geometry
geometric transformation
homothety
ratio

[hide="My Solution....Please check"]Let the medians from $ A,B,C$ meet $ BC,CA,AB$ at $ D,E,F$. Let $ DE$ and $ DF$ meet $ MN$ in $ Y$ and $ X$ respectively.
From Menelaus in $ \triangle ABE$,
\[ \frac{BM}{MA} \cdot \frac{AN}{NE} \cdot \frac{EG}{GB}\equal{}\minus{}1\]
\[ \Longrightarrow \frac{BM}{MA}\equal{}\frac{\minus{}2NE}{AN}\]
Similarly,Applying Menelaus in $ \triangle AFC$,
\[ \frac{CN}{NA}\equal{}\frac{\minus{}2MF}{AM}\]
Adding, we get
\[ \frac{CN}{NA}\plus{}\frac{BM}{MA}\equal{}\frac{\minus{}2MF}{AM}\plus{}\frac{\minus{}2NE}{AN}\]
From this the signs are readily checked to be the required ones.
So it remains to prove
\[ \frac{NE}{AN}\plus{}\frac{MF}{AM}\equal{}\frac{1}{2} ...(in \ magnitude)\]
For this, applying midpoint theorem in $ \triangle ABC$, we get $ DF\parallel{}AC$ and $ DE\parallel{}AB$
\[ \Longrightarrow \frac{NE}{AM}\equal{}\frac{NY}{NM} \ and \ \frac{MF}{AM}\equal{}\frac{MX}{MN}  \ \ (Thales' \ theorem)\]
Therefore, it reduces to proving
\[ NY\plus{}MX\equal{}\frac{1}{2}MN \iff XY\equal{}\frac{1}{2}MN\]
For this we note that $ \triangle ABC$ and $ \triangle DEF$ are homothetic with $ G$ as the centre of homothety and $ 2: 1$ the mapping ratio.
\[ NG\equal{}2GX \ and \ MG\equal{}2GY\]
Adding these two together proves the required statement...
$ Q.E.D.$ :wink:
[/hide]
Please post some answers....mine I feel is too long..... :o

Please...I need a better solution for this problem.... :|

BTW....Source=Kiran Kedlaya Geometry Unbound.. :wink:

No one interested? Please answer.... :o  :(

$ AN.MB\plus{}AM.NC \equal{} AM.AN$
Divide both side by $ AM.AN$, we get:
$ \frac{MB}{AM} \plus{} \frac{NC}{AN} \equal{} 1$
Let the median from $ A$ meets $ BC$ at $ P$. Draw $ BD$,$ CE$ parallel to $ MN$ ($ D,E \in AP$). By Thales theorem:
$ \frac{MB}{AM} \plus{} \frac{NC}{AN} \equal{} \frac{GD}{AG} \plus{} \frac{EG}{AG} \equal{} \frac{GD\plus{}EG}{AG} \equal{} \frac{GD\plus{}GP\plus{}EP}{AG}$
Notice that $ DP\equal{}EP$ because $ \frac{DP}{EP}\equal{}\frac{BP}{PC}\equal{}1$
So: $ \frac{MB}{AM} \plus{} \frac{NC}{AN} \equal{} \frac{GD\plus{}DP\plus{}GP}{AG}\equal{}\frac{2GP}{AG}\equal{}1$
Problem's solved :lol:

[quote="Unidranix"]$ AN.MB \plus{} AM.NC \equal{} AM.AN$
Divide both side by $ AM.AN$, we get:
$ \frac {MB}{AM} \plus{} \frac {NC}{AN} \equal{} 1$
Let the median from $ A$ meets $ BC$ at $ P$. Draw $ BD$,$ CE$ parallel to $ MN$ ($ D,E \in AP$). By Thales theorem:
$ \frac {MB}{AM} \plus{} \frac {NC}{AN} \equal{} \frac {GD}{AG} \plus{} \frac {EG}{AG} \equal{} \frac {GD \plus{} EG}{AG} \equal{} \frac {GD \plus{} GP \plus{} EP}{AG}$
Notice that $ DP \equal{} EP$ because $ \frac {DP}{EP} \equal{} \frac {BP}{PC} \equal{} 1$
So: $ \frac {MB}{AM} \plus{} \frac {NC}{AN} \equal{} \frac {GD \plus{} DP \plus{} GP}{AG} \equal{} \frac {2GP}{AG} \equal{} 1$
Problem's solved :lol:[/quote]
Thanks very much Unidranix...I've checked your solution and its correct....and very elementary too!! :lol:  :D

There's another solution for this problem using the area.
Notice that two triangle have the same height then the ratio of their areas is the ratio of their bases.
 :) Good luck

[quote="Poincare"][quote="Rijul saini"]if a line thru the centroid $ G$ of $ ABC$ meets $ AB$ in $ M$ and $ AC$ in $ N$ then prove that $ AN.MB \plus{} AM.NC \equal{} AM.AN$ both in magnitude and sign.[/quote]

Whoa. Wait a second. Let AN=NC=y, AM=MB=z, let the spot where the median hits BC be O. so,

$ (AN)(MB) \plus{} (AM)(NC) \equal{} (AM)(AN)$ is the same as
$ yz \plus{} yz \equal{} yz$
This is defnitely not true, unless I don't know what the "." means or I'm screwing up the centroid with something else.[/quote]

 :arrow: Hey poincare, $ AN\equal{}NC$ and $ AM\equal{}MB$ can't both be true cause if it's true, then $ MN$ doesn't go through the centroid, $ MN$ will go through the midpoint of the median from $ A$ to $ BC$.

[quote="Unidranix"][quote="Poincare"][quote="Rijul saini"]if a line thru the centroid $ G$ of $ ABC$ meets $ AB$ in $ M$ and $ AC$ in $ N$ then prove that $ AN.MB \plus{} AM.NC \equal{} AM.AN$ both in magnitude and sign.[/quote]

Whoa. Wait a second. Let AN=NC=y, AM=MB=z, let the spot where the median hits BC be O. so,

$ (AN)(MB) \plus{} (AM)(NC) \equal{} (AM)(AN)$ is the same as
$ yz \plus{} yz \equal{} yz$
This is defnitely not true, unless I don't know what the "." means or I'm screwing up the centroid with something else.[/quote]

 :arrow: Hey poincare, $ AN \equal{} NC$ and $ AM \equal{} MB$ can't both be true cause if it's true, then $ MN$ doesn't go through the centroid, $ MN$ will go through the midpoint of the median from $ A$ to $ BC$.[/quote]
Not trying to be stubborn or anything but: http://en.wikipedia.org/wiki/File:Triangle.Centroid.svg

[quote="Poincare"][/quote][quote="Unidranix"][quote="Poincare"][quote="Rijul saini"]if a line thru the centroid $ G$ of $ ABC$ meets $ AB$ in $ M$ and $ AC$ in $ N$ then prove that $ AN.MB + AM.NC = AM.AN$ both in magnitude and sign.[/quote]

Whoa. Wait a second. Let AN=NC=y, AM=MB=z, let the spot where the median hits BC be O. so,

$ (AN)(MB) + (AM)(NC) = (AM)(AN)$ is the same as
$ yz + yz = yz$
This is defnitely not true, unless I don't know what the "." means or I'm screwing up the centroid with something else.[/quote]

 :arrow: Hey poincare, $ AN = NC$ and $ AM = MB$ can't both be true cause if it's true, then $ MN$ doesn't go through the centroid, $ MN$ will go through the midpoint of the median from $ A$ to $ BC$.[/quote][quote="Poincare"]
Not trying to be stubborn or anything but: http://en.wikipedia.org/wiki/File:Triangle.Centroid.svg[/quote]

Ok, poincare, If $ AN=NC$ and $ AM=MB$, then $ M,N,G$ wont be collinear(the problem gives that line $ MN$ go through the centroid $ G$), because it divides the median into two equal parts but the centroid divides the median into two parts which their ratio is $ 2: 3$. It's too basic and besides, you can see and read it here [url]http://www.mathopenref.com/trianglecentroid.html[/url](in the centroid facts)
Hope you get it and we should end this topic
Bye :P

Another approach: let $\{D\}\in AG\cap BC$ (i.e. midpoint of BC) and $\{E\}\in MN\cap BC$.
Apply Menelaos for $\triangle ABD$ with the transversal $\overline{EMG}$ and for $\triangle ADC$ with transversal $\overline{EGN}$, add the two necessary ratios, i.e. $\frac{BM}{MA}+\frac{CN}{NA}$, keep in mind that $AG=2GD$ and $BE+CE=2DE$ and finally get $\frac{BM}{MA}+\frac{CN}{NA}=1$, done.

Best regards,
sunken rock

[color=darkred][b][u]PP[/u].[/b] Let $ABC$ be a triangle . For the points $M\in AB$ , $N\in AC$ and $P\in MN$ denote the point

$D\in AP\cap BC$ . Prove that $\boxed{\frac {MB}{AM}\cdot DC +\frac{NC}{AN}\cdot BD=\frac {PD}{AP}\cdot BC}$   both in magnitude and sign.[/color]

[b][u]Proof[/u].[/b] Denote $\{\begin{array}{c}
X\in MN\cap BC\\\\
Y\in MN\ ,\ AY\parallel BC\end{array}\|$ . Therefore, $\{\begin{array}{c}
\frac {MB}{AM}=\frac {XB}{AY}\\\\
\frac {NC}{AN}=\frac {XC}{AY}\end{array}|$ $\begin{array}{cc}
\ \ \odot\ DC\\\\
\ \ \odot\ BD\end{array}\|\bigoplus\ \implies$ $\frac {MB}{AM}\cdot DC +\frac{NC}{AN}\cdot BD=$

$\frac {XB\cdot DC+XC\cdot BD}{AY}=\frac {(XD+DB)\cdot DC+(XD+DC)\cdot BD}{AY}=$ $\frac {XD\cdot(DC+BD)}{AY}=$ $\frac {XD}{AY}\cdot BC=$ ${\frac {PD}{PA}\cdot BC}$ .

[b][u]Particular cases[/u].[/b] $\{\begin{array}{cccccccc}
1\blacktriangleright & P:=G & \implies & \frac {MB}{AM} & + & \frac{NC}{AN} & = & 1\ .\\\\
2\blacktriangleright & P:=I & \implies & b\cdot \frac {MB}{AM} & + & c\cdot \frac{NC}{AN} & = & a\ .\end{array}\|$ [b]See[/b] [url=http://www.artofproblemsolving.com/blog/38873][color=darkred][b]here[/b][/color][/url].

[b][u]Remark[/u].[/b] $P\in XY$ means that the point $P$ belongs to the line $XY$
and  $[XY]$ is the notation of a segment. In upper proof $XY$ means $\overrightarrow{XY}$ .

lemma:L is the line through G.from A,B,C draw perpendicular to line L and the feet of prependiculars are X,Y,Z Respectively.so we have AX=BY+CZ
solution:
we have to prove AM(AN-NC)=AN(MB)and we know AX=BY+CZ .
AM/MB=AN/(AN-NC) and we have AM/MB=AX/BY,AN/(AN-NC)=AX/(AX-CZ)=AX/BY
so
AM/MB=AN/(AN-NC)


ind the last 1000 digits of the number
   N= $ {1\plus{}50\plus{}50^2\plus{}50^3\plus{}/dot\plus{}50^999}$

[quote="Mewto55555"][quote="sura33"]ind the last 1000 digits of the number
   N= $ {1 \plus{} 50 \plus{} 50^2 \plus{} 50^3 \plus{} /dot \plus{} 50^999}$[/quote]

Did you mean $ N \equal{} 1 \plus{} 50 \plus{} 50^2 \plus{} ... \plus{} 50^{999}$?

If so, we use some clever manipulations using the fact that its a geometric series.

$ N \equal{} 1 \plus{} 50 \plus{} 50^2 \plus{} ... \plus{} 50^{999}$

$ 50N \equal{} 50 \plus{} 50^2.... \plus{} 50^{1000}$

Subtracting,

$ 49N \equal{} 50^{1000} \minus{} 1$

$ N \equal{} \frac {50^{1000} \minus{} 1}{49}$

We only have to look at the last 1002 digits of the numerator.

Thus we have $ N \equal{} \frac {2500....00 \minus{} 1}{49} \equal{} \frac {2499999999999999...}{49}$ where there are 1000 9's on top.

I am stumped.[/quote]
[hide="hint"]50^1000 is not 25000...0000[/hide]

This is true, but if we are only worried about the last 1000 digits of N, we don't need to worry about more than the last 1002 digits of the numerator (because 10^1002/49>10^1000)

Oops.

Maybe you could try rewriting it as (100/2)^n, I'm not sure where that would lead though.

Anyways, I calculated out the first 8 or so digits, and they didn't seem to follow any pattern.

Maybe he meant to find the remainder $ \pmod{1000}$? If so, here it is:

[hide="My Solution"]
Using the formula for geometric series, we obtain

$ N \equal{} \frac {50^{1000} \minus{} 1}{49}\implies 49N \equal{} 50^{1000} \minus{} 1.$

Now we have $ \lambda(1000) \equal{} 100$ where $ \lambda(n)$ is the Carmichael function. So we have

$ 49N\equiv 50^{1000} \minus{} 1\equiv 0 \pmod{1000} \implies N \equiv \boxed{0} \pmod{1000}.$

Hmm. That doesn't seem right. I definitely made a mistake somewhere.
[/hide]

$ 50^1000 \equiv 0 \implies 50^1000\minus{}1 \equiv \minus{}1 (mod 1000)$

You forgot to subtract the 1  :wink:

[quote="sunehra"]Maybe he meant to find the remainder $ \pmod{1000}$? If so, here it is:
[/hide][/quote]

Hmm.. If it's $ \pmod{1000}$, we only have to compute the first 3 terms.


Let ABC be a triangle, M the midpoint of BC, C1 and B1 are the feet of the perpendipular on AB and  AC. H is the intersection of the perpendipular in the triangle AB (i forgot how its called in English). Let D be the intersection of the lines B1C1 and BC.
Prove that AM is perpendipular on DH.

It has been posted before as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110887]Triangle, perpendicular lines[/url], by [b][size=100]Yimin Ge[/size][/b].

Kostas Vittas.


geometry
circumcircle
area of a triangle
geometry proposed

A triangle is said te be harmonic if $a^2+b^2+c^2=4R^2$, ($R=$circumradius).
Show that $ABC$ is harmonic if and only if the circumcircle and the nine point circle are orthogonal.

The condition of orthogonality is equivalent to $ON^2=R^2+\frac{R^2}4\iff 4ON^2=5R^2$, where $R$ is the circumradius, and $O,N$ are the circumcenter and nine-point center respectively.

However, $4ON^2=OH^2$, and $OH^2=9R^2-(a^2+b^2+c^2)$. This shows that $4ON^2=5R^2$ is equivalent to $a^2+b^2+c^2=4R^2$, and we're done...

It easy to prove that [b]the area of a triangle $\triangle ABC$ is equal to that one of its orthic triangle iff $\triangle ABC$ is obtuse-angled and harmonic.[/b]


inequalities
vector
inequalities proposed

At the Internet site of Kvant journal

http://kvant.mccme.ru/1993/06/gif/93_06-31.gif

I have found an inequality (problem M1385-year 1993) with a three pages fascinating (I hope) solution.

I suspect it is very nice but I cannot understand it because I don't know Russian language.

I am asking for your help to translate it from Russian or to find it in the English version of Kvant journal.

Thank you very much for your help.

I'm just translating the problem:

[b]M1385.[/b] Let ABC be an arbitrary triangle. Show that

[b](a)[/b] for every equilateral triangle $A_1 B_1 C_1$, the inequality

$\displaystyle A_1 A^2 + B_1 B^2 + C_1 C^2 \geq \frac{AB^2 + BC^2 + CA^2}{6} - 2 \sqrt3 S_{ABC}$

holds;

[b](b)[/b] there exists a equilateral triangle $A_1 B_1 C_1$ for which equality holds in the above inequality.

  Darij

Darij,thank you very for your translation.

I think it is an interesting geometric inequality to solve by vector method.


You are true! Inductive is natural  :blush:


dar mosalas $ABC$ $L,M,N$ be tartib vasat haye $BC,AC,AB$ hastand .az noghteye delkhahe $P$  daroone mosalas amoodhaye $PD,PE,PF$ ra be tartib bar $BC,AC,AB$ mikeshim .
farz konid :
$(EF,MN)=X$
$(FD,LN)=Y$
$(DE,LM)=Z$

Sabet konid $DX,EY,FZ$ roye yeki az noghate barkhorde dayerehaye mohitie $\triangle LMN,\triangle DEF$ hamresand . :D  :D

[img]http://image59.webshots.com/459/3/60/4/2892360040097546774grBAaN_ph.jpg[/img]

hichki nemikhad ino hal kone.

kheili tooope.


inequalities trigonometry function integration calculus derivative limit

Prove that $ \sin x<x<\tan x$ for $ 0<x<\frac{\pi}{2}$

Do we have any simpler method rather than drawing a circle :?:

Since for $ x\equal{}0$ we have equality, we can just compare derivations, so we get:

$ \cos x < 1 < \frac{1}{\cos^2x}$

$ \cos^3 x < cos^2x < 1$

and this is true since $ |cos x| < 1$ on the interval $ \left[0,\frac{\pi}{2}\right]$.

Hm.. Don't know what you mean. I meant, if at the beginning all three functions are equal, then if the rate of change of first one is always smaller of the rate of change of the second that is always smaller that the rate of change of third, then the initial functions will satisfy the same inequality. If you like, I can say this:

Since we have:

$ \cos x < 1 < \frac {1}{\cos^2x}$

we have

$ \int^t_0\cos x\,dx < \int^t_0\,dx < \int^t_0 \frac {dx}{\cos^2x}$

$ \sin x < x < \tan x$

Would this do!?

Let $ f(x)\equal{}sinx\minus{}x$

$ f'(x)\equal{}cosx\minus{}1\equal{}\minus{}2sin^{2}\frac{x}{2}<0$ for $ 0<x<\frac{\pi}{2}$

$ f(x)$ is decreasing function on the interval $ 0<x<\frac{\pi}{2}$

i.e. for $ x>0$

$ f(x)<f(0)$

$ sinx\minus{}x<sin0\minus{}0$

$ sinx<x$

Similarly,take $ g(x)\equal{}tanx\minus{}x$

Is that the desired proof? It is not until that the inequality is proved, we understand the differentiation of $ \sin x$ is $ \cos x$, thus your proof is circlar reasoning.

The picture I attach here is the circle I've talking about.

It shows that $ \forall x\in\left(0,\frac{\pi}{2}\right),\sin x<x<\tan x$.

And also implies that $ \lim_{x\rightarrow 0}\frac{\sin x}{x}\equal{}1$ by using Squeezing Theorem. :D


What's Pascal's Triangle?

We discover Pascal's Triangle and its relationship to combinations.

hello, see here
http://mathworld.wolfram.com/PascalsTriangle.html
Sonnhard.

[quote="Waffle"]No - he knows what Pascal's triangle is. This is a place to discuss the Alcumus video. Or maybe I'm misinterpreting...?[/quote]

I think this forum is to ask questions, like if you don't understand why they did something in the video.  I'm pretty sure Mr. Rusczyk knows what Pascal's Triangle is  :P

Hi...

Excuseme. I go here just to say this triangle is Khayyam-Pascal's triangle and Khayyam is former than Pascal write this triangle (Omar Khayyam lived in 10 century ago and hi is a Iranian scientist and...). of course Pascal worked on this triangle and presentation some properties of this triangle. You can search "Omar Khayyam triangle" and see who is he and what is his triangle.
[url]http://en.wikipedia.org/wiki/Omar_Khayyam[/url]
You can see Khayyam in wikipedia and see Mathematician topic.
and you can search more...

By...


geometry
number theory unsolved
number theory

Is there an infinite number of pairs of distinct primitive pythagorean triples (a,b,c) & (x,y,z) such that ab=xy? (i.e. the triangles have the same area)


geometry
circumcircle
cyclic quadrilateral
projective geometry
geometry proposed

Let $ ABCD$ be cyclic quadrilateral with circumcircle $ (O)$.Let $ I;K$ be points such that $ \hat{IAD}=\hat{IBC}=\hat{KBA}=\hat{KCD}=90^{o}$.
Prove that the points $ I;K;O$ are collinear.

[color=purple][size=134][b]Very nice problem ![/b][/size][/color]
[quote="chien than"][color=darkred]Let $ A,B,C,D$ be four points of the circle $ (O)$. Define the points $ E,I,K$ for which $ \{\begin{array}{c}E\in AC\cap BD\\\ AI\perp AD\ ,\ BI\perp BC\\\ BK\perp BA\ ,\ CK\perp CD\end{array}$. Prove that $ \{I,K\}\subset OE$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the diameters $ [AA']$, $ [CC']$ and $ [DD']$ of the circle $ (O)$. Observe that $ \{\begin{array}{c}O\in AA'\cap CC'\cap DD'\\\ I\in AD'\cap BC'\\\ K\in BA'\cap CD'\end{array}$.

Apply the [b]Pascal's theorem[/b] to the hexagons $ \{\begin{array}{cccccc}1\blacktriangleright & \boxed{ACC'BDD'}& \mathrm{\ \ \ : \ \ \ }& \{\begin{array}{c}E\in AC\cap BD\\\ O\in CC'\cap DD'\\\ I\in C'B\cap D'A\end{array}\| & \implies & I\in OE\\\\ 2\blacktriangleright & \boxed{ACD'DBA'}& \mathrm{\ \ \ : \ \ \ }& \{\begin{array}{c}E\in AC\cap DB\\\ K\in CD'\cap BA'\\\ O\in D'D\cap A'A\end{array}\| & \implies & K\in OE\end{array}\|$ $ \implies$ $ \{I,K\}\subset OE$.[/color]


calculus
integration
limit
trigonometry
function
real analysis
real analysis solved

Compute

 \lim   (n.cos(n). \int _[0;1]y^n.cos(ny)dy + n.sin(n). \int _[0;1]y^n.sin(ny)dy )

I bet the following is the greatest stupidity of all times:
   Using that cosn(x-1)=cosncosnx+sinn*sin nx, we find that the expression is lim n*integral(0 to 1) (x^n*cos[n(1-x)])dx. Make t=1-x. We have to find lim n*integral (0 to 1) (1-t)^n*cos ntdt.Take a small eps>0. Take t=1-x^1/n. We have that n*integral (eps to 1) (1-t)^n*cos ntdt= integral ( eps to 1) x^1/n*cos(n(1-x^1/n))dx, which tends to integral (eps to 1) cos(ln x)dx when n tends to infinity ( x^1/n tends to 1 and n(1-x^/n) tends to lnx. So, lim n*integral (eps to 1) (1-t)^ncosntdt tends to integral (eps to 1) cos(lnx) dx for any eps>0. So, the limit is integral (0 to 1) cos(lnx)dx. But by integration by parts we find that integral (a to b) cos(lnx))dx= bcos(lnb)+bsin(lnb)-acos(lna)-asin(lna). Take a to tend to 0 and b=1 and I think you will obtain 1.

A = \int _[0;1]cos(lnx)dx change variable lnx=y, x=e^y
A=  \int _]-oo;0][cosy.e^ydy again y = -z
A=  \int _[0;+oo(cosz.e^(-z)dz = 1/2
The limit is 1/2

Here  my solution
With trigonometry formulae

a(n)=  n.\int _[0;1]y^n.cos(n(y-1))dy change variable y= -z/n + 1

a(n) =  \int _[0;n] (1 - z/n)^n.cosz dz

a(n) =  \int _[0;+oo[ Fn(z) dz where

Fn(z) = (1-z/n)^n.cosz.C_[0;n](z)

with C_[0,n](z) is 1 if z \in [0,n], 0 if z not in [0,n], C_[0,n] is the caracteristic function on [0,n]

Fn converge simply to e^(-z).cosz
|Fn(z)|  \leq e^(-z) integrable on R+

Henri Lebesgue dominated convergence theorem gives

lim a(n) = \int _[0;+oo[  \lim Fn(z)dz= \int _[0;+oo(cosz.e^(-z)dz = 1/2


Let $x_1, x_2 ..., x_{n+1}$ be positive real numbers such that $x_1+x_2 + ... + x_n = x_{n+1}$ Prove that:

\[\displaystyle\sum_{i=1}^{n} \sqrt{x_i(x_{n+1} - x_i)} \le \sqrt{\sum_{i=1}^{n} x_{n+1}(x_{n+1} - x_i)}\]

i got somewhere with this problem.... but couldnt seem to finish it off

use cauchy in pretty much the only way possible?

Actually, just use Cauchy like this: $\sqrt{a_1}+\sqrt{a+_2}+...+\sqrt{a_n}\leq\sqrt{n(a_1+a_2+...+a_n)}$ ;)

 Then by some calculations you will get an expected inequality and then apply Cauchy again.

 P.S. I remembered that I solved this problem in the 9-th grade while preparing for Olympiad. ;)

Cauchy is the way to go.

When is it ever not? :D

oh wow i didnt think about using cauchy cuz this problem was in kiran kedlaya power means section -_____-

but yea thanks for the help

From Cauchy, we have

$(a_1^2+\ldots+a_n^2)(1+\ldots+1)\geq (a_1+\ldots+a_n)^2$ which is just

$\frac {a_1^2+\ldots+a_n^2}{n}\geq \left(\frac {a_1+\ldots+a_n}{n}\right)^2$

which you should recognize as exactly the RMS-AM inequality.  So it is a power-means question, in a sense, although the power-mean inequality in the general case is far stronger than Cauchy.


calculus
integration
trigonometry
analytic geometry
calculus computations

Evaluate $ \int \int_D x^2e^{ \minus{} x^2 \minus{} y^2}dxdy,\ \ \ D \equal{} \{(x,\ y)\in\mathbb{R}^2|0\leq y\leq x\}$.

Substitute $ x\equal{}r\cos\theta, y\equal{}r\sin\theta$ and use the change of variables theorem.  Use Fubini, substitute $ u\equal{}r^2,v\equal{}\minus{}\frac{e^{\minus{}r^2}}{2}$ and use integration by parts.  I believe, if I have not made any computational mistakes, that the answer should be $ \frac{\pi\plus{}2}{32}$.  I have to go to school now.

I forgot to multiply an integral by 2 when evaluating $ I_2$, but that is the solution I had.

[quote="hjbrasch"]Change of variables (polar coordinates) yields $ I: \equal{} \int\int_Dx^2e^{ \minus{} (x^2 \plus{} y^2)}dydx \equal{} \int_0^{\frac {\pi}{4}}\cos^2\varphi d\varphi\cdot\int_0^{\infty}R^3e^{ \minus{} R^2}dR$. For the evaluation of the two integrals occurring as factors we perform an appropriate product integration, then

$ I_1: \equal{} \int_0^{\frac {\pi}{4}}\cos^2\varphi d\varphi \equal{} \sin\frac {\pi}{4}\cos\frac {\pi}{4} \plus{} \int_0^{\frac {\pi}{4}}\sin^2\varphi d\varphi \equal{} \frac {1}{2} \plus{} \frac {\pi}{4} \minus{} \int_0^{\frac {\pi}{4}}\cos^2\varphi d\varphi$ which implies $ I_1 \equal{} \frac {2 \plus{} \pi}{8}$ and

$ I_2: \equal{} \int_0^{\infty}R^3e^{ \minus{} R^2}dR \equal{} \int_0^{\infty}Re^{ \minus{} R^2} \equal{} \frac {1}{2}$

and putting all together we have $ I \equal{} I_1I_2 \equal{} \frac {2 \plus{} \pi}{16}$[/quote]

That's right. :lol:


Consider all functions $f:\mathbb{N}\to\mathbb{N}$ satisfying $f(t^2 f(s)) = s(f(t))^2$ for all $s$ and $t$ in $N$.  Determine the least possible value of $f(1998)$.

http://www.kalva.demon.co.uk/imo/isoln/isoln986.html
Let f(1) = k. Then f$ (kt^2) = f(t)^2$ and $ f(f(t)) = k^2t$. Also $ f(kt)^2 = 1.f(kt)^2 = f(k^3t^2) = f(1^2f(f(kt^2))) = k^2f(kt^2) = k^2f(t)^2$. Hence $ f(kt) = k f(t)$.

By an easy induction $ k^nf(t^{n + 1}) = f(t)^{n + 1}$. But this implies that k divides f(t). For suppose the highest power of a prime p dividing k is a > b, the highest power of p dividing f(t). Then a > b(1 + 1/n) for some integer n. But then na > (n + 1)b, so $ k^$n does not divide $ f(t)^{n + 1}$. Contradiction.

Let $ g(t) = f(t)/k$. Then $ f(t^2f(s)) = f(t^2kg(s)) = k f(t^2g(s) = k^2g(t^2g(s))$, whilst s $ f(t)^2 = k^2s f(t)^2$. So $ g(t^2g(s)) = s g(t)^2$. Hence g is also a function satisfying the conditions which evidently has smaller values than f (for k > 1). It also satisfies g(1) = 1. Since we want the smallest possible value of f(1998) we may restrict attention to functions f satisfying f(1) = 1.

Thus we have f(f(t) = t and f(t^2) = f(t)^2. Hence f(st)^2 = f(s^2t^2) = f(s^2f(f(t^2))) = f(s)2f(t^2) = f(s)^2f(t)^2. So f(st) = f(s) f(t).

Suppose p is a prime and f(p) = m·n. Then f(m)f(n) = f(mn) = f(f(p)) = p, so one of f(m), f(n) = 1. But if f(m) = 1, then m = f(f(m)) = f(1) = 1. So f(p) is prime. If f(p) = q, then f(q) = p.

Now we may define f arbitarily on the primes subject only to the conditions that each f(prime) is prime and that if f(p) = q, then f(q) = p. For suppose that $ s = p_1^{a_1}...p_r^{a_r}$ and that $ f(p_i) = q_i$. If t has any additional prime factors not included in the $ q_i$, then we may add additional p's to the expression for s so that they are included (taking the additional a's to be zero). So suppose $ t = q_1^{b_1}...q_r^{b_r}$.Then $ t^2f(s) = q_1^{2b_1 + a_1} ...q_r^{2b_r + a_r}$ and hence $ f(t^2f(s) = p_1^{2b_1 + a_1} ...p_r^{2b_r + a_r} = s f(t)^2$.

We want the minimum possible value of f(1998). Now $ 1998 = 2.3^3.37$, so we achieve the minimum value by taking f(2) = 3, f(3) = 2, f(37) = 5 (and f(37) = 5). This gives $ f(1998) = 3.2^35 = 120$


inequalities
vector
abstract algebra
linear algebra
linear algebra unsolved

Notation: $M_{m \times n }(K) =$ set of all m x n matrices in K.

1. Let $A,B\in M_{n \times n}(\mathbb{R})$, $AB=0$. Prove that $rank A+rank B \leq n$.

2. Let $A\in M_{m\times n}(K)$, $rank A=n$. Prove that set
$W_{A}= \{ B\in M_{n \times m}(K) | BA = I_{n}\}$ is coset in $M_{n\times m}(K)$. What is dimension of $W_{A}$?

1.By the inequality of  [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sylvester&t=43691]Sylvester[/url] $rank(A)+rank(B)\leq rank(AB)+n$ .

Yes, I know it that way, but it is from Linear Algebra exam for computer scientists, so I wonder if there is any elementary way to do it.

for 1, think of $A$ and $B$ as linear transformations on an $n$-dimensional vector space.  Then $AB=0$ means that the image of $B$ is contained in the kernel of $A$.

Thus the dimension of the image of $B$ is less than or equal to the dimension of the kernel of $A$.

Hence by the Rank-Nullity Theorem, $rank(A)\le n-rank(B)$.


After looking at the solution, how do you know that TD and TE are congruent?

http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10B_Problems/Problem_20

Please keep all discussion in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1376256#1376256]original thread.[/url] Merged and locked.


calculus
integration
limit
real analysis
calculus computations

For $ a>1$, find $ \lim_{n\to\infty} \int_0^a \frac{e^x}{1\plus{}x^n}dx.$

Hi :) !!!

LET $ a > 1$

than: $ \int_0^a \frac {e^x}{1 \plus{} x^n}dx \equal{} \int_0^1 \frac {e^x}{1 \plus{} x^n}dx \plus{} \int_1^a \frac {e^x}{1 \plus{} x^n}dx$

let $ g_n(x) \equal{} \frac {e^x}{1 \plus{} x^n}$ it's oubvious that $ g_n$ are continue in $ [0;a]$ than measurables and:

$ \forall x \in [0;1] : |g_n(x)|\leq e^x$ and $ \lim_{n \to \plus{} \infty} g_n(x) \equal{} e^x$.

$ \forall x \in [1;a] : |g_n(x)| \leq \frac {e^x}{2}$ and $ \lim_{n \to \plus{} \infty} g_n(x) \equal{} 0$

than th. of Lebesgue give that:

$ \lim_{n \to \plus{} \infty}\int_0^a \frac {e^x}{1 \plus{} x^n}dx \equal{} \int_0^1e^xdx \equal{} e \minus{} 1$

and thanks :)

thx,and is the answer above right  ,  kunny???

Well, I have given fully worked out part with the help of DCT, the idea is same as that of Mathema .

Given $ \int_{0}^{a} \frac {e^{x}}{x^{n} \plus{} 1}.dx, a > 0$

Well consider,

$ \frac {e^{x}}{x^{n} \plus{} 1} \equal{} e^{x}(1 \plus{} x^{n})^{ \minus{} 1}, x \in [0,1]$

$ \equal{} (1 \plus{} x \plus{} \frac {x^{2}}{2!} \plus{} \frac {x^{3}}{3!} \plus{} ....)(1 \minus{} x^{n} \plus{} x^{2n} \minus{} ....)$

$ \equal{} (e^{x} \minus{} x^{n}(e^{x}) \plus{} x^{2n}(e^{x}) \minus{} ...)$

$ \lim_{n \to \infty}e^{x}(1 \minus{} x^{n} \plus{} x^{2n} \minus{} ...) \equal{} e^{x}, x \in [0,1]$

  $ \lim_{n \to \infty}\frac {e^x}{x^{n} \plus{} 1} \equal{} e^{x}$

$ \lim_{n\to \plus{} \infty}\int_{0}^{1}e^{x}dx \equal{} e \minus{} 1$

For $ x \in [1,a] , \lim_{n \to \infty}\frac {e^{x}}{x^{n} \plus{} 1} \equal{} 0$

$ \lim_{n\to\infty}\int_{0}^{a}\frac {e^{x}}{1 \plus{} x^{n}}dx. \equal{} e \minus{} 1$


Prove that for all $ n\geq 3$, any integer between 1 and $ n!\minus{}1$, exclusive, is the sum of distinct factors of $ n!$.

[hide="Hint"]Induct. And be careful.[/hide]

[quote="solafidefarms"]Prove that for all $ n\geq 3$, any integer between 1 and $ n! \minus{} 1$, exclusive, is the sum of distinct factors of $ n!$.

[hide="Hint"]Induct. And be careful.[/hide][/quote]
Is there a typo in the problem?


ratio
algorithm
combinatorics unsolved
combinatorics

A graph, each edge is associated with two numbers a and b, now given a start vertex s, find a path with highest sum(a)/sum(b) that starts and ends both at s. Note that each edge can appear in the path only once.

Ok, I think what you are asking for is an algorithm....And I tend to believe that in the general case, this problem is NP-complete....Maybe the graph has some certain properties???Complete? However, I am not sure that would help..... : :huh:

Is there an existing graph theory problem similar to this one?
If so, maybe we would get lucky to find the solution.

Hi!
I would kindly ask you, if possible, to tell us the source of the problem, and if possible, the restrictions on the graph. That would be the maximum number of vertices and/or edges.....I think that would be of help. Although it could be NP-complete, it could still have a solution better than finding all cycles.....

The source of this problem is.
A person wants to select a route to biking with the maximum efficiency, i.e., the maximum elevation gain/distance.
Each road segment is identified by a start point, end point, elevation gain and distance.
All elevation gains are positive. A descent elevation has a gain of 0.
The person starts and comes back at the same place and no segment(in either direction) in his route can be used twice.
The algorithm should be take care of a few thousand road segments.


Ο Ράλλης μου ζήτησε να βάλω τα θέματα της εξωγήινης (  :huh:  ) ολυμπιάδας λίγα λίγα και δεν μπορούσα ν'αρνηθώ  :mad:  (για διαφόρους λόγους).

Για αρχή λοιπόν να τα δύο πρώτα:

1)Λύστε στους ακεραίους την εξίσωση  $xy(x^{2}+y^{2})=2z^{4}$ .

2)Λύστε στους ακεραίους την εξίσωση $x^{n}+y^{n}= 1994$ με $n\geq 2$.

Υ.Σ. Λύσεις γίνονται δεκτές μέχρι και 31 Ιουλίου ,οπότε και θα δημοσιευθούν άλλα προβλήματα για λύση.
Οι νικητές θα παραλάβετε τα βραβεία από τον [color=green]RAKAR[/color] (για πληροφορίες στείλτε στο e-mail του)  !  :rotfl:

πολυ καλα Στελιο και Ραλλη.οι ασκησεις ειναι πολυ ωραιες.τι βραβειο θα παρουμε?  :lol:  :D

Ego pantow den dino tipota.......
Stelio esi an hes dose kati....

RALLIS :D

Λοιπόν για την πρώτη :
Υποθέτουμε οτι ΜΚΔ(x,y)=1 (giati?)  και στη συνέχεια παίρνουμε περιπτώσεις στις οποίες στην πρώτη ο x είναι άρτιος και ο y περιττός και στην άλλη και οι δύο .Στη συνέχεια προκύπτουν κάποιες εξισώσεις οι οποίες δεν έχουν λύση και τελικά η μόνη λύση είναι η (x,y,z)=(1,1,1)

Gia τη δεύτερη αν x,y >0 και $x\geq 1$ βρίσκουμε ότι $n\leq 11$ και στη συνέχεια εξετάζουμε απλά περιπτώσεις για $n\leq 11$ χωρίς να έχουμε πολλές δοκιμές αφού τα x,y φράσονται

Μαλλον δεν πολυδουλευτηκε η πρωτη...
Παραθετω μια πολυ κομψη λυση (οχι δικια μου).......
Η σχεση $xy(x^{2}+y^{2})=2z^{4}$ γινεται

$(x-y)^{4}+(2z)^{4}=(x+y)^{4}$ οποτε απο την αποδειξη του Wiles του τελευταιου θεωρηματος του Fermat παιρνουμε $x=y$με λυση την$(m,m,m)$και την$(m,m,-m)$ειτε$z=0$και βγαινουν οι λυσεις $(0,m,0)$ και$(m,0,0)$
Aπο το βιβλιο 360 problems for mathematical contests

ΡΑΛΛΗΣ :D

Ράλλη αν δεν κάνω λάθος είναι $(x+y)^{4}-(x-y)^{4}=(2z)^{4}$   :!:  αλλά πάλι το ίδια λογική ακολουθείται οπότε πολύ ωραία λύση..........
Βάλε άλλες δύο τώρα....... :P

Oντως εχεις δικιο,τωρα ομως το διορθωσα....
Για το δευτερο καμια ιδεα?? Δεν ειναι δυσκολο απλως θελει λιγη δουλιτσα...
Αντε λιγη βοηθεια για εκκινηση...(n,x,y)=(2,+-25,+-37) ειναι λυση....Για n>2 ομως??

ΡΑΛΛΗΣ :D

Αν είδες παραπάνω είχα γράψει λίγα πράγματα για την δεύτερη .
Τέλος πάντων βάζω τη λύση ....
προφανώς αν $n \geq 2$, τότε $x,y$ περιττοί .

Λόγω συμμετρίας υποθέτουμε $x > y$ ($x > 0$)

Αν $n$ άρτιος , χρειάζεται μόνο να εξετάσουμε την περίπτωση  $n=2$ όπου
$x^{2}+y^{2}= 1994$,
Τότε  $32 \leq x \leq 44$, και εξετάζοντας παίρνουμε
$(x,y,n) ={(-37,25,2), (-37,-25,2), (-25,37,2), (-25,-37,2), (25,37,2), (25,-37,2), (37,25,2), (37,-25,2)}$

Αν $n$ περιττός, $(n \geq 3)$.
τότε είτε $x+y = 2$ ή $x+y = 1994$  ή $x+y=997$
Aν $x,y > 0$, σε αυτή την περίπτωση δεν έχουμε λύση ,επειδή e $x^{n}+y^{n}\geq x^{3}+y^{3}\geq (x+y)^{2}$, άρα ο $x+y$ μπορεί να είναι μόνο $2$, αλλά τότε
 $x = y = 1$ άτοπο. .

Τώρα αν $y < 0, n \geq 3$ περριτός
έστω  $k =-y$, και θεωρούμε πλεον την εξίσωση
 $x^{n}-k^{n}= 1994, k > 0, x \geq k$    $(1)$
Για  $y =-1$ δεν έχουμε λύση , άρα $k \geq 2$
και προφανώς  $x^{n}-k^{n}\geq (x-k)^{2}$
επειδή $x,y,k$ περριτοί θα είναι $x-k = 2$

Αντικαθιστώντας $x = k+2$ στην  $(1)$  παίρνουμε
$2k^{n-1}+4k^{n-2}+\cdots+2^{n}= 1994$
$k^{n-1}+2k^{n-2}+\cdots+2^{n-1}= 997$   $(2)$
$997 = LHS \geq 2^{n-1}\cdot n$, που σημαίνει  $n \leq 7$
$997 = LHS \leq 2^{n-1}\cdot n$,που σημαίνει $k \geq 9$   $(3)$

i) $n = 3$
Από $(2), k^{2}+2k+4 = 997$ δεν έχουμε λύση.

ii) $n = 5, n = 7$
Από $(2), 997 = LHS > k^{n-1}\geq k^{4}$.
Άρα , $k \leq 5$. αλλά αυτό έρχεται σε άτοπο με την  $(3)$
Άρα δεν υπάρχουν άλλες λύσεις

Υ.Γ. Πάνε τα χεράκια μου....... :)

κάτσε ρε φίλε
οκ με την λύση σου στο 1ο όμως για να χρησιομοποιήσεις το λήημα του wiles θέλει απόδιξη 10000 σελίδων μέσες άκρεσ :rotfl:  :rotfl:  :rotfl:  οπότε πές έτσι αυθαίρετα επειδή κάποιος έγκυρος το είπε

Για το α ερωτημα το ιδιο εκανα και γω.Την ελυσα θετωντας x+y=a kai x-y=b οποτε κετεληξα στην εξισωση a^4=b^4+(2z)^4 που συμφωνα με τον Wiles η παραπανω εξισωση δεν εχει λυση στους θετικους ακεραιους.

ρε παιδια η περιπτωση που εχουμε την a^4+b^4=c^4  αποδεικνυεται και ευκολα με ενα λημμα 2 σελιδων που αρκετοι απο εδω (το ελληνικο) μπορουν να αποδειξουν.συγκεκριμενα οτι η α^4+b^4=x^2 δεν λυνεται. οποιος θελει και μπορει ας δωσει την δικη του λυση και αρα του θ, του willes για 4ν

Χάρις την ιδέα του Κώστα!  :wink:

[color=darkblue][size=150][b]Λήμμα:[/b][/size][/color] Δεν υπάρχουν $a,b,c \in \mathbb Z^{*}$ που να είναι λύσεις της εξίσωσης $a^{4}+b^{4}=c^{2}$.

[color=darkblue][b]Απόδειξη:[/b][/color]
Έστω ότι υπάρχουν $a,b,c \in \mathbb Z^{*}$ τέτοια ώστε $a^{4}+b^{4}=c^{2}$.
Χωρίς βλάβη της γενικότητας υποθέτουμε ότι οι αριθμοί $a^{2},b^{2},c$ είναι πρώτοι μεταξύ τους,οπότε
επειδή αποτελούν πυθαγόρεια τριάδα θα υπάρχουν $x,y$ τέτοια ώστε:
$a^{2}=2xy$
$b^{2}=x^{2}-y^{2}$
$c=x^{2}+y^{2}$ .
Έτσι προκύπτει άλλη μια πυθαγόρεια τριάδα,αφού $b^{2}+y^{2}=x^{2}$ ,άρα υπάρχουν $k,l$ με $(k,l)=1$ τέτοια ώστε:
$y=2kl$
$b=k^{2}-l^{2}$
$x=k^{2}+l^{2}$ .
Οπότε ισχύει ότι $a^{2}=2xy=4kl(k^{2}+l^{2})$ . Όμως $(k,l)=1$ άρα και $(kl,k^{2}+l^{2})=1$ ,συνεπώς οι αριθμοί $kl$ και $k^{2}+l^{2}$ είναι τέλεια τετράγωνα.
Έστω λοιπόν ότι $k^{2}+l^{2}=p^{2}$. Αυτό όμως οδηγεί στην ύπαρξη μίας ακόμη πυθαγόρειας τριάδας.
Επειδή τώρα ισχύει $p^{2}=x<c\leq c^{2}$ ,η ίδια διαδικασία μας οδηγεί σε συνεχώς "μικρότερες" πυθαγόρειες τριάδες (άτοπο).
Συνεπώς δεν υπάρχουν $a,b,c \in \mathbb Z^{*}$ που να είναι λύσεις της εξίσωσης $a^{4}+b^{4}=c^{2}$ .

[color=darkblue][size=150][b]Θεώρημα:[/b][/size] [/color]Δεν υπάρχουν $x,y,z \in \mathbb Z^{*}$ τέτοια ώστε $x^{4n}+y^{4n}=z^{4n}$ .

[color=darkblue][b]Απόδειξη:[/b][/color]
Έστω ότι υπάρχουν $x,y,z \in \mathbb Z^{*}$ τέτοια ώστε $x^{4n}+y^{4n}=z^{4n}$ ,τότε αυτόματα προκύπτει μια λύση στην εξίσωση
$a^{4}+b^{4}=c^{2}$ για
$a=x^{n}$
$b=y^{n}$ και
$c=z^{2n}$ (άτοπο λόγω του λήμματος).
Άρα ισχύει το θεώρημα.

Υ.Γ. Εφαρμόζοντας το παραπάνω θεώρημα για $n=1$ εύκολα λύνεται η πρώτη άσκηση της Εξωγήινης( :huh: ) Μαθηματικής Ολυμπιάδας (ΕΜΟ) χωρίς τη χρήση του τελευταίου θεωρήματος του Fermat.

Stelios...

Μπραβο Στελιο, ωραια αποδειξη. :)

nice!!! se prolabe omos allos!  :wink:
abros mallon den katalaba kala... gia na xreisimopoiisoume ena theorima prepei na parathetoume (ara kai na jeroume) tin apodeiji kathe fora?

Όταν λες με πρόλαβε άλλος ,εννοείς τον Fermat ή τον Wiles;;;;  :P  :wink:

Θανο για αυτο που ρωτας προφανως καθε βοηθητικη προταση-λημμα που δε θεωρηται γνωστη(πχ οτι καθε ακεραιος ισουται με το γινομενο πρωτων αριθμων ειναι γνωστο) προφανως την αποδεικνυεις,αλλα αυτο πρεπει να το πουν και  παιδια που εχουν παει σε ΒΜΟ,ΔΜΟ και εχουν πειρα

και στους 2 :lol:
Το ΤΘΦ δε θεωρείται γνωστό?????

τι να σου πω.οχι απλα ειναι γνωστο αλλα το ξερουν και αλλοι εκτος μαθηματικων.αλλα δεν ειναι πολυ trendy χωρις αποδειξη  :rotfl:

φίλε κώστα υπερβάλλεις λίγο με την λέξη trendy στα μαθηματικά δεν ισχύει το τι ειναι trendy αλλά το τι είναι ΣΩΣΤΟ ΜΑΘΗΜΑΤΙΚΑ κι μόνο αυτό
χαίρεται

φαντασου οτι στην αρχη ειχα γραψει glamour, :D  εχεις δικιο αστειο ηταν.

κώστα σου εφιστώ την προσοχή στο θέμα του ΣΤΑΡΤ είναι πολύ επικύνδυνα τοποθυετημένος οσον αφορά την κριτική κι πανεύκολά καποιος θα μπει στον κόπο να ξεκατινιαστεί όπως εγω με τον ανώνυμο αν θυμάσαι εγω δεν εχω προβλημα να το κάνω απλώς για σένα το λέω επειδη εισαι φίλος μου

Παιδια δεν ξερω αν συμφωνησεται  αλλα η δικη μου αποψη ειναι να προσπαθουμε να μαθαινουμε αποδειξεις των θεωρηματων που με καμποση προσπαθεια να ειναι εφικτο(δεν αναφερομαι μονο στο θεωρημα του Fermat η Wiles)Εξαλλου μην ξεχνατε οτι πολυ καλες τεχνικες και γνωσεις υψηλου επιπεδου χρησιμ.για την αποδειξη καποιου ''καλου''θεωρηματος.Πολλες εξαλλου ασκησεις κατασκευαζονται απο τεχνασματα και γνωσεις που περιεχονται σε ενα θεωρημα. :wink: Οσον αφορα για την συζητηση  σας θα επιμεινω σε αυτο που γραφω ακριβως απο κατω. :wink:

Δεν είπα ότι δεν είναι καλό, πόσο μάλλον περιττό, να μαθαίνουμε την απόδειξη αλλά απλώς αναρρωτιώμουν αν το ΤΘΦ δεν είανι αρκούντως διάσημο ώστε να χρησιμοποιούται χωρίς απόδειξη. Δεν είχα αμφιβολία μέχρι τώρα αλλά η μάλλον έντονη παρέμβαση (έως επίπληξη) του ambros προκάλεσε την απορία μου. Θα με ενδιέφερε να ακούσω και τη δική του τοποθέτηση αν φυσικά το επιθυμεί. Παρόλο που το παρακάτω δεν είναι εντός θέματος του συγκεκριμένου topic θα ήθελα να μάθω, μιας και το ανέφερε ο ambros, η συζήτηση αυτή έχει διαγραφεί ή έγινε μέσω pm? Γιατί δεν τη βρίσκω στο forum...

γεια χαρα Θανο! για ποια συζυτηση λες?

Φίλε μου πολύ το παρεξήγησες καθότι δεν είναι επίπληξη αυτό που ειπα ελίναι μια ξεκάθαρη κι χωρίς αμφιβολία τοποθέτηση πάνω στο θέμα τώρα για τρην τοποθέτηση που ζητάς η γνώμη μου ειναι ότι μερικά θεωρήματα (γνωστά η δυσαπόδεικτα)δεν χρειάζονται απόδειξη όπως ΤΘΦ η θεώρημα χορδής κι εφαπτομένης όσο για την συζητητση μιλάω για υτήν που α΄νοιξε ο στάρτ κι είναι επικυνδυνη
Σε ευχαριστώ  κι είμαι πάντα ανοιχτός κι στην διάθεση όλων για το κάθε τι που μας αφορά
Χαιρετώ

εννούσα την παλιά συζήτηση στην οποία αναφέρθηκε ο abros Κώστα, αυτή με τον ανώνυμο


inequalities
linear algebra
matrix
calculus
derivative

Solution.
Let $f(a,b,c)=\sqrt{\frac{a}{b+c}}$. Consider the Hessian Matrix
\[H(f) = \begin{bmatrix}\frac{\partial^{2}f}{\partial a^{2}}& \frac{\partial^{2}f}{\partial a\,\partial b}& \frac{\partial^{2}f}{\partial a\,\partial c}\\ \\ \frac{\partial^{2}f}{\partial b\,\partial a}& \frac{\partial^{2}f}{\partial b^{2}}& \frac{\partial^{2}f}{\partial b\,\partial c}\\ \\ \frac{\partial^{2}f}{\partial c\,\partial a}& \frac{\partial^{2}f}{\partial c\,\partial b}& \frac{\partial^{2}f}{\partial c^{2}}\end{bmatrix}\]
Since $H(f)$ is negative definite (I believe the product of the eigenvalues is negative) for $a,b,c\in (0,1)$, $f$ is concave.

By Jensen's inequality (multivariable),

\[\sum_{cyc}cf(a,b,c)\leq (a+b+c)f\left(ab+bc+ac,ab+bc+ac,ab+bc+ac\right)=\frac{1}{\sqrt{2}},\]

as required.$\blacksquare$

Masoud Zargar

Are you allowed to cite multivariable jensen?

[quote="boxedexe"]What do you meant by allowed?  In an Olympiad?  Most likely, yes.[/quote]

Yes, like use it without a proof on an olympiad. I was wondering because it seems to trivialize some inequalities with only one application.

There's a catch, though: the [u]exact[/u] criterion of the convexity of a multiargument function. Is the positivity of second partial derivatives [i]necessary and sufficient[/i]? Calculus experts should be consulted.

Well, I'm not a Calculus expert...almost never studied it fully, but you could consider the Hessian Matrix.:P By the way, I think that the positivity of the second partial derivatives is not sufficient, since I believe that Mathematics is usually not a tautology.

Whew. I think I have a solution...

[hide]
Cauchy!
\[\left( \sum_\text{cyc}\frac{xy}{\sqrt{xy+yz}}\right)^{2}\leq ([xy+xz]+[yz+xy]+[zx+yz])\left( \sum_\text{cyc}\frac{x^{2}y^{2}}{(xy+yz)(xy+xz)}\right)\]
We need to show that the right hand side is less than or equal to $\frac{1}{2}(x+y+z)^{2}$. We make the substitution
\[s=x+y+z,\ t=xy+yz+zx,\ u=xyz\]
The right hand side is equal to
\begin{eqnarray*}2t\sum_\text{cyc}\frac{xy}{(s-x)(s-y)}&=&2t\left( \frac{xy(s-z)+yz(s-x)+zx(s-y)}{(s-x)(s-y)(s-z)}\right) \\ &=&\frac{2t(st-3u)}{st-u}\\ \end{eqnarray*}
So our inequality is equivalent to
\[s^{3}t-4st^{2}-s^{2}u+12tu\geq 0\]
We look to Schur, and obtain the identity
\[s^{3}-4t+9u=\sum_\text{cyc}x(x-y)(x-z)\]
by setting $x^{3}$ coefficients equal and noting that zeros occur at $x=y=1,\ z=0$ and $x=y=z=1$. So we rewrite as
\[t(s^{3}-4st+9u)-u(s^{2}-3t)\]
AAAAAHHHH!!! THAT'S A MINUS SIGN!!! WE NEED INSANE SUM OF SQUARES!!!!
The above expression is one-half of
\[\sum_\text{cyc}[(xy+yz+zx)(x+y-z)-xyz](x-y)^{2}\]
\[=\sum_\text{cyc}(x^{2}y+x^{2}z+y^{2}x+y^{2}z-z^{2}x-z^{2}y)(x-y)^{2}\]
We need to show that this expression is nonnegative. WLOG let $x\geq y\geq z$ because this expression is symmetric. Then only the $(y-z)^{2}$ term has a chance of being negative, since the $x^{2}(y+z)$ keeps the other weights nonnegative. So we must [b]eat[/b] this troublesome term, and the $(x-z)^{2}$ term is very hungry:
\[(x^{2}y+x^{2}z+y^{2}x+y^{2}z-z^{2}x-z^{2}y)(x-y)^{2}\]
\[+2(xz^{2}+yz^{2})(y-z)^{2}\]
\[+(x^{2}y+x^{2}z+z^{2}x+z^{2}y-y^{2}x-y^{2}z)(x-y)(x+y-2z)\]
[b]YUMMMMM!!!!!!!![/b]  :icecream:
Every line above is nonnegative, so we're done.
[/hide]

[quote=pohoatza]For pozitive numbers with $x+y+z=1$ we have the following ineq:

$\sum_{cyc}{\frac{xy}{\sqrt{xy+yz}}\leq \frac{1}{\sqrt{2}}}$

[hide="solution with jensen"]first we see that for the ponders $x,y,z$ and $f(a)=\sqrt{\frac{a}{s-a}}$ the inequality is to strong, so we try to find stronger ponders, in order to apply jensen:

so we rewrite :

$\sum_{cyc}{\frac{xy}{\sqrt{xy+yz}}= \sum_{cyc}{(\frac{x+y}{2}\cdot\sqrt{\frac{4x^{2}y}{(x+y)^{2}(x+z)})}}}$

consider now $f(a)=\sqrt{a}$ which is concave, and apply jensen :
$\sum_{cyc}{(\frac{x+y}{2})\cdot f(\frac{4x^{2}y}{(x+y)^{2}(x+z)})}\leq f(\sum_{cyc}{\frac{2x^{2}y}{(x+y)^{2}(x+z)}})$ , obvious the ponders are : $\frac{x+y}{2}$, $\frac{y+z}{2}$, $\frac{z+x}{2}$

$\Longleftrightarrow LHS \leq f(\sum_{cyc}{\frac{2x^{2}y}{(x+y)^{2}(x+z)}})$ , where $LHS =\sum_{cyc}{\frac{xy}{\sqrt{xy+yz}}}$

so it remains us to prove that :

$\sqrt{(\sum_{cyc}{\frac{2x^{2}y}{(x+y)^{2}(x+z)}})}\leq \frac{1}{\sqrt{2}}$

$\Longleftrightarrow \sum_{cyc}{\frac{2x^{2}y}{(x+y)^{2}(x+z)}}\leq \frac{1}{2}$ , which is true by expanding : $\Longleftrightarrow \sum{x^{3}y}+\sum{xy^{3}}\geq 2\sum{x^{2}y^{2}}$

so we are done.[/hide][/quote]

The Jensen wasn't written in right way


every squre of a $m\times n$ grid is colored either black or white, with equal probability each.  Find the probability that every row and column has at least one black square.

uhh yea im almost positive this is wrong but yea....

[hide]

the probability of choosing a row to be completely blank and choosing two columns to be completely blank is $\binom{m}{1} \binom{n}{2} \Big(\frac{1}{2}\Big)^{mn}$

we can do this with all the other cases, which will ultimately give us:

$1 - \Big( \binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n}\Big)\Big(\binom{m}{0} + \binom{m}{1} + \ldots + \binom{m}{m}\Big)\Big(\frac{1}{2}\Big)^{mn}$

But this is not entirely correct so we have to do some tweaking

first off, ther cant be a case of $\binom{m}{0}\binom{n}{0}$, for otherwise we wouldnt be choosing any rows or columns

also, any case with $\binom{m}{m}$ and $\binom{n}{n}$ is not technically valid, since if all the columns are chosen to be blank, then the rows are automatically blank as well, and visa versa

since we're subtracting these cases even though we shouldn't even include them, we have to add back:

$\binom{n}{0}\binom{m}{0} \Big(\frac{1}{2}\Big)^{mn} + \binom{n}{n}\Big( \binom{m}{0} + \binom{m}{1} + \binom{m}{2} + \ldots + \binom{m}{m}\Big) \Big(\frac{1}{2}\Big)^{mn} + \binom{m}{m}\Big(\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n}\Big)\Big(\frac{1}{2}\Big)^{mn}$

then one final adjustment to be made is we've added $\binom{n}{n}\binom{m}{m}\Big(\frac{1}{2}\Big)^{mn}$ twice so we subtract it once, hence our answer is...

$1 - (2^{n})(2^{m})\Big(\frac{1}{2}\Big)^{mn} + \Big(\frac{1}{2}\Big)^{mn} + (2^n)\Big(\frac{1}{2}\Big)^{mn} + (2^m)\Big(\frac{1}{2}\Big)^{mn} - \Big(\frac{1}{2}\Big)^{mn}$

$= \Big(\frac{1}{2}\Big)^n + \Big(\frac{1}{2}\Big)^m$

heh... ne1 else willing to give it a shot  :P cuz this answer seems like it would be way too small as $m$ or $n$ got bigger...

[/hide]

yeah...
[hide]
for 2*3 the probability is 25/64[/hide]

i don't know if this is a good way to count, but consider if the main diagonal is black, and the rest of the squares are gray, any switching of the rows and or columns produces a new arrangement, the gray could be any color, and we would have a desired pattern, but this massively overcounts, so maybe not...


Find the minimum value of
$ \frac{(x\plus{}\frac{1}{x})^6\minus{}(x^6\plus{}\frac{1}{x^6})\minus{}2}{(x\plus{}\frac{1}{x})^3\plus{}(x^3\plus{}\frac{1}{x^3})}$
 for $ x>0$.


The Good, The Bad and The Ugly get into a pistol duel under unusual terms. They draw lots and determine the order in which they start firing (who first, who second and who third). They also decide to use the same type of pistol and bullets. Then,they take up their positions at the corners of an equilateral triangle. They come to an agreement that each of them will fire single shots, in turn, and continue in the same cyclic order until two of them are dead. At his turn, the one who is firing may aim wherever he pleases. It is also well known that The Bad always hits his target, The Ugly hits his target three out of four turns and The Good manages only one out of
two. Assume that all of them adopt the best strategy in the duel and no one is killed by an arbitrary shot not intended for him. Who has the best chance to survive?
What are the exact survival probabilities of the three men?
How will your answer differ if the triangle is not equilateral and, in turns, they start
shooting bullets at regular intervals of time? Explain.

This could go to Puzzles&Brainteasers. Mods, please react!


You need to have the package fancyhdr in the folder.

If you just downloaded latex and have not used it yet, it probably is not a good idea to try to learn to typeset and type up all your solutions from now until midnight.

It is probably a better idea just to scan your solutions if you have them written down and e-mail them as attachments.

Or you could type them in word and print and scan them.

[quote="Max300"]I just downloaded the things necessary for LaTeX. I downloaded both MikTex and LaTeX. Then I went on the USAMTS site and downloaded all 3 things that need to be put in the folder. However, when I compiled it I got 19 errors and 1 warning. Some of the errors were Package Fancyhdr Warning: \headheight is too small (12.0pt) : Make it at least 13.59999;
then this one was there more than once among the 19 errors \begin{document} ended by \end{solution}. Also, another one that appeared more than once was Environment solution undefined.

What did I do wrong?!  HELP!!!! I have to submit my solutions today for Round 3   :([/quote]

fancyhdr.sty and usamts.tex have to be in the same folder as your solution file

I do have fancyhdr in the same folder as usamts.tex and the solutions1.tex folder. So I don't know whats wrong.

I'm familier with Latex. I typed up Round 2 and sent the solutions in. It is just that the computer that I had Latex on got messed up and my dad is fixing it so I have to download it on the one I'm using.

[quote="Max300"]this one was there more than once among the 19 errors \begin{document} ended by \end{solution}.[/quote]1) Check to make sure you didn't accidentally delete the \end{document} line.

2) Make sure all your files for round 3 are in the same folder, including the ones you downloaded that aren't "part" of your solution.

Not sure what else it could be.

[quote="Max300"]I haven't deleted any lines, I checked for them all. The only things I have altered from the original form are my name, id, the round number, and the username.

Also, I tried the Hello, World program and it works just fine so I'm sure there is something wrong with one of the files or I am missing something and doing it wrong.

I have 3 files in a folder, and those 3 are fancyhdr.sty, solutions[1].tex, and usamts[1].tex[/quote]Try it with everything unaltered and see if there are still error messages.

If this is taking up too much of your time and you haven't typed solutions yet, remember that you don't need the USAMTS solution form. Any document will do, as long as you put your USAMTS info in it.  :)

try renaming "usamts[1].tex" to "usamts.tex".

I know somewhere an "/include{usamts}" is called, so it will look for a file called "usamts.tex"

I will keep thinking though... and I might look at the downloadable files...

good luck ^_^

oh, yeah that's definitely the problem. you can't change the filenames...

[quote="Mathlete9366"]try renaming "usamts[1].tex" to "usamts.tex".
[/quote]

OMG THANK YOU!!!!!!!!!!

did i tell you i <3 you btw?

also thanks everybody for all the advice and help, i really appreciate it


function
linear algebra
matrix
algebra
domain
limit
complex analysis

I came across the following problem when trying to solve a question in complex analysis:

Suppose that you have an infinite matrix (indexed by $ \mathbb{N} \times \mathbb{N}$) of complex functions, all of which are analytic on a certain domain $ A \subseteq \mathbb{C}$. Assume that each row in this matrix converges pointwise to an analytic function on $ A$. Assume also that each column in this matrix converges uniformly (on $ A$) to some analytic function. Let $ f_1$ denote the analytic function obtained by taking the limit along the first column, let $ f_2$ be the analytic function obtained by taking the limit along the second column and so on. Does $ \lim_{n \to \infty} f_n$ necessarily exist and is it necessarily analytic?

The answers to what is written (but not necessarily to what was meant) are "No" and "No". Just put $ f_{i,j} \equal{} 0$ when $ i < j$ and $ f_{i,j} \equal{} f_i$ when $ i\ge j$. Every matrix like that (with analytic on $ A$ functions $ f_i$) satisfies the conditions but there are absolutely no restrictions on the behaviour of $ f_i$.

Now, if the uniform convergence of $ f_{i,j}(z)$ to $ f_j(z)$ as $ i\to\infty$ means uniform in [i]both[/i] $ z$ and $ j$, the story is completely different and the answers are "Yes" and "Yes".

Ah, too bad. The above problem came up when I tried to construct a counterexample to the following proposition:

Given an absolutely convergent series of analytic functions (on some common domain A), the limit of the series must also be analytic.

But now I'm not sure if this proposition is false.


geometry
rotation
analytic geometry
geometry unsolved

Point $ P$ is inside an equilateral triangle $ ABC$ such that $ PA\equal{}3$, $ PB\equal{}4$ and $ PC\equal{}5$. Find the area of the triangle.

Rotate the triangle ABP 60 degs around B, A maps C and P goes to A'. The triangle BA'P is equilateral, while PCA' is the one constructed with the segments PA, PB and PC as sides.
Rotate the triangle BPC 60 degs around C, B maps A, P goes to B'; the triangle CPB' is equilateral, APB' the one constructed with the segments PA, PB and PC as sides.
Finally, rotate the triangle ACP 60 degs around A, C maps B', P goes to C'; the triangle APC' is equilateral, BPC' the one constructed with the segments PA, PB and PC as sides.

Thus the hexagon AC'CA'CB' has the area twice the one of ABC, as constructed. (*)
The triangles PA'C, PB'A and PC'B are equal, their area being 6 (they are right-angled, as their sides are 3-4-5). (**)
The triangles APC', BPA' and CPB' are equilateral, their sides being, of course, PA, PB and PC respectively, hence their total area will be [(PA^2 + PB^2 + PC^2)sqrt(3)]/4 = 12.5 sqrt(3)(***)

Hence 2S = 18 + 12.5 sqrt(3), that is, S = 9 + 6.25 sqrt(3).

Best regards,
sunken rock

This could be one of those few applications of the [url=http://mathworld.wolfram.com/TripolarCoordinates.html]tripolar coordinates[/url].

In the equilateral triangle $ \triangle ABC,$ the tripolar coordinates of $P(x,y,z)$ satisfy the relation:

$ a^2(x^2 \plus{} y^2 \plus{} z^2) \plus{} x^2y^2 \plus{} y^2z^2 \plus{} x^2z^2 \minus{} (x^4 \plus{} y^4 \plus{} z^4) \minus{} a^4 \equal{} 0$

Substituting $(x,y,z)=(3,4,5)$ yields the biquartic equation

$ a^4 \minus{} 50a^2 \plus{} 193 \equal{} 0 \Longrightarrow  a^2 \equal{} 25 \plus{} 12\sqrt {3} \Longrightarrow [\triangle ABC] \equal{} \frac {_{25}}{^4}\sqrt {3} \plus{} 9.$


limit
real analysis
real analysis unsolved

Let the sequence $(x_{n})_{n\geq 1}$ with $\lim_{n\rightarrow\infty}{(x_{n+2}+2x_{n+1}+x_{n})}=0$.  Show that $\lim_{n\rightarrow\infty}{\frac{x_{n}}{n^2}}=0$

Take $y_n=(-1)^nx_n$.  We have that
\[ \lim_{n\rightarrow\infty}{(x_{n+2}+2x_{n+1}+x_{n})}=0\Rightarrow \lim_{n\rightarrow\infty}(-1)^n{(x_{n+2}+2x_{n+1}+x_{n})}=0\Rightarrow \lim_{n\rightarrow\infty}{(y_{n+2}-2y_{n+1}+y_{n})}=0 \]

 Aplying Cesaro-Stolz twice we get:$0=\lim \frac{y_n}{n^2}=\lim \frac{x_n}{n^2}$.

I used the fact that $a_n\to 0\Leftrightarrow |a_n|\to 0\Leftrightarrow (-1)^na_n\to 0$.

xirti,I think you may reverse the "what is known" and "what we should to prove"...

[quote="xirti"]:rotfl: yes.. but if you red the solution backwards it was Ok :rotfl: Now it is edited
.[/quote]
Oh,sorry,I reversed... :blush:


$\frac{\overline{AB}}{\overline{BC}}\approx\overline{0.BCA}$

Find $A$, $B$, and $C$.

Masoud Zargar

Woah, first I thought you were talking about line segments and then I thought the digits were repeating and then I realized they were concatenations. Just use $AB_{10}$. :D

I had just seen this notation used in JBMO, so I used it. :)  By the way, this is an IQ test type problem, which was basically expecting guessing, however, I found a mathematical way, so do not guess.  ;)

[quote="boxedexe"]Anyone want to try this?[/quote]

well...I still didn't get the notation...can you explain a little??

$\overline{AB}=(AB)_{10}$, which is one number with A teh first digot and B teh second.  The same goes with BC.  $\overline{0.BCA}=(0.BCA)_{10}$ whcih is one number.

well, using normal notation:

$\frac{10A+B}{10B+C} = 10^{-1}B + 10^{-2}C + 10^{-3}A$
$10A+B = B^2 + 10^{-1}(2BC) + 10^{-2}(AB+C^2) + 10^{-3}(AC)$

we can immediately get $AC \equiv (\bmod 10)$.  I'm not good on number theory so I'll start guessing numbers.  having $A$ and $C$ be 0 sounds good because all the decimal places become zero.  so $A = 0$, $B = 1$, $C = 0$

That is one solution, even though I was looking for some non zero digits.  There is a solution where A,B, and C are non zero.

The only result I've found that even remotely qualifies as "approximate" is

[hide="this"]$\frac{35}{59}\approx 0.593$, and admittedly I've written a program for it. Have no idea how to approach approximate "equalities" with mathematical strictness.[/hide]

I think the decimal is repeating, so 9,9,9 works exactly. (I might be wrong, not too sure about his notation.)

Since the overline extends before the point, I'd say he just wanted to make sure that the RHS is not understood as a product.

The overline in the periodic decimal numbers normally NEVER extends before the point.

Ah, true enough. Yeah, that would clear that up (I thought maybe it was an accident?)  :roll:

[hide="HINT"]Try finding an equation with two terms that their difference is 0.  Then you may use number theory to find the best suitable A, B, and C that maximize one of the terms.[/hide]


induction
inequalities unsolved
inequalities

Okey,relax,relax :lol:
Here is a problem in IRan MO 1999,round 1
It is not hard,and my hint for you is useing induction


superior algebra
superior algebra unsolved

Q1.Find all prime elements in ring of Gaussian Integers.
Q2.How do we prove that there exist infinite number of primes of form 4k+3 using ring of Gaussian integers?

For Q2, you don't need the Gaussian integers. Assume there are finitely many primes of the form 4k+3. Multiply them together. If the result is of the form 4k+1, add 2, and if it's of the form 4k+3, add 4. Is the result divisible by any of these primes? And what happens if the result only has prime factors of the form 4k+1?

However, to prove that there are infinitely many primes of the form 4k+1, you do need the Gaussian integers ;)

Also, Q1 is a quite classic and nontrivial result.

The other standard choice for Q2 is to use $ 2p_1 p_2 ... p_n \plus{} 1$ to get the contradiction.  You don't need the Gaussian integers for $ 1 \bmod 4$ either:  take $ (2 p_1 p_2 ... p_n)^2 \plus{} 1$.

This looks like homework.  What have you done on Q1?  What do you know about factorization in the Gaussian integers?

[quote="t0rajir0u"]You don't need the Gaussian integers for $ 1 \bmod 4$ either:  take $ (2 p_1 p_2 ... p_n)^2 \plus{} 1$.[/quote]

How does this work?

$ p | x^2 \plus{} 1 \Leftrightarrow p \equiv 1 \bmod 4$.  This is not hard to see:  if $ x^2 \equiv \minus{}1 \bmod p$ then $ x$ is an element of order $ 4$, which exists if and only if $ 4 | p\minus{}1$.  And in fact this is usually a lemma that is used to solve Q1 and describe factorization in the Gaussian integers, not the other way around.


Let 62ab427 be a multiple of 99, find the digits a and b.

[quote="humblemastermind"][hide="solution"] For divisibility by 99, you need divisibility by both 11 and 9. So, $6+a+4+7-2-b-2=13+a-b$ must be divisible by 11. Also, $6+2+a+b+4+2+7=21+a+b$ must be divisible by 9. Since both $a$ and $b$ must be less than 10, you know (from the first expression - it must be 11) that $a-b=2$. Additionally, you know (From the second expression - it must be 27) that $a+b=6$. Solving, we get that $a=4$ and $b=2$. [/hide][/quote]
[hide]
supposed to be $b-a=2$ not $a-b=2$
so $6242427$ doesnt work, but $6224427$ works
[/hide]

[quote="humblemastermind"][hide="solution"] For divisibility by 99, you need divisibility by both 11 and 9. So, $6+a+4+7-2-b-2=13+a-b$ must be divisible by 11. Also, $6+2+a+b+4+2+7=21+a+b$ must be divisible by 9. Since both $a$ and $b$ must be less than 10, you know (from the first expression - it must be 11) that $a-b=2$. Additionally, you know (From the second expression - it must be 27) that $a+b=6$. Solving, we get that $a=4$ and $b=2$. [/hide][/quote]

[hide]Why can't the second expression be 36?[/hide]

[quote="Ignite168"]Why can't the second expression be 36?[/quote]
Assume that the second expression = 36, then a + b = 15.
Since b - a = 2, both the digits a, b must be either odd or even, then a + b must be even. Hence a + b can't be 15.


Here, you can tell about your oldest books. So far, the oldest in my family is a 181-year-old book, published in 1825. It's called "Schoolmaster's Assistant: Improved and Enlarged; being a plain and practical system of Arithmetic: Adapted to the United States." It's the 61st edition, by Nathan Daboll.

How about you?

[quote="tim1234133"]In one of the maths classrooms at my school we found some textbooks from the 1790's- and they still had all that stuff written in them by bored students that modern books do (On page 3: "Don't look at page 74", on page 74: "Haha, I knew you would, you turd")[/quote]

They used the word turd like that back then?  Interesting...

I can't remember the exact the words they used but that was the essence of it.

[quote="tim1234133"]In one of the maths classrooms at my school we found some textbooks from the 1790's- and they still had all that stuff written in them by bored students that modern books do (On page 3: "Don't look at page 74", on page 74: "Haha, I knew you would, you turd")[/quote]
This sounds incredibly unlikely seeing as books were hand-written and very expensive.

They weren't hand written (printing was invented well before the end of the eighteenth century), and my school was fairly rich in the past (still, but less so, now).

I was amazed that they were written in as well, but they clearly were. Most of the books in that cupboard were from the 19th century, the oldest was from 1793, I think.

So the youngest one on this topic was one from 1825.


Let $a_{1},...,a_{n}$ be positive numbers with product $1$. Show that there is $k$ ($1\leq k\leq n$) such that ${a_{k}a_{k+1}^{2}(a_{k+1}^{3}+2})\geq 3$, where $a_{n+1}=a_{1}$

[quote="RIP"]Excuse me for my bad English[/quote]
Never! :lol:


its easy to see that the equation $ x^2\equal{}2^x$ only has 3 roots. 2 positive ones r $ 2,4$ . how abt the negative one ?

yeah i know that appro too. any representation

Can't think of one.  Maybe if you do a Taylor approximation, you can get closer but otherwise, no idea.

Wouldn't it involve finding the inverse of xlnx?

$ x_3 \equal{} \minus{}2\frac{LW(\frac12ln(2))}{ln(2)}$

where $ LW(x)$ is this function:
[url]http://mathworld.wolfram.com/LambertW-Function.html[/url]


1. Courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof. What could be the weight in grams of a package for which the charge is $ \$$1.55?

A. 1155
B. 1145
C. 1040
D. 950
E. 259

[hide]Let $ x$ be the number of additional 100 grams or part thereof and $ y$ be the weight of the package.
$ .65+.1x=1.55$
$ x=9$
So, $ 1050<y\le1150$.
Looking at our answer choices, B. is the right answer.[/hide]

wait, but if there are 9 additional weights of 100, than when u add it all up, doesn't the actual weight of the package come out to be 1150...that's what i got.

There is no answer choice 1150.. but 1145 is a perfectly legal answer.

[quote="mountain.dew"]Read the question again please. It says "...or part thereof" which means that anything less than 100 grams would still add 10 cents.[/quote]
That's what I thought it meant...


search
inequalities unsolved
inequalities

Prove that ${a \over 1+bc}+{b \over 1+ac}+{c \over 1+ab}\leq 2$ for all $a,b,c \in [0,1]:$.

maybe $a,b,c\in[0,1]$.we can prove that $\frac{a}{1+bc}\leq\frac{2a}{a+b+c}$.
In fact,if $a=0$,this is correct,
just to prove $a+b+c\leq2+2bc$,$(1-a)+bc+(1-b)(1-c)\geq0$,it's obvious that it's correct.
Likewise,$\frac{b}{1+ac}\leq\frac{2b}{a+b+c}$,$\frac{c}{1+ab}\leq\frac{2c}{a+b+c}$,
Add them!

yes of course it was some bug. Soo fast response!!!

2005, 56th polish MO. Use [i]search[/i].

Jin, thank you for the solution. It is very beautiful

another proof:Wlog $a \ge b \ge c$
so:\[ {a \over 1+bc}+{b \over 1+ac}+{c \over 1+ab}\leq \frac{a+b+c}{1+bc} \]
so we should to prove:
  \[ \frac{a+b+c}{1+bc} \le 2 \]
\[ \iff a+b+c \le 2+2bc \]
\[ \if bc+(1-b)(1-c)+1-a \ge 0 \]
obvious. :)