geometry
rectangle
trigonometry
power of a point
radical axis
geometry unsolved

ON AB, BC of square ABCD, E, F are constructed so BE=BF, let BN the altitude jof triangle BCE prove angle DNF is a right angle

Click on the picture for a bigger view

Let BN meets DA at G, the right $\triangle BCE \cong \triangle ABG$ are congruent by ASA, AG = BE = BF. Then CDGF is a rectangle, let (O) be its circumcircle. Let M be midpoint of BC. Let a parallel to AB through O meet BC at N. Then $ON = \frac{AB}{2},$ $NM = \frac{BC}{2}-\frac{BC-BF}{2}= \frac{BF}{2}= \frac{AG}{2},$ The right $\triangle ABG \sim \triangle NOM$ are centrally similar, because $\frac{NO}{AB}= \frac{NM}{AG}= \frac{1}{2}$ and $NO \parallel AB,\ AG \parallel NM,$ hence $OM \parallel BG \equiv BN.$ Since $BN \perp CN,$ N lies on a circle (M) with diameter BC intersecting (O) at C. OM is their center line, their radical axis is a normal through C to OM, i.e. to BN, i.e., their radical axis is CN, hence N is their other intersection. DF is a diameter of (O), $\angle DNF$ is right.

Well done yetti  :)

[quote="yetti"]Let BN meets DA at G, the right $\triangle BCE \cong \triangle ABG$ are congruent by ASA, AG = BE = BF. Then CDGF is a rectangle, let (O) be its circumcircle. [/quote]

Since $\angle GNC = \angle GFC =90^{\circ}$ point $N$ must be on $(O)$. So $\angle DNF = \angle DGF=90^{\circ}$.

So we don't need this:

[quote="yetti"] Let M be midpoint of BC. Let a parallel to AB through O meet BC at N. Then $ON = \frac{AB}{2},$ $NM = \frac{BC}{2}-\frac{BC-BF}{2}= \frac{BF}{2}= \frac{AG}{2},$ The right $\triangle ABG \sim \triangle NOM$ are centrally similar, because $\frac{NO}{AB}= \frac{NM}{AG}= \frac{1}{2}$ and $NO \parallel AB,\ AG \parallel NM,$ hence $OM \parallel BG \equiv BN.$ Since $BN \perp CN,$ N lies on a circle (M) with diameter BC intersecting (O) at C. OM is their center line, their radical axis is a normal through C to OM, i.e. to BN, i.e., their radical axis is CN, hence N is their other intersection. DF is a diameter of (O), $\angle DNF$ is right.[/quote]

 :wink:

[color=green]A trigo solution[/color][hide]
Let $\angle CNF=\delta$ and $\angle ECB=\beta$
$\frac{CF}{FB}=\frac{CF}{NF}\frac{NF}{FB}=\frac{\sin \delta}{\sin \beta}\frac{\cos \beta}{\cos \delta}= \frac{\tan \delta}{\tan \beta}$
so $\tan \delta = \frac{CF}{FB}\tan \beta= \frac{CF}{FB}\frac{EB}{BC}=\frac{CF}{BC}=\tan \angle CDF$
so $\angle CNF = \angle CDF$ and $CFND$ is circumscribed.[/hide]

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 7 Miniatures géométriques sur un carré  p. 4
Sincerely
Jean-Louis

My solution:
We have $\angle DCE= \angle CEB= \angle CEN$
Also,
$\frac{DC}{CN}= \frac{EC}{CN}= \frac{EB}{EN}= \frac{EF}{EN}$
Thus, $\triangle BCN \sim \triangle ENF$, so $\angle DNC= \angle ENF$, so $\angle DNF= \angle ENC=90$.

From $\Delta BCE$ we have $BC^2=CN\cdot CE\iff CN\cdot CE=CD^2$ i.e. $\Delta CDN\sim\Delta CED\implies\angle CND=\angle CDE$, but $\angle CDE=\angle AED$, and $\angle AED=\angle CFD$ from $\Delta AED\cong\Delta CFD$, so $\angle CND=\angle CFD$ and $CFND$ is cyclic, done.

Best regards,
sunken rock

Oof, I just complex bashed it.

[hide=WARNING!]This stunt is performed by a professional idiot. Do not try this at home.
[hide]I am not joking, it actually gets very ugly.
[/hide]
[/hide]

Actually I think when you bash it that it is somewhat faster, I solved it using standard synthetic geometry and bashing.

Enough talking, let's get to the solution.

We fix $a=1$ and $b=i$, $c=-1$, $d=-i$, or in other words we fix them to be the $4-th$ roots of unity.
Obviously we have that $f=\overline{e}$.
We now want to calculate $n$.
Since $C,N,E$ are colinear, we have that:
$$\frac{c-n}{\overline{c-n}}=\frac{c-e}{\overline{c-e}}$$
when we plug in the values we have and we plug in $\overline{e}=f$ we get that:
$$n=\frac{1+f}{1+e}n+\frac{f-e}{1+e}$$
From the prependicularity condition we have that:
$$\frac{b-n}{\overline{b-n}}=-\frac{c-e}{\overline{c-e}}$$
when we plug in everything we have gotten so far we get that:
$$n=\frac{(i-1)(f-e)}{2(1+f)}$$
and we have that:
$$\overline{n}=\frac{(i+1)(f-e)}{2(1+e)}$$

So we got to show that:
$$\frac{d-n}{\overline{d-n}}=-\frac{f-n}{\overline{f-n}}$$
but this is seen when we plug everything we got.

Thus we have that:
$$DN \perp NF$$

$\dfrac{EN}{NB}=\dfrac{EB}{BC}=\dfrac{EF}{BD}$, $\angle CEF=\angle NBD$

$\implies \triangle ENF\cong\triangle BND$
$\implies \angle ENF=\angle BND$
$\implies \angle FND=\angle ENB=90^\circ$
 :help:


geometry
Euler
radical axis
geometry open

[b]Definition. [/b]The point $Q$ is the complement of the point $P$ with respect to $\triangle ABC$ having $G$ as centroid if $Q, G, P$ are collinear in this order and $PG = 2GQ$. Then $P$ is the anticomplement of the point $Q$ with respect to $\triangle ABC$.

I was finding some results on plane geometry and found this definition... What about this, is there any good propiety about complement????

thank u

José Carlos

[quote="gemath"]Can you post some your results ?? and we can exchange, I had found some properties of this definition. :)  :)[/quote]

I haven't found any interesting result, could you share your results?

Ok hi hucht, sorry because I reply lately, we have more accurate definition :The point $Q$ is the complement of the point $P$ with respect to the $\triangle ABC$ having $G$ as centroid if $2\vec{GQ}+\vec{GI}=\vec{0}$
special: $G,H,O$ and Euler line.
My result: radical center of three escribed circle is Q, and incircle centre is I then $2\vec{GQ}+\vec{GI}=\vec{0}$ I think this is a nice result, I have some other results and can you share your result ??
Nice to meet you hucht  :)  :)

Dear Mathlinkers,
a nice message writing by Darij Grinberg about "the isotomcomplement Theory"
http://tech.groups.yahoo.com/group/Hyacinthos/message/6423
Sincerely
Jean-Louis


geometry
circumcircle
trigonometry
angle bisector
geometry solved

$ABC$ is an acute-angled triangle. Let $D$ be the point on $BC$ such that $AD$ is the bisector of $\angle A$. Let $E, F$ be the feet of perpendiculars from $D$ to $AC,AB$ respectively. Suppose the lines $BE$ and $CF$ meet at $H$. The circumcircle of triangle $AFH$ meets $BE$ at $G$ (apart from $H$). Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled.

The problem is wrong and I do not see how to fix it. The only non-trivial right angle that I can see is $\angle BGD = 90^\circ$, but the triangle $\triangle BGD$ is already constructed.

Anyway, the intersection H lies on the A-altitude of the triangle $\triangle ABC$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41286]Another geometry question[/url] in the high school section for a proof). Let K be the foot of this A-altitude. Denote a = BC, b = CA, c = AB. Since K is the altitude foot and D the angle bisector foot,

$BK = c \cos \widehat B,\ \ BD = \frac{ca}{b + c},\ \ BK \cdot BD = \frac{c^2a}{b + c} \cos \widehat B$

The power of the point B to the circumcircle (P) of the triangle $\triangle AFH$ is

$BH \cdot BG = BA \cdot BF = BA \cdot BD \cos \widehat B = \frac{c^2a}{b + c} \cos \widehat B$

It follows that

$BH \cdot BG = BK \cdot BD,\ \ \frac{BG}{BD} = \frac{BK}{BH}$

Thus the triangles $\triangle BKH \sim \triangle BGD$ having the common angle at the vertex B are similar by SAS, which means that the angle $\angle BGD = \angle BKH = 90^\circ$.

[quote]Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled.[/quote]

It should have been BG, GE, BF instead of BG, CE, BF. Problem corrected now.  :)

Thanks, orl. The above solution then continues:

Since $\angle DGB = \angle DGE = 90^\circ$ is right,

$DG^2 = BD^2 - BG^2 = DE^2 - GE^2$

The right angle triangles $\triangle ADF \cong \triangle ADE$ are trivially congruent by ASA, because AD bisects the angle $\angle EAF \equiv \angle CAB$, hence DF = DE and

$BD^2 - BG^2 = DF^2 - GE^2$

$BD^2 - DF^2 = BG^2 - GE^2$

Since $DF \perp AB$, the triangle $\triangle BDF$ is right and $BD^2 - DF^2 = BF^2$, so

$BF^2 = BG^2 - GE^2$

$BF^2 + GE^2 = BG^2$

BG would be the hypotenuse of the right angle triangle constructed from BF, GE, BG.

I just wanted to add a synthetic proof of < BGD = 90° (that's what Yetti proved in post #2):

At first, we apply the result of http://www.mathlinks.ro/Forum/viewtopic.php?t=41286 and http://www.mathlinks.ro/Forum/viewtopic.php?t=5460 to see that the point of intersection H of the lines BE and CF lies on the altitude of triangle ABC issuing from the vertex A, so that $AH\perp BC$.

We will use directed angles modulo 180°.

Let G' be the orthogonal projection of the point D on the line BE. Then, < BG'D = 90°. On the other hand, < BFD = 90°, since $DF\perp AB$. Thus, the points G' and F lie on the circle with diameter BD; hence, < BG'F = < BDF. But since $DF\perp AB$, we have < (AB; DF) = 90°, and since $AH\perp BC$, we have < (AH; BC) = 90°. Thus,

  < HG'F = < BG'F = < BDF = < (BD; DF) = < (BD; AB) + < (AB; DF) = < (BD; AB) + 90° = < (BD; AB) + < (AH; BC) = < (BD; AB) + < (AH; BD) = < (AH; AB) = < HAF.

Hence, the point G' lies on the circumcircle of triangle AFH. In other words, the point G' is the point of intersection of the circumcircle of triangle AFH with the line BE different from the point H. Thus, the point G' coincides with the point G. Hence, < BG'D = 90° becomes < BGD = 90°, and we are done.

  darij

I have a nice solution to prove that BGD=90 and it is dedicated to Darij (my figure is different from yetti  :(  I have AC>AB)
Let the altitude from A intersects BC at K .
Then from the cyclic FKDA we have that $BF\cdot BA=BK\cdot BD$  . But from the power point theorem we have that
$BH\cdot BG=BF\cdot BA$  so $BH\cdot BG=BK\cdot BD$ so the triangles BHK and BGD are similar and we are done  :)

PS I hope you like it


Problem: Let $ a$ and $ n$ be positive integers such that $ a < n!$. Prove that $ a$ can be represented as a sum of at most $ n$ distinct divisors of $ n!$.

Outline of book's solution
[hide]
Suppose it holds for $ n$. For $ a \leq n!$, rewrite it as $ a\equal{}b(n\plus{}1)\plus{}c$ with $ 0\leq c\leq n$. Then $ b\leq n!$, so $ b$ can be represented as a sum of at most $ n$ distinct divisors of $ n!$: $ d_1, d_2,...,d_k$ ($ k\leq n$). Hence $ a \equal{} d_1(n \plus{} 1) \plus{} ... \plus{} d_k(n \plus{} 1) \plus{} c$, where $ d_1(n \plus{} 1), ..., d_k(n \plus{} 1), c$ are $ k \plus{} 1$ distinct divisors of $ (n \plus{} 1)!$.
[/hide]

Question: how do you know that it's the only way to represent $ a$? The keyword "at most" is not proved here.

You don't need any information about the other representations of $ a$.  The problem asks to prove that there exists a representation with at most $ n$ divisors, and there is.  Problem solved.

But then what's the difference between "there exists a representation with $ n$ divisors" and "there exists a representation with [b]at most [/b]$ n$ divisors "?

The first condition is stronger.  We can't say that there exists a representation with exactly $ n$ divisors; for example, $ 2$ cannot be represented as a sum of $ 3$ divisors of $ 3!$.

Oh, I see where I misunderstood, thanks a lot.


The equation y = x^2 + 2ax + a represents a parabola for all real values
of a.
(a) Prove hat each of these parabolas pass through a common point
and determine the coordinates of this point.
(b) The vertices of the parabolas lie on a curve. Prove that this curve
is a parabola and find its equation.

I got part a) correct (-1/2, 1/4)
for part b) the formla for vertex =-b/2a and i noticed that for a=n, the vertex=-a. So i'm not sure how this a parabola.

Then i tried to complete the square and arrived at (x+a)^2+(a-a^2). This makes more sense but why didn't the first approach work.

And if (a-a^2) is the vertex, how do I find the equation of the parabola

The vertex is [-a,a-a^2] in parametric coordenates.To find a-a^2 you must put x=-a in the equation of the parabola.So x=-a or a=-x that put in a at parametric coord. we obtain [x,-x^2-x].So the equation is y=-x^2-x


If Popeye buys 120 truckloads of spinach at $8000 per truckload, and he sells all of it for a million dollars, how many dollars profit will he make?

[hide]
$120 \cdot 8000=960000$
that is how much he bought it for.  Since he sells it:
$1000000-960000=40000 $
40000 becomes his profit[/hide]


[quote="Silverfalcon"]Let $f$ be the function defined by $f(x) = ax^2 - \sqrt{2}$ for some positive $a$. If $f(f(\sqrt 2)) = - \sqrt 2$ then $a =$

$\text{(A)} \ \frac{2 - \sqrt 2}{2} \qquad \text{(B)} \ \frac{1}{2} \qquad \text{(C)} \ 2 - \sqrt 2 \qquad \text{(D)} \ \frac{\sqrt 2}{2} \qquad \text{(E)} \ \frac{2 + \sqrt 2}{2}$[/quote]

[hide]
$f^2(x)=a(ax^2-\sqrt{2})^2-\sqrt{2}$
$a(2a-\sqrt{2})^2-\sqrt{2}=-\sqrt{2}$
$a(2a-\sqrt{2})^2=0$
Since $a$ is positive,
$(2a-\sqrt{2})^2=0$
$a=\frac{\sqrt{2}}{2}$
[/hide]

[hide="Solution"]$f(f(\sqrt{2}))=f(2a-\sqrt{2})=a(2a-\sqrt{2})^2-\sqrt{2}=-\sqrt{2}\implies a(2a-\sqrt{2})^2=0\implies 2a-\sqrt{2}=0\implies a=\frac{\sqrt{2}}{2}$[/hide]
Masoud Zargar


superior algebra
superior algebra unsolved

Show that if $\chi$ is a character af the finite group $G$, then $\chi(g)$ is an algebraic integer for all $g\in G$.

I mean that $\chi$ is a homomorphism $G\to\mathbb{C}^{*}$. If i am true then $(\chi (g))^{|G|}=\chi (g^{|G|})=\chi (e)=1$, so $\chi (g)$ is a root of $x^{|G|}-1=0$ and we are done!

In this context, a character is a map of the type $G\to\mathbb C,g\mapsto\mathrm{tr}\,\rho(g)$ for some representation $\rho\colon G\to\mathrm{GL}_{n}(\mathbb C)$. For $n=1$, this is characters in the sense of homomorphisms $G\to\mathbb C^\times$. owk


Can anybody recommend a decent geometry rendering program, preferably freely downloadable for windows? I have Microsoft's Visio on my computer right now, but I don't really know if it's that useful.  How is LaTeX just by itself for drawing?

you might wanna try geometer's sketchpad. They have a nice demo version, student licences, and the only bad thing about the demo is that you can't rate stuff :-( anyway it's a good tool.

[i]Posted from a Sony Ericsson K700i[/i]

If you try using just LaTeX for drawing diagrams, you'll spend a lot of time looking like this:  :?  then like this :( then like this  :mad: then finally like this   :no:.

Draw the diagrams elsewhere, then include them as described in the Pictures section of our LaTeX pages.

[quote="rrusczyk"]If you try using just LaTeX for drawing diagrams, you'll spend a lot of time looking like this:  :?  then like this :( then like this  :evil: then finally like this  :cry: .

[/quote]

You have a great sense of humor, but what's wrong with 'evil' and 'cry' icons?

[quote="paganinio"]You have a great sense of humor, but what's wrong with 'evil' and 'cry' icons?[/quote]They were from the old emoticons pack, so they were deprecated. We don't need evil emoticons (aside from this case :P) on the site.

I'd go for Geometer's Sketchpad. Can't realli remember how much I paid, but it's a good deal. *nods*

The student version is somewhat limited in that you need to have the CD in the drive every time you use it. I think there are a few other features missing. Apparently the full version is $120 (although I paid nothing  :lol: ).

If you purchase a student license (I got one at IMO 2001 :D) you can use [url=http://www.daemon-tools.cc/dtcc/portal/index.php]daemon-tools[/url] along with an image (.iso, .cue, .bin) of your Sketchpad CD so that you won't ocupy your CD drive with the Sketchpad cd all the time.

[quote="MithsApprentice"]Eh...I'm personally not too keen on always having a virtual drive. Besides, isn't that illegal?[/quote]No, I think it's not illegal (it says right so on the first page of that software) if you, of course, bought (or have been given ;) ) a license for that software.

Ah interesting. Thanks for the info on that.

What do you think about [url=http://www.chartwellyorke.com/cabri.html]Cabri Geometry II Plus[/url]? I use it and I like it pretty much. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38410]Here[/url] is an examle.


geometry
inequalities proposed
inequalities

if $a,b,c$ are the sides and $S$ the area of triangle $ABC$ then show that:
$\frac{1}{a^2+b^2+4c^2}+\frac{1}{b^2+c^2+4a^2}+\frac{1}{c^2+a^2+4b^2}\le \frac{\sqrt{3}}{8S}$

[quote="Albanian Eagle"]if $a,b,c$ are the sides and $S$ the area of triangle $ABC$ then show that:
$\frac{1}{a^2+b^2+4c^2}+\frac{1}{b^2+c^2+4a^2}+\frac{1}{c^2+a^2+4b^2}\le \frac{\sqrt{3}}{4S}$[/quote]

i think the best version is

[quote] if $a,b,c$ are the sides and $S$ the area of triangle $ABC$ then show that:
$\frac{1}{a^2+b^2+4c^2}+\frac{1}{b^2+c^2+4a^2}+\frac{1}{c^2+a^2+4b^2}\le \frac{\sqrt{3}}{ 8 S}$[/quote]

isn't it ?? :)

[hide="if so...."]then it is a very very nice problem  ;) [/hide]

 :lol:

yes sorry for my mistake
I edited!
now try.

Well, this is a very nice problem.I really liked it!  :lol:

Note that $\sum_{cycl} \frac{1}{a^2+b^2+4c^2}=\sum \frac{1}{4(m_a^2+m_b^2)}\le \sum \frac{1}{8m_a \cdot m_b} \frac{}{} \frac{\sqrt{3}}{8S}$

due to a well-known ineq : $\sum \frac{1}{m_a \cdot m_b} \frac{}{} \frac{\sqrt{3}}{S}$ which i posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=69002_post6]here[/url]

(two nice slutions exist)

 :)

By AM-GM we have that,

$\frac{1}{a^2+b^2+4c^2}+\frac{1}{b^2+c^2+4a^2}+\frac{1}{c^2+a^2+4b^2} \le \frac{1}{\sqrt[3]{4{(abc)}^2}}$

but we know that,

$S=\frac{ab.sin \angle{C}}{2}=\frac{bc.sin \angle{A}}{2}=\frac{ac.sin \angle{B}}{2} \implies {(abc)}^2=\frac{8S^3}{sin \angle{A}.sin \angle{B}.sin \angle{C}}$ therefore,

$\frac{1}{a^2+b^2+4c^2}+\frac{1}{b^2+c^2+4a^2}+\frac{1}{c^2+a^2+4b^2} \le \frac{\sqrt[3]{sin \angle{A}.sin \angle{B}.sin \angle{C}}}{2\sqrt[3]{4}.S}$

$\le \frac{1}{3}\left(\frac{sin \angle{A}+sin \angle{B}+sin \angle{C}}{2\sqrt[3]{4}.S}\right) \le \frac{\sqrt{3}}{2}\left(\frac{1}{2\sqrt[3]{4}.S}\right) =\frac{\sqrt{3}}{8S}$ by Jensen's inequality.


Take any number. Reverse it and add it to the original .
If required; repeat this process until the final sum is palindromic.

For example,
$47+74 = 121$

$58+85 = 143$
$143+341 = 484$

For some numbers, the number of steps may be more.

Is it possible to say that for any given number , this can always be reached ; given no limit on the number of steps allowed ?

All positive integers except the "[url=http://en.wikipedia.org/wiki/Lychrel_number]Lychrel Numbers[/url]" are believed to hold this property. The most well-known of the Lychrel Numbers is 196.


This problem is similar to hanoi puzzle
The rule is same.
i) only one disk may be moved at a time
ii) no disk may be placed on top of a smaller disk

There are three rods $ A,B,C$
(disks size are same as their numbers, $ 1<2<3<\cdots$ i.e. bigger number, bigger size)

1. There are $ 2n$ disks. We divide disks in two groups; odd and even.
The odd number disks are on $ A$, and the even number disks are on $ B$ with smaller number are upper than bigger number
($ ex.$ $ 1,3,5,7$ are on $ A$ , $ 2,4,6,8$ are on $ B$ from top to bottom)
We want to move disks to be on $ C$ in order $ 1,2,3,\cdots , 2n$ from top to bottom.
How many moves are needed?

2. There are $ 3n$ disks. We divide disks in three group; $ 1,2,3$ ($ mod$ $ 3$)
Rods $ A,B,C$ contains group $ 1,2,3$ separatively.
We want to move disks to be on $ C$ in well order. (just same as upper problem)
How many moves are needed?

Both questions can be solved by the following method:  let $ a_{n}$ be the number of moves necessary to move $ n$ disks in the given initial configuration so that all are on stick $ A$, and likewise $ b_{n}$ and $ c_{n}$.  Then analyze the solutions as follows: if the largest disk begins on stick $ i$ and needs to end up on stick $ i$, you might as well pretend it's not there and reduce to a simpler case.  If the largest disk is on stick $ i$ and needs to end up on stick $ j\neq i$, it will move exactly once.  Before it moves, all of the other disks must be on the stick $ k\not\in\{i, j\}$.  This then reduces to a simpler case of the same problem combined with a case of the conventional Towers of Hanoi problem.  This gives you a bunch of recurrence relations that you can solve by conventional means.

As an aside, every position in the Towers of Hanoi puzzle can be represented by a number written in ternary, where if the disk numbered $ i$ is on the destination stick we make the $ i$th digit from the right of the number a zero, and we arbitrarily label the other two sticks $ 1$ and $ 2$ and number the associated digits appropriately.  (Leading zeros may occur but don't matter: if the largest disk is already on the destination stick, it may as well not exist.)  I wonder if there is any simple way to relate the number associated to a state to the number of moves necessary to complete the puzzle from that state.


Where would I voice your complaint if i feel the grader has made a mistake correcting my work?

the subforum ""grading issues" are for protests for your grades.  The link is at the top of the USAMTS forum (I think I'll protest, too)

Please make sure to read all the rules regarding grading issues before filing a protest.


Hey, I found this site very recently and I've looked through it...awesome stuff.  At my school I would be considered a decent math student, but pretty much all I know is the stuff inside typical algebra and geometry books; I haven't had much experience with higher level thinking unfortunately.  I was amazed at the level of difficulty in algebra and geometry...  Some of my friends say algebra and geometry are easy, yet the problems I saw on this site seem impossible...  I'm planning to really study all of this stuff, enter some math competitions when I get better, and see where my hard work will take me.

However, I'm stuck with these problems as of now. :(

1. |2x+1|=|x-8|

I've tried writing it four different ways for x>0 or x<0, and I [i]think[/i] I got the right answer, but I'm not sure.

2. Write |-5x^2y^3| two ways.

Tried it, but got confused...

3. A somewhat long question...  You are about to take a test that contains questions of type A worth 4 points and of type B worth 7 points.  You must answer at least 5 of type A and 3 or type B, but time restrict answering more than 10 of either type.  In total, you can answer no more than 18.  How many questions of each type must you answer, asuming all of your answers are correct, to maximize your score?  What is the maximum score?

My math teacher told us to use linear combinations, and I came up with this to start (all < or > signs actually less than or equal to/greater than or equal to):

5<A<10
3<B<10
A+B<18
Score=4A+7B

However, this is as far as I got...

Thanks a lot for your time.  :)

Just remember that if you have an $|x|$, your value for x is either x or -x.
So in setting up a relationship like $|x| = |y|$
You have the possibilities
$x = y$
$-x = y$
$x = -y$
$-x = -y$
So can you do 1 and 2 now?

In number 3, which type of question will yield the most points?  You want to include as many of those as possible and it ought to fall into place.

[quote="brady00n"]

1. $|2x+1|=|x-8|$

I've tried writing it four different ways for x>0 or x<0, and I [i]think[/i] I got the right answer, but I'm not sure.[/quote]

Don't try it with $x>0$ or $x<0$ because it doesn't lead to anywhere. Try to set the whole thing inside the absolute value $\ge 0$ or $< 0$.

For here, you can set $x\ge 8$ and $x<8$. What will $|2x+1|$ and $|x-8|$ be in these two cases?

[quote="brady00n"]2. Write $|(-5)x^2y^3|$ two ways.

Tried it, but got confused...[/quote]

Look at the number inside the absolute value. What can you see right away?

$x^2$ is always $\ge 0$ right? So the only thing which determine $(-5)x^2y^3$ is $y$.

Set two cases when $y\ge 0$ and $y<0$

[quote="brady00n"]3. A somewhat long question... You are about to take a test that contains questions of type A worth 4 points and of type B worth 7 points. You must answer at least 5 of type A and 3 or type B, but time restrict answering more than 10 of either type. In total, you can answer no more than 18. How many questions of each type must you answer, asuming all of your answers are correct, to maximize your score? What is the maximum score?

My math teacher told us to use linear combinations, and I came up with this to start (all < or > signs actually less than or equal to/greater than or equal to):

$5\le A\le 10$
$3\le B\le 10$
$A+B\le 18$
Score=$4A+7B$

However, this is as far as I got..[/quote]

To minimize $4A+7B$, easy right? $A=5$ and $B=3$ will give you $41$

To maximize $4A+7B$: Because we can want as many B as possible (B is worth more than A), so $B=10$ and $A=8$ will you give $102$.

Hope it helps.

here is a little thing that i thought of for absolute values that might help you understand what you are doing, lets say you have
$|a|=|b|$, square both sides,
$|a|^2=|b|^2$ notice that the absolute value symbols are usless here, so
$a^2=b^2$, rearranging,
$a^2-b^2=0$ factoring
$(a+b)(a-b)=0$ using the zero product property,
$a=-b$ or $a=b$
hopefully, you see what you should do with the problems now

also when you have absolute value signs, it does not matter if the inside is positive or negative, just the magnitude matters, so like |5|=|-5|, similiarly, you have $|-5x^2y^3|=|5x^2y^3|$ (the reason is that it will be positive no matter what, so the positive or negative symbols do nothing, more rigirously, we have
$|a| = |-1*-a|=|-1||-a|=|-a|$


A three digit number 2A4 is added to 239, yielding a sum of 5C3. Find the largest possible value of A such that 5C3 is divisible by 3.

A. 1   B. 3    C. 4   D. 7  E. 8

Looking at the addition, I think $ A\plus{}4\equal{}10\plus{}C$.

$ 5C3$ divisible by $ 3$ means $ 8\plus{}C$ is divisible by $ 3$ so $ C\equal{} 1, 4$ or $ 7$ and $ A\plus{}4\equal{}11, 15$ or $ 17$.

But $ A<9$ so $ A\equal{}7$.


pigeonhole principle
trigonometry
inequalities
triangle inequality

[size=150]Week 1[/size]

Let $f(x)=\frac{x}{3x-2}$. Show that if $n$ has $d$ digits, then $f(n)=\frac{n}{10^{d}}$ if and only if $n$ is an integer and $n=\underbrace{33\cdots 33}_{(d-1) \text{ threes}}4$ in base 10.
[hide="my solution"]
$f(n)=\frac{n}{3n-2}$ by definition, so $f(n)=\frac{n}{10^{d}}$ iff $3n-2=10^{d}$. $n=\lceil 333...3.3333\rceil$ etc $n=\underbrace{33\cdots 33}_{(d-1) \text{ threes}}4$
[/hide][hide="Week 1 Solution"]
[quote="matt276eagles"]Here's my proof:

The desired condition is $f(n)=\frac{n}{10^{d}}$.  By the definition of $f$, this condition is equivalent to $\frac{n}{3n-2}=\frac{n}{10^{d}}$.  Rearranging, we have $3n-2=10^{d}$, or $n=\frac{10^{d}+2}{3}$.  We only need to show that if $n$ has $d$ digits, then $n=\frac{10^{d}+2}{3}$ if and only if $n$ is an integer and $n=\underbrace{333\cdots 333}_{(d-1)\text{ threes}}4$.

For any given value of $d$, the equation $n=\frac{10^{d}+2}{3}$ is linear and has exactly one solution for $n$.  We will proceed by induction.

If $d=1$, then we have $n=4$.

For some value of $k$, assume that the equation $n=\frac{10^{k}+2}{3}$ has the solution $n=\underbrace{333\cdots 333}_{(k-1)\text{ threes}}4$.

Now, consider the equation $n=\frac{10^{k+1}+2}{3}$.  We have $n=\frac{10(10^{k})+2}{3}=\frac{9(10^{k})+10^{k}+2}{3}=3(10^{k})+\underbrace{333\cdots 333}_{(k-1)\text{ threes}}4=\underbrace{333\cdots 333}_{k\text{ threes}}4$.

The induction is complete.  Therefore, if $n$ has $d$ digits, then $f(n)=\frac{n}{10^{d}}$ if and only if $n$ is an integer and $n=\underbrace{333\cdots 333}_{(k-1)\text{ threes}}4$.  QED.[/quote]

(All the solutions were good, there's no real reason I picked this one, so don't be sad :D)
[/hide]

[size=150]Week 2[/size]

Compute, with proof, all primes $p$ such that $2^{p}+p^{2}$ is also prime.
[hide="my solution"]For low cases besides the trivial ones, it seems like $2^{p}+p^{2}$ is divisible by three. Let's prove this:
Assume $p>3$. Then $p\equiv \pm 1 (3) \implies p^{2}\equiv 1 (3)$.
Also, $p$ is odd, so $2^{p}\equiv (-1)^{p}\equiv-1 (3)$. Thus $2^{p}+p^{2}\equiv 0 (3)$ if $p>3$. Checking $p=2,3$ yields $p=3$ as the only solution to the equation.[/hide][hide="Week 2 Solution"][u][b]Week 2 Solution:[/b][/u]

[quote="tarquin"]First we check for $p=2$. $p=2 \Rightarrow 2^{p}+p^{2}=8$ is not a prime.

Now suppose $p \not = 3$ is a prime. Then $2^{p}+p^{2}\equiv-1+1 \equiv 0 \mod 3$ using Fermat's theorem. We verify that $2^{3}+3^{2}= 17$ is a prime and see that $3$ is the only solution.[/quote][/hide]

[size=150]Week 3[/size]

You have three piles of stones containing 5, 49, and 51 stones. You can join any two piles together into one pile, and you can divide a pile with an even number of stones into two piles of equal size. Is it possible to eventually have 105 piles each with one stone?
[hide="my solution"]Combine the 5 and 51 piles, then split the 56 piles down to 8 sevens. Combine a seven with the 49 pile, and split down to 8 more sevens giving us 15 piles of seven. From here it is unsolvable; so there is no solution is my answer...[/hide][hide="Week 3 Solution"][u][b]Week 3 Solution[/b][/u]

[quote="jrshoch"]First, if all of the current piles have a common prime factor other than $2$, let's call it $n$, then the possible resulting piles must be multiples of $n$.  This follows logically from the fact that if the piles have $a_{1}n$, $a_{2}n$, $a_{3}n$... stones, the only possible results from adding the piles will be multiples of $n$, and splitting the piles in half cannot eliminate $n$ as a factor, since $n$ is prime and not $2$, and is therefore odd.

So if the piles reach a situation in which they all share such a common prime factor $n$, it would be impossible for the the piles to all be split into piles of $1$, because all resulting piles must be multiples of $n$.

The only possible first steps are adding the piles containing $5$ stones and $49$ stones, $5$ and $51$ stones, or $49$ and $51$ stones, because all the piles are odd.  This would result of piles of $54$ and $51$, $49$ and $56$, or $5$ and $100$, respectively.  These pairs have common prime factors $3$, $7$, and $5$, respectively, which means, by the previous argument, that they cannot be divided into only piles of $1$.  And since this covers all possible situations, it follows that the piles containing $5$, $49$, and $51$ stones cannot be added or divided into only piles of $1$.[/quote][/hide]

[size=150]Week 4[/size]

Let $\mathcal{S}$ be a subset of $[0,1]$ with [u]total[/u] length greater than $\frac{1}{2}$. Prove that there are two elements of $\mathcal{S}$ that differ by exactly $0.1$.
[hide="my solution"]If we can prove that three exist elements farther than .1 unit apart and there exists elements closer than .1 apart, then by continuity there exist elements exactly .1 apart. But these are both trivial to prove (pidgeonhole into 10 regions to prove there are two elements less than, and all elements must be in a segment of length .2 for greater than to fail).
[/hide][hide="Week 4 Solution"]Solution: Consider partitioning the interval $[0,1]$  into the pairs

$\{\alpha, \alpha+\frac{1}{10}\}$

for $\frac{k}{5}\le \alpha \le \frac{k}{5}+\frac{1}{10}$ with $0 \le k \le 4$ . By the given condition, we can have at most one value from each of these pairs so at most half of the total length, and since the total length of the interval is 1 , the maximum length of $S$ is $\frac{1}{2}$ . $QED$.[/hide]

[size=150]Week 5[/size]

Prove that, for all values of $\theta$, (provided that $\tan\theta$ and $\sec^{2}\theta$ are defined), \[2\tan\theta \leq \sec^{2}\theta\] [hide="my solution"]$\sin^{2}\theta+\cos^{2}\theta=1\implies \tan^{2}\theta+1=\sec^{2}\theta$ (divide thru by $\cos^{2}\theta$) \[2\tan\theta \leq \sec^{2}\theta \equiv 2\tan\theta \leq \tan^{2}\theta+1\] which is just AM-GM on $\tan^{2}\theta$ and $1$
[/hide][hide="Week 5 Solution"][quote="matt276eagles"]Since the maximum of the function $f(x)=\sin x$ is $1$, we have $\sin 2\theta\le 1$ for all values of $\theta$.  Using the double angle identity for sine, we obtain $2\sin\theta\cos\theta\le 1$.

Dividing by $\cos^{2}\theta$, which is positive for all values of $\theta$ such that $\tan\theta$ and $\sec^{2}\theta$ are defined, we have $\frac{2\sin\theta}{\cos\theta}\le\frac{1}{\cos^{2}\theta}$.  By the definitions of tangent and secant, we have $2\tan\theta\le\sec^{2}\theta$, which is the desired result.[/quote][/hide]

[size=150]Week 6[/size]

Let $a$, $b$, and $c$ be fixed natural numbers such that, for every positive integer $n$, there is a triangle whose sides have length $a^{n}$, $b^{n}$, and $c^{n}$.   Prove that these triangles are not scalene.[/quote]
[hide="my solution"]Assume no two sides are equal. Then relabel c as the strictly largest side. We now show that for large enough $k, c^{k}>a^{k}+b^{k}$ (which means the triangle inequality no longer holds). This is equivalent to $1>(\frac{a}{c})^{k}+(\frac{b}{c})^{k}$, which is obviously true because $\frac{a}{c},\frac{b}{c}<1$.[/hide][hide="Week 6 Solution"][quote="Jiang"]Let $a>b>c$ and $a<b+c$. For there to be a triangle for every positive integer n, $a^{n}<b^{n}+c^{n}$ must be true for all n. Dividing the inequality by $b^{n}$, we get $(\frac{a}{b})^{n}<(\frac{c}{b})^{n}+1$. Because $\frac{a}{b}>1$, as n gets larger, $(\frac{a}{b})^{n}$ also gets larger without limit. While because $\frac{c}{b}<1$, as n gets larger, $(\frac{c}{b})^{n}$ approaches more and more to 0. So the inequality would eventually prove to be false. If either two or all of the sides are same, it is obviouse that they will always be the same for any n. A scalene triangle must have all of its sides in different lengths, so it would not be possible.[/quote][/hide]

[size=150]Week 7[/size]

$f=x^{3}-x-1$
Prove that $f$ has [i]only[/i] one real root $r \in \left(1, \frac{3}{2}\right)$
Prove that if $a$ is a root and $a \notin \mathbb{R}$ then $|a| \in \left( \sqrt{\left(\frac{2}{3}\right)}, 1\right)$

They tried to turn the thread into a proof marathon, but it didn't work
I will volunteer to take turns posting problems with someone... I hope there is interest!

Your week 3 solution is incomplete. Just because you found a case in which you did not reach the desired ending does not mean it is not possible at all.

Yes, I didn't think of how to finish that before I looked at the answer... now its pointless to change what I wrote


Find all functions $ f: \mathbb{N}\rightarrow\mathbb{R}$ such that for all $ x,y,z\in\mathbb{N}$,
\[f(xy)\plus{}f(xz)\geq f(x)f(yz)\plus{}1.\]

this is quite interesting question
$ f(xy) \plus{} f(xz) >\equal{} f(x)f(yz)\plus{}1$
hence we get
$ f(0)\plus{}\frac{1}{f(0)}<\equal{}2$
but we know
$ f(0)\plus{}\frac{1}{f(0)}>\equal{}2$
hence we can say
$ f(0)\equal{}1$
now, put $ x\equal{}y\equal{}z\equal{}1$
we get
$ f(1)[2\minus{}f(1)]>\equal{}1$
which again gives
$ (f(1)\minus{}1)^2\equal{}0$
that is $ f(1)\equal{}1$
now,
put y=0
$ f(0)\plus{}f(xz)>\equal{}f(x)f(0)\plus{}1$
solving this we get
$ f(xz)>\equal{}f(x)$..........................(1)
put x=1
$ f(z)>\equal{}f(1)$............................(2)
also put z=1 in (1)
we $ f(0)>\equal{}f(x)$......................(3)
combining (2) and (3)
we get
$ f(1)<\equal{}f(x)<\equal{}f(0)$
that is
$ 1<\equal{}f(x)<\equal{}1$
hence we get
$ f(x)\equal{}1$

In this question $ \mathbb{N}$ is a set of natural numbers,i.e $ \mathbb{N}\equal{}\{1,2,3\dots\}$.So you can't consider $ f(0)$.

[quote="Erken"]Find all functions $ f: \mathbb{N}\rightarrow\mathbb{R}$,such that for all $ x,y,z\in\mathbb{N}$:
$ f(xy) \plus{} f(xz)\geq f(x)f(yz) \plus{} 1$.[/quote]

Let $ P(x,y,z)$ the property $ f(xy) \plus{} f(xz)\geq f(x)f(yz) \plus{} 1$

$ 1)$ $ P(1,1,1)$ gives $ 2f(1)\geq f(1)^2\plus{}1$, and so $ 0\geq f(1)^2\minus{}2f(1)\plus{}1\equal{}(f(1)\minus{}1)^2$ and so $ \boxed{f(1)\equal{}1}$

$ 2)$ $ P(x,1,1)$ gives $ 2f(x)\geq f(x)f(1)\plus{}1\equal{}f(x)\plus{}1$ and so $ \boxed{f(x)\geq 1\: \forall x\in\mathbb{N}}$

$ 3)$ $ P(x,x,x)$ gives $ 2f(x^2)\geq f(x)f(x^2)\plus{}1$ and so $ f(x^2)(f(x)\minus{}2)\leq \minus{}1$ and so $ \boxed{f(x)<2\: \forall x\in\mathbb{N}}$

$ 4)$ Let then $ a\equal{}\sup(\{f(n),n\in\mathbb{N}\})$ the least upper bound of $ f(\mathbb{N})$. $ a$ exists since $ f(\mathbb{N})\subseteq[1,2)$ and $ 2\geq a\geq 1$
Let then $ a>\epsilon>0$ and $ p$ such that $ 0\leq a\minus{}f(p)<\epsilon$
$ P(p,p,1)$ gives $ f(p^2)\plus{}f(p)\geq (f(p))^2\plus{}1$. And, since LHS is less tha $ a\plus{}a$ : $ (f(p))^2\plus{}1 \leq 2a$ and so $ (a\minus{}\epsilon)^2\plus{}1\leq 2a$
And so $ (a\minus{}1)^2<2\epsilon a \minus{}\epsilon^2<4\epsilon$ $ \forall \epsilon\in(0,1)$
And so $ a\equal{}1$
And so $ \boxed{f(x)\equal{}1\: \: \forall x\in\mathbb{N}}$


geometry
trigonometry
function
conics
parabola
analytic geometry
quadratics

Find the area bounded by the curve $ C$ defined by $ x(t) \equal{} \cos{t}\sin{t}$ and $ y(t) \equal{}\minus{} \sin{t}\sin{3t}$.

That can't be the correct picture.

A small test: let $ t$ be a small positive number. Then all of its trignometric functions are positive and the point should lie in the first quadrant very close to the origin. In fact, we can say that for small positive $ t,$ the curve is approximated by the parabola $ (t,3t^2).$ Which isn't what you drew.

As for the area - I've always liked the Green's Theorem based formula: for a simple closed counterclockwise oriented curve $ C,$ the area enclosed is
\[ A \equal{} \frac12\oint_C \minus{} y\,dx \plus{} x\,dy
\]
That will be a fairly messy thing to compute here, but it can be computed.

Edit: Now the picture makes sense.

Actually it's not bad -- just upside down ;)  (And now fixed.)

I don't like polar coordinates, and these equations are not that hard to de-polar-ize:  we can rewrite the defining equations as $ x \equal{} \frac {1}{2} \sin 2t$ and $ y \equal{} \frac {1}{2} (\cos 2t \plus{} \cos 4t)$.  This means $ 4\left((y \minus{} 1 \plus{} 2x^2)^2 \plus{} x^2\right) \equal{} 1$, and we can solve this as a quadratic equation for $ y$.  To actually integrate, we have to find the points where one of the solutions crosses the $ x$-axis, which involves solving a quartic that's really just a quadratic in $ x^2$.

hello, i have plotted the curve with Maple, i think your picture must be upside down.
Sonnhard.

my apologies, $ y \equal{} \minus{} \sin{t}\sin{3t}$
Why jump all the way to greens? Can we not partition the range $ t\in (0,\pi)$ appropriately and apply $ \int^{t_2}_{t_1} y\text{ }dx$ as the area bounded by the x-axis and the curve $ C$? The visualization is difficult, but I could certainly draw the series of rectangles with mathematica.
Let's assume we do not partition $ (0,\pi)$, and simply compute $ \int^{0}_{\pi} y\text{ }dx \equal{} \frac {\pi}{4}$. Is this result correct?

[b]edit:[/b] Yes, Dr. Sonnhard it is. I flipped the original curve so that it would more closely resemble the star trek logo.


Seven distinct points are identified on the circumference of a circle. How many different triangles can be formed if each vertex must be one of these 7 points?

since no 3 lie on a straight line, the number is simply $ \binom{7}{3}\equal{}35$

For more information, go to the entry on Combinations in my blog. Link's in my signature.  :wink:

[quote="vallon22"]for those of u who don't know what $ \binom{7}{3}$ means, it basically just means......

with any number $ x$ and $ y$, $ \binom{x}{y} \equal{} \frac {x!} {(y!)((x \minus{} y)!)}$

idk how clear that was[/quote]

using this definition, $ x$ and $ y$ must be non-negative integers.  Note that $ 0!\equal{}1$.

There is also the form that looks uglier but is still useful:
$ \binom{x}{y}\equal{}\frac{x(x\minus{}1)(x\minus{}2)(x\minus{}3)...(x\minus{}y\plus{}1)}{y!}$.


limit
real analysis
real analysis unsolved

Denote:\[F(z)=\prod_{n=0}^{\infty}\left(1-z^{2^{n}}\right),|z|<1\]
Prove that $F(a)$ is transcendental for every algebraic number $a$,with $|a|<1$.

I know it has been proved by Mahler,but I can't find the proof.Can somebody offer a proof or a link.Thanks

[quote="zhaobin"]Denote:
\[F(z)=\prod_{n=0}^{\infty}\left(1-z^{2^{n}}\right),|z|<1 \]
Prove that $F(a)$ is transcendental for every algebraic number $a$,with $|a|<1$.

I know it has been proved by Mahler,but I can't find the proof.Can somebody offer a proof or a link.Thanks[/quote]

I don't know if it helps, but recall that for $|z|<1,$

\[h(z)=\prod_{n=0}^{\infty}\left(1+z^{2^{n}}\right)=\lim_{N\to\infty}(1+z)(1+z^{2})(1+z^{4})(1+z^{8})\cdots\left(1+z^{2^{N}}\right)\]
\[=\frac1{1-z}\lim_{N\to\infty}(1-z)(1+z)(1+z^{2})(1+z^{4})(1+z^{8})\cdots\left(1+z^{2^{N}}\right)\]
\[=\frac1{1-z}\lim_{N\to\infty}(1-z^{2})(1+z^{2})(1+z^{4})(1+z^{8})\cdots\left(1+z^{2^{N}}\right)\]
\[=\frac1{1-z}\lim_{N\to\infty}(1-z^{4})(1+z^{4})(1+z^{8})\cdots\left(1+z^{2^{N}}\right)\]
\[\vdots\]
\[=\frac1{1-z}\lim_{N\to\infty}\left(1-z^{2^{N+1}}\right)\]
\[=\frac1{1-z}\]

and hence

\[|F(z)| = \left|\prod_{n=0}^{\infty}\left(1-z^{2^{n}}\right)\right| = \prod_{n=0}^{\infty}\left|1-z^{2^{n}}\right| \leq \prod_{n=0}^{\infty}\left(1+|z|^{2^{n}}\right)=\frac1{1-|z|}\]

Thank you...
But I don't see if there is any hope from your hint :(

I had started a topic on this some time ago, look here: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=irrationality&t=121584

The above is not hint at all...

I can show you a proof if no one else becomes interested in this problem -- it is a really beautiful and surprisingly simple (for transcendence theory) method. But again, let's see first if anyone else is interested, so that I don't spoil the fun.

My suggestion: try to prove first that $f(2/3)$ is irrational. Once you manage to do this, you should have no problem in generalizing your arguments to show the general fact. Now, the problem is difficult without a hint and quite simple after the hint, but... ok, I will give a hint if no one else posts in the next couple of days. And a solution then, if the hint is insufficient. :)

Vess,Can you post the solution now?
I even can't prove $f(2/3)$ is irrational...
Thank you


One is that
In m = n if $m = e^n$

By the way, e is the value of $\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} +...$

In general, In($e^x$) = x is a special case of the theorem $\log_b(b^x) = x$

I don't get natural logrithms too well either.

[quote="4everwise"]My textbook covers a lot of things that can be done with $log_{10}$, but does not cover $log_e$ that well. Posting of formulas involving the natural base e in logarithms would be most appreciated.[/quote]It looks like you're just asking for formulas that involve the natural logarithm function, or $\ln{x}$ or $\exp{x}$. There are hundreds upon hundreds of equations you could find involving this function, so I don't really know what you're looking for. Here's a few I found on MathWorld, althought it's probably not what you're looking for at all. \[\ln(1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\ldots.\]\[\ln\frac{1+x}{1-x}=\frac{2x}{1-\frac{x^2}{3-\frac{4x^2}{5-\frac{9x^2}{7-\frac{16x^2}{9-\ldots}}}}}\]

[quote="nolachrymose"][quote="236factorial"]By the way, e is the value of $\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} +...$[/quote]

Actually, you made a little mistake.
$e = \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$[/quote]

Oh yeah, sorry  :oops:
That reminds me of another topic. *to be posted soon*


Anybody knows the lowest score needed for the Open math in order to qualify for the camp?

No, I got a 73 and made it. I'm thinking that It'll be mid 60's.

i had a 72, got an invitation, but didn't apply :( cause i just got the letter last monday and i had too much homework

you make it if u have mid 60s? WTF WHERE MY INVITATION IM RIGHT ON MID 60. -.-

meh i got high 60's but no invitation
i guess the cutoff this year was low 70's.. maybe 70 or 71?

I got 67 and I made it.  :D Am I probably the lowest scorer?

[quote="zjzjimbo"]Well，first of all, you have to be in grades 8-10 in order to be invited..
I think the cutoff is mid-60s.[/quote]

umm.. i'm in gr. 11, and got invited.. i think they invite anyone who qualifies in their list :p

That's strange. Both my invitation and [url]http://www.cms.math.ca/MathCamps/indexe[/url] say you need at least 2 years remaining in high school (ie 8-10).

OOOHHHHH LMFAO

srry guys, i was thinking about the CANADA/USA Math Camp  :blush:

yea, i went to the "national math camp" last year, it was pretty awesome :D

Its next week!
It should be lots of fun.
By the way, what is the level of the material covered compared to the seminar?


Find all real numbers $ x$ such that \[ \sqrt{x\plus{}1}\plus{}\sqrt{x\plus{}3} \equal{} \sqrt{2x\minus{}1}\plus{}\sqrt{2x\plus{}1}.\]

[quote="ZetaX"]Find all real numbers $ x$ such that
\[ \sqrt {x \plus{} 1} \plus{} \sqrt {x \plus{} 3} \equal{} \sqrt {2x \minus{} 1} \plus{} \sqrt {2x \plus{} 1}.
\]
[/quote]
[hide]
The difference in the radicals on both sides is two...also $ x\ge \frac{1}{2}$
So they should only be equal when $ x\plus{}1\equal{}2x\minus{}1$ and $ x\plus{}3\equal{}2x\plus{}1$...or $ x\equal{}2$ ...[/hide]

[hide="In other words"] $ f(x)\equal{}\sqrt{x\minus{}1} \plus{} \sqrt{x\plus{}1}$ is a strictly increasing function, so $ f(a)\equal{}f(b) \implies a\equal{}b$

Here, $ f(x\plus{}2)\equal{}f(2x) \implies x\plus{}2\equal{}2x \implies x\equal{}2$[/hide]

[quote="ZetaX"]Find all real numbers $ x$ such that
\[ \sqrt {x \plus{} 1} \plus{} \sqrt {x \plus{} 3} \equal{} \sqrt {2x \minus{} 1} \plus{} \sqrt {2x \plus{} 1}.
\]
[/quote]

Rationalization of both sides gives $ \sqrt{x\plus{}3}\minus{}\sqrt{x\plus{}1}\equal{}\sqrt{2x\plus{}1}\minus{}\sqrt{2x\minus{}1}$.

Adding both sides of these equations, we have $ 2\sqrt{x\plus{}3}\equal{}2\sqrt{2x\plus{}1}$, yielding $ x\equal{}2$.

Very nice, kunny!  :)

[hide="More traditionally"] Write as $ \sqrt{x \plus{} 1} \minus{} \sqrt{2x \plus{} 1} \equal{} \sqrt{2x \minus{} 1} \minus{} \sqrt{x \plus{} 3} \implies$
$ 2 \sqrt{(x \plus{} 1)(2x \plus{} 1)} \equal{} 2 \sqrt{(2x \minus{} 1)(x \plus{} 3)} \implies$
$ 2x^2 \plus{} 3x \plus{} 1 \equal{} 2x^2 \plus{} 5x \minus{} 3 \implies$
$ x \equal{} 2$.

We substitute to verify. [/hide]

How does $ \sqrt{x \plus{} 1} \minus{} \sqrt{2x \plus{} 1} \equal{} \sqrt{2x \minus{} 1} \minus{} \sqrt{x \plus{} 3}$ imply $ 2 \sqrt{(x \plus{} 1)(2x \plus{} 1)} \equal{} 2 \sqrt{(2x \minus{} 1)(x \plus{} 3)}$?

Squaring gives $ 3x \plus{} 2 \minus{} 2 \sqrt {(x \plus{} 1)(2x \plus{} 1)} \equal{} 3x \plus{} 2 \minus{} 2 \sqrt {(2x \minus{} 1)(x \plus{} 3)}$.  The key observation is that $ (x \plus{} 1) \plus{} (2x \plus{} 1) \equal{} (2x \minus{} 1) \plus{} (x \plus{} 3)$.  (This is actually the same key observation that motivates the other solutions.)


floor function
calculus
calculus computations

Dear all;

Can anyone provide assistance in the solving of the question below?

1) Does $ \sum{\frac{\sqrt{n\plus{}1}\minus{}\sqrt{n}}{\sqrt{n}}}$ converge? Justify your answer.

2) Use Dirichlet's test to show that $ \frac{1}{1}\plus{}\frac{1}{2}\minus{}\frac{1}{3}\minus{}\frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}\minus{}\frac{1}{7}\minus{}\frac{1}{8}$+... converges.

Thank You.

hello, to your first question. Write $ \frac{\sqrt{n\plus{}1}\minus{}\sqrt{n}}{\sqrt{n}}$ in the form $ \frac{1}{\sqrt{n^2\plus{}n}\plus{}n}$. It is easy to prove that $ \frac{1}{\sqrt{n^2\plus{}n}\plus{}n}>\frac{1}{3n}$ holds for $ n \geq1$ and $ \sum_{n\equal{}1}^\infty \frac{1}{3n}$ does diverge so your sum does also diverge.
Sonnhard.

Thank You Dr. Sonnhard. May I know how do you know you need to compare it with $ \frac{1}{3n}$? Like how you see you need to compare it with $ \frac{1}{3n}$

For my original answer before I posted here, I let $ An$ be the original series and $ Bn \equal{} \frac{1}{\sqrt{n\plus{}1}(\sqrt{n\plus{}1}\plus{}\sqrt{n\plus{}1})}$ Since Bn diverge, so is $ An$.

[quote="Symbian"]May I know how do you know you need to compare it with $ \frac {1}{3n}$? Like how you see you need to compare it with $ \frac {1}{3n}$[/quote]
After that algebraic manipulation, we can see that the terms "look like" $ \frac1{2n}$ for large $ n.$ We could use exactly that, and call it the Limit Comparison Test, or we could do what Dr. Graubner did, which is to do a one-sided comparison to a constant times $ \frac1n$ where the constant is smaller than $ \frac12.$ That's where $ \frac13$ came from - just some convenient constant that is smaller than $ \frac12.$ He could have said $ \frac1{5n}$ or $ \frac1{\pi n}$ - the $ 3$ was just arbitrary.

Fully understood. Thanks Kent Merryfield.

Anyone would like to have a go on Part 2 of my question?

I see two ways for showing it:

1. Split the series in two: $ \sum \frac {( \minus{} 1)^{n \plus{} 1}}{2n}$ and $ \sum \frac {( \minus{} 1)^{n \plus{} 1}}{2n \minus{} 1}$ and then it's easy to apply Leibniz' test (which is a consequence of Dirchlet's test).

2. Apply the Leibniz test on the original series: can you find the general term? The general term is $ ( \minus{} 1)^\alpha\frac {1}{n}$ (find the alpha!).

[hide]Alpha can be given as $ \lfloor \frac{n\minus{}1}{2}\rfloor$[/hide]

As the sequence $ \frac {1}{n}$ is monotone decreasing and its limit is zero, and the partial sums of $ \sum( \minus{} 1)^\alpha$ is bounded, the conclusion follows.


logarithms
algorithm
function
analytic geometry
search

How to solve it? In the formu i found that it is: BFS+deque.

[url]http://acm.tju.edu.cn/toj/showp2505.html[/url]

but how?

[hide="an idea"]
The rocks break stuff up into at most 8 regions. So we could BFS stuff, and keep track of where the two rocks are and which region the person is in.

In this case the only real problem is finding some fast way to update the regions with very little memory. Perhaps some modification of Union-Find would allow this in $O(\log (N))$, though it is questionable as to whether $8N^{4}\log (n)$ runs in time.
[/hide]

can you please clear it a little bit more.

have you considered that, a stone can be pushed more than one time?

Yes.

The idea is that we keep track of where each stone is, and what range of movement the person has. The problem is how to quickly update the range of movement. I was just sharing my thoughts on the problem, I have not yet solved it either.

Another optimization, though somewhat messy, is to compress all regions that are significantly wide that they could never be blocked by the stones into a single node and treat the entire structure simply as a graph. This could probably speed up your worst-case performance by a factor of about 16 (as effectively the possible nodes drops from 2500 to 625), but I can think of no remotely good way to code it in a reasonable amount of time.

Another idea:

Find all accessible position for the rock A.
For each (new) position of A find all accessible positions of the rock B ...
Repeat until you can reach the treasure

With "accessible positions" I mean, where we can move this rock without move any other rocks.
For example, the accessible positions of the rock in (2,5) are
(3,5), and you are on (2,5)
(4,5), and you are on (3,5)

The above algorithm can be done easily. We will need:

[color=brown]
[u][b]1st function[/b][/u]
function Reachable(mx,my,x,y)
returns true if we can reach the position (x,y) starting from (mx,my) without move any rock[/color]

[color=green]
[u][b]2nd function[/b][/u]
The function that finds the accessible positions works like this:
Let (ax,ay) be the position of the rock A
Find an adjacent cell of the rock A, [u]for example[/u] the cell (ax-1,ay)
If this cell is free:
If this is reachable:
If the symmetric cell, (ax+1,ay) is free:
Then move the rock to (ax+1,ay) - so you go to the position (ax,ay)
This position is labeled as accecible (you may keep and your current position I think)[/color]

Both the functions are recursive.
To avoid checking already checked positions, when we check the position (x,y) we put the value CHECKED[x][y]=1 in the array CHECKED
[color=red]This works for the Reachable function, but I'm not sure if we can use it for the Accessible_Positions function[/color]

[color=blue]
[b]P.S.[/b]
Forgive me if something above is confusing, or if I wrote something which is previously written, but I haven't deal with such problems (when I started thinking about this problem, I didn't know how to find the reachable positions)

If we have more than 2 rocks the problem becomes more complicated because, after moving the rock A we don't know which rock we should move so we must check all the possible cases

And what is this "BFS" thing?[/color]

That is interesting, but I think runtime is still a problem here. However, going off of your idea, I have another idea that should run fairly quickly:

For the BFS, always move the same block as you previously moved, unless such a move would change the range of motion of the other block, in which case you expand from both.

This should run fairly quickly, as although it is still theoretically O(N^6), I think the runtime should in reality be much faster, as you branch only when movement is inhibited or freed in some way, and if we can optimize the "find reachable positions" functon then it could be even faster.

well keep in mind these types of cases:

[code]
.T#.
....
#AB.
S.#.
[/code]

where the person starts from S, pushes A, then pushes B, then again A and then to T.

Yeah, that would be covered because the range of motion for B chances (as now it is possible to push it).

I still cannot understand what is [b]BFS[/b]

Well, I didn't like my previous suggestion. I was adversely influenced by http://www.games4brains.de/sokobluejavaenglisch.php

[hide="Here is another suggestion"]
We won't find all possible destinations of the rock. We'll move it one step at a time.

The position of the treasure and the wall-blocks are constant.

Let $M=(mx,my)$ be the coordinates of the player
$R_{i}= (x_{i},y_{i})$ the coordinates of the rock number-$i$
$R[\ ]$ the array of all the rock-coordinates

You have to solve the problem [color=blue]Task (M,R)[/color]

[u]In the procedure [color=blue]Task[/color]:[/u]

- Choose a rock
- Choose an adjacent cell
- If possible, move the rock one step only.

Let M' be the new position of the player and R' the new array of the rocks. Then you have to solve the problem [color=blue]Task(M',R')[/color]
- Continue until you can reach the treasure

I think that this algorithm is faster than my previous post, because we move a rock one step each time.

Again we have to mark the checked positions to avoid endless loops.
And this is the hard part...[/hide]

BFS = Breadth First Search

Basically you keep a queue of states, every time you take something out of the queue, if it doesn't complete the problem, you take all not-already-visited states that can be reached from the current state, and put them in at the end of the queue.

Thank you! :)

I googled "Breadth First Search" and I found interesting things.

Well, I wrote a program in JavaScript, that finds quickly [u]one[/u] solution, not the faster.
Accurately, when it finds a sequence of moves that lead to the treasure, it stops.
If I change it to search for all the possible moves, it will become too slow.
[color=brown](anyway, if we use other language, the program will be faster; I simply prefer JavaScript because it's easy and nice)[/color]

Also, I didn't check the ability to go back to the entrance.

I'm still trying to find ways to avoid meaningless moves in my algorithm

well i think, it can be solved by BFS. just think a bit. using BFS is enough, deque is not needed


conics
hyperbola
ellipse
trigonometry
function
algebra proposed
algebra

Show that the distance between a point on the hyperbola $xy=5$ and a point on the ellipse $x^{2}+6y^{2}=6$ is at least $\frac{9}{7}$.

Denote $M(a,\frac{5}{a})(a>0)$ and $N(\sqrt{6}\cos{\alpha},\sin{\alpha})(0\leq\alpha \leq \frac{\pi}{2})$. We only need prove $MN\geq \frac{9}{7}$. Or , prove $f(a,\alpha)=(\sqrt{6}\cos{\alpha}-a)^{2}+(\frac{5}{a}-\sin{\alpha})^{2}\geq\frac{81}{49}\forall a>0\forall \alpha\in [0,\frac{\pi}{2}]$.

Now, i think I can fix $\alpha$, use analysis with function $g(a)=f(a,\alpha)$,...


An integer $ n>101$ is divisible by $ 101$. Suppose that every divisor $ d$ of $ n$ with $ 1<d<n$ equals the difference of two divisors of $ n$. Prove that $ n$ is divisble by $ 100$.

Let $ \frac{n}{101}\equal{}\frac{n}{k}\minus{}\frac{n}{m}$, were $ 101,k,m, n/101,n/k.n/m$ are divisors of n.
Then $ 101k\equal{}m(101\minus{}k)$. Because $ 101$ is prime $ 101\minus{}k|k\to k\equal{}100\to 100|n$.


There are n triangles of positive area that have one vertex A(0,0) and the other two vertices whose coordinates are drawn independently with replacement from (0,1,2,3,4).  Compute n

[hide="solution"]We need to find the number of possible coordinates for the other two vertices of the triangle, and the coordinates are chosen from the first 5 whole numbers. There are $ 5*4\equal{}20$ possible coordinates, and we need to choose two of them. $ \binom{20}{2}\equal{}\boxed{190}$[/hide]

Consider the pair $ (1, 1)$ and $ (2, 2)$, then modify your solution :)

What does this mean??

[quote="firecricket91"]of positive area [/quote]

 :)

If lets say B is at (3,2) and C is at (2,3).

Is that traingle different than a triangle that has C at (3,2) and B and (2,3)?

[hide]
First these points are not coincident, so in this $ 5\times 5$ grid, with $ (0,0)$ alreday chosen, we have $ \binom{24}{2}\equal{}276$ triangles.
However, we also need to eliminate the cases when the three points are collinear. Within this grid, among all the rays started from origin, we have three rays with four lattice points and two rays with two lattice points. Thus, we need to subtract $ 3\binom{4}{2}\plus{}2\binom{2}{2}\equal{}20$ triangles, so $ n\equal{}256$.

[/hide]

When it means n trianlges does it mean the vertices can be interchanged?

Because it gave A as (0,0) doenst that kinda imply the verticies matter

A triangle doesn't care what you name its vertices.


$ ABC$ is a triangle. We draw a circle that tangents to $ BC$ and the extensions of $ AB,CA$. Similarly, draw two other circles (tangent to $ CA$ and $ AB$ externally). Let the centers be $ I_a,I_b,I_c$. Prove that $ AI_a,BI_b,CI_c$ are the altitudes of triangle $ I_aI_bI_c$.

I think you are meaning the excircles.
[hide]If so the center of any of them are equidistant from the 3 sides.So lies on the angle bisector of the opposite angle.And the centers lie on the external bisectors of the angles at the vertexes (same equidistant argument).And the result follows from that the internal and external bisectors create a right angle between them.[/hide]

Dear Mathlinkers,
this problem comes from Jules Mention.
Sincerely
Jean-Louis


Given a function $ f$ continuous on the closed interval $ [0,1]$, let $ 0 \leq f(x) \leq 1$ whenever $ 0 \leq x \leq 1$. Prove that there exists some $ c \in (a,b)$ such that $ f(c) \equal{} c.$

Consider $ g(x) \equal{} f(x) \minus{} x$. Note that $ g(0) \geq 0$ and $ g(1) \leq 0$. Hence, there is $ c \in (0,1)$ such that $ g(c) \equal{} 0$ or $ f(c) \equal{} c$.

This problem gives rise to a very difficult conjecture.

This is a consequence of the intermediate value theorem.

Every continuous and bounded function on a given interval must "cross" the function y = x at some point on that interval.


Find all integers$ (b,c)$ such that
$ \frac{2bc}{b\plus{}c\plus{}1\minus{}bc}$ is an integer

I have got stuck here solving a problem of United Kingdom '95

Oni.

Set $ x \equal{} b \minus{} 1$,$ y \equal{} c \minus{} 1$ .....$ xy \minus{} 2/2(x \plus{} y \plus{} 3) \equal{} > |xy \minus{} 2| < \equal{} 2|x \plus{} y \plus{} 3|$

I can't understand ur solution. Plz write it fully...

$ b \plus{} c \plus{} 1 \minus{} bc/2bc \equal{} > bc \minus{} b \minus{} c \plus{} 1 \minus{} 2/2bc \equal{} > (b \minus{} 1)(c \minus{} 1) \minus{} 2/2bc \equal{} > (b \minus{} 1)(c \minus{} 1) \minus{} 2/2(b \minus{} 1)(c \minus{} 1) \plus{} 2(b \plus{} c \minus{} 1) \equal{} > (b \minus{} 1)(c \minus{} 1) \minus{} 2/2[(b \minus{} 1)(c \minus{} 1) \minus{} 2] \plus{} 2(b \minus{} 1) \plus{} 2(c \minus{} 1) \plus{} 4 \plus{} 2 \equal{} > (b \minus{} 1)(c \minus{} 1) \minus{} 2/2(b \minus{} 1) \plus{} 2(c \minus{} 1) \plus{} 6$ set $ x \equal{} b \minus{} 1$ , $ y \equal{} c \minus{} 1$ so $ xy \minus{} 2/2(x \plus{} y \plus{} 3) \equal{} > |xy \minus{} 2| < \equal{} 2|x \plus{} y \plus{} 3| \equal{} > .....$


the whole objective is to present a complex mathematical concept or some sort, in a visual presentation.

i've checked out some proof without words as an idea..

any suggestions?

A nice theorm for visual presentation.. hmm...  the more complex it is the higher marks earned!

RM>=AM>=GM>=HM with two variables.  It can be done in a semi-circle.

[quote="miraculouspostmaster"]A few years ago I did a project on Lewis Carroll's algorithm for finding the day of the week for any given date (using the numbers of the day, year, and month).  Quite a useless concept, obviously, but an interesting one.  I wrote some computer programs and did a powerpoint presentation.[/quote]

wow I do too did a presentation on that, and it may be useless, but it impresses others.

[quote="miraculouspostmaster"]A few years ago I did a project on Lewis Carroll's algorithm for finding the day of the week for any given date (using the numbers of the day, year, and month).  Quite a useless concept, obviously, but an interesting one.  I wrote some computer programs and did a powerpoint presentation.[/quote]
i'd like to know how that works.

You should do a study on infinity.

Show properties and paradoxes regarding the idea.

Explain its importance in math.

I read something in a thread recently about how a point from (0,0) to (0,1) has an infinite number of points. However, so does a point from (0,0) to (2,0)

It is easy to see that the latter has a larger length, but both contain an infinite number of points. I believe someone said that showed how you can have different (quantities of?) infinite.

Interesting to me...

[url=http://www.cs.usyd.edu.au/~kev/pp/TUTORIALS/1b/carroll.html]Day of the Week[/url]

Maybe you can do a project on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12191]Visual Calculus[/url]?

heres something deep...
how does the euclidian method of solving for lowest common denominator work...

I think that's a bit tough for your average high schooler... :)


algebra
polynomial
algorithm
number theory
Euclidean algorithm

Let $ p(x) \equal{} x^{2}\plus{}bx\plus{}c$, where b and c are integers. If $ p(x)$ is a factor of both $ x^{4}\plus{}6x^{2}\plus{}25$ and $ 3x^{4}\plus{}4x^{2}\plus{}28x\plus{}5$, find $ p(1)$.

Answer

By dividing $ x^{4}\plus{}6x^{2}\plus{}25$ by $ x^{2}\plus{}bx\plus{}c$ using polynomial division we get remainder $ (bc\minus{}b(6\minus{}c\plus{}b^{2}))x\plus{}25\minus{}(6\minus{}c\plus{}b^{2})c$.

This remainder must be zero for all $ x$

Hence $ bc \equal{} b(6\minus{}c\plus{}b^{2})$ and $ 25 \equal{} (6\minus{}c\plus{}b^{2})c$

If $ b \equal{} 0$ then $ 25 \equal{} (6\minus{}c)c$

$ \Rightarrow c^{2}\minus{}6c\plus{}25 \equal{} 0$

This equation has no real roots and contradicts the fact that $ c$ is an integer.

Hence $ b\not \equal{} 0$ and we can divide the first equation through by b.

$ c \equal{} 6\minus{}c\plus{}b^{2}\Rightarrow b^{2}\equal{} 2c\minus{}6$

$ c^{2}\equal{} 25\Rightarrow c \equal{}\pm 5$

But $ b^{2}> 0$ hence $ c \equal{} 5$.

$ \Rightarrow b \equal{}\pm 2$

Dividing $ 3x^{4}\plus{}4x^{2}\plus{}28x\plus{}5$ by $ x^{2}\plus{}bx\plus{}5$ gives remainder $ (15b\plus{}28\minus{}b(3b^{2}\minus{}11))x\plus{}5\minus{}5(3b^{2}\minus{}11)$

This remainder must be zero for all $ x$:

$ \Rightarrow 15b\plus{}28 \equal{} b(3b^{2}\minus{}11)$

We know that $ b \equal{} 2$ or $ b \equal{}\minus{}2$ and by simple substitution we find that $ b \equal{}\minus{}2$

Hence $ p(x) \equal{} x^{2}\minus{}2x\plus{}5\Rightarrow p(1) \equal{} 4$

I think this is a little easier...
[hide="Solution without division"]Let $ q(x)\equal{}x^{4}\plus{}6x^{2}\plus{}25$ and $ r(x)\equal{}3x^{4}\plus{}4x^{2}\plus{}28x\plus{}5$
If $ p(x)$ divides both of these for all x, then $ p(x)$ also divides $ 3q(x)\minus{}r(x)$...
$ 3q(x)\minus{}r(x)\equal{}14x^{2}\minus{}28x\plus{}70\equal{}14(x^{2}\minus{}2x\plus{}5)$
It follows that $ p(x)\equal{}x^{2}\minus{}2x\plus{}5$, so $ p(1)\equal{}4$[/hide]

Basically we do the euclidean algorithm with polynomials.

$ (x^{4}\plus{}6x^{2}\plus{}25,3x^{4}\plus{}4x^{2}\plus{}28x\plus{}5)$

$ \equal{}(x^{4}\plus{}6x^{2}\plus{}25,(\minus{}14)x^{2}\plus{}28x\minus{}70)$

$ \equal{}(x^{4}\plus{}6x^{2}\plus{}25,x^{2}\minus{}2x\plus{}5)$

$ \equal{}(2x^{3}\plus{}x^{2}\plus{}25,x^{2}\minus{}2x\plus{}5)$

$ \equal{}(5x^{2}\minus{}10x\plus{}25,x^{2}\minus{}2x\plus{}5)\equal{}x^{2}\minus{}2x\plus{}5$

That is to say, there is literally a process that you can follow to solve this problem.


In Triangle ABC we have AB=25, BC=39, and AC=42. Points D and E are on AB and AC respectively, with AD=19 and AE=14. What is the ratio of the area of triangle ADE to the area of quadrilateral BCED?

A. 266/1521

B. 19/75

C. 1/3

D. 19/56

E. 1

I don't know if there is an easier way of doing this, but if you use the formula $\frac{1}{2}ab \sin{C}$.

[hide]Area of $\triangle{ADE}$=$\frac{1}{2}\sin A(19)(14)=133\sin A$.

Area of $\triangle{ABC}$=$\frac{1}{2}\sin A(25)(42)=525\sin A$.

$\frac{[ADE]}{[ABC]}=\frac{19}{75}\Longrightarrow \frac{[ADE]}{[BCED]}=\frac{19}{56}=\boxed{D}$.[/hide]

[hide]Recall that the ratio of the areas of triangles with equal height, is equal to the ratio of their bases.
$[ADC]/[ABC]=19/25$, $[ADE]/[ADC]=14/42=1/3$, $\implies [ADE]/ABC]=19/75 \implies [ADE]/[BDC]=19/(75-19)=19/56, \boxed{D}$.[/hide]


Select four numbers from the table, selecting exactly one from each row and exactly one from each column. What is the sum of the four numbers that produce the least possible product?
\begin{tabular}{|c|c|c|c|}\hline 3&5&7&14\\\hline 15&8&6&4\\\hline 2&9&10&11\\\hline 13&12&15&1\\\hline\end{tabular}

$ \begin{tabular}{|c|c|c|c|}\hline 3&5&7&14\\\hline 15&8&6&4\\\hline 2&9&10&11\\\hline 13&12&15&1\\\hline\end{tabular}$

we should try to choose numbers that are small, so their product is small
i would choose the numbers 1, 2, 5, and 6
their product is 60
and their sum is $ \boxed{14}$


geometry
ratio
incenter
area of a triangle

An equilateral triangle and square are both inscribed in the same circle. Find the ratio of the areas of the triangle to the square.

Let the diameter of the circle be $d$.

Then the square has a diagonal $d$. Also $d^2=s^2+s^2$ by Pythagorean where $s$ is a side of the square, so $s=\frac{\sqrt{2}d}{2}$. Hence, its area, $s^2$, is $\frac{d^2}{2}$.

Consider a triangle within the equilateral triangle formed by radii from two vertices and the side between them. The triangle is $30^\circ-30^\circ-120^\circ$ since the radii are also angle bisectors of the vertices. Drawing a perpendicular down from the incenter, we have a $30^\circ-60^\circ-90^\circ$ triangle formed by the perpendicular, the radius, and half a side of the equilateral triangle. Since the hypotenuse is $\frac{d}{2}$, the half side of the equilateral triangle is $\frac{\sqrt{3}d}{4}$.
The area of a triangle is $\frac{\sqrt{3}s^2}{4}$ where $s$ is the length of a side, so the area of this triangle is $\frac{\sqrt{3}\left(2*\frac{\sqrt{3}d}{4}\right)^2}{4}=\frac{3\sqrt{3}d^2}{16}$.

Thus, the ratio is $\frac{3\sqrt{3}}{8}$.

Okay, so first you have a circle and let's pick $1$ to be its radius. Therefore the equilateral triangle will be equivalent to $3$ iscocles triangles with the two equal sides of $1$. Since those three iscoceles triangles are composed of two $30,60,90$ triangles each thus, the area of all the triangles is $1/2(3^{1/2}/2)(3)$, thus the area of the equilateral triangle is $(3(3^{1/2}))/4$. Now, the square will be composed of a diagonal of $2$ which is the diameter of the circle, thus each side of the square will be $2^{1/2}$ so the area would be $2$, thus the ratio would be $(3(3^{1/2}))/8$.


In how many different ways can four students stand in a straight line if two of the students refuse to stand next to each other?

Complementary counting: $ 24\minus{}12\equal{}12$.

i get the 24 part, but why are you subtracting 12?

One way of thinking of it is to first find out how many total combinations there are then subtract the ones we don't want, which are where the two are next to each other. So in total there are obviously $4! = 24$ different arrangements without the restriction. To deal with the restriction, we let the group of two students that don't want to be next to each other be a "superblock" which we will call G, and the other 2 students will each be a unique S. So we have to find the number of arrangements of $GS_1S_2$. This is going to be ${3!} \times 2! = 12.$


modular arithmetic
number theory theorems
number theory

Solve in positive integers the follwing equation:

     $ 2^x+3^y=z^3 $ ;)

[quote="cezar lupu"]Solve in positive integers the follwing equation:

     $2^x+3^y=z^3$ ;)[/quote]

This simple but nice equation seems so hard. In fact it's not hard to derive what $(x,y)$ in positive integers so that $2^x+3^y$ is square!

For this equation may be the hardest part is in the case $y=1$. In fact for $y>1$ here is some thoughts of mine : Assume that $y>1$, It's easy to check that $z^3\equiv0,\pm1\pmod{7}$ and  $z^3\equiv0,\pm1\pmod{9}$. So firstly works on modulo $9$, then $x$ must be divisible by $3$. Now work on modulo $7$ then $y$ must be divisible by $3$. So by FLT this contracts!

[quote=pluricomplex]Now work on modulo $7$ then $y$ must be divisible by $3$. So by FLT this contracts![/quote]
y can be 5(mod 6)
also i dont get how FLT is used.

$y>1\implies 3|x \implies 3^y=z^3-2^{3x_1}=(z-2^{x_1})(z^2+...)\implies z-2^{x_1}=1,3$
$y=1\implies 2^x=z^3-1=(z-1)(z^2+z+1)\implies z-1=1$
no solns in either case.


The roots of $ x(x^2\plus{}8x\plus{}16)(4\minus{}x)\equal{}0$ are:
$ \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 0,4 \qquad\textbf{(C)}\ 0,4,\minus{}4 \qquad\textbf{(D)}\ 0,4,\minus{}4,\minus{}4 \qquad\textbf{(E)}\ \text{none of these}$

Using the zero-product property, the roots of any polynomial set to zero is the solution for x of each individual factor set to zero. Thus, the roots are:

[b]$ x\equal{}0$[/b]

$ x^2\plus{}8x\plus{}16\equal{}0$
$ (x\plus{}4)^2\equal{}0$
$ x\plus{}4\equal{}0$
[b]$ x\equal{}\minus{}4$[/b]

$ 4\minus{}x\equal{}0$
$ \minus{}x\equal{}\minus{}4$
[b]$ x\equal{}4$[/b]

Thus, the answer is $ \fbox{C}$.

[quote="r15s11z55y89w21"]Using the zero-product property, the roots of any polynomial set to zero is the solution for x of each individual factor set to zero. Thus, the roots are:

[b]$ x \equal{} 0$[/b]

$ x^2 \plus{} 8x \plus{} 16 \equal{} 0$
$ (x \plus{} 4)^2 \equal{} 0$
$ x \plus{} 4 \equal{} 0$
[b]$ x \equal{} \minus{} 4$[/b]

$ 4 \minus{} x \equal{} 0$
$ \minus{} x \equal{} \minus{} 4$
[b]$ x \equal{} 4$[/b]

Thus, the answer is $ \fbox{C}$.[/quote]

The answer is (D), -4 is a double root


probability
logarithms
expected value
advanced fields
advanced fields unsolved

I've asked myself while queueing behind some kids today:

They are collecting stickers for their album. Stickers are contained in chocolate bars, one in each bar. Suppose that they have to collect N stickers to fill the album. Estimate the number of chocolates they have to buy to achieve that.

This looked to me like an easy question, but I haven't really studied probability, so I can't measure how difficult it is... :ewpu:

If you have $k$ stickers already, the probability that the next one will be one you don't have is $\frac {n-k}n$. The number of bars you need to find the $k+1$st distinct sticker follows the geometric distribution with that characteristic probabilty $\frac {n-k}n$; this has mean $\frac n{n-k}$. Adding these, the expected number you need is $n\cdot\sum_{j=1}^n\frac1j= n\ln n+O(n).$

Since these geometric distributions are independent, we can also add variances: the variance is $\sum_{j=1}^n\frac{n^2}{j^2}=\frac{\pi^2}6n^2+O(n).$


real analysis
topology
real analysis unsolved

[quote="eugene"]It may have already been posted, i don't remember exactly
Given $\epsilon >0$, construct an open set $E \in [0,1]$ which is dense in $[0,1]$, such that $m(E)=\epsilon$.
$m$ stands for Lebesgue measure on $\mathbb{R}$.[/quote]Let $\mathbb Q \cap [0,1] =\{x_1,x_2,\dots\}$. Set $E(\delta) = \cup_{k\ge 1} I(x_k, \frac{\delta}{2^k})$ where $I(x,\epsilon) = (0,1)\cap (x-\epsilon,x+\epsilon)$.  This is open, contains a dense subset and hence is dense, and has measure $m(E(\delta)) \le \delta$. Also, $m(E(1)) = 1$. Now show that  $m(E(\delta))$ is continuous in $\delta$ and use the intermediate value theorem.

How to show continuity of $m(E(\delta))$ in $\delta$?

@ math1601: First note that $E_\delta \subset E_\eta$ when $\delta \le \eta$ hence $m(E_\delta)$ is increasing in $\delta$. Write $I_{k\delta} :=I(x_k,\delta/2^k)$ and again suppose that $\delta\le \eta$. Then $E_\eta\setminus E_\delta =\cup_k I_{k\eta}\setminus \cup_k I_{k\delta}\subset \cup_k (I_{k\eta}\setminus I_{k\delta}) $ and $m(E_\eta\setminus E_\delta)\le \sum_k m( I_{k\eta}\setminus I_{k\delta})=\sum_k {{\eta-\delta}\over {2^k}}=\eta-\delta$. Hence $m(E_\delta)$ is indeed continuous in $\delta$.


projective geometry
geometry
angle bisector
geometry proposed

In $\triangle ABC$, let $M,N$ be points on side $BC$ such that $AM,AN$ are the angle bisector and the median, respectively. $P$ is on $AM$ such that $BP$ is perpendicular to $AM$. $BP$ meets $AN$ at $S$. Show that $MS$ is parallel to $AB$.

Let $D$ be the point such that $ABDC$ is a parallelogram. Then, applying Desargues Theorem to triangles $AMC$ and $BSD$, it is sufficient to prove that $PN$ and $AC$ (and $BD$) are parallel. We have to prove that $\frac{AP}{CN}= \frac{AM}{CM}$. This can be easily checked using the following substitutions :
$AP = c \times cos\frac{\alpha}{2}$, $cos\frac{\alpha}{2}= \sqrt{\frac{(a+b+c)(b+c-a)}{4bc}}$, $CN = \frac{a}{2}$, $AM = \sqrt{bc(1-(\frac{a}{b+c})^{2})}$, $CM = \frac{ab}{b+c}$

Benjamin

PS : Huyền Vũ, how do you see immediately that $PN//AC$ ?

It's should be $1+\frac{AT}{PN}=\frac{AC}{PN}-1$. Note that: $PN=\frac{1}{2}CT$ and in his (her) proof $AB<AC$

[quote="Huyền Vũ"]to benjamin: because triangle BAT is isosceles so P is the midpoint of BT[/quote]

Thanks !  :lol:

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=60661 .

  darij


On the "login" page, registred should be spelled registered.

In context:

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Let ABCD is the convex quadrilateral. M, N, P, Q(not in ABCD) are the points such that $ \Delta AMB,\Delta BCN,\Delta DCP,\Delta QAD$ are the isosceles triangles and $ \angle AMB \equal{} \angle BCN \equal{} \angle DPC \equal{} \angle DAQ \equal{} \ 120^0$. H, I, K ,J is respectively the mid-points of AM, CN, CP, AQ.Prove that
$ IJ \equal{} \sqrt 3 HK$ :lol:


[quote="rnwang2"]15. Let $f$ be a function for which $f(x/3)=x^{2}+x+1$. Find the sum of all values of $z$ for which $f(3z)=7$.

18. In year $N$, the $300^{\text{th}}$ day of the year is a Tuesday. In year $N+1$, the $200^{\text{th}}$ day is also a Tuesday. On what day of the week did the $100^{\text{th}}$ day of year $N-1$ occur?[/quote]

18.
[hide]We know that there are either 65 or 66 days remaining in the year, so we get 265 or 266 has to be divisible by 7.  266 is divisible by 7, so there are 366 days in the year N, which means that there are 365 days in year N-1, so this means that we have to go back 300+265 days and that is 465 days backwards, which is really 5 days backwards which is Thursday.[/hide]


videos
Alcumus
probability
expected value

[quote="zheng"]For some reason all the videos in which it is under the topic of expected value won't work even though alcumus says they are available. It says "You do not have access to this video" :( .Please help.[/quote]

It may be a problem in the site. Or it might be a problem that you blocked part of the site?


calculus
integration
LaTeX
absolute value
calculus computations

What is $ I$?

$ \text{\LaTeX}$ note:

use:

\int for $ \int$

^ for powers

_ for indexes

Also, clicking on other peoples $ \text{\LaTeX}$ inputs shows you the code, and I don't think it's hard to understand from that how stuff is done.

Is this what you mean?

$ \int_{ \minus{} 1}^0|3x \minus{} 1|\,dx$

General advice for dealing with absolute values in calculus: the absolute value is a piecewise defined function. Make that explicit, and split the integral up every time we change formulas.

By the way, there's a typo in the title- and "Integral" doesn't say much about the content of the problem. More specific titles are greatly preferred.

The next step in the solution: rewrite the integral without any absolute values.


geometry
inradius
calculus
geometry unsolved

I want to fit as large a circle as possible inside a triangle whose side lengths are 2, 2, and 2$ k$. Find $ k$

Let $ x,y,z > 0$ such that $ x \plus{} y \equal{} 2,y \plus{} z \equal{} 2,z \plus{} x \equal{} 2k$. Then $ x \equal{} k,y \equal{} 2 \minus{} k,z \equal{} k$. Therefore $ r^2 \equal{} \frac {xyz}{x \plus{} y \plus{} z}\equal{} \frac {k^2(2 \minus{} k)}{2 \plus{} k}$, where $ r$ denotes the inradius of the triangle. Note that $ f(k) \equal{} \frac {k^2(2 \minus{} k)}{2 \plus{} k}$ implies $ f'(k) \equal{} \frac {2k(4 \minus{} 2k \minus{} k^2)}{(2 \plus{} k)^2}$. Therefore $ f'(k) \equal{} 0$ implies $ 4 \minus{} 2k \minus{} k^2 \equal{} 0$, since $ k > 0$. This has the positive root $ \sqrt 5 \minus{} 1$, which is the desired value of $ k$ that maximizes $ r$.


There is  a <n> surfaces such that degrees of each head is 4, This shape has <m> surfaces that those are triangle (maybe there is some surfaces that those aren't triangle).
find minimum of <m>.

Can anyone make sense of this post?  "Head" is obviously supposed to be "vertex" -- is "surface" supposed to be "face" (i.e. of a planar graph) or just a 3-cycle?

I think it means a planar graph such that each vertex has degree $ 4$, and it is asked for the minimum of triangular faces.
Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?p=962198#962198

Pierre.

I can't speak English well.  :blush:

Thanks pbornsztein, That's right.  :thumbup:
Please take me a solution...


ratio
limit
integration
function
floor function
logarithms
real analysis

Theorem:

Suppose $ a_n$ is a sequence of positive numbers such that $ \lim_{n\to\infty}a_n^{1/n}\equal{}L$ for some $ 0<L<\infty.$ Then $ \lim_{n\to\infty}\frac{a_{n\plus{}1}}{a_n}\equal{}L.$

This theorem shouldn't be that hard to prove starting with the idea that for sufficiently large $ n,$ $ (L\minus{}\epsilon)^n<a_n<(L\plus{}\epsilon)^n.$

Now, all we have to do is let $ a_n\equal{}\int_X|f|^n$ and quote a theorem we should know, namely that $ \lim_{n\to\infty}\left(\int_X|f|^n\right)^{1/n}\equal{}\|f\|_{\infty}.$

As near as I can tell, the only purpose served by the hypothesis that $ \mu(X)<\infty$ is to assure that $ \int_X|f|^n<\infty$ for all $ n.$ We could just assume that (or that for sufficiently large $ n$) separately.

[quote="perfect_radio"]Let $ \mu$ be a positive measure on $ X$ such that $ \mu (X) < \infty$; let $ f \in L^\infty$ s.t. $ \| f \|_\infty > 0$.
Prove that $ \lim_{n \to \infty} \frac {\int_X |f|^{n \plus{} 1}}{\int_X |f|^n} \equal{} \| f \|_\infty$.[/quote]

You may use distribution function to solve  :) .

[quote="Kent Merryfield"]Theorem:

Suppose $ a_n$ is a sequence of positive numbers such that $ \lim_{n\to\infty}a_n^{1/n} \equal{} L$ for some $ 0 < L < \infty.$ Then $ \lim_{n\to\infty}\frac {a_{n \plus{} 1}}{a_n} \equal{} L.$[/quote]
Unfortunately, this only holds in the other direction :(
(because e.g. let $ a_n$ be a sequence such that $ \frac {a_{n \plus{} 1}}{a_n} \equal{} L$ for virtually all $ n$, except for some rare $ n$ for which it is $ \frac {L}2$)

As for a proof, I think I found one in the mean time.
(I used the converse of the theorem you speak and some magic fact about the sequence)

Boy, did I screw that one up, big time.

perfect_radio is right. For a more concrete version of his example (with $ L \equal{} 1$) let $ a_n \equal{} 2^{ \minus{} \lfloor \log_2n\rfloor}.$

I see my mistake, and it's a stupid one. I got to $ \frac {a_{n \plus{} 1}}{a_n}\le(L \plus{} \epsilon)\left(\frac {L \plus{} \epsilon}{L \minus{} \epsilon}\right)^n$ and somehow (without writing anything down, which was a mistake) thought I saw $ e^{2\epsilon/L}$ in that second factor. No. It's just something which goes to $ \infty$ and $ n\to\infty$ and thus provides no control.

So now it's back to doing what bookworm_vn suggested, looking at the distribution function.

That is, if $ \lambda_f(\alpha) \equal{} \mu(\{|f| > \alpha\}),$ then $ \int_X|f|^n \equal{} \int_0^{\|f\|_{\infty}}n\alpha^{n \minus{} 1}\lambda_f(\alpha)\,d\alpha.$

(And the proof is still not obvious - there's work to do.)

[hide="What I had in mind"]
We know that $ \log a_n$ is convex, so $ \log a_{n\plus{}1} \minus{} \log a_n$ is increasing, and thus has a limit $ L$.
That means $ \lim_{n \to \infty} \frac{a_{n\plus{}1}}{a_n} \equal{} e^L$. This indeed implies $ \lim_{n \to \infty} \sqrt[n] {a_n} \equal{} e^L$.
But we already know that limit, so that's that.
[/hide]

Or you can use Cauchy $ \frac{\int_{X}|f|^{n\plus{}1}}{\int_{X}|f|^{n}} \geq \frac{\int_{X}|f|^{n}}{\int_{X}|f|^{n\minus{}1}}$ to see that  the sequence has a limit $ \leq \| f\|_\infty$ and Kent's theorem is valid now.


calculus
integration
algebra unsolved
algebra

:D Hi to all !!  I found a good problem in the book..

This is a problem..

___________________________
If [tex]n[/tex]  [tex]\succeq {6}[/tex], then show that the equation

[tex]\frac {1}{x_1 ^2}+\frac {1}{x_2 ^2}+...+\frac {1}{x_n ^2} =1[/tex]

have integral solutions.
___________________________

Quite good problem.... Don't you want to challenge?

 :D  :challeging is beautiful !

And the book is '250 problems in elementary number theory' by W. Sierpinski? :D

Pierre.

Especially impressive how to manually find for $n=8$: (2,2,2,2,3,7,14,21).


A 4x6x9 cuboid consists of 216 unit cubes. The game is about dividing the cuboid with Vertical/horizontal slices and taking away one of the remaining cuboids. So lets say for example the first player divides the cuboid with a vertical slice leaving 1x6x9 and taking away the rest. The second player divides and takes away and so on. The winner is the one who makes the last slice and take-away, which is leaving one unit cube.

Let´s say you start first. Is there a strategy that guarantees you a win?

[hide]If the cuboid is (x,x,1) or (x,1,x) or (1,x,x) this is a lost situation, since in this case, the strategy to win is to maintain the square for the opponent. So players must create such situation for the opponent to win.

So from the beginning, players must avoid creating any square face, since creating a square face, the opponent may create one of the above mentioned cuboid for him whence he lose. Furthermore, if he create a side with length 1 without having any square face during his turn, he lose for the opponent will be able to create a square for him.

Thus the (2,3,4) situation is a lose situation. This imply that (2,5,6) is a lose situation and in turn imply (3,5,7) lose situation.

(3,5,7) lose imply (4,6,7) lose. First move is to create (4,6,7) [/hide]

See [url=http://en.wikipedia.org/wiki/Nim]Nim[/url]; the game mentioned is equivalent to having three piles of $ l\minus{}1, w\minus{}1, h\minus{}1$ objects.


superior algebra
superior algebra unsolved

Prove that there is no continuous isomorphism from $ O_4$ to $ O_{3,1}$.

Where:
$ O_4$ be the group of orthogonal matrices in $ GL_4(\mathbb{R})$, that is $ QQ^t\equal{}I_4$
$ O_{3,1}$ the [i]Lorentz[/i] group, that is $ \{P\in  GL_4(\mathbb{R}) | P^tI_{3,1}P \equal{} I_{3,1} \}$, where $ I_{3,1}$ is the $ 4 \times 4$ identity with the element in the lower right corner $ \minus{}1$ instead of $ 1$.

In fact there is no isomorphism at all. $ O_{3,1}'$ has index 4 and $ O_4'\equal{}SO_4$ has index 2.

With respect to which group are those indexes?

There are not many groups here which could it be  :wink:

I mean the indices $ |O_{3,1}: O_{3,1}'|\equal{}4$ and $ |O_4: O_4'|\equal{}2$


linear algebra
matrix
linear algebra unsolved

I find this one interesting: (surely already posted ?)

Suppose that $ f: \text{M}_n\left(\mathbb{C}\right) \to \text{M}_n\left(\mathbb{C}\right)$ is a linear operator which maps every invertible matrix onto an invertible matrix. Show that $ f$ conserves the rank.

I posted this one as Dieudonne theorem. Actually, it is a theorem due to Purves and Marcus. It uses 4 different ideas. The one who finds them is really good.  :D

OK, here are 4 steps to get to a solution
1. $ \mbox{rank}(M) \leqslant n\minus{}1$ iff $ \exists P \in \mbox{GL}_n(\mathbb{C})$ s.t. $ \forall \lambda \in \mathbb{C}$, $ P\minus{}\lambda M \in \mbox{GL}_n(\mathbb{C})$ (hint: $ M$ is equivalent to an upper triangular matrix with $ 0$s on its main diagonal).
2. $ f^{\minus{}1}(\mbox{GL}_n(\mathbb{C}) \subset \mbox{GL}_n(\mathbb{C})$
3. Let $ M \in \mathcal{M}_n(\mathbb{C})$. $ \exists P \in \mbox{GL}_n(\mathbb{C})$ s.t. $ P\minus{}\lambda M \in \mbox{GL}_n(\mathbb{C})$ for exactly $ \mbox{rank}(M)$ values of $ \lambda$. Hence $ \mbox{rank}(f(M)) \geqslant \mbox{rank}(M)$ for all $ M \in \mathcal{M}_n(\mathbb{C})$.
4. $ f \in \mbox{GL}(\mathcal{M}_n(\mathbb{C}))$, hence  $ \mbox{rank}(f(M)) \equal{} \mbox{rank}(M)$

P.S.: I hope that this was not asked at Polytechnique without any hint :huh:


trigonometry
limit
logarithms
calculus
calculus computations

How would you evaluate
\[\lim_{x\to\infty}(\ln{(2\cosh{x})}-x\tanh x) \]
easiest?

Observe that

$\ln (2\cosh{x}) = \ln (e^{x}+e^{-x}) = x+\ln (1+e^{-2x})$

and $\ln (1+e^{-2x}) \to 0$ as $x \to \infty$. So if we can evaluate $\lim_{x\to\infty}x(1-\tanh x)$, then we have the answer. But this is quite easy, because

$x(1-\tanh x) = x\left(1-\frac{e^{2x}-1}{e^{2x}+1}\right) = \frac{2x}{e^{2x}+1}$.

So the limit is 0.


geometry
geometric transformation
rotation

Bonjour,

j'ai vu l'exercice suivant à l'oral de l'X:

étant donnés 1000 points dans le plan (non alignés), montrer qu'il existe une droite qui les sépare en deux ensembles de 500 points chacun. ;)

Amitiés,

Michael

La première chose c'est de prouver l'existence d'une droite D qui partage le plan en deux ensembles dont l'un contient tous les 1000 points. (facile)
Ensuite sur cette droite sélectionne un point A tel que tous trois points parmis les 1001 points dont on dispose ne soient pas alignés (un tel point existe car le nombre de droites déterminées par deux points pris parmis les 1000 points est fini, donc le nombre d'intersections de ces droites avec D est fini.
Maintenant par rotation continue de D autour de A on obtiendra la droite voulue.

voila ce que je pense: note $E$ l'ensemble des droites qui contiennent au moins 2 points de l'ensemble. Il n'y en a qu'un nombre fini.
prend un droite qui n'est pas dans cette ensemble. Alors toute droite parallele à cette droite ne contient qu'au maximum un point. La conclusion est alors immédiate non ?

On peut même imposer des conditions plus fortes sur la répartition des points :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=discrete+continuity&t=34319

Pierre.

un ami fait un TIPE sur le sujet suivant.
Dans un espace de dimension n, si on définit correctement ce qu'est une répartition de masse alors on montre qu'étant donné 2 repartitions de masse il existe un hyperplan qui coupe ces 2 répartitions chacune en 2 sous répartition égale. L'exo initial en est un cas particulier !

[quote="alekk"]un ami fait un TIPE sur le sujet..[/quote]

Est-ce que son TIPE est disponible sur le web ?

voilà ce que je propose moi : il existe une droite traverssant les 2n points et ne contenant aucun de ces points ,  notons p,q le nombre de points de part et d'autre par rapport à cette droite , si p<q , on peut trouver un point de cette droite et une droite passante par ce dernier point de telle sorte que l'espace inter-droite contient un seul point des q points sinon le nombre de ces points serait immediatement infini , en construit ainsi à base d'une droite ne contenant aucun de ces points une droite coupant les 2n points en deux parties ayant réspectivement p+1 et q-1 points , en itérant le procédé un nombre fini de fois on obtient ce qu'on veut


find all intiger$x,y$ which that:
$(x^2-y^2)^2=1+16y$

This one from ARO, pr.1, 9 grade. I participated in it :) It is the simpliest problem from the set ;)

Amir2!
Please, read http://www.mathlinks.ro/viewtopic.php?p=140098 . The title "nice" is out of law! You should name problems properly in order to understand what this problem about.

Myth

I think this is a Russian problem, although I'm not sure.

Note that $(x,y)$ is a solution iff $(-x,y)$ is one. If $x=0$, we have $y^4=16y+1$ which implies that $y \mid 1$, but $y=1$ isn't a solution. So we will assume $x$ to be positive. Also note that $y$ must be non-negative.

If $x=y$, we have $0=16y+1$, so either $x>y$ or $x<y$. If $x>y$, take $r=x-y$, then we have
\[16y+1=(2ry+r^2)^2 \geq (2y+1)^2=4y^2+4y+1 \\
12y \geq 4y^2 \\
3 \geq y.\]
So we check that $y=0$ gives $x^4=1$, so $x=1$; $y=1, 2$ that have no solutions because the RHS is not a square, and we see that $y=3$ gives the solution $x^2-9=7, x=4$.

So now assume that $x<y, r=y-x$, then we have
\[16x+16r+1=(2rx+r^2)^2=4r^2x^2+4r^3x+r^4 \geq 4x^2+4x+r^4 \\
4x(3-x)+r(16-r^3)+1 \geq 0 \]
So if $r \geq 3$, $x \leq 3$. Then we check that $x=1$ gives $y^4-2y^2+1=16y+1$, $y^3-2y=16$, $y(y^2-2)=16$, which has no solutions. $x=2$ gives $y^4-8y^2+16=16y+1$, so $y \mid 15$, so we can check the divisors of 15 to see there are no solutions. If $x=3$ it's easy to prove there are no solutions, also. So we have to check $r=1, r=2$. $r=1$ yields $x=4, y=5$, and $r=2$ gives $4 \mid 1$, absurd. So all solutions are $(1, 0), (-1, 0), (4, 3), (-4, 3), (4, 5), (-4, 5)$.


There is a very boring proof where you compare the matrices entry-by-entry.  Try to interpret the matrices as linear transformations for a second proof.


geometry
geometric transformation
geometry proposed

Let $ \Delta ABC$ be a triangle and $ P$ be a point on $ BC$. Let $ E,F$ be the projections of $ P$ on $ AC,AB$, respectively ($ E \in AC,F \in AB$). Find the locus of the midpoint of $ EF$.

It's going to be a line, and you can find it by putting $ P\equal{}B$ and $ P\equal{}C$, and then drawing a line between the two different midpoints of $ EF$.

My solution:

Let $ S( \triangle ) $ be the area of $ \triangle $ .
Let $ D $ be the midpoint of $ EF $ .
Let $ Y, Z $ be the projection of $ B, C $ on $ CA, AB $, respectively .

Since
$ S(\triangle DYC)+S(\triangle DBZ)=\frac{1}{2} \cdot [S(\triangle FYC)+S(\triangle EBZ)] $

$ =\frac{PC}{BC} \cdot S(\triangle BYC)+\frac{BP}{BC} \cdot S(\triangle ZYC)+\frac{PC}{BC} \cdot S(\triangle YBZ)+\frac{BP}{BC} \cdot S(\triangle CBZ) $

$ =\frac{1}{2} \cdot S(BCYZ) $ ,

so $ D $ lie on the Newton line of complete quadrilateral $ \{BC, CY, YZ, ZB \} $ ,
hence the locus of the midpoint of $ EF $ is a line passing through the midpoint of $ BY, CZ $ .

Circle $\odot(PEAF)$ with diameter $\overline{PA}$ cuts $BC$ again at the foot $D$ of the A-altitude. Angles $\angle DEF=\angle DAB$ and $\angle DFE=\angle DAC$ are constant $\Longrightarrow$ all $\triangle DEF$ are similar. Thus if $M$ is the midpoint of $\overline{EF},$ then $\angle FDM$ is constant and $\tfrac{DF}{DM}=k$ is constant $\Longrightarrow$ locus of $M$ is the line $\ell$ image of $AB$ under spiral similarity with center $D,$ rotational angle $\angle (DM,DF)$ and coefficient $k.$ Making $P \equiv B$ and $P \equiv C,$ then $\ell$ is just the line through the midpoints of the altitudes issuing from $B$ and $C.$


Let a,b,c are positive real numbers.Prove that:
$ \sum a^2 \plus{}2abc \plus{} 1 \ge 2\sum ab$

[quote="bigbang195"]Let a,b,c are positive real numbers.Prove that:
$ \sum a^2 \plus{} 2abc \plus{} 1 \ge 2\sum ab$[/quote]
It is well-known inequaltiy.
Among the numbers $ (1\minus{}a),(1\minus{}b),(1\minus{}c)$ there are two of them with the same sign.let us say $ (1\minus{}b)(1\minus{}c)\geq 0$.
we have:
$ {a}^{2}\plus{}{b}^{2}\plus{}{c}^{2}\plus{}2abc\plus{}1\minus{}2\left(ab\plus{}bc\plus{}ca \right)\equal{}{\left(a\minus{}1 \right)}^{2}\plus{}{\left(b\minus{}c \right)}^{2}\plus{}2a\left(1\minus{}b \right)\left(1\minus{}c \right)\geq 0$
So we have Q.E.D.Equality occurs for $ a\equal{}b\equal{}c\equal{}1$ :)

[quote="trungk42sp"][quote="bigbang195"]
Among the numbers $ (1 \minus{} a),(1 \minus{} b),(1 \minus{} c)$ there are two of them with the same sign.let us say $ (1 \minus{} b)(1 \minus{} c)\geq 0$.
[/quote][/quote]
why !!!!!!!
vì sao có cái này hả anh  :lol: anh add nick em đi


function
vector
complex numbers
complex analysis
complex analysis unsolved

[hide="1."]
Let $ f \equal{} u \plus{} iv$ be an analytic function in the unit disk, with $ u, v$ its real, respectively, imaginary part.
Assume that there are non-zero constants $ a, b, c,$ so that
\[ au \plus{} bv \plus{} c \equal{} 0
\]
in the disk.
What can be said about $ f?$[/hide]

[hide="2."]
Find all analytic functions $ f$ in the unit disk satisfying
\[ n^{\minus{}5/2} \leq \left| f\left( \frac{1}{n} \right) \right| \leq 2 n^{\minus{}5/2}.
\]
[/hide]

For 1, I'm not sure if this is what you're look for, but both $ u$ and $ v$ are harmonic functions.

[quote="JRav"]What do you mean, "elaborate"?  It's a pretty straightforward statement.  The real and imaginary parts of an analytic functions are harmonic.  Your textbook should cover harmonic functions or it's not doing a good job.[/quote]

well, he's probably confused by your statement, which doesn't appear to help in the slightest.

a better idea (i think) is to differentiate the relation between $ u$ and $ v$ and combine this with the cauchy-riemann equations.

I don't think my statement is ambiguous, assuming you know what harmonic means.  Based on blahblahblah's suggestion, I'll show you what I mean:
By CR,
$ \frac{\partial u}{\partial x}\equal{}\frac{\partial v}{\partial y}$, $ \frac{\partial v}{\partial x}\equal{}\minus{}\frac{\partial u}{\partial y}$.
Then if $ f\equal{}u\plus{}iv$,
we can write
$ v_{xx}\equal{}\minus{}u_{yx}\equal{}\minus{}u_{xy}\equal{}\minus{}v_{yy}$
Hence $ v_{xx}\plus{}v_{yy}\equal{}0$.  We can do the same thing for $ u$ to show that it is harmonic as well.

well, it is kind of a given that $ u$ and $ v$ are harmonic.  The condition $ au \plus{} bv \plus{} c \equal{} 0$ actually gives you a lot more (differentiate it with respect to $ x$ and $ y$ to get two equations, use the cauchy riemann equations to get two linear equations in two unknowns ($ u_x$ and $ u_y$), and finally find what $ u_x$ and $ u_y$ have to be).

Thank you. So I get that $ f$ is a constant. Is that correct? Does anyone know how to do #2?

Just ask yourself what can be the first non-zero term in the Taylor expansion of $ f$ at $ 0$.

From df/dx = df/idy, we find that u_x+ iv_x = -iu_y + v_y <=> u_x + iu_y = v_y + iv_x
Differentiating the given relation and plugging into vector components, a(u_x+iu_y) + b(v_x+iv_y) = 0 <=> a(v_y+iv_x)+b(v_x+iv_y)=0
As a,b are non-zero, this means that the vectors $ (v_y\plus{}iv_x),(v_x\plus{}iv_y)$ are multiples of each other, so $ v_x\equal{}\pm v_y$ and similarly for $ u$. Then, v_xx=v_yx=v_xy=v_yy or v_xx=-v_yx=-v_xy=v_yy, which contradicts v_xx+v_yy=0 unless v_xx=v_yy=0. So, v_x is constant, etc., and we come to the conclusion that all the functions f that work are: \[ f \equal{} (a\plus{}bi)(x\pm y) \plus{}(c\plus{}di)\]


Δίνονται οι θετικοί αριθμοί $ a_1,a_2,...,a_n$ ώστε $ a_1a_2...a_n\equal{}1$ . Να αποδείξετε ότι

$ (2\plus{}a_1)(2\plus{}a_2)....(2\plus{}a_n)\geq3^n$

Η λύση μου
[hide]Παίρνουμε AM-GM για το $ 2 \plus{} a_1 \equal{} 1 \plus{} 1 \plus{} a_1$ και έχουμε ότι $ 2 \plus{} a_1\geq3\sqrt [3]{a_1}$  . Ομοίως και για τα υπόλοιπα και στη συνέχεια πολλαπλασιάζουμε κατά μέλη και έχουμε
$ (2 \plus{} a_1)(2 \plus{} a_2)....(2 \plus{} a_n)\geq3^n\sqrt [3]{a_1a_2...a_n}\Rightarrow (2 \plus{} a_1)(2 \plus{} a_2)....(2 \plus{} a_n)\geq3^n$ So QED :lol:
[/hide]

Απλός και ωραίος :!:

Μια πιο άκομψη λύση θα ήταν με επαγωγή.

Ακόμα πιο κομψά: Holder


algebra
polynomial
superior algebra
superior algebra unsolved

An ideal $ I$ in $ R[x_1,x_2,...,x_n]$ is called a $ homogeneous$ $ ideal$ if whenever $ p \in I$ then each homogeneous component of $ p$ is also in $ I$.
Prove that an ideal is a homogeneous ideal if and only if it may be generated by homogeneous polynomials.

$ \Rightarrow$ is clear. $ \Leftarrow$: assume $ I$ is generated by the homogeneous polynomials $ (h_j)$ and $ p \in I$. if $ p \in I$, say $ p \equal{} \sum_j f_j h_j$, the homogeneous components of $ p$ are obviously generated by $ h_j$, i.e. they lie in $ I$.

[quote="Bradley"]Why is $ \Rightarrow$ clear??[/quote]

Because we may homogenize any generating set.