25 people sit around a circular table. Three of them are chosen randomly. What is the probability that at least two of the three had been sitting next to each other.

Solution: (i don't understand why this is wrong)

The sample space is 25C3

1st case: considering 2 people sitting next to each other. We can set one person in any of the seats. This can be done in 25 ways. For the next person to be beside the first person, we have 2 choices for his seat, to the left and to the right. Lastly, because the 3rd person can't be beside the 2 of them, we only have 21 seats for him (25-2-2). For a total of 25x2x21 ways

2nd case: considering all of them next to each other. We can set one person in any of the seats. This can be done in 25 ways. For the next person to sit beside him, we have 2 ways. For the third person to sit next beside them, we have 2 ways (left and right). So, this is for a total of 25x2x2

Now, i don't know, after adding, the result is 1/2.but the correct answer is 11/46. Where's my mistake?

There are 25 groups of three consecutive seats at a table with 25 seats.  You counted each such grouping exactly once in your denominator (the $ \binom{25}{3}$) and exactly four times in your numerator (the $ 25 \cdot 2 \cdot 2$).  Similar problem in the other case.


real analysis
advanced fields
advanced fields unsolved

what does this symbol mean: $A\Delta B$?

The question was ($\nu$ is lebesgue measure): if $\nu(A\Delta B)=0,$ show that $\nu(A)=\nu(B)$. The question must be completely trivial, but $A\Delta B$ can mean a lot of things... can anyone tell me this?

Thanks!

I think it is the symmetric difference $A\Delta B=(A\setminus B)\cup (B\setminus A)$

Maybe I'm wrong, but isn't \[ A \Delta B = (A \backslash B) \cup (B \backslash A)  \] for all sets $A$ and $B$? Not sure... (Our Algebra professor uses the symbol that way)

Ok... It seemed most logic, since elseway the answer would be more than 8 characters long.  :roll:

Thanks. :)


Square an integer greater than or equal to 3, and subtract 1. What is the probability that the resulting value is a prime number?

$ x^3 \minus{} 1 \equal{} (x \minus{} 1)(x^2 \plus{} x \plus{} 1)$
If $ x \minus{} 1 \equal{} 1 \ \text{i.e.} \ x \equal{} 2$, $ x^3 \minus{} 1 \equal{} 7\in\text{prime}$.
If $ x^2 \plus{} x \plus{} 1 \equal{} 1 \ \text{i.e.} \ x \equal{} \minus{} 1, 0$, $ x^3 \minus{} 1 \equal{} \minus{} 2, \minus{} 1\not\in\text{prime}$.

@Kouichi Nakagawa:  It's "square an integer", not cube an integer

So we can express what the problem is saying in algebra:

$ x^2 \minus{} 1$

We know that $ x^2 \minus{} 1$ factored is $ (x \plus{} 1)(x \minus{} 1)$.  Thus, the probability is $ \boxed{0}$ because it is the product of two integers which are not equal to $ 1$.

Sorry.
I seem to have misunderstood anything.

I think you missed a number.

We know that $x^{2}-1=(x+1)(x-1)$

Since the number is prime, one of the factors is equal to 0.

$x-1=1$

$x=2$

$x+1=1$

$x=0$

Therefore, we have a $\frac{1}{100}$ probability.

However, since the problem says to square an integer greater than or equal to 3, the probability and answer is 0.

Hmm wait i'm confused, the problem I reviewed said :I
n the set of integers ${1,2,3,...98,99,100}$, what is the probability that a randomly selected number when squared and 1 subtracted from it will form a prime number.

Yes I agree with Grawr, this is what I comprehended too maybe we might be doing this wrong

@above: maybe problem was changed?

I would say this answer is 0, since this is a probability in an infinite sequence of integers.

Yes, in FTW!, the problem asks for probability from integers 1~100.

According to this problem, the probability is 0.
[hide = Why]
We can simplify the problem to whether $x^2-1$ is prime or not
This is expanded into $(x+1)(x-1)$ The only way for that to be prime is if $x-1 = 1$, therefore, the only possibility is when $x=2$

However, since the problem is telling that the integer is greater than or equal to 3 the probability is $\boxed0$,  as 2 is less than 3.
[/hide]

However, there is a similar problem on FTW
[quote=titanium7]The problem I got states, "Choose an integer from the set {1,2,3,...99,100}. Square it, then subtract 1. What is the probability that the resulting value is a prime number?".[/quote]

In this case, the answer is[hide = Answer (Why is above)] $\boxed{1/100 }$ One successful possibility (2) out of a hundred possibilities [/hide]

I believe that this problem has been edited in the database. I think that's why


abstract algebra
group theory
superior algebra
superior algebra unsolved

let $ G$ be a simple group of order $ \geq 2$ , prove that :

$ Z(Aut(G))\equal{}\{e\}$ iff $ G$ is non-abelian

 :)

For $ z\in Z(Aut(G))$ and $ \forall g,h\in G$ we let $ h$ act as an inner automorphism on $ g$ and use the commutativity with $ z$, i.e. $ (g^z)^{h^z}\equal{}(g^h)^z\equal{}(g^z)^h$ which is equivalent to $ h^{\minus{}1}h^z\in Z(G)\equal{}\{e\}$ since $ G$ is simple and $ g$ arbitrary. But this implies $ \forall h: \,h^z\equal{}h\Longrightarrow z\equal{}id_G$

Thank you for answer

This is my 1st I see the symbol $ h^z$ ?

what do u mean by the power .

 :roll:

$ h^z$ is the image of $ h$ under the automorphism $ z$.

For the reverse, the fundamental theorem of finite abelian group may be useful :
Suppose on the contrary that G is abelian then from the fundamental theorem of finite abelian group, we can get:
$G \cong \mathbb{Z}_{{p_1}^{\alpha_1}}....\mathbb{Z}_{{p_n}^{\alpha_n}}$
and then, check that $Z(Aut(\mathbb{Z}_{{p_1}^{\alpha_1}}....\mathbb{Z}_{{p_n}^{\alpha_n}})) \neq {e}$


geometry
logarithms
modular arithmetic
3D geometry
number theory

I need to do each of these problems in less than 20 seconds:

1.How many digits are there in $ 2^{100}$?
2.What is the coefficient of $ x^{99}$ in the expansion of $ (x \plus{} 1)(x \plus{} 2)(x \plus{} 3) ....(x \plus{} 100)$
3.If $ (4^{x} \plus{} 4^{\minus{}x}) \equal{} 23$, find $ ((2^{x}) \plus{} (2^{\minus{}x}))$
4.Factorize completely: $ (x^2)\minus{}(y^2)\minus{}(z^2) \plus{} (2yz)$
5.Given A and B are two sets such that A = {2,3,4,5,0,-1} and (A intersection B )= { 8,10}.Find (A union B)
6.Find the lateral surface area of a cone.

[hide="1)"]
calcbash  :)  giving us 23.
And as AIME15 said: Log 2 gives us .301
Log 2^100=30.1 (approximated)[/hide]

1: 31

2: 5050

3: 5

4: (x+y-z) (x-y+z)

5: If A = {2,3,4,5,0,-1} it is impossible to have: (A intersection B )= { 8,10}.
Intersection of A has to be a part of A.

You gave no info on set B. So (A union of B) can be anything that contains A.

6: in any geometry book

I have the answers.. :wink: ..I wanted solutions :)

I didnt make these problems up...so I think the people who made them made a mistake :maybe:

The last one asks how to derive the formula

Numero 4.

Let guy be x, thing be y, and that one dude be z.

Now, we have

$ \text{guy}^2 \minus{} \text{thing}^2 \minus{} (\text{that one dude})^2 \plus{} 2(\text{thing})(\text{that one dude})$

We note that the negative sum of the squares of a thing and that one dude added to twice the produce of that one dude and a thing is merely the negative of the square of the difference of the thing and that one dude. I.e. for all y,z, $ \minus{} y^2 \minus{} z^2 \plus{} 2yz \equal{} \minus{} (y \minus{} z)^2$

Our expression becomes:

$ \text{guy}^2 \minus{} (\text{thing} \minus{} \text{that one dude})^2$

And by the Differencio of Squares thingamadoozle.

$ (\text{guy} \minus{} \text{thing} \plus{} \text{that one dude})(\text{guy} \plus{} \text{thing} \minus{} \text{that one dude})$

Substituting back,

$ (x \minus{} y \plus{} z)(x \minus{} y \plus{} z)$

:rotfl:

Oh, nobody's answered #2 yet?

An $ x^{99}$ term is created when we take the $ x$ from $ 99$ of the binomials and the constant from the others...thus, the answer is $ 1\plus{}2\plus{}3\plus{}\dots\plus{}100\equal{}\boxed{5050}$.

[quote="AIME15"]Oh, nobody's answered #2 yet?

An $ x^{99}$ term is created when we take the $ x$ from $ 99$ of the binomials and the constant from the others...thus, the answer is $ 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} 100 \equal{} \boxed{5050}$.[/quote]

Hm...is that a formula? What would it be for $ x^{98}$?

Then, it'd be taking 2 of the constant terms - $ 1(2\plus{}3\plus{}4\plus{}\dots\plus{}100) \plus{} 2(3\plus{}4\plus{}5\plus{}\dots\plus{}100) \plus{} 3(4\plus{}5\plus{}6\plus{}\dots\plus{}100) \plus{} \dots \plus{} 99(100)$.

Interesting :) ...How did you get it? :P

I'm still a little confused about $ \#2$, and also if how do you do this problem [b]as fast as possible[/b]?

Find the largest multiple of $ 17$ less than $ 1000$.

Also this: Find the largest multiple of $ 17$ less than $ 1000$, such that it is one less than a multiple of $ 8$?

[quote="generalmotors"]
Find the largest multiple of $ 17$ less than $ 1000$.
[/quote]
[hide="This is how I would approach it."]
If you simply divide $ 1000$ by $ 17$, you would get $ 58$ with a remainder of $ 14$.  This means that the largest multiple of $ 17$ less than $ 1000$ is $ 14$ less than $ 1000$, or $ \boxed{986}$.
It would look like this if expressed in modular arithmetic (solving for $ x$):
\[ 1000\equiv 14\pmod{17}\]

\[ x\equiv 0\pmod{17}\]
\[ 1000 \minus{} 14\equiv 14 \minus{} 14\equiv 0\equiv x\equiv \boxed{986} \pmod{17}\]
[/hide]

Know your multiplication tables.

Start with $ 850,$ which is divisible by $ 17.$ Then add the largest multiple of $ 17$ less than $ 150,$ which happens to be $ 136,$ so the answer is $ 850\plus{}136\equal{}986.$

K..thanks so far....I just need a little explanation for $ 2$, and one for $ 6$

6. The lateral surface area of a cone is $ \pi r l$ where $ l$ is the slant height of the cone.

So for example, if we are given a cone with a radius of 4 and a slant height of 5, we would do $ \pi \times 4\times 5 \equal{} 20\pi \approx 62.82 cm^2$

[quote="tiger21"][quote="generalmotors"]I just need a little explanation for $ 2$, and one for $ 6$[/quote]

You can understand $ 2$ easily if you read more on Binomial Theorem.

As for explanation for $ 6$:

The lateral surface area of a cone = $ (2\pi r)/(2\pi l) * \pi l^2 \equal{} \pi rl$[/quote]

Isn't this what I just said?  :wink:

[quote="AIME15USAMO"]
Isn't this what I just said? [/quote]

No. He/she wants to know how you get the result.
He/she wants to know how the formula is derived.

I added some extra formula if someone who understands the geometry background of it, he/she should know how the final formula is derived. He/she should understand why the area is $ \pi r l$, not $ 2\pi rl$, not $ \pi r^2l$, not $ \pi r l^2$.


gravitational force conservates angular momentum -> if i throw a projectile with horizontal speed $ \overline{v_x}$, it's angular momentum must be conserved -> $ \frac{d(\overline{r} \times m \overline{v})}{dt}\equal{}0$ -> $ \overline{v_y}\times \overline{v_x}\equal{}\overline{0}$, which is false

where is the mistake?

$ \frac{d \overline{L}}{dt} \equal{} \frac{d (\overline{r} \times m \overline{v})}{dt} \equal{} \frac{d \overline{r}}{dt} \times m \overline{v} \plus{} \overline{r} \times m\frac{d \overline{v}}{dt} \equal{} m (\overline{v} \times \overline{v} \plus{} \overline{r} \times \frac{d \overline{v}}{dt})$

$ \equal{} \overline{0} \plus{} \overline{r} \times m \overline{a} \equal{} \overline{r} \times \overline{F}$

which is zero when: 1) the resultant force is zero; 2) the force applied to a body is a central force, as happens with satelites and planets, in which the central force is the gravitational force.


LaTeX
arithmetic series
geometric series

1.) Find the sum of the following series

3 + 5 + 7 +...+99

I know it is an arithmetic series and increases by 2. I just need help to find n and then find the sum.

2.) Find the sum of the following series.

3 + 6 + 12 +...+3072

Again, I know it is a geometric series and each term is multiplied by 2, but need help to find n and then the sum.

Thanks in advance!

[hide]
1)
$n=100/2-1=49$
Traditional Way:
$(99+3)*\frac{49}{2}=\fbox{2499}$
My Way:
$50^{2}-1=\fbox{2499}$

2)
$3*2^{x}=3072$
$2^{x}=1024$
$x=10$
$11$ terms
YIKES!
I forgot the formula for finding the sum of finite geometric sequences
Someone PLZ help!

[/hide]

$S_{n}= \frac{a(1-r^{n})}{(1-r)}$

Where $a$ is the first term and $r$ is the common ratio.

Okay I have a couple questions.

1.)  What formula did you use to find n in both problems?

2.)  the formula that max_tm gave to find the sum I thought is for if r < 1.  In this case we have r > 1.  I thought the formula was

S_n = (a_1(r^n - 1))/(r - 1)

I am not sure?  Thanks for your help!

1)
For number one I just found how many odd numbers there were from
$1$
to
$99$
You add $1$ to each term
So $1$ becomes $2$
and $99$ becomes $100$
You divide each term by two
You get numbers
$1$
to
$50$
$50-1=49$
I didn't use a formula
but it would be easy to come up with one (even though understanding it would be better for this situation)

For Number 2
I figured out how many times you had to multiply by $2$ in order to get to $3072$
Again I didn't use a formula
but it would be easy to come up with one (even though understanding it would be better for this situation again)

2)
It doesn't matter
They're both right

Edit:
except max_tm has a formula that works for any number

[quote="rd5493"]or u could do it this way

[hide]First, make 1) say $1+3+5+...+99$

Now, the answer will be a square number.  From $3-99$ there are $49$ odd numbers.  Now $49^{2}=2401$ and $2401-1=2400$.
So the answer to the first question, i believe, is 2400.  Am I right?[/hide]

EDIT: Aznwithabrian, what did i do wrong, i thought i had the right answer.[/quote]

no
[hide]
you can't just do $49^{2}$ because it's not the first $49$ odd numbers
[/hide]

oh, thanx for clearing that up  :lol:

ps, is ne1 elses latex down, i cant seem to type with it anymore


[hide]Ok. The first requirement--place the first 3 listed in the order noted--
IV               I                    III

The second requirement--place "II" in the only place it can go--
IV      II       I                    III

The third requirement--well there are two places "V" can go, but the fourth requirement states it can not be third. So placing it into the correct order you get
[b]IV      II       I        V        III[/b][/hide]

[quote="cellular_mem"]Julius Caesar wrote the Roman Numberals I, II, III, IV and V in a certain order from left to right.  He wrote I before III but after IV.  He wrote II after IV but before I.  He wrote V after II but before III.  If V was not the third numeral, in what order did Caesar write the five numberals from left to right?[/quote]

He wrote them from left to right.

<___<

Just for people who need it in arabic numerals...[hide]42153[/hide] and their superiors...[hide]IV II I V III[/hide]:winner_second:

[hide] Oh...I might want to add I have it backwards and that is why I don't have a  :winner_first:...[/hide]

[hide="answer!"]IV, II, I, V, III[/hide]


modular arithmetic
induction
function
number theory
totient function

is 4^237 congruent to 4 mod12?

is phi(1000)=400 and is phi(2401)=640

can someone verify this for me please?

thanks

$ 4^237$ is congruent to $ 4$ modulo $ 12$.  Here is why:

$ 4 \equiv 4 \pmod{12}$
$ 4^2 \equiv 4 \pmod{12}$
$ 4^3  \equiv 4 \pmod{12}$
$ \Downarrow$
$ 4^n \equiv 4 \pmod{12}$

Thanks a lot for confiming, I just couldn't believe that was that easy, you know when you are used to solve hasrd problems , you always over think.  :lol: I had the same solution.

By the way, do you think that I might need a proof for the 4^n congruent to 4 mod 12?

You can definitely prove it with induction (sort of what AIME15 did)...

$ 4^1\equiv 4\pmod{12}$, obviously.
Assume for arbitrary $ k\in\mathbb{Z}^ \plus{}$ that $ 4^k\equiv 4\pmod{12}$. Then, multiplying both sides by 4, you have $ 4^{k \plus{} 1}\equiv 16\equiv 4\pmod {12}$.
So $ 4^n\equiv 4\pmod {12},\;n\in\mathbb{Z}^ \plus{}$, Q.E.D.

As for your other questions about Euler's totient function:

$ \varphi(1000) \equal{} \varphi(2^3\cdot 5^3) \equal{} 1000\left(1 \minus{} \frac {1}{2}\right)\left(1 \minus{} \frac {1}{5}\right) \equal{} 400$
$ \varphi(2401) \equal{} \varphi(7^4) \equal{} 2401\left(1 \minus{} \frac {1}{7}\right) \equal{} 2058$

Oh, for future reference, this is [b]modular arithmetic[/b], not abstract algebra  :lol:


number theory
least common multiple
number theory proposed

How many solutions has the following equality:
$ LCD(a,b\plus{}1)\equal{}LCD(b,a\plus{}3)$, where $ a$ and $ b$ are positive integers < $ 10^6$.

[quote="aviateurpilot"]what's $ LCD$???[/quote]

$ LCD$=lowest common denominator  :wink:

if it's $ LCD$ that then number of solution is $ (10^{6}-1)^2$
if it's $ lcm$ we note $ lcm(x,y)=(x,y)$

we suppose that $ c=(a,b+1)=(b,a+3)$
$ b,b+1|c\Rightarrow b(b+1)|c\Rightarrow \exists k: \ (a,b+1)=kb(b+1)\Rightarrow b|a$
if $ gcd(a,3)=1$ then:
$ a,a+3|c\Rightarrow a(a+3)|c\Rightarrow \exists k': \ (b,a+3)=k'a(a+3)\Rightarrow a|b$
then a=b but we get $ c=a(a+1)=a(a+3)$ (impossible)

then $ a=3a'$ for some $ a'$.
$ 3a',3a'+3|c\Rightarrow 3a'(a'+1)|c\Rightarrow \exists k": \ (b,3a'+3)=3k"a'(a'+1)\Rightarrow a'|b$
and we have $ b|a=3a'$ so $ b=a'$ because $ b=3a'=a$ is not solution
and we get $ c=(3b,b+1)=(b,3b+3)\iff gcd(b,3)=gcd(b+1,3)=1$

then $ (a,b)\in\{(3n,n): \ n\in[|1,\frac{10^6-1}{3}|];\ n\equiv 1[3]\}$
and $ |\{(3n,n): \ n\in[|1,\frac{10^6-1}{3}|];\ n\equiv 1[3]\}|=\frac{10^6-1}{9}$

[quote="ZetaX"]The least common denominator of integers is always $ 1$  :wink:
You rather may want to use $ \mathrm{lcm}$, meaning "least common multiple"  :wink:[/quote]

Thank you Zetax.
Yap, I meant $ LCM$. Sorry, I confused the words...


On a four-by-four grid, points are 1 unit apart horizontally and vertially.  Find the probably that two randomly selected points P and Q are less than 2.5 units apart.

The AoPS Volume 2 book has a similar problem.  I'm not great at probability problems so I'll try to give advice.  The fact that it is a grid will really help you see things.  Draw a picture of the situation in terms of boxes and it should help.  I hope that was of some help.

im not very good at probablitiy questions either so answer/solution might be very wrong

[hide]Ok all points that are less than 2.5 points away from P (we will say that this is the first one picked) will be inside a circle of radius 2.5 cm so the area of this circle is $ (2.5^2)\pi$ and then we put this over 16 so our answer is $ \boxed{\frac{6.25\pi}{16}}$[/hide]

By points, he means the 16 points on the grid. Not arbitrary ones.

ok so the answer would be [hide]$ \frac{21}{32}$[/hide]

Solution:

[hide]Ok so we know that the probablitiy of any point on the grid being picked as point P is 1/16 so we can examine each of three cases

P is a corner point -- there would be a 7/16 probablitiy that Q would satisfy the requirements and there are 4 corner points so our probablitiy is $ \frac{1}{16} \times \frac{7}{16} \times 4 \equal{} \frac{7}{64}$

P is an edge point -- there would be a 10/16 probability that Q would satisfy the restraints and there are 8 edge points so our probablitly is $ \frac{1}{16} \times \frac{10}{16} \times 8 \equal{} \frac{20}{64}$

P is in the middle -- there would be a 15/16 probablity that Q would satisfy the restraints and therea are 4 middle points so our probability is $ \frac{1}{16} \times \frac{15}{16} \times 4 \equal{} \frac{15}{64}$

If we add all of these us we have $ \frac{7}{64} \plus{} \frac{20}{64} \plus{} \frac{15}{64} \equal{} \boxed{\frac{21}{32}}$[/hide]


geometry circumcircle inradius inequalities trigonometry

Let ABC be a right triangle. If R is the circumradius and r the inradius, show that

[tex]
R \geq (1 + \sqrt{2})r
[/tex]

[hide]

Lemma: c >= (a+b)*:rt2: / 2.

Proof:

The inequality is homogeneous. Reduce to c = 1.

Then a = sin :theta:, b = cos :theta: for some :theta:.

RHS = (sin :theta: + cos :theta:) :rt2: / 2

       = (sin :theta:cos pi/4 + sin pi/4 cos :theta:)

        = sin(:theta: + pi/4) <= 1 by the range of sin.

[tex]\blacksquare[/tex]

So we have

c >= (a+b) * :rt2: / 2.

And :rt2: / 2 = (1+:rt2:)/(2+:rt2:), so

(2+:rt2:)c >= (a+b)(1+:rt2:)

=> c >= (a+b-c)(1+:rt2:)

=> c/2 >= ((a+b-c)/2)(1+:rt2:)

Since it is a right triangle, c/2 = R, and r = (a+b-c)/2, so

R >= (1+:rt2:)r, as desired.

[/hide]

Another proof (Let's make it more complicated ):

[hide]

[tex]r/{(4R)}=\sin{(A/2)}\sin{(B/2)}\sin{(C/2)}=\sin{(A/2)}\sin{(B/2)}\sin{(\pi/4)}[/tex]

[tex]=\displaystyle{{1\over2}\sin{\pi\over4}(\cos{{A-B}\over2} - \cos{{A+B}\over2})}[/tex]

[tex]=\displaystyle{{\sqrt{2}\over4}\cos{{A-B}\over2} - {1\over4}}[/tex]

[tex]\leq\displaystyle{{\sqrt{2}-1}\over4}}[/tex]

[/hide]

See here for a proof of the identity [tex]\sin{A/2}\sin{B/2}\sin{C/2}=r/4R[/tex]

[quote="fuzzylogic"]The lemma can be proved without trignometry, it is a simple case of Cauchy-Schwarz:

[tex]\sqrt{2}\sqrt{a^2+b^2} \ge a+b[/tex] (or simply equivalent to) [tex](a-b)^2\ge0[/tex][/quote]

It's all good.

It's also interesting to note that cauchy takes the form

[tex] ||w|| \; ||v|| >= w \cdot v = ||w||\; ||v||\; \cos \, \theta, [/tex]
where :theta: is the angle made by w and v.

So the approaches are remarkably related.


analytic geometry
graphing lines
slope
trigonometry
geometry
geometric transformation
reflection

We have a slope of incline angle $\alpha$ height $h$ and mass $M$ which can slide freely on a horizontal frictionless table. We put a small mass $m_{1}$ on top of the slope, and attach a string of length $h$ and negligible mass to it. To the other end of the string we attach a mass $m_{2}$ which is laying on the table. after how much time will the two masses be the closest to each other? (all friction is negligible)

I got a messy answer  :mad: . For the following values, I get for time: $1,83~ \rm seconds$.

$m_{1}=7~\rm kg \\ m_{2}=1~\rm kg \\ M=20~\rm kg \\ \alpha=30^\circ \\ h=1~\rm m$

Do you get the same result?  :D

When substituted into my answer, I get $t \approx 0.3\text{s}$, unless I've made a mistake...

Here's what I got...

$t=\sqrt{\frac{h}{g}\frac{M(m_{1}+m_{2})\sin\alpha}{m_{1}^{2}\cos^{2}\alpha+m_{1}M\sin^{2}\alpha-m_{2}M\sin\alpha}}$

Plugging in those numbers, i got 0.36s... I think

Well, my solution isn't correct, because for $\alpha = \pi/2$ you have to get \[t^{2}= \frac{h}{g}\frac{m_{1}+m_{2}}{m_{1}-m_{2}},\] which I don't get :(

I've found the mistake, and my new response is
\[t = \sqrt{\frac{h(1+\sin\alpha)}{2g}\frac{m_{1}+m_{2}-\frac{m_{1}^{2}\cos^{2}\alpha}{m_{1}+m_{2}+M}}{m_{1}\sin\alpha-m_{2}}}. \]

I forgot a factor of $2$ in the expression. Sorry, I'll edit that.

:ewpu:

This somehow reflects my feelings at this point... I've discovered more errors in my solution. Here's the (hopefully) correct one:
\[t = \sqrt{\frac{h}{g}\frac{m_{1}+m_{2}-\frac{m_{1}^{2}\cos^{2}\alpha}{m_{1}+m_{2}+M}}{m_{1}\sin\alpha-m_{2}}}. \]

[b]Thaakisfox[/b], please respond! I'm not changing my mind about this one, so  :huuh:

Yes Djole that is the correct answer :D
nice problem right? :P

[b]Djole[/b] or [b]Thaakisfox[/b]: can you post your solution :?:

Edit: OK, I found the solution myself (conservation of energy, conservation of momentum and some simple integrating). Does anyone know a solution using laws of dynamics :?:

I did it with lagrangians, but it shouldn't be too hard to do it by Newton's laws. I'll try to post that soon, please bear with me...

Usings Newton laws I do not get the same result as you. I am very probable doing something wrong. I'll wait for your solution [b]Djole[/b].  :D Thanks.

I don't get the same solution for the accelerations as well. [b]Thaakisfox[/b], do you know the solution using Newton's laws? The correct accelerations are \[a_{1}= g \frac{m_{1}\sin\alpha-m_{2}}{m_{1}+m_{2}-\frac{m_{1}^{2}\cos^{2}\alpha}{M+m_{1}+m_{2}}}\] and \[a = a_{1}\frac{m_{1}\cos\alpha}{M+m_{1}+m_{2}},\] where $a$ is the horizontal acceleration of the incline and $a_{1}$ is the vertical acceleration of $m_{2}$.

Not quite :maybe: Better wait for [b]Thaakisfox[/b] and see if he has a solution using Newton's laws...

Ok I have some time now, so I will write the solution using newtons-laws.
lets fix a coordinate system to the table. $(x,y)$. clearly the two masses will be the closest when half of the string $\frac{h}{2}$ will be on the slope and half of it hanging.(I think everyone can prove this :D ).
let $N$ be the normal force by the slope on $m_{1}$, $T$ the tension in the string.Let $a_{1x}$ , $a_{1y}$ be the components of acceleration of $m_{1}$, $A$ be the acceleration of the slope, and $a_{2y}$ be the acceleration of $m_{2}$.
So the equations of motion:
$N\sin \alpha-T \cos \alpha =(M+m_{2})A$                  $(1)$
$N\sin \alpha-T\cos \alpha = m_{1}a_{1x}$                $(2)$
$m_{1}g-N\cos \alpha-T\sin \alpha = m_{1}a_{1y}$     $(3)$
$T-m_{2}g = m_{2}a_{2y}$                                        $(4)$

now we have to find some link between the accelerations.
Lets look at $m_{1}$. while it moves downwards towards the floor by $s_{1y}$ it moves sideways by $s_{1x}$ with respect to the table. But it the slope moves by $S$ with respect to the table, so $m_{1}$ moves $s_{1x}+S$ with respect to the slope sideways. Since the string is non elastic, we must have:
$\tan \alpha =\frac{s_{1y}}{s_{1x}+S}$
since the initial velocity of everything was $0$ and the forces are constant, hence the accelerations are constant, so the latter ratio after time $t$
$\tan \alpha =\cfrac{\frac12 a_{1y}t^{2}}{\frac12 At^{2}+\frac12 a_{1x}t^{2}}= \frac{a_{1y}}{a_{1x}+A}$
 So:
$a_{1y}=(a_{1x}+A) \tan \alpha$                            $(5)$
and we also know that since the string is non elastic, the movement of $m_{1}$ with respect to the slope is the same as the vertical movement of $m_{2}$. hence:
$s_{1y}=s_{2y}\sin \alpha$
So from here just as we did before:
$a_{1y}=a_{2y}\sin \alpha$                                   $(6)$
now the $T$ time we are looking for can be written as:
$T=\sqrt{\frac{2 \cdot h/2}{a_{2y}}}=\sqrt{\frac{h}{a_{2y}}}$  $(7)$
so we need to find $a_{2y}$
Now solving $(1)$, $(2)$, $(3)$ , $(4)$, $(5)$ and $(6)$ for $a_{2y}$ (this is "just" algebra :P ) we get:

$a_{2y}=\frac{(m_{1}\sin \alpha-m_{2})(m_{1}+m_{2}+M)}{m_{1}(m_{2}+M)+m_{1}^{2}\sin^{2}\alpha+m_{2}(m_{1}+m_{2}+M)}g$

Now plugging this into $(7)$ we get:

$T=\sqrt{\frac{m_{1}(m_{2}+M)+m_{1}^{2}\sin^{2}\alpha+m_{2}(m_{1}+m_{2}+M)}{(m_{1}\sin \alpha-m_{2})(m_{1}+m_{2}+M)}\cdot \frac{h}{g}}$

which is the same as djole's answer. (of course after some rearrangement) :D

That's similar to what I've tried to do, except I worked in the non-inertial frame of the moving incline. I didn't get the right answer, though -- I have no idea why :( Anyway, go lagrangians ;)

You probably screwed the calculation. Its really long solving that system of 6 equation. it took me about 3-4 pages :D

Yeah, probably :( But I did it twice and I got the same result, so I'm not so sure...

[quote="Thaakisfox"]$N\sin \alpha-T \cos \alpha =(M+m_{2})A$                  $(1)$[/quote]
I believe that this is the equation of motion for the horizontal movement of the slope and $m_{2}$. Then why is $T \cos \alpha$ there - does the string exert any force on the slope :?:

Yes, it does, through the tension force acting on the pulley which is a part of the incline.

:w00tb: Of course, now I see it.  :rotfl: That was the mistake in my solution. Now I get a correct answer solving the problem in the non-inertial frame of the moving incline...

Would I be too much of a pain if I asked you to post that solution, please? I really don't see where I've gone wrong...

And if anyone wants I can post my lagrangian solution...

Let $N_{1}$ be the normal force exerted by the slope on $m_{1}$, $N_{2}$ - the normal force exerted by the slope on $m_{2}$, $T$ the tension in the string, $a$ the acceleration of the slope and $a'$ - vertical acceleration of $m_{2}$.
In the inertial frame of reference we have got the equation of motion for the slope:
$N_{1}\sin\alpha-N_{2}-T\cos\alpha=Ma$ $(1)$
In the non-inertial frame of the moving incline we have got the following equations (force components normal to the slope acting on $m_{1}$ and $m_{2}$ must balance each other):
$N_{1}+m_{1}a\sin\alpha=m_{1}g\cos\alpha$ $(2)$
$N_{2}=m_{2}a$ $(3)$
And the equations of motion for $m_{1}$ and $m_{2}$:
$m_{1}g\sin\alpha+m_{1}a\cos\alpha-T=m_{1}a'$ $(4)$
$T-m_{2}g=m_{2}a'$ $(5)$

By comparing $a'$ from $(4)$ and $(5)$ we get $T$.
Putting $T$, $N_{1}$ from $(2)$ and $N_{2}$ from $(3)$ into $(1)$ we can find $a$:
$a=\frac{m_{1}g\cos\alpha(m_{1}\sin\alpha-m_{2})}{(M+m_{2})(m_{1}+m_{2})+m_{1}(m_{1}\sin^{2}\alpha+m_{2})}= \frac{m_{1}g\cos\alpha(m_{1}\sin\alpha-m_{2})}{(M+m_{2}+m_{1})(m_{1}+m_{2})-m_{1}^{2}\cos^{2}\alpha}$
Now adding equations $(4)$ and $(5)$ we get:
$m_{1}g\sin\alpha+m_{1}a\cos\alpha-m_{2}g=(m_{1}+m_{2})a'$
And finally putting $a$ into it we get:
$a'=\frac{g}{m_{1}+m_{2}}\frac{(m_{1}\sin\alpha-m_{2})(M+m_{2}+m_{1})(m_{1}+m_{2})}{(M+m_{2}+m_{1})(m_{1}+m_{2})-m_{1}^{2}\cos^{2}\alpha}= \frac{g(m_{1}\sin\alpha-m_{2})}{m_{1}+m_{2}-\frac{m_{1}^{2}\cos^{2}\alpha}{M+m_{2}+m_{1}}}$
The rest is obvious...

Thanks a lot! I'll check it out as soon as possible, and I'll try to post my solution soon as well.

My equations are exactly the same as yours. Too bad I suck at algebra :(

But here's my solution. Let $x$ be the coordinate of the lower right corner of the incline, and let $y$ be the height of $m_{2}$ above the horizontal. The coordinates and velocity components of the bodies are: \[M: \quad (x, 0), \quad (\dot x, 0);\] \[m_{2}: \quad (x, y), \quad (\dot x, \dot y);\] \[m_{1}: \quad (x-y\cos\alpha, h-y\sin\alpha), \quad (\dot x-\dot y\cos\alpha,-\dot y\sin\alpha).\]

The kinetic energies are now easy to computate; the total kinetic energy of the system is \[K = \frac{M}{2}\dot x^{2}+\frac{m_{2}}{2}(\dot x^{2}+\dot y^{2})+\frac{m_{1}}{2}[(\dot x-\dot y\sin\alpha)^{2}+\dot y^{2}\cos^{2}\alpha].\] The potential energy of the system is \[V = m_{2}gy+m_{1}g(h-y\sin\alpha),\] so the lagrangian is \[\mathcal L = \frac{M+m_{1}+m_{2}}{2}\dot x^{2}+\frac{m_{1}+m_{2}}{2}\dot y^{2}-m_{1}\dot x\dot y\cos\alpha-m_{2}gy+m_{1}gy\sin\alpha.\]

The E-L equations are \[\frac{\text d}{\text dt}\left(\frac{\partial \mathcal L}{\partial \dot x}\right) = \frac{\text d\mathcal L}{\text dx}\] and \[\frac{\text d}{\text dt}\left(\frac{\partial \mathcal L}{\partial \dot y}\right) = \frac{\text d\mathcal L}{\text dy},\] which yields \[\ddot x = \ddot y \frac{m_{1}\cos\alpha}{M+m_{1}+m_{2}}\] and \[\ddot y = g \frac{m_{1}\sin\alpha-m_{2}}{m_{1}+m_{2}-\frac{m_{1}^{2}\cos^{2}\alpha}{m_{1}+m_{2}+M}},\] and now it is trivial, as [b]Thaakisfox[/b] noted.


suppose (C) is a circle and P is a fixed point in the exterior of it. let L be a variable line wich goes through P, and cuts

(C) in points B,A respectively. M is the midpoint of arc AB.and N is a point on (C) , in which MN is diameter.

let C be a point on chord AB in which :

                                                     PA:PC=PC:PB

prove that:  angle MCN is fixed.

[color=red][Moderator edit: This problem has also been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=47265 .][/color]

Let's label the given circle (O) (rather than (C)), because the defined point C is not the center of the given circle. The condition $\frac{PA}{PC} = \frac{PC}{PB}$ means $PC^2 = PA \cdot PB$, which is the power of the point P to the circle (O). Hence, PC is the tangent length from the point P to the circle (O). Let S, T be the tangency points of the tangents to (O) from P. The point C is the intersection of a circle (P) with radius PS = PT with the chord AB. Let the tangents PS, PT meet the circle (P) at points U, V different from S, T, so that SU, TV are diameters of the circle (P). Then the angles $\angle SCU = \angle TCV = 90^\circ$ are right and PC is the common median of the right angle triangles $\triangle SCU, \triangle TCV$ from the common vertex C. The triangles $\triangle MSN, \triangle NTM$ on the diameter MN are also right, with the medians OS, OT - radii of the circle (O). Since $US \equiv PS \perp OS,\ CP \perp MN$, the angles $\angle MOS = \angle SPC$ are equal. As a result, the right angle triangles $\triangle SCU \sim \triangle MSN$ are similar, having equal angles between the hypotenuse and the right angle median. Likewise, since $VT \equiv PT \perp OT,\ CP \perp MN$, the angles $\angle NOT = \angle TPC$ are also equal and the right angle triangles $\triangle TCV \sim \triangle NTM$ are also similar. From the similar isosceles triangles $\triangle CPU \sim \triangle SON$, the angles $\angle MNS = \angle PCU$ are equal. Since the lines $PC \perp MN$ are perpendicular, it follows that the lines $NS \perp UC$ are also perpendicular. But since the angle $\angle SCU = 90^\circ$ is right, this implies that the point $C \in NS$. Likewise, we can deduce that the point $C \in MT$. Since the points $U, V \in (P)$ are fixed and $C \in (P)$, the angle $\angle UCV$ is constant. But

$\angle UCV = \angle PCU + \angle PCV = \angle PUC + \angle PVC \equiv \angle MUC + \angle NVC =$

$= \angle SNM + \angle TMN \equiv \angle CNM + \angle CMN = 180^\circ - \angle MCN$

which means that the angle $\angle MCN$ is also constant. Even simpler, since the points $S, T \in (P)$ are fixed and $C \in (P)$, the angle $\angle SCT = \angle MCN$ is constant.

This looks like to use some projective transformations again. I'l try later.

Hmmmm....

Thinking....

number 1,did u know that it has a solution by pole and polar and harmonic?

yetti ,did u know that MS and NT meet each other on the circumcircle of triangle SVU,in point K

,inwich CK is diameter?


induction
number theory proposed
number theory

The sequence $ (L_n)$ is given by $ L_0 \equal{} 2, L_1 \equal{} 1$ and $ L_{n \plus{} 1} \equal{} L_n \plus{} L_{n\minus{}1}$ for $ n \geq 1$. Prove that if a prime number $ p$ divides $ L_{2k}\minus{} 2$ for $ k \in \mathbb{N}$, then $ p$ also divides $ L_{2k \plus{} 1} \minus{}1$.

[hide]
I will prove the identity

$ (L_{2k} \minus{} 2)(L_{2k \plus{} 2} \plus{} 2) \equal{} (L_{2k \plus{} 1} \minus{} 1)^2$, from which the result follows.

First, $ L_{n}L_{n \plus{} 2} \minus{} L_{n \plus{} 1}^2 \equal{} ( \minus{} 1)^n(5)$. I will prove this identity by induction on $ n$:

$ L_{n}L_{n \plus{} 2} \minus{} L_{n \plus{} 1}^2 \equal{} ( \minus{} 1)^n(5)$
$ (L_{n \plus{} 2} \minus{} L_{n \plus{} 1})L_{n \plus{} 2} \minus{} L_{n \plus{} 1}(L_{n \plus{} 3} \minus{} L_{n \plus{} 2}) \equal{} ( \minus{} 1)^n(5)$
$ L_{n \plus{} 2}^2 \minus{} L_{n \plus{} 1}L_{n \plus{} 3} \equal{} ( \minus{} 1)^n(5)$
$ L_{n \plus{} 1}L_{n \plus{} 3} \minus{} L_{n \plus{} 2}^2 \equal{} ( \minus{} 1)^{n \plus{} 1}(5)$.

This identity is analogous to the Fibonacci identity $ F_{n}F_{n \plus{} 2} \minus{} F_{n \plus{} 1}^2 \equal{} ( \minus{} 1)^{n \plus{} 1}$.

Now we have that

$ L_{2k}L_{2k \plus{} 2} \minus{} L_{2k \plus{} 1}^2 \equal{} 5$
$ (L_{2k} \minus{} 2)(L_{2k \plus{} 2} \plus{} 2) \plus{} 2L_{2k \plus{} 2} \minus{} 2L_{2k} \plus{} 4 \minus{} L_{2k \plus{} 1}^2 \equal{} 5$
$ (L_{2k} \minus{} 2)(L_{2k \plus{} 2} \plus{} 2) \plus{} 2L_{2k \plus{} 1} \plus{} 4 \minus{} L_{2k \plus{} 1}^2 \equal{} 5$
$ (L_{2k} \minus{} 2)(L_{2k \plus{} 2} \plus{} 2) \equal{} L_{2k \plus{} 1}^2 \minus{} 2L_{2k \plus{} 1} \plus{} 1$
$ (L_{2k} \minus{} 2)(L_{2k \plus{} 2} \plus{} 2) \equal{} (L_{2k \plus{} 1} \minus{} 1)^2$.

Hence if a prime divides $ L_{2k} \minus{} 2$ then it divides $ L_{2k \plus{} 1} \minus{} 1$.

[/hide]

Note that $ L_n\equal{}\phi^n\plus{}(\minus{}\phi)^{\minus{}n}$ where $ \phi$ is the golden ratio. Thus
\[ L_{2n}\minus{}2\equal{}(\phi^n\minus{}\phi^{\minus{}n})^2\text{ and }L_{2n\plus{}1}\minus{}1\equal{}(\phi^n\minus{}\phi^{\minus{}n})(\phi^{n\plus{}1}\plus{}\phi^{\minus{}(n\plus{}1)})\]
The result follows, as the first implies $ p|\phi^n\minus{}\phi^{\minus{}n}$


function
complex analysis
real analysis
real analysis unsolved

Consider the entire function $\prod_{n\geq 0}{(1+x^{2^n}+x^{2^{n+1}})}=a_0+a_1x+a_2x^2+...$. Prove that $ (a_i, a_{i+1})=1$ for all $i$ and the sequence $(\frac{a_{n+1}}{a_n})$ is a enumeration of all rational numbers.

Nice problem.
[hide="my solution"]Let $f(x)=\prod_{n=0}^{\infty}(1+x^{2^n}+x^{2^{n+1}})=\sum_{i=0}^{\infty}a_{i} x^{i}$, we know that $f(x)=(1+x+x^2)f(x^2)$, compare the coefficients in both sides, we get two equations:
$\{\begin{array}{rll}a_{2n}&=&a_{n}+a_{n-1}\\a_{2n+1}&=&a_{n}\end{array}$.
    Thus $(a_{i},\ a_{i+1})=1$ can be proved easily by induction.
    As for the second conclusion, what we need to prove is $\forall (P,\ Q)=1$, we can find some $m$, s.t. $a_{m}=P,\ a_{m+1}=Q$, assume $P>Q$ first, from the equations we know $m$ must be even, $m=2k$(if $P<Q,\ m$ must be odd), now we need some $k$, s.t. $a_{k-1}=P-Q,\ a_{k}=Q$, similar to the successive division, we have successive subtraction here, finally we need only find some $k$, s.t $a_{k}=1,\ a_{k+1}=1$, but we know $a_{0}=a_{1}=1$, Q.E.D.[/hide]


ina graph with 2n vertices ,prove that there existe two vertices who has even degrees

I can immediately think of several different counterexamples for any $ n$.  What do you mean?


Ok this problem isn't hard or anything, but it was a homework problem. The answer in the back of the book said that that question had an impossible solution (there is no answer), but me and a couple of classmates solved the problem and found a solution. We showed it to the class, and everyone agreed with it and didn't see anything wrong with our solution. Who is right?

Word for word from book : The face diagonals of a rectangular box are 2, 3, and 6. Find the diagonal of the box.

I got the answer 7 root 2 all over 2. (the square root of 49/2)

Name the three sides of the prism x, y, and z.
Therefore,
$ x^2 \plus{} y^2 \equal{} 2^2$
$ z^2 \plus{} y^2 \equal{} 3^2$
$ x^2 \plus{} z^2 \equal{} 6^2$
Adding them up gives:
$ 2(x^2 \plus{} y^2 \plus{} z^2) \equal{} 49$
$ x^2 \plus{} y^2 \plus{} z^2 \equal{} \frac {49}{2}$
And since the space diagonal of a rectangular prism is $ \sqrt {x^2 \plus{} y^2 \plus{} z^2}$, the space diagonal of this rectangular prism is...

$ \frac {7}{\sqrt {2}}$

Which simplifies to...

$ \frac {7\sqrt {2}}{2}$

Unfortunately, you forgot to check that $ x, y, z$ exist.  It turns out that we must have $ y^2 \equal{} \frac{49}{2} \minus{} 36 \equal{} \minus{} \frac{23}{2}$.  ;)

aww man..... i thought i had finally proved the book wrong and therefore we didnt have to do bookwork anymore. crap

way to go torajirou

Geomety for Enjoyment and Challenge by McDougal Littell/Houghton Mifflin. Why?


I think most of the scores should be out now, and I'm just curious how AoPs places against the rest of the nation.

Yeah, it belongs there, but I don't think I'm allowed to make polls in that forum.


induction
inequalities
number theory unsolved
number theory

Let $E$ be the set of numbers of the form $2^{a}3^{b}$, with $a,b$ non-negative integers. A positive integer $n$ is called [i]appalling[/i] if it can be written as the sum of elements of $E$, such that for any two terms of the sum, one does not divide the other.

a) Prove that every integer $n > 0$ is [i]appalling[/i].

b) Prove that for any $k > 0$ there exists a positive integer $n$ such that $n$ can be written in $k$ ways in the form explained above.

Has it been posted before?

a) Done by induction, firstly we can obviously assume that $6\not |n$. Let $k,m$ be positive integers defined by $3^{k}\leq n < 3^{k+1}$ and $m=c$ or $m=c-1$ so that $2^{c}\leq n-3^{k}< 2^{c+1}$ and $3|n-2^{m}$.
Let $\sum 2^{a_{i}}3^{b_{i}}$ be a representation of $\frac{n-3^{k}-2^{m}}{6}$ so that $n=3^{k}+2^{m}+\sum 2^{a_{i}+1}3^{b_{i}+1}$. I claim that this is desired representation of $n$. All the summands $2^{a_{i}+1}3^{b_{i}+1}$ meets the conditions of course, and we are to check that $k>b_{i}+1$ and $m>a_{i}+1$, but that can be easily shown by proof ad absurdum - if you suppose that one of the reversed inequalities hold then you obtain contradiction with the way we've choosen $k,m$.

They are obviosly.
a)$n=1+1+...+1$ (n- times), ($1=2^{0}*3^{0}$).
$n=\sum_{i}2^{a_{i}}$, were $a_{i}$ are distinct 2-adic expression.
b) If $n>3^{k}$ we have at least k expressions
$n=3^{l}+m_{l}, \ 1\le l\le k,$ were $m_{l}=\sum_{i}2^{a_{i,l}}$.

this is very similar with one problem that i see in tournament of towns!  :)


LaTeX
trigonometry
algebra
polynomial
geometry
3D geometry

solve for x.  16cos^2x.sin^3x = 2sinx-sin5x
if this was in latex would my code be? 16\cos^\theta\sin^\theta=2sin\theta-sin5\theta..? help.i am new
dont take cos to the power 2x. it is cosx squared. and similarly for sin x. {{this is the consequence of no hand in latex}}

$ \text{\LaTeX}$-ifying it makes it \[ 16\cos(x)\cdot\sin(3x)=2\sin(x)-\sin(5x)\]

here's the code: [code]\[16\cos(x)\cdot\sin(3x)=2\sin(x)-\sin(5x)\][/code] (the [code]\[ and \][/code] center the expression)

We want every term to be in terms of $ a\sin(x)b\cos(x)$ to make it easier (in other words, none of the $ \sin(5x)$ stuff).

\[ 16\cos(x)\sin(2x+x)=2\sin(x)-\sin(3x+2x)\]

I know how to approach this (using double angle bashing and $ \sin(\alpha+\beta)$) but I don't know how to do it without having it get too messy.

The solutions aren't pretty. The problem is solved by expanding the expression to an 8th degree polynomial in $ \cos{x}$ which can't be solved by a human. Mathematica gives all the roots as a nested-radical mess.

hello, write your equation in the following form
$ \sin(x)(16\cos^4(x)\plus{}64\cos^3(x)\minus{}12\cos^2(x)\minus{}16\cos(x)\minus{}1)\equal{}0$.
Sonnhard.

i am sorry there is a mistake. its actually cos square x and sin cube x. or sin^3.x and cos^2.x. there is a mistake in the sin. now i think its solvable

so it's
\[ 16\cos^2(x)\sin^3(x) = 2\sin(x) - \sin(5x)?
\]
I'm not sure. Are you sure it's supposed to be $ \sin(5x)?$ It would get very messy :maybe:

It would be great if you learned $ \text{\LaTeX}$ :)

$ 16\cos^2x\sin^3x\equal{}4\sin x\cdot\left(2\sin x\cos x\right)^2\equal{}$ $ 4\sin x\sin^22x\equal{}$ $ 2\sin x\cdot 2\sin^22x\equal{}$ $ 2\sin x\cdot(1\minus{}\cos 4x)\equal{}$ $ 2\sin x\minus{}2\sin x\cos 4x\equal{}$

$ 2\sin x\minus{}(\sin 5x\minus{}\sin 3x)\equal{}$ $ 2\sin x\plus{}\sin 3x\minus{}\sin 5x$ . In conclusion, our equation is equivalently with $ \sin 3x\equal{}0$ , i.e. $ x\in \frac {\pi}{3}\cdot\mathbb Z$ .


function
number theory proposed
number theory

Find all functions $ f$ satisfiying: $ f$:  $ \mathbb{N^*} \minus{} > \mathbb{N^*}$

and :

$ \forall x,y \in \mathbb{N^*}: x^2 \plus{} f(y) / y \plus{} f^2(x)$

/ means :"divides"

some one knows where this problem was proposed, please ?

1)$ f(1)\equal{}1$
2) Take $ x\equal{}1$ then
$ 1\plus{}f(y)|y\plus{}1$ so  $ y\geq f(y)$
3)Take $ y\equal{}1$ then
$ x^2\plus{}1|f(x)^2\plus{}1$ so $ f(x)\geq x$
It mean that $ f(x)\equal{}x$


calculus
derivative
trigonometry
calculus computations

how does one calculate the partial derivative of

xcos(y) with respect to x and y?

When calculating the partial with respect to one variable, the other one is to be treated as a constant.  So

$\frac{d}{dx}\left( x \cos y \right) = \cos y$
$\frac{d}{dy}\left( x \cos y \right) =-x \sin y$


Demonstrati ca daca $A \in M_n(R)$ cu $n \in 2N+1$ si $A^n=I_2$ atunci exista $y \in R^*$ a.i. $A=yI_2$

Cauta in solved problems algebra. Am demonstrat ca problema e falsa.


x.y.z = 1

$ x\plus{}\frac{1}{z}\equal{}5$ and $ y\plus{}\frac{1}{z}\equal{}29$

find $ z\plus{}\frac{1}{y}$

[quote="epsilon07"]x.y.z = 1

$ x \plus{} \frac {1}{z} \equal{} 5$ and $ y \plus{} \frac {1}{z} \equal{} 29$

find $ z \plus{} \frac {1}{y}$[/quote]
$ z\equal{}\frac 1{xy}$ and so $ x\plus{}xy\equal{}5$ and $ y\plus{}xy\equal{}29$

So $ y\equal{}\frac{5\minus{}x}x$ and $ z\equal{}\frac 1{5\minus{}x}$ and $ \frac{(5\minus{}x)(x\plus{}1)}x\equal{}29$

So $ x^2\plus{}25x\minus{}5\equal{}0$ and $ x\equal{}\frac{\minus{}25\plus{}\epsilon\sqrt{645}}2$ where $ \epsilon\equal{}\pm 1$

And so $ z\plus{}\frac 1y\equal{}\frac{x\plus{}1}{5\minus{}x}\equal{}\frac{\minus{}40\pm 3\sqrt{645}}{145}$

[quote="epsilon07"]x.y.z = 1

$ x \plus{} \frac {1}{z} \equal{} 5$ and $ y \plus{} \frac {1}{z} \equal{} 29$

find $ z \plus{} \frac {1}{y}$[/quote]

If x,y,z = 1... means that they all are positive integers, . . . EDIT:  The user did not correctly
type x*y*z = 1, $ x \cdot y \cdot z \equal{} 1, or xyz \equal{} 1$.

then $ \frac 1z$ would be a positive integer.

And the variables' values could be x = 4, y = 28, and z = 1
to satisfy both equations.

Then $ z \plus{} \frac 1y \equal{}$

$ 1 \plus{} \frac {1}{28} \equal{}$

$ \frac {29}{28}$.

[quote="Arrange your tan"] If x,y,z = 1... means that they all are positive integers,[/quote]

No, it is "$ x\cdot y\cdot z\equal{}1$" and this means $ x,y,z$ are three real numbers whose product is $ 1$.

[quote="epsilon07"]x.y.z = 1

$ x \plus{} \frac {1}{z} \equal{} 5$ and $ y \plus{} \frac {1}{z} \equal{} 29$

find $ z \plus{} \frac {1}{y}$[/quote]
epsilon07,

just type the condition as:

xyz = 1.

Do not use periods for multiplication.

Something tells me that $ y+\frac{1}{z}$ should be $ y+\frac{1}{x}$. My solution is based on that assumption.
[hide="Solution"]
Multiply the three expressions (substituting $ z+\frac{1}{y}=n$), then substitute and manipulate (cleverly of course):\begin{align*}\left(x+\frac{1}{z}\right)\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)&=5\cdot29n\\xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}&=145n\\1+1+\left(x+\frac{1}{z}\right)+\left(y+\frac{1}{x}\right)+\left(z+\frac{1}{y}\right)&=145n\\2+5+29+n&=145n\\36&=144n\\n&=\boxed{\frac{1}{4}}.\end{align*}[/hide]

[quote="Arrange your tan"][quote="epsilon07"][quote="varunrocks"]I think this is an AIME problem.[/quote]

as I know x.y.z means x, y and z are real numbers and their products are 1 as navier said...[/quote]

As you know now, don't use periods for multiplication.  Period.[/quote]
To be honest that is quite unfair. There are a lot of places around the world that use the period as a multiplication sign.

[quote="Arrange your tan"]Their ways are immaterial here.[/quote]  As much as I'd like to pursue the bizarre mindset implied by this sentence, I think it would be more productive just to remind everyone that Arrange your tan is fundamentally a troll.  I strongly recommend just ignoring its posts -- among other things, it appears to have only minimal ability to understand the intent of a variety of kinds of communication, and becomes aggresive and repetitive when the intended content is explained to it.  (Just look through a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=324591]random[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1750021#1750021]selection[/url] of its [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1750009#1750009]recent[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1726384#1726384]posts[/url], for example.)  Eschew!

[quote="Arrange your tan"][quote="ckck"]To be honest that is quite unfair.  [...][/quote]

It's not unfair; you use our conventions here.  Their ways are
immaterial here.  You don't use commas for periods and/or
periods for multiplication.  It is analogous to not typing
out problems here in a language other than English.  You
adapt to the convention here.[/quote]
"our"? Although AoPS is in America, there are many users that are non-american that have different perspectives in writing math. You can't just bash how they do it seeing as they have been writing math like that for almost their entire lives. Look at some other forums and you'll see that other people use the period as multiplication signs. I don't see you bashing them about it.

[quote="ckck"]
"our"? Although AoPS is in America, [1]there are many users that are non-american that have different perspectives in writing math. [2]You can't just bash how they do it seeing as they have been writing math like that for almost their entire lives. [3]Look at some other forums and you'll see that other people use the period as multiplication signs. [4]I don't see you bashing them about it.[/quote]

[1] Stop changing the subject by referring to non-Americans
when it is about *conforming* to our ways.
[2] That is immaterial as to how *they* have been doing it,
and the longer they have been doing doesn't add anything
to it being immaterial.
[3] No, I don't have to look at other forums.  Two wrongs,
or multiple wrongs as you have indicated, don't make it right.
[4] There is no logic there.  There are a "thousand fires" of
ignorance out there, and I can only put out a relative selective
few.   You "don't see?"  I have made corrections before on
certain ones, but you could not have noticed all of the posts in
all the posts in all of the forums to detect them.  And you not
seeing my posts is *your* own limitation.  It is never an argument
that if you don't see something (or claim to have not seen
something) that it hasn't been done and/or it isn't being done.

I advise you to not try to continue to defend your off-topic
point.  Your immediate history of emotionalism keeps you from
making certain sufficient logical statements here where it matters.

--------------------------------------------------------------------------------

Just for the sake of an illustration:

1) What is another equivalent expression for 2.3^5.4?

Is it $ (2)[3^{(5.4)}]?$

Or is it $ [(2.3)^5](4)?$

Or is it $ (2)[3^5](4)?$

Or is it $ (2)[3^{[(5)(4)]}]?$

Or is it $ [(2)(3)]^{5.4}?$

Or is it $ [(2)(3)]^{[(5)(4)]}?$

Or is it $ [(2)(3)]^5(4)?$

Or is it just $ (2.3)^{(5.4)}?$

[quote="Arrange your tan"]1) What is another equivalent expression for 2.3^5.4?[/quote]The difference between between your example and his is that x.y doesn't exist as a number. The only meaning it can have is $ x*y$. On the other hand, 2.3^5.4 has multiple meanings.

Also, can you stop attacking people personally? If you want to continue the debate, it would be best if you don't take any cheap shots.

[quote="JBL"]I strongly recommend just ignoring its posts ... becomes aggresive and repetitive ... Eschew![/quote]  Should have listened to me ;)

(Also, how come no one has pointed out yet that this website is the union of an American and European website, i.e., even if it ever makes sense to think of websites as being in a country, it makes no sense in the case of [i]this particular[/i] website.  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=326161]Here's[/url] the mathlinks.ro version of this thread, for example.)

[quote="quattayangel"][quote="pco"][quote="epsilon07"]x.y.z = 1

$ x \plus{} \frac {1}{z} \equal{} 5$ and $ y \plus{} \frac {1}{z} \equal{} 29$

find $ z \plus{} \frac {1}{y}$[/quote]
$ z \equal{} \frac 1{xy}$ and so $ x \plus{} xy \equal{} 5$ and $ y \plus{} xy \equal{} 29$

So $ y \equal{} \frac {5 \minus{} x}x$ and $ z \equal{} \frac 1{5 \minus{} x}$ and $ \frac {(5 \minus{} x)(x \plus{} 1)}x \equal{} 29$

So $ x^2 \plus{} 25x \minus{} 5 \equal{} 0$ and $ x \equal{} \frac { \minus{} 25 \plus{} \epsilon\sqrt {645}}2$ where $ \epsilon \equal{} \pm 1$

And so $ z \plus{} \frac 1y \equal{} \frac {x \plus{} 1}{5 \minus{} x} \equal{} \frac { \minus{} 40\pm 3\sqrt {645}}{145}$[/quote]

There is something that doesn't make sense to my mind. Can you explain more about how you got $ z \plus{} \frac 1y \equal{} \frac {x \plus{} 1}{5 \minus{} x} \equal{} \frac { \minus{} 40\pm 3\sqrt {645}}{145}$[/quote]

I proved two lines above that $ y \equal{} \frac {5 \minus{} x}x$  and $ z \equal{} \frac 1{5 \minus{} x}$ and so $ z \plus{} \frac 1y \equal{}\frac 1{5 \minus{} x}\plus{}\frac x{5 \minus{} x}$ $ \equal{} \frac {x \plus{} 1}{5 \minus{} x}$

And since I proved too that $ x \equal{} \frac { \minus{} 25 \plus{} \epsilon\sqrt {645}}2$ where $ \epsilon \equal{} \pm 1$, you just have to plug this value in $ \frac {x \plus{} 1}{5 \minus{} x}$ in order to get the result $ \frac { \minus{} 40\pm 3\sqrt {645}}{145}$

For what step do you need more precision ?

What is the deal of using different kind of way to express multiplication? Move on people
Can any here answer my question that I post before?

[quote="quattayangel"]What is the deal of using different kind of way to express multiplication? Move on people[/quote]
No interest. Just ignore.

[quote="quattayangel"]Can any here answer my question that I post before?[/quote]
I did


algebra
binomial theorem
binomial coefficients
superior algebra
superior algebra unsolved

Let R be a commutative ring with characteristic p, for prime p.Then show that $ (a\plus{}b)^p\equal{}a^p\plus{}b^p$ for $ a, b \in R$.

[quote="HilbertThm90"]... notice something about the coefficients.[/quote]

[hide][b]If[/b] $ k \in \{1, 2 \ldots, p \minus{} 1\}$ [b]then[/b] $ p |\binom{p}{k}$. [b]Since[/b] $ \mathrm{char(R)} \equal{} p$[b], this  implies in turn that[/b] $ \binom{p}{k} \equal{} 0$ [b]whenever that[/b] $ k \in \{1, 2 \ldots, p \minus{} 1\}.$  [b]Done.[/b][/hide]

[quote="HilbertThm90"]Eh. You don't even need to check small cases. Write it out with the binomial theorem and notice something about the coefficients.[/quote]
Obviously the question is pretty easy if you already know the answer. :P  But it's possible to make the conjecture about prime divisors of binomial coefficients through just a little pencil and paper work. (the difficulty of proving it, on the other hand, depends on the OP's knowledge and experience)


trigonometry
calculus
calculus computations

Please help, I got stuck in this problem of my homework.
[img]http://img204.imageshack.us/img204/9561/001cp9.jpg[/img]

Find the deviation of the direction (relative to the forward direction) of the resultant force which these engines exert on the rocket.

$ f_x(\theta) \equal{} 760 \plus{} 473\cos \theta$
$ f_y(\theta) \equal{} 473\sin \theta$

angle of deviation will be $ \tan^{ \minus{} 1} \left( \frac {473\sin \theta}{760 \plus{} 473\cos \theta}\right)$


Okay, here is the update. Use d, f, h, j, space, up, down, left, right


group theory
abstract algebra
number theory
greatest common divisor
Ring Theory
superior algebra
superior algebra unsolved

Hi all,

  I came across the following proof for Chebotarev Density Theorem for $ K(\zeta)|K$, where $ K$ is a number field, and $ \zeta$ is an $ m$-th root of unity, $ A$ the ring of integers of K. But I have some issues in parts of the proof for which I need some help.

  It is clear that $ L \equal{}K(\zeta)$ is a finite Galois extension of $ K$ and $ Gal (L|K)$ is isomorphic to $ Gal (Q(w)|Q(\zeta) \cap K)$ which is a subgroup of $ Gal(Q(\zeta)|Q)$. By restriction, for every $ \sigma \in Gal (L|K)$, there exists $ r$, $ 1 \le r < m$, with gcd $ (r, m) \equal{} 1$, such that $ \sigma$ is identified with $ \sigma_r \in Gal(Q(\zeta)|Q)$, where $ \sigma_r$ is the automorphism sending $ \zeta$ to its $ r$-th power

  Let $ A_r$ = {$ P$ prime ideal in $ K$ such that the Frobenius automorphism $ (\frac {L|K}{P})\equal{} \sigma_r$} and $ P_f$ the set of prime ideals $ P$ in $ K$ with inertial degree $ f$ over $ Q$.

  The author claims that $ P \in A_r \cap P_1$ iff $ N(P) \equiv r (mod m)$ and $ P$ has inertial degree 1 over $ Q$.

  While I certainly believe the claim is true, I guess the proof supplied by the author is not correct. Does anyone know the correct proof of the claim above?

PS If $ P \in A_r \cap P_1$ then on the one hand $ \frac {L|K}{P} (\zeta) \equiv  \zeta^p  (mod P)$ [I think this is wrong here, it should be $ Q$ for a prime in $ L$ lying above $ P$] where $ p \equal{} N(P)$. On the other hand, $ (\frac {L|K}{P} (\zeta) \equal{} \zeta^r$, so $ \zeta^r \equal{} \zeta^p (mod P)$ [again I believe it is $ mod Q$]. If $ p \not\equiv r (mod m)$, then $ P$ [or is it $ Q$] divides $ A(1 \minus{} \zeta^{p\minus{}r}) \equal{} A(1\minus{} \zeta)$ which in turn divides $ Am$. This is a contradiction as $ P$ is not ramified.

I think some parts of the proof above are incorrect and there are also gaps.

Can anyone help?


analytic geometry
graphing lines
slope
calculus

the [b]equilibrium point[/b] of a particle is the position at which the net force acting on it is 0 right?

if so then on a PE (of a an isolated system in which the acting force is conservative) vs Position (of a particle in that system) curve would this point be [b]the x value of the coordinate at which the slope of the curve is 0[/b] because this means the PE of the system has no rate of change and that means no energy is being transferred from the system's PE to the particle's KE or vice versa, which means there is no conservative force doing work on the particle to account for either of these energy transfers, which means the net force of the particle is zero and it is at equilibrium.

is this reasoning correct?

Right.

In terms of calculus, $ \vec{F} \equal{} \minus{} \frac {\partial U}{\partial x} \mathbf{\hat{x}}$, or in the general case, $ \vec{F} \equal{} \minus{} \nabla U$, where $ \nabla U$ means the [url=http://en.wikipedia.org/wiki/Gradient]gradient[/url] of $ U$.

This assumes that all energy is either stored as spatially dependent potential energy or as kinetic energy. In the general case, the relation is just $ \nabla \mathcal{H} \equal{} 0$, where $ \mathcal{H}$ is the total energy.


calculus
real analysis
real analysis unsolved

:(  :(  :(  :(  :(

consider a population taht grows according to a linear growth model.  the initial population is Po=8 and the population in the 10th generation is P10=38.

a) find the common difference d.

b) find P50

c) give an explicit description of the population sequence

how do I solve this problem

This isn't calculus and doesn't belong in this part of the forum. Your previous post of the same problem has been moved to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64748]here,[/url] which is where you're more likely to have it answered.


[quote="sinajackson"]For positive reals $ a,b,c\in\mathbb{R}^{\plus{}}$ where $ abc \equal{} 1$ Prove that:

$ 2\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\minus{}\left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)\leq3$[/quote]
It's very known.
By Schur and Muirhead we obtain: $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}3 \equal{}$
$ \equal{} a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}3\sqrt [3]{a^{2}b^{2}c^{2}}\geq\sum_{cyc}(\sqrt [3]{a^{4}b^{2}}\plus{}\sqrt [3]{a^{4}c^{2}})\geq2(ab\plus{}ac\plus{}bc).$


search
number theory unsolved
number theory

Find all odd integers $n > 1$ such that, whenever $a$ and $b$ are coprime
divisors of $n$, the number $a+b-1$ is also a divisor of $n$.

Can you say more clearly? I think is problem is difficult, and I have discovered no clue.

Searching "coprime divisors" in this subforum works very well, even the title agrees with yours...


My friend bet me that this problem couldn't be solved. It's easier to try and solve this in Microsoft Excel as it recalculates formulas for you. Here's what to post in Excel:

A1: ?
A2: ?
A3: ?
A4: ?
A5: ?
B1: =SUM(20*A1)-A2-A3-A4-A5
B2: =SUM(10*A2)-A1-A3-A4-A5
B3: =SUM(5*A3)-A1-A2-A4-A5
B4: =SUM(2*A4)-A1-A2-A3-A5
B5: =SUM(1*A5)-A1-A2-A3-A4

Now, what would the values of A1 to A5 have to be to make B1 to B5 result in whole numbers (ie. numbers greater than or equal to 0)? The only condition is that A1 to A5 had to be greater than or equal to 1. He said that it's a circular equation and that the values couldn't ALL be whole numbers without it resulting in at least ONE negative number (With the exception of 0). I disagree, I'm sure there MUST be a solution. Help!

From the requirement that each of the $b's$ be non-negative, we get the following inequalities in the $a's$:

$a_5\ge a_1+a_2+a_3+a_4$
$2a_4\ge a_1+a_2+a_3+a_5\ge 2(a_1+a_2+a_3)+a_4\Rightarrow a_4\ge 2(a_1+a_2+a_3)$
$5a_3\ge a_1+a_2+a_4+a_5\ge 2(a_1+a_2+a_4)+a_3\ge 6(a_1+a_2)+5a_3\Rightarrow 0\ge6(a_1+a_2)$

Now since the $a's$ are non-negative also, we get $a_1=a_2=0$.

But now $b_1=20a_1-a_2-a_3-a_4-a_5\le0$ with equality iff all of the $a_i$ are zero, and hence all of the $b_i$ are zero, so the only way that each of the $a_i$ and the $b_i$ can be non-negative is if they are all zero.

Also, this contradicts the stronger assumption that all of the $b_i$ could be positive with each of the $a_i$ not less than one.

It's come to my attention that I may have worded my terminology incorrectly as I have not dealt with these terms for a number of years. What I meant to say about the conditions is this:

The A values must be natural numbers (1, 2, 3...), and the resulting equations, the B values, must be whole numbers (0, 1, 2...).

I don't know if that changes the results of the first reply.


logarithms
limit
real analysis
real analysis unsolved

Intuitively the second one is $\frac{1}{2}$, the third one is 2.

[quote="uldis"]first can be solved without taking logarithms, namely, it is a simple geometric progression[/quote]

That's true. Multiplying the eqation by $1-\frac{1}{2^{2}}$ gives the answer. :D

All three of these telescope- you can find the partial sums or products exactly.


MATHCOUNTS
Support
AMC
AIME
greatest common divisor

I also got 440, but I think when it says equivalent common fraction, it means its most simplified form

It was 341, because it is smaller. I had 440 :mad:

[quote="footballrocks41237"]If I remember correctly, unless otherwise specified, Common Fraction [i]always[/i] means a simplified fraction on MathCounts tests.[/quote]yeah...22/40 can still be simplified further, making it an uncommon fraction

lol @ uncommon fraction.

Is that actually the right term? I thought it was like... unsimplified or unreduced or w/e. No clue.

[quote="binonunquineist"]Oh, haha, Bobby and I both put 440 so we decided to protest and I think we got it approved  :lol: so they counted both 341 and 440 right in Texas[/quote]

So what did you get at states Kevin?

lol, i don't get how saying 440 is the answer is a valid argument...

jorian

It isn't, for in the problem it clearly said that it has to be simplified  I thought.

They also in the example said 154/253=14/23, which is simplified completely, so I don't know.

[quote]Monica reduced 154/253 to a common fraction by canceling the middle 5s to get 14/23.[/quote]
We can infer by the wording of the question that 154/253 is NOT intended to be defined as a common fraction, but that 14/23 IS intended to be defined as a common fraction.  Furthermore, this implied definition in the question is consistent with the explicit definition printed on the "Forms of Answers", which all State-level competitors should have reviewed, and which may have been printed on the back of the Sprint Round, itself.

[quote="Forms of Answers"]A "common fraction" is to be considered a fraction in the form $\pm\frac{a}{b},$ where $a$ and $b$ are natural numbers and $\text{gcf}(a,b)=1.$[/quote]

[quote="binonunquineist"]Oh, haha, Bobby and I both put 440 so we decided to protest and I think we got it approved  :lol: so they counted both 341 and 440 right in Texas[/quote]
The judging panel at each local or state competition does have the final say in what is accepted at that competition.  But, by the definition implied in the question, and the definition printed in the "Forms of Answers", I don't think 22/40 qualifies as a common fraction.

The posted answer key doesn't accept 440 as an answer.  Florida and Kentucky (perhaps others) took the contest after Texas.  Competitors from those states haven't reported that they accepted 440 as an answer.

If Texas actually accepted 440 as an answer, I think the judges made a mistake.  But, of course, I don't know what argument the coach used when submitting the protest.  Your judges *are* the final arbiters in your jurisdiction.

The Mathcounts Handbook has a vocabulary list of over 200 mathematical terms.  It defines none of them.  It is left to students and teachers to use reference books, (dictionaries, math textbooks, etc.), for definitions of these terms.

All the refernces I've checked define a common fraction as an expression using a horizontal or diagonal bar to indicate the division of an integer by an integer.  None of them indicate that the fraction is to be simplified or in lowest terms.  Thus 154/253 and 22/40 are both common fractions.

The Mathcounts Handbook contains a page on Forms of Answers which also appeared on the back cover of the Sprint Round.  These are exactly what the title says.  They are for the benefit of graders working in the scoring room.  They are not definitions of mathematical terminology.  They are rules for how certain answers are to be written.

Problem 25 asked for the greatest possible value of x.  The answer is an integer so the answer form for a common fraction doesn't apply.  To find an "implied" definition for a common fraction which removes Mathcounts from the mainstream of mathematics and then try to support it with a non-definition that doesn't apply isn't logical.

I'm happy to see that several students found 440 as the  answer.  It's sad that they didn't trust their own reasoning and failed to appeal a poorly written problem.  The authors could easily have worded the problem in such a way that 341 would have been the answer.  They failed to do so.  The student's role is to answer the problem as it is written.  It shouldn't be a test of the student's ability to be a mind reader.

While students at the national competition are given opportunities to protest answers after each round, students in many state and chapter MC competitions have no option to protest.  Texas is probably one of a very small number of states that permit either coaches or students to see the answers on the day of the contest.  North Carolina does not provide mechanisms for protest.

Most state coordinators are unwilling to chance compromising the integrity of the contest in other states by allowing protests - even if the answers are released under highly supervised conditions.

Does it take mind reading ability to understand what a common fraction is?

Before the contest, moderators are supposed to read general instructions. I definitely remember a SPECIFIC definition of common fraction, which included the words "simplified".

You know, it would have just been easier for the test-writers to not allow such a predicament to arise in the first place.  They could've just picked some other numbers that didn't create such a controversy.

Easy to say, but it is hard to catch every single mistake.  I think Mathcounts does a good job of catching almost all of them.

to Zmastr:

Yes, they certainly could have done that.  It would have been nice if they had.

to Fanatic:

I believe you recall the reading of a required "answer form for a common fraction", not the definition of a "common fraction".  There is a difference.

Mathcounts wants it simplified for ease of grading.  That doesn't mean that it ceases to be a common fraction when it isn't simplified.  It just isn't what they want to see in the grading room.

The AIME avoids this kind of confusion by writing conditions about fractions and radicals into each individual problem.[/img]

Here is the quote again from the MathCounts General Instructions:

[quote="MathCounts"]A "common fraction" is to be considered a fraction in the form $\pm \frac{a}{b}$, where $a$ and $b$ are natural numbers, and $\mathrm{GCF}(a, b) = 1$.[/quote]

Here MathCounts is clearly defining the term "common fraction", not merely "the required answer form for a common fraction".  According to MathCounts, $\frac{2}{4}$ is not a common fraction.  That's one of the first things I explain to my students.

You might not like the MathCounts definition, but that's a different story.

Is it your position that, according to MathCounts, $\frac{2}{4}$ is a common fraction?

Ravi B

I have read that quote many times.  It is taken from a page titled "Forms of Answers".

I note that you have also chosen to omit part of the quote.  It includes the line, "In some cases the term "common fraction" is to be considered a fraction of the form A/B where A and B are algebraic expressions and A and B do not share a common factor."

It's my position that when a Mathcounts problem ends with a sentence that says, "Express your answer as a common fraction.", you should refer to this RULE in writing your answer.

Mathcounts knows that a fraction containing algebraic expressions is not considered to be a "common fraction" by the mathematics community.  Why is it included in this rule?  Because Mathcounts also wants these fractions to be simplified.  It allows the problem writers to end problems with, "Express your answer as a common fraction.", thus putting this rule into effect and avoiding the longer and more detailed statements found on the AIME.

There are many rules on that page.  There's a rule about the form of an answer that involves a radical.  It is not a definition of a radical; just another rule about an answer form.

As I noted in my first post on this topic the Mathcounts Handbook includes a list of over 200 vocabulary words and defines none of them.  Common fraction happens to be in that list.  Based on the nonexisting Mathcounts' definition I wouldn't have a clue about 2/4.  The mathematical community does consider it to be a common fraction.

Your assumption that I might not like the Mathcounts' definition is silly.  Mathcounts would have to write a definition before I could decide to dislike it.

Our disagreement hinges on whether the quoted sentence is a definition or not.  In my opinion, the structure of the sentence [A "common fraction" is considered to be...] is precisely that of a definition.  I'll let others decide whether the sentence is a definition as I assert.

That the definition appears on a page called "Form of Answers" is irrelevant; what is relevant is what the sentence says.  MathCounts has used the term "common fractions" beyond just asking for the form of an answer.  For example, here is a problem from a recent handbook:

[b]2002-03 Handbook, Warm Up 5, Question 8:[/b] $P$ and $Q$ are whole numbers such that $0 < P < 10$ and $0 < Q < 10$.  How many common fractions $\frac{P}{Q}$ exist if $\frac{1}{2}< \frac{P}{Q}< 1$?

The answer provided is 13, which means that MathCounts excludes the fractions $\frac{4}{6}$, $\frac{6}{9}$, and $\frac{6}{8}$.

You assert that the math community has an agreed-upon definition of common fraction that counts $\frac{2}{4}$ as a common fraction.  I was a mathematician for many years, and we never used the term common fraction.  If you asked 100 mathematicians to define common fraction, I doubt a third of them would come up with your definition.  Because common fraction is not an agreed-upon term, MathCounts is free to define it as desired.  I'm sure there are other sources that use your definition, but on a MathCounts test, the MathCounts definition applies.

You are right that the Form of Answers page does not define the word radical, but instead states rules about what form of radicals is acceptable.  My point was that common fraction is treated differently from other terms on that page; it is one of the few terms actually defined, based on the structure of the relevant sentences.  That's why, when MathCounts asks for a radical, it specifies "Express your answer in simplest radical form", but when it asks for a fraction, it merely says "Express your answer as a common fraction".  In the latter case, MathCounts doesn't say "simplified common fraction", because, by its definition, common fractions are in simplified form.

You point out that MathCounts page provides two sentences for common fractions, one for fractions of integers, the other for fractions of algebraic expressions.  Those are two separate definitions, overloading the term "common fraction".  It's similar to how mathematicians define the term "greatest common divisor" twice: once for integers, and later for polynomials.  That there are two definitions doesn't mean that neither counts as a definition; the definition that applies depends on context.

Those of you who have watched this Forum over the years know that I believe when a question is worded ambiguously, the ambiguity should be resolved in favor of the students.

I do not see *any* ambiguity in this question.  From the information I have seen, Texas is the only State in which an alternate answer was accepted.  As I said in my previous post, "If Texas actually accepted 440 as an answer, I think the judges made a mistake."  Their ruling is final.  If some of you benefited from their ruling, be thankful for the gift.

Ravi has very eloquently explained his position.  I am in complete agreement with Ravi's analysis.

Alfalfa:  Your definition is inconsistent with the wording of the problem.  [See my previous post.]  I cannot believe that any seasoned MATHCOUNTS Coach would not drill their State Competitors with the MATHCOUNTS definition of "common fraction" or would teach their students a definition that is inconsistent with the MATHCOUNTS definition!

On the other hand, there are valid reasons why a coach might have challenged this question.  (Time constraints are near the top of the list.)  I don't question the motive of the coach or the decision of the judges.  In the heat of the competition, decisions must be made quickly.

Not to sound rude, but I don't think we should just debate over this controversial issue. I understand that in this scenario, we are debating over a MathCounts issue, which makes it more understandable. However, we should focus more on suitable answers fundamentally pertaining to the [b]Actual problem![/b]

fanatic, you recall a specific definition that claimed that common fractions, if stated as such in a problem, should "always be simplified"?  REALLY? I looked and my coach looked and neither of us saw. I agee wholeheartedly with alfalfa's reasoning - 440 was the only correct answer based on the wording of the problem as far as I could tell. Please tell me and others on this site where you saw it said "simplified"?

[quote="Zmastr"]25. Monica reduced 154/253  to a common fraction by canceling the middle 5s to get 14/23 . This answer is correct, but the procedureis incorrect. Kia’s example worked withMonica’s incorrect procedure, too; the fraction 242/x also reduced to its equivalent common fraction when the identical middle digits of the two three-digit integers were canceled out. What is the greatest possible value of x?

I got 440.  If you reduce 242/440, the answer is 11/20, and by taking out the middle 4's, you get 22/40, which is also 11/20.  How is this not the answer?[/quote]
It says equivalent common fraction, meaning that $ \frac{22}{x}$ MUST be in simplest terms.

I like to think that you're a genius and that I suck at mathcounts beyond belief (in comparison to other contests).

[quote="stevenmeow"][quote="binonunquineist"]Oh, haha, Bobby and I both put 440 so we decided to protest and I think we got it approved  :lol: so they counted both 341 and 440 right in Texas[/quote]
our finest moment  :D  :D  :D
but anyway, we led
46-binonunquineist
45-stevenmeow
40-umm shuld i tell?
40-not sure of username

so i like to think our luck would not make too much difference[/quote]

True, but it would have for 5th-10th written. (This was last year, right?) I was 5th and, according to an early estimate from the 10th place person, we got the same score of 38. It probably helped me that I got 7/10 of the last 10 on sprint (including #30!) right.

I am envious...but then again I bombed the cd, so it's really my own fault.

EDIT: The estimate is most likely off. I'm talking about after-round chat, where he thought he had a 24/14. I suppose I kinda bombed the written too.

[quote="stevenmeow"][quote="binonunquineist"]Oh, haha, Bobby and I both put 440 so we decided to protest and I think we got it approved  :lol: so they counted both 341 and 440 right in Texas[/quote]
our finest moment  :D  :D  :D
but anyway, we led
46-binonunquineist
45-stevenmeow
40-umm shuld i tell?
40-not sure of username

so i like to think our luck would not make too much difference[/quote]

maybe it would not be changing the top two students who went to nationals , but it probably affected the other two. It even affects who participated in the countdown round. but of course this is all in the past. we need to look this year now. Good luck to everyone in the state competition


Find a matrix whose determinant, when set equal to zero, represents the equation of a circle passing through the points $(-3, 1)$, $(2, 4)$, $(5,-2)$.

I'm not sure if PenguinIntegral will like me doing this but could someone please post a solution to this problem? Thanks in advance :)

[quote="eryaman"]Hmm, wasn't this problem in AoPS Volume 2 in the Determinents/Cross Products chapter or something?  It looks really familiar.[/quote]
Yes, it was.

looks like a good question. must have some deep ties between geometry and matrices

for some reason, i felt like writing up a solution in my blog...look here:

http://www.artofproblemsolving.com/Forum/weblog.php?w=313

[quote="236factorial"][quote="eryaman"]Hmm, wasn't this problem in AoPS Volume 2 in the Determinents/Cross Products chapter or something?  It looks really familiar.[/quote]

I've never seen it in there, if I remember correctly.[/quote]
I missed it the first time I looked too (:blush:). It's in the ch.11: Cross products/determinants, pg.161, question no. 172.
Thanks for the solution Altheman :)


The measure of angle NOA is also 78 since it is a vertical angle.

So the intercepted arc NA has measure of 78.

By the inscribed angle theorem. The measure of angle NGA is half the measure of the intercepted arc NA.

So, the measure of angle NGA is 39.


graph theory
combinatorics unsolved
combinatorics

In a group of 2n+1 people; each person has minimally n friends.

Proof someone exists that all the others are his friends.

It seemed to be a wrong problem, I think.
Let n=2 and they form a Hamilton ring exactly.

this is russia 2001 i think... the problem states that in a group of $2n+1$ for every group of $n$ persons there is one that knows all of them. then there exists a person that knows everybody in the group.
prove that the graph has a triangle(not hard). then from this it easy to construct a $K_{n+1}$. So you are left with a complete graph in $n+1$ vertices and other $n$ vertices. There exists a vertex $v$ adjacent to all the other $n$ vertices, and this vertex has to be in the $K_{n+1}$. So $v$ is adjacent to $2n$ vertices.

you see suppose that when n=2 we can have  C5 which implies that there is nobody who wants to know all the others !!  another point is that the question which is stated in Russia 2001 is different to this question i think !!

[quote="manuel"]oops, i thought he meant the russian problem  :blush:[/quote]
what do you mean ?

[quote="manuel"]i  thought he wanted to post the russian problem and he wrote it incorrectly...[/quote]
but i think instead of $n$ he should have written $n+1$  :huh:

[quote="grobber"]It still doesn't work. We can easily find examples of graphs with $7$ vertices, for example, in which each vertex has degree $\ge 4$, but no vertex has degree $6$. I really think kharazmi wanted to post the Russian problem that Manuel solved.[/quote]
yes you are right !!  :blush:  :P


Let $ a,\ b,\ c$ be positive real numbers. Prove the following inequalities.

(i) $ \boxed{(8a \plus{} 5b \plus{} 5c)^3 \plus{} (5a \plus{} 8b \plus{} 5c)^3 \plus{} (5a \plus{} 5b \plus{} 8c)^3\geq 1944(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$

(ii) $ \boxed{(8a \plus{} 5b \plus{} 5c)(5a \plus{} 8b \plus{} 5c)(5a \plus{} 5b \plus{} 8c)\geq 972(a \plus{} b \plus{} c)\sqrt [3]{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers. Prove the following inequalities.

(i) $ \boxed{(8a \plus{} 5b \plus{} 5c)^3 \plus{} (5a \plus{} 8b \plus{} 5c)^3 \plus{} (5a \plus{} 5b \plus{} 8c)^3\geq 1944(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$
[/quote]
It's very easy, kunny! Even I have proved it. :lol:
$ \sum_{cyc}(8a \plus{} 5b \plus{} 5c)^3\geq\frac{1}{3}\sum_{cyc}(8a \plus{} 5b \plus{} 5c)\sum_{cyc}(8a \plus{} 5b \plus{} 5c)^2\equal{}$
$ \equal{}\sum_{cyc}6a\sum_{cyc}(114a^2\plus{}210ab)\geq\sum_{cyc}6a\sum_{cyc}324ab\equal{}1944(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc).$

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers. Prove the following inequalities.

(i) $ \boxed{(8a \plus{} 5b \plus{} 5c)^3 \plus{} (5a \plus{} 8b \plus{} 5c)^3 \plus{} (5a \plus{} 5b \plus{} 8c)^3\geq 1944(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$[/quote]
by holder we have
$ (8a \plus{} 5b \plus{} 5c)^3 \plus{} (5a \plus{} 8b \plus{} 5c)^3 \plus{} (5a \plus{} 5b \plus{} 8c)^3\ge 3(\frac{18(a\plus{}b\plus{}c)}{3})^3\equal{}648(a\plus{}b\plus{}c)^3$

we need prove that $ 648(a\plus{}b\plus{}c)^3\ge 1944(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$

$ \Leftrightarrow (a\plus{}b\plus{}c)^2\ge 3(ab\plus{}bc\plus{}ca)$ which is true :wink:

[quote="arqady"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers. Prove the following inequalities.

(i) $ \boxed{(8a \plus{} 5b \plus{} 5c)^3 \plus{} (5a \plus{} 8b \plus{} 5c)^3 \plus{} (5a \plus{} 5b \plus{} 8c)^3\geq 1944(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$
[/quote]
It's very easy, kunny! Even I have proved it. :lol:
$ \sum_{cyc}(8a \plus{} 5b \plus{} 5c)^3\geq\frac {1}{3}\sum_{cyc}(8a \plus{} 5b \plus{} 5c)\sum_{cyc}(8a \plus{} 5b \plus{} 5c)^2 \equal{}$
$ \equal{} \sum_{cyc}6a\sum_{cyc}(114a^2 \plus{} 210ab)\geq\sum_{cyc}6a\sum_{cyc}324ab \equal{} 1944(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc).$[/quote]

Sorry, I can't understand your solution.

To bunhiacovski:

I don't know Holder's inequality. :P

[quote="kunny"]

Sorry, I can't understand your solution.

[/quote]
For non-negative $ x,$ $ y$ and $ z$ the following inequality holds:
$ x^3\plus{}y^3\plus{}z^3\geq\frac{1}{3}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2)$ and let $ x\equal{}8a\plus{}5b\plus{}5c,$ $ y\equal{}...$

[quote="kunny"]Is that Chevyshev's inequality?[/quote]
It's one of the options.
$ (8a \plus{} 5b \plus{} 5c)(5a \plus{} 8b \plus{} 5c)(5a \plus{} 5b \plus{} 8c)\equal{}\sum_{sym}(100a^3\plus{}645a^2b\plus{}227abc)\geq$
$ \geq648(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)\geq 972(a \plus{} b \plus{} c)\sqrt [3]{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}.$

[quote="kunny"]
To bunhiacovski:

I don't know Holder's inequality. :P[/quote]
$ (a\plus{}b\plus{}c)(x\plus{}y\plus{}z)(m\plus{}n\plus{}p)\ge (\sqrt[3]{axm}\plus{}\sqrt[3]{byn}\plus{}\sqrt[3]{czp})^3$

$ (1 \plus{} 1 \plus{} 1)(1 \plus{} 1 \plus{} 1)(\sum (8a \plus{} 5b \plus{} 5c)^3)\ge (\sum(8a \plus{} 5b \plus{} 5c))^3$

$ \Leftrightarrow 9(\sum (8a \plus{} 5b \plus{} 5c)^3)\ge [18(a \plus{} b \plus{} c)]^3 \equal{} 5832(a \plus{} b \plus{} c)^3$

$ \Leftrightarrow \sum (8a \plus{} 5b \plus{} 5c)^3\ge 648(a \plus{} b \plus{} c)^3$

[quote="arqady"][quote="kunny"]Is that Chevyshev's inequality?[/quote]
It's one of the options.
$ (8a \plus{} 5b \plus{} 5c)(5a \plus{} 8b \plus{} 5c)(5a \plus{} 5b \plus{} 8c) \equal{} \sum_{sym}(100a^3 \plus{} 645a^2b \plus{} 227abc)\geq$
$ \geq648(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)\geq 972(a \plus{} b \plus{} c)\sqrt [3]{abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}.$[/quote]

That's all Greek to me!


probability
calculus
geometry
integration
function
probability and stats

I'm just curious. I had this problem where I have the pdf for X.

f(x)= 6x(1-x) for 0<x<1  and 0 otherwise.

Y=X^3
Find the pdf of Y.
I know that you find the cdf of why and differentiate it etc. etc.

I'm just wondering and I need someone to correct me. The pdf of Y is supposed to tell you the probability that Y=y for some. Why is it not true that P(Y=y)=P(X=y^1/3), in other words, why can't I just plug y^1/3 into the pdf of X?

Thanks!

For a continuous distribution, $ P(X \equal{} x) \equal{} 0$ for all $ x.$ So the pdf is [i]not[/i] telling you what $ P(X \equal{} x)$ is.

Go back to your first calculus class and all those problems you worked finding area, and volume (and work, and all the other things in the "applications of integration" chapter). You were always taking slices of things, and saying things like this: at height $ y$, this object has cross-sectional area $ A(y),$ so we think about a thin slab of thickness $ dy$ which then has volume $ dV \equal{} A(y)dy.$ You can think of that as the volume between $ y$ and $ y \plus{} dy.$ Then multiply by a density, multiply by the height lifted to get work, and so on.

That's a pretty informal argumentation style - we'd have to do certain other things to make it rigorous. But let's stay with that argumentation style.

In that same informal sense, what does the pdf $ f(x)$ mean? It means that $ P(x\le X\le x \plus{} dx) \equal{} f(x)\,dx.$

Now, we have $ Y,$ which is a certain (one-to-one differentiable) function of $ X,$ and we want to find the pdf of $ Y.$ Thinking of that in terms of probabilities, one infinitesimal probability should equal another infinitesimal probability, so we set
\[ f_Y(y)\,dy \equal{} f_X(x)\,dx\]
We can then solve that for $ f_Y(y) \equal{} f_X(x)\cdot\frac {dx}{dy}.$

So for $ Y \equal{} X^3,$ we have $ x \equal{} y^{1/3}$ and $ \frac {dx}{dy} \equal{} \frac13y^{ \minus{} 2/3}.$

We should then have that $ f_Y(y) \equal{} 6y^{1/3}(1 \minus{} y^{1/3})\frac13y^{ \minus{} 2/3} \equal{} 2(y^{ \minus{} 1/3} \minus{} 1)$ for $ 0 < y < 1$ and zero otherwise.

=====

Compared to finding the cdf, transforming the cdf, and differentiating, this method is significantly shorter but (for most students) significantly more confusing. If you're going to be confused by it, going through the cdf is safer. Try it that way here to confirm this answer.

What you're missing in your comment is the change of variables adjustment given by $ \frac {dx}{dy}.$ In higher dimensions, that would be a Jacobian.


calculus
geometry
rotation
geometric transformation

<NO CALCULUS, USE ONLY GEOMETRY, BECAUSE LOOKING FOR DIFFERENT ANSWER>

  There are two overlapping circles A and B, such that if one sixth is drawn in circle A,

and rotated around 90 degrees, it fits exactly over that of circle B.  If the area of circle

A is exactly that of circle B, and the area is $\frac{x^2}{4}$, the point of rotation

being labeled C, the other vertex of the sixth of circle A being D, that of B being E,

FIND THE AREA OF THE FIGURE OBTAINED BY THE BOUNDARIES OF ARC DC, ARC CE,

AND THE STARIGHT LINE DE.

[quote="mgao"] one sixth is drawn in circle A, [/quote]

One-sixth of what?   :huh:

The way I read the problem, I think it looks something like this:
[img]http://i9.photobucket.com/albums/a70/littlekasper/circles.jpg[/img]

[hide]First, let the center of the top circle be O.  Draw triangle ECD, and that is equal to the area of the question since the areas of the lunes(?) of arc DC and arc CD are equal.  Also, triangle ECD is a right isoceles triangle because EC=DC and DCE=90.  But ECO is a equilateral triangle, and CO = $\sqrt{\frac{x^2}{4}}/\pi=\frac{x}{2\pi}$  So the area is ${\frac{x^2}{8\pi^2}}$   ?[/hide]

What program do you use to make those drawings?  If it's paint, you will make me feel dumb...

Use Autodesk Inventor!  :lol:
[size=59][color=white]Serious overkill.[/color] [/size]   :D


geometry
incenter
trigonometry
angle bisector

The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P,$  and its radius is 21.  Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.

[hide=" Dios mio! Most straightforward #12 ever....... "]Let the point of tangency of the circle with $AC$ be denoted as $Q$, and the point of tangency of the circle with $AC$ be denoted as $R$. So $AP=AQ=23$, $BP=BR=27$, and $CQ=CR=x$, where $x$ is some variable we are trying to solve for.

We will utilize the equation $rs=[ABC]$, $r$=radius of incircle, $s$=semiperimeter, and $[ABC]$= area of triangle $ABC$.

We have $r=21$, $s=50+x$. Now $[ABC]=\sqrt{(50+x)(x)(27)(23)}$ by our good friend Heron.

$rs=[ABC]$
$\Rightarrow (21)(50+x)=\sqrt{(50+x)(x)(27)(23)}$
$\Rightarrow (441)(50+x)^2=(50+x)(x)(27)(23)$
$\Rightarrow (441)(50+x)=621x$
$\Rightarrow (441)(50)+441x=621x$
$\Rightarrow (441)(50)=180x$
$\Rightarrow (441)(5)=18x$
$\Rightarrow (49)(5)=2x$
$\Rightarrow x=122.5$
So we are trying to find the perimeter of the triangle, $p$.
$p=2s=2(50+122.5)=\fbox{345}$[/hide]

$PA=s-a=x=23,\ PB=s-b=y=27,\ r=21\ (c=x+y).$

$(s-a)(s-b)(s-c)=sr^2\Longrightarrow s[(s-a)(s-b)-r^2]=(s-a)(s-b)c\Longrightarrow$

$s=\frac{(s-a)(s-b)c}{(s-a)(s-b)-r^2}\Longrightarrow s=\frac{xy(x+y)}{xy-r^2}\Longrightarrow$

the perimeter $p$ of the triangle $ABC$ is $p=\frac{2xy(x+y)}{xy-r^2}.$ [b]Answer:[/b] $p=345.$

Hmm.... I just realized we can meow use angle bisector...... anyone care to try it? I'm feeling kinda lazy now.

Sounds interesting, I can't find a way to do it though, care to explain? :?

[quote="pontios"][hide="Answer"]$172,5$[/hide][/quote]

Actually, this was $s$, the semi-perimeter

[hide="BAM...trig!"]
Let the incenter be denoted $I$.  It is commonly known that the incenter is the intersection of the angle bisectors of a triangle.  So let $\angle ABI =\angle CBI =\alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$

We have that

\begin{eqnarray*} \tan \alpha &=& \frac{21}{27}\\ \tan \beta &=& \frac{21}{23}\\ \tan \gamma &=& \frac{21}x. \end{eqnarray*}

So naturally we look at $\tan \gamma.$  But since $\gamma = \frac \pi2 -(\beta+\alpha)$ we have

\begin{eqnarray*} \tan \gamma &=& \tan\left(\frac \pi2 -(\beta+\alpha)\right)\\ &=& \frac 1{\tan(\alpha+\beta)}\\ \Rightarrow \frac{21}x &=& \frac{1-\frac{21\cdot 21}{23\cdot 27}}{\frac{21}{27}+\frac{21}{23}} \end{eqnarray*}

Doing the algebra, we get $x=\frac{245}2.$

The perimeter is therefore $2\cdot\frac{245}2 + 2\cdot 23 + 2\cdot 27 =\boxed{345}.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=654[/img][/hide]

My method was exactly joml88's, but I'm not sure about the angle bisector approach suggested by 1234567890... any ideas?

[quote="PhireKaLk6781"]I just tried using angle bisector, but I ended up with $\boxed{154}$.

I set up the ratios in respect to $\angle{A}$, called the point of tangency on $\overline{AC}$ $Q$, and called $CQ$ $x$.

$\frac{50}{27}=\frac{23+x}{x}$
$23x=621$
$x=27$

$2(23)+4(27)=154$.[/quote]

Doesn't work. The point of tangency of the incircle to the side of the triangle is (almost always) not the angle bisector.

Oh.

As is often the case, I'm being stupid.

I found it rather simple that we equate the inradius*perimeter/2=Heron's.

But how would one solve this problem if they do not know Heron's?

Let the other tangent length be $x$. Then the side lengths are $23+x, 27+x, 50$, so the area is $\sqrt{(50+x)(x)(23)(27)}$ by Heron. Then $a=rs$ gives $\sqrt{(50+x)(x)(23)(27)}=21(50+x)$, or $621x(50+x)=441(50+x)^2$, equivalent to $621x = 441(50+x) \implies 180x = 441 \cdot 50 \implies x = \frac{245}{2}$. The perimeter is $23 \cdot 2 + 27 \cdot 2 + \frac{245}{2} \cdot 2 = \boxed{345}$.

Trigo bash: tanA/2tanB/2 +tanB/2tanC/2+tanC/2tanA/2=1 , solving we get 2x=245 >> perimeter =345

Let unknown side has length as $x$, Assume three sides of triangles are $a,b,c$, the area of the triangle is $S$.

Note that $r=\frac{2S}{a+b+c}=21,S=1050+21x$

$\tan\angle{\frac{B}{2}}=\frac{7}{9}, \tan\angle{B}=\frac{63}{16}$. Use trig identity, knowing that $1+\cot^2\angle{B}=\csc^2\angle{B}$, getting that $\sin\angle{B}=\frac{63}{65}$

Now equation $(x+27)*50*\frac{63}{65}*\frac{1}{2}=1050+21x; x=\frac{245}{2}$, the final answer is $245+100=345$


板凳符号，看了真想坐上去，虽然不知是干什么用的。

[quote="paganinio"]板凳符号，看了真想坐上去，虽然不知是干什么用的。[/quote]
连乘。


It might have been posted before....but I can't find it anywhere in this site. Hope someone can help:

$a, b, c$ are the sides of a triangle. Prove:
$a^2(b + c - a) + b^2(c+a-b) + c^2(a+b-c) \leq 3abc$

 :)


geometry
geometric transformation
rotation

I. There are approximately two billion children (persons under 18) in the world. However, since Santa does not visit children of Muslim, Hindu, Jewish or Buddhist religions, this reduces the workload for Christmas night to 15% of the total, or 378 million (according to the Population Reference Bureau).
At an average (census) rate of 3.5 children per house hold, that comes to 108 million homes, presuming that there is at least one good child in each.

II. Santa has about 31 hours of Christmas to work with, thanks to the different time zones and the rotation of the earth, assuming he travels east to west (which seems logical). This works out to 967.7 visits per second. This is to say that for each Christian household with a good child, Santa has around 1/1000th of a second to park the sleigh, hop out, jump down the chimney, fill the stockings, distribute the remaining presents under the tree, eat whatever snacks have been left for him, get back up the chimney, jump into the sleigh and get on to the next house.

Assuming that each of these 108 million stops is evenly distributed around the earth (which, of course, we know to be false, but will accept for the purposes of our calculations), we are now talking about 0.78 miles per household; a total trip of 75.5 million miles, not counting bathroom stops or breaks. This means Santa's sleigh is moving at 650 miles per second --- 3,000 times the speed of sound. For purposes of comparison, the fastest man-made vehicle, the Ulysses space probe, moves at a poky 27.4 miles per second, and a conventional reindeer can run (at best) 15 miles per hour.

III. The payload of the sleigh adds another interesting element. Assuming that each child gets nothing more than a medium sized Lego set (two pounds), the sleigh is carrying over 500 thousand tons, not counting Santa himself. On land, a conventional reindeer can pull no more than 300 pounds. Even granting that the "flying" reindeer could pull ten times the normal amount, the job can't be done with eight or even nine of them--- Santa would need 360,000 of them. This increases the payload, not counting the weight of the sleigh, another 54,000 tons, or roughly seven times the weight of the Queen Elizabeth (the ship, not the monarch).

IV. 600,000 tons traveling at 650 miles per second creates enormous air resistance --- this would heat up the reindeer in the same fashion as a spacecraft re-entering the earth's atmosphere. The lead pair of reindeer would absorb 14.3 quintillion joules of energy per second each. In short, they would burst into flames almost instantaneously, exposing the reindeer behind them and creating deafening sonic booms in their wake.

The entire reindeer team would be vaporized within 4.26 thousandths of a second, or right about the time Santa reached the fifth house on his trip.

Not that it matters, however, since Santa, as a result of accellerating from a dead stop to 650 m.p.s. in .001 seconds, would be subjected to centrifugal forces of 17,500 g's. A 250 pound Santa (which seems ludicrously slim) would be pinned to the back of the sleigh by 4,315,015 pounds of force, instantly crushing his bones and organs and reducing him to a quivering blob of pink goo.

Therefore, if Santa did exist, he's dead now.

Santa doesn't visit homes in Africa or Asia cause they have no souls.

He doesn't visit homes in South America because there are no good children there.

Lastly, he gets his friends to visit homes in Europe (Papa Noel, Pere Noel, etc.)

So he only has to do homes in the US.

Get your facts straight.

[quote="chess64"]So in South America Christmas is celebrated on June 25th? :P[/quote]

LOL... I would think that in South America and Australia, they would have something different for Christmas (ever heard of White Christmas in Australia? :P)

[quote="Ars"]no there are good children in South America but they are visited at there winter so there you go[/quote]

No, they're not visited cause there are no good children there.

[quote="h_s_potter2002"][quote="Ignite168"]Santa doesn't visit homes in Africa or Asia cause they have no souls.

He doesn't visit homes in South America because there are no good children there.[/quote]

That sounds pretty racist to me.[/quote]

It's not racism, it's the truth. Even if it wasn't true (which it is) what's racist about it?

The calculations above fail to account for the fact that the average number of children per household may vary by religion.  Although I suppose if assuming the houses are evenly distributed across the planet is an acceptable error, this probably is too.

Santa is a supernatural existence. He defies the laws of nature. No further questions?

[quote="bubka"]You are assuming there is only one santa. I would rather think along the lines of a huge team santa. There are thousands of fresh recruits every year. All santa-look-alikes are invited, and hardly any one ever refuses.

By the way in Africa and Asia there are too few christian children and no snow and other sissy stuff, so santa does not visit these continents. It would be a waste of tiem even if they had souls. And just because Latin America is full of revolutionaries, it does not mean any joker clad in red can slip through. So santa dares not step there. But I am not sure abouth the Europe part. The children there are mostly dumb enough to pass as Americans. So if I were Santa, I wouldn't depend on papa noel and send a part of my team instead.[/quote]
it snows in asia...i've seen it myself.
it may not snow in some countries, but a large portion of asia receives snow

when it was winter, it snowed just like it snows in the united states.
asia has 253 million christians, while north america has only 236 million christians. so there are actually more christians in asia than north america.
it might not seem that there are more christians in asia because asia's total population is about 8 times that of north america's population, so asia seems to have 8 times less christians.

edit-sorry i just noticed that this thread is 1 month old.


Let $a,b,c$ be positive integers with $b$ odd.
	Define the sequence $\{a_n\} $ as follows: $a_0 = a$, and for $n \in \mathbb{N}$,

$ \{ a_n \} =\frac{1}{2}a_n $ if  $a_n$ is even
or
$ \{ a_n \} = a_n +b$  otherwise.

(a)	Show that $ a_n \leq b $ for some $ n \in \mathbb{N}$.
(b)	Show that the sequence $\{a_n\} $ is periodic from some point onwards.

a)

if u(n) is odd and > b then u(n+2)=(u(n)+b)/2 < u(n)
if u(n) is even and > b then u(n+1)=u(n)/2 < u(n)

So it's possible to build a strictly decreasing sub-sequence of u(n) to go below b.

b)

Clear since there are a finite number of integer <= b, the sequence must take a previously reached value <= b and then become periodic.