trigonometry
geometry
parallelogram
circumcircle
geometry solved

Let $ABC$ be a triangle and $AB=BC=CA$, and let $P$ be a point in the $int(ABC)$. If $\alpha= \angle BPC$,$\beta =\angle CPA$ and $\gamma = \angle APB$ prove that:

          $\frac{PA}{sin(\alpha-\frac{\pi}{3})}= \frac{PB}{sin(\beta-\frac{\pi}{3})}= \frac{PC}{sin(\gamma-\frac{\pi}{3})}.$

A direct consequence of a known "theorem", which states that, given a point $P$ in an arbitrary triangle, the expression $\frac{AB\cdot PC}{\sin(\angle APB-\angle ACB)}$ is a constant.

Then can someone post the proof of this theorem? Thanks.

Let $\triangle ABC$ be arbitrary. On the segments PA, PB, construct triangles $\triangle APX,\ \triangle PBY$ directly similar to $\triangle ABC,$ the vertices X, Y on the same side of the broken line APB as C. The triangles $\triangle APB \sim \triangle AXC \sim \triangle CYB$ are directly similar by SAS. Considering $\triangle ABC$ as a degenerate quadrilateral ABCC, according to the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109435]"prove that it is parallelogram"[/url], PXCY is a parallelogram. Its sides are

$YC = PX = PA \cdot \frac{BC}{AB},\ \ XC = PY = PB \cdot \frac{CA}{AB}$

and they form angles $\angle YPC = \alpha-\angle A,\ \angle XPC = \beta-\angle B$ with the diagonal PC. By the sine theorem

$\frac{\sin(\alpha-A)}{\sin(\beta-B)}= \frac{PX}{PY}= \frac{PA}{PB}\cdot \frac{BC}{CA},\ \ \ \frac{a \cdot PA}{\sin(\alpha-A)}= \frac{b \cdot PB}{\sin(\beta-B)}$

where a = BC, b = CA. If the triangle $\triangle ABC$ is equilateral, the triangles $\triangle APB \cong \triangle AXC \cong \triangle CYB$ are directly congruent by SAS and the parallelogram PXCY with the sides YC = PX = PA, XC = PY = PB is obvious without a reference to another problem.

It's trivial using some basic properties of pedal triangles:

[color=blue][b]Theorem 1 (pedal triangle formulas).[/b] If P is a point in the plane of a triangle ABC, and X, Y, Z are the orthogonal projections of this point P on the lines BC, CA, AB, then $YZ=AP\cdot\sin A$, $ZX=BP\cdot\sin B$, $XY=CP\cdot\sin C$. Also, if the point P lies inside the triangle ABC, we have < ZXY = < BPC - A, < XYZ = < CPA - B, < YZX = < APB - C.[/color]

[i]Proof.[/i] We will only regard the case when the point P lies inside the triangle ABC and the points X, Y, Z lie inside the segments BC, CA, AB. All other cases can be considered in analogous manner.

Since < AYP = 90° and < AZP = 90°, the points Y and Z lie on the circle with diameter AP. Thus, < AZY = < APY and < PYZ = < PAZ. By the sine law in triangle AYZ, we have $\frac{YZ}{\sin\measuredangle YAZ}=\frac{AY}{\sin\measuredangle AZY}$. Since < AZY = < APY, we have $\frac{AY}{\sin\measuredangle AZY}=\frac{AY}{\sin\measuredangle APY}$, and in the right-angled triangle APY, we have $\frac{AY}{\sin\measuredangle APY}=AP$. Thus, $\frac{YZ}{\sin\measuredangle YAZ}=AP$, so that $YZ=AP\cdot\sin\measuredangle YAZ=AP\cdot\sin A$. Similarly, $ZX=BP\cdot\sin B$ and $XY=CP\cdot\sin C$.

Since < PYZ = < PAZ and similarly < PYX = < PCX, we have

  < XYZ = < PYZ + < PYX = < PAZ + < PCX = < PAB + < PCB
            = (180 - < APB - < PBA) + (180° - < CPB - < PBC)          (by the sum of angles in the triangles APB and CPB)
            = (360° - < APB - < CPB) - (< PBA + < PBC)
            = < CPA - < CBA = < CPA - B,

and similarly < YZX = < APB - C and < ZXY = < BPC - A. Theorem 1 is proven.

By the sine law in triangle XYZ, we have $\frac{YZ}{\sin\measuredangle ZXY}=\frac{ZX}{\sin\measuredangle XYZ}=\frac{XY}{\sin\measuredangle YZX}$. Using Theorem 1, this becomes $\frac{AP\cdot\sin A}{\sin\left(\measuredangle BPC-A\right)}=\frac{BP\cdot\sin B}{\sin\left(\measuredangle CPA-B\right)}=\frac{CP\cdot\sin C}{\sin\left(\measuredangle APB-C\right)}$. Now, by the sine law in triangle ABC, we have $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Multiplying these two equations, we get

$\frac{AP\cdot\sin A}{\sin\left(\measuredangle BPC-A\right)}\cdot\frac{a}{\sin A}=\frac{BP\cdot\sin B}{\sin\left(\measuredangle CPA-B\right)}\cdot\frac{b}{\sin B}$
$=\frac{CP\cdot\sin C}{\sin\left(\measuredangle APB-C\right)}\cdot\frac{c}{\sin C}$,

what simplifies to $\frac{AP\cdot a}{\sin\left(\measuredangle BPC-A\right)}=\frac{BP\cdot b}{\sin\left(\measuredangle CPA-B\right)}=\frac{CP\cdot c}{\sin\left(\measuredangle APB-C\right)}$. This is exactly the assertion of Erik. in his post #2.

  darij

A very nice and also helpful solution. :wink:  Thank you, Darij.

[color=darkblue]I"ll construct ([b]similarly to Yetti[/b]) a triangle which is similarly to the [b]generalized Pompeiu's triangle for the point[/b] $P$[b] w.r.t. the triangle[/b] $ABC$,
i.e. the triangle $XYZ$ for which $YZ=a\cdot PA$, $ZX=b\cdot PB$, $XY=c\cdot PC$, where $BC=a$, $CA=b$, $AB=c$.

[b]Proof.[/b] Let $R$ be the point for which the line $PB$ separates the points $A$, $R$ and $PBR\sim ABC$.

Therefore, $\boxed{\ m(\widehat{BPR})=A\ }$, $\boxed{\ m(\widehat{BRP})=C\ }$, $m(\widehat{PBR})=B$ and $\frac{PB}{AB}=\frac{BR}{BC}=\frac{RP}{CA}$ $\Longrightarrow$ $\widehat{ABP}\equiv\widehat{CBR}$ and $\boxed{\ b\cdot PB=c\cdot PR\ }\ .$

Thus, $ABP\sim CBR$, i.e. $\frac{AB}{CB}=\frac{BP}{BR}=\frac{PA}{RC}$ and $m(\widehat{CRB})=\gamma$, $\boxed{\ a\cdot PA=c\cdot RC\ }\ .$ Look at the triangle $CPR$ :

$\blacktriangleright$ $m(\widehat{CPR})=\alpha-A\ ,\ m(\widehat{RCP})=\beta-B\ ,\ m(\widehat{PRC})=$ $\gamma-C\ \ ;\ \ \frac{CR}{a\cdot PA}=\frac{RP}{b\cdot PB}=\frac{PC}{c\cdot PC}\ .$

Apply the Sinus' theorem in the triangle $CPR\ : \ \boxed{\ \frac{a\cdot PA}{\sin (\alpha-A)}=\frac{b\cdot PB}{\sin (\beta-B)}=\frac{c\cdot PC}{\sin (\gamma-C})\ }\ .$

[b]Remark.[/b] The area of the [b]generalized Pompeiu's triangle[/b] is equally to $[ABC]\cdot p_{w}(P)$,
where $p_{w}(P)$ is the power of the point $P$ w.r.t. the circumcircle $w$ of the $\triangle ABC\ .$[/color]


geometry
trigonometry
AMC 10
area of a triangle
AMC

Solutions to 2005 AMC 10 Contest B ??

I can't figure out how to solve
questions 9, 12,13,14,15,16 and 20.

Could someone please help me ??

They only posted the correct answer on the
 AMC website.

9) Total of 6*6, or 36 possibilities cuz both die have 6 faces. Only way sum will be odd is if one of the numbers is odd and the other is even. It can be 2 and 5. There are 2 choices for each for 2*2 = 4. Also 1/3 and 4/6 works, and there are 4 possibilities on each die, so 4*4 = 16.

(4+16)/36 = 20/36 = 5/9

12) Only way for the product to be prime is if one of the numbers is prime and the rest are 11's. There are only 3 primes under 6. And they can be on any of the 12 die and the rest are 1's, so there are 3*12 = 36 ways. The total number of possibilities is obviously 6^12
and (6^2)/(6^12) = 1/(6^10)

13) First find out how many multiples of 3 are under 2005, and that is 2005/3 and only take the integer part of it, so we get 688. Do the same with 4 to get 501. Now we want to subtract the multiples of 12 that are in common to both and there are 167 multiples of 12. However we want none of the numbers to be a multiple of 12, so subtract 167 again.
668+501 - 167 - 167 = 835

14) Since M is the midpoint, MC is 1. Since C is the midpoint, CD is 2. And we are given that ABC is equilateral, so angle ACB is 60 degrees and MCB is supplementary, so it's 120 degrees. A formula to determine the area of a triangle is (a*b*sinC)/2, where a and b are lengths of any 2 sides and C is the angle between them. So we get (2*1 * sin120)/2 = sin120
If you've taken trig, you'll know that that is rt3/2, or if you don't evaluate it on your calculator and firgure out which answer it is from there.

15) Easier to write out all possibilities if you don't get it. The number in the parantheses is the number of these combinations there are:
10 + 10 = 20 (1)
20 + 1 = 21 (4)
20 + 5 = 25 (4)
20 + 10 = 30 (4)
20 + 20 = 40 (1)
add them up to get 14
since we're considering 10 and 20 the same as 20 and 10, there are 8C2 combinations or 28 of them. 14/28 = 1/2

20) There basically are 5! different numbers possible. Since there will be the same number of each number of the digits in each place (like tens, or hundreds) and there are 5 digits so, there will be 24 of each digit being summed in each place. so you can add up all the digits and multiply it by 24 to get the value that'll be summed in each place. (1+3+5+7+8)24 = 24*24 = 576. Since it'll be summed in 5 places, you can multiply it by 11111, which is 6399936. Now you can divide by the total number of numbers, or 5! to get 53332.8.

[quote="djshowdown2"]9
[hide]You can find the median by just imagining there are 100 students, then finding the 50th student.  10 got 70, 25 got 80 so thats 35, 20 got 85, so the middle value must be 85.
To find the mean, first find out what percent got 95 but subtracting all the percentages by 100%, yielding 30% recieved a 95.  No you can just multiply 10%*70 + 25%*80 + 20%*85 + 15%*90+ 30%*95 = 86

86-85 = [b]1[/b]

if you don't follow this then you could always just make a list of 20 students, but that takes a bit longer.[/hide]

13 - hint, solve for x1, solve for x2, solve for x3
hint 2 - [hide]logarithms[/hide]
answer-[hide] solving for x1 you get log base 4 of 5, which with change of base theorem can be written as log base 2 of 4 divided by log base 2 of 5.  Similatrly x2 = log base 2 of 6 / log base 2 of 5, x3 = log base 2 of 7 / log base 2 of 6 ,... so when you multiply them together, everything cancels out except for log base 2 of 128 / log base 2 of 4 which = 7/2[/hide]

20-
[hide]When you're minimizing the sum of suares of two numbers, such that sum of the two numbers remains constant, the minimum will be when the two numbers are equal, because of the shape of the x^2 graph (increases faster as a number gets larger (or smaller on the negative side).  So in this case since -7-5-3-2+2+4+6+13 = 8, the minimum would normally occur at 4^2 + 4^2 = 32.  However, with the given numbers you can't get a sum of 4, so the next best thing is 3^2 + 5^2 = 34.[/hide]

I got 16 too, but I'm gonna have to rework it really quick.[/hide][/quote]
really hate to post 2 times in a row, but methinks you did the AMC 12B problems...

Thanks everyone.

I still don't quite understand question#20.

How you come up with 24 digits and multiplied by 11111 ?

Could you please explain a bit more ??

If you tell me what problem 20 is, i might be able to help.


calculus
integration
function
abstract algebra
advanced fields
advanced fields unsolved

Consider the space of continuous functions $ f(u)$ on the interval $ [0,1]$ and continuous functions $ A(u,v)$ on the square $ 0\leq u,v \leq 1$ and the integral transform:

$ A\cdot f \equal{} \int_{0}^{1} A(u,v) f(u) \, du$

With $ A(u,v)\equal{}u\plus{}v$, determine the image of the operator explicitily. Determine its nonzero eigenvalues and describe its kernel in terms of the vanishing of some integrals.

Denote the space of all affine mappings (i.e. mappings of the form $ av \plus{} b$) from $ [0,1]$ by $ A[0,1]$.
The image of the operator is $ A[0,1]$.
[hide="proof"]Indeed, $ A.f \equal{} \int\limits_0^1uf(u)du \plus{} v\int\limits_0^1f(u)du$ belongs to $ A[0,1]$. Conversely, let $ g(v) \equal{} av \plus{} b$ in $ A[0,1]$, then we can find $ f$ of the form $ \alpha u \plus{} \beta$ which satisfies $ a.f \equal{} g$.[/hide]
For the eigenvalues, let $ \lambda\not \equal{} 0$, we want to find $ f$ such that $ A.f \equal{} \lambda f$. Thus, $ f$ is of the form $ au \plus{} b$. We then substitute $ f(u) \equal{} au \plus{} b$ back to the equation of eigenvectors, and find $ a$, $ b$.
For the kernel, it is the set of all $ f$ such that $ A.f \equal{} 0$, hence the set of all $ f$ such that $ \int\limits_0^1uf(u)du \equal{} \int\limits_0^1f(u)du \equal{} 0$


Suppose $ F(x) \equal{} \frac {1 \minus{} \sqrt {1 \minus{} 3x}}{5x} \equal{} a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} ... \plus{} a_nx^n \plus{} .....$
Find $ a_n$
Could tell me some thinhs about fruction "GeneratingFunctions"
Thank you  :)

[quote="pco"][quote="shortlist"] Could tell me some thinhs about fruction "sinh"[/quote]

You already asked. : http://www.mathlinks.ro/Forum/viewtopic.php?t=325055

I already answered.

It seems you did not read, or understand my answer. :(
If you did not understand, just say.
If you did not read, you'll certainly never read this one too ... :([/quote]
I am sorry ,
it is "GeneratingFunctions"
Could you find $ a_n$?


While replying to another thread today, I noticed that text within the code tag displays differently within the standard thread view and in the "thread review" below the posting window. Specifically, it's significantly smaller in the normal view than in the review. (Previewing a post gives the normal view)
It's a monospace font in both versions, which is good. I like the larger size better, for readability.
Is this intended behavior?

This applies cross-browser for me; both Firefox and Safari.

[code]Testing...
Monospace.[/code]

Yes that is the intended behavior, we can't change that any time soon.


geometry
geometric transformation
reflection
conics
parabola

A projectile is launched from the very center of the floor of a rectangular room which is 40 feet wide with a very high ceiling. The projectile hits the wall at a height exactly 10 feet above the floor, reflects off this wall, hits the opposite wall, and reflects again, finally landing back exactly where it was launched, without hitting the ceiling (the projectile travels along parabolic segments, not straight lines). When the projectile is at its highest point, how high above the floor is it?

[hide] it travels a total of 80 feet (ie 20 to the wall, 40 back across, and 20 back to where it started), so assuming that it losses no speed in the bounce, we can map its height as a parabola, so let x be how far it has gone horizonatlly:

y = k(x^2 - 80x)  since it has zeros at 0 and 80 feet

now since it hits the wall 20 feet away at 10 feet high:

10 = k (400 - 1600)
10 = k (-1200)
k = -1/120

now to find the vertex of y = -1/120(x^2 - 80x)
y = -1/120(x^2 - 80x + 1600) + 1600/120
y = -1/120(x - 40)^2 + 40/3
so the max is at x = 40, y = 40/3[/hide]


Here are two questions that were proposed to 12th grade students, 1st year university students, beginning teachers and experienced teachers, in the context of a study that was made in my country whose purpose was to ascertain the level of chemical knowledge of people.

1. We can say that the [i]solubility product constant[/i], $ K_{s}$, is...

A - ... the value of the dissociation constant of any solid.
B - ... the product of the concentrations of the ions in equilibrium, dividing by the concentration of the solid present in the aquous solution.
C - ... the inverse of B.
D - ... the product of the concentrations of the ions in equilibrium with the solid in the aquous solution.

2. The [i]molar solubility[/i] of a low soluble salt...

A - ... is the molar concentration (of the solid that does not dissolve) in a saturated solution.
B - ... is the molar concentration (of the solid that does dissolve) in a solution.
C - ... is the molar concentration (of the solid that does dissolve) in a saturated solution.
D - ... is the quantity of salt that does dissolve in a certain volum of liquid.
E - ... none of the previous choices.


How does the "Intellectual Whirlwind" round work,exactly? What kinds of questions do they ask (an example would be helpful)?

Also, for the Synergy round last year, the test on the website doesn't have the triangles to put together in a square-- anyone know what the triangles were? (I'm trying to work through a mock contest with my team)

Thankd

Hi.  I'm glad you're coming this year.  I'm the problem coordinator so perhaps I can be of help.

"Intellectual Whirlwind" will be a fast round of quick questions.  They can range from algebra, trig, geometry, probability, math history, and anything in between.  However, they're not meant to take long to solve.  For example, we might ask:

Which is bigger, $ 10^{100}$ or $ 100^{10}$?
Which mathematician is on the Fields Medal?
How many internal degrees are in a pentagon?
(Not saying we will ask these questions, just examples of the types of questions)

Don't worry about that triangle problem.  I had the unfortunate task of grading that problem last year and there were problems with it, namely that there were multiple solutions.

I'm looking forward to the tournament this year.


geometry
circumcircle
conics
ellipse
trigonometry
parabola
geometry unsolved

In following figure, $ AB$ is an arbitrary chord of circle($ O$), tangent to circle($ I$).
Prove that the locus of the circumcenter of triangle $ ABO$ is ellipse with its focus $ O$.

I got the fact that the locus of the circumcenter of triangle $ ABI$ is a circle with center $ O$, recently.  :blush:
May I ask you a favor ?  [b]synthetic[/b] proofs as possible,please~
Thanks for reading my post.

Let (O), (I) be 2 arbitrary circles with radii R, r (not necessarily the circumcircle and incircle of some triangle), let d = OI, let M be arbitrary point on (I), $ \phi \equal{} \angle MOI.$ Let tangent to (I) at M cut (O) at A, B, and let N be foot of perpendicular from O to AB. $ ON \equal{} IM \minus{} OI \cdot \cos \widehat{MOI} \equal{} r \minus{} d \cos \phi.$ Tangents to (O) at A, B intersect at pole K of AB with respect to (O). Inversion in (O) takes the line AB into the circumcircle (C) of the triangle OAB and N to K, $ OK \equal{} \frac {R^2}{ON}.$ The circumcenter C of the triangle OAB is midpoint of OK, $ OC \equal{} \frac {R^2}{2 \cdot ON} \equal{} \frac {R^2}{2(r \minus{} d \cos \phi)},$ which is polar equation of a conic with focus O and excentricity $ \epsilon \equal{} \frac {d}{r}.$ Ellipse if d < r (O is inside (I)), parabola if d = r (O is on (I)), and hyprebola if d > r (O is outside of (I)).

I always admire your proof , yetti  :blush: thanks


What programmes are there available for those people in the summer between high school and college? (Are there any?) Preferably ones which admit international students.

Cheers

SimonM

[quote="t0rajir0u"]That depends on what kind of program you're talking about.  As far as I'm aware, all of the major math camps admit such students.[/quote]

Something similar to RSI in that the mathematics will be of a 'high level' and there is a focus on research.

I shall look into them, thanks


$ \tan \frac{B}{2} \equal{} \frac{r_b}{p}$ so we can costruct easly the angle B, now let a line r and a point $ D \in r$ and $ E \in r$; let a point F s.t. $ \angle FDE\equal{} \angle B$. A parallel line to DE that have distance $ h_a$ from DE meet FD on G, so $ \overline {GD} \equal{} \overline {AB}$. Then the circle with center A and radius $ GB$ meet the halfline AB on B. The circle with center B and radius p meet AB on K, s.t. K is on the same part of A. K is the tangency point of the B-excenter with AB, so costruct the b-excenter and the tangent line from B to it ($ \neq AB$) meet the tangent line from A to it ($ \neq AB$) on C.


How many non-empty subsets containing only prime numbers
does the set $ \{1, 2, 3, 4, 5, 6, 7\}$ have?

You only care about the prime numbers, so consider the set $ \{2,3,5,7\}$. It has $ 16$ subsets, but $ 1$ is empty, so there are $ 15$ subsets total.


For the game show $ \emph{Who Wants To Be A Millionaire?}$, the dollar values of each question are shown in the following table (where K = 1000).
\[ \begin{scriptsize}
\begin{tabular}{rc@{\hspace{.06in}}c@{\hspace{.06in}}c@{\hspace{.06in}}
                 c@{\hspace{.06in}}c@{\hspace{.06in}}c@{\hspace{.06in}}
                 c@{\hspace{.06in}}c@{\hspace{.06in}}c@{\hspace{.06in}}
                 c@{\hspace{.06in}}c@{\hspace{.06in}}c@{\hspace{.06in}}
                 c@{\hspace{.06in}}c@{\hspace{.06in}}c}
Question & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14
& 15 \\
Value & 100 & 200 & 300 & 500 & 1K & 2K & 4K & 8K & 16K & 32K & 64K & 125K &
250K & 500K & 1000K \\
\end{tabular}
\end{scriptsize}
\vspace{\minus{}2pt}
\]
Between which two questions is the percent increase of the value the smallest?
\\ \\
$ \text{(A)}\ \text{From 1 to 2} \qquad \text{(B)}\ \text{From 2 to 3} \qquad \text{(C)}\ \text{From 3 to 4} \\
\text{(D)}\ \text{From 11 to 12} \qquad \text{(E)}\ \text{From 14 to 15}$ \\ \\
Select the most appropriate letter.

(A) is 100% increase.
(B) is 50% increase.
(C) is 66% increase.
(D) is close to 100% increase.
(E) is 100% increase.

So the answer is B.


induction
invariant
pigeonhole principle
geometry
3D geometry
combinatorics unsolved
combinatorics

In the land of Xanadu there are 2^n countries.  The roads connecting adjacent countries form the edge of a n-hypercube with the countries as vertices.  At your disposal are 2^n armies stationed at various countries.  A single army is not strong enough to invade an adjacent country.  However if two armies attempt to invade an adjacent country, they will succeed, but incur such heavy losses, that they must regroup into a single army.  Prove that is is possible to invade any given country irregardless of how your armies are initially stationed.

Can two armies from two different countries $A,B$ invade a country $C$ if $C$ is adjacent to both $A$ and $B$, or do the two armies have to come from the same country?

Sorry about the wording - they must come from the same country.  Hopefully this is better:

In the land of Xanadu there are 2^n countries. The roads connecting adjacent countries form the edge of a n-hypercube with the countries as vertices. At your disposal are 2^n armies stationed at various countries. A single army is not strong enough to invade an adjacent country. However if two armies, attempt to invade an adjacent country, they will succeed, but incur such heavy losses along the heavily fortified road that they must regroup into a single army. Prove that is is possible to invade any given country irregardless of how your armies are initially stationed.

Hmm...
It is obvious, isn't it? By induction.

No - it is not obvious.  What are you trying to induct on?

I was thinking about number of dimensions.
I think, I will write it tommorow (if grobber will not be so fast :) )

If you consider 2 dimensions, if you have 2 armies at (0,0) and 1 at (0,1) and 1 at (1,0), how do you invade (1,1)? Are you allowed to move armies between adjacent countries ? If this moving is allowed, then the induction on n becomes more possible.

But moving the armies is the whole point: that's how yu invade a country, you move the armies there. In this case, take the two armies from $(0,0)$ and put one army (one gets wasted) on $(1,0)$, for example. You now have two armies on $(1,0)$, and, with them, you can invade $(1,1)$.

Take every possible path from the furthest point to the point you want to invade of length $n$. Give a value to the point on the path which is equals to $\frac{1}{2^n}$, where n is the distance. Note that if the value of a path is $1$, there is a way to get an army to the desired spot, because every invasion keeps the total sum invariant, and the only way that we cannot find any legal moves is if there is 1 army on every other spot except the desired spot, which is less than $1$)

However, the total value of each point will be $1$ (because a point $k$ steps away will be on $2^k$ paths).  This means the sum total over all paths is the number of armies, because each give a value of 1, which is $2^n$. Furthermore, there are $2^n$ paths, so by pigeonhole, at least 1 path must have a value of 1. QED

[quote]Take every possible path from the furthest point to the point you want to invade of length n. Give a value to the point on the path which is equals to $\frac{1}{2^n}$, where n is the distance. Note that if the value of a path is 1, there is a way to get an army to the desired spot, because every invasion keeps the total sum invariant, and the only way that we cannot find any legal moves is if there is 1 army on every other spot except the desired spot, which is less than 1.
[/quote]
This is correct and a good insight.

However it does NOT follow that there is a path with value 1.  For example take

3 on 110
3 on 011
2 on 101

with a target country of 000.

Note that no path has value 1.  Also the claim that there are 2^n paths does not seem to be correct.

Hmm...
Can anybody say me whether we allowed move armies between two adjacent invaded countries without losses? :?
If it is true than due to my solution it is sufficient to have $2^{n-1}+1$ armies.

You still need to "invade" an occupied country with 2 troops, losing 1 troop along the way.

No, I don't think we're allowed to do that, Myth. I don't think this is a regular invasion :), i.e. every time we want to move from a country to another one, we lose troops. Thjat's what I think, anyway.

kueh, if I understand correctly what you are saying, I think the number of paths is $n!$, and that's whay that doesn't work. I might be wrong though.. :?

[quote="grobber"]kueh, if I understand correctly what you are saying, I think the number of paths is $n!$, and that's whay that doesn't work. I might be wrong though.. :?[/quote]

Yes, and the value of each point ranges. The furthest point has value $\frac{n!}{2^n}$, the second furthest has value $\frac{(n-1)!}{2^{n-1}}$.... the kth furthest point has value $\frac{(n-k-1)!k!}{2^{n-k}}$. So maybe the proof can be modified... or strengthened with induction... I am still unsure of the details. One idea is we make all moves to maximise the total value over all paths. Because some moves actually increase the value. But some moves reduce the value. But if we can have a lower bound, it may be done, but it is complicated.

Ok!
My solution works for both variants:
1. We are allowed to move armies between two adjacent countries without losses.
Then answer is $2^{n-1}+1$ and it can't be reduced.

2. We are not allowed to move armies between two adjecent without losses.
Then answer is, indeed, $2^{n}$.
Ops... One more question: if we have three armies and want to invade adjacent country can we move all three armies at the same time ($3\to 2$)?

Just to be absolutely sure that I understand you correctly:

* The only way of moving armies into another country is
by sending TWO of them along THE SAME edge. One
of them is lost by doing this.

* If I have one army on (0,1) and one army on (1,0),
then I cannot use them to invade country (1,1).

Is this correct?

=====================================================
If yes, then the invariant of kueh can not work out,
since then it is possible to make fatal mistakes.

Beginning: 4 armies at (0,0); rest empty.
Move: 2 armies to (1,0); one is lost.
Move: 2 armies to (0,1); one is lost.
Situation: one army on (0,1) and one army on (1,0).
Game over!

[quote="test20"]Myth wrote:
>  kueh doesn't make such moves as you described.

Oh boy.  That wasn't my claim...

I claimed that an invariant of the type proposed by kueh cannot
work, since it does not distinguish between good situations and
GAME OVER situations.[/quote]

That's not true either. Once you get a path of value 1, we follow it as planned. In this example, we -already- have a path of value 1, hence, voila.

I assume that the only feasible atomic move is:
  Move two armies simultaneously from one vertex to an adjacent vertex.
  By doing this, one loses one of these two armies.

=================================================================
AUXILIARY RESULT.
Consider the game played on the path 0-1-2-3-...-n. The goal is
to get one army onto vertex n through a sequence of atomic moves.
Let x_i denote the number of armies on vertex i.

CLAIM:
The goal can be reached if and only if sum(i=0..n) x_i*2^i >= 2^n.

PROOF of claim:
Any atomic forward move leaves the sum unchanged (since two
times 2^i is equal to one times 2^(i+1)).
As long as there is no army on vertex n, we can make at least one
feasible atomic move: Otherwise, every vertex 0,1,...,n-1 contains
at most 1 army and vertex n contains no army, and the sum would
be at most 2^n-1.
=================================================================

[Edited; incorrect/useless pieces have been removed]

Ok, I know I am very stupid these days.
Why do you wrote there are $2^n$ x-paths? I thought there are $n!$ such paths :?

> Why do you wrote there are  x-paths? I thought there are  such paths

Because I am an idiot.

(There indeed are n! such paths, one for every ordering of bit-positions in the string).

[quote="Myth"]So, we no need to read your proof?  ;)[/quote]

well its just my idea restated/paraphrased/rewritten.

[quote]Just to be absolutely sure that I understand you correctly:

* The only way of moving armies into another country is
by sending TWO of them along THE SAME edge. One
of them is lost by doing this.

* If I have one army on (0,1) and one army on (1,0),
then I cannot use them to invade country (1,1).

Is this correct?
[/quote]
Yes this is corrrect.  As previously noticied, avoiding "bad" moves is definitely an issue that new ideas need to address.

It probably is not possible to settle this problem through invariants that solely
depend on paths.  Consider the following example:
========================================================
On the 3-dimensional cube, there are 5 armies in one fixed corner x, and
1 army in each of the three corners adjacent to x. The goal is to move one
army into the corner diametrical to x.
========================================================
In this example, every successful schedule must split the five armies into
two groups and move them along separate paths.

Consider an n-hypercube with 2^n countries.
Assume that there are altogether A armies in these countries.
Assume that B of the countries contain an odd number of armies.
==================
Claim 1:
If A>=2^n, then we can eventually move one army to the goal country.

Claim 2:
If A+B>2^{n+1}, then we can eventually move two armies to the goal country.
==================

Both statements are proved simultaneously by induction on n.
We divide the n-hypercube H into two (n-1)-hypercube H_1 and H_2, where
H_1 contains the goal vertex. For i=1,2, the hypercube H_i contains A_i
armies, and has B_i countries with an odd number of armies.
The proof is then easily done by some case distinctions.

(For instance for Claim 1: If A_1>=2^(n-1), the inductive hypothesis works.
Else if B_2<=A_1, move (A_2-B_2)/2 armies from H_2 to their immediate
neighbors in H_1. Then apply the inductive hypothesis to H_1. Etc.)


Triangle ABC is such that AB=12 cm, AC=13 cm, and BC=14 cm. The bisector of the interior angle BAC intersects BC in a point M. What is the length of the segment BM? Give your answer as a common fraction in lowest terms.

[hide]By the Angle Bisector Theorem, AC/CM=AB/BM. Replacing AC and AB with the given values gets 13/CM=12/BM. CM+BM=14, so solving both equations gets CM=182/25 and [b]BM=168/25[/b].[/hide]


linear algebra
matrix
algebra
polynomial
quadratics
abstract algebra
absolute value

Let $ a \in ( \minus{} 1;1)$ be a real number and $ A \in \mathcal{M}_2{(\mathbb{R})}$ a matrix with the propriety that : $ \det \left(A^4 \minus{} aA^3 \minus{} aA \plus{} I_2 \right) \equal{} 0$.

Compute $ \det{A} \equal{} ?$

[hide="Answer"]$ \det{A} \equal{} 1$ :P [/hide]

Use the fact that if $ \lambda$ is an eigenvalue of $ A$ then $ P(\lambda)$ is an eigenvalue of $ P(A)$.

Yes, but we have to prove that if $ z$ is a root of the equation $ x^4\minus{}ax^3\minus{}ax\plus{}1\equal{}0$, then $ |z|\equal{}1$ :)

Let's try to factor the polynomial $ p(x) \equal{} x^4\minus{}ax^3\minus{}ax\plus{}1$ as

$ x^4 \minus{} ax^3 \minus{} ax \plus{} 1 \equal{} (x^2 \minus{} ux \plus{} 1)(x^2 \minus{} vx \plus{} 1) \equal{}$

$ \equal{} x^4 \minus{} (u\plus{}v) x^3 \plus{} (uv\plus{}2) x^2 \minus{} (u\plus{}v) x \plus{} 1$.

This works if we can make $ u\plus{}v \equal{} a$ and $ uv\equal{}\minus{}2$, namely $ u$ and $ v$ are the two roots of $ t^2 \minus{} a t \minus{} 2 \equal{} 0$. The discriminant of this quadratic is $ a^2\plus{}8$ which is always positive, so $ u$ and $ v$ exist and are real.

Therefore, the roots of $ p(x)$ are just the roots of $ x^2 \minus{} u x \plus{} 1$ and $ x^2 \minus{} v x \plus{} 1$. Let's looks at $ x^2 \minus{} u x \plus{} 1$ (the same argument obviously works for the other one, too). This real polynomial has two roots whose product is $ 1$, so if they are non-real they have absolute value $ 1$ (since they are conjugate). So it's enough to show that $ u^2 \minus{} 4 \leq 0$. But $ u \equal{} \frac{1}{2}(a \pm \sqrt {a^2\plus{}8})$, and

$ \minus{}1 < a < 1$

$ 8 < a^2\plus{}8 < 9$

$ 2 < \sqrt{a^2\plus{}8} < 3$

$ 1 < a \plus{} \sqrt{a^2\plus{}8} < 4$

$ \frac{1}{2} < u < 2$

so indeed $ u^2 < 4$.

Hi turcas_c,
you can assume $ A\in\mathcal{M}_n(\mathbb{R})$.
If $ n$ is odd then $ A$ doesn't exist.
in $ n$ is even then necessarily $ det(A)\equal{}1$. Moreover $ A$ is similar to a matrix in this form:
$ diag(U,\cdots,U,V,\cdots,V)$ where $ U\equal{}\begin{pmatrix}0&\minus{}1\\1&u\end{pmatrix},V\equal{}\begin{pmatrix}0&\minus{}1\\1&v\end{pmatrix}$ and $ u,v$ are as in the randomgraph's post.

Now, how about if we generalize the statement:

If $ \det \left( A^{2n} \minus{} a \cdot A^{2n \minus{} 1} \minus{} a \cdot A \plus{} I_2 \right) \equal{} 0$, prove that $ \det A \equal{} 1$. Where $ a \in ( \minus{} 1;1)$.

This is equivalent to prove that if $ z$ is a root of the equation $ x^{2n} \minus{} ax^{2n \minus{} 1} \minus{} ax \plus{} 1 \equal{} 0$, then $ |z| \equal{} 1$. This is more interesting :). I'm looking forward to your solutions.

Let $ z\in\mathbb{C}$, with $ f(z)=0,z=r(cost+isint)$. Then
$ z^{2n-1}=\frac{az-1}{z-a}=\frac{arcost-1+iarsint}{rcost-a+irsint}$
And we pass to modules:
$ r^{2n-1}=\sqrt{\frac{(arcost-1)^2+(arsint)^2}{(rcost-a)^2+(rsint)^2}}\leftrightarrow \\
r^{4n-2}=\frac{a^2r^2-2arcost+1}{r^2-2arcost+a^2}\leftrightarrow
r^{4n-2}-1=\frac{(a^2-1)(r^2-1)}{\fbox{+}}\\\Rightarrow sgn(r^{4n-2}-1) =-sgn(r^2-1)$,
but we know that $ sgn(r^{4n-2}-1)=-sgn(r^2-1)$, wich means that $ r=1=|z|$.


linear algebra
matrix
linear algebra unsolved

Show that if A is matrix mxn such $ A*X=0$ (null matrix) where X is a matrix nx1, then $ A=0$
My question is why X can't be the null matrix?

Probably because the originally-posed question had more words in it, e.g. "for all $ X$" or some such thing.  As stated, there are many possibilities (including those for which neither $ A$ nor $ X$ is the zero matrix).


I twice today (once earlier this afternoon, and once just now at 7:12 UTC -5) had the experience of clicking on a link and going to a page that said "Critical error: could not connect to database."  It cleared up both times within a few seconds.

We're working on it. Let us know if you experience that error anytime in the following days.

That happened to me sometime around 12:00 PM PST today also.


probability
rotation
symmetry
geometry
geometric transformation

[hide]P(at least 2 girls sit next to each other) = 1-P(no 2 girls sit next to each other).

There are 2 arrangements of boys / girls such that no two girls sit next to each other (gbgbgbgbgb and bgbgbgbgbg), and there are $ 5!\equal{}120$ ways to seat the boys / girls in each arrangement, for a total of $ 2\cdot5!\cdot5!$ arrangements.  There are $ 10!$ total ways to seat the children, so the probability is $ 1\minus{}\frac{2\cdot5!\cdot5!}{10!}\equal{}125/126$[/hide]

[hide="not sure, buta correction:"]
aren't gbgbgbgb... and bgbgbg... the same because it's a circular table?[/hide]

Wait.. I'm not sure either now.

b-flat didn't correct for over counting of orders that are the same when rotated for the total number of arrangements so does solution still work?

Actually... If we counted gbgbgbgbgb and bgbgbgbgbg the same we would only have have $ 5! \times 5!$ arrangements. If we also corrected for over counting for rotational symmetry for the table, would we get $ \frac {10!}{10} \equal{} 9!$ total arrangements, giving us an entirely different answer?

Since it is circular, the rotations need to be taken into account.
For an $ n$ spots in the circle, divide by $ n$ to discount the rotations.

Gah; I made a ton of errors yesterday!

Because the table is circular, there are only $ 9!$ total arrangements.  There is also only one arrangement pattern in terms of boys / girls (they must alternate).  This makes the answer:

[hide]$ 1\minus{}\frac{5!\cdot5!}{9!}\equal{}121/126$[/hide]

[quote="mihail911"][b]A class contains 5 boys and 5 girls. For the class banquet, they select seats at random around a circular table that seats 10. What is the probability that some two girls will sit next to one another?[/b][/quote]
[hide]There are $ 9!$ ways to seat the girls and boys around the circular table. Let us first seat the girls around the table leaving one seat between two girls. Since they are seated in a circular fashion, this can be done in $ 4!$ ways. Having seated the girls, the 5 boys can be seated in $ 5!$ ways. Hence the probability of seating them such that no two girls are next to each other is $ \frac{4! \times 5!}{9!} \equal{} \frac{1}{126}$. So the probability of seating them such that some two girls are seated together is $ 1 \minus{} \frac{1}{126} \equal{} \frac{125}{126}$.[/hide]

[quote="aufha"]mthiyaga, why is the number of ways to seat the boys different than the ways to seat the girls? There are five of each.[/quote]
[hide]Once you seat the 5 girls in $ 4!$ ways  in the circular table, the boys can only be seated between the girls. So we don't need to worry about the over counting that occurs in a circular permutation. Since there are now 5 seats and 5 boys, we can seat the boys in $ 5!$ ways.[/hide]


Also, the mean of the numbers in an arithmetic sequence is the same as the median... so the answer is 11.


For n variables $ a_1,a_2,\cdots,a_n >0$ such that $ (n\minus{}1)\sum_{i\equal{}1} a_i \equal{}\sum_{i \ne j} a_i a_j$.
Let $ x_i\equal{}\sum_{j\equal{}1} a_j \minus{}a_i$
Prove:$ \sum_{i\equal{}1}^n \frac{1}{1\plus{}x_i} \le 1$.

EV or SMV works here, zhaobin. All we need to do is to prove it when n-1 variables are equal. Surely, I dont like this kind of solution, so I will think more.

I suppose you are right,hungkhtn.(Because I am not very famililar with the two methods)
But My proof only use Cauchy :D

You are right, Bin. Cauchy works here, and it is actually nicer.

Very beautiful problem,

Thank you,hungkhtn.
But seems noone here want to see the proof,so I am lazy to post that :D

Oh,You solution is right and nice,Thank you. :)


topology
conics
hyperbola
geometry
geometric transformation
rotation
real analysis

Compute trajectory starting at (x0,y0) for each of the following:

a) (x   y)' = (-x  y)

b) (x   y)' = (y   x)

Are these two systems homeomorphic, diffeomorphic, or lineomorphic?  Contruct the homeo/diffeo/lineo morphism and show that it maps trajectories in one system to trajectories in the other.

The trajectories are $ xy \equal{} k$ for (a) and $ x^2 \minus{} y^2 \equal{} k$ for (b). Both are hyperbolas, the only difference is rotation by $ 45$ degrees (and scaling by $ \sqrt {2}$ or something like that).

This is how the question should read... does this change your answer at all?

Compute the trajectory starting at
\[ (x_0, y_0)
\]

for each of the following:

a)
\[{ \left( \begin{array}{c} x \\
y \end{array} \right)' = \left( \begin{array}{c} - x \\
y \end{array} \right)}
\]
b)
\[{ \left( \begin{array}{c} x \\
y \end{array} \right)' = \left( \begin{array}{c} y \\
x \end{array} \right)}
\]
Are these two systems homeomorphic, diffeomorphic, or lineomorphic?  Construct the homeo/diffeo/lineo - morphism and show that it maps trajectories in one system to trajectories in the other.


:clap:  :wallbash:  :starwars:
hello plz prove this inequality

we have that:
$ x\plus{}y\plus{}z\minus{}xy\minus{}xz\minus{}yz\equal{}1\minus{}2xyz$
so we only should to prove:
$ 1\minus{}2xyz \geq \frac{3}{4}$
or
$ \frac{1}{8} \geq xyz$

1 case:
$ x\plus{}y\plus{}z \geq \frac{3}{2}$
then:
$ xyz\equal{}(1\minus{}x)(1\minus{}y)(1\minus{}z) \leq (\frac{3\minus{}x\minus{}y\minus{}z}{3})^3 \leq \frac{1}{8}$

2 case:
$ x\plus{}y\plus{}z \leq \frac{3}{2}$
then:
$ xyz \leq (\frac{x\plus{}y\plus{}z}{3})^3 \leq \frac{1}{8}$


Prove that each integer number exept zero can be presented in that form: $(3x-1)(2y-1)$, where $x, y$ are integers

Do you mean all composite numbers?  All primes have only two factors, and only $2y-1$ can be expressed as the factor $1$, and obviously not all primes can be written in the form $3x - 1$...

But I ment that $x$ and $y$ are integer numbers, so $3x-1$ can be equal to $-1$ and then all odd number and also prime can be expressed as
$2y-1$

[hide]If the number is odd, then we can prove it easily in the manner outlined above by dondigo.

If it is even and not a multiple of three, then we can construct it as follows :

Let us call the number we wish to express in the given form $n$.  If $n\equiv 4\mod{6}$, then we can set $y$ at 1 and $x$ at $\displaystyle \frac{n-1}{3}$, which we can see is always an integer if we represent $n$ as $4+6j$.

If $n\equiv 2\mod{6}$, then we can set $y$ at 0 and $x$ at $\displaystyle \frac{-n-1}{3}$, which we can show is always an integer in the same manner.

If $n$ is an even multiple of three, then we can express it as $a\cdot 3^{b}$, where $a$ and $b$ are integers and $a$ is an even non-multiple of three.  Then we can set $2y-1$ to plus or minus $3^{b}$, whichever suits us better, and revert to the abovementioned method.

Q.E.D.[/hide]

Hint: Click only if you want to peek at the answer....

[hide]You can write any integer  as $2^nq$  where $q$ is odd.
You can always write $q$ as $2y-1$
You can write either $2^n$ (when n is odd)  or $-2^n$ (when n is even) as $3x-1$
so factor it as $(\pm 2^n) \cdot (\pm q)$  choosing the right sign for $q$ depending on  n is even  or odd , and you are done...

[hide](Of course you could also have said that any number can be wirtten as $(\pm 3^m)(\pm p)$  (p is not divisible by 3)
 $3^m$ is always odd so it can be written as$2y-1$ and  ether $p$ or $-p$ could be written as $3x-1$ )[/hide][/hide]


Prive that if the fuction $ f: R^{ \plus{} } to R^{\plus{}}$ such that $ f(\frac {x \plus{} y}{2}) \plus{} f(\frac {2xy}{x \plus{} y}) \equal{} f(x) \plus{} f(y)$ for all postive real $ x,y$ then $ 2f(\sqrt {xy}) \equal{} f(x) \plus{} f(y)$ for all positive real $ x,y$

[quote="pco"][quote="math10"]Prive that if the fuction $ f: R^{ \plus{} } to R$ such that $ f(\frac {x \plus{} y}{2}) \plus{} f(\frac {2xy}{x \plus{} y}) \equal{} f(x) \plus{} f(y)$ for all postive real $ x,y$ then $ 2f(\sqrt {xy}) \equal{} f(x) \plus{} f(y)$ for all positive real $ x,y$[/quote]

Are you sure that there are no supplementary constraints ?
For example, this problem is rather simple if there is the continuity constraint.[/quote]
I changed the conditions of the problem

[quote="pco"][quote="math10"] I changed the conditions of the problem[/quote]

:)

So you mean you made an error in the first problem and now, you correctly stated the problem you got from your book / contest / friend / class ...

And you are sure there is a solution.

OK[/quote]
Link: http://www.mathlinks.ro/viewtopic.php?t=241518
I have read this problem in a document of web The IMO Compendium  :lol: .
I really want to know more solution for this problem  :D

I dont understand your link. In this link, the function is continuous. In such a case, as I previously said, your request is easy to solve.
If the function is not supposed to be continuous, I dont know for the moment how to prove the required result.
And I'm not sure that changing $ \mathbb R$ in $ \mathbb R^\plus{}$ helps a lot.


You can't, in general. Take $ \mathbb R_{\geq0}$, the set of nonnegative reals and $ \mathbb R$. If there was an order preserving bijection $ f: \mathbb R_{\geq 0} \to \mathbb R$, then $ f(0)$ would be a least element of $ \mathbb R$, contradiction to the fact that there is no least real number.

[quote="bharathhm"]Consider two [i]uncountable[/i] sets $ A$ and $ B$, having the same cardinality. Both of them are ordered. How do we construct an [i]order- preserving bijection[/i] between A and B?
By order-preserving, I mean, if a,b $ \in A$ and  x,y $ \in B$ are their images respectively, then a>b iff x>y.

I have one way of constructing such a bijection. But, unfortunately, it assumes a distance function (more precisely, [i]continuity[/i]) defined on both the sets $ A$ and $ B$[/quote]
I assume here that $ A$ and $ B$ are subsets of the real line.
Then clearly the same metric $ \rho$ which exists on the real line gets induced on them.if they were not subsets of real line one can always construct a metric using the definition of order that exists on the sets.
Now $ B$ being a subset of real line the basis for topology on it are balls.Consider a ball $ B(r,\epsilon)$ around $ r$.if there were an order preserving bijection  $ f$ then clearly $ f^{ \minus{} 1}(B(r,\epsilon)) \equal{} B(f^{ \minus{} 1}(r),$min$ (f^{ \minus{} 1}(r \plus{} \epsilon) \minus{} f^{ \minus{} 1}(r),f^{ \minus{} 1}(r) \minus{} f^{ \minus{} 1}(r \minus{} \epsilon)))$.Which is also an open set in $ A$.Hence the function $ f$ is not only continuous but also establishes an homeomorphism between the sets.Hence the existence of such a function forces the topological equivalence of the two sets.
if $ \epsilon$ is such that $ r \plus{} \epsilon$ does not belong to the set then we can choose any number lying outside the ball belonging to the set


induction
combinatorics unsolved
combinatorics
bipartite graph
graph theory

Let a square $ 100$ cells $ 10*10$. Each step we colour $ 4$ unit squares at intersection of $ 2$ rows and
$ 2$ colums satisfy at least $ 1$ cells hasn't coloured before. What is the MAXIMUM number of steps to colour all cells of big square?

I have found that the maximum number is $ 81$.
But seem nobody like this problem?

Can you prove it is 81? I have 2 constructions for 81, but can't prove it's the maximum

I prove that: If let square $ n*n$ then the maximum number of step is $ (n\minus{}1)^{2}$ by use induction. I prove that $ S(n)\leq S(n\minus{}1)\plus{}2n\minus{}3$.

For a $ n*n$ board how do you know it first involves coloring $ (n\minus{}1)*(n\minus{}1)$ grid and then the rest of the squares?

I solve its by that way:
Denote cell row $ i$ coloum $ j$ by $ (i;j)$. If we colour $ 4$ cells $ (1;j),(1;j'),(i;j),(i;j')$ and $ 4$ cells $ (1;j),(1;j'),(k;j),(k;j')$ then we can't colour $ 4$ cells $ (i;j),(i;j'),(k;j),(k;j')$. Then number of steps to colour by this way is equal number of steps to colour by below way: colour $ 4$ cells $ (i;j),(i;j'),(k;j),(k;j')$ and $ 4$ cells $ (1;j),(1;j'),(i;j),(i;j')$. It mean we can colour the square $ (n\minus{}1)*(n\minus{}1)$ first, then colour the rest. Now it is very easy to use induction!
I thinks it is correct, but my english can make you don't understand my idea!


Assume that  $A,B,C,D$ are real numebrs and $\varepsilon >0 .$
[b]1)   Prove that   $f(x)=x^5+Ax^4+ \left(\frac{4A^2}{7}+\varepsilon\right) x^3 +Bx^2+Cx+D$ has at least two complex roots. [/b]
[b]2) If   $f(x)=x^5+Ax^4 +\beta x^3 + Bx^2 +Cx+D$ has only real roots,  prove/or disprove
\[ \left\{ \begin{array}{lcl} \displaystyle (AC - \beta B)^2 & < & \displaystyle 4(\beta^2 -AB)(B^2-\beta C) \\ \displaystyle (\beta D-BC)^2 & < & \displaystyle 4(B^2-\beta C)(C^2-BD) \end{array}\right.  \]
[/b]

For the first one: Suppose that all the roots are real then by Viete's formula:

$0<\sum x_ i^2=(\sum x_i)^2-2\sum x_ix_j=A^2-\frac{8A^2}{7}-2\epsilon<0$ contradiction, hence there are at least two complex roots.

For the second it seems like Viete's formula will be handy as well but I don't dare to expand it all  :P

[quote="Megus"] ... the second it seems like Viete's formula will be handy as well but I don't dare to expand it all  ...[/quote]
The second question is  much deep : the assertoin may be generalized in the following way:
[color=green] [b] 2) If   $f(x)=x^n+\sum\limits_{k=1}^{n}a_kx^k \; ,\; n\ge 4,$ has only real roots,  then  for $1\le i\le n-3$
\[ \left| \begin{array}{cc} a_i & a_{i+1} \\ a_{i+1} & a_{i+2} \end{array}\right| \left|\begin{array}{cc} a_{i+1} & a_{i+2}\\ a_{i+2}&a_{i+3} \end{array}\right| > 0  \]
and moreover
\[ \left|\; \begin{array}{|cc|} a_i & a_{i+1} \\ a_{i+2}&a_{i+3} \end{array}\; \right| < 2 \sqrt{ \left| \begin{array}{cc} a_i & a_{i+1} \\ a_{i+1} & a_{i+2} \end{array}\right| \left|\begin{array}{cc} a_{i+1} & a_{i+2}\\ a_{i+2}&a_{i+3} \end{array}\right| }\; .  \]
[/b][/color]


http://www.breitbart.com/article.php?id=080218195400.xhq81wua&show_article=1

I just thought you guys/girls might want to know.


induction
limit
number theory proposed
number theory

Here's a cute little problem (I suppose it should be awfully easy :)):

Let $(a_n)_{n\ge 1},(b_n)_{n\ge 1}$ be two sequences of natural numbers s.t. $a_{n+1}=na_n+1,\ b_{n+1}=nb_n-1,\ \forall n\ge 1$. Show that the two sequences can have at most a finite number of terms in common.

An easy induction shows that for each $n\geq 2$,
\[a_{n}=(n-1)!(a_{1}+\sum_{k=0}^{n-1}\frac{1}{k!})\]
\[b_{n}=(n-1)!(b_{1}-\sum_{k=0}^{n-1}\frac{1}{k!})\]
If $b_{1}<e$, $b_{n}$ is negative for $n$ large enough, and as $a_{n}>0$ the sequences have only a finite number of terms in common. Suppose now that $b_{1}>e$.
Suppose that $a_{n}$ and $b_{n}$ have infinitely many terms in common. As both sequences have infinite limits, we know that , for each $N$, there are infinitely many couples $(p,q)$ such that $p,q\geq N$ and $a_{p+1}=b_{q+1}$ ie:
\[\frac{a_{1}+s_{p}}{b_{1}-s_{q}}=\frac{q!}{p!}\].
with $s_{n}=\sum_{k=0}^{n}\frac{1}{k!}$. $\lim s_{n}=e$. Suppose that $l=\frac{a_{1}+e}{b_{1}-e}>1$ (we can not have $l=1$...). We can find $N> l+1$ such that (*)
\[\forall p,q\geq N, l+1>\frac{a_{1}+s_{p}}{b_{1}-s_{q}}>1\]
If $p,q \geq N$ and $a_{p+1}=b_{q+1}$, we have
\[l+1>\frac{q!}{p!}>1\]
But as soon as $\frac{q!}{p!}>1$, we have $\frac{q!}{p!}\geq\frac{(p+1)!}{p!}=p+1>N>l+1$, which contradicts (*). A similar argument works if $a_{1}+e<b_{1}-e$.
Is there a nicer approach?

That is exactly what I did. Why would you think it's not nice? :? :)

Well :-) in fact it is not so ugly :-) At the beginning, I thought that I could reach a solution easily finding different equivalents for $a_{n}$ and $b_{n}$, but then I found the general expressions and once you see this the solution is not so far is it?   :P  In my view, the idea that $\frac{p!}{q!}>1$ implies $\frac{p!}{q!}>q+1$ is the nicest part of the exercise  :blush:


for $ x,y,z > 0$, such that $ 2xyz \equal{} 3x^2 \plus{} 4y^2 \plus{} 5y^2$.Find min:

$ A \equal{} 3x \plus{} 2y \plus{} z$
 :P

hello, could it be that you meant $ 2xyz\equal{}3x^2\plus{}4y^2\plus{}5z^2$?
Sonnhard.

I got $ 9\sqrt [3]{45}$.

[hide="Solution"]Applying AM-GM to $ 3x^2 \plus{} 4y^2 \plus{} 5y^2$, we have $ \frac {3x^2 \plus{} 4y^2 \plus{} 5y^2}{3}\ge \sqrt [3]{60x^2y^2z^2}$.

Since $ 3x^2 \plus{} 4y^2 \plus{} 5y^2 \equal{} 2xyz$, we have $ 8x^3y^3z^3\ge27\cdot60x^2y^2z^2$.

Thus, $ xyz\ge \frac {27\cdot15}{2}$.

Now applying AM-GM to $ 3x \plus{} 2y \plus{} z$, we have $ \frac {3x \plus{} 2y \plus{} z}{3}\ge \sqrt [3]{6xyz}$.

Since $ xyz\ge\frac {27\cdot15}{2}$, we have

$ \frac {3x \plus{} 2y \plus{} z}{3}\ge \sqrt [3]{6xyz}\ge\sqrt [3]{\frac {6\cdot27\cdot15}{2}}$.

Hence, we have $ 3x \plus{} 2y \plus{} z\ge9\sqrt [3]{45}$.

Answer: $ 9\sqrt [3]{45}$[/hide]

Meh, did I do something wrong? :(

EDIT: Darn. :stink:

I don't believe that's attainable. The minimum value you stated ($ 27 \times 15/2$) was attained when $ 3x^2 \equal{} 4y^2 \equal{} 5z^2$ (by the way, there's a typo in your first line :P). Call that value $ k$. Then $ x \equal{} \sqrt {\frac {k}{3}}$, $ y \equal{} \frac {\sqrt{k}}{2}$, and $ z \equal{} \sqrt {\frac {k}{5}}$, so $ \frac {k^{3/2}}{\sqrt {30}} \equal{} 3k$, so $ k \equal{} 30$.

However, $ 9 \sqrt [3]{45}$ is attainble only when $ 3x \equal{} 2y \equal{} z$ as well, which isn't true for the given values.

I think it is $ 36$, when $ x \equal{} y \equal{} z \equal{} 6$...
 :huh:

[quote="fesha3t"]for $ x,y,z\in R$, such that $ |x|\plus{}|y|\plus{}|z|\equal{}1994$.Find min:
$ B\equal{}|x\minus{}1|\plus{}|y\minus{}2|\plus{}|z\minus{}3|$[/quote]

It's $ 1988$, if I am not mistaken.


Many of our students take multiple classes at a time, particularly during the summer.

All the classes are transcripted, meaning that you can later review everything that happens in a class.  Therefore, as long as you're willing to review later, missing classes is ok.

All of our courses provide extra problems on the message board.  The subject classes also provide two thorough challenge sets for homework.  (The Problem Series classes do not have challenge sets, but have plenty of message board problems.)

[quote="bluemoom"]I am thinking to take two classes at same time this summer, is it doable (in terms of work load )?

Does the class give homework or reading material?

I will be out of country for four weeks, and during that time, I may not find internet access , is it a way for me to catch up with those missed class?

Thanks[/quote]

I've done up to three or four classes at a time. Works great. Very doable, even if you're planning on missing loads of classes. I generally attend 8 or 9 of per 12 classes just b/c of the amount of stuffs that tend to overlap into those class times. It usually doesn't cause too much of a hassle. There's always time to go back over the material at a later date (e.g. Summer or winter vacation, thanksgiving, spring break, etc).

Thanks for your reply, I think I know what should I do this summer now


I hope I'm in the appropriate forum.  I'm trying to help my girlfriends brother with his homework.  He neglected to bring the math book home with the necessary formual, and I certainly don't remember any high school math... or the math I learned last week in college  :blush:

I hope someone can help me, here's the problem.

__________________________________________
Senior class members were asked to name their favorite subject.  The results of the survey of 220 students are listed in the table below. (sorry for the crude table)

[b]Subject/Number[/b]: English: 30 | Social Studies: 40 | Science: 50 | Math: 35 | Foreign language: 20 | Gym: 10 | No Preference: 35 |

Then the question is: [b]What is the probability that a randomly chose student did not say math?[/b]

I don't really need the answer, but I do need the formual in order to solve several other questions based on this table.

Err... $1 - \frac{35}{220}$.

In this case each student has one and only one choice, so $P_{math} + P_{notmath} = 1$ and the result follows.


calculus
integration
function
limit
geometry
calculus computations

Maybe try to apply L'Hopital Rule :?  but i'm not sure the second limit satisfy the condition to apply that.
What you mean
[quote]... a function whose restricetion to $[a,x]$ is integrable for every $x>a$...
[/quote]

I don't get it.

It is pretty straightforward.

For large $x$, $f(x)$ is very close to $L$ and hence the integral of $f$ is also close to $L$, because the 'bad parts', e.g, $\int^b_a f(x)dx$ for $b$ close to $a$ cannot be too big (by the hypothesis of integrability).

Writing it rigorously as follows.

For $\epsilon > 0$, there exists $X>0$ such that $|f(x)-L|<\epsilon/3$ for $x>X$. Then for $x>X$, we write

$\frac{1}{x}\int_a^x f(x)dx-L = \frac{1}{x}\int_a^X f(x)dx + \frac{1}{x}(\int_X^x f(x) dx-L(x-X))-L\frac{X}{x} = S_1(x) + S_2(x) + S_3(x)$.

$S_1(x)$ goes to 0 as $x\to \infty$ since $\int_a^X f(x)dx$ is a constant now. $S_3(x)\to 0$ as $x\to\infty$, too.

Then there exists $X_1>X$ such that $|S_1(x)|<\epsilon/3$ and $|S_3(x)|<\epsilon/3$ for all $x>X_1$.

Now let us look at
$|S_2(x)| = |\frac{1}{x}(\int_X^x f(x) dx-L(x-X))| = |\frac{\int_X^x (f(x)-L) dx}{x}| \leq \frac{\int_X^x |f(x)-L| dx}{x} < \frac{\frac\epsilon 3(x-X)}{x} < \frac \epsilon 3$

Hence for $x>X_1$,

$|\frac{1}{x}\int_a^x f(x)dx-L| \leq |S_1(x)| + |S_2(x)| + |S_3(x)| < \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3= \epsilon$

Could anyone tell me more about in detail  " It is pretty straightforward.For large $x$, $f(x)$ is very close to $L$ and hence the integral of $f$ is also close to $L$, because the 'bad parts', e.g, $\int^b_a f(x)dx$ for $b$ close to $a$ cannot be too big (by the hypothesis of integrability)",  recent post by "blahblahblah"

And What Method did "liyi" use in his post?

Thx alot.

blahblahblah was attempting to be intuitive, saying that $f(x)=L+$ (some early stuff that we can swamp out in the long run) + (something that's small as $x\to\infty.$)

Putting in a few more details, but sticking to blahblahblah's plan:

$\frac1x\int_a^xf(t)\,dt=\frac1x\int_a^xL\,dt+\frac1x\int_a^x(f(t)-L)\,dt$

By direct calculation, $\frac1x\int_a^xL\,dt=\frac{(x-a)L}x\to L$ as $x\to\infty.$

So all we have to do is to show that the leftover stuff goes to zero.

Since $f(x)\to L$ as $x\to\infty,$ we can find $b$ such that $|f(t)-L|<\epsilon$ for $t\ge b.$

So $\left|\frac1x\int_a^x(f(t)-L)\,dt\right|\le\frac1x\int_a^x|f(t)-L|\,dt$

$\le\frac1x\int_a^b|f(t)-L|\,dt+\frac1x\int_b^x\epsilon\,dt.$

The first part of this is just some constant times $\frac1x,$ so it goes to zero.

The second part of this is less than $\epsilon.$

liyi's full argument is just this argument stated with even more care and formality. Because there are three pieces to estimate, he used an $\epsilon/3$ argument.

[quote="ABM"]Need some help with the following. Thx

Suppose that $a$ and $L$ are real numbers and $f: [a,\infty]\rightarrow R$ is a function whose restricetion to $[a,x]$ is integrable for every $x>a$

Show that if $\lim_{x\to\infty}f(x)=L$ then $\lim_{x\to\infty}\frac{1}{x}\int_{a}^{x}f(t)dt=L$[/quote]

well one way to think of it is that the second expression is essentially the average value of the area [a,x] (the a becomes insignificant as x approaches infinity) , and the limit will essentially also be that average value, i know that is not rigirous, but it makes logical sense

But I still don't get it, did I miss something? :?  in the liyi's post he get
for $x> X_1$ then $\left | \frac{1}{x} \int_{a}^{x} f(t) dt \right | < \epsilon$ , that's mean
$\lim_{x \rightarrow \infty} \frac{1}{x} \int_{a}^{x} f(t) dt = 0$, but in the problem we should prove $\lim_{x \rightarrow \infty} \frac{1}{x} \int_{a}^{x} f(t) dt = L$

For [b]Kent[/b]:
$\frac{1}{x} \int_{b}^x \epsilon \, dt = \frac{1}{x} \epsilon (x-b)$ then for $x \rightarrow \infty$ it tends to $\epsilon$ , why this prove the terms tends to $0$ ?

Thanks

[quote="Beginner"]But I still don't get it, did I miss something? :?  in the liyi's post he get
for $x> X_1$ then $\left | \frac{1}{x} \int_{a}^{x} f(t) dt \right | < \epsilon$ , that's mean
$\lim_{x \rightarrow \infty} \frac{1}{x} \int_{a}^{x} f(t) dt = 0$, but in the problem we should prove $\lim_{x \rightarrow \infty} \frac{1}{x} \int_{a}^{x} f(t) dt = L$
[/quote]
Typo. I've fixed it.

We're trying to prove the following: given any $\epsilon>0,$ (no matter how small), we can find $M$ (large enough) so that if $x>M,$ then

$\left|\frac1x\int_a^xf(t)\,dt-L\right|<\epsilon.$

As for the line of mine you quoted and are questioning: we picked $b$ after we already knew $\epsilon.$ It really is sufficient just to prove that that term is small. If I were to go back and flesh out the proof with more details, I would actually have $\frac{\epsilon}3$ in that term rather than $\epsilon,$ and then I will be able to claim that that term is $<\frac{\epsilon}3.$

Once I prove that each of the other two terms is $<\frac{\epsilon}3,$ using, if necessary, a condition that $x$ be large enough, I will be done with the proof.

My intention with that post was to find a middle ground between blahblahblah's extremely informal, all-motivation with no details post and liyi's strictly formal, less motivated post. I wasn't trying to provide a complete formal proof but rather to , reach a point from which a person with ordinary skills and experience in constructing epsilon-delta proofs should be able to provide the details needed to finish it.


geometry
geometric transformation
reflection
circumcircle

ABC is an acute-angled triangle. D is the reflection of A in BC, E is the reflection of B in AC, and F is the reflection of C in AB. Show that the circumcircles of DBC, ECA, FAB meet at a point and that the lines AD, BE, CF meet at a point.

1) "who can prove" is NOT a title for a problem. An appropriate title would be, for instance, "Reflections of A, B, C in BC, CA, AB and concurrent lines/circles".

2) If you copy a problem from Kalva, then please mention the source (in the "Subject description" field). Here it's ASU 1973 problem 9 (found by googling for the exact text of your post).

3) This problem is more appropriate for the Pre-Olympiad forum (though it's ASU). Therefore, moved.

4) The ASU has nothing to do with the MathLinks contest (this concerns your other post).

  darij

...Do you just want a proof that there exists a unique orthocenter for any triangle?  That would do it.


ABC is right-angled at A.  D is a point on AB such that CD = 1.  AE is the altitude from A to BC.  If BD = BE = 1, what is the length of AD?

[quote="ashrafmod"][color=darkred]$ABC$ is right-angled at $A$. $D$ is a point on $AB$ such that $CD = 1$.
$AE$ is the altitude from $A$ to $BC$. If $BD= BE=1$, what is the length of $AD\ ?$[/color][/quote]
$AC^{2}+AD^{2}=CD^{2}\Longrightarrow$ $AC^{2}+x^{2}=1$ $\Longrightarrow$ $AC^{2}=1-x^{2}\ .$
$AB^{2}+AC^{2}=BC^{2}$ $\Longrightarrow$ $(1+x)^{2}+1-x^{2}=(1+y)^{2}$ $\Longrightarrow$ $(1+y)^{2}=2 (1+x)\ .$
$AB^{2}=BC\cdot BE$ $\Longrightarrow$ $(1+x)^{2}=(1+y)\cdot 1$ $\Longrightarrow$ $1+y=(1+x)^{2}\ .$
Therefore, $(1+y)^{2}=(1+x)^{4}$ $\Longrightarrow$ $2(1+x)=(1+x)^{3}$ $\Longrightarrow$ $\boxed{\ x=\sqrt [3] 2-1\ }\ .$


[b]FIRST PRIZE[/b] and the Sun Life Financial Cup goes to
[b]Dong Uk (David) Rhee[/b], McNally Composite High School Edmonton, AB;

[b]SECOND PRIZE[/b] goes to
[b]Yufei Zhao[/b], Don Mills Collegiate Institute, Toronto, ON;

[b]THIRD PRIZE[/b] goes to
[b]Shawn Eastwood[/b], Canadian International School, Singapore.

[b]HONOURABLE MENTIONS[/b] were awarded to:
[b]Alan Guo[/b], O'Neill Collegiate and Vocational Institute, Oshawa, ON,
[b]Kent Huynh[/b], University of Toronto Schools, Toronto,  ON,
[b]Viktoriya Krakovna[/b], Vaughan Road Academy, Toronto,  ON,
[b]Alexander Remorov[/b], Waterloo Collegiate Institute, Waterloo  ON,
[b]Thomas Tang[/b], A.Y. Jackson Secondary School, North York,  ON.

[quote="Elyot"]A huge congrats to David and Yufei!  Awesome job!

I guess the Penguins really got owned this year!!![/quote]

They picked up the lead on the USAMO, though. And on Euclid. But total penguinization was indeed not accomplished. And they are even refusing to penguinize the Seminar! Blasphemy ...

[quote="lightrhee"][quote="YawningLion"]Are the APMO winners posted anywhere as well? I think Australia posted theirs long ago, but Canada is still being secretive. I got just 14/35, so I'm not opting for anything anyway, but it's interesting how people did. Will the results be posted eventually, or will they forever remain a mystery?[/quote]
Didn't they send you an email with your result? I got mine(20/35, and that's 7-9th :o ).[/quote]

Yup, they sent an email, so I know only my result, but have no basis for comparison. Anyway, good job! I suppose you were one of those people who did try the 'scary' #5 (I didn't). And congrats for the CMO 1st - you rock!


Let $ n\geq 3$ be an integer and let $ x_0,x_1,..,x_{n \plus{} 1}$ be nonegative real number such that $ x_0 \equal{} x_1 \equal{} 0,x_n \equal{} x_{n \plus{} 1} \equal{} 1$ .Prove that there exist $ j\in \{1,2,...,n\}$ for which:$ |x_{j \plus{} 1} \plus{} x_{j \minus{} 1} \minus{} 2x_j|\geq \frac {4}{n^2}$


[quote="Vegetius"]Prove that for any positive real numbers $a,b,c$ we have the inequality:
      \[ (1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a}) \geq 2(1+\sum \sqrt[3]{\frac{bc}{a^2}})  \][/quote]

Let's multiply the both parts of inequality by $abc>0$
So we have to prove $(a+b)(a+c)(b+c) \ge 2(abc+\sqrt[3]{ab^4c^4}+\sqrt[3]{a^4bc^4}+\sqrt[3]{a^4b^4c}) \Leftrightarrow$
$2abc + (a^2b+a^2c+b^2a+b^2c+c^2a+c^b) \ge 2(abc+\sqrt[3]{ab^4c^4}+\sqrt[3]{a^4bc^4}+\sqrt[3]{a^4b^4c}) \Leftrightarrow$
$\sum_{ciclyc}(a^2b+a^2c) \ge 2(\sqrt[3]{ab^4c^4}+\sqrt[3]{a^4bc^4}+\sqrt[3]{a^4b^4c})$

Let's note
$a^2b+2c^2a \ge 3\sqrt[3]{a^4bc^4}$
$c^2a+2b^2c \ge 3\sqrt[3]{ab^4c^4}$
$b^2c+2a^2b \ge 3\sqrt[3]{a^4b^4c}$
$b^2a+2c^2b \ge 3\sqrt[3]{a^4bc^4}$
$c^2b+2a^2c \ge 3\sqrt[3]{a^4bc^4}$
$a^2c+2b^2a \ge 3\sqrt[3]{a^4b^4c}$
We get the required inequality by summarizing this ones.
V:)

I have another solution for Vegetius's problem, and I think it's shorter :)
Note that $\prod_{cyc} (1+\frac{a}{b})+1 = (a+b+c)(\frac1a+\frac1b+\frac1c)$(1)
By AM-GM $(a+b+c) \geq 3\sqrt[3]{abc}$, so
$\frac23(a+b+c)(\frac1a+\frac1b+\frac1c) \geq 2\sum \sqrt[3]{\frac{bc}{a^2}}$(2)
and $\frac13(a+b+c)(\frac1a+\frac1b+\frac1c) \geq 3$ (by AM-GM)(3)
From (2),(3) and (1) we get the answer.

[quote="darkmaster"][quote="arqady"]$(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a}) \geq 2(1+\sum \sqrt{\frac{bc}{a^2}})$ is stronger. ;)[/quote]
And easier, just expand LHS and then AM-GM.[/quote]
Therefore an initial inequality even easier, since
$\sum \sqrt{\frac{bc}{a^2}}\geq\sum \sqrt[3]{\frac{bc}{a^2}}$ has a negative level of complexity. :P

How can you obtain $\sum \sqrt{\frac{bc}{a^2}}\geq\sum \sqrt[3]{\frac{bc}{a^2}}$?, I don't think it is short.

[quote="darkmaster"]How can you obtain $\sum \sqrt{\frac{bc}{a^2}}\geq\sum \sqrt[3]{\frac{bc}{a^2}}$?[/quote]
This is meurhead's inequality.  ;)
[quote="darkmaster"] I don't think it is short.[/quote]
If $(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a}) \geq 2(1+\sum \sqrt{\frac{bc}{a^2}})$ then $(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a}) \geq 2(1+\sum \sqrt[3]{\frac{bc}{a^2}}).$


Let $ a$ and $ b$ be nonnegative real numbers.

(a) If $ 2a^2\plus{}b^2\equal{}2a\plus{}b$, then $ 1\minus{}ab\ge \frac {a\minus{}b}{3}$;

(b) If $ a^3\plus{}b^3\equal{}2$, then $ 3(a^4\plus{}b^4)\plus{}2a^4b^4\le 8$.

[quote="Vasc"]Let $ a$ and $ b$ be nonnegative real numbers.

(a) If $ 2a^2 \plus{} b^2 \equal{} 2a \plus{} b$, then $ 1 \minus{} ab\ge \frac {a \minus{} b}{3}$;
[/quote]
$ 2a \plus{} b \equal{} a^2 \plus{} a^2 \plus{} b^2 \geq \frac {(2a \plus{} b)^2}{3} \Rightarrow 2a \plus{} b \leq 3$

We have to prove that :

$ 1 \minus{} ab\ge \frac {a \minus{} b}{3} \Leftrightarrow (3a \minus{} 1)(3b \plus{} 1) \leq 8 \Leftrightarrow (3b \plus{} 1)(6a \minus{} 2) \leq 16$

But,this is easy from AM-GM because :

$ (3b \plus{} 1)(6a \minus{} 2) \leq \frac {( 3.(2a \plus{} b) \minus{} 1 )^2 }{4} \leq 16$

Done !

[quote="Vasc"]Let $ a$ and $ b$ be nonnegative real numbers.

(a) If $ 2a^2 \plus{} b^2 \equal{} 2a \plus{} b$, then $ 1 \minus{} ab\ge \frac {a \minus{} b}{3}$;

(b) If $ a^3 \plus{} b^3 \equal{} 2$, then $ 3(a^4 \plus{} b^4) \plus{} 2a^4b^4\le 8$.[/quote]

(b) Applying the AM-GM Inequality, we have
$ 3a^4 \le a^3\plus{}a^3\plus{}a^6,3b^4 \le b^3\plus{}b^3\plus{}b^6,$
and
$ 2a^4b^4\le \frac{2}{3}a^3b^3(a^3\plus{}b^3\plus{}1)\equal{}2a^3b^3.$
Adding up these three inequalities, we deduce that
$ 3(a^4\plus{}b^4)\plus{}2a^4b^4 \le 2(a^3\plus{}b^3)\plus{}(a^6\plus{}b^6\plus{}2a^3b^3)\equal{}8.$

Nice your solutions, Can_hang. The second is almost the same as my solution.

[quote="rachid"][quote="Vasc"]Let $ a$ and $ b$ be nonnegative real numbers.

(a) If $ 2a^2 \plus{} b^2 \equal{} 2a \plus{} b$, then $ 1 \minus{} ab\ge \frac {a \minus{} b}{3}$;
[/quote]
The following more general ( and very easy ) inequality holds:

If $ a,b,c \geq 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} a \plus{} b \plus{} c$,then :

$ 2 \minus{} c(a \plus{} b) \geq \frac {a \plus{} b \minus{} 2c}{3}$

For $ a \equal{} b$ we get the above inequality.

[quote="Vasc"]Nice your [b]solutions[/b], Can_hang. The second is almost the same as my solution.[/quote]
The first solution is mine,isn't it ?! :maybe:  :P[/quote]

We can easly get $ a \plus{} b \plus{} c\le3$

ineq is equivalent to

$ 2 \minus{} c(a \plus{} b)\geq \frac {a \plus{} b \plus{} c \minus{} 3c}{3}$ and so it remains to prove that

$ 2 \minus{} c(a \plus{} b)\ge 1 \minus{} c$ which is equivalent to $ 1\ge c(a \plus{} b \plus{} c) \minus{} c^2 \minus{} c$ which is true since we have

$ c^2 \plus{} c \plus{} 1 \ge 3c\ge c(a \plus{} b \plus{} c)$

[quote="rachid"]
[quote="Vasc"]Nice your [b]solutions[/b], Can_hang. The second is almost the same as my solution.[/quote]
The first solution is mine,isn't it ?! :maybe:  :P[/quote]
Sorry, Rachid. Indeed, the first solution is yours. Nice solution and generalization, and nice solution of Evariste-Galois to generalization :lol:

how about exponnent  generalisation ??  let $ a,b\geq 0$  find $ k$ such that :
$ a^k\plus{}b^k\equal{}3$  ==>  $ 3(a^{k\plus{}1}\plus{}b^{k\plus{}1})\plus{}2(ab)^{k\plus{}1} \geq 8$


If two numbers are selected randomly without replacement from the set $ \{1,2,3,4,5,6\}$, what is the probability that their product will be greater than 15? Express your answer as a common fraction.

There are 18 ways to choose the numbers: $ \frac {6 \times 5}{2}$. Of these, $ 3 \times 6, 4 \times 5, 4 times 6, and 5 \times 6$ work. Thus the answer is $ \boxed{\frac {4}{15}}$.

There are $ 6$ choices for the first number, then $ 5$ for the second, so there are $ 6\cdot 5$ ways to choose the two numbers without replacement.  However, this counts each arrangement $ 2!$ times, so we need to divide to correct for this, so there are $ \frac{6\cdot 5}{2}\equal{}15$ ways to choose the numbers.

The pairs that work are $ 3\cdot 6$, $ 4\cdot 5$, $ 4\cdot 6$, and $ 5\cdot 6$, so the probability is $ \boxed{\frac{4}{15}}$


function
algebra
domain
real analysis
calculus
integration
analytic geometry

Let $ f: \mathbb R \to \mathbb R$ be a function, two times derivable on $ \mathbb R$ for which there exist $ c\in\mathbb R$ such that
\[ \frac { f(b)\minus{}f(a) }{b\minus{}a} \neq f'(c) ,\] for all $ a\neq b \in \mathbb R$.

Prove that $ f''(c)\equal{}0$.

The image of the function $ g(a,b)\equal{}\frac{f(a)\minus{}f(b)}{a\minus{}b}$ defined for $ a\ne b$ being an interval (connected subset of the real line) and $ f'(c)$ not being in this image, it follows that we may assume that $ f'(c)<g(a,b)$ for all $ a\ne b$. But then $ f'(c)\leq f'(x)$ for all $ x$ and so $ c$ is a minimum point of $ f'$. Of course, this can be written in 11-th grade vocabulary.  :D

The set of pairs $ (a,b)$ such that $ a\ne b$ is a connected subset of the plane and the function $ g$ is continuous on this domain, thus its image is a connected subset of the line, thus an interval. I agree however that this is not a solution of an 11-th grade student, but that's how life is.  :D I will not be amazed if in a few years I see complex analysis, Lebesgue integration and other such stuff at RMO. It's quite à pity, it gives huges advantages to some people.  :(

[quote="harazi"]But then $ f'(c)\leq f'(x)$ for all $ x$ and so $ c$ is a minimum point of $ f'$. [/quote]
Why for all $ x$?

Well, $ f'(c)\leq g(a,x)$ for all $ a\ne x$ and now make $ a$ close to $ x$ and use the definition of derivative.

Well, I said however a very stupid thing: the function $ g$ should be defined on the set of pairs $ (a,b)$ such that $ a<b$ to ensure that its domain is connected. Of course, all the rest works with this modification, I don't know how I could write such a stupid thing.  :D

Well, when I created this problem, I was thinking to the obvious geometric meaning: if $ f''(c) \ne 0$, then $ f$ is strictly concave up or down on some neighbourhood of the point $ c$, thus one can draw a close enough parallel to the line tangent at $ c$ to the function's graph that intersects the graph in two points $ (a,f(a))$ and $ (b,f(b))$. The slope of that tangent would be $ \frac{f(b)\minus{}f(a)}{b\minus{}a}$, equal to the slope of the tangent at $ c$, that is,$ f'(c)$.

One of above solutions uses that $C=\mathbb{R}^2-\{(a,a) \mid a \in \mathbb{R}\}$ is connected, but why $C$ is connected ? $C$ has clearly two connected components. However with this spirit we can also solve the problem, extend $g(a,b)=\frac{f(a)-f(b)}{a-b}$ to whole plane defining $g(a,a)=f'(a)$ and now $g$ is continuous on whole plane and do similar thing.

[quote=Svejk]I tried to solve it harazi's way but I didn't get maximum  since I didn't manage to prove that $ g(a,b)\minus{}f'(c)$ has the same sign for all $ a,b$.Can you be give me more detailes please  :lol: ?The official solution is based on the fact that the function $ g(x)\equal{}f(x)\minus{}f'(c)\cdot x$ is injective ,hence monotonic.[/quote]

  where did you find the official solution?


limit
trigonometry
algebra proposed
algebra

Let $n\in\mathbb{N}^*$. Prove that \[ \lim_{x\to 0}\frac{ \displaystyle (1+x^2)^{n+1}-\prod_{k=1}^n\cos kx}{ \displaystyle x\sum_{k=1}^n\sin kx}=\frac{2n^2+n+12}{6n}.  \]

Using $\boxed{\lim_{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta ^2}}=\frac{1}{2},$ we have

${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}=\frac{1}{12}n(n+1)(n+2),$ without using Landau sign $o.$

Thus, by $\lim_{x\rightarrow 0} \frac{(1+x^2)^{n+1}-1}{x^2}=\lim_{t\rightarrow 0} \frac{(1+t)^{n+1}-1}{t}$

$=\lim_{t\rightarrow 0}(1+t+t^2+\cdots +t^{n})=n+1,$

$\lim_{x\rightarrow 0} \sum_{k=1}^n \frac{\sin kx}{kx}\cdot k=1+2+\cdots+n=\frac{n(n+1)}{2}.$

Thus

$\lim_{x\to 0}\frac{ \displaystyle (1+x^2)^{n+1}-\prod_{k=1}^n\cos kx}{ \displaystyle x\sum_{k=1}^n\sin kx} =\lim_{x\rightarrow 0} \frac{\frac{(1+x^2)^{n+1}-1}{x^2}+\frac{\displaystyle 1-\prod_{k=1}^n\cos kx}{x^2}}{\displaystyle \sum_{k=1}^n \frac{\sin kx}{kx}\cdot k}$

$=\frac{n+1+\frac{n(n+1)(n+2)}{12}}{\frac{n(n+1)}{2}}=\frac{2n^2+n+12}{6n}.\ Q.E.D.$


inequalities
advanced fields
advanced fields unsolved

Let D be an infinite set of cardinality d. Let E be the set of all partial orders on D. Then $ |E| \leq 2^d$.

The other inequality is easy so this implies that $ |E| \equal{} 2^e$.

If you are asking for a proof of $ |E|\le 2^{|D|}$, here a [hide="proof:"]$ E$ is a subset of the power set $ P(D\times D)$ of $ D\times D$, thus $ |E|\le |P(D\times D)|\equal{}2^{|D\times D)|}\equal{}2^{|D)|}$.[/hide]

there are even $ 2^d$ well-orderings:
http://www.mathlinks.ro/Forum/viewtopic.php?t=190094


At each lattice point of a finite grid paper, we draw an arrow parallel to one of the sides of the paper (no arrows on the boundary can point outwards.) Show that there exist two neighboring points (horizontally, vertically, or diagonally) which the arrows point to opposite directions.

Lemma: If you color the lattice points of a grid red and blue, there is either a red path from the bottom to the top, or a blue path from the left to the right, where you are allowed to move diagonally.

If you're familiar with the game of hex, this lemma is similar to the lemma that says there can't be a draw in that game.  The proof of the lemma is also similar.

Sketch of proof of lemma: Consider the right hand border together with all blue points connected to it by a path.  If this region is connected to the left hand border, we're done.  If not, by following the boundary of this region we can find a path from the bottom to the top.

In your problem, we color the up/down arrows red, and the right/left arrows blue.  If there's a red path from top to bottom, it must start with a down arrow on top (since no arrows point outwards) and end with a up arrow on bottom (ditto), so it changed somewhere along the way between two adjacent squares in the path.  The same holds true for a left/right blue path.


5 green balls and 2 red balls together weigh 10 pounds, and 1 green ball and 4 red balls together weigh 7 pounds.  If all red balls weigh the same amount and all green balls weigh the same, then what is the weight of 8 red and 8 green balls together?

Let $ x$ be the weight of a green ball, and let $ y$ be the weight of a red ball.
We have \[ 5x \plus{} 2y \equal{} 10\\
1x \plus{} 4y \equal{} 7\]
Add these together to get \[ 6x \plus{} 6y \equal{} 17\]
Now multiply by $ \frac{4}{3}$ to get \[ 8x \plus{} 8y \equal{} \boxed{\frac{68}{3}}\]


Dear all,

The online application for Mathcamp 2005 is now available here:

https://www.mathcamp.org/application

Please note that if you are a Mathcamp alum seeking to return to Mathcamp 2005, then this is NOT the page for you.  We'll have the alumni signup page ready soon (the site for new applicants was our first priority).

California is not in the drop down menu for where I live.[img]http://users.pandora.be/eforum/emoticons4u/sad/1011.gif[/img]

When is the deadline for the mathcamp application?

The application deadline is April 27.

Thanks for pointing out that Califorina is missing.  Alabama is also missing, and Arkansas is there twice; there are also some problems with the list of countries.  We're not quite sure how that happened, since we copied the lists directly from other places, but we'll take care of it asap.

Neither the link here nor the one on Mathcamp website is working any more.
I was planning on submitting my quiz online tomorrow...Will it be back up by then?
It's a little too late to submit via snail mail.

Thank you,

[quote="litoxe2718281828"]Neither the link here nor the one on Mathcamp website is working any more.
I was planning on submitting my quiz online tomorrow...Will it be back up by then?
[/quote]

It should be back up now.


$ x,y,z,t$ positive number
$ (x\plus{}y\plus{}z)^{t}\plus{}(x\plus{}z\plus{}t)^{y}\plus{}(x\plus{}y\plus{}t)^{z}\plus{}(x\plus{}y\plus{}z)^{t}>3$

[quote="zaya_yc"]$ x,y,z,t$ positive number
$ (x \plus{} y \plus{} z)^{t} \plus{} (x \plus{} z \plus{} t)^{y} \plus{} (x \plus{} y \plus{} t)^{z} \plus{} (x \plus{} y \plus{} z)^{t} > 3$[/quote]
Try $ x\equal{}y\equal{}z\rightarrow0^\plus{}.$ :wink:


Solution :   [hide]$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+.......................+\frac{1}{n\times(n+1)}= 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{n}-\frac{1}{n+1}= \frac{n}{n+1}$
$\Rightarrow$ $\frac{n}{n+1}= \frac{2\times\sqrt{5^{n}-4^{n}}}{9}$

if $n$ is odd we have  $gcd( 2\times\sqrt{5^{n}-4^{n}}, 9 ) = 1$ and $n+1=9$ and
$n = 2\times\sqrt{5^{n}-4^{n}}$ thus we must have: $5^{n}-4^{n}= 16$ but this is not possible.

so $n$ is even. if $n$ is even then $9 | 5^{n}-4^{n}$ , so we can write:

$\frac{n}{n+1}= \frac{2\times\frac{\sqrt{5^{n}-4^{n}}}{3}}{3}$ $\Rightarrow$ $2\times\frac{\sqrt{5^{n}-4^{n}}}{3}= 2$ $\Rightarrow$ $5^{n}-4^{n}= 9$ and $n=2$.
In other case if $81 | 5^{n}-4^{n}$ then we must have $n = 0$ that is not possible, so the only answer is $n=2$.[/hide][/hide]

at the person who last posted on this topic: please hide your solution.

[hide="solution"]The left hand side is equal to $\frac{n}{n+1}$ so $\frac{n}{n+1}=\frac{2\sqrt{5^{n}-4^{n}}}{9}$. When the RHS gets bigger, the LHS gets closer to one. We need to find an n such that the RHS is greater than one. Therefore, $5^{n}-4^{n}$ has to be greater than $\frac{81}{4}$. That can be achived by n=3. Therefore, n has to be less than or equal to 2. We test for two, and it works. \[\boxed{\bf{2}}\] [/hide]


induction
limit
integration
real analysis
real analysis solved

We have the sequence difiened like so:
a1=2,  a_(n+1)=2^(an/(n+1)).
1) Prove that it converges.
11)compute the limit lim(n-->oo) an;
111)compute the limit lim(n-->oo) an^n;
1111)compute the limit lim(n-->oo) n(an-1);

How do you prove that it converges? :?
cheers! :D

just for convergence we can prove by induction that an is decreasing. Suppose that an < an-1 .

Then  an / n+ 1 < an-1 / n and raising to the power of 2, we get an+1 < an and since (an)n is positive, we obtain that it is convergent.

It's now obvious that an is convergent to 1=20.

a_n-->1
(a_n)^n = 2^(a_(n-1)) -->2 when n->+oo
Since a_(n-1)ln2/n -->0
a_n = exp(a_(n-1)/n.ln2) = 1+ a_(n-1)/n.ln2+ a_(n-1).ln2/(2n)+o(a_(n-1)/n) (*)
n(a_n-1) = a_(n-1).ln2+  \varepsilon (n)
 \lim n(a_n-1)= ln2 using (*) we can get an aymptotic expansion
a_n = 1 +ln2/n + ln2/(2n) + o(1/n)
 \cap  \cap  \emptyset  \cup \beta  \int  \cap  \emptyset  \lambda


There are 2 non-congruent truangles. Each triangle has side legnths of 10cm ad 12cm, and the measure of the angle opposite the 10cm side is 45 degrees in each triangle. In square centimeters, wat is the sum of the areas of these two triangles?

This problem needs trig. boo. No mathcounts problems require it and it isn't allowed in this forum.

[quote="daermon"]aren't there an infinite number of answers?  :huh:[/quote]
No. The SSA property only applies if you know if the triangle is obtuse or acute.

This involves trig, so I am not going to solve it.


vector
superior algebra
superior algebra unsolved

Show that if we have a field then axiom $ x\plus{}y\equal{}y\plus{}x$ follows from the other axioms.

Expand $ (1\plus{}x)(1\plus{}y)$ in two ways.

For further discussion see my prior post [1]. In a nutshell:  a semiring with 1 and cancellative addition must have commutative addition. Such semirings are simply subsemirings of rings (e.g. $ \mathbb{N} < \mathbb{Z}$) because any commutative cancellative semigroup embeds canonically into a commutative group, its group of differences. See [1] for definitions, examples, and references.

--Bill Dubuque

[1] sci.math, 29 Nov 2005, Commutative axiom unnecessary in vector spaces
http://google.com/group/sci.math/msg/86f109f83a36f628
http://google.com/groups?selm=y8zhd9vvlav.fsf%40nestle.csail.mit.edu


Compute the value of $ 3t\plus{}4q$ where $ t$ is the number of triangles and $ q$ is the number of quadrilaterals in the tiling shown.

[asy]draw((2,3)--(6,1)--(0,1)--(2,3)--(6,3)--(6,0)--(0,0)--(0,3)--(2,3));[/asy]

More in-depth:
there are 3 triangles, they should be obvious
there are 5 quadrilaterals (two trapezoids each consisting of 2 triangles, and 3 rectangles)
so $ 3(3)\plus{}4(5)\equal{}\boxed{29}$


induction
trigonometry
calculus
calculus computations

Find the Taylor serie for
[url=http://imageshack.us][img]http://img148.imageshack.us/img148/1558/int9jk7.gif[/img][/url]

By http://mathworld.wolfram.com/TaylorSeries.html we need only find $f^{(n)}(x)$. By induction on $n$, we have $\cos^{(n)}x=\cos{(-x+n\cdot\frac{\pi}{2})}$.


does anyone know where i can find some amc problems?

Seeing how there are now 4 topics asking for help with AMC prep within the first 7 topics on this page, I propose that someone make a sticky.

There actually is a sticky...the AMC FAQ's thread.


If there are 3 toys.

One for a baby, one for a boy and one for a teenager.

However the parents managed to give the wrong presents to all of them.

How many possibilites are there?

I guess, by simple counting possibilities there are 2.

How about if there were $n$ toys and they were all given to the wrong person.

How many ways are there for this to happen?

This is known as a derangement and is denoted by \[ D_n = n!(\frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}}...... + ( - 1)^n \frac{1}{{n!}})  \]

See
[url=http://mathworld.wolfram.com/Derangement.html]mathworld.wolfram.com/Derangement.html[/url] for more information


[quote="Maths_KTL"]Who can send me IMO 2006 shortlisted problems?
dastan_ktl@mail.ru[/quote]

This has nothing to do with Geometry. Besides, nobody is going to send you the shortlist problems of 2006 before the next IMO without further authentification on your side.

  Darij


My 5-year-old brother picked up a huge calculus book and was attempting to read it seriously. Now who can come up with a cuter thing than that?

... maybe he saw you reading it and tried to make the same you did? :D

Wow....  Awwwwwwww....   :lol:   I wonder if (should I not say his name?) remembers Denver?  I doubt he remembers me.

omg that's SOOOOOOOOOOOOOOOOOOOOOOOOOO cute!!! were you reading the book right before? little kids are soooo cute!!

I was [i]not[/i] reading the book, which is the darndest thing.


conics ellipse parameterization geometry calculus algebra polynomial

my friend , when i was 15.
His problem is a bit easier
For k = 2 .
But i can't solve with k= 2 ...
Try k=2 first ...

I know this problem for $k=1$ :) My teacher gave it to us when I was 11 :D

ok , k = 1 is very easy
but for k = 2 i can't do ............up to now ...
may be .. it is very hard ...

If I remember correctly, Descartes introduced a generalization of the ellipse, called Descartes's Oval. It's the locus of all the points satisfying $MA+kMB=c$ for some constant $c$ and fixed points $A,B$ and parameter $k$. I think the answer to this question is: construct all the ovals with $A,B,k$ being the given ones and $c$ going through the reals. There is one which touches the line $d$ in a point. That's the point we are looking for (I think :)).

Pretty nasty, huh? :D

It leads to 4-th power equation. I have great doubts it has roots, which can be expressed in square radicals.

I read a note in Descartes's "Geometry" a few seconds ago. Indeed, the locus of the point $M$ for which $\alpha MA+\beta MB=c$ is an "oval" called Descartes's Oval (today; he didn't actually name it like that when he introduced it :)).

As far as I am concerned, I'm pretty convinced the point we are looking for is not constructible by ruler and compasses, but I don't think the problem asks for a construction.

[quote="grobber"]I read a note in Descartes's "Geometry" a few seconds ago. Indeed, the locus of the point $M$ for which $\alpha MA+\beta MB=c$ is an "oval" called Descartes's Oval (today; he didn't actually name it like that when he introduced it :)).[/quote]
yes , i know it
it is in our lesson ...  in vietnam ,we learn it when we are at 12th class .
But .. the problem is find the minimum of MA+kMB ....

Did you also read my previous post? If $k>0$, the ovals are similar to an ellipse (they do not go to infinity). Among all the ovals, there is one which touches the line. The point of tangency is the one you are looking for. If $k<0$, on the other hand, I don't know..

Like Myth said, the equations involved have a rather high degree (even though I haven't actually computed anything :)).

Another question: does your problem include the case $k<0$?

[quote="InToTheRainBow"]Given 2 point A and B ; 1 line d ; a number k
Find the point M on the line d satisfied that MA + kMB min ...[/quote]

This problem is a real thorn in my side. I am 95% sure that it cannot be solved by compass and ruler, but the remaining 5% should be checked by an algebraist. What I can say now is that for positive k, the problem of finding a point M on the line d such that $MA+k\cdot MB$ is minimum is equivalent to the Fermat-Snellius problem about light refraction between two homogeneous media (at least, it is equivalent in the case when the points A and B lie on different sides of the line d - but the other case can be easily reduced to this one ;) ).

Here is some data I calculated with my CAS for direct use:

Let U and V be the orthogonal projections of the points A and B on the line d; we use directed segments on the line d. Then, a = AU, b = BV and s = UV are given values, and k is a given value as well. What we have to find is the x = UM such that $MA+k\cdot MB$ is minimal. After a bunch of algebra and calculus, with the values a = 3, b = 2, s = 1 and k = 2, we see that x is a root of the polynomial $3x^4-6x^3+35x^2-72x+36$. If you succeed to show that the roots of the polynomial are not compass and ruler constructible, then it will follow that the problem is not solvable with compass and ruler for k = 2. Since some people might think that negative k would behave differently, I also checked a = 3, b = 2, s = 1 and k = -2, asking for the maximum of $MA+k\cdot MB$ instead of the minimum, of course, and got x as a root of the same polynomial $3x^4-6x^3+35x^2-72x+36$ as in the above (but another root!). Now a Galois check would solve the problem, right?

[b]EDIT.[/b] [url=http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=29857]Thanks to Dr. Sonnhard Graubner[/url], the problem is now solved. The roots of the polynomial $3x^4-6x^3+35x^2-72x+36$ cannot be constructed by compass and ruler, so that the original problem is generally unsolvable with compass and ruler (neither for positive k, nor for negative k).

  Darij


Do you know how to prove than the subset of roots of unity of all order is dense in the complex unity circle?

[quote="t0rajir0u"]This is equivalent to showing that $ \mathbb{Q}$ is dense in $ \mathbb{R}$ (why?) or, more specifically, that $ \mathbb{Q}/\mathbb{Z}$ is dense in $ \mathbb{R}/\mathbb{Z}$.[/quote]

ok, if I understood well, you took the image of $ \mathbb{Q}$ under the continuous and onto map $ \exp{(2i\pi*x})$. Nice. :). Thanks.

It still seems to work with only the $ {2^n}$-roots of unity, do you see the reason?

The set of rational numbers of the form $ \frac{k}{2^n}$ is also dense in $ \mathbb{R}$, and this is not much harder than the first proof; both are essentially because of the Archimedean property.  There is a slightly more general statement:  if $ (a_n)$ is a strictly increasing sequence of natural numbers, then the set of rational numbers of the form $ \frac{k}{a_n}$ is dense in $ \mathbb{R}$.

[quote="t0rajir0u"]The set of rational numbers of the form $ \frac {k}{2^n}$ is also dense in $ \mathbb{R}$, and this is not much harder than the first proof; both are essentially because of the Archimedean property.  There is a slightly more general statement:  if $ (a_n)$ is a strictly increasing sequence of natural numbers, then the set of rational numbers of the form $ \frac {k}{a_n}$ is dense in $ \mathbb{R}$.[/quote]

Thanks very much for the knowledge.


On each of the 2006 cards a natural number is written. Cards are placed arbitrarily in a row. 2 players take in turns a card from any end of the row until all the cards are taken. After that each player calculates sum of the numbers written of his cards. If the sum of the first player is not less then the sum of the second one then the first player wins. Show that there's a winning strategy for the first player.

Let $S_1$ be the sum of the cards with even indexes and $S_2$ be the sum of the cards with odd indexes.
If $S_1>S_2$ then at his move the first player should pick the card with even index, otherwise he should chose the one with odd index. Of course, in this way he always wins.

This was also a problem in the second QEDMO, but was also posted here on the forum several times ago (so not a very new problem).


number theory
prime numbers
number theory proposed

Is there a prime number $ p$, for which the least positive integer $ d$ satisfying $ p|2^{d} \minus{} 1$ also satisfies $ p^2|2^{d} \minus{} 1$?

They called Wieferich primes. Known only 2 ($ p<10^{15}$) Wieferich primes $ p\equal{}1093,3511$.

Thank you Rust! I only needed to know if there exists such prime number. It solves Germany TST 2006 problem 3. Namely, if you remember that problem, those prime numbers can't be represented in the given form.

If it's the problem I think it is, then $ 23$ can't be represented either, as $ 2$ has order $ 11$ $ \mod 23$ and $ \mod 89$, thus both can only occure together. One has to do some work using Zigmondy again to finish it from here. Here I'm talking about the general products of numbers of type $ \frac{2^a\minus{}2^b}{2^c\minus{}2^d}$, so we allow an arbitrary number of such factors (for that one, you need to use a bit more, too).

Working on this problem, i came across the statement

p divides the numerator of (1/1+1/2+1/3+........1/(p-1)/2) . Will anyone comment on this statement, or even characterize such primes ?


(a). Kaoticity (Of course you already know me :))
(b). During the summer... Hmm.  Almost anytime is good.

u already know my sn...starting from july would be nice

coolcam2433 - Friday during the summer would be optimum, and monday nights.

I'm interested but I don't know if I'll have time.

u know who i am and you know im interested, just that ill be off from mid july to end of august

i don't even have AIM, how do you get it?

Sounds like fun.  I guess any time is good for me.   My sn is highwind331

[quote="Lucky707"]Over the summer, I'd like to host sessions of MAFIA over AOL INSTANT MESSENGER.

I actually hosted one after the AOPS TRIVIA INVITATIONAL on FRIDAY and it went pretty well. Even better when we hit 8 people, which was our peak at the time.

I think with more than 8 people, it would be a roaring success.

I need to know:

(a) who is interested (leave your AIM here if you are or message me: I'm "hezgotzbeenz")
(b) what times/dates would be good[/quote] HEY YOU'RE COPYING SOMEONE

Lol i'm sorry that I had to go... otherwise we could have played longer. Anyways,, i'm "TJ Valthonis" (Yeap you know me already).

[quote="mathbomberII"] HEY YOU'RE COPYING SOMEONE[/quote]

Um... what? I'm not copying anyone.

Or if you want to be literal, I'm unknowingly copying someone who unknowingly copied someone else who unknowingly copied someone else who ....

yeah, you get the point.

So I have:

Lucky
DarkKnight
GoBraves
WindSlicer
phatso *Maybe*
Chigr
Secret Asian
TJ

That's eight ... that's not bad. Let's get more people! C'mon, sign up everyone! Then we'll figure out a time from all this! Don't be shy!

I'll play. Weekends would be the best; I'm gone at various times during the weekdays, so...

I think you know mine... but FYI, it is es7abl1sh3d.

I am available sometimes.

I'm interested too (sn: tocolorthesky).
I don't have a specific time, but I do have stuff to do at random times during the day, but most of the times I can probably work around.

Cmon hez, don't leave me out here.  :)  You know my SN.

I've never played in one of the mafia games yet.. seems like fun i was just too lazy to read through all the rules... i could try once i guess. Im not home from July 12-sometime around early august or late july (probably be back by august 5). If u need to contact me.. click on my "AIM" button under my post  :) .

FYI - to get aim go to AIM.com. MSN and other messengers are better by themselves, but AIM is much more popular because of its connection to AOL and compatibility w/ it.

Haha that was fun online mafia. I was mafia like 4 times in a row though.. it was strange. First time I was the only citizen without any special ability.. =( But it was fun! And one time I figured out who mafia was even though I was only an SCV!  :D  Uhh so yeah I'm gone till next Saturday afternoon so Iono if I'll play next time.

Well yesterday was fun, so when's the next one?

Hahaha

Dear Lucky,

I am writing this, pleading unto you that you may bestow forgiveness upon me. I know my actions deserve beheading but I have heard that you are a lenient lord and forgive those when they beg it.

:roll:*rolls eyes*:roll:

lol.

Sorry!  :D

[quote="Lucky707"]Thanks,

still waiting for 10 more letters of apology.[/quote]

Haha. My sincere apologies Lucky. I couldn't resist killing you as serial killer.

Well, I guess I'm excused from writing a letter since, at the last minute, I couldn't make it. When are you thinking the next one will be?

Dear Lucky,

I never voted you out first. Sorry I couldn't stop other people from being stupid and doing it though.

Yay, that made me happy.

What's a good time for everyone?

[quote="Lucky707"]We'll hold the next one once everyone writes me a written letter of apology for voting me first every time and promises to not do it too often ever again :p[/quote]

HAHAHAHA. yeah right :D its not like you aren't after me all the time on the first day, too, im just repaying the favor ;) im up for more online mafia, weekends would work best, but weekday nights late im free too (or afternoon, this is all converted to EST too).

I'm pretty sure that I won't be able to go, since I'm at camp right now, but is there a set date and time for Online Mafia yet?

How do you play online mafia?  I have played it with cards but it was much simpler. :?:

Online Mafia was fun last time, this weekend would be perfect.

I be playing Chat Mafia with the people from MafiaScum ...

11PM EST every Monday. That starts today.

If you'd like to join, send me an IM.

Of course, you need to be on your best, tip-top behaviour ... if you aren't, I won't be sending you any more invites from now on.


Let be $ x,y$ real numbers and let be $ m,n$ real nonnegative constants numbers , such that $ mx^2\plus{}ny^2>0$ . Find the values of :

$ E\equal{}\frac{x\plus{}y}{\sqrt{mx^2\plus{}ny^2}}$

[hide="Vector solution"]$ E \equal{} \frac {\langle x\sqrt {m},y\sqrt {n}\rangle\cdot\left\langle{1\over\sqrt {m}},{1\over\sqrt {n}}\right\rangle}{|\langle x\sqrt {m},y\sqrt {n}\rangle|}\implies |E|\leqslant\sqrt {{1\over m} \plus{} {1\over n}}$. The extremes are reached for $ {x\sqrt {m}\over y\sqrt {n}} \equal{} \frac {1/\sqrt {m}}{1/\sqrt {n}}\implies mx \equal{} ny\implies x \equal{} an\land y \equal{} am, a\in\mathbb{R}$[/hide]

Note on $ m,n$ being non-negative: if either of them is zero, then $ E$ is unbounded. Hence both should be taken positive.


Solve in the set of the real numbers the equation :

$ \sum_{i\equal{}1}^na_i^{2i}\equal{}2\sum_{i\equal{}1}^ni\cdot a_i\minus{}n^2\ ,\ (\forall)n\in \mathbb{N}$

[hide]$ \sum_{i \equal{} 1}^n [1 \plus{} (2i)(a_{i} \minus{} 1)] \equal{} \sum_{i \equal{} 1}^n [1 \plus{} (a_i \minus{} 1)]^{2i}$, bernoulli's implies $ a_i \equal{} 1,\ 1 \le i \le n$.[/hide]


function
logarithms
algebra
floor function

what is the greatest integer function mean? and what is the purpose of it [/quote][/code][/list][/list]

The Greatest Integer Function tells the greatest integer less than or equal to a number.  One common use of the function is to tell how many digits are in a number.  For example:

Let [x] denote the greatest integer function. The number of digits in x (d(x)) are given by
[tex]
\displaystyle
d(x) = 1 + [\log_{10} x]
[/tex]

This is also sometimes known as "floor."

A similar function, is the least integer function, or ceiling function.  It gives the smallest integer greater than or equal to a number.


set theory
real analysis
real analysis solved

Prove that: for any set $A$, $\O \subset A$.

The axiom of the empty set is:

$\exists x \forall y: y\notin x$.

It says there exist an object $x$ that no other set belongs to $x$.
It is called empty set.

reductio ad absurdum:
suppose that fails, then there is a set A such that the empty set is not its subset: then there must be an element of the empty set (already a contradiction with its definition) which is not contained in A. Contradiction reached. We didn't use that element except its existence which is actually important, not that it's not an element of A.
The standart Zermelo-Fraenkel axiomatic set theory system that i've studied does not include an axiom for the empty set, its existence and uniqueness are derived as follows:
there is an axiom that states that there must be at least one set, another axiom (axiomatic scheme to be precise) states that for every set A, and every predicate (FO-described of course) there exist the set which contains only those elements of A that satisfy the predicate. So we must take the set from the first axiom and use the prediacate x <> x for example to derive a set with the property that no set belongs to it and convince ourselves that this set is only one: this can be concluded from the axiom of volume: two sets are considered the same iff ( every set is an element of the first set iff it is an element of the second one)