left[\frac {n \plus{} 2}{2^2}\right] \plus{} ... \plus{} \left[\frac {n \plus{} 2^k}{2^{k \plus{} 1}}\right] \plus{} ...
\]

I assume the square brackets denote the floor function?

Edit: ... and I assume also that you mean $ \left\lfloor \frac {n \plus{} k}{2^{k \plus{} 1}} \right\rfloor$? Or is it $ \left\lfloor \frac {n \plus{} 2^k}{2^{k \plus{} 1}} \right\rfloor$?

[quote="worthawholebean"]I assume the square brackets denote the floor function?[/quote]

Greatest integer function!

[hide="Hint"] Let $ S_n$ be the sum in question and compute $ S_{n\plus{}1} \minus{} S_n$. [/hide]

[hide="Solution"]Let $ n\equal{}a_m2^m\plus{}a_{n\minus{}1}2^m\minus{}1\plus{}...\plus{}a_0$ ; $ a_m\equal{}1$
\[ a_1,...,a_{m\minus{}1} \in {0,1}\]
If $ k>m$ $ \boxed{\left[\frac{n\plus{}2^k}{2^{k\plus{}1}}\right]\equal{}0}$
If $ k\equal{}m$ $ \boxed{\left[\frac{n\plus{}2^k}{2^{k\plus{}1}}\right]\equal{}1\equal{}a_m}$
For $ k<m$ we have the cases:
[color=red][b]1)[/b][/color]
$ a_k\equal{}0$ $ \implies$ $ n\plus{}2^k\equal{}a_m2^m\plus{}...\plus{}a_{k\plus{}1}2^{k\plus{}1}\plus{}2^k\plus{}a_{n\minus{}1}2^{k\minus{}1}\plus{}...\plus{}a_0$
\[ \left[\frac{n\plus{}2^k}{2^{k\plus{}1}}\right]\equal{}a_m2^{m\minus{}k\plus{}1}\plus{}a_{m\minus{}1}2^{m\minus{}1\minus{}k\minus{}1}\plus{}...\plus{}a_{k\plus{}1}.\]
[color=red][b]2)[/b][/color]
$ a_k\equal{}1$ therefore $ \left[\frac{n\plus{}2^k}{2^{k\plus{}1}}\right]\equal{}a_m2^{m\minus{}k\minus{}1}\plus{}a_{m\minus{}1}2^{m\minus{}1\minus{}k\minus{}1}\plus{}...\plus{}a_{k\plus{}1}\plus{}1$

In both cases we have:$ \left[\frac{n\plus{}2^k}{2^{k\plus{}1}}\right]\equal{}a_m2^{m\minus{}k\minus{}1}\plus{}a_{m\minus{}1}2^{m\minus{}k\minus{}2}\plus{}...\plus{}a_{k\plus{}1}\plus{}a_k$
\[ \sum_{k\equal{}0}^{\infty} \left[\frac{n\plus{}2^k}{2^{k\plus{}1}}\right]\equal{}a_m(2^{m\minus{}1}\plus{}2^{m\minus{}2}\plus{}...\plus{}2\plus{}1)\plus{}a_m\plus{}a_{m\minus{}1}(2^{m\minus{}2}\plus{}2^{m\minus{}1}\plus{}...\plus{}1)\plus{}a_{m\minus{}1}\plus{}...\plus{}a_2(2\plus{}1)\plus{}a_2\plus{}a_1\plus{}a_1\plus{}a_0\equal{}a_m(2^m\minus{}1\plus{}1)\plus{}a_{m\minus{}1}(2^{m\minus{}1}\minus{}1\plus{}1)\plus{}...\plus{}a_2(2^2\minus{}1\plus{}1)\plus{}a_1(1\plus{}1)\plus{}a_0\equal{}a_m2^m\plus{}a_{m\minus{}1}2^{m\minus{}1}\plus{}...\plus{}a_1 \cdot 2\plus{}a_0\equal{}n\]
 And this can be proved with induction![/hide]


[quote="Virgil Nicula"][hide="Here are two short and complete solutions."][b]The first method.[/b]
$2\left(\cos^2x+\cos^22x+\cos^23x\right)=2\Longleftrightarrow$
$2\cos^2x+(1+\cos 4x)+(1+\cos 6x)=2\Longleftrightarrow$
$2\cos^2x+\cos 4x+\cos 6x=0\Longleftrightarrow$
$2\cos^2x+2\cos 5x\cos x=0\Longleftrightarrow$
$\cos x(\cos x+\cos 5x)=0\Longleftrightarrow$
$\cos x\cos 2x\cos 3x=0\Longleftrightarrow$
$\boxed {x\in \pi Z+A,\ \mathrm {\ where\ }A=\left\{\frac{\pi}{6}\ ,\ \frac{\pi}{4}\ ,\frac{\pi}{2}\ ,\ \frac{3\pi}{4}\ ,\ \frac{5\pi}{6}\right\}}\ .$
[b]The second method.[/b]
Denote $z=\cos x+i\cdot \sin x$, $x\in [0,2\pi )$. Thus, $z\cdot \overline z=1$ and for any $n\in N$,
$z^n=\cos nx+i\cdot \sin nx$. Therefore, $\boxed {\ \cos nx=\frac{z^{2n}+1}{2z^n}\ }\ .$
The our equation becomes: $\left(\frac{z^2+1}{2z}\right)^2+\left(\frac{z^4+1}{2z^2}\right)^2+\left(\frac{z^6+1}{2z^3}\right)^2=1\Longleftrightarrow$
$z^{12}+z^{10}+z^8+2z^6+z^4+z^2+1=0\ \ \boxed {z^2=t}\Longleftrightarrow$
$t^6+t^5+t^4+2t^3+t^2+t+1=0\ \ \boxed {t+\frac 1t=u}\Longleftrightarrow$
$u(u^2+u-2)=0\Longleftrightarrow u\in \{-2,0,1\}\Longleftrightarrow t\in \left\{-1,\pm i,\frac{1\pm i\sqrt 3}{2}\right\}\Longleftrightarrow$
$\boxed {z\in \left\{\pm i,\pm \frac{\sqrt 2}{2}(1\pm i),\pm \frac 12(\sqrt 3\pm i)\right\}}$.[/hide][/quote]
[b]Remark.[/b] \[ \boxed {\ z^2=a+bi\ ,\ a,b\in R\ ,\ b\ne 0\ ,\ \rho=\sqrt{a^2+b^2}\ } \] \[ \Downarrow\Downarrow\Downarrow \] \[ \boxed {\ z_{1,2}=\pm\left(\sqrt{\frac{\rho +a}{2}}+i\cdot \mathrm {sgn}b\cdot \sqrt{\frac{\rho-a}{2}}\right)\ } \]

$z_{1,2}$ are the roots of the equation $Z^2=a+bi$, i.e. the $2$-roots of the complex number $a+bi$. Use for the above substitution $z^2=t$.


You may use the estimation $ \frac{n^2 \minus{} n}{n^2 \sqrt{n^2 \plus{} 2n \plus{} 1}} \leq \frac{n^2 \minus{} k}{n^2 \sqrt{n^2 \plus{} k}} \leq \frac{n^2}{n^2 \sqrt{n^2}}.$

Thank you, but I got a question: how do you make those estimations?

btw/ in case ppl are wondering, the answer is $ \minus{}1$. at least that is what i get using riemann summation. :|


geometry
rotation
function
algebra
system of equations

I found these Problems Sets from [url]http://www.chem.utoronto.ca/IChO.Ontario/problems/english/englishprob.htm[/url] and I can't find any answers so I thought it might be good if we solve them because I have solved some of them but I am not sure if I got them correct. So just feel free to post answers or thoughts on these and try to keep it in order (so set 1,2,3.....).

Thanks

P.S. They are not very hard.

Problem Set 1, Problem 1

"1. Element X reacts with oxygen to produce a pure sample of X2O3. In an experiment, it is found that 1.0000 g of X produces 1.1596 g of X2O3. If the atomic weight of oxygen is known to be 15.9994 g/mol, what is the atomic weight of X?
(a) 20.70 (b) 66.85 (c) 100.2 (d) 150.4 (e) 168.9"

X reacts with oxygen according to the chemical equation

$4X+3O_{2}\longrightarrow 2X_{2}O_{3}$

So, the number of moles of X, $N_{X}$, that react is double of number of moles of oxide produced, $N_{P}$. But the number of moles is given by the quotient of mass and molar mass. So:

$\frac{1.0000g(X)}{M_{X}}= 2\frac{1.1596g(X2O3)}{2M_{X}+3M_{O}}$

where $M_{X}$ and $M_{O}$ are the molar masses of X and oxygen. Solving this we get $M_{X}= 150.4 g/mol$, and the correct answer is (d).

Problem Set 1, Problem 2

"2. An ore contains 1.34% of the mineral argentite, Ag2S, by weight. How many grams of this ore would have to be processed in order to obtain 1.00 g of pure solid silver, Ag?
(a) 74.6 g (b) 85.7 g (c) 107.9 g (d) 134.0 g (e) 171.4 g"

Whatever method is used to reduce the sulphide to the metal, for example

$Ag_{2}S+2H^{+}\longrightarrow 2Ag+H_{2}S$,

we always have that, from 1 mol of Ag2S we get 2 mol of Ag. As 1.00 g of silver correspond to 9.27 mmole, then we must use 4.64 mmole of sulphide, which corresponds to a mass of 1.15 g. Now, in 100 g of ore there is 1.34 g of pure sulphide, so the mass of ore that contains 1.15 g of Ag2S is 85.8 g. The correct answer is (b).

Problem Set 1, Problem 3

"3. When pentane, C5H12, is burned in excess oxygen, the products of the reaction are CO2(g) and H2O(l). In the balanced equation for this combustion

C5H12(g) + a O2(g) → b CO2(g) + c H2O(l)

The coefficient a should be

(a) 16 (b) 12 (c) 11 (d) 8 (e) 6."

From the stoicheometry of the reaction we must have:

$b = 5; 12 = 2c; 2a = 2b+c$.

Solving this system of equations we get b = 5, c = 6, and a = 8. So the correct answer is (d).

Problem Set 1, Problem 4

"4. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The weight percentage of carbon in cortisone is 69.98%. What is the molecular weight of cortisone?
(a) 176.5 (b) 252.2 (c) 287.6 (d) 312.8 (e) 360.4."

As the weight percentage of carbon in cortisone is 69.98%, that means that in 100 g of cortisone we will have 69.98 g of carbon. So in M g of cortisone (where M is the molecular weight) we will have

$m_{C}= \frac{69.98M}{100}g$

of carbon. But, as each molecule of cortisone has 21 atoms of carbon, then this mass must also equal (knowing that the molar mass of carbon is 12.011 g/mol) $m_{C}= 21\cdot12.011$. Equating these two relations we get M = 360.4 g/mol.
The correct answer is (e).

Problem Set 1, Problem 5

"5. Which atom has the same number of neutrons as 85Rb?

(a) 85Kr (b) 87Rb (c) 85Sr (d) 86Sr (e) 86Kr."

An atom with an atomic number one unit greater and a massa number also one unit greater will have the same number of neutrons. So the answer is (d).

Problem Set 1, Problem 6

"6. An aqueous solution of which of the following compounds will not conduct
electricity ?
(a) CH3OH (b) Ba(OH)2 (c) KOH (d) Ca(CH3COO)2
(e) All of compounds (a)-(d) will conduct electricity."

Compounds (b)-(c) are ionic, and so in aquous solution will dissociate in ions. Compound (a), methanol, is a molecular compound that will not ionize appreciably in solution. So the answer is (a).

Problem Set 1, Problem 7

"7. The first ionization energy of an element is the energy required to remove one electron from a gaseous atom of that element, that is, it is the energy required for the reaction X(g) → X+(g) + e− where X stands for any element. Which of the following elements would you expect to have the lowest first ionization energy?
(a) Mg (b) Rb (c) Li (d) Ca (e) Be."

The first ionization energy decreases down a group and increases along a period (with some exceptions). So Rubidium is the atom, from the ones given, with the lowest first ionization energy. Answer (b).

Problem Set 1, Problem 8

"8. The name of the SO32− ion is
(a) sulfite (b) sulfate (c) sulfide (d) sulfurate (e) bisulfide."

Answer (a), sulfite.

$5.$ Which atom has the same number of neutrons as $^{85}Rb$?
$(a)^{85}Kr \qquad (b)^{87}Rb \qquad (c)^{85}Sr \qquad (d)^{86}Sr \qquad (e)^{86}Kr$
[hide="Problem Set 1, Problem 5"]
So the order on he table is Krypton, Rubidium, Strontium if I remember correctly with Krypton a noble gas of course.
Let $Rb$ have $x$ protons so this isotope has $85-x$ neutrons. The answer is either $^{86}Sr$ or $^{86}Kr$ which have $86-(x+1),86-(x-1)$ neutrons respectively so the answer is $(d)$
[/hide]

Problem Set 1, Problem 9

"9. The combustion of butane, C4H10, yields only CO2(g) and H2O. A sample of butane gas at 23.8°C and 753.4 mm Hg occupies a volume of 1.85 L. How many litres of CO2(g), measured at the same temperature and pressure, will be formed on complete combustion of this C4H10?
(a) 7.40 L (b) 9.25 L (c) 12.0 L (d) 16.7 L (e) 24.0 L."

The chemical equation is

$2C_{4}H_{10}+13O_{2}\longrightarrow 8CO_{2}+10H_{2}O$.

Since we are dealing with gases we will naturally approximate their behaviour with the ideal gas law. Now, as from 2 moles of butane we get 8 of carbon dioxide, and from the gas law

$n = \frac{PV}{RT}$

we conclude that $V(CO_{2}) = 4V(C_{4}H_{10}) = 7.40 L$. The correct answer is (a).

Problem Set 1, Problem 10

"10. At 27°C and 1.00 atm, the density of a gaseous hydrocarbon is 1.22 g/L. The hydrocarbon is (a) CH4 (b) C2H4 (c) C2H6 (d) C3H8 (e) C3H6."

From the perfect gas law we have $M = \frac{dRT}{P}$, where M is the molar mass and d the density. Substituting by the given values, and using $R = 0.0820578 L atm K^{-1}mol^{-1}$, we get M = 30.05 g/mol. So the correct answer is (c).

Problem Set 1, Problem 11

"11. A mixture of 50.0 mL of ammonia and 60.0 mL of oxygen gas reacts according to the following equation: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g).
If all gases are at the same temperature and pressure, and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced ?
(a) 48.0 mL (b) 60.0 mL (c) 72.0 mL (d) 75.0 mL (e) 110 mL."

Consider ideal gas behaviour. Avogadro's principle tells us that, under the same conditions of temperature and pressure, the number of moles of a gas is directly proportional to its volume. So, for a given volume of oxygen, the volume of ammonia needed to react completely with the oxygen is given by

$V(NH3) = \frac{4V(O2)}{5}$.

In the present case, for V(O2) = 60.0 mL we would need V(NH3) = 48.0 mL. But we use 50.0 mL of ammonia, so the limiting reagent is oxygen and reaction continues until all the oxygen is consumed. So, if 5 moles of O2 produce 6 moles of water vapour, then the volume of water produced will be

$V(H2O) = \frac{6V(O2)}{5}= 72.0 mL$

The correct answer is (c).

Problem Set 1, Problem 12

"12. What is the partial pressure of SO2 if 100 g of O2 are mixed with 100 g of SO2, and the total pressure is 600 mm Hg ?
(a) 500 mm Hg (b) 400 mm Hg (c) 300 mm Hg (d) 200 mm Hg
(e) 100 mm Hg."

We have

$P(SO_{2}) = x(SO_{2})P$

where $x(SO_{2})$ is the mole fraction of SO2 and P is the total pressure. From the molar masses of the gases we get 100 g (O2) = 3.12 mol and 100 g (SO2) = 1.56 mol, and so $x(SO_{2}) = 0.333$. Finally, $P(SO_{2}) = 0.333\cdot600 mm Hg = 200 mm Hg$. The correct answer is (d).

Problem Set 1, Problem 13

"13. A 0.20-mol sample of a hydrocarbon CxHy yields, after complete combustion with excess O2 gas, 0.80 mol of CO2 and 1.0 mol of H2O. The molecular formula of CxHy is (a) C4H10 (b) C4H8 (c) C4H5 (d) C8H16 (e) C8H10."

In the complete combustion of an hydrocarbon CxHy, for every mole of CxHy we obtain x moles of CO2 and y/2 moles of H2O. Since the problem tells us that from 0.20 mol of CxHy we get 0.80 mol CO2 and 1.o mol H2O, then we must have x = 4 and y = 10. The answer is (a).

Problem Set 1, Problem 14

"14. A sample of gas is kept in a closed container of fixed volume at 250 K and 400 mmHg pressure. If the gas is heated to 375 K, its pressure increases to 600 mm Hg. By what factor will the average speed of the molecules increase ?
(a) 1.22 (b) 1.50 (c) 2.25 (d) 2.00
(e) There is not enough information to determine this."

According to the kinetic model of gases, the average speed of molecules of gas is directly proportional to the square root of absolute temperature. So,

$\frac{v_{2}}{v_{1}}= \sqrt{\frac{T_{2}}{T_{1}}}= \sqrt{\frac{375K}{250K}}= 1.22$.

The correct answer is then (a).

Problem Set 2, Problem 16

"16. On the basis of the relative strengths of intermolecular forces, predict the order of increasing boiling points of the following substances. [N.B. CH3CHO is acetaldehyde or ethanal (IUPAC)]
(a) CH3CH3 < CH3CHO < CH3CH2OH
(b) CH3CH2OH < CH3CH3 < CH3CHO
(c) CH3CHO < CH3CH2OH < CH3CH3
(d) CH3CH2OH < CH3CHO < CH3CH3
(e) CH3CH3 < CH3CH2OH < CH3CHO."

The CH3CH3 molecules (ethane) are non polar, and so between them there are weak London interactions. The acetaldehyde molecules are polar, but they cannot form hydrogen bonds between them: so the only intermolecular forces operating here are dipole-dipole interactions. Now, between the molecules of ethanol relatively strong hydrogen bonds can and do operate. So the order of increasing boiling point is ethane < ethanal < ethanol, and the correct answer is (a).

Problem Set 3, Problem 31

"31. Ksp for lead iodate, Pb(IO3)2, is 3.2 × 10−14 at a given temperature. What is the solubility of Pb(IO3)2 in mole litre–1?
(a) 1.8 × 10–7 (b) 3.6 × 10–7 (c) 2.0 × 10–5 (d) 4.0 × 10–5 (e) 8.0 × 10–5."

The solubility equilibrium is

$Pb(IO_{3})_{2}(s)+aq \rightleftharpoons Pb^{2+}(aq)+2IO_{3}^{-}(aq)$.

So we have $[Pb^{2+}] = S$ and $[IO_{3}^{-}] = 2S$ where S is the molar solubility. So $K_{sp}= [Pb^{2+}][IO_{3}^{-}]^{2}= 4S^{3}$ from which we get $S = 2\times10^{-5}mol/L$.

Problem Set 3, Problem 32

"32. A solution has the following molar concentrations: [Cl–] = 1.5 × 10–1, [Br–] = 5.0 ×10–4, [CrO42–] = 1.9 × 10–2. A solution of AgNO3 (100% dissociated) is added to the above solution, drop by drop. Which silver salt will precipitate first ?
Data: Ksp(AgCl) = 1.5 x 10–10; Ksp(AgBr)= 5.0 x 10–13; Ksp(Ag2CrO4) = 1.9 x 10–12
(a) AgCl (b) AgBr (c) Ag2CrO4 (d) AgCl and AgBr together
(e) None of the above"

A given salt will start to precipitate as soon the quotient of reaction equals Ksp, and the salt that will precipitate first is the one that requires the lower concentration of silver ions in solution. The relevant equilibria are:

$AgCl(s)+aq \rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq), K_{sp}= [Ag^{+}][Cl^{-}]$
$AgBr(s)+aq \rightleftharpoons Ag^{+}(aq)+Br^{-}(aq), K_{sp}= [Ag^{+}][Br^{-}]$
$Ag_{2}CrO_{4}(s)+aq \rightleftharpoons 2Ag^{+}(aq)+CrO_{4}^{2-}(aq), K_{sp}= [Ag^{+}]^{2}[CrO_{4}^{2-}]$.

For AgCl we have: $[Ag^{+}] = \frac{K_{sp}}{[Cl^{-}]}= 1.0\times10^{-9}M$.
For AgBr: $[Ag^{+}] = \frac{K_{sp}}{[Br^{-}]}= 1.0\times10^{-9}$.
For Ag2CrO4: $[Ag^{+}] = \sqrt{\frac{K_{sp}}{[CrO_{4}^{2-}]}}= 1.0\times10^{-5}$.
So AgCl and AgBr will precipitate first and at the same time. The correct answer is (d).

Problem Set 3, Problem 33

"33. BaSO4 is a compound that aids radiographic investigation of the gastrointestinal (GI) tract because it is opaque to radiation and only slightly soluble (Ksp = 1.1 x 10–10). Suppose that you had added some solid BaSO4 to pure water to obtain a saturated solution and wished to reduce the Ba2+ concentration to a minimum to prepare it for ingestion by a patient uncommonly allergic to Ba2+. Which procedure might be recommended ?
Data: BaSO4(s) → Ba2+(aq) + SO42–(aq) is endothermic
(a) Heat the system
(b) Add sufficient H2SO4 to give a [H+] = 1.0 molar
(c) Add more solid BaSO4
(d) Reduce the amount of saturated solution over the solid
(e) Add enough Na2SO4 to give a [Na+] = 10–2 molar."

(a) As the solubilization process is endothermic, increasing the temperature will favour the direct reaction, so more solid will dissolve and increase the concentration of Ba2+.

(b) The equilibria that describe the ionization of sulphuric acid in solution are:
$H_{2}SO_{4}(aq)+H_{2}O(l) \longrightarrow HSO_{4}^{-}(aq)+H^{+}(aq), K very large$
$HSO_{4}^{-}(aq)+H_{2}O(l) \rightleftharpoons SO_{4}^{2-}(aq)+H^{+}(aq), K small$

So the concentration of sulphate ions due to sulphuric acid will be small but the concentration of H+ will be high (1.0 M as stated in the problem), and these H+ will reduce the concentration of SO42-in solution due to

$SO_{4}^{2-}(aq)+H^{+}\rightleftharpoons HSO_{4}^{-}(aq)$

which will shift the solubility equilibrium of BaSO4 to the right, dissolving more solid and increasing the concentration of Ba2+.

(c) Adding more solid will of course favour the direct reaction, dissolving more solid and increasing the concentration of Ba2+.

(d) The concentration of ions in solution will remane the same, so this will have no effect.

(e) Adding a salt with a common ion with the solid will shift the equilibrium to the left and diminish the concentration of Ba2+. This is the common ion effect.

The correct answer is then (e).

Problem Set 3, Problem 34

"34. A 10.00g sample of Ca(OH)2 (Ksp = 4.7 x 10−6) is added to 250.0mL of pure water. Determine the pH of the solution after just enough HCl is added to completely dissolve the Ca(OH)2. Assume that there is no change in the solution volume during the addition of HCl.
(a) 10.50 (b) 11.70 (c) 12.11 (d) 11.47 (e) 10.89."

10.00 g of Ca(HO)2 correspond to 0.135 mol, so when the solid is completely dissolved the same number of moles of Ca2+ will exist in solution, that, with a volume of 0.250 L, correspond to a concentration of 0.54 M. In the end there is no solid but the solubility equilibrium still holds, and we can use it to compute the equilibrium concentration of hydroxide:

$[HO^{-}] = \sqrt{\frac{K_{sp}}{[Ca^{2+}]}}= 2.95\times10^{-3}M$.

This corresponds to a pOH of 2.53 and so a pH of 11.47 - answer (d).

Problem Set 3, Problem 35

"35. The bicarbonate ion, HCO3–, can act as a base by hydrolyzing in aqueous solution. Given only Ka1(H2CO3) and Kw, what is Kb(HCO3–)? Note: Kb(HCO3–) is K for the reaction of HCO3– acting as base: HCO3– + H2O ↔ H2CO3 + OH–
(a) Ka1 x Kw (b) Ka1/Kw (c) Kw/Ka1 (d) Ka12/Kw
(e) There is insufficient data as Ka2(H2CO3) must be given."

We have $H_{2}CO_{3}(aq) \rightleftharpoons HCO_{3}^{-}(aq)+H^{+}(aq), K_{a1}$, and so

$HCO_{3}^{-}(aq)+H^{+}(aq) \rightleftharpoons H_{2}CO_{3}(aq), \frac{1}{K_{a1}}$
$H_{2}O(l) \rightleftharpoons H^{+}(aq)+HO^{-}(aq), K_{w}$.

Adding these two equilibria we get:

$HCO_{3}^{-}(aq)+H_{2}O(l) \rightleftharpoons H_{2}CO_{3}(aq)+HO^{-}(aq), \frac{K_{w}}{K_{a1}}= K_{b}$.

The correct answer is then (c).

Problem Set 3, Problem 36

"36. What is the pH of a 1.0 molar sodium acetate, CH3COONa, solution, given that Ka(CH3COOH) = 1.8 x 10–5 and Kw = 1.0 x 10–14?
(a) 0.0 (b) 2.4 (c) 4.8 (d) 7.0 (e) 9.4."

We have:

$CH_{3}COONa(s)+aq \longrightarrow CH_{3}COO^{-}(aq)+Na^{+}(aq)$
$CH_{3}COO^{-}(aq)+H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq)+HO^{-}(aq), K = \frac{K_{w}}{K_{a}}$.

The equilibrium concentrations will be [CH3COOH] = [HO-] = x, [CH3COO-] = 1.0 - x. But, as Kw/Ka is quite small, and the initial concentration of acetate is relatively large, than we can make the approximation [CH3COO-] = 1.0 in the equilibrium. So:

$\frac{x^{2}}{1.0}= \frac{K_{w}}{K_{a}}= 5.6\times10^{-10}\Rightarrow  [HO^{-}] = x = 2.4\times10^{-5}\Rightarrow  pOH = 4.6 \Rightarrow  pH = 9.4$.

The correct answer is (e).

Problem Set 3, Problem 42

"42. Which of the following statements describes Markovnikov's rule?
(a) In the electrophilic addition of H–Cl to an alkene, the chlorine atom becomes
attached to the carbon which has the most hydrogen atoms.
(b) In the nucleophilic addition of H–Br to an alkene, the major product is formed from the more stable carbocation intermediate.
(c) In the electrophilic addition of H–Br to an alkene, the major product is formed from the more stable radical intermediate.
(d) The transition state for an endothermic process will closely resemble the product.
(e) In the electrophilic addition of H–Cl to an alkene, the chlorine atom becomes
attached to the carbon which has the least hydrogen atoms."

The Markovnikov's rule states that in the electrophilic addition of a compound A-B to an alkene, where A have a partial or total positive charge and B a partial or total negative charge, A will bond to the carbon atom of the double that possess the greatest number of hydrogen atoms, and B to the other atom. So the correct answer is (e).

Problem Set 3, Problem 43

"43. Betamethasone is a drug widely prescribed for the treatment of rheumatoid arthritis. How many degrees of unsaturation are present within the molecule?
(a) 3 (b) 4 (c) 7 (d) 8 (e) 10."

Every double bond and every ring contributes each with one degree of unsaturation, so counting them on the molecule we get the answer (d), 8.

Problem Set 3, Problem 44

"44. The number of different constitutional isomers generated by radical
monochlorination of 1,3-diethylcyclopentane is
(a) 4 (b) 5 (c) 9 (d) 12 (e) 18."

The radical chlorination is rather unselective, so every hydrogen on the molecule will be chlorinated. Noting that 1,3-diethylcyclopentane has a plane of simmetry, then the number of monochlorinated isomers is 5 - answer (b).

Problem Set 3, Problem 45

"45. The correct IUPAC name for the following compound is
(a) (3E)-3,7-dimethyl-1,3,6-octatriene
(b) (2E,5E)-2,6-dimethyl-2,5,7-octatriene
(c) (1Z,3E,6E)-3,7-dimethyl-1,3,6-octatriene
(d) (2Z)-2,6-dimethyl-2,5,7-octatriene
(e) (3E,6E)-3,7-dimethyl-1,3,6-octatriene."

The compound given has three double bounds, but only the middle one can have cis/trans isomery. We also start to number the carbon chain from the right in order to give the lowest numbers for the double bounds, and we have two methyl groups as substituents in carbons 3 and 7. The geometry of the central double bond is trans or E, so the correct answer is (a).

Problem Set 3, Problem 46

"46. A chiral compound, C5H8, on catalytic hydrogenation yields an achiral compound, C5H10. What is the original chiral compound?
(a) 1-methylcyclobutene (b) 3-methylcyclobutene
(c) 1,2-dimethylcyclopropene (d) cyclopentene
(e) 3-ethylcyclopropene."

From the compounds given, the only one that have a chiral carbon is 3-methylcyclobutene, so the answer is (b).

Problem Set 3, Problem 47

"47. Which of the following compounds has an infra red absorption band at 3300 cm–1 but no band between 1650 and 1850 cm–1 ?
(a) CH3COOH (b) CH3OCH3 (c) CH3CONH2 (d) CH3CH2OH (e) CH3COCH3."

There are three bonds that can have an IR absorption at 3300 cm-1: O-H of alcohols, N-H of amines and amides, and C-H of terminal alkynes. However, for the amide we would also have a carbonyl absorption near 1650 cm-1, so the correct answer (d).

Problem Set 3, Problem 48

"48. Which alkyl halides upon dehydrohalogenation give alkenes that do not show
cis-trans isomerism?
i) 2-chloropentane iv) 1-chloro-2-methylpropane
ii) 2-chlorobutane v) 2-chloro-2-methylpropane
iii) chlorocyclohexane
(a) i, ii and iv
(b) iii, iv and v
(c) i, iii and iv
(d) ii, iii and v
(e) ii, iii and iv."

For an alkene to not show cis-trans isomerism it must have two equal groups attached to at least one of its carbons. So the correct answer can only be (iii), (iv), and (v) - (b).

Problem Set 3, Problem 49

"49. What is the relationship between these two structures?
(a) identical compounds (b) enantiomers (c) diastereomers
(d) cis/trans isomers (e) constitutional isomers."

The structures are the same compound, since for example the structure on the right, if rotated by 180º can then be exactly superposed to the one on the left. The answer is (a).

Problem Set 3, Problem 50

"50. Which one of the following statements is false?
(a) The bond order of C2– is 2.5.
(b) The predominant source of strain energy in a planar cyclopentane is the deviation
from tetrahedral geometry at carbon.
(c) The correct IUPAC name for the following compound is
(E)-3,5,5-trimethyl-3-heptene:
(d) Halogenation of an alkene takes place with anti stereochemistry
(e) The following species can be described as having aromatic properties:

(a) True (the configuration of C2- would be $\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{2}\pi_{2p}^{2}\sigma_{2p}^{1}$, and $bond order = \frac{number of bonding electrons-number of antibonding electrons}{2})$.
(b) False.
(c) True.
(d) True.
(e) True.

Problem Set 3, Problem 39

"39. Sodium thiosulfate, Na2S2O3·5 H2O (commonly known as "hypo" from an old name, sodium hyposulfite) is used extensively in photography and volumetric analysis. Iodine reacts quantitatively with sodium thiosulfate in water and sodium tetrathionate is formed: 2 Na2S2O3 + I2 → Na2S4O6 + 2 NaI
What are the oxidation numbers of sulfur in the thiosulfate ion, S2O3
2– and the tetrathionate ion, S4O62–?
S2O32– S4O62–
(a) –2 –2.5
(b) –0.5 –1
(c) +0.5 +1
(d) +2 +2.5
(e) +3 +3."

For the thiosulfate ion we have 2x -6 = -2 => x = +2. For the tetrathionate ion we have 4x - 12 = -2 => x = +5/2. So the correct answer is (d).

Problem Set 3, Problem 40

"40. One of the steps in the manufacture of nitric acid from ammonia, NH3, involves the disproportionation (self-oxidation and self-reduction) of nitrous acid, HNO2, according to the following unbalanced equation:
a HNO2 → b NO + c HNO3 + d H2O
What are the stoichiometric coefficients when the equation is properly balanced ?
a b c d
(a) 1 1 1 1
(b) 2 2 1 1
(c) 5 2 3 1
(d) 6 6 3 3
(e) None of the above."

Let's start by computing the oxidation number of nitrogen, x, in each compound: HNO2, x = +3; NO, x = +2; HNO3, x = +5. So the semi equation of reduction is

$HNO_{2}\longrightarrow NO$,

and the semi equation of oxidation is

$HNO_{2}\longrightarrow HNO_{3}$.

Now, to balance semi equations of reduction or oxidation we follow the following steps:

1) Balance the number of atoms that undergo red. or ox. in each side of the equation.
2) Add the appropriate number of electrons, according to step 1, to the left if reduction and to the right if oxidation.
3) Balance the number of charges so as to have the same number in each side: with H+ if we are in acid medium and with HO- in alcaline medium.
4) Balance the number of hydrogen and oxygen atoms by adding enough molecules of water to the appropriate side of the equation.
5) If necessary, multiply each semi equation by a number so as to have the same number of electrons transferred in both.
6) Add the two semi equations.

Following these steps we get:

$HNO_{2}+e^{-}+H^{+}\longrightarrow NO+H_{2}O$
$HNO_{2}+H_{2}O \longrightarrow HNO_{3}+2e^{-}+2H^{+}$.

Now multiply the first one by 2 and sum:

$3HNO_{2}\longrightarrow HNO_{3}+2NO+H_{2}O$.

So the correct answer is (e).

Problem Set 3, Problem 41

"41. Calcium is the fifth most abundant element in man. The Ca2+ content in the blood of an average adult is 0.90–1.15 mg per 10 cm3 blood. In a typical analysis, a 10.0 cm3 blood sample from an adult male patient was diluted to 100 cm3. From a 10.0 cm3 portion of this diluted sample, the Ca2+ was precipitated as CaC2O4. After this precipitate had been dissolved in H2SO4, it required 1.00 cm3 of 0.00500 normal (1.00 x 10–3 mole litre–1) KMnO4 to titrate to an end point. What is the [Ca2+] expressed as milligrams of Ca2+ per 10.0 cm3 of blood? The equation for the titration reaction is:
5 CaC2O4 + 2 KMnO4 + 8 H2SO4 → K2SO4 + 2 MnSO4 + 5 CaSO4 + 10 CO2 + 8 H2O
(a) 0.100 (b) 1.00 (c) 2.00 (d) 3.20 (e) 10.0."

The number of moles of KMnO4 that reacted is $n(KMnO4) = 1.0\times10^{-3}L \cdot 1.0\times10^{-3}mol/L = 1.0\times10^{-6}mol$. According to the titration reaction, 5 moles of calcium oxalate react with 2 moles of KMnO4, so the number of moles of CaC2O4 that reacted is $n = 2.5\times10^{-6}mol$, which is also the number of moles of Ca2+ present in the 10 cm3 portion of the diluted sample of 100 cm3. As the concentration is the same, then the number of moles of Ca2+ present in the diluted sample of 100 cm3 is $n(Ca^{2+}) = \frac{100cm^{3}\cdot2.5\times10^{-6}mol}{10cm^{3}}= 2.5\times10^{-5}mol$. But this is also the number of moles initially present in the blood sample of 10 cm3. Using the molar mass of calcium, we finally get that the concentration of calcium ions is $[Ca^{2+}] = 1.00 mg/10mL$ of blood. The correct answer is then (b).

Problem Set 4, Problem 51

"51. 1.00 mole of an ideal gas at 0oC confined initially to a volume of 22.4 litres is first allowed to expand isothermally against a constant external pressure of 0.50 atm until equilibrium is attained. The gas is then compressed isothermally and reversibly back to a volume of 22.4 litres. The total enthalpy change ΔH of the gas as a result of the two processes is:
(a) 11.2 L atm (b) 15.5 L atm (c) 4.3 L atm (d) −4.3 L atm (e) 0.0 L atm."

As H is a state function, then for every cyclic process we must have ΔH = 0. As the problem describes a cyclic process, then the correct answer is (e).


How to change the line spacing to double space? Thank you

Put [code]\usepackage{setspace}
\doublespacing[/code]in the preamble - before \begin{document}


geometry
perimeter
ratio
geometry proposed

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

Let a triangle ABC . M , N , P are the midpoints of BC, CA, AB .
a) $d_1, d_2, d_3$ are lines throughing M, N, P and dividing the perimeter of triangle ABC into halves . Prove that : $d_1, d_2, d_3$ are concurrent at K .
b) Prove that : among the ratios : $\frac{KA}{BC}, \frac{KB}{AC}, \frac{KC}{AB}$, there exists at least one ratio $\geq \frac{1}{\sqrt3}$.

See [url=http://groups.yahoo.com/group/Hyacinthos/messages/9627?expand=1&viscount=8]Hyacinthos messages #9627, #9628, #9629 and some later messages[/url]. Since some people have troubles with displaying these pages, here is a copy of the relevant contents:

[quote="Hyacinthos message #9540"]
[b]From:[/b] "orl_ml" (Orlando Döhring)
[b]Subject:[/b] a ratio

On the sides of triangle ABC take the points M_1, N_1, P_1 such
that each line MM_1, NN_1, PP_1 divides the perimeter of ABC in two
equal parts (M, N, P are respectively the midpoints of the sides BC,
CA, AB).

1/ Prove that the lines MM_1, NN_1, PP_1 are concurrent at a point K.

2/ Prove that among the ratios KA/BC, KB/CA, KC/AB there exist at
least a ratio which is not less than 1/?ã3.

[/quote]

[quote="Hyacinthos message #9627"]
[b]From:[/b] "ben_goss_ro" (aka Grobber)
[b]Subject:[/b] Re: a ratio

--- In "Hyacinthos", "orl_ml" (Orlando Döhring) wrote:
> On the sides of triangle ABC take the points M_1, N_1, P_1 such
> that each line MM_1, NN_1, PP_1 divides the perimeter of ABC in two
> equal parts (M, N, P are respectively the midpoints of the sides BC,
> CA, AB).
>
> 1/ Prove that the lines MM_1, NN_1, PP_1 are concurrent at a point K.
>
> 2/ Prove that among the ratios KA/BC, KB/CA, KC/AB there exist at
> least a ratio which is not less than 1/?ã3.

(1): The intersection point is the centroid of the perimeter of
triangle ABC. It's the well-known Spieker center (X(10)). The fact
that it's the centroid of the perimeter is a good-enough argument to
show the concurrence: the centroid of the perimeter of ABC must lie
on MM_1, on NN_1 and on PP_1, so they must be concurrent.

What's the ratio for (2)? I see gibberish, but maybe it's my
browser's fault.

[/quote]

[quote="Hyacinthos message #9628"]
[b]From:[/b] Darij Grinberg
[b]Subject:[/b] Re: a ratio

Dear "ben_goss_ro" (= grobber at MathLinks?),

In Hyacinthos message #9627, you wrote:

>> (1): The intersection point is the centroid of
>> the perimeter of triangle ABC. It's the
>> well-known Spieker center (X(10)). The fact
>> that it's the centroid of the perimeter is a
>> good-enough argument to show the concurrence:
>> the centroid of the perimeter of ABC must lie
>> on MM_1, on NN_1 and on PP_1, so they must be
>> concurrent.

This puzzles me a bit. In fact, the lines MM_1,
NN_1, PP_1 do bisect the perimeter of triangle
ABC, but the cevians of the Nagel point (X(8))
do it as well; however, the former lines pass
through the centroid of the perimeter of
triangle ABC, whereas the latter ones don't!
I guess this is a consequence of my bad
physics knowledge (in fact, must a centroid of
a perimeter always lie on every perimeter
bisector?), but I am still searching a better
explanation why X(10) is the centroid of the
perimeter, while X(8) is not.

>> What's the ratio for (2)? I see gibberish,

Actually it's because Orlando used a non-ASCII
sign. If I understand him correctly, he means
"[...] a ratio which is not less than
1/sqrt(3)". Actually, the problem is quite
simple; K doesn't need to be the Spieker point
in order to have this satisfied.

We will prove that for every point K in the
plane of triangle ABC, at least one of the
ratios KA / BC, KB / CA, KC / AB is not less
than 1/sqrt(3).

Well, assume the contrary: Let all ratios
KA / BC, KB / CA, KC / AB be less than
1/sqrt(3). If the sidelengths BC, CA, AB are
denoted by a, b, c, then we have

  KA^2 < a^2 / 3,
  KB^2 < b^2 / 3,
  KC^2 < c^2 / 3.

Hence,

  KA^2 + KB^2 + KC^2 < (a^2 + b^2 + c^2) / 3.

But the well-known Leibniz formula gives

  KG^2 = (KA^2 + KB^2 + KC^2) / 3
            - (a^2 + b^2 + c^2) / 9,

where G is the centroid of triangle ABC.

[Actually, this formula is also known in the
form

  KG^2 = (KA^2 + KB^2 + KC^2 - GA^2 - GB^2 - GC^2) / 3.

But this is equivalent to the above, as some
formulas for the length of a median show.]

Since KG^2 >= 0, we thus have

  KA^2 + KB^2 + KC^2 >= (a^2 + b^2 + c^2) / 3,

contradicting our assumption. Hence, at
least one of the ratios KA / BC, KB / CA,
KC / AB must be >= 1/sqrt(3). Qed..

  Sincerely,
  Darij Grinberg[/quote]

[quote="Hyacinthos message #9629"]
[b]From:[/b] "ben_goss_ro" (aka Grobber)
[b]Subject:[/b] Re: a ratio

--- In "Hyacinthos", Darij Grinberg wrote:
> Dear "ben_goss_ro" (= grobber at MathLinks?),

> This puzzles me a bit. In fact, the lines MM_1,
> NN_1, PP_1 do bisect the perimeter of triangle
> ABC, but the cevians of the Nagel point (X(8))
> do it as well; however, the former lines pass
> through the centroid of the perimeter of
> triangle ABC, whereas the latter ones don't!
> I guess this is a consequence of my bad
> physics knowledge (in fact, must a centroid of
> a perimeter always lie on every perimeter
> bisector?), but I am still searching a better
> explanation why X(10) is the centroid of the
> perimeter, while X(8) is not.
>

Hi! Yes, I am grobber (:)).

About the first part of the problem:
I apologize for not completing it. No, not all the lines which bisect
the perimeter pass through Spieker's point, but these ones do because
they also pass through the midpoints of the sides, and here's why:

Let's look at MM_1. Assume AB<AC, and take X on [AC] s.t. AB=CX. It's
obvious that M_1 is the midpoint of AX. Now look at
the "quadrilateral" ABCX and let's try to find the centroid of its
perimeter. First of all, we take the centroid G_1 of the system
formed by the sides AX and BC. To do this we take the midpoints of AX
and BC (which are M_1 and M respectively), and G_1 must lie on MM_1
s.t. MG_1*BC=M_1G_1*AX (basic rod equilibrium physics). Now take G_2
to be the centroid of the system formed by the sides AB and XC. We do
pretty much the same as before, taking G_2 on the line connecting the
midpts of AB and XC (call them S and T respectively) s.t.
G_2X*AB=G_2B*XC, only this time AB=XC, so SG_2=BG_2. This means that
G_2 is also the centroid of the system of points A,B,X,C and that
it's also situatd on the line connecting the midpts of AX and BC,
namely MM_1.

We have 2 things: G_1 is on MM_1 and G_2 is on MM_1, but G_1 and G_2
are the centroids of 2 disjoint parts of our system (AB, BC, CX, XA)
the union of which is the entire system, so the centroid of the whole
system must lie on G_1G_2=MM_1, so the centroid of the perimeter of
ABC is on MM_1. I'd like to stress that G_1 is never equal to G_2,
because that would mean that AX=BC iff AC-AB=BC, which is clearly
false (triangle inequality). X(10) is on MM_1 so it must, of course,
be on NN_1 and PP_1 as well, so we're done.

[/quote]

  Darij


geometry
vector
analytic geometry
Euler
circumcircle
ratio
geometry solved

prove that for $n\geq 3$ if an n-simplex has an orthocenter and if the point coincides with centroid or cercumcirle then simplex is regular.

(All multiplication of vectors here is the usual dot product)
Lemma: The n-simplex $X_1X_2X_3 \dots X_{n+1}$ has an orthocenter in its interior if and only if a congruent simplex $A_1A_2A_3 \dots A_{n+1}$ can be embedded into $\Re^{n+1}$ with each $A_i$ on a coordinate axis but not at the origin.

Proof: If $A_1A_2A_3 \dots A_{n+1}$ has $A_i=a_iE_i$, where $E_i$ is the i-th unit vector in the standard basis of $\Re^{n+1}$, let $H$ be the projection of $0$ onto the hyperplane containing the simplex. Since $(A_i-0) \cdot (A_j-0) = 0$ for all distinct $i,j$, $(A_i-0) \cdot (A_j-A_k) =0$ for any distinct $i,j,k$ and $(A_i-H) \cdot (A_j-A_k)= (0-H) \cdot (A_j-A_k)=0$ by the projection, making $H$ the orthocenter of $A_1A_2A_3 \dots A_{n+1}$.
WLOG, suppose each $a_i>0$. Then the equation of the hyperplane is $\frac{x_1}{a_1}+\frac{x_2}{a_2}+\dots + \frac{x_{n+1}}{a_{n+1}}=1$ and  $H$ is $\frac1B(\frac1{a_1},\frac1{a_2}, \dots ,\frac1{a_{n+1}})$, where $B=\frac1{a_1^2}+\frac1{a_2^2}+ \dots +\frac1{a_{n+1^2}}$. Since H has positive coordinates, it must be in the interior of the simplex.

If $X_1X_2X_3 \dots X_{n+1}$ has an orthocenter $H$, we have $(X_i-H)\cdot (X_j-X_k)=0$ for any distinct $i,j,k$. This gives $(X_i-H)\cdot (X_j-H) = (X_i-H)\cdot (X_k-H)=-C$, where C is the same for any triple of distinct indices $i,j,k$. Consider right triangle $X_iX_jF_i$, where $F_i$ is the foot of the altitude from $X_i$. $H$ is in the interior of segment $X_iF_i$, so $\angle X_jHX_i$ is obtuse and $(X_i-H)\cdot (X_j-H)$ is negative, making $C$ positive.
Then
$(X_i-X_j)^2=((X_i-H)-(X_j-H))^2 =$
$ (X_i-H)^2-(X_i-H)\cdot (X_j-H)+(X_j-H)^2-(X_i-H)\cdot (X_j-H) = $
$(X_i-H)^2+C+(X_j-H)^2+C$.
Let $a_i=\sqrt{(X_i-H)^2+C}$ and $A_i=a_iE_i$ as above. We have $A_iA_j=X_iX_j$ for all $i,j$, so $X_1X_2X_3 \dots X_{n+1}$ and $A_1A_2A_3 \dots A_{n+1}$ are congruent.

Problem. Assume that a simplex has an orthocenter $H$ which coincides with its centroid $G$. The centroid is in its interior, so we can embed the simplex as in the lemma. The vector $H-0=G-0=\frac1{n+1}(a_1,a_2,\dots ,a_{n+1})$ is orthogonal to the hyperplane $\frac{x_1}{a_1}+\frac{x_2}{a_2}+\dots + \frac{x_{n+1}}{a_{n+1}}=1$, so $a_1^2=a_2^2= \dots =a_{n+1}^2$ and the simplex is regular.
 As a side note, it is trivial to establish from here the generalization of the Euler line- the circumcenter $O$ is the projection of the point $\frac12(a_1,a_2,\dots ,a_{n+1}) $. The ratio of distances $\frac{HG}{GO}$ is $\frac2{n-1}$, so the same result would hold if we started with $H$ and $O$ coinciding.


topology
advanced fields
advanced fields unsolved

Hello,

let
$ M \equal{} \left \{([x],[y]) \in \mathbb{RP}^n \times \mathbb{RP}^m | n \leq m, \sum_{i\equal{}0}^n x_i y_i \equal{} 0 \right \}$

Prove M is a differntial manifold of dimension $ n\plus{}m\minus{}1$.

I would like to write M as $ f^{\minus{}1}(y)$ for $ y \in$ some manifold Y and "good" map f, but I can't find such objects.
any help?

thx

Set $ f \equal{} \sum_{i \equal{} 0}^n x_i y_i$


Hi!  I'm going to be a senior, and I'm currently in AP Bio.  I know I want to major in bio in college (either molecular or cellular).  Should I take AP Chem or AP Environmental Science next year?

Thanks!

I suggest AP Chem. Chemistry is fundamental to the understanding of biological systems.


There are 100 seats on an aeroplane. Each passenger boarding must sit in their own seat unless it is occupied in which case they may choose a seat at random. The first passenger on is a little old lady who sits in the first seat she comes to. What is the probability that the last passenger gets to sit in their own seat?

I keep coming up with either 1/2, 1/100 or 100!. Any ideas please?

I remember having solved this problem a while ago. If you don't want me to ruin it for you, well, try this:

-manually calculate it for smaller values of $n$ (here $n=100$);
-try to find a recurrence between the cases for different $n$;
-induction. ;)

Let the plane have n seats and let $p_n$ be the probability that the last passenger goes to his own seat. Assume that the old lady is assigned seat #1 and passengers #2, #3, ..., #n are assigned seats #2, #3, ..., #n.  The old lady chooses the seat #2 with probability $\frac 1 n$ and some other seat with probability $\frac {n - 1}{n}$. If she selects seat #2, then passenger #2 is exactly in the position of the old lady in a (n-1)-seat airplane and the probability of the last passenger getting his own seat (under the condition that she chose seat #2) is $p_{n-1}$. If she selects some other seat, then we can ignore passenger #2 and his seat, consider an (n-1)-seat airplane and again, the probability of the last passenger getting his own seat (under the condition that she did not select seat #2) is $p_{n-1}$. Hence,

$p_n = \frac 1 n p_{n-1} + \frac{n - 1}{n} p_{n-1} = p_{n-1}$

Thus the probability $p_n$ does not depend on the number of seats in the airplane. It is trivial to see that in a 2-seat  plane, the probability $p_2 = \frac 1 2$ and consequently, $p_n = \frac 1 2$ for any $n \ge 2$. :P


AMC
USA(J)MO
USAMO
USAJMO
AMC 10
AMC 10 A
AMC 12

Hi,
I just have a question about how to qualify for USAMO or USAJMO

I am taking AMC10A through my high school, and AMC12B through another math organization

For example, if I get a 150 on 10A and 138 on 12B and a 9 on AIME (which I will take through my high school) will it still be possible for me to make USAMO, or will I only be able to make USAJMO because I took 12B through at a different place than I took AIME?

It's irrelevant where you take the tests

On the other hand, the AMC has been known to lose track of people who take the tests in multiple locations. If something like this happens and you wind up on the wrong list, contact them to correct it.

Fill in the same name and home address on every answer form.  We lose track of people when they take contests at different locations using nickames, initials, or if the answer form is returned with incomplete information.

Steve Dunbar
MAA Director, American Mathematics Competitions


Well , we know that a number is divisible by $3$ iff the sum of the digit is divisible by 3. We see that $1+2+3+4+5+6=21$ is divisible by three . Now you want to make a five digit numbers . Obviously the number you are going to give up is either 3 or 6 (or else the sum wont be divisible by 3) .

If you give up 3 , then the number of way of such arrangement is $5!=120$ . If it is not divisible by 12 , the number should be not divisible by 4 also . This means there are 1/4 of them is divisible by 4 . So those not divisible by 4 is 3/4 . Hence the total number not divisible by 4 is $120\cdot 3/4 = 90$ . But you can also give up " 6 " .Hence th enumber of ways is $90\cdot 2=180$ .So the probability is

$\frac{180}{240}=\frac{3}{4}$  :)

EDITTED : Thanks for pointing out FMako  :)

[quote="FMako"]I disagree, assuming I understand what I'm doing.

I agree that you have to remove only a $3$ or $6$.

Case 1: $1,2,4,5,6$

Total combinations: $5! = 120$

Total combinations that are divisible by $12$, since it is divisible by $3$ already, that means it is divisible by $4$.
Therefore you only need to look at the last two digits of the number.
Endings of $12, 52, 24, 64, 16, 56$ satisfy divisibility by $4$,
therefore the total combinations divisible by $4$ are $3! \cdot 6 = 36$.

Case 2: $1,2,3,4,5$

Total combinations: $5! = 120$

"": $12,32,52,24$ Thus the total combinations divisible by $4$ are $3! \cdot 4 = 24$

Thus the probability that the number is divisible by $3$ but not $12$ is $\frac{240-60}{240} = \frac{3}{4}$

I hope I did this right.[/quote]

Why divide by 240 ? The total number of ways of writing 5 digit number from 6 number is $6P5=720$  :) For the 240 , you only included the case which you didnt take 3 or 6. But what about if you didnt take 1,2,4,6 ?  :P There are another 480 cases .

[quote="FMako"][quote="eminem"]this is from IIT JEE 96

a number of 5 distinct digits is written down at random using the digits 1,2,3,4,5 and 6 . What is the chance that the no. is divisible by 3 but not by 12?[/quote]

The question asks for the probability that the number formed is not divisible by 12, if the number is divisible by 3.

I hope that's what the question means at least.[/quote]

Oh , I got what you mean :) . I read the first sentence only and forget the back one . Then I agree with you that it is 3/4  :P


Mykolka the numismatist possesses $ 241$ coins of the total amount $ 360$ tughriks (the Mongolian currency), where the value in tughriks of each coin is an integer. Does it necessarily follow that these coins can be divided into three heaps of equal amount?


Jenny and Alex flip four fair coins each. What is the probability that they get the same number of tails?

[hide="Generalization"]
\[ P(n)= \sum_{k=0}^{n}\left( \frac{ \binom{n}{k}}{2^{n}}\right)^{2}= \frac{\sum \binom{n}{k}^{2}}{4^{n}}= \boxed{ \frac{ \binom{2n}{n}}{4^{n}}}\]
Note that $ \sum_{k=0}^{n}\binom{n}{k}^{2}= \binom{2n}{n}$ by the [url=http://mathworld.wolfram.com/Chu-VandermondeIdentity.html]Vandermonde Convolution Formula.[/url]

This gives $ P(4)=\frac{35}{128}$.[/hide]
[b]winluo2[/b]: Check your calculations in the beginning.


Hopkins requires you to buy all your textbooks. At minimum, they would cost around $\$$300 and that's if you buy all your books used, at the maximum and buying them new, they could cost you well over $\$$500 and also, the weight of your backpack for all your homework is crazy! I'm like stunting my growth with the weight of my backpack.

Also, the amount of homework you get is ridiculous.

[quote="Phelpedo"]You guys are freshman. You don't know anything about homework.[/quote]

Freshman at Hopkins' Homework > Seniors at Hamden High Homework

kthx.

[quote="Phelpedo"]No duh seniors never have much homework. Freshman at Hopkins < most people, depending on courses.[/quote]

Seniors should have the most homework.

And any student > most people.

kthx.

No no. The homework hierarchy goes like this: Senior $\leq$ Freshman < Sophomore < Junior. Why don't seniors have any homework? BECAUSE THEY"VE ALREADY DONE COLLEGE APPLICATIONS AND DON"T CARE ABOUT ANYTHING ANYMORE.

Also, when I said Freshmen @ Hopkins < most people (upperclassment), I was talking about homework load.

kthx.

Except you're just generally assuming that. It's not true at Hamden High, it's not true at Hopkins.

Ok. So Hamden high guys get no homework and Hopkins freshman get a little homework.

Ummmm... That is DEFINITELY true at Hopkins.

Ok the "that" is quite out of context now so I don't know what you're referring to. However, I will assume it to be Ignite's statement. Anyhow, I don't think you can really say until you have a better frame of reference (I mean come on, you get like 30 minutes of HW every night in middle school).  4-5 hours a night for freshman? Not unusual at all.

Well you can expect around 4 hours a week per subject.

Up to a maximum of 7 subjects

28 hours a week

[quote="worthawholebean"]For me, it generally ends up about

6h for English
4h for Math
2h for the other 3

So 16h of homework

And most students take 5 subjects.[/quote]

I think most students take 6.

I don't think he's including non-academic subjects.

No actually many people take 6 academic subjects

5 ACADEMIC - foreing language, social studies, sci, MATH, la

[quote="worthawholebean"]If you count MG as "many people", let it be known he is not a representative sample.

From freshman year onward, not many people take 6 academics.[/quote]
I was talking about in the Jschool.

How do you expect me to know how many courses he takes?

[quote="worthawholebean"]You know him. How many courses do you think he's taking?[/quote]
Ummm......fifteen and a half?

fifteen in a half....... you've gotta be kiding me.. :0

Is he taking two arts?

I'm thinking of taking two sciences in my junior or senior year.

Or he might still be taking both Spanish and Latin

I don't really know. Last year I had one class with him but not this year.

See, MG has time to play Ping Pong. I told you study halls aren't very necessary. I should've took Chinese and Spanish and an art and the other 5 courses.

I think this whole heavy book bag, and lot's of hw thing isn't that big of a problem. Hopkins is a very academic school. So you should expect to keep alot of hw and heavy books to carry home each day. :)

Hopkins, is a hard school and academic's are a major so heavy books and lots of hw.. shouldn't be a bir problem to you. Since you applied and learn alot about the school..

:


analytic geometry
number theory proposed
number theory

Given a Cartesian coordinate system in the plane, a point with coordinates (x; y) is called a grid point if and only if his coordinates x and y are integers. Does there exist a circle with exactly five grid points lying on its periphery?

  Darij

This is a problem that needs some spirit of observation ;]
There are two ideas that may bring you to its solution:
1) The equation sin^2(x)+cos^2(x)=1 has only four integer solutions;
2) If a grid point lies on the periphery then its symmetric for the center of the circle also lies on the periphery of the circle.

1)One knows that the parametrical equations of the circle are :
x=a+Rcos(x)
y=b+Rsin(x),  where (a,b) are the coordinates of the center of the circle, and R is its radius.
But cos(x) and sin(x) are integer only for four points on the circle, so there exist only 4 grid points on the periphery of the circle. (this may be difficult to understand for R irrational, but when one looks at the circle and at the coordinates, everything becomes clear :))

2) This is a solution only for circles with grid center points :
 Let M(c,d) be a point on the periphery of the circle, then its symetrical point M' (according to the center of the circle) is also grid, because its coordinates are (2a-c,2b-d). So there can be only an even number of grid points on the periphery of the circle, and because 5 is even there doesn't exist a circle with such a property.

This solution came to my mind quickly, but its generality is hard to prove.

__________________________________________________________

[SYSTEM OF A DOWN]
"Our days are never coming back" (Highway Song) ;]

This is a theorem of A. Schinzel, which in fact states that for any n there exist a circle with has exactly n laticial point on it.

@ Palosh : Hmm... in fact, there does exist a circle through 5 grid points! Of course, its center is not a grid point; it is not even a point with coordinates (x/2; y/2) where x and y are integers.

  Darij

Valentin, could you tell me that theorem... and Darij what is then the complete solution of this problem?
Thanx...
:):):)

______________________
"Our days are never coming back..." (Highway Song -> System of a Down)

The complete solution is quite straightforward once you have the explicite answer: The circle with equation $x^{2}+\left( y+\frac{7}{4}\right) ^{2}=\left( \frac{25}{4}\right) ^{2}$ passes through the points (0; -8), (5; 2), (-5; 2), (6; 0) and (-6; 0) and through no other lattice points. About the general theorem, I think there is [url=http://mathworld.wolfram.com/SchinzelsTheorem.html]some information on the Mathworld site[/url].

  Darij

Thanx a lot...
Nice problem though i've such a mistake!
 :):):)

       Palosh alias Downito
_____________________________
"Our days are never coming back... " (Highway Song -> System of a Down)


Do you think if your parents should be invited to join AoPS site?  Give your reason please.

for the most part, parents would learn quite a bit if they were in AoPS. my mom was the one who invited ME to AoPS however.

Aren't there alredy parents browsing the AoPS forums, along with the teachers?

well...I guess. But mine already is.  Hi Dad! :)

I think he should because he's a math professor, and this would be fun for him, but he hasn't joined yet.


AMC USA(J)MO USAMO geometry circumcircle induction

I was totally confuzzled by the problems. I suck at proofs :(

How do you tackle this:

2008 USAMO #2: Let ABC be an acute, scalene triangle and let M, N, and P be the midpoints of BC, CA, and AB, respectively. Let teh perpendicular bisectors of AB and AC intersect ray AM in points D and E respectively, and let lines BD and CE intersect in point F, inside triangle ABC. Prove that points A, N, F, and P all lie on one circle.

I tried drawing the figure as described and drew the circumcenter of ABC. I tried to make a circle with AO as a diameter. Knowing that angles APO and ANO are right angles, they form two semicircles so the four points and O must form a circle.

Do I have good enough progres to at least get SOME credit for the problem?

here is another one too:

4. Let P be a convex polygon with n sides, n ¸ 3. Any set of n¡3 diagonals of P that do not
intersect in the interior of the polygon determine a triangulation of P into n¡2 triangles.
If P is regular and there is a triangulation of P consisting of only isosceles triangles, ¯nd
all the possible values of n.

The problems are under "Contests" so you should not repost them.

But anyways, #2 is tricky but not too far-fetched.
#4 uses more intuition than geometry knowledge.

To be honest, your #2 would probably not earn any points... it's pretty obvious that any quads with opposing right angles are always cyclic, and that doesn't demonstrate enough.

Try #1. It's much, much simpler imo, and doesn't need anything strange or obscure. Try to find one such sequence (since it just says to prove that one exists); play around with the numbers, preferably without a calculator... and then

[hide]

use induction[/hide]

But you can't use a calc for USAMO.
 :maybe:


I get the feeling there may be other schools added.

like I said in the mc forum, I doubt this is from this year.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=675930#675930
Then again I may be wrong.

It is definately from this year because AMC of last year was in 2005, and the title says 2006, so it's from this year.  I need to stop repeating myself.

yay i got first at the place i took it and its on the honor roll :D

Same here! Perfect at Kennedy Junior High School, Lisle, IL.
What's weird is that our school is in Lisle, but it's a Naperville school.

After the latest update, we are on the list.


[quote="i_like_pie"][hide]I sort of saw the solution immediately the first time I solved this, so I don't have a very good logical explanation.

One of the digits is either 5 or 6. 6 doesn't work, so one of the digits (the last one) is 5. The first one must be a 1, which leaves the last one to be 4. The number is $\boxed{145}$.[/hide][/quote]
Read the question again ;)

[quote="i_like_pie"][hide]I sort of saw the solution immediately the first time I solved this, so I don't have a very good logical explanation.

One of the digits is either 5 or 6. 6 doesn't work, so one of the digits (the last one) is 5. The first one must be a 1, which leaves the last one to be 4. The number is $\boxed{145}$.[/hide][/quote]
Just wondering, why does one of the digits have to be five or six?

[quote="xscapezaer"]Find a three digit number of the form $xyz$ such that the number $xyz = x!+y!+z!$, where $x$ and $y$ and $z$ are all between 4 and 10, inclusive.

In other words, find a three digit number, whose digits are between 4 and 10 inclusive, s.t. it equals the sum of the factorials of its digits.[/quote]

is 10 a digit? :rotfl:

Oops... I didn't see they had to be bewteen 4 and 10.

[hide="Solution"]The maximum possible value of any of the digits is $5$ because $6!=720$, which means one of the digits has to be $7$, but $7!=5040$.

The only possible digits are $4$ and $5$, but the sum of the factorials of any combination of those digits will always be less, so there is $\boxed{\texttt{no solution}}$.[/hide]
[quote="davidlizeng"]the largest possible way to get xyz is 10*10*10 or 1000
since 6!=720 and 7!=5040, the numbers x, y, z could at most be 6!
then just use 6! and work your way down.[/quote]
It can't be 6! because then one of the digits is 7.

[quote="xscapezaer"]Find a three digit number of the form $xyz$ such that the number $xyz = x!+y!+z!$, where $x$ and $y$ and $z$ are all between 4 and 10, inclusive.

In other words, find a three digit number, whose digits are between 4 and 10 inclusive, s.t. it equals the sum of the factorials of its digits.[/quote]

The original poster means $\overline{xyz}= x!+y!+z!$, where $\overline{xyz}$ is the number with digits $x,y,z$, as can be inferred from the second paragraph. ;) So yeah, 145.

[quote="chess64"]The original poster means $\overline{xyz}= x!+y!+z!$, where $\overline{xyz}$ is the number with digits $x,y,z$, as can be inferred from the second paragraph. ;) So yeah, 145.[/quote]
[quote="xscapezaer"]Find a three digit number of the form $xyz$ such that the number $xyz = x!+y!+z!$, where $x$ and $y$ and $z$ [b]are all between 4 and 10, inclusive.[/b][/quote]
Read my second post (the one above your post).

It is [hide]no solution.

A digit can obviously not be above 6, since 7! is over 1000. If one of the digits is 6, then the hundreds digit would be at least 7, which is not possible. Since 3X4!=72, one of the numbers mst be 5. It is obvious that this is not possible by looking at the remaining possibilities.[/hide]


calculus
derivative
limit
function
calculus computations

find the limit as x->4 of (4^x - x^4) / (x-4).

We are not allowed to use l'hopital's rule.

Any help is appreciated. Thanks!

Are you allowed to use the definition of the derivative?
(For the given limit this amounts to the same as using L'Hopital.)

In any case you should try to break up the fraction into a sum of two terms
that are difference quotients and deal with them seperately.

yes, we are allowed to use the definition of the derivative.

I tried using algebra to try and cancel terms so that i could substitute 4 in, but i havent come up with anything. I also tried multiplying by the conjugate, using exponentials, etc.

Any tips? :)

You're going to have to quote your knowledge of derivatives. $ \lim_{x\to 4}\frac{4^4\minus{}x^4}{x\minus{}4}$ is doable algebraically, but you need to know the derivative of the exponential for $ \lim_{x\to 4}\frac{4^x\minus{}4^4}{x\minus{}4}$.

is it basically just differentiating $ \frac{4^{x} \minus{} 4^{4}}{x \minus{} 4}$ and then substituting the value 4?

but differentiating that would give a denominator of (x-4)^2 which would give a denominator of 0 when substituting 4.

I'm still stuck :(

$ \lim_{x\to 4}\frac {4^x \minus{} 4^4}{x \minus{} 4}$  [i]is[/i] a derivative -- can you recognize the function that it's a derivative of (and the point at which it's the derivative)?

oops, its $ \lim_{x\to 4}\frac{4^{x}\minus{}x^{4}}{x\minus{}4}$ not $ \lim_{x\to 4}\frac{4^{x}\minus{}4^{4}}{x\minus{}4}$

if i let y = $ \frac{4^{x}\minus{}x^{4}}{x\minus{}4}$ , and i differentiate w.r.t. x, and substitute it in at point 4, i get 240/0. Does that mean the limit is infinity?

No, it means that you don't know how to recognize a limit as the definition of a derivative.  What's the definition of the derivative of the function $ f(x)$ at the point $ x \equal{} a$?

(Also, jmerry and I both know that the problem we're suggesting to you is not exactly the same as the problem you asked; but it's an intermediate step in solving your problem.  Once you've figured out our problem, you'll begin to see why.)

$ \frac{240}{0}$? $ 4^4$ is $ 256$, not $ 16$.

It has to be a $ \frac{0}{0}$ form, because it's a derivative.

As for the different limits, there are two pieces in my post #4. Their sum is the original limit, and each is simpler than that original.

$ \lim_{x\to 0}\frac{f(a\plus{}m)\minus{}f(a)}{m}$   ? I think thats right. Sorry if I'm a bit slow, I'm really bad at limits. Taking calculus in college and learning derivatives without learning limits in high school is a terrible experience.

I did realise that I could add them together and get the limit, but I can't find the derivative. I'm going around in circles. I can't seem to apply anything I've learnt in my calc class to this. I think I'm missing a huge chunk of the concept here.

For the way you wrote that, you wanted to take $ \lim_{m\to 0},$ not $ x.$

But you might also want to consider this alternate (but algebraically equivalent) form:

$ f'(a) \equal{} \lim_{x\to a}\frac {f(x) \minus{} f(a)}{x \minus{} a}.$

[quote="tennischan"]Sorry if I'm a bit slow, I'm really bad at limits.[/quote]  No need to apologize :)

Kent made a helpful suggestion for another way to write the same expression.  Now, the question becomes: can you recognize the limit you're given as the derivative of some function?  This is a pattern-matching exercise: you have to come up with the right choice of $ f(x)$ and $ a$ so that the definition of the derivative of $ f$ at $ a$ ends up looking like the limit you're trying to compute.

Okay, so I tried something I thought of, and I added the limits together by comparing the definition of the derivative for both parts and finding out what f(x) for each part was, and then finding f'(x).

So I found the derivative of 4^x and the derivative of x^4, which are 4^x (ln 4) and 4x^3 respectively.

Adding them together, I got 256 ln 4 + 256 = 256 (ln 4 + 1).

Is that right?

Almost: are you sure adding the two limits was what you wanted to do?  After you fix this, you'll be done; for purposes of your own education, though, you should definitely make sure you understand [i]why[/i] you did each step in this process and practice doing other examples so that you don't forget the important ideas  :)

$ \frac {4^x \minus{} x^4}{x \minus{} 4} \equal{} \frac {4^x \minus{} 4^4}{x \minus{} 4} \minus{} \frac {x^4 \minus{} 4^4}{x \minus{} 4}$.

And apply definition of derivative.

oops, its 256(ln 4 -1)

I get it now, thanks so much for all your help! :)