inequalities
trigonometry
geometry
rearrangement inequality
inequalities unsolved

Let $ABC$ be a triangle. Prove that: \[\frac{h_{a}}{m_{b}}+\frac{h_{b}}{m_{c}}+\frac{h_{c}}{m_{a}}\le 3\]
where $m_{a},\,m_{b},\,m_{c}$ denote the lenghts of medians to the side $BC,\,CA,\,AB$ respectively, $h_{a},\,h_{b},\,h_{c}$ denote the lenghts of altitudes to the side $BC,\,CA,\,AB$ respectively.

You can write $h_{a}=c\sin B=\frac{bc}{2R}=\frac{abc}{2Ra}$ and permutations thereof.
From this, it follows that $a\geq b\geq c$ implies $h_{a}\leq h_{b}\leq h_{c}$.
By Apollonius' Theorem, you can write:
$m_{a}^{2}=\frac{2b^{2}+2c^{2}-a^{2}}{4}=\frac{a^{2}+b^{2}+c^{2}}{2}-\frac{3a^{2}}{4}$. So $m_{a}\leq m_{b}\leq m_{c}$.
For obvious geometry reasons $h_{a}\leq m_{a}$. Now, by the rearrangement inequality:
$\frac{h_{a}}{m_{b}}+\frac{h_{b}}{m_{c}}+\frac{h_{c}}{m_{a}}\leq \frac{h_{a}}{m_{a}}+\frac{h_{b}}{m_{b}}+\frac{h_{c}}{m_{c}}\leq 1+1+1=3$
Equality occurs when all the "$\leq$"s are "=", ie the triangle is equilateral!

[quote="Inquisitor"]You can write $h_{a}=c\sin B=\frac{bc}{2R}=\frac{abc}{2Ra}$ and permutations thereof.
From this, it follows that $a\geq b\geq c$ implies $h_{a}\leq h_{b}\leq h_{c}$.
By Apollonius' Theorem, you can write:
$m_{a}^{2}=\frac{2b^{2}+2c^{2}-a^{2}}{4}=\frac{a^{2}+b^{2}+c^{2}}{2}-\frac{3a^{2}}{4}$. So $m_{a}\leq m_{b}\leq m_{c}$.
For obvious geometry reasons $h_{a}\leq m_{a}$. Now, by the [color=blue]rearrangement inequality[/color]:
$\frac{h_{a}}{m_{b}}+\frac{h_{b}}{m_{c}}+\frac{h_{c}}{m_{a}}\leq \frac{h_{a}}{m_{a}}+\frac{h_{b}}{m_{b}}+\frac{h_{c}}{m_{c}}\leq 1+1+1=3$
Equality occurs when all the "$\leq$"s are "=", ie the triangle is equilateral![/quote]
Are you sure?


trigonometry
calculus
integration
MATHCOUNTS
USAMTS
real analysis
AMC

Well, I'm not very advanced compared to many of the people in this community.  I will be going into 10th grade and will take pre-calc honors.

-interesting_move

[color=white]Not advanced?  [i]Not[/i] advanced??  I'm only going to 8th grade and will take trig!  Hope that made you feel a wee bit more advanced...

That said and done, I'll just make this comment: Other than me and an assortment of other users, most are probably much older/advanced.  Right, the Great Rusczyk?[/color]

Don't know about how advanced everyone on the site is, but you're certainly all way ahead of where I was at your age.  When I was in 8th grade, I was taking...  Algebra I.

Never even saw integration in a classroom in high school (learned that on my own for the BC Calc exam).

And I had to walk to school uphill 6 miles both ways through 6 feet of snow.  Y'all have it easy now.

9th
Calc AB
AMC10/12/AIME/USAMO, Mathcounts (last year), Mandelebrot (next year), USAMTS (next year)

P.S. High-five, Rep123max.  Same here last year.  8th precalc!  WTG.  You two, jojo, etc.

[quote="rrusczyk"]And I had to walk to school uphill 6 miles both ways through 6 feet of snow.  Y'all have it easy now.[/quote]

It's amazing that all sorts of people say that. My father used to try to convince me that downhill was only invented in 1970 and that downhill both ways is now being perfected in laboratories.

I'm going to 12th and last grade here in Sweden. Don't know what maths I will read in high school that year because the classes here suck, but I will attend to two classes at a time in Stockholm University (haven't decided which ones yet), which my mentor fixed kindly for me. Right now I'm reading Principles of Mathematical Analysis by Walter Rudin and An Introduction to the Theory of Numbers by Ivan Niven on my own.

Quote:Not advanced? Not advanced?? I'm only going to 8th grade and will take trig! Hope that made you feel a wee bit more advanced...

Actually, Trig is very advanced for an 8th grader; You'll probably be taking Calc when you get into 10th grade.  Therefore, I'm not advanced.[/hide]

[quote]Not advanced? Not advanced?? I'm only going to 8th grade and will take trig! Hope that made you feel a wee bit more advanced...[/quote]

Actually, Trig is very advanced for an 8th grader; You'll probably be taking Calc when you get into 10th grade.  Therefore, I'm not advanced.

-interesting_move

9th grader (freshman) taking accelerated precalc/trig and teaching myself calculus.

Middle School: Mathcounts, Mu Alpha Theta, AMC, and Math League.  And took algI/algII and geometry.

And in high school I hope to make the USAMO and do the Mandelbrot and MAYBE ARML.

In mathcounts...as a seventh grader I got second at chapter and 38th at state I think.  And in eighth grade I got first at chapter, fourth at state, and 59th at nationals.  Just to clear that up.

Oh yes, I forgot to post the math competitions I've been in.  I was in Mathcounts from 7th (11th regional, 39th state) to 8th grade(1st regional, 4th state), and I competed in the AMCs this year.

-interesting_move

See, in my school system every things integrated.  See, i'm going into 9th grade next year and will take Math AB, its the most difficult class that my school will offer.  I took the Math A in 8th grade, also the most advanced offered wich really sucks cuz it was way easy.  So you guys are actually all advantaged compared to me.  I have to learn things on my own by way of AoPS and pass tests.  Did mathcounts all through out middle school, but dont know what math copetitions are at my highschool.

9th

Calc AB or BC... not sure yet

I guess I should compete in math competitions, but I don't get that much opportunity...

[quote="mathfanatic"]9th
Calc AB
AMC10/12/AIME/USAMO, Mathcounts (last year), Mandelebrot (next year), USAMTS (next year)

[/quote]

if ur going into 9th, im [i]exactly[/i] the same.  :D

last year:

10th grade: Calc BC, amc, aime

next year: 11th: Calc III and IV at local college, independent study number theory, AP stat, and oh, AP comp. sci.

[quote="Ripper1"]Hehe, you guys all think you aren't advanced.  In 8th grade, I was in "Advanced" Algebra 1 and next year I am doing Calc AB as a senior which is the advanced class at my high school.  For math contests, in 8th grade I placed third in the Serra High School Contest(local high school) .  After that, I didn't care much for math for the next couple of years, during my junior year(last year), I did AMC, scored in the high 80's(Pretty bad, but it was my first and I had absolutely no experience in high school math contests), and I did fairly well on the last 3 California Math leagues. Heh, I didn't take the first 3 and scored in the top 5 of my school.  I also participated in BAMM which is another local math contest and I placed honorable mention in the Cryptography contest!! Haha, that is my math experience, I'm getting better :D .[/quote]

Cool beans, Jack.

I'm sorry, my name is Phil, not Jack. Why the sarcasm?

No, no, no sarcasm.  I'm serious.  Sounds good.  And you're not not advanced.  I know people who are taking Trig as Seniors.

And I didn't think your name was Jack  :lol:  But your name is "Ripper1" and so I was thinking of Jack the Ripper.  BTW, what does your name mean?

9th
Algebra II/Trig, and Precalc.
AMC10/AIME/USAMO (hopefully)
USAMTS
Mathcounts (Last three years)

I would have liked to take Calc now, but my school would not let me jump that far.

Oh, heh sorry about that then, usually when people say phrases such as those, sarcasm is usually implied.  Anyways, the name Ripper1 has been my handle for hmm 8-9 years.  My dad was playing a game on the PC, called Ripper, which is about Jack the Ripper, it was sort of a "cyberpunk" themed game.  I really liked the plot of the game but I wasn't allowed to play it because it was too violent but for some reason I was allowed to play Doom and Duke Nukem 3D(Don't ask me how that works), anyways, when trying to play online(on my 28.8 kbps modem!!) someone had taken the name Ripper, so Ripper1 it was! I was quite an original 4th grader wasn't I?

I remember my dad playing Doom when I was little...it was pretty cool for me, considering how old I was at the time.  I even got to switch the weapons for him when he told me to...wow! :mrgreen: Ah, bringing back old memories...


I can't make it :(

Youth group on wednesdays from 5:30-7:00 CST

(And I would also be interested in it being administered a little bit early)

i am all for all it.
but probably posting it a bit earlier in the day is better/

Are you guys done? I'm leaving on vacation on Monday, out of the country, so if I could take it this weekend, that would be cool.

Thanks, Max. I may be able to do it after I leave anyway, but early makes it easier.

hey, how does it work?

I've never taken a mock AMC before on AOPS.

Do you just upload the file and let us download it, then we send you back the answers?

or do you put the questions in a post?

im gonna be leaving  sunday and wont be coming back until friday... is there some way i can still take it?

cool. i'll take it. wednesday shouldn't be a problem for me.

I'm in. but what is it in Pacific Time? (because i'm in California, and i have no knowledge what so ever about time zone)

When it's 5 PM CST, it is 6 PM EST, and 3 PM PST.

and not to leave the mountain time zone out, it would be 4:00  :D

The test has been finalized so, as Rep123Max said, you may take it tomorrow as early as noon (central time I assume) [i]if you absolutely cannot make the official time.[/i]  Please PM him about requests to take it early as I start school tomorrow and won't have much time to respond.

Also, there will be a solution guide.  It's almost done (I have 3 or 4 problems left).  It will be available only after the official administration of the test when the results of the test are posted.

Have a good day :)


i'm new to the inequality thing and am stuck on a particular question:

x^2+|2x+3|[u]>[/u]6

any help would be much appreciated

-lizzie

Sketch the graphs of $y_{1}=-x^{2}+6$ and $y_{2}=|2x+3|$ and find the range of $x$ which can be valued such that $y_{1}\leq y_{2}$. The point is that solving an inequality means that examining which gaphs locates above or below the other one.

[hide="solution"]first assume that $2x+3\ge 0$ i.e. $x\ge-\frac{1}{2}$ then our inequality becomes
$x^{2}+2x+3\ge 6$
$\iff (x+1)^{2}\ge 4$
so $x\ge \pm 2-1$
since $x\ge-\frac{1}{2}$, we have $x\ge 1$      $(1)$

now asume that $2x+3<0$ so $x^{2}-2x-3\ge 6$
in the same way as above we get $x\ge \pm \sqrt{10}+1$
since $x<-\frac{3}{2}$ we get $x\ge-\sqrt{10}+1$      $(2)$

combining $(1)$ and $(2)$ we conclude that $x\ge 1$ or $-\sqrt{10}+1\le x<-\frac{3}{2}$[/hide]

nayel, your answer is incorrect. The right answer is $x\leq 1-\sqrt{10},\ 1\leq x$.

first , i think u should notice about the well-known but important things in that kind of problem
$|A|=A$ when $A \geq 0$ and $|A|=-A$ when $A \leq 0$
so $|2x+3|=2x+3$ when $x\geq \frac{-3}{2}$ and $|2x+3|=-2x-3$ when $x\leq \frac{-3}{2}$
i think now u can do the rest


we don't have the conditions to use the karamata inequality,can you post your solution math boy


If p is prime then p divides ((p -2)! -1)

It's just Wilson's theorem:

If $p\in\mathbb{P}$, then $(p-2)!\equiv 1\pmod p$.  The result then follows.

Masoud Zargar

But, is there another way to prove it without using Wilson's Theorem?

It [u]is[/u] Wilson's theorem ;)
To prove it:
For any factor $a \neq 1$ in the product $1 \cdot 2 \cdot ... \cdot (p-2)$, there is an $b \neq a$ with $ab \equiv 1\mod p$. Now all these pairs kill each other out of the product, so it's $1 \mod p$.


for all non-negative real a,b,c,d which satisties a+b+c+d=4,

prove that
a^2bc + b^2cd + c^2da + d^2ab ≤ 4

[quote="profcool"]for all non-negative real a,b,c,d which satisties $ a\plus{}b\plus{}c\plus{}d\equal{}4$,

prove that
$ a^2bc \plus{} b^2cd \plus{} c^2da \plus{} d^2ab\leq 4$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960


geometry
geometric transformation
reflection
circumcircle
parallelogram
trigonometry
linear algebra

2.In the triangle $ABC$,$O$-is the circumcentre ,$A',B',C'$-are such  a point that $CA'=CA$ and $AA'\bot{BC}$,$CB'=CB$ and $BB'\bot{AC}$,$BC'=BC$ and $CC'\bot{AB}$,
$A_1\in{OA'}\cap{BC}$,$B_1\in{OB'}\cap{AC}$,$C_1\in{OC'}\cap{AB}$.
Prove that the lines $AA_1,BB_1,CC_1$ intersecting in the same point. $\clubsuit$

[quote="Tiks"]... $A_1\in{OA'}\bot{BC}$, $B_1\in{OB'}\bot{AC}$, $C_1\in{OC'}\bot{AB}$ ...[/quote]
This should be $A_1\equiv{OA'}\cap{BC},$ $B_1\equiv{OB'}\cap{AC},$ $C_1\equiv{OC'}\cap{AB}.$ :bruce:

A' is a reflection of A in BC. Hence the triangle $\triangle AA'A_1$is isosceles with $\angle AA'A_1 = \angle A'AA_1.$ Let $A_0, B_0, C_0$ be the midpoints of BC, CA, AB. Let D be the foot of the A-altitude and $O_0$ the circumcenter of the triangle $\triangle AC_0B_0$ centrally similar to the triangle $\triangle ABC$ with similarity center A and similarity coefficient $\frac 1 2.$ Since D is the reflection of A in $C_0B_0$ and $AD = \frac{AA'}{2},$ the triangles $\triangle AA'O \sim \triangle ADO_0$ are also centrally similar with the same similarity center and coefficient. Let N be the reflection of $O_0$ in $C_0B_0,$ the perpendicular bisector of AD. The triangles $\triangle DNA \cong \triangle ADO_0$ are then oppositely congruent and the triangles $\triangle DNA \sim \triangle AA'O$ oppositely similar. Consequently,

$\angle A'AN \equiv DAN = \angle AA'O \equiv \angle AA'A_1 = \angle A'AA_1$

This means that the lines $AN \equiv AA_1$ are identical. But obviously, N is the circumcenter of the medial triangle $\triangle A_0B_0C_0 \cong \triangle AC_0B_0,$ i.e., the center of the 9-point circle (N) of the triangle $\triangle ABC,$ and $N \in AA_1.$ In exactly the same way, we can show that $N \in BB_1, N \in CC_1,$ so that N is the concurrency point of these 3 lines.

BTW, this problem also has an easy trigonometric solution.

[quote="yetti"][quote="Tiks"]... $A_1\in{OA'}\bot{BC}$, $B_1\in{OB'}\bot{AC}$, $C_1\in{OC'}\bot{AB}$ ...[/quote]
This should be $A_1\equiv{OA'}\cap{BC},$ $B_1\equiv{OB'}\cap{AC},$ $C_1\equiv{OC'}\cap{AB}.$ :bruce:
[/quote]
Sorry $yetti$ :blush: ,I have edited my post  :) .

@yetti: Sorry, but how can someone even think of a trigonometric solution, when there are so beautiful elementary solutions? :roll:

I solved it this way:

If we reflect $OC'$ in $AB$, then the line $OC'$ becomes $CC_1$, hence the problem is equivalent to the following:

[quote]Let $ABC$ be a triangle with circumcentre $O$. $X,Y,Z$ are the reflections of $O$ in $BC, CA, AB$ respectively. Then $AX, BY$ and $CZ$ are concurrent.[/quote]

We have $\angle ZAC + \angle ACX = 90-\angle C + \angle A + 90 - \angle B + 180 - 2\angle A = 180$ and also $AZ = OA = OC = CX$, hence $ACXZ$ is a parallelogram. Hence $AX \cap CZ$ bisects $AX$ and $CZ$.
By the same reasoning $AYXB$ is a parallelogram and $AX\cap BY$ bisects $AX$ and $BY$.
The result follows
qed

[quote="Yimin Ge"]@yetti: Sorry, but how can someone even think of a trigonometric solution, when there are so beautiful elementary solutions? :roll: [/quote]

That's because trigomometry is a general method and you guys keep underestimating it all the time. I do not remember using 9-point circle center in my life, except for fun. On the other hand I use trigonometry all the time for processing experimental data. It is not just the distance-angle calculations, for example, think of an orthogonal matrix of a rotation. Or, Bessel functions $ J_{n \plus{} 1/2}(x)$, which pop-up in numerous problems, can be expressed by trigonometric functions: $ J_{ \plus{} 1/2}(x) \equal{} \sqrt {\frac {2}{\pi x}} \sin x,\ J_{ \plus{} 3/2}(x) \equal{} \sqrt {\frac {2}{\pi x}} \left(\frac {\sin x}{x} \minus{} \cos x\right), ...,$ or spherical harmonics $ Y_{lm}(\theta, \phi) \sim P^{(m)}_l(\cos \theta) e^{\text i m \phi}$ (where $ P^{(m)}_l(z)$ are derivatives of Legendre polynomials), which likewise pop-up in numerous problems, etc. Below is an image of ion beam obtained by passing the beam through a lattice mask, with microchannel detectors and a fluorescent screen behind, the image taken by a digital camera as a huge bitmap. That's after I filtered out the noise. Given this image, I had to calculate the basic parameters of the beam (the image comes every second or so) and also devise a way to calibrate the system. I surely did not use the 9-point circle or a cyclic quadrilateral to do that.


Let $ f\left(n\right)$ be the number of walks from $ \left(0,0\right)$ to $ \left(n,n\right)$ using steps of $ \left(0,1\right)$, $ \left(1,0\right)$, and $ \left(1,1\right)$.  Show that

\[ \sum_{n\geqslant 0}x^{n}f\left(n\right)\equal{}\frac{1}{\sqrt{1\minus{}6x\plus{}x^{2}}}.\]

The generating function implies a recursion; one approach would be to determine this recursion and then try to show that $ f(n)$ satisfies that recursion.

The other is to consider the function $ f(m, n)$ that gives the number of walks from $ (0, 0)$ to $ (m, n)$ and compute $ \sum f(m, n) x^m y^n$, and then isolate the terms you want from here.  (There are a few very natural ways to do this; take advantage of orthogonality.)


Suppose $ n$ is not multiple of $ 3$. Find all integer solutions of
\[ (a^{2}-bc)^{n}+(b^{2}-ac)^{n}+(c^{2}-ab)^{n}=1\]


trigonometry
geometry
circumcircle
trapezoid
geometry solved

A convex hexagon ABCDEF is inscribed in a circle such that AB = CD = EF and diagonals AD;BE;CF are concurrent. Let P be the intersection of AD and CE. Prove that $\frac{CP}{PE}=\left(\frac{AC}{CE}\right)^2$.

I may be wrong, but this looks like 50 times of Sine Law... :lol:

I think, this problem was posted to forum some time ago.

[quote="MysticTerminator"]my solution was similar to [url]http://www.kalva.demon.co.uk/usa/usoln/usol943.html[/url][/quote]

That's a nice solution! Here is the mine:

Denote by Q the point of intersection of the diagonals AD, BE, CF. Consider the segments AB, CD and EF as chords in the circumcircle of the hexagon ABCDEF; since these chords are equal in length, their chordal angles are also equal. In other words, < AEB = < CAD = < ECF. This rewrites as < QEA = < QAC = < QCE.

Now, if you know triangle geometry, you can conclude from this that the point Q is the [url=http://mathworld.wolfram.com/BrocardPoints.html]first Brocard point[/url] of triangle ACE, and thus, the equation $\frac{CP}{PE}=\left(\frac{AC}{CE}\right)^2$ follows from the following property of the first Brocard point of a triangle:

[b]Theorem 1.[/b] If Q is the first Brocard point of a triangle ABC, so that < QBC = < QCA = < QAB, and if the line BQ meets the line CA at a point Y, then $\frac{CY}{YA}=\left(\frac{a}{b}\right)^2$, where a = BC, b = CA and c = AB are the sidelengths of triangle ABC.

[i]Proof of Theorem 1.[/i] After the Sine Law in triangles CYB and AYB, we have

$CY=BC\cdot\frac{\sin\measuredangle CBY}{\sin\measuredangle BYC}$;
$YA=AB\cdot\frac{\sin\measuredangle YBA}{\sin\measuredangle AYB}$.

But since < BYC + < AYB = 180°, we have sin < BYC = sin < AYB, so that

$\frac{CY}{YA}=\frac{BC\cdot\frac{\sin\measuredangle CBY}{\sin\measuredangle BYC}}{AB\cdot\frac{\sin\measuredangle YBA}{\sin\measuredangle AYB}} = \frac{BC\cdot\sin\measuredangle CBY}{AB\cdot\sin\measuredangle YBA} = \frac{a\cdot\sin\measuredangle QBC}{c\cdot\sin\measuredangle QBA}=\frac{a}{c}\cdot\frac{\sin\measuredangle QBC}{\sin\measuredangle QBA}$.

Since < QBC = < QCA, we thus have

$\frac{CY}{YA}=\frac{a}{c}\cdot\frac{\sin\measuredangle QCA}{\sin\measuredangle QBA}$.

Now, from the Sine Law in triangles CQA and AQB, we have

$\sin\measuredangle QCA = \sin\measuredangle CQA \cdot \frac{AQ}{CA}$;
$\sin\measuredangle QBA = \sin\measuredangle AQB \cdot \frac{AQ}{AB}$.

Hence,

$\frac{\sin\measuredangle QCA}{\sin\measuredangle QBA} = \frac{\sin\measuredangle CQA \cdot \frac{AQ}{CA}}{\sin\measuredangle AQB \cdot \frac{AQ}{AB}} = \frac{\sin\measuredangle CQA \cdot AB}{\sin\measuredangle AQB \cdot CA} = \frac{\sin\measuredangle CQA \cdot c}{\sin\measuredangle AQB \cdot b} = \frac{c}{b}\cdot \frac{\sin\measuredangle CQA}{\sin\measuredangle AQB}$.

But finally, since < QCA = < QAB, we have < QCA - < QAB = 0°, but the sum of angles of triangle CQA yields

  < CQA = 180° - < QCA - < CAQ = 180° - < QCA - (< CAB - < QAB)
             = 180° - < CAB - (< QCA - < QAB) = 180° - < CAB - 0° = 180° - < CAB = 180° - A,

so that sin < CQA = sin A, and similarly, sin < AQB = sin B. Thus,

$\frac{\sin\measuredangle QCA}{\sin\measuredangle QBA} = \frac{c}{b}\cdot \frac{\sin\measuredangle CQA}{\sin\measuredangle AQB} = \frac{c}{b}\cdot \frac{\sin A}{\sin B} = \frac{c}{b}\cdot \frac{a}{b}$,

where we have used the sine law for triangle ABC. Consequently,

$\frac{CY}{YA}=\frac{a}{c}\cdot\frac{\sin\measuredangle QCA}{\sin\measuredangle QBA} = \frac{a}{c}\cdot\left(\frac{c}{b}\cdot \frac{a}{b}\right) = \left(\frac{a}{b}\right)^2$,

completing the proof of Theorem 1.

Okay, indeed, I have used the sine law quite often...

  Darij

[quote="paul_mathematics"]A convex hexagon ABCDEF is inscribed in a circle such that AB = CD = EF and diagonals AD;BE;CF are concurrent. Let P be the intersection of AD and CE. Prove that $ \frac {CP}{PE} \equal{} \left(\frac {AC}{CE}\right)^2$.[/quote]
Sorry to revive an old topic, but I have a rather odd solution to this problem:
[hide="Solution"]
Let the point of concurrency be $ Q$.  Notice that $ AB \equal{} CD \equal{} EF$ and quadrilaterals $ ABCD$, $ CDEF$, and $ EFAB$ are all cyclic, so they are also all isoceles trapezoids.  This means that $ AF\parallel EC$, $ BC\parallel AD$, and $ ED\parallel FC$.  Now, notice that $ \angle QAC \equal{} \angle QFD$ and $ \angle AQC \equal{} \angle FQD$, so $ \triangle AQC\sim \triangle FQD$.  This means that $ \frac {AC^2}{FD^2} \equal{} \frac {[AQC]}{[FQD]}$.  Notice that because of isoceles trapezoid $ EFCD$, we see that $ FD \equal{} EC$, so $ \frac {AC^2}{EC^2} \equal{} \frac {[AQC]}{[FQD]}$.  Now, since $ FQ\parallel ED$, we have that $ [FQD] \equal{} [QEF]$.  Since $ EQ\parallel AF$, we also have that $ [QEF] \equal{} [AQE]$.  Thus, $ \frac {AC^2}{EC^2} \equal{} \frac {[AQC]}{[FQD]} \equal{} \frac {[AQC]}{[AQE]} \equal{} \frac {PC}{PE}$, as desired. [/hide]


Prove that if $ M$ is an $ n\times n$ matrix that can be written in the form $ M\equal{}\begin{pmatrix}A&B\\O&C\end{pmatrix}$, where $ A$ and $ C$ are square matrices, then $ \det(M)\equal{}\det(A)\det(C)$.

Prove that $ \delta: M_{2\times 2}(F)\rightarrow F$ is a 2-linear function if and only if it has the form $ \delta(A)\equal{}A_{11}A_{22}a\plus{}A_{11}A_{21}b\plus{}A_{12}A_{22}c\plus{}A_{12}A_{21}d$ for some scalars $ a,b,c,d\in F$.

For the first, one approach is to use elementary row operations to make $ C$ upper triangular, then use the rows of $ C$ to eliminate every entry of $ B$.

Does anyone have a way for the second one?

It would help if you defined the terms precisely. I'm not sure what "2-linear" means in this context, since it's not written as a function of two arguments.


Let $a,b,c$ positive integers with sum $1$. Prove that:
   $\frac{\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{b}}{1-\sqrt{b}}+\frac{\sqrt{c}}{1-\sqrt{c}}\geq \frac{3+3\sqrt{3}}2$

How can $a+b+c=1$ if all the numbers are positive integers?

I am really sorry :oops:  a,b,c are just positive numbers  :blush:

It is well known this inequality.Anyway was posted on the forum.I'll solve it.Note $a=x^{2},b=y^{2},c=z^{2}$.From Cauchy we have $\sum_{cyc}\frac{x}{1-x}\geq\frac{(\sum x^{2})^{2}}{\sum x^{3}-\sum x^{4}}$.It remains now to prove $\sum x^{3}-\sum x^{4}\leq\frac{\sqrt{3}-1}{3}$.Now from $(\sum x^{4})(\sum x^{2})\geq (\sum x^{3})^{2}$ we are done because if we note $s=\sum x^{3}$ we have to prove $3s^{2}-1+\sqrt{3}-3s\geq 0$ so we have $(s\sqrt{3}-1)(s\sqrt{3}+1-\sqrt{3})\geq 0$.We know $s\sqrt{3}\geq 1$,qed. :lol:

Of just note that $f(x)=\frac{1}{\sqrt{x}(1-\sqrt{x})}$ is convex since $f''(x)\geq 0$.  Then use Jensen's inequality.


I have a die. It has n faces labeled 1-n. Prove that if I roll a finite number of times, I will roll all numbers on the die.

Um, it can't be proven, I don't think.  Or else I'm misunderstanding.

Not true.  The probability that you only ever roll one face is finite.

[quote="lotrgreengrapes7926"]Is this solvable??
I have a die. It has $n$ faces. Find the probability that if I roll it $k$ times, I will roll all faces of the die.[/quote]

Yeah this should be solvable. You have $n^{k}$ as your denominator, because you roll it $k$ times.

Then you must roll every side of the die, which means $n!$ on top,  and you can roll the rest $k-n$ rolls anything you like, so they are all$n's$, so you only need to see what order the $n$ faces can come up on, which would be $kCn$, so the whole thing is:

$\frac{n!n^{k-1}kCn}{n^{k}}$

but I'm almost sure this is wrong because I did not account for overcounting. Nevertheless, I think it is solvable as long as $k \geq n$.


geometry
analytic geometry
geometric transformation
reflection
angle bisector
geometry proposed

In a triangle $ ABC$,we consider points $ B',C'$ on $ AB,AC$ such that $ BB' \equal{} CC'$.Let $ D$ be the intersection of lines $ BC'$ and $ B'C$.Prove that the symmetric point of $ D$ with respect to the midpoint $ M$ of $ BC$ lies on the angle bisector of $ \angle A$.

It suffices to show that $ D$ lies on the anticomplement of the bisector of $ \angle BAC,$ i.e.  the $ A_0$ bisector of the anticomplementary triangle $ \triangle A_0B_0C_0.$ Let $ k \equal{} BB' \equal{} CC'.$ Using barycentric coordinates we have

$ B' \ (k: c \minus{} k: 0) \ , \ C' \ (k: 0: b \minus{} k) \ , \ BC' \cap CB' \equiv D \ (k: c \minus{} k: b \minus{} k)$

Eliminating $ k$ from the coordinates of $ D,$ the  locus of $ D$ is $ (b \minus{} c)x \plus{} by \minus{} cz \equal{} 0,$ which is the anticomplement of  the A-angle bisector $cy \minus{} bz \equal{} 0.$

Denote by$ P$ the symmetric point of $ A$wrt$ M.$
$ BC'\cap CP \equal{}S$.
We have :
$ \frac{DB}{DS}\equal{}\frac{BB'}{CS}\equal{}\frac{CC'}{CS}\equal{}\frac{PB}{PS}$
$ \equal{}>D$lies on the angle bisector of the angle $ BPS$.
$ \equal{}>E$lies on the angle bisector of the angle $ BAC$.
 :)

Let $ D'$ be the reflection of $ D$ across $ M$
Since $ BC'//D'C$ and $ CB'//D'B$ , we have $ [BD'B']\equal{}[BD'C]\equal{}[D'CC']$ and hence, $ D'$ lies on the bisector of $ \angle BAC$


function
symmetry
LaTeX
calculus
integration
real analysis
real analysis unsolved

And sorry for the pic, I don't know how to use index , so i attached a pic below this chart.

The qestion is to evaluate the function that is shown on the pic.

Hint: Don't expend it but use symmetry instead.

Anyone? Thanks for helping.[/i][/quote]

When writing something in latex, put what you want to write between two dollar signs (the symbol above the 4 in your keyboard).. For example, to write your integral in latex type \int^1_{-1}(e^x-e^{-x})\,dx between those symbols:

$ \int^{1}_{-1}(e^{x}-e^{-x})\,dx$

(roll your mouse over the latex to see the code used to write it).
Now, if that is the integral you want then its value is zero because $ f(x) = e^{x}-e^{-x}$ is an odd function.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 is our useful LaTeX FAQ page.

:)    Thanks for helping me out guys~~    I Appreciate It~!


Okay, this is a pretty lame game, but you guys might think otherwise. Basically I will start the game by saying something, and then the next person will add to the sentence, by saying what I said then 'because' and then say why it happened.

Example;

Wicked said "Wicked didn't go to school today."

Next person said "Wicked didn't go to school today, because he didn't want to."

Next person said "Wicked didn't go to school today, because he didn't want to, because he thought ninjas would attack him."

Next person said "Wicked didn't go to school today, because he didn't want to, because he thought ninjas would attack him, because they had heard of his amazing math skills, and they wanted to see if he could fight as well as he could solve math problems."

Next person said "Wicked didn't go to school today, because he didn't want to, because he thought ninjas would attack him, because they had heard of his amazing math skills, and they wanted to see if he could fight as well as he could solve math problems, because if he couldn't then he needed body guards."

Next person said "Wicked didn't go to school today, because he didn't want to, because he thought ninjas would attack him, because they had heard of his amazing math skills, and they wanted to see if he could fight as well as he could solve math problems, because if he couldn't then he needed body guards, because the cops were always busy eating donuts."

Sorry, I got a bit carried away. But you get the idea.

Start with this;

Wicked's computer blew up.

Wickedestjr's computer blew up, because ernie said so.  THE END.

[quote="james4l"]Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, while a nuke blows up the entire world, causing mass destruction, however people survived because they didn't end their sentences with periods

james4l and 33286, you need to use BECAUSE!

skip james4l and 33286.

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, [b]because Bob Barr is a fairy,[/b]

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died.,

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died, because I saw Medusa, because I have 20-20 vision.

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died, because I saw Medusa, because I have 20-20 vision because I am just skilled like that.

skip 33286 b/c i wanted to continue.

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died, because I saw Medusa, because I have 20-20 vision because I am just skilled like that, because I am better than Wickedestjr at a lot of things, so I must be skilled, after all Wicked is just plain awesome.

My last post wasn't meant as part of the game...

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died, because I saw Medusa, because I have 20-20 vision because I am just skilled like that, because I am better than Wickedestjr at a lot of things, so I must be skilled, after all Wicked is just plain awesome because I can't think of a way to continue this.

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died, because I saw Medusa, because I have 20-20 vision because I am just skilled like that, because I am better than Wickedestjr at a lot of things, so I must be skilled, after all Wicked is just plain awesome because I can't think of a way to continue this, because it's sort of dying, :(

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died, because I saw Medusa, because I have 20-20 vision because I am just skilled like that, because I am better than Wickedestjr at a lot of things, so I must be skilled, after all Wicked is just plain awesome because I can't think of a way to continue this, because it's sort of dying, because Wicked's computer blew up,

Wicked's computer blew up, because ernie said so, because he is able to say things and then they will happen (except the end), because McCain is creepy, because he is stupid, because Barack Obama is really stupid, because ernie should be president, because obama is an idiot and mccain is awesome, because Bob Barr is even more awesome, because Bob Barr is a fairy,because I just died, because I saw Medusa, because I have 20-20 vision because I am just skilled like that, because I am better than Wickedestjr at a lot of things, so I must be skilled, after all Wicked is just plain awesome because, I can't think of a way to continue this, because it's sort of dying, because Wicked's computer blew up, because crazychinesemaniac shot a missile at it, because he had an existential breakdown, because he knew that the Simpsons would be cancelled one day, because


$ f: [0;\infty)\rightarrow [0;\infty)$ is bounded on the interval $ [0, 1]$ and satisfies $ f(x) f(y)\leq x^2f(\frac{y}{2}) \plus{} y^2f(\frac{x}{2})$ for all $ x, y$. Show that $ f(x)\leq x^2$.

[hide="Solution"]$ y\equal{}x$: $ f(x)^2\leq 2x^2f(\frac{x}{2})\Leftrightarrow f(x)\leq x\times \sqrt{2f(\frac{x}{2})}$
If $ x\equal{}0$ then $ f(0)\equal{}0$
If $ x\neq 0$ then $ \frac{f(x)}{x^2}\leq\sqrt{\frac{1}{2}\times\frac{f(\frac{x}{2})}{(\frac{x}{2})^2}}$
With $ g(x)\equal{}\frac{f(x)}{x^2}$ then $ g(x)\leq \sqrt{\frac{g(\frac{x}{2})}{2}}\leq ...\leq \sqrt[2^n]{\frac{g(\frac{x}{2^n})}{2^{2^n\minus{}1}}}$
With $ n$ is enough large then $ \frac{x}{2^n}<1$ then exists $ A>0: f(\frac{x}{2^n})<A\Leftrightarrow g(\frac{x}{2^n})<\frac{A.2^{2n}}{x^2}$
It mean $ g(x)\leq \lim_{n\rightarrow\infty}\sqrt[2^n]{\frac{A.2^{2n}}{2^{2^n\minus{}1}.x^2}}\equal{}1$
Then $ f(x)\leq x^2\forall x$[/hide]


function
floor function
algebra proposed
algebra

1. Given positive numbers a,b,c,d and that cd=1.
   Show that the closed interval [ab,(a+c)(b+d)] contains an integer square.

2. [tex]f,g:R\rightarrow R[/tex] are functions such that [tex]f(x+g(y))=2x+y+5[/tex] for all x,y. Find an explicit representation of [tex]g(x+f(y))[/tex].

OK ok next time I will come up with a difficult one.


blogs
superior algebra
superior algebra unsolved

Is there somebody who know where can I find a solution manual to Galois Theory by Ian Stewart.

I think the solution manual does not exists. I tried searching in Google and in Amazon. Why not writing one yourself?

[quote="HilbertThm90"]Stewart wrote a Galois Theory book !?  :ewpu: (I just wrote a huge blog post complaining about his calc text).[/quote]
Not the same Stewart...


percent
probability
algebra
binomial theorem

If a ten question True/False test is given and 60% is a passing grade, waht isyour percent chance, to the nearest whole percent, of passing if you guess on each question?

I got the same answer as Scrambled; here's how I did it:

[hide]Using the Binomial Theorem for Probability,

Prob. of getting 6 right = 10C6 x (.5)^10 = 210/1024

Prob. of getting 7 right = 10C7 x (.5)^10 = 120/1024

Prob. of getting 8 right = 10C8 x (.5)^10 = 45/1024

Prob. of getting 9 right = 10C9 x (.5)^10 = 10/1024

Prob. of getting 10 right = 10C10 x (.5)^10 = 1/1024

So prob. of getting a passing grade = 210/1024 + 120/1024 + 45/1024 + 10/1024 + 1/1024 = 386/1024 = 193/512.[/hide]

Quote:How could you have answered this question by only computing the probability of getting exactly 50%?

One could just [hide]find the probability of getting 50%, subtract it from 1, and divide the result by 2.[/hide] This is a much better way of solving the question than just brute-forcing it, because you don't have to calculate all the separate probabilities. Thanks to Mr. Rusczyk for suggesting this approach.

ooooooooo. This probability problem is just like rolling a pair of dice in games like craps.

[hide] since 1024 is the total number of combinations,

(10C6)/1024

(10C7)/1024

(10C8)/1024

(10C9)/1024

(10C10)/1024

Add them all together to get 386/1024

Simplify to 193/512[/hide]

Quote:I don't understand why you divide the result by 2 after computing the possibility of getting 50%....

Here's why:

[hide]If you let the probability of getting 50% be x and the probability of getting any of 60%, 70%, 80%, 90%, or 100% (all the passing grades) be y, the probability of getting any of 0%, 10%, 20%, 30%, or 40% also equals y (because P(0%) = P(100%), P(10%) = P(90%), etc.). Since we've covered all possible outcomes, we have x + 2y = 1, so y = (1-x)/2.[/hide]

oops. forgot that you had to use a percent, not just a fraction. why do i always screw up on this stuff?


What is the number of 0-1 strings of length n such that there aren't 2 consecutive ones ?

Let $a_n$ be a desired number. Then $a_n=a_{n-1}+a_{n-2}$, since the number of sequences started with 0 is $a_{n-1}$ and the number of sequences started with $0$ is $a_{n-2}$.

BTW, it is a basic property of Fibonachi's base representation.


Prove that every positive integer not exceeding n! can be expressed as the sum of at most n distinct positive integers each of which is a divisor of n!.

a "strong" induction on $n$ will do it I think.

Let's suppose that the result is true for $k = 1 .. n-1$
Let's take $k$ between $1$ and $n!-1$
If $k \leq m!$ with $m < n$ we use inductiuon hypothesis.

If $(n-1)! < k \leq n!$ then $k = n!/q + l$ then apply induction hypothesis to $l < (n-1)!$


linear algebra
matrix
parameterization
algebra
polynomial
invariant
linear algebra open

Let u be a real number,v be a non real complex number and w be its conjugate.
Let A,B,C be real (n,n) matrices which are simultaneously diagonalizable over C:
A~diag(ai), B~diag(bi), C~diag(ci) where, for all i, (ai,bi,ci) is a permutation of (u,v,w).
I think that necessarily n is even.
Is it true?

(Sorry, for the long proof, but I have written it down in a brute force manner. I guess someone will easily spot the essential argument and incorporate it into a more efficient proof :idea: . In particular, some aspects shown are not relevant for the "$ n$ is even conclusion")

Since $ tr(X)\in\mathbb{R}$ for $ X\in\{A,B,C\}$ we must have that $ v$ and its conjugate $ \bar{v}$ occur with the same multiplicity as eigenvalues. Performing a basis change in $ \mathbb{R}^{n}$ we can therefore assume that with $ k$ copies of $ S(v): \equal{}\left(\begin{array}{cc}Re(v) & Im(v)\\ \minus{}Im(v) & Re(v)\end{array}\right)$ the matrix $ A$ has the following form

$ A\equal{}\left(\begin{array}{cccc}uI_{n\minus{}2k}& & &\\ & S(v) & &\\ & &\ddots &\\ & & & S(v)\end{array}\right)$

Note, that for $ T\equal{}\left(\begin{array}{cc}1 & 1\\ \minus{}i & i\end{array}\right)$ we have $ T^{\minus{}1}S(v)T\equal{}\left(\begin{array}{cc}\bar{v}& 0\\ 0 & v\end{array}\right)\equal{}: N(v)$.

Any matrix $ X\in M_{n}(\mathbb{R})$ represented in the (identical) block form $ X\equal{}\left(\begin{array}{cccc}X_{00}& X_{01}& & X_{0k}\\ X_{10}& X_{11}& & X_{1k}\\ & &\ddots &\\ X_{k0}& X_{k1}& & X_{kk}\end{array}\right)$

commutes with $ A$ iff for all $ i,j\equal{}1,\ldots,k: uX_{0j}\equal{}X_{0j}S,uX_{j0}\equal{}SX_{j0},SX_{ij}\equal{}X_{ij}S$ holds (and $ X_{00}$ arbitrary). Direct calculation using $ Im(v)\not\equal{}0$ yields that this is equivalent to

$ X_{ij}\equal{}S(x_{ij}),X_{j0}\equal{}^{t}X_{0j}\equal{}0_{2\times (n\minus{}2k)}$ with suitable parameters $ x_{ij}\in\mathbb{C}$, which we collect in the matrix $ X^{*}\in M_{k}(\mathbb{C})$

Setting $ U\equal{}\left(\begin{array}{cccc}uI_{n\minus{}2k}& & &\\ & T & &\\ & &\ddots &\\ & & & T\end{array}\right)$ we get $ U^{\minus{}1}AU\equal{}\left(\begin{array}{cccc}uI_{n\minus{}2k}& & &\\ & N(v) & &\\ & &\ddots &\\ & & & N(v)\end{array}\right)$ and for any matrix $ X$ commuting with $ A$ we compute $ U^{\minus{}1}XU\equal{}\left(\begin{array}{cccc}X_{00}& & &\\ & N(x_{11}) & & N(x_{1k})\\ & &\ddots &\\ & N(x_{k1}) & & N(x_{kk})\end{array}\right)$. Note, that this implies for the characteristic polynomial:

$ f_{X}(t)\equal{}\det(tI_{n}\minus{}X)\equal{}f_{X_{00}}(t)\cdot f_{X^{*}}(t)\cdot f_{\bar{X^{*}}}(t)$

Since $ B,C$ commute with $ A$  we can use the above description of matrices and the symbols $ B_{00},b_{ij},B^{*}$ (resp. $ C_{00},c_{ij},C^{*}$) to denote the respective parameters. Since $ A\plus{}B\plus{}C\equal{}(u\plus{}2 Re(v))I_{n}$ holds (invariant under basis transformations !), we get:

$ B_{00}\plus{}C_{00}\equal{}2 Re(v)I_{n\minus{}2k},\forall i\not\equal{}j\equal{}1,\ldots, k: b_{ii}\plus{}c_{ii}\equal{}u\plus{}\bar{v},b_{ij}\equal{}\minus{}c_{ij}\Longleftrightarrow\\ B_{00}\plus{}C_{00}\equal{}2 Re(v)I_{n\minus{}2k},C^{*}\equal{}(u\plus{}\bar{v})I_{k}\minus{}B^{*}$

If $ B_{00}$ (resp. $ C_{00}$) has the eigenvalue $ u$, then $ C_{00}$ (resp. $ B_{00}$) would have the eigenvalue $ 2Re(v)\minus{}u\not\in\{v,\bar{v}\}$. Therefore, all eigenvalues of the matrices $ B_{00},C_{00}$ must lie in $ \{v,\bar{v}\}$. Since $ tr(B_{00})\in\mathbb{R}$ they must both occur with the same multiplicity. Therefore $ n\minus{}2k\equal{}rk(B_{00})$ is even and $ n$ must be even, too (as a necessary criterion).

That an even $ n$ is sufficient for the existence of some $ A,B,C$, can be seen from the example $ A\equal{}uI_{n},B\equal{}diag(S(v),\ldots,S(v)),C\equal{}diag(S(\bar{v}),\ldots,S(\bar{v}))$

the essential and short argument is as follows:

Clearly, $ \dim Eig(A,v)\equal{}\dim Eig(A,\bar{v})$ holds, because of $ Im(v)\not\equal{}0$ and $ A$ being a real matrix. Therefore $ n\equiv\dim Eig(A,u)\bmod{2}$ is valid. Since $ B,C$ commute with $ A$, they leave the eigenspace $ V\equal{}Eig(A,u)$ invariant. $ B$'s restriction on $ V$ cannot have the eigenvalue $ u$, so $ B$ has only the eigenvalues $ v,\bar{v}$ on $ V$. Since $ V$ is spanned by real eigenvectors, $ B$ must have the eigenvalues $ v,\bar{v}$ with the same multiplicity, i.e. $ \dim V$ is even and therefore $ n$, too.


2^u=3^y=6^-z
find 1/u+1/y+1/z

i want the cmplete solution

Try taking logarithms and see where that gets you.  And don't be so rude in demanding complete solutions.  People will solve as much as they want/can and may want you to actually try it for yourself.

I tried by logarit but have'n got any way. Who can slove this problem? But I will try to do it.

$ 2^{u}\equal{}3^{y}$ gives $ \frac{1}{y}\equal{} \frac{\log{3}}{\log{2}}\times\frac{1}{u}$
$ 2^{u}\equal{}6^{\minus{}z}$ gives $ \frac{1}{z}\equal{} \minus{}\frac{\log{6}}{\log{2}}\times\frac{1}{u}\equal{}\minus{}\frac{\log{3}\plus{}\log{2}}{\log{2}}$

Therefore,
$ \minus{}\frac{1}{u}\equal{}\frac{1}{y}\plus{}\frac{1}{z}$
$ \frac{1}{u}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}0$

Oh! First akshayarora9 post problem not true. If you change so, it's easy.


induction
arithmetic sequence
algebra proposed
algebra

Let positive numbers $a_1, a_2, ..., a_{3n}$ $(n \geq 2)$ constitute an arithmetic progression with common difference $d > 0$. Prove that among any $n + 2$ terms in this progression, there exist two terms $a_i, a_j$ $(i \neq j)$ satisfying $1 < \frac{|a_i - a_j|}{nd} < 2$.

Without loss of generality, suppose $i<j$.

Then $|a_i-a_j|=(j-i)d\Rightarrow\frac{|a_i-a_j|}{nd}=\frac{j-i}{n}$, and the given condition $1<\frac{|a_i-a_j|}{nd}<2$ becomes $i+n<j<i+2n$.

It's pretty easy to show by induction that if the given set contains $k$ elements of the AP satisfying the condition, then there must be $n+k-2$ other elements from the AP which would violate the condition if they were added to the set.  Letting $k=n+1$ proves the desired result.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=326543#p326543]http://www.mathlinks.ro/Forum/viewtopic.php?p=326543#p326543[/url] for another proof.


hi,
theres a wording problem i cannot solve, Im not sure how to write out the equation from the problem.

this is the question:
Bus company charges a half of its normal far for concession card holders. If company takes $ 2600 on a certain route and carries 900 passengers, then how many of those passengers were concession card holders if the normal far was$4.

All i know so far is to
Let X be number of concession card holders. Can anyone tell me what is the equation?been workin on it for a while. I will know what to do if i know the equation. I know its suppose to be an easy question but i feel so dumb not knowing it.

thank u

What's the [Unparseable or potentially dangerous latex formula.Error: 6: 944x15] thingy?
I can't tell what the problem is.

[quote="ludasky"]Bus company charges a half of its normal far for concession card holders. If company takes $ \$$2600 on a certain route and carries 900 passengers, then how many of those passengers were concession card holders if the normal far was $ \$$4.[/quote]
Dollar signs on this board cause it to enter Latex math mode, which allows for typesetting mathematics but is not so nice to normal next. To type a dollar sign, use [code]$ \$ $[/code]

$ x$ is non-card holders, $ y$ is card holders

$ 4x\plus{}2y\equal{}2600$
$ x\plus{}y\equal{}900$

[quote="facis"]$ x$ is non-card holders, $ y$ is card holders

$ 4x \plus{} 2y \equal{} 2600$
$ x \plus{} y \equal{} 900$[/quote]

A method following facis' incomplete solution without algebra that I learnt in middle school (still useful now)

Assume all 900 passengers are not card holders. Then ther bus company would have collected $ 900 \times 4 \equal{} 3600$ dollars. But the company collected $ 3600 \minus{} 2600 \equal{} 1000$ dollars less, so these extra 1000 dollars must have been caused by some of the passengers being card holders. Thus the number of card holders is $ 1000/2 \equal{} 500$.


[quote="Nameless"][quote="anthony19891019"]1/2*(3,0,0) + (1,1,1)  and (2,1,0), which bigger?[/quote]
I think it should be $\frac{1}{3}$(3,0,0)[/quote]
sorry, i only want to compare these two... i only want to know 1/2*(3,0,0) + (1,1,1)  and (2,1,0), which is bigger..
can you give me a satisfied solution? thanks..

[quote="Soarer"]do you mean $(3,0,0)+(1,1,1) \ge 2(2,1,0)$? If yes, then it is just schur's inequality[/quote]

thankyou, soarer..

actually i want to know 0.5*(3,0,0) + (1,1,1) > (2,1,0)...

but now it becomes trivial..


superior algebra
superior algebra unsolved

Assume $k$ is a field that has no extension of the form $K/k$ with $[K: k]=p$, $p$ fixed prime. Is there a division algebra $D$ such that $\dim_{k}D=p^{n}$ for some $n$ and $Z(D)=k$?

If $k$ is such a field, then $\mathrm{char}\,{k}= 0$ and $D^{\otimes p}\neq M_{m}(k)$.

edit: and every element of $D$ has order $p$, so it is contained in $k$, so $D = k$.

uhm..
actually, i don't think $\text{char}k$ has to be $0$: just take a prime $q$ and the field $F = \mathbb{F}_{q^{p^\infty}}: = \bigcup_{n}\mathbb{F}_{q^{p^{n}}}$. $F$ has no extension of degree $p$, but has $\text{char}F = q$.

anyway, the only other thing i can say is that $k$ has to be infinite, since there's a theorem (wedderbrun's theorem, maybe?) that states that every finite division ring is a field...

[quote="ziuuz"]$D^{\otimes p}\neq M_{m}(k)$[/quote]
Is that just your conjecture or do you have a proof for it? If so, please post it  :wink:


trigonometry
inequalities
function
inequalities solved
geometry

[quote="hipsishopsis"][hide]
First note that if one of the angles is $\ge \frac{\pi}{2}$ radians then the left side is $\ge 0$ and right side is $\le 0$.

So let's assume they are all in the range $0 < A,B,C < \frac{\pi}{2}$.

$(1-\cos A)(1-\cos B)(1-\cos C) \ge \cos A \cos B \cos C$

On the given range this is equivalent to:
$(\frac{1}{\cos A}-1)(\frac{1}{cos B}-1)(\frac{1}{\cos C}-1) \ge 1$.

Now note that as $0 < \cos x < 1$ when $0 < x < \frac{\pi}{2}$ every term on the left side is $\ge 1$.
[/hide][/quote]

That last part doesn't work; it only guarantees each term nonnegative.
[hide]
It is true that we can reduce this to acute triangles, because if right/obtuse,
\[(1-\cos A)(1-\cos B)(1-\cos C)\geq 0\geq \cos A\cos B\cos C\]
Otherwise, let $x=\cos A,y=\cos B,z=\cos C$. Observe that one of $(2x-1)(2y-1),(2y-1)(2z-1),(2z-1)(2x-1)$ must be nonnegative (since they are three numbers with nonnegative product). WLOG let $(2x-1)(2y-1)\geq 0$. Now the inequality is equivalent to
\[(1-x)(1-y)\geq (xy+(1-x)(1-y))z\]
From $\cos (A-B)\leq 1$, we have $\cos (A-B)+\cos (A+B)\leq 1+\cos (A+B)$, which becomes $z\leq 1-2xy$. We can now multiply by the positive $xy+(1-x)(1-y)$. It remains to prove
\[(1-x)(1-y)\geq (xy+(1-x)(1-y))(1-2xy)\]
Guess what? It turns out that their difference is $xy(2x-1)(2y-1)$ (that's what motivated the WLOG in the first place), so we're done.
[/hide]

[quote="scorpius119"][quote="hipsishopsis"][hide]
First note that if one of the angles is $\ge \frac{\pi}{2}$ radians then the left side is $\ge 0$ and right side is $\le 0$.

So let's assume they are all in the range $0 < A,B,C < \frac{\pi}{2}$.

$(1-\cos A)(1-\cos B)(1-\cos C) \ge \cos A \cos B \cos C$

On the given range this is equivalent to:
$(\frac{1}{\cos A}-1)(\frac{1}{cos B}-1)(\frac{1}{\cos C}-1) \ge 1$.

Now note that as $0 < \cos x < 1$ when $0 < x < \frac{\pi}{2}$ every term on the left side is $\ge 1$.
[/hide][/quote]

That last part doesn't work; it only guarantees each term nonnegative.
[/quote]

Oops, you're right - careless thinking here :)

[quote="hien"]Let $A,B,C$ be three angles of triangle $ABC$. Prove that:
$(1-\cos A)(1-\cos B)(1-\cos C)\ge\cos A\cos B\cos C$[/quote]
$(1-\cos A)(1-\cos B)(1-\cos C)\ge\cos A\cos B\cos C\Leftrightarrow$
$\Leftrightarrow(a+b-c)^{2}(a+c-b)^{2}(b+c-a)^{2}\geq(a^{2}+b^{2}-c^{2})(a^{2}+c^{2}-b^{2})(b^{2}+c^{2}-a^{2}).$
Let $\Delta ABC$ is acute angle triangle and use the following inequality:
$(a+b-c)^{2}(a+c-b)^{2}\geq(a^{2}+b^{2}-c^{2})(a^{2}+c^{2}-b^{2})$ and similar.

hmm... if $xyz \ge 8$ does that imply $(x-1)(y-1)(z-1) \ge 1$ for $x,y,z > 1$?

if so then the inequality follows from $\cos{A}\cos{B}\cos{C}\le \frac{1}{8}$ which can be shown by Jensen's.

EDIT: never mind, the first assumption does not hold - consider $x = 7, y = z = 2\sqrt{\frac{2}{7}}$

$\bullet$ If triangle $ABC$ is obtuse then the proposed inequality is true.

$\bullet$ If triangle $ABC$ is acute, then we have:
\[\prod\cos A\le\prod (1-\cos A)\iff\prod\cos A(1+\cos A)\le\prod(1-\cos^{2}A)\\ \iff 8\prod\cos A\prod\cos^{2}\frac{A}{2}\le\prod\sin^{2}A\iff\prod\cot\frac{A}{2}\le\prod\tan A\\ \iff\sum\cot\frac{A}{2}\le\sum\tan A\]
which is trivial.

[quote="colorfuldreams"][quote="arqady"]
$(a+b-c)^{2}(a+c-b)^{2}\geq(a^{2}+b^{2}-c^{2})(a^{2}+c^{2}-b^{2})$ and similar.[/quote]

How you prove  this without expand?[/quote]

they are sides of triangle ...

$(a^2-(b-c)^2)^2\ge a^4-(b+c)^2(b-c)^2$
$2(b-c)^2(b^2+c^2-a^2)\ge 0$

but in acute inequality we have $a^2+b^2-c^2\ge 0$

就是余弦定理了

Upon expansion, the inequality is equivalent to

$1-\sum\cos A+\sum\cos A\cdot\cos B\ge2\prod\cos A$.

Making use of the following identities: $\begin{cases} \sum\cos A=1+\frac{r}{R} \\ \\ \sum\cos A\cdot\cos B=\frac{s^2+r^2}{4R^2}-1 \\ \\ \prod\cos A=\frac{s^2-(2R+r)^2}{4R^2}\end{cases}$

the inequality is further equivalent to

$1-\left(1+\frac{r}{R}\right)+\left(\frac{s^2+r^2}{4R^2}-1\right)\ge2\left(\frac{s^2-(2R+r)^2}{4R^2}\right)$

$\iff \frac{s^2+r^2-4Rr-4R^2}{4R^2}\ge \frac{2s^2-8R^2-2r^2-8Rr}{4R^2}$

$\iff s^2\le4R^2+4Rr+3r^2$

which is the well-known Gerretsen's Inequality.

Note that $(1-\cos A)(1-\cos B)(1-\cos C)\ge 0$ so we only need to consider the case of triangular  ABC tip.  Inequality to prove rewritten into:
\[\dfrac{1-\cos A}{\cos A}.\dfrac{1-\cos B}{\cos B}.\dfrac{1-\cos C}{\cos C}\ge 1\\\Leftrightarrow \tan A.\tan B.\tan C\ge\cot\dfrac{A}{2}.\cot\dfrac{B}{C}.\cot\dfrac{C}{2}\] On the other hand, in the triangle  ABC any, we always have: \[\tan A+\tan B+\tan C=\tan A.\tan B.\tan C\\\cot\dfrac{A}{2}+\cot\dfrac{B}{2}+\cot\dfrac{C}{2}=\cot\dfrac{A}{2}.\cot\dfrac{B}{2}.\cot\dfrac{C}{2}\]   Therefore, we should point out that: \[\tan A+\tan B+\tan C\ge\cot\dfrac{A}{2}+\cot\dfrac{B}{2}+\cot\dfrac{C}{2}\] At the function  $f(x)=\tan x$,with $x\in \left(0;\dfrac{\pi}{2}\right)$ We have: $f''(x)=-\sin x,\forall x\in\left(0;\dfrac{\pi}{2}\right)$ therefore, $f(x)$  convex on the account $x\in\left(0;\dfrac{\pi}{2}\right)$ Using the inequality Jensen , we have: \[\tan A+\tan B\ge 2\tan\dfrac{A+B}{2}=2.\cot\dfrac{C}{2}\;\;\;\;\;\;\;\;\;(1)\] Similarly \[\tan B+\tan C\ge 2\tan\dfrac{B+C}{2}=2.\cot\dfrac{A}{2}\;\;\;\;\;\;\;\;\;(2)\] \[\tan C+\tan A\ge 2\tan\dfrac{C+A}{2}=2.\cot\dfrac{B}{2}\;\;\;\;\;\;\;\;\;(3)\] Czech side under the side of the inequality (1), (2) and (3) we obtain issues need to prove . The problem is proven. Mark equality occurs if and only if the triangle  ABC were.

[quote=hien]Let $A,B,C$ be three angles of triangle $ABC$. Prove that:
$(1-\cos A)(1-\cos B)(1-\cos C)\ge\cos A\cos B\cos C$[/quote]
[url=http://artofproblemsolving.com/community/c6h131123p743688]The inequality is equivalent to[/url]
Let $a,b,c>1$ be such that $a+b+c=abc$. Prove that$$(a^{2}-1)(b^{2}-1)(c^{2}-1)\leq 8.$$

This inequality is stronger:
$ (1-cosA)(1-cosB)(1-cosC)\ge (2-\frac{2r}{R})cosAcosBcosC $
See "Improvement of a Problem from American Mathematicai Monthly ",
in Mathematical Reflections 2 (2011), an article written by Wei-dong Jiang and Mihály Bencze.
The proposed inequality is equivalent to
$ s^2\le 4R^2+4Rr+r^2+\frac{Rr^2}{R-r}  $

[quote=sqing][quote=hien]Let $A,B,C$ be three angles of triangle $ABC$. Prove that:
$$(1-\cos A)(1-\cos B)(1-\cos C)\ge\cos A\cos B\cos C$$[/quote]
Let $A,B,C$ be three angles of  acute-angled triangle $ABC$. Prove that:
$$(1-\cos A)(1-\cos B)(1-\cos C)\geq \frac{3\cos A\cos B\cos C}{2+8\cos A\cos B\cos C}.$$
[size=50]Middle school mathematics research (China Nanchang)No.3(2019)[/size][/quote]

In any triangle,have

\[(1-\cos{A})(1-\cos{B})(1-\cos{C})(1+\cos{A}\cos{B}\cos{C})\geq \frac{9}{8}\cos{A}\cos{B}\cos{C}\]
<=>

\[4\,{R}^{2}+4\,Rr+3\,{r}^{2}-2\,{\frac {{r}^{2} \left( R-2\,r \right)
 \left( R+2\,r \right) }{ \left( 3\,R-2\,r \right)  \left( 3\,R+2\,r
 \right) }}\geq{s}^{2}\]
<=

\[4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{\frac {{r}^{2} \left( R-2\,r \right) }{R-
r}}\geq {s}^{2}\]

In any triangle,prove

\[ \left( 1+\cos \left( 2\,A \right)  \right)  \left( 1+\cos \left( 2\,B
 \right)  \right)  \left( 1+\cos \left( 2\,C \right)  \right)  \left(
1-\cos \left( 2\,A \right) \cos \left( 2\,B \right) \cos \left( 2\,C
 \right)  \right) +{\frac {9}{8}}\,\cos \left( 2\,A \right) \cos
 \left( 2\,B \right) \cos \left( 2\,C \right)\geq 0 \]

equality hold when $a=b=\frac{\sqrt{2}}{2}c$

In any triangle,prove that

\[ \left( 1+\cos \left( 4\,A \right)  \right)  \left( 1+\cos \left( 4\,B
 \right)  \right)  \left( 1+\cos \left( 4\,C \right)  \right)  \left(
1-\cos \left( 4\,A \right) \cos \left( 4\,B \right) \cos \left( 4\,C
 \right)  \right) +{\frac {9}{8}}\,\cos \left( 4\,A \right) \cos
 \left( 4\,B \right) \cos \left( 4\,C \right) \geq 0\]

equality hold when $a=b=kc,k\in \{\frac{1}{3}\,\sqrt {3},\frac{1}{2}\,\sqrt {4+2\,\sqrt {2}},\frac{1}{2}\,\sqrt {4-2\,\sqrt {
2}}\}$


very beautiful
$ a,b,c$ are prime number
find $ a,b,c$ such that:
$ b\minus{}a,c\minus{}a,c\minus{}b$ are prime numbers also
(with solution)

[quote="JRav"]$ a \equal{} 2$, $ b \equal{} 5$, $ c \equal{} 7$[/quote]
what about the solution ?

Here's a quick note (explanation of JRav's solution):

[hide]obviously c>b>a

Also, if all three are odd, all three new primes are even, hence 2, nut it would mean a=b, so it isn't ok.
That's why one of them is 2. It is obviously a (2 is the smallest prime!). Then, c-b=2, but note that b-2 is also prime, so you have three odd consecutive twins (c,b,b-2): only such example is 3,5,7. (easy to prove, because one of them has to be divisible with 5, hence it has to be 5).[/hide]


If you have found transcripts for other classes useful, you should find the AIME Problem Seminar transcripts useful as well.  I'll leave it up to you to judge the value of that much AIME problem solving transcript.

The primary source of problems will be old AIMEs, but there will be AIME-like problems from other sources.

I'm going to the Rice tournament, but I'll still probably sign up for the AIME Special Seminar.  You can always look at the transcripts; it's not like the AIME is the next day and you wont have time to look at them.  Plus, it's preparation for the AIME!  Everybody could use that  :lol:

I did not qualify for AIME this year.

However, I'm pretty sure that I'll qualify AIME next year either by USAMTS or AMC so would it be a good thing if I take the seminar this year and maybe look at it and see what I can understand and save for next year when I'm qualified?

Also, are the problems from this year and the next year will be different? Because if they're same, there's no way I'm taking now and then both. ;)

Thanks for the reply.


Let $K$ be a given real number. Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$ such that that continuity is not demanded at $ x=K$ for $f$ and \[ f\left( {\frac {{x \plus{} y}} {{1 \plus{} (xy/K^2 )}}} \right) \equal{} f(x)f(y),\] for all reals $x$ and $y$ such that $xy\neq -K^2$.

[quote="Xaenir"]Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying:$ f\left( {\frac {{x \plus{} y}} {{1 \plus{} (xy/K^2 )}}} \right) \equal{} f(x)f(y)$ ,$ \forall x,y$[/quote]

Notice that this equality can never be verified $ \forall x,y$ since it cant be verified for $ (K,\minus{}K)$
If we consider that the problem is modified by adding $ \forall x,y$ [b][u]such that[/u][/b] $ xy\neq \minus{}K^2$, then :

$ y\equal{}\minus{}x\notin\{\minus{}K,\plus{}K\}$ $ \implies$ $ f(0)\equal{}f(x)f(y)$ and so $ f(x)\equal{}c$ and so $ c\equal{}c^2$ and only two solutions :

$ f(x)\equal{}0$
$ f(x)\equal{}1$

Obviously $ xy \ne  \minus{} K^2$
I can't understand how you got $ f(0)\equal{}f(x)f(y)$
Anyway , I got the answer $ \left( {\frac{{K \plus{} x}}
{{K \minus{} x}}} \right)^a$

[quote="Xaenir"]Obviously $ xy \ne \minus{} K^2$[/quote]
OK, better to write it.
:)
[quote="Xaenir"]
I can't understand how you got $ f(0) \equal{} f(x)f(y)$[/quote]

nvm, it's a mistake ...  :blush:

[quote="Xaenir"]Obviously $ xy \ne \minus{} K^2$
I can't understand how you got $ f(0) \equal{} f(x)f(y)$
Anyway , I got the answer $ \left( {\frac {{K \plus{} x}} {{K \minus{} x}}} \right)^a$[/quote]

This solution is not continuous for $ x \equal{} K$.

This solution is not defined for $ |x| > |K|$ and $ a\notin\mathbb Q$

Some other solutions exist (including all continuous).

I'll post my solution in the next 24h.

[quote="Xaenir"]Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying:$ f\left( {\frac {{x \plus{} y}} {{1 \plus{} (xy/K^2 )}}} \right) \equal{} f(x)f(y)$ ,$ \forall x,y$[/quote]

Obviously $ f(x)\equal{}0$ $ \forall x$ and $ f(x)\equal{}1$ $ \forall x$ are solutions (and are the only constant solutions).
Let us look now for non constant solutions.

We'll show that no non constant solution exist (no non constant solution is continuous for $ x\equal{}K$). So I'll consider that [u][b]there is a second error in the problem statement[/b][/u] and that continuity is not demanded for $ x\equal{}K$

First get rid of $ K$ (I suppose $ K>0$) :

$ f(\frac {Kx \plus{} Ky} {1 \plus{} (KxKy/K^2 )}) \equal{} f(Kx)f(Ky)$ $ \implies$ $ f(K\frac {x \plus{} y} {1 \plus{} xy}) \equal{} f(Kx)f(Ky)$

Let $ g(x)\equal{}f(Kx)$ : $ g(\frac {x \plus{} y} {1 \plus{} xy}) \equal{} g(x)g(y)$

1) [u][b]Some preliminary results[/b][/u] :
$ y\equal{}1$ and $ x\neq \minus{}1$ $ \implies$ $ g(x)g(1)\equal{}g(1)$ and so either $ g(1)\equal{}0$, either $ g(x)\equal{}1$ $ \forall x\neq \minus{}1$. So $ g(1)\equal{}0$
$ y\equal{}\minus{}1$ and $ x\neq 1$ $ \implies$ $ g(x)g(\minus{}1)\equal{}g(\minus{}1)$ and so either $ g(\minus{}1)\equal{}0$, either $ g(x)\equal{}1$ $ \forall x\neq 1$. So $ g(\minus{}1)\equal{}0$

If $ g(x_0)\equal{}0$ for some $ |x_0|\neq 1$, then, using $ y\equal{}\frac{z\minus{}x_0}{1\minus{}x_0z}$, we get $ f(x_0)f(y)\equal{}f(z)$ and so $ f(x)\equal{}0$ $ \forall x\neq \frac{1}{x_0}$ and so (since we are studying non constant solutions) $ g(x)\neq 0$ $ \forall x\notin \{\minus{}1,\plus{}1\}$

Then, $ \forall x,y\in\mathbb R^*$ :

$ g(x)g(\frac{1}{y})\equal{}$ $ g(\frac {x \plus{} \frac{1}{y}} {1 \plus{} \frac{x}{y}})$ $ \equal{}g(\frac {1 \plus{} xy} {x \plus{} y})$

$ g(y)g(\frac{1}{x})\equal{}$ $ g(\frac {y \plus{} \frac{1}{x}} {1 \plus{} \frac{y}{x}})$ $ \equal{}g(\frac {1 \plus{} xy} {x \plus{} y})$

So $ g(x)g(\frac{1}{y})\equal{}g(y)g(\frac{1}{x})$, which implies $ \frac{g(x)}{g(\frac{1}{x})}$ $ \equal{}\frac{g(y)}{g(\frac{1}{y})}$ $ \equal{}c$

So $ g(x)\equal{}c\cdot g(\frac{1}{x})$ $ \forall x\neq 0$ and since then $ g(\frac{1}{x})\equal{}c\cdot g(x)$, we get $ c\equal{}\plus{}1$ or $ c\equal{}\minus{}1$.

So :
either $ g(x)\equal{}g(\frac{1}{x})$ $ \forall x\neq 0$
either $ g(x)\equal{}\minus{}g(\frac{1}{x})$ $ \forall x\neq 0$

2) [u][b]Computation of non constant $ g(x)$ for $ |x|<1$[/u][/b] : $ g(x)\equal{}\left(\frac{1\plus{}x}{1\minus{}x}\right)^b$ $ \forall x\in(\minus{}1,\plus{}1)$
Consider $ x,y\in(\minus{}1,1)$ : $ \exists! u,v\in\mathbb R$ such that $ x\equal{}\tanh(u)$ and $ y\equal{}\tanh(v)$ Then :

$ g(\tanh(u\plus{}v))\equal{}g(\tanh(u)g(\tanh(v))$ $ \forall x,y\in \mathbb R$ and, since $ g(\tanh(x))$ is continuous and non constant :

$ g(\tanh(x))\equal{}e^{ax}$ $ \forall x\in\mathbb R$

$ \implies$ $ g(x)\equal{}e^{a\tanh^{[\minus{}1]}(x)}$ $ \forall x\in(\minus{}1,\plus{}1)$

$ \implies$ $ g(x)\equal{}\left(\frac{1\plus{}x}{1\minus{}x}\right)^b$ $ \forall x\in(\minus{}1,\plus{}1)$ (with $ b\equal{}\frac{a}{2})$

Notice here that this function has no limit when$ x\to \plus{}1$ and so the function cant be continuous in $ 1$.

[u][b]So the only continuous solution are constant solutions $ g(x)\equal{}0$ and $ g(x)\equal{}1$ $ \forall x$[/u][/b]

In case of second error in the problem statement (pleaaaase be more rigorous in your statements) allowing the function not to be continuous at $ \plus{}1$, let's go on :)

3) All solutions for $ g(x)$
We have as mandatory conditions for non constant solutions :
$ g(x)\equal{}\left(\frac{1\plus{}x}{1\minus{}x}\right)^b$ $ \forall x\in(\minus{}1,\plus{}1)$

But $ g(\frac{1}{x})\equal{}\epsilon\cdot g(x)$ (with $ \epsilon\equal{}1$ or $ \epsilon\equal{}\minus{}1$) and so :

$ g(x)\equal{}\epsilon\left(\frac{1\plus{}\frac{1}{x}}{1\minus{}\frac{1}{x}}\right)^b$ $ \equal{}\epsilon\left(\frac{x\plus{}1}{x\minus{}1}\right)^b$ $ \forall x\notin[\minus{}1,\plus{}1]$ and so :

Case $ \epsilon\equal{}\plus{}1$
$ g(x)\equal{}\left|\frac{1\plus{}x}{1\minus{}x}\right|^b$ $ \forall x\notin\{\minus{}1,\plus{}1\}$
$ g(1)\equal{}0$
$ g(\minus{}1)\equal{}0$
The continuity at $ x\equal{}1$ cant be obtain.
The continuity at $ x\equal{}\minus{}1$ may be obtained iff $ b>0$
And it's easy to check that this solution matches the original equation.

Case $ \epsilon\equal{}\minus{}1$
$ g(x)\equal{}\left(\frac{1\plus{}x}{1\minus{}x}\right)^b$ $ \forall x\notin\{\minus{}1,\plus{}1\}$ but then we need $ b\in\mathbb N$ in order to have power of negative values meaningful (I dont want to discuss about $ b\in\mathbb Q)$
$ g(1)\equal{}0$
$ g(\minus{}1)\equal{}0$
The continuity at $ x\equal{}1$ cant be obtain.
The continuity at $ x\equal{}\minus{}1$ may be obtained iff $ b>0$
And it's easy to check that this solution matches the original equation.

4) [u][b]Synthesis : All solutions of the required <modified> problem[/b][/u]  :
Modified problem :
Find all functions $ f: \mathbb{R} \to \mathbb{R}$, continuous in $ \mathbb R\backslash\{K\}$ and satisfying:$ f\left( {\frac {{x \plus{} y}} {{1 \plus{} (xy/K^2 )}}} \right) \equal{} f(x)f(y)$ , (K>0) $ \forall x,y$ such that $ xy\neq \minus{}K^2$

Solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}1$ $ \forall x$

$ f(x)\equal{}\left|\frac{K\plus{}x}{K\minus{}x}\right|^b$ $ \forall x\neq K$ and for any $ b\in\mathbb R\plus{}$
$ f(K)\equal{}0$

$ f(x)\equal{}\left(\frac{K\plus{}x}{K\minus{}x}\right)^n$ $ \forall x\neq K$ and for $ n\in\mathbb N$
$ f(K)\equal{}0$


Im am currently doing a project that has to be based on the history of maths and need to do a 6000 word project on a certain topic of this.  I have always been intersted in astronomy but dont know that much about it.  I would appreciate it if you could think of any project titles that come under mathematical astronomy and its history for me to research (a topic big enough to do 6000 words on)

Thank you for any input

6000 words? :o Wow...

Perhaps one of the topics you could discuss could be how the mathematical formulas for finding planets--or predicting where they will be--have changed, from using the Titius-Bode Law to using more accurate math to using computers, etc., along with discussing other "updates" to astrophysics (or physics as a whole), such as general relativity.

Good luck with your project. :)

[quote="deanj2k"]and need to do a 6000 word project on a certain topic of this.  [/quote]
How do you count equations? :)
Anyway, how much time do you have?
There are numerous things you can do if you have a semester or so (in a place with reasonably clear skies). Some Astronomy books (at the beginner undergraduate level) have a section on experimental and/or observational projects. [i]Astronomy: Principles and Practice[/i] by A.E. Roy and D. Clarke is a good example, but it is out of print (it may be available in libraries). You can also get excellent ideas from past problems from the practical round of the National/International Astronomy olympiads. You may club together four or five similar olympiad problems to get to the required word limit. I can look up some of these if you are interested.(Experiments with a single pole and an open space with lots of sun can alone provide several pages of data)
Good Luck!

I think you could talk a bit about early experimental calculations, like ptolemy etc. and then maybe you could talk about newtons gravitational law, and then about keplers laws, how kepler got them from observational data, and then derive them using newtons law. If you need, I have a really simple derivation for all of them, just let me know and I can tell you. Then about newer methods for observing now planets,e.g how uranus, neptune was discovered. you could mention the n-body problem, and also show lagranges part solution to the 3-body problem, the mathematics are quite simple for it. and then you could talk about anomalies, like the earths tilted axis, and mercuries perihelium displacement, and here you could mention GR which explains it, you could also mention the einstein equations etc. and evolution of stars, galaxies, black holes etc

I hope I could give you some ideas :)

Good like for your project :)

hey first of all best of luck and then i think u should make a project on space time distortion and its chromatic aberration using VLT'S

This topic is kind of too old to revive.  :wink:


logarithms
calculus
calculus computations

Basic rule for handling exponents: if you have $a^b$ where both $a$ and $b$ are important, then rewrite it as $a^b=e^{b\ln a}.$ (In fact, this is the useful definition of what $a^b$ means, anyway.)

So $\frac{d}{dx}x^x=\frac{d}{dx}e^{x\ln x}=e^{x\ln x}\frac{d}{dx}(x\ln x) = x^x(\ln x+1).$

Here's a related question:

Let $u=u(x)$, $v=v(x)$ be everywhere differentiable functions.  Find

$\frac {d}{dx}u^v$

I learnt a shortcut, but its not rigourous - treat u as a constant, differentiate, treat v as a constant, differentiate, and add the two. but it is equivalent i think.

kueh is claiming that

$\left(u^v\right)'=vu^{v-1}u'+u^v\ln u\,v'$

Doing it in a more transparent way:

$\left(u^v\right)'=\left(e^{v\ln u}\right)'=e^{v\ln u}(v\ln u)' = e^{v\ln u}\left(v'\ln u+v\frac{u'}u\right)$

$=u^vv'\ln u + u^vv\frac{u'}u=u^vv'\ln u + vu^{v-1}u'.$

Yes, it is equivalent.

[quote="sky9073"]Supposedly very easy, but I don't get it.
Why is d/dx (x^x)= x^x (ln x +1)?[/quote]
Set $y = x^{x}$ therefore:

\begin{eqnarray*}y &=& x^{x}\hfill \\ \ln y &=& x\ln x \hfill \\ \frac{d}{{dx}}\ln y &=& \frac{d}{{dx}}x\ln x \hfill \\ \frac{1}{y}\cdot y' &=& \ln x+x \cdot \frac{1}{x}\hfill \\ \frac{1}{y}\cdot y' &=& 1+\ln x \hfill \\ y' &=& x^{x}(1+\ln x) \hfill \\ \end{eqnarray*}


probability
calculus
derivative
number theory

I'm participating in the Mu Division this year.
For those of you who have done this before, can you explain to me how each test (individual, cipher, topic,etc.)is scored and how people generally do on them. Thanks.

Scoring this year is 4 for right, 1 for blank, and 0 for wrong.

I cannot tell you how people do, because it differentiates depending on the test, and also the amount of disputes. You just have to talk around after the test.

BTW, what school are you from?

http://www.mualphatheta.org/National_Convention/MAO_2005_Math_Events.htm

^^ That should answer most of your questions about scoring and what not.

For Mu, the most difficult tests in my opinion would be Number Theory and Probability while Sequences and Series and Limits and Derivatives have been known to be hard (FL State MAO L & D: 92(Rickards whoo), 2nd place 72...probably the hardest L & D ever)

I'm only a Theta so I can't exactly say anything about Mu with alot of assurity.

Everyone does well on their tests haha, there will be very few tests where everyone gets a low score.

If you have any more questions, feel free to ask


function
real analysis
real analysis unsolved

Given non-compact subset $ K$ of $ \mathbb{R}$, could anybody show me how to construct a unbounded continuous map defined on K to  $ \mathbb{R}$?

thanks

There are two ways a subset of $ \mathbb{R}$ can fail to be compact: either it's unbounded, or it's not closed. Construct an example for each.

thanks

I know for the case K is unbouned
but for K is not closed, then there exists a sequence $ x_{n}$  in K which has no limit point.
I believe f must be $ f : K\minus{}\minus{}\minus{}>\mathbb{R}$ and $ f(x_{n})\equal{}n$  how can I prove that $ f$ is continuous in case  K may be very bad

Don't look at it from the perspective of a sequence. Look at that limit point $ L$ which isn't in $ K$. Can you construct a function that blows up there, so that any sequence $ x_n$ converging to $ L$ makes $ f(x_n)$ diverge?

Okie

$ f : K\minus{}\minus{}\minus{}\minus{}>\mathbb{R}$  and $ f(x)\equal{}\frac{1}{x\minus{}L}$ works, doesn't it ?


I will speak persain:

Masalehaye Marhaleye sevome olampiyade IRAN sale 1384  ro mitonid inja peyda konid:(Albate ye serishon kheli ajib ghariban , :D  :D ,mesike dar tarhe masale,tarahan kheli abtekar be kharj dadan)just kidding :P

http://www.mathlinks.ro/Forum/viewtopic.php?t=49899

Kheili jalebe!  Man say mikonam haleshoon konam. :)

Masoud Zargar

salam bar hameye bache khafanaye irooni.daste arman joon dard nakone ke sabab kheir shod va ye forum ham be ma dadan.
 :D

من سعی میکنم که صورت فارسی سوالات را بفرستم

سلام بر همه ی ایرانیان عزیز :D  با با دست مریزاد ...آفرین...ای ولا ... و سلام ویژه خدمت امید خان  ;)

khoda vakili soalahaye marboot be ham resish kheily ghashange. ;)

Tabestoon ke vaghtam azad shod in kar ra mikonam.


Does anyone know how I can contact Anya Katsevich? She was 6th in National written round.

Also see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276439

Why do you want to contact her?

BTW, does anyone know if Gene Katsevich, who will be going to blue mop this year, is Anya's brother?

Same city, only two USAMO qualifiers ever with that last name... probably.

As for the original question: No stalking.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $f$ be a polynomial over a field $K$, and let $L: K$ be a Galois extension of $K$ which is also a splitting field extension for $f$. Then we have:

$|G(L: K)|=dim[L: K]$

where $G(L: K)$ is the group of automorphisms from $L$ into $L$ which left $K$ fixed.

In other words, if $f$ is a separable polynomial of $F[X]$ and $E$ is a splitting field for $f$, then $|Aut(E/F)|=[E: F]$.

hm..
say $\displaystyle f(x) = \prod_{k=1}^{\partial f}(x-\alpha_k)$ in some algebraic closure $\bar{F }$ of $F$.
since $E = F(\alpha_1, \dots, \alpha_{\partial f})$, there exist $\beta \in E$ such that $E = F(\beta)$ (by primitive element theorem).
if $\mu \in F[x]$ is the minimal polynomial of $\beta$ over $F$, then we have both $\left|\text{Gal}(E/F)\right| = \partial \mu = [E: F]$.


Points $ M$ and $ N$ are the midpoints of sides $ PA$ and $ PB$ of $ \triangle PAB$. As $ P$ moves along a line that is parallel to side $ AB$, how many of the four quantities listed below change?

(a) the length of the segment $ MN$
(b) the perimeter of $ \triangle PAB$
(c) the area of $ \triangle PAB$
(d) the area of trapezoid $ ABNM$

[asy]
draw((2,0)--(8,0)--(6,4)--cycle);
draw((4,2)--(7,2));
draw((1,4)--(9,4),Arrows);
label("$A$",(2,0),SW);
label("$B$",(8,0),SE);
label("$M$",(4,2),W);
label("$N$",(7,2),E);
label("$P$",(6,4),N);[/asy]

$ \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

[hide="solution"] Only the perimeter changes. I thought this was kind of obvious so its $\boxed{B}$.[/hide]

[hide]From the given information, we get AB=2MN.

a) doesn't change because AB is constant.

b) changes (it's the smallest when its an isosceles, and increases as it stretches).

c) doesn't change since the base (AB) and the distance from AB to the line with point P (altitude) are equal

d) remains equal the distance from MN to AB remains the same, as well as MN and AB.

Thus, only one quantity changes. [/hide]


ratio
geometric series
number theory
relatively prime
algebra
system of equations

An infinite geometric series has sum $2005$. A new series, obtained by squaring each term of the original series, has $10$ times the sum of the original series. The common ratio of the original series is $\frac{m}{n}$ where $m$ and $n$ are relatively prime integers. Find $m+n$.

[hide="solution"]well using the infinite geometric series formula where $a$ is the first term we have

$\frac{a}{1-r}=2005$
and
$\frac{a^2}{1-r^2}=20050$

from here we see that

$a=(2005)(1-r)$
and that
$a^2=(2005)(10)(1-r)(1+r)$

we then see that $a^2=(a)(10)(1+r)$
so we now have

$a=(1-r)(2005)$
and
$a=10+10r$

setting these equal we find that $r=399/403$

$399+403=802$[/hide]

[hide="Solution"]First, we must translate this into the language of mathematics. If $a$ is the first term and $r$ is the ratio, we have $a+ar+ar^2+\cdots=2005$ and $a^2+a^2r^2+a^2r^4+\cdots=10*2005$.

Applying the formula for an infinate geometric series, $\frac{a}{1-r}=2005$ and $\frac{a^2}{1-r^2}=10*2005$.

Notice that if we factor the second equation, we obtain $\left( \frac{a}{1-r}\right) \left( \frac{a}{1+r} \right) = 10*2005$ and $\frac{a}{1+r}=10$. This then gives us $a=10(r+1)=10r+10$.

But from our original geometric series, we get $a=2005(1-r)=2005-2005r$

Solving this system of equations, we find that $r=\frac{399}{403}$, and our answer is $802$.[/hide]

Let this sequence be:
$$\implies a + ar^2 + a^3 + \ldots$$

So the sum of the series is, by infinite geometric series sum formula,
$$\implies \frac{a}{1 - r}$$

Let's look at the second series.

The second series basically is every term squared of the first series. Hence our second series is:
$$\implies a^2 + a^2r^2 + a^2r^4 + \ldots$$

So the sum of the series is, by infinite geometric series sum formula,
$$\implies \frac{a^2}{1 - r^2}$$

We know that the second series has $10$ times the sum of the original series. Hence:
$$\frac{\frac{a^2}{1 - r^2}}{\frac{a}{1 - r}} = \frac{a}{1 + r} = 10$$

Hence $a = 10(1 + r)$.

We know that the sum of the original series is $2005$.

Hence, we have:
$$\frac{10(1 + r)}{1 - r} = 2005$$
$$10(1 + r) = 2005 - 2005r$$

Rearranging gives that:
$$r = \frac{399}{403}$$

Hence $m + n = 399 + 403 = \boxed{802}$!

edit: lol, continuously switching tabs, every tab open with a problem :)

You can just make a system of equations, and solve with either substitution or elimination.

p1: Let the common ratio of the original sequence be $r$ and let the first term be $a$. Then $\frac{a}{1-r} = 2005, \frac{a^2}{1-r^2} = 2005 \cdot 10$. The first equation gives $\frac{a^2}{(1-r)^2} = 2005^2$, so $\frac{(1-r)^2}{1-r^2} = \frac{10}{2005} \implies r = \frac{399}{403}$. The answer is $399+403 = \boxed{802}$.


calculus
integration
real analysis
real analysis unsolved

Define $ f(x) \equal{} \frac {x^x}{\Gamma(x)\, e^x}$ and $ a_n \equal{} \int_0^1 x^n\, f(x)\, f(1 \minus{} x)\, dx$ .

Can you show that $ (a_n)_{n\ge 0} \equal{} \left(\frac {1}{2^3\cdot 3},\frac {1}{2^4\cdot 3},\frac {73}{2^7\cdot 3^2\cdot 5},\frac {11}{2^8\cdot 5},\frac {5^3\cdot 29}{2^{10}\cdot 3^4\cdot 7},\frac {5^2\cdot 13\cdot 17}{2^{11}\cdot 3^4\cdot 7},\frac {1571\cdot 3331}{2^{15}\cdot 3^5\cdot 5^2\cdot 7},...\right)$

is a sequence of rational numbers?

Hi :) !!

than i think that your question $ \Large \sim$ a this question:

$ a_n \equal{} \frac{1}{e} \int_0^1 \frac{x^{x\plus{}n\plus{}1}(1\minus{}x)^{1\minus{}x}}{B(x;1\minus{}x)}dx$

prove that: $ \forall n \in \mathbb{N} : a_n \in \mathbb{Q}$.

that's OK?????

and thanks

Yes, my question is equivalent with:

Is $ a_n\equal{}\frac{1}{e}\int_{0}^{1}\frac{x^{x\plus{}n\plus{}1}(1\minus{}x)^{1\minus{}x}}{B(x,1\minus{}x)}\, dx$ rational for all $ n\in\Bbb{N}$ ?


probability
articles
analytic geometry
inequalities
geometry
function

A point (x, y) in the coordinate plane satisfies the inequalities
−6 < x < 6 and −6 < y < 6, where x and y are real numbers. Find the probability that the point
(x, y) is less than 4 units from the origin.

[hide]Well if you draw two concentric circles, one of which has a radius of 4 and the other with a radius of 6, you could use the areas, but the circle with a radius of 6 doesn't include the curcumference area, aka the line drawn around a circle. So if you want area less than r. you just take pi R^2= 36pi and then subtract the string manifested as the line around the empty space inside the circle. It can be a rectangle(since 2d space) with length of 2pi r and width of w. This denotes all the possible sums of x and y. Now for the smaller circle, do the same thing and obtain 16pi -8pi(m)   so this yields [16pi-8pi(m)]/36pi-16pi(m) and then factor out stuff, leaving (4-2m)/(9-4m) but if one takes the limit as m approaches zero, than the result is just 4/9[/hide]

P.s. I'm sorry, should this be in basics?

[hide="Hint"]You would have a square with vertices on (6,6), (-6,6), (6,-6), and (-6,-6). Then, you would have a circle with radius 4. The probability is the area of the circle/area of the square.[/hide]

I'm sorry, but i don't understand why it would be a square. Given that x and y are independent, you could have a point (5,5) for example(which also means that the square doesn't represent all possible coordinates). Even though it doesn't count for a sucessful outcome, it's still a possible one.

Yes. You have a square with vertices on (6,6), (-6,6), (6,-6), (-6,-6), since you could have a point within that and satisfy the equation

[b]adrenalinefix36[/b] — all the points inside the square, but not on its border, satisfy the inequalities $ \minus{} 6 < x < 6$; $ \minus{} 6 < y < 6$.

[hide="Solution"]The area of the square would be $ [6 \minus{} ( \minus{} 6)]^{2} \equal{} 12^{2} \equal{} 144$, and the area of the circle would be $ 4^{2}\pi \equal{} 16\pi$. The probability, as [b]#H34N1[/b] suggested, would then be $ P(x) \equal{} \frac {16\pi}{144} \equal{} \frac {\pi}{9}$. It is important to note that the point cannot be on the circumference of the circle, or on the sides of the square, so there is an infinitesimal difference between the calculated probability and the actual probability — but, since the difference is so small, we can dismiss it.[/hide]

"infinitesimal difference," metafor?  The problem is considering 2d spaces, and, since lines and other 1 dimensional paths have no area, they do not affect the probability.  Perhaps this is what you meant, but the calculated probability is correct either way.

@adrenalinefix36: the "string" that you want to wrap around the circle only has length, not area, and thus does not affect the answer.

To illustrate the reason for this, consider the following question:  If Jack chooses a real number between 0 and 1 (inclusive or not, makes no difference) and each number has an equal chance of being chosen, what is the probability that Jack chooses .5?

There is 1 way to have a success and an infinite number of possibilities, so the answer is $ 1/\infty \equal{} 0$.

If we instead try to do this by using the probability distribution function, here $ f(x) \equal{} 1 \forall (x | 0 < x < 1)$, we are left trying to find the area under the curve at $ x \equal{} .5$.  But there is no area under a point, so our answer is again $ 0$.

The issue that you guys are worrying about is the same idea, just in a different form.

Thanks for much for the help. Honestly I was really tired last night, and didn't understand lol. O well, good days and bad days(or in my case nights); thanks though! At facis, I was fallaciously adding dimension to a line giving it area in plans to later define it as irrelevant. It seemed correct at a point in time when i was very tired but not now.

Yes, that's what I meant, [b]facis[/b]; sorry if I confused anyone. :P


I was using Koma's script's scrbook as my document class, and after I included the asymptote package the following warning occured:

Package scrkbase Warning: \ifundefined was redefined
(scrkbase)                at the document preamble.
(scrkbase)                If you did not redefine it by yourself, have a
(scrkbase)                a look at the description of this command at
(scrkbase)                the manual of every package you are using.

And I didn't see my picture in the final product.  I'm guessing that maybe they both uses the same variable ifunderfined, but have no idea how to even fix it.

Any help is appreciated.  Man, I know how to code, but I don't know anything about the system...

[b]Either[/b] (from comp.text.tex newsgroup)
[i]supp-pdf.tex loads supp-mis.tex that also defines \ifundefined.
Update supp-pdf.tex. The new one (2006-09-02) doesn't load
supp-mis.tex anymore. [/i]

[b]Or[/b]
Use Asymptote to make the picture then include the graphic in your tex file.

OK I fixed it by getting the new versions of the files listed above.

You can get it here:
[url]http://osdir.com/ml/tex.metapost/2006-10/msg00011.html[/url]

And for future reference, try to name your latex file in one word and not ending in a number, because that causes the Asymptote to not able to save a pdf containing the picture, and thus the failure...(because it tries to use a filename similar to your tex file name)


geometry
circumcircle
trigonometry
geometry solved

Let $P$ be a point on the circumcircle of $\triangle ABC$. Let $L$, $M$ and $N$ the ortogonal projections of $P$ on sides $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$. Let $\overline{PL}=l$, $\overline{PM}=m$, $\overline{PN}=n$ and $a,\ b,\ c$ be the sides of $\triangle ABC$, then prove that $a\cdot m\cdot n=b\cdot l\cdot n+c\cdot l \cdot m$

How it can be proved without trig?

Thanks.

i guess you could check this out
http://www.math.uci.edu/~mathcirc/math194/lectures/inscribed/node2.html

Yeah thanks.., but in the proof was used trig..., that's what I'm not lookin' for...

Does anybody have a proof without trig.?

[quote="MathTeX"]prove that $a\cdot m\cdot n=b\cdot l\cdot n+c\cdot l \cdot m$[/quote]

This is equivalent to $\frac{a}{l}=\frac{b}{m}+\frac{c}{n}$, or, with a proper convention for [b]signed[/b] distances, to $\frac{a}{l}+\frac{b}{m}+\frac{c}{n}=0$. This is Theorem 2 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=157240#157240]http://www.mathlinks.ro/Forum/viewtopic.php?t=24945 post #3[/url]. There I give a proof without trigonometry.

  Darij


geometry
3D geometry
pyramid
ARML
rectangle
inequalities
triangle inequality

Source: ARML 2004 Team
10. Pyramid $EARLY$ has rectangular base $EARL$ and apex $Y$, $ER = \sqrt{1105}$, and $\overline{EY}\perp\overline{RY}$. If $EA$, $AR$, $EY$, $AY$, $RY$, and $LY$ are distinct integers, compute the area of the smallest face of the pyramid.

[hide]1105=$4^{2}$+$33^{2}$=$9^{2}$+$32^{2}$=$12^{2}$+$31^{2}$=$23^{2}$+$24^{2}$.
Therefore, the smallest face would be a 4-9-12 triangle.  By Heron's theorem, its area is $\sqrt{(25/2)(17/2)(7/2)(1/2)}=5/4\sqrt{119}$.
This however, skips over the series of reasons on why this works.
Draw two semicircles so that the diagonals are diameters.  This means the center of the rectangle is the center of both of these semicircles.  Any triangle that has two points on the rectangle's opposite sides and the last point on the corresponding semicircle is right.  These two semicircles intersect since they share the same center and are just halves of great circles.  This forces EYR and AYL to be right angles.  Once that is reached, one just needs to hunt for the smallest possible triple and confirm that it satisfies the triangle inequality, which it does.
4-9-12
33-9-31
33-12-32
4-32-31[/hide]


In triangle ABC prove

abc +(a-b)(b-c)(c-a) = 4Rr(aCosC +bCosA + cCosB)

R is the circumradius,r the inradius.

Since the area of triangle ABC can be expressed both as $\frac{abc}{4R}$ and as sr, where $s=\frac{a+b+c}{2}$, we get $\frac{abc}{4R} = sr$, and thus abc = 4Rsr. Hence, $4Rr=\frac{abc}{s}$.

Also, by the Cosine Law, $\cos C = \frac{a^2+b^2-c^2}{2ab}$ and consequently, $a\cos C = \frac{a^2+b^2-c^2}{2b}$. Similarly, we can find b cos A and c cos B. This all simplifies your trigonometric identity into

$abc+\left(a-b\right)\left(b-c\right)\left(c-a\right)=\frac{abc}{s}\left( \frac{a^2+b^2-c^2}{2b}+\frac{b^2+c^2-a^2}{2c}+\frac{c^2+a^2-b^2}{2a} \right)$.

Now, this equation can be easily proven by direct calculation (it's easier than you would expect). I wouldn't expect a geometric proof since I don't see a geometric significance of the terms involved...

  Darij

Indeed, recalling a couple of formulas and juggling around the algebraic expression will do it. :)  :D  :)


I was thinking of starting an online ARML meeting session during the course of the year. If you are interested in joining, please post here!

if we can get enough people, i'll be glad to do it

That'd actually be great if you could join, Howard. We already have approximately 12 or so people willing to attend the meetings. Please email me at [url]karthik.narayan.0@gmail.com[/url] or please post here the following information:

Name
AIM Screen Name
Preferred time of choice

We will be conducting the meetings through AIM, so as far as TeX is concerned, we might just have everyone open up AoPS's TeXer website or something.  :wink:

If anybody else is interested in joining, please follow the format above. The current "most preferred time" is Sunday at 1:00 PM (Eastern Time).

Well my name and AIM screenname are both available in my profile, BUT

Name: Howard
AIM screenname: captcha008

As far as time is concerned, I won't be able to do 1 pm on Sundays for a few weeks, at least until my SAT class is finished (around the October SAT). I would be willing to do it anytime Saturday night, my Sundays are kind of full and too much homework on weeknights.

Alright, that's cool. I'm pretty sure that we can accomodate for that  :) I'll add you to the list.

I am so thoroughly in.

Barry
aviairboy
Sunday or Wednesday, whenever. Saturdays except for tournaments and October (marching band), but I can deal.


geometry
inequalities
trigonometry
inequalities unsolved

Prove that in every triangle ABC with sides a,b,c and area F, the following inequalities hold
a)$(ab+bc+ca)\sqrt{\frac{abc}{a^{3}+b^{3}+c^{3}}}\ge 4F$,
b)$8R(R-2r)\ge (a-b)^{2}+(b-c)^{2}+(c-a)^{2}$

Ok I will help you but it's pity that nobody want to help me [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=126559]here[/url] !  :(  :(
ok now we well know
$yz\sin^{2}A+zx\sin^{2}B+xy\sin^{2}C\le \frac{(x+y+z)^{2}}{4}(*)$ for all $x,y,z$
now let $x=bc,y=ca,z=ab$ we will get
$(ab+bc+ca)^{2}\ge 4(\sum a^{2}bc\frac{a^{2}}{4R^{2}})\Leftrightarrow (ab+bc+ca)\sqrt{\frac{abc}{a^{3}+b^{3}+c^{3}}}\ge 4F$
$8R(R-2r)=8R^{2}-16Rr=8R^{2}-\frac{8abc}{a+b+c}\ge \sum (b-c)^{2}\Leftrightarrow 8R^{2}\ge \sum (b-c)^{2}+\frac{8abc}{a+b+c}= \frac{(\sum (b-c)^{2})(a+b+c)^{2}+8abc(a+b+c)}{(a+b+c)^{2}}= \frac{2\sum (a+b)(a+c)a^{2}}{(a+b+c)^{2}}\Leftrightarrow 4R^{2}(a+b+c)^{2}\ge \sum (a+b)(a+c)a^{2}$
which is true by from $(*)$ we let $x=b+c,y=c+a,z=a+b$
 :)


[all the initial square roots are always positive, so no domain restrictions off the bat]

$2x^{2}+x+5=x^{2}-3x+13+x^{2}+x+1-2\sqrt{(x^{2}-3x+13)(x^{2}+x+1)}$

$3x-9=-2\sqrt{(x^{2}-3x+13)(x^{2}+x+1)}$ [x<3]

$4x^{4}-8x^{3}+35^{3}+94x-29=0$
$(2x^{2}-7x+29)(2x^{2}+3x-1)=0$

the first has no real roots, the second has two real roots,
$x=\frac{-3\pm\sqrt{17}}{4}$

both of which are solutions

[quote="mathgeniuse^ln(x)"]How did you factor that Altheman?[/quote]

if i were to do it by hand, i would take (2x^2+ax+b)(2x^2+cx+d) and equate coefficients, and hope for a nice solution, (or (4x^2+..)(x^2+...))

but i just plugged into a calculator

Thanks Altheman.

I get it now.  If it weren't to have worked, would you have done 4x^2 and x^2

Wow, that is smart.

How did you figure out to do those two polynomials for the equation?

You want to find some constants $(m,n)$ such that $m(2x^{2}+x+5)+n(x^{2}+x+1) = x^{2}-3x+13$. (Because then you would just have to substitute $2x^{2}+x+5$ and $x^{2}+x+1$ to get a nice equation.) For the $x^{2}$-term you get $2a+b = 1$, for the $x$ you get $a+b =-3$, so $1 = 2a+b = a-3$, so $a=4$ and $b =-7$. Check it for the constant term. We're lucky!  :D So you get the nice identity \[4(2x^{2}+x+5)-7(x^{2}+x+1) = x^{2}-3x+13.\]


This was a problem i was doing while preparing for the putnam. but im sure its not above the Pre-Olympiad level.
Let $a,b,c$ be distinct integers. Does there exist a polynomial with integer coefficients that cyclically permutes these values in the sense that $P(a)=b,$ $P(b)=c,$ and $P(c)=a$? Explain.
However the site did not have a solution for this so i am wondering if anyone could provide me with one. thanks.

polynomial with integer coeff. means you can say $x-y|P(x)-P(y)$. That should be enough to complete the problem.


There are $ 90$ contestants in a mathematics competition. Each contestant gets acquainted with at least $ 60$ other contestants. One of the contestants, Amin, state that at least four contestants have the same number of new friends. Prove or disprove his statement.

The statement is true.

Since everyone knows between 60 and 89 of the other people (30 different possibilities), if there are not four people with the same number of acquaintances then there must be three people who know $ k$ other people for each integer $ k\in[60,89]$. But now, sum over all 90 people the number of other people they know, and we get $ 3\sum_{i\equal{}60}^{89}i\equal{}6705$. But since this counts each acquaintance between two people exactly twice, it must be an even number - a contradiction. To avoid contradiction, some four people must have the same number of acquaintances.


calculus derivative function inequalities trigonometry algebra polynomial

Let $f:R->R$ a $C^{\infty}$ class function whose first derivative is 1 in 0 and $|f^{(p)}(x)|\leq 1$ for all $x$ and all $p$. Then prove $ f(x)=sinx$ for all $x$. I tried everything and no proof.  :(

Did you try to express $f$ as a Fourier series?  Because the derivatives are bounded by a uniform constant, and using Bessel's inequality, we might be able to show that all the Fourier coefficients vanish except for $\sin x$, $\cos x$, and the constant.

Anyway, it's just an idea.  I don't claim to have a proof.

I tried, but I failed.  I imagine that after all one needs Fourier series, but I don't think that it's some direct consequence of some classic estimations.

<ok, wrong>

[hide="old post"]Hmm... if even harazi can't do this, then I am wrong. But where? :)

Let's first assume $f$ can be represented in a finite number of sines and cosines, we then have: $f(x) = \sum_{n=0}^k a_n \sin(nx) + b_n \cos(nx)$. Since for any $n>1$ we have the $p$th derivative has sine/cosine coefficients $n^p\cdot a_n$ and $n^p\cdot b_n$, the limitedness of each derivative for all $x$ gives us that $a_n=0, b_n=0, \forall n>1$, so $f=a\sin(x)+b\cos(x)+c$. Due to the conditions we have $f=sin(x)$.

Now we have proven it for any finite series of $f$. We know (*I think we know right?*) that you can arbitrarily approximate $f$ with a finite fourier series $f_n$. So since the function should be $\sin(x)$, for any finite number of terms, then $f(x)=sin(x)$. To see this, assume $g(x)=f(x)-\sin(x)$, then we can for every point $x$ and positive real $\epsilon$ we can always find $n$ so that $g_n(x)<\epsilon$[/hide]

I think you are not right with the approximation argument. Indeed, let $\epsilon>0$ and $ g_n$ a trigonimetric polynomial which approximates $f$ with error at most $\epsilon$ on $[0,1]$. What guarantees you that $g_n$ verifies the same conditions as $f$ refering to its derivatives so that to deduce that $ g_n(x)=sinx$?

Man, I really don't undersatnd your claim. How can you prove that the function that you say it approximates very well our initial function $f$ must be $sinx$ or must tend to $sinx$. This is what I don't understand. I mean, here is what I understood: you proved that any trigonometric polynomial which is a solution of our problem must be $sinx$. Next, you use Weierstrass theorem to approximate $f$ very well with trigonometric polynomials, but then you say all these polynomials must be $sinx$. Why? How do you deduce that?

Peter, I don't think this approach works.  For example, let $f(x) = \cos(x/2)$.  This function obeys the bounded derivatives condition.  But we can check that none of its Fourier cosine coefficients vanish.

Here by Fourier series, I assume we are restricting f to between $-\pi$ and $\pi$ (and then periodically extending beyond that).  I think part of the problem is that this periodic extension is not guaranteed to be differentiable or even continuous.

I think I now got what you meant, harazi... when I differentiate p times, the upper limit is dependent on p, so it's not bounded in fact, so it's not the same case for $f_n$ as for an f with finite series. :blush:

Indeed ravi, as it's not bounded, the coefficients don't vanish. So it's only valid for the case where we have a finite series. (which is pretty trivial)

The problem has already been posted (see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24773[/url]) though no solution was given in that thread. Here is a small hint. Suppose that $f$ is bounded on the real line and the Fourier transform of $f$ is supported by $[-1,1]$. Show that $f'(0)\leqslant \sup_{\mathbb R}|f|$ and the equality is attained only for $f(x)=\sin x$.

I apologize for the absolutely stupid question, but what does the fact that the Fourier transform of f is supported by $[-1,1]$ mean?

I think I can answer your question, harazi, but then I will add my own dumb question.  The support of a function is the set of points where the function is nonzero.  So fedja's condition means that the support of the Fourier transform of f is a subset of [-1, 1].  In other words, the Fourier transform vanishes outside of [-1, 1].  That makes some sense, because if f had a nonzero term corresponding to say $\cos(1.5x)$, then by taking derivatives, we would get an unbounded sequence.

My dumb question: I thought to take the Fourier transform (integral) of a function $f$ you had to know something like the integral of $|f|$ is finite.  We don't know something like that, do we?  Do we first "truncate" the function?  Does the function $\sin x$ have a Fourier transform?
</end of dumb questions>

[quote="Ravi B"]
My dumb question: I thought to take the Fourier transform (integral) of a function $f$ you had to know something like the integral of $|f|$ is finite.  We don't know something like that, do we?  Do we first "truncate" the function?  Does the function $\sin x$ have a Fourier transform?
[/quote]
You can take the Fourier transform of any tempered distribution, in particular, of any function that grows not faster than polynomially. Of course, you'll get a tempered distribution rather than an ordinary function but you can always do some approximations. In our particular case, we know that $f$ is entire and grows not faster than $e^{|z|}$, which, if we use its boundedness on the real line, can be immediately improved to $e^{|\text{Im\,}z|}$ (this estimate can also be easily obtained directly using the Taylor series at the point $\text{Re\,}z$). Of course, the Fourier transform of $f$ cannot be taken in the classical sense, but we can approximate $f$ by $f_\delta(z)=f(z)\left(\frac{\sin\delta z}{\delta z}\right)^2$, which is integrable on the real line and grows not faster than $e^{(1+2\delta)|\text{Im\,}z|}$. Then the Fourier transform of $f_\delta$ is supported by $[-1-2\delta, 1+2\delta]$ (If I remember it right, the corresponding statement is called the Paley-Wiener theorem) and we can play with $f_\delta$ instead of $f$ and then pass to the limit as $\delta\to 0$.

The function $\sin x$ has a Fourier transform in this generalized sense, which is a linear combination of 2 Dirac $\delta$-measures at the points $1$ and $-1$.

Sorry for posting in such an old topic but I have a question for fedja ? how does one know that $f$ is the restriction of an analytic function ? (to apply Paley theorem etc ..)

Just use the Taylor estimates to show that the power series of $f$ converges to $f$.