ratio
geometry
3D geometry
number theory
modular arithmetic
area of a triangle
Heron\u0027s formula

Since the last one was locked, I have decided to make a new marathon. I don't want this one to be locked, so I am posting some new guidelines.

[b]1. Very Important! Be sure it's classroom math level! I can't stress this enough![/b]
If you are not sure if it's appropriate, please, please, private message it to me and I will give you my opinion. Do NOT post any problems you are unsure about. And, chances are, if you have doubts, it's not appropriate.

[b]2. Explain the solutions![/b]
Don't just give an answer! Explain it thoroughly!

[b]3. Post a new problem after solving one![/b]
After you solve a problem, make sure you post a new one! See Rule #1.

On that note, here's the first problem!

If 2 yings equal 4 yangs, and 2 yangs equal 1 yung, how many yings equal 8 yungs?

[hide="Here is my unique way of solving it"]
Since 2 yings equal 4 yangs, we can set up the ratio $ \frac {\text{ying}}{\text{yang}} \equal{} \frac 24 \equal{} \frac 12$.

Similarly, 2 yangs equal 1 yung, so we can set up another ratio $ \frac {\text{yang}}{\text{yung}} \equal{} 2$. Thus, multiplying $ \frac 12$ by $ 2$, we see that the resulting ratio of yings to yungs is 1. Thus 8 yings equal 8 yungs.[/hide]

New problem:
Find the value of ten quarters.

[hide="Major Hint"]
A quarter is 25 cents.[/hide]

[hide="uh... this better not be a trick question"]25*10= 250 cents or 2.50 dollars[/hide]
While generally not difficult for people who use AoPS forums, you'd be surprised how many middle schoolers, and even high schoolers, would have difficulty with this problem.

[hide]
Since the letter $ b$ is included, and the product includes $ 0$, by the multiplicative identity, the whole product will turn out to be $ \boxed{0}$.[/hide]

New problem:
Find 15 + 88.

[quote]New problem:
Find 15 + 88.[/quote]
Something like this may be too elementary, but hey, I guess we could have elementary level people here :)

Umm...I guess it's hard to say anything more than 15+88=103... :maybe:

Anyways...

NEW PROBLEM:

If there are 7 people, each with 6 bags, each with 6 vegetables, each with 5 seed pods, each with 4 seeds, how many seeds does each person have?

To solve this problem, we multiply.

So:

4 seeds * 5 pods * 6 veggies * 6 bags * 7 people =

$ 4*5*6*6*7 \equal{} 5040$

This is the total number of seeds that these 7 people have. To find how much each person has, divide by 7 and get $ 720$. So, each person has 720 seeds.

Or, a smarter way, just multiply $ 4*5*6*6$ and get 720, since we are only finding for one person, not all.

New Problem: 2^2005 = 2^2004 + 2^n  Find n.

do you mean times? if not, this problem isn't classroom level and the answer isn't an integer.

just so you all know, modular arithmetic isn't classroom math level.

Edit: It is classroom math level.

Wow... you are so used to saying "mc level" that you say put it where it should say "classroom math level". :rotfl:

Anyway,

[hide]
$ 2^{2005} \equal{} 2 \cdot 2^{2004} \equal{} 2^{2004} \plus{} 2^{2004}$. Thus $ n \equal{} 2004$.[/hide]

New problem:
Find the difference between the cube of the square of 6859 and the square of the cube of 6859. Express your answer in simplest form.

0!

x^2^3=x^3^2
x^6=x^6
that was on state cd this year :)

An isosceles triangle with a vertex angle of 60 degrees has a base of 2. Find the area.

@yongyi, oops :D

Levans, I'm just wondering if you changed my post around a little. I looked back to my problem, and the problem was erased and it said [quote="someone"]While generally not difficult for people who use AoPS forums, you'd be surprised  how many middle schoolers, and even high schoolers, would have difficulty with this problem.[/quote] instead. I thought someone hacked my account for a moment. Please just tell me in case.7

P.S. There are some elementary people here; I'm in 5th grade.

If any angle of an isosceles triangle is 60 degrees, wouldn't that just make it an equilateral triangle?

So with sides of 2 (I'm gonna use Heron's Formula here),

$ s\equal{}\frac{2\plus{}2\plus{}2}{2}\equal{}\frac{6}{2}\equal{}3$

therefore

$ A\equal{}\sqrt{3(3\minus{}2)(3\minus{}2)(3\minus{}2)}\equal{}\sqrt{3(1)^3}\equal{}\sqrt{3(1)}\equal{}\sqrt{3}$

New problem: Do this without Heron's Formula...

[hide="Classroom Style - Memorization"]
In a typical classroom, you would just be taught that the [b]formula[/b] for the area of an equilateral triangle of side length $ s$ is $ \frac {s^2\sqrt 3}4$. Thus plugging in, we see that $ A \equal{} \sqrt{3}$.[/hide]

New problem:
Using the Heron's Formula, find the area of a 6-7-8 triangle. Do [b]not[/b] round. Leave your answer in simplest radical form.

[hide]"All the integers" include 0, so the final product will be $ 0$.

However, if the question is asking $ 0 \cdot \infty$, then there is no answer.
Similarly, how many negative numbers are there? The sum of all the numbers excluding 0 will oscillate between $ \infty$ and $ \minus{}\infty$, so there is no well-defined answer to your question.[/hide]

New problem:
If a green smiley is found in an array of 20 smileys, how many smileys are expected to be found in an array of 240 smileys?

To solve this problem, we can write a proportion.

$ \frac{1}{20} \equal{} \frac{ x }{240}$

We cross-multiply:

$ 240*1\equal{} 20*x$

$ 240 \equal{}20x$

$ 12 \equal{} x$

So, 12 green smileys.  :D

New Problem:

Cheryl has $ x$ amount of marbles. She has $ 12$ more than Scott, who has $ \frac{1}{2}$ $ x$ marbles. How many marbles do Cheryl and Scott have all together?

[hide]

so x-12=1/2x

1/2x=12
x=24

so 24+12=36.
[/hide]

anyways, you guys got my problem wrong.

anyways, here it is again

An isosceles triangle with a vertex angle of 60 degrees has a
base of 2. Find the area. Remember what the vertex angle is.

the sum of the angles in a triangle is 180, and the base angles in a isosceles triangle are congruent

let a base angle equal b
b+b+60=180
2b+60=180
2b=120
b=60

so all angles in the triangle are 60 degrees

the formula for the area of an equilateral triangle is $ \frac{s^2 \cdot \sqrt{3}}{4}$

plz dont tell me i have to prove that

so the area is $ \frac{2^2 \cdot \sqrt{3}}{4}$
$ \frac{4 \cdot \sqrt{3}}{4}$
$ \sqrt{3}$

i assume there are no units so...

$ \sqrt{3}$ units

Please do not create any more marathons in Classroom Math - they're never going to last and it's really a waste of effort.


Suppose$a, b$are natural numbers such that:
$p=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$
is a prime number. What is the maximum possible value of$p$?

We will try to find an odd prime $p$. Let $\gcd (a,b)=d$ then $a=da_1,b=db_1$ where $\gcd (a_1,b_1)=1$. We have $p= \frac b4 \sqrt{\frac{2a-b}{2a+b}}$.

[b]Case 1.[/b] If $p \mid 2a_1-b_1$ then since $\gcd (2a_1-b_1,2a_1+b_1) \mid 2^k$ so $p^2 \mid 2a_1-b_1$. let $2a_1-b_1=p^2k$ with $k \in \mathbb{N}$ then $\sqrt{\frac{b^2k}{(p^2k+2b_1)16}}=1$ or $k= \frac{32b_1}{b^2-16p^2}$. If there exist prime $q \mid b_1, 2 \nmid q$ and $q \mid b^2-16p^2$, this follows $p=q$. If $v_p(b_1)=1$ then $p^2 \mid b^2-16p^2$, a contradiction. Thus, $p^2 \mid b_1$. Let $b_1=p^x2^y$ then $k= \frac{32p^{x-2}2^y}{d^2p^{2x-2}2^{2y}-16}$. Since $x \ge 2$ so $(dp^{x-1}2^y-4)(dp^{x-1}2^y+4) \mid 32 \cdot 2^y$. We find $dp^{x-1}2^y=12$ so $p=3,x=2,2^yd=4$ and $2^7\mid 32 \cdot 2^y$ so $y \ge 2$. Thus, $d=1,y=2,p=3,x=2$ so $b=db_1=36$. This follows $k=1$. Hence, $2a_1=45$, a contradiction.

 Thus, there doesn't exist such $q$, which follows that $b^2-16p^2=2^x \mid 32b_1$. Note that $2a_1=b_1+p^2k$ so $\gcd (k,b_1) \mid 2^x$. Also, if $2 \mid b_1$ then $2 \nmid a_1$. This means if $v_2(b_1)=1$ then $4 \mid k$. or if $v_2(b_1)=2$ then $v_2(k)=1$. Thus, we always have $b^2-16p^2 \mid 64$. This follows that $4 \mid b, b=4b_2$ so $b_2^2- p^2\mid 4$. We couldn't find any odd prime $p$ in this case.

[b]Case 2.[/b] If $p \mid b$ then since $\gcd (2a_1-b_1,2a_1+b_1) \mid 2^k$ so $2a_1-b_1=2^x, 2a_1+b_1=2^y$. This gives us $b_1=2^{x-1}\left(2^{y-x}-1 \right)$ so $\frac{d \cdot 2^{x-1} \left( 2^{y-x}-1 \right)}{4 \cdot 2^{(y-x)/2}}=p$.
Hence, if $p \mid d$ then $2^{y-x}-1=1$ so $y-x=1$, a contradiction since $2 \mid y-x$. If $p \mid 2^{y-x}-1$ then $d=1$ and $2+(y-x)/2=x-1$ or $3x-6=y$. Thus, $2^{y-x}-1=2^{2x-6}-1=4^{x-3}-1$. Hence, $p=3,x=4,y=6$.

Thus, $p=3$ is the answer.  $\blacksquare$

[quote=shinichiman]Then since $\gcd (2a_1-b_1,2a_1+b_1) \mid 2^k$ so $2a_1-b_1=2^x, 2a_1+b_1=2^y$.
[/quote]
Unfortunately, this conclusion is not justified and indeed not correct since actually there is a solution for $p=5$.


Steve has one quarter, two nickels and three pennies. Assuming no items are free, for
how many different-priced items does Steve have exact change?

1) 1 penny
2) 2 pennies
3) 3 pennies
4) 1 penny + 1 nickel
5) 1 penny +2 nickels
6) 2 pennies +1 nickel
7) 2 pennies + 2 nickels
8) 3 pennies +1 nickel
9) 3 pennies + 2 nickels
10) 1 quarter
11) 2 quarters
12) 1 penny + 1 quarter
13) 2 pennies +1 quarter
14) 3 pennies + 1 quarter
15) 1 penny + 1 nickel +1 quarter
16) 1 penny + 2 nickels + 1 quarter
17) 2 pennies +1 nickel + 1 quarter
18) 2 pennies + 2 nickels + 1 quarter
19) 3 pennies +1 nickel + 1 quarter
20) 3 pennies +2 nickel + 1 quarter
21) 1 quarter + 1 nickel
22) 1 quarter + 2 nickels
23) 1 nickel + 1 quarter
24) 2 nickels + 1 quarter

Quicker solution

4 possibilites for # of pennies x 3 for # of nickels x 2 for # of quarters - 1 for when you have 0 of each=23

Solutions #21 and #23 are the same...

22 and 24 are also the same, and 11 is invalid, but you forgot 1 nickel and 2 nickel

this is vanilla minecraft therefore coins do not exist and steve has nothing


from Mandelbrot November 1992:

8. Find the length of the median of a trapezoid whose diagonals are perpendicular segments with lengths 7 and 9.

Pretty easy... or am I missing something?

[hide]The diagnols being perpedicular implies it is a rhombus, and using thing the Pythag. Thm, we see the side length is $\frac{\sqrt{130}}{2}$.
Now the area of the trapezoid is $9*7=63$

So $63=bh=\frac{\sqrt{130}}{2}h$

So $h=\frac{63\sqrt{130}}{65}$[/hide]

it doesn't have to be a rhombus(and it's not in this problem).

But a rhombus fits, as two sides are parallel, making it a trapezoid.

isn't a trapezoid definitionally only 1 pair of parallel sides? http://mathworld.wolfram.com/Trapezoid.html

Hmm, well I thought a rectangle was a trapezoid, too, right?

Rectangles are not trapezoids, as the definition of a trapezoid is a quadrilateral with one pair of parallel sides.

If a trapezoid did have 2 pairs of parallel sides, it wouldn't be a trapezoid.  :D

Hope I didn't mess this one up. ;)

If a trapezoid is orthodiagonal, the sum of its sides squared is equal to its diagonals squared. The proof is hard to explain, but I'll post it if somebody asks.

Let our median be x so that $(2x)^2=7^2+9^2=130$.

Our answer is thus $\boxed{\sqrt{\frac{65}{2}}}$

If you're still online, is this right? :)

well, i am not quite sure what the right answer is.

The answer book says $\frac{\sqrt{30}}{2}$ and I get $\frac{\sqrt{130}}{2}$

The way to prove the thing you were talking about though:

Label the vertices of the trapezoid $ABCD$ where $AB$ and $CD$ are the two paralel sides. Translate $BD$(a diagonal) onto the vertice $C$ so you have a parallelogram. Label the point $B$ is translated onto $B'$ $AB' = AB + DC$ and $B'CA = 90^{\circ}$, by the pythagorean therem then we have $AC^2 + B'C^2 = (AB + BC)^2 \implies AC^2 + BD^2 = (AB + DC)^2$

I think median refers to the average of the two parallel sides.

[quote="agolsme"]well, i am not quite sure what the right answer is.

The answer book says $\frac{\sqrt{30}}{2}$ and I get $\frac{\sqrt{130}}{2}$[/quote]I get the same answer as you. I just didn't rationalize the denominator in $\boxed{\sqrt{\frac{65}{2}}}$  ;)

I think the book is wrong and that we're right. :D:P

[hide="hint"]suppose it is TRAP, TR parallel AP

extend TR beyond R to S so RS=AP, RSAP is a parallelogram...

there is a problem just like this in the AoPS books...
[/hide]

[quote="pkothari13"][quote="4everwise"]
Let our median be x so that $(2x)^2=7^2+9^2=130$.

Our answer is thus $\boxed{\sqrt{\frac{65}{2}}}$[/quote]

Why did you relate the median to the diagonals in that? Don't you have to use the sides instead?[/quote]Sorry for the brief solution.

I let the median be x, so that the length of the sides would be 2x. :D

Let $ABCD$ be a trapezoid ($AB\parallel CD$). I note the middlepoints $M$, $N$ of the diagonals $AC$, $BD$ respectively, $AB=a$, $CD=b$, $AD=m$, $BC=n$, $AC=c$, $BD=d$ and the length $l=\frac 12 (a+b)$ of the median
($2MN=|a-b|$). From the Euler's relation (in a convex quadrilateral)  $a^2+b^2+m^2+n^2=$ $c^2+d^2+4\left(\frac{a-b}{2}\right)^2$ results $m^2+n^2+2ab=c^2+d^2\ (1)$. But $AC\perp BD$ $\Longleftrightarrow$  $a^2+b^2=m^2+n^2\ (2)$. From the relations $(1)$ and $(2)$ we obtain $\overline {\underline {\left| \ l=\frac 12 \sqrt{c^2+d^2}\ \right| }}$. If $c=7$ and $d=9$ then $l=\frac 12\sqrt{130}$.

[u]Remark.[/u] Better they choose $c=24$ and $d=10$ ! Then $l=13$.


I have trouble forming an explanation to the following problems:

a. does every 3 by 3 matrix have an inverse under matrix addition? [b]explain your reasoning[/b].

b. does every 2 by 2 matrix have an inverse under matrix multiplication? [b]explain your reasoning[/b].

[hide="a"] Yes.  Negate each entry. [/hide]
[hide="b"] No.  The determinant is multiplicative, so the inverse matrix of a matrix needs to have determinant the inverse of the determinant of the original matrix.  But what if the determinant is $0$? [/hide]

[quote="jeez123"][quote="t0rajir0u"][/hide]
[hide="b"] No.  The determinant is multiplicative, so the inverse matrix of a matrix needs to have determinant the inverse of the determinant of the original matrix.  But what if the determinant is $0$? [/hide][/quote]

what do you mean by that? give me an example perhaps? thanks![/quote]
[hide="An example is..."]

Try the matrix

[3 9]
[1 3]

The determinant is $0$, so it has no inverse.[/hide]


quadratics
function
Gauss
number theory solved
number theory

I kind of have a proof for this, but it's rather long and ugly:

Prove that if p = 8k+3 then there is an x s.t. p|x^2+2 and if p=8k+5 then there is no such x.

Easy problem for someone who knows quadratic residues.
Hard for the rest.

(1) (2/p) = (-1)^[(p^2-1)/8]
(2) (-1/p) = (-1)^[(p-1)/2]
(3) the Legendre symbol is a multiplicative function

These 3 things are all you need, and the toughest is the first one. This is the one I have an ugly proof for. I used the fact that Z[2^1/2] is a principal ring (I don't know if this is the term in English) and stuff like that. You can forget about the initial problem. The idea of this topic is this: does anyone have a nice (preferably elementary, but I doubt it) proof for (1)?

Yes, Gauss has  :D.
You should find an elementary proof for that in any respectable NT book.

Could you post one (if you have he time, of course)? Unfortunatly, I don't seem to have any respectableNT books around right now... :D  :D

Probably the most direct proof of (1) comes from Gauss' lemma, which states that if we have an integer m and an odd prime p with (m,p)=1, let r be the number of integers in the set {m, 2m, 3m, ... , (p-1)/2*m} whose least residues modulo p are negative, then (m|p) (Legendre symbol) = (-1)^r.

Now, we want to find when (2|p), so we look at the set of integers {2, 4, 6, ... , p-1} and finding the number of integers in this set whose least residues mod p are negative is equivalent to seeing how many even integers are in the interval [(p+1)/2, p-1] it follow that r = 2s when p = 8s + 1, r = 2s + 1 when p = 8s + 3, r = 2s + 1 when p  = 8s + 5, and r = 2s + 2 when p = 8s + 7, then from Gauss' lemma we see that (2|p) = 1 when p = 1 or -1 mod 8, and (2|p) = -1 when p = 3 or -3 mod 8.  This coincides exactly with (-1)^(p^2-1)/8 which is 1 when p = 1 or -1 mod 8 and -1 when p = 3 or -3 mod 8 so our identity is established.

Thanks. I did eventually manage to find a reasonably good NT book. I'll post a proof for Gauss' Lemma (proof which was found by Gauss, not me
:D)  so that this problem can be moved to its proper place (we have enough unsolved ones already  :D ):

Let ~ mean "congruent modulo p". im=q_i*p+r_i for i from 1 to (p-1)/2 with ri>0. Let's assume we have u numbers among the r_i's which are
p/2 (call them b_1, b_2, .., b_v).

We have \pi a_i *   \pi b_j =  \pi r_i ~  \pi im = m (p-1)/2 * [(p-1)/2]!. (*)

Let's assume a_i=p-b_j for some i and j. This means that p|(i+j)m => p|(i+j) which is false because i,j<p/2. This means that a_1, a_2, .., a_u, p-b_1, p-b_2, .., p-b_v are the residues 1, 2, 3, .., (p-1)/2 (not necessarily in this order, of course). This, in turn, means that
 \pi a_i * (-1) v  \pi b_j ~ [(p-1)/2]! (**)

By combining (*) and (**) we find m (p-1)/2 ~ (-1) v (1)

Euler's criterion (easy enough to prove) states that
(m/p) ~ m (p-1)/2  (2)

From (1) and (2) we get what we want: (m/p) ~ (-1) v


NEST (for admission to NISER and CBS-DAE) and the ISI entrance test are on the same day, 10 May. Does anyone have an idea why they've done this?

can u plz provide some more information on NEST????(or related website)

National Entrance Screening Test ([url]http://nest.niser.ac.in/[/url])

got it! thanks!!! very imp. exam!!!! BOL to all aops ers  (TO U ALSO!!)

Where does NEST rank among the universities in India? Also where are univerisities like IISc with respect to global standings. And is it possible to transfer to univs like MIT/UC Berkley after joining integrated M.Sc in the IISc?

IISc is in worlds top 50!! :)

2-3 yrs back it was 37th
Now mayb better :oops:

So if say...you get into IISc is it worth leaving IITs? Especially if your final aim is going into research and doing M.Sc. from US? :huh:

I wont leave India 99.99999%

search for a "Research abroad" thread in GD by ashwath

And yes,IISc in for post graduate studies only--so u cant do nething there now :P

I forgot to add....IIT is in Top 10...so dont leave it 4 IISc  :rotfl:

Thanks bro!
P.S. Why are you offline in gtalk?  :huh:

NEST is a pretty easy exam. It is mostly below AIEEE level but the questions in the general section are really tough. You need to have a really good funda of science to score well there.
I guess, since you guys are really so smart, you shouldn't get below 99%ile.

[hide]After your results come out, you'll be asking why the hell you considered joining NISER anyways! :lol: [/hide]

[quote="rituraj007"]

I forgot to add....IIT is in Top 10...so dont leave it 4 IISc  :rotfl:[/quote]

Was that a joke?
So there are less than 10 better research institutes in the world than IIT ..(Which IIT?)

I dont think IISc is in the top 50 of the world...
But definitely the best in India... ( I tend to include IISc under TIFR as do many people... So we can say IISc/TIFR)

But of course it is among the very few institutes in India which have some global standing in terms of research


calculus
integration
function
inequalities
real analysis
real analysis unsolved

Let $ \{f_n\}$ be a uniformly bounded sequence of functions which are Riemann-integrable on $ [a,b]$, and put
\[ F_n(x) \equal{} \int_a^x f_n(t)dt, (a \le x \le b)\]
Prove that there exists a sequence $ \{F_{n_k}\}$ which converges uniformly on $ [a,b]$.

Let $ M > 0$ be a uniform bound of $ f_n$. Then for any $ \epsilon > 0$, there is $ \delta \equal{} \epsilon / M$ such that $ |x \minus{} y| < \delta$ implies $ |F_n(x) \minus{} F_n(y)| \equal{} |f_n(c)| |x \minus{} y| < \epsilon$ for all $ n$. This means $ F_n$ is equicontinuous.

Clearly, $ F_n$ is pointwise bounded by the inequality $ |F_n(x)| \leq \int_{a}^{x} |f_n(x)| \, dx \leq M(x\minus{}a)$.

Then by Theorem 7.25, $ F_n$ has a uniformly convergent subsequence.

$ \{f_n\}$ is a totally bounded subset of $ L^1[a,b]$. Since $ L^1$ is complete, it follows that there is a subsequence $ f_{n_k}$ converging to $ f \in L^1[a,b]$ (not nec. Riemann-integrable, but definitely approximable by Riemann-integrals, so this could be stated in terms of riemann integration). Let $ F(x) \equal{} \int_a^x f(t) \, dt$. Then $ |F_{n_k}(x) \minus{} F(x)| \leq \int_a^b |f_{n_k}(t) \minus{} f(t)| \, dt \to 0$ as $ n \to \infty$


give yourself a pat on the back if you [color=white]post[/color] spam here

we're officially second in posting (~90% spam) in North America (and first place is Canada). YAY

In a way, we're spammier than Canada too.
Canada's population is about 32 million, compared to Bay Area's 7.2 million. Their forum has 8108 posts compared to our 4041.

Therefore Canada has 1 post for every $ 32000000/8108 \equal{} 3946.719$ people living in it
And the Bay Area has 1 post for every $ 7200000/4041 \equal{} 1781.737$
And $ 1781.737 < 3946.719$

Dudes like canada is tank.
We have a massive post count

MAN WE USED TO LEAD THE SPAM CHARGE BUT THEN OUR FORUM GOT ALL LAME

okay this topic is pointless
the spam in this topic is, amazingly, spammier than the others
the other spam is off-on-tangent spam while this is just random spam
im locking it


induction
graph theory
combinatorics solved
combinatorics

In the space there are 8 points that no four of them are in the plane. 17 of the connecting segments are coloured blue and the other segments are to be coloured red. Prove that this colouring will create at least four triangles. Prove also that four cannot be subsituted by five.

Remark: Blue triangles are those triangles whose three edges are coloured blue.

Come on guys! The first part of this problem is a trivial consequence of my proposed (but still unsolved :( ) problem :
http://www.mathlinks.ro/viewtopic.php?t=5820

In another hand, as stated, the second part is not true, as we may prove (that's always 'my' problem) that for any bicoloring of the edges of the complete graph $K_8$, there will be at least 8 monochromatic triangles.
Is the problem about only blue triangles?

Pierre.

Ok, that's what I thought : We only count blue triangles, as proved by the following exemple :
Consider the complete graph with 8 vertices $A_1,A_2,A_3,A_4,B_1,B_2,B_3,B_4$.
Now color the edge $A_iB_j$ blue for all $i,j$ and color $A_1A_2$ blue too. The others are red.
We then have 17 blue edges and exactly 4 blue triangles, which are $A_1A_2B_i$ for $i=1,2,3,4$ (but more than 4 red triangles too, which does not contradict my previous post).
This solves the second part of the problem.

Now, let's turn to the first part. The following is not very elegant but....

Given a coloring with exactly 17 blue edges, there are exactly 11 red edges.
Let's consider the graph whose vertices are the 8 points but deleting the red edges. Since there are 17 blue edges in the initial graph, the sum of the degrees of our new graph is $2 \times 17 = 34 = 4\times 8 + 2$.
From the pigeon-hole principle, it follows that there is a vertex with degree at least 6 or at least two vertices with degree 5 in this new graph.

- If there is a vertex with degree 7, say $A$ : It means that, in the initial coloring, $A$ is connected to each of the other vertices. There are 10 other blue edges left, and each one of them, say $BC$ gives a blue triangle $ABC$. These triangles are pairwise distincts, so we are done in this case.

- If there is a vertex, say $A$, with degree 6 :
Let $A_1,...,A_6$ be the vertices connected with $A$ by a blue edge, and $B$ such that $AB$ is red.
There are $ \frac {6 \times 5} {2} = 15$ edges of the form $A_iA_j$, and at most 10 are red, thus we have at lest 5 blue edges among the $A_iA_j$, and each of them gives a different blue triangle with $A$. Thus, the conclusion holds in that case.

- If there is no vertex with degree at least 6. Then, from above there are at least two vertices, say $A,B$ with degree 5.

1) If $AB$ is blue :
then there are two red edges from $A$ and two others from $B$. It left 7 red edges.
Moreover, $A,B$ both are connected with blue edges to at least two common other vertices.
a) If they have 4 common vertices connected with blue edges, then we are done.
b) If they have exactly 3 common vertices, say $C,D,E$ conneceted with blue edges : Then we have 3 blue triangles, which are $ABC,ABD,ABE$.
If $CD,DE$ or $EC$ is blue, we are done. Otherwise, they are all red, which left only 4 remaining red edges not already determined.
Let $M$ (resp. $P$) be the fourth vertex connected with $A$ (resp. $B$) with a blue edge. Among the edges $MC,MD,ME,PC,PD,PE$ at least two are blue, which gives at least two more blue triangles, and we are done.
c) If they have only $C,D$ as common other vertices connected with blue edges : We then have two blue triangles which are $ABC$ and $ABD$.
There are exactly $ \frac {4 \times 3} {2} - 1 = 5$ edges other than $CD$ which connect two vertices connected with $A$ with a bue edge. Similarly, we have 5 such edges for $B$. And, in our case, these 10 edges are pairwise distincts. Since there are only 7 red edges left, at least 3 of these 10 edges are blues, giving at least 3 mor blue triangles, and we are done again.

2) If $AB$ is red :
Then, there is one red edge from $A$ (resp. $B$) other than $AB$.
Thus, there are 8 red edges left, and $A,B$ are connected with blue edges to at least 4 common vertices.
a) If they are connected with blue edges to 5 common vertices. There are 10 edges between these 5 points, then at least 2 are blue. It gives at least 4 blue triangles, and we are done in that case.
b) If they are connected to 4 common vertices, say $C,D,E,F$, with blue edges :
There are $ \frac {5 \times 4} {2} = 10$ edges which connect two of the 5 vertices adjacent to $A$ with blue edges. Some of them connect vertices adjacent to $B$ with blue edges, but there are 4 more edges connecting two vertices adjacent to $B$ with blue edges (if $P$ is connected to $B$ but not to $A$, with a blue edge then $PC,PD,PE,PF$ are these four edges). It leads to a total of 14 edges, and among them at least 6 are blue. Each of these blue edges connect two vertices adjacent to either $A$ or $B$ with blue edges, and then gives 6 pairwise distinct blue triangles.
This ends the proof.

Pierre.

in fact it can be prove the more general statement: if we have 2n points and n^2+ 1 blue segments then we have at least n blue triangles.

the prof is by induction: for n=2 is obvious.

now for n there is at least one blue triangle(acording to Turan theorem), ABC and let x,y,z be their blue degree. we have 2 cases

I x+y+z <=3n+1 then we can choose two of them(let,s say A and B) such that
x+y<=2n so we have 2n-1 edges which are incident with one of A or B. let us remove A and B, and in the new graph we have 2(n-1) points, al lest n^2+1-(2n-1) =(n-1)^2+1 blue segments, so according to the induction at lest n-1 triangles + ABC=n , so we are done

II x+y+z>=3n+2 and let n0,n1,n2,n3 be the number of points P such that P is connected to exactly 0,1,2 or 3 of A,B,C. then we have that n0+n1+n2+n3=2n-3 and n0+n1+2n2+3n3 >= 3n+2-6 = 3n-4 and the number of triangles is at least 1+ n2+3n3= 1+n1+2n2+3n3-n1-n2 >= 1+3n-4-n0-n1-n2-n3= 3n-3-(2n-3)= n

@orl: please correct it: the triangels are required to be blue and please change it away from _the_ plane ;)


In the figure below, the shaded region is formed by drawing two parallel segments which connect the midpoints of congruent squares. Each square has side length 1 centimeter. What is the number of square centimeters in the shaded region? Express your answer as a common fraction.

[asy]size(4cm,4cm);

pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,a,b,c,d;

A=(0,1);
B=(1,1);
C=(1,2);
D=(0,2);

draw (A--B--C--D--A);

E=(1,0);
F=(2,0);
G=(2,1);
H=(1,1);

draw(E--F--G--H--E);

I=(2,1);
J=(3,1);
K=(3,2);
L=(2,2);

draw(I--J--K--L--I);

M=(1,2);
N=(2,2);
O=(2,3);
P=(1,3);

draw(M--N--O--P--M);

a=(1/2)*C+(1/2)*D;
b=(1/2)*F+(1/2)*G;
c=(1/2)*I+(1/2)*J;
d=(1/2)*M+(1/2)*P;

fill((a--b--G--c--d--C--cycle),gray);[/asy]

So you just move the extra triangles that go into the outer squares so that 2 of them complete the square in the middle, and the other 2 for a quarter square.
$ 1\plus{}\frac14$=$ \frac54$


geometry
trapezoid
ratio
trigonometry
incenter
circumcircle
geometry proposed

Triangle $ABC$ is right angled at $A$; $D,E,F$ are the circumcentres of triangles $ABC,ABD,ACD,$ respectively.

1. Let $M$ be the midpoint of $EF$. Prove that $MD\perp BC$.

2. If $O=EF\cap AD$ prove that
\[\frac{OE}{OF}=\left(\frac{AB}{AC}\right)^{2}\]

1. Let $G,\,H$ be the projections of $E,\,F$ on $BC$
We have: $D$ is the midpoint of $BC$ so $D$ also is midpoint of $GH$
On the otherhand, $EFHG$ is a trapezium and $M$ is the midpoint of $EF$ so $DM\parallel EG\parallel FH\Longrightarrow DM\perp BC$

2. Let $K$ be the projection of $A$ on $BC$
We have: $\triangle ABC\sim\triangle DEF$ and $DO$ is the altitude of triangle $DEF$
So $\frac{OE}{OF}=\frac{KB}{KC}=\left(\frac{AB}{AC}\right)^{2}$

[quote="April"]We have: $\triangle ABC\sim\triangle DEF$[/quote]

Find the ratio:$\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$ in terms of the sides or angles of triangle $ABC$.

Happy birthday to you, [b]nayel[/b]! And thanks for your problem.
Here is my solution.

Use the notations above, we have: $\frac{AB}{DE}=\frac{AK}{DO}=\frac{b\sin C}{\frac{1}{2}AD}=\frac{b\sin C}{\frac{1}{4}BC}=\frac{4b\sin C}{a}$

[quote="April"]Let $I_{1},\,I_{2}$ be the incenters of triangle $ABK,\,ACK$ respectively.
Prove that: the circumcicles of $\triangle ABI_{1},\,\triangle ACI_{2}$ are tangent at $A$.[/quote]

thank you April for your nice problem.  :) i'm not so good at geometry, so let me think about it for a while...

okey, here's my solution. Let $AP$ be the bisector of $\angle BAC$. I conjecture that $AP$ is the common tangent of the circumcircles of $\triangle AI_{1}B$ and $\triangle AI_{1}C$. Denote by $O_{1}, O_{2}$ the circumcentres of these two circles respectively. we have $\angle AI_{1}B=90^{o}+\frac{1}{2}\angle AKB=135^{o}$.

we also have $\angle AO_{1}B=2(180^{o}-\angle AI_{1}B)=90^{o}$
so $\angle O_{1}AB=45^{o}$
thus $\angle PAO_{1}=\angle PAB+\angle O_{1}AB=45^{o}+45^{o}=90^{o}$

in the same way it can be shown that $\angle PAO_{2}=90^{o}$

so we conclude that $O_{1}, A$ and $O_{2}$ are collinear and we can rest. :D

may i see your solution?

Yes, and here is my solution  :D
[color=indigo][b] Solution.[/b] Denote: $I$ be the incenter of triangle $ABC,$ then we have:
$\angle I_{1}BA+\angle I_{1}AB=\angle I_{1}AB+\angle IAI_{1}=45^\circ\Longrightarrow \angle IBA=\angle IAI_{1}$
So $\triangle IAB\sim\triangle II_{1}A\Longrightarrow AI^{2}=II_{1}\cdot IB$
Thus $AI$ is the tangent of circumcircle of triangle $ABI_{1}$
Similar, we have: $AI$ is the tangent of circumcircle of triangle $ACI_{2}$
And hence: $AI$ is the common tangent of the circumcircles of $\triangle AI_{1}B$ and $\triangle AI_{2}C$[/color]


[quote="kyyuanmathcount"]Mr. Crawford,

The AIME Problem Series class will begin in June, but that is kind of too early for me.  However, if this is the last one offered before next year's AIME, I might have to take this one.  Please let me know.  Thanks.[/quote]

We will set the fall schedule this week.  It will [i]very [/i]likely include one of the AIME Problem Series classes.


logarithms
number theory open
number theory

I've learned from Hardy and Wright that $ \sum_{p\leq n}\frac{1}{p} \equal{} \ln\ln n \plus{} O(1)$.
Do anyone know a similar result or an asymptotic (not necessarily tight)
upper bound for the generalized sum
$ \sum_{p^{\alpha}\leq n} \frac{1}{p^{\alpha}}$
(where the sum ranges over the powers of primes) ?!?

$ \sum_{p^{\alpha}\leq n} \frac{1}{p^{\alpha}}  \equal{} \ln\ln n \plus{} O(1)$ too.

Reason:
$ \sum_{p\leq n}\frac{1}{p}$ and $ \sum_{p^{\alpha}\leq n} \frac{1}{p^{\alpha}}$ differe by some proper powers of primes. So it suffices to show that $ \sum_{m\equal{}2}^\infty \sum_{n\equal{}2}^\infty \frac{1}{m^n}< \infty$ (if converging, this sum converges absoluetly, and the the sum of powers of primes is a sub-sum).
But we have a telescoping sum
\[ \sum_{m\equal{}2}^\infty \sum_{n\equal{}2}^\infty \frac{1}{m^n} \equal{} \\
\equal{} \sum_{m\equal{}2}^\infty \frac{1}{m^2\minus{}m} \equal{} \\
\equal{} \sum_{m\equal{}2}^\infty \frac{1}{m\minus{}1}\minus{}\frac{1}{m} \equal{} \\
\equal{} 1.\]


Hi,
Is possible include a .rtf file on a latex document?

Thanks.

No - how could latex know what this proprietary format is all about or deal with formatting? Convert to text first then use that.

You could use this utility. http://www.tug.org/utilities/texconv/pctotex.html


How does $\displaystyle \sqrt{3+2\sqrt2}$ simplify to $\displaystyle 1+\sqrt2$?

I did it like this.

$\displaystyle \sqrt{3+2\sqrt2} = a+b$
$\displaystyle 3+2\sqrt2=a^2+2ab+b^2$

This is the part I don't understand.

$\displaystyle 3=a^2+b^2$
and $\displaystyle 2\sqrt2=2ab$
Then solve.

Why does $\displaystyle a^2+b^2=3$?

i dont think 3 = a 2 + b 2 because 2 :sqrt: 2 doesnt = 2ab, 2 :sqrt: 2 can equal either a 2  or b 2

Well, by writing $3+2\sqrt{2}$ as a square we have $3+2\sqrt{2}=2+2\sqrt{2}+1=(1+\sqrt{2})^2$.

In your equation, we have $3+2\sqrt{2}=a^2+b^2+2ab$.  Since the coeficient of $\sqrt{2}$ is $2$, it makes sense that $a*b=2/\sqrt{2}$, and thus what's left must equal $3$, so we have $a^2+b^2=3$.
This is because we know that either $a$ or $b$ must be a radical thus and $a^2+b^2$ cannot be, so it must equal $3$.
Let me know if this helps.

You should look up viete equalities as this is a similar concept.

[quote="tarquin"]Umm
[hide]To make it a little easier let $\sqrt {3 + 2 \sqrt 2 } = \sqrt a + \sqrt b$. Squaring both sides $3 + 2 \sqrt 2 = a + 2 \sqrt {ab} + b$. And the numbers that work are 1 and 2... which makes it simplify to $1 + \sqrt 2$. Kind of cheap though because I just picked "numbers that work".[/hide][/quote]

well from your solution, we know that $ab = 2$, $a+b=3$. So we do this:

$ab = 2$, $a = \frac{2}{b}$.

substitute that into $a+b=3$, we have:

$b+\frac{2}{b} = 3$
$b^2 - 3b + 2 =0$
$(b-2)(b-1) = 0$, $b=2 \text{ or } 1$

so then a = 1 or 2. So the number $\sqrt {3+2\sqrt2}$ reduces to: $\sqrt{a} + \sqrt{b} = 1+\sqrt{2}$

i dont get it, how do we know a = 1 and b = 2

From what xxreddevilzxx wrote, we see that $a=\frac{2}{b}$ and that $b=1 \text{ or } 2$ therefore the two pairs of answers $(a,b)$ are $(1,2)$ and $(2,1)$, the final answer is $\sqrt{a}+\sqrt{b}$, so it doesn't matter which pair we take.

You can see from the first equation that $a=\frac{2}{b}$ and from the last that $b=1 \text{ or } 2$.
If $b=1$, then $a=2$.
If $b=2$, then $a=1$.

So the "two" final answers are $1+\sqrt{2}$ and $\sqrt{2}+1$ which are the same.


Call a number "$n$-consecutable" if it can be expressed as the sum of $n$ positive consecutive integers. What is the smallest number that is: 3-consecutable, 7-consecutable, 37-consecutable AND 73-consecutable?

If the desired number is $x$ and the first of $n$ consecutive numbers is $k$, then $x=k+(k+1)+\dots+(k+n-1)=\frac{k+(k+n-1)}{2}\cdot n=n(k+\frac{n-1}{2})$

Hence we have

$x=3(k_1+1)$
$x=7(k_2+3)$
$x=37(k_3+18)$
$x=73(k_4+36)$

Since all the given constants are prime numbers, it's obvious that $x$ must be their common multiplier, and the least such number is $56721.$

[hide]
Let $x$ be such a number. Then, for integers $a, b, c, d$,
$x=(a-1)+a+(a+1)=(b-3)+\dots+(b+3)=(c-18)+\dots+(c+18)=(d-36)+\dots+(d+36)$
$x=3a=7b=37c=73d$

Obviously, $3, 7, 37, 73$ must all be factors of $x$, so the smallest value of $x=(3)(7)(37)(73)=\boxed{56721}$
[/hide]


Find all possible solutions to the equation: $w!=x!+y!+z!+u!$ , where $w,x,y,u,z$ are positive integers.

[b]Answer for reference[/b]
[hide]WLOG, assume $x\leq y\leq z\leq u$. Then the answers are (w,x,y,z,u) = (3,1,1,2,2) and (4,3,3,3,3) only.[/hide]

[hide="The important step"]If $w > 4$ then $w! > 4(w - 1)!$ and the equation can have no solutions, since each of $x, y, z, u$ must be less than $w$.[/hide]


Source: ARML 2001

In triangle $ABC$, $AB=AC=1$, $BC=k$, and $\frac{\tan B}{\tan A}=2001$. Compute $k^2$.

[quote]kimby_102 wrote:Source: ARML 2001

In triangle $\ABC, AB=AC=1, BC=k$, and =2001. Compute $\ k^2$[/quote]

"and=2001" ? ;)

I think he meant $\frac{\tan{B}}{\tan{A}} = 2001$

Ok, assuming that $\frac{tan B}{tanA} = 2001$,
[hide]
Let $\angle BAC = \angle ACB = a$. Therefore, $\angle BAC = 180-2a$.
$\frac{sina}{cosa} * \frac{cos(180-2a)}{sin(180-2a)} = 2001$

Since $cos(180-2a) = -cos2a$ and $sin(180-2a) = sin2a = 2sinacosa$, the original equation becomes

$\frac{-cos2a}{2cos^2a} = 2001$. In addition, $-cos2a = 1-2cos^2a$, so solving for $cos^2a$ gives

$cos^2a = \frac{1}{4004}$.

Now, using the cosine law, we get:
$k^2 = 1+1-2(-cos2a)$
$k^2 = 2+2(2cos^2a-1)$
$k^2 = 4cos^2a$.
Therefore, $k^2 = \frac{1}{1001}$.
[/hide]


AMC
AIME
calculus
integration
trigonometry
ratio
trig identities

11. Let x= (cos 1+cos 2+cos 3+....+cos 43+cos 44)/(sin 1+sin 2+ sin 3+ ... + sin 43 + sin 44). What is the greatest integer that does not exceed 100x?  (all trig. terms are in degrees, not radians)

Can someone provide a plain and easy to understand explanation of the answer?

I hope someone else does not reply while I type up this solution :

[hide]

Good to know:

cos(a)*sin(b) = (-sin(a-b)+sin(a+b))/2

sin(a)*sin(b) = (cos(a-b)-cos(a+b))/2

Let x =  :Sigma: (i = 1 to 44) cos(i), y =  :Sigma: (i=1 to 44) sin(i)

Multiply each sum by catalyst equal to 1 (sin(0.5)/sin(0.5) = 1)

x = sin(0.5)/sin(0.5)*:Sigma: (i = 1 to 44) cos(i) = :Sigma: (i = 1 to 44) (-sin(i-0.5)+sin(i+0.5))/(2sin(0.5)). This telescopes, and we obtain

x = (-sin(0.5)+sin(44.5))/(2sin(0.5)).

y = sin(0.5)/sin(0.5)*:Sigma: (i = 1 to 44) sin(i) = :Sigma: (i = 1 to 44) (cos(i-0.5)-cos(i+0.5))/(2sin(0.5)). This telescopes, and we obtain

y = (cos(0.5)-cos(44.5))/(2sin(0.5)).

x/y = (-sin(0.5)+sin(44.5))/(cos(0.5)-cos(44.5)) (the denominators cancel) = (2cos(22.5)sin(22))/(2sin(22.5)sin(22)) = cot(22.5).

Using half angle formula, 100*cot(22.5) = 100sqrt((2+sqrt(2))/(2-sqrt(2)) = 100sqrt((6+4sqrt(2))/2) = 100sqrt(3+2sqrt(2)) = 100sqrt(2)+100  :approx: 241.

[/hide]

Sayoonara.

The top is pretty close to the integral of cos(x) dx from 0 to 45.  The denominator is close to the integral of sin(x) dx from 0 to 45.  So you get

[ :rt2: /2] / [1 -  :rt2: /2].

Then you can get the answer.

Hmm. I wonder if there is a reason (other than luck) that the integral approximation works out to the exact answer.

I will not put this in spoiler because it is too hard to understand without latex.

Solution:

Let our expression be y.  First, we know that

[tex]
\displaystyle\cos x = \sin (90-x)
[/tex]
Thus we turn our expression into

[tex]
\displaystyle y = \frac{\sin 46 + \sin 47 + \cdots + \sin 89}{\sin 1 + \sin 2 + \cdots + \sin 44}
[/tex]

We note that 46 = 1+45, 47 = 2+45 etc.  Thus we use the fact that
[tex]
\displaystyle\sin (a+b)=\sin a \cos b + \sin b \cos a
[/tex]

to find that

[tex]
\displaystyle\sin (x+45) = \frac{\sqrt{2}}{2}(\sin x + \cos x)
[/tex]

Our expression becomes

[tex]
\displaystyle\frac{\sqrt{2}}{2} \cdot \frac{\sin 1 + \sin 2 + \cdots + \sin 44 + \cos 1 + \cos 2 + \cdots + \cos 44}{\sin 1 + \sin 2 + \cdots + \sin 44}
[/tex]

[tex]
\displaystyle y = \frac{\sqrt{2}}{2} \cdot \left(1 + \frac{\cos 1 + \cos 2 + \cdots + \cos 44}{\sin 1 + \sin 2 + \cdots + \sin 44}}\right)
[/tex]

[tex]
\displaystyle= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}y
[/tex]

Solving for y gives [tex]y = \frac{\sqrt{2}}{2 - \sqrt{2}}[/tex].  We know that the square root of two is about 1.4142, so doing the arithmetic and multiplying by 100, we find that the answer is 241.

If you didn't see or remember the trig identities that have been used here, there's another place to start. The numerator and denominator of this expression are, respectively, the real and imaginary parts of a finite geometric series with ratio equal to e^(i:pi:/180). Everyone should know about finite geometric series:
sum = ((first term) - (next term after the last))/(1 - ratio).

This observation doesn't make the answer immediate - there's still quite a bit of complex number arithmetic to be done, and done quite carefully - but you're now working on a calculation that you know should work.

Good Hint: [hide]Multiply both the numerator and denominator by sin .5 :^o:.  From there, use sum-product identities.[/hide]  Ok oops everyone beat me to it.

And sadly I got that answer but since I didn't know the value of :rt2: to a few decimal places, I'd have gotten that wrong anyways.


$ \ce{C6H6} \plus{} \ce{CH2(Cl)CH2CH2COOH} \plus{}\ce{HF}   \plus{} \Delta$

What kind of reaction is this? Are you sure about the reagents?

Is the answer a 2 ring system ?
I think so because HF is used for cyclising and i think the first reagent is used for chlorination.

yes madness( :D ) u r right .
please post the complete mechanism

If the person of have made this problem intended to expect alpha-tetralone as the product, then I think he/she have chosen a not very good example.

[quote="Pardesi"]yes and the product even more weird[/quote]

I believe the mechanism that your book or madness will propose will be even more weird.

[quote="Madness"]and i think the first reagent is used for chlorination.[/quote]

What reagent?

The reagent I was referring to was CH2(Cl)CH2CH2COOH and HF.
Its a bit like Friedel and crafts alkylation .
First CH2(Cl)CH2CH2COOH and HF form the carbocation CH2+CH2CH2COOH .
I think it rearranges to form CH3CH+CH2COOH .
Then it attacks the benzene ring to form C6H5CHCH2COOH .
                                                               |
                                                               CH3
Then here HF acts as an acid and gives a H+
Amazingly ( I think this is what will happen as I vaguely remember it as an IIT coaching test question )
the COOH becomes O=C+ .
Then the compound cyclises to form a 5 membered ring on the phenyl ring
****____ **CH3
***/**O*\__|
***\____/***\
*******\____\
*******||
*******O
Sorry for the *stars didn't know another way to draw the rings
Another alternative maybe as Carcul says $ \alpha$ tetralone for which we could propose a similar mechanism ( Sorry if its weird but my name is MADNESS !!!!!!!  :wink: )

Madness can u plese use Mchem installed here all u have to do is read [url=http://www.mathlinks.ro/viewtopic.php?mode=attach&id=11410]this[/url] first thwo-three pages and enclose all codes within dollar symbol

Well my answer was
($ \ce{CH(CH3)CH2CO}$)cyclic attached to
$ \ce{C6H6}$

no there is no branching in the ring attached exxcept for the keto part.
whence i beleive there is no rearrangemnet in the first step :rotfl:

Then it must be $ \alpha$ tetralone as Carcul said .
Then rearrangement occurs after the linear chain is attached to the benzene ring.
 :maybe:

[quote="Madness"]First CH2(Cl)CH2CH2COOH and HF form the carbocation CH2+CH2CH2COOH.[/quote]

And what is the mechanism of that amazing carbocation formation reaction?

[quote="madness"]
****____ **CH3
***/**O*\__|
***\____/***\
*******\____\
*******||
*******O
Sorry for the *stars didn't know another way to draw the rings
[/quote]

Worry not.  Within the next couple days, we should have a structure-drawing Asymptote package installed on this forum.

Thanks Gregx007, that would be very nice, specially for drawing organic structures and mechanisms when necessary.

I THINK AN ACYLIUM ION IS FORMED INITIALLY BY ELiMINATION OF WATER NOT Cl, WHICH UNDERGOES FRIEDEL CRAFTS REACTION.

I CAME ACROSS THE SAME QUESTION IN ONE OF THE BOOKS. AND THE FINAL PRODUCT WAS A BICYCLIC RING WITH A KETO GROUP WITH NO BRANCHING( ALPHA TETRALONE)(THE MECH WASNT GIVEN THO). SO I WAS WRONG ABOUT ACYLIUM ION BEING FORMED FIRST. CL- IS REMOVED FIRST TO GIVE CARBOCATION AND THE REACTION PROCEEDS LIKE ANY NORMAL FRIEDEL CRAFTS REACTION LASTLY HYDROXYL GROUP IS ELIMINATED AS WATER TO GIVE A CYCLIC COMPOUND. IN THE FIRST STEP F- ACTS AS A NUCLEOPHILE TO REMOVE CL-. I KNOW CL- ISNT SUCH A GOOD LEAVING GROUP BUT I GUESS, ITS ABILITY IS ENHANCED BY ANCHIMERIC ASSISTANCE DUE TO NEIGHBOURING CARBOXYL GROUP. I DUNNO MUCH ABOUT SYMPHORIA N STUFF SO YOU BETTER FIND OUT FROM SOMEONE WHO IS IN A BETTER POSITION TO ANSWER

Varunarasu can u please stop using capitals it's almost like u r yelling answers at us  :huh:
Thanks

First of all we should know that HF is a good dehydrating agent.
 It will react with the -COOH group and eliminate the -H2O to form -C+=O (Acylium ion)
which undergoes the friedel crafts reaction.(This is one of the steps in the Haworth synthesis of preparation of naphthalene and other fused rings.)

After that the reaction proceeds.

[quote="VARUNARASU"]IN THE FIRST STEP F- ACTS AS A NUCLEOPHILE TO REMOVE CL-.[/quote]

What an amazing reaction!

[quote="VARUNARASU"]I KNOW CL- ISNT SUCH A GOOD LEAVING[/quote]

Chloride ion is a very good leaving group, since it is a very weak base.

[quote="Chemrock"]First of all we should know that HF is a good dehydrating agent.[/quote]

Which is also amazing, since HF is a weak acid (unless we use a ~100% solution, but then we would also be in danger).

[quote="Chemrock"]After that the reaction proceeds.[/quote]

And what exactly happens to chlorine?

hey carcul, i know f- is a poor nucleophile and cant replace chlorine otherwise. dats why i said i think the carboxyl group also plays a part due to anchimeric assistance. im not sure abt dat but if you can think of a better mechanism plz post it.
[quote]Which is also amazing, since HF is a weak acid (unless we use a ~100% solution, but then we would also be in danger). [/quote] n wat does acidity have to do with its dehydrating ability?? :?:
n who cares if we are in danger, we are not actually gonna carry out the reaction to find out the product


Find all solutions (x,y) to:

6*Binomial[x,3]=Binomial[y,x-3]

(where y>=x-3 and x>=3).

Is Binomial(a,b) the same thing as C(a,b)?

[hide]

Expanding that gives x! = y(y-1)...(y-x+4)

In other words, we want n! = a product of n-3 consecutive (positive) integers.

Now if we look at n(n-1)...(5)(4), which is n-3 consecutive integers, its product is n!/6, a bit too small. The next one up is this times (n+1)/4, the one after that you also multiply by (n+2)/5, etc.

So we can change the problem into finding n such that (n+1)/4 * (n+2)/5 * (n+3)/6 * ... as much as we want, = 6.

Just taking the first term gives n=23 works. Taking the first two terms shows any other n>=10 won't work (first two terms make product too big, and increasing the number of terms can only increase it).

So all we have to do is check n from 4 to 9 by hand - very easy to do, you just keep multiplying by the next term until you get to 6 or higher as it always increases. In the end there are four solutions, which lead to (x,y) as

(4,24), (6,10), (7,10), (23,24).

[/hide]

See the 1990 AIME, problem #11. :D

The brute-force works pretty well, but one would hope there is a more combinatorial way...  Do you know of a combinatorial solution, fairyofwind?  I'd be interested to see it.


What is the units digit of $ 23^{45} \plus{} 67^{89}$?

Since $ 45 \equiv 1\pmod{4}$ and $ 89 \equiv 1\pmod{4}$, we have $ 3\plus{}7$, or $ \boxed{0}$.

We only need to look at the units digits. The tens digits do not affect the end units digit. So basically we have \[ 3^{45} \plus{} 7^{89}\] Now \[ 3^1 \equal{} \boxed{3}\]\[ 3^2 \equal{} \boxed{9}\]\[ 3^3 \equal{} \boxed{7}\]\[ 3^4 \equal{} \boxed{1}\] So, for every 4 consecutive exponents, the unit digits repeat. Let's do this for 7, too. \[ 7^1 \equal{} \boxed{7}\]\[ 7^2 \equal{} \boxed{9}\]\[ 7^3 \equal{} \boxed{3}\]\[ 7^4 \equal{} \boxed{1}\] So the remainder when 45 is divided by 4 is 1; the units digit for the first term is 3. When 89 is divided by 4, the remainder is also 1. The units digit for the second term is 7. Add them together to get $ \boxed{0}$.


real analysis
calculus
integration
function
algebra
domain
polynomial

Is the space of signed measures on the Lebesgue-measurable subsets of $ [0,1]$ separable under the (ba) total variation norm?
http://en.wikipedia.org/wiki/Ba_space

If $ h_n \rightarrow h$ under the total variation norm, do their integrals (of any continuous function on $ [0,1]$) converge as well?

Thanks.

It's not separable.

What I'm going to say is not in any sense a proof, but it's a plausibility argument. For Banach spaces, especially Banach spaces that include functions on familiar domains (like $ [0,1],$ separability (which is having a countable dense [i]subset[/i]) tends to go hand in hand with having a "nice" dense [i]subspace.[/i]

If you take a variety of "nice" subspaces of functions on $ [0,1]$ - step functions, continuous functions, polynomials, etc. - and find their closures under the total variation norm, you don't get all of the signed measures, you only get the absolutely continuous ones - that is, $ L^1[0,1].$ We do know that $ L^1$ is itself separable, but it's a non-trivial closed subspace of $ M.$

As I said, that is in no sense a proof.

[quote]If $ h_n \rightarrow h$ under the total variation norm, do their integrals (of any continuous function on $ [0,1]$) converge as well?[/quote]
Yes. $ M$ is the dual space of $ C,$ and the total variation norm on $ M$ is the induced norm on the dual space. Your question is effectively to ask if norm convergence implies weak* convergence, which it does.

An uncountable discrete set: point masses on each point of $ [0,1]$. That rules out separability.


So I've been trying to do math contest prep in a more organized fashion from January on; I pick a few chapters from AoPS vol 2 at at a time and work through them, jumping back and forth to make sure that I'm not just doing one subject area for a week or two weeks straight. Do you guys have any specific way you like to organize yourself or is it more of a do what you want when you want it?

I actually try to focus on just one subject for a while, then move on.  I don't know if thats the best idea, but it works for me as long as I go back and do problems on previous topics regularly.


function
number theory proposed
number theory

Let a function f:N--->N be called "cautious" if |f(n)-f(n+1)|  2001 for all
 n \in  N. Suppose f,g,h are cautious functions such that f(n)g(n)= h(n) for all n \in  N. Prove that there exists a constant C such that
min{f(n), g(n)}<C for all n

-to my understanding this problem was propsed, but never included because of its difficulty :? I may be wrong however

Yeah, I consider this one a super-hard problem. I remember spending lots of time trying to solve it, with no success.

yeah, harazi do you know if anyone has solved this problem yet? I don't think so, because esp. for rookies this is an extremely hard problem and it is probably is a hard problem for veterans also ( although it is a beautiful problem :D)

No, I asked quite lots of people, but they couldn't solve it.


Where could you find images for most of the problems more specifically

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433599#p433599
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433604#p433604
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433703#p433703

I'm trying to practice 2006. How can I view the 2006 pictures? I can't seem to do so.  :(

So does that mean they are completely inaccesible? That sucks, as Geometry is probably my worst area and I was trying to improve it. While #13 is impossible without the diagram, it seems easy. The other ones I cannot even start to visualize.


Tennyson is building a circular corral for his pet dinosaur.  To form the corral he pounds 20 posts equally spaced around the circumference.  Each post contributes 6 inches to the circumference and the fencing between adjacent posts contributes 8 feet to the circumference.  To the nearest whole number, what is the number of square feet in the area of the corral?  Express your answer to the nearest integer.

[hide]20*6 = 120 so the posts make 10 feet. The number of fences will be one less then the number of posts so it is 19 * 8 = 152 feet. Combined it is 162 feet circumfernce. d = c/ :pi:  so the diameter is about  51.6 feet. Radius = 25.8. The area is 2091 square feet.[/hide]

[hide]

20 posts and 20 wires

10 feet and 160 feet

170 feet is the circumference

so 85/pi feet is the diameter,

pi(85)^2/pi^2

7225/pi

CALCULATOR'D

2300 square feet

Where did Tennyson get that dinosaur? I want one of those too  .

A+Math, it's a circle. There will be exactly as many wires as posts.

[/hide]


inequalities
quadratics
graph theory
combinatorics unsolved
combinatorics

Determine the maximal number of edges of graph with $n$ vertices that doesn't contain a 4-cycle.

Ok, let $d_1,d_2...d_n$be the degrees of the graph, The idea is to double count trios of vertices $a,b,c$ such that $ab$ and $bc$ are edges. Call this number $N$.

If we fix $a,c$ then there is at most another vertex $b$ to whom both of them are conected, (otherwise we would get a four cycle) so we get $N\leq \binom{n}{2}$.

On the other hand, fixing $b$, we get $\binom{d_b}{2}$ pair $(a,c)$ conected to $b$ so $N=\sum_{i=1}^n\binom{d_i}{2}$.

These obseravtions take us to our inequality:

\[ n(n-1)\geq \sum_{i=1}^n d_i^2-d_i\geq \frac{(\sum_{i=1}^nd_i)^2}{n}-\sum_{i=1}^nd_i  \]

So let $e$ be the number of edges. Our inequality becomes then a quadratic one in $e$:

\[ 0\geq 4e^2-2ne-n^2(n-1)  \]

Which occurs only if:

\[ e\leq \frac{n+n\sqrt{4n-3}}{4}  \]

Which is a good bound, however in the practice, we do not apply Cauchy, but try to maximize $d_1+d_2...+d_n$ with the given restrictions (maximun depends on the regularity of the sequence).

Do you need exact answer? It seems to be a very difficult question.

Well, this is a generalisation of a problem that was given in the second round of our national olympiad but I was able to find solution only for case proposed at the olympiad.
Anyway, very good solution, Pascual2005, though it's a bit standart, but nevertheless very good, 10x :)

[quote="Pascual2005"]

$\sum_{i=1}^n d_i^2-d_i\geq \frac{(\sum_{i=1}^nd_i)^2}{n}-\sum_{i=1}^nd_i$

[/quote]

I'm sorry, but I don't get that part..  :oops:

Heh, yeah, how could I miss that one..
Thanks! ;)

Here, one should read :
[quote="Pascual2005"]
Which occurs only if:

\[ e \leq \left \lfloor \frac{n+n\sqrt{4n-3}}{4} \right \rfloor  \]
[/quote]

And, for $n = 5$, it yields $e \le 6$, hence we can easily see that we can have the equality.

Just to add my "grain of salt" to this beautiful proof!  :P

Hm this seems slightly non-rigorous, but:

Assuming n>=4,

There exist n*(n-1)*(n-2)*(n-3) ordered sets of four vertices. You must not have an edge between at least one of the 4 successive elements in each ordered set.

However, each edge belongs to 4*(n-2)*(n-3) ordered sets of four vertices.

Thus there must be at least n*(n-1)/4 edges missing from the complete graph of $n$ vertices.

There are n*(n-1)/2 edges, so there can be at most $\frac{n*(n-1)}{4}$ edges with no 4 cycles in a graph.

All that remains is to find an equality case (which I haven't found yet).

EDIT: This doesn't agree with the actual solution, where did I screw up in logic?

[quote="ThAzN1"]I didn't read it all, but you seemed to have proved a weaker upper bound (so it's true but doesn't say very much) Can we get equality?[/quote]

Except for n = 5 it doesn't work.


Find a closed formula for the recursion a_n=6_(n-1)-9_(n-2) where a_2=4 and a_1=1.

Enjoy


For each positive integer k, define the sequence {an} by
a[size=75]0[/size] = 1 and a[size=75]n[/size] = k[size=75]n[/size] + (-1)^n a[size=75]n[/size]-1 for each n > or = 1
Determine all values of k for which 2000 is a term of the sequence.

I've solved it but my solution is based on a "boring fact".

[hide]$ a_{4n} \equal{} 4n k  \plus{}  1$

$ a_{4n\plus{}1} \equal{}  k  \minus{}  1$

$ a_{4n\plus{}2} \equal{} (4n\plus{}3) k  \minus{} 1$

$ a_{4n\plus{}3} \equal{} 1$[/hide]

I was lazy to prove that by induction (easy but boring).

Then is easy to see that

[hide]$ k\in\{2001, 667, 87, 23, 3\}$.[/hide]

What does $ k_n$ mean? Isnt $ k$ a constant?
If it is, then I have
$ a_0\equal{}1$
$ a_1\equal{}k\minus{}1$
$ a_2\equal{}k\plus{}k\minus{}1\equal{}2k\minus{}1$
$ a_3\equal{}k\minus{}2k\plus{}1\equal{}\minus{}k\plus{}1$
$ a_4\equal{}k\plus{}k\minus{}1\equal{}2k\minus{}1$
$ a_5\equal{}k\minus{}2k\minus{}1\equal{}\minus{}k\plus{}1$
....
Therefore to let a term equal 2000, our only postive k is $ 2001$

I'll use $ \text{\LaTeX}$.

For each positive integer k > 1, deﬁne the sequence $ \{a_n\}$  by

$ a_0 = 1 \qquad a_n = kn + ( - 1)^n \ a_{n - 1}\quad n\geq 1$

Determine all values of k for which 2000 is a term of the sequence.

See my hints.


pigeonhole principle
geometry
real analysis
real analysis unsolved

So I know (b/c our prof told us) that this question is supposed to be solved by the PHP.

Let T be a triangle such that T can be placed inside the unit square (sides of length one) with the center of the square not contained in the triangle. Prove that one of the sides of T must have a length of less than 1.

Saying that the center isn't in the triangle means that we can draw a line through the center, and have all three vertices of the triangle on the same side. We have three triangle vertices; if we can force two into some set $ S$, their distance is at most the diameter of $ S$. For Pigeonhole to work with this, we need that $ S$ is one of two pieces.
OK, can we further divide that half of the square (sliced through the center at an arbitrary angle) into two pieces, each of diameter at most $ 1$?

[hide="Hint"]Slice with an orthogonal line to the one given by jmerry, also passing through the center[/hide]

I think I see where you are going in that you're trying to say that 3 vertices must be in three separate parts (since we just split the square into four part) but if they can only be in two otherwise they contain the center. This makes intuitive sense to me but I have no idea how you'd even start to go about proving it. Most of my proof experience comes from set theory...I haven't done many intense geometry proofs.


k is a nonnegative integer and the integers \[ a_{1},a_{2},a_{3},...,a_{n}\]
give at least 2k different remainders on division by n+k . prove that among the \[ a_{i}\] there are some whose sum is divisible by n+k


Given that both $ p$ and $ p \plus{} 1$ are prime numbers, what is the least positive composite number that is not divisible by $ p$ nor $ p \plus{} 1$ ?

The only two consecutive numbers that are prime are $ 2$ and $ 3$.

The first composite number that is not divisible by either $ 2$ or $ 3$ is $ 5 \cdot 5 \equal{} \boxed{25}$


geometry
3D geometry
sphere
invariant
geometric transformation
rotation

What does it mean for a finite set of points to be [b]uniformly distributed[/b] over the surface of a sphere?

Is that expression somehow related to the usual [b]UNIFORM DISTRIBUTION[/b] that we encounter at school?

Got any books, articles, links, etc., that might be of some aid TO HELP me learn these type of topics?

Thanks in advance for your replies.

In this case, uniform means invariant under rotations; the probability of a region is the same as the probability of any rotation of that region.
For the uniform distribution on an interval, it means invariance under translation mod $ 1$.
Uniform distributions naturally arise given compact groups, and compact spaces they act transitively on.

[b][color=red]Thanks for taking the time to reply, Sir.[/color][/b]

Wonder whether you'd NOW let me know where you got all that 411 from. Probabilities, rigid motions, compactness, groups: it all sounds like a party to me!  :D

[url=http://en.wikipedia.org/wiki/Haar_measure]Haar measure[/url]. Probability is frequently the same thing as measure theory on spaces of total measure $ 1$, after all.
It's really just obvious to me; uniform means that the probability is proportional to the standard measure for the set. That allows you to talk about uniform distributions on balls or other irregularly shaped sets in $ \mathbb{R}^n$, which aren't covered nicely by a transitive group action.


MATHCOUNTS
ratio
number theory
relatively prime

i have no idea how to do this problem please tell me how and give me answer

When 3.583 (the three in the thousanths digit is recurring goes on forever i dont know how to put the bar) is expressed as the ratio of 2 relatively prime integers a and b what is a + b?

PLEASE HELP QUICKLY THANKS A LOT IF YOU DO

Try not to overuse exclamation points.

[hide]$ 3.58\overline{3}\equal{}\frac{39775}{11100}$

$ 39775\plus{}11100\equal{}50875$[/hide]

Overuse of exclamation points is the sign of a diseased mind.
[hide]It is $ 3.58+\frac{1}{100}\cdot{\frac{1}{3}}$ which is $ \frac{1075}{300}=\frac{43}{12}$.
So 55.[/hide]
i_like_pie, those aren't relatively prime.

[quote="i_like_pie"]Try not to overuse exclamation points.

[hide]$ 3.58\overline{3} \equal{} \frac {39775}{11100}$

$ 39775 \plus{} 11100 \equal{} 50875$[/hide][/quote]
which method did you use to get that fraction?
The first fraction I get (before simplifying) was 3225/900

ok..
.583 with a line over the numbers= 583/999 so 3  583/999
i guess you would simplify the fraction?

add on to last comment by me
and add the numerator and denominator?

[quote="LadyKn1ght"]To repeat what other people say, don't use excessive exclamation points; they are not necessary to get people to look at your topic :wink: . (I removed the exclamations, btw.)[/quote]
That's absolutely true; mathirons, overusing exclamations can sometimes hurt you from solving a math problem you're stuck on. If you just calm down you can eventually do it yourself without help.

[quote="xpmath"]Overuse of exclamation points is the sign of a diseased mind.
[hide]It is $ 3.58 + \frac {1}{100}\cdot{\frac {1}{3}}$ which is $ \frac {1075}{300} = \frac {43}{12}$.
So 55.[/hide]
i_like_pie, those aren't relatively prime.[/quote]

That's one nice way of looking at it!

Nice! :lol:

Let $ F\equal{}3.58\overline{3}$.  Then $ 100F\equal{}358.\overline{3}\equal{}358\dfrac{1}{3}\equal{}\dfrac{1075}{3}$.  Dividing by 100 we get that $ F\equal{}\dfrac{1075}{300}\equal{}\dfrac{43}{12}$.  $ 43\plus{}12\equal{}\boxed{55}$

[quote="DonkeyKong"][quote="funmath"]Let $ F \equal{} 3.58\overline{3}$.  Then $ 100F \equal{} 358.\overline{3} \equal{} 358\dfrac{1}{3} \equal{} \dfrac{1075}{3}$.  Dividing by 100 we get that $ F \equal{} \dfrac{1075}{300} \equal{} \dfrac{43}{12}$.  $ 43 \plus{} 12 \equal{} \boxed{55}$[/quote]

Let's not try to solve it again..
This question has been already solved by xpmath...[/quote]
Right. Thats true, DK.

Mods plz lock this.

[quote="ProtestanT"][quote="i_like_pie"]Try not to overuse exclamation points.

[hide]$ 3.58\overline{3} \equal{} \frac {39775}{11100}$

$ 39775 \plus{} 11100 \equal{} 50875$[/hide][/quote]
um...how did u get those numbers?[/quote]
That is not the correct answer.. so it really doesn't matter..

Mods, can you lock this?

It does help to know what he's thinking...because those number look a bit familiar.

He probably used a method similar to the other two answers.

Most likely:
\begin{eqnarray*}
(3.58\overline{3}\times 10000) + (3.58\overline{3}\times 1000)+(3.58\overline{3}\times 100)&=&(35833+3583+358)+3\times 0.\overline{3}\\
&=&39774+1\\
\end{eqnarray*}

So $ 3.58\overline{3}=\frac{39775}{10000+1000+100}$

He just forgot to simplify the fraction to $ \frac{43}{12}$, but the two fractions are equal.

Btw, mathirons, here's what to type to make "$ 3.58\overline{3}$":

[code]$3.58\overline{3}$[/code]

So, the problem has been solved three times. Topic is now locked.


im not going to state (im a back up) but kansas's is on March 8.

PA's is like on March 29th or something, all i know is that its the last week end of spring break!

I'm pretty sure it's March 8th for MA as well

Virginia = March 15

eh... this topic is kind of pointless.

I'm not going to get top 20 unless everyone else fails, so don't worry.

right...

I'm not going to make a prediction cause that makes weird things happen

I'm not gonna make nats unless I miraculously become a middle schooler again...

[quote="ZhangPeijin"]Meh, I ain't gonna make it to Nats unless something miraculously amazing happens.
Like everyone gets sick.[/quote]

theres poisonous stuff everywhere. :rotfl:

yeah PA is the 29th http://www.pspe.org/mathcounts_schedule2008.shtml

[quote="infinity4ever"]You guys need some self esteem.  :D

Anyways, it really doesn't make much a difference when peoples' state competitions are... everybody takes the same test anyways.  :huh:[/quote]
Well actually it does because in some states you could get like a 39 and be #1, while in others you could get a 39 and be tied for 15th place.

Oklahoma's state competition is on March 8th.

GA's is on March 15th at the Atlanta Zoo.

Wisconsin is on March 1st... do I really have to repeat what xpmath, Zhangp, or all the other (their statements apply to me as well) or does everyone get the picture?

TX= March 29

I won't make it to nats.
ra5249 and stevenmeow probably will.

Wait, so it's actually on the 8th?
Or are you using reverse psychology?
I have to check on website now...

???
I was talking about MA state competition...
What are you talking about?

[quote="Moldytape"]I guess ra5249 will use cheap, underhanded, illegal tactics to get to nats because the last time I heard that he is in 9th grade now. And I wish States was in March 9th because then I get to miss tennis/chinese school/piano. :([/quote]
What he said.

Wait. Who's ra5249 then? I'm so confused.

[quote="frost13"][quote="illusiat"]MA state competition is on March 9

[quote="Ihatepie"]I heard that state sprint round gets harder as the years go bye. Everything else, including nats just jumps around.[/quote]
Not really; 2006 was easier than 2005 and 2005 and 2007 were about the same[/quote]

You're from Acton right?  If you guys show up March 9th that will make it easier for my team to repeat since the rest of us will be there March 8th when they hold the competition:-)  Now if only we can trick Diamond to showing up on the 9th as well.  Hey Moldytape, states is on the 9th now ;-)

Frost[/quote]

hmm, never say that side of you before...

I've heard there's going to be some bad weather in MA and that states will be moved to the 15th :P


There is a post-rating bug. I have two pictures taken at the same time. Windows Vista, AoPS skin, Firefox 3

I really don't understand those pictures. How about you try to explain what you think is going wrong.

[quote="eggylv999"]When i rated the post spam, its rating went up.[/quote]Someone might have rated the post at the same time you were.

Seriously, who would give to that post a rating higher than 1?
Sorry; if this is spam, rate it as just 1.

This has happened to me before also.  Strangely, it only has happened with mewto55555's posts.


Let $ a,b,c > 0$, then
$ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}\geq\frac{9}{4}\plus{}(\frac{a\plus{}b}{c}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{b\plus{}c}{a})\minus{}\frac{9(ab\plus{}bc\plus{}ca)}{4(a^2\plus{}b^2\plus{}c^2)}$

[quote="filomen"]Let $ a,b,c > 0$, then
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq\frac {9}{4} \plus{} (\frac {a \plus{} b}{c} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {b \plus{} c}{a}) \minus{} \frac {9(ab \plus{} bc \plus{} ca)}{4(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]

not true...

try $ a\equal{}b\equal{}c\equal{}1$ .

[quote="filomen"]Let $ a,b,c > 0$, then
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq\frac {9}{4} \plus{} (\frac {a \plus{} b}{c} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {b \plus{} c}{a}) \minus{} \frac {9(ab \plus{} bc \plus{} ca)}{4(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]

$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq\frac {9}{4} \plus{} \frac {1}{2} (\frac {a \plus{} b}{c} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {b \plus{} c}{a}) \minus{} \frac {9(ab \plus{} bc \plus{} ca)}{4(a^2 \plus{} b^2 \plus{} c^2)}$

$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq 7\plus{} \frac {1}{2} (\frac {a \plus{} b}{c} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {b \plus{} c}{a}) \minus{} \frac {7(ab \plus{} bc \plus{} ca)}{(a^2 \plus{} b^2 \plus{} c^2)}$

the inequality holds

[quote="filomen"]Let $ a,b,c > 0$, then
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq\frac {9}{4} \plus{} (\frac {a \plus{} b}{c} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {b \plus{} c}{a}) \minus{} \frac {9(ab \plus{} bc \plus{} ca)}{4(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]

Sorry... Correct

$ \frac{a^{2}}{b^{2}}\plus{}\frac{b^{2}}{c^{2}}\plus{}\frac{c^{2}}{a^{2}}\geq\frac{9}{2}\plus{}(\frac{a}{c}\plus{}\frac{c}{b}\plus{}\frac{b}{a})\minus{}\frac{9(ab\plus{}bc\plus{}ca)}{2(a^{2}\plus{}b^{2}\plus{}c^{2})}$


I know there are a lot of subjects about it, but I didn't see a proof using this  inequality.

It was already on the forum... You just have to look for it.

Here's the [url=http://www.math.auckland.ac.nz/olympiad/Rear.pdf]link[/url] (example 5).

I didn't understand the solution...I think I have another one...thanks!

ok I noticed that the sum of the cofficients of this expression equal to $n^{n}$  $(x_{1}+x_{2}+...+x_{n})^{n}$, for example $(a+b)^{2}=a^{2}+2ab+c^{2}$ and the sum is $4=2^{2}$. now, hod do I prove it, and may I can use it to prove AM-GM by rearrangement inequality? thanks. :)
because there is the same number of numbers on both sides of the inequality we can maybe use rearrangement..?


Prove that :
 for all x>0 then x 2 + \pi x +15/2 \pi sinx >0

discuss for three cases:
(i) x \in (0, \pi]
(ii) x \in (\pi, 3/2 \pi) (hint: x-\pi \in (0,\pi/2) ==> sin(x-\pi) < x-\pi
(iii) x \in [3/2 \pi, +\infty)

u can use the derivative.. it's easier i think


how do you produce the letters  QED with the central 'E' dropped just like \TeX{} \[ \text{\TeX}{} \]

I have seen it done but have never seen how its done.

Like this $Q\!_{E}\!D$? Though many people use the qed symbol $\square$ or $\blacksquare$ (or similar) automatically added in the proof environment.

using the E as a subscript just doesn't have the same effect.  THX anyway.

$Q\!_\text{\normalsize E}\!D$ is the best I can do.


I have created a card tictactoe. A lay out 9 random cards in a 3*3 square.  I tell them the object is to pick 3 in a row that are red-red-red, or black-black-black.

The 9 cards are faced down.

What is the probability of them losing when 2 cards are faced up?

1?

Or none at all?

(there are 5 red cards and 4 black cards by the way)

Well, there are three possible cases for this:
1. Both face-up cards are red
2. Both are black
3. There is 1 of each

Choosing those 2 cards randomly, the probability of the first case occurring is $\left(\frac{5}{9}\right)\left(\frac{1}{2}\right)=\frac{5}{18}$.  The probability of the second case is $\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)=\frac{1}{6}$, and the probability of the third case is $\left(\frac{5}{9}\right)\left(\frac{1}{2}\right)+\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)=\frac{5}{18}+\frac{5}{18}=\frac{5}{9}$.

When the first case is true, the probability of losing is $\frac{4}{7}$, when the second case is true, it is $\frac{5}{7}$, and when the third case is true, it is $1$.  Multiplying these values by the probabilities of each case occuring, we get $\frac{10}{63}$, $\frac{5}{42}$, and $\frac{5}{9}$.  Adding these together, we find that the probability of losing is $\boxed{\frac{5}{6}}$.

I hope this is what you meant by this question.  If I misinterpreted it, could you please clarify?  Asking what the probability is when two cards are face up is pretty ambiguous...

[quote="white_horse_king88"]Yes. Much needs to be clarified.
Are your friends 'smart' pickers. Or is this random. That affects the answer... along with other things.
Also Dan, I think you simplified the cases too much (also I think the conditional probabilities for your first two cases are reversed...).[/quote]

No, the conditional probabilities are correct.  There are 5 red cards and 4 black out of 9 to start with...

Simplified the cases too much?  How so?

I saw this as an entirely different question, with the face up cards revealed before the pick begins.

That part I said about the conditional prob. being reversed is nonsense. I misread, sorry.


calculus integration logarithms function geometry rectangle trigonometry

[quote="misan"]show :D

$ \int_{0}^\infty\;\frac{\textbf dx}{\left(x^{2}+\frac{\pi^{2}}{4}\right)\cdot\cosh x}\;=\;\boxed{\frac{2\ln 2}{\pi}}$[/quote]
I didnt solve this (yet), but I realized something really really interesting.

I was going to do the more generalized integral:
$ \int_{0}^{\infty}\frac{dx}{(a^{2}+x^{2})\cosh x}, \ ~ a>0$

By considering the function $ f(z) = \frac{1}{(a^{2}+z^{2})\cosh z}$ on the rectangle with verticies $ -R,+R,-R+\pi i, R+\pi i$.

Now look at the poles:
1)$ z^{2}+a^{2}= 0$ so $ z=ai$ in the rectangle.
2)$ \cosh z = 0$ so $ z=\pi i/2$.

Now the interesting thing is that Misan used the special case when $ a=\pi i/2$ so we dont get a simple pole, rather a double pole. Maybe this is important in the solution.  :maybe:

I think this is a generalization:

Let $ a>0$ be a positive real number. Then

$ I=\int_{0}^{\infty}\frac{1}{(x^{2}+a^{2})\cosh x}dx=\frac{1}{a}\sum_{m\geq 0}^{\infty}\frac{(-1)^{m}}{m+1/2+a/\pi}$.

When $ a=\frac{\pi}{2}$ we get that our integral equals $ \frac{2\ln 2}{\pi}$.
First show that

1) $ \int_{0}^{\infty}e^{-ay}\cos xydy =\frac{a}{a^{2}+x^{2}}$,   $ a>0$.

and

2) $ \int_{0}^{\infty}\frac{\cos xy}{\cosh x}dx=\frac{\pi}{2\cosh \pi y/2}$.

We have

$ I=\frac{1}{a}\int_{0}^{\infty}\int_{0}^{\infty}e^{-ay}\cos xy dy\frac{1}{\cosh x}dx=\frac{1}{a}\int_{0}^{\infty}e^{-ay}\left(\int_{0}^{\infty}\frac{\cos xy}{\cosh x}dy\right)dy$

$ =\frac{\pi}{a}\int_{0}^{\infty}\frac{e^{-ay}}{e^{\piy/2}+e^{-\piy/2}}dy=\frac{\pi}{a}\sum_{m\geq 0}^{\infty}(-1)^{m}\int_{0}^{\infty}e^{-y(a+\pi/2+\pi m)}dy$

$ =\frac{1}{a}\sum_{m\geq 0}^{\infty}\frac{(-1)^{m}}{m+1/2+a/\pi}$.

Perhaps the sum can be expressed in terms of well-known functions....
I hope I made no computational mistakes.


geometry
incenter
trigonometry
angle bisector
geometry proposed

Point $D$ is on side $AC$ of triangle $ABC$. Let $I$ be the incenter of triangle $CBD$ and $I'$ be the incenter of triangle $CBA$. Is it true that $CI<CI'$?

It is true.
$CI=\frac{1}{2}(BC+AC-AB)/\cos (\frac{C}{2}), CI'=\frac{1}{2}(BC+CD-BD)/ \cos (\frac{C}{2})$
So $CI>CI' \iff BC+AC-AB>BC+CD-BD \iff AD+BD>AB$.
This result is trivial.

[hide="Another approach"]
$I,I'$ are both on the angle bisector of $\angle BCA$
$I$ is on the angle bisector of $\angle CBD$
$I'$ is on the angle bisector of $\angle CBA$

$D$ is in the interior of $AC\Rightarrow $

the ray $BD$ is in the interior of $\angle CBA \Rightarrow $

$\widehat{CBD}< \widehat{CBA}\Rightarrow $

$\frac{\widehat{CBD}}{2}< \frac{\widehat{CBA}}{2}\Rightarrow $

$\widehat{CBI}< \widehat{CBI'}\Rightarrow $

the ray $BI$ is in the interior of $\widehat{CBI'}\Rightarrow $

$I$ is in the interior of $CI'\Rightarrow $

$CI<CI'$[/hide]


[quote="tuandokim"]Prove that:if x,y,z>0 and x+y+z=1 then
$ 9(xyz)^2 \plus{} \sum(xy)^2 \plus{} 2xyz\sum (xy) \geq 2xyz$[/quote]
It is equivalent to $ \sum_{sym}(x^4y^2\minus{}x^4yz\plus{}x^3y^3\minus{}2x^3y^2z\plus{}x^2y^2z^2)\geq0,$ which is obvious.

Arqady, does the expression you wrote represent 6 sums corresponding to xy, xz, yx, yz, zx, and zy?

dgreenb801, I think we can say so.
$ \sum_{sym}(x^4y^2\minus{}x^4yz\plus{}x^3y^3\minus{}2x^3y^2z\plus{}x^2y^2z^2)\equal{}$
$ \equal{}x^4y^2\plus{}x^4z^2\plus{}y^4x^2\plus{}y^4z^2\plus{}z^4x^2\plus{}z^4y^2\minus{}2(x^4yz\plus{}y^4xz\plus{}z^4xy)\plus{}6x^2y^2z^2\plus{}$
$ \plus{}2(x^3y^3\plus{}x^3z^3\plus{}y^3z^3)\minus{}2(x^3y^2z\plus{}x^3z^2y\plus{}y^3x^2z\plus{}y^3z^2x\plus{}z^3x^2y\plus{}z^3y^2x).$


function
real analysis
real analysis unsolved

Does there exist an injective continuous function $ f: \mathbb{R}^2 \rightarrow \mathbb{R}$ ?

Probably not because it seems hard to find one :)

You are right. It is a well known fact that such a function doesn't exist.

Theorem:
Let $ f: \mathbb{R}^2\rightarrow \mathbb{R}$. If $ f$ is continuous in both variables (separately) then $ f$ is not an injection.

proof:
Suppose WLOG that $ f(0,0) \equal{} a > b \equal{} f(1,0)$. Because $ f$ is continuous in the first variable we can find $ c$ so that $ f(c,0) \equal{} \frac {a\plus{}b}{2}$ (consider the continuous function $ f(x,0)$)
Now simply consider the function $ g(y) \equal{} f(c,y)$ which is also continuous by the properties of $ f$.
we have $ b < g(0) < a$ which implies that for a neighborhood of $ 0$ , $ U_0$ we will have $ b < g(y) < a$ for all $ y\in U_0$.
thus there exists a $ c'\not \equal{} 0$ so that $ g(c') \equal{} f(x,0)$ for some $ x\not \equal{} c$
This contradicts the injectivity of $ f$.

 :)

Now another question for you:
Does there exist such an injective function if we only ask that it be continuous in one of its variables?

[quote="Albanian Eagle"]
Suppose WLOG that $ f(0,0) \equal{} a > b \equal{} f(1,0)$. Because $ f$ is continuous in the first variable we can find $ c$ so that $ f(c,0) \equal{} \frac {1}{2}$ (consider the continuous function $ f(x,0)$)
[/quote]

Don't you mean $ \frac{a\plus{}b}{2}$ instead of $ 1/2$ ?


search
number theory proposed
number theory

Find solution in positive integer of equation:

\[
x^2  + y^2  = z^{2l + 1}  + z
\]

wth l is also a positive integer

$z(z^{2l}+1)$ is a sum of two squares [b]iff[/b] $z$ is a sum of two squares.
We can take $(x,y,z)=(t(t^2+u^2)^l-u,t+u(t^2+u^2)^l,t^2+u^2)$  ($u,t$ integers), wich give an example...

[quote="blang"]$z(z^{2l}+1)$ is a sum of two squares [b]iff[/b] $z$ is a sum of two squares.[/quote]
Could you explain it in more details, please?:)

A number is the sum of two squares iff every prime divisor of the form $4k+3$ appears with an even exponent. This is true for $z^{2l}+1$, which is a sum of two squares, so it's true for $z(z^{2l}+1)$ iff it's true for $z$.

You are so fast, grobber :(
I was writing reply with the same phrase too ;)

[quote="grobber"]A number is the sum of two squares iff every prime divisor of the form $4k+3$ appears with an even exponent.[/quote]

Could you prove this please?

I think it has been discussed quite a few times on ML. The necessity follows from the fact that if a prime of the form $4k+3$ divides $a^2+b^2$, then it divides both $a$ and $b$. The sufficiency follows from the fact that each prime of the form $4k+1$ can be written as the sum of to squares and the product of sums of two squares is also a sum of two squares.

Sorry I don't give the proofs, but, as I said, it's been discussed before. You could also search for "sum of two squares" on Google, for example. You'll find lots of stuff I could not explain properly :).

[quote="Myth"]I see we have an expert in Number Theory now ;)[/quote]

As far as I know him, I think you are right. :)

Pierre.

I would like to be an expert in number theory. :D


1.Comment on why the relative rate of the reaction
$(n-C_{3}H_{7})_{3}N+CH_{3}I \rightarrow (n-C_{3}H_{7})_{3}N^{+}(CH_{3})I^{-}$

in $n-(C_{6}H_{14})$:$CHCl_{3}$ is $1: 13000$

2.Why does the reaction between $n-Butyl Chloride$ and $n-Octyl Alcohol$ in a solution of aqueous $NaOH$ containing little amount of quaternery Ammonium salt give a high yield(approx. $95$ Percentage) of $n-Butyl$ $n-Octyl Ether$ as compared to the standard procedure in which the ethers are prepared by Williamson's synthesis?

In 2, the quaternary ammonium salt acts as a phase-transfer catalyst.

sir can you be more specific as to what the two phases are here.i am actually confused by the one of the methods used in williamson's synthesis where concentrated aqueoud $NaOH$ is used .now say the raection proceed between the above two compounds.Well as in most textbooks the mechanism written is first formation sodium alkoxide then the alkoxide attacking the halide and the entire thing proceeding via $S_{N}2$ .But my doubt is if the sodium ion can attack the alcohol then why can't the hydroxyl part attack the halide and result in nucleophillic displacent of the halide ion and since the solution is concentrated so the effect of water though exisit but still there is an high probability that the halide will be converted to an alcohol Can you clarify my doubt :maybe:

The two phases are: 1) the aqueous phase containing the NaHO; 2) the organic phase containing the alkyl chloride and a significant proportion of the alcohol. What might happen is the following: some of the alcohol may dissolve in the aquous phase because of the equilibrium

$ROH+HO^{-}\rightleftharpoons RO^{-}+H_{2}O$.

Don't forget that the quaternary ammonium salt is not only ionic but also lipophilic because of the four alkyl chains surrounding the nitrogen. So, because of the equilibrium that occurs in the aqueous phase,

$R_{4}N^{+}X^{-}+RO^{-}\rightleftharpoons R_{4}N^{+}RO^{-}+X^{-}$,

then the alcoxide ion can travel to the alkyl chloride organic phase due to the lipophilicity of quaternary ammonium salt,

$R_{4}N^{+}RO^{-}(aqueous phase) \rightleftharpoons R_{4}N^{+}RO^{-}(organic phase)$.

Now the alcoxide is very reactive because it is weakly solvated, and reaction with the chloride is faster. In the aqueous phase, the alcoxide would be surrounded by a layer of water molecules, and any reaction that could occur in the surface of contact between the two phases would be slow.
After the reaction the quaternary ammonium salt can return to the aqueous phase and repeat the cycle.

[quote="Pardesi"]But my doubt is if the sodium ion can attack the alcohol then why can't the hydroxyl part attack the halide and result in nucleophillic displacent of the halide ion and since the solution is concentrated so the effect of water though exisit but still there is an high probability that the halide will be converted to an alcohol. Can you clarify my doubt[/quote]

Because the hydroxide ions are not soluble in the alkyl chloride organic phase, and acid-base reaction with the alcohol is perhaps favoured.

Also, the standard procedure for the Williamson Ether Synthesis is to first make the alcoxide, by adding a Group 1 metal or metal hydride to the alcohol, and then add the alkyl halide to the alcoxide solution: this way any elimination side reaction is less probable because of the low concentration of alkyl halide.

thank you sir but just one more doubt since here the alcohol($\n-Hexyl anlcohol$)
has the majority part as the organic part can we nelect this effect.now because of it's huge organic part it can also show lipohillic behaviour and may be  even soluble in the halide :maybe:

Yes, you are right: your alcohol have low solubility in water and so should be soluble in the alkyl chloride. But I think the acid base equilibrium with the hydroxide ions (the first equilibrium in my previous post) helps to increase its solubility in the aqueous phase.

Thank you  :) .what about the first one :?:

Perhaps because chloroform molecules are polar and hexane molecules are non-polar: the first ones are thus more appropriate for stabilizing the ions that form in the reaction.


group theory
abstract algebra
superior algebra
superior algebra theorems

How can I show that $ \mathbb{Q}$ is not a free group?

Any two elements are commensurable, so there are no subgroups which are free or free abelian on more than one generator. The whole group isn't cyclic, so it's not free on one generator, and thus it's not free or free abelian.


(n)  1,  2,    3,
(m)7,  17,  31,

What is the relationship between n and m as an equation?

If you want a single solution, one is

31(x-1)(x-2)/2+7(x-2)(x-3)/2-17(x-1)(x-3) The reason why this works should be apparent

Just after  general solution describing n & m. Can't seem to see it  fro some reason.

Only one of those terms is non-zero for 1,2,3, so then I just weighted the term effected by the number by a suitable amount. There are an infinite number of ways you can have a relationship between n and m, there is no one solution.

If I were you, I would use a quadratic

[hide="Click to see the quadratic"]$m=2n^{2}+4n+1$[/hide]

rofler's answer simplifies to yours.  It's a technique known as Lagrange Interpolation.


conics
ellipse
linear algebra
matrix
calculus
calculus computations

1. Use Lagrange multipliers to find the point closest to the origin on the intersection of the planes $ x \plus{} 2y \plus{} z \equal{} 5$ and $ 2x \plus{} y \minus{} z \equal{} 1$.

OK. So we apply the method of Lagrange multipliers and get the 3 equations:

$ 2x \equal{} \lambda \plus{} \mu$
$ 2y \equal{} 2\lambda \plus{} \mu$
$ 2z \equal{} \lambda \minus{} \mu$

The problem is I don't know how to work with this system of equations.

2. Find the points closest to and farthest from the origin on the ellipse 2x^2+4xy+5y^2=1.

Again we apply the method of Lagrange multipliers and obtain:

$ 2x \equal{} \lambda \left( 4x \plus{}4y \right)$

$ 2y \equal{} \lambda \left( 10y \plus{}4x \right)$

Again I am not really sure how to work with this system.

(First question)
Your system is the equation of a third plane. Intersect it with the first two.
The problem would probably be easier with single-variable techniques- find and parametrize the line of intersection.

(Second question)
That system is an eigenvalue problem: for what $ v\equal{}\begin{bmatrix}x\\y\end{bmatrix}$ and $ \lambda$ is $ \lambda Av\equal{}2v$, where $ A\equal{}\begin{bmatrix}4&4\\4&10\end{bmatrix}$? For most $ \lambda$, this system has only the trivial solution $ v\equal{}0$, which does not correspond to any points on the ellipse. For two particular values of $ \lambda$, the space of solutions $ v$ is a line, which intersects the ellipse in two opposite points each. These lines come from eigenvectors of $ A$, and the $ \lambda$ (which don't really matter) are related to eigenvalues. How do you find them? Set up the matrix $ \lambda A\minus{}2I$ of coefficients, and look for where its determinant is zero.

[quote="jmerry"]
(Second question)
That system is an eigenvalue problem: for what $ v \equal{} \begin{bmatrix}x \\
y\end{bmatrix}$ and $ \lambda$ is $ \lambda Av \equal{} 2v$, where $ A \equal{} \begin{bmatrix}4 & 4 \\
4 & 10\end{bmatrix}$? For most $ \lambda$, this system has only the trivial solution $ v \equal{} 0$, which does not correspond to any points on the ellipse. For two particular values of $ \lambda$, the space of solutions $ v$ is a line, which intersects the ellipse in two opposite points each. These lines come from eigenvectors of $ A$, and the $ \lambda$ (which don't really matter) are related to eigenvalues. How do you find them? Set up the matrix $ \lambda A \minus{} 2I$ of coefficients, and look for where its determinant is zero.[/quote]

The eigenvectors of that matrix satisfy 2x=y or 2y = -x. We plug that into the constraint equation to obtain to $ (\pm \frac{1}{\sqrt{30}}, \pm\frac{2}{\sqrt{30}})$ and $ (\pm \frac{2}{\sqrt{5}},\mp \frac{1}{\sqrt{5}})$ so the minimum distance is $ \frac{1}{10}$ and the maximum distance is $ \frac{3}{5}$.

I will work on the first one now.

[quote="jmerry"](First question)
Your system is the equation of a third plane. Intersect it with the first two.
The problem would probably be easier with single-variable techniques- find and parametrize the line of intersection.
[/quote]

OK. I figured this one out also. My answer checks out with the back of the book so I won't post my solution.


Prove the following for all $ n\in \mathbb{N}$:

\[ 1\binom{n}{1} \plus{} 2\binom{n}{2} \plus{} \ldots \plus{} (n\minus{}1)\binom{n}{n\minus{}1} \plus{} n\binom{n}{n} \equal{} n\cdot 2^{n\minus{}1}\]

[hide="hint"] Add $ 0\dbinom{n}{0}$. Then, write the sum backwards and add it to the original.[/hide]

[hide="Method 1"] Let $ F(x) \equal{} (1 \plus{} x)^n \equal{} \sum_{k \equal{} 0}^{n} x^k {n \choose k}$.  Then

$ F'(x) \equal{} n(1 \plus{} x)^{n \minus{} 1} \equal{} \sum_{k \equal{} 0}^{n} kx^{k \minus{} 1} {n \choose k} \implies$
$ F'(1) \equal{} n \cdot 2^{n \minus{} 1} \equal{} \sum_{k \equal{} 0}^{n} k {n \choose k}$. [/hide]
[hide="Method 2"] The LHS is the number of ways to pick a committee of $ 1$ or more people from a group of $ n$ people and then to select a president from that committee.  We can do this by picking the President first and then toggling whether the $ n \minus{} 1$ other people are in or not in the committee, which gives the RHS. [/hide]

Nice solutions! :)

I'll post my own later today.

[hide="yet another solution"]

Let S equal the sum given, and rewrite it as:

$ S \equal{} \sum_{k\equal{}1}^n{k*{n \choose k}} \equal{} \sum_{i\equal{}1}^{n}{\sum_{j\equal{}i}^n{n\choose j}}$
$ \equal{} \sum_{i\equal{}1}^n{\left(2^n \minus{} \sum_{j\equal{}0}^{i\minus{}1}{n\choose j}\right)}$
$ \equal{} n*2^n \minus{} \sum_{i\equal{}1}^n{\sum_{j\equal{}0}^{i\minus{}1}{n \choose j}}$

Notice that the sum on the RHS of the last equation equals $ S$ because $ {n \choose i} \equal{} {n \choose n\minus{}i}$ so the summation $ \sum_{j\equal{}0}^{i\minus{}1}{n \choose j}$ equals $ \sum_{j\equal{}n\minus{}i\plus{}1}^n{n \choose j}$ (you can check all the summations correspond).

Therefore $ S \equal{} n*2^n \minus{} S$ and it follows that $ S \equal{} n*2^{n\minus{}1}$.

[/hide]

[b]Follow-up:[/b]  Compute $ \sum_{k \equal{} 0}^{n} {k \choose r} {n \choose k}$.  (This problem is the special case $ r \equal{} 1$.)

[hide="Note"] Both of the techniques I used generalize immediately to this problem.  The other techniques given so far do not ;) [/hide]

[url=http://www.initdotd.com/~nikita/LaTeX/CombinatorialProof.pdf]Proof of Original Problem[/url]


Please post organic problem on Elimination and Substitution reactions

Browse this forum, there are plenty of problems.


modular arithmetic
number theory unsolved
number theory

Suppose $ \{a_{1},a_{2},\ldots,a_{k}\}$ is a complete residue system modulo $ k$, where $ k$ is a prime.  Prove that for each integer $ n$ and each nonnegative integer $ s$ there exists a congruence of the form $ n\equiv\sum_{j \equal{} 0}^{s}b_{j}k^{j}\pmod{k^{s \plus{} 1}}$, where each $ b_{j}$ is one of the $ a_{i}$.

This is nonsense as written.  Was $ k^k$ intended to be $ k^j$?  In that case, think about the base $ k$ expansion of $ n$.

Sorry, typo.  Should be $ k^{j}$.  I was thinking the base-$ k$ representation but then how would I choose the coefficients.  Zero isn't necessarily in my system, is it?

No, but a multiple of $ k$ is.  To get a feel for what's going on here you might want to read up on [url=http://en.wikipedia.org/wiki/Ternary_numeral_system]ternary representation[/url], which is a special case.

Here's what I came up with: by Euclid's division lemma and the basis representation theorem, we can write
$ n\equal{}kq\plus{}r$, $ 0\leq r<k$.  We can write $ q\equal{}b_{s}k^{s}\plus{}b_{s\minus{}1}k^{s\minus{}1}\plus{}\cdots\plus{}b_{1}k\plus{}b_{0}$, so $ n\equal{}b_{s}k^{s\plus{}1}\plus{}b_{s\minus{}1}k^{s}\plus{}\cdots\plus{}b_{0}k\plus{}r\Rightarrow n\minus{}(b_{s\minus{}1}k^{s}\plus{}\cdots\plus{}b_{0}k\plus{}r)\equal{}b_{s}k^{s\plus{}1}$
$ \Rightarrow n\equiv \sum_{j\equal{}0}^{s}b_{j}k^{j}\pmod{k^{s\plus{}1}}$.
But this doesn't depend on $ k$ being prime, or the complete residue system.


geometry
trapezoid
analytic geometry
quadratics
Pythagorean Theorem

Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart.  A tall circular building 100 feet in diameter is centered midway between the paths.  At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart.  Let $t$ be the amount of time, in seconds, before Jenny and Kenny can see each other again.  If $t$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

[hide]Let $J_1$ and $K_1$ be Jerry's and Kenny's starting point respectively and $J_2$ and $K_2$ be there ending points. Draw the radius from the center of the circle, lets call it $O$, to $J_2K_2$, and another parallel to $J_1J_2$. Call the intersections of the two radii with $J_2K_2$, $L_1$ and $L_2$ respectively. Drop a perpendicular from $J_1$ onto $K_1K_2$. Call the intersection of this perpendicular with $K_1K_2$, $L_3$. Draw the radius from $O$ to $JK$. Call the intersection with $J_1K_1$, $L_4$

We have $\triangle OL_1L_2 \sim \triangle J_2L_3K_2$. If we let $J_1J_2 = x$ then it is easy to see that $K_1K_2 = 3x$ and $L_3K_2 = 2x$. Let $L_1L_2 = m$. We thus have

\[ \frac{L_1L_2}{OL_2} = \frac{K_2L3}{L_3J_2} = \frac{2x}{200} = \frac{m}{50} \Rightarrow m = \frac x2 \]

Applying Pythagorean Theorem yields

\[ OL_2 = \sqrt{50^2 + \frac{x^2}{4}} \]

Since $L_2L_4$ is midway between $J_1J_2$ and $K_1K_2$ we have

\[ \frac{3x+x}{2} = L_2L_4 = 50 + \sqrt{50^2 + \frac{x^2}{4}} \Rightarrow x= \frac{160}{3} \]

$160 + 3 = \fbox{163}$[/hide]

[quote="nsato"]Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart.  A tall circular building 100 feet in diameter is centered midway between the paths.  At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart.  Let $ t$ be the amount of time, in seconds, before Jenny and Kenny can see each other again.  If $ t$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?[/quote]

At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart
Does this mean they are on the same line if draw L perpendicular to their path?

[b]cdymdcool:[/b] Yes, if I'm understanding you correctly.

[hide="Solution"]
Look at the diagram (all lengths are the distances from the two marked points next to it on the segment it labels); most of it is easy to see, using the properties of the medians of trapezoids gives all the lengths parallel to the lines of length $ t$ and $ 3t$ (since all five points on he bottom are $ 50$ apart from their nearest neighbors), and the fact that the tangents from a point to a circle are equidistant. Now, from the Pythagorean Theorem, $ 100^2\plus{}t^2\equal{}(4t\minus{}100)^2\implies t^2\equal{}(4t\minus{}100)^2\minus{}100^2\equal{}4t(4t\minus{}200)$
$ \implies t\equal{}4(4t\minus{}200)\equal{}16t\minus{}800\implies15t\equal{}800\implies t\equal{}\boxed{\frac{160}3}$.
[/hide]

Another way would be to coordinate bash: Find the equation of the line to be: $y=\frac{-100}{t}x+300$, plug it into $(x-50)^2+(y-100)^2=50^2$, create a quadratic in x and set $b^2-4ac=0$ as at the point of tangency, there is only one intersection, and solve for a.

Draw the two segments parallel to to the paths that are tangent to the circle. This forms a trapezoid with the circle as its inscribed circle. Then, sum of lengths of opposite sides are same of the new trapezoid and wala.


[quote="greatestmaths"]let f is a function such that:

for all ((x;y) in RxR)   f(x+y) = f(x) f(y)

1) prove that this function isn't surjective.

2)prove if  f is injective <==> f(0)=1 and for all (x in R) f(x) isn't 1[/quote]

Take $ x\equal{}y\equal{}\frac{a}{2}$ then $ f(a)\equal{}f(\frac{a}{2})^{2}\geq 0$. So $ y\equal{}f(x)$ has no solution if $ y<0$ and so $ f(x)$ is not surjective.

Take $ y\equal{}0$  and so $ f(x)\equal{}f(x)f(0)$ and so either $ f(0)\equal{}1$, either $ f(x)\equal{}0$ $ \forall x$ and then $ f(x)$ is not injective. So, if $ f(x)$ is injective, $ f(0)\equal{}1$.
Then, since $ f(0)\equal{}1$ and $ f(x)$ is injective, $ f(x)\neq 1$ $ \forall x\neq 0$

1, Put $ x \equal{} y \equal{} 0$, then $ f(o) \equal{} 0$ or $ f(0) \equal{} 1$.

-Case 1 . $ f(0) \equal{}0$ .

Put $ x \equal{} 0$, then $ f(y) \equal{} 0,\forall y\in\mathbb{R}$ then $ f$ isn't surjective .

-Case 2. $ f(0) \equal{} 1$ .

Put $ x \equal{}\minus{}y$ then $ f(y)f(\minus{}y) \equal{} 1,\forall y\in\mathbb{R}(1)$.

So if there exist $ t$ such that $ f(t) \equal{} 0$, put $ y \equal{} t$ in $ (1)$ we have contracdiction . :)

2, From 1, we have if $ f(0) \equal{} 1$ then $ f(x) \equal{} 0,\forall x\in\mathbb{R}$ so $ f$ isn't injective .

If there exist $ f(t) \equal{} 1,t\not \equal{} 0$ , put $ y \equal{} 1$ then $ f(x\plus{}1) \equal{} f(x)$ so so $ f$ isn't injective .

We will prove that if $ f(0) \equal{} 1,f(y)\not \equal{} 1,\forall y\in\mathbb{R}$ then $ f$ is injective .

Consider $ x_{1}\not \equal{} x_{2}$

Let $ x_{1}\equal{} x_{2}\plus{}a,a\not \equal{} 0$

Then $ f(x_{1}) \equal{} f(a).f(x_{2})\not \equal{} f(x_{2})$

We're done


Let $ f$ be a holomorphic function on the unit disk. If the image of $ f$ is also the unit disk and $ f'$ is nowhere zero, then is $ f$ injective?


function
floor function
limit
calculus
calculus computations

Let f(x) be the greatest integer less than or equal to x.
compute the lim x->0  xf(1/x).

Because $x-1<\left\lfloor x\right\rfloor = f(x) \le x$ we have $\frac{1}{x}-1 <f \left( \frac{1}{x}\right) \le \frac{1}{x}$

It's $1-x < x \cdot f \left( \frac{1}{x}\right) \leq 1$  therefore $\lim_{x \to 0 }x \cdot f \left( \frac{1}{x}\right) = 1$  .


[quote="sri340"]Chocolate costing 10 dollars per pound is mixed with nuts costing 4 dollars per pound to make a mixture costing 6 dollars per pound. What fraction of the mixture’s weight is chocolate? Express your answer as a common fraction.[/quote]

Taking a random case that fulfills the given conditions, we get 1/3 lb chocolate for 2/3 lb nuts.  (10/3+8/3=6.)  Because they add up to one pound, the fraction of the mixture's weight that is chocolate is 1/3.

If there are x lb chocolate, y lb nuts, then...

10x+4y=6(x+y)
10x+4y=6x+6y
4x=2y
x/y=1/2
x/(x+y)=1/3

The logic between steps 4 and 5 is this (if you don't know it yet): we have 1 part chocolate for 2 parts nuts, so we have 1 part chocolate for 2 parts nuts and 1 part chocolate, or 1 part chocolate for 3 parts mixture.

Cough. Not 1/6.

The issue with the above solution is that the mixture costs 6 dollars, but is not exactly one pound.

Whoops.   :blush:   -Proceeds to edit post-

Please explain your motivation for guessing 1/3 and 2/3?
 :rotfl: