geometry
3D geometry
pyramid
vector
analytic geometry
trigonometry
inradius

30-60-90 triangles are my favorite, but I think that's pretty common... I'll explain anyway

In a triangle with angles 30,60, and 90 degrees,
The short leg = x
The hypotenuse = 2x
The long leg = ¬/3 * x

[quote="kyyuanmathcount"]Square any number from 25 to 50 (assuming you have memorized the squares up to 25).

$(50-n)^2+100(N-25)$.

I discovered that while looking at charts of squares up to 50  :P .[/quote]
Squaring any number of the form $\bar{n5}$.
$\bar{n5}^2=\bar{\{[n(n+1)]25\}}$.
So $15^2=\bar{\{[1(1+1)]25\}}=225$

Masoud Zargar

you are given two sides of a triangle and the included angle where the two sides are a and b and the included angle is c. the area of the triangle is {a*b*sin(c)}/2

Masoud, if you use \overline instead of \bar, I think you will get what you want.

E.g. $\overline{ \{ a+b+c\}}$.

[size=75]have you guys ever seen that one trick for squaring numbers that end with 5? you take the number in the tens place and multipy it by the number immediately following it and then add 25

for 45^2...

4 x 5=20

answer=2025

well, i thought it was interesting...
[/size]

Formula for a frustrum (cut-off pyramid)

$\displaystyle \frac 13 h ( \text {Area of Base 1} + \sqrt{\text {Area of Base 1} \times \text {Area of Base 2}} + \text {Area of Base 2})$

biffanddoc, if you want to actually evaluate a large number $\bmod 11$ you have to make sure that the ones digit is given a positive sign ($10^0 \equiv +1 \bmod 11$).

But anyway, here's a trick:  The Shoelace Theorem.  The area of a "simple" region (a region whose defining lines do not cross each other) defined by the lines connecting the points $(x_1, y_1), (x_2, y_2), ... (x_n, y_n)$ (in clockwise or counterclockwise order) is given by

$A = \frac{1}{2} \left(x_1 y_2 + x_2 y_3 + ... + x_n y_1 \right) - \frac{1}{2} \left( x_2 y_1 + x_3 y_2 + + x_1 y_n... \right)$

(This is simply half of the cross products of the vectors given by the lines.)  To evaluate this product quickly, write the points in order in a column, and then write the first point after the last one.  Then start from the x-coordinates and draw diagonal lines down to the next y-coordinate and evaluate those products - then start from the y-coordinates and draw diagonal lines down to the x-coordinates and evaluate those products - and compute one half the difference.  The lines create a "shoelace" pattern, hence the name.  :)

The area of a triangle is $K=\frac{abc}{4R}$, where $a,b,c$ are the side lengths and $R$ is the circumradius. This is easily proven with the extended law of sines and $K=\frac{ab\sin C}{2}$.
I'm sure a lot of people might know this one already, but when was the last time you actually used it? I used it on only one contest problem in my life, in an unofficial contest with many other solutions. It's a really pretty formula, but it's almost completely obsolete.

Well, as long as we're talking about area formulas...

Denote $r$ to be the inradius of a triangle with side lengths $a, b, c$ and $r_a, r_b, r_c$ to be the exradii of the excircles tangent to $a, b, c$, respectively.  Denote $s = \frac{a + b + c}{2}$ to be the semiperimeter of the triangle.  Denote $K$ to be the area of the triangle.  Denote $R$ to be the circmradius of the triangle.  Then:

$K = rs$
$K = r_a(s - a), r_b(s-b), r_c(s-c)$
$K = \sqrt{r r_a r_b r_c}$
$K = \sqrt{s (s-a)(s-b)(s-c) }$
$4R = r + r_a + r_b + r_c$ (not an area formula, but cool enough :) )

And if we have a cyclic quadrilateral, K^2 = (s-a)(s-b)(s-c)(s-d).

Edit: wow, the shoelace theorem looks extremely usefull for computational geometry. I'll have to remember that one.

Oh, here's another extremely awesome coordinate geometry trick.

Pick's Theorem:  Given a region bounded by lines (again, none of the lines cross) between lattice points such that the perimeter/boundary of the region contains $B$ lattice points and the interior of the region contains $I$ lattice points, the area of the region is $A = I + \frac{B}{2} - 1$.

(This is often actually used to find $I$ rather than $A$ :) )

That's actually on the MathCounts CD as one of the most essential lessons or something like that.
My coach showed it to us on Friday morning..

[quote="mbabbitt2003"]I think that people should post ideas and tricks. For one, square 999 (without using a calculator).  It isn't hard. Let 999 be n. I will prove that $(n-1)(n+1)+1=n^2$.  Multiply the binomial out and add one. $(n-1)(n+1)=n^2-1$  TA-DAA  :D  NOW square 999 without using a calculator.[/quote]

What about  : $(1000-1)^2$?  :P ($= 1000^2 - 2*1000 + 1$)

any cool stuff in algebra and numbertheory?

for $999^2$
i just do $1000^2-1999$
is that easiest way?

I suppose a lot of people who compete know the Rational Roots Theorem, but it really doesn't get the exposure it deserves.

If $P(x)=a_0+a_1x+...+a_nx^n$ and $\frac{p}{q}$ is a root of $P(x)=0$ with p and q integral, then $p|a_0$ and $q|a_n$.

Also, here are four very similar theorems about trapezoids, one of which you should already know:
Suppose you have a trapezoid with bases of length a and b, and you draw in a median of length m.
1. If the median divides the trapezoid into two trapezoids of equal height, then m is the arithmetic mean of a and b.
2. If the median divides the trapezoid into two trapezoids of equal area, then m is the quadratic mean of a and b.
3. If the median divides the trapezoid into two similar trapezoids, then m is the geometric mean of a and b.
4. If the median passes through the intersection of the diagonals, then m is the harmonic mean of a and b.

I have only seen these come up twice in competition.  Once they asked explicitly for m in terms of a and b in one case, and the other was at ARML last year.  I tried to solve the trapezoid problem (individual #4, I think; the one with the intersection of the diagonals) using part 4 here and some very very nasty algebra, without success.  That took about 9 minutes.  Of the other minute, I spent half of it solving #3 and the other half using a special case of #4 to get a quick answer.

Edit: Oops, read post below.  Thanks for the correction, t0rajir0u.

[quote="Sly Si"]
If $P(x)=a_0+a_1x+...+a_nx^n$ and $\frac{p}{q}$ is a root of $P(x)=0$ with p and q integral, then $p|a_0$ and $q|a_n$.[/quote]

This statement requires qualification.  If $a_0, a_1, ... a_n \in \mathbb{Z}, a_n \neq 0$, and $(p, q) = 1$, only then can you say $p | a_0$ and $q | a_n$.

Incidentally, this theorem provides a very simple proof of the irrationality of $\sqrt{2}$.  :)

Euler's Totient Function:  $\phi(n)$ denotes the number of integers $1 \le k \le n$ such that $(n, k) = 1$.  It can be calculated quickly by writing $n = p_1^{a_1} p_2^{a_2} ...$, then $\phi(n) = (p_1 - 1) p_1^{a_1 - 1} (p_2 - 1) p_2^{a_2 - 1} ...$.

Euler's Totient Theorem:  For integers $a, p$ such that $(a, p) = 1$, then $a^{ \phi(p) } \equiv 1 \bmod p$.  When $p$ is prime this becomes Fermat's Little Theorem, since $\phi(p) = p - 1$.

Very useful for number theory.  :)

(Note:  $\phi(n)$ is not the smallest exponent that satisfies the equality!  That assumption cannot be made.)

in 3rd grade I created a formula for finding pythogorean triplets, but it probably has been invented.

for $x$ is a positive odd integer:

square $x$ and divied the resulting number by two. now that number is .5 more and .5 less than whole numbers. those two whole numbers and $x$ are pythogorean triplets

Here's Herons formulas for finding areas of a triangle.
s is the semi perimeter.
a,b and c are sides.
$\sqrt(s(s-a)(s-b)(s-c))$
For an inscribable quadrilateral,
s is still the semi perimeter.
a,b,cand d are sides.
$\sqrt(s-a)(s-b)(s-c)(s-d)$
I hope you will notice that the first formula is equivilant to the second only if d=0

You can square any number that ends in 5 easily-if the number is n5, then you square n, add n, add two zeros to the end of that number, and then add 25.   Example, $25^2,2^2+2=6$(add two zeros),$600$(add 25) $=625$.  This works for hundreds also.  for the form $xn5^2$ =$(xn)^2+xn+25$, another example, $115^2=100(11^2+11)+25=100(132)+25=13225$

Actually this post reminds me of the FOrmulas post in the Mathcoutns forum.

Hey, I have an AWESOME math trick that nobody has ever taught you before! Basically, what you do is you take the fixed, sharp end of a compass and you poke your finger with it. If you do it hard enough, it will eventually turn red!


Using a balance scale, it was determined that an algebra book
and a geometry book together balance a trigonometry book,
and two trigonometry books balance three geometry books.
How many geometry books are needed to balance four algebra
books?

Here, we will refer to the algebra book as "a", the geometry book as "g", and the trigonometry book as "t". a + g = t,  2t = 3g. We can see that 2a + 2g = 2t.
Since 3g = 2t, 2a = g, so 4a = 2g, so the answer is 2.


Let $n\in\mathbb{N}$ and let $f,g$ are two permutations of set $X=\{1,2,...,n\}$. We say $f$ and $g$ have common point if there exist $k\in X$ such that $f(k)=g(k)$.
a)If $m>\frac{n}{2}$, prove that there are $m$ permutations $f_{1},f_{2},...,f_{m}$ of $X$ such that for every permutation $f$ of $X$, there is an index $i$ so that $f$ and $f_{i}$ have common points.
b)Prove that for $m\leq \frac{n}{2}$, we can't find permutations $f_{1},f_{2},...,f_{m}$ of $X$ satisfying the above condition.

(a) Let the permutations be $f_{1}=1,2,...m..,f_{2}=2,3...m,...,f_{3}=3,4...m,1,2...$ and so on(the elements with indices greater than m are arbitrary). Now if f does not have a common point with no $f_{i}$ then all elements in f which are$\leq m$ have indices greater than m from here $m\leq n-m$ contradiction.
For(b) apply Hall's theorem.


Let m,n are different positive integers.Prove that exist z complex number with [size=150]|z|=1 and |1-z^m+z^n|>2 [/size]

hi,
write $ z \equal{} e^{i\phi}$ , with $ \phi \equal{} \frac{\pi}{3(m\plus{}n)}$.

How did you get it?
I think its not solution...
try for wxample m=1 n=3...


Let $ ABC$ be a triangle with $ \angle BAC\equal{}60^{\circ}$. The incircle of $ ABC$ is tangent to $ AB$ at $ D$. Construct a circle with radius $ DA$ and cut the incircle of $ ABC$ at $ E$. If $ AF$ is an altitude, prove that $ AE\ge AF$.

[hide] Call $ J,K$ are tangent points of incircle $ (I)$ with $ CA,BC$.
From $ \widehat{BAC}\equal{}60^0$ then $ AD\equal{}DJ\equal{}JA$ and $ \widehat{DIJ}\equal{}120^0$
Hence $ \widehat{DEJ}\equal{}60^0$, it mean $ DE\equal{}EJ\equal{}JD$
Now $ I\in AE$
But $ A$ is outside of incircle $ (I)$ then $ \widehat{AKE}\geq 90^0\geq \widehat{AEK}$, it mean $ AE\geq AK\geq AF$[/hide]


Let $X$, $Y$, $Z$ be points on the sides $BC$, $CA$ ,$AB$ respectively of triangle $ABC$. Prove that:  $\frac{1}{area(AYZ)}+\frac{1}{area(BZX)}+\frac{1}{area(CXY)}\geq \frac{3}{area(XYZ)}$

Right now, I can prove the following

Let $M, N, P$ be the intersection of $YZ$ and $BC$, $ZX$ and $CA$, $XY$ and $AB$. The point $X$ is defined as "good" to $YZ$ if the condition $(XB-XC)(MB-MC) \leq 0$ satisfies. Then your inequality is true if $X, Y, Z$ are good to $YZ, ZX, XY$.

I will think of the general case later.

There is another way is using Balkan inequality, But I don't post it, it isn't belong to me :D.


calculus
derivative
function
calculus computations

For $ \vec{x}\equal{}(x,y,z)\in\mathbb{R}^3\minus{}\{0\}$ and $ t\in\mathbb{R}$ let $ F(\vec{x},t)\equal{}r^{\minus{}1}g(ct\minus{}r)$, where $ c$ is a constant, $ g$ is a $ C^2$ function of one variable, and $ r\equal{}|\vec{x}|$. Show that $ \partial^2_x F\plus{}\partial^2_y F\plus{}\partial^2_z F\equal{}c^{\minus{}2}\partial^2_t F$.


Hey everybody!
I need a joke to start my math team campaign speech for VP. Ideas please!!!
Thank you so much!

say something along the lines of "vote for me cuz i'm cool"

i guarantee you that people will find it funny


A word consists of n letters from the alphabet {a, b, c, d}. A word is complicated if it has 2 consecutive identical blocks of letters (e.g. bacacb, acbbc and bcabcab are complicated but cabdab is not). For each n, find a closed formula / or a recursion for the number of complicated words with n letters.

It's indeed messy.
The idea of recursion souds simple:
Any complicated word of length n can be written in exactly one way as follows:
Take a simple word with length m, take its last subword with length k, repeat it (m+k<=n), then add last letters as you wish.
But it's hard to prove. I'm not sure.
Please tell me if you know. Is this solution correct?

Note: A version of this problem was at Romanian TST 2003( no one solved it, I think). It's already on this site

i think that is correct and easy to prove:
consider a complicated word of length k. consider the first string of m letters such that this string is complicated, but the string without the last letter of m is "simple". this string must contain a latter sequence of, say, n letters which form identical blocks. so any complicated word of any length can be written in that way. Further it is unique; to see this, suppose you can do it in an extra way. Then the first letters in that word contains the substring described above, so it is not simple. So it is in fact true i think. (sorry i didn't explain very well  :maybe: ).

here is my solution:
[hide]Let f(n) denote number of complicated words of length n.

Consider a k-lettered complicated word. its first k-1 letters should also form a complicated word. Thus to get a c-word of lenth k,
we can add any of the 4 letters to the right of an arbitary (k-1) lettered c-word except the case when we create a repeated  string of the form:

 $ \boxed{***\cdots*} \boxed {a_1 a_2 a_3 a_4 \cdots a_r} \boxed {a_1 a_2 a_3 a_4 \cdots a_{r \minus{} 1}} ar$

where there are $ (k \minus{} 2r) *'s$ .

Now we find how often can we have a repeated string of length r.

For r> $ [{\frac {k}2}]$ , the answer is 0.

for $ \boxed{1 \le r \le [{\frac {k}2}]}$

 answer is $ f(k \minus{} r)$. this is because there is a bijection between set of c-words of length (k-r),
and of $ \b{in \minus{} complicated}$ k-lettered words that have exactly two repeated strings of letter $ r$ at the end.

SO,  $ f(k) \equal{} 4 f(k \minus{} 1) \minus{} \sum{ f(k \minus{} r)}$, 2nd sum being taken over $ r \equal{} 1 to [{\frac {k}2}]$[/hide]

It's wrong -- your sequence gives $ f(5) \equal{} 252$ when in fact $ f(5) \equal{} 264$.  I don't know precisely what the error is, though.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246318

(The original solution offered in this thread is also wrong: we generate some simple strings in multiple ways, e.g., starting with abcab as your simple word leads to both abcabab... and abcababcab....)


Let a,b,c>0 such that a+b+c \leq 3. Prove the following inequality:
   9(1/a+1/b+1/c)-3 \geq 8(a+b+c)/ \sqrt abc.

1/a + 1/b + 1/c  \geq 9/(a+b+c) \geq 3 (1)
We denote :
x= \sqrt (bc/a)
y= \sqrt (ac/b)
z= \sqrt (ab/c)
Hence
1/a + 1/b + 1/c  \geq (a+b+c)/ \sqrt (abc)
<=> x+y+z \geq yz+xz+xy (xy+yz+xz \leq 3) (2)
If x+y+z \geq 3 =>(2) is obvious
If x+y+z \leq 3
We denote:
m=x/3
n=y/3
t=z/3
=> m+n+t \leq 1
=> m+n+t \geq (m+n+t) 2  \geq 3(mn+nt+mt)
=> (2) is right
(1) (2) => q.e.d


Construct a circle with center $O$ of radius $20$, with a fixed point $A$ and a variable point $P$ on the circle. Let the midpoint of $OP$ be $M$, and the midpoint of the perpendicular from $P$ to $OA$ be $N$. The area bounded by the locus of the midpoint of $MN$, as $P$ moves around the circle, is $\textit{A}$. Find $\lfloor \textit{A}\rfloor$.

[hide]
let $O$ be at the origin and
let $A$ be at $(20,0)$

let the location of $P$ be referred to by an angle $\theta$
$P$ is at $(20 cos\theta , 20 sin\theta )$

it follows that
$M$ is at $(10 cos\theta , 10 sin\theta)$
and (since OA is the x axis)
$N$ is at $(20 cos\theta , 10 sin\theta)$

the locus of their midpoints is the parametric graph
$(15 cos\theta , 10 sin\theta)$
this is the graph of an ellipse with $a = 10$ and $b = 15$
$A = \pi(10)(15) = 150\pi$
$\lfloor{A}\rfloor = 471$
[/hide]


[quote="13375P34K43V312"][/url]http://www.addictinggames.com/theidiottest.html[/url]

How many times does it take you to beat it? It's harder than it looks. It took me 3 tries.[/quote]

Ya i have done that before, pretty fun.

I screwed up only once. wasn't ready 4 that green button...

i missed that light blue because what they thought was light blue was cyan

grr, that stupid "Push the green button quick!"

I was being dumb for the first one.  I got Idiot Test 2 in one try.

I was like...hm...is this a square?  Beats me...

I tried 28 times and I still can't get past border-line idiot. Now I feel stupid since I worked hard to try to prove that I'm not an idiot, so I'm going to go cry myself to sleep.

got it in 3 tries but the first two times was not concentrating
pressed blue four times, and missed the smallest red button
but the 'quick!' didn't beat me :)

Yeah that got me, I saw like 20 rectangular shapes and it said to click the rectangle. wtf. Obviously AoPSers are superior to the game maker at geometry.

yeah I didn't get it when I Saw it, I was like omg

nice game though

I totally failed at that game :mad: . After the 4th try, I got bored and gave up :rotfl:

I don't blame you. That's what this guy did also from our school.

It's bad for my self-confidence being called an idiot by my computer. :(

It took me 2 tries, I messed up the first time when I thought the computer was lagging

[quote="weiquan"]Haha you can get through the whole thing without following instructions except for the "push the green button quick!"[/quote]

What do you mean?

[quote="moogra"][quote="weiquan"][quote="zeb"]hmm... right click play?[/quote]

yeah.. which goes to show he doesn't know enough about Flash actionscript...[/quote]

lol i guess. Right Click Forward works for a bit, but fails at don't clikc the button. How would you do it?[/quote]

How would I do what? All the author had to do was add one line of script to remove the menu, and another to disable keyboard shortcuts... Anyway I can't remember now, but I think I used right-click play, and besides the game is just silly...


conics
ellipse
analytic geometry
superior algebra
superior algebra solved

We have an ellipse with the extremities A,A',B,B' and let M(x,y) be a variable point on it.Let M1(x,0) and M2(0,y).Find out the geometric locus of the intersection of BM1 and AM2.

I made some calculations and obtained an equation which indicates an order 4 curve,but in the book i found the problem says its an hiperbola.

I used the fact that for every M on the ellipse there is a t such that
M(x,y)=M(a cost, b sint) where OA=a and OB=b.
Using this I got to a description for the point of intersection of the 2 lines by (x(t),y(t))=(a cost(1-sint)/(1-sint*cost), b sint(1-cost)/(1-sint*cost)).

I don't think this is an hiperbola. Notice that as M aproaches A,A',B or B', the intersection point of AM2 and BM1 also aproaches A,A',B or B' respectively. So the curve passes through A,B,C,D and it is not something the hiperbola does, or does it? Anyway, notice that the curve leaves the interior of the ellipse only for pi<t<3pi2.

Yeah, that's what i got too. Th only difference is that I tried seeing what the locus is for a circle (I looked at th ellipse as being the projection of a circle on a plane). The coordinates I got were the same as amfulger's, the only differenc being that a=b (his notation). I plugged this into Mathematica and got a really strange shape. It's definitely not a hyperbola. First of all, it looks different for the 4 different quarters of the circle.

[quote="amfulger"]I got to a description for the point of intersection of the 2 lines by (x(t),y(t))=(a* cost(1-sint)/(1-sint*cost), b* sint(1-cost)/(1-sint*cost)).
[/quote]

Because that curve does not have a nice geometrical form, it cannot be named nicely, so my description is the best we can get.
I guess it will not hurt my conscience to move this topic to solved  :D .


inequalities
geometry
perimeter
inequalities proposed

Consider $ABC$ a triangle with perimeter $1$ and sides lenghts $a, b, c$. Find the maximum value of $E=(a+b)^c   (b+c)^a  (c+a)^b$ ;)

We  can  assume  [tex]a\geq b\geq c[/tex]  so  its   easy  to find that [tex]ln(b+c)\leq ln(c+a)\leq ln(a+b)[/tex].Use  Chebychev  to  find  :
[tex]3(cln(a+b)+aln(b+c)+bln(c+a))\leq(a+b+c)(ln(a+b)+ln(b+c)+ln(c+a))=ln((a+b)(b+c)(c+a))   (1)[/tex]

By  am-gm  we  have :
[tex](a+b)(b+c)(c+a)\leq[\frac{2(a+b+c)}{3}]^3=\frac{8}{27}    (2) [/tex]

Use  (1) , (2)  to find [tex]E\leq\frac{2}{3}[/tex]. The  equality  holds  if-f  [tex]a=b=c=\frac{1}{3}[/tex]


The incircle of $\triangle ABC$ of radius $r$ is tangent to the sides $AB,BC,AC$ at  $N,Y,H$ respectively. Denote the distances from the points $N,Y,H$ to the sides $BC,AC,AB$ by $d_N,d_Y,d_H$ respectively. Prove that
$\sum_{cyclic} \frac{\sqrt d_H}{\sqrt d_Y+\sqrt d_N}(\frac{1}{d_Y}+\frac{1}{d_N}) \geq \frac{2}{r}$


Problem 1. For $ a,b,c > 0$ such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 1$. Prove that:

$ \frac {a^2 \plus{} b}{\sqrt {a \plus{} b}} \plus{} \frac {b^2 \plus{} c}{\sqrt {b \plus{} c}} \plus{} \frac {c^2 \plus{} a}{\sqrt {c \plus{} a}} \geq \frac {4}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1}$

Tran Cong Dieu

I hope it will be solve by everyone

Problem 2. For constant $ k,l \geq 1$ and $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:

$ \sqrt {\frac {a}{a \plus{} kb} \plus{} l} \plus{} \sqrt {\frac {b}{b \plus{} kc} \plus{} l} \plus{} \sqrt {\frac {c}{c \plus{} ka} \plus{} l} \geq \sqrt {3\frac {a \plus{} b \plus{} c}{k \plus{} \sqrt [3]{abc}}\plus{}9l}$

Tran Cong Dieu

[quote="vipCD_A1"]Problem 1. For $ a,b,c > 0$ such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 1$. Prove that:

$ \frac {a^2 \plus{} b}{\sqrt {a \plus{} b}} \plus{} \frac {b^2 \plus{} c}{\sqrt {b \plus{} c}} \plus{} \frac {c^2 \plus{} a}{\sqrt {c \plus{} a}} \geq \frac {4}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1}$
[/quote]
We have
\[ LHS \ge \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} a \plus{} b \plus{} c}{\sqrt {a \plus{} b \plus{} c}} \ge 2\sqrt {a^2 \plus{} b^2 \plus{} c^2}
\]
It suffices to show that
\[ \sqrt {a^2 \plus{} b^2 \plus{} c^2} \ge \frac {4}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 1}
\]
which is true since $ a^2 \plus{} b^2 \plus{} c^2 \ge a^4 \plus{} b^4 \plus{} c^4 \equal{} 1$.
Equality holds if and only if $ a \equal{} 1,b \equal{} c \equal{} 0$ or any cyclic permutations. :)

[quote="vipCD_A1"]Problem 2. For constant $ k,l \geq 1$ and $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:

$ \sqrt {\frac {a}{a \plus{} kb} \plus{} l} \plus{} \sqrt {\frac {b}{b \plus{} kc} \plus{} l} \plus{} \sqrt {\frac {c}{c \plus{} ka} \plus{} l} \geq \sqrt {3\frac {a \plus{} b \plus{} c}{k \plus{} \sqrt [3]{abc}} \plus{} 9l}$

Tran Cong Dieu[/quote]
It's wrong for $ a\equal{}\sqrt{2},b\equal{}1,c\equal{}0,k\equal{}l\equal{}1$.


After the examination, one of my friends made a suggestion:
"Start the exam from the last question. That way, if you don't know how to do the hard question, you won't disappointed that much. On the other hand, if you start from the first question and you don't know how to do it, you are screwed for the rest of the three hours because you don't even know how to do the EASIEST question."
well.. it was nice joke after three hours of AIME.. but it made sense eventhough the first few questions were easy..  :rotfl:

Though he makes a valid point, I like to get some of the easier questions "under my belt" before I get discouraged by the harder ones.

[quote="nat mc"]I would do mine starting at number 1 and go on, doing the easy ones, and skipping ones with bad computation or the ones I get stuck on. I go to the end and then start again with the ones I haven't done and work them all through.[/quote]

So the order for me would be: 1,2,3,4,5,6,7,8,9,10,11,12,13,14, 15, 7,8,9,10,11, 12, 13, 14, 15, 7,8,9,10 11, 12, 13, 14, 15, 7,8,9,10,11, 12, 13, 14, 15, 11, 12, 13, 14, 15...........................
An infinite series! :)  :(


prove sum of the five angles of a star is $180$.

suppose the name of this pentagon is$ABCDE$.

I should prove that: $\angle CAD+\angle BEC+\angle ADB+\angle ECA+\angle DBE=180$ $(1)$

It is very easy to prove it:

Suppose $AC$ intersects $BE,BD$ at points $F,G$ respectively.

In triangle $AGD$, $\angle CAD+\angle ADB=\angle AGB$  $(2)$

In triangle $EFC$, $\angle\angle BEC+\angle ECA=\angle CFB$  $(3)$

If we sum the equalities $(2),(3)$ we have:

$\angle CAD+\angle ADB+\angle\angle BEC+\angle ECA=\angle AGB+\angle CFB$

if we add $\angle DBE$ to this equality :

$\angle CAD+\angle ADB+\angle\angle BEC+\angle ECA+\angle DBE=\angle AGB+\angle CFB+ \angle DBE$

but the three angles in right hand of the equality, are the angles of triangle $BFG$, and sum of

them is equal to $180$.

There for the left hand of the equality should also be equal to $180$ :

$\angle CAD+\angle ADB+\angle\angle BEC+\angle ECA+\angle DBE=180$

and the problem is solved.

I didn’t want to solve it this way.i really want to solve it by inversion.

………………………………………………………………………………………………………...

the inversive proof:

I will invert my shape round point $A$ and an arbitrary radius.

Suppose $B',C',D',E'$ are the inversive form of points $B,C,D,E$ after inversion.

And suppose $C',D',$ are in one side of $B'E'$, (the other cases have similar ways)

We know that: $\angle B'AD'+\angle D'AE'=\angle B'AE'$

Then: $\angle B'AD'=\angle D'AE'-\angle B'AE'$

Then:$180-\angle B'AD'=180-\ angleB'AE'+\angle D'AE'$

finally: $\angle AB'D'+\angle AD'B'=\angle AE'B'+AB'E'+\angle D'AE'$

$\angle AB'D'+\angle AD'B'-\angle AE'B'-AB'E'-\angle D'AE'=0$

In triangle $AC'E'$, $\angle C'AD'+\angle AE'C'+\angle AC'E'+\angle D'AE'=180$

Then I add $180$ to the recent equality.

$\angle AB'D'+\angle AD'B'-\angle AE'B'-AB'E'- \angle D'AE'+\angle C'AD'+\angle AE'C'+\angle AC'E'+\angle D'AE'=180$

and it shows us that:

$\angle ADB+\angle ABD-\angle ABE-\angle AEB- \angle DAE+\angle CAD+\angle ACE+\angle AEC+\angle DAE=180$

and again it shows us, that :

$\angle CAD+\angle BEC+\angle ADB+\angle ECA+\angle DBE=180$

and every thing is ok.

should only me prove it?

is there any one with a different solution.

may be this problem is so easy, which have only two ways.

one, angle chasing. and the other inversion. ;)

[quote="JBL"]It is much easier than either of your solutions -- use the fact that an inscribed angle is half the intercepted arc.[/quote]

yes it is easier than that one.

good idea ;)

[hide]The five triangles on the sides of the pentagon have a total sum of interior angles $5\times 180$ degrees. We want to subtract the base angles, which represent two sets of exterior angles of the pentagon.
$5\times 180-2\times360=180$ degrees.[/hide]

[quote="Altheman"]the star is not necessarily inscribed in a circle...this proof is easy if you use symmetry[/quote]

is pianofort solution synethic?

and $Ashegh$, does inversion solve all problems?

Not at all...

only about $99$ pecent :rotfl:

Ah, Al, you're absolutely right -- I assumed the star was regular.  How about this, then: suppose we have a standard $2n + 1$-pointed star.  (By standard, I mean that if the vertices are labelled 1, 2, 3, ..., 2n + 1, then vertex 1 is connected to vertex n + 1 and vertex n + 2.)  Find the sum of the pointy angles in this case.

I should note that pianoforte's solution works here, but it also relies on the regularity of the star.  It needs further arguments as to why the pentagon exists in general, or what to do if it doesn't.

Here's one solution that covers this problem:
[hide]If a polygonal path winds around, always in the same orientation (clockwise or counterclockwise), the sum of the exterior angles of the path will necessarily equal the number of loops multiplied by 360 degrees.  Thus, for a simple polygon, the sum of the exterior angles is always 360 degrees.  For a pentagonal star, it's 720 degrees because there are two loops.  In general, a standard star of $2n + 1$ vertices has $n$ loops, so there are $360n$ degrees of exterior angle and $180(2n +1)$ total degrees of interior angle + exterior angle.  By subtraction, we're done.[/hide]


analytic geometry
graphing lines
slope
algebra proposed
algebra

Determine:
$min_{p,q \in R}max_{x \in (1,2)}(px^{2}+qx-1)^{2}$.
In other words determine the adequate values for the reals $p,q$ in order to minimize the maximum value for $(px^{2}+qx-1)^{2}$ when $x \in (1,2)$.

There could be no maximum in the open interval $(1,2)$ (only a lowest upper bound), so I suppose the problem was meant to be with a closed interval:
$\min_{p,q \in R}\max_{x\in[1,2]}(px^{2}+qx-1)^{2}$.

I'm working with the equivalent condition
$\min_{p,q \in R}\max_{x\in[1,2]}f(x)$,
$f(x)=\mid px^{2}+qx-1\mid$.

Since $f$ is differentiable, $\max_{x\in[1,2]}f(x)=\max\{f(1),f(2),f(m)\mid f'(m)=0,m\in[1,2]\}$.
$f(1)=\mid p+q-1\mid$.
$f(2)=\mid 4p+2q-1\mid$.
$m=\frac{-q}{2p}\in[1,2]$ $\to$ $f(m)=\mid \frac{-q^{2}}{4p}-1\mid \le \max_{-4p\le q\le-2p}\{\mid-p-1\mid ,\mid-4p-1\mid \}$.

We should study every case.
If I'm not wrong, the solution is when $f(1)=f(2)=0$, $f(m)=\frac{1}{8}$, when $p=\frac{-1}{2}$, $q=\frac{3}{2}$.

[quote="lordWings"]If I'm not wrong, the solution is when $f(1)=f(2)=0$, $f(m)=\frac{1}{8}$, when $p=\frac{-1}{2}$, $q=\frac{3}{2}$.[/quote]

Wrong - just shift your line with the same slope $p=\frac{-1}{2}$ slightly slow for a smaller $q$ so...?

Yes, it was an error. I assumed the balance $f(1)=f(2)$ was enough, when it clearly wasn't.

In fact, the local optima will be achieved with the balance $M=f(1)=f(2)=f(m=\frac{3}{2})$.
Solving the equations, there are four local optima,
$M=1$, when $p=0$, $q=0$.
$M=\frac{1}{17}$, when $p=\frac{-8}{17}$, $q=\frac{24}{17}$.
$M=\frac{1}{5}$, when $p=\frac{-4}{5}$, $q=2$.
$M=\frac{-1}{11}$, when $p=\frac{-4}{11}$, $q=\frac{14}{11}$.

Among them, the absolute minimum is $M=\frac{1}{17}$, when $p=\frac{-8}{17}$, $q=\frac{24}{17}$.


Give $ a,b,c \ge 0$
Prove
$ \frac{a}{b}\plus{} \frac{b}{c}\plus{} \frac{c}{a}\plus{} \frac{kabc}{a^3\plus{}b^3\plus{}c^3} \ge \frac{k\plus{}9}{3}$
and $ k \equal{} 9m^2$.m is root of equation $ m^3 \plus{} 3m^2 \minus{} 1 \equal{} 0$ and $ 0 < m < 1$
We have $ k > 2,54$.For $ k \equal{} \frac {9}{4}$,posted by me and $ k \equal{} \frac {5}{2}$,posted by canhang_2007.

With $ k \equal{}\frac {8}{3}$, I had had a nice proof by Am-Gm. The best constant $ k$ ~ $ \frac {11}{4}$
But with $ k \equal{} \frac {11}{4}$, I have no solution for it  :ninja:

@nhocnhoc or canhang_2007[hide]anh oi!,em hoi? chut'em muon' viet 1 file pdf voi' dau de la "tuyen? tap. cac' bai` bat' dang thuc",em muon' lay' 1 so' bai tren mathlinks,vi' du. lay bai` cua cac' anh chang han,the em co phai duoc su dong y' cua tac' gia? khong a.[/hide]


[quote="afaouzi"]Find all functions f : [u]R[/u] to [u]R[/u] such that:

For all x and y real numbers we have:   [b]f(x+f(y))=y+f(x)[/b]

Oops! I discovered that there is a same problem ( answered) for integer numbers.

So: Find all functions f : [u]R[/u] to [u]R[/u] such that:

For all x and y real numbers we have:   [b](f(x+f(y)))²=(y+f(x))(x+y)[/b][/quote]
Let a be: f(0)
x=0 $f(f(y))^2=(y+a)y$............(*)
x=y=0  $f(a)^2=0 \Rightarrow f(a)=0$
x=0, y=a $a^2=2a^2$ therefore a=0
x=0 $f(f(y))^2=y^2$
If f(t)=0, therefore $t^2=f(f(t))^2=f(0)^2=0$ therefore t=0
So $t \neq0 \Rightarrow f(t) \neq0$
y=0 $f(x)^2=xf(x)$
If $x \neq0$ then f(x)=x
Hence $f(x)=x; \forall x \in \mathbb R$ since f(0)=0

$f(x+f(y))=y+f(x)$
Setting $y=0$ we wil get $f(x+f(0))=f(x) \rightarrow f(0)=0$
so from here $f(x)=x$

Its not so simple as you wrote:
Solutions are $f(x)=x$ & $f(x)=-x$
Take $x=0$ then $f(f(y))=y+f(0)$ so if $f(y)=f(t)$ then $y=t$ this means that $f$ is one to one.
Take $y=0$ then $f(x+f(0))=f(x)$ then $x+f(0)=x$ so $f(0)=0$
Put $t=f(y)$ we have $f(x+t)=f(x+f(y))=f(x)+y=f(x)+f(t)$ so $f$ verify causchy's equation and thus $f(x)=f(1)x$ for all rationals.
Now from $f(f(y))=y$ we conclude that $f$ is a surjection as it's one to one then $f$ is a bijection
So $f$ is strictly increasing or strictly decreasing and finaly $f(x)=f(1)x$ for all reals.
Plug this in the original equation gives $f(1)^2=1$


Fnd all the positive integral roots of the equation mx=m+10x,for integral values of m. :lol:

[hide]You can rewrite the equation as $x=\frac{m}{m-10}$. Since x is positive, m>10.

When m=11, x=11.

When m=12, x=6

When m=15, x=3

When m=20, x=2.

We can't go any higher, because m>m-10, meaning m>1. The smallest integer greater then 1 is 2.[/hide]

[color=green][size=75]mod edit: hide solutions![/size][/color]

You also need to show there are no solutions for $m<0$, in which case $x$ would be positive as required.

[hide]Suppose $m<0$. Either $x=1$ and we have $m=m+10$, which is impossible, or $x>1$. Then $mx<m\Longrightarrow{m}x-m<0<10x\Longrightarrow{m}x<m+10x$, a contradiction.[/hide]

How about this(pretty same solution)ps-i dont know how to do the gide stuff.

x=m/(m-10)
x=1+10/(m-10)

since x is +ve,therefore m-10 must be a divisor of 10.
SO,m-10=1,2,5 or 10.
then do the rest :D


Let $ ABC$ be a triangle with angle $ A$ obtuse. Let $ Q$ be a point other than $ A,B,C$ on the circumcircle of triangle $ ABC$, on the same side of chord $ BC$ as $ A$, and let $ P$ be the other end of a diameter through $ Q$. Let $ V,W$ be the feet of the perpendiculars from $ Q$ onto $ CA,AB$ respectively. Prove that triangles $ PBC,AVW$ are similar.

We have: (with $ V$ inside $ AC$ and $ W$ outside $ AB$, other case is similar!)
$ \angle{WAV}\equal{}180^0\minus{}\angle{BAC}\equal{}\angle{BPC}$
And $ \angle{WVA}\equal{}\angle{WQA}\equal{}90^0\minus{}\angle{WAQ}\equal{}90^0\minus{}\angle{BCQ}\equal{}\angle{BCP}$ then OK!


A census-taker knocks on a door, and askes the woman inside how many children she has and how old they are.
        "I have three daughters, their ages are whole numbers, and the product of the ages is 36," says the mother.
        "That's not enough information," responds the census-taker.
        "I'd tell you the sum of their ages, but you'd still be stumped."
        "I wish you'd tell me something more."
        "Okay, my oldest daughter Annie likes dogs."
What are the ages of the daughters?

This is a classic problem that seems impossible but it's not.
[hide="Hint"]"I'd tell you the sum of their ages, but you'd still be stumped."
 ;) This is an important line. ;)  [/hide]

This problem is already in the Puzzles and Brainteasers subforum in Other Topics.


inequalities
geometry
function
circumcircle
perimeter
area of a triangle
Heron\u0027s formula

In a triangle $\Delta ABC $ let $\;a,\;b,c\; , \;R,\;r\;$ the usual elements.

[b]Questions:[/b]

[b]i)[/b]  Prove that there exists a  triangle  $\;A_1B_1C_1\;$ having the  length-sides $\;\sqrt{a},\;\sqrt{b},\;\sqrt{c}$;

[b]ii)[/b] Show that $\;\Delta A_1B_1C_1\;$ is an acute triangle;

[b]iii)[/b] Find area $ F_1 $ of $\;\Delta A_1B_1C_1\;$  as a function of   $\;R\;$ and $\;r$;

[b]iv)[/b]  If $\;R_1\;$ is the radius of circumcircle of  $\;\Delta A_1B_1C_1\;$ prove that
\[\frac{R}{R_1^2}\ge \sqrt{3}\; .\]

i) We must prove $ \sqrt{a} + \sqrt{b} > \sqrt{c} $
which is equivalent to $  a+b+2\sqrt{ab} > c $ ( obviously true )

ii) To prove $ A1B1C1 $ is an acute triangle we must prove $ b+c>a $ (obviously true)

iii) Use Heron's formula we have
$ 16F1^2=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c}) $
so $ 16F1^2=2(ab+bc+ca)-a^2-b^2-c^2 $
Use the indentity $ ab+bc+ca=p^2+4Rr+r^2 $ we have $ F1=\frac{\sqrt{r(4R+r)}}{2} $

iv) $ \sqrt{abc}=4R1F1 $
so $ abc=16R1^2F1^2=4R1^2r(4R+r) $
Replace $ abc=4pRr $ (p denotes perimeter) we have $ \frac{R}{R1^2}=\frac{4R+r}{p} $
Therefore, all we have to do is proving $ \frac{4R+r}{p} \geq\sqrt{3} $ which is equivalent to $ tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2} \geq\sqrt{3}  $


modular arithmetic
quadratics
pigeonhole principle
number theory proposed
number theory

let $ f(x) \equal{} x^{2} \minus{} a$ . Prove that, for any given a such that $ a \equal{} 1(mod8)$, and any given integer n>2, there exist integers t such that f(t) is fully divisible by $ 2^{n}$
Also, there exist 2 such integers t incongruent mod $ 2^{n}$

One possible solution:

If $ x^2\equiv y^2\pmod{2^n}$ (x and y are odd integers) then $ x\equiv y \pmod{2^{n-1}}$ or $ x\equiv -y \pmod{2^{n-1}}$

To prove this you just have to write $ x^2-y^2=(x+y)(x-y)$ and since $ (x+y) + (x-y)=2x$ is divided exactly by 2, then one between $ x+y$ and $ x-y$ is divided exactly by 2, so the other must be divided by $ 2^{n-1}$ (because the product is divided by $ 2^n$).

So if $ x^2-a$ has a solution modulo $ 2^n$ then it has at most 4 solutions. There are $ 2^{n-1}$ odd residue class modulo $ 2^n$ therefore there are at least $ 2^{n-3}$ odd quadratic residue modulo $ 2^n$

Since if an odd number is $ %Error. "unequiv" is a bad command.
1 \pmod 8$ it can't be a quadratic residue $ \pmod 8$, so it can't be a quadratic residue $ \pmod{2^n}$ for $ n\ge 3$

If we suppose (by absurd) that there exist an integer $ a\equiv 1\pmod 8$ that is not a quadratic residue $ \pmod{2^{n-1}}$ then there are at most $ 2^{n-3}-1$ odd quadratic residue, which is a contradiction, because we proved that there are at least $ 2^{n-3}$ quadratic residue.

I hope that I helped you.

Hello, thankyou for your concern.
I've got some problems understanding your argument. How did you come to the conclusion that there are atleast $ 2^{n\minus{}3}$ odd quadratic residues mod $ 2^{n}$ ?

Also, isn't it true that the congruence $ x^{2}\equal{}a(mod2^{n})$ where a is odd, posseses solutions iff a=1(mod8) for every n>2
More generally, for odd a the congruence $ x^{n}\equal{}a(mod2^{m})$ has exactly one solution if n is odd and in the case n is even, it has $ 2^{q\plus{}1}$ solutions or no solution according as $ a\equal{}1(mod2^{q\plus{}2})$ or not where $ 2^{q}\equal{}(n,2^{m\minus{}2})$

This is a special case of [url=http://en.wikipedia.org/wiki/Hensel%27s_lemma]Hensel's lemma[/url].

Yes t0rajir0u. I built this problem using Hensel's lemma. I was expecting solutions not employing this result

[quote="lajanugen"]I've got some problems understanding your argument. How did you come to the conclusion that there are atleast $ 2^{n \minus{} 3}$ odd quadratic residues mod $ 2^{n}$ ?[/quote]

Because mod $ 2^{n}$ there are $ 2^{n\minus{}1}$ odd residues. If there were less than $ 2^{n \minus{} 3}$ odd quadratic residues, then (by pigeonhole, or Dirichlet principle, I don't know which one is the correct name) there are 5 odd residues that give the same residue when you square them. That is to say you have $ a_1,a_2,a_3,a_4,a_5$ (which are distinct mod $ 2^{n}$) such that $ a_1^2\equiv a_2^2\equiv a_3^2\equiv a_4^2\equiv a_5^2\pmod{2^n}$

This is absurd, because you can have maximum four of them (as I proved in my earlier post).

Is it clear now?

Btw, I think that you have to be a bit careful when using Hensel's lemma here, because $ 2|f'(x_0)$ but I think that you can easily fix this problem.

[quote="piever"]Btw, I think that you have to be a bit careful when using Hensel's lemma here, because $ 2|f'(x_0)$ but I think that you can easily fix this problem.[/quote]
The general lemma just requires $ v_p(f(x_0)) > 2 \cdot v_p(f^\prime(x_0))$.


vector
linear algebra
linear algebra unsolved

Let $F$ be a finite field of size $q$, and let $V$ be an $n$-dimensional vector space over $F$. Show that there are
\[ \frac{(q^n-1)\cdots (q-1)}{(q^k-1)\cdots (q-1)(q^{n-k}-1)\cdots (q-1)} \]
subspaces of $V$ of dimension $k$.

consider all pairs (W,g)
where W is a k dimensional subspace, and g is a basis for W

now count all possible pairs

you can do it in two ways : for every subspace W there $(q^k-1)\cdots (q-1)$ possibilities

however you can also count all basissen of a k dimensional subspace, and for every such basis there is exactly one subspace of dimension k of course

there, that should give you the number of all W's


Here is a simple game that I made up. I wouldn't be surprised if some variation already exists, but there are none that I know of.

The game is for two players and rules are as follows:

-Each player writes down an ordered triple of nonnegative real numbers (x, y, z) such that x + y + z = 1.
-Both players reveal their triples.
-The player whose triple "defeats" the other player's triple wins.

A triple (x1, y1, z1) is said to "defeat" another triple (x2, y2, z2) if any two of the following are true:
* x1 > x2
* y1 > y2
* z1 > z2

The greater-than sign ">" can be used to represent this relation. However, note that the relation is not transitive.

Examples:

(0.66, 0, 0.33) > (0.33, 0.66, 0)
(0.33, 0.66, 0) > (0, 0.33, 0.66)
(0, 0.33, 0.66) > (0.66, 0, 0.33)

(0.33, 0.33, 0.33) > (0.8, 0.1, 0.1)
(0.8, 0.1, 0.1) > (0.5, 0.5, 0)
(0.5, 0.5, 0) > (0.33, 0.33, 0.33)

"<>" denotes that neither triple defeats the other

(0, 0.5, 0.5) <> (0, 0.4, 0.6)
(0.33, 0.33, 0.33) <> (0.33, 0.33, 0.33)

"t" refers to a triple (x, y, z)
"tn" refers to (xn, yn, zn)

Some properties that are easy to prove, e.g.:

1. Given any triple (x, y, z), there exists a third triple (a, b, c) that defeats it.
2. Given any two triples (x1, y1, z1) and (x2, y2, z2), there exists a triple (a, b, c) that defeats both

Here are some conjectures I am working on:

-Given the "symmetric set" of a triple (a, b, c): {(a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a)}, there exists a triple (d, e, f) that defeats more than half of the elements in the symmetric set.

-Given a finite set of triples {(x1, y1, z1), (x2, y2, z2),... (xn, yn, zn)} there exists a triple (a, b, c) that defeats more than half of the elements in the set.

-There exists an infinite set of triples such that no triple defeats more than half of the elements in the set. (If possible, prove this conjecture by describing an example of such a set.)

Define "collection" as an unordered list of elements that can contain duplicates of the same element. One "collection" is equivalent to another if it contains the same number of each element for each "element class" in either collection. For example, [1, 2, 2, 3] is a collection which is equivalent to [3, 2, 1, 2], but not to [1, 2, 3]; all three collections contain only the element classes {1, 2, 3}. There is probably an official term for this; please inform me of it if you know!

-Given an infinite collection of triples containing a finite number of distinct element classes [t1, t1, t1... t2, t2, t2... ... tn, tn, tn...], there exists a triple (a, b, c) that defeats more than half of the elements in the collection.

-There exists an infinite collection of triples such that no triple defeats more than half of the elements in the collection. (If possible, prove this conjecture by describing an example of such a collection.)

For what it's worth, I find it helpful to visualize the triples as points on the equilateral triangle in the xyz plane. ((0.33, 0.33, 0.33) is the center of the triangle; (1, 0, 0), (0, 1, 0), (0, 0, 1) are the corners.)

Any help on these conjectures, or further analysis of this game would be deeply appreciated! And feel free to ask any clarifying questions.

EDIT: Should this be in the Strategy Games subforum? My apologies if this is the case!

EDIT: As some of you may have noticed, I have been constantly adding to this post as I have been trying to clarify my ideas on paper. Apologies for any inconvenience.

[quote="snarles"]For what it's worth, I find it helpful to visualize the triples as points on the equilateral triangle in the xyz plane. ((0.33, 0.33, 0.33) is the center of the triangle; (1, 0, 0), (0, 1, 0), (0, 0, 1) are the corners.)[/quote]

This is interesting.  For a given point, try to graph the locus of points that defeat it and the locus of points it can defeat.

White = the triple t (x, y, z)
Yellow = the set of all points u, where t > u
Black = the set of all points s, where t < s

Playing around with this, you see visually why the triples (1, 0, 0), (0, 1, 0), (0, 0, 1) can't beat anything, and why (0.33, 0.33, 0.33) would be rather strong against an opponent who randomly picks any triple in the triangle.

PS: I love how you can use attachments in these forums.  :)


linear algebra
matrix
linear algebra unsolved

If $ A \equal{} \left[\begin {array}{cc} \minus{} 12 & 14 \\
\minus{} 27 & 27 \end {array} \right]$ then find $ A^n \equal{} ?$ when $ n\in N$

$ \left[\begin {array}{cc} 2 & 0 \\
0 & 27 \end {array} \right] ^{n} \equal{} \left[\begin {array}{cc} 2^n & 0 \\
0 & 27^n \end {array} \right]$

(But I wonder how eigenvalues of $ 6,9$ can yield a matrix with $ 2$ and $ 27$ diagonal values after diagonalisation.) :lol:

$ A \equal{} PA'P^{ \minus{} 1} \Rightarrow A^n \equal{} PA'^{n}P^{ \minus{} 1}$
You know all that is needed to calculate RHS. P is the matrix formed by writing two linearly independent column eigenvectors (in order of eigenvalue taken in diagonal matrix).

[quote="phhj"][quote="QuyBac"]Read http://www.mathlinks.ro/viewtopic.php?t=266247
 $ A$ have eigenvalues are $ 6,9$  diagonalize it ![/quote]
If diagonalize it $ \left[\begin {array}{cc} 2 & 0 \\
0 & 27 \end {array} \right]$  . Then how to solve? Could you explain?  :maybe:[/quote]

It seems to me that you have performed some (arbitrary) row or column operations to obtain the diagonalised form. I can see two column operations on the given matrix that get you "your" result of diagonalisation.

But diagonalising is just not about making it a diagonal matrix, its about finding a diagonal matrix [b]similar[/b] to the original matrix.
[url]http://en.wikipedia.org/wiki/Diagonalizable_matrix[/url]

I could see instantly that $ \begin{bmatrix}2 & 0 \\
0 & 27\end{bmatrix}$ was not similar to the original matrix.

How "instantly"? Similarity of matrices preserves the [b]trace[/b] of the given matrix and it preserves the [b]determinant[/b] of the given matrix. (Similarity preserves other things as well, although for the $ 2\times 2$ case, there aren't really any other things - determinant and trace are the whole show for that case.)

The trace of the original matrix is $ \minus{} 12 \plus{} 27 \equal{} 15.$ The trace of the proposed matrix is $ 2 \plus{} 27 \equal{} 29.$ That's not the same, so they aren't similar. Note also that the determinant of the original matrix is $ ( \minus{} 12)27 \plus{} 14\cdot27 \equal{} 2\cdot 27 \equal{} 54.$ (The incorrect proposed matrix did have the correct determinant but not the correct trace.) Hence the eigenvalues $ r$ and $ s$ (the elements that will be on the diagonal of the diagonal matrix) must be numbers such that $ r \plus{} s \equal{} 15$ and $ rs \equal{} 54.$ In other words, what QuyBac said.


This is a slight adjustment from a recent AIME problem:

[i]f[/i] is a complex-valued function on the complex numbers such that  [i]f[/i](x) = (a + bi)z, where  a and b  are real and |a + bi| = 6.  If  [i]f[/i](z)  is always equidistant from 0 and z, what is the value of b^2?

Hmm, which AIME is this from? I don't want to do it if its on a past AIME that I have and that I want to take in a real exam condition.

This problem is similar to #9 on the 1999 AIME.

Hmm, this one was quite simple for a #9. The solution (in spoiler):

[hide]f(z) = (a+bi)z = az + bzi. If f(z) is always equidistant from 0 and z, then az = (1/2)z <=> a = 1/2. Noting that |a + bi| = 6 and substituting, we have |1/2 + bi| = (1/2)^2 + b^2 = 6 <=> b^2 = 23/4.[/hide]


$ a,b,c,d,e,f$ real number . System solve:

$ 4a\equal{}(b\plus{}c\plus{}d\plus{}e)^{4}$
$ 4b\equal{}(c\plus{}d\plus{}e\plus{}f)^{4}$
$ 4c\equal{}(d\plus{}e\plus{}f\plus{}a)^{4}$
$ 4d\equal{}(e\plus{}f\plus{}a\plus{}b)^{4}$
$ 4e\equal{}(f\plus{}a\plus{}b\plus{}c)^{4}$
$ 4f\equal{}(a\plus{}b\plus{}c\plus{}d)^{4}$

Clearly all are positive.
Assume $ a$ largest(not strictly).
$ 4a\geq 4f\implies (b \plus{} c \plus{} d \plus{} e)^4\geq (a \plus{} b \plus{} c \plus{} d)^4\implies e\geq a$
Thus all were equality. $ a \equal{} f \equal{} e$
$ 4e\geq 4d\implies (f \plus{} a \plus{} b \plus{} c)^4\geq (e \plus{} f \plus{} a \plus{} b)^4\implies c\geq e$
Thus all were equality. $ e \equal{} d \equal{} c$
Similarly $ a \equal{} b \equal{} c$
Thus $ a \equal{} b \equal{} c \equal{} d \equal{} e \equal{} f$
Then clearly $ a \equal{} b \equal{} c \equal{} d \equal{} e \equal{} f \equal{} 0$ or $ a \equal{} b \equal{} c \equal{} d \equal{} e \equal{} f \equal{} 1$

P.S. I think I did something wrong. It looked too easy.


How would you make a border around the page?

The [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/fancybox.html]fancybox package[/url] will do this.

I've downloaded the package but I don't understand the command for framing the page. Please help?

The commands are on page 17 of the documentation (= page 16 at the top of the page).
Simple code such as [code]\fancypage{\fbox}{}[/code] puts a box round the text on the page but excludes headers and footers
[code]\fancypage{}{\fbox} [/code] puts a box around the page [i]including[/i] the header and footer.

Of course fancybox is capable of much more complicated effects than that, such as the framing of page 17 itself, but \fancypage should get you started.


(From ARML 1982)
A normal 6-sided die is thrown until 6 comes up.  What is the probability that 1, 2, 3, and at least one of 4 or 5 all come up before the 6?
i.e. a sequence of 341226 works but 1312222222226 does not.

[hide]I have $\frac{1}{27}\displaystyle\left(\frac{1}{6} + \frac{5}{6}*\frac{1}{6} + \frac{5}{6}^2 * \frac{1}{6} + \ldots \right)$, which is $\frac{1}{27}$.

I'm probably wrong though.[/hide]

That's not right.  What puzzled me is that the provided solution does not have an infinite series, but mine and yours did (my answer was wrong, too).

Well, I can't really see the gap in my reasoning yet.

Can you post the solution so I can compare? Though, I fear the solution may be on a totally different train of thought.

Here's my reasoning.

The probability that 1, 2, 3 and/or 4/5. Come before 6.

SO 1/6 * 1/6 * 1/6 * 1/3 * 24 (Different arrangements) is the probability that that happens. So that is 1/27.

Now the probability of the six coming right after that is 1/27 * 1/6. And the probability that a 6 comes after another digit after that is 1/27 * 5/6 * 1/6. No? So on and so forth.

Well what you're saying...that's only considering the possibilities when the 1, 2, 3, and 4/5 are in the first 4 rolls.  For instance, you're not considering the possibility 122356, because the 5 is not in the first four rolls.

Perhaps they trivialized it without losing generality?

Can anyone explain where the numbers came from?

The probability of getting ****6 is 2^4*3/6^5=1/162
The probability of getting *****6 is 5!(9)/6^6 (the 9 is because you could get 11234, 12234, 12334, 12344, 12345, 12355, 12335, 12235, 11235)=5/216
The probability of ******6 is 6!(31)/6^7=165/3^5*2^3
I don't know if there's a pattern...i think i calculated it out correctly...if there's a pattern, then you can find the overall sum..

[quote="Wumbate"][hide="Solution"]We are interested in first occurence sequences of the types: ••••6 or ••••4,6 or ••••5,6.  There are 5!+2(4!) of these.  Since there are 6! equally probable first occurence sequences, the desired probability equals $\frac{5!+2(4!)}{6!}=\frac{7}{30}$[/hide]
Explain?[/quote]
In this problem, repeating a number that you have already seen does not change your "state".  Thus the problem is equivalent to the following: Given a permutation of the numbers 1 to 6, what is the probability that the 6 comes after the 1, 2, and 3, and at least one of the 4 and 5.

There are two disjoint cases: (1) The 6 is the last number, or (2) The 6 is the second-to-last number with a 4 or 5 as the last number.  The first case has probability $1/6$.  The second case has probability $1/15$.  Adding both probabilities up gives $7/30$.


Find m such that the following equation has solution $ x \in (\minus{}1;3)$ :
$ (\minus{}x^2\plus{}4x\plus{}2)^2\plus{}2(\minus{}x^2\plus{}4x\plus{}2)\plus{}2m\minus{}1\equal{}0$

I don't think so .Please check your solution !

[quote="searcher44"]-47/2<m<1 :blush:  :blush:  :blush:[/quote]

You have alomost solved, the answer is $ \minus{}\frac{47}{2}\leq m<1$.

[quote="kunny"][quote="searcher44"]-47/2<m<1 :blush:  :blush:  :blush:[/quote]

You have alomost solved, the answer is $ \minus{} \frac {47}{2}\leq m < 1$.[/quote]
How do you solve it ?

The correct answer is $ \minus{}\frac{47}{2}\leq m \leq 1$

 :lol:


ratio
Asymptote
geometry
probability
perpendicular bisector

I had this question in a island wide competition/camp

The [hide="answer(as I remember it)"]
The ratio as I remember is 2:1
[/hide]

Please tell me if im right.

Actually, in this case it's $ 1: 2$, because of the order. And you need to prove your solution.

[quote="Temperal"]Let $ A,B$ be arbitrary points on a circle with center $ O$. Let point $ C$ be a point on the circle such that $ CO$ is the perpendicular bisector of $ AB$. Find the ratio $ \angle ACB: \angle AOB$.

[asy]import cse5; size(150); pen p=fontsize(10); pair O=(0,0), A=O+dir(60), B=O+dir(0), P=O+2/3*dir(30), C=O+dir(210); D(CR(O,1)); D(MP("A",D(A),NNE,p)--MP("O",D(O),WSW,p)--MP("B",D(B),ENE,p)); D(MP("A",D(A),NNE,p)--MP("C",D(C),WSW,p)--MP("B",D(B),ENE,p));[/asy][/quote]
[hide]
tonypr is right.  We will prove that $ \angle AOB \equal{} 2\angle ACB$ if C is anywhere.  Okay, so now, extend CO to reach the circle at P.  Say that $ \angle ACP \equal{} \alpha$ and $ \angle OCB \equal{} \beta$.  Since O is the center of our circle, we have that $ OA \equal{} OC$ and $ OC \equal{} OB$.  So, triangles $ OAC$ and $ OAB$ are isosceles.  Thus, we have that $ \angle A \equal{} \angle OCA \equal{} \alpha$ and $ \angle B \equal{} \angle OCB \equal{} \beta$.  Thus, we have that $ \angle AOP \equal{} \angle A \plus{} \angle OCA \equal{} \alpha \plus{} \alpha \equal{} 2 \alpha$.  Similarly, we have that $ \angle BOP \equal{} \angle B \plus{} \angle OCB \equal{} \beta \plus{} \beta \equal{} 2 \beta$.  So, we have that $ \angle AOB \equal{} \angle AOP \plus{} \angle BOP \equal{} 2 \alpha \plus{} 2 \beta \equal{} 2(\alpha \plus{} \beta) \equal{} 2(\angle ACO \plus{} \angle BCO) \equal{} 2\angle ACB$.  Thus, it is proven. [/hide]

[hide="My Solution"]
Not sure if this has been done already, and I'm actually really bad at geometry, so bear with me.

[asy]import cse5; size(5cm); pen p=black; pair O=(0,0), A=O+dir(60), B=O+dir(0), P=O+2/3*dir(30), C=O+dir(210); D(CR(O,1)); D(MP("A",D(A),NNE,p)--MP("O",D(O),NNW,p)--MP("B",D(B),ENE,p)); D(MP("A",D(A),NNE,p)--MP("C",D(C),WSW,p)--MP("B",D(B),ENE,p)); pen g = red+solid; D(A--B,g); D(O--C,g);[/asy]

Adding line segments at $ AB$ and $ OC$ (shown in red) we get three isosceles triangles - isosceles because two of their sides are also radii of the circle. Call the identical angles in each isosceles $ a$, $ b$ and $ c$, in the triangles $ \triangle AOB$, $ \triangle OBC$ and $ \triangle AOC$ respectively. Since the three isosceles also tessellate to make one larger (non-isosceles) triangle, $ a + a + b + b + c + c = 180$ or $ 2a + 2b + 2c = 180$.

However this does not involve $ \angle AOB$, so this is not useful in its current form. Due to the fact $ \triangle AOB$ is isosceles however, we can write $ a$ as $ \frac {180 - \angle AOB}{2}$. The algebra now follows:

\begin{align*}
2\left(\frac {180 - \angle AOB}{2}\right) + 2b + 2c &= 180\\
180 - \angle AOB + 2b + 2c &= 180\\
2b + 2c &= \angle AOB\\
b + c &= \frac {\angle AOB}{2}\\
\end{align*}

However, $ b + c$ is $ \angle ACB$ which we wanted to find in ratio to $ \angle AOB$. So the ratio is $ \boxed{1: 2}$ and we are done.
[/hide]

Both are correct.

Extension:

Suppose $ O$ is a just a random point on the circle, with everything else being fixed as it is now. Find the probability that the ratio $ \angle ACB: \angle AOB$ is less than $ 1: 2$.

EDIT: Sorry, AB is 4 units, $ \angle AOB \equal{} 30^{\circ}$.

[quote="Temperal"]Extension:

Suppose $ O$ is a just a random point on the circle, with everything else being fixed as it is now. Find the probability that the ratio $ \angle ACB: \angle AOB$ is less than $ 1: 2$.[/quote]
We need more information.  If $ C$ is on minor arc $ AB$, then $ \angle ACB$ could or might not be less than $ \frac {\angle AOB}{2}$.

So, we need the length of $ AB$ and $ \angle AOB$.


inequalities
geometry
trigonometry
trig identities
Law of Cosines

Try witth aalll your heart , i mean BRAIN

hint:use R<=2r and heron's formula for area

at last i have got a solution :D  :spider: errr..thanks anorithdialga

By the looks of it no one else is interested in the problem. no one else will solve it.

so can u post the solution for me ,? plz? :D

Let $ \angle ABC\equal{}2B$ , $ \angle BAC\equal{}2A$  and $ \angle ACB\equal{}2C$.
It`s easy to prove that $ 4.R .senA .sen B .sen C\equal{}r$ holds for every triangle ABC.
$ \implies \frac{r}{R}\equal{}4.sen A .sen B .sen C$
By the law of cosines we have that:
$ a^2\minus{}b^2\minus{}c^2\equal{}\minus{}2.b.c.cos2A$

$ 2a^2\minus{}(b\minus{}c)^2\equal{}a^2\plus{}b.c(1\minus{}cos2A)\equal{}8R^2.sen^2A(2 .cos^2A \plus{}2 .sen2B .sen2C)$

Then we have to prove that:
$ 32R^3.senA.sen B.sen C\sqrt{(2 .cos^2A \plus{}2 .sen2B .sen2C)(2 .cos^2B \plus{}2 .sen2A .sen2C)(2 .cos^2C \plus{}2 .sen2B .sen2A)}\geq \frac {r}{R}(a\plus{}b)(b\plus{}c)(a\plus{}c)$

$ \Leftrightarrow (2 .cos^2A \plus{}2 .sen 2B .sen 2C)(2 .cos^2B \plus{}2 .sen2A .sen2C)(2 .cos^2C \plus{}2 .sen2B .sen2A)\geq [(sen 2A \plus{}sen 2B)(sen 2B \plus{}sen 2C)(sen 2C \plus{}sen 2A)]^2$

But we can prove that $ 2.cos^2 A\geq sen^2 2C \plus{} sen^2 2B$

It true that:
$ \frac{\sqrt2}{2} \geq sen A$ because the triangle is acute.

$ \implies \sqrt2\geq 2 .cos (C\plus{}B)$ because $ A\plus{}B\plus{}C\equal{}90^o$

Suppose that $ B>C$

$ \implies \sqrt2.sen (B\minus{}C) \geq sen 2B \minus{}sen 2C$

$ \implies 2.sen^2 (B\minus{}C) \geq (sen2B \minus{}sen2C)^2$

$ \implies 2.sen^2 (C\plus{}B) \geq sen^2 2C \plus{} sen^2 2B$

Then:
$ (2 .cos^2A \plus{}2 .sen2B .sen2C) \geq (sen^2 2C \plus{} sen^2 2B \plus{}2 .sen2B .sen2C)\equal{}(sen 2C \plus{}sen 2B)^2$

And the inequality is proved.


Prove that if $ a,b,c>0$ then $ (1\plus{}a^3)(1\plus{}b^3)(1\plus{}c^3) \geq (1\plus{}abc)^3$
Generalize.

$ (1\plus{}a_1^n)(1\plus{}a_2^n)...(1\plus{}a_n^n) \ge (1\plus{}a_1a_2...a_n)^n$ which is true by holder  :wink:

OK ,Dimitris proof is correct , but let me give a proof by AM-GM for your problem.

By AM-GM

\[ \frac{1}{a^3 \plus{}1} \plus{}\frac{1}{b^3 \plus{}1 } \plus{}\frac{1}{c^3 \plus{}1} \geq \frac{3}{\sqrt[3]{(a^3 \plus{}1)(b^3 \plus{}1)(c^3\plus{}1)}}\]
\[ \frac{a^3}{a^3 \plus{}1} \plus{}\frac{b^3}{b^3 \plus{}1} \plus{}\frac{c^3}{c^3 \plus{}1} \geq \frac{3abc}{\sqrt[3]{(a^3 \plus{}1)(b^3\plus{}1)(c^3 \plus{}1)}}\]

Summing we have :

\[ 3 \geq \frac{3(abc\plus{}1)}{\sqrt[3]{(a^3 \plus{}1)(b^3 \plus{}1)(c^3 \plus{}1)}}\]

\[ \implies (1\plus{}a^3)(1\plus{}b^3)(1\plus{}c^3) \geq (1\plus{}abc)^3\]

In the same way you can generalise it .

Thanks Dimitris and enndb0x for the very nice solution.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Find a group $ G$ that is not an Abel group, and for any $ a\in G$, we have $ a^3\equal{}e$.

(So according to the Sylow Theorem we know $ G\equal{}3^l$; and it's obvious that $ l\neq 1,2$. So how about $ l\equal{}3$?)

And if we replace $ 3$ with $ 4$, the following 2 groups satisfy the conditions.

$ D_4\equal{}\left\{ \langle a,b \rangle: \ a^4\equal{}b^2\equal{}e,\ (ab)^2\equal{}e \right\}$
$ H\equal{}\{\pm 1,\ \pm i,\ \pm j,\ \pm k\}$

Take a Sylow 3-subgroup P of GL(3,3), the group of invertible 3×3 matrices with entries from a field of three elements.  For example, take P to the subgroup of GL(3,3) consisting of upper triangular matrices with 1s on the diagonal.  Such a group is also called a "Heisenberg group" and has presentation < x,y,z : x^3 = y^3 = z^3 = 1, yx = xyz, zx = xz, zy = yz >.

Indeed, and we can show that the Heisenberg group is isomorphic to $ G \equal{} (\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z})\rtimes \mathbb{Z}/3\mathbb{Z}$.

So we found a non Abelian subgroup of $ \mathrm{GL}_3(\mathbb C)$ of exponent $ 3$ and order $ 3^3$.

Follow up question: For which $ m > 2$ and $ n > 1$ can we find a non Abelian subgroup of $ \mathrm{GL}_n(\mathbb C)$ of exponent $ m$ and order $ m^n$?

I think you may have confused exponent with order.

For every integer n > 2, there is a non-abelian finite group G with x^n = e for all x in G.  If n is a power of 2, then G=D8 works and if p>2 is a prime dividing n, then the Sylow p-subgroup G of GL(3,p) works,

Your definition of S seems to be intended to omit those n such that there is no non-abelian finite group of order n.

For followup question 2, you probably want to look at restricted burnside groups.  They are the largest finite groups of given exponent and given number of generators.  The groups you are looking for are exactly their quotient groups.  However, even for n=7 and 2 generators, the restricted burnside group has order 7^20416, so even understanding 2-generated groups for n=7 at any level of detail is likely beyond the capabilities of the human mind.

We are right JackSchmidt, I had two ideas in mind, and in the end what I wrote didn't make much sense.
I edited above, it should make sense now :blush:


hello, are ppl from the us allowed to like talk here, or do u talk in hindi?-- i don't know a lot of hindi, only english and bengali.
ppl like u are smart -how do u get so smart?

yes ofc we all make friends here not segregate each other and u r right they r genuises present here(plzplzplz exclude me)like ritraj007,pardesi,choronz etc.
even i dont know there secret
and ofc we talk in hindi but sometimes..............
for further details u contact our mods.........
till then bye!

so u do talk in hindi, kewl......... lucky u.
i bet ur smart too, many ppl in america think asians r way too smart.
i believe so too.i know a little hindi, like kese ho meand how are u......i think...right???
i dunno if i even spelled that right...do ppl in ur school think it's good to b smart???

why not, u can talk here

the dominating language here is tamil but no problem about that

asians are smart but it is not their merit
they have to face such competition that makes them smart(or force them to be)

COOL SKAND
REALLY COOOOOOOLL!
AND BY THE WAY ITS NOWAY TO WORRY ON LANGUAGE
U JUST COME HERE AND RELAX!
AT LEAST THATS WAT MY AIM FOR COMING HERE IS!
BYE!

this is INDIA
where ppl just do it 4 fun!!!!!

You guys speak tamil??? I used to know a lot of tamil when I was 10, but now I can only understand it not speak.

i don't  :(
but most of the people here do
check out metathread,and earlier posts
you'll find some tamil

ummmmmm, things do seem different here, can i ask u somthing?what happens if you go off math topic, the admin in the us, or even in aops don't like "SPAM"  :spam: but there's this one music post and sorry if i'm wrong but that's not mathy....don't u care if somethings a bit off topic. i mean not to be too much of a goodie goodie, but its a habit i guess.....i was just wondering.......

here things r just cool !
so no hyping up!

yay!!!!  :lol:  :thumbup:
do any of you live in the northeast? that's where my cousins and relatives live.

remove the east from it and there is where i live!

but the place here i live is nearer to North East in comparison with others'
but i think none of us lives there(north east)

hey skand my place is not thaaaaaat far from urs!"
am i right !" :lol:  :o

ah so it is in the end rt-08 that finds it out. GOOD JOB! never heard of it? oh well! it ROX. [b]Disney [/b]is way too close to it -- don't know why u don't know of it!!!!

ofc we kno now and that's not bad at all :D  :D  :rotfl:

okay--so..........i don't think i have anything to say

i dont know where skand?
i havent chatted with my friend for quite  a long trime
hey crazy u know bout intel ISEF

OH MY GOSH, is it a science fair, i think i was in it last year. died!!! at regionals.... 2nd place so close....

i am here :D  :D

you may have got a chance to meet rt_08 there (this year)
btw,what is procedure?i mean to participate

u were in US regionals i suppose??

and rt ,who's sponsoring u ,IRIS or what??? :?:

well to be precise for IRIS it is national botanical research institute at lucknow and birbal sahni institute of paleobotany alnog with indian institute of toxicology research and central drug research institute and lucknow university ofc...........
for intel it is IRIS and iit's and iisc along with iiser
bye!

my teacher's sponsored me and ya us regionals. the math sector had only a few people it's wasn't that competitive, i didn't get any money-- math sectors don't i was upset i dunno if I'm doing one this year.

@rt_08

don't forget i am also sponsoring you :rotfl:

MY BLOG HAS A NEW POST!!!! I ENCOURAGE U TO VISIT!!!


search
number theory unsolved
number theory

Find all nonnegative integers $ a,b,c$ such that $ 2^a\plus{}2^b\plus{}2^c$ is a perfect square.

the case that $ a,b,c$ are even numbers,is discussed here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=160202[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1116561944&t=163569[/url]

Let $ a\ge b\ge c$.
0. If $ a\equal{}b\equal{}c$, then $ 2^a\plus{}2^b\plus{}2^c\equal{}3*2^c$ is not square.
1. If $ a\equal{}c\plus{}1,b\equal{}c$, then $ 2^a\plus{}2^b\plus{}2^c\equal{}2^{c\plus{}2}$ is square if and only if $ c$ is even.
2. If $ a>c\plus{}1,b\equal{}c$ and $ 2^a\plus{}2^b\plus{}2^c\equal{}2^{c\plus{}1}(1\plus{}2^{a\minus{}c\minus{}1})$ is square, then c is odd and $ k\equal{}a\minus{}c\minus{}1\ge 3$ and $ 1\plus{}2^k\equal{}x^2$ is square. It give $ x\minus{}1\equal{}2^m,x\plus{}1\equal{}2^l,k\equal{}l\plus{}m,2^l\minus{}2^m\equal{}2\to m\equal{}1,l\equal{}1,x\equal{}3$ unique solution.
Solution is (c+4,c,c) - c is odd.
3. $ a\equal{}b>c$ give unique solution $ (c\plus{}2,c\plus{}2,c)$ c -even.
Let $ a>b>c$, then c must be even and $ b\ge c\plus{}3$. Let $ y\equal{}2^{c/2}\plus{}2^{b\minus{}c/2\minus{}1}$, then $ y^2\equal{}2^c\plus{}2^b\plus{}2^k, k\equal{}2b\minus{}c\minus{}2$,
therefore $ x^2\minus{}y^2\equal{}2^a\minus{}2^k\ge 0$. If $ a\plus{}c\equal{}2b$ c even we get solution $ x\equal{}y$ or $ (a,b,c)\equal{}(2(m\plus{}k),2m,2(m\minus{}k))$.
4. $ a>k$ hadnot solution.

[quote="Rust"]4. $ a > k$ hadnot solution.[/quote]
How do you rule out solutions like $ 2^9 \plus{} 2^4 \plus{} 2^0 \equal{} 23^2$?

Yes. It is true only if $ a$ is even, when $ (2^{a/2})^2<2^a\plus{}2^b\plus{}2^c<(2^{a/2}\plus{}1)^2$.
Therefore my solution is not complete for case $ a>b>c$ a - odd, c - even. (we can consider only c=0, b>=3), then
$ 2^{2k\plus{}1}\plus{}2^m\plus{}1\equal{}x^2,3\le m\le k$.


Find number of $x$ such that $\lfloor \frac{x}{3}\rfloor+\lfloor \frac{x}{4} \rfloor+\lfloor \frac{x}{5} \rfloor =x$

obviously $x\in Z$ then use $x-1<\lfloor{x}\rfloor \leq x$.... :lol:
Interesting! :)


[quote="Pain rinnegan"]Let be $ a_1,a_2,...,a_n\ge 0$ . Prove that :

$ \frac {(1 \minus{} a_1)(1 \minus{} a_2)...(1 \minus{} a_n)}{(1 \plus{} a_1)(1 \plus{} a_2)...(1 \plus{} a_n)}\ge \frac {1 \minus{} (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)}{1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n}$[/quote]
Try to use the mathematical induction. :wink:


real analysis
function
real analysis solved

Consider $A$ the algebra of sequences $ (a_n)$  of real numbers with the property that the radius of convergence of  the series $ a_0+a_1x+...+a_n x^n+...$ is at least $R>0$ (which is fixed) (obviously, the product of convolution for sequences is adopted as product in $A$). Find all morphisms from $A$ to the algebra of real numbers.

They are the evalutaions in some $t$ with $|t|<R$.

Let $\varphi$ be the homomorphism, and let's identify the algebra $\mathcal A$ with the algebra of power series itself, instead of the representing sequences. We can also assume WLOG that $R=1$.

Clearly, $\varphi(1)=1$, and put $\varphi(1_{\mathbb R})=t$ ($1$ is the constant function $1$ and $1_{\mathbb R}$ is the identity function $x\mapsto x$; I'm trying to show that $\varphi(g)=g(t),\ \forall g\in\mathcal A$). For any $\alpha\in\mathbb R$ with $|\alpha|\ge 1$, there is a function $g\in\mathcal A$ s.t. $(\alpha-1_{\mathbb R})g=1$, so we conclude that $(\alpha-t)\varphi(g)=1$. In particular, $t\ne\alpha$, meaning that $|t|<1$. This shows that $g(t)$ makes sense for every $g\in\mathcal A$.

For every $g\in\mathcal A$, we have $g-g(t)=(1_{\mathbb R}-t)h\ (*)$ for some $h\in\mathcal A$, and if we apply $\varphi$ to $(*)$ we get $\varphi(g)=g(t)$, as desired.

I hope it's Ok.


[quote="linj"]1.The sides of triangle ABC lie on the lines:
3x+4y=0
4x+3y=0
x=3
Let (h,k) be the center of the circle inscribed in triangle ABC.
Find h+k.[/quote]
Do anyone know if I can upload a geometry figure. such as a triangle.

here's how you do it
XD too lazy to actually calculate

[hide]find the coordinates of the intersections to be
(0,0) (3,-4) (3,-9/4)
use 2r=a*b*c/area of the triangle to find the radius to be 25/4
then set up equation (x-m)^2+(y-n)^2=(25/4)^2, use the 3 coordinates to solve for m and n
[/hide]


On a quiz show, you must answer 8 consecutive questions. (Assume that for each question, your probability of answering correctly is 1/2.) One correct answer is worth 1 point. If you get 2 in a row, the second is worth 2 points. If you get 3 in a row, that question is worth 3 points, and so on. An incorrect answer is worth 0 points, and the next question's value starts over at 1. So for example:

If you get questions 2,3,4, 6 and 8 correct, you recieve 1+2+3 + 1 + 1 = 8 total points.

What is the probability that you will score at least 9 points?

$A) \frac{57}{256} B) \frac9{32} C) \frac{35}{128} D) \frac{13}{32} E) \frac12$

[hide]This isn't that hard.
If you scored 8 right, that's ${8\choose 0}/256$ or 1/256 chance of that happening.  Obviously, that must have scored beyond 9 points.
If you scored 7 right, that's ${8\choose 1}/256$ or 8/256 chance of that happening.  Of course, that must have also scored beyond 9 points.
If you scored 6 right, that's ${8\choose 2}/256$ or 28/256 chance of that happening.  No matter how the wrong answers are placed, you can't score lower than 9 points.
Now, let's work the other way.  If you missed all 8 of them, you couldn't have scored 9 points.  Likewise if you missed 7, 6, or 5.  If you miss 4, the only way to score beyond 9 points is to get 4 right in a row.  There's 5 ways to arrange that.  The chances of that happening is 5/256.
Now, we deal with the chances when you scored 5 correct.  The only patterns that yield 9+ points is scoring 3 in a row followed by 2 in a row, 4 in a row with one more, or if I scored 5 in a row.  There are 4 ways to get 5 in a row.  To get four in a row but not five in a row has 3+2+2+2+3 ways or 12 ways.  To get three followed by 2 in a row is 3+2+1+1+2+3=12.  The total is 12+12+4=28/256.  We add up the possibilities to get:
$(1+8+28+28+5)/256$ or $70/256=35/128$ which is C).[/hide]


Let [tex]g(z) = a_n z^n + \cdots + a_1 z + a_0[/tex] be a polynomial of degree > 0 with complex coefficients such that [tex]g(0)=0[/tex]. Show that there exists a complex number [tex]z_0[/tex] such that [tex]\displaystyle \left| g(z_0) - 1\right| < 1[/tex].

Please do not use [b]Fundamental Theorem of Algebra[/b] to prove it since this will lead to an elementary proof of the [b]Fundamental Theorem of Algebra[/b] (I will show how later).

i'm not too familiar with the notation, and probably couldn't solve the problem even if i knew what it said, but just for the sake of brining me into the loop, could you explain what the problem says (and what the notation means)?

I would recommend that you give a low-notation restatement of the problem -- I don't think I would have known what some of those things meant before taking college-level math classes.

*Edit: Ah, I see Nosepuppy beat me to the punch!

Anyhow, let me restate it:

Given a polynomial g with complex coefficients such that g(0) = 0, prove that there is some complex number z0 such that |g(z0) - 1| < 1.

[quote="JBL"]I would recommend that you give a low-notation restatement of the problem -- [/quote]
Point well taken. I have copied your restatement. Thanks.

Not too rigourous :

[hide]

f(z) = z^n * P(z) with P(0) != 0 and n != 0. P(z) = a0  + z*Q(z)

f(r*e^(i*a)) = a0 * r^n * e^(i*n*a)  (1+  r/a0 * Q(r*e^(i*a))

For r small enough, |f(r*e^(i*a))| < 1 and  f(r*e^(i*a)) takes all the values of the circle (0, r^n). One of these values has argument such as f(r*e^(i*a)) is a positive integer. This value works.[/hide]


Euler
geometry
circumcircle
ratio
analytic geometry
geometry proposed

Let a triangle $ABC$ and its circumcircle .Let $S$ be an arbitary point on the circle .Let $K$, $L$, $M$ the centres of the Euler circles of the triangles $SAB$ , $SBC$ and $SCA$ .Prove that the triangles $KLM$ and $ABC$ are similar and find also the ratio of their areas .

I think you mean $KLM$ if so
on the complex plane let $0$ be the center of the circle and $a,b,c ,s$ be the coordinates of $A,B,C,S$ the coordinates of $K,L,M$  are $\frac{a+b+s}{2},\frac{a+c+s}{2},\frac{b+c+s}{2}$ so the ratio of similarity is $\frac{1}{2}$

[quote="Albanian Eagle"]I think you mean $KLM$
[/quote]

Yes I meant $KLM$ ...
Your solution is same as mine .I believe it's hard to find another solution .
Bravo for solving the problem  :D

An elementary proof of this problem is the follows: ( In my drawing AB = 9.0, BC = 12.0, AC = 11.0, SB = 5.0, where S, on the arc BC not containing the vertex A of ABC ).

     Let K', L', M', the orthocenters of the triangles SAB, SBC, SCA respectively, which have the same circumcenter O, with the given triangle ABC.

     So, we have that KM//K'M' with (KM)/(K'M') = 1/2, (1) and KL//K'L' with (KL)/(K'L') = 1/2, (2) and LM//L'M' with (LM)/(L'M') = 1/2. (3).

    From the perpendicularities of the lines CM', K'B, to the same line AS, we have that CM'//K'B, (4)  and we will prove that K'M' //= BC.

     It is easy to prove that the quadrilateral AK'SM', is cyclic ( from the perpendicularities of the lines AB, K'S and SB, AK', we have that <AK'S = 180º - <ABS = <ACS and <AM'S = 180º - <ACS = <ABS ).

     So, <M'CB = <M'CS - <BCS = <M'AS - <BAS = <M'KS - <BK'S = <BK'M'

     =>  <M'CB = <BK'M', (5) and from (4), (5), => K'M' //= BC, (6).

     Similarly we can prove that K'L' //= AC, (7) and L'M' //= AB, (8).

     From (1),(2),(3),(6),(7),(8), we conclude that the triangles KLM and ABC are similar, with similarity ratio = 1/2.

     Kostas Vittas.


trigonometry
function
trig identities
tangent half angle

If given tanx=1/2

I need to find the following:

1. sin (2x)
2. cos (2x)
3. sin (4x)
4. cos (6x)

how can I do this without a calculator? Any help is appreciated, thank you.

You know that $ \sin x / \cos x \equal{} 1/2$ and that $ \sin ^2 x \plus{} \cos ^2 x \equal{} 1$. Now, you can find $ \sin x$ and $ \cos x$.

Now, you can use the sin and cos addition formulas to find sin and cos of (x+x)=2x.
From there, you can use sin and cos addition formulas to find sin and cos of (2x+2x)=4x
Finally, you can use cos addition formula to find cos of 6x

You can find the addition formulas for sin and cos [url=http://www.artofproblemsolving.com/Wiki/index.php/Trig_identity#Angle_Addition.2FSubtraction_Identities]here[/url], or in your textbook  :)

Let $ t \equal{} \tan x$.  The [i]tangent half-angle identities[/i] state that

$ \cos 2x \equal{} \frac{1 \minus{} t^2}{1 \plus{} t^2}$
$ \sin 2x \equal{} \frac{2t}{1 \plus{} t^2}$

which wouldn't be that interesting (they're nice formulae, but it's more useful to be able to derive them for basic identities) except that $ \sin 2x, \cos 2x$ are simultaneously rational [i]if and only if[/i] $ t$ is.  In other words, the above two functions rationally parameterize the circle.  :)


Is there a formula to find out what $ ord_{n}2$ for a certain $ n$ is ($ ord_{n}m$ is the smallest integer $ k$ such that $ m^{k}\equiv 1(mod n)$). If not, is there a reason why $ ord_{n}2$ cant divide $ n$ in certain cases? or general formulas for $ ord_{n}m$? Please feel free to post any related topics or formulas to orders and stuff.

This is, as far as I know, an open question.  Such a formula would resolve [url=http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots]Artin's conjecture[/url], which is open.

Of $ \mathrm{ord}_n(2)|n$ then you get $ n|2^n\minus{}1$, so this: http://www.mathlinks.ro/viewtopic.php?t=150382 .


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

What is the possible order of a matrix in $GL_{3}(Z/7Z)$? Is there a method for computing this for $GL_{n}(Z/pZ)$, even for small values of $n$ (not $n=1$  :D ).

It is a group. How many elements are in this group?

E=GL3(Z/7Z). If A is in E and order(A)=n then the irreducible factors of its minimal polynomial are divisors of x^n-1.
  The following is a Maple calculation; we give the p-values s.t. there exists an irreducible polynomial which is a divisor of x^p-1 and is not a divisor of x^m-1 with m<p; we give, in the left side, an example of such a polynomial.
degree 1:
x-1->1, x+1->2, x-2->3, x-3->6
degree 2:
x^2+1->4, x^2+3*x+1->8, x^2+2->12, x^2+x-1->16, x^2+x+4->24, x^2-x+3->48
degree 3:
x^3+3->9, x^3+2->18, x^3+2*x-1->19, x^3+2*x+1->38, x^3+x-1->57
Remark: for each preceding polynomial P of degree d we can find a (d,d) matrix s.t. its characteristic polynomial is P (think of companion matrices).
    diff(x^n-1)=n*x^(n-1); thus if 7 is not a divisor de n then A is diagonalizable over K, an algebraically closed extension of Z/7Z.
    If A is not diagonalizable let A=D+N where D is its diagonalizable part and N its non zero nilpotent part (DN=ND); A^7=D^7 and if 7 is not a divisor of k then A^k is not diagonalizable. The minimal polynomial of A is (x-a)^3 or (x-a)^2*(x-b) with a<>b; in the first case a is in Z/7Z; in the second case it has a multiple root and is reducible; we deduce that a,b are in Z/7Z; thus the possible orders of A are:
7*1,7*2,7*3,7*6.
  Conclusion: the possible orders are:
1,2,3,6
lcm(u,v) where u is in degree 1 list and v is in degree 2 list.
9,18,19,38,57
7,14,21,42
Remark: the maximal order is 57.

Rectification :
The degree 3 list is incomplete;
You must add:  x^3+x+1->114, x^3+3*x+5->171,
x^3+3*x+2->342
Thus the maximum order is 342.

The following maple program calculates the degree 3 list;
(time: 3")
> l:=vector(20):
> k:=0:
> for a from 0 to 6 do
> for b from 0 to 6 do
> for c from 1 to 6 do
> p:=z^3+a*z^2+b*z+c:
> if (Irreduc(p) mod 7)=true then
>  for i from 4 to 343 do
>      if (Rem(z^i-1,p,z) mod 7)=0 then
>          test:=0:
>          for j from 1 to k do
>          if i=l[j] then test:=1:break: fi:
>          od:
>          if test=0 then print(a,b,c,i);k:=k+1:l[k]:=i: fi:
>     break:
>     fi:
>     if i=343 then print(alert,a,b,c); fi:
>  od:
> fi:
> od:od:od:

In my last post the maple program uses implicitly the following results:
Over the field Fq, the product of all irreducible polynomials of degree p s.t. p is a divisor of n is:x^(q^n)-x.
Thus all non zero elements of the algebraic closure of Fq are roots of the unity.
  Moreover the maximal order of the companion matrix of an irreducible polynomial of degree n is inferior to q^n-1.
   Here q=7;  if n=3 this maximal order is <=
7^3-1=342; if n=2 this maximal order is <=7^2-1=48.


Just started to learn limits :) Please tell me if this is correct.

$f: \mathbb{R}_{+}^{*}\to \mathbb{R}_{+}^{*}$

$f(x_{1})=\frac{1}{x_{1}}$

$\lim_{x_{1}\to  \infty}f(x_{1}) = 0$

Second, can you help me find the limits of this functions and please explain your computations?

$f(x_{2})=x_{2}+3$

$\lim_{x_{2}\to  5}f(x_{2}) =$ ?

$f(x_{3})=x_{3}\sqrt{x_{3}}$

$\lim_{x_{3}\to  \infty}f(x_{3}) =$ ?

All are defined on $\mathbb{R}^{*}_{+}$

I might not be right, but isn't the limit of $f(x_{2})$ as $x_{2}$ approaches $5$ simply $5+3=8$?
As for the limit of $f(x_{3})$ as $x_{3}$ approaches infinity, it's $\infty$, is it not? Since both terms tend to infinity.

i'm in agreement with davidyko.
if you can see why the first example would tend to 0, then the others should be...basic

Well, the limit is 0 because when it tends to infinity the values get smaller and smaller, i see that. But i wouldn't be able to solve a more complicated limit since i don't really know the steps.

Let's try another one, $f(x_{4})= \sqrt{x_{4}}$
$\lim_{x_{4}\to 5}f(x_{4}) = \sqrt{5}$ ?

and $f(x_{5})=\frac{1}{x_{5}(x_{5}+1)}$

$\lim_{x_{5}\to 1}\frac{f(x_{5})-0.5}{x-1}=-500$ ?

I repeat, all are defined on [img]http://www.mathlinks.ro/Forum/latexrender/pictures/7e0da82fcdf1387bb35d774456ac6ace.gif[/img]

Edit: for the last one, i gave $x_{5}$ some values until i found that.

look at the graphs of these functions.
that should help you understand it
later there are limits that can't be done through just straight substitution. algebraic manipulation is required and sometimes you just have to graph it. other times right and left hand limits don't meet and you don't have a limit. then there are corners and cusps. are you studying from a textbook?

that last limit. I'm not understand why there's an x sub 5 and then just an x

kevyy, that was my mistake, it is supposed to be $x_{5}$ instead of $x$.

I took the basic definitions of limits from a book, i don't have much examples and that's why i make mistakes.

[quote="ckck"][quote="7h3.D3m0n.117"]Mods, please move this to a calculus board, thanks.[/quote]
Yea, thats just what I was thinking. This is wayy too advanced for the HSB forum.[/quote]

Yeah my knowledge on limits is severely limited so i couldn't asnwer any  :D


function
calculus
integration
trigonometry
limit
complex analysis
calculus computations

Prove that \[ \text{sech}(x) \equal{}4\sum_{k \equal{} 0}^{\infty}{( \minus{} 1)^{k}}\frac {(2k \plus{} 1)\pi}{(2k \plus{} 1)^2 \pi^2 \plus{} 4x^2}\]
This is need to prove [url=http://www.mathlinks.ro/viewtopic.php?t=208660]misan's topic[/url].

I'm just thinking out loud here so don't go nuts if it's wrong;

That sort of looks like a Fourier series so could it be computed that way?

Maple can't simplify that series, so i tried numericaly calculate it with different variables of $ x$ and found that both functions are equal not only in point $ e$:

$ f(x) \equal{} 4\sum_{k \equal{} 0}^{\infty}{( \minus{} 1)^{k}}\frac {(2k \plus{} 1)\pi}{(2k \plus{} 1)^2 \pi^2 \plus{} 4x^2},\;g(x) \equal{} \text{sech}(x)$

$ f(1) \equal{} 0.64805427$
$ g(1) \equal{} 0.64805427$

$ f(6) \equal{} 0.00495747$
$ g(6) \equal{} 0.00495747$

$ f(e) \equal{} 0.13140388$
$ g(e) \equal{} 0.13140388$

Alsow if it is true we can generalize misan's integral $ \int _ 0^\infty \; \frac { 2\cos \; 2a x} {\cosh \; \pi x } \; \; \textbf dx \; \equal{} \; \text{sech} \; a$
I'll think about Fourier :)

I found a solution. This formula can be proved using partial fraction decomposition. Here is theorem we should use:

Consider a function $ f(z)$, whose only singularities in the finite part of the plane are simple poles $ a_1,\;a_2,\;a_3,\;\dots$, where $ |a_1|\leq |a_2|\leq|a_3|\leq\dots$. Let $ b_1,\;b_2,\;b_3,\dots$ be residues at these poles, and let it be possible to choose a sequence of circles $ C_m$ (the radius of $ C_m$ being $ R_m$) with center at $ O$, not passing through any poles, such that $ |f(z)|$ is bounded on $ C_m$. Suppose further that $ R_m\to\infty$ as $ m\to\infty$ and that the upper bound of $ |f(z)|$ on $ C_m$ is itself bounded as $ m\to\infty$; so that, for all points in the circle $ C_m$, $ |f(z)| < M$, where $ M$ is independent of $ m$.

Then, if $ x$ be not a pole of $ f(z)$, since the only poles of the integrated are the poles of $ f(z)$ and the point $ z \equal{} x$, we have
\[ \frac {1}{2\pi i}\int_{C_m}\frac {f(z)}{z \minus{} x}dz \equal{} f(x) \plus{} \sum_{r}\frac {b_r}{a_r \minus{} x}
\]
where summation extends over all poles in the interior of $ C_m$.

But
\[ \frac {1}{2\pi i}\int_{C_m}\frac {f(z)}{z \minus{} x}dz \equal{} \frac {1}{2\pi i}\int_{C_m}\frac {f(z)}{z}dz \plus{} \frac {x}{2\pi i}\int_{C_m}\frac {f(z)}{z(z \minus{} x)}dz \equal{}
\]

\[ \equal{} f(0) \plus{} \sum_{r}\frac {b_r}{a_r} \plus{} \frac {x}{2\pi i}\int_{C_m}\frac {f(z)}{z(z \minus{} x)}dz
\]
if we suppose the function $ f(z)$ is analytic at the origin.

Now as $ m\to\infty$, $ \int_{C_m}\frac {f(z)}{z(z \minus{} x)}dz$ is $ O(R_{m}^{ \minus{} 1})$, and so tends to zero as $ m$ tends to infinity.

Therefore, making $ m\to\infty$, we have
\[ 0 \equal{} f(x) \minus{} f(0) \plus{} \sum_{n \equal{} 1}^{\infty}{b_n\left\{\frac {1}{a_n \minus{} x} \minus{} \frac {1}{a_n}\right\}} \minus{} \lim_{m\to\infty}\frac {x}{2\pi i}\int_{C_m}\frac {f(z)}{z(z \minus{} x)}dz
\]
i.e.
\[ f(x) \equal{} f(0) \plus{} \sum_{n \equal{} 1}^{\infty}{b_n\left\{\frac {1}{x \minus{} a_n} \plus{} \frac {1}{a_n}\right\}}
\]
which is an expansion of $ f(x)$ in rational fractions of $ x$; and the summator extends over all the poles of $ f(x)$.

Our topic in on the page 136 of the book [url=http://rapidshare.com/files/196322311/Whittaker_K.T.__Watson_G.N._A_course_of_modern_analysis__CUP__1935__T__609s__MCet_.djvu.html]Whittaker K.T., Watson G.N. A course of modern analysis (CUP, 1935)(T)(609s)[/url].

Using Fourier:

$ \sinh (bx)\equal{}\frac{2\sinh (b\pi)}{\pi}\sum_{n\equal{}1}^{\infty}(\minus{}1)^{n\plus{}1}\frac{n\sin (nx)}{b^2\plus{}n^2}$

Therefore

$ \frac{\pi}{2}\cdot\frac{\sinh(bx)}{\sinh(b\pi)}\equal{}\sum_{n\equal{}1}^{\infty}(\minus{}1)^{n\plus{}1}\frac{n\sin (nx)}{b^2\plus{}n^2}$

Since the series is valid for $ \minus{}\pi\leq x\leq \pi$, we may replace $ x$ by $ \pi\minus{}x$, and obtain

$ \frac{\pi}{2}\cdot\frac{\sinh(b\pi\minus{}bx)}{\sinh(b\pi)}\equal{}\sum_{n\equal{}1}^{\infty}\frac{n\sin (nx)}{b^2\plus{}n^2}$

Adding the two series, and nothing that all term with $ n$ even cancel, we obtain the result

$ \frac{\pi}{4}\cdot\frac{\cosh(\frac{1}{2}b(\pi\minus{}2x))}{\cosh(\frac{1}{2}b\pi)}\equal{}\sum_{n\equal{}0}^{\infty}\frac{(2n\plus{}1)\sin (2n\plus{}1)x}{b^2\plus{}(2n\plus{}1)^2}$

and in particular, if $ x\equal{}\frac{\pi}{2}$

$ \text{sech}\left(\frac{1}{2}b\pi\right)\equal{}\frac{4}{\pi}\sum_{n\equal{}0}^{\infty}(\minus{}1)^{n}\frac{2n\plus{}1}{b^2\plus{}(2n\plus{}1)^2}$

Maybe you're fully aware, but your first solution is borrowing from complex analysis.  I happen to like solutions using complex analysis.

And I see that you used my Fourier idea ( :D ).

[b]JRav[/b], :D i study complex analysis myself a little, but my knowledges are superficial, this theorem is in that book too, but I do not understand to the end how it works - contours, etc. Now i even don't remember how to calculate residue of a function, need to look to the book.


Because of the little Fermat's theorem, $a^{p-1}\equiv 1$ mod $p$, and so the sum is $\equiv 2004$ mod p. And so sum$\equiv 2004-L_p\equiv 0$ mod $p$..(look behind, man :) )

Awww, you're right, Anyway, we can use the equation $2004-L_p=0$ (mod $p$), where $L_p=[\frac{2004}{p}]$ or Lagrange-De puissin


No one interested in My hard inequality?

Let me post the more beutiful form:Given $x_{i}>0,\sum_{i=1}^{n}{x_{i}}=1,$ prove that $\sum_{i=1}^{n}{\sqrt{x_{i}^{2}+x_{i+1}^{2}}\leq 2-\frac{1}{\sum_{i=1}^{n}{\frac{x_{i}^{2}}{x_{i+1}}}+\frac34}}$
I can't edit the orginal post.....

Still no one interested in it?I think the second inequality is beautiful and strong. :blush: There is a weaker one:
Given $x_{i}>0,\sum_{i=1}^{n}{x_{i}}=1,$ prove that $\sum_{i=1}^{n}{\sqrt{x_{i}^{2}+x_{i+1}^{2}}\leq 2-\frac{1}{\sum_{i=1}^{n}{2\frac{x_{i}^{2}}{x_{i+1}}}}}$

I am so sad no one is intereseted in my problem :(
for the second one :
$x_{i}+x_{i+1}-\sqrt{x_{i}^{2}+x_{i+1}^{2}}\geq \frac{x_{i}x_{i+1}}{2(x_{i}+x_{i+1})}$
Please have a try on the problem in the fifth post.It is very strong, I think.

[i]You are really intelligent , Shalex

  Very nice Solution

  I have thought this problem is unsolvable

  Congratulation[/i] :lol:

[i]  The ineq of China is always beautiful[/i]

 [i] What a pity,  there  is no ineq in VietNam TsT 2008[/i]


Prove that there is no polynomial $f(x)$ with integer coefficients such that $f(7)=11$ and $f(11)=13$.

[hide]Use $a-b|f(a)-f(b)$ for $a,b\in\mathbb{Z}$.[/hide]

does integer coeficients mean : a_1,a_2,.....,a_n £ N

if that's the case
[hide]
if f(x) is 2sd degre or higher than 2sd degre

f(x)>=x^2

if f(x) is constant then f(11)=f(7)=k
if f(x) is a 1st degre polynom then

7a+b=11
11a+b=13

4a=2==>a=1/2 impossible cause a £ N

then there is no polynomial [img]http://www.mathlinks.ro/Forum/latexrender/pictures/3/e/0/3e03f4706048fbc6c5a252a85d066adf107fcc1f.gif[/img] with integer coefficients such that  [img]http://www.mathlinks.ro/Forum/latexrender/pictures/5/2/0/520b6ddb059e6dd80afa1a86de0ad9f0b2471e78.gif[/img]  and [img]http://www.mathlinks.ro/Forum/latexrender/pictures/d/d/8/dd88222a07130aa39401938e987dc4b2042d554a.gif[/img][/hide]


Here is my solution
  We can rewrite the ineq like this
  a/b+b/a+d/c+c/d+d^2/b+c^2/a+b^2/c+a^2/d>=4(1+E)
  But a/b+b/a+d/c+c/d>4
  And  \sum a^2/d> = \sum a>=(abcd)^(1/4)=E
  Adding the 2 inequalities we get what we want

  Hope the solution is corect!


we had 2 tsts
These are the problems :
1st Day

[b]Problem 1[/b]

Let the triangle [i][b]ABC[/b][/i] with[u][i] BC [/i][/u]the smallest side. Let [u][i]P[/i][/u] on [u][i](AB)[/i][/u] such that angle [u][i]PCB[/i][/u] equal angle [u][i]BAC[/i][/u]. and [u][i]Q[/i][/u] on side [i][u](AC) [/u][/i]such that angle [u][i]QBC[/i][/u] equal angle [u][i]BAC[/i][/u]. Show that the line passing through the circumenters of triangles [u][i]ABC[/i][/u] and [i][u]APQ [/u][/i]is perpendicular on [i][u]BC[/u][/i].

[b]Problem 2[/b] (very easy :))

Show that:
 a) there exist infinitely many positive integers with the from [i]3p+1[/i]  p - positive integer that can pe represented as m :^3: -n :^3: with m,n positive integers
 b) there exist infinitely many positive integers of the form [i]5q+1[/i], q - positive integer that can be written as
  k :^5: -l :^5: with k,l positive integers.

[b]Problem 3[/b]
Let [i]a1, a2, a3, ... , an [/i] positive real numbers. Let [i]S = a1 +a2 +a3 + ... +an [/i], [i]p=a1*a2*...*an[/i]; Show that 2 :^n: :sqrt: p  <= [i]1+ s/1! + s 2 /2! + s :^3: /3! + ... + s :^n: .

[b]Problem 4 [/b]
Let the function [i]f:N* --> N*[/i] with [i]f(1) = 1[/i] and that verifies:
 1) [i]3*f(n)*f(2n+1) = f(2n)*[1+3*f(n)];[/i]
2) [i] f(2n) < 6f(n), [/i], n positive integer.
Find all solutions of [i]f(k)+f(l) = 293, k<l. [/i]
                                                                2nd Day
[b]Problem 1 [/b]

Let the acuteangled triangle ABC, CF - altitude with F on side AB. BM - median, M on (CA) such that CF=BM.
Then the angle MBC is equal to angle FCA if and only if ABC is equilateral

[b]Problem 2[/b]
Let n be a positive integer and [i]x1, x2, . . . xn, [/i]positive real numbersthat verify :
       [i]   1/x1 +1/x2 + ... +1/xn = n. [/i]
Find the minimal value of
           [i]  x1+ x2 :^2: /2 + x3 :^3: /3 + ... +xn :^n: /n. [/i]

[b]Probelm 3[/b]
Let p be aprime number and [i]a[/i] and [i] n [/i] positive nonzero integers. Prove that
if 2 :^p: + 3 :^p: = a :^n: then [i] n=1 [/i].
[b] Problem 4[/b]
Let the functions [i]fm(x) = (m :^2: + 1)*x :^2: + 3mx + m :^2: -1, gm(x)=m :^2: x :^2: + mx - 1, where m real parameter.
Show that for any function h (2nd grade) that satisfies gm(x)<=h(x)<=fm(x) any real x, there exists  :lambda in [0, 1] that verifies: h(x) =  :lambda fm(x) + (1- :lambda )gm(x) any real x.

[quote="TheMaskedMagician"]Can someone help post these on the forums and I will add them to the contest page.[/quote]
I posted $1$,$4$,$5$ and $7$ problem. Hopefully someone will finish others
$1)$ [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=602387[/url]
$4)$ [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=602391[/url]
$5)$ [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=602388[/url]
$7)$ [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=602392[/url]


The largest integer such that seven times that number is less than 100, is

$ \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 16$

[hide]$ n\equal{}\left\lfloor\frac{100}{7}\right\rfloor\equal{}14\ \Rightarrow\ \mathrm{C}$[/hide]

[hide]Let the integer you are looking for be $ n$. Then, $ 7n<100 \Rightarrow n<\frac{100}{7} \approx 14.2857$. The greatest integer that satisifes the inequality is $ 14 \Rightarrow \boxed{C}$.[/hide]


quadrilateral $ B$ with coordinates \[ W(7,4), X(9,2), Y(4,2), and Z(2,4)\] is reflected about its side, $ WX$, to form quadrilateral A. Find the greatest distance between a point on quadrilateral $ A$ and a point on quadrilateral $ B$.

[hide="hint"]
The distance between Y and Z'
[/hide]


Tyler is half the age of his father and a quarter the age of his grandfather. If their total ages amount to $ 154$ years, how old is Tyler's grandfather?

[hide="Solution"] Call Tyler's age $ x$. We then have $ x\plus{}2x\plus{}4x\equal{}7x\equal{}154 \implies x\equal{}22$. Since Tyler is $ 22$ years old, his grandfather is $ 22\times{4}\equal{}\boxed{88}$ years old. [/hide]


I'm a sophomore this year in high school and I have doubled up on math in order to take precalculus this year. This means I am taking both Algebra II and Precalculus. However, my algebra class is accelerated (highest level) while my precalculus class is normal. I had to take the normal class because the accelerated was packed over the limit by like 10 kids. Will colleges look down upon my not taking the accelerated precalculus course?

(btw, I am self learning the topics not covered in normal precalc. I have taught myself series and sequences (but not the taylor series) and I am going to learn on my own polar coordinates.)

Don't worry about that - if you're strong at math (and it sounds like you are), you'll ace the AP Calc exam when the time comes.  Develop your problem solving skills and do well on the AMC/AIME test, and you'll really have nothing to worry about as far as convincing schools you know your math.


find all triplets  of positive integer $ (k,l,m)$ with sum $ 2010$ and for which the system has real roots

[size=150]$ \begin{array}{*{20}{c}} {\frac {x}{y} \plus{} \frac {y}{x} \equal{} k} \\
{\frac {y}{z} \plus{} \frac {z}{y} \equal{} l} \\
{\frac {z}{x} \plus{} \frac {x}{z} \equal{} m} \\
\end{array}$[/size]

[quote="imedeen"]find all triplets  of positive integer $ (k,l,m)$ with sum $ 2010$ and for which the system

[size=150]$ \begin{array}{*{20}{c}} {\frac {x}{y} \plus{} \frac {y}{x} \equal{} k} \\
{\frac {y}{z} \plus{} \frac {z}{y} \equal{} l} \\
{\frac {z}{x} \plus{} \frac {x}{z} \equal{} m} \\
\end{array}$[/size][/quote]

$ \frac{x}{y}\equal{}a$ , $ \frac{y}{z}\equal{}b$ , $ \frac{z}{x}\equal{}c$ ---> $ abc\equal{}1$

$ a\plus{}\frac{1}{a}\equal{}k$

$ b\plus{}\frac{1}{b}\equal{}l$

$ c\plus{}\frac{1}{c}\equal{}m$---> $ k^{2}\plus{}l^{2}\plus{}m^{2}\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{c^{2}}\plus{}6$

$ klm\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{c^{2}}\plus{}2$---->

$ k\plus{}l\plus{}m\equal{}2010$
$ k^{2}\plus{}l^{2}\plus{}m^{2}\equal{}klm\plus{}4$--> ...


A circular garden is enlarged so that the new diameter is twice the
old diameter. What is the ratio of the original area to the enlarged
area? Express your answer as a common fraction.

The ratio of the areas is the square of the ratio of any two corresponding lengths, so our answer is $ \left(\frac12\right)^2\equal{}\boxed{\frac14}$.


advanced fields
advanced fields unsolved

Evaluate $ 1\cdot 1!+2\cdot 2!+3\cdot 3!+\ldots+n\cdot n!$.

$ k.k!=(k+1)!-k!$, now use telescopic sum.

Is there other means of solving it using combinatorics?? By the way what is telescopic sum<it seems new to me>? thnx

No problem, but i think that, this problem (and some other) of you must post in Olympiad box or High School box. Please don't post them in College Playground  box.  :wink:


integration
parameterization
calculus
calculus computations

Find the length of a curve defined by $y=(1-{x^{\frac{2}{3}})^\frac{3}{2}}$ where $\frac{\sqrt{2}}{4}\le{x}\le{1}$

Some tweaks: In symmetric form, this curve has equation $x^{2/3}+y^{2/3}=1$. That gives us a parametric form $x=t^{3/2},y=(1-t)^{3/2}$, and arclength is usually nicer in parametric terms. The half of the curve you're looking at is $\frac12\le t\le 1$.

[hide="Solution"]$s=\int_{\frac12}^{1}\sqrt{\left(\frac32 t^{1/2}\right)^{2}+\left(-\frac32 (1-t)^{1/2}\right)^{2}}\,dt$
$s=\int_{\frac12}^{1}\sqrt{\frac94}\,dt=\frac34$.
This was rigged to be nice. The natural parametrization is almost an arclength parametrization already.[/hide]


inequalities
geometry
3D geometry
perimeter
circumcircle
USAMTS

[b]Prove that the cube of the perimeter of a triangle is greater than or equal to 108 times the product of its area and circumradius[/b]

Let $ p$ be the perimeter, $ T$ the area, $ R$ the circumradius and $ a,b,c$ the sides.
Just remember that $ p\equal{}a\plus{}b\plus{}c$, and $ T\equal{}\frac{abc}{4R}$

$ p^3\equal{}(a\plus{}b\plus{}c)^3\ge (3\sqrt[3]{abc})^3\equal{}27abc\equal{}108\frac{abc}{4}\equal{}108T\cdot R$

By AM.GM.

 :)

[quote="Temperal"]This, I recall, another one from AoPS 2.[/quote]
good that you remember it  :)
i liked this one a lot

I personally liked the USAMTS $ \sqrt{n}\le \$sqrt[n]{n!}$ one better (a bit harder than this one), but each to his own.

I still haven't done the Inequalities section :wink:
UH..I remember when I first got the book, and was just looking around, that I saw this one.


the circle with center C and radius P meet one sice of the angle C on E and the other on F. The perpendicolar line from E to CE meet the perpendicolar line from F to CF on K. The circle with center K and radius KE is the C-excircle $ \Gamma_C$. Call $ \Gamma$ the circle with center C and radius $ h_c$. Then the side BC is one of the 2 internal tangent line of $ \Gamma$ and $ \Gamma_C$.


geometry
geometric transformation
reflection
circumcircle

Given an acute triangle ABC with orthocenter H, consider the reflection of H with respect to each side. Prove that the three reflected points lie on the circumcircle of ABC.

Oh! It's very easy.

You can see my picture.

We have $ \angle H_1CB \equal{} \angle BCH$, but $ \angle BCH \equal{} \angle BAH$, (because $ \angle BCH \plus{} \angle ABC \equal{} \angle BAH \plus{} \angle ABC \equal{} 90^0$).

$ \Longrightarrow \angle BAH \equal{} \angle H_1CB \Longrightarrow$ $ A, B, H_1, C$ lie on a circle  $ \Longrightarrow$ lie on the circumcircle of $ \triangle ABC$

Suggestions we get $ H_2, H_3$ lie on the circumcircle of $ \triangle ABC$

Then OK.


trigonometry
calculus
integration
calculus computations

Calculate the integral

\[ \int\limits_{0}^{1}\left(\sin x\plus{}\sin \frac{1}{x}\right)\frac{dx}{x}.\]

Insight: $ \frac{dx}{x}$ is the Haar measure for the multipicative group $ (0,\infty).$

Upshot: $ \int_0^1\left(\sin \frac1x\right)\frac{dx}{x}\equal{}\int_1^{\infty}(\sin x)\,\frac{dx}{x}.$

Hence the integral didilica is asking for is $ \int_0^{\infty}\frac{\sin x}{x}\,dx$

- which is pretty well known, and has been calculated a few times on this site (including by me).