geometry
Locus
circle
Locus problems
IMO
IMO 1992

In the plane let $\,C\,$ be a circle, $\,L\,$ a line tangent to the circle $\,C,\,$ and $\,M\,$ a point on $\,L$. Find the locus of all points $\,P\,$ with the following property: there exists two points $\,Q,R\,$ on $\,L\,$ such that $\,M\,$ is the midpoint of $\,QR\,$ and $\,C\,$ is the inscribed circle of triangle $\,PQR$.

Let $C$ be tangent to $L$ at $X$ and $O$ be the center of $C$. Let $R$ be one of points that has the problems property. Let $P,Q$ be the vertices of triangle which has the problems property. Let $X'$ be on $L$ such that $MX=MX'$.Consider the THe circle $C'$ such that be tangent to $PQR$ externally . and let $OX$ intersects $C$ at $Y$ again. we know $P$ is center of homothety which convert $C$ to $C'$ and we know point $Y$ will conwert to $X'$ in this homothety . so $P,Y,X'$ are collinear and since $Y,X'$ are constant , so the locus of $P$ is aline passes through $Y,X'$ .

Let the point of tangency of $C$ with $l$ be $T$, and the reflection of $T$ across the centre of the circle be $S$, so that $ST$ is a diameter of $C$. Let the intersection of $PS$ and $QR$ be $X$. Then by a well known lemma, $QT=RX$, and therefore $TM=MX$. Since $T,M$ are fixed, $X$ must be the reflection of $T$ through $M$, and hence the locus of $P$ is the line $SX$ such that $P$ is on the other side of $S$ from $X$. It is easy to check that this is true.

Let line L be tangent to the circle at X. Point M is fixed, then so is the reflection of X over M, say Z. The circle itself is fixed then so is the point diametrically opposite X, say Y. It is well known that P,Z,Y are collinear. And since line XY is fixed, all points P lie on this line.

[asy]
size(8cm);
defaultpen(fontsize(10pt));

pair P = dir(120), Q = dir(210), R = dir(330), I = incenter(P,Q,R), T = foot(I,Q,R), M = (Q+R)/2, S = Q+R-T, Tt = 2I-T;

dot("$P$", P, dir(45));
dot("$Q$", Q, dir(270));
dot("$R$", R, dir(270));
dot("$T$", T, dir(270));
dot("$M$", M, dir(270));
dot("$S$", S, dir(270));
dot("$T'$", Tt, dir(240));

draw(Tt--Tt+dir(S--Tt)*1.2, red, Arrow);
draw(Q--P--R, blue);
draw(2Q-T--2R-S, Arrows);
draw(incircle(P,Q,R));
draw(T--Tt--S, blue+dashed);
[/asy]

Let $T$ be the point where $C$ touches $L$, $T'$ be the point diametrically opposite $T$ on $C$, and $S$ be the reflection of $T$ over $M$. We claim that the locus is a ray with endpoint $T'$ pointing toward the direction from $S$ to $T'$, not including $T'$.

It is well known that in order to satisfy the given conditions, we must have $P$, $T'$, and $S$ collinear (consider a homothety at $P$ that sends $C$ to the $P$-excircle, where $S$ is the point where the excircle touches $\overline{QR}$). Thus $P$ must lie on line $\overline{T'S}$. However, if $P$ is closer to $S$ than $T'$ or $P = T'$, it is impossible to draw the tangents $\overline{PQ}$ and $\overline{PR}$ such that $C$ is the incircle.

To show that all points in the locus works, pick any $P$ and let its tangents to $C$ meet $L$ at $Q$ and $R$. Since the midpoint $M'$ of $\overline{PQ}$ is the unique point such that $T$ is the reflection of $S$ over $M'$, we must have $M'=M$, completing the proof. $\blacksquare$

Let the point $T$ be the tangency point, and $E$ be the reflection of $T$ over $M$. Then, let $D$ be the antipode of $T$.

We claim that the locus is the points beyond $D$ on the ray $ED$. We first show such points work. It is well known that $PD\cap L$ is the excenter of $\triangle PQR$ (proof by homothety). Then, the midpoint of QR is the midpoint of $T$ and $PD\cap L = E$ which is $M$, so we're done.

We now show that only points on this ray work. Firstly, points on the other side of $\overline{DE}$ fail since $D$ must be in the interior of $\triangle PQR$. Then, for $P\notin DE$, note that $PD\cap L= E'$ is still the excenter of $\triangle PQR$. Then, the midpoint of $QR$ is the midpoint of $TE'$ which is not equal to the midpoint of $TE$ since $E\neq E'$, so these points don't work.

Thus, we 've shown that all points on the ray work, and all points not on the ray don't work, so we're done. $\blacksquare$.

Seems easy for even P1.
Let $A$ be center of $C$ and $B=L\cap C$ and $B'$ be reflection of $B$ over $M$ and $B_1$ be antipode of $B$ . So, because of " Diameter of incircle Lemma" we know that $P$ should be on line $MB_1$. So, locus of $P$ is $\overrightarrow{\rm B_1X}$ where $X$ is any point on  $\overrightarrow{\rm MB_1}$ such that $B_1$ lies between $M$ and $X$. The question "Why all points on this ray work?" can be easily answered like this : Take any point $P$ on this ray and draw tangents from $P$ to $C$ and intersect them with $L$ at $Q,R$. Then again because of "Diameter of Incircle Lemma" $M$ is midpoint of $QR$. So we are done!

Let $X=\mathcal{C}\cap\ell,$ $X'$ the antipode of $X$ with respect to $\mathcal{C},$ and $Y$ the reflection of $X$ over $M.$ We claim that $P$ lies on $\overline{X'Y}$ beyond $X'.$ Since $\mathcal{C}$ is the incircle of $\triangle PQR,$ we know $P$ must lie on $\overline{X'Y}$ by the Diameter of an Incircle lemma. Moreover, with this lemma, we can draw the tangents from $P$ (on $\overline{X'Y}$ beyond $X'$) to $\mathcal{C}$ to intersect $\ell$ at $Q$ and $R,$ noting that $MQ=MR.$ $\square$


G is a simple bipartite graph. Let $ v$ be the vertices and $ e$ the edges of the graph. Prove that: $ e \le \frac{{v^2 }}{4}$

Let's set $ V$ the set of all vertices. And $ V_1$, $ V_2$ the sets of vertices for the bipartition. $ v_1$ and $ v_2$ the cardinality of those sets.

At most each point of $ V_1$ is connected with all the points of $ V_2$, we wolud have $ v_1v_2$ edges. so

$ e\le v_1v_2\le \frac{(v_1\plus{}v_2)^2}{4}\equal{}\frac{v^2}{4}$.

As an extra: Find the max number of edges when $ v$ is odd, and the same conditions of above.

 :)


What is the meaning of the term CUBE ROOT OF UNITY
and what does $ \omega$ represent???

$ 1,\omega,$ and $ \omega^{2}$ are the three cube roots of unity. State the values of $ \omega^{3}$ and $ 1+\omega+\omega^{2}$. Hence simplify each of the expressions $ (1+3\omega+\omega^{2})^{2}$ and $ (1+\omega+3\omega^{2})^{2}$ and show that their sum is -4 and their product is 16

Roots of unity are the roots of the polynomial $ z^{n}=1$. When graphed on an Argand Diagram, they spread evenly around a unit circle to give a regular $ n-gon$. So cube roots of unity represent the roots to $ x^{3}=1$.

So in your case $ \omega = cis \frac{\pm 2 \pi}{3}$ so $ \omega^{3}=1$ and $ 1+\omega+\omega^{2}= 0$.

I assume your first expression is $ (1+3\omega+\omega^{2})^{2}= (2 \omega)^{2}= 4 \omega^{2}$

$ (1+\omega+3 \omega^{2}) = (2 \omega^{2})^{2}= 4 \omega^{4}= 4 \omega$

So their:

$ \text{SUM}= 4(\omega+\omega^{2}) =-4$

$ \text{PRODUCT}= 16 \omega^{3}= 16$.

ok i understand it now.

[url=http://img71.imageshack.us/my.php?image=71661118es5.png][img]http://img71.imageshack.us/img71/5051/71661118es5.th.png[/img][/url]

For this one? Do i Solve for z?

You could solve for $ z$. But as $ z$ is a third root of unity, it is a solution to $ 0 = z^{3}-1 = (z-1)(z^{2}+z+1)$. Therefore, either $ z=1$ or $ z^{2}+z+1 = 0$. So the two values are $ 1^{2}+1+1 = 3$ and $ 0$.


algebra
polynomial
inequalities
calculus
derivative
function
complex analysis

Find all complex polynomials $P(z)$ such that the sequence
\[
\Big\{P^{(n)}=\underbrace{P\circ P\circ\cdots\circ P}_{n\,\,\text{terms}}\Big\}_{n=1}^\infty
\]
converge uniformly on compact subsets.

On all compact subsets of $\mathbb{C}$ ¿

Then:

If the degree of $P$ is higher than $1$, there is some $r \in \mathbb{R}$, so that for all $z,|z|>r$ the inequality $|P(z)|>2|z|$ holds, so it doesn't even converge pointwise for such numbers.
So $P(z)=az+b$.

If $a=1$ one easy sees that $b=0$ and this gives a uniformly convergent sequence.

If $a \neq 1$:
Let be $x=\frac{b}{1-a}$ the solution of $ax+b=x$.
Then $P(z)-x = az+b-ax-b=a(z-x)$, so $|P(z)-x|=|a||z-x|$, so the distance to $x$ multiplies by $a$, so if:

$|a|>1$: the sequence diverges pointwise for all $z \neq x$

$|a|<1$: the sequence converges pointwise to $x$ for all $z$. But it also converges uniformly on compact subsets, since every compact subset is contained in a compact subset of the form $B_r(x)= \{ z | |z-x| \leq r \}$ for some $r \in \mathbb{R}$. And its easy to see that $|P^{(n)}(z)-x| = |z-x||a|^n \leq r|a|^n $ for all $z$ from that set.

$|a|=1$: Then
$|P^{(n)}(z) - P^{(n-1)}(z) |= |P^{(n)}(z) -x - (P^{(n-1)}(z)+x)| =$
$ = |a^n(z-x) - a^{n-1} (z-x)| = |a^{n-1}||z-x||a-1|=|z-x||a-1|$
so when $z \neq x$, then it doesn't converge pointwise since the absolute difference is constant and not $0$.

Very nice solution ZetaX!  :)

Here is another problem from UMCP:

(UMCP 2004) Let $f^{(n)}$ denote the $n$'th (complex) derivative of $f$. Find all entire function $f$ such that $\{f^{(n)}\}_{n=0}^\infty$ converges uniformly on all compact subsets of $\mathbb{C}$.


logarithms
floor function
factorial
geometry
3D geometry
MATHCOUNTS
function

[quote="kheg_k"]in order to replace R49917S's 24 and to supplement Klebian's 24 in case a 24 is particularly tough, i have created another 24. to get rules for the game, u can look up Klebian's 24's rules.

numbers: 6,9,6,5

new numbers will be posted a full day after a solution unless another solution is posted.[/quote]

$(6*9)-(6*5)$

[quote="kheg_k"]in order to replace R49917S's 24 and to supplement Klebian's 24 in case a 24 is particularly tough, i have created another 24. to get rules for the game, u can look up Klebian's 24's rules.

numbers: 6,9,6,5

new numbers will be posted a full day after a solution unless another solution is posted.[/quote]

$(9-6-5+6)!$

3,4,2,1

there are many for this one

points
ch1n353ch3s54a1l          1
TZ                                   1

(4+2) * (3+1)

[size=59]Are you just like using 24 game cards? (cuz this seems like a 1 dot)[/size]

4! * (3/[2+1])
And for my 24, for rules and playing, look in my signature.

$1\times2\times3\times4$ (I know it's just a backwords. Just felt like posting it)

I can't believe this is our third 24 in about 2 months. I quit on mine because Kleiban made his own one which took all the attention off mine. And now you made this one since sometimes his are too hard.

or you could go the other way: -3+4!+1+2

ch1n353ch3s54a1l 1
TZ 1
nonie 1
biffanddoc 1
klebian 1
slamchess 1
easyas3.14159... 1
X_wingredleader   1

new numbers:

2;7;2;3

sorry.. i think i just made up my own rules.
ok here is another one:
EDIT: sorry  :blush: i messed up again
i double-checked this one:
$7,31,255,2006$

EDIT2: another one with primes:
$3,7,19,23$

EDIT3: $3,4,5,10$ (i hope no one has done this before)

i was beginning to wonder if i had killed it :blush:
[quote="rem"]$3*4*\frac{10}{5}=24$[/quote]i was thinking of: $4^{5}-10^{3}$
[quote="rem"]$(23-19)!\times (7-3!)=24$[/quote]$\sqrt{19\cdot(23+7)+3!}$

[quote="rem"]Has anybody solved this one yet:
$7,31,255,2006$
 :maybe:[/quote]
ill post the solution if nobody has solved it within a few days.

EDIT:[hide="solution"]$(2006-7\cdot(31+255))!$[/hide]

Has this topic died?

For the first one:

$(3+2-1^{456})!$

$\sqrt{456+1(2+3)!}$

[quote="minsoens"]nice job SplashD
but the solution i first thought of is this: $\frac{115-19}{\sqrt{16}}$
hmm.. anyway, the numbers $1,15,16,19$ correspond to the letters A, O, P, S on the alphabet  :)[/quote]
You cant do ${\sqrt{16}}$ unless you have a 2.

can someone post all of the rules of 24 like what symbols you can use with what numbers so we don't have these conflicts?

*looks on Wikipedia*
The 24 Game is a mathematics game in which the object is to find a way to manipulate four numbers so that the end result is 24. Each set of four numbers is found on a game card. [b]In the original version, addition, subtraction, multiplication, or division could be used[/b] to make four digits from one to nine equal 24. For an example card with the numbers 4,7,8,8, a possible solution is the following: 8/8=1, 7-1=6, 6×4=24. There are currently nine official variations of the 24 Game. Robert Sun invented the 24 Game in 1988 and sold it through his company, Suntex International Inc. The tournament-style competition 24 Challenge is based on the 24 Game.

The original version tells you only to use addition, subtraction, multiplication, and division.  No square roots.  But we don't always have to follow the original rules, do we?

when I started this topic in the classrom math forum a while ago, I decided to use the rules of Klebian's 24 (with a few exceptions of scoring)

[b]Order of operations ARE required, and use parentheses and the like whenver necessary.
You may use exponents, add, subtract, multiply, divide, and each as many times as you like (i. e. you can do 27 - 1 -1 -1 as a solution). Square roots can be used, but you will need the number in front of the radical to do it.
And yes, you may use factorial.
Also, you can make 2 digit numbers. [/b]
You may only use each number once.

[quote="minsoens"]$10,132,213,321$
its pretty easy.. don't think too hard[/quote]
Yeah:
$\frac{132+321-213}{10}$


1 1 1 =6
2 2 2 =6
3 3 3 =6
4 4 4 =6
5 5 5 =6
6 6 6 =6
7 7 7 =6
8 8 8 =6
9 9 9=6
you are asked to put operators so that the expressions are true. you are allowed to use different operators on each line. i will just show the solution for the two most obvious ones.
2+2+2=6
6+6-6=6

3 times 3 -3=6

5 divided by 5 +5 =6

6 times 6 divided by 6=6

7-7 divided by 7=6

r u allowed to use operators other than add, subtract, times, and divide???

you are allowed to use any operator except  the ones with numbers like for example
powers, or logaritms of a certain base, or radicals of a certain degree.

can they be 3 in base 2 or something like that???

[quote="Geehoon"]I think he meant

$(\sqrt{8+\frac{8}8})! = 6$[/quote]
Yes, I meant that. Sorry :blush: .

congratulations! you got it!
myself when i got the test had some problems with the line of 1's cause i just couldn't think of using the "!" operrator. :)


Determine the triples of integers $(x,y,z)$ satisfying the equation \[x^{3}+y^{3}+z^{3}=(x+y+z)^{3}\]

[quote="neelnal"]
yx^2+xy^2+xz^2+zx^2+yz^2+zy^2=0
[/quote]

how'd you get that?

shouldn't it be like cyclic sum of xy^2 + 6xyz = 0?

[quote="andyciup"]Use the identity \[(x+y+z)^{3}-(x^{3}+y^{3}+z^{3})=3(a+b)(b+c)(c+a)\] and it is easy from here..[/quote]I knew I had to factor that somehow, but didn't find the factorization.  :blush:

[hide="Solution"]We start by simplifying. \begin{eqnarray*}0= (x+y+z)^{3}-(x^{3}+y^{3}+z^{3})&=& 3(x^{2}y+x^{2}z+xy^{2}+y^{2}z+xz^{2}+yz^{2}+2xyz) \\ &=& 3(x+y)(y+z)(x+z) \end{eqnarray*} This statement is true when $x=-y,-z$ or $y=-z.$ Hence, the solutions are $(x,y,z)=(m,-m,n),(m,n,-m),(n,m,-m)$ where $m$ and $n$ can be any integer. [/hide]

[quote="neelnal"][quote="Hokkage"][quote="neelnal"]This was a USAMO problem before was it not?

yx^2+xy^2+xz^2+zx^2+yz^2+zy^2=0

So, (0,0,0) is a solution.

We can also realize that plugging in z=0, we get that
all (n,-n,0) are solutions as are (0,n,-n) and (-n,0,n).[/quote]This doesn't prove that there aren't any other solutions, you've just found the trivial ones.[/quote]

actually, factor out yx^2+xy^2+xz^2+zx^2+yz^2+zy^2=0

i assumed that everyone would infer that step easily.[/quote]

The problem is that your expansion is incorrect, so your solution is not complete, as shown by the real solution by 4everwise.


geometry
perpendicular bisector
geometry solved

[quote="Thaakisfox"]In the $ABCD$ convex quadrilateral $ABD \angle =ACD \angle$. The midpoints of $BC$ and $AD$ sides are $E$ and $F$ respectivaly. The normal projection of the intersecting point of $AC$ and $BD$ diagonals on the $AB$ and $CD$ lines are $G$ and $H$. Prove that $EF \bot GH$.[/quote]

"Convex quadrilaterals ABCD satisfying < ABD = < ACD" are better known as "convex cyclic quadrilaterals ABCD", and the assertion of your problem can be extended as follows: Prove that EF is the perpendicular bisector of the segment GH.

This extension of your problem was proved at
http://www.mathlinks.ro/Forum/viewtopic.php?t=39107 and
http://www.mathlinks.ro/Forum/viewtopic.php?t=55177 .

  darij


I was wondering if there was any easy way to produce exact sequences in Latex.  For example, if I had a diagram with:

$ L \xrightarrow{\varphi} M \xrightarrow{\theta} N$

$ L' \xrightarrow{\varphi'} M' \xrightarrow{\theta'} N'$

And then the maps $ \alpha : L \rightarrow L'$, $ \beta : M \rightarrow M'$, $ \gamma : N \rightarrow M'$.

I want this to look something like

[code]
L ------> M ------> N
|         |         |
|         |         |
|         |         |
v         v         v
L'------> M' -----> N'
[/code]

I don't know how to solve your problem, but I will move this thread into the $ \text{\LaTeX}$-section ;)

Ah, thank you.  Whenever I visit this forum, I go straight into the College Playground, I never even knew there was a Latex section...  :blush:

[quote="Ricky"]I was wondering if there was any easy way to produce exact sequences in Latex.[/quote]The diagxy package will do this. See [url=http://sixthform.info/steve/wordpress/?p=28]Commutative Diagrams[/url] for examples and link to the documentation.

What a fantastic package, thank you very much.


USAMTS
AMC
USA(J)MO
USAMO
AIME
number theory

[quote="mustafa"]How hard is the USAMO compaired to the USAMTS?[/quote]

I vaugely remember somewhere where it says USAMTS is AIME level

And yet you get a month to do the USAMTS, and you can research stuff? If the USAMTS was AIME level, wouldn't you get a lot more perfect scores?

USAMTS is [i]far[/i] easier than USAMO.  There's no comparison.  I guess you could say it required about AIME-level heuristic ability but is more in-depth, but it isn't really like the AIME at all.  Plus, if you have all (or most) of the mathematical prerequisites used by the AMC, AIME, USAMO, etc., there isn't really much "research" you can do that will help you—it's mostly for people who don't know all that, so that they can find well-known theorems (e.g., Fermat's Little Theorem, which is part of standard knowledge but excluded from most school curricula) that will help them.

USAMTS has some really trivial problems occasionally, you can't really compare them to like MC or AMC because those are tests with strict time limits, and it takes some time to do them though they are trivial (such as 1.1 from year 18 and year 17).

And there are proof questions, which can be comparable to olympiads from countries who don't do extraordinarily well at IMO, easier olympiads.

However, some are alot like AIME problems, like 18.1.5, which I could definitely see as an AIME #9 or 10.

I figured that much, but I would think that the problems would be harder then AIME problems. I guess that the proofs make it harder.....

So it ranges from AIME 1-5 to a bit harder than the hardest AIME problems/easy olympiad problems?

You could say that...but the "olympiad" problems would show up on like...regional olympiads (bamo, sdmo, leningrad).

It's generally very hard to compare the difficulty of contests with radically different time/problem values. With a month for each problem set, the USAMTS is at the extreme end of this scale.

That said, a student who can reliably qualify for the USAMO shouldn't have any trouble solving USAMTS problems. That student should do the USAMTS (and other proof contests) to practice writing skills.

But if you get a 100 on USAMTS, it doesn't necessarily mean you will even get to AIME. I agree that it is much easier than USAMO, although, for me, it is mainly because we have over a month for 5 problems. :D

[quote="lotrgreengrapes7926"]But if you get a 100 on USAMTS, it doesn't necessarily mean you will even get to AIME.[/quote]
Um, no, because you can qualify for the AIME via USAMTS, remember? :wink:

ALL the usamts problems are #5-#15 AIME difficulty, and maybe 3 or 4 annually are USAMO difficulty, but overall, you can't compare USAMO and USAMTS because of time constraints, etc.


hello i am interested in physics olympiad problems .the first stage is nsep.
can i find question papers of previous years.i have found nsep 2000 on net.
can anyone help me

[quote="AndersoN"]What is 'nsep' ?![/quote]

it is the national standard examination in phyics conducted in india
i have found NSEP 2000

http://www.hbcse.tifr.res.in/olympiads/olyphy/NSEP2kpdf

here is one of the paper 2000 year.i am finding other papers too
especially 2005


quadratics
modular arithmetic
number theory
Perfect Squares
system of equations
IMO
IMO 1986

Let $d$ be any positive integer not equal to $2, 5$ or $13$. Show that one can find distinct $a,b$ in the set $\{2,5,13,d\}$ such that $ab-1$ is not a perfect square.

see $(mod.16)$ the numbers. the quadratic residues $(mod.16)$ are $0$,$1$,$4$,$9$.
suppose that $2d-1$,$5d-1$,$13d-1$ are perfect squares.
for $2d-1\equiv 0,1,4,9 (mod.16)$, we have $d \equiv 1,5,8 or 13 (mod.16)$.
if $d$ is $1$ or $13$, then $13d - 1$ is not one of these values.
If $d$ is $5$ or $9$, then $5d - 1$ is not one of these values.

contradiction.  :)

Assume $ 2d,5d,13d$ are all perfect squares. So $ d$ is divisible by $ 2,5,13$. Let $ d\equal{}130n$. Note that $ d/2,d/5,d/13$ are perfect squares, let the values be $ p^2,q^2,r^2$ respectively. So $ 65n\equal{}p^2,26n\equal{}q^2,10n\equal{}r^2$. Multiply the three equations, we get $ 130^2n^3\equal{}(pqr)^2$. Therefore $ n$ is a perfect square. But this implies that $ 2d\equal{}260n$ is not a perfect square. Contradiction.

[quote="Johan Gunardi"]Assume $ 2d,5d,13d$ are all perfect squares. So $ d$ is divisible by $ 2,5,13$. Let $ d \equal{} 130n$. Note that $ d/2,d/5,d/13$ are perfect squares, let the values be $ p^2,q^2,r^2$ respectively. So $ 65n \equal{} p^2,26n \equal{} q^2,10n \equal{} r^2$. Multiply the three equations, we get $ 130^2n^3 \equal{} (pqr)^2$. Therefore $ n$ is a perfect square. But this implies that $ 2d \equal{} 260n$ is not a perfect square. Contradiction.[/quote]

We have to assume $ 2d\minus{}1, 5d\minus{}1, 13d\minus{}1$ are all perfect squares. :huh:

it is IMO1986 first prob..and it is also exist in pen
[url]http://www.mathlinks.ro/viewtopic.php?t=150420[/url]

[quote="Complex_Ninja"]Assume $ 2d\minus{}1, 5d\minus{}1, 13d\minus{}1$ are perfect squares
Then $ (2d\minus{}1)(5d\minus{}1)(13d\minus{}1) \equal{} x^2$ for some $ x \ge 0$
$ (2d\minus{}1)(5d\minus{}1)(13d\minus{}1) \equal{} 130 d^3\minus{}101 d^2\plus{}20 d\minus{}1$
$ 130 d^3\minus{}101 d^2\plus{}20 d\minus{}1 \equiv \minus{}d^2\minus{}1 \bmod 5$
The residues $ \bmod$5 are $ {1,4}$
Then $ \minus{}d^2\minus{}1 \equiv \minus{}2, 0 \bmod 5$, but since this equals $ x^2$ the residues$ \bmod 5$, have to match. Contradiction.[/quote]

I dont see the contradiction in your argument.

-2 mod 5 is not equal to 1 or 4 mod 5 and 0 mod 5 is not equal to 1 or 4 mod 5. since x^2 is a perfect square it is either 1 or 4 mod 5. but in this case it is -2 which is also 3 or 0 mod 5. so contradiction.

edit: oops. dont know how i missed that 0 is not a quadratic residue mod 5. sorry. :blush:

I don't mean to revive this topic, but I was wondering if this was a valid proof (I'm pretty sure it is, but just in case I missed something...):

[hide="My Solution"]Temporarily assume $2d-1, 5d-1,$ and $13d-1$. All square numbers are congruent to either $1$ or $0$ mod $4$.

Notice that $2d-1$ must be odd, since $2d$ is even. This means it is congruent to $1$ mod $4$, and subsequently, $2d$ is congruent to $2$ mod $4$. This means $d$ is odd. Now let $5d-1 = y^2$ and $13d-1 = z^2$. Then $z^2 - y^2 = 8d$. $y^2$ and $z^2$ are both congruent to $0$ mod $4$ since $5d-1$ and $13d-1$ are even. This allows us to write $y^2 = 4m^2$ and $z^2 = 4n^2$. So we have $4n^2 - 4m^2 = 8d \rightarrow n^2 - m^2 = 2d $. Now, since $2d$ is even, $n^2$ and $m^2$ must be both even or both odd. Either ways, $n^2 - m^2 \equiv 0 mod 4$ since it is either $1$ mod $4 - 1$ mod $4$ or $0$ mod $4 - 0$ mod $4$. This means that the LHS is divisible by $4$. However, the RHS is only divisible by $2$ since $d$ is odd. Thus, there is a contradiction, and not all of them may be squares.[/hide]

Perfectly correct, flare. If you will take a look at the link mentioned above your post (pointing to the problem in PEN), you will see a solution very much along your line.

[hide="Solution"]
Say there does exist some $d$, with $2d-1=a^2, 5d-1=b^2, 13d-1=c^2$. Then we see that $a$ is odd, and thus $2d\equiv 2\mod 8$. and so $d\equiv 1\mod 4$. Let $d=4k+1$, Then we see that:
\[8k+1=a^2, 20k+4=b^2, 52k+12=c^2\]
 So $2|b, 2|c$, and let $b=2b_1, c=2c_1$. Then the equations are:
\[8k+1=a^2, 5k+1=b_1^2, 13k+3=c_1^2\]
So $a^2+b_1^2+1=c_1^2$. But $a^2\equiv 1\mod 4$, so $b_1^2+2\equiv c_1^2\mod 4$, clearly impossible. So no such $d$ exists.
[/hide]

To the contrary let us assume that $13d-1,5d-1,2d-1$ are squares and let
$13d-1=k^2$....(1)
$5d-1=l^2$.......(2)
$2d-1=m^2$.....(3)

(1) $\Rightarrow 13d \equiv 1,2\pmod{4} \Rightarrow d \equiv 1,2\pmod{4}$.

$d \equiv 1\pmod{4} \Rightarrow d=4k+1$ for some $k \in \mathbb{Z}$.Then (2) $\Rightarrow 20k+4=4(5k+2)=l^2 \Rightarrow (\frac{l}{2})^2 \equiv 3\pmod{5}$.Contradiction!

$d \equiv 2\pmod{4} \Rightarrow d=4k+2$ for some $k \in \mathbb{Z}$.Then (3) $\Rightarrow m^2 \equiv 3\pmod{8}$.Contradiction!

We have acheived a contradiction!

Taking $A=a^2, B=b^2, C=c^2$ in your system,[b] it does not easily solve [/b]to give $A=B=C=-1$. The system is indeterminate (compute its determinant, and see it is null). The general solution is $C=X, B = \dfrac {5C - 8} {13}, A = \dfrac {10C - 55} {65}$, and[b] indeed [/b]it has the [b]particular[/b] solution $A=B=C=-1$ for $X=-1$, but it also has infinitely many others (in rationals), and you have to show that none yields $A,B,C$ perfect squares.

Assume that there is exist $d$ such that for every 2 distinct $a,b \in \{2, 5, 13, d\}$, $ab-1$ is perfect square.
So there are exist $a,b,c\in\mathbb{N}$ such that $2d-1=a^2$, $5d-1=b^2$, $13d-1=c^2$.
Clearly see that $a$ is odd so $2d-1\equiv 1\pmod 4$ that means $d$ is odd too. So, $b,c$ is even.
There are exist $r,s\in\mathbb{N}$ such that $2^r\parallel b$ and $2^s\parallel c$
If $r<s$ then $2^{2r}\parallel b^2-c^2$. Because $2^3\parallel 8d$ and $b^2-c^2=8d$. So $2r=3$ that is impossible.
If $r>s$ then $2^{2s}\parallel b^2-c^2$. In the same way, we get $2s=3$ that is impossible too.
If $r=s$ then assume that $b=2^rm,c=2^rn$ when $m,n$ is odd. So $8d=b^2-c^2=2^{2r}(m^2-n^2)$
Because $m^2\equiv n^2\equiv 1 \pmod 8$. So $8\mid m^2-n^2$. That means $2^{2r}$ is odd so $r=0$ that is impossible. Contradiction.

[quote=supercomputer]Is this valid:
[hide="solution"]
Considering mod 4, we find that $d\equiv1\pmod{4}$.  Now, let $d=4k+1$.  We have $\{2d-1,5d-1,13d-1\}=\{8k+1,4(5k+1),4(13k+3)\}$. Considering mod 4 again, we have a contradiction because for all possible residues of $d$ mod 4, at least one of the numbers $\{8k+1,5k+1,13k+3\}$ is not 0 or 1.
[/hide][/quote]

I don't think this works because the first term can be $1\pmod{4}$ and the other two can be $0\pmod{4}$ in the set $\{8k+1,4(5k+1),4(13k+3)\}$.

Providing with a very beginner solution (which I solved in my Art Period)  :D :

[b][u][size=150]SOLUTION[/size][/u][/b]

[hide=Claim 1]The square of any number is always of $4k$ or $8k+1$ type.[/hide]

[hide=Proof]
Let $a=2n+1$ be an odd number.

Then,

$a^2=(2n+1)^2=4n^2+4n+1$

$a^2=4n(n+1)+1$

$n(n+1)$ is divisible by $2!=2$

So,$a^2=4(2k)+1=8k+1$

Now, Let $b=2n$ be an even number.

Then,

$b^2=(2n)^2=4n^2=4k$[/hide]

This claim is only used in the whole solution.

We note down all possiblities of $ab-1$ where $a,b \in (2,5,13,d)$:

$(9,64,25)$ and $(2d-1,5d-1,13d-1)$

The first set is a perfect square. Let us have a look at the second set.

Assuming all the elements of the second set are perfect squares, we get:

$2d-1=a^2$     $(\clubsuit)$

$5d-1=b^2$     $(\bigstar)$

$13d-1=c^2$    $(\spadesuit)$

We know the LHS of $(\clubsuit)$ is an odd number.

So, $2d-1=8k+1\Rightarrow d=4k+1$

Replacing values of $d$ in $(\bigstar)$ and $(\spadesuit)$, we get:

$20k+4=4(5k+1)$ and $52k+12=4(13k+3)$.

Thus, $(5k+1) \equiv (k+1) \equiv {0,1}\pmod{4}$.

so, from $(\bigstar)$, we get $k \equiv {-1,0}\pmod4$.

Similarly, $(13k+3) \equiv (k-1) \equiv {0,1}\pmod{4}$.

so, from $(\bigstar)$, we get $k \equiv {1,2}\pmod4$.

But, these two cannot hold together for any $d\in\mathbb{N}$.

So, our assumption is wrong.

$\boxed{Q.E.D}$

This problem has been the bane of my existence for far too long.  I guess it's better late than never....

It suffices to show that at least one of $2d-1$, $5d-1$, or $13d-1$ is not a perfect square.  Suppose for the sake of contradiction that all of them are.  First observe that since $2d-1$ is a perfect square, $d$ must be odd (else $2d-1\equiv 3\pmod 4$).  Furthermore, $d$ divides each of $a^2 + 1$, $b^2 + 1$, and $c^2 + 1$; thus $d\equiv 3\pmod 4$ is impossible as well.  It follows that $d\equiv 1\pmod 4$.

So let $d = 4k + 1$.  Substitution implies that
\[
\frac{5d-1}4 = \frac{5(4k+1) - 1}4 = 5k + 1\quad\text{and}\quad \frac{13d-1}4 = \frac{13(4k+1) - 1}4 = 13k + 3
\]
are both perfect squares.  But now observe that these two numbers differ by $8k + 2$, which is an integer that is $2\pmod 4$; thus these two numbers cannot simultaneously be perfect squares, and we are done.

[hide=My (somewhat bashy) Solution]
We reframe the problem statement as follows
[Quote=reframation] Let $d$ be a positive integer not equal to $2,5,13$. Prove that $2d-1, 5d-1, 13d-1$ cannot all be perfect squares.[/quote]
It is easy to check that both the problem statements are equivalent.
Assume the contrary, that $2d-1,5d-1,13d-1$ are all perfect squares.
Let $2d-1=(2a+1)^2$. This gives
$d=2a^2+2a+1$
$5(2a^2+2a+1)-1=10a(a+1)+4$ is a perfect square, say equal to $4b^2$
This gives $a(a+1)=\frac25(b^2-1)$
Similarly,
$13(2a^2+2a+1)-1=26a(a+1)+12$ is a perfect square, say equal to $4c^2$.
This gives $a(a+1)=\frac{2}{13}(c^2-3)$
$\frac{2}{5}(b^2-1)=\frac{2}{13}(c^2-3)\implies 5c^2-13b^2=2$
As perfect squares are congruent to $0,1$ modulo $4$, we find that this equation has no solutions, a contradiction. Therefore, our assumption is incorrect. Hence, atleast one of $2d-1, 5d-1, 13d-1$ is not a perfect square.

[hide=Timepass][b]Solution.[/b] Assume the contrary and let
\begin{align*}
2d-1 &=a^2\\
5d-1 &=b^2\\
13d-1 &=c^2 \\
\end{align*}Note that if $b,c$ have same parity. We claim that both are even. Assume the contrary, then working $\pmod{4}$ gives $d-1\equiv 1 \pmod{4}$ so $d\equiv 2$. But then $$a^2=2d-1\equiv 2\cdot 2-1\equiv 3 \pmod{4}$$Contradiction. Therefore, $b,c$ are even. Then again, taking 2nd equation $\pmod{4}$ we get $d\equiv 1\pmod{4}$. Let $d=4\ell+1$ and $b=2b^*$ and $c=2c^*$ then all three reduces to
\begin{align*}
8\ell+1 &=a^2 \\
5\ell+1 &=(b^*)^2 \implies \ell+1 \equiv (b^*)^2 \pmod{4}~(\clubsuit)\\
13\ell+3 &=(c^*)^2 \implies \ell+3 \equiv (c^*)^2 \pmod{4}~(\spadesuit)\\
\end{align*} Clearly, $\ell\in\{0,3\}$ modulo $4$. However, if $\ell\equiv 0$ then $(\spadesuit)$ gives contradiction, otherwise $\ell\equiv 3$ then again $(\spadesuit)$ gives contradiction. Consequently, there exists no such $a,b,c$ sastisfying above systems which finishes the problem. $\blacksquare$[/hide]:whistling:

[Hide = The Longest solution to the simplest problem] FSOC Let $2d-1=A^2, 5d-1=B^2, 13d-1 =C^2$

$(A^2, B^2)=(3,d+1) = 1$ since if the GCD is $3$, it implies a contradiction $\implies (A,B)=1$

$(B^2,C^2)=(8,d+3)=4,1 = x$

If $x=4, B=2p, C=2q, (p,q) = 1$. Notice that $d$ has to be odd for the conditions to hold.
We get $(C+B)(C-B) = 8d \implies (q+p)(q-p)=2d$. But if q is odd and p is even or vice versa $(q+p)(q-p)$ is odd, whereas if both are odd then $(q+p)(q-p)$ is divisible by $4$, contradiction. Therefore we are left with the case $(B^2,C^2) =1$. But this is a contradiction in itself since $B, C$ are even so the gcd must be $\geq 4$.
[/hide]

QED

We will show that amongst $2d-1,5d-1,13d-1$ there is a number that is not a perfect square. For the sake of contradiction assume that there doesn't, and let \[p^2 = 2d-1;\;q^2 =5d-1;\;r^2 = 13d-1\] we then have that \[\frac{p^2+1}{2} = \frac{q^2+1}{5}\iff 5p^2+5 = 2q^2+2\iff 2q^2-5p^2=3\] However, running through the fact that $x^2\in\{0,1\}\mod{4}$ we have that this is impossible. Thus we have a contradiction and we are done.

[quote=DrYouKnowWho]we then have that \[\frac{p^2+1}{2} = \frac{q^2+1}{5}\iff 5p^2+5 = 2q^2+2\iff 2q^2-5p^2=3\] However, running through the fact that $x^2\in\{0,1\}\mod{4}$ we have that this is impossible. Thus we have a contradiction and we are done.[/quote]

Something must be all wrong here, since $(p,q)=(1,2)$ actually is a solution.

I did with mod $3$. I am not sure of it though
[hide= Solution] If $ab-1$ is not a perfect square then $ab-1 \equiv 2 (mod 3)$ $\implies$ $ab \equiv 0 (mod 3)$
Therefore let's say $d \equiv 0 (mod 3)$. Hence $3d,5d,13d$ satisfy the condition and we are done :)
[/hide]

Let's assume that ab-1 is a square
We know that all perfect squares are either of form 4k or 8k+1
(i) If ab-1 =4k then ab=4k+1
    But this isn't possible since for a=2, ab is even
(ii)If ab-1=8k+1 then ab=8k+2
    This implies d is odd as 2d=2(4k+1)
    But then 5d ,13d are odd, making ab is odd
Hence we have a contradiction in each case, our assumption is wrong
Therefore we can always have d in the set for which ab-1 is not a perfect square.

Was just wondering if this solution works... can anyone figure out any faults or shortcomings?

Suppose not.

Then $2d-1$, $5d-1$, and $13d-1$ are all perfect squares.

We note that the quadratic residues $\pmod {16}$ are $0,1,4,9$.

Case 1: $d$ is even.
Then $2d-1\equiv 3\pmod4$, and thus not a perfect square.

Case 2: $d$ is odd.
Then $5d-1$ and $13d-1$ are both even perfect squares. Since they are distinct $\pmod {16}$, either $8d\equiv 4\pmod {16}$ or $8d\equiv 12\pmod{16}$. Both are not possible as they both imply $8\nmid 8d$.

Suppose FTSOC that $2d-1,5d-1,$ and $13d-1$ are all perfect squares; therefore, they are quadratic residues modulo $4.$ Then, $2d-1\equiv 0,1\pmod{4}$ or $d\equiv 1,3\pmod{4}.$ Also, $5d-1\equiv d-1\equiv 0,1\pmod{4}$ or $d\equiv 1,2\pmod{4}.$ Hence, $d\equiv 1\pmod{4},$ so let $d=4d'+1.$ We can then let $b^2=5d-1=20d'+4$ and $c^2=13d-1=52d'+12.$ Since $2\mid b,c,$ let $2b_1=b$ and $2c_1=c,$ and we know that $b_1^2=5d'+1$ and $c_1^2=13d'+3.$ Subtracting yields $$c_1^2-b_1^2\equiv 8d'+2\equiv 2\pmod{8}$$ which is not a difference of quadratic residues. $\square$

Suppose $2d-1, 5d-1, 13d-1,$ are squares. Then they are congruent to $0$ or $1$ in modulo $4.$ Notice that $2d-1$ implies $d$ is odd.
Setting $d=2k+1\implies 10k+4=d^2\implies b,k$ is even.
Setting $k=2h \implies 5k+1$ is a square.
Similarly we get that $13h+3$ is a square.
We can easily get a contradiction from the difference of two squares, $$(13h+3)-(5h+1)=8h+2. \text{Q.E.D.}$$

Here is my solution: [url]https://calimath.org/pdf/IMO1986-1.pdf[/url]
And I uploaded the solution with motivation to: [url]https://youtu.be/JS6xuvdYd5Y[/url]

here we go:

FTSOC assume $\exists$ $a,b \in \{2,5,13,d\}$ for which $ab-1$ is a perfect square, then:\\

$\alpha^2=2d-1$\\

$\beta^2=5d-1$\\

$\gamma^2=13d-1$\\

now $\alpha^2 \equiv 1 \pmod 4 \implies 2d-1=4m+1 \implies d=2m+1$ which is odd\\

now plugging this $d$ back into the $\beta^2$ and $\gamma^2$ we get $\beta^2=10m+4$ and $\gamma^2=26m+12$ now clearly this gives us that $\beta$ and $\gamma$ are even so set $\beta=2x$ and $\gamma=2y$\\

so we get $4(x^2-y^2)=8d \implies x^2-y^2 \equiv 2 \pmod 4$ which is a contradiction! as there is no $(x,y) \in \mathbb{Z}$ s.t. $x^2-y^2 \equiv 2 \pmod 4$\\

and hence we get $\exists$ $a,b \in \{2,5,13,d\}$ such that $ab-1$ is not a perfect square $\blacksquare$

Suppose FTSOC that there exists $d$ such any pair produces a perfect square. Then all three of $2d-1$, $5d-1$, $13d-1$ are perfects squares. We must have $2d-1\equiv 1 \pmod 4\Rightarrow d$ is odd, then $5d-1$ is even, and so $5d-1\equiv 0 \pmod 4 \Rightarrow d\equiv 1 \pmod 4$. Let $d=4k+1$ for some integer $k$. Inserting it into $5d-1$ and $13d-1$, we get that $20k+4$ and $52k+12$ are perfect squares, respectively, and they should remain such after being divided by $4$, i.e. $5k+1$ and $13k+3$ are perfect squares, but $$5k+1\equiv 0,1 \pmod 4\Rightarrow k\equiv 0,3 \pmod 4$$ $$13k+3\equiv 0,1 \pmod 4\Rightarrow k\equiv 1,2\pmod 4$$which is a desired contradiction.

Assume for the sake of contradiction that there exist $k_1^2=2d-1$, $k_2^2=5d-1$, $k_3^2=13d-1$ for positive integers $k_1$, $k_2$, $k_3$. Since squares are only 0, 1, 4 mod 8, we must have $2d-1 \equiv 1 \pmod 8 \implies d \equiv 1 \pmod 4$. Let $d=4a+1$. Substituting, we get $k_1'^2=8a+1$, $k_2'^2=5a+1$, $k_3'^2=13a+3$ for some positive integers $k_1'$, $k_2'$, $k_3'$. Then, note that $k_3'^2-k_2'^2=8k+2 \equiv 2 \pmod 4$, but this is impossible since squares are only 0, 1 modulo 4, contradiction. $\blacksquare$

I did differently from of all you guys, let me show:
FTSOC, suppose that all the combinations we try give us a perfect square. Firstly, if none of $a$ or $b$ is $d$, obviously give us all perfect squares:
$ab-1=\begin{cases} 2\cdot 5-1= 3^2 \\2\cdot13-1= 5^2 \\ 5\cdot 13-1= 8^2\end{cases}$ are all perfect squares.
Now, assume WLOG, $b=d$. Then:
$ab-1= \begin{cases}
    2d-1= q_1^2&\text{(1)}\\5d-1=q_2^2&\text{(2)}\\13d-1=q_3^2&\text{(3)}
\end{cases}$,  where $q_1, q_2,q_3\in\mathbb{Z^*_+}$
Now, multiply (1) by $65$, (2) by $26$ and (3) by $10$:
$\begin{cases}
    130d-65= 65q_1^2\\130d-26=26q_2^2\\130d-10=10q_3^2
\end{cases}$

So, $130d=65(q_1^2+1)= 26(q_2^2+1)= 10(q_3^2+1)$.
In particular, $26(q_2^2+1)= 10(q_3^2+1)\implies 13(q_2^2+1)= 5(q_3^2+1)$
As $13\nmid 5\implies q_3^2+1= 13\implies q_3^2= 12\implies q_3\not\in\mathbb{Z}$, contradiction!
Therefore, our supposition is false and we can always choose $a,b\in\{2,5,13,d\}$ such $(ab-1)$ is not a perfect square.


inequalities
function
search
Cauchy Inequality

1. $ x,y,z$ are (+) real nember such that $ x \plus{} y \plus{} z \ge 6$.

Prove that: $ \frac {{x^3 }}{{y \plus{} z}} \plus{} \frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{z^3 }}{{x \plus{} y}} \ge 6$

2.  $ x,y,z$ are (+) real nember such that $ x \plus{} y \plus{} z \ge a$. ($ a$ is (+) real number)

Find the minximum of $ A_n \equal{} \frac {{x^n }}{{y \plus{} z}} \plus{} \frac {{y^n }}{{x \plus{} z}} \plus{} \frac {{z^n }}{{x \plus{} y}}$ with $ n$ is natural number and $ n \ne 0$.

Nice problems!  :D
[hide="1"]By Hölder:
 $ (1 \plus{} 1 \plus{} 1)((y \plus{} z) \plus{} (x \plus{} z) \plus{} (x \plus{} y))(\frac {{x^3 }}{{y \plus{} z}} \plus{} \frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{z^3 }}{{x \plus{} y}})\geq (x \plus{} y \plus{} z)^3 \Leftrightarrow \frac {{x^3 }}{y \plus{} z} \plus{} \frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{z^3 }}{{x \plus{} y}}\geq \frac {(x \plus{} y \plus{} z)^2}{6}\geq 6$
Equality holds iff $ x \plus{} y \plus{} z \equal{} 6$ and $ x \equal{} y \equal{} z$, so $ x \equal{} y \equal{} z \equal{} 2$[/hide]

[hide="2"]Again by Hölder:
 $ (1 \plus{} 1 \plus{} 1)^{n \minus{} 2}((y \plus{} z) \plus{} (x \plus{} z) \plus{} (x \plus{} y))(\frac {{x^n }}{{y \plus{} z}} \plus{} \frac {{y^n }}{{x \plus{} z}} \plus{} \frac {{z^n }}{{x \plus{} y}})\geq (x \plus{} y \plus{} z)^n \Leftrightarrow A_n\geq \frac {a^{n \minus{} 1}}{3^{n \minus{} 2}\times 2}$
Equality holds iff $ x \equal{} y \equal{} z \equal{} \frac {a}{3}$[/hide]

Alternate Method:
[hide="Solutions"]
Define $ S\equal{}x\plus{}y\plus{}z$, and $ f(x)\equal{} \dfrac{x^3}{S\minus{}x}$ Therefore, $ f(x)$ is a convex function.

Hence, by Jensen We have:
$ f(x) \plus{} f(y) \plus{} f(z) \ge 3 f\left (\dfrac{x\plus{}y\plus{}z}{3} \right) \equal{} 3\left(\dfrac{\dfrac{S^3}{27}}{S \minus{} \dfrac{S}{3}} \right) \equal{} \dfrac{S^2}{6} \ge 6$

Number 2 follows similarly from Jensens, except $ f(x) \equal{} \dfrac{x^n}{S\minus{}x}$
This implies that: $ f(x) \plus{} f(y) \plus{} f(z) \ge 3 f\left (\dfrac{x\plus{}y\plus{}z}{3} \right) \equal{} 3\left ( \dfrac{ \dfrac{S^n}{3^n}}{\dfrac{2S}{3}}\right) \equal{}\dfrac{ S^{n\minus{}1} }{2(3^{n\minus{}2})} \ge \dfrac{a^{n\minus{}1} }{2(3^{n\minus{}2})}$[/hide]

and so we are done. :D

I agree, these are really nice!

[quote="Jarle10"][hide][quote="Nawahd Werdna"]Alternate Method:
[hide="Solutions"]
Define $ S \equal{} x \plus{} y \plus{} z$, and $ f(x) \equal{} \dfrac{x^3}{S \minus{} x}$ Therefore, $ f(x)$ is a convex function.[/quote][/hide][/hide]

Can you explain why you can do this? Isn't S a function of x as well, don't you treat it as a constant?[/quote]

I treat $ S$ as a constant because we can choose an arbitrary (constant) value $ k$, such that $ S \equal{} k \ge 6$ and then just vary $ x$, $ y$, and $ z$ to obtain the $ S$. Therefore, since we can choose any $ S$ (greater than $ 6$), we can choose any $ x$, $ y$, and $ z$, and the inequality will be true. So, in this case, treating $ S$ as a constant is ok.  :)

Possibly you may consider the following.
$ n \in \mathbb{N}$, simplify that
\[ E_n \equal{} \frac{a^n}{(a\minus{}b)(a\minus{}c)} \plus{} \frac{b^n}{(b\minus{}c)(b\minus{}a)} \plus{} \frac{c^n}{(c\minus{}a)(c\minus{}b)}\]

Example
$ E_0 \equal{} 0$
$ E_1 \equal{} 0$
$ E_2 \equal{} 1$
$ E_3 \equal{} a\plus{}b\plus{}c$

This identical equation is well-known relations called the fractional expression of Euler.
But, I was not able to find the URL which introduced this to.
Sorry  :(

[quote="Jarle10"][quote="Jorge Miranda"]Nice problems!  :D
[hide=1]By Hölder:
 $ (1 + 1 + 1)((y + z) + (x + z) + (x + y))(\frac {{x^3 }}{{y + z}} + \frac {{y^3 }}{{x + z}} + \frac {{z^3 }}{{x + y}})\geq (x + y + z)^3$[/quote]
Hi, I have only seen Hölders inequality with 2 factors, here you use 3. Can you explain why this is possible, and if it holds in general?

That is,

$ \sum (a_i b_i c_i ) \leq (\sum a_i^p)^{\frac {1}{p}}(\sum b_i^q)^{\frac {1}{q}}(\sum c_i^r)^{\frac {1}{r}}$, where $ {\frac {1}{p}} + {{\frac {1}{q}} + {\frac {1}{r}} = 1}$ ? Or perhaps with n factors?[/quote]

Yeah, it can be generalized to n factors  :D
See here http://www.mathlinks.ro/viewtopic.php?search_id=2062434983&t=102983 or here http://www.mathlinks.ro/viewtopic.php?search_id=974757376&t=211273

[quote="L_Euler"]1. $ x,y,z$ are (+) real nember such that $ x \plus{} y \plus{} z \ge 6$.

Prove that: $ \frac {{x^3 }}{{y \plus{} z}} \plus{} \frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{z^3 }}{{x \plus{} y}} \ge 6$
[/quote]cs and rearrangement: $ LHS\equal{}\sum_{cyc}\frac{x^2}{\frac{y}x\plus{}\frac{z}x}\ge\frac{(x\plus{}y\plus{}z)^2}{\sum_{sym}\frac{x}y}\ge\frac{(x\plus{}y\plus{}z)^2}{\sum_{sym}\frac{x}x}\equal{}\frac{6^2}{6}\equal{}6$

Q1: Use Cauchy Inequality, we have:

$ \begin{array}{*{20}c} {\frac {{x^3 }}{{y \plus{} z}} \plus{} \frac {{y \plus{} z}}{2} \plus{} 2 \ge 3.\sqrt [3]{{\frac {{x^3 }}{{y \plus{} z}}.\frac {{y \plus{} z}}{2}.2}} \equal{} 3x} \\
{\frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{x \plus{} z}}{2} \plus{} 2 \ge 3.\sqrt [3]{{\frac {{y^3 }}{{x \plus{} z}}.\frac {{x \plus{} z}}{2}.2}} \equal{} 3y} \\
{\frac {{z^3 }}{{y \plus{} x}} \plus{} \frac {{y \plus{} x}}{2} \plus{} 2 \ge 3.\sqrt [3]{{\frac {{z^3 }}{{y \plus{} x}}.\frac {{y \plus{} x}}{2}.2}} \equal{} 3z} \\
\end{array}$

So $ \begin{array}{l} \frac {{x^3 }}{{y \plus{} z}} \plus{} \frac {{y \plus{} z}}{2} \plus{} 2 \plus{} \frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{x \plus{} z}}{2} \plus{} 2 \plus{} \frac {{z^3 }}{{y \plus{} x}} \plus{} \frac {{y \plus{} x}}{2} \plus{} 2 \ge 3(x \plus{} y \plus{} z) \\
\leftrightarrow \frac {{x^3 }}{{y \plus{} z}} \plus{} \frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{z^3 }}{{y \plus{} x}} \ge 3(x \plus{} y \plus{} z) \minus{} 6 \minus{} (x \plus{} y \plus{} z) \\
\leftrightarrow \frac {{x^3 }}{{y \plus{} z}} \plus{} \frac {{y^3 }}{{x \plus{} z}} \plus{} \frac {{z^3 }}{{y \plus{} x}} \ge 2(x \plus{} y \plus{} z) \minus{} 6 \ge 6 \\
\end{array}$

"=" is happened when $ x \equal{} y \equal{} z \equal{} 2$.

Q2: Do you find the analogous solution with Q1?  :D  :D  :D

[quote="Jorge Miranda"]
Yeah, it can be generalized to n factors  :D
See here http://www.mathlinks.ro/viewtopic.php?search_id=2062434983&t=102983 or here http://www.mathlinks.ro/viewtopic.php?search_id=974757376&t=211273[/quote]

Thanks alot! But I am a bit sceptical to the notation $ a_{ij}$, shouldn't it be $ a_{i,j}$? So one could seperate - for example for $ a_{123}$ -
$ a_{1,23}$ from $ a_{12,3}$.

Am I right presenting it in this way:
$ \sum_j \frac {1}{p_j} \equal{} 1, \ a_{i,j},p_j > 0$
$ \sum^n_i (\prod^k_j a_{j,i}) \leq \prod^k_j (\sum^n_i a_{j,i}^{p_j})^{\frac {1}{p_j}})$ ?

Can we guess: "=" is happended when $ x \equal{} y \equal{} z \equal{} \frac{a}{3}$?


[quote="GameBot"]A regular polygon has sides of length 5 units and an exterior angle of 120 degrees. What is the perimeter of the polygon, in units?[/quote]

If the exterior angle is 120 degrees, then the number of sides is $ \frac {360}{x} \equal{} 120 \implies x \equal{} 6$. Thus the perimeter is $ 5 \times 6 \equal{} 30$.

EDIT: Oh yeah, because 360/120 isn't 6...

[quote="isabella2296"][quote="GameBot"]A regular polygon has sides of length 5 units and an exterior angle of 120 degrees. What is the perimeter of the polygon, in units?[/quote]

If the exterior angle is 120 degrees, then the number of sides is $ \frac {360}{x} \equal{} 120 \implies x \equal{} 6$. Thus the perimeter is $ 5 \times 6 \equal{} 30$.[/quote]

What?? The exterior angle is 120 degrees, so the interior is 60, and there are 3 sides, giving a perimeter of 15.


integration
calculus
calculus computations

Shouldn't this always equal 1?

$ \Gamma(z)^{\minus{}1} \int_0^\infty \frac{x^{z\minus{}1}}{e^x\plus{}1}\mathrm{d}x$

as it simplifies...

$ \Gamma(z)^{\minus{}1} [\minus{}\Gamma(z,\infty) \plus{} \Gamma(z, 0)] \equal{} \Gamma(z)^{\minus{}1}\Gamma(z) \equal{} 1$

[quote="JRav"]I don't quite understand your notation in the second part.[/quote]

Ahh, I think I found my mistake. I was thinking that $ \int \frac{x^{z\minus{}1}}{e^x\plus{}1}\mathrm{d}x\equal{} \minus{}\Gamma(z,x)$, when actually $ \int \frac{x^{z\minus{}1}}{e^x}\mathrm{d}x \equal{} \minus{}\Gamma(z,x)$. The +1 threw me off.


geometry
incenter
conics
hyperbola
analytic geometry
function

$P$ is on the right part of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $F_1$, $F_2$ are the focuses. Find the abscissa of the incentre of $\triangle{PF_1F_2}$.

[hide][img]http://mathworld.wolfram.com/images/eps-gif/HyperbolaFoci_750.gif[/img]
[color=red](in the sketch, the point P is on the left part)[/color]

$F_1F_2=2c$

$PF_1-PF_2=2a$

The incircle $(I,r)$ touches the sides $F_1F_2,PF_1,PF_2$ respectively at the points $A,B,C$.

$2a=PF_1-PF_2=(PB+BF_1)-(PC+CF_2) = BF_1-CF_2 = AF_1-AF_2$

$AF_1-AF_2=2a \Rightarrow$ The point $A$ satisfies the equation of the hyperbola

And $AF_1>AF_2$ so $A$ is on the right part. Also $A \in F_1F_2$ so the cordinates are $A(a,0)$

For any position of $P$ (such that $P \not \equiv A$) we have $IA \perp F_1F_2$, hence the abscissa of $I$ is fixed ($a$)[/hide]

Correct.

BTW, what did you use to draw the hyperbola?

Nothing!  :D  he he...

If you see the url of the picture, it is from mathworld

By the way, there are programs for geometry which allow to draw conics or , at least,  some locus.

Also, you can draw it with a program which makes the graph of a function

Thanks.
Can you recommend a programme to draw conics? I only have the geometer's sketchpad.


Hey, I'm having some trouble with this problem:

(−x2y2+4y3)dx+(4xy2−2y4)dy=0

It's a non-exact equation which can be made exact by finding an integrating factor. I found the integrating factor u=-2y using (Nx-My)/M, yet when I multiply each side by this, my My and Nx aren't equal. I must be making an error somewhere?

Thanks,
Matt

This is hard to read without some markup. Are those exponents; did you mean $ (-x^2y^2+4y^3)dx+(4xy^2-2y^4)dy=0$?

The usual convention for exponents in plain text is to use the ^ symbol. This is also the code in $ \text{\LaTeX}$, the pretty math here.

Sorry, and yes, that's exactly what I meant. I was scattered as it was for an assignment I was struggling to finish. Well, the assignment was due and I didn't receive credit for the problem, but I'm still very interested to find my error. Any help would be greatly appreciated!

You may assume an integrating factor $ I \equal{} x^{a}y^{b}$

In this case, you will find that $ I \equal{} \frac {1}{y^{2}}$ is a valid integrating factor.

Now your DE becomes $ \left( \minus{} x^{2} \plus{} 4y\right) dx \plus{} \left( 4x \minus{} 2y^{2}\right) dy \equal{} 0,$ which is exact.

Upon solving, you will obtain the solution

$ \minus{} \frac {1}{3}x^{3} \plus{} 4xy \minus{} \frac {2}{3}y^{3} \equal{} C$

Hope this explains.


When a number is reduced by $ 40\%$, the result is 36.  What is the original number?

Reduced by $ 40\%$ is the same as $ 60\%$ of the original number.  Then
$ .6x\equal{}36\Rightarrow x\equal{}60$.


find the coordinates of the points of intersection of the graphs of the equations y=abs(2x)-2 and y=-abs(2x)+2

[hide]10-10[/hide]

[hide]solving for x and y, (1,0) , (-1,0)?[/hide]

Yeah, that's what I get.

btw you can make the absolute value sign by hitting Shift+\.


Given an inscribed convex quadrilateral with sides of length m,n,p,q, taken in order around the quadrilateral,diagonals of length d and d',
prove that  sqrt(mp+nq)<=(1/2).(d+d')

It is direct consequence of Ptolemei theorem: $mp+nq=dd'$.


geometry
circumcircle
analytic geometry
graphing lines
slope
trigonometry
power of a point

Let $ABC$ be an isosceles triangle with $AB=AC$. Let $D$ be a point of $BC$ such that $BD=2DC$. At $AD$ we take a point $P$ such that angle$BPD=A$angle. Prove that angle$A=2DPC$ angle

Silouan, do you have the solution or not. This problem reminds me something. I think I have seen the solution somewhere.
  You can make a pdf file and a figure. Send it!
Babis

Take the point $H$ on $AB$ such that $AH=\frac{AB}{3}$

$\frac{AH}{AB}=\frac{DC}{BC}\Rightarrow  HD \parallel AC$, hence $\angle BHD=\widehat A \Rightarrow  \angle BHD=\angle BPD$.

Since $H,P$ are on the same side (half-plan) of $BC$ we have that $B,D,P,H$ are cyclic.

Let $C_1$ the circumcircle of $BDPH$.

Now bring a perpendicular $DF$ from $D$ to $BC$, where $F \in AC$

If $M$ the midpoint of $BC$, then $MD =\frac{MC}{3}$

and from parallel lines $AM,DF$ we get that

$AF=\frac{AC}{3}$. Also, $\angle DFC = \frac{1}{2} \widehat{A}$

Now we construct the circle $C_2$ with diameter $CF$.

$\angle CDF=90^0 \Rightarrow  D \in C_2$

The power of $A$ w.r.t. $C_1$ and $C_2$ respectively is  $AH \cdot AB$ and $AF \cdot AC$

But $AH=AF$ and $AB=AC \Rightarrow$

$A$ is on the radical axis of $C_1$ and $C_2$.

we have also that $D \in C_1 \cap C_2$

So the radical axis is the line $AD$.

But $P \in AD$ and $P \in C_1 \Rightarrow  P \in C_2$

So, $F,P,D,C$ are cyclic, hence the segment $DC$ is shown under equal angles from $P$ and $F$, that is $\angle DPC = \angle DFC = \frac{1}{2} \widehat A$

Nice. i have another solution. I will post it as soon as possible.
I take this problem from mr Stergiou book Mathematical Olympiads A lyceum page 155.

Indeed it is very nice

Hi everyone. :lol:
Lets give the solution silouan mentioned(from a book).
We extend $AD$ to $D$ and we take a point $E$ such that $PE=PB$. Then the

 triangle $PBE$ is isosceles  and similar to $ABC$. So $ACEB$ is inscribanle and so

angle$AEC$=angle$ABC$ , so $ED$ is bisector at the triangle $EBC$. So

$\frac{EB}{EC}=\frac{DB}{DC}=2$ so $EB=2EC$. Now if $M$ is the midpoint of $BE$

then $ME=\frac{EB}{2}=EC$ . So the triangles $PEM$ and $PEC$ are equal.So

angle$BPE$=angle$2MPE$=angle$2EPC$ so angle$A$=angle$2DPC$.

Really nice problem.  :lol:

[color=darkred][size=117]Very nice the problem and the its Pontios' proof ![/color][/size]

[quote="silouan"]$\{\begin{array}{c}\underline{\underline{\triangle ABC\ ,\ AB=AC}}\\\\ D\in (BC)\ ,\ BD=2\cdot DC\\\\ P\in (AD)\ ,\ m(\widehat{BPD})=A\end{array}\|$ $\implies$ $m(\widehat{DPC})=\frac{A}{2}$.[/quote]
[color=darkblue][b][u]Pontios' proof.[/u][/b] Denote the point $H$ for which $DH\parallel AC$. Thus, $AH=\frac{b}{3}$ and $\widehat{BHD}\equiv \widehat{BAC}\equiv\widehat{BPD}$ $\implies$ $\widehat{BHD}\equiv\widehat{BPD}$ $\implies$  the quadrilateral $BHPD$ is cyclically $\implies$ $AH\cdot AB=AP\cdot AD$ $\implies$ $\boxed{AP\cdot AD=\frac{b^{2}}{3}}\ \ (1)$. Denote the point $F\in (AC)$ for which $FD\perp BC$. Thus, $AF=\frac{b}{3}$, i.e.  $\boxed{AF\cdot AC=\frac{b^{2}}{3}\ \ (2)}$. From the relations $(1)$ and $(2)$ obtain that $AP\cdot AD=AF\cdot AC$, i.e. the quadrilateral $CDPF$ is cyclically. Therefore, $\widehat{DPC}\equiv\widehat{DFC}$. But $m(\widehat{DFC})=\frac{A}{2}$. In conclusion, $m(\widehat{DPC})=\frac{A}{2}$.

[size=117][b]Here are two ugly, but instructive proofs.[/b][/size]

[b][u]An analytical proof.[/u][/b] Let $a$ be a given positive number and suppose w.l.o.g. $\{\begin{array}{c}A(0,a)\ ,\ B(-3,0)\ ,\ C(0,3)\ ,\ D(1,0)\\\\ y=m(\widehat{ABP})=m(\widehat{CAD}\end{array}$

and $s(XY)$- the slope of the line $XY$. Thus, $\boxed{\tan\frac{A}{2}=\frac{3}{a}}$ and $\tan B=\frac{a}{3}$. Therefore, $\{\begin{array}{c}s(AD)=-a\\\\ s(AC)=-\frac{a}{3}\end{array}\|$ $\implies$

$\tan y=|\frac{s(AC)-s(AD)}{1+s(AC)\cdot s(AD)}|=\frac{2a}{a^{2}+3}$ $\implies$ $\boxed{\tan y=\frac{2a}{a^{2}+3}}$. Obtain $s(BP)=\tan (B-y)=\frac{\tan B-\tan y}{1+\tan B\cdot\tan y}=\frac{\frac{a}{3}-\frac{2a}{a^{2}+3}}{1+\frac{2a^{2}}{3(a^{2}+3)}}$ $\implies$

$s(BP)=\frac{a(a^{2}-3)}{5a^{2}+9}$. Thus, $\{\begin{array}{ccc}BP & \mathrm{\ : \ }& y=\frac{a^{2}-3a}{5a^{2}+9}\cdot (x+3)\\\\ AD & \mathrm{\ : \ }& y=-a\cdot (x-1)\end{array}\|$ $\implies$ $P\ : \ \{\begin{array}{c}x=\frac{a^{2}+9}{3(a^{2}+1)}\\\\ y=\frac{2a(a^{2}-3)}{3(a^{2}+1)}\end{array}\|$ $\implies$ $s(CP)=\frac{\frac{2a(a^{2}-3)}{3(a^{2}+1)}}{\frac{a^{2}+9}{3(a^{2}+1)}-3}$ $\implies$  $s(CP)=\frac{3-a^{2}}{4a}$.

Therefore, $\tan\widehat{CPD}=$ $|\frac{s(CP)-s(AD)}{1+s(CP)\cdot s(AD)}|=$ $|\frac{\frac{3-a^{2}}{4a}+a}{1-\frac{3-a^{2}}{4}}|=$ $\frac{3(a^{2}+1)}{a(a^{2}+1)}=\frac{3}{a}$ $\implies$ $\tan \widehat{CPD}=\frac{3}{a}=\tan\frac{A}{2}$ $\implies$ $m(\widehat{CPD})=\frac{A}{2}$.

[b][u]A trigonometrical proof.[/u][/b] $\{\begin{array}{c}B=90-\frac{A}{2}\\\ m(\widehat{DPC})=x\\\ m(\widehat{ABP})=m(\widehat{DAC})=y\end{array}\|$ $\implies$ $\{\begin{array}{c}m(\widehat{DAB})=A-y\\\ m(\widehat{DBP})=90-y-\frac{A}{2}\\\ m(\widehat{DCP})=90+y-x-\frac{A}{2}\end{array}$. Therefore,

$1\blacktriangleright \triangle BAC\ : \ 2=\frac{DB}{DC}=\frac{AB}{AC}\cdot \frac{\sin (A-y)}{\sin y}=\frac{\sin A-\cos A\cdot\tan y}{\tan y}$ $\implies$ $\boxed{\tan y=\frac{\sin A}{2+\cos A}}\ \ (1)$.

$2\blacktriangleright \triangle BPC\ : \ 2=\frac{DB}{DC}=$ $\frac{PB}{PC}\cdot\frac{\sin A}{\sin x}=$ $\frac{\cos(\frac{A}{2}+x-y)\cdot\sin A}{\cos(\frac{A}{2}+y)\cdot\sin x}=$ $\frac{\sin A}{\sin x}\cdot\frac{\cos(\frac{A}{2}+x)+\sin(\frac{A}{2}+x)\cdot\tan y}{\cos\frac{A}{2}-\sin\frac{A}{2}\cdot\tan y}$ $\implies$

$2\sin x\cdot(\cos\frac{A}{2}-\sin\frac{A}{2}\cdot\tan y)=\sin A\cdot[\cos(\frac{A}{2}+x)+\sin(\frac{A}{2}+x)\cdot\tan y]$ $\implies$ $\boxed{\tan y=\frac{2\sin x\cdot \cos\frac{A}{2}-\sin A\cdot\cos(\frac{A}{2}+x)}{2\sin x\cdot\sin\frac{A}{2}+\sin A\cdot\sin(\frac{A}{2}+x)}}\ \ (2)$.

The relations $(1)$ and $(2)\implies$ $\frac{2\sin x\cdot \cos\frac{A}{2}-\sin A\cdot\cos(\frac{A}{2}+x)}{2\sin x\cdot\sin\frac{A}{2}+\sin A\cdot\sin(\frac{A}{2}+x)}=\frac{\sin A}{2+\cos A}$ $\Longleftrightarrow$ $\frac{2\cos\frac{A}{2}\cdot[\sin x-\sin\frac{A}{2}\cdot\cos(\frac{A}{2}+x)]}{2\sin \frac{A}{2}[\sin x+\cos\frac{A}{2}\cdot\sin (\frac{A}{2}+x)]}=\frac{2\sin\frac{A}{2}\cdot\cos\frac{A}{2}}{2+\cos A}$ $\Longleftrightarrow$

$\frac{\sin x-\sin\frac{A}{2}\cdot\cos(\frac{A}{2}+x)}{\sin x+\cos\frac{A}{2}\cdot\sin(\frac{A}{2}+x)}=\frac{1-\cos A}{2+\cos A}$ $\Longleftrightarrow$ $\frac{3\sin x-\sin (A+x)}{3\sin x+\sin (A+x)}=\frac{1-\cos A}{2+\cos A}$ $\Longleftrightarrow$ $\frac{6\sin x}{2\sin (A+x)}=\frac{3}{1+2\cos A}$ $\Longleftrightarrow$

$\sin (A+x)=\sin x\cdot (1+2\cos A)$ $\Longleftrightarrow$ $\sin (A+x)=\sin x+\sin (A+x)-\sin (A-x)$ $\Longleftrightarrow$ $\sin (A-x)=\sin x$ $\Longleftrightarrow$ $x=\frac{A}{2}$, i.e. $m(\widehat{DPC})=\frac{A}{2}$.[/color]

[size=117][color=darkred][b]Remark.[/b] These proofs are hardly to please, but "the last wins" and the all is O.K.[/color][/size]

[quote][size=134][color=darkred][b][u]Generalization.[/u][/b] $\{\begin{array}{c}\underline{\underline{\triangle ABC}}\\\\ D\in (BC)\ ,\ \frac{DB}{DC}=\frac{c(b+c)}{b^{2}}\\\\ P\in (AD)\ ,\ m(\widehat{BPD})=A\end{array}\|$ $\implies$ $m(\widehat{DPC})=\frac{A}{2}$. For $b=c$ obtain the proposed problem ![/color][/size][/quote]

[color=darkblue][b][u]Remark.[/u][/b] If $DB=DC$, i.e. $b^{2}=c(b+c)$, then obtain the my proposed problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=57122 [/color]


What is the slope of the line determined by any two solutions
to the equation $ \frac{2}{x}\plus{}\frac{3}{y} \equal{} 0$? Express your answer as a common fraction.

Should I just do the algebra or is it intuitive?

Algebra is the only way I see how to do this.

Multiply both sides by $ xy$.

\[ 2y \plus{} 3x \equal{} 0 \]

Now set this up in slope-intercept form.

\[ y \equal{} \minus{}\frac{3}{2}x \]

Clearly, the slope is $ \boxed{\minus{}\frac{3}{2}}$.

This is simply the graph of the line of $ y\equal{}\frac{\minus{}3}{2} x$ with a hole at x=0.

Alteratively one can plug in numbers

at x=1 2/1+3/y=0 3/y=-2 y=-3/2

at x=2 1+3/y=0 so y=-3

(1,-3/2) (2,-3) so by the sllope formula $ \frac{\frac{\minus{}3}{2} \plus{}3}{1\minus{}2}\equal{}\frac{\minus{}3}{2}$


A,B and C are positive angles with sum 4pi/3

Find bounds on the value of (CosA)^2 +(CosB)^2 +(CosC)^2

British MO Year ?

... I don't think it's that simple. $\cos x$ isn't convex or concave on $[0,4\pi /3]$ and even if it was, it would only give one bound. I'm sure there is a trick such that Jensen can be used.


modular arithmetic
number theory
number theory proposed

Let $ A\equal{}\{n: 1 \le n \le 2009^{2009},n \in \mathbb{N} \}$ and let $ S\equal{}\{n: n \in A,\gcd \left(n,2009^{2009}\right)\equal{}1\}$. Let $ P$ be the product of all elements of $ S$. Prove that \[ P \equiv 1 \pmod{2009^{2009}}.\]
[i]Nanang Susyanto, Jogjakarta[/i]

[quote="Raja Oktovin"]Let $ A \equal{} \{n: 1 \le n \le 2009^{2009},n \in \mathbb{N} \}$ and let $ S \equal{} \{n: n \in A,\gcd \left(n,2009^{2009}\right) \equal{} 1\}$. Let $ P$ be the product of all elements of $ S$. Prove that
\[ P \equiv 1 \pmod{2009^{2009}}.\]
[i]Nanang Susyanto, Jogjakarta[/i][/quote]
Easy  :)
We have: if $ x\in S$ then $ 2009^{2009} \minus{} x \in S$
So: $ P\equiv ( \minus{} 1)^{\frac {|S|}{2}}$
We easy prove that $ |S| \equal{} \phi (2009^{2009}) \equal{} 4k$
So problem had proven.

this problem can be found here : PEN D12

Generalised Wilson's theorem states that: $\prod_{1\le x\le n, \gcd(x,n)=1} x\equiv 1\pmod{n}$ for $n\ne 4, p^{\alpha}, 2p^{\alpha}$ for odd prime $p$. Since $2009$ is composite, the result follows.


a,b,c- positive real numbers then prove that:
1/(a(b+1))+1/(b(c+1))+1/(c(a+1))>=3/(abc+1)


parameterization
conics
parabola
analytic geometry
function
trigonometry
calculus

can you explain why we consider the parametric form of equation for parabola etc
We introducing the parameters in equation is said to be a parametric form..what is the signifigance of these parameter?????give examples ..

         thank you

The parametric form of a curve allows you to see what the particle (usually, a curve is thought of as the path of some particle in space) does with respect to time.  The expression $ (x(t),y(t))$ tells you what each of the coordinates are doing independently as a function of time.  So while $ x^{2}\plus{}y^{2}\equal{}1$ and $ (\cos{t},\sin{t})$ both represent the same circle, the latter is in parametric form and allows us to see that both the $ x$ and $ y$ directions are oscillating as time moves forward.  Similarly, $ (\cos{t},\sin{t},t)$ is a helix.  How do we know this?  Well as I just pointed out, the first two coordinates trace out a circle.  The third coordinate tells us that as time moves forward, the particle simply moves upward; i.e., a helix.

we consider only the time as a parameter....

There is no generic "significance" of the parameter. It's just a tool that lets us express more curves in more ways than other forms.


[quote="Chocoboom"]In an isoceles triangle $ABC$, $AB = AC$. A line is drawn from $A$ to a point $D$ on line $CB$. Another line is drawn from $D$ to a point $E$ on line $AC$ so that $AD = DE$. If $\angle BAD = \alpha$ then find $\angle CDE$ in terms of $\alpha$[/quote]
[hide="Answer"] Is the answer $2\alpha-360$?[/hide]

[quote="lotrgreengrapes7926"][hide="Answer"] Is the answer $2\alpha-360$?[/hide][/quote]
Since $0<\alpha<180^\circ$, that gives a negative value...

Yeah, I just found a mistake in my solution.  :blush:

I don't think a definite relationship exists. There are many relationships depending on the degree measure of $\angle BAC$. For example, with trial and error, if $\angle BAC = 100$, then the sum of $a + \angle CDE$ is a constant 60 degrees. If the measure of $\angle BAC$ is 110 degrees, then the sum $a +\angle CDE$ is a constant 75 degrees with increments of 15 for each 10 degree increase of $\angle BAC$. This makes sense because clearly, $\angle BAC>60$ because you won't be able to construct a triangle that satisfy the condition of this problem if $\angle BAC \leq 60$.


algebra
polynomial
calculus
binomial theorem
calculus computations

i need to prove that:
n
sum (nCk)^2=2nCn
k=0

i tried using (1+x)^2n=(1+x)^n(1+x)^n

but i got frustrated, can someone please help me on this.

thanks in advance.

The coefficient of $x^n$ in $(1+x)^{2n}$ is (think binomial theorem)
\[ \binom{2n}{n}  \]
The coefficient of $x^n$ in $(1+x)^n(1+x)^n$ is
\[ \sum^n_{i=0}\binom{n}{i}\binom{n}{n-i}  \]
but since
\[ \binom{n}{n-i}=\binom{n}{i}  \]
we get
\[ \binom{2n}{n}=\sum^n_{i=0}\binom{n}{i}^2  \]

[quote="alon"]is this argument rigourous enough?[/quote]

Since two polynomials are equal iff the respective coefficients are equal, I would have to say "yes".

[quote="perfect_radio"][quote="kunny"]We can prove that in point of view combinatrics. :lol:[/quote]

Of course we can. It's the proof of blahblahblah translated into the language of combinatorics: $2n$ objects partitioned into $2$ groups of $n$. Picking $n$ of the total $2n$ can be done in $\binom{2n}{n}$ ways; else we take $0$ from the first group, $n$ from the second; then $1$ from the first and $n-1$ from the second etc.

@alon, polynomials can be treated as a sequence $(a_n)_{n \geq 0}$ which is eventually $0$. (the coefficient's sequence)

Thus, having the coefficients equal also means that they have the same number of terms.

BTW, what's this doing in the CALCULUS section?[/quote]
i'm asking here because the source of my question is a calculus book.

[quote="alon"][quote="blahblahblah"][quote="alon"]wait a minute, but polynomial equality needs not only coefficients equality but also terms equality.
e.g, 1+2x+x^2>1+2x
(bernouli inequality).

what i want to say is that also the number of terms has an impact here, is it not?[/quote]

I really have no clue what you're trying to say, but two polynomials with equal coefficients are equal.[/quote]
what i'm trying to say is, that take for example:
x^2+2x^3
and x^2+2x^3+x^4

the second polynomial is greater than the former although the coeeffients are identical.[/quote]

The coefficients are not identical; $x^2+2x^3$ is really the same as $x^2+2x^3+0x^4$, whence it is clear that they have different coefficients in the $x^4$ term.


Your answer is absolutely true.
I think you have seen this problem before.

Yes, I've seen and solved it before. Does that actually matter? :)


Merhaba Arkadaşlar,

Türkiye de Matematik Dergisi düşüneniz var mı?
Beraber çıkarabiliriz. Fakat sistemli çalışma gerekli... Vebol sabır.Bu işi en iyi yapan Sayın Ali Nesin Beydir.Matematik DÜnyası Dergisi  11000 satıyor. Bu müthiş bir durum...

Bende şu an fakülteden bağımsız bir Matematik dergisi çıkarmaya çalışıyorum.Şimdiden 60 sayfa tamam gibi...Yazar kadrosunda sürporiz isimler var...
Kendine özgü çalışmaları olanlar ban özel mesaj atabilir...
Sevgiyle kalın...

ihsan YÜCEL

Derginin kapak konusu [color=blue][b]FRAKTALLAR[/b][/color]!...
Hazırlayan Kemal Ilgar Eroğlu(Washington üniv. doktora öğrencisi)

merhaba
foruma bugün kayıt oldum
yazdıklarınızı okudum ve size hak veriyorum
ben bu işte varım...

Ihsanbey sizi tebrik ediyorum! Bende size katilma shansim var mi?
Dergii hangi dille yapmaya dusunuyorsunuz? Lutfen sorularima cevap verseydiniz iyiyi olurdu. Benimde bir kac projeler vardi onlarida kime gondermem lazim etc

saygilarimla
davron latipov

Sevgili Devran,
Öncelikle eğer kendine özgün çalışmaların varsa ve bizimle paylaşırsan bundan büyük memnuniyet duyarız. Kadromuzda çeşitli üniv. öğretim üyeleri ve üniv. öğrencileri var. Sen 1 ağustosa kadar çeşitli düzenlemelerini yaparak word formatında yazını ihsan_einstein@yahoo.com adresine yollayabilirsin. Fakat projen hakkında kısa bir bilgi sunarsan aynı e-posta adresine daha iyi olur.
Teşekkür ediyorum ve kolay gelsin diyorum.

Not: Danışma kurulunda TOBB üniv Yard.Doç.Dr.Hüseyin Merdan bulunmaktadır. Ayrıca dergi çıktığı takdirde Matematik DÜnyası Dergisinde duyurusu da yapılacaktır.
...

Sevgili Arkadaşlar,
Sizedaha öncede bahsetmiş olduğum gibi dergimiz görücüye çıkmış bulunmaktadır:)) Fakat bazı mali sorunlardan dolayı şu an ilk sayıyı matbaa basımı, daha sonraki sayıları E-Dergi formatında internet ortamından paylaşım olacağını söylemek istiyorum.
Şu an son aşamlarını tamamlıyoruz. Ümit ediyorum ki sizlerin özgün çalışmalara sahip olabilecek potansiyelin var olduğunu söylemeye gerek duymuyorum.Bu nedenle özellikle 'olimpiyatçı' arkadaşların paylaşımını da bekliyorum. Örneğin konu sıkıntısı çekenler için şöyle izah edebiliriz. Örneğin bazen bir soru farklı ufuklara yelken açabilmeyi sağladığından bu o kadar da zor olmadığını göreceksiniz. Amerikan Matematical Monthly dergisinin sorularından bambaşka ve özgün yeni projelere imza atabilen arkadaşlarımı tanıyorum.Neden olmasın? Şu an Türkiye de sadece Matematik Dünyası Dergisi süreli oalrak çıkmaktadır.
Kolay gelsin. :)


analytic geometry
geometry
circumcircle
incenter
search
graphing lines
slope

Well right now im getting a C in grade 10 academic math and we have a test on tuesday :?

ive been crusing throguh my notes and ahev even looked around a couple of websites but im having problems either finding a decent explanation or undertsanding it

so if you guys have a couple of minutes to spare maybe you can tell me how to find these?(from a triangle)or even just give me the link of a website that could explain it to me? :?:

Find the altitude,median or right bisector of a triangle; for example A triangle with points D(-4,-7) E(8,5) F(2,-3), find the median/altidue/right bisector.

Find the circumcentre,centroid, or orthocenter.

And by help me i mean show me how to solve an equation like this:find the circumcentre/centroid/orthocenter of the triangle defined by D(5,1) E(-2,0) and F(4,8)

and lastly questions like
Points P(1,8) and Q(9,2) are end points of a diameter of a circle, and R(8,9) is another point on the same circle. Find the coordinates of the centre of the circle.

I know this is alot but im begging anyone out there, i need your help! :o

just explain 1 or 2 or just give me a link, anything helps

thank you for your future help :rotfl:

[quote="Telmob"]Points P(1,8) and Q(9,2) are end points of a diameter of a circle, and R(8,9) is another point on the same circle. Find the coordinates of the centre of the circle.[/quote]
well if PQ is a diameter then its midpoint would be the center of the circle, wouldn't it?
to make sure, use the distance formula to see if the distance from the center to R is the same as to P and Q

[quote="Telmob"]Find the circumcentre,centroid, or orthocenter.

[/quote]

A link could be [url=http://www.mathworld.com]mathworld[/url]

alright thx, i now understand the circle part but anyone can help me with median,altitude and right bisector?

i know i need them to understand the circumcenter,centroid or orthocenter :huh:

woooooooooooooooow, i just went to mathworld and that explanation just blew me away

it didnt help  me at all and confused the bejesus out of me :rotfl:

I don't know what exactly you're having trouble with, but I'll try and explain the basics

Median: line from vertex of triangle to midpoint of side opposite. The three medians are concurrent at a point called the centroid. Also, the medians are divided by each other in 1 to 2 ratios.

Angle Bisector: line drawn separating an angle into two equal angles, touching the opposite side. The three angle bisectors are concurrent at the incenter, which can be used to inscribe a circle within the triangle. If you drop a line segment from the incenter to any of the three sides of the triangle and form a right angle, the line segment is the radius of the incircle, or inradius.

Perpendicular Bisector: line draw perpendicular to a side of the triangle that goes through the midpoint of the side. The three perpendicular bisectors are concurrent at the circumcenter, which is the same distance from each of the three vertices of the triangle. In acute triangles, the circumcenter is within the triangle. In right, it lies on the hypotenuse, the midpoint of it actually. In obtuse, it lies outside of the triangle. Because the vertices are of equal distance from the circumcenter, the distance from any vertex to the circumcenter is the radius of the circle with center at the circumcenter and who has points on each vertex. This radius is called the circumradius.

Altitude: basically the height of the triangle if you were to set it upon any side. In acute triangles, it goes through the triangle. In right, 2 altitudes are the legs, the other goes through the triangle somewhere. In obtuse, 2 altitudes will fall outside the triangle and the other inside. The three altitudes are concurrent at the orthocenter. Honestly, I don't know of any purpose for the orthocenter. Search for it.

I had to draw a ton of pictures before I finally understood these terms (do not draw equilateral triangles, these will confuse you because many of the above mentioned are the same). Hope I helped.

lol not exactly i know what all these shapes are but in the test im going to have questions like:

A triangle has vertices P(-12,6), Q(4,0) and R (-8,-6). Find the coordinates of the circumcenter

i have to find out the equations or the actual answer.

Like i know what the distance formula or slope formula is but i dont know where to put them and what to add to what and all that.

i want to know the formulas to find the circumcenter or altitude.

sry im not being clear enough and im very great full of the help you guys have given me :lol:

but if you still have some time please help me

Oh, I see. I'm ok at finding lengths, but when it comes to finding the coordinates, I'm awful. Maybe someone else has heard of these formulas and can help us both.

Ok... so you know how to derive a line right?  Based on two coordinates.  In order to find the orthocenter of a triangle take one of the coordinates, make a line passing through that point, perpendicular to the other side.  Call that altitude x just for now.  Now take another point, and do the same to find altitude y.  Then you find the intersection of x and y...

Centroid:  You simply find the average of all the coordinates...

alright im so happy

today i went to class and the teacher said that we had an entire period to just review :roll: i love her now

well i asked her how to get the altitude,median or right bisector

i didnt ask her about the circumcenter,centroid or othocenter but maybe you guys can help me

and by help i mean now i know that to get median i need to get the mid-pt,then the slope, then cross multipy and turn into an equation :lol:

to find the centroid, i was pretty sure u found the x and y values that satisfied each equation. so find the line made between the actual vertex of the triangle to the midpoint of the opposite side, and do the same for another point, than solve the equation...i did tenth grade math a long time ago lol...but im pretty sure thats how you do it. ill post on the other stuff later if no one else does.

ok, centroid is the easiest: take the average of all the x and y coordinate (if you know physics, which I doubt it, centroid is the center of mass when mass is uniformly distributed, i.e. the weight for the coordinates are all equal.)

So lets say you have a triangle with vertices  A(0, 0),  B(1, 2) C(5, 1) then you have:

$\frac{0 + 1 + 5}{3} = 2$

$\frac{0 + 2 + 1}{3} = 1$.

so the coordinate of the centroid is (2, 1).


Let p(x) be the sum of all the positive divisors of an integer x (including x and 1), without repeat divisors.

I know that p(8)=15, but is there any other x such that p(x)=15?

Yes. The sum of divisors of $p_{1}^{i}p_{2}^{j}...$ is the product of ths sums of $p_{1}^{i}$, $p_{2}^{j}$, etc. This means that the specific sums are 3 and 5 or 15. 15 leads to $2^{3}$ (1, 3, 5, 7, 11, 13 don't work) and $5$ leads to no primes.

Get that?


algebra
polynomial
LaTeX
algebra proposed

The case $n=3$
The roots of $x^3 -x^2-0,5x+a_3=0$ are $x_1\leq x_2 \leq x_3 $ with

(1) $x_3-x_1=\sqrt(10/3)$

Relation with the roots and the coefficients give
(2) $x_1+x_2+x_3=1$
(3) $x_1x_2+x_2x_3+x_3x_1=-0,5$
(4) $x_1x_2x_3=-a_3$

(1),(2),(3) give a second order equation with unknown $x_1$, solve it to get
$x_1$, with (1) we get $x_3$, (2) for $x_2$.Finally (4) give $a_3$

I don't think this works for larger n. I think this can be settled using Cauchy-Schwarz inequality., I will think about it.

A wonderful problem. But the Latex seems to be angry with me. $| 2^n $ and $ |f(x)| $.


logarithms
calculus
integration
calculus computations

What is approximate value of the $ 1^1 2^2 3^3 \dots n^n$?

$ \ln\left(\prod_{k\equal{}1}^nk^k\right)\equal{}\sum_{k\equal{}1}^nk\ln k.$

That will lead us to investigate the integral $ \int_1^nx\ln x\,dx.$

So our first very crude statement will be that this product behaves something like $ n^{n^2/2}e^{\minus{}n^2/4}.$

The level of crudeness is the same as saying that $ n!$ behaves like $ n^ne^{\minus{}n}.$ It's not detailed enough to be an asymptotic approximation, but it's a start, and it can be refined in part by just taking a little more care in estimating how the integral approximates the sum.

Just playing around with this empirically, it looks like $ n^{\frac {n^2 \plus{} n}2}e^{ \minus{} \frac {n^2}4}$ does pretty well.

Just to be silly, here's another way to come up with a similar approximation:
\begin{align*} P & : = \prod_{i = 1}^n i^i \\
& = \frac {(n!)^n}{\prod_{i = 1}^{n - 1} i!} \\
& \approx \frac {(n^ne^{ - n})^n}{\prod_{i = 1}^{n - 1} i^i e^{ - i}} \\
& = \frac {n^{n^2} e^{ - n^2}}{P \cdot n^{ - n} \cdot e^{ - n(n - 1)/2}}, \end{align*}
so
\[ P^2 \approx n^{n^2 + n} e^{ - (n^2 + n)/2}
\]
which gives an approximation not that different than Kent's.  If we'd used the more precise $ i! \sim c\cdot i^{i + 1/2} e^{ - i}$, we would get out
\[ P^2 \approx c \cdot \frac {n^{n^2 + 3n/2 + 1/2} e^{ - n(n + 1)/2}}{\sqrt {n!}} \approx \sqrt {c} \cdot n^{n^2 + n + 1/4} e^{ - n^2/2}
\]
(assuming no  errors on my part :)).
Edit: false assumption ;)

If I use a spreadsheet to add up $ \sum_{k\equal{}1}^nk\ln k,$ I get that for $ n\equal{}500,$ $ \ln P\approxeq 715880.431$ while $ (n^2\plus{}n\plus{}1/4)(\ln n)/2\minus{}n^2/4\approxeq 715880.441.$

That was using the last line in JBL's post.

Original answer to our product:

$ P\approx A n^{n^2/2 + n/2 + 1/12}e^{ - n^2/4}$

where $ A$ - Glaisher-Kinkelin Constant $ \approx 1.2824271\dots$

$ A = e^{1/12 - \zeta'( - 1)} = (2\pi e^{\gamma - \zeta'(2)/\zeta(2)})^{1/12}$

[url=http://mathworld.wolfram.com/Glaisher-KinkelinConstant.html]If interesting...[/url]

My solution:

By Euler's summation formula: $ \sum_{1\leq k < n}f(k) = \int_{1}^{n}f(x)dx + \sum_{k = 1}^{m}{\frac {B_k}{k!}(f^{(k - 1)}(n) - f^{(k - 1)}(1))} + R_{mn}$, where $ R_{mn} = \frac {( - 1)^{m + 1}}{m!}\int_{1}^{n}{B_{m}(\{x\})f^{(m)}(x)dx}$. And for even $ m: \; \left|\frac {B_m(\{x\})}{m!}\right|\leq\frac {|B_{m}|}{m!} < \frac {4}{(2\pi)^m}$.

$ \sum_{1\leq k < n}k\ln k = \left(\frac {1}{4} + \frac {n^2\ln n}{2} - \frac {n^2}{4}\right) + \frac { - 1/2}{1!}(n\ln n - 0) + \frac {1/6}{2!}(\ln n + 1 - 1) + \dots =$

$ = \frac {1}{4} + \ln\left(n^{n^2/2 - n/2 + 1/12}\right) - \frac {n^2}{4} + \sum_{k = 3}^{m}{\frac {( - 1)^{k - 1}B_k}{k!} \left( \frac {(k - 3)!}{n^{k - 2}} - (k - 3)! \right) + R_{mn} =}$

$ = c + \ln\left(n^{n^2/2 - n/2 + 1/12}\right) - \frac {n^2}{4} - \frac {B_4}{2\cdot 3\cdot 4n^2} - \dots - \frac {B_{2t}}{(2t - 2)\cdot (2t - 1)\cdot (2t)n^{2t - 2}}$$ + O\left(\frac {1}{n^{2t}}\right)$

And:

$ P = n^{\frac {n^2}{2} - \frac {n}{2} + \frac {1}{12}}\cdot\exp \left(c - \frac {n^2}{4} - \frac {B_4}{2\cdot 3\cdot 4n^2} - \dots - \frac {B_{2t}}{(2t - 2)\cdot (2t - 1)\cdot (2t)n^{2t - 2}} + O\left(\frac {1}{n^{2t}}\right)\right)$

Only I don't understand why i got $ - n/2$ instead of $ + n/2$ in the power of $ n$? Kent or JBL, can you help me? Looks right because $ B_1 = - \frac {1}{2}$ :(


geometry
number theory
relatively prime
geometry open

Show that there are infinitely many noncongruent triangles which satisfy the following conditions:
i) the side lengths are relatively prime integers;
ii)the area is an integer number;
iii)the altitudes' lengths are not integer numbers.

Can someone post a solution to this nice problem?? Probably Pohoata himself if possible.
I have worked with it in vain as I do not know how to relate altitudes not being integers with this problem.

take the triangle ABC.
let AD be the altitude from A with D on BC and p,q,r,s four positive integers which are pairwise coprime.
let's take AD=2pqrs,BD=$ (pr)^{2}-(qs)^{2} $,CD=$ (pq)^{2}-(rs)^{2} $.
that means AB=$ (pr)^{2}+(qs)^{2} $,AC=$ (pq)^{2}+(rs)^{2} $.
now we only have to take s<p because AB,BC,CA are the sides of a triangle.
if exists d>2 which divides $ r^{2}+q^{2} $ and $ p^{2}-s^{2} $ and d<$ r^{2}+q^{2} $ and d<$ p^{2}-s^{2} $  and ($ q^{2}-r^{2}$,$ p^{2}-s^{2} $)=($ q^{2}-r^{2} $,$ q^{2}+r^{2} $)=1 then (AB,BC,CA) is d.
the altitudes from B,C are only rational numbers. they don't are integers.
now if we take the triangle which is similar to ABC with the sides BC/d,AC/d,AB/d we obtain a triangle which satisfies the problem.
if you see any mistake tell me.

Let $a=9t(5t+2)+5$, $b=16t(5t+2)+5$, $c=25t(5t+2)$ the sides of triangle $\triangle ABC$. Let $S$ the area of $\triangle ABC$ $\Longrightarrow$ by Heron theorem we get:
$$S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{9.16.25.(5t+1)^2.(5t+2)^2}=60(5t+1)(5t+2)$$
$\Longrightarrow$ $h_a=\frac{120(5t+1)(5t+2)}{9t(5t+2)+5}$, $h_b=\frac{120(5t+1)(5t+2)}{16t(5t+2)+5}$, $h_c=\frac{120(5t+1)(5t+2)}{25t(5t+2)}$, if we choose the integer positive $t\equiv 106\pmod {210}$ it so easy note that $a,b,c$ are pairwise coprime and $h_a,h_b,h_c$ are not integers and $S$ is a integer.


This is the official invitation to the Columbus High School Invitational Mathematics Tournament. Attached documents include the invitation letter and the registration form. If you wish to attend, please print the registration form and send it to Mrs. Franklin at the school address indicated on the invitation letter.
Please remember that individuals are welcome to come!
Thanks for the time, and let me know if you have any problems downloading the files.
Mark

Download Details: Click on the link to download the registration form. In order to get a larger version of the invitation letter, right-click on the image and select "save target as." After selecting a location, save the file, and then open the file with Windows Picture and Fax Viewer. From here you should be able to print a full-size version of the letter. Ask if you need any assistance!


In order to increase the probability of correct transmission of a message over a noisy channel, a repetition code is often used . Assume that the message consist of a single bit, and the probability of correct transmission on a single trial is $ p$. [b]With a repetition code of rate $ 1/n$[/b] the message is transmitted $ n$ number of times and a majority voter is used for decoding. Assuming $ n\equal{}2m\minus{}1 , m\equal{}1,2,3....,$ determine the error probability $ P_e$ of a repetition code as a function of $ m$.

[hide="what i have understood "]
the message will only be decoded when $ m$ or more messages out of $ n$ reach the destination correctly.
but i am confused with the working of repetition code and also what that bold part means. [/hide]

I assume this means you're using a $ (n,1)$ [url=http://en.wikipedia.org/wiki/Repetition_code]repetition code[/url]. The question is asking what the probability is that more repetitions will be transmitted incorrectly than correctly, given $ n$ repetitions.

Wouldn't it just be
\[ P_e \equal{} \sum_{k \equal{} m}^{2m \minus{} 1}{2m \minus{} 1 \choose k}p^k(1 \minus{} p)^{2m \minus{} 1 \minus{} k}
\]
or am I misunderstanding the question?  If not, then I think any closed form expression of the above will necessarily involve hypergeometric functions.  Mathematica thinks so, too.  Specifically, Mathematica returns
\[ P_e \equal{} 2^{2m\minus{}1}(1\minus{}p)^{m\minus{}1}p^m \frac {\Gamma \left(m \plus{} \frac {1}{2}\right)}{\Gamma(m \plus{} 1)\sqrt {\pi}} \ _2F_1\left(1,1 \minus{} m,1 \plus{} m,\frac {p}{p \minus{} 1}\right)
\]

after going through the link,
i think the answer should be

  $ P_e \equal{} \sum_{k \equal{} 0}^{m\minus{}1}{2m \minus{} 1 \choose k}p^k (1 \minus{} p)^{2m \minus{} 1 \minus{} k}$

because if more than $ m$ bits reach the output correctly, then the majority voter will give the correct output.


Use the binomial theorem to express (x+y)^7 in expanded form:

Thanks

Ryan

Wouldn't you just... use the biomial theorem?

I don't think that you're supposed to post this kind of question here, but...

$x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35 x^3y^4 + 21x^2y^5 + 7xy^6 + y^7$

[quote="joml88"]Hmm, why was this locked?[/quote]I unlocked it. Somebody probably thought it was a homework question (which it probably was), but heh.

You can re-lock it if you want.  :D


Hi I was reading the proof of the inverse of matrix and came upon these lines:

AB=BA=In
AC=CA=In
Now AB=In
=>C(AB)=CIn
=>(CA)B=C (BECAUSE OF ASSOCIATIVE PROPERTY) *
=>InB=C
=>B=C

I.E The inverse of a matrix is unique. *

(1)(2)Please can anybody explain the lines marked with asterisks, including what is the meaning of unique?

Also, I need help with these lines regarding inverse;
A(adjA)=(adjA)A=|A|In
A{1/|A|*(adjA)}={1/|A|*(adjA)}A=In

(3)Also when we find the inverse by 1/|A|*(adjA), what happened to A itself?

Please make yourself more clear and use $\LaTeX$.

$\LaTeX$ is a program to make mathematical expressions. You can find how to use it by clicking "LaTeX" at the left. You can download a program to your computer but you don't need to to use it on the forums. Just enclose your LaTeX code in dollar signs. For example,

[code]$7x=17y^2$[/code]

produces

$7x=17y^{2}$[/code]

$\LaTeX$'d

Where $I_{n}$ is the $n \times n$ identity matrix, and $B$ and $C$ are (possibly distinct) inverses of matrix $A$
$AB=BA=I_{n}$
$AC=CA=I_{n}$
Now $AB=I_{n}$
$C(AB)=CI_{n}$
$(CA)B=C$ by associativity
$I_{n}B=C$
$B=C$

[quote="sidk47"]$(CA)B=C$ by associativity[/quote]
Matrix multiplication is associative. For any matrices $A$, $B$, and $C$ of any size provided that $AB$ and $BC$ are both defined, then the product $AB$ multiplied by $C$ is always the same as $A$ multiplied by $BC$. It sounds simple and self-evident, but some operations such as subtraction are not associative.

[quote="sidk47"]The inverse of a matrix is unique. *[/quote]
Take for example the square root of 4. You don't get a unique answer from this; it can either be $2$ or $-2$. There are many other mathematical operations that don't get unique answers. Here, we are showing that when one talks about the inverse of a matrix, there's no ambiguity whatsoever. There can only be one answer (if there is one at all). This isn't true when talking about things like the square root.

[quote="sidk47"]$A(\text{adj }A)=(\text{adj }A)A=|A|I_{n}$
$A\frac{1}{|A|\cdot(\text{adj }A)}=\frac{1}{|A|\cdot(\text{adj }A)}A=I_{n}$ [/quote]
I don't understand what you need help with. You know what the adjoint ($\text{adj}$) is, right? This is just an identity that makes it easier to find the inverse. I have even less of an idea of what you mean by your third question.

Thank you very much for your help and patience.

To clarify my third question, it is:

(3) Why isn't the inverse found by $A*(1/|A|*adjA)$, instead of just ${1/|A|*adjA}$, so my question is what happened to A when finding the inverse.

Thanks once again.

$A$ times $\frac{1}{|A|\cdot\text{adj }A}$ gives the identity matrix. That means by definition, the second operand $\frac{1}{|A|\cdot\text{adj }A}$ is the inverse. What your third equation says is not $A \cdot \frac{1}{|A|\cdot\text{adj }A}= A^{-1}$, but rather $A \cdot \frac{1}{|A|\cdot\text{adj }A}= I_{n}$. Because the inverse is unique, i.e. there is only one matrix $A^{-1}$ such that $AA^{-1}= I_{n}$, we have that $\frac{1}{|A|\cdot\text{adj }A}= A^{-1}$

Thank you very much once again for your time, patience and kindness. I understand clearly now. Thanks once again.


Solve for nonnegative reals: let $n$ be a positive integer:

$x_{1}+x_{2}^{2}+...+x_{n}^{n}=n$
$x_{1}+2x_{2}+...+nx_{n}=n(n+1)/2$

By AM-GM : $k-1+x_{k}^{k}\geq kx_{k}$. That is all :D

thank you, and I am sorry that I accidentally put this into the combinatorics section.


The problem committee of a mathematical olympiad prepares some variants of the contest. Each variant contains 4 problems, chosen from a shortlist of $ n$ problems, and any two variants have at most one problem in common.
(a) If $ n \equal{} 14$, determine the largest possible number of variants the problem committee can prepare.
(b) Find the smallest value of $ n$ such that it is possible to prepare ten variants of the contest.


If we denote by h the altitude in triangle DEF corresponding to side EF, we have area of DEF = EF*h/2. Then denote r=(area of DEF)/(EF^2)=h/(2*EF). Because AFDE is cyclic, we get EF/h=cot(BAD)+cot(CAD), so r is constant for a fixed D. So it's enough to prove the inequality r<=1/4*AD^2 (which is the ineq we want to prove) for any points E and F , as long as we have a fixed D and AFDE is cyclic. Then we can take points E and F such that DE perpendicular to AC and DF perpendicular to AB. Then AD is the diameter of the circumcircle of AFDE, so EF^2/4*AD^2=(sinA)^2/4, so we have to prove that area DEF <= area ABC*(sinA)^2/4 whenever de perpendicular to AC and DF perpendicular to AB. Because DE*AC+DF*AB=2*S (S is area of ABC), we find by AM-GM that the largest area of such a triangle DEF is achieved when D is the midpoint of BC. If we calculate in this case we find equality in our inequality, so we proved what we wanted. Hope it's correct.

That looks right to me!

And forgive me for being a little slow, but how did you get that EF/h=cot(BAD)+cot(CAD)?

Let P be the point of intersection between h and EF (the projection of D on EF). Then PF/h=cotDFE=cotBAD because angles BAD and DFE are equal (AEFD is cyclic). You do the same for PE/h and you add them.

[quote="grobber"]angles BAD and DFE are equal (AEFD is cyclic). [/quote]
could u prove that?


[quote="GameBot"]Seven students set their lunch bags on a table while they went to wash their hands. When they return, each randomly grabs one of the seven identical bags. What is the probability that exactly six students will grab the bag that was originally theirs?[/quote]

If exactly six students grab the bag that was originally theirs, then the only bag left will belong to the 7th student. Thus 7 students will have the bag that was originally theirs. However, the problem says EXACTLY six, so the probability must be 0.


algebra
polynomial
number theory proposed
number theory

Find all polynomials $ p\in\mathbb Z[x]$ such that $ (m,n)\equal{}1\Rightarrow (p(m),p(n))\equal{}1$

Suppose $ P(x) \equal{} x^tQ(x)$, with $ Q(0) \not \equal{} 0$. Then $ (m, n) \equal{} 1 \Rightarrow (Q(m), Q(n)) \equal{} 1$. Taking a large prime $ m$, we see that either $ Q$ is constant or there is a prime $ p$ such that $ p|Q(m)$. If $ p \equal{} m$ then $ m|Q(m) \Rightarrow m|Q(0)$, witch is avoided for $ m$ large enough. Then $ p \not \equal{} m$. So $ (m, m \plus{} p) \equal{} 1$ but $ Q(m \plus{} p) \equiv Q(m) \equiv 0  \text{ } mod \text{ } p$, contradiction. Thus, $ Q$ is constant and $ P \equal{} \pm x^t,  \text{ } t \ge 0$.

Notice that for all $n\in\mathbb{Z}$, if there exists a prime dividing $P(n)$ that doesn't divide $n$, we look at $(n+p,n) = 1\implies (P(n+p),P(n)) = 1$. This is absurd. Hence all the primes dividing $P(n)$ also divides $n$.
Now if say $P$ is not constant, then by schur's there exists infinitely primes dividing $P(n)$ for some $n$, hence by the above observation we see that there then exists infinitely many primes dividing $P(0)$, implying that $P(0)  =0$.
We may proceed this argument on $Q(x) = P(x)/x$ and so on, hence we obtain $P(x) = \pm x^t$ for nonnegative integers $t$.

A small thought: what would the answer be if the condition is $(P(m),P(n))=1\Rightarrow (m,n)=1$?


function
real analysis
real analysis unsolved

1) Let $ f\colon {\mathbb R}\to {\mathbb R}$ be a continuous function such that  $ f(x \plus{} 2\pi) \equal{} f(x)$ and $ f(x \plus{} 2\pi \omega) \equal{} f(x)$ for any $ x,$ $ \omega \in {\mathbb R}\backslash {\mathbb Q}.$ Prove that $ f$ is constant

The upper formulation was too generous. It should have been

2) Let $ f\colon {\mathbb R}\to {\mathbb R}$ be a continuous function such that  $ f(x \plus{} 2\pi) \equal{} f(x)$ and $ f(x \plus{} 2\pi \omega) \equal{} f(x)$ for any $ x$ and $ \omega \in {\mathbb R}\backslash {\mathbb Q}$ fixed. Prove that $ f$ is constant

Let me try:

if $ y\minus{}x\not\equal{}2k\pi$

take $ \omega\equal{}\frac{1}{2\pi}(y\minus{}x)$

we have $ f(x\plus{}2\pi\omega)\equal{}f(y)\equal{}f(x)$

if $ y\minus{}x\equal{}2k\pi$

we have

$ f(y)\equal{}f(x\plus{}2k\pi)\equal{}f(x\plus{}2(k\minus{}1)\pi)\equal{}...\equal{}f(x)$

so If $ y \in {\mathbb{R}}\backslash{\mathbb{Q}}$

we get the conclusion

because $ f(x)$ is a continuous function, so we get the conclusion

Your proof is nice and good for 1). The actual problem is 2) . Sorry for 1) which has generated the misunderstanding.

[quote="Xingyuan"]Let me try:....[/quote]

A periodic non-constant continuous function has a period of minimal length $ \lambda$, such that any other period's length is an integer multiple of $ \lambda$ (elementary fact).

Therefore $ \omega \equal{} \frac {2\pi\omega} {2\pi} \in \mathbb{Q}$, contradiction.


A polynomial

$P(x)=c_{2004}x^{2004}+c_{2003}x^{2003}+...+c_1x+c_0$
has real coefficients with $c_{2004}\not=0$ and 2004 distinct complex zeroes $z_k=a_k+b_ki$, $1\leq k\leq 2004$ with $a_k$ and $b_k$ real, $a_1=b_1=0$, and

$\sum_{k=1}^{2004}{a_k}=\sum_{k=1}^{2004}{b_k}$.
Which of the following quantities can be a nonzero number?

$\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k=1}^{2004}{a_k} \qquad \text {(E)}\sum_{k=1}^{2004}{c_k}$

[hide]Since $a_{1}=b_{1}=0$, we have $z_{1}=0$.  Then $P(x)$ is divisible by $x$, and $c_{0}=0$.

Complex roots occur in conjugate pairs, so there is an even number of complex roots.  $b_{2}b_{3}\cdots b_{2004}$ is non-zero only if all $2003$ of the other roots are complex, which is impossible.  Therefore, $b_{2}b_{3}\cdots b_{2004}=0$.

The sum of the roots is $\sum_{k=1}^{2004}(a_{k}+b_{k}i)=-\frac{c_{2003}}{c_{2004}}$, which is real.  Therefore, $\sum_{k=1}^{2004}b_{k}=0$.  Then $\sum_{k=1}^{2004}a_k=0$ also.  Therefore, the sum of the roots is $0$, and $c_{2003}=0$.

Therefore, all answer choices except E are equal to $0$.  Notice that $P(1)=\sum_{k=1}^{2004}c_{k}$.  Therefore, $\sum_{k=1}^{2004}c_{k}=0$ iff $1$ is a root, which is not necessarily true.  The answer is E.[/hide]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=416108#p416108]here[/url] for other discussions on this exact same problem...


geometry
geometric transformation
reflection
geometry unsolved

Let $ ABC$ be a triangle, and $ P$ a variable point in its plane. Let $ L$, $ M$, $ N$ be the orthogonal projections of the point $ P$ on the lines $ \left(BC\right)$, $ \left(CA\right)$, $ \left(AB\right)$.

Find the point $ P$ for which the value of $ BL^2 \plus{}CM^2 \plus{}AN^2$ is minimum.

I think P must be the circumcenter. I don't have time to post a complete soln right now (Internet is EXPENSIVE during the day  :D ), but I used Carnot's thm.

[color=red][Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=42446 for this problem.][/color]

[color=red][Moderator edit: This post is related to another problem, namely to finding the minimum of $PL^2+PM^2+PN^2$.][/color]

It's well known point. So-called Lemuan point of triangle. This point is the point of intersection of three simedians of triangle.

We need to find minimum of x^2+y^2+z^2 if ax+by+cz=2S (we consider
directed distances). It is obvios that minimum is when x:a=y:b=z:c= u.

If  distances from M to BC, CA, AB are ua, ub, uc then M_1 --- the reflection of M for bissector of \angle A, lyes on median of ABC, since
S(AM_1C)=S(AM_1C) , ie M lies on simedian.


pigeonhole principle
inequalities
graph theory
combinatorics unsolved
combinatorics

At a meeting there n people. For each subset of 6 people you will find two people who know each other. Prove that there is at least one subset of 6 people who all know each other.

a.) n = 512 (iterated application of the pigeonhole principle)
b.) n = 252 (elementary estimate)
c.) n = 102 (Ramsey: R(6,6,2))

Hmmm...and what if no one knows someone in this meeting?
Clearly, it sounds like a ramsey problem but I think you forgot something in the statement.

Pierre.

Hi Pierre,

is there still any missing information ?

Are you sure about the n= 102?
Seems to me that 102 =< R(6,6) =< 165, but the exact value of R(6,6) is not known.

Let n >= R(6,6).
Consider the people as the vertices of the complete graph K_n and 2-color the edges of K_n by red if the two people known each other and blue otherwise.
The value of n ensures that there is a monochromatic K_6. But, from the statement of the problem, there is no blue K_6 (since among each group of 6, there is two people who know each  other). Thus there is a red K_6 and we are done.

So, from the initial remark, I think that we may say that n = 165 works, but...

Pierre.

69 \leq R(6,6,2) \leq 102 - maybe it is not up-to-date anymore, it is from 'Ramsey Theory by Graham and Rothschild, 1980'

Well, are we talking about the same number (see my 'solution' of the problem)? Is 'your' R(6,6,2) the same as 'my' R(6,6)'?

I've got 'Ramsey theory' from Graham, Spencer and Rothschild (and it is really a wonderful book on the subject), Second edition (1989), and they say that 102 =< R(6,6) =< 178.
The upper bound has been improved to 165 in 1993 (or 1994).

Pierre.

Pierre! Why did you not publish solution for a) or b)?  :(

Because the Ramsey bound is better, so that I did even thought about the other problems.

Pierre.

I do not understand what you mean. Since the Ramsey bound is better a) and b) are solved too. If you want a more constructive bound than the optimal knwon, which may be non-trivial (I don't know), you may use the well-known inequality (which is one of the way to prove Ramsey's theorem) :
\[R(a,b) \leq R(a-1,b) + R(a,b-1)\]
with strict inequality when the two numbers of the RHS are even.

If I made no mistake in my computations, and since $R(5,3) = 14$ and $R(4,4) = R(6,3)=18$ it leads to $R(6,6) \leq 222$, which is still better than $252$.

Pierre.

You can write it in more details, then we will close topic.

Let me write it, to close the topic :)
Let $R(a,b)$ denote the minimal number of vertexes a graph must have, so that when coloring its edges with 2 colors, we have either a $K_a$ or a $K_b$. Take any vertex from a graph with at least $R(a-1,b)+R(a,b-1)+1$ vertexes. Then, we have either $R(a-1,b)$ edges of one color, or $R(a,b-1)$ edges of another color, and in the particular subgraphs, we will have, by the meaning of $R(a,b)$, one complete monocolor subgraph of the needed kind, and also we may add to it the initial vertex, because of the colors of its edges.
Too clumsy, I think :?

[quote="DusT"]LTake any vertex from a graph with at least $R(a-1,b)+R(a,b-1)+1$ vertexes.[/quote]
I think you mean $R(a-1,b)+R(a,b-1)$, because if sum of two numbers is $R(a-1,b)+R(a,b-1)-1$ then it is impossible that we simultaneously have first of them is less than $R(a-1,b)$ and second is less than $R(a,b-1)$.

Yes, you are right, sorry. My mistake  :?