Let $ABC$ be a triangle in which $AB =AC$ and $\angle CAB = 90^{\circ}$. Suppose that $M$ and $N$ are points on the hypotenuse $BC$ such that $BM^2 + CN^2 = MN^2$. Prove that $\angle MAN = 45^{\circ}$.

[hide="Some idea..."]
Take a point $P$ outside $\triangle ABC$ such that $PM=BM$ and $PN=CN$. Since $PM^2+PN^2=MN^2$, we know that $\angle MPN=90^\circ$.

If we could show $AP=AB=AC$, then $\triangle ABM\cong\triangle APM$ and $\triangle ACN\cong\triangle APN$. From here, it is easy to see that $\angle MAN=\angle MAP+\angle NAP=\frac{1}{2}\cdot 90^\circ=45^\circ$.[/hide]

[u]A analytical solution.[/u]

$A(0,0),\ B(1,0),\ C(0,1),\ M(m,1-m),\ N(n,1-n),\ 0<n<m<1$.

$MB^2+NC^2=MN^2\Longleftrightarrow n^2+(m-1)^2=(m-n)^2\Longleftrightarrow \overline {\underline {\left| 2m(1-n)=1\right| }}.$

The [b]g[/b]radients of the lines $AM,\ AN$ are $g(MA)=\frac{1-m}{m},\ g(NA)=\frac{1-n}{N}$

and $\tan (\angle MAN)=\mod \frac{g(MA)-g(NA)}{1+g(MA)g(NA)}=$ $\mod \frac{\frac{1-m}{m}-\frac{1-n}{n}}{1+\frac{(1-m)(1-n)}{mn}}=$

$\mod \frac{m-n}{-[2m(1-n)]+m+1-n}=\mod \frac{m-n}{-1+m+1-n}=1\Longrightarrow$

$\tan (\angle MAN)=1\Longrightarrow m(\widehat {MAN})=45^{\circ}.$

[u]A metrical solution.[/u]

I suppose (w.l.o.g.) $AB=AC=1\ (BC=\sqrt 2)$ and I note $m(\widehat {MAN})=m,$

$BM=y,\ MN=x,\ BN=z\ (\underline {x^2=y^2+z^2},\ BC=\underline {x+y+z=\sqrt 2})$.

$[ABC]=[ABM]+[BMN]+[ACN]\Longrightarrow$

$\frac 12 =\frac{y\sqrt 2}{4}+\frac{z\sqrt 2}{4}+\frac 12\cdot AM\cdot AN\cdot \sin m\Longrightarrow 2\cdot AM\cdot AN\cdot \sin m=$

$2-(y+z)\sqrt 2=2-(\sqrt 2 -x)\sqrt 2=x\sqrt 2\Longrightarrow \underline {\overline {\left| AM\cdot AN\cdot\sin m=\frac{x}{\sqrt 2}\right| }}.$

$AM^2=1+y^2-y\sqrt 2,\ AN^2=1+z^2-z\sqrt 2,$

$x^2=AM^2+AN^2-2\cdot AM\cdot AN\cdot\cos m\Longrightarrow$

$\underline {x^2}=2+\underline {(y^2+z^2)}-(y+z)\sqrt 2-x\sqrt 2\cot m\Longrightarrow$

$(\sqrt 2 -y-z)\sqrt 2=x\sqrt 2 \cot m\Longrightarrow x\sqrt 2=x\sqrt 2\cot m$

$\Longrightarrow cot \ m=1\Longrightarrow m=45^{\circ}.$

[quote="Rushil"] [color=darkred]Let $ ABC$ be a $ A$ - right and isosceles triangle. Consider two points $ \{M,N\}\subset (BC)$ . Prove that $ BM^2 \plus{} CN^2 \equal{} MN^2\ \Longleftrightarrow\ m(\widehat {MAN}) \equal{} 45^{\circ}$ . [/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. that $ a \equal{} 1$ . Thus,

$ \left\{\begin{array}{ccccccc} \boxed {\ \begin{array}{c} x \equal{} m(\widehat {BAM}) \\
\ u \equal{} \tan x\end{array}\ } & \implies & \frac {MB}{MC} \equal{} \tan x & \implies & MB \equal{} \frac {\tan x}{1 \plus{} \tan x} & \implies & \boxed {\ BM \equal{} \frac {u}{1 \plus{} u}\ } \\
 \\
\boxed {\ \begin{array}{c}y \equal{} m(\widehat {CAN}) \\
\ v \equal{} \tan y\end{array}\ } & \implies & \frac {NC}{NB} \equal{} \tan y & \implies & NC \equal{} \frac {\tan y}{1 \plus{} \tan y} & \implies & \boxed {\ CN \equal{} \frac {v}{1 \plus{} v}\ } \\
 \\
MN \equal{} BC \minus{} MB \minus{} NC & \implies & ******* & \implies & MN \equal{} 1 \minus{} \frac {u}{1 \plus{} u} \minus{} \frac {v}{1 \plus{} v} & \implies & \boxed {\ MN \equal{} \frac {1 \minus{} uv}{(1 \plus{} u)(1 \plus{} v)}\ }\end{array}\right\}$ $ : \ \implies$

$ \frac {BM}{u(1 \plus{} v)} \equal{} \frac {CN}{v(1 \plus{} u)} \equal{} \frac {MN}{1 \minus{} uv} \equal{} \frac {1}{(1 \plus{} u)(1 \plus{} v)}$ . Therefore, $ \boxed {\ BM^2 \plus{} CN^2 \equal{} MN^2\ }$ $ \Longleftrightarrow$ $ u^2(1 \plus{} v)^2 \plus{} v^2(1 \plus{} u)^2 \equal{} (1 \minus{} uv)^2$ $ \Longleftrightarrow$

$ (1 \plus{} u)^2\cdot v^2 \plus{} 2u(1 \plus{} u)\cdot v \minus{} (1 \minus{} u^2) \equal{} 0$ $ \Longleftrightarrow$ $ (1 \plus{} u)\cdot v^2 \plus{} 2u\cdot v \plus{} u \minus{} 1 \equal{} 0$ $ \Longleftrightarrow$ $ (1 \plus{} v)[(1 \plus{} u)\cdot v \minus{} (1 \minus{} u)] \equal{} 0$ $ \Longleftrightarrow$

$ \frac {u \plus{} v}{1 \minus{} uv} \equal{} 1$ $ \Longleftrightarrow$ $ \frac {\tan x \plus{} \tan y}{1 \minus{} \tan x\tan y} \equal{} 1$ $ \Longleftrightarrow$ $ \tan (x \plus{} y) \equal{} 1$ $ \Longleftrightarrow$ $ x \plus{} y \equal{} 45^{\circ}$ $ \Longleftrightarrow$ $ \boxed {\ m(\widehat {MAN}) \equal{} 45^{\circ}\ }$ . See and[/color]  [url=http://www.mathlinks.ro/viewtopic.php?search_id=372486077&t=85445][color=red][b]here[/b][/color][/url] .

[quote="10000th User"][quote="frt"]If we could show $ AP \equal{} AB \equal{} AC$[/quote]New challenge :!:  frt never showed this step... is frt right? :P[/quote]

Can anyone supply a proof of this claim?

BM^2=AM^2-AB^2+√2 BM×AB ,〖CN〗^2=〖AN〗^2-〖AC〗^2+√2 CN×AC ,
〖MN〗^2=〖AM〗^2+〖AN〗^2-2AM×AN×〖cos∠〗AMN  so we have                                                                            √2 CN×AC+√2 BM×AB-BC^2=-2AM×AN×〖cos∠〗AMN                                                                                →BC ×MN/2=AM×AN×〖cos∠〗AMN,2 area of ∆AMN=AM×AN×〖cos∠〗AMN
So  cos ∠ AMN=sin∠ AMN

Take two identical pieces of paper, square-shaped, perfectly overlapping; pick them with a needle in the center and rotate the upper one by an angle $\alpha$. You will obtain an octogon - the overlapping area of the two squares and eight congruent right-angled triangles, the corners of the squares, non overlapping. Each of these triangles is the mirror image of its two neighbouring ones; all of them are seen from the center all-around the 'horizon', i.e. $360^\circ$, hence each of them is seen at an angle of $45^\circ$, done.

Best regards,
sunken rock

Wasn't this also a romanian tst problem?

   We are given that BM^2+CM^2=MN^ and AB^2+AC^2=BC^2.  Hence AB^2+AC^2=BM^2+MN^2+CN^2+2BM.MN+2MN.CN+2BM.CN
FURTHER,2AB^2=2MN^2+2BM.CN+2BM.MN+2MN.CN THAT IMPLIES, AB^2=MN^2+BM.MN+MN.CN+BM.CN that implies, AB^2=BN.CM
from which we get AB/BN=CM/AB. AB=AC hence AC/BN=CM/AB and∠ ABN=∠ACM FOR BOTH being 45degrees. So
∆ABN|||∆ACM. Hence∠ AMC=∠BAN and ∠ANB=∠ MAC. NOW
∠BAN+∠MAC-∠MAN=90 and1/2{∠AMN   + ∠ANM    +∠MAN}=90
∠BAN   +∠MAC   -∠MAN   =1/2 {∠AMN   +∠ANM   +∠MAN}  that implies ∠MAC+∠BAN=3∠MAN implies  ∠MAC   +∠BAN   +∠MAN    =2∠MAN
IMPLIES 2∠MAN=90 means ∠MAN=45  which we wanted to show :-D

It's hard to beleive that this problem is the generalization of $RMO$ $2013$ $P5$...(What else do you expect from indians :P)
Both these problems can be solved easily using Rotation about $A$ about $90$deg. that maps $B$ to $C$


inequalities
integration
inequalities proposed

If $ A(x):= 6x^3+72x^2+360x+720 $ and $ B(x):=\ds x^2+6x+12  ,
$ prove/or disprove that for each $ x \in (0,4) $ there exists $ \varepsilon(x)\in (0,1)$ such that
\[
e^x=\frac{A(x)+\varepsilon(x)x^2B(x)}{A(-x)+\varepsilon(x)x^2B(-x)}\;
.
\]


geometry
3D geometry
probability
MATHCOUNTS

3: A 4x4x4 wooden cube is painted on all six faces and then cut into 64 unit cubes.  One unit cube is randomly selected and rolled.  What is the probability that a painted face shows up on top of the rolled cube?  Express your answer as a common fraction.  Our answer was 1/4.  I don't really know how we got it cuz David got the answer and told us or me and we wrote it down so you guys will have to ask him about it and he'll post it sometime soon....i hope...i'll tell him 2marrow.

6: Nanette has $ 48 to spend to build a fence along three sides of htr rectangular garden.  The fourth side is along a garage and no fence is needed there.  The fence materials cost$3 per foot. if the width and lengeth of the garden are each a whole number of feet, what is the greatest number of square feet Nanette can have for the area of her garden?  try to figure it out on your own be4 looking at the answer to #6. that is if you didnt get the answer to 6 lol

[hide]Well i did this problem so i think i still remember how to do it.  ok so first i drew a diagram to help me solve it. the garage didnt have a side so i left it blank.  and then i did 48/3 which = 16 so if one side is longer than the others, how much can she spend of that? well i thought that if one side is longer than the rest, then she would have to spend more on that side than the others and i came up with 4s for the shorter side and 8 for the longer side and then the area was 32ft2 which made the price add up to exactly 48 dollars ($ 12 for one of the short sides and$24 for the long side. 24+24=48 :) )[/hide]

[quote="themathmaster770"]3: A 4x4x4 wooden cube is painted on all six faces and then cut into 64 unit cubes.  One unit cube is randomly selected and rolled.  What is the probability that a painted face shows up on top of the rolled cube?  Express your answer as a common fraction.  Our answer was 1/4.  I don't really know how we got it cuz David got the answer and told us or me and we wrote it down so you guys will have to ask him about it and he'll post it sometime soon....i hope...i'll tell him 2marrow.
[/quote]

I am another David, so I probably qualify to post the solution.. :lol:

Since it's 4*4*4 cube, we know that there are 64 cubes, which means they have 384(64*6) sides(faces).
Now we know the total # of possibility of choosing 1 side= $ \ 384$
Then we try to crack the problem.
Cubes with only 3 sides painted would be= 4(corners) so 4*3=$ \ 12$ faces total.
Cubes with only 2 sides painted would be= 24(sides - corners) so 24*2=$ \ 48$faces total.
Cubes with only 1 side painted would be= 24, so 24*1= $ \ 24$

Then we add them up. $ \ 12 \plus{} 48 \plus{} 24 \equal{} 84$
That's the number of sides that are painted, So...
$ \frac {84}{384} \equal{} \frac {7}{32}$
..... wait.. Oh..no..I got this answer last time.. Ah!! What did I do wrong?
I figured it out how to do it after the competition..(did the addition wrong..)What the heck!! :ewpu:  I can't think straight...
What did I do wrong????  :help:

i got 1/196

so the correct answer is 1/4????
explain plz

[quote="spiders_hate_chestnut"]:D  it took me and my dad to get this right

lets call the 4x4x4 square the 'big square' and each of the 64 squares 'little squares.

64 little cubes right?
and each little cube has 6 sides (we dont care about if its painted or not right now).
so 64x6=384
there are 384 sides total.

and now look at the painted sides.
64=4x4x4, which means the large cube is cut into cubes that are 1 unit lenghth/width/heigth.
so one side of the large square equals 16 sides of the little square.
therefore, 16x6 for the total of painted sides of the little squares.
which equals 96.
so now, 96/384
which is the total painted sides of the little squares/just the total sides of the little squares.
simplify to 1/4

SOLUTION: 1/4[/u][/quote]
Hmmm... Can you explain where 96 came out from with more details??
I don't get that part. :(

[quote="spiders_hate_chestnut"]u get the 384 part, right

anyway
the 4x4x4 cube (the big cube) is cut into 64 little cubes
if u do the math, each of the little cube's dimensions are 1x1x1
now, one painted side of the big cube is 16sq unit (4x4)
16sq unit is just one side of the cube
and a cube has 6 sides
so 16x6=96

and so on....

if u hav problem understanding me
im not surprised
cus i kno i have problem with that kind of stuff
ask for jennifer in mathcounts[/quote]

Hmm.... I think I get it.
So what you are saying is that since the outer parts of big cube(as you call it) are painted and there are 16 painted surface on each 6 sides.. 16*6= 96.. Right??

Let's see if I can help any (I am really good at making everything look muddier!)

What we want to find is $ \dfrac{the.number.of.sides.painted}{the .total.number.of.sides.available}$

So, the sides painted:
--The sides will be cut into 16 squares on each side when the cube is cut into 64 cubes, making 16 new faces showing on each side of the larger cube when the larger cube is cut into 64 smaller cubes.
--There are 6 sides of the original cube, so the number of faces that are painted on the sides of the 64 smaller cubes will be 6 x 16 = 96
--the total number of faces available on 64 smaller cubes is 6 faces x 64 cubes (384 faces), thus giving us
$ \dfrac{96}{384} \equal{} \dfrac{1}{4}$

Does that help DK?


Let $ ABCDEF $ be a convex insribed hexagon (O, R) . $ AB=CD=EF=R $
Let $ M, N , P $ be the midpoints of $ BC , DE , FA $ . Prove that : $ MNP $ is an equilateral triangle

Solved here: http://www.mathlinks.ro/Forum/viewtopic.php?t=16908

  Darij


inequalities
function
inequalities unsolved

Let $ a,b,c\in R\plus{}$ prove that
1.
$ \frac {1}{a(1 \plus{} b)} \plus{} \frac {1}{b(1 \plus{} c)} \plus{} \frac {1}{c(1 \plus{} a)}\geq\frac {3}{1 \plus{} abc}$
2.
$ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c})\geq \frac {9}{1 \plus{} abc}$
I don't know which function can be used here..... help me plz  :blush:

1. \[ \sum_{cyc}\frac{1}{a(1\plus{}b)} \geq \frac{3}{1\plus{}abc} \leftrightarrow
\sum_{cyc}\frac{1\plus{}abc}{a(1\plus{}b)} \geq 3\]
\[ \leftrightarrow \sum_{cyc}\frac{(1\plus{}a)\plus{}ab(1\plus{}c)}{a(1\plus{}b)} \geq 6\]
which is obvious by am-gm.
[hide="@rose-joker"]ข้อนี้โคตรง่ายเลยอะครับ[/hide]

2. \[ \sum_{cyc}\frac{1}{a\plus{}1} \equal{} \sum_{cyc}\frac{\frac{1}{a}}{1\plus{}\frac{1}{a}}\]
\[ \geq 3\sqrt[3]{\frac{1}{abc}}\frac{1}{\sqrt[3]{(1\plus{}\frac{1}{a})(1\plus{}\frac{1}{b})(1\plus{}\frac{1}{c})}}\]
\[ \geq \frac{9}{\sqrt[3]{abc}(3\plus{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})}\]
thus \[ LHS \geq \frac{9}{\sqrt[3]{abc}(\frac{3}{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}})\plus{}1} \geq \frac{9}{\sqrt[3]{abc}(\sqrt[3]{abc}\plus{}1)}\]
it's remain to prove that $ \frac{9}{\sqrt[3]{abc}(\sqrt[3]{abc}\plus{}1)} \geq \frac {9}{1 \plus{} abc}$
which is easy.

[quote="Rose_joker"]Let $ a,b,c\in R \plus{}$ prove that

2.
$ (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(\frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c})\geq \frac {9}{1 \plus{} abc}$
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145655

......Can you please use Jensen inequality for these 2 problems..... That's what i want here  :) Thx a lot


An ordered pair $(m,n)$ of non-negative integers is called "simple" if the addition $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$.

[hide="Solution"]Let's examine each of the digits seperately. There are $2$ ways the numbers can sum to thousands digit of $1$. The thousands digit of one of the numbers is $1$, while the thousands digit of the other is $0$. Similarly, there are $5$ ways to obtain a hundreds digit of $5$, $10$ ways to obtain a tens digit of $9$, and $3$ ways to obtain a units digit of $2$. Therefore, our answer is $(2)(5)(10)(3)=300$.[/hide]

so units digit it has to be 0 2 1 1 or 2 0 so that is 3 way
ten digit can be 0 9 ... 9 0 and that is 10 way
hundred digit is 0 4 ... 4 0 so 5 way
thousand is 0 1 or 1 0
so answer is 3*10*5*2=300

Combo AIME grind: Consider each of the digits alone. To get units digit $2$ we can have $1+1, 2+0, 0+2$, so $3$ ways. To get tens digit $9$ we have $10$ ways. To get hundreds digit $4$ we have $5$ ways. To get thousands digit $1$ we have $2$ ways. The answer is $3 \cdot 10 \cdot 5 \cdot 2 = \boxed{300}$.

Pretty, "simple". To get the units digit, we have $3$ ways, for the tens, $10$ ways, hundreds $5$ ways, thousands $2$ ways. So $3\cdot 10 \cdot 5 \cdot 2=\boxed{300}.$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group.   Show that if $ H \lneq G$ then $ \bigcup_{g \in G} gHg^{\minus{}1} \subsetneq G$


The sequence of real numbers $a_1,a_2,\dots$ is defined as follows: $a_1=56$ and $a_{n+1}=a_n-\frac{1}{a_n}$ for $n\ge 1$. Show that there is an integer $1\leq{k}\leq2002$ such that $a_k<0$.

Solution by Daniel Vaz, member of the Portuguese IMO team in 2007 and 2008:

Suppose there isn't $ k$ with $ 1\leq k \leq 2002$ for which $ a_k<0$.

First, observe that $ a_n$ is decreasing for $ 1\leq n \leq 2002$, because $ a_{n\plus{}1}\equal{}a_n\minus{}\frac{1}{a_n} \leq a_n$, since $ a_n\geq 0$.

Now, note that for $ n\geq1$, $ a_n\leq a_1 \equal{}56$, so $ a_{n\plus{}1}\leq a_n \minus{}\frac{1}{56}$, and from this we easily get that $ a_{57}\leq 56\minus{}\frac{56}{56}\equal{}55$.
Now in a similar manner we obtain $ a_n\leq 55$ for $ n\geq 1\plus{}56$, and then get $ a_{57\plus{}55}\leq 54$, and continuing this we get $ a_{1597}\equal{}a_{1\plus{}56\plus{}55\plus{}...\plus{}1} \leq 56\minus{}56\equal{}0$, contradiction, as desired.

In fact we proved that $ a_k\leq 0$ for some $ k$, with $ 1\leq k\leq 1957$, which is something a bit stronger  :)

[quote="Jorge Miranda"]Now, note that for $ n\geq1$, $ a_n\leq a_1 \equal{} 56$, so $ a_{n \plus{} 1}\leq a_n \minus{} \frac {1}{56}$, and from this we easily get that $ a_{57}\leq 56 \minus{} \frac {56}{56} \equal{} 55$.[/quote]

I thought that if we substitute $ 56$ for $ n$ we get $ a_{57} \leq a_{56} \minus{} 1/56 \leq 56 \minus{} 1/56.$

No, you don't plug it in like that, but instead use it repeatedly (or induction :P), like this:
$ a_{57}\leq a_{56} \minus{}\frac{1}{56} \leq a_{55}\minus{}\frac{2}{56}\leq \dots \leq a_1\minus{}\frac{56}{56}\equal{}55$

All the other times are analogous, of course  :)


Hi!

My name is Frederic Vanhove, and I'm also a Belgian student (unlike Gery however, I'm from the northern half so I speak Dutch)

I participated in the IMO 2002 in Glasgow. Because several reasons, one of which happened to be no training ( flemish government didn't wish to offer one Euro for us :( ) , and many others, the score for me wasn't too fantastic to say the least...

Anyway, lots more can be found concerning myself but also last IMO on my site http://users.pandora.be/frederic_bel6/
i have written journals about every day and i have about 45 pictures too

it is embarassingly outdated, so about myself :

after the IMO 2002, I enrolled at the university of Ghent to study civil engineering, but that didn't seem to be well fit for me

i quickly discovered that pure mathematics should have been my original choice

the result : i didn't make it trough the first year because of psychology, economy, chemistry,... but I do lose about three to five courses next year  :D , so it should be more fun

(i am now one of those guys who strongly believe that pure maths is the place to be)

well, that's all i can think of too say right now

Hello,

I went to see your web page. On the Maths section you are asking about generalizations of IMO problems, I have the proof of generalization of problem 2 from IMO 2001 - inequality. I downloaded it some time ago from somewhere, I don't know where it can be found, but I put it here: http://www.geocities.com/beata_stehlikova/index.htm

Beata


Let $a,b,c$ be three positive real numbers such that $a+ b +c =1$.
Let $\lambda = min \{ a^3 + a^2bc , b^3 + b^2 ac , c^3 + ab c^2 \}$
Prove that the roots of $x^2 + x + 4 \lambda = 0$ are real.

Super straightfoward way:

---

Real roots condition is $\frac{1}{16} \ge \lambda$

Notice that $b^3 + b^2ac \ge a^3 + a^2bc \Leftrightarrow (b-a)(a^2 + b^2 + ab + abc) \Leftrightarrow b \ge a$.

Wlog, $a$ is the smallest of $(a,b,c)$.  Fix $a$.  It fixes $b+c$, which fixes the maximum of $bc$ as $(\frac{1-a}{2})^2$.

So we just need $a^3 + a^2(\frac{1-a}{2})^2 \le \frac{1}{16}$

becoming $a(a+1) \le 1/2$, which is true in $a \in [0, 1/3]$.

We have to prove that the roots of the equation x^2 + x + 4ƛ = 0 are real, if the roots of this equation are real then b^2 - 4ac ≥ 0, so we get 1 - 16ƛ ≥ 0.          Therefore ƛ ≤ 1/16. Hence we have to prove that ƛ ≤ 1/16 .
  And now a^3 + a^2bc = a^2(a + bc)  which is equal to a^2(1-b-c+bc) which is then equal to a^2(1-b)(1-c). Similarly for b^3 + b^2ca and
c^3 + c^2ab we get that b^3 + b^2ca = b^2(1-c)(1-a) and c^3 + c^2ab = c^2(1-b)(1-a). On multiplying a^2(1-b)(1-c) with b^2(1-c)(1-a) and c^2(1-b)(1-a) we get
(a(1-a))^2(b(1-b))^2(c(1-c))^2. So (a(1-a))^2(b(1-b))^2(c(1-c))^2 ≤ (1/16)^3. Since ƛ is the minimum of these three hence, we prove that ƛ ≤ 1/16

[quote=homersimpson] So $(a(1-a))^{2} (b(1-b))^{2} (c(1-c))^{2} \le (\frac{1}{16})^{3}$.[/quote]

Please explain how u got this ?

It is simple.. straight forward use of De Gua's rule (not to be confused with De Gua's theorem)

 De Gua's rule - The rule that if, in a polynomial equation ƒ(x) = 0, a group of r consecutive terms is missing, then the
                            equation has at least r imaginary roots if r is even, or the equation has at least r + 1 or r- 1imaginary
                            roots if r is odd (depending whether the terms immediately preceding and following the group have like
                            r unlike signs).


inequalities
geometry
email
inequalities unsolved

While I was solving a problem relating areas in a triangle I arrived to  have to prove that

$[4,1,0] + [2,2,1] \geq [3,2,0] + [3,1,1]$

I am really new with Muirhead inequality and I think that $[4,1,0]$ majorizes both $[3,2,0]$ and $[3,1,1]$, but $[2,2,1]$ does not,

a) Can this inequality be proven with Muirhead?

b) If not, I think that this inequality is true, isn't it?

c) How can we prove or disprove the inequality?

Thank you!

That inequality was related with the sides $a$, $b$, $c$ of a triangle

I have made the change
$x=-a+b+c$, $y =a-b+c$, $z=a+b-c$

Then, using Mathematica, I have arrived to $[3,2,0] \geq [2,2,1]$ that is true by Muirhead

Please, can anybody check if this is all right?

Thank you,

[quote="pacoga"]

$[4,1,0] + [2,2,1] \geq [3,2,0] + [3,1,1]$

a) Can this inequality be proven with Muirhead?

b) If not, I think that this inequality is true, isn't it?

c) How can we prove or disprove the inequality?

Thank you![/quote]
a) directly, no.
b)I think you are right.
I used very ugly and dibily method:
Let $a\geq b\geq c$ and $f(a)=[4,1.0]+[2,2,1]-[3,2,0]-[3,1,1].$
Then $f'''(a)=24a(b+c)-6b^2-6c^2-6bc\geq0.$
Hence, $f''(a)\geq f''(b)=4b^3+8b^2c-4bc^2-2c^3\geq0.$
Hence, $f'(a)\geq f'(b)=c(b-c)(2b^2+2bc-c^2)\geq0.$
Hence, $f(a)\geq f(b)=bc(b-c)^2(b+2c)\geq0.$ :)
I am sure, that existing much nicer proof. ;)

Sorry that I didn't reply here earlier, since I didn't see this thread. In fact, I have already replied to Pacoga by email, now I post the proof here for the sake of completeness.

In fact, the inequality $\left[4,1,0\right]+\left[2,2,1\right]\geq\left[3,2,0\right]+\left[3,1,1\right]$ is equivalent to

$a\left(b+c-a\right)^2\left(b-c\right)^2+b\left(c+a-b\right)^2\left(c-a\right)^2+c\left(a+b-c\right)^2\left(a-b\right)^2\geq 0$,

and thus it holds for any three nonnegative reals a, b, c (not necessarily being sidelengths of a triangle).

  Darij

[quote="arqady"]Very nice, darij! The next is a present for you from me.

$a,b,c$ are real numbers.
 Prove that $[6,2,0]-2[6,1,1]-2[4,4,0]+2[4,3,1]-2[3,3,2]\geq0.$ ;)[/quote]

Not true

counter example a=b=c=1

[quote="11235813"][quote="arqady"]
$a,b,c$ are real numbers.
 Prove that $[6,2,0]-2[6,1,1]-2[4,4,0]+2[4,3,1]-2[3,3,2]\geq0.$ ;)[/quote]
Not true

counter example a=b=c=1[/quote]
My dear, I did mean to
$a,b,c$ are real numbers. Prove that
$\sum_{cyc}(a^6b^2+a^6c^2-2a^6bc-2a^4b^4+2a^4b^3c+2a^4c^3b-2a^3b^3c^2)\geq0.$ ;)

[quote="arqady"][quote="11235813"][quote="arqady"]
$ a,b,c$ are real numbers.
 Prove that $ [6,2,0] \minus{} 2[6,1,1] \minus{} 2[4,4,0] \plus{} 2[4,3,1] \minus{} 2[3,3,2]\geq0.$ ;)[/quote]
Not true

counter example a=b=c=1[/quote]
My dear, I did mean to
$ a,b,c$ are real numbers. Prove that
$ \sum_{cyc}(a^6b^2 \plus{} a^6c^2 \minus{} 2a^6bc \minus{} 2a^4b^4 \plus{} 2a^4b^3c \plus{} 2a^4c^3b \minus{} 2a^3b^3c^2)\geq0.$ ;)[/quote]
What is your solution?

Sunjee, it's just a test whether you own a CAS or not. ;)

$ \sum_{\text{cyc}}\left(a^6b^2 \plus{} a^6c^2 \minus{} 2a^6bc \minus{} 2a^4b^4 \plus{} 2a^4b^3c \plus{} 2a^4c^3b \minus{} 2a^3b^3c^2\right)$
$ \equal{} \left(b \minus{} c\right)^2\left(c \minus{} a\right)^2\left(a \minus{} b\right)^2\left(a \plus{} b \plus{} c\right)^2$,

what is obviously $ \geq 0$.

  darij


inequalities
induction
inequalities solved

Let {x} be the fractional part of x. Prove that \sum^{n 2 }_{k=1} {k 1/2 } \leq 1/2 * (n 2 - 1)


Find the maximum value of $ \frac {15x^2}{2} \plus{} 4xy$ for two real numbers $ x, y$ where $ x^2 \plus{} y^2 \equal{} 2$.

Find the maximum value of $ \frac {1}{1 \plus{} |x|} \plus{} \frac {1}{1 \plus{} |y|} \plus{} \frac {1}{1 \plus{} |z|}$ for real numbers $ x,y,z$ where $ 3x \plus{} 2y \plus{} z \equal{} 1$.

Find the minimum value of $ \frac {x^2 \plus{} 2y^2 \plus{} 2x \minus{} 6y \plus{} 5}{xy}$ where $ x,y$ are positive integers.

Find the maximum value of $ 26x^3 \plus{} 27xyz$ for real numbers $ x, y, z$ where $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$

Find the minimum value of $ \frac{(x\plus{}y^2)}{\sqrt{x^2\minus{}4}\plus{}\sqrt{y^2\minus{}9}}$ for real numbers $ x, y$ where $ x>2, y>3$

For your information, these problems are neither my homework nor from on-going competitions or anything.

hello, we have $ \frac{15x^2}{2}\plus{}4xy \le 16$ for all real $ x,y$ with $ x^2\plus{}y^2\equal{}2$. The equal sign holds if $ x\equal{}\minus{}4\sqrt{\frac{2}{17}},y\equal{}\minus{}\sqrt{\frac{2}{17}}$.
Sonnhard.

This is completely unhelpful. Please post a solution from now on. I've noticed that you never post a solution. Please do so in the future.

Bump...

Anybody?

Please help  :(

If nobody answers, mods please move this thread to Intermediate topics or Pre-olympiad.

Thanks :)

In the first one substitute $ \frac {x}{\sqrt {2}} = sin \frac {a}{2} \quad \frac {y}{\sqrt {2}} = cos \frac {a}{2}$

Hence
$ 15\frac {x^2}{2} + 4xy = 15sin^2 \frac {a}{2} + 8 sin\frac {a}{2} cos \frac {a}{2} = \frac {15}{2} (1 - cos a) + 4 sin a = \frac {15}{2} - \frac {15}{2} cos a + 4 sin a = \frac {15}{2} + \sqrt {(\frac {15}{2})^2 + 4^2} \left( - \frac {15}{2\sqrt {(\frac {15}{2})^2 + 4^2} } cos a + \frac {4}{\sqrt {(\frac {15}{2})^2 + 4^2}} sin a \right)$

There exists $ \alpha \in \mathbb{R}$ , that $ cos(a - \alpha) = \left( - \frac {15}{2\sqrt {(\frac {15}{2})^2 + 4^2} } cos a + \frac {4}{\sqrt {(\frac {15}{2})^2 + 4^2}} sin a \right)$

$ \frac {15}{2} + \sqrt {(\frac {15}{2})^2 + 4^2} cos(a - \alpha) \le \frac {15}{2} + \sqrt {(\frac {15}{2})^2 + 4^2}=16$

I think there isn't a typo in calculations and my method works here, I hope so...  :roll:

Okay, mods please move this to Pre-Olympiad. :(

Thanks.

Remember, you can use your forced conditions to simplify your expression!

[hide="Solution to first problem"]$ \frac {15 x^2}{2} \plus{} 4xy \equal{} \frac {15 x^2 \plus{} 8xy}{2} \equal{} \frac {16x^2 \plus{} 8xy \plus{} y^2 \minus{} 2}{2} \equal{} \frac {(y \plus{} 4x)^2 \minus{} 2}{2}$

Clearly, this is maximized when we maximize $ y \plus{} 4x$, so the problem boils down to maximizing $ y \plus{} 4x$ given $ x^2 \plus{} y^2 \equal{} 2$, which is the same as finding the positive value $ k$ for which $ y \equal{} k \minus{} 4x$ is tangent to the circle $ x^2 \plus{} y^2 \equal{} 2$, which is just similar triangles.

You get $ k \equal{} \sqrt {34}$, giving an answer of $ 16$, as expected.[/hide]

In number 2, think about the graph of $ \frac{1}{1\plus{}|x|}$. Given that at least one of $ x,y,z$ must be positive from the given, is it ever optimal to use negative numbers? This is an important thing to consider because if negative numbers are suboptimal, we can limit our domain and simplify the problem.

[quote="FantasyLover"]Bump again...

Oh wow, I should've posted this on Olympiad Forum...[/quote]

I don't see the necessity of bumping your thread once a day.


abstract algebra
binomial coefficients
superior algebra
superior algebra unsolved

Hi, I'm a 1st year Maths student from the UK doing a module on 'Finite Mathematics' and having problems with the following question on a homework assignment.

Show that

$ \sum_{m\equal{}0}^{500} \ (\minus{}1)^m \binom{1000}{2m} \equal{} 2^{500}$

I've just started learning about binomial coefficients and the only thing I know about them is that $ (x\plus{}y)^n \equal{} \sum_{j\equal{}0}^n \binom{n}{j}x^{n\minus{}j}y^{j}$

So I tried to use this to tackle the problem, and rewrote $ 2^{500}$ as $ (3\plus{}(\minus{}1))^{500}$ (so that $ y$ could be -1)

It wasn't long before it dawned on me that I was barking up the wrong tree.

Would anybody be able to help me out as to which tree I should be barking at to tackle this question?

Try to consider $ (x\plus{}i)^{1000} \plus{} (x\minus{}i)^{1000}$.
By the way, this should be in combinatorics forum.

Thanks for the tip

Sorry for posting in the wrong place


How many times does an hourhand,minute-hand and second-hand of a clock met together in a day ?

[quote="prototypex24"]24?[/quote]

I dun know how it works when deal with hour-minute-second hands but for the case where you only consider hourhand and minute-hand , it woulb be less than 24 times ..

[quote="Cyberman"]It's more than 24, I forget the equation for this though.

at 6:33 the hands may all touch each other.[/quote]

I doubt it would be more than 24 times . Since if we only consider the case for hourhand and minute-hand , they both meet together 22 times a day (because theres no coincide at 11+hr) So i tihink it should be less than 22 times or maybe 22  :?

[quote="236factorial"]I think only at noon and midnight. Twice?[/quote]

Ya , thats what i think before . But you see , the minute-hand only moves $\frac{1}{60}$ step when the second-hand moves $1$ step . So the space between would be small enough to be ignored and hence we can say that the amount of coincide of hour-minute-second hands is actually the same as the amount of coincide of hour-minute hand which is 22 times . Could that be true  :roll:

Hourhand=h
minutehand=m
secondhand=s
we divide the hour circle to 3600.when our h goes 1 step m goes 12 step and s goes 720 step.the journey around the circle ends in 3600 unit times because h goes there.h and m meet 3600/11
unit times.m and s meet $3600/708$ unit times equal to $300/59$.$(3600/11)*k$ is different from $(300/59)*n$ up to 3600,k and n are positive integers.When these numbers equal to each other h and m and s meet.They equal each other at 3600 and 7200.In a day this comes twice.

[quote="gunes"]Hourhand=h
minutehand=m
secondhand=s
we divide the hour circle to 3600.when our h goes 1 step m goes 12 step and s goes 720 step.the journey around the circle ends in 3600 unit times because h goes there.h and m meet 3600/11
unit times.m and s meet $3600/708$ unit times equal to $300/59$.$3600/11$ and $300/59$ have no meeting place in $0<x<3600$ and the times meet at 3600.In a day this comes twice.[/quote]

sorry but I cant understand what this sentence means $%Error. "arrow" is a bad command.
$ $3600/11$ and $300/59$ have no meeting place in $0<x<3600$ and the times meet at 3600.

[quote="236factorial"][quote="shyong"]So the space between would be small enough to be ignored and hence we can say that the amount of coincide of hour-minute-second hands is actually the same as the amount of coincide of hour-minute hand which is 22 times . Could that be true  :roll:[/quote]

You can't ignore that small space. It's still something  ;)[/quote]

Theorically we would say there are still intervals in between , but practically , we cant see the difference with our bare eyes , can we ? I mean , by looking at a clock , the movement of $\frac{1}{60}$ step is really too small  :P

[quote="236factorial"]I can, but I have to stare at the clock  :o[/quote]

$\text{lol}$  :rotfl:  :rotfl:

Then I guess my clock is too small for me to do so .... Anyway , concerning the problem , if the time peroid starts from 12midnight to another 12midnight , then it would have $2$ coincide .But isnt a peroid of one day starts from 12:01midnight to another 12midnight ? If it is so , then the answer would be only $1$  :o  :o

[quote="236factorial"]Not quite. I think the day starts at midnight sharp. Even if it started on 12:00:00.000.....00001, there would still be two coincides: one at noon, and one at the way end of the day, [i]just[/i] before the next day (12:00:00.0000...0001) starts.[/quote]

Oh ya you are right  :P I keep thinking that theres only one coincide at 12midnight per day and forgot about the 12 noon  :blush:  :blush:  :blush:

The reason I said more than 24 is because I think I've seen this problem before at totse.com in the mad scientist section.
The question back there was: How many times will a clock ring when the all hands H,M,S touch each other within a 24 hours period?
I think it was an SAT/ACT question.

6:32:32 or so. The gray stuff is a shadow i think..

[img]http://img322.imageshack.us/img322/2307/time0pv.jpg[/img]

No, they're still a little off, by only like 3 degrees or so.


analytic geometry
probability
algebra
function
domain
conics
parabola

I never looked up the answers for last year's AMC, so I decided to take it again, but I didn't get the following ones:

18. If n has 5 digits (with the leftmost digit nonzero) and n = 100q + r, for how many values of n does 11|q+r ?

20. If 5 A's, 5 B's, and 5 C's are placed in a 15-letter arrangement such that there are no A's in the first five letters, no B's in the next 5 letters, and no C's in the next five letters, how many arrangements are possible?

22. On a coordinate place, A starts at (0,0) and either goes up one or right one each step, while B starts at (5,7) and either goes down one or left one each step. If A and B move simultaneously, what is the probability that they meet?

(I thought that they could meet at 6 different points after 6 steps, one for each of x=0,1,2,3,4,5 , so the probability would be 6/6^2 = 1/6, but that reasoning seems to be wrong.)

25. If f(x) = Sqrt(ax^2+bx), how many real values of a exist such that there is at least one positive value of b for which both the domain and range of f(x) are the same set?

Thanks.

For 22, wouldn't the total possibilities... be 2^6?

But I think we only want to see the number of possibilities at the 6th step, since that is the only step at which A and B can meet.

But don't you need to take into account the probability that A and B are at places that they CAN meet on the first step?

for 18, is r supposed to be between 0 and 100?

confuted:

But even though A and B don't meet after one step, they have the potential to meet later on, right? So we don't know whether they will meet yet.

MysticTerminator:

Yeah, q is the quotient and r is the remainder, forgot to specify.

20. we have x A's in the second set of five letters. we thus have 5-x A's in the third set of five letters and 5-x C's in the second set of five letters. thus we have x C's in the first set of five letters, x B's in the third set of five letters, and 5-x B's in the first set of five letters. x can here run from 0 to 5. For a given x, the number of ways we can arrange the letters is C(5,x)^3. We could try to derive a combinatorial identity for this or something, or we could just sum over x: 1+125+1000+1000+125+1=2252.

18. 11|q+r is equivalent to 11|100q+r or 11|n. You want the number of 5-digit multiples of 11. I'll let you take it from here.

domain: ax^2+bx:ge:0 => 0 and -b/a are the roots. depending on the value of a (positive or negative) the domain will differ. but you can see it's either [-b/a, 0] or the opposite of that.

range: if a is positive, we have a parabola that opens upward, so the range will be [0, :inf:). This does not correspond to either of the possible domains (a closed interval, or the union of two intervals). So therefore, a is negative, and the domain is [0, blah] where blah is what we get when we plug in the vertex. So, the vertex is -b/2a, we plug it in to get sqrt(a(-b/2a)^2+b(-b/2a))=sqrt(b^2/4a-b^2/2a)=b/2sqrt(-a). This has to be equal to -b/a, so -a=2sqrt(-a) => a^2=4a => two values, a=0 and a=4. However, a has to be negative. So no values.

MyticTerminator:

18. Bah, missed that.

20. Nice.

22. Where did you get C(12,5) from?

25. Hmm, the answer is 2...

22. C(12,5) is the number of paths of length 12 that have 5 right-moves and 7 up-moves.

25. yeah, a=0 works for one. i'm figuring out where i messed up.

25. okay let me just start over. the range clearly contains no negative points. therefore, the domain cannot contain any negative points, and x(ax+b) must therefore be strictly negative on when x is negative. Thus, ax+b must be positive for any negative x. We clearly see that a:le:0. We note that a=0 is a solution (where both domain and range are [0, :inf:) for any b), and forget about it temporarily. So now we have a strictly negative. To simplify things, we replace a everywhere by -c, where c is positive. So now, f(x)=sqrt(bx-cx^2). The domain is the closed interval with endpoints 0 and b/c. Let z=b/2c, i.e. the vertex of the parabola that we're taking the square root of. The range will be the closed interval from 0 to f(z)=sqrt(z(b-cz))=sqrt((b/2c)(b-b/2))=sqrt(b^2/4c). So this thing is b/c. We set them equal and take squares to get b^2/4c=b^2/c^2. So, either b=0 (which we can discard) or c=4. So we get a=-4 as another solution. I must've messed up on my signs last time. Bah i really need to start using paper...

22. Hmm, so C(12, 5)/2^12 is the probability that A will go to the point (5, 7) (the initial position of B). Why is this the same probability as the probability that the moving points A and B meet?

25. wait a minute:

[quote]25. okay let me just start over. the range clearly contains no negative points. therefore, the domain cannot contain any negative points, and x(ax+b) must therefore be strictly negative on when x is negative. Thus, ax+b must be positive for any negative x. We clearly see that a<=0.[/quote]

Ok, so f(x) >= 0, so for both the range and domain to be the same, x >= 0. Since ax^2 + bx = x(ax+b) >= 0 for f(x) to be real, ax+b must be greater than or equal to 0, right? So why is a<=0?

12) Just try to slowly build up the sequence of numbers based on which red numbers divides which blue numbers.

17) For this one, I just let A be the origin of the plane and wrote equations for the two circles. I then found the intersection points.

[quote="Fierytycoon"]
18. If n has 5 digits (with the leftmost digit nonzero) and n = 100q + r, for how many values of n does 11|q+r ?

22. On a coordinate place, A starts at (0,0) and either goes up one or right one each step, while B starts at (5,7) and either goes down one or left one each step. If A and B move simultaneously, what is the probability that they meet?
[/quote]

i think these two were the ones youre still having trouble with? oh well

18)notice that n=100q+r=99q+q+r. so then its divisible if and only if n is so find all of those

22)its just a bijection so C(12,5)/2^12

[quote="hello"]How about the geometry problem 17. CAn someone help?
Also problem 12 with the red and blue?[/quote]

can you post those problems for us?

The ones I was still a bit confused about were 22 and 25.

Syntax Error:

Bijection? Ok..so what does that tell us? (I'm not very used to the terms bijections, injections, and surjections, so I may be missing something obvious here)

bijection is just another term for one to one correspondence

heres an example from confuted's going away contest

[quote]The game of Chomp is played with a 5 x 7 board. Each player alternately takes a bite out of the board by removing a square any and any other squares above and/or to the left of it. How many possible subsets of the 5 x 7 board (including the original board and the empty set) can be obtained by a sequence of bites? [/quote]

you can find a one to one correspondence with the number of paths going up and left or whatever those directions should be, so then its just C(12,5), because it includes 7 step and 5step things

edit: basically for you know there are 12c5 steps from 0,0 to 5,7 up or right. the bijection is between the where they meet etc. and there are 2^12 pairlike thingies, so then its just c(12,5)/2^12

more explaining?

i dont have the contest; would someone post questions#12 and 17?

Well, a bijection has to be onto as well, no?

Anyways, so the bijection is between

- the number of possible paths between (0,0) and (5,7)

and

- the number of ways in which A and B can meet

But why can we establish this bijection?

http://mathworld.wolfram.com/Bijection.html

basically you want to find another set where you can pair off

12) Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6.  She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card.  What is the sum of the numbers on the middle three cards?

17) Square ABCD has sides of length 4, and M is the midpoint of segment CD.  A circle with radius 2 and center M intersects a circle with radius 4 and center A at points P and D.  What is the distance from P to segment AD.

first ones pretty much just divisibility and knowing what the ends should be. the second one was just assinging a coordinate plain

12. the two bad red cards are 4 and 5. the only multiples they have among the blue cards are 4 and 5. thus they must be on either end. now, the red 3 must be surrounded by the blue 3 and the blue 6. it is clear that these three are in the middle, so the ans. is 12. the complete order is R4 B4 R2 B6 R3 B3 R1 B5 R5 if you really care.

17. bah

this is something i never really understood... what tips you off in a problem that makes you think to put it on the coordinate plane?

Put things on a coordinate plane if you don't know how to solve them in other ways.

it years for it too, 2 circles intersecting, midpoints, gravy

[quote]basically you want to find another set where you can pair off[/quote]

Yeah, I know, but I'm saying that I don't quite see exactly how

- the number of possible paths between (0,0) and (5,7)
and
- the number of ways in which A and B can meet

pair up with one another. I guess I could form some hazy intuitive argument, but...

[quote]Just draw in the routes each point is taking.. if they meet, it must form a line going from 0,0 to 5,7..[/quote]

Oh..that was simple.  :lol:

Still confused on #25 :(

Since f(x) = Sqrt(...), the range must be nonnegative, so the domain must be nonnegative. Since ax^2+bx=x(ax+b) must be nonnegative and x>=0, then ax+b>=0. If b is positive, I don't get why a <= 0.

if a>0 then range of f(snort) contains negative number things, but its not supposed to, then you go through some stuffs ending up with 2:sqrt:-a=-a, thus two values of a=0 and a=-4 work

[quote]if a>0 then range of f(snort) contains negative number things[/quote]

How does it do that? If both x and b are nonnegative, then a>=0 yields x(ax+b)>=0, no?

[quote]I think we're talking about when x is negative, and the result must be undefined, ie negative inside the square root too..[/quote]

But if x is negative, the domain and range would not be the same because the range is nonnegative, so why are we considering it?

[quote]Are you considering that the letters are not distinct from one another? And aren't they dependent, so why is it (5 C x) ? I'm not sure, shouldn't it just be (5 C x)?[/quote]

You must choose x A's, B's, or C's in each set of 5 letters, and since there are three such sets, it is C(5,x)^3.

[quote]Also, how come it's 2^12 instead of 4^12, since there are 4 ways at each step for both pairs to move, unless I don't understand something here.[/quote]

As I understand it, a downward or leftward motion of B is equivalent to an upward or rightward motion of A.

Oh, I get what you all were saying about #25 now. You were saying that if x<0, x(ax+b)<0, and so etc... Hmm, for some reason I didn't see that yesterday. Anyways, thanks to everyone who helped.

Thanks Fiertytycoon for 17 I finally got it.


Εστω $ x,y>0$ και $ x^2\plus{}y^3\geq x^3\plus{}y^4$. Να αποδείξετε ότι:
$ x^3\plus{}y^3\leq 2$

 :)

Ανέβασε τα θέματα της μεσογειάδας. Αν όχι στείλε τα στο email που σου έστειλα στο κινητό.

Για κάποιο περίεργο λόγο (αν και υποψιάζομαι γιατί) το αντίστοιχο post του Κώστα (Protonios) εξαφανίστηκε ξαφνικά.

η μεσογειαδα πραγματοποιείται σε διαφορετικές ώρες απο χώρα σε χώρα και τα θέματα ειναι τα ίδια...γιαυτό μεχρί να τελειώσει ο διαγωνισμός παν-μεσογειακώς δν βγαίνουν τα θέματα...υπομονή...

Όντως πολύ περίεργη ανισότητα ρε μιμ!
Λοιπόν αυτό που έκανα εγώ είναι το εξής (δυστυχώς έχει πολλές περιπτώσεις αλλά νομίζω ότι δουλεύει):
Είναι προφανές ότι αν $ x,y>1$ τότε παίρνουμε άτοπο από τη δοσμένη. Άρα μας μένει να εξετάσουμε την περίπτωση και οι 2 μεταβλητές να είναι το πολύ 1, οπότε η ανισότητα γίνεται προφανής και τις περιπτώσεις η μία να είναι μεγαλύτερη του 1 και η άλλη μικρότερη του 1 (σε περίπτωση που μια από τις δύο μεταβλητές είναι ίση με 1 η ανισότητα είναι πάλι προφανής)
Περίπτωση 1: $ x>1>y$
Τότε γράφουμε τη δοσμένη ως εξής:
$ y^3(1\minus{}y) \geq x^2(x\minus{}1) \implies \frac{x\minus{}1}{1\minus{}y}\leq \frac{y^3}{x^2}$.
Επίσης γράφουμε τη ζητούμενη ως εξής:
$ x^3\minus{}1\leq1\minus{}y^3 \iff (x\minus{}1)(x^2\plus{}x\plus{}1) \leq (1\minus{}y)(y^2\plus{}y\plus{}1)$ και θέλουμε να δείξουμε ότι $ \frac{x\minus{}1}{1\minus{}y} \cdot \frac{x^2\plus{}x\plus{}1}{y^2\plus{}y\plus{}1}\leq1$. Άρα σύμφωνα με τα παραπάνω αρκεί να δείξουμε ότι
$ \frac{y^3}{x^2}\cdot \frac{x^2\plus{}x\plus{}1}{y^2\plus{}y\plus{}1}\leq1$ που ισοδυναμεί με $ y^3x^2\plus{}y^3x\plus{}y^3 \leq x^2y^2\plus{}x^2y\plus{}x^2$ που είναι προφανής (κάθε όρος στο αριστερό μέλος είναι μικρότερος από τον αντίστοιχο στο δεξί!)

Περίπτωση 2: $ y>1>x$

Όμοια με την Περίπτωση 1

Άρα σε κάθε περίπτωση ΙΣΧΥΕΙ (<-Μιμ αυτό σε σένα πάει  :P ) και με ισότητα αν και μόνο αν $ x\equal{}y\equal{}1$

Ηλίας

Δεν πάμε καλά μου φαίνεται περιπτώσεις ... περιπτώσεις ... just joking . Πρωτότυπη λύση Hλία. Here is mine

$ x^3 \plus{} y^3 \equal{} 2 \minus{} 3\left(x^2 \plus{} y^3 \minus{} x^3 \minus{} y^4\right) \minus{} (x \minus{} 1)^2(2x \plus{} 1) \minus{} (y \minus{} 1)^2\left(3y^2 \plus{} 2y \plus{} 1\right)\leq 2.$

re paidia sorry pou epemvaino alla auth den einai mia polu oraia alla pasignosti anisotita ????

[quote="Nick Rapanos"][quote="Athinaios"]yep! ΚΛΑΣΙΚΕΣ-ΝΕΕΣ ΑΝΙΣΟΤΗΤΕΣ , Στεργιου-Σκομπρης, σελ 30   :P[/quote]

ok...leo ki ego...
an uparxei phgh sto vivlio, kalo tha itan na ti vazame..

giati i anisotita arxika emfanistike se kapoio diagonismo (pou den thumamai pou akrivos) kai meta egine anarpasth...
gia rosiki murizei pantos ee?[/quote]
Εξαιρετικη και δυσκολη κατα τη γνωμη μου ανισοτητα...
Eιναι οντος Ρωσια....
Δειτε και εδω http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1450237#1450237

Γεια χαρά. Έπεσα πάνω στο topic και σκέφτηκα να γράψω ακόμα μια λύση. Έστω $x\leq1$ και $y\geq1$ τότε η δοσμένη γράφετε ως $y^4-y^3\leq x^2-x^3$ οπότε διαιρώντας διαδοχικά με $y$ την αριστερή πλευρά και με $x$ την δεξιά έχουμε
$y-1\leq y^2-y\leq y^3-y^2\leq y^4-y^3\leq x^2-x^3\leq x-x^2\leq1-x$
Τώρα απ αυτές διαλέγουμε τις
$y^3-y^2\leq  x^2-x^3\Rightarrow x^3+y^3\leq x^2+y^2$
$y^2-y\leq  x-x^2\Rightarrow x^2+y^2\leq x+y$
$y-1\leq  1-x\Rightarrow x+y\leq 2$
Η περίπτωση $1\leq x$ και $y\leq 1$ βγαίνει όμοια ενώ οι άλλες περιπτώσεις είναι τετριμμένες όπως ανέφερε και ο κύριος salonikios


In $\triangle ABC$, let $P$ be the point on the bisector of angle $B$ such that $CP$ is perpendicular to $BP$. Let $M$ be the midpoint of $AC$. Prove that $MP||AB$

[hide]Let $CP$ intersect line $AB$ in $S$. The $\Delta BCS$ is isosceles $BC=BS$ so $P$ is the midpoint of $CS$. As $M$ is the midpoint of $CA$ $PM$ is paralel to $AS$.[/hide]


Decide whether there exists a function f: N --> N such that

f2003(n) = 2n

for all positive integers n.

Here f2003 means f(f(f(...))) [2003 times].

I think such a function does exist, but I wonder if I've managed to define it correctly :?

First of all, we have f(2n)=2f(n), so it's enough to define the function on odd numbers.

We take the first 2002 odd numbers: 1, 3, .., 4005. We take f(1)=3, f(3)=5, .., f(4003)=4005. Then necessity guides us: we must have f(4005)=2, f(2)=6 etc. We're done with this group of odd numbers. We take the next group of 2003 odd numbers: 4007, 4009, .., 8011 and we do what we did with the previous group: f(4007)=4009, f(4009)=4011, .., f(8009)=8011, and then f(8011)=2*4007=8014 and so on. We do this for every group of 2003 consecutive odd numbers.

Does it work?

Yes, you're correct. I think it's more challenging to find a bijective function satisfying those criteria

Are you proposing it? I mean, do you have a solution?

Just an observation: It's obvious that the idea above works for f(k)(n)=2n, no matter what k is. I think we can find functions f s.t. f(k)(n)=n2 for all natural n, and I'm not sure right now, but I think the problems can be related to each other.


For $ a,b,c$ real numbers such that $ a\plus{}b\plus{}c\equal{}1$ prove

$ \sum \frac{1}{1\plus{}a^2} \leq \frac{27}{10}$

[quote="manlio"]For $ a,b,c$ real numbers such that $ a \plus{} b \plus{} c \equal{} 1$ prove

$ \sum \frac {1}{1 \plus{} a^2} \leq \frac {27}{10}$[/quote]
We can use my theorem for this inequality, manlio. :)
The inequality is equivalent to
\[ \sum \frac{a^2}{a^2\plus{}(a\plus{}b\plus{}c)^2} \ge \frac{3}{10}\]
By the Cauchy Schwarz Inequality, we get
\[ \sum \frac{a^2}{a^2\plus{}(a\plus{}b\plus{}c)^2} \ge \frac{(a^2\plus{}b^2\plus{}c^2)^2}{a^4\plus{}b^4\plus{}c^4\plus{}(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)^2}\]
It suffices to prove that
\[ 10(a^2\plus{}b^2\plus{}c^2)^2 \ge 3(a^4\plus{}b^4\plus{}c^4\plus{}(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)^2)\]
\[ \Leftrightarrow 4\sum a^4 \plus{}14\sum a^2b^2\minus{}6\sum a^3b\minus{}6\sum ab^3 \minus{}6abc\sum a \ge 0\]
From this, we get $ m\equal{}4>0,n\equal{}14,p\equal{}g\equal{}\minus{}6$ and
\[ 3m(m\plus{}n)\minus{}p^2\minus{}pg\minus{}g^2 \equal{}108>0\]
Now, using my theorem, the inequality is proved.
We have done. :),

[quote="silouan"][quote="can_hang2007"]
Now, using my theorem, the inequality is proved.
[/quote]

Which is the theorem ? :maybe:[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?t=199301]Here is it[/url] :)

Thank you very much, can_hang2007. :)

Does it exists a proof using tangent line tecnique?

I really don't remember. :blush:

I have seen the solution use targent line .But I can prove it by cauchy-schwarz :D
<=>$ \sum \frac{a^2}{a^2\plus{}(a\plus{}b\plus{}c)^2} \ge \frac{3}{10}$
We have :$ (a\plus{}b\plus{}c)^2 \le ( |a|\plus{}|b|\plus{}|c|)^2$
Setting :$ x\equal{}|a|;y\equal{}|b|;z\equal{}|c|$
Assume that :$ x\plus{}y\plus{}z\equal{}1$
By cauchy-schwarz ,we need to prove :
$ 10(x^2\plus{}y^2\plus{}z^2)^2 \ge 3(x^2\plus{}y^2\plus{}z^2)\plus{}3\sum x^4$
<=>$ 17q^2\minus{}11q\minus{}6r\plus{}2 \ge 0$
By Am-Gm we have :$ r \le \frac{q}{9}$
$ LHS \ge 17q^2\minus{}11q\minus{}\frac{2q}{3}\plus{}2 \equal{}\frac{2}{3}(3q\minus{}1)(17q\minus{}6) \ge 0$
Done ;)

[quote="silouan"]yes there exist one line solution with tangent line .

$ \frac {1}{1 \plus{} a^2}\leq\frac {27}{50}(2 \minus{} a)$[/quote]

No, you will not complete this one if you using only tangent line  :wink: Note that:
\[ \frac {1}{1 \plus{} a^2}\leq\frac {27}{50}(2 \minus{} a)\]
\[ \iff \frac{(3x\minus{}1)^2(4\minus{}3x)}{50(1\plus{}x^2)}\ge 0\]
It is not always holds for $ x\ge \frac{4}{3}$. We don't have $ x\le 1$ because x,y,z are [b]real [/b] numbers.

The following is also holds:
\[ \sum \frac{a}{a^2\plus{}1}\le \frac{9}{10},\forall a,b,c\in \mathbb{R},a\plus{}b\plus{}c\equal{}1\]

[quote="manlio"]For real numbers $ a,b,c$ such that $ a \plus{} b \plus{} c \equal{} 1$, prove that

$ \sum \frac {1}{1 \plus{} a^2} \leq \frac {27}{10}.$

Does it exists a proof using tangent line tecnique? [/quote][quote="mitdac123"]We have $ (a \plus{} b \plus{} c)^2 \le ( |a| \plus{} |b| \plus{} |c|)^2$.

Setting $ x \equal{} |a|,y \equal{} |b|,z \equal{} |c|$.

Assume that $ x \plus{} y \plus{} z \equal{} 1$.[/quote][quote="silouan"]yes there exist one line solution with tangent line .

$ \frac {1}{1 \plus{} x^2}\leq\frac {27}{50}(2 \minus{} x).$[/quote][quote="zaizai-hoang"]
$ \iff \frac {(3x \minus{} 1)^2(4 \minus{} 3x)}{50(1 \plus{} x^2)}\ge 0.$

It is holds for $ x\le 1$ (because $ x,y,z$ are [b]nonnegative[/b] numbers). :wink:
[/quote]If natural number $ n\geq 2$ and real numbers $ a_1, a_2,\cdots,a_n$ satisfy $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$, then

$ \sum_{i \equal{} 1}^{n} \frac {1}{1 \plus{} a_{i}^2} \leq \frac {n^3}{n^2 \plus{} 1},$

with equality if and noly if $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n \equal{} \frac {1}{n}.$

[hide="Proof."]$ \sum_{i \equal{} 1}^{n} \frac {\left(\sum_{k \equal{} 1}^{n}a_{k}\right)^2}{\left(\sum_{k \equal{} 1}^{n}a_{k}\right)^2 \plus{} a_{i}^2}\leq\sum_{i \equal{} 1}^{n} \frac {\left(\sum_{k \equal{} 1}^{n}|a_{k}|\right)^2}{\left(\sum_{k \equal{} 1}^{n}|a_{k}|\right)^2 \plus{} a_{i}^2}\equiv\sum_{i \equal{} 1}^{n} \frac {\left(\sum_{k \equal{} 1}^{n}x_{k}\right)^2}{\left(\sum_{k \equal{} 1}^{n}x_{k}\right)^2 \plus{} x_{i}^2}.$

Without loss of generality, we assume $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n \equal{} 1$. When $ n\geq 3$,

$ \frac {1}{1 \plus{} x_i^2} \plus{} \frac {2n^3}{(n^2 \plus{} 1)^2}\left(x_i \minus{} \frac {1}{n}\right) \minus{} \frac {n^2}{n^2 \plus{} 1}$

$ \equal{} \minus{} \frac {(n^2 \minus{} 1 \minus{} 2nx_i)(1 \minus{} nx_i)^2}{(n^2 \plus{} 1)^2(1 \plus{} x_i^2)}\leq 0\Longrightarrow$

$ \sum_{i \equal{} 1}^{n}\frac {1}{1 \plus{} x_i^2}\leq\sum_{i \equal{} 1}^{n}\left[\frac {n^2}{n^2 \plus{} 1} \minus{} \frac {2n^3}{(n^2 \plus{} 1)^2}\left(x_i \minus{} \frac {1}{n}\right)\right] \equal{} \frac {n^3}{n^2 \plus{} 1}.$

When $ n \equal{} 2$,

$ \frac {(a \plus{} b)^2}{(a \plus{} b)^2 \plus{} a^2} \plus{} \frac {(a \plus{} b)^2}{(a \plus{} b)^2 \plus{} b^2} \minus{} \frac {8}{5}$

$ \equal{} \frac { \minus{} (a \minus{} b)^2(a^2 \plus{} b^2)}{(2a^2 \plus{} 2ab \plus{} b^2)(a^2 \plus{} 2ab \plus{} 2b^2)} \leq 0\Longrightarrow$

$ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2}\leq\frac {8}{5}$ with $ a \plus{} b \equal{} 1$.[/hide]
For nonnegative numbers $ x_1, x_2,\cdots,x_n$, the reverse inequality (due to Ting-Yan Tian)

$ \sum_{i \equal{} 1}^{n} \frac {1}{1 \plus{} x_{i}^2} \geq n \minus{} \frac {1}{2}$

holds for $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n \equal{} 1$.

[hide="Proof."]$ \frac {1}{1 \plus{} x_i^2} \minus{} 1 \plus{} \frac {x_i}{2} \equal{} \frac {x_i(1 \minus{} x_i)^2}{2(1 \plus{} x_i^2)}\geq 0\Longrightarrow$

$ \sum_{i \equal{} 1}^{n} \frac {1}{1 \plus{} x_{i}^2} \geq\sum_{i \equal{} 1}^{n} \left(1 \minus{} \frac {x_i}{2}\right) \equal{} n \minus{} \frac {1}{2}.$[/hide]

[quote="zaizai-hoang"]
The following is also holds:
\[ \sum \frac {a}{a^2 \plus{} 1}\le \frac {9}{10},\forall a,b,c\in \mathbb{R},a \plus{} b \plus{} c \equal{} 1
\]
[/quote]

See also
http://www.mathlinks.ro/viewtopic.php?t=169923

[quote="Ji Chen"]
If natural number $ n\geq 2$ and real numbers $ a_1, a_2,\cdots,a_n$ satisfy $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$, then

$ \sum_{i \equal{} 1}^{n} \frac {1}{1 \plus{} a_{i}^2} \leq \frac {n^3}{n^2 \plus{} 1},$

with equality if and noly if $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n \equal{} \frac {1}{n}.$

[hide=Proof.]$ \sum_{i \equal{} 1}^{n} \frac {\left(\sum_{k \equal{} 1}^{n}a_{k}\right)^2}{\left(\sum_{k \equal{} 1}^{n}a_{k}\right)^2 \plus{} a_{i}^2}\leq\sum_{i \equal{} 1}^{n} \frac {\left(\sum_{k \equal{} 1}^{n}|a_{k}|\right)^2}{\left(\sum_{k \equal{} 1}^{n}|a_{k}|\right)^2 \plus{} a_{i}^2}\equiv\sum_{i \equal{} 1}^{n} \frac {\left(\sum_{k \equal{} 1}^{n}x_{k}\right)^2}{\left(\sum_{k \equal{} 1}^{n}x_{k}\right)^2 \plus{} x_{i}^2}.$

Without loss of generality, we assume $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n \equal{} 1$. When $ n\geq 3$,

$ \frac {1}{1 \plus{} x_i^2} \plus{} \frac {2n^3}{(n^2 \plus{} 1)^2}\left(x_i \minus{} \frac {1}{n}\right) \minus{} \boxed{\frac {n^2}{(n^2 \plus{} 1)^2}}$

$ \equal{} \minus{} \frac {(n^2 \minus{} 1 \minus{} 2nx_i)(1 \minus{} nx_i)^2}{(n^2 \plus{} 1)^2(1 \plus{} x_i^2)}\leq 0\Longrightarrow$

$ \sum_{i \equal{} 1}^{n}\frac {1}{1 \plus{} x_i^2}\leq\sum_{i \equal{} 1}^{n}\left[ \boxed{\frac {n^2}{(n^2 \plus{} 1)^2}} \minus{} \frac {2n^3}{(n^2 \plus{} 1)^2}\left(x_i \minus{} \frac {1}{n}\right)\right] \equal{} \frac {n^3}{n^2 \plus{} 1}.$
[/quote]

I proved your generalization by the same way but there's a little mistake in your solution:

all $ \boxed{\frac {n^2}{(n^2 \plus{} 1)^2}}$ should be replaced by $ \boxed{\frac {n^2}{(n^2 \plus{} 1)}}$ :wink:

$ \frac {1}{1 \plus{} x_i^2} \plus{} \frac {2n^3}{(n^2 \plus{} 1)^2}\left(x_i \minus{} \frac {1}{n}\right) \minus{} \frac {n^2}{n^2 \plus{} 1}$

$ \equal{} \minus{} \frac {(n^2 \minus{} 1 \minus{} 2nx_i)(1 \minus{} nx_i)^2}{(n^2 \plus{} 1)^2(1 \plus{} x_i^2)}\leq 0\Longrightarrow$

I think this is wrong as $ x_i$ can be a very great positive number. :wink:

[quote="Ji Chen"]
If natural number $ n\geq 2$ and real numbers $ a_1, a_2,\cdots,a_n$ satisfy $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$, then

$ \sum_{i \equal{} 1}^{n} \frac {1}{1 \plus{} a_{i}^2} \leq \frac {n^3}{n^2 \plus{} 1},$

with equality if and noly if $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n \equal{} \frac {1}{n}.$

[/quote]

  generalization:
If natural number $ n\geq 3$ and real numbers $ a_1, a_2,\cdots,a_n$ ,satisfy $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$, $ 0 < \lambda \le \frac {{n^2 }}{{2n \plus{} 1}}$,then
\[ \sum\limits_{i \equal{} 1}^n {\frac {1}{{1 \plus{} \lambda a_i^2 }}} \le \frac {{n^3 }}{{{n^2} \plus{} \lambda }}
\]

[quote="scpajmb"]If natural number $ n\geq 3$ and real numbers $ a_1, a_2,\cdots,a_n$ satisfy $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$, $ 0 < \lambda \le \frac {n^2 }{2n \plus{} 1},$  then

$ \sum_{i \equal{} 1}^n {\frac {1}{{1 \plus{} \lambda a_i^2 }}} \le \frac {{n^3 }}{{{n^2} \plus{} \lambda }}$.[/quote]Good generalized orientation, scpajmb. :)

For all real numbers $ a_1, a_2,\cdots,a_n(n\geq2)$ such that $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$,

the above inequality holds if and only if $ 0 \leq \lambda \leq \frac {n^2(n \minus{} 1)}{n^2 \minus{} n \plus{} 1}.$

With equality if $ \lambda \equal{} \frac {n^2(n \minus{} 1)}{n^2 \minus{} n \plus{} 1},a_1 \equal{} \frac {n \minus{} 1}{n},a_2 \equal{} \cdots \equal{} a_n \equal{} \frac {1}{n(n \minus{} 1)}.$


I just finished precalculus, but my school doesn't teach vectors and limits, do I have to be familiar with these two topics before I move on to Cal AB?  If so, when can I get the handouts and resources of these two topics?

You won't need to know vectors to understand the material offered in AB calculus. Also, an intuitive notion of limits is covered in AB, which is all you'll need. So no, you won't need either of those things to do AP calculus, though I'd suggest learning stuff about vectors just for the sake of learning it.

For Calculus AB, you don't need vectors, but once you move past single variable calculus you'll want a solid foundation in vectors. It's good to get comfortable with dot products and cross products in particular before taking any multivariable calculus.

i dont think we used vectors until multi. calc.

ill move this to the calc forum

As for limits- the standard Chapter 1 in a calculus textbook is limits. It is essential to calculus, but it's also part of the course.


Let n be a positive integer. Prove that n is power of an odd prime if and only if
\[ n \minus{} 1 \equal{} min\{k > 0: k(k \plus{} 1) \vdots 2n\}
\]
:)

My solution:

If $ n$ is a power of a prime then $ n\minus{}1$=$ min\{k>0: 2n|k(k\plus{}1)\}$ is obvious. The other direction follows quickly if you just write $ n\equal{}ab$, where $ a,b>1$ and $ gcd(a,b)$=1 (clearly possible given the condition), then use the fact that you can get $ x,y$ such that $ ax\plus{}by$=1, and $ 0<x<b$.


calculus
integration
logarithms
trigonometry
calculus computations

$ \int \frac{x^{2}\plus{}7}{x^{4}\minus{}x^{2}\plus{}4}dx$

$ \equal{}\frac{1}{4}\int \left( \frac{7\plus{}\sqrt{5}x}{x^{2}\plus{}\sqrt{5}x\plus{}2}\plus{}\frac{7\minus{}\sqrt{5}x}{x^{2}\minus{}\sqrt{5}x\plus{}2}\right) dx$

$ \equal{}\allowbreak \frac{\sqrt{5}}{8}\ln \frac{x^{2}\plus{}\sqrt{5}x\plus{}2}{x^{2}\minus{}\sqrt{5}x\plus{}2}\plus{}\frac{3\sqrt{3}}{4}\allowbreak \tan ^{\minus{}1}\frac{4\sqrt{3}x}{3\minus{}2\sqrt{15}}\plus{}C$


Prove that the triangle $ABC$ is right-angled if and only if
\[\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}-\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\frac{1}{2}\]

$2s=a+b+c$, then $\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$, and $\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$

our condition is then $2\sqrt{(s^{3})(s-a)(s-b)(s-c)}-2(s-a)(s-b)(s-c)=abc$

now...i have a feeling...that if we get rid of that radical, we will get a polynomial (that is 6th degree in any one variable), that will factor into:

$k*(a^{2}-b^{2}-c^{2})(b^{2}-c^{2}-a^{2})(c^{2}-a^{2}-b^{2})$

I wonder if this had work. I reduced the problem down to proving that

$(sinA+sinB+sinC)-(cosA+cosB+cosC)=1$ holds iff $\triangle ABC$ is a right-angle triangle.

Now, assume WLOG, $C \geq B \geq A$

Case 1: $C=90^{\circ}$

Case 2: $C<90^{\circ}$

Case 3: $C>90^{\circ}$


geometry
incenter
trigonometry
analytic geometry
angle bisector
similar triangles

Given $ \Delta ABC$ and a ruler in which you [i]are[/i] allowed to measure, but  no compass, construct (or locate) the points $ D%Error. "inBC" is a bad command.
,E%Error. "inCA" is a bad command.
,F%Error. "inAB" is a bad command.
$ such that the incircle is tangent to $ \Delta ABC$ at $ D,E,F$.

Prove that $ AD,BE,CF$ are concurrent. Their point of concurrency is called the Gergonne (Gorgonne?) Point.

Extention: with the same ruler, construct the incenter of the triangle.

Finding the points:
[hide]
Let our sides have lengths a, b, and c, so that side a is opposite A, etc.

Construct G, the foot of the angle bisector from A to BC, by measuring out $ \frac{ab}{b\plus{}c}$ units from B on BC.  This is $ \frac{ac}{b\plus{}c}$ units from C. These distances satisfy the angle bisector theorem. Similarly, on AC and AB, mark H and J, the feet of the angle bisectors from B and C to AC and AB, respectively, so that $ CH \equal{}\frac{bc}{a\plus{}c}$ and $ AJ \equal{}\frac{ac}{a\plus{}b}$.

Connect AG, BH, and CJ and you have the incenter I of ABC where all three segments intersect each other.

Now, with a calculator and/or a lot of patience, you could use Heron's to find the area of one of the small triangles, say, BIC. You know BI and BC, so the sine of IBC is twice the area over (BI)(BC). Now use similar triangles and SOHCAHTOA to find the altitude from I to BC, which is D. Repeat with the other two triangles to find E and F.

I don't know Ceva or anything else about concurrency, so I won't be able to prove the concurrency without major major coordinate bashing.[/hide]

Simpler way to find D, E, and F:

[hide="Some Lengths..."]We know that $ AE\equal{}AF\equal{}s\minus{}a$, $ BF\equal{}BD\equal{}s\minus{}b$, and $ CD\equal{}CE\equal{}s\minus{}c$.  Just measure and we're done.  Concurrency is proven by Ceva's.[/hide]

@FightTheTide: Ceva's states that if $ D$ is on $ BC$, $ E$ is on $ AC$, and $ F$ is on $ AB$, then $ AD$, $ BE$, and $ CF$ are concurrent iff:
$ \frac{AE}{EC}\cdot\frac{CD}{DB}\cdot\frac{BF}{FA}\equal{}1$


I can't find out who the moderator to the India forum is.  Can any one please help?  Thanks.

There doesn't appear to be a moderator for that forum. If there's something you'd like deleted, contacting an administrator would be a good place to start.

How could a local community find a moderator ?We need a moderator for the Greek Community  :maybe:

You could pm Valentin and ask him to assign one

[quote="mysmartmouth"][b]Rushil [/b]is the moderator, I believe[/quote]
and now it's me  :wink:

[quote="Ramanujam297"]I think the morderator is Gouthom[/quote]

yes [b] now [/b] the mod is Goutam, but this thread is 5 years old :) so they had different mods before.

mathman did you notice that there was a 5 year period between posts?
EDIT: My mistake.

dude  you had to edit it so that you wouldnt make a fool of yourself.....

[quote="ahaanomegas"]Dude, this thread is hilarious! It's been running for $8$ years now, with almost no useful post.[/quote]

Well!

[quote="9001BigJ"]Well well, hello there![/quote]

[quote="math26"]Whoops did not see that...[/quote]

Lol, this is so funny, hahahahahaaaaaa, I'm gonna fall over with laughter

That is not needed as well, and mods please lock.
I will now stop.

... maybe something called the report abuse button is useful.

[quote="LIONBLAZE333"]I agree.
math26, maybe if you post less, people won't jump at the chance to insult you.[/quote]
Add on to that: If you post hurtful comments, people are definitely going to insult you! :P


function
modular arithmetic
number theory
least common multiple
geometry
geometric transformation
reflection

What does the Lambda function state? Does it have any application for mods? Cuz I know that the phi function also says that if a is coprime to n, that is, $ \gcd(a, n) \equal{} 1$, then $ a^{\varphi{n}} \equiv 1 \mod{n}$ .

The function you are referring to is the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael Function[/url]. This function denotes the smallest $ n$ such that $ a^n \equiv 1 \pmod{m}$. It tends to be more useful when finding the units digit in high moduli.

"The first 26 values of λ(n) for n = 1, 2, 3, ... are 1, 1, 2, 2, 4, 2, 6, 2, 6, 4, 10, 2, 12, 6, 4, 4, 16, 6, 18, 4, 6, 10, 22, 2, 20, 12, ... "
If I'm not misinterpreting the definition of the function I believe $ \lambda(3) \equal{} 1$ since $ 4^1\equiv1\pmod3$ with $ (4,3) \equal{} 1$.
Trivially lamda=1 for every positive integer that is coprime to n...... this leads me to assume the def. in wiki left out the condition that the positive integer must be smaller than n... (not sure if this is the condition though as i'm not familiar with this function)

I think you overlooked the condition that the power $ n$ must be the lowest power such that it leaves a remainder of 1 for [b]every[/b] integer $ a$ that is coprime to $ m$.

Actually i didn't overlook... i misunderstood the def. i thought that by every it meant from the set of all a that from coprime to n - that is, we could choose the a that would yield the minimum power instead of every a in the set must satisfy a^lamda congruent modulo n..... (oh well i even misunderstood things like this on a lot of problems, perhaps b/c english is still not my native...:( and it usually takes me longer to understand problems than others and sometimes even misunderstand - disadvantage in math competitions... i know)

So whats the closed form for that function?

One way to define the Carmichael function is recursively, as follows:

$ \lambda(1) \equal{} \lambda(2) \equal{} 1, \lambda(4) \equal{} 2, \lambda(2^n) \equal{} 2^{n\minus{}2}, n \ge 3$,
For all odd primes $ p$, $ \lambda(p^k) \equal{} (p\minus{}1) p^{k\minus{}1}$,
$ \gcd(m, n) \equal{} 1 \Rightarrow \lambda(mn) \equal{} \text{lcm}(\lambda(m), \lambda(n)).$

The second and third conditions basically mean "use Euler's theorem and apply CRT," while the first condition reflects the fact that there are no primitive roots $ \bmod 2^n$ for $ n \ge 3$.

So to find the value of the Carmichael function for any integer $ n$, simply prime factorize that integer and apply the multiplicative rule that t0rajir0u stated.

[quote="mathwizarddude"]How to prove those properties? Also, what are some advantages of using the lambda function over the totient function? It seems the totient function would do no worse than the lambda function in reducing powers, so when to use the lambda function?[/quote]

What's $ \lambda(10000)$?

What's $ \varphi(10000)$?

[quote="mathwizarddude"]How to prove those properties?[/quote]
The third property is equivalent to CRT.  The first and second properties are equivalent to the following standard theorem and its proof can be found in any good number theory text:

[b]Theorem:[/b]  $ (\mathbb{Z}/m\mathbb{Z})^{*}$ is cyclic if and only if $ m \equal{} 1, 2, 4, p^k, 2p^k$ for $ p$ an odd prime and $ k \ge 1$.  If $ m \equal{} 2^k, k \ge 3$ then $ (\mathbb{Z}/m\mathbb{Z})^{*}$ is the product of a cyclic group of order $ 2$ and a cyclic group of order $ 2^{k \minus{} 2}$.

[quote="mathwizarddude"] Also, what are some advantages of using the lambda function over the totient function?[/quote]
$ \lambda(n)$ divides $ \varphi(n)$ and is strictly smaller in almost every case.  In grn_trtle's example, while $ \varphi(10000) \equal{} 4000$ it is easy to compute that $ \lambda(10000) \equal{} 500$.  So if you need to find the last four digits of something, you really want to use Carmichael's theorem instead of Euler's theorem.

Hmm looking back I'm actually getting $ \lambda(10000)\equal{}1500$:
$ \lambda(2^4)\equal{}2^2\equal{}4$
$ \lambda(5^4)\equal{}3*5^3$

$ [4,3*5^3]\equal{}4*3*5^3\equal{}20*3*5^2\equal{}20*75\equal{}1500$... so what's wrong? (also 1500 does not divide 4000)

Here's your error: $ \lambda(5^{4})\equal{}5^{4 \minus{} 1}(5 \minus{} 1) \equal{} 4\cdot 5^3 \not\equal{} 3\cdot5^3$. The least common multiply turns out to be $ 500$, with your error fixed, as t0rajirou stated before.

Thanks for catching that!! (making way tooooo many dumb mistakes... even copying off the def... *sigh*)

By the way, I don't really see how to prove the last part using CRT... would you mind proving it for me? Thanks.

Sorry to bother again, but regarding [url]http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_4_2006-2007_Problems/Problem_7[/url]
of finding $ 3^{3^{3^3}}\mod1000$ I believe there's a mistake in the solution: $ 3^{27}\equiv 3^7\equiv 12\mod25$, not 18. Here's my solution - could someone please kindly check if my answer is correct (using any means whatsoever, like Mathematica)? If so, that would be great (and replace the original solution in the Wiki if anyone feels to do it)...
[hide="Solution"]$ \phi(1000) \equal{} 400$ We want to find $ 3^{27}\mod400$. $ \lambda(2^4\cdot5^2) \equal{} 20$, and we find $ 3^7 \equal{} 729\cdot3\equiv187\mod400$.
Now we want to find $ 3^{187}\mod1000$. $ \lambda(1000) \equal{} 100$ and so we want to find $ 3^{87}\mod1000$
Since $ 3^6 \equal{} 729$, we find $ 3^9\equiv683$ and $ 3^{10}\equiv 49\mod1000$. And now we want to find $ 3^7\cdot49^8\mod1000$. $ 49^2\equiv401;401^2\equiv801;801^2\equiv601$ and $ 3^7\equiv187$ and $ 187\cdot601 \equal{} 389\mod1000$ So the answer is $ \boxed{389}$ Carmichael function is sooo useful in this problem!! :lol: [/hide]
BTW on the same Mock AIME could somone please check if the answer to #11 is $ 152$? Thanks.

You don't even need Mathematica, you could use [url=http://www.wolframalpha.com/]Wolfram Alpha[/url]. But for some reason, it won't do the problem directly (maybe it doesn't have that much precision). And according to Wolfram Alpha, you are right about the fact that $ 3^{27} \equiv 12\pmod{25}$.

I just quickly substituted 7625597484987 for $ 3^{3^3}$ (which I calculated using a calculator), and Wolfram alpha can do this:

http://www.wolframalpha.com/input/?i=3%5E7625597484987

Which returns the last three digits as 387, and not 389.

Edit: I updated the solution on http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_4_2006-2007_Problems/Problem_7#Solution so it should be correct now. Also...

mathwizarddude, you have the correct solution, except that $ 187\cdot601 \equal{} 387\pmod{1000}$, not 389 :P The Carmichael function can be useful, but making arithmetic errors aren't!


Let $p$ be a prime number for which there exists a positive integer $n$ such that $n \mid p-1$ and $p \mid n^3-1$. Prove that $4p-3$ is the square of a integer.

It was also given in MMO this Year. :lol:


vector
geometry
circumcircle
cyclic quadrilateral
geometry unsolved

PROBLEM (Lokman GÖKÇE): $ABCD$ is a cyclic quadrilateral that the diagonal $AC$ perpendicular to $BD$ at point $E$. The gravity centers of $\triangle ABC,\triangle BCD, \triangle CDA, \triangle DAB$ are $G_{1},G_{2},G_{3},G_{4}$, respectively.If the circumcenter is $O$, prove that:
$\vec{OG_{1}}+\vec{OG_{2}}+\vec{OG_{3}}+\vec{OG_{4}}=2\vec{OE}$

Let $H$be the orthocenter of triangle $ABC$

On one hand,
$\vec{OG_{1}}+\vec{OG_{2}}+\vec{OG_{3}}+\vec{OG_{4}}=1/3*(\vec{OA}+\vec{OB}+\vec{OC})+1/3*(\vec{OB}+\vec{OC}+\vec{OD})+1/3*(\vec{OC}+\vec{OD}+\vec{OA})+1/3*(\vec{OD}+\vec{OA}+\vec{OB})=\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=\vec{OH}+\vec{OD}$

On the other hand,
$|EH|=2RCos<BAC*Cos<BCA=2R*Cos<BDC*Cos<ADB=2R*Sin<ACD*Cos<CAD=|ED|$.

Hence,
$\vec{OG_{1}}+\vec{OG_{2}}+\vec{OG_{3}}+\vec{OG_{4}}=\vec{OH}+\vec{OD}=2\vec{OE}$

Let $M,N$ be mid-points of $[BD],[AC]$.$ENOM$ is a rectangle.We will show that $\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=2.\vec{OE}$.
Since $\vec{OA}+\vec{OC}=2.\vec{ON}$ and $\vec{OB}+\vec{OD}=2.\vec{OM}$, $\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=2.(\vec{ON}+\vec{ON})=2.\vec{OE}$.

Lokman GÖKÇE


Sa se determine functile bijective $f:[0,1]\to[0,1]$ care verifica relatia :$\frac{1}{2}[f(x) + f^{ - 1} (x)] = x,\forall x \in [0,1].$

Yes of course I would be very happy if you would do this.

I know no one appreciate my help. :(

Forget...

Come on, Myth, I just joked.  :( I am sorry. :(  Of course we appreciate your help.

It is not funny for me to see it again and again...  :(

To see what, Myth? I do not remember to have said anything bad in the near past.

Humble opinion: that is not very useful.

Moubinool could you explain me how you found f(x)=x.

$x,f(x),f(f(x))...f^{(n)}(x)$ form an arithmetical progression which can't be bounded, unless $x=f(x)$

oops, acuma mi-am dat seama ca trebuia pe romana :))
$x,f(x),f(f(x))....f^{(n)}(x)$ formeaza o progresie aritmetica care nu e marginita decat daca ratia e $0$, deci $x=f(x)$.
 :lol:

However,Myth how was your solution like?

Myth that is not a big probleme,because I have learn a lot of english in the last years.


In how many different ways can 10 be written as a sum using only the numbers 1, 3 and 5?
Note that 3 + 3 + 3 + 1 and 1 + 3 + 3 + 3 are two different ways that should be included.

I think this might be $ 47$:

Using two $ 5s$: $ 1$ way: $ 5\plus{}5$.

Using one $ 5$ and no $ 3s$: $ 6$ ways: $ 5, 1, 1, 1, 1, 1$ and permutations.

Using one $ 5$ and one $ 3$: $ 12$ ways: $ 5, 3, 1, 1$ and permutations.

Using no $ 5s$ and no $ 3s$: $ 1$ way: $ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$.

Using no $ 5$ and one $ 3$: $ 8$ ways: $ 3, 1, 1, 1, 1, 1, 1, 1$ and permutations.

Using no $ 5s$ and two $ 3s$: $ 15$ ways: $ 3, 3, 1, 1, 1, 1$ and permutations.

Using no $ 5s$ and three $ 3s$: $ 4$ ways: $ 3, 3, 3, 1$ and permutations.


How many positive integers have cube roots less than $ 1 \plus{} \sqrt{2}$?

[hide="solution"]
The number $ n$ satisfies \[ \sqrt[3]{n}<1+\sqrt{2}.\] Cubing both sides gives \[ n<1^3+3\cdot1^2\sqrt{2}+3\cdot1(\sqrt{2})^2+(\sqrt{2})^3=7+3\sqrt{2}.\]
Now is when knowing approximations of common square roots comes in handy! We (should) know (learn if you don't) that $ \sqrt{2\approx1.414.}$ Plugging that back into our original expression gives the approximation \[ 7+3\times1.414=7+4.242=11.242.\] The number of positive integers $ n$ satisfying that is $ \boxed{11}.$
[hide="list of common square roots and their approximations"]
\begin{align*}\sqrt{2}&\approx1.414\\\sqrt{3}&\approx1.732\\\sqrt{5}&\approx2.236\\\sqrt{10}&\approx3.162\end{align*}[/hide][/hide]

Cubing both sides actually gives $ 7 \plus{} 5\sqrt{2}$, not $ 7 \plus{} 3\sqrt{2}$. Hence, the answer is actually $ \boxed{14}$.

[quote="RoFlLoLcOpT"]
Now is when knowing approximations of common square roots comes in handy! We (should) know (learn if you don't) that $ \sqrt {2\approx1.414.}$
[/quote]

I think you mean $ \sqrt{2}\approx1.414$.


Μπορεί κάποιος να με βοηθήσει με το εξής πρόβλημα;

Αποδείξτε ότι αν [b]σ(Ν) = 2Ν + 1 [/b] , τότε ο [b]Ν[/b] είναι τετράγωνο ενός περιττού ακεραίου.

[b]Φώτη πρώτα απ'όλα νομίζω ότι δεν χρειάζεται $ \sigma(N) \equal{} 2N\plus{}1$ αλλά γενικά $ \sigma(N) \equal{}$ περιττός γενικά.

Θα ξεκινήσω αναφέροντας μερικά χρήσιμα πράγματα που υπάρχουν στο "backround".
[/b]

[b][color=blue][u]Ορισμός 1[/u][/color]
Μία μη μηδενική συνάρτηση $ f:\mathbb{N}\to\mathbb{R}$ ονομάζεται πολλαπλασιαστική εαν για κάθε $ m,n\in\mathbb{Z}$ με $ (m,n) \equal{} 1$, ισχύει
f(mn)=f(m)f(n).

[color=red]
[u]Πόρισμα 1[/u][/color]
Για κάθε πολλαπλασιαστική συνάρτηση $ f$ έχουμε $ f(1) \equal{} 1$.

[color=red]
[u]Πόρισμα 2[/u][/color]
Έστω $ f$ μία πολλαπλασιαστική συνάρτηση και $ a_{1},\ldots,a_{n}$ θετικοί ακέραιοι, πρώτοι μεταξύ τους ανά δύο. Τότε
$ f(a_{1}\cdots a_{n}) \equal{} f(a_{1})\cdots f(a_{n})$. (Απόδειξη με επαγωγή).

[color=red]
[u]Πόρισμα 3[/u][/color]
Έστω $ f$ μία πολλαπλασιαστική συνάρτηση και $ n\in\mathbb{N}$ με $ n > 1$. Αν $ n \equal{} p_{1}^{a_{1}}\cdots p_{r}^{a_{r}}$ είναι η ανάλυση του $ n$ σε (διακεκριμένους) πρώτους παράγοντες, τότε $ f(n) \equal{} f(p_{1}^{a_{1}})\cdots f(p_{r}^{a_{r}})$.

[color=blue][u]Ορισμός 2[/u][/color]
Εαν $ k\in\mathbb{N}$ θεωρούμε τη συνάρτηση $ \sigma_{k}$ που ορίζεται από τη σχέση

$ \sigma_{k}(n) \equal{}\sum_{d\mid n}d^{k}$ για κάθε $ n\in\mathbb{N}$.

Παρατήρηση: Η συνάρτηση $ \sigma$ είναι απλά η $ \sigma_{1}$ και συμβολίζει το άθροισμα των διαιρετών του $ n$ ενώ η συνάρτηση $ \sigma_{0}$ την συμβολίζουμε συνήθως με $ \tau$ και παριστάνει το πλήθος των διαιρετών του $ n$.

[color=brown]
[u]Πρόταση 1[/u][/color]
Η συνάρτηση $ \sigma_{k}$ είναι πολλαπλασιαστική.

[color=brown][u]Πρόταση 2[/u][/color]
Εαν $ n$ ένας φυσικός με $ n\geq 2$ και $ n \equal{} p_{1}^{a_{1}}\cdots p_{r}^{a_{r}}$ η ανάλυση του $ n$ σε (διακεκριμένους) πρώτους παράγοντες τότε από το Πόρισμα $ 2$ έχουμε $ \sigma_{k}(n) \equal{}\sigma_{k}(p_{1}^{a_{1}})\cdots\sigma_{k}(p_{r}^{a_{r}}) \equal{}\prod_{i \equal{} 1}^{r}(1\plus{}p_{i}^{k}\plus{}\cdots\plus{}p_{i}^{ka_{i}})$.
Επομένως εαν $ k \equal{} 0$ τότε
$ \sigma_{0}(n) \equal{}\tau(n) \equal{} (a_{1}\plus{}1)\cdots (a_{r}\plus{}1)$ (Πολύ χρήσιμος τύπος εαν θέλουμε να βρίσκουμε το πλήθος των διαιρετών ενός αριθμού)

ενώ εαν $ k\geq 1$ τότε
$ \sigma_{k}(n) \equal{}\prod_{i \equal{} 1}^{n}\frac{p_{i}^{k(a_{i}\plus{}1)}\minus{}1}{p_{i}^{k}\minus{}1}$.

[u]Επιστροφή τώρα στην άσκηση που ζητάς:[/u] (συμβολίζω το δικό σου $ N$ με $ n$ για ευκολία δική μου)
Καταρχήν σύμφωνα με τα παραπάνω, εαν $ n \equal{} p_{1}^{a_{1}}\cdots p_{r}^{a_{r}}$, η ανάλυση του $ n$ σε πρώτους παράγοντες τότε $ \sigma(n) \equal{}\sigma_{1}(n) \equal{}\prod_{i \equal{} 1}^{r}(1\plus{}p_{i}\plus{}\cdots p_{i}^{a_{i}})$.

Αφού ο $ n$ δεν είναι άρτιος, άρα κανένας από τους $ p_{i}$ ($ i \equal{} 1,\ldots, r$) δεν είναι ο $ 2$. Άρα είναι όλοι περιττοί.

Σκέψου λοιπόν να υπήρχε κάποιος εκ των $ a_{j}$ που να ήταν περιττός. Tότε το αντίστοιχο άθροισμα $ 1\plus{}p_{j}\plus{}\cdots p_{j}^{a_{j}}$
 θα ήταν αριθμός άρτιος, άρα ολόκληρο το γινόμενο θα ήταν αριθμός άρτιος, άτοπο γιατί είναι περιττός.

Συνεπώς  $ a_{i}$ είναι άρτιος για κάθε $ i \equal{} 1,\ldots, r$, και ας υποθέσουμε $ a_{i}\equal{} 2a_{i}'$.

Τότε προφανώς $ n \equal{}\left(p_{1}^{a_{1}'}\cdots p_{r}^{a_{r}'}\right)^{2}$, δηλαδή τέλειο τετράγωνο.

Ελπίζω να σε διαφώτισα λιγάκι. Για εξάσκηση μπορείτε να δοκιμάσετε τις εξής:

[size=150][color=red]ΑΣΚΗΣΗ 1[/color] [/size]

Ο φυσικός $ \tau(n)$ είναι περιττός, αν και μόνο αν, ο $ n$ είναι τέλειο τετράγωνο και

[size=150][color=red]ΑΣΚΗΣΗ 2[/color] [/size] (Ευθύ και αντίστροφο της άσκησης που πρότεινε ο Φώτης)

Ο φυσικός αριθμός $ \sigma(n)$ είναι περιττός, αν και μόνο αν, ο $ n$ είναι τέλειο τετράγωνο ή διπλάσιο τέλειου τετραγώνου ακεραίου.

Αλέξανδρος[/b]

[quote="cretanman"]

[b]Αφού ο $ n$ δεν είναι άρτιος,[/b] άρα κανένας από τους $ p_{i}$ ($ i \equal{} 1,\ldots, r$) δεν είναι ο $ 2$. Άρα είναι όλοι περιττοί.

[/quote]

Αλέξανδρε καταρχήν σε ευχαριστώ για τον κόπο σου.Αν καταλάβα καλά  παίρνεις ως δεδομένο ότι ο $ N$ είναι περιττός , ενώ αυτό ζητείται να αποδειχθεί.Ισως δεν ήμουν σαφής στην εκφώνηση της άσκησης αλλά ζητειται να αποδειχθεί οτι ο $ N$ είναι και περιττός και τέλειο τετράγωνο.Ισως πάλι να το απέδειξες κάπου και να μη το κατάλαβα....

Στο περιττός είναι που κολλάω και εγώ...κατα τα άλλα έχω κάνει τις ίδιες σκέψεις με σένα.

Τελος πάντων ευχαριστώ και πάλι.

Για την 1 απλά χρησιμοποίησε τον τύπο για το τ(ν) και για να είναι περιττός πρέπει όλοι να είναι περιττοί άρα τα ρ1, ρ2 άρτια άρα τέλειο τετράγωνο. :wink:

Για τη 2 είπε τα περισσότερα ο cretanman. Είναι απλές και έχουν την ίδια φιλοσοφία. :wink:

Μπορούμε να αποδείξουμε ότι ο $ N$ είναι περιττός ως εξής:
Έστω ότι $ N\equal{}2^{p}q^{2}$ όπου $ q$ περιττός. Θα δείξουμε ότι $ p\equal{}0$.
Η σχέση $ \sigma(N)\equal{}2N\plus{}1$ γράφεται τώρα $ (2^{p\plus{}1}\minus{}1)\sigma(q^{2})\equal{}2^{p\plus{}1}q^{2}\plus{}1$, απ' όπου έπεται ότι $ 2^{p\plus{}1}\minus{}1/q^{2}\plus{}1$. Αυτό όμως είναι δυνατό όταν $ 2^{p\plus{}1}\minus{}1\equiv1\pmod{4}$, που ισχύει μόνο όταν $ p\equal{}0$.

Α ωραία. Σε ευχαριστώ!

Φώτη είχες δίκιο! Πριν λίγο επέστρεψα από το χωριό και το είδα. Όταν την είχα δει μου θύμισε την γνωστή άσκηση που έγραψα και γι'αυτό δεν καλοδιάβασα την εκφώνηση. Νόμιζα ότι ήταν εκείνη. Όμως ο Μιχάλης κάλυψε αυτό το κενό με την απόδειξή του και τον ευχαριστώ πολύ. Είναι αρκετά γνωστές τεχνικές αυτές με τις οποίες αντιμετωπίζουμε τέτοια προβλήματα χωρίς όμως να είναι πάντα τετριμμένες.

Αλέξανδρος


I got admitted to Stony Brook.
yeah. I know this school can't compare to what everyone is discussing about in this forum(e.g. MIT). But it's most likely where I will go.
After I did some research, the SBU's math club website is dead. Last problem of the month was in 2007 and the only page I can find something about Putnam was talking about a practice in 2006.
These make it seems like math is just courses + seminars and everyone just doesn't care about math if it can't earn them credits.
Also someone told me in those Math 3XX courses, there is usually less than 10 people in it. I mean, less than 10 people in courses like introduction to analysis? This make me wonder about the ability of SBU's math department.

Anyone here ever in Stony Brook can comment on it?

Although I live on LI, I don't attend Stony Brook.  I always thought of SB as more of a biology school (I could be completely wrong, I didn't even apply there).  But how good a department is depends on how much you can get out of it.  If all you want is to take the classes and get a degree, then it probably won't matter where you go.  I would get into contact with people who are at SB and ask them for their opinions.

Why are you concerned that there are only 10 people in an analysis class?  Most people at a university don't take analysis.  In my analysis course, there were 11 people total.  And it was great.

If you do go there next year, then there should be an active math club by that time. I have a friend who is a student there, and he is trying to restart the math club. Apparently it died out not too long ago when most of its active members graduated.

I don't know anything about the undergraduate experience there, but Stony Brook's math department is pretty good for research, especially geometry.  (And thanks to Jim Simons, it has a lot of dough, too.)  Or a better way to say it is that, in comparison to Stony Brook's general reputation as a university, it's math department's reputation is significantly better.  If I were to choose a public university in New York to attend on the basis of math alone, I would choose Stony Brook without question.

Small classes are usually a good thing, because it means more individualized attention.  Of course, if the other students are weak, that is not a good thing, but at least it would mean better opportunities to interact with faculty.

[quote="yenlee"]If I were to choose a public university in New York to attend on the basis of math alone, I would choose Stony Brook without question.[/quote]  My basic impressions are similar to yours, with the unhelpful additional comment that the quoted statement should probably depend on whether or not one considers Cornell to be a public university.

Cornell is a private university. However, Cornell has seven colleges, four of which are entirely privately funded and three of which are funded in part by the state of New York.

[b]Entirely privately funded:[/b]
College of Arts and Sciences -- [i]The Department of Mathematics is in here[/i]
College of Engineering
College of Architecture, Art, and Planning
School of Hotel Administration

[b]Funded in part by the state of New York:[/b]
College of Agriculture and Life Sciences
College of Human Ecology
School of Industrial and Labor Relations

Indeed, if you plan to attend Cornell for math, you'll be in the school of Arts & Sciences, which, as TZF pointed out, is private.  Hence you'll be paying full price.

[quote="yenlee"]

Small classes are usually a good thing, because it means more individualized attention.  Of course, if the other students are weak, that is not a good thing, but at least it would mean better opportunities to interact with faculty.[/quote]

Indeed Jim Simons from hedge fund Renaissance Technologies was chairman of the mathematics department at Stony Brook University. [url=http://www.bloomberg.com/apps/news?pid=20601109&sid=aq33M3X795vQ&refer=exclusive]Here[/url] is more information about his person. He is also funding [url=http://www.sns.ias.edu/csb/]The Simons Center for Systems Biology, Institute for Advanced Study, Princeton University.[/url]


Let x,y,z be postive numbers s.t. $xyz=1$. Prove that \[\frac{xy}{x^5+xy+y^5}+\frac{yz}{y^5+yz+z^5}+\frac{zx}{z^5+zx+z^5}\leq1\]

I met it several times. Anyway it is beautiful.
Show $ \frac {xy}{x^{5}+xy+y^{5}} \leq  \frac {z}{x+y+z} $

or $xy(x+y) \leq z(x^{5}+y{5})$ or $ x^{2}y^{2}(x+y) \leq (x^{5}+y^{5})$ that is clear from AM-GM.


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How to prove that there are infinitely many primes $ 1$ mod $ 3$ ?
I will try to imitate the prove of Euclides.
Suppose there is a finite number of primes $ 1$ mod $ 3$.
Define $ N$=$ 3*P_1*P_2*..*P_n\plus{}1$
then we have $ N$=$ 1$ mod $ 3$ and $ N$ is a solution of:
$ x^2\plus{}x\plus{}1$=$ 0$ mod $ 3$
Which means that: $ N$ is a solution of $ x^3$=$ 1$ mod $ 3$.
But Now I have to consider a prime number $ p$ which is a divisor of $ x$ and consider its unit group of order $ p\minus{}1$.
I don't know how to get further! Can anyone help me?

Consider the quadratic residues that a prime congruent to $ 2 \bmod 3$ cannot have and write $ N$ so you can derive a contradiction.

I don't think that there is any way to make the original proof work.  Surely all of $ N$'s factors could be $ 2\ (mod\ 3)$.  For example: $ 3\times7\times13\plus{}1\equal{}2\times137$.

Maybe this is what t0rajir0u was implying.

I mean the original proof doesn't work (although it will for primes congruent to $ 2 \bmod 3$) and you need to choose a better $ N$.  :)

[hide="In particular, use the fact that..."] $ \minus{}3$ is never a quadratic residue of a prime congruent to $ 2 \bmod 3$. [/hide]

I am not sure whether this can be done, correct me if I am wrong.

Suppose we had finite number of primes of form $ 3k\plus{}1$ . Then we can find a $ n\in \mathbb{N}$ so that every prime $ >n$ is of form $ 3k\plus{}2$. Consider the product $ P$ of all the primes $ \leq n$ except $ 3$.

case (i) If there are even number of primes of form $ 3k\plus{}2 \leq n$ select the smallest prime $ >n$ (which is obviously of the form $ 3k\plus{}2$ by assumption), call it $ p$. We have $ pP\minus{}1$ of form $ 3K\plus{}1$ which is not divisible by any prime less than it, contradiction.

case(ii) If there are odd number of primes of form $ 3k\plus{}2 \leq n$ , then we have $ P\minus{}1$ of form $ 3k\plus{}1$ which is not divisible by any prime less than it, contradiction.

If I've read your solution right, neither case excludes the possibility that the number you've come up with has prime factors larger than $ n$.

Here is the proof to which I am hinting:  suppose there are a finite number of primes congruent to $ 1 \bmod 3$ and let $ P$ be their product.  Then $ N \equal{} P^2 \plus{} 3$ cannot have a prime factor congruent to $ 2 \bmod 3$ (because of the fact I have previously cited); contradiction.

:)

Alternately, the same proof can be done using $ N \equal{} P^2 \plus{} P \plus{} 1$.  Note that $ 4N \equal{} (2P \plus{} 1)^2 \plus{} 3$.

There is no point beating around the bush.
I'll prove, that there are infinitely many primes $ \equiv 1\pmod{n}$ for all $ n\in\mathbb{N}$.
This is hardly more than an exercise in most algebra textbooks
and uses only some basics about [url=http://en.wikipedia.org/wiki/Cyclotomic_polynomial#Cyclotomic_polynomials]cyclotomic polynomials[/url].
The idea is to find primes $ p$, for which the field $ \mathbb{F}_p\equal{}\mathbb{Z}/p\mathbb{Z}$ contains some primitive $ n$-th root of unity.

Assume $ p_1,\ldots,p_m$ are all primes $ \equiv 1\pmod{n}$ and take some $ N\in\mathbb{N},N>2$ with $ np_1\cdots p_m\mid N$.
Then $ |\Phi_n(N)|\equal{}\prod_{k\equal{}1,\ldots,n\atop \gcd(k,n)\equal{}1}|N\minus{}e^{2k\pi i/n}|\ge (N\minus{}1)^{\varphi(n)}>1$.
So we find some prime $ p|\Phi_n(N)$, and then $ p|N^n\minus{}1$ gives $ p\nmid N$,
and therefore $ p\nmid n$ and $ p\not\equal{}p_k$ for $ k\equal{}1,\ldots,m$.
This leaves to prove $ p\equiv 1\pmod{n}$, i.e. $ n|p\minus{}1$.
(look how this generalizes Euklids proof)

The coset $ \zeta: \equal{}N\plus{}p\mathbb{Z}\in \mathbb{F}_p$ is a root of the polynomial $ \Phi_n(X)|X^n\minus{}1\in\mathbb{F}_p[X]$.
This gives $ \zeta^n\equal{}1$, and if also $ \zeta^d\equal{}1$ for some $ d|n,d<n$,
then $ \zeta$ would be a multiple root of $ (X^d\minus{}1)\Phi_n(X)|X^n\minus{}1$,
and is therefore also root of the derivative $ {d\over dX}(X^n\minus{}1)\equal{}nX^{n\minus{}1}\in \mathbb{F}_p[X]$.
But then $ p\nmid n$ gives $ \zeta\equal{}0$ and the contradiction $ p\mid N$.

So we get $ \zeta$ of order $ n$ in the muliplictive group $ \mathbb{F}_p^\times$ of order $ p\minus{}1$.
And then [url=http://en.wikipedia.org/wiki/Lagrange%27s_theorem_%28group_theory%29]Lagrange[/url] gives $ n|p\minus{}1$.

Obviously all prime divisors of $ P(x)\equal{}\frac{x^3\minus{}1}{ x\minus{}1}$ are congruent to $ 1$ modulo $ 3$,hence it is enough to prove that the set
$ P\equal{}\{ p\in\mathbb{P},$such that $ p$ divides $ \frac{x^3\minus{}1}{ x\minus{}1}$ for some $ x\in\mathbb{N}\}$ is infinite.
But it is obvious,because $ P(p_1p_2\dots p_n)$ is pairwise coprime with $ p_1,p_2,\dots,p_n$,where $ p_i$ are prime divisors of $ P(x)$.

A search for cyclotomic polynomials would have revealed everything necessary  :wink:

I gave a proof with cyclotomic polynomials [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=155461]some time ago[/url], but I think my last step is a little messy.  I didn't notice how a generalization of Euclid's proof could be used, olorin - thanks :)

[quote="t0rajir0u"]I mean the original proof doesn't work (although it will for primes congruent to $ 2 \bmod 3$) and you need to choose a better $ N$.  :)

[hide="In particular, use the fact that..."] $ \minus{} 3$ is never a quadratic residue of a prime congruent to $ 2 \bmod 3$. [/hide][/quote]

If $ P\equal{}11$ then $ 5^2\equal{}3$ mod $ 11$ and $ 11 \equal{}2$ mod $ 3$.
if   $ P\equal{}13$ then $ 4^2\equal{}3$ mod $ 13$ and $ 13 \equal{}1$ mod $ 3$.
So there is some  $ p$ congruent to $ 2 \bmod 3$ or $ 1 \bmod 3$ such that $ p$ is a quadratic residue.

$ 3$ and $ \minus{}3$ are two different things  :wink:

[hide]

Suppose that there are a finite number of primes of the form $ 3n+1$, say $ \alpha_{1}, \alpha_{2},..., \alpha_{k}$. Then $ P = \alpha_{1}*...*\alpha_{k} = 3T + 1$ for some $ T > 1$. (can show by induction easily)

Consider $ \beta = (2P)^{2} + 3$. $ \beta$ is of the form $ 3n+1$ so it can not be prime ($ \beta > \alpha_{i}$). Let $ q$ be a prime dividing $ \beta$. Since $ \beta$ is odd, $ q \neq 2$. If $ q \equiv 0 (mod 3)$ then we obtain a contradiction since $ (2P)^{2} \equiv 1 (mod 3)$. If $ q \equiv 1 (mod 3)$ then we have a contradiction since $ q$ can not be one of the $ \alpha_{i}$ and then we would have found another prime of the form $ 3n+1$.

So $ q \equiv 2 (mod 3)$. $ \beta = 4(P^{2})+3 \equiv 4(3m+1) + 3 \equiv 12m+7 \equiv 7 (mod 12)$ for some positive integer $ m$. Suppose that $ q \equiv 1 (mod 4)$. Then $ (\frac{-1}{q}) = 1$. Note that $ (2P)^{2} \equiv -3 (mod q)$. Since $ (\frac{-3}{q}) = (\frac{-1}{q})(\frac{3}{q})$ and $ (\frac{-3}{q}) = (\frac{(2P)^{2}}{q}) = 1$, it follows that $ (\frac{3}{q}) = 1$. By quadratic reciprocity ($ q \equiv 1 (mod 4)$) we have that $ (\frac{q}{3}) = 1$. But the only quadratic residue $ mod 3$ is 1 and $ q \equiv 2 (mod 3)$, giving a contradiction.

Suppose on the other hand that $ q \equiv 3 (mod 4)$.  Since $ q \equiv 2 (mod 3)$ this means that $ q \equiv 11 \equiv -1 (mod 12)$. If $ q$ is a prime dividing $ \beta$ than it must be of the form $ 4k+3$ since none exist of the form $ 4k+1$. $ \beta$ must be a product of primes that are congruent to $ -1 (mod 12)$. Thus we can write $ \beta = q_{1}*...*q_{r}$ where $ q_{i} \equiv -1 (mod 12)$. If $ r$ is odd then $ \beta \equiv -1 (mod 12)$. If $ r$ is even then $ \beta \equiv 1 (mod 12)$. But $ \beta \equiv 7 (mod 12)$ giving a contradiction either way.

Thus $ \beta$ must have a divisor which is a prime of the form $ 3n+1$ and is not one of the $ \alpha_{i}$. This implies that there are an infinite number of primes of the form $ 3n+1$.

[/hide]

Or, i guesss, you can just say:"  By dirichlet's theorem."  Because this is a corollary of dirichlet's theorem.

But how unsatisfactory is that?  See my discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=203526]here[/url].  (By the way, does anyone know if the answer to the question I raise is affirmative?)

Here's a proof that there are infinitely many primes congruent to $ 1 \bmod{p}$

Since $ (a)$ is a subgroup of the multiplicative group $ \mod{a^p \minus{} 1}$, it follows that $ p | \phi{(a^p \minus{} 1)}$ by Lagrange's Theorem (this is a classic problem from Herstein). Suppose there are finitely many primes congruent to $ 1 \bmod{p}$, say all of them are $ p_1, p_2, ..., p_n$. Then let $ a \equal{} p p_1 p_2 ... p_n$. It follows that $ a^p \minus{} 1$ does not have any of them as prime factors, but $ a^p \minus{} 1$ must have a prime factor congruent to $ 1 \bmod{p}$ since $ p | \phi{(a^p \minus{} 1)}$, meaning that there is another prime congruent to $ 1 \bmod{p}$, and we are done.


linear algebra
matrix
geometry
linear algebra solved

I have tried filling in just random 2x2 matrices for the following problem, but I always get that they are equal.
The problem is:

Find 2x2 matrices for A and B such that (A+B)^2 DOES NOT EQUAL A^2 + 2AB + B^2

Please help me get on the right thinking path to start this instead of just randomly picking matrices

Try with 2 matrixes which don't commute.

First of all I want to apologize for posting this question in the wrong forum...I didn't realize it until it was too late :(

In response to the suggestion of using matrices that don't commute:  Could you give me an example of two by two matrices that don't commute.  I guess I am missing that part of the concept and that is what is making it hard for me.  I thank you for any help you can give in advance.

All $2x2$ matrices that commute have the property $B=\alpha A+\beta I_2$ . The choices that you must make are obvious now.

If you note Eij the canonic base of the set of matrix n*n.
Then you have Eij*Ekl=0 if j<>k else =Eil
It should be easy to find a counter-example.

I tried some rational numbers and was able to get it to work out so that (A+B)^2 DID NOT = A^2 + 2AB  + B^2.  I also checked my answers in my graphing calculator to make sure I did it correctly.  Now I have a new problem I can't figure out.  I have worked it out on three sheets of paper.  I need help on it, but I need it told to me in very elementary terms.  I am preparing to teach high school geometry, so linear algebra is killing me!

My new problem will be posted in unsolved problems this time.  Please help.  I appreciate any help I receive.

[/b]

Okay.

So you want matrices for which $(A+B)^2\not=A^2+2AB+B^2$.

By definition, $(A+B)^2=A^2+AB+BA+B^2$.  So $(A+B)^2\not=A^2+2AB+B^2$ if and only if $AB\not=BA$.

Let $A=\left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$ and $B=\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$

What are $AB$ and $BA$?

[b]

Thank you so much for asking me what are AB and BA.

So, I end up with (A + B)^2 ?=? A^2 + 2AB + B^2

(A+B)^2 =
[1 2
 0 0]

and

A^2 + 2AB + B^2 =
[1 3
 0 0]

(A + B)^2 DOES NOT EQUAL A^2 + 2AB + B^2 :lol: [/b]


Visto che l'OliForum è chiuso posto qua per parlare dei Giochi di Archimede di quest'anno...gli ultimi per me  :(

Ecco la mia griglia.

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