geometry
trapezoid
trigonometry
geometry unsolved

[quote="Tiks"][quote="shobber"]In $\triangle{ABC}$, $S_{ABC}=1$. $D$, $E$ are two points of $AC$ and $AB$.  $P= BD \cap CE$ such that $\frac{S_{BCDE}}{S_{BPC}}=\frac{16}{9}$. Find the max value of $\triangle{DEP}$ when $D$ and $E$ moves on $AC$ and $AB$.[/quote]
Hi,$shobber$.
I think I have a solution of this interesting problem.
I note $\frac{AD}{DC}=\gamma$ and $\frac{AE}{EB}=\beta$
then we have  that $\frac{S_{ADE}}{S_{ABC}}=\frac{\beta\gamma}{(\beta+1)(\gamma+1)}$=>$S_{BCDE}=\frac{(\beta+\gamma+1)^2}{(\beta+1)(\gamma+1)}=\frac{16}{9}$
also we have that $\frac{S_{BCDE}}{S_{BPC}}=\frac{BD*CE}{BP*CP}=\frac{(\beta+\gamma+1)^2}{(\beta+1)(\gamma+1)}$=>$S_{BPC}=\frac{1}{\beta+\gamma+1}$
then $\frac{S_{BPC}}{S_{DPE}}=\frac{BP*CP}{PD*PE}=\frac{(\beta+1)(\gamma+1)}{\beta\gamma}$=>$S_{DEP}=\frac{\beta\gamma}{(\beta+1)(\gamma+1)(\beta+\gamma+1)}$
So we have that $\frac{(\beta+\gamma+1)^2}{(\beta+1)(\gamma+1)}=\frac{16}{9}$
and we must find $max(\frac{\beta\gamma}{(\beta+1)(\gamma+1)(\beta+\gamma+1)})$ after long Algebra I proved that $\frac{\beta\gamma}{(\beta+1)(\gamma+1)(\beta+\gamma+1)}$ is max  when $\gamma=\beta=\frac{1}{3}$ so answer is $\frac{1}{18}$,I hope that is true :? ,(if my algebra isn't bad :(  :D ) ;)[/quote]
Is my answer true? :?
If yes,can you find nice solution for laste parte :?
[quote]So we have that $\frac{(\beta+\gamma+1)^2}{(\beta+1)(\gamma+1)}=\frac{16}{9}$
and we must find $max(\frac{\beta\gamma}{(\beta+1)(\gamma+1)(\beta+\gamma+1)})$[/quote] ;)

Here is my solution(?):

  At first we have to establish when (PED) is max. It is evident that

  (PED) increases by taking away area from the flanks BDP and CEP.

   This increase stops when flanks are trimmed to the  same area,

   i.e. when BDEC is a trapezoid with  point P on the A-median.

   Now the problem  is a pis o' cake.

   Allow me to take a 'useless break' here.

   Our old friend Yetti will be delighted to learn that  the 4 points

   (you all know which ones I'm talking about )  are still in harmonic range 3:1:2,

  that his sturdy proof withstood the relentless test of time

   and like the inradii problem, his wonderful results will be used and I'm not afraid

   to say used again by future generations.

   see         http://www.mathlinks.ro/Forum/topic-55062.html

     where he brought plenty of transparency into the mirky question and finally put to

    rest  all lingering doubts on the subject.

   After due respect is paid, let's return and complete the problem, for what's left

   is though very little,  yet I hope very arithmetically educational.

   Thus, the area of PED is 1 unit and area of BEDC is 16 units, area of ADE  is 2 units.

   This makes  area of ABC 18 units.     PED = 1/18 of ABC.

   And it is pretty much in line with the results in the previous posts.

  Thank you.

   M.T.

[quote="Virgil Nicula"]Denote $\frac{EA}{x}=\frac{EB}{1}=\frac{AB}{x+1}$, $\frac{DA}{y}=\frac{DC}{1}=\frac{AC}{y+1}$, where $x\ge 0$, $y\ge 0$.
From the Menelaus' theorem applied to the transversal $\overline {BPD}$ to the $\triangle ACE$
we obtain: $\frac{BE}{BA}\cdot \frac{DA}{DC}\cdot \frac{PC}{PE}=1\Longrightarrow \frac{PE}{y}=\frac{PC}{x+1}=\frac{CE}{x+y+1}$. Therefore,

$[PED]=\frac{PE}{EC}\cdot [EDC]=\frac{y}{x+y+1}\cdot \frac{CD}{CA}\cdot [AEC]=$
$\frac{y}{x+y+1}\cdot \frac{1}{y+1}\cdot \frac{AE}{AB}=\frac{xy}{(x+1)(y+1)(x+y+1)}\Longrightarrow$
$\boxed {[PED]=\frac{xy}{(x+1)(y+1)(x+y+1)}}\ .$

$[BPC]=\frac{CP}{CE}\cdot [BEC]=\frac{x+1}{x+y+1}\cdot \frac{BE}{BA}=\frac{1}{x+y+1}\ .$
$[BCDE]=1-[AED]=1-\frac{AE}{AB}\cdot \frac{AD}{AC}=1-\frac{xy}{(x+1)(y+1)}=\frac{x+y+1}{(x+1)(y+1)}\ .$
$\frac{[BCDE]}{[BPC]}=\frac{16}{9}\Longrightarrow \frac{(x+y+1)^2}{(x+1)(y+1)}=\frac{16}{9}\Longrightarrow \boxed {3(x+y+1)=4\sqrt{(x+1)(y+1)}}\ .$

$\boxed {u=x+1\ge 1,\ v=y+1\ge 1}\Longrightarrow$$3(u+v-1)=4\sqrt{uv}\ge 3(2\sqrt{uv}-1)\Longrightarrow \sqrt{uv}\le \frac 32$
$[PED]=\frac{3(u-1)(v-1)}{4uv\sqrt{uv}}=\frac{3[uv+1-(u+v)]}{4uv\sqrt{uv}}\le \frac 34 \cdot \frac{uv+1-2\sqrt{uv}}{uv\sqrt{uv}}=$
$\frac 34\cdot \frac{(\sqrt{uv}-1)^2}{uv\sqrt{uv}}$. Denote $\boxed {t=\sqrt{uv}}\in \left[1,\frac 32\right]$.
Thus, $[PED]\le \frac 34 f(t)$, where $\boxed {f(t)=\frac{(t-1)^2}{t^3}}\le f\left(\frac 32\right)=\frac{2}{27}$.
Therefore, $\boxed {[PED]\le \frac{1}{18}}$ with equality if and only if $\boxed {x=y=\frac 12}$.

[b]Generalization.[/b] Let $\triangle ABC: \ [ABC]=1\ ;\ D\in (AC)\ ,\ \frac{DA}{DC}=y>0\ ;$

$\ E\in (AB)\ ,\ \frac{EA}{EB}=x>0\ ;\ P\in BD\cap CE\ .\ [BCDE]=$ $k^2\cdot [BPC]\ ,\ k\in (1,2)\Longrightarrow$

$\boxed {[DEP]\le \frac{(2-k)(k-1)^2}{k}}\le \frac{k^2}{27}$ with equality if and only if $\boxed {x=y=\frac{k-1}{2-k}}\ .$[/quote]
So my answer is true! :lol:

[quote="Virgil Nicula"]Welcome, Tiks, to http://www.mathlinks.ro ! [u]And[/u] so my answer is true ! Understand you what say Armpist in the last reply w.r.t. my dear friend Yeti ? I didn't.[/quote]

  Hello Virgil !

  Never mind what I say about anyone, they are my personal views which are not as

  important as a  solution to a specific problem, as for this particular case I gave credit to

  Yetti, who is a dear friend of both you and me,  for a wonderful and timely

  proof of a fact I used  as a centerpiece in the  my solution.

  BTW, what is your educated opinion on that? Did you understand it?

  Do not be affraid to put forward some constructive criticism, but please

  do not volunteer your opinions of me (such as 'badly', 'goodly',etc),

  but rather on my solutions(?).

  Thank you.

  M.T.

Dear Armpist, I wish read only your nice solutions and no your opinions about X, Y a.s.o. There are on Mathlinks some sections where you can talk over anything or using the private message to X, Y a.s.o. Such I learned from Darij, such Darij suggest me, because and I made sometimes the same "mistake" (in my opinion). I am sorry for these annoying words ! What means BTW ? [i]Thank you.[/i]
Come this leisurely and can help me with some ideas at the own problem from the last reply at http://www.mathlinks.ro/Forum/viewtopic.php?t=70753

[quote="Virgil Nicula"] What means BTW ?

Come this leisurely and can help me with some ideas at the own problem from the last reply at http://www.mathlinks.ro/Forum/viewtopic.php?t=70753[/quote]

BTW -  is short for   'by the way'

I have to decline your offer with topic 70753, for it is not in Geometry, and I do not

trust several things, among them trigonometry, electricity and both principles of

thermodynamics. They are unreliable, in my opinion.

  Thank you.

  M.T.


so, here is problem 2.43:

 [geogebra]31adb1f2ffa1a846e4c54f240c7802c1b91c0ae0[/geogebra]
FC is parallel to BD. Angle DBE is 2x and angle FCD=4x and angle CAB = 3x. find x (or angle C). So, the book has the solution that 3x=angle CAB=angle EAB. And then you have $ 5x \equal{} 180$ and $ x \equal{} 36$ BUT what i'm saying is by vertical angles or something like that angle DBE= angle FBA then $ 6x \equal{} 180$ and $ x \equal{} 30$, contradiction. please help :)

Here FC is NOT parallel to BD
just join the lines and see  :)

i meant BC  :blush:  :blush: no wonder i fail :noo:

Even if there is a mistake (I haven't looked through your post yet), it shouldn't be *too* surprising. Think about how long the book is. Even if you have many people proofreading it, there's still a chance that there are mistakes. I'm actually impressed by the lack of mistakes.

By the way, I've found a few mistakes in AoPS books too. The first one that I can remember off the top of my head was something in chapter 14 (?) of Volume 2.

Um...Poincare, that also doesn't work, because just looking at the names tells you that they share an endpoint...

Do you mean AB?  :lol:

JUST THE ONES THAT LOOK PARRALEL OMG!! :D

Which I would say is FC and AB, as they are connected, and appear to be parallel...

[quote="nick42"]are you sure you haven't made any typos?

cuz some of those angles you typed don't look right.[/quote]
yea, i meant BCD= 4x and yes you are correct fishythefish.


Let $x$ a real number, $-1 \le x \le 1$ ,$0 \le arccosx \le \pi$ and $n$ a positive whole number. Prove that $f_{n}(x)=cos(narccosx)$ can be expressed by a polinom $P(x)$, n degree, which the coeffient of the term of maximum degree is $2^{n-1}$.

Sorry about my english!

[quote="santosguzella"]Let $x$ a real number, $-1 \le x \le 1$ ,$0 \le arccosx \le \pi$ and $n$ a positive whole number. Prove that $f_{n}(x)=cos(narccosx)$ can be expressed by a polinom in $P(x)$, n degree, which the coeffient of the term of maximum degree is $2^{n-1}$.

Sorry about my english![/quote]

What is $P(x)$?

[quote="santosguzella"]$P(x)$ is a polynomial which you must prove that $f_{n}(x)$ can be expressed. ok?[/quote]
I can't understand, It is $x$. This is what you need, i think  :wink:
http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html


geometry
circumcircle
trigonometry
trapezoid
Pythagorean Theorem
trig identities
Law of Cosines

6 points $A$, $B$, $C$, $D$, $E$, and $F$ are concyclic.
Also, $AB=BC=CD=2$ and $DE=EF=FA=4$.

Find the area of hexagon $ABCDEF$.

this can be done with trig...i will write the solution after i am done with my hw

If we denote the center of the circumcircle as $O$,

we find that $\angle AOB=\angle BOC=\angle COD=36$

and that $\angle DOE=\angle EOF=\angle FOA=72$

Therefore, we can find the radius by law of cosines.  From this, we use the fact that $Area_{triangle}=\frac{1}{2} a b cos(\theta)$ to find the total earea of the hexagon.

how do you know those angles, just because the chord is twice as long, does not mean that it intercepts double the arc and with those angles, $(36 + 72)3 = 324$ which is not correct since there must be 360 degrees in a circle

look at the diagram attatched, i did not name it correctly so bear with me
AC and AD are drawn perpendicular to the respective chords
since they are perpendicular, they biscect the chord
so DB = $\frac{2}{2} = 1$, and DC = $\frac{4}{2} = 2$
Let $AD = r$, then by the pythagorean theorem, $AC = \sqrt{r^2 - 4}$ and $AB = \sqrt{r^2-1}$
since half of the chords are 2 and half of the chords are 4, then if we pair up 2 different chords, they intercept an arc of 120 degrees
since $\angle CDB$ is inscribed in the circle and intercepts 2 of these pairs, $\angle CDB = \frac{240}{2} = 120$
by the law of cosines, $(BC)^2 = 2^2 + 1^2 - 2(2)(1)\cos{120}$ so $BC = \sqrt{7}$
since $\angle ABD = \angle ACD = 90$ the sum of opposite angles equals 180 so quadrilateral $ABDC$ is cyclic
since it is cyclic, by ptolemy's theorem,
$(1)(\sqrt{r^2 - 4}) + (2)(\sqrt{r^2 -1}) = (r)(\sqrt{7})$
solving yields $r = \sqrt{\frac{28}{3}}$
using the pythagorean theorem gives us then $AC = \frac{4\sqrt{3}}{3}$ and AB = $\frac{5\sqrt{3}}{3}$
using $\frac{bh}{2} = A_{\triangle}$
${Area = 3(\frac{2\frac{5\sqrt{3}}{3}}{2} + \frac{4\frac{4\sqrt{3}}{3}}{2}) = \boxed{13\sqrt3}}$

Nice work, Altheman. That is correct. :yup:

Does anyone have a different solution?
I took a different approach, so I'll post my solution later if no one else posts it.

[hide="hint for another approach"]
If we connect each of the six points to the center of the circle, we obtain 6 triangles. Play around with them a bit...
[/hide]

another better approach is using more of ptolemy's theorem, that lets you find the chord between the long and short sides, and that is pretty easy too, just there is a few systems of equations that need to be solved, and you need to use the law of cosines, once you have that, there are 2 isosoles trapezoids that are easy to find the area of

That's a nice one also, but how about this?
[hide]
If we connect each of the six vertices to the center of the circle, we get 6 triangles (as mentioned in the previous hint).

Consider another cyclic hexagon $HIJKLM$ such that $HI=JK=LM=2$ and $IJ=KL=MH=4$. Connect each of these six points to the circumcenter. Again we obtain 6 triangles.

Now compare these two hexagons. Both of them consist of exactly the same triangles, so they have the same area.

Extend $IJ$, $KL$ and $MH$ so that these three lines intersect each other and form an equilateral triangle. We can observe that the large equilateral triangle has side length 8. The area of hexagon is calculated by subtracting the area of 3 small equilateral triangles from the area of the large equilateral triangle.

$\therefore\text{Area}=16\sqrt{3}-3\sqrt{3}=\boxed{13\sqrt{3}}$[/hide]

No need for trig :)

wow, that is better than i did, i thought of rearranging the sides would not change the area, but i did not really pursue it at all


This is the question:

Exist a prime $p$ such that $\gcd(h,p^{h-1}+p^{h-2}+...+1)$ = 1,  $\forall$h$\geq$1?

Bye, Uber

$h=2$ gives necesarily $\gcd(2,p+1)=1$, so $p=2$, but the last one is not possible, too (e.g. by $h=6$).

It is wrong. Exist infinely numbers h with property $\phi (h)|h$. For suth h we have $(h,\frac{p^h-1}{p-1})\not=1$ if $(p-1)\not|h$.

ok... it was to start:

let $d$ a divisor of  $p$-1 (p is always a prime, an odd prime).  Obviously: $p^{d^n}\equiv1(mod d^n) \forall n$.
Conversely: Esists a prime p such that if h$|p^h$-1 then h=$d^n$, for some d divisor of p-1 and n integer?

Bye, Uber

It is wrong. Let $h|(p+1),h>2$ and even, then h is not divisor p-1. However $h|(p^h-1)$.
If $h\not=2^k$, then $h\not=d^m,d|(p-1)$. It give for all p not be Mersen primes.
For Mersen primes we chose divisor $h|(p^k-1),k|h,h\not|(p-1)^m$.


A sequence of positive integers is constructed as follows. If the last digit of $ a_n$ is greater than $ 5$, then $ a_{n \plus{} 1}$ is $ 9a_n$. If the last digit of $ a_n$ is $ 5$ or less and $ a_n$ has more than one digit, then $ a_{n \plus{} 1}$ is obtained from $ a_n$ by deleting the last digit. If $ a_n$ has only one digit, which is $ 5$ or less, then the sequence terminates. Can we choose the first member of the sequence so that it does not terminate?

Claim: For all $ n$ we have $ \min\{a_{n\plus{}1},a_{n\plus{}2}\}<a_n$ or the sequence terminates latest at step $ n\plus{}2$.

Proof: If the sequence does not terminate before $ n\plus{}2$, let $ C$ denote the last digit of $ a_n$. Clearly, we can assume $ C>5$ and $ a_n>9$, otherwise $ \min\{a_{n\plus{}1},a_{n\plus{}2}\}<a_n$ holds and we are finished. For $ C>5$, the number $ a_{n\plus{}1}$ has exactly the same last digit as $ 9C$, which is smaller than $ 5$ and greater than zero. This implies the inequality $ a_{n\plus{}2}\leq\frac{a_{n\plus{}1}\minus{}1}{10}\equal{}\frac{9a_n\minus{}1}{10}\leq a_n\minus{}1$ and the claim is verified.

Since $ a_n\geq 1$ holds, if the sequence has not terminated before, we conclude from the claim and the decreasing trend behavior, that any sequence will terminate.


Please help me solve this gravitation problem :

A particle of mass equal to mass of the earth is projected by an external agency so that there is no recoil in earth when particle is projected. In this case escape velocity of the particle will be ?
( g is acceleration due to gravity on the surface of the earth and R = radius of earth )

[quote="_MM_"]
there is no recoil in earth when particle is projected

[/quote]

That makes it easier!

If there's no recoil, just use conservation of energy.

Yes, you should conserve energy But what should the final energy be ?

I was thinking that as the particle is also of same mass as that of earth we can't neglect gravitational force between them and thus earth also gains some velocity....

Could you please work it out

[hide="ans"]
$ \sqrt{ 4gR }$
[/hide]

Could someone please help me..I still haven't got the answer

Am i wrong in my concept somewhere  :(

The earth too acquires a velocity $ \minus{}v$ due to momentum conservation.


Vieta
algebra
polynomial
quadratics
SFFT
special factorizations

[quote="ProblemSets"]x + y +2xy = 7
2(x+y) + xy = 8

solve for x,y[/quote]
[hide] Say that $x+y=a$ and $xy=b$.  Thus, we have that
\[a+2b=7 \]
and
\[2a+b=8 \]
Adding our equations together, we obtain that
\[3a+3b=15\Rightarrow a+b=5 \]
We can find that $a=3, b=2$.  So, we have that $x+y=3, xy=2$.  Thus, $y=3-x$ and $y=\frac{2}{x}$.  So, we have that
\[3-x=\frac{2}{x}\Rightarrow 3x-x^{2}=2\Rightarrow x^{2}-3x+2=0\Rightarrow x=1,2 \]
Plugging this back in, we  get that $y=2$ if $x=1$ or $y=1$ if $x=2$.  So, our possibilities are $\boxed{\text{(1,2) and (2,1)}}$[/hide]

[quote="ProblemSets"]x + y +2xy = 7
2(x+y) + xy = 8

solve for x,y[/quote]

[hide="or...."]Let $x+y=S$ and $xy=P$.

$S+2P=7$ and $2S+P=8$.

Multiply the second equation by 2 and subtract the first equation.

$3S=9$ so $S=3$ which means $P=2$.

By vieta's theorem, $x$ and $y$ are the solutions to $m^{2}-3m+2=0$.  The solutions are $1$ and $2$.

$\therefore (1,2) (2,1)$[/hide]

[quote="Aznwithabrain"][quote="jli"]

By vieta's theorem, [/quote]

What is Vieta's Theorem?[/quote]

For jlis question, vietas therom wasnt neccessary, he just basically factored..vietas theorm just deals the witht he coefficiants in terms of the roots (product, sum, alternating sums..) cant explain it very well..

[quote="Temperal"][hide]Let $x+y=n$ and $xy=t$. We obtain $n+2t=7$ and $2n+t=8$
Thus, $n=3$; $t=2$.

Therefore, $x+y=3$, $xy=2$. We can turn this into the quadratic $x^{2}-3x+2=0$
Thus, $(x,y)=(1,2)or(2,1)$.[/hide][/quote]

Well that's exactly the same thing as before, no need to post it again.

[hide="Alternate Solution"]
Subtract the first equation from the second to get $x+y-xy=1$, or $xy-x-y+1=0$. By Simon's Favorite Factoring trick this becomes $(x-1)(y-1)=0$, so either $x=1$ or $y=1$. Plugging these back into $x+y+2xy=7$ yields that $x=1\Rightarrow y=2$ and $y=1\Rightarrow x=1$, hence we have the solutions $(1,2),(2,1)$.

Of course, this method will not work every time, I got lucky. But substitution is an important tool for one to have.[/hide]

Vieta's theorem can be derived by multiplying:

$(x-a_{1})(x-a_{2})...(x-a_{n})$

$x^{n}-(a_{1}+a_{2}+...+a_{n})x^{n-1}+(P)x^{n-2}-\cdots+a_{1}a_{2}\cdots a_{n}$

P is the symmetric sum taken two at a time

[quote="13375P34K43V312"]
...Well that's exactly the same thing as before, no need to post it again....
[/quote]
Se the other thread in which you reprimanded me for my answer to this.

Anyways, most threads have dozens of congruent answers to the same thread. You're missing the point of these forums.

Don't go off topic again, 13375P34K43V312.

A little different approach by substituting immediately:

$x+y+2xy = 7$
$2(x+y)+xy = 8$

[hide]First equation:
$2xy+x+y = 7$
$y(2x+1)=7-x$
$y=\frac{7-x}{2x+1}$

Plug into second equation:
$2(x+\frac{7-x}{2x+1})+x(\frac{7-x}{2x+1})=8$
$2x+2\frac{7-x}{2x+1})+x(\frac{7-x}{2x+1})=8$
$2x(2x+1)+2(7-x)+x(7-x)=8(2x+1)$ Assuming $2x+1$ is not $0$
$3x^{2}-9x+6=0$ Simplify.
$x^{2}-3x+2=0$
$(x-2)(x-1)=0$
$x=1 \text{ and }x=2$

Plugging back into original equation:
$(1,2) \text{ and }(2,1)$[/hide]


geometry
geometric transformation
rotation
ratio
geometry proposed

The circle $ (C_1)$ and $ (C_2)$  intersect at distinct points $ A, B$. An arbitrary line through $ A$ intersects again $ (C_1)$ at point $ C$ and $ (C_2)$ at point $ D$. Let $ M$ be an arbitrary point on the segment $ CD$. The parallel line to $ BC$ from $ M$ intersects the segment $ BD$ at point $ K$, the parallel line to $ BD$ from $ M$ intersects the segment $ BC$ at point $ N$. The perpendicular line from $ N$ to $ BC$ intersects the arc $ BC$ of $ (C_1)$ which does not contain $ A$ at point $ E$. The perpendicular line from $ K$ to $ BD$ intersects the arc $ BD$ of $ (C_2)$ which does not contain $ A$ at point $ F$. Prove that $ \angle EMF \equal{} 90^0$.

i don't know where exactly but i saw this problem in this forum,
anyway ,i' ll give some hints,
[hide="hint"]
to prove it u must prove first
$ \Delta FKD \sim \Delta ECN,\Delta ENB \sim FKB$.
then after proving those similarities just see that a rotation centered at $ E$ and tranforming $ C$ to $ B$ and a rotation centered at $ F$ and transforming $ B$ to $ D$ is equal to a rotation centered at $ M$ and transforming $ C$ to $ D$.
to prove this use the ratios in the similarities and also the fact that $ \frac{1}{2} \angle ECB \plus{}  \frac{1}{2} \angle BFD\equal{}90^\circ$.   [/hide]

$ \bullet$ From $ MK\parallel BC$ and $ MN\parallel BD,$ based on the [b]Thales theorem[/b], we have that $ \frac {KB}{KD} \equal{} \frac {MC}{MD} \equal{} \frac {NC}{NC}$ $ ,(1)$

Let be the point $ X\equiv (O_{1})\cap EN$ and it is easy to show that the triangles $ \bigtriangleup FBD,\ \bigtriangleup XCB$ are similar, because of $ (1)$ and $ \angle BFD \equal{} \angle CAB \equal{} \angle CXB$ and $ FK\perp BD$ and $ XN\perp CB.$

So, from this similarity we have that $ \angle BFK \equal{} \angle CXE \equal{} \angle CBE$ $ ,(2)$

From $ (2)$ we conclude that the right triangles $ \bigtriangleup KBF,\ \bigtriangleup NEB$ are similar and then, we have that $ \frac {KF}{NB} \equal{} \frac {KB}{NE}$ $ \Longrightarrow$ $ \frac {KF}{KM} \equal{} \frac {NM}{NE}$ $ ,(3)$

$ \bullet$ From $ (3)$ and because of $ \angle MKF \equal{} 90^{o} \plus{} \angle KMN \equal{} \angle ENM$, we conclude the similarity of the triangles $ \bigtriangleup MKF,\ \bigtriangleup ENM$ and then, we have that $ \angle KFM \equal{} \angle NME$ $ ,(4)$

It is easy to show now, that $ \angle EMF \equal{} \angle NME \plus{} \angle NMF \equal{} \angle NMF \plus{} \angle KFM \equal{} 90^{o}$ and the proof is completed.

Kostas Vittas.


Let $n$ and $k$ be positive integers. There are given $n$ circles in the plane. Every pair of them intersect in two distinct points, and all points of intersection are pairwise distinct. Each intersection point must be coloured with one of $n$ distinct colours and each colour must be used at least once. Exactly $k$ distinct colours must occur on each circle. Find all values of $n \geq 2$ and $k$ for which such a colouring is possible.


All these are pretty basic, but I forget how to do them

simplify:
1) (e^x-3)/(e^x-4)
2) (e^x)/(e^1-x)

Solve each equation for x(I think ur supposed to use logs, not positive though):
5^3x=5^4x-2
7^x^2=7^2x+3
(1-x)^5=(2x-1)^5
2xe^-x=0
(x^2)(e^x)-5xe^x=0

Help is greatly appreicated, and im more intrested in the steps to solving these problems than the answers

When you have only the same base on each side of the equation then you can set the exponents equal to each other.

$\frac{e^{x-3}}{e^{x-4}} = e^{(x-3)-(x-4)} = e$

$7^{x^2} = 7^{2x+3} \Rightarrow x^2 = 2x +3 \Rightarrow x = -1, 3$

$2xe^{-x} = 0$  For this, note that $e^t$ is always positive for all real $t$, thus $x=0$.

$x^2e^x - 5xe^x = 0 \Rightarrow x^2-5x = 0 \Rightarrow x = 0, 5$

You should be able to do the rest based on those.

I didnt understand what you meant,
So are you saying they are equal because they cant be.

The general mathematical convention is to do exponents from top to bottom, i.e.

$2^{3^4}=2^{81}, not 2^{12}$

So $x^{x^x} \neq x^{2x}$

thx, mathnerd314 this thing was really annoying me.

[quote="shadysaysurspammed"]I'm not too clear how mathnerd got 2^81.could someone tell me?[/quote]
$2^{3^4}$
$=2^{81}$
because $3^4=81$


complex numbers
real analysis
real analysis unsolved

suppose z= a+bi, w = c+di. Define z<w if a<c, and also if a=c but b<d. Prove that this turns the set of all complex numbers into an ordered set. Does this oredered set have the least-upper-bound property?

THXXX

Does $ \{x\plus{}yi\mid x<0\}$ have a least upper bound? Does $ \{x\plus{}yi\mid x\le 0\}$ have a least upper bound?

Note the careful phrasing of the problem statement: $ \mathbb{C}$ is a field, and this turns it into a totally ordered set - but this order does not make $ \mathbb{C}$ an ordered field.


function
logarithms
integration
complex analysis
analytic geometry
complex analysis unsolved

Find all functions analytical in $\mathbb{C}\setminus\{0\}$ such that the image of any circle with center 0 belongs to some circle with center 0. (Here circle is a line.)

remark that if $0$ is a pole, $\alpha z^{n}f(z)$ is an entire function (for $n$ large enough and $\alpha \in \mathbb{C}$ well chosen) that stabilizes the unit circle and one has already proved that such functions are of the form $C z^{k}$, $k \geq 0$. So if $0$ is a pole there are only functions of the form $C z^{k}$, $k \in \mathbb{Z}$. Remains to prove that $0$ is always a pole ..

Note that $\log |f|$ is a radial harmonic function outside of $0$, and all such functions can be classified (as solutions of a certain ordinary differential equation).

A generalization of the mean-value property of harmonic functions says that if $u$ is harmonic in an annulus around $0$, then $\int_{0}^{2\pi}u(re^{i\theta})\mbox d\theta$ is a linear function of $\ln r$ (when $u$ is harmonic, this linear function becomes the constant $2\pi\cdot u(0)$). If we apply this to $u=\ln|f|$, we find a relation of the form $|f(z)|=C|z|^\alpha\ (*)$ for some constants $C>0,\alpha\in\mathbb R$.

It's clear from $(*)$ that there are small enough neighborhoods of $0$ which $f$ does not map onto dense subsets of the plane, and this is enough to conclude that $0$ is not an essential singularity (a holomorphic function maps every neighborhood of an isolated essential singularity onto a dense subset of the plane).

(edited; see alekk's comment below)

I'm not sure that if $u$ is harmonic in an annulus around $0$ then
$g(r)=\int_{0}^{2 \pi}u(r, \theta) d \theta$ is a linear function of $r$. For example if
$u(r, \theta)=\ln(r)$, $g(r)= 2 \pi \ln(r)$.

In polar coordinate, the laplacian is:
$\Delta (u) = \partial_{r,r}u+\frac{1}{r}\partial_{r}u+\frac{1}{r^{2}}\partial_{\theta, \theta}u$
Hence:
$g'(r)=\int_{[0; 2 \pi]}\partial_{r}u(r, \theta) d \theta$
$g^{''}(r)=\int_{[0; 2 \pi]}\partial_{r,r}u(r, \theta) d \theta =-[\int_{[0; 2 \pi]}\frac{1}{r}\partial_{r}u(r, \theta)+\frac{1}{r^{2}}\partial_{\theta, \theta}u(r, \theta) d \theta]$
$g^{''}(r) =-[\int_{[0; 2 \pi]}\frac{1}{r}\partial_{r}u(r, \theta) d \theta]=-\frac{1}{r}g'(r)$
thus
$r g^{''}(r)+g'(r)=0$
this is why
$g'(r)=\frac{\lambda}{r}$
$g(r)=\lambda \ln(r)+K$
The conclusion is the same: if $f$ is analytic with the hypothesis in the problem, $u=\ln(|f|)$ is radial
(and well defined because $|f(z)| \neq 0$ for $z \neq 0$) and harmonic. Hence
$g(r)= 2 \pi u(r) = 2 \pi ln(|f(z)|) = \lambda \ln(r)+K$ so
$|f(z)|=C |z|^{\alpha}$, as Grobber mentioned above.

It was a typo. I meant a linear function of $\ln r$. You could have seen that from the fact that I drew the conclusion $|f(z)|=C|z|^\alpha$ :).

yep , I knew it was a typo  :lol:
anyway the natural next question is:
what are the analytic functions $f: \mathbb{C}^{*}\to \mathbb{C}$ that stabilize the unit circle.

[quote="alekk"]what are the analytic functions $f: \mathbb{C}^{*}\to \mathbb{C}$ that stabilize the unit circle.[/quote]
Define $z^{*}=1/\overline{z}$. The analytic functions that stabilize the unit circle are precisely those for which $f(z^{*})=f(z)^{*}$. [hide="This is why"]because $f(z^{*})^{*}=f(z)$ on the circle, and both functions are analytic[/hide] You can create them in the following way: given any analytic function $g\colon \mathbb{C}^{*}\to \mathbb{C}$, let $f(z)=g(z)+g(z^{*})^{*}$. In fact, this gives all $f$s that stabilize the circle. [hide="This is why"]because the Laurent expansion of $f$ is symmetric[/hide]


Let $ ABC$ be a triangle $ B',C'$ be midpoint of $ AC,AB$, $ I_a$ is A-excenter, let $ M,N$ be the points on ray AB,AC such that $ BM\equal{}BC\equal{}CN$ the lines through $ M$ and perpendicular $ I_aC'$ and through $ N$ and perpendicular $ I_aB'$ intersect at $ A''$ prove that $ AA''\perp BC$.

[url=http://www.mathlinks.ro/viewtopic.php?t=249089]See more here[/url]  :D

Maybe this problem will be true for all point on line $ OI_a$, I inspired from [url=http://www.mathlinks.ro/viewtopic.php?t=249089]here[/url]  :| !


geometry
rectangle
quadratics
algebra
quadratic formula

1.For any positive integer n let s(n) denote the number of ordered pairs (x,y) of positive integers for which
$ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{n}$
Find all positive integers n such that s(n)=5

2.Solve in positive integers
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}\frac{3}{5}$

3.Determine all real x such that $ x^{n}\plus{}x^{\minus{}n}$ is an integer.

4.If r is a mXn rectangle formed from unit squares, find the number of unit squares containing a segment of one diagnol of R.

5.Find the number of ways in which nine digits 1,2,3,....9 can be arranged in a 3X3 square array so that the determinant of this array is an odd number.

[quote="hell_ever"]1.For any positive integer n let s(n) denote the number of ordered pairs (x,y) of positive integers for which
$ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{n}$
Find all positive integers n such that s(n)=5

2.Solve in positive integers
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {3}{5}$

3.Determine all real x such that $ x^{n} \plus{} x^{ \minus{} n}$ is an integer.

4.If r is a mXn rectangle formed from unit squares, find the number of unit squares containing a segment of one diagnol of R.

5.Find the number of ways in which nine digits 1,2,3,....9 can be arranged in a 3X3 square array so that the determinant of this array is an odd number.[/quote]
One solution for number 2 is 3, 6, 10.

[quote="hell_ever"]1.For any positive integer n let s(n) denote the number of ordered pairs (x,y) of positive integers for which
$ \frac {1}{x} + \frac {1}{y} = \frac {1}{n}$
Find all positive integers n such that s(n)=5

2.Solve in positive integers
$ \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = \frac {3}{5}$

3.Determine all real x such that $ x^{n} + x^{ - n}$ is an integer.

4.If r is a mXn rectangle formed from unit squares, find the number of unit squares containing a segment of one diagnol of R.

5.Find the number of ways in which nine digits 1,2,3,....9 can be arranged in a 3X3 square array so that the determinant of this array is an odd number.[/quote]

[hide="1"]
\begin{align*}
\frac 1x + \frac 1y &= \frac 1n \\
(x+y)n &= xy \\
n^2-(x+y)n+xy &= n^2 \\
(n-x)(n-y)=n^2
\end{align*}
This implies that $ n^2$ has exactly 5 divisors, then n has exactly 3 divisors and has to be of the form $ p^2$ with p prime.
[/hide]

[hide="Approach for 2"]
From the equation it's obvios that $ 2\le \min(x,y,z)\le 5$, so WLOG $ x=\min(x,y,z)$ and we just have to look at each case x=2,...,5 separately.
[/hide]

For #3, would $ n$ just fall under any value as an element of the Reals?
If so, then [hide]The only solution I can see that will always equal an integer upfront for all values of $ n$ is when $ x \equal{} { \plus{} \minus{} }1$, which for $ x \equal{} 1 \minus{} > 1^n \plus{} 1^{ \minus{} n} \minus{} > 1 \plus{} 1 \equal{} 2$. For $ x \equal{} \minus{} 1$ the solution would equal $ 2$ or $ \minus{} 2$, depending on the value of $ n$. There might be more but I'm not certain yet, this is just what I'm thinking upfront. I might be wrong, but I'm not sure-yet-if there's any other real value when itself plus its reciprocal, under any power, besides $ n \equal{} 0$ (which would mean that all values of $ x$ satisfy this) would equal an integer.[/hide]

[quote="Patterns_34"]I'm not sure if there's any other real value when itself plus its reciprocal, under any power, besides $ n \equal{} 0$ (which would mean that all values of $ x$ satisfy this) would equal an integer.[/quote]

$ \frac{k \plus{} \sqrt{k^2 \minus{} 4}}{2} \plus{} \frac{2}{k \plus{} \sqrt{k^2 \minus{} 4}} \equal{} \frac{k \plus{} \sqrt{k^2 \minus{} 4}}{2} \plus{} \frac{k \minus{} \sqrt{k^2 \minus{} 4}}{2} \equal{} k$

The value of $ x$ in question is real for $ k \ge 2$.

However, I believe the correct question is to determine the real values of $ x$ such that $ x^n \plus{} \frac{1}{x^n}$ is an integer [b]for every positive integer $ n$.[/b]

Well if it works for n=1 it will work for all n.

[hide="Proof"]
Set $ s_n \equal{} x^n \plus{} x^{ \minus{} n}\Rightarrow s_{n \plus{} 2} \equal{} s_1s_{n \plus{} 1} \minus{} s_n$, so if $ s_1\in\mathbb{N}$ all $ s_i\in\mathbb{N}.$$ (s_0\equal{}2\in\mathbb{N})$
[/hide]

[quote="bertram"]Well if it works for n=1 it will work for all n.

[hide="Proof"]
Set $ s_n \equal{} x^n \plus{} x^{ \minus{} n}\Rightarrow s_{n \plus{} 2} \equal{} s_1s_{n \plus{} 1} \minus{} s_n$, so if $ s_1\in\mathbb{N}$ all $ s_i\in\mathbb{N}.$
[/hide][/quote]
So don't we have infinitely many x, just pick a k and use the quadratic formula on $ x^{2}\minus{}kx\plus{}1\equal{}0$, and you have 2 working values of x.


calculus
function
analytic geometry
graphing lines
slope
real analysis
real analysis solved

Function f is differentiable on [a,b] and f'(a)=f'(b). show that there exists \ksi \in [a,b] such that f'(\ksi) = (f(\ksi)-f(a)) / (\ksi-a).

I'll make a statement here, which I'm going to use without proof:

First of all, let's generally denote by X the point (x,f(x)). I say that in the given conditions there is a point c in (a,b) s.t. C is on the line AB. Try to prove it; it's not that hard. The new problem I want to prove is this: there is a  \lambda in (c,b) s.t. the tangent at  \lambda to the graph of f passes through A. This is equivalent to the fact that there is x in (c,b) s.t.

(x-a)f'(x)-f(x)+ka+f(b)-kb = 0, where k is the slope of AB (and CB as well). But if we divide the left part by (x-a)^2 we get g'(x), where g(x)=[f(x)-(kx+f(b)-kb)]/(x-a) and g:[c,b]. It's easy to see that g has value 0 in c and b, so, by Rolle's thm, there is a pt x ( \lambda ) in (c,b) s.t. g'(x)=0, which is what we want. Hope it's clear enough...

Great! How did the idea come into your mind? I mean, how did u find to consider the function (x-a)f'(x)-f(x)+ka+f(b)-kb and then divide it by (x-a)^2?

Well, after you solve a few of these they become quite standard. The only problem is finding the right auxillary function. You know, joust like in the standard proof for Rolle's, Lagrange's and Cauchy's theorems.


geometry
circumcircle
perpendicular bisector
geometry unsolved

As shown in the following figure, point $ P$ lies on the circumcicle of triangle $ ABC.$ Lines $ AB$ and $ CP$ meet at $ E,$ and lines $ AC$ and $ BP$ meet at $ F.$ The perpendicular bisector of line segment $ AB$ meets line segment $ AC$ at $ K,$ and the perpendicular bisector of line segment $ AC$ meets line segment $ AB$ at $ J.$ Prove that

\[ \left(\frac{CE}{BF} \right)^2 \equal{} \frac{AJ \cdot JE}{AK \cdot KF}.\]

$ KB \equal{} AK$ and $ CJ \equal{} JA$ $ \Longrightarrow$ $ \angle FKB \equal{} 2 \angle A \equal{} \angle CJE.$ In addition,

$ \angle BFK \equal{} \angle BFC \equal{} \angle C \minus{} \angle CBF \equal{} \angle C \minus{} \angle CBP \equal{} \angle C \minus{} \angle CAP \equal{}$
$ \equal{} (\angle C \minus{} \angle A) \plus{} (\angle A \minus{} \angle CAP) \equal{} \angle BCJ \plus{} \angle PAB \equal{} \angle BCJ \plus{} \angle PCB \equal{}$
$ \equal{} \angle PCJ \equal{} \angle ECJ.$

Triangles $ \triangle BFK \sim \triangle ECJ$ are therefore similar, having equal angles $ \Longrightarrow$

$ \frac{EC}{BF} \equal{} \frac{JE}{KB} \equal{} \frac{CJ}{FK},\ \ \left(\frac{EC}{BF}\right)^2 \equal{} \frac{JE}{KB} \cdot \frac{CJ}{FK} \equal{} \frac{JE \cdot JA}{AK \cdot FK} \equal{} \frac{AJ \cdot JE}{AK \cdot KF}.$


I am about to rgister for Ijntermediate Trig, but I just  was looking through the problem set and I could do 5 out of 8 problems, and Counting: 6 out of 10 problems. My question is: will either of the classes still be good for me? There will be harder problems in each of the classes, right?
And does AoPS V.2 counting/probability chapters cover generating functions and other stuff to the same detail as the Intermediate class? What about AcOPS? Thanks!

Never mind. I decided to not take those classes. Sorry to bother you all.


modular arithmetic
limit
number theory unsolved
number theory

Let $ S(n)$ be defined as the the sum of all digits of $ n$. Prove that if we pick any $ 2$ consecutive numbers, there always exists a number such that it can be performed as $ n\plus{}S(n)$.

[quote="Bugi"]The question is unclear...

1) performed=expressed?

2) where did n come from?[/quote]

The question is :
Show that $ \forall n\in\mathbb N$, $ \exists p\in\mathbb N$ such that either $ n\equal{}p\plus{}S(p)$, either $ n\plus{}1\equal{}p\plus{}S(p)$

[quote="mathVNpro"]Let $ S(n)$ be defined as the the sum of all digits of $ n$. Prove that if we pick any $ 2$ consecutive numbers, there always exists a number such that it can be performed as $ n \plus{} S(n)$.[/quote]

Consider the sequence $ a_n\equal{}n\plus{}S(n)$

If $ n\neq 9\pmod{10}$, $ a_{n\plus{}1}\equal{}a_n\plus{}2$
If $ n\equal{}9\pmod{10}$, $ a_{n\plus{}1}<a_n$
And since $ a_n>n$, $ a_n$ may be as great as we want

So the set $ \{a_n\}$ contains $ 2$, and contains arbitrarily great numbers and no gap greater than $ 2$.
Hence the result.

can you please explain it a little bit clearer, dear [b]pco[/b]?

[quote="pco"][quote="mathVNpro"]Let $ S(n)$ be defined as the the sum of all digits of $ n$. Prove that if we pick any $ 2$ consecutive numbers, there always exists a number such that it can be performed as $ n \plus{} S(n)$.[/quote]

Consider the sequence $ a_n \equal{} n \plus{} S(n)$

If $ n\neq 9\pmod{10}$, $ a_{n \plus{} 1} \equal{} a_n \plus{} 2$
If $ n \equal{} 9\pmod{10}$, $ a_{n \plus{} 1} < a_n$
And since $ a_n > n$, $ a_n$ may be as great as we want

So the set $ \{a_n\}$ contains $ 2$, and contains arbitrarily great numbers and no gap greater than $ 2$.
Hence the result.[/quote]

[quote="mathVNpro"]can you please explain it a little bit clearer, dear [b]pco[/b]?[/quote]

Sure. Hope this will be clearer :

$ a_n\equal{}n\plus{}S(n)$

If the rightmost decimal digit of $ n$ is different from $ 9$, $ S(n\plus{}1)\equal{}S(n)\plus{}1$ and so $ a_{n\plus{}1}\equal{}a_n\plus{}2$
If the rightmost $ p>0$ last decimal digits of $ n$ are all $ 9$, $ S(n\plus{}1)\equal{}S(n)\plus{}1\minus{}9p$ and so $ a_{n\plus{}1}\equal{}a_n\plus{}2\minus{}9p<a_n$
So $ a_{n\plus{}1}\leq a_n\plus{}2$ $ \forall n$

$ a_1\equal{}2$
$ a_n>n$ and so $ \lim_{n\to\plus{}\infty}a_n\equal{}\plus{}\infty$

So $ \forall m\geq 1$ $ \exists p>0$ such that $ a_p\leq m\plus{}1< a_{p\plus{}1}$ (else, either $ a_k> m\plus{}1$ $ \forall k$, which is impossible since $ a_1\equal{}2\leq m\plus{}1$, either $ a_k\leq m\plus{}1$ $ \forall k$ which is impossible since $ \lim_{n\to\plus{}\infty}a_n\equal{}\plus{}\infty$).

And since $ a_{p\plus{}1}\leq a_p\plus{}2$, we have $ a_p\leq m\plus{}1< a_p\plus{}2$ and so either $ m\equal{}a_p$, either $ m\plus{}1\equal{}a_p$
Q.E.D.


trigonometry
geometry
trig identities
Law of Sines

Right triangle ACD is constructed w/ C being the right angle.  The measure of CAD is 30 degrees.  Triangle BEF is inscribed and is equilateral, w/ E touching CD, B touching AD, and F touching AC.  The measure of segments CE and DE are both one unit.
I don't think this can be approached with mass points.
Can anybody help?

1) :roll:
2) :!:
3) :oops:

Generally you've got some important point of concurrency or a couple lengths/ratios for mass points, so I don't see it here either.

[hide="A triggy solution"]
Call some angle $\alpha$, I chose $FEC$.  Then angle chase a little to get $EFC = 90^\circ - \alpha$ and $DBE = \alpha$.  Then law of sines says $\frac{1}{\sin{\alpha}} = \frac{s}{\sin{60^\circ}}$ and $\frac{1}{\sin{(90^\circ- \alpha)}} = \frac{1}{\cos{\alpha}} = \sec\alpha = \frac{s}{\sin{90^\circ}}$.

Solve for $s$ and equate or just divide one by the other to cancel the $s$ so you get $\tan\alpha = \frac{\sqrt3}{2}$.  Pythagorean identity to get $\sec\alpha$ which is equal to $s$ which ends up being $\frac{\sqrt7}{2}$.
[/hide]

Solve for the sides of the equilateral triangle!

mgao, did you see djshowdown2's solution???


algebra
polynomial
vector
linear algebra
linear algebra unsolved

Suppose $M\in M_n(F)$ and $f,g$ are two polynomials prime to each other. Prove that if $A=f(M),B=g(M)$. Then for each vector $v$ if $ABv=0$ then there exist 2 vectors $v_1,v_2$ which $Av_1=Bv_2=0$ and $v=v_1+v_2$

Since $f$ and $g$ are coprime there exists polynomials $m,k$ such that $mf+kg=1$. So $m(M)f(M)+k(M)g(M)=I$ in other words $m(M)A+k(M)B=I$, which means that $m(M)Av+k(M)Bv=v$ and now we can take $v_1=k(M)Bv,v_2=m(M)Av$ then $Av_1=Av_2=0$ and $v=v_1+v_2$ .
Actually the more general result holds which can be proved in the same way as above was presented:
If $P_1,P_2,...,P_k$-are pairwise coprime polynomials and $M \in L(E)$-linear operator on the $\mathbb{K}$-vector space $E$, then
$\bigoplus_{i=1}^k kerP_i(M)=ker(P_1P_2 \cdots P_k)(M)$.


calculus
integration
geometry
real analysis
real analysis unsolved

reading how to calculate a path integral(or line integral,or cubilinear integral
or however you may call it)
i got lost allong the way.
ok
so here is the integral
$I=\int_{r}(x+y)dx (y-x)dy$ where is supposed to be a square
(the square is aligned so that it sits in first quadrant and 2 of it sides are on the
axes) wich is walked in positive way(counterclockwise) of side $\sqrt{2}$
first of all i don't get the parametrisation they give here.
for $AB$ they give parametrisation $x=t,y=1-t$ with $0\leq t \leq 1$
isnt that parametrisation for the first bisector or the diagonal of the sqare ?
ok,so they go on calculating the for the first side of the square and they
get $I=I_{1}+I_{2}+I_{3}+I_{4}$(now every of those for I's are for every side
of the square).
and they get $I_{1}=\int_{1}^{0}dx+\int_{0}^{1}2y-1dy =-1$
(why is the first integral with 1 and 0 switched ??)
(for the second integral having as I states $y-xdy=1-2t$ ? but i see they
write $2y-1dy$ , why is this ?)
analogous to all the other sides,and conclusion is that $I=-4$
ok,but the area of the square should have been $2$ .
i guess i would have gotten the area if $I$ were equal to $\int_{r}xdx+ydy$ ?
what should i use to get the area enclosed by the sides of the square ?
what is the interpretation of $I$ as written in the statement ?

how about applying Green's theorem?

$I = \int_{r}(x+y) \, dx+(y-x) \, dy = \int \! \! \! \! \int_{\text{inside }r}\left( \frac{ \partial(y-x)}{\partial x}-\frac{ \partial(x+y)}{\partial y}\right) \, dxdy$

oh uhm , so

$I=\int_{0}^{1}\int_{0}^{1}(y-x-2 )dx dy =-4$

that works well.
the problem is , my book has yet to cover the green theorem(it is 26 pages away from where
i am now)

ok,for the other questions ?


algebra
polynomial
number theory
relatively prime

[quote="kunny"]For an integer $ a$, find the root such that $ 4x^3 \minus{} (a \minus{} 2)x \minus{} (a \plus{} 4) \equal{} 0$ has positive rational root which is not integer.[/quote]I think what was meant to be asked was “Find an integer value of $ a$ that will make this equation have a root that is a positive rational number, but is not an integer.”
In that case, I have a solution. Although it is not an elegant one, it may be what was expected.
[hide]A positive, non-integer rational root would be $ m/n$ where $ m$ and $ n$ are relatively prime positive integers, and $ n>1$.
We know that, if there is such a root, $ n$ is a factor of $ 4$ $ (2$ or $ 4)$, and $ m$ is an odd factor of $ (a\plus{}4)$.
Because the coefficients are integers, $ n\equal{}4$ will not work.
Substituting the rational root, $ x\equal{}m/2$, we get
$ m^3\minus{}m(a\minus{}2)\minus{}2(a\plus{}4)\equal{}0$, or
$ m(m^2\minus{}a\plus{}2)\equal{}2(a\plus{}4)$
Since $ m$ is not even, $ (m^2\minus{}a\plus{}2)$ must be even, which makes $ a$ and $ (a\plus{}4)$ odd.
Making $ m\equal{}a\plus{}4$ with $ m^2\minus{}a\plus{}2\equal{}2$ yields no solutions.
Then $ m$ must be just a factor of the composite odd integer $ (a\plus{}4)$.
Chosing $ a\plus{}4<0$ would make the whole polynomial positive for $ x>0$, so it must be $ a\plus{}4>0$.
We can try the composite odd integers in order: $ a\plus{}4\equal{}9$, $ a\plus{}4\equal{}15$, and so on.
Luckily $ a\plus{}4\equal{}9$  $ (a\equal{}5)$, with $ m\equal{}3$,  results in a solution.
The answer is $ a\equal{}5$.
With $ a\equal{}5$, $ x\equal{}3/2$ is a positive rational root which is not an integer.[/hide]

Assume it factors as $ (2x \plus{} b)(2x^2 \minus{} bx \plus{} c)$ (since the $ x^2$ cancels out).
The x coefficient is 6 greater than the constant term, so we obtain
$ b^2 \plus{} bc \minus{} 2c \plus{} 6 \equal{} 0$.
$ 10 \equal{} (2 \minus{} b)(b \plus{} c \plus{} 2)$
b must be negative for the root to be positive
Searching the factors of 10, we have for (b,c) (-3,3) and (-8,7)
So we have
$ (2x \minus{} 3)(2x^2 \plus{} 3x \plus{} 3)$ with root $ 3/2$.
and
$ (2x \minus{} 7)(2x^2 \plus{} 8x \plus{} 7)$ with root $ 7/2$
Of course there are a few other polynomials where the roots are negative.


Sarah, on a motorcycle, starts at the back of 2 km train as its front enters a 4km tunnel. Both travel at a constant speed and she exits the tunnel just as the train is entirely in the tunnel. When the front of the train emerges from the tunnel, Sarah turns and heads back toward the train. How many meters from the tunnel does Sarah meet the front of the train?

[hide="Solution"]The train travels $ 2$ kilometers as Sarah travels $ 6$ kilometers, so Sarah travels three times as quickly as the train. Therefore, when the train travels $ 4$ kilometers, Sarah has traveled $ 12$ kilometers, or $ 6$ extra kilometers from the front of the tunnel. Therefore, she travels $ 4500$ meters as the train travels $ \boxed{1500}$ meters from the front of the tunnel.[/hide]


The following inequalities a) and b) are well known.
    Let $ABC$ be a triangle and $x, y, z$ are real numbers. Show that
       a) $\sum yzcosA \leq \frac{x^2+y^2+z^2}{2}$
       b) $\sum yzsin^2A \leq \frac{(x+y+z)^2}{4}$
 In one of my posts, I have shown that
      c) $\sum yzsinA \leq \frac{(x+y+z)^2}{2\sum tg{\frac{A}{2}}}$
 Here is my problem in this post :
      Let $ABC$ be given. Maximize ${\frac{\sum yzsin2A}{(x+y+z)^2}}$ (here again, $x, y, z$ are reals)

It's really nice but not hard.Max=$\frac{1}{2\sqrt3}$.Here is my synthetic proof:
Put $A'=\pi-2a$,etc... then we need to prove:$\frac{\sum yzsinA'}{(\sum x)^2}\leq \frac{1}{2\sqrt3}$ where $A'+B'+C'=\pi$
We will prove that:
$\sum x^2 +2\sum xy(1-2sin\frac{\pi}{3}sinA')\geq 0$
But we have:
${L.H.S\geq \sum x^2+2\sum xy*cos(\frac{2\pi}{3}-A'})\geq 0$
which follows from the following result:
$\sum x^2+2\sum xy*cosM\geq 0$ where $M+N+Q=\pi$
And we are done!! :D

Let $ABC$ be an acute-angled triangle and $x, y, z$ are real numbers. Show that$$\frac{x^2}{cosA}+\frac{y^2}{cosB}+\frac{z^2}{cosC}\geq 4(yzsin\frac{A}{2}+zxsin\frac{B}{2}+xysin\frac{C}{2})$$


How many ways are there of placing 20 identical balls into 21 baskets?

Each basket can hold any number of balls.

Thanks

[hide="the method"]Ah, a classic question.

What we need to do is to make a bijection to something that is easier to count. Instead of putting balls into baskets, let us imagine that the balls are lined up in a line, and we're putting dividers between them. Let's take a smaller example, with 5 balls and 4 baskets. An example of an arrangement of balls in baskets:

o||ooo|o
(note that there is 1 less divider than there are baskets)

this would represent putting 1 ball into the first basket, 0 balls in the second, 3 balls in the third, and 1 ball in the fourth. Notice that every arrangement of 5 o's and 3 |'s will give a different arrangement of balls. However, it is easy to count how many ways you can arrange o's and |'s! It is just $ \frac {8!}{3!\cdot 5!} \equal{} \binom{8}{3}$.

(It's very similar to the question, how many ways can you rearrange the letters in "BANANA")

I trust you will be able to extend this to 20 balls and 21 baskets.[/hide]

Thankyou for your help.

Here is the method applied to my question -

20 balls and 21 baskets --> 20 balls 20 dividers

There are $ 40!$ ways of arranging the balls and dividers.

All the balls are identical so we divide by $ 20!$ to get rid of repeated cases. Same goes for the dividers.

Final number of arrangments = $ \frac {40!}{20!\cdot20!} \equal{} 137846528820$

Then you're asking for the number of partitions of 20 into 21 or fewer parts, which just reduces to asking for the number of partitions of 20.  There is no simple closed formula for the partition function, but (as with many other nice sequences) it can be found in the [url=http://www.research.att.com/~njas/sequences/Seis.html]OEIS[/url].  (The partitions are sequence [url=http://www.research.att.com/~njas/sequences/A000041]A000041[/url].)

On another note, is there any easy way of calculating the number of partitions of a number without listing them (a recurrence or something)?

Point of note, recently i read that an exact formula has been found, and it proceeded to give a gigantic formula involving lots of cos's and arccos's as well as some log and exp thrown in. Have a look for it, it was in some obscure journal. You wouldnt use it in a problem though  :lol:

Recently?  Rademacher's formula was found in 1937.  :)

Edit:  Well, Rademacher's is a convergent series, so if you mean a single closed form, that would be fantastic.

Yup  :)  a single (huge) formula. Ive been racking my brains to try remember where it was, but just cant  :(


symmetry
algebra
polynomial
ARML
complex numbers

[quote="AndrewTom"]Solve for complex numbers $ x$,

$ 2x^{8} \plus{} 6x^{7} \minus{} 44x^{6} \plus{} 24x^{5} \minus{} 90x^{4} \plus{} 24x^{3} \minus{} 44x^{2} \plus{} 6x \plus{} 2 \equal{} 0$[/quote]

Let $ y\equal{}x\plus{}\frac{1}{x}$ the equation becomes:

$ y^4\plus{}3y^3\minus{}26y^2\plus{}3y\plus{}1\equal{}0$

Let $ z\equal{}y\plus{}\frac{1}{y}$ the equation becomes:

$ z^2\plus{}3z\minus{}28$, so $ z\equal{}4$ or $ \minus{}7$,

$ y^2\minus{}4y\plus{}1\equal{}0$ or $ y^2\plus{}7y\plus{}1\equal{}0$

$ x^2\minus{}(2 \pm \sqrt{3})x\plus{}1\equal{}0$  or  $ x^2\plus{}\frac{7 \pm 3 \sqrt{5}}{2}x\plus{}1\equal{}0$

$ x\equal{}\frac{2 \pm \sqrt{3} \plus{}\sqrt{3 \pm 4 \sqrt{3}}}{2}$, $ \frac{2 \pm \sqrt{3} \minus{} \sqrt{3 \pm 4 \sqrt{3}}}{2}$,

$ \frac{\minus{}7 \mp 3 \sqrt{5} \plus{}\sqrt{78 \pm 42 \sqrt{5}}}{4}$, $ \frac{\minus{}7 \mp 3 \sqrt{5} \minus{}\sqrt{78 \pm 42 \sqrt{5}}}{4}$

Thanks stephencheng.

An aquaintance made this up and I thought it was nice (but felt he could have worked backwards from some nicer roots).

Nice solution, but I'm wondering how you thought to make that substitution!
To see that it can work you must expand out a power of 8! Is there something about the problem that allows you to see the substitution? thanks

The symmetry in the coefficients suggests this approach. Sometimes, even a polynomial equation without this symmetry can be manipulated to make this approach work. Can't think of an example off the top of my head right now, but I seem to remember there was one at ARML a few years ago. A tiebreaker question I think...

Two questions:

A) How did you figure out that factorization?

B) What now?

By the way, the ARML problem that I mentioned in my post above was tiebreaker 1 in 2007. It wasn't really related to this problem as much as I had thought though.

hello, the both factors can be written in the form
$ x^2\plus{}\frac{1}{x^2}\minus{}4\left(x\plus{}\frac{1}{x}\right)\plus{}3\equal{}0$
$ x^2\plus{}\frac{1}{x^2}\plus{}7\left(x\plus{}\frac{1}{x}\right)\plus{}3\equal{}0$
Sonnhard.

[quote="gauss202"]Two questions:

A) How did you figure out that factorization?[/quote]

He uses Maple extensively.

[quote="gauss202"]The symmetry in the coefficients suggests this approach.[/quote]
More precisely, the following is true.

[b]Lemma:[/b]  Let $ P(x)$ be a polynomial of degree $ n$ with nonzero constant term, and let $ Q(x) \equal{} x^n P \left( \frac{1}{x} \right)$.  Then the roots of $ Q$ are the reciprocals of the roots of $ P$.

[b]Corollary:[/b]  If $ P(x) \equal{} Q(x)$ (i.e. the polynomial is symmetric in its coefficients), then the roots of $ P$ come in pairs $ r_i, \frac{1}{r_i}$ and $ P$ is a polynomial in $ x \plus{} \frac{1}{x}$ (up to multiplication by $ x$).


geometry
incenter
circumcircle
geometry solved

A,B,C are three points which divide circle O to three arcs.X is a unfixed point on the arc AB.$I_1$ is the incentre of $\triangle{CAX}$,$I_2$ is the incentre of $\triangle{CBX}$
Prove the circumscribed circle of $\triangle{XI_1I_2}$ meets circle O at a fixed point.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=19777 .

  darij


superior algebra
superior algebra unsolved

Let $ K$ be a finite extension of $ \mathbb{Q}$. Prove that $ K$ contains only finitely many roots of unity.

If not, can the set of $ n$ such that $ K$ contains a primitive $ n$-th root of unity, be bounded?


[quote="Art of Owna"]How many occuences of the digit 0 are there in the list of numbers 1,2,3,...9999,10000?[/quote]

Would it be
[hide]$1000$ (units digit is 0)
+
$1000-9$ (tens digit is 0)
+
$1000-99$ (hundreds digit is 0)
+
$1000-999$ (thousands digit is 0)
=
$2893$?
(I just did it quickly - not sure)[/hide]

Extension

If I said how many times 0 occurs from 1,2,3..2006,2007, would the answer be [hide]506[/hide]

i just want to make sure i can do these "easy" counting problems for the AIME


inequalities
Euler
inequalities unsolved

Let $ a,b,c$ are 3 triangular side. Prove that
$ \frac{a^3\plus{}b^3}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{b^3\plus{}c^3}{c^2\minus{}ca\plus{}a^2}\plus{}\frac{c^3\plus{}a^3}{a^2\minus{}ab\plus{}b^2}\geq 2(a\plus{}b\plus{}c)$

[quote="Toxic_Frog"]Suppose $ a \ge b \ge c$ we have :
$ b \plus{} c > a \Rightarrow (b \minus{} c)(b \plus{} c) \ge a(b \minus{} c) \Rightarrow a^2 \plus{} b^2 \minus{} ab \ge c^2 \minus{} ca \plus{} a^2$ with a similar way we will have $ c^2 \minus{} ca \plus{} a^2 \ge b^2 \plus{} c^2 \minus{} bc$ so :
$ {1 \over a^2 \plus{} b^2 \minus{} ab} \le {1 \over a^2 \plus{} c^2 \minus{} ac} \le {1 \over b^2 \plus{} c^2 \minus{} bc}$

and we have $ a^3 \plus{} b^3 \ge a^3 \plus{} c^3 \ge b^3 \plus{} c^3$

Now using the right part of  Hardy's ineq simply gives the wanted result.[/quote]
I don't understand

he used rearrangement and the fact that $ \frac{a^3\plus{}b^3}{a^2\minus{}ab\plus{}b^2}\equal{}a\plus{}b$.

However, I do not think that we are done yet because the inequality is not symmetric (it is only cyclic)

we have to consider the case where $ a\le b\le c$ or $ a\ge c\ge b$

first, we show $ \sum\frac {b^3}{b^2 \minus{} bc \plus{} c^2}\geq \sum a$(well know).

Let $ A \equal{} \sum\frac {b^3}{b^2 \minus{} bc \plus{} c^2},B \equal{} \sum\frac {c^3}{b^2 \minus{} bc \plus{} c^2}$. $ A \minus{} B \equal{} \sum\frac {b^3 \minus{} c^3}{b^2 \minus{} bc \plus{} c^2} \equal{} \sum b \minus{} c \equal{} 0$.

Hence $ A \equal{} \frac {A \plus{} B}{2} \equal{} \sum\frac {b^3 \plus{} c^3}{2(b^2 \minus{} bc \plus{} c^2)}\geq \sum\frac {b^3 \plus{} c^3}{b^2 \plus{} c^2}\geq \sum\frac {b \plus{} c}{2} \equal{} \sum a$

 It is suffice to show $ \sum\frac {a^3}{b^2 \minus{} bc \plus{} c^2}\geq \sum a$

$ \sum\frac {a^3}{b^2 \minus{} bc \plus{} c^2}\geq \frac {1}{2}\sum\frac {a^3}{b^2 \plus{} c^2} \equal{} \frac {1}{2}\sum a\frac {a^2}{b^2 \plus{} c^2}$.

$ a\geq b$ iff $ \frac {a^2}{b^2 \plus{} c^2}\geq \frac {b^2}{c^2 \plus{} a^2}$.

Hece $ \sum a\frac {a^2}{b^2 \plus{} c^2}\geq \frac {1}{3}(\sum \frac {a^2}{b^2 \plus{} c^2})(\sum a)\geq \frac {1}{3}\cdot\frac {3}{2}\sum a \equal{} \frac {1}{2}\sum a$

[quote="jacques"]The last proof is wrong. It contains a mistake among a computation[/quote]
you are right.

[quote="Leonhard Euler"]

 It is suffice to show $ \sum\frac {a^3}{b^2 \minus{} bc \plus{} c^2}\geq \sum a$

[/quote]
By Cauchy-Schwarz inequality we have
$ \sum\frac {a^3}{b^2 \minus{} bc \plus{} c^2} \equal{}\sum\frac {a^4}{a(b^2 \minus{} bc \plus{} c^2)} \geq \frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum a(b^2 \minus{} bc \plus{} c^2)}$
It suffits to prove
 $ (a^2\plus{}b^2\plus{}c^2)^2 \ge (a\plus{}b\plus{}c)(\sum a(b^2 \minus{} bc \plus{} c^2))$
which is equivalent to
 $ a^4\plus{}b^4\plus{}c^4\plus{}abc(a\plus{}b\plus{}c) \ge \sum a(b^3\plus{}c^3)$ (Schur)

[quote="Leonhard Euler"]first, we show $ \sum\frac {b^3}{b^2 \minus{} bc \plus{} c^2}\geq \sum a$(well know).

Let $ A \equal{} \sum\frac {b^3}{b^2 \minus{} bc \plus{} c^2},B \equal{} \sum\frac {c^3}{b^2 \minus{} bc \plus{} c^2}$. $ A \minus{} B \equal{} \sum\frac {b^3 \minus{} c^3}{b^2 \minus{} bc \plus{} c^2} \equal{} \sum b \minus{} c \equal{} 0$[/quote]
You are wrong Euler,notice that $ b^3\minus{}c^3\equal{}(b\minus{}c)(b^2\plus{}bc\plus{}c^2)$ :wink:

[quote="Algadin"]Let $ a,b,c$ are 3 triangular side. Prove that
$ \frac {a^3 \plus{} b^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3 \plus{} c^3}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^3 \plus{} a^3}{a^2 \minus{} ab \plus{} b^2}\geq 2(a \plus{} b \plus{} c)$[/quote]

It is a difficult problem.

$ L\minus{}R\equal{}\frac{\sum_{cyc}a^5(a\minus{}b)(a\minus{}c) \plus{}(b^2a\plus{}ca^2\plus{}bc^2\minus{}abc)(b^3c\plus{}ba^3\plus{}c^3a\minus{}b^2c^2\minus{}b^2a^2\minus{}a^2c^2)}{(b^2\minus{}bc\plus{}c^2)(c^2\minus{}ac\plus{}a^2)(a^2\minus{}ab\plus{}b^2)}$

So, we only need to prove  $ b^3c\plus{}ba^3\plus{}c^3a \geq b^2c^2\plus{}b^2a^2\plus{}a^2c^2$ (1)

Since $ a,b,c$  are 3 sides of the triangle. We have $ a\equal{}x\plus{}y$ , $ b\equal{} y\plus{}z$,  $ c\equal{}z\plus{}x$

(1) becomes $ y^3x\plus{}yz^3\plus{}zx^3 \geq xy^2z\plus{}yz^2x\plus{}x^2yz$
Divide both side by xyz
$ \frac{y^2}{z}\plus{} \frac{z^2}{x} \plus{} \frac{x^2}{y}  \geq x\plus{}y\plus{}z$
Good problem !

[quote="Algadin"]Let $ a,b,c$ are 3 triangular side. Prove that
$ \frac { a^3\plus{} b^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3 \plus{} c^3}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^3 \plus{} a^3}{a^2 \minus{} ab \plus{} b^2}\geq 2(a \plus{} b \plus{} c)$[/quote]
This is my solution:
we have $ \frac{ a^3\plus{} b^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3 \plus{} c^3}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^3 \plus{} a^3}{a^2 \minus{} ab \plus{} b^2}\geq 2(a \plus{} b \plus{} c)$
$ \Leftrightarrow\frac{ a^3\plus{} b^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3 \plus{} c^3}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^3 \plus{} a^3}{a^2 \minus{} ab \plus{} b^2}\geq\sum {\frac{b^3\plus{}c^3}{b^2\minus{}bc\plus{}c^2}}$
$ \Leftrightarrow\sum {\frac{a^3}{b^2\minus{}bc\plus{}c^2}}\ge\sum {\frac{c^3}{b^2\minus{}bc\plus{}c^2}}$(*)
we may assum that :$ a\geq b\geq c$ so, $ a^3\geq b^3\geq c^3$ and $ \frac{1}{b^2\minus{}bc\plus{}c^2}\geq\frac{1}{c^2\minus{}ca\plus{}a^2}\geq\frac{1}{a^2\minus{}ab\plus{}c^2}$(notice that $ a\plus{}b>c,b\plus{}c>a,c\plus{}a>b$ )
So,(*) obviously true by rearangement,done!

It is the same thing that Toxic_Frog did

sorry, but I think you (ToanIneq) are wrong because we don't assume that $ a \geq b \geq c$... It is easy to see it. :D

This is my proof.
We can assume that a=max{a,b,c}.
Let $ X \equal{} a^2 \minus{} ab \plus{} b^2, Y \equal{} a^2 \minus{} ac \plus{} c^2, Z \equal{} c^2 \minus{} cb \plus{} b^2$. We have show that
$ \frac {(a \plus{} b)X}{Z} \plus{} \frac {(c \plus{} b)Z}{Y} \plus{} \frac {(a \plus{} c)Y}{X}\geq 2(a \plus{} b \plus{} c)$
*If $ a\geq b\geq c$, it is easy to show that $ X\geq Y\geq Z$. Then rewrite the inequality in the form (by use expand Abel)
$ (b \minus{} c)\frac {X}{Z} \plus{} (a \minus{} b)(\frac {X}{Z} \plus{} \frac {Y}{X} \minus{} 2) \plus{} (c\plus{}b)(\frac {X}{Z} \plus{} \frac {Y}{X} \plus{} \frac {Z}{Y} \minus{} 3)\geq 0$
Easy to check it out (by apply AM_GM inequality).
*If $ a\geq c\geq b$, we have $ Y\geq X\geq Z$, the inequality is writen as
$ (c \minus{} b)\frac {Y}{X} \plus{} (a \minus{} c)(\frac {Y}{X} \plus{} \frac {X}{Z} \minus{} 2) \plus{} (c\plus{}b)(\frac {Y}{X} \plus{} \frac {X}{Z} \plus{} \frac {Z}{Y} \minus{} 3)\geq 0$
And similarly we have it is true. :D

[quote="thegod277"]This is my proof.
We can assume that a=max{a,b,c}.
Let $ X \equal{} a^2 \minus{} ab \plus{} b^2, Y \equal{} a^2 \minus{} ac \plus{} c^2, Z \equal{} c^2 \minus{} cb \plus{} b^2$. We have show that
$ \frac {(a \plus{} b)X}{Z} \plus{} \frac {(c \plus{} b)Z}{Y} \plus{} \frac {(a \plus{} c)Y}{X}\geq 2(a \plus{} b \plus{} c)$
*If $ a\geq b\geq c$, it is easy to show that $ X\geq Y\geq Z$. Then rewrite the inequality in the form (by use expand Abel)
$ (b \minus{} c)\frac {X}{Z} \plus{} (a \minus{} b)(\frac {X}{Z} \plus{} \frac {Y}{X} \minus{} 2) \plus{} (c \plus{} b)(\frac {X}{Z} \plus{} \frac {Y}{X} \plus{} \frac {Z}{Y} \minus{} 3)\geq 0$
Easy to check it out (by apply AM_GM inequality).
*If $ a\geq c\geq b$, we have $ Y\geq X\geq Z$, the inequality is writen as
$ (c \minus{} b)\frac {Y}{X} \plus{} (a \minus{} c)(\frac {Y}{X} \plus{} \frac {X}{Z} \minus{} 2) \plus{} (c \plus{} b)(\frac {Y}{X} \plus{} \frac {X}{Z} \plus{} \frac {Z}{Y} \minus{} 3)\geq 0$
And similarly we have it is true. :D[/quote]
[color=blue]You 're wrong[/color]

hmm...  can't we use weighted AM -GM here.


Show that for every positive interger $n$  there exist one number which, wirtten in base $10$, has $n$ digits all equal to $1$ or $2$  and is divisible by $2^{n}$

You can probably do it by showing that there is a number which is a string of 2's followed by n 0's that has the same residue mod n as a smaller string of 1's followed by n 0's.

[hide="Proof"]For $n=1$ we have $N=2$, for $n=2$ we have $N=12$. Assume that for some $n$ we can find the desired $N$. We'll prove that a $1$ or $2$ can always be appended to the beginning of $N$ so that the new number is divisible by $2^{n+1}$

Indeed, the two new numbers are $10^{n}+N$ and $2\cdot 10^{n}+N$. If we put $N=M\cdot 2^{n}$, which is the inductive presumption, we can divide those numbers by $2^{n+1}$ to obtain

$\frac{10^{n}+N}{2^{n+1}}=\frac{10^{n}+M\cdot 2^{n}}{2^{n+1}}=\frac{5^{n}+M}{2}$

$\frac{2\cdot 10^{n}+N}{2^{n+1}}=\frac{2\cdot 10^{n}+M\cdot 2^{n}}{2^{n+1}}=\frac{2\cdot 5^{n}+M}{2}$

Since $M$ is either even or odd, exactly one of those quotients is an integer, which proves both the inductive step and the uniqueness of $N$ for the given $n$. The resulting sequence is $2,12,112,2112, 22112,122112,2122112,22122112,\dots$[/hide]

Doesn't this seem strangely similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=336189#p336189]2003 USAMO #1[/url]?

[quote="Phelpedo"]The solution does.[/quote]

Are you implying I'm a Xerox machine? Behave, boy.

[quote="Gyan"]Small nitpick - [quote]
Since $M$ is either even or odd, exactly one of those quotients is an integer, which proves both the inductive step [b]and the uniqueness of[/b] $N$ [b]for the given[/b] $n$. The resulting sequence is $2,12,112,2112, 22112,122112,2122112,22122112,\dots$ [/quote]
There is no uniqueness, for example, if nothing else,  for n=2, you have 12,112,212,1111112 etc.[/quote]

Read the problem carefully: For every given $n$, the desired number must have $n$ digits - that makes it unique.

Right, yes I did not read the problem carefully.


[quote="moldovan"]Each diagonal of a convex pentagon is parallel to one side of the pentagon. Prove that the ratio of the length of a diagonal to that of its corresponding side is the same for all five diagonals, and compute the ratio.[/quote]
[b][color=brown]If th side lenght is l the ratio will be
 l / cos 36
 right or wrong!![/color][/b]


geometry
circumcircle
trigonometry
radical axis
geometry unsolved

the triangle ABC is inscribed in a circle center O. The tangent of the circle through A meet BC at D. AB, AC meet DO at E,F,respectively. M,N are the midpoints of the segments AB, AC, respectively. Prove that  the lines FM, EN, AO have a common point.

We denote the points $ B'\equiv AB\cap NO,$ $ C'\equiv AC\cap MO$ and it is easy to show that the point $ O,$ is the orthocenter of the triangle $ \bigtriangleup AB'C'.$

Let $ S$ be, the intersection point of the line segments $ B'C',$ $ MN$ and we denote the points $ E,$ $ F,$ on $ AC,$ $ AB$ respectively, as their intersection points from the line segment $ SO.$

It is easy to prove that the point $ P\equiv FM\cap EN,$ lies on $ AO,$ because of the concurrency of the line segments $ MN,$ $ EF,$ $ B'C',$ at the point $ S.$

We denote as $ D,$ the intersection point of the sideline $ BC$ of $ \bigtriangleup ABC,$ from the line segment $ SO$ and it is enough to prove that $ DA\perp AO$ $ \Longrightarrow$ $ DA\parallel B'C'.$

Based on the below [b][size=100]Lemma[/size][/b], we have $ MK\parallel AL$ $ ,(1)$ where $ K\equiv B'C'\cap AO$ and $ L\equiv BC\cap MO.$

From $ (1)$ $ \Longrightarrow$ $ \frac {OA}{OK} \equal{} \frac {OL}{OM}$ $ ,(2)$ and from $ MN\parallel BC$ $ \Longrightarrow$ $ \frac {OL}{OM} \equal{} \frac {OD}{OS}$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {OA}{OK} \equal{} \frac {OD}{OS}$ $ \Longrightarrow$ $ DA\parallel SK\equiv B'C'$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ AD,$ $ BE,$ $ CF$ be, its altitudes. We draw the circle $ (H),$ centered at the orthocenter $ H$ of $ \bigtriangleup ABC,$ with radius $ HA,$ which intersects $ AC,$ $ AB,$ at points $ E',$ $ F',$ respectively. Prove that $ AP\parallel DF,$ where $ P\equiv CF\cap E'F'.$[/color][/size][/b]

I will post here later the easy proof of the above [b][size=100]Lemma[/size][/b], I have in mind.

A happy new year to all, Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ AD,$ $ BE,$ $ CF$ be, its altitudes. We draw the circle $ (H),$ centered at the orthocenter $ H$ of $ \bigtriangleup ABC,$ with radius $ HA,$ which intersects $ AC,$ $ AB,$ at points $ E',$ $ F',$ respectively. Prove that $ AP\parallel DF,$ where $ P\equiv CF\cap E'F'.$[/color][/size][/b][/quote]
[b][size=100]PROOF[/size][/b]. - Because of the midperpendiculars $ HE,$ $ HF,$ of the segments $ AE',$ $ AF'$ respectively, we have $ EF\parallel E'F'$ $ ,(1)$

Because of $ (1)$ and cyclic quadrilaterals $ AEHF,$ $ AEDB,$ $ BDHF,$ we conclude $ \angle APF \equal{} \angle F'PF \equal{} \angle EFP \equal{} \angle EAD \equal{} \angle EBD \equal{} \angle DFP$

$ \Longrightarrow$ $ \angle APF \equal{} \angle DFP$ $ \Longrightarrow$ $ AP\parallel DF$ and the proof is completed.

Kostas Vittas.

Denote $ K \in AO \cap (O), S \in EF \cap MN$.
We have $ ABKC$ is harmonic quadrilateral, so $ A(AKBC) \equal{} \minus{} 1 \Rightarrow O(ASMN) \equal{} \minus{} 1 \Rightarrow A(OSMN) \equal{} \minus{} 1$, which means that $ AO,EN,FM$ are concurrent

Here is another solution

Let $ ON$ meet $ AB$ at $ X$ and $ ON$ the $ AC$ at $ Y$  and let $ S$ be the intersection of $ XY,\ MN$
I will show that $ S,\ O,\ D$ are collinear finishing the proof .

Consider the symmedian $ AK$ of the triangle $ ABC$ .It also symmedian of $ AMN$ . It is easy to show that $ (C,K,B,D)=-1$
Now we have that $ XY,\ MN$ are antiparallel so AK passes through the midpoint $ Q$ of $ XY$ . So $ AQ\equiv AK$ is the polar of $ S$ with respect the circumcirlce of MNYX . So $ SO\bot AQ$ . But $ AK$ is the polar of $ D$ with respect $ (O)$ so $ DO\bot AK$ so  $ S,\ O,\ D$ are collinear finishing the proof  :)

Hope you like it

A not so nice way, using poles/polars and then a quick application of the Law of Sines...

Let the tangents to the circumcircle from $ B$, $ C$, and $ A$ meet pairwise at $ X$, $ Y$, and $ Z$, so that $ X$ is opposite $ A$, $ Y$ opposite $ B$, and $ Z$ opposite $ C$.  Create lines through $ Y$ and $ Z$ parallel to $ CA$ and $ AB$, respectively.  Let these lines intersect at $ T$.  Construct lines through $ Y$ and $ Z$ parallel to $ AX$.  Let the line from $ Y$ and the first line from $ Z$ intersect at $ R$, and let the line from $ Z$ parallel to $ AT$ and the line from $ Y$ parallel to $ CA$ intersect at $ S$.  Okay.

$ ZR$ is the polar of $ M$, $ YS$ is the polar of $ N$
The intersection of $ YZ$ with the line at infinity is the pole of $ AO$.

The poles of $ AB$ and $ AC$ are $ Z$ and $ Y$, and the pole of $ DO$ is the intersection of $ AX$ with the line at infinity.  Therefore, the polars of E and F are lines $ ZS$ and $ YR$.

So then we have $ S$ is the pole of $ EN$, $ R$ is the polar of $ FM$.  It then suffices to show $ RS\parallel{}YZ$.  We will show $ RSZY$ is a parallelogram.  We will show the diagonals bisect each other.

Notice $ m\angle{TZY} \equal{} m\angle{ZAB} \equal{} m\angle{C}$.  Simiarly, $ m\angle{TYZ} \equal{} m\angle{B}$.  Also, $ m\angle{SZT} \equal{} m\angle{BAX}, m\angle{ZST} \equal{} m\angle{CAX}$.

So $ \frac {ST}{ZT} \equal{} \frac {\sin{\angle{BAX}}}{\sin{\angle{CAX}}}$
But $ AX$ is a symmedian.
$ \equal{} \frac {BA}{CA} \equal{} \frac {YT}{ZT}$
$ \implies ST \equal{} YT$

Similarly, $ RT \equal{} ZT$.  And we're done.

EDIT: A synthetic/harmonic proof.  Let $ AX \cap DO\equal{}P$.  Clearly $ DO \perp AX$, as $ AX$ is the polar of $ D$.

It suffices to show that if $ EF \cap MN\equal{}Q$, then $ O$ and $ Q$ are harmonic conjugates with respect to $ M$ and $ N$.  Define $ B', C'$ as Vittasko did ($ NO \cap AB$ and $ MO \cap AC$).  We know that if $ MN$ and $ B'C'$ intersect at $ Q'$, then $ AO, AQ', AB',$ and $ AC'$ are harmonic lines.  Therefore, it suffices to show $ Q\equal{}Q'$: namely, that $ MN, EF,$ and $ B'C'$ concur.

As $ O$ is the orthocenter of $ \triangle{AB'C'}$, notice $ B'C'NM$ is cyclic.
As $ m\angle{APO}\equal{}m\angle{AMO}\equal{}m\angle{ANO}\equal{}90^{\circ}$, notice $ A, M, N, P, O$ are concyclic.
As $ m\angle{XPO}\equal{}m\angle{XBO}\equal{}m\angle{XCO}\equal{}90^{\circ}$, notice $ X, B, C, P, O$ are concyclic.  As $ 180^{\circ}\minus{}m\angle{OC'C}\equal{}m\angle{MC'A}\equal{}90\minus{}m\angle{A}\equal{}m\angle{OXC}$, notice $ C'$ is on this circle.  So is $ B'$.

Therefore, because $ MNOP$, $ OPB'C'$, and $ B'C'NM$ are cyclic, the three lines $ MN, PO,$ and $ B'C'$ must concur at the radical center of these three circles.

[quote="caothujjj"] [color=darkred]Let $ C(O)$ be the circumcircle of the triangle $ ABC$ .Denote $ D\in BC\cap AA$ , $ E\in AB\cap DO$ , $ F\in AC\cap DO$ and the midpoints $ M$ , $ N$ of the segments $ AB$ , $ AC$ respectively. Prove that  the lines $ FM$ , $ EN$ , $ AO$ are concurrently.[/color] [/quote]
[color=darkblue][b][u]A metrical proof.[/u][/b] $ O\in EF\ \Longrightarrow\ \frac {EB}{EA}\cdot b\cdot\cos B \plus{} \frac {FC}{FA}\cdot c\cdot\cos C \equal{} a\cos A\ \ (1)$ . Apply

the [b]Menelaus[/b]' theorem to the transversal $ \overline {DEF}$ in $ \triangle ABC\ : \ \ \frac {DB}{DC}\cdot\frac {FC}{FA}\cdot\frac {EA}{EB} \equal{} 1$ and

$ \frac {DB}{DC} \equal{} \frac {c^2}{b^2}\ \ \Longrightarrow\ \ c^2\cdot\frac {FC}{FA} \equal{} b^2\cdot\frac {EB}{EA}\ \ (2)$ . From the relations $ (1)$ , $ (2)$ obtain :

$ \frac {FC}{b\cdot\cos A} \equal{}$ $ \frac {FA}{c} \equal{}$ $ \frac {b}{c \plus{} b\cdot\cos A} \equal{}$ $ \frac {2\cdot\left|FA \minus{} \frac b2\right|}{c \minus{} b\cdot \cos A} \equal{}$ $ \frac {2\cdot FN}{a\cdot\cos B}$ .

$ \frac {EB}{c\cdot\cos A} \equal{}$ $ \frac {EA}{b} \equal{}$ $ \frac {c}{b \plus{} c\cdot\cos A} \equal{}$ $ \frac {2\cdot\left|EA \minus{} \frac c2\right|}{b \minus{} c\cdot\cos A} \equal{}$ $ \frac {2\cdot EM}{a\cdot\cos C}$ .

Thus,  $ \frac {OE}{OF} \equal{} \frac {AE}{AF}\cdot\frac {\cos C}{\cos B}$ $ \implies$ $ \frac {OE}{OF}\cdot\frac {NF}{NA}\cdot\frac {MA}{ME} \equal{}$ $ \frac {AE}{AF}\cdot\frac {\cos C}{\cos B}\cdot \frac {2\cdot NF}{b}\cdot\frac {c}{2\cdot ME} \equal{}$

$ \frac {AE}{ME}\cdot\frac {NF}{AF}\cdot\frac cb\cdot\frac {\cos C}{\cos B} \equal{}$ $ \frac {b}{a\cdot\cos C}\cdot\frac {a\cdot \cos B}{c}\cdot\frac {c\cdot\cos C}{b\cdot\cos B} \equal{} 1$ . In conclusion,

$ \frac {OE}{OF}\cdot\frac {NF}{NA}\cdot\frac {MA}{ME} \equal{} 1$ , i.e. (the [b]Ceva[/b]'s theorem) the lines $ FM$ , $ EN$ , $ AO$ are concurrently.[/color]

Dear Virgil, can you please show how this relation comes?

[quote="Virgil Nicula"] $ O\in EF\ \Longrightarrow\ \frac {EB}{EA}\cdot b\cdot\cos B \plus{} \frac {FC}{FA}\cdot c\cdot\cos C \equal{} a\cos A\ \ (1)$ . [/quote]


inequalities
geometry
3D geometry
tetrahedron
ratio
function
vector

If $ a\equal{}b\to\infty$, $ c\in\mathbb{R}$ then

 $ (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\equal{}(2a\plus{}c)(\frac{2}{a}\plus{}\frac{1}{c})\to\infty$

I'm sorry but [i]c[/i] cannot be arbitrary here.

In fact to satisfy that [i]1/a[/i], [i]1/b[/i] and [i]1/c[/i] are the sides of a triangle, [i]c[/i] must tend to infinity as [i]a[/i] and [i]b[/i] do.

This makes the term in the second bracket tend to zero and your quoted expression does not necessarily tend to infinity.

Thanks for your interest though.

Steve

A fascinating problem, especially one of its generalizations (see below).

I will prove $ M \equal{} 6 \plus{} 2\sqrt {5}\approx10.472$.
We can suppose wlog $ a\ge b\ge c$ and $ ac \equal{} 1$, cuz dividing $ a,b,c$ by $ \frac {1}{\sqrt {ac}}$ doesn’t change $ F(a,b,c): \equal{} (a \plus{} b \plus{} c)(\frac1a \plus{} \frac1b \plus{} \frac1c)$
Then the problem is maximising $ F(a,b,c) \equal{} (a \plus{} \frac1a)^2 \plus{} (a \plus{} \frac1a) (b \plus{} \frac1b) \plus{} 1$, with conditions $ a \minus{} \frac1a\le b$ and $ \frac1b\ge a \minus{} \frac1a$. From both conditions we get $ a^2\le a \plus{} 1$, i.e. $ a\le\phi: \equal{} \frac {\sqrt {5} \plus{} 1}2$. Note that for fixed $ a$, F has its minimum at $ b \equal{} 1$.
[b]Case 1: [/b]$ a\le\sqrt {2}$. Then $ F(a,b,c)\le F(a,a,c)\le 10$.
[b]Case 2: [/b]$ \sqrt {2}\le a\le \phi$. Then
$ F(a,b,c)\le F(a,\frac {a}{a^2 \minus{} 1}, \frac1a ) \equal{} 2(\frac {a^4}{a^2 \minus{} 1} \plus{} 1) \le2(\frac {\phi^4}{\phi^2 \minus{} 1} \plus{} 1) \equal{} 2(\phi^3 \plus{} 1) \equal{} 2(3 \plus{} \sqrt {5}).$
(Note that $ \frac {a^4}{a^2 \minus{} 1}$ has its minimum at $ a \equal{} \sqrt {2}$.)

Equality occurs for $ (a,b,c) \equal{} ( \phi,1, \phi^{ \minus{} 1})$ Note that for this extremal case the triangle (a,b,c) is degenerated, as well as the one with sides $ (\frac1a,\frac1b,\frac1c)$ (which is in fact the same).

Several [b][i]generalizations [/i][/b]are possible, i.e. searching the maximum of $ F: \equal{} F(a_1,\dots,a_n) \equal{} \sum a_i\sum\frac1{a_i}$ under certain conditions:
[i][b](1)[/b]	any three of $ a_1,\dots,a_n$ and any three of $ \frac1{a_1},\dots,\frac1{a_n}$ must form a triangle.
[/i]
The same method as above yields that $ (a_1,\dots,a_n) \equal{} ( \phi,1 ,\dots,1, \frac1{\phi} )$ is extremal
[i][b](2)[/b]	More generally : given $ k\ge3$, require that any $ k$ of $ a_1,\dots,a_n$ and any $ k$ of $ \frac1{a_1},\dots,\frac1{a_n}$ must form a $ k$-gon.

[/i]Almost the same thing. $ (a_1,\dots,a_n) \equal{} ( x,1, \dots,1, \frac1x)$ is extremal, where $ x \equal{} k \minus{} 2 \plus{} \frac1x$.

More interesting is the following involving higher dimensions :) :D
[i][b](3)[/b]	Let $ d\ge2$ and $ n \equal{} \frac {d(d \plus{} 1)}{2}$. Require that $ a_1,\dots,a_n$ as well as $ \frac1{a_1},\dots,\frac1{a_n}$ are edges of a $ d$-dimensional simplex.
[hide] Note that this simplex won’t be uniquely determined in general. If e.g. $ d \equal{} 3$ and $ (a_1,\dots,a_n) \equal{} (1,1,1, \sqrt {3},\sqrt {3},\sqrt {3})$, there is one tetrahedron with base triangle of sides $ (1,1,1)$, having a thorn-like shape, and another one with base triangle of sides $ (\sqrt {3},\sqrt {3},\sqrt {3})$, which is ‘flat’, i.e. degenerated with height $ 0$. [/hide][/i]
Let’s call a sequence $ (a_1,\dots,a_n)$ [i] self-reciprocal [/i] if $ a_1a_n \equal{} a_2a_{n \minus{} 1} \equal{} \cdots \equal{} a_na_1$. Note that for such a sequence, the sequence $ (\frac1{a_1},\dots,\frac1{a_n})$ yields the same simplexes as $ (a_1,\dots,a_n)$.
For $ d \equal{} 3$, I conjecture $ 36\le F\le 22 \plus{} 10\sqrt {5}\approx 44.36$, where the maximum is yielded by the six edges contained in the golden ratio constellation: $ (1, \phi, \phi, \phi^2, \phi^2, \phi^3)$ as illuatrated. It may be possible to prove this in a similar way as above, but this probably is awkward, distinguishing several cases.
Note that like in the original $ d \equal{} 2$ case (and in the first two generalizations), this is self-inverse and the tetrahedron is the most degenerated possible, with all points not only in a plane, but on a line.
For $ d > 3$, I don’t think there exists a constellation with $ d \plus{} 1$ points on a line such that the reciprocals of the distances form a $ d$-simplex. No clue what may be the shape of a maximal constellation. For $ d\equal{}4$, it may be the above plus a point on its side such that $ xy\equal{}\phi^3$

Prove or disprove that a maximal constellation is always self-reciprocal... :)

[quote="hexomino"]Let [i]a[/i],[i]b[/i] and [i]c[/i] be the lengths of the sides of a triangle such that the numbers [i]1/a[/i], [i]1/b[/i] and [i]1/c[/i] are also the lengths of the sides of a triangle.

What is the maximum possible value of (a+b+c)(1/a +1/b + 1/c)?

In terms of an inequality problem I want to be able to write

9 <= (a+b+c)(1/a + 1/b + 1/c) <= M

At the moment, I can show that M <= 11 but I'm sure this isn't the best bound.

Thanks

Steve[/quote]
Assume that $ a \ge b \ge c$, then we have $ a \le b\plus{}c$ and $ c \ge \frac{ab}{a\plus{}b}$. Consider the function
\[ f(c)\equal{}(a\plus{}b\plus{}c)(\frac{1}{a} \plus{}\frac{1}{b}\plus{}\frac{1}{c})\]
We have
\[ f'(c) \equal{}\frac{1}{a}\plus{}\frac{1}{b} \plus{}\frac{1}{c} \minus{}\frac{a\plus{}b\plus{}c}{c^2} \equal{}\frac{(a\plus{}b)(c^2\minus{}ab)}{ abc^2} \le 0\]
Hence,
Case 1. If $ a \minus{}b \le \frac{ab}{a\plus{}b}\Leftrightarrow a \le \frac{1\plus{}\sqrt{5}}{2}b$, then we have
\[ f(c) \le f( \frac{ab}{a\plus{}b})  \equal{}(a\plus{}b\plus{}\frac{ab}{a\plus{}b})( \frac{1}{a}\plus{}\frac{1}{b} \plus{}\frac{a\plus{}b}{ab}) \equal{} \frac{ 2(a^2\plus{}3ab\plus{}b^2)}{ ab} \le 6\plus{}2\sqrt{5}\]
(since $ b \le a \le \frac{\sqrt{5}\plus{}1}{2}b$)
Case 2. If $ a \minus{}b \ge \frac{ab}{a\plus{}b}\Leftrightarrow a \ge \frac{\sqrt{5}\plus{}1}{2}b$, then we have
\[ f(c) \le f(a\minus{}b) \equal{}2a  (\frac{1}{a} \plus{}\frac{1}{b} \plus{}\frac{1}{a\minus{}b} ) \equal{}2\plus{}\frac{2a}{b} \plus{}\frac{2a}{a\minus{}b} \equal{}g(a)\]
\[ g'(a) \equal{}\frac{2}{b} \minus{}\frac{2b}{(a\minus{}b)^2} \le \frac{2}{b} \minus{}\frac{2b}{(2b\minus{}b)^2}  \equal{}0\]
Hence $ g(a) \equal{}\le g(\frac{\sqrt{5}\plus{}1}{2}b) \equal{}6\plus{}2\sqrt{5}$.
:)

Thanks for the solutions guys!

Also, spanferkel, nice generalisations!
 :)

Cheers

Steve

[quote="hexomino"]Thanks for the solutions guys!

Also, spanferkel, nice generalisations!
 :) [/quote]
But the best one remains to be proved  :maybe:

[quote="spanferkel"]
For $ d > 3$, I don’t think there exists a constellation with $ d \plus{} 1$ points on a line such that the reciprocals of the distances form a $ d$-simplex. No clue what may be the shape of a maximal constellation. For $ d \equal{} 4$, it may be the above plus a point on its side such that $ xy \equal{} \phi^3$
[/quote]
In fact, for $ d \equal{} 4$, this is simply solved by $ x\equal{}\phi^2\equal{}\phi\plus{}1$ and $ y\equal{}\phi$, so the constellation would be part of the usual pentagram pattern (=pentagon with its diagonals), and $ 100\le F\le M_4: \equal{}4\phi^7\equal{}58\plus{}26\sqrt{5}\approx 116.14$.
For $ d>4$, there cannot be more points added to this in a plane if we still want to obtain a constellation where the $ \frac1{a_i}$ are also feasible edge lenghts.

But based on this, I would conjecture that an analogous thing may well be the maximal pattern for higher dimensions:
For $ d\equal{}5$, add to this a vertex in $ \mathbb R^3$ with the same distances from the initial "golden ratio" segment, and at a distance of $ \phi^{\frac32}$ from the other vertice. The edge of length $ \phi^{\frac32}$ is "self-reciprocal", and $ 225\le F\le M_5: \equal{}(1\plus{}2\sqrt{5}\phi^{\frac52})^2\approx 252.59$.
Generally, for any $ d\ge4$, instead of this one edge, we would have $ (d\minus{}3)$ points forming a $ (d\minus{}4)$-dimensional regular simplex of edge length $ \phi^{\frac32}$, in addition to the four "golden ratio" points as above, and this all happens in $ \mathbb R^{d\minus{}2}$ .  :lol:  :D

I’ve just solved the $ d\equal{}3$ case. The extremal constellation (more precisely the limit case) is a 1-dimensional degenerate tetrahedron, but not the above... Just note that $ (t,t,t,\frac1t,\frac1t,\frac1t)$ are edges of a tetrahedron for each $ t>0$. So $ F\equal{}9(t\plus{}\frac1t)^2$ has no maximum. :) But for $ d>3$, a similar argument doesn’t necessarily apply, :?: at least not for $ d\equal{}4$.

For $ d\equal{}4$, starting with the edges $ (t,t,t, 1,1,1,1,\frac1t, \frac1t,\frac1t)$, we find that the maximal feasible $ t$ for this is the biggest solution of $ 3t^6\minus{}12t^4\plus{}4\equal{}0$, which is $ t\approx1.978$, yielding $ F\approx 131.124$, a value much bigger than the one I suggested above. The associated constellation is a 3-dimensional degenerated 4-simplex, in fact a tetrahedron with edges $ (t,t,t,\frac1t,\frac1t,\frac1t)$ as above plus a central point which is at distance 1 from its 4 vertices.

Inspired by this, I suggest two more generalizations:

[b](4)[/b] This extends (3) the same way as (2) extends (1): Let $ n>k: \equal{}\frac{d(d\plus{}1)}{2}$ and require that any $ k$ of $ a_1,\dots,a_n$ and any $ k$ of $ \frac1{a_1},\dots,\frac1{a_n}$ must form a $ d$-simplex.
Even the simplest case $ d\equal{}3, n\equal{}7$ is already rather complicated. Starting with $ (t,t,t, 1,\frac1t, \frac1t,\frac1t)$ for example,  we find that the maximal feasible $ t$ for this is the biggest solution of $ t^4\minus{}t^2\minus{}\sqrt{3}t\minus{}1\equal{}0$, which is $ t\approx1.58$, yielding $ F\approx 58.35$. But nothing proves of course that the extremal pattern is of this type.

[b](5)[/b] Note that for a given set of 6 edge lengths, there are up to 30 non-isomorphic tetrahedrons (because there are 15 ways of selecting 3 pairs of opposite edges, and each of those yields 2 tetrahedrons). So we can formulate a much sharper condition: require that not only at least one, but all 30 possible ways of labeling a tetrahedron with edges $ a_1,\dots,a_6$ and all 30 ways of labeling one with $ \frac1{a_1},\dots,\frac1{a_6}$ are valid. (and likewise for higher dimensions :) ).

[hide]By the way, this condition is stronger than just requiring that for each ‘coloring’ (=labeling) of the edges of the complete graph $ K_{d\plus{}1}$ with the values $ a_1,\dots,a_{\frac{d(d\plus{}1)}{2}}$, the triangle inequalities must be fulfilled in all the triangles of $ K_{d\plus{}1}$.   [/hide]

Possibly for $ d\equal{}3$, the maximum is already achieved for $ (\phi, \phi, \phi, 1,1,1)$, yielding $ F\equal{}9\phi^3\approx 38.124$. Note that there are 3 non-isomorphic labelings for the tetrahedron in this case, only one of which yields a degenerate one.

what's about maximum of $ (x^2\plus{}y^2\plus{}z^2)(\frac{1}{x^2}\plus{}\frac{1}{y^2}\plus{}\frac{1}{z^2})$ with $ x\plus{}y\plus{}z\equal{}3$
and  $ (x^n\plus{}y^n\plus{}z^n)(\frac{1}{x^n}\plus{}\frac{1}{y^n}\plus{}\frac{1}{z^n})$

[quote="gagaga"]what's about maximum of $ (x^2 \plus{} y^2 \plus{} z^2)(\frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^2})$ with $ x \plus{} y \plus{} z \equal{} 3$
and  $ (x^n \plus{} y^n \plus{} z^n)(\frac {1}{x^n} \plus{} \frac {1}{y^n} \plus{} \frac {1}{z^n})$[/quote]Try $ x\to0$  :rotfl:

But here is a similar problem, inspired by the original one, only this one is ‘less homogenous’. :D Again, it allows for lots of generalizations:

(a)	For $ n \equal{} 3$, under the condition that [b]each permutation [i][in terms of the variables $ a,b,c$][/i] of $ (a,b, \frac1c)$ and of $ (\frac1a,\frac1b,c)$ yields a triangle[/b], prove $ F(a,b,c): \equal{} (a \plus{} b \plus{} c)(\frac1a \plus{} \frac1b \plus{} \frac1c)\le 6 \plus{} 2\sqrt {5}\approx10.472$. (In fact, exactly as in the original case, $ (a,b,c) \equal{} ( \phi,1, \phi{ \minus{} 1})$ is again an extremal vector.)

(b)	Similarly for $ n \equal{} 6$, under the condition that each permutation [...] of $ (a,b,c, \frac1d, \frac1e, \frac1f)$ yields 6 edges of a feasible tetrahedron, prove or disprove that $ \max F \equal{} 20 \plus{} 8\sqrt {6}\approx 39.596$ with an extremal vector $ (\frac {\sqrt {3} \plus{} 1}{\sqrt {2}},\frac {\sqrt {3} \plus{} 1}{\sqrt {2}},1,1,1,1)$. :)

(c)	etc. etc. 	... :D


See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]

The number $123456789$ is written on the blackboard. At each step it is allowed to choose its digits $a$ and $b$ of the same parity and to replace each of them by $\frac{a+b}{2}.$ Is it possible to obtain a number larger then
a)$800000000$; b)$880000000$ by such replacements?

8 can be obtained only with 9+7 or 8+8. So we want to transform the first digit into a 7 or an 8. In the first case the operations are 1+3, 2+4, 3+5, ..., 6+8, 9+7, and we obtain 822345678. The second case can't occur because to maximize the number we must add consecutive integer starting from the left. So 822345678 is also the maximum we can obtain.

[quote="Andreas"]The second case can't occur because to maximize the number we must add consecutive integer starting from the left. [/quote]Why? :?
[quote="Andreas"] So 822345678 is also the maximum we can obtain.[/quote]??? :?


function
algebra
system of equations
algebra proposed

Find all triples $\left(x,\ y,\ z\right)$ of integers satisfying the following system of equations:

$x^3-4x^2-16x+60=y$;
$y^3-4y^2-16y+60=z$;
$z^3-4z^2-16z+60=x$.

Equations are equivalent to :
\begin{eqnarray*} (x-4)^2(x+4)=y+4 \\ (y-4)^2(y+4)=z+4 \\ (z-4)^2(z+4)=x+4\end{eqnarray*}
we can see that $x=y=z=-4$ leads to a solution so we can suppose that none of $x,y,z$ equals $-4$. Hence above equations are equivalent to the followings.

\begin{eqnarray*} (x-4)^2=\frac{y+4}{x+4} \\ (y-4)^2=\frac{z+4}{y+4} \\ (z-4)^2=\frac{x+4}{z+4}\end{eqnarray*}

so

$(x-4)^2(y-4)^2(z-4)^2=1$, hence $(x-4)^2=(y-4)^2=(z-4)^2=1$ so we conclude after all:

$\{x,y,z\} \in \{3,5\}$ or $x=y=z=-4$

[quote="Megus"]$\{x,y,z\} \in \{3,5\}$[/quote]

Almost. There is also the solution $\left(x,\ y,\ z\right)=\left(-4,\ -4,\ -4\right)$. Guess you will quickly see where you dropped it...

  Darij

how were u able to factor that so easily? it would have taken me forever to get that, assuming i got it at all. :(

$s_k=x^k+y^k+z^k$ and add them all together

$s_3-4s_2-17s_1+60=0$, this is the newton's sum for

$r^3-4r-17r+60=0=(r-5)(r-3)(r+4)$, the rest is trivial

[quote="Megus"]$(x-4)^2=(y-4)^2=(z-4)^2=1$...[/quote]

Substituting this into the first set of equations, we get $x=y=z$, hence this

[quote="Megus"]$\{x,y,z\} \in \{3,5\}$[/quote]

is not correct. The solutions are $(3,3,3), (5,5,5), (-4,-4,-4).$

I think Megus means too.What about for x,y,z reals???

[quote="Nima Ahmadi Pour"]if $f(x)=x^3-4x^2-16x+60$ then $f'(x)=3x^2-8x-16$ and we simply find that $f'(x)>0$ for all $x$.[/quote]

(Edit:) It seems I forgot to check both your work and mine.  :blush:

(Keeping as a sidenote:) I once saw someone (was it Vornicu?) post the obvious generalization, if $f(x)$ is strictly increasing function and we have the system

$f(x) = y$
$f(y) = z$
$f(z) = x$

It follows that $x = f(f(f(x))) \implies x = f(x)$ and then $y = f(x), z = f(f(x))$.

how $f(x)=3x^2-8x-16=(3x+4)(x-4)$ is increasing for all $x$ ?? I cant see it  :( Someone explain please ?

[quote="Nima Ahmadi Pour"]Another solution (also for real numbers) :
Just see that if $f(x)=x^3-4x^2-16x+60$ then $f'(x)=3x^2-8x-16$ and we simply find that $f'(x)>0$ for all $x$.
So $f(x)$ is an increasing function.
Suppose that two of the $x,y,z$ are not equal for example $x\neq y$.
We have two cases :
1) $x>y$ : Then $f(x)>f(y)$, it means $y>z$ and $f(y)>f(z)$ that is $z>x$ but then we have $x>y>z>x$ which is contradiction.
2) $x<y$ : Similar as above we arrive at $x<y<z<x$ which is also contradiction.
So we must have $x=y=z$ but then we must find the roots of $f(x)=x$, as mentioned earlier they are $3,-4,5$.
So the only solutions are $x=y=z\in\{3,-4,5\}$.[/quote]
Your solution is just a litle incorrect :) .
this problem was proposed in Armenian TST and I solved it in this way,here is my solution.
We have that $f'=3x^2-8x-16=(3x+4)(x-4)$,so if $x\in{(-\infty;-4)\cap{(4;\infty)}=I}$

then $f$ is increasing function ,now that isn't hard to see that if $x\in{I}$=>$f(x)\in{I}$=>$f(f(x))\in{I}$=>if $f(x)>x$=>\[ f(f(x))>f(x)>x \]=>\[ f(f(f(x)))>f(f(x))>f(x)>x \]=>\[ f(f (f(x)))>x \],and with same way if $f(x)<x$=>$f(f(f(x)))<x$=>if $x\in{I}$=>$f(x) =x$=>$x=5$,also we need to find out are ther any other answers when
 \[ x={\set{-4;- 3;-2;-1;0;1;2;3;4}} \] and we will see that when $x=3$ or $x=-4$=>$y=z=3$ or $y=z=-4$ ;)


limit
real analysis
real analysis unsolved

$\|f_k\|_p \leq M$ holds for all $k$. $2<p<\infty$ and $\lim_{k\to\infty} \|f_k-f\|_{p/2}=0.$
Show that $\lim_{k\to\infty}\|f_k-f\|_2 = 0.$

For $p < 4$ we use the result which dickchimney proved here?
http://www.mathlinks.ro/Forum/viewtopic.php?t=25007

and for $p \geq 4$, $\lim_{k\to\infty} \|f_k-f\|_2 \leq \lim_{k\to\infty} \|f_k-f\|_{p/2}$ can be used.

I had an approach later.

Write $2 = 2\lambda+2(1-\lambda)$ where $\lambda$ satisfies
\[\frac{2\lambda}{\frac{p}{2}}+\frac{2(1-\lambda)}{p}=1\]
Then apply $\mbox{H\\"older}$...

[quote="liyi"]I had an approach later.

Write $2 = 2\lambda+2(1-\lambda)$[/quote]
That works great! Very nice tip but I wonder why$||f_k||_p \leq M$ was not used.


Determine all polynomials $f (x)$ with real coeffcients that satisfy
\[f (x^{2}-2x) = f^{2}(x-2)\]
for all $x.$

[quote="N.T.TUAN"]Determine all polynomials $f (x)$ with real coeffcients that satisfy
\[f (x^{2}-2x) = f^{2}(x-2) \]
for all $x.$[/quote]

Let $g(x)=f(x-1)$, then $g((x-1)^{2})=g^{2}(x-1)$ and $g(x^{2})=g^{2}(x)$

Then, if $z$ is a complex nonzero root of $g(x)$, then $z^{2}$ is also a root and, since we have a finite number of roots, all nonzero roots have their modulus equal to 1.
But then, if $e^{ix}$ $(x\in[0,2\pi))$ is a complex root, $e^{i\frac{x}{2}}$ is also a root, and so are $e^{i\frac{x}{2^{k}}}$ $\forall k\in\mathbb{N}$, and $x$ must be 0 (else we have an infinite number of roots). But, if $1=e^{0*i}$ is a root, so is $-1=e^{i\pi}$ and $e^{i\frac{\pi}{2^{k}}}$ $\forall k\in\mathbb{N}$.

So $g(x)$ can only have zeroes roots and we have $g(x)=ax^{n}$. This necessary condition works if $a^{2}=a$, so if $a=0$ or $a=1$

The solutions to the initial problem are :
$f(x)=0$
$f(x)=(x+1)^{n}$ $\forall n\in\mathbb{N}\cup\{0\}$

--
Patrick


I know most people here is in high school, but does anyone know who else is gonna be at state?

MOI! See ya there, lucy! Pizitz is going down! :D Not really, just kidding, Pizitz is one of the best schools there. You'll see. :)

Btw, are you the only girl on your team? Last year, I was the only girl AND the only seventh grader. But I scored the highest too.  :lol:

[quote="LuCy4EvA"]haha, okay, so maybe i'm not the only girl... but I am the only SEVENTH GRADE GIRL.... and the ONLY SEVENTH GRADER....

wait, what school are you from? what chapter? If you mind, what's your name? I'm guessing you are a girl in eighth grade... I really want to find someone on AOPs that's gonna be at state.... I'm gonna look for you... and anyone that's from state and is on AOPs.....

I think i"m the only person from my school on AOPs.... and my other friend who only posted 2 times.....

haha, wow, you were in seventh grade and scored the highest? wow, that's incredible... all the rest of the people from Pizitz is wayyyyy too smart..... I can barely keep up with them....[/quote] last year i was the only seventh grader, the only girl, AND the only seventh grade girl...and the only one with some common sense, if you ask me, lol. Yeah, I am an eighth grader, so I don't have another chance next year :( . Though I scored the highest last year, it's not that incredible when you think about it, because I won not because I was good (at States, I found that I had a LONG way to go) but because the other people in my chapter weren't that good.

[color=red]STATES IN 48 HOURS!!!! [/color]*starts biting on nails*  :? It's kinda weird thinking about it...in less than 48 hours I'll know if I qualify. Are you nervous?

no... not really, since i have no idea what it is like, which is the thing i like about mathcounts, i don't get nervous like usual... but i still want to do good though, just to beat an eight-grader... haha, wait, what school r u from, what chapter? is the people in your chapter hard? wait, are y'all driving to birmingham? it's lucky that state is like a 20 minute drive from where i live...

I think i am the youngest in the chapter (i am a 5th gradr)

zomg this forum has been reborn after like a year

wow way to revive a 2 and 1/2 year old topic.... challenger is looking really strong this year

Wow thanks! (I'm on challengers team.) I think...... state competition will be FIERCE!   Anyone here from Huntsville chapter? Lets revive this entire forum! Its so boooooooooring.

How was that fierce? You destroyed me...

Oh well... Anyone know what high school does for math competitions?

AMC and AIME stuff. i had 2 take those. anyway, what place did u get at state MC?

what were the results? who got into countdown and what teams got 1st 2nd and 3rd???????? WHAT IS ALABAMA'S TEAM??????

[quote="chestnut"]i got first :lol:
2 people from challenger got 2nd and 3rd
and weasly (Simmons) got 4th

i got 2nd on cd. it wuz unofficial though. the person who got 3rd got 1st

pizitz 1st  :clap:
some other school 2nd
challenger 3rd[/quote]

Tehe. Weasly, as in ron weasly.

Me= 3rd written = 1st cd.

My coach figures we bombed team round...

ANYONE GOING 2 GRISSOM TOURN. THIS SAT.? I think I'm gonna bomb because all I've been thinking about is MC...

Dude you just made national mathcounts. You should place in the top 5 at least if not 1st at Grissom. I mean it's not all that different. Me, on the other hand, will be lucky to get 20th what with Vestavia and Hoover competing

I'll be there. hope to see you there!

a link that tells you the places for top 12 in alabama. Mr. Crawford, the speaker at state MC, runs that website

here is the link
 [/url]http://www.mistacademy.com/almathcounts.htm#2008state
if it doesn't work, copy paste time!!

If you can get 3rd at state mc, you came get top 5 in some regular MT tourn.
I on the other hand, make stupid mistakes. :wallbash_red:
A lot of stupid mistakes.

And this is the arml webpage
 [url]http://www.mistacademy.com/alabamaarml.htm[/url]

Yeah,  I hope! Last year I didnt even place in pre-alg and this year I'm in alg! Plus chestnut beats me [b][u][i]everywhere [/i][/u][/b](ooh, triple threat. Italics, bold, and underlined.)remember randolph 6th grade????

[quote="lifesapain"]Haha Sid was that your first post? Your cousin told me you prepared for it 24/7 in an OCD kind of way. Good job though. Hey how did Challenger only get 3rd when they had 2 people in cd and a 3rd person got 12th? Did you guys do bad in team or was it that 4th guy? And Im pretty sure that Wesley kid from Simmons got 4th written last year too...[/quote]

evil_ed good for you!!!! first post!!! I knew you should have gotten first place ( :wink: )...

Our coach (im a challengerian) says she figures we may have BOMBED team round but Matthew Crawford saw the scores and said the top 3 were within the "slimmest of margins"...

Wesley got 5th  (according to evil_ed)last year.... anyone know if he has an aops account cuz then we would have a full aops team!!!!!!!

[quote="chestnut"]I left early for some piano comp...
 so I don't really know what happened @ Randoph.

Hey, u beat me in cd, and got a gift card and the title.[/quote]

Hm coulda sworn remembering standing next to you and being the only ones left.... but whatever! My memoory is leaving me as I cram my small brain with math...

Yes. The gift card is going towards my camera fund (which if I get before nats I will have with me EVERYWHERE but testing unless there's another one of those "No recording devices" thing......)

gah.......ME got 6TH CURSED BY LIFESAPAIN!!!!!!!!!!!!!

chestnut did get top 5 (gj! btw the pizitz girls (the only one whose name I know is Shanna Liu) were asking me questions like my last name and my score on the (grissom) test..... (I think they graded mine wrong bcuz shanna said she got an 84.1 or something and I think i got a 81.3 and shanna got like 11th or something and I got 6th...))

I thought we had first place team darn! (even with like 9 out of the top 10 being from pizitz)

[quote="chestnut"]yeah, i remember shanna talking 2 me after the compitition. she said somthing like "is that the person who's going to mathcouts with you and a bunch of other stuff."[/quote]

whoa we are on at the same time freaky

THE BANNER IS GOING TO ROCK!

I'm making the team "banner" that we're gonna hold on stage (maybe). I've spent at least 4 hours on it, and it ROCKS. (i hope). U'll see it at nats.

you need to post a picture some time  :P

I should be done with it tonight.... however whilst (fun word) I wait for my parents to make a decision about letting me get a new camera, i will be forced to use my non-broken one. picture quality may be shabby.

the picture i was going 2 use is 2 big or somting like that

dynamo, were on againg @ the same time!

[img]http://bp0.blogger.com/_klL-NR6348w/SBvIASN-33I/AAAAAAAAABU/2kMkNxqDCE8/s1600-h/P5020100.JPG[/img]

There you go. I hope you like it. I worked 7+ hours on it ARGGGGH!

EDIT: If you don't see the image, go to my blog at [url]http://www.dynamosblog.blogspot.com/[/url].