linear algebra
matrix
algorithm
vector
algebra
polynomial
invariant

I wonder if there is an algorithm to transform any matrix such as

$A = \left[\begin{array}{ccc} A_{1,1} & A_{1,2} & A_{1,3} \\
A_{2,1} & A_{1,2} & A_{2,3} \\ A_{3,1} & A_{3,2} & A_{3,3}
\end{array}\right]$

into its companion form

$B = \left[\begin{array}{ccc} 0 & 1 & 0 \\
0 & 0 & 1 \\ b_1 & b_2 & b_3
\end{array}\right]$

Could this be done effectively for matrices of any order?  :oops:

The naive approach is to start with a random row vector $v$, and take basis $v,vA,vA^2,\dots$, terminating when we reach a vector which is a linear combination of the previous. This basis is then used to transform the matrix; if $X$ has rows $v,vA,\dots$, $XAX^{-1}=B$ This almost always works; if the minimal polynomial is the characteristic polynomial, it works if the initial vector is not in some finite collection of proper subspaces. If the field is infinite, this is almost certain. If the minimal polynomial is of smaller degree than the characteristic polynomial, there is no chance; in that case, the form must include several diagonal blocks.

This method has no guarantee of being numerically decent.

A problem is:  "What sort of matrices can be transformed into companion form"?
There is an explicit algorithm for matrices to be transformed into Jordan canonoical form. Are there explicit algorithm to transform matrices into thier companion form as well?  :?:

[quote="bchui"]"What sort of matrices can be transformed into companion form"? [/quote]
A $n\times n$ matrix $A$ is similar to one in companion form if and only if its minimal polynomial $p$ has degree $n$

If you're using a standard of "explicit" that allows the standard arguments for Jordan form, I'll just add to my previous algorithm that we choose $v$ outside the invariant subspaces of $q(A)$ for each $q|p$. This is always possible, so we do it.

The whole thing is really very easy, as long as you start in the right place.


vector
inequalities
trigonometry
geometry proposed
geometry

Let ABC be a triangle having I is the incenter.
  Prove that (IA+IB+IC)^2 >= 4r(4R+r).

[quote]We have 4r(4R+r)= 2(ab+bc+ca)-a^2-b^2-c^2[/quote]

why? Can you explain it.

Can you have a solution without sin cos tan

[quote="treegoner"]4r(4R+r)= 2(ab+bc+ca)-a^2-b^2-c^2=4[(p-a)(p-b)+(p-b)(p-c)+(p-c)(p-a)][/quote]

This can be simplified a bit: if we involve the exradii $r_{a},$ $r_{b},$ $r_{c}$, then a well-known result of Steiner tells us that $4R=r_{a}+r_{b}+r_{c}-r$, and we also know $rr_{a}=\left( p-b\right) \left( p-c\right) $, $rr_{b}=\left( p-c\right) \left( p-a\right) $, $rr_{c}=\left( p-a\right) \left( p-b\right) $, so that we have $4r\left( 4R+r\right) =4r\left( r_{a}+r_{b}+r_{c}\right) =4\left(
rr_{a}+rr_{b}+rr_{c}\right) =4\left( \left( p-b\right) \left( p-c\right)
+\left( p-c\right) \left( p-a\right) +\left( p-a\right) \left( p-b\right)
\right) $. But what I don't understand is the following observation:

[quote="treegoner"]4[IA.IBcos(A/2)cos(B/2)+...]<=(IA+IB+IC)^2[/quote]

Can you explain it to me? Thanks!

  Darij

We have cos(A/2)cos(B/2)=sin((B+C)/2)sin((C+A)/2)<=sin^2((A+B+2C)/4)=sin^2(X).
   Similarly we have XYZ is a triangle.
   but we have sum 4xysin^2Z<=(x+y+z)^2
   I think it is a very cool inequality.  :cool:

Can we put the vector on IA;IB;IC to have your last inequality?
We have:
$\sum 4\cdot IA\cdot IB\cdot \cos \frac{A}{2}\cos \frac{B}{2} =\sum 2\cdot
IA\cdot IB\cdot \left( \cos \frac{A+B}{2}+\cos \frac{A-B}{2}\right)  \\
\leq -\sum 2\cdot IA\cdot IB\cdot \cos \frac{\pi +C}{2}+2\sum IA\cdot IB$.
We have $\left( \overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\right)^{2}\geq 0$
$\Longleftrightarrow \sum IA^{2}\geq -\sum 2\cdot IA\cdot IB\cos \frac{\pi +C}{2}$
So $\sum 4\cdot IA\cdot IB\cdot \cos \frac{A}{2}\cos \frac{B}{2}\leq \sum
IA^{2}+2\sum IA\cdot IB=\left( IA+IB+IC\right) ^{2}$ :)

Thanks, actually both of your proofs are cool :-)

  Darij


On hypotenuse $ AB$ of right triangle $ ABC$ are taken points $ K$ and $ L$ such that $ AK \equal{} KL \equal{} LB$. Find angles of triangle $ ABC$, if $ CK \equal{} \sqrt {2}CL$

Midpoint $ O$ of $ AB$ is the circumcenter, $ R \equal{} OA \equal{} OB \equal{} OC$ is the circumradius, $ \phi \equal{} \angle COB < 90^\circ.$ $ O \equal{} (0, 0),$ $ A \equal{} ( \minus{} R, 0),$ $ B \equal{} ( \plus{} R, 0),$ $ K \equal{} ( \minus{} R/3, 0),$ $ L \equal{} ( \plus{} R/3, 0),$ $ C \equal{} (R \cos \phi, R \sin \phi).$

$ R^2[(\cos \phi \plus{} 1/3)^2 \plus{} \sin^2 \phi] \equal{} CK^2 \equal{} 2\ CL^2 \equal{} 2R^2[(\cos \phi \minus{} 1/3)^2 \plus{} \sin^2 \phi]$

$ \cos \phi \equal{} \frac {5}{9},\ \ \sin \widehat{ABC} \equal{} \cos \widehat{CAB} \equal{} \cos \frac {\phi}{2} \equal{} \sqrt {\frac {1 \plus{} \cos \phi}{2}} \equal{} \frac {\sqrt 7}{3}$

I do not remember the name of the following theorem, so whoever knows it, please, communicate, thanks:

For a point $\{D\}\in (AB)$, the hypothenuse of a right-angled triangle, there exists following relation:
$BC^2\cdot AD^2+AC^2\cdot BD^2=CD^2\cdot AB^2$; apply it for $K$ and $L$, getting: $9CK^2=a^2+4b^2, \ 9CL^2= 4a^2+b^2$, where $a=BC, b=AC$; now easily get $\tan(\widehat{BAC})=\sqrt{\frac{3}{7}}$, a.s.o.

Best regards,
sunken rock

Let $AK=KL=LB=x, CL=y, CK=y\sqrt{2},BC=a$.

Then in $\triangle BCK$ segment $CK$ is a median, hence $CL^2={BC^2+CK^2\over 2}-{BK^2\over 4}\iff y^2={a^2+2y^2\over 2}-{4x^2\over 4}\iff a=x\sqrt{2}$.

Hence $b=\sqrt{(3x)^2-a^2}=x\sqrt{7}$

Now $\angle BAC=\arctan\sqrt{3\over 7}$


I'm confused as to when resonance is necessary to describe a Lewis structure. I'll use an example from a recent test:

The question asked which structure needed resonance to accurately describe it and two of the choices were $ \text{NO}_{2}^{\minus{}}$ and $ \text{ClO}_{4}^{\minus{}}$. When I drew the structures I found that both of them exhibit resonance but why is it that the first one NEEDS resonance to accurately describe it?

Also, on the subject of resonance, my textbook doesn't do the greatest of jobs describing it. Does it mean that when a structure has resonance, what does that say about the bond length of each specific bond? Does it mean that each one can vary or just that all of the lengths are the same?

A common mistake people make when thinking about resonance is they assume that the structure of say $ \ce{NO2\minus{}}$ rapidly interconverts between its two forms.

THIS IS WRONG!

The correct way to picture molecular bonding is as a superpositioning of all the possible resonance structures but each structure does not necessarily contribute as much to the overal form. By the way don't both structures need resonance to be described accurately?

The ressonance concept arises as a result of the natural limitation of Lewis structures: in Lewis notation, electron pairs are represented by small lines, between atoms if they are bonding, or on the atoms if they are non-bonding. However, this representation makes the electrons' position artificially static, localized in a given region of space. However, there are plenty of situations where we know that the electrons are free to move between three or more nuclei. But how can we continue to represent that situation with Lewis structures? The answer is ressonance and ressonance structures. Each of those structures represent a possible localization for electrons, but none is the true picture of the molecule.

Remember, science is not fact. Chemistry is not 100% correct. Resonance was established as a result of some molecules that could be drawn in multiple Lewis structures. In reality, the actual geometry of the molecule is the [i]average[/i] of the total amount of Lewis diagrams drawn. For example, in $ \ce{SO2}$, the bond lengths between the sulfur and the oxygen are the average of the lengths of two Lewis diagrams. Experiments have been carried out that prove that the bond lengths between the sulfur and oxygen are all the same, and that neither is shorter or longer than the other.


let a,b,c be the length of the sides of triangle ABC
prove that:
$ \frac{1}{(a\plus{}b\plus{}c)^2}\plus{}\frac{1}{(ab\plus{}bc\plus{}ca)^2}\geq \frac{1}{3(a^2b\plus{}b^2c\plus{}c^2a)}\plus{}\frac{1}{3(ab^2\plus{}bc^2\plus{}ca^2)}$

[quote="nguoivn"]It's an old and nice problem. Can_hang had a nice proof for it. But I can't find his link. Sorry  :wink:[/quote]
can you ask him??thanks
I really want to know the solution of this inequality.

[quote="tuandokim"]let a,b,c be the length of the sides of triangle ABC
prove that:
$ \frac {1}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{(ab \plus{} bc \plus{} ca)^2}\geq \frac {1}{3(a^2b \plus{} b^2c \plus{} c^2a)} \plus{} \frac {1}{3(ab^2 \plus{} bc^2 \plus{} ca^2)}$[/quote]
$ \frac {1}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{(ab \plus{} bc \plus{} ca)^2}\geq \frac{2}{(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)}$ and easy SOS. :wink:

[quote="arqady"][quote="tuandokim"]let a,b,c be the length of the sides of triangle ABC
prove that:
$ \frac {1}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{(ab \plus{} bc \plus{} ca)^2}\geq \frac {1}{3(a^2b \plus{} b^2c \plus{} c^2a)} \plus{} \frac {1}{3(ab^2 \plus{} bc^2 \plus{} ca^2)}$[/quote]
$ \frac {1}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{(ab \plus{} bc \plus{} ca)^2}\geq \frac {2}{(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}$ and easy SOS. :wink:[/quote]
Can you tell me more about your solution???thanks :blush:

[quote="tuandokim"][quote="arqady"][quote="tuandokim"]let a,b,c be the length of the sides of triangle ABC
prove that:
$ \frac {1}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{(ab \plus{} bc \plus{} ca)^2}\geq \frac {1}{3(a^2b \plus{} b^2c \plus{} c^2a)} \plus{} \frac {1}{3(ab^2 \plus{} bc^2 \plus{} ca^2)}$[/quote]
$ \frac {1}{(a \plus{} b \plus{} c)^2} \plus{} \frac {1}{(ab \plus{} bc \plus{} ca)^2}\geq \frac {2}{(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}$ and easy SOS. :wink:[/quote]
Can you tell me more about your solution???[/quote]
Of cause!
We need to prove that
$ \frac {2}{(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}\geq\frac {1}{3(a^2b \plus{} b^2c \plus{} c^2a)} \plus{} \frac {1}{3(ab^2 \plus{} bc^2 \plus{} ca^2)}.$
But $ \frac {2}{(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}\geq\frac {1}{3(a^2b \plus{} b^2c \plus{} c^2a)} \plus{} \frac {1}{3(ab^2 \plus{} bc^2 \plus{} ca^2)}\Leftrightarrow$
$ \Leftrightarrow6\sum_{cyc}(a^4bc \plus{} a^3b^3 \plus{} a^2b^2c^2)\geq\sum_{cyc}(a^2b \plus{} a^2c)\sum_{cyc}(a^2b \plus{} a^2c \plus{} abc)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}( \minus{} a^4b^2 \minus{} a^4c^2 \plus{} 4a^3b^3 \plus{} 4a^4bc \minus{} 5a^3b^2c \minus{} 5a^3c^2b \plus{} 4a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow \minus{} \sum_{cyc}(a^4b^2 \minus{} 2a^3b^3 \plus{} a^2b^4) \plus{} \sum_{cyc}(c^3a^3 \minus{} c^3a^2b \minus{} c^3ab^2 \plus{} c^3b^3) \plus{}$
$ \plus{} 2\sum_{cyc}(a^4bc \minus{} a^3b^2c \minus{} a^2b^3c \plus{} b^4ac) \minus{} 2\sum_{cyc}(a^3bc^2 \minus{} 2a^2b^2c^2 \plus{} ab^3c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2( \minus{} a^2b^2 \plus{} c^3(a \plus{} b) \plus{} 2abc(a \plus{} b) \minus{} 2abc^2)\geq0.$
Let $ a\geq b\geq c$ and $ S_c \equal{} \minus{} a^2b^2 \plus{} c^3(a \plus{} b) \plus{} 2abc(a \plus{} b) \minus{} 2abc^2.$
Hence, $ S_a\geq0,$ $ S_b\geq0$ and $ (a \minus{} c)^2\geq(a \minus{} b)^2.$
Id est, $ \sum_{cyc}(a \minus{} b)^2S_c\geq(a \minus{} c)^2S_b \plus{} (a \minus{} b)^2S_c\geq(a \minus{} b)^2\left(S_b \plus{} S_c\right).$
Thus, we need to prove that $ S_b \plus{} S_c\geq0.$
But $ S_b \plus{} S_c \equal{} \minus{} a^2c^2 \plus{} b^3(a \plus{} c) \plus{} 2abc(a \plus{} c) \minus{} 2ab^2c \minus{}$
$ \minus{} a^2b^2 \plus{} c^3(a \plus{} b) \plus{} 2abc(a \plus{} b) \minus{} 2abc^2 \equal{}$
$ \equal{} a(b^3 \plus{} c^3) \plus{} b^3c \plus{} c^3b \plus{} 4a^2bc \minus{} a^2b^2 \minus{} a^2c^2\geq$
$ \geq a^2(b^2 \minus{} bc \plus{} c^2) \plus{} b^3c \plus{} c^3b \plus{} 4a^2bc \minus{} a^2b^2 \minus{} a^2c^2 \equal{}$
$ \equal{} b^3c \plus{} c^3b \plus{} 3a^2bc\geq0.$ Done! :)


integration
trigonometry
calculus
vector
function
derivative
analytic geometry

Let $ \bold{F} \equal{} \minus{} 150 \bold{j}$

(a) If $ C$ is the straight line segment $ y \equal{} x$ from $ P(100,100)$ to $ Q(0,0)$ find the work.

So $ x \equal{} x, \ y \equal{} x, \ 0 \leq x \leq 100$

$ \minus{} \int_{0}^{!00} ( \minus{} 150 \bold{j}) \cdot (\bold{i} \plus{} \bold{j}) \equal{} 15000$

(b) $ C$ is the circular arc $ x \equal{} 100 \sin t, \ y \equal{} 100 \cos t$, $ P(0,100)$ to $ Q(0,0)$. I think this is a typo and it should be $ Q(100,0)$. Find the work.

So $ \int_{0}^{\pi/2} ( \minus{} 150 \bold{j}) \cdot (100 \sin t \bold{i} \plus{} 100 \cos t \bold{j}) \equal{} \minus{} 15,000$

(c) $ C$ is the parabolic arc $ y \equal{} \frac {x^2}{100}$ from $ P(100,100)$ to $ Q(0,0)$. Find the work.

$ x \equal{} x, \ y \equal{} \frac {x^2}{100}$

$ \int_{0}^{100} ( \minus{} 150 \bold{j}) \cdot (x \bold{i} \plus{} \frac {x^2}{100} \bold{j}) \equal{} \minus{} 66666.6$

Are these correct?

Well first $ W \equal{} \int_a^b \bold{F} \cdot \bold{r'(t)} ~dt$.

$ \bold{(a)}$ So thus your first integral is correct if that $ !$ is a typo for $ 1$.

$ \bold{(b)}$ This one I believe you got a little mixed up on.  We'll get:

$ \bold{r}(t) \equal{} \bold{<} 100 \sin t, 100 \cos t  \bold{>}$
$ \implies \bold{r}'(t) \equal{} \bold{<} 100 \cos t, \minus{} 100 \sin t  \bold{>}$

Thus: $ W \equal{} \int_0^{\frac{\pi}{2}} \bold{<} 0, \minus{}150 \bold{>} \cdot \bold{<} 100 \cos t, \minus{} 100 \sin t  \bold{>} ~dt$

$ \bold{(c)}$ The whole $ \bold{r}'(t)$ thing also needs fixing in this one as well.  Also, we are going from $ x\equal{}100$ to $ x\equal{}0$ so flip your bounds.

Thanks.

I get $ 15000$ for all of them (number 2 its negative).

This is a conservative force. The work should be path-independent. That is, we only need to know the endpoints of the path, not the intermediate details, in order to calculate the work.

Can you use a differential form to evaluate this?

Maybe a 1-form: $ A \ dx \plus{} B \ dy \plus{} C \ dz$?

The distinction between a vector field and a 1-form is one of notation, not substance. Changing the notation won't change anything of any importance about the computations. I called it a conservative field; I could just as easily have called it an exact differential form. The meaning is the same.

(Technical definitions: A vector field $ v$ is an operator from scalar-valued functions to scalar-valued functions which behaves like a derivative (linear at each point, product rule). It naturally corresponds to a vector-valued function, with the vectors tangent to the underlying space. A $ 1$-form is an operator from vector fields to scalar-valued functions, which naturally corresponds to a "covector"-valued function.)

Actually, there is a difference between vector fields and 1-forms: the spaces they live in are "dual" to each other. They are (locally) isomorphic, but that isomorphism depends on a choice of coordinates. This distinction between vectors and covectors gets blurred in the standard first treatments of multivariable calculus, but it shows up if you try to apply coordinate transformations- it's covariant for one and contravariant for the other.

That said, this $ F$ is a 1-form, not a vector field. The things that we integrate, that have the Stokes family of theorems for them, are forms.

An example of what I was talking about in the first paragraph: What happens if we scale our coordinate system- the point which was represented by $ (x,y,z)$ is now represented by $ (2x,2y,2z)$? The vector field is doubled (it should give the same result when applied as a derivative to a function, now stretched out over twice the coordinate range) and the 1-form is halved (it should give the same result when integrated over a curve, now covering twice the coordinate range).

And now- I'm not really contradicting Kent. We have to do our computations in coordinates anyway, and on $ \mathbb{R}^n$ we have a standard basis to relate everything to. If we want to write these things down, we need to use some sort of vector notation, and duplicating that of vector fields works pretty well. You're not even using "real" vector fields much anyway.


hey the NSO results are out but we dont remember the roll no how do we check last time we used to enter the school code???now what could we do??? :blush:

[schoolcode]2M[yourcode]

Anyways I got AIR222.... Not a big deal but I guess I'll get a minimum of 250Rs. :D

Was too bored to go to school.. So din write nso.. :D

i got air 55 :)
but i heard theres gonna be a 2nd round also  :|
is that true

yeah ur rite shreyas. 250 bucks indeed :D

i got 'AIR' 304  :wink:

@anitashu: yes, there is gonna be a 2nd round.. and jus 4 the record- the AIR rank one in the second round gets 51 grand!!
:w00tb:

hey wats d eligibility for 2nd round...
im AIR 500...
n when is it likely to be??

@abishek: Top 500 is eligibility. But I'm not sure if you've either missed the bus or just hopped onto it. ie: Dunno if AIR 500 is included. :(

Anyways, anyone here writing NCO 2nd Level. Cause it clashes with Fiitjee open test?

Got AIR 562 . Am shocked  :o
Did the paper terribly for words ( got 35 or 36  :maybe: )
Expected rank less than 2000.
Now am sad that if i had got 2 more questions would have got thro.  :(
Anyways wanna write cyber and then paper 2 of fiitjee ( mostly paper 2 will be tougher )

@pardesi : Did you write nco and nso?

yes and the [b]most[/b] miserable performance 1000+ :rotfl:
none from our school qualifies this time and last time we had got the best school medal :o


induction
inequalities unsolved
inequalities

[quote="pohoatza"]
I think that in $RHS$ is $\frac{(a_{1}+a_{2}+...+a_{n})^{3}}{27}$?[/quote]

Yes! I have mistakes. :wink:


Find all values of $ a, b, x$ and $ y$ that satisfy the simultaneous equations:

$ a\plus{}b\equal{}1$

$ ax\plus{}by\equal{}\frac{1}{3}$

$ ax^{2} \plus{} by^{2} \equal{}\frac{1}{5}$

$ ax^{3}\plus{}by^{3}\equal{}\frac{1}{7}$.

Do you calculate Gaus's formula for $ \int_0^1f(x)dx?$.
You can calculate as \[ by^2(y\minus{}x)\equal{}\frac 17 \minus{}\frac x5,\]
\[ by(y\minus{}x)\equal{}\frac 15 \minus{}\frac x3,\]
\[ b(y\minus{}x)\equal{}\frac 13 \minus{}x.\]
Then \[ y\equal{}\frac{3\minus{}5x}{5(1\minus{}3x)}\equal{}\frac{15\minus{}21x}{21\minus{}35x}\to 35x^2\minus{}30x\plus{}3\equal{}0,...\]

Hi rust. Could you explain in more detail, please: I don't know what Gauss's formula for $ \int_{0}^{1} f(x)\ dx$ is, but your method looks like it might be quite interesting?

I used

$ (ax^{n}\plus{}by^{n})(x\plus{}y) \equal{} (ax^{n\plus{}1}\plus{}by^{n\plus{}1}) \plus{}xy(ax^{n\minus{}1}\plus{} by^{n\minus{}1})$

(which may also have a name).

Are you out there Rust?

Or could someone else explain?

hello, solving this system i have got
$ a_1\equal{}\frac{1}{2}\plus{}\frac{1}{36}\sqrt{30}$
$ b_1\equal{}\frac{1}{2}\minus{}\frac{1}{36}\sqrt{30}$
$ x_1\equal{}\frac{3}{7}\minus{}\frac{2}{35}\sqrt{30}$
$ y_1\equal{}\frac{3}{7}\plus{}\frac{2}{35}\sqrt{30}$
or
$ a_2\equal{}\frac{1}{2}\minus{}\frac{1}{36}\sqrt{30}$
$ b_2\equal{}\frac{1}{2}\plus{}\frac{1}{36}\sqrt{30}$
$ x_2\equal{}\frac{3}{7}\plus{}\frac{2}{35}\sqrt{30}$
$ y_2\equal{}\frac{3}{7}\minus{}\frac{2}{35}\sqrt{30}$
i don't want to type all my calculations from my sheet of paper in this editor.
Sonnhard.

Thanks Sonnhard. They're the same as my answers. I was looking for some other ways to solve the problem, and I'm still confused by Rust's mention of Gauss's formula. I like Rust's method of solution but I can't see what Gauss's formula has to do with it; perhaps because I don't know what Gauss's formula is. Can anyone shed some light on this?

Has anyone figured out what Rust is alluding to with $ \int_{0}^{1} f(x)\ dx$?


I was reasearching a little bit of [url=http://mathworld.wolfram.com/FigurateNumber.html]Figurate Numbers[/url], such as square numbers, pentagonal numbers, etc.  It is possible to find the formulas and the first few numbers of each of these types fairly easily. However, I was just wondering, what exactly is an r-gonal number? For example, what would the first few r-gonal numbers be? Is it possible to write a formula to determine the nth r-gonal number?

[quote="ZetaX"]the n-th r-gonal number is $ n \cdot \left( \frac {(r-2)(n-1)}{2}+1 \right) $[/quote]

And what exactly is a r-gonal number? Is that formula the same thing as:

[hide]$\frac{1}{2}r[(n-2)r-(n-4)]$[/hide]

Where $r$ represents the number of sides, and $n$ is the $n$th r-gonal number?

You can define a m-th r-gone by dividing every side of an r-gone into m segments, giving their endpoints as the 'm-th r-gone points'. Now if you set the first n 'm-th r-gones' (m from 0 to n) 'inside' each other, having two edges and one vertice in common, you get the n-th r-gonal number as the number of different 'r-gone-points'.
(sorry for my bad explanation  :( )


hmm
I dunno how to make polls
but vote how many posts will get in this topic before it gets locked?

it would be really sad if a moderator was the first to look at this and locked it immediately...

and that's why I don't think that this is a very good idea...

Well...on the top next to "message body", it should say "poll", which would allow you to insert a poll.

Yay! $ \ge2$

If this thread is to survive, the moderator must have a secondary goal besides locking the thread. Perhaps he wants to make our poll as inaccurate as possible. If the moderator has just the one goal, we are doomed no matter what, so we may as well assume that this is not the case. We must therefore be extremely careful in voting so that the moderator will not lock the thread for fear of us winning.

This is very important! Think before you vote!

here is a question: can mods look at the results of a poll BEFORE they vote? After all they are mods.

everyone can
and this is just spam
i think ill lock it

this is pretty tricky

its like game theory omg

I think the amount of posts needed to [b]arouse[/b] the attention of a mod is on the order of $ \lim_{x\to\infty} x$

Try making a TWSS joke out of that!

-wonders who unlocked this- just out of curiosity and such.


Somebody told me recently that in a "show that" problem, you could use the result and work backwards as your reasoning, but in a proof problem you had to come to your conclusion without what was asked to prove in the question. This seems wrong to me as styles of proof such as contradiction and mathematical induction [b]require[/b] you to use what was to prove as part of the proof. But that lead me to thinking; is there a difference between these types of problems? Is it just in the level of rigour?

In a proof, you [i]can[/i] work backwards from the requested statement, but you then have to show that all of your steps are reversible. If something is necessary for something else (that is, $p\implies q$) it may not be true that it's sufficient ($q\implies p$). Showing the reversibility of your steps is -very- important if you plan on working backwards.


Let $x,y$ two real numbers which are not both $0$. Prove that

$\dfrac{x+y}{x^2-xy+y^2}\leq\dfrac{2\sqrt 2}{\sqrt{x^2+y^2}}$.

Please I want from the moderators to send all the problems to Olympiad recousrces.
I will put all the for problems from 2004.

Hi silouan:
  I think it is an easy inequality.  :D
   1>If $x+y<0$ then $LHS<0<RHS$
   2> If $x+y>0$ then square all of them we have to solve the new inequality:
$V(x,y)=8(x^2-xy+y^2)^2-(x^2+y^2)(x^2+2xy+y^2)\geq0$
$V(x,y)= (x-y)^2(7x^2-4xy+7y^2)\geq0$
  Done! :!:

Siouan, your inequality follows immediately from this easy one:
$2(x^{3}+y^{3})\geq (x+y)(x^{2}+y^{2})$ which is a direct consequence of Chebyshef. ;) , but it can be also proved by calculations. very easy. ;)


Given a natural number $n.$ Generate new numbers by the following mapping $\Phi.$ Erase the last digit and add to this number 4 times the last digit, e.g. $\sigma = 1995 \mapsto 199 + 4 \cdot 5 = 219 = \tau.$

a.) Determine starting numbers $\sigma$ for which after a finite number $k$ of compositions of $\Phi,$ $(\Phi^{(k)}),$ a number $\tau$ occurs several times. Does there $\exists$ a starting numbers $\sigma$ for which after a finite number applications of $\Phi,$ $(\Phi^{(k)}),$ only the same number $\tau$ occurs (called one-cycle).

b.) Let $1001 = \tau$ for a certain $\Phi^{(k)}, k \geq 1.$ Prove that there does not exist any $\tau$ for a certain $\Phi^{(k)}$ so that $\tau$ is a prime.

Edited for error:

[hide]a) For the first part, any number, after repeated applications of Phi, will eventually lead to a cycle.  Consider any four digit number.  When you erase the last digit of the number, it decreases by at least 100, and when you multiply the former last digit by 5 and add it to the new number, you add at most 45.  Therefore, any application of Phi to a four digit number will result in a smaller number (the same argument holds for numbers with more digits; for an n digit number, an application of Phi will decrease the number by at least (10)^(n-2) and increase it by at most 45).  Thus, if we have a number of four or more digits, we will eventually get a number with three or fewer digits.  When we have a number with three or fewer digits, two things can happen.  Either an application of Phi will lead to another number with three or fewer digits, or we get a number with four or more digits (I know, this second one is impossible, but I'm not sure how to prove it quickly), either of which will lead to another number with three or fewer digits.  Either way, we'll eventually run out of such small numbers, and we'll have to repeat some.  For the second part, application of Phi to 26 will result in 26, thus there is a starting number (49) that will result in a one-cycye.[/hide]

I'm not sure about part b)...

b) goes as follows:

1001 -> 104 -> 26 -> 26 -> ... furher the row always evolves 26-26-26-... so what is the problem? :?

Or do you mean we have to run back and check that 1001 cannot come from any primes, regardless of how far you return?

[quote="orl"]
b.) Let $1001 = \tau$ for a certain $\Phi^{(k)}, k \geq 1.$ Prove that there does not exist any $\tau$ for a certain $\Phi^{(k)}$ so that $\tau$ is a prime.[/quote]

At first I have to say that you cannot start the sequence with $1001$ as you have to look for $ \tau$ and not $ \sigma.$ $ \tau$ cannot occur before the second element in the sequence (first element is $ \sigma$) of your numbers, e.g. you can choose $(9971, 1001, 104, 26, \ldots).$ And this case would mean that you chose $k =1$ and I said $k \geq 1.$ So you have to consider all previous numbers in the sequence too.  :)

part b:

[hide]The transformation 10a+b -> a+4b is equivalent to multiplication by 4 in modulo 39, i.e. 4(10a+b)=40a+4b=a+4b mod 39. Since 1001=26 mod 39, which is divisible by 13, any number in a sequence containing 1001 will be divisible by 13. The only prime in such a sequence is 13. But 1001 and 13 cannot be in the same sequence, for the following reason:

13 can't come after 1001, since 1001->104->26->26..., which does not contain 13.

1001 cant come after 13, since 13->13->..., which doesnt contain 1001.[/hide]

Btw, this also happens to be a divisibility test for 13; it retains divisibility by 13, though any remainders will get distorted by the x4 part.

For example, we all know 1001=7x11x13
1001-->100+4x1=104-->10+4x4=26, which is a multiple of 13, hence 1001 is a multiple of 13


[quote="AIME15"]The radius of the circle is $ 6\sqrt{2}$, so the diagonal is $ 6\sqrt{2}$, so the side of the square is $ 6$ and its area is $ \boxed{36}$.[/quote]

Just to add a little to what AIME said:

Now that you know what the radius of the circle is, you can use it to find the area of the square, so (6sqrt2)^2 divided by 2 is 72/2 which is 36!!! The formula to find the area of the square is diagonal^2 divided by two.

You can just divide by 2pi. It works every time for a problem like this

[quote=GameBot]A square is drawn inside a circle so that one vertex lies at the center of the circle and another vertex lies on the circumference. The are of the circle is $ 72\pi$ square inches. How many square inches are in the area of the square?

[asy]size(50);
draw(Circle((0,0),5));
draw((-5,0)--(5,0));
draw((0,5)--(0,-5));
draw((3.5,0)--(3.5,-3.5)--(0,-3.5));[/asy][/quote]

you made a typo


trigonometry
function
inequalities
algebra proposed
algebra

A calculator treats angles as radians. It initially displays 1. What is the largest value that can be achieved by pressing the buttons cos or sin a total of 2001 times? (So you might press cos five times, then sin six times and so on with a total of 2001 presses.)

It is easy to show maximum is $max=cos(sin(sin \dots 1)\dots )$ It give approch $cos(\sqrt{\frac 3n})=1-\frac{3}{2n}+O(n^{-2}), \ n=2000.$

Yes. But formula $max=1-\frac{3}{2n}+O(n^{-2})$ works.

Yes you are right.But I think for this problem we needn't to estimate the value.

After some years thinking and computing :D , I will post a solution:

Thinking in Dynamic Programming, we will define the sequences $ M_k$ and $ m_k$ as the maximum and the minimum value of the values of pressing sequence, with $ k$ presses.

We can assume all values are in the first quadrant (between 0 and $ \pi/2$).

As $ sin$ and $ cos$ are monotonic (increasing and decreasing, respectively) functions, we can say that $ \{M_{k\plus{}1},m_{k\plus{}1}\} \in \{sin(M_k),sin(m_k),cos(M_k),cos(m_k)\}$.

The idea is to put restrictions in this fact.

These are easily verifiable facts about $ sin$ and $ cos$:

1 - $ sin(x) > cos(y) \Leftrightarrow x\plus{}y > \pi/2$
2 - $ sin(x) < cos(y) \Leftrightarrow x\plus{}y < \pi/2$
3 - $ sin(x) > sin(y) \Leftrightarrow x>y$
4 - $ cos(x) > cos(y) \Leftrightarrow x<y$

If $ M_k\plus{}m_k > \pi/2$ then $ M_{k\plus{}1}\equal{}sin(M_k)$ and $ m_{k\plus{}1}\equal{}cos(M_k)$
If $ M_k\plus{}m_k < \pi/2$ then $ M_{k\plus{}1}\equal{}cos(m_k)$ and $ m_{k\plus{}1}\equal{}sin(m_k)$

So, we conclude that $ M_k^2\plus{}m_k^2\equal{}1$ for $ k \geq 1$.

But using Power Mean Inequality, we cn assure that $ M_k\plus{}m_k \leq \sqrt{2} \leq \pi/2$, and so we can use the formula $ M_{k\plus{}1}\equal{}cos(m_k)$ and $ m_{k\plus{}1}\equal{}sin(m_k)$ for any $ k \geq 1$.

Now the problem is almost obvious :)


Let the two base angles of a triangle be $A$ and $B$, with $B$ larger than $A$.  The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$, with $C_2$ adjacent to side $a$.  Then:
(A) $C_1 + C_2 = A + B$
(B) $C_1 - C_2 = B - A$
(C) $C_1 - C_2 = A - B$
(D) $C_1 + C_2 = B - A$
(E) $C_1 - C_2 = A + B$

[hide]The altitude to the base makes a 90 degree angle, making the other two angles in the triangles inside ABC euqal each other:

C1 + A = C2 + B.

We need to simplify it, since that isn't one of the answer choices.

C1-C2 = B-A, or B.[/hide]


Six cards with numbers 1, 1, 3, 4, 4, 5 are given. We are drawing 3 cards from 6 given cards one by one and are forming a three-digit number with the numbers over the cards drawn according to the drawing order. Find the probability that this three-digit number is a multiple of 3. (The card drawn is not put back)

$\textbf{(A)}\  \frac {1}{5} \qquad\textbf{(B)}\  \frac {2}{5} \qquad\textbf{(C)}\  \frac {3}{7} \qquad\textbf{(D)}\  \frac {1}{2} \qquad\textbf{(E)}\ \text{None}$

The number of all outcomes is $ \frac {6!}{2!2!} \equal{} 180$

The number of favorable outcomes is:

1. Digits $ 114$ - either one of the two fours can be drawn, and the drawn digits can be ordered in $ 3$ ways, hence $ 2\cdot 3 \equal{} 6$

2. Digits $ 135$ - either one of the two one's can be drawn, and the drawn digits can be ordered in $ 6$ ways, hence $ 2\cdot 6 \equal{} 12$

3. Digits $ 144$ - either one of the two one's can be drawn, and the drawn digits can be ordered in $ 3$ ways, hence $ 2\cdot 3 \equal{} 6$

4. Digits $ 345$ - either one of the two fours can be drawn, and the drawn digits can be ordered in $ 6$ ways, hence $ 2\cdot 6 \equal{} 12$

Total: $ 6 \plus{} 12 \plus{} 6 \plus{} 12 \equal{} 36$

Probability: $ {36\over 180} \equal{} {1\over 5}$

Farenhajt, are you sure the total number of outcomes is 180? That seems a little high to me. I think you calculated the number of ways to form 6 digit numbers instead of 3 digit numbers.

You're right, I should really go to sleep :roll:

Designate the digits thus: $ 1, 1', 3, 4, 4', 5$. Then drawing a 3-digit number is non-repetitive variation, so there's $ 6\cdot 5\cdot 4 \equal{} 120$ outcomes.

Nothing will change in the count of favorable cases, they will only split, like $ 11'4 \plus{} 11'4'$ etc.

But then the result is $ {36\over 120} \equal{} {3\over 10}$, so the answer is $ E$...?

Okay, now I agree with your denominator but I disagree with your numerator (according to the way you are counting). For example, 114 could be either 11'4, 1'14, 11'4', or 1'14' (plus all the permutations of those).

Grrrrr!

Now all the digits are distinct, hence each possibility yields $ 6$ orders, and there are $ 8$ possibilities, hence $ P\equal{}{48\over 180}\equal{}{4\over 15}$

And if this is also wrong, please just give the solution (it's 4:50 am here).

I get $ {48 \over 120} \equal{} {2 \over 5}$, which is answer choice $ B$

You see what I'm doing? Writing $ 180$ instead of $ 120$ in the denominator... Tragic.


This isn't a typical AMC type problem but the idea is quite classic and useful if you haven't seen it before.

(Me)

$ \displaystyle \frac{1}{\sqrt{2} \plus{} \sqrt{3}} \plus{} \frac{1}{\sqrt{3} \plus{} \sqrt{4}} \plus{} \frac{1}{\sqrt{4} \plus{} \sqrt{5}} \plus{} \frac{1}{\sqrt{5} \plus{} \sqrt{6}} \equal{} 2 \sin k (\tan m \minus{} \tan k)$

If $ 0 < k,m < \frac{\pi}{2},$ then find $ k\plus{}m$.

$ \displaystyle \text{(A)} \ \frac{5\pi}{12} \qquad \ \text{(B)} \ \frac{\pi}{2} \qquad \text{(C)} \ \frac{7\pi}{12} \qquad \text{(D)} \ \frac{2\pi}{3} \qquad \text{(E)} \ \frac{5\pi}{6}$

[hide]Answer: $ (\text{C}) \frac{7\pi}{12}$

Multiply the numerator and denominator of each fraction by the conjugate (?) of the denominator ($ \sqrt {2} \minus{} \sqrt {3}$, $ \sqrt {3} \minus{} \sqrt {4}$, etc.)

This will result in denominator of $ \minus{} 1$ for all of the fractions, which we may now add together.
We now find that $ \frac {\sqrt {6}}{2} \minus{} \frac {\sqrt {2}}{2} \equal{} \sin{k}(\tan{m} \minus{} \tan{k}) \equal{} \sin{k}\tan{m} \minus{} \sin{k}\tan{k}$
From here I assumed that $ \sin{k}\tan{k} \equal{} \frac {\sqrt {2}}{2}$ which gave me $ k \equal{} \frac {\pi}{4}$. Substituting this value for $ k$ in $ \sin{k}\tan{m} \equal{} \frac {\sqrt {6}}{2}$ gave me $ m \equal{} \frac {\pi}{3}$

$ k\plus{}m\equal{}\frac{7\pi}{12}$[/hide]

[quote="Altheman"]There is obviously not a unique answer to this...[/quote]

I know, I realized it after I wrote the problem so I decided not to put on the contest but like, the way I did it was:

[hide="Essentially same"]
After simplifying the fractions, it came down as $ \sqrt{6} \minus{} \sqrt{2} \equal{} \sqrt{2} (\sqrt{3} \minus{} 1) \equal{} 2 \frac{\sqrt{2}}{2} (\sqrt{3} \minus{} 1) \equal{} 2 \sin \frac{\pi}{4} (\tan \frac{\pi}{3} \minus{} \tan \frac{\pi}{4})$ [/hide]

Hmm, I wonder if it's possible to find ALL the $ (k,m)$ pairs that satisfy this value at all? I mean, the numbers will have to be rounded but would there be a way to do this? Multiplying $ \sin k$ only gives $ \frac{\sqrt{6} \minus{} \sqrt{2}}{2} \equal{} \frac{\sin m \sin k}{\cos m} \minus{} \frac{\sin^2 k}{\cos k}$ which doesn't really help anything though...

Of course, that was just a question though.

[quote="Silverfalcon"]$ \frac {\sqrt {6} \minus{} \sqrt {2}}{2} \equal{} \frac {\sin m \sin k}{\cos m} \minus{} \frac {\sin^2 k}{\cos k}$[/quote]
\[ \equal{} 2\sin{15^{\circ}}
\]

Dunno if that's any help...

Wait, I don't get it.. but then again I'm not exactly creative...

So i simplified it down to $ \frac{\sqrt{6} \minus{} \sqrt{2}}{2} \equal{} sin(tan(k)\minus{}tan(m))$

but... how do I make this...work?

You probably misread the question. The the argument of the sine function does not call the tangent function -- you have sin(stuff) times tan(stuff)


group theory
abstract algebra
superior algebra
superior algebra unsolved

In simple group A_5, what is number of subgroups of order 12?

There are 5 - namely, the copies of $ A_4$ given by the permutations that fix $ k$, for $ 1\leq k\leq 5$.

To show there are no more...

[hide]suppose $ H$ is a subgroup of order 12. $ H$ contains an element of order 3 (a 3-cycle). But $ H$ cannot contain two 3-cycles which are independent (that is, one not generating the other). For if $ H$ contained two such cycles, one of the following must hold about the sets of elements moved by the cycles:

- They would intersect in 1 element. WLOG let the cycles be $ (123),(145)$. Then $ (123)(145)\equal{}(14523)$, a 5-cycle, which could not be in $ H$.

- They would intersect in 2 elements. WLOG let the cycles be $ (123),(124)$. These cycles generate all even permutations fixing 5, contrary to the assumption that $ H$ was not a copy of $ A_4$.

So $ H$ contains exactly one subgroup of 3-cycles. WLOG let this subgroup be $ <(123)>$. If $ K$ is the subgroup of $ A_5$ fixing 1, then we have
\[ |H\cap K|\equal{}\frac{|H\parallel{}K|}{|HK|}\equal{}\frac{12^2}{|HK|}\geq \frac{12^2}{60}\equal{}\frac{12}{5}>2\]
But also $ H\cap K$ is a subgroup of both $ H$ and $ K$, so cannot contain a 3-cycle. Our only option is that $ H\cap K$ is the subgroup $ <(23)(45),(24)(35)>$ of $ K$. In particular, $ (24)(35)\in H$. We find that $ H$ contains $ (123)(24)(35)\equal{}(12435)$ which is a 5-cycle, a contradiction. So $ H$ must be one of the 5 copies of $ A_4$.[/hide]

There's probably a shorter/easier way to do this...


What is the product of the numbers $ 1,2,3,4,5,6,7,8,9,$ and 0?

This is $ 0$ because $ 0$ times anything is $ \boxed{0}$.


Let $\log_{3x+4}(2x+1) = a$ and $\log_{2x+1}(3x+4) = b$. We then have $ab = 1$ and $4 = \log_{3x+4}((2x+1)^{2})+\log_{2x+1}((2x+1)(3x+4)) = 2a+1+b$, so $3 = 2a+b$, i.e. $b^{2}-3b+2 = 0$. So we have $b = 1$ or $b = 2$.

When $b = 1$, we obtain $\log_{2x+1}(3x+4) = 1$, i.e. $3x+4 = 2x+1$. But then $x =-3$, which means that $3x+4$ and $2x+1$ or both negative, which makes both logarithms undefined. Therefore, no solution when $b = 1$.

When $b = 2$, that means $\log_{2x+1}(3x+4) = 2$, i.e. $(2x+1)^{2}= 3x+4$, so $0 = 4x^{2}+x-3 = (x+1)(4x-3)$. For $x =-1$ we have the same undefined logarithms, so the only solution is $x = \frac{3}{4}$.

[quote="ashrafmod"]$ \log_{3x \plus{} 4}(4x^{2} \plus{} 4x \plus{} 1) \plus{} \log_{2x \plus{} 1}(6x^{2} \plus{} 11x \plus{} 4) \equal{} 4$[/quote]
$ \Leftrightarrow \log_{3x \plus{} 4}(2x \plus{} 1)^2 \plus{} \log_{2x \plus{} 1}((3x \plus{} 4)(2x \plus{} 1)) \equal{} 4$
$ \Leftrightarrow 2\log_{3x \plus{} 4}(2x \plus{} 1) \plus{} \log_{2x \plus{} 1}(3x \plus{} 4) \plus{} \log_{2x \plus{} 1}(2x \plus{} 1) \equal{} 4$
$ \Leftrightarrow 2\log_{3x \plus{} 4}(2x \plus{} 1) \plus{} \log_{2x \plus{} 1}(3x \plus{} 4) \equal{} 3$
Set $ \log_{3x \plus{} 4}(2x \plus{} 1) \equal{} t$
We have:
$ 2t \plus{} \frac {1}{t} \equal{} 3$
Or: $ (2t \minus{} 1)(t \minus{} 1) \equal{} 0 \Leftrightarrow t \equal{} 1 V t \equal{} \frac {1}{2}$
$ t \equal{} 1: \log_{3x \plus{} 4}(2x \plus{} 1) \equal{} 1 \Leftrightarrow x \equal{} \minus{} 3$,impossible
$ t \equal{} \frac {1}{2}: \log_{3x \plus{} 4}(2x \plus{} 1) \equal{} \frac {1}{2} \Leftrightarrow (2x \plus{} 1)^2 \equal{} (3x \plus{} 4)$
$ \Leftrightarrow 4x^2 \plus{}x\minus{}3 \equal{} 0 \Leftrightarrow (4x\minus{}3)(x\plus{}1)\equal{}0 \Leftrightarrow x\equal{}\frac{3}{4}$
Then equatiom has one root: $ x\equal{}\frac{3}{4}$

__________
Editted!

[quote="queen"][quote="Kurt Gödel"] = , i.e. $ b^{2} \minus{} 3b \plus{} 2 \equal{} 0$. [/quote]

How did you derive  that  :huh:[/quote]
From $ ab\equal{}1$, we have $ a\equal{}\frac{1}{b}$.  Substituting this into $ 3\equal{}2a\plus{}b$ gives us $ b^2\minus{}3b\plus{}2\equal{}0$ after multiplication by $ b$.

[quote="JBL"][quote="NguyenDungTN"]$ t \equal{} \frac {1}{2}: \log_{3x \plus{} 4}(2x \plus{} 1) \equal{} \frac {1}{2} \Leftrightarrow 2x \plus{} 1 \equal{} (3x \plus{} 4)^2$[/quote]

Nice work, but this step is wrong.  You also could have established the incorrectness of your solution by checking to see that the value K.G. gave is in fact a root.[/quote]
I'm sorry! I've editted my solutions!


real analysis
limit
real analysis unsolved

Let $ limsup|a_n|^{\frac{1}{n}}\equal{}R$ where $ R\in[0,\infty]$.

Why $ limsup|na_n|^{\frac{1}{n\minus{}1}}\equal{}limsup|na_n|^{\frac{1}{n}}$??

$ \lim \frac{n}{n\minus{}1} \equal{} 1$, does that help?


inequalities
number theory
prime numbers

Find all pairs of natural numbers $(n,k)$ for which

$(n+1)^{k}-1 = n!$.

if $n>3$ and $n+1$ was composite then we would have $n+1 \mid n!$ so we get that $n+1 \mid 1$...
so we have $n\leq 3$ or $n+1$ is a prime number...
so assume that $n+1$ is prime,then we have $n!=(n+1)^{k}-1=n[(n+1)^{k-1}+...+1]$
so we get that $(n-1)!=(n+1)^{k-1}+...+1$ now if $n>4$ then its a composite number,thus we would have:
$(n+1)^{k-1}+...+1 \equiv k (\bmod n)$ and also $n \mid (n-1)!$ so we must have $n \mid k$ so $k\geq n$ but one can see that if $n\geq 2$:
$(n+1)^{k}\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions...
so the remain case is $n\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$...

[quote="BaBaK Ghalebi"]if $n>3$ and $n+1$ was composite then we would have $n+1 \mid n!$ so we get that $n+1 \mid 1$.[/quote]

Can you explain this further?  :)

maybe the folowing explanation would make it clear...
well if $n+1$ was composite then we would have $n+1=pq$ s.t $\gcd (p,q)=1$ and $p,q>1$ so we get that $p,q<n+1$ so $p,q\leq n$ so we get that:
$i)p \mid n!$
$ii)q \mid n!$
now from $i),ii)$ and $\gcd (p,q)=1$ we get that $pq \mid n!$ so $n+1 \mid n!$...
now in the equation $(n+1)^{k}-1=n!$ we have $n+1 \mid RHS$ so we get that $n+1 \mid LHS$ so $n+1 \mid 1$...

[quote="BaBaK Ghalebi"][quote="cincodemayo5590"][quote="BaBaK Ghalebi"]well if $n+1$ was composite then assume that $p^{\alpha}\mid n+1$ s.t $\alpha>1$ [/quote]

Alright; everything made sense except for why you can assume that $\alpha >1$. It doesn't seem like this treats the case of something like, $n+1=6=2 \cdot 3$. Then, you wouldn't have $2p \leq p^{\alpha-1}<n \iff 4 \leq 1 <5$. Unless I'm missing something, it doesn't look like you treat the cases where $n+1 = \prod p_{i}$; only cases where $n+1$ is prime or is composed of primes with higher powers than $1$.[/quote]
sorry it was a typo...
I edited my post,I think its correct now...[/quote]

OK, that makes more sense now. Thanks. :)

Well yeah, my proof is pretty much the same.

For $n>4$ it is clear that $n$ muc be even so let $n=2m$ and so $n!= (2m)! = (2m).(2m-1)!$ so it is clear that $n^{2}| n!$. So $n^{2}| (n+1)^{k}-1$. Using binomial expansion we see that if $n^{2}| (n+1)^{k}-1$ then $n|k$ and so $k \geq n$.

Therefore $(n+1)^{k}\geq (n+1)^{n}> (n-1)!+1$. Now all we have to do is check for $n=1,2,3,4$.

[quote="me@home"]Everything is fine until here
[quote="BaBaK Ghalebi"]now if $n>4$ then its a composite number,thus we would have:
$(n+1)^{k-1}+...+1 \equiv k (\bmod n)$ and also $n \mid (n-1)!$ so we must have $n \mid k$ so $k\geq n$ but one can see that if $n\geq 2$:
$(n+1)^{k}\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions...
so the remain case is $n\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$...[/quote]
Then I think your english gets in the way of the rest of the proof, at least for me... :blush:[/quote]
well we assumed that $n+1$ is prime and also if $n>4$ then $n$ must be composite (because there doesnt exist two consecutive prime numbers greater than $4$) so $n$ is a composite number so $n \mid (n-1)!$ now from the equation $(n-1)!=(n+1)^{k-1}+...+1$ we get that $n \mid LHS$ so $n \mid RHS$ but $RHS \equiv 1+1+...+1 \equiv k (\bmod n)$ so we get that $n \mid k$ so $n\leq k$,but we know that the folowing inequality is true for $n\geq 2$:
$(n+1)^{k}\geq (n+1)^{n}>(n-1)!+1$ contradicting the equality of the problem...
so we have $n\leq 4$ so we only need to check these cases...
I hope its clear now...  :maybe:


search
advanced fields
advanced fields unsolved

Can anyone versed in Physics help me with this problem?.

"Two slabs, each one inch thick, each have one surface at 0 degrees and the other at 100 degrees. At t=0, they are stacked with their 100 degree faces together and then the outside surfaces are held at 100 degrees. Find u(x,t) for t>0".

Can someone help?.  Should this problem contain u_0 and u_f

Thanks,
Cody

i think you need $u(x,0)$. Then you just have to find a solution to the heat diffusion equation, which can be done with fourier expansion. Try to search solution of the form $u(x,t)=f(x)g(t)$, i think (not sure) it will help.

I think the initial condition is that the temperature varies linearly within each block from 0 at the outside edge to 100 at the inside edge- then we introduce a discontinuity by holding the outside edges at 100. The system will of course decay toward a uniform temperature of 100.

another pb: find the evolution of the heat on a ring .

Because there are two slabs, u_0 is piecewise.  One for x in (0,1) where the temperature goes from 0 to 100 and one for x in (1,2) where the temperature gores from 100 to 0.  This gives u_0-u_f=100x-100 and 100-100x.


Determinati solutiile sistemului:

   2006^x=2005+y
   2006^y=2005+z
   2006^z=2005+x

$a>0\ ,\ x\in R\ : \ f(x)=(a+1)^x\ \ (\uparrow),\ g(x)=x+a\ \ (\uparrow)\Longrightarrow$ \[ \boxed {\ \{ \begin{array}{ccc} f(x) & = & g(y)\\ f(y) & = & g(z)\\ f(z) & = & g(x) \end{array} \ }  \] [u]The first case[/u] : $\boxed {\ x\le y\ }\Longrightarrow f(x)\le f(y)\Longrightarrow g(y)\le g(z)\Longrightarrow$

$\boxed {\ y\le z\ }\Longrightarrow f(y)\le f(z)\Longrightarrow g(z)\le g(x)\Longrightarrow \boxed {\ z\le x\ }\Longrightarrow x=y=z\ .$

[u]The second case[/u] : $x>y\Longrightarrow \ldots \Longrightarrow x>y>z>x\Longrightarrow x>x$, what is false.

Therefore, $\boxed {\ x=y=z\ \ ;\ \ (a+1)^x=x+a\ }.$ Denote $b=a+1>1$.

$b\ln b<1\Longrightarrow 0<x_1<x_2=1\ ;\ b\ln b=1$$\Longrightarrow x_0=1\ ;\ b\ln b>1\Longrightarrow 1=x_1<x_2\ .$

Thank you for this solution.

Here is my solution: If 2006^x - 2005=f(x), we have:

f(x)=y
f(y)=z
f(z)=x.
 So f(f(f(x)))=x.  But f is monotone, so f(x)=x (a known problem), and we have x=y=z.  We obtain that only (1,1,1) is solution.

In my solution, the function $f(x)=(a+1)^x-x-a$ isn't always monotonely; this point depends of the basis $a\ !$


calculus
derivative
function
trigonometry
topology
real analysis
real analysis unsolved

[quote="xirti"]Prove or (give a counterexample ) that for a derivative function there exists an interval $(a,b)$ such that $f$ is bounded on $(a,b)$.[/quote]
$f(x)=\sqrt{x}$
$a=0,b=1$

Give a counterexample. not an example of function bounded on an interval. every continuous function is bounded on some intervals. I asked to prove that for a function there exists [b]an[/b] interval such that f is bounded on it. Or, if this statment is false you may give a [b]counter[/b]example (an example of derivative function unbounded an every interval).

I assume that the question is whether the derivative $f=g'$ of an [i]everywhere[/i] differentiable function $g$ can be unbounded on every open interval. Then the answer is "no". Indeed, $f$ is a pointwise limit of a sequence of continuous functions $f_n(x)=n\left(g\left(x+\frac 1n\right)-g(x)\right)$. Let $F_N=\{x\in\mathbb R\,: \, |f_n(x)-f_m(x)|\le 1\text{ for all }n,m>N\}$. Each $F_N$ is closed and their union is the entire real line. Hence, by the Baire theorem, one of $F_N$ contains an open interval. The rest should be clear. Actually, a slight modification of this reasoning shows that $f$ must even have a lot of points of continuity.

Note carefully what fedja said: there must be [b]some[/b] interval on which $g'$ is bounded (and some interval on which $g'$ is continuous.) That certainly allows $g'$ to be discontinuous at some points and unbounded in some neighborhoods.

With that said, the example of $g(x) =\sqrt{x}$ on $(0,1)$ is not really what we're looking for. For every $a\in(0,1),$ there is a neighborhood of $a$ on which $g'$ is bounded. And it won't do to try to put this example on $[0,1)$ (or, for that matter, to use $\sqrt[3]{x}$ on the whole line) because then the function will not be differentiable at zero.

But here's a member of the classic family of examples:

Let $g(x)=\begin{cases}x^2\sin\left(\frac1{x^{10}}\right)&x\ne0\\0&x=0\end{cases}.$

This function is differentiable everywhere (even at zero) but the derivative is not bounded in any neighborhood of zero. In fact, $g'$ is not (absolutely) integrable in any neighborhood of zero.

Great solution fedja. I was even trying to find an counterexample :D  I have a question: It is wrong if I work with
$G_N=\{x\in \mathbb{R}: |f_n(x)-f(x)|\le 1 ,\forall n\ge N\}$ instead $F_N$ on the same method? It seems more natural to me.

The problem with that approach: we show that $F_N$ is closed by using the fact that both $f_m$ and $f_n$ are continuous for all pairs $m,n$; $F_N$ is an intersection of closed sets.

You don't know that $f$ is continous, so your $G_N$ sets are not obviously closed.

Random notational fact: $F$ usually stands for a closed set, $G$ for an open set.


calculus
integration
calculus computations

[hide]Performing the substitution $ x^a \equal{} u$ we have

$ I(a) \equal{} \int_0^1 \frac{1}{\sqrt{1 \minus{} x^a}}\,dx \equal{} \frac{1}{a} \int_0^1 u^{\frac{1}{a} \minus{} 1} (1\minus{}u)^{\frac12 \minus{}1}\,du$

$ \equal{} \frac{1}{a} B\left(\frac{1}{a}, \frac12\right) \equal{} \frac{1}{a} \frac{\Gamma \left(\frac{1}{a}\right) \Gamma \left(\frac12\right)}{\Gamma \left(\frac{1}{a} \plus{} \frac12\right)} \equal{} \frac{\sqrt{\pi}}{a } \frac{\Gamma \left(\frac{1}{a}\right)}{\Gamma \left(\frac{1}{a} \plus{} \frac12\right)}.$[/hide]


Guess the age based on the discussion below...

A: "Write down my date of birth in the DDMMYYYY format.  My age this year multiplied by DD and MM gives you YYYY."

B: "That's not good enough."

A: either DD or MM or both could be single-digit numbers or double-digit numbers.

B: "I'm still clueless."

A: "My son will join my birthday celebration this year."

B: "I'm still clueless."

A: "Both DD and MM are prime numbers."

B: "I got it!"

What is A's date of birth based on above dialogue?

[quote="amcavoy"]$x\implies\textrm{year}$

$y\implies\textrm{month}$

$z\implies\textrm{day}$

Then we have

$\left(2006-x\right)yz=x\implies x=\frac{2006yz}{yz+1}$

and $y\in\left\{2,3,5,7,11\right\}$, $z\in\left\{2,3,5,7,11,13,17,19,23,29,31\right\}$.  So now we have $2006=2\cdot 17\cdot 59$ so $yz=1,16,58$.  The only one that works here is $58=2\cdot 29$, so the date of birth is $\boxed{29/02/1972}$.[/quote]

hmm.... does this mean that the date of birth in this year is 29th Feb??

but 2006 is not a leap year?

right? what are we missing...

anyone...

February 29th, 1972 existed.  In fact, it was a Tuesday.  Remember that we aren't talking about 2006 here.


If the equation  $ ax^2\plus{}bx\plus{}c \equal{} 0$ ,with $ a , b , c$ non zero digits, has only one integer root, find all triples $ (a,b,c)$

Oops, I read it to mean only one root, which happened to be an integer T_T. It seemed a bit easy...


Prove there there exists positive integers $ a_1, a_2,..., a_ n$ such that $ |a_i\minus{}a_j|$ divides $ a_i\plus{}a_j$ for all distinct $ i$ and $ j$.

[quote="stevenmeow"]so you mean for all n?
it is easy to prove if say n=2 or 3[/quote]

yes, using induction


integration
algebra
function
domain
calculus
real analysis
real analysis unsolved

Suppose that $ f,g \in L^p([0,1])$ for $ 1<p$ finite. Assume that for all $ y \in (0,\infty)$

$ |\left\{g(x) > y\right\}| \leq \frac{1}{y} \int_{g(x) > y}f(x)dx$

Show that $ |g|_p \leq q |f|_p$, where $ q$ is the conjugate of p.

There seems to be a tacit assumption here that both $ f$ and $ g$ are nonnegative functions. I'll keep that assumption; without it, we'd have a number of absolute values in the statement but it would otherwise work the same way. It also doesn't matter at all that the domain is $ [0,1].$ The domain could be any measurable $ E,$ of finite or infinite measure.

A piece of exponent arithmetic that we'll use later: $ \frac{p}{p\minus{}1}\equal{}q.$

$ \|g\|_p^p\equal{}\int_0^{\infty}py^{p\minus{}1}|\{g(x)>y\}|\,dy \le\int_0^{\infty}py^{p\minus{}1}\int_{\{g(x)>y\}}f(x)\,dx\,dy$

Of course we use Tonelli. Why else write a double integral?

$ \equal{}\int_Ef(x)\int_0^{g(x)}py^{p\minus{}2}\,dy\,dx\equal{}\frac{p}{p\minus{}1}\int_Ef(x)g(x)^{p\minus{}1}\,dx$

Now we use Hölder. Note that $ (g(x)^{p\minus{}1})^q\equal{}g(x)^p.$

$ \le \frac{p}{p\minus{}1}\|f\|_p\left(\int_Eg^p\right)^{\frac{p\minus{}1}p}\equal{}q\|f\|_p\|g\|_p^{p\minus{}1}.$

If $ \|g\|_p<\infty$ (which we assumed), we can then divide $ \|g\|_p^{p\minus{}1}$ to get $ \|g\|_p\le q\|f\|_p.$


Let $A,B$ - real matrices $n \times n$. Suppose that there exist pairwise distinct $t_1, ..., t_{n+1} \in \mathbb{R}$ such, that $A+t_iB$ are nilpotent matrices. Show that $A$ and $B$ are nilpotent.

I remember this problem was discussed many times on Mathlinks. Also it was on IMC 1995, you can find it from the http://www.imc-math.org.

Oh sorry, I know the solution, I was hoping there are more than one solution to this problem.


geometry
incenter
circumcircle
geometry unsolved

Let $ I$ and $ O$ denote respectively the incenter and the circumcenter of a triangle $ ABC$. Given that $ \hat{AIO} \equal{} 90$. Prove that the area of the triangle $ ABC$ is less than $ \frac {3\sqrt {3}}{4}AI^2$

[url]http://www.mathlinks.ro/viewtopic.php?t=168836&search_id=79333260&start=20[/url]
You should know you was posting many problems of M&Y.
Lock this topic!


algebra
polynomial
trigonometry
Vieta
trig identities

Let $ a < b < c$ and $ a,$ $ b$ and $ c$ are roots of the equation: $ x^3 \minus{} 3x \plus{} 1 \equal{} 0.$ Prove that
\[ \frac {a}{b \plus{} 1} \plus{} \frac {b}{c \plus{} 1} \plus{} \frac {c}{a \plus{} 1} \equal{} \minus{} 3
\]

[hide]Note from the coefficient of $ x^{2}$ that $ a \plus{} b \plus{} c \equal{} 0$. Then $ a\plus{}c \equal{} \minus{}b$, $ b\plus{}c\equal{}\minus{}a$, $ a\plus{}b\equal{}\minus{}c$, so it suffices to show that:

\[ \frac{1}{a\plus{}1} \plus{} \frac{1}{b\plus{}1} \plus{} \frac{1}{c\plus{}1} \equal{} 0\]

Note that $ P(x\minus{}1) \equal{} (x\minus{}1)^{3} \minus{} 3(x\minus{}1) \plus{} 1$ has roots $ a\plus{}1, b\plus{}1, c\plus{}1$. So the polynomial with reverse coefficients has the roots $ 1/(a\plus{}1), 1/(b\plus{}1), 1/(c\plus{}1)$. So we must show that the $ x^{2}$ coefficient of the polynomial with reverse coefficents is $ 0$; the $ x$ coefficient of $ P(x\minus{}1)$ is $ 0$. This is easy: The $ x$ coefficient is simply $ 3\minus{}3\equal{}0$ (from binomial expansion), as desired.[/hide]

[quote="vishalarul"][hide]Note from the coefficient of $ x^{2}$ that $ a \plus{} b \plus{} c \equal{} 0$. Then $ a \plus{} c \equal{} \minus{} b$, $ b \plus{} c \equal{} \minus{} a$, $ a \plus{} b \equal{} \minus{} c$, so it suffices to show that:
\[ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \equal{} 0
\][/hide][/quote]

How do you get that?  I can't figure out how to get from the given equation to this equation.

Yeah, i was wondering that too.

All I get is (2-c)/(b+1)+(2-b)/(a+1)+(2-a)/(b+1)=0.

Sorry, I messed up (and I can't edit that post anymore...)

[hide="Ideas"]It turns out that indeed $ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \equal{} 0$. I thought we had to prove that

$ \frac {a \plus{} c}{b \plus{} 1} \plus{} \frac {a \plus{} b}{c \plus{} 1} \plus{} \frac {b \plus{} c}{a \plus{} 1} \equal{} \minus{} 3$

(which should be true, since $ a \plus{} b \plus{} c \equal{} 0$)

However, this is indirectly helpful. It remains to show that

$ \frac {c}{b \plus{} 1} \plus{} \frac {a}{c \plus{} 1} \plus{} \frac {b}{a \plus{} 1} \equal{} 0$

But I bet that's just as hard as the original problem. Well, if we expand this, we need to show that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \plus{} a^2 b \plus{} b^2 c \plus{} c^2 a \equal{} 0$

Since

$ 0 \equal{} (a \plus{} b \plus{} c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \plus{} (ab \plus{} bc \plus{} ca) \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \minus{} 3$

we need to show that $ a^2 b \plus{} b^2 c \plus{} c^2 a \equal{} \minus{} 3$. If we look at $ (a \plus{} b \plus{} c)^3 \equal{} 0$ and expand,

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a^2 b \plus{} b^2 c \plus{} c^2 a \plus{} a b^2 \plus{} b c^2 \plus{} c a^2) \plus{} 6abc \equal{} 0$

From $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(STUFF) \equal{} 0$, we get $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc \equal{} \minus{} 3$,  so we need to show that:

$ a b^2 \plus{} b c^2 \plus{} c a^2 \equal{} 6$

Let $ X \equal{} a^2 b \plus{} b^2 c \plus{} c^2 a$ and $ Y \equal{} a b^2 \plus{} b c^2 \plus{} c a^2$. Notice that:

$ X \plus{} Y \equal{} ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \equal{} \minus{} 3abc \equal{} 3$

And now, I'm stuck... I'm not sure how to incorporate $ a < b < c$ into this...

EDIT: $ Y \minus{} X \equal{} (b \minus{} a)(c \minus{} b)(c \minus{} a)$, but that doesn't really help...

EDIT2: Here's some actual progress. $ \minus{}X\equal{}a/b\plus{}b/c\plus{}c/a$ and $ \minus{}Y\equal{}b/a\plus{}c/b\plus{}a/c$. Mutiplying these two together and expanding,

$ XY \equal{} 3\plus{}(ac/b^2\plus{}ab/c^2\plus{}bc/a^2) \plus{} (b^2/ac\plus{}c^2/ab\plus{}a^2/bc)$

The second sum, $ (b^2/ac\plus{}c^2/ab\plus{}a^2/bc)$ is easy: put a common denominator $ abc\equal{}\minus{}1$:

$ (b^2/ac\plus{}c^2/ab\plus{}a^2/bc) \equal{} \minus{}(a^3\plus{}b^3\plus{}c^3) \equal{} 3$

The first sum is more annoying. We put a common denominator, and we wish to find $ a^3b^3\plus{}b^3c^3\plus{}c^3a^3$. Noting that this equals $ [(a^3\plus{}b^3\plus{}c^3)^2\minus{}(a^6\plus{}b^6\plus{}c^6)]/2 \equal{} \minus{}24$ (well, at least it should: a formal proof requires newton's sums, which aren't that hard to compute). So FINALLY, $ XY \equal{} 3 \plus{} 3 \minus{} 24 \equal{} \minus{}18$

So $ (X,Y)\equal{}(\minus{}3,6)$ or $ (6,\minus{}3)$. To figure out which one we use $ Y\minus{}X \equal{} (b\minus{}a)(c\minus{}b)(c\minus{}a)>0$ (it's not that hard to prove; just plug in and factor) since $ a<b<c$.

So $ X<Y$, implying that $ X \equal{} \minus{}3$. Which means we're done.[/hide]

This problem is a lot harder than I thought. :blush: (I really, really, hope I didn't make any mistakes...)

Hints

Use Trig Substitutions to explicitly calculate a,b, and c.

Then simplify that expression to prove to prove something else, and then use trig identities and Vieta's sums to get to the core.

$ a\plus{}b\plus{}c\equal{}0$ is due to the fact that a cubic polynomial $ a_3x^3\plus{}a_2x^2\plus{}a_1x\plus{}a_0$ is equal to $ a_3[x^3\minus{}(r_1\plus{}r_2\plus{}r_3)x^2\plus{}(r_1r_2\plus{}r_1r_3\plus{}r_2r_3)x\minus{}r_1r_2r_3]$, where $ r_1$, $ r_2$, and $ r_3$ are the roots of the polynomial. In this case $ a$, $ b$, and $ c$ are the roots of the polynomial. Therefore, vishalarul is correct.

[hide="Trig Method"]Expanding,

$ \sum_{cyc} a(a\plus{}1)(c\plus{}1) \equal{} \minus{}3(a\plus{}1)(b\plus{}1)(c\plus{}1)$
$ \iff$
$ \sum_{cyc} a^{2}c \plus{} ac \plus{} a^{2} \plus{} a \equal{} \minus{}3(abc \plus{} ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c \plus{} 1)$
$ \iff$
$ \sum_{cyc} a^{2} c \plus{} (a\plus{}b\plus{}c)^{2} \minus{} (ab\plus{}bc\plus{}ca) \plus{} (a\plus{}b\plus{}c) \equal{} \minus{}3 (abc\plus{}ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c \plus{} 1)$
$ \iff \text{Vieta}$
$ \sum_{cyc} a^{2} c \equal{} \minus{}3 \minus{} 3(\minus{}1\minus{}3\plus{}1) \equal{} 6$
$ \iff$
$ \sum_{cyc} a^{2} c/abc \equal{} \minus{}6$
$ \iff$
$ \sum_{cyc} a/b \equal{} \minus{}6$

Since the roots of $ x^{3} \minus{} 3x \plus{} 1 \equal{} 0$ are $ a \equal{} 2\cos 160^{\circ}$, $ b\equal{}2\cos 80^{\circ}$, $ c\equal{}2\cos 40^{\circ}$, we need to show:
$ \cos 160^{\circ}/\cos 80^{\circ} \plus{} \cos 80^{\circ}/\cos 40^{\circ} \plus{} \cos 40^{\circ}/\cos 160^{\circ} \equal{} \minus{}6$

We use double angle formula for cosine:

$ 2\cos 80^{\circ} \minus{} 1/\cos 80^{\circ} \plus{} 2\cos 40^{\circ} \minus{} 1/\cos 40^{\circ} \minus{} 2\cos 20^{\circ} \plus{} 1/\cos 20^{\circ} \equal{} \minus{}6$
$ \iff$
$ 2\cos 80^{\circ} \minus{} 1/\cos 80^{\circ} \plus{} 2\cos 40^{\circ} \minus{} 1/\cos 40^{\circ} \plus{} 2\cos 160^{\circ} \minus{} 1/\cos 160^{\circ} \equal{} \minus{}6$
The sum of the roots is $ 0$, so we need to show that twice the sum of the reciprocal of the roots is $ 6$. ($ 1/\cos 40^{\circ} \plus{} 1/\cos 80^{\circ} \plus{} 1/\cos 160^{\circ} \equal{} 6$). But this is clearly true: the polynomial with negative roots is the polynomial with reverse coefficients: $ x^{3} \minus{} 3x^{2} \plus{} 1$. The sum of the roots of this polynomial is $ 3$, so twice the sum of the roots (i.e. $ 1/\cos 40^{\circ} \plus{} 1/\cos 80^{\circ} \plus{} 1/\cos 160^{\circ}$) is indeed $ 6$.[/hide]
That was pretty long. I wonder if there's a shorter way to do it.
[hide="And is there a way to prove that..."]the roots of $ x^{3} \minus{} 3x \plus{} 1$ are $ \cos 40^{\circ}$, $ \cos 80^{\circ}$, and $ \cos 160^{\circ}$?[/hide]
which was the key fact used in the proof?

[quote="vishalarul"]the roots of $ x^{3} \minus{} 3x \plus{} 1$ are $ \cos 40^{\circ}$, $ \cos 80^{\circ}$, and $ \cos 160^{\circ}$?
which was the key fact used in the proof?[/quote]
No! With them this is nicer. For example:
Let $ a < b < c$ and  $ a,$ $ b$ and $ c$ are roots of the equation: $ x^3 \minus{} 6x \plus{} 1 \equal{} 0.$  Prove that
\[ \frac {a}{b \plus{} 1} \plus{} \frac {b}{c \plus{} 1} \plus{} \frac {c}{a \plus{} 1} \equal{} \minus{} \frac {5 \plus{} \sqrt {93}}{4}
\]

That is so random, but there must be some nice way of getting it.


[size=150]Next Question[/size]

Suppose that we are to be presented with $n$ distinct prizes in sequence.  After being presented with a prize we must immediately decide whether to accept it or reject it and consider the next prize.  The only information we are given when deciding whether to accept a prize is the relative rank of that prize compared to ones already seen.  Suppose that once a prize is rejected it is lost, and that our objective is to maximize the probability of obtaining the best prize.  Assuming that all $n!$ orderings of the prizes are equally likely, what is the probability that this strategy will select the best prize?

Why did you post the same question twice?

sry, you can delete this one....I had clicked the wrong link :oops:


I know that you cannot take the AIME A and then say I want to not count that and take the AIME B for sure... I don't know about the skipping thing, but it sounds reasonable.

AIME & USAMO Registration

There is no additional registration fee for the American Invitational Mathematics Examination, unless you choose to take the second sitting of the exam. An additional administrative/shipping fee is charged, with a minimum of $25.00 for the first 10 students and $50.00 for over 10 students. This is explained more thoroughly in the AMC 10/12 and AIME Teacher's Manual.

Shouldn't this apply, such that I could register for the B?

[quote="spang"]Shouldn't this apply, such that I could register for the B?[/quote]

Not if you already took the first AIME this year, as explained in the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04AIME/html/04-04aime-text.html]AIME manual[/url]: [b]Under no circumstances can a student take both AIME's.[/b] So, no, no one can annull an AIME score from today's test just to take the test on the second date.

[quote="spang"]I was wondering if we could null the AIME A score and intead count the AIME B score to be official? Also, is it legit to have skipped AIME A to take AIME B (@ another testing location)?[/quote]

um, you can skip the AIME a to take the AIME b, but in general the B is harder and it's not considered very legit.

[quote="z9R4C3"][quote="WarpedKlown1335"]Well umm...this isn't very uplifting to someone here who's taking the AIME II.  LoL., wish me luck.[/quote]
heh, sorry.  but you didn't skip the first one on purpose, right?  good luck!  :)[/quote]

heck...there's a lot of us that'll need good luck for this test!!

The first AIME would have been during my spring break so I couldn't take it.

[quote]It doesn't matter which is harder, because for the USAMO they take a number of people from each test proportional to the number who took each one. So, if the B is a lot harder, they'll still take the top 1% on the B.[/quote]

Could someone knowledgable comment more on this, because as far as I know, the USAMO cutoff index has remained constant for both AIME 1 and AIME 2 takers in the past few years (ie the cutoff was 226 for everyone last year).

Really? So what do they do when one test is obviously easier than another?

I presume that if a serious discrepancy between the tests should present itself, they would do something about it.  However, that seems rather improbable, especially given that the average score still hovers consistently at a 3 or below, right?


geometry
geometric transformation
reflection

Suppose we have an equilateral triangle with vertices at points $(0, 0), \ (10, 0), \text{ and }(5, 5\sqrt{3})$. If a point $(a,b)$ is chosen within the triangle, what is the shortest distance that the point can travel so that it touches all three sides? (Touching a vertex is the same as touching 2 of the sides.) Is there a "best" point that minimizes this distance? What if the point is outside of the triangle?

[hide=" :maybe:.. "]
Just a guess..but can you start at the vertex and make your way down to the opposite side thus touching all three sides and its $5\sqrt{3}$
[/hide]

[hide]Something tells me it is just the altitude of $5\sqrt{3}$[/hide]

That's what I'm thinking also. Can anyone prove it?

What about the general case?

I think the general is that you take the vertex opposite the largest side and draw the altitude and that is the answer.

heres an attempted proof that the altitude is the minimal distance, but its not the only path with that length

Clearly we must start on a side, otherwise would just be stupid.  Simply reflect the triangle over the edge other than the one you are startng at, particularly the one your starting point is closer to, so then go from whatever point you start on to the opposite edge of the reflected triangle, the minimal length of this is the perpendicular of 5sqrt(3).  Then relfect the part of the path on the reflected triangle back on to the original, and it has the same distance.


im really confused over this problem.
Can anybody please help or advise please?

=========================
Mark each statement true or false
1. { x | x = A } = {A}
2. { x | x $\in$ A } = A
3. { x | {x} $\subset$ A } = {A}
4. { x | {x} $\subset$ $\varnothing$ } = $\varnothing$
5. { x | {x} = A } = {A}
6. { x | {x} $\in$ A } = A
=========================

Thank you very much for your advise ^^!

First, we don't solve homework problems here (at least, not immediate homework problems), secondly, try posting it in some high-school or pre-olympiad forum. As you can see, this is the advanced forum, so probably nobody will bother to answer you.

[quote="Ointer"]how do you know???
could anyone give me some suggestions????[/quote]

Let sets $A$ and $B$. Do you know defined of $A=B$ ?  :huh:


Find all functions f:$ R \to R$ such that
$ f(x^3\minus{}y)\plus{}2y(3f^2(x)\plus{}y^2)\equal{}f(y\plus{}f(x))$ for all $ x;y \in R$

$ P(x,0): f(x^3)\equal{}f(f(x))$----------------(1)
Now we prove $ f(x)$ is injective. Suppose it is not true and $ f(a)\equal{}f(b)$ for $ a\neq b$.
$ P(a,y)\minus{}P(b,y): f(a^3\minus{}y)\equal{}f(b^3\minus{}y)$. So $ f(x)$ is periodic with period $ a\minus{}b$.
$ P(x,a\minus{}b): f(x^3)\plus{}6(a\minus{}b)f(x)^2\plus{}2(a\minus{}b)^3\equal{}f(f(x))\equal{}f(x^3)$. This implies
$ 3f(x)^2\plus{}(a\minus{}b)^2\equal{}0$, a contradiction with $ f(x)\in R$.
Now formula (1) implies $ f(x)\equal{}x^3$.

$ y: \equal{} \frac {x^3 \minus{} f(x)}{2}$

[b]Done[/b]

 :!:  :w00t:  :P

[hide="BTW"]I Hope you meant $ f^2(x)\equal{}(f(x))^2$ and not $ f(f(x))$. [/hide]


analytic geometry
geometry
parallelogram
trigonometry
absolute value
trig identities
Law of Cosines

Given a triangle $ ABC$, let $ M$ be mid point of $ BC$ such that $ AB\equal{}4,\ AM\equal{}1$. Find the possible least measure of angle $ \angle{BAC}$.

[hide="my solution..."]
Let $ A$ be the center of a circle of raidus $ 1$. Now $ M$ is a point on that circle and the midpoint of $ BC$.
Now we can see that $ \angle BAC$ is minimised then $ \overline{BC}$ is tangent to the cirlce

$ \Rightarrow Min: \angle BAC \equal{} 2\arccos \left(\frac{1}{4}\right) \sim 151.045...$[/hide]

[hide]
Let's place triangle $ ABC$ in coordinate system such that $ A (0,0)$, $ B (4,0)$ and $ C$ lies in II. quadrant. Let $ M$ lie on circle $ x^2 + y^2 = 1 \implies M (sin \phi,cos \phi)$. Let $ x_0$ and $ y_0$ be the coordinates of point $ C$. Since $ M$ is midpoint of $ BC$, $ sin \phi = \frac {x_0 + 4}{2}$ and $ cos\phi = \frac {y_0}{2} \implies x_0 = 2(sin\phi - 2), y_0 = 2cos\phi$. Straight line $ AC$ has equation $ y = \frac {y_0}{x_0}x \implies y = \frac {cos\phi}{sin\phi - 2}x$. The angle we are looking for is the angle which this line closes with positive x-axis, so $ tg\alpha = k = \frac {cos\phi}{sin\phi - 2}$. This angle will achieve minimum for $ |k|_{max}$ (because $ C$ is in II. quadrant). Let $ f(\phi) = \frac {cos\phi}{sin\phi - 2}$. Now let's find $ f'(\phi)$ in order to find extremes of the function. Since $ f'(\phi) = \frac {2sin\phi - 1}{(sin\phi - 2)^2}$ which equals $ 0$ for $ sin\phi = \frac {1}{2} \implies f(\phi)_{min} = - \frac {1}{\sqrt {3}}$ (notice that maximum has same absolute value since $ f(\phi)$ is a trigonometric function). Using $ f(\phi) = tg\alpha \implies \boxed{\alpha_{min} = 150^{\circ}}$.[/hide]

[hide="or"]
By parallelogram theorem,we have
$ AB^2\plus{}AC^2\equal{}2(AM^2\plus{}BM^2)\Longrightarrow 16\plus{}b^2\equal{}2(1\plus{}\frac{a^2}{4})\Longrightarrow a^2\equal{}2b^2\plus{}28$.
and by law of cosines and AM-GM we have
$ \cos\angle BAC\equal{}\frac{16\plus{}b^2\minus{}a^2}{8b}\equal{}\frac{\minus{}b^2\minus{}12}{8b}\equal{}\minus{}\frac{1}{8}\left(b\plus{}\frac{12}{b}\right)\le \minus{}\frac{2}{8}\sqrt{b\cdot \frac{12}{b}}\equal{}\minus{}\frac{\sqrt{3}}{2}$,
equality occurs when $ b\equal{}2\sqrt{3}$.

Hence the least is $ \boxed{150^\circ}$
[/hide]


Find the condition of $ f(x)$to this equation is true
$ lim_{x\to 0}(\frac {1}{x}(1 - \frac {1}{f(x)}) = lim_{x\to 0}(\frac {1-\frac {1}{L}}}{x}$
where $ L = lim_{x\to 0} f(x)$
And can you show your genaral one

Do you mean:

$ lim_{x \to 0} \left ( \frac{1\minus{}\frac{1}{f(x)}}{x} \right ) \equal{} lim_{x \to 0} \left( \frac{1 \minus{} \frac{1}{L}}{x} \right )$, where $ L : \equal{} lim_{x \to 0} f(x)$?

For each $ L \neq 1$ the RHS is 'equal' to negative or positive infinity I would say. Writing such an equation down is kind of meaningless in those cases.
So you would have to have $ lim_{x \to 0} f(x) \equal{} 1$. We then get the inderminate form $ \frac{0}{0}$ on the LHS, so we can try to apply l'Hopital to get:

$ lim_{x \to 0} \left ( \frac{1\minus{}\frac{1}{f(x)}}{x} \right ) \equal{} lim_{x \to 0} \left ( \frac{\frac{1}{(f(x))^2}f^`(x)}{1} \right )
\equal{} lim_{x \to 0} \frac{f^`(x)}{(f(x))^2},$ which would have to be $ 0$ to fit the RHS.

I thus come to the following two conditions:

- $ lim_{x \to 0} f(x) \equal{} 1$

- $ lim_{x \to 0} \frac{f^`(x)}{(f(x))^2} \equal{} 0$

Did I make any sense in the above?

[quote="Tyl"]

- $ lim_{x \to 0} f(x) \equal{} 1$

- $ lim_{x \to 0} \frac {f^`(x)}{(f(x))^2} \equal{} 0$

[/quote]
Thank you,I am too


How many integers satisfy $ \minus{}4 \le 2x\plus{}1 \le 6$?

Subtracting 1 from all,

$ \minus{}5 \le 2x \le 5$

Dividing by 2

$ \minus{}2.5 \le x \le 2.5$

so x=-2,-1,0,1,2 or 5 vlues.


Solve in $ \mathbb{R}$ the equation $ \left| x \minus{}\frac{\pi}{6} \right|\plus{}\left| x\plus{}\frac{\pi}{3} \right|\equal{}\arcsin \frac{x^3\minus{}x\plus{}2}{2}.$


induction
real analysis
real analysis unsolved

Define a sequence by $ n(n\minus{}1)a_{n} \equal{} (n\minus{}1)(n\minus{}2)a_{n\minus{}1}\minus{}(n\minus{}3)a_{n\minus{}2}$, $ n \ge 2$, where $ a_0 \equal{} a_1$ are any real numbers. Prove that $ \displaystyle\sum_{n\equal{}0}^{\infty}a_{n}$ converges to $ a_{0}e$.

[hide="answer"] Note that $ \frac {1}{(n \minus{} 2)!} \equal{} \frac {1}{(n \minus{} 3)!} \minus{} \frac {n \minus{} 3}{(n \minus{} 2)!}$, so by induction it's easy to show $ a_n \equal{} \frac {a_0}{n!}$, and the sum is $ a_0\sum_{n \equal{} 0}^\infty{\frac {1}{n!}} \equal{} a_0 e$ [/hide]


geometry
3D geometry
probability
expected value
probability and stats

a funny one I was asked yesterday during a job interview:
there is a ant walking on the edges of a cube. Every time she arrives at a vertex of the cube, she chooses randomly another edge, and continues to walk on this edge. Suppose that it takes $ 1$ minute to travel one edge: what is the average amount of time it takes for the ant to go from one vextex to the opposite one ?

I'm surprised I wasn't able to find this already on the forum anywhere.

[hide]Let $ A$ be your starting vertex, call all three of the vertices adjacent to $ A$ $ B$, call all three of the vertices adjacent to the $ B$ $ C$.  Let $ a$ be the expected number of steps from $ A$, $ b_A$ be the expected number of steps from one of the $ B$ if you've just arrived from $ A$, $ b_C$ be the expected number of steps from one of the $ B$ if you've just arrived from $ C$, and $ c$ the expected number of steps from one of the $ C$.  Then it's easy to see that
\begin{align*} a & = 1 + b_A \\
b_A & = 1 + c \\
b_C & = 1 + a/2 + c/2 \\
c & = 1 + b_C/2. \end{align*}
Solving this system gives $ a = 6, b_A = 5, b_C = 6$ and $ c = 4$.

That's pretty good savings compared with a "dithering" ant that chooses from among the three edges randomly, regardless of where it came from.[/hide]


Steal the avatar!
http://www.artofproblemsolving.com/Forum/images/avatars/1678273634461505c0c3a30.gif
[img]http://www.artofproblemsolving.com/Forum/images/avatars/1678273634461505c0c3a30.gif[/img]
Supporters:
footballrocks41237

Please don't, this is retarded.
The point of avatars are to make the posts more distinguishable or whatever, if a bunch of people have same avatars, I actually have to check their username before reading their posts, which gets annoying

Dude, everyone should just steal my avatar. It's so creative.  :D

I use avatars as a way to express my general mood. But no one can ever figure out what each one means.

w00t yeah i stole it too cause i was bored XD


How many positive integers $ n$ satisfy $ 200 < n^2 < 900$?

The smallest number that can be $ n$ is $ 15$ because $ 15^2\equal{}225$. The largest number that can be $ n$ is $ 29$ because $ 29^2\equal{}841$. Therefore, the numbers that can be $ n$ are $ 15\minus{}29$, which is a total of $ 15$ numbers


Hello everyone,

I am having some trouble with the following two problems pertaining to exponents, and I would appreciate any help or hints!

Thank you!

---

1. Let $ a$ and $ b$ be real numbers with $ a > 1$ and $ b > 0$. If $ ab = a^b$ and $ \frac{a}{b} = b^3$, find and b. (Hint: Use substitution or multiply the equations side by side)

First off, could anyone please explain what is meant by multiplying the equations side by side?

[u]My Work[/u]:

Since $ \frac{a}{b} = b^3$, $ a = b^4$.

I substituted this into the other equation:

$ b^{4}(b) = (b{}^4){}^b$

$ b^5 = b^{4b}$

$ 5 = 4b$

$ b = \frac{4}{5}$.

However, the provided answer for b is 0.5 (I wanted to get the right answer for $ b$ before continuing to find $ a$.Where might I have erred?)

2. Let $ f(x) = 2^{kx} + 9$ where $ k$ is a real number. If $ f(3): f(6) = 1: 3$, find the value of $ 2^{3k}$.

Since $ f(6) = f(3) x 3$:

$ \frac{2^{6k} + 9}{2^{3k} + 9} = 3$

$ \frac{2{}^{3k}{}^2 + 9}{2^{3k} + 9} = 3$

Let $ y = 2^{3k}$:

$ \frac{y^2 + 9}{y + 9} = 3$

$ y^2 + 9 = 3y + 27$

$ y^2 - 3y - 18 = 0$

$ (y - 6)(y + 3) = 0$

$ y = -3, 6$

The provided answer gives $ y = 2^{3k} = 6$. Here, I am unsure of how I can rule out -3 as a solution for y.

[/i]

This place:
5 = 4b
b is 5/4, NOT 4/5   :)

And about second. It's impossible to get negative value, when you are involuting positive number!


Determine all real numbers $ r$ such that there is precisely one pair $ (x,y)$ of real numbers satisfying the conditions :

$ 1. y\minus{}x\equal{} r$
$ 2.  x^2 \plus{}y^2 \plus{} 2x \le 1$.

The values are $ r \in \{ \minus{}1, 3 \}$; for $ r \in (\minus{}1, 3)$ there are two solutions, while for $ r \in (\minus{}\infty, \minus{}1) \cup (3, \plus{}\infty)$ there are none.
$ x^2 \plus{} y^2 \plus{} 2x \equal{} (x\plus{}1)^2 \plus{} y^2 \minus{}1$, so the second relation describes the points inside the closed disk $ D$ of center $ (\minus{}1,0)$ and radius $ \sqrt{2}$, while the first describes the line $ \ell$ given by $ y \equal{} x\plus{}r$, parallel to the first bisector. The unique solution is obtained when $ \ell$ is tangent to $ D$.

hello, the discriminant of the quadratic equation
$ x^2\plus{}x(r\plus{}1)\plus{}\frac{r^2\minus{}1}{2}\equal{}0$ must be zero, this is the case if
$ \left(\frac{r\plus{}1}{2}\right)^2\equal{}\frac{r^2\minus{}1}{2}$
$ \minus{}r^2\plus{}2r\plus{}3\equal{}0$ with the solutions
$ r\equal{}3$ or $ r\equal{}\minus{}1$
Sonnhard.


Dear Math Enthusiast,

The 2006 Four-by-Four Competition took place on Thursday, February 9th. This competition consists of ten rounds in which teams of four are challenged to solve four problems in three minutes. The first-place winners in each division are:

School: Vestavia Hills High School, Vestavia Hills, AL

Middle School Team:
     Jack Dickerson
     Kevin Shutzberg
     Max LaVictoire
     Sam Kallman
     from The Davis Academy, Atlanta, GA

"Rock" Division Team:
     Alan Chou
     Peter Cheng
     Kathryn Brewer
     Rebecca Mai
     from Grissom High School, Huntsville, AL

"Paper" Division Team:
     Linda Liu
     William Dobny
     Stephanie Wang
     Zane Blanton
     from Parkview High School, Lilburn, GA

"Scissors" Division Team:
     Amir Talebi
     Eric Chatterjee
     Elliot Lee
     Dianna Wu
     from Eastlake High School, Sammamish, WA

The recognized scores in each division were:

Schools: 83-210
Middle School Teams: 19-40
"Rock" Division Teams: 60-64
"Paper" Division Teams: 61-72
"Scissors" Division Teams: 68-86

Complete results for the competition (the top 25 school and the top 5 teams in each division) are available on our website; take a look to see who in your area you can congratulate!

[quote="tom_clymer"]
"Scissors" Division Team:
     Amir Talebi
     Eric Chatterjee
     Elliot Lee
     Dianna Wu
     from Eastlake High School, Sammamish, WA
[/quote]

yay :)

a nice energy boost before the AIME :)

[quote="tom_clymer"]Dear Math Enthusiast,

"Scissors" Division Team:
     Amir Talebi
     Eric Chatterjee
     Elliot Lee
     Dianna Wu
     from Eastlake High School, Sammamish, WA
[/quote]

Good to know I won!
Wait, I never did this...

-Eric Chatterjee

Hello.

In case there is any confusion arising from the previous post, I just wanted to let everyone know that I (Eric Chatterjee, Eastlake High School Class of 2007, Sammamish, WA), am the one who took part in the team that won the Scissors Division of Four by Four.

Hope that this is now clear to everyone viewing this topic. :)

According to my research, there are actually 3 of us. You in WA, me in OH and another "Eric Chatterjee" somewhere in Germany. Know any others? :)

Based on my search, I believe there is a director/producer by this name.  ;)


advanced fields
advanced fields unsolved

$ E$ is a fibration over $ B$ with fiber $ F$, and $ F\minus{}>E\minus{}>B$ is an exact sequence,
if $ F$ is simply connected , then $ \pi(E)\equal{}\pi(B)$. Is there such a thm? in which book can
I find the proof?

[url]http://en.wikipedia.org/wiki/Homotopy_group#Long_exact_sequence_of_a_fibration[/url]

take that exact sequence and apply it. they probably also give references.


I have a question about midpoints.  I am sure it is a simple solution but I am having difficulty solving.  P Q R S ant T are on line k such that Q is the midpoint of PT, R is the  midpoint of QT and S is the midpoint of RT.  If PS=9, then what is PT?

This is how far I have gotten
P________Q____R__S__T
QT=PT/2
RT=QT/2
St=RT/2
I know that PQ=QT or PQ=PT/2
Now I am stuck.

Any help would be appreciated and thanks :D

[quote]This is how far I have gotten
P________Q____R__S__T
QT=PT/2
RT=QT/2
St=RT/2
I know that PQ=QT or PQ=PT/2
[/quote]

Now make use of these statements to form an expression for PS:

QT = PQ = 2QR = 4RS.

PS = PQ + QR + RS
     = 7RS

So, RS = PT / 8 = 9 / 7

So, PT = 72 / 7

Alternatively,

PS = PQ + QR + QS
     = PT / 2 + PT / 4 + PT /8
     = 7PT / 8

Again, 7PT / 8  = 9
                7PT = 72
                  PT = 72 / 7

Use your equality statements to form an expression for a portion of PT and solve for that portion.  Then use the relationship of that portion to PT to solve for PT.

Easy!
[hide] As you know:
  ST = 1/2 RT =1/4 QT = 1/8 PT
or  ST = 1/7 PS = 1/7 * 1/9 = 1/63
=> PT = 9 + 1/63 = 568/63[/hide]
 Maybe I am wrong. I am not sure! :(

[hide="Answer to dholstein"]
Alright $PQ=\frac{1}{2}PT, QR=\frac{1}{4}PT, RS=ST=\frac{1}{8}PT$. Let's make the value of $PT=x$ so all the named segments are the named fraction of $x$. Now $PS=PQ+QR+RS=\frac{x}{2}+\frac{x}{4}+\frac{x}{8}=\frac{7x}{8}$. So we set $\frac{7x}{8}=9=\frac{7x}{8}=\frac{72}{8}$ which we can make $7x=72$, so $x=\frac{72}{7}=PT$[/hide]

[hide="Correction for Chocolate milk"]
In the second line of your work you have $PS=(\frac{1}{7})(\frac{1}{9})$. However, you are supposed to instead have $9(\frac{1}{7})$ because $9$ is 7 times the amount of $ST$, which is $\frac{9}{7}$ then add that to $9$ to obtain $\frac{72}{7}$[/hide]

[hide="Correction for mustafa"]
In the last step you are supposed to divide $\frac{9}{(\frac{7}{8})}$ which is basically $(\frac{8}{7})(9)=\frac{72}{7}$[/hide]

Whoops, lol I think i'm too tired to do math right now :oops:


Let $ ABCD$ be a quadrilateral , $ X$ and $ Y$ be the mid points of $ AC$ and $ BD$ respectively. The lines through $ X$ and $ Y$ respectively parallel to $ BD$ and $ AC$ meet at $ o$. Let $ P,Q,R,S$ be the mid points of $ AC,BC,CD,DA$ respectively. Prove that $ [APoS]\equal{}\frac{1}{4}[ABCD]$.

Because of $ (APS) \plus{} (CQR) \equal{} \frac {(ABD)}{4} \plus{} \frac {(CBD)}{4} \equal{} \frac {(ABCD)}{4},$ it is enough to prove that $ (OPS) \equal{} (CQR)$ $ ,(1)$

Because of now, the triangles $ \bigtriangleup OPS,\ \bigtriangleup CQR,$ have $ PS \equal{} \frac {BD}{2} \equal{} QR,$ it is enough to prove that their altitudes through $ O,\ C$ respectively, are congruent.

But it is true, because of $ XM \equal{} KN$ $ ($ easy to prove from $ MC \equal{} MK$ and $ XA \equal{} XC$ $ ),$
where $ K\equiv AC\cap BD,\ M\equiv AC\cap QR,\ N\equiv AC\cap PS$ and the proof is completed.

Kostas Vittas.


trigonometry
limit
function
integration
real analysis
real analysis solved

I still hope somebody will solve the nested root problem. Meanwhile, let me post one more. It is not really hard but I like it anyway...

Find $\sum_{n\geq 1}\frac1{2^n}\tan \frac1{2^n}$.

I think that if we also add the term corresponding to $n=0$, which is more natural anyway :), we get $1-2\cot 2$, but I'm not that sure. I might have made some computational errors.

First of all, we develop $x\tan x$ in a Taylor series around the origin: $x\tan x=\sum_{n\ge 1}\frac{(-1)^{n-1}\cdot 2^{2n}\cdot (2^{2n}-1)\cdot B_{2n}}{(2n)!}\cdot x^{2n}$. After this, we fix $n$, and sum up all the terms corresponding to $n$ as $x$ runs through the powers of $\frac 12$ from $0$ to $\infty$. That ugly $2^{2n}-1$ cancels out, and a $2^{2n}$ appears, so the series looks like $\sum_{n\ge 0}\frac{(-1)^{n-1}\cdot 4^{2n}\cdot B_{2n}}{(2n)!}\ (*)$.

Now, since $x\cot x=\sum_{n\ge 1}\frac{(-1)^n\cdot B_{2n}}{(2n)!}\cdot (2x)^{2n}$ (for $x\in[-\pi,\pi]$), this matches $(*)$, if we take $x=2$ and multiply it by $-1$.

Is it Ok?

[Edited]

Very nice, grobber! You lost $+1$ somewhere (I leave it to you to find the exact spot) because the answer for summation including $0$ is $1-2\cot 2$ (and without $n=0$ it is $1-\cot 1$). Otherwise it's right. I had a slightly different solution in mind but congratulations anyway :lol:. Did you try the nested root problem?

Nope, I didn't try it, but I don't think it's the sort of problem I like to try really hard to solve :). As for this one, it doesn't look like I'll be finding that lost $1$ any time soon :D.

In general
\[
\displaystyle\sum_{n=0}^{+\infty}\frac{1}{2^n}\tan\frac{x}{2^n}=\frac{1}{x}-2\cot 2x
\]
following another way like this:
Applying the identity $\cot x =2\cot 2x+\tan x \quad \textrm{for}\quad x\not=k\frac{\pi}{2}\quad k\in\mathbb{Z}$  and the fact that if $\displaystyle\lim_{n\to+\infty}a^{-n}f(b^{-n}x)=l(x)$
then
\[
\displaystyle\sum_{n=0}^{+\infty}a^{-n}g(b^{-n}x)=\frac{l(x)-af(bx)}{c}
\]
where a,b,c are non zero costant and the functions f and g satisfy the condition
$f(x)=af(bx)+cg(x).$
In our case $f(x)=\cot x,\quad g(x)=\tan x,\quad a=2,\quad b=2\quad\textrm{and}\quad c=1.$
Remember, also, that
\[
\displaystyle\lim_{n\to+\infty}\frac{1}{2^n}\cot \frac{1}{2^n}=\frac{1}{x}
\]

I meant that I don't like grobber's approach, since it is too ugly, though it leads to the answer ;) (without any offences). Your solutions is very nice. I was expecting it, but I haven't in mind that identity.

I figured out where I lost the $1$ :). The series for $x\tan x$ starts from $n=1$, so in the end, we "almost" get the series for $x\cot x$, without the initial term, which is $1$. When we multiply it by $-1$, the $1$ we were missing becomes an extra $1$. I'll edit it.

Edit: no offence taken, Myth :)

Some words about first sum: $S(x)=\sum_{n\geq0}\frac{1}{2^n}tan(\frac{x}{2^n}) $. Integrating by parts we may obtain that $\int S(x)dx=\sum_{n\geq0}(-ln(cos(\frac{x}{2^n})))=-ln(\frac{sin(2x)}{x})=ln(x)-ln(sin(2x))$, so we obtain $S(x)$
(sorry for not being very precise)

[quote="Moubinool"]Find this one $\sum_{n\geq 0}\frac{1}{2^{2k}}\tan^2(\frac{x}{2^k})$[/quote]
By integrating we get $\sum_{k\geq 0}\frac{1}{2^k}\left(\tan(x/2^k)-x/2^k\right)=\sum_{k\geq 0}\frac{1}{2^k}\tan\frac{x}{2^k}-\frac{4x}{3}$.


modular arithmetic
number theory unsolved
number theory

Find all positive integer solutions $ (a,b,c,n)$ of the equation: $ 2^n\equal{}a!\plus{}b!\plus{}c!$.

[quote="moldovan"]Find all positive integer solutions $ (a,b,c,n)$ of the equation: $ 2^n \equal{} a! \plus{} b! \plus{} c!$.[/quote]

Wlog say $ a\geq b \geq c$

If $ c\geq 3$, $ 3$ divides RHS but not LHS. So $ c<3$
1) $ c\equal{}1$
Then $ b\equal{}1$ and $ a>1$ else RHS would be odd while LHS even. So $ 2^n\equal{}a!\plus{}2$
If $ n\equal{}1$, this equation has no solution
If $ n\equal{}2$, we get the solution $ 2^2\equal{}2!\plus{}1!\plus{}1!$
If $ n>2$, $ a<4$ else $ LHS\equal{}0\pmod 4$ while $ RHS\equal{}2\pmod 4$. And we get the solution $ 2^3\equal{}3!\plus{}1!\plus{}1!$

2) $ c\equal{}2$
Then $ 2^n\equal{}a!\plus{}b!\plus{}2\geq 6$ and so $ n\geq 3$
If $ b\geq 4$, then $ LHS\equal{}0\pmod 4$ while $ RHS\equal{}2\pmod 4$. So $ b<4$

2.1) $ c\equal{}2$ and $ b\equal{}2$
Then $ 2^n\equal{}a!\plus{}4$.
If $ a\geq 4$, $ RHS\equal{}4\pmod 8$ and so $ n<3$ and so no solution
We just have to test $ a\in[2,3]$ and no solution.

2.2) $ c\equal{}2$ and $ b\equal{}3$
Then $ 2^n\equal{}a!\plus{}8$
If $ a\geq 8$, $ RHS\equal{}8\pmod 16$ and so $ n<4$ and so no solution
We just have to test $ a\in[3,7]$ and two solutions :
$ 2^5\equal{}4!\plus{}3!\plus{}2!$
$ 2^7\equal{}5!\plus{}3!\plus{}2!$

So 4 solutions :
$ 2^2\equal{}2!\plus{}1!\plus{}1!$
$ 2^3\equal{}3!\plus{}1!\plus{}1!$
$ 2^5\equal{}4!\plus{}3!\plus{}2!$
$ 2^7\equal{}5!\plus{}3!\plus{}2!$


number theory
relatively prime
Diophantine equation
IMO Shortlist

(a) Let $ n$ be a positive integer. Prove that there exist distinct positive integers $ x, y, z$ such that

\[ x^{n\minus{}1} \plus{} y^n \equal{} z^{n\plus{}1}.\]

(b) Let $ a, b, c$ be positive integers such that $ a$ and $ b$ are relatively prime and $ c$ is relatively prime either to $ a$ or to $ b.$ Prove that there exist infinitely many triples $ (x, y, z)$ of distinct positive integers $ x, y, z$ such that

\[ x^a \plus{} y^b \equal{} z^c.\]

[quote="orl\]
(b) Let $ a, b, c$ be positive integers such that $ a$ and $ b$ are relatively prime and $ c$ is relatively prime either to $ a$ or to $ b.$ Prove that there exist infinitely many triples $ (x, y, z)$ of distinct positive integers $ x, y, z$ such that
\[ x^a \plus{} y^b \equal{} z^c.
\][/quote]

See it here  http://www.mathlinks.ro/viewtopic.php?t=150638

My solution.
Exist u, v, w such that: $ u \plus{} v \equal{} w$ and:
$ u \equal{} \prod_i^n p_i^{\alpha_i}$;
$ v \equal{} \prod_i^n p_i^{\beta_i}$;
$ w \equal{} \prod_i^n p_i^{\gamma_i}$;
By chinese remainder theorem, exist $ \delta_i, i \equal{} 1,..n$  such that:

$ a\mid\delta_i \plus{} \alpha_i$
$ b\mid\delta_i \plus{} \beta_i$
$ c\mid\delta_i \plus{} \gamma_i$

Choose $ z \equal{} \prod_i^n p_i^{\delta_i}$, This problem proven