Well, you did, for one.  As did I.  I'm hoping for an interesting semi-final exam.  The quarter-final exam this year was disappointingly easy.

[quote="Benjamin Hu"]I was hoping to get in, but I didn't; all I know is that at least one person from my school got in.
I agree that the quarter-final exam was a bit simple this year.[/quote]
I am stupid, so I didn't get in.
Even though I got owned, I thought the problems were pretty bad, having extremely straightforward solutions.

I got in. The problems were easy, so I expected a high cutoff. It wasn't high enough to screw me over for messing up problem 3 though (I couldn't remember the formula $ C \equal{} \dfrac{\epsilon_0 A}{d}$).

I got in. I got problems 1, 2, and most of 4. I didn't attempt 3.

Just confirming suspicions, for number 3 all you had to do was find the difference in the stored energy in the capacitor right? Putting the plate in would halve the energy or something (I don't quite remember the problem,  but my impression was the plate would be the same size as one of the plates of te capacitor).

I first got #4, then #2. All I put for #3 was $ W \equal{} \minus{} q \Delta V$ and hoped that might be useful. I did #1, but might have misread the question because I got $ 1.7 \times 10^6$ solar masses.

Made it..

I got #4a and messed up b (which wasted me a bunch of time).  Got #1a, and about half of b (b/c of the time I wasted on #4).  Got all of #2.  For number, #3, couldn't remember the capacitance formula (like Mij), but I still put down the observation that serialk11r made.

Problem #3 defined $ g$ as the "acceleration due to gravity."
With up as the positive direction, does $ g\equal{}\minus{}9.8$ m/s or $ \plus{}9.8$ m/s? By convention, isn't $ g$ just the magnitude without the sign? I had $ g$ be positive and thus labeled the force due to gravity be $ \minus{}mg$. The other semifinalist at my school had $ g$ be negative and used $ F_g\equal{}mg$. Which is the correct one?
I feel as if I should already know this...

[quote="hillarryous"]Problem #3 defined $ g$ as the "acceleration due to gravity."
With up as the positive direction, does $ g \equal{} \minus{} 9.8$ m/s or $ \plus{} 9.8$ m/s? By convention, isn't $ g$ just the magnitude without the sign? I had $ g$ be positive and thus labeled the force due to gravity be $ \minus{} mg$. The other semifinalist at my school had $ g$ be negative and used $ F_g \equal{} mg$. Which is the correct one?
I feel as if I should already know this...[/quote]

I've always used g as equaling +9.8 as does my physics teacher.

same.. I would use $ g\equal{}9.8 m/s^2$ only.. However, don't confuse with g and a because $ a\equal{}\minus{}g\equal{}\minus{}9.8 m/s^2$

does anyone know when were allowed to discuss the tesT?


geometry
perimeter
parallelogram
trigonometry
combinatorics unsolved
combinatorics

Fedja used matches to put down the equally long sides of a parallelogram whose vertices are not on a common line. He figures out that exactly 7 or 9 matches, respectively, fit into the diagonals. How many matches compose the parallelogram's perimeter?

By cosine law

$ 2a^2\plus{}2a^2cosY\equal{}81$
$ 2a^2\minus{}2a^2cosY\equal{}49$

$ 4a^2\equal{}130$ ? This question cannot be right.

Does this question even belong in combinatorics?

[color=red]the answer is 22[/color]

We call $ a,b$ the two diagonals, with $ a \equal{} 7$ and $ b \equal{} 9$. $ c,d$ are the sides of the parallelogram.

[color=blue]it is well known that $ 2c^2 \plus{} 2d^2 \equal{} a^2 \plus{} b^2$. [/color]
So we have that $ c^2 \plus{} d^2 \equal{} 65$.

$ 9^2 < (c \plus{} d)^2$ because $ 9,c,d$ are sides of a triangle.

[color=blue]it is well known that $ x^2 \plus{} y^2\geq 2xy$[/color]

$ 9^2 < (c \plus{} d)^2 \equal{} c^2 \plus{} 2cd \plus{} d^2\leq c^2 \plus{} (c^2 \plus{} d^2) \plus{} d^2 \equal{} 2(c^2 \plus{} d^2) \equal{} 130 < 12^2$

So now we have two cases 1) $ c \plus{} d \equal{} 10$ and 2) $ c \plus{} d \equal{} 11$

[color=blue]1)[/color]

$ 4cd \equal{} 2(c \plus{} d)^2 \minus{} 2(c^2 \plus{} d^2) \equal{} 2\cdot 10^2 \minus{} 130 \equal{} 70\Longrightarrow cd \equal{} \frac {35}{2}$ but $ c$ and $ d$ must be integers.

[color=blue]2)[/color]

this case is obvious, and we find that $ 2(c \plus{} d) \equal{} 22$


Hi, could someone help me with a couple of questions with finding roots (and variables) in a polynomial.
The question is:

The cubic equation 2x^3 − x^2 + x − 1 = 0 has roots a, b and c. Find

a) a + b +c
b) ab + ac  + bc
c) a^2 + b^2 + c^2

Ok i may have just answered my own question here. Here is how i think it can be done.
First, find a root (for e.g. 1, 2 or whatever it is) which satisfies the equation so it equals 0.  If the root is 1 (just take it as an example). Then use long division to divide the entire question by (x-1) ...This will eventually give an equation at the end after the long division. Ok lets pretend the equation it comes down to is

2x^2 + 4x + 6

So then in order to do a + b + c .....i would just do 2 + 4 + 6 ...which is 12. ...is that how you do it?? If you don't understand, could someone do it for me ...but i'm pretty sure i'm correct, just not 100% sure. Thanks

The sum, product, etc. of the roots of a polynomial can be found by looking at the coefficients like so: if you have the cubic $(x-a)(x-b)(x-c)$, when this is factored out it becomes $x^3-(a+b+c)x^2+(ab+bc+ac)x-abc$. Now it's pretty easy to find what you need to find.

[quote="coolio2005"]Hi, could someone help me with a couple of questions with finding roots (and variables) in a polynomial.
The question is:

The cubic equation $2x^3-x^2 + x - 1 = 0$ has roots $a$, $b$ and $c$. Find

a) $a + b +c$
b) $ab + ac + bc$
c) $a^2 + b^2 + c^2$[/quote]
[hide]
Well...  part a) and b) can be done with Vieta's Theorem,
You can look up this site for the Theorem: http://mathworld.wolfram.com/VietasFormulas.html .

Answer for both a) and b) are $\frac{1}{2}$

For c) $(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$
Plug in  the values of a+b+c and ab+ac+bc.
[/hide]

Is the method i've said also correct?
Just finding the root of the equation, divide using long division, grab the equation which you get after long division, and look at the co-efficients from there?

so if the equation turns out to be 2x^2 + 5x + 9

a = 2
b = 5
c = 9

and just do your calculations from there?

Nope, that's not right. Try using a factored form and expanding it out.

If you could find one particular root (which won't be easy to do in many cases), you could then use that to factor the rest of the equation. However, it won't always factor out into linear terms, in which case it'll be better to use the relations between roots and coefficients.

[quote] So if the equation turns out to be 2x^2+5x+9

a=2
b=5
c=9
[/quote]

Those don't work as roots, the coefficients aren't the roots of the equation.

But your method is hard if the roots are not integers.
In other words, what if you can't find the root?

[quote="coolio2005"]My way does work, but yes, if a root cannot be found easily, then the long method is required. Thanks guys![/quote]

Hmm? The method you outlined does not work, the coefficients are not simply the roots. Plus, the method the others have outlined here is much faster  :)

[url=http://upal.ca/adeel/final_paper.pdf]Unfinished paper about polynomials...[/url] I'll try to finish it sometime this month.


[quote="lowest_reynolds"]Hey guys.  Ready for Thursday?[/quote]
Thurrrr...sday? How come I remember it as next Wednesday? Anyone please tell me I am wrong!

[quote="martha"]
Thurrrr...sday? How come I remember it as next Wednesday? Anyone please tell me I am wrong![/quote]

Haha, my bad.  Yeah it's Wednesday.

[quote]
Haha, my bad. Yeah it's Wednesday.
[/quote]

lol

[quote="Hellstar"]I guess its because we don't have much homework from school. So for me, it's either math or fooling around.[/quote]
Which grade are you?

[quote="martha"]
Which grade are you?[/quote]

11

[quote="Hellstar"]

11[/quote]
wow Grade 11 and not yet have homework.... sweet

gr 12, still dont do homework
lots of time on my hands
yet still no math practice
guess ill need a lot of luck, althoguh i did try to do some math this year, so i hope ive improved enough
good luck

I find these contests are a bit like DDR.  If you've ever played, once you become somewhat good (at getting high scores) you feel the difference between A, AA, and AAA scores come out to [b]luck[/b].  But if you watch people with ALOT of [b]practice[/b], they can consistently get AA/AAA.

[quote="TheDreamer"][quote="Sunny"][quote="Hellstar"]Making last ditch efforts. Increased math time to 4hrs/day (personal high). Dunno if it will do any good though seeing as I only got 118 on Cayley last year.[/quote]

4 hrs a day?!  I envy you the time![/quote]

Yes, how is it possible to have so much time!?

Anyways, good luck all..I know I'll need it; haven't had ANY time for math at all.[/quote]

I got more time, even though I'm in grade 12!!
My school rarely gives us homework. And when it does, it's usually optional. :D My quizes and tests are easy since most of them are multiple choice, and I'm [b]VERY[/b] good at guessing. :rotfl:


geometry
3D geometry
sphere
geometric transformation
dilation

[quote="shobber"]True or false: If the electric field strength is $0$, then the electric potential is also $0$.[/quote]
[color=blue]I don't think it's true because we know that $\large \vec{E}=-\vec{gradV}$
If $E=0$ then we can't affirm that $V=0$ but vice versa, if $\large V=0 \Longrightarrow E=0$
[/color]

yeah, think about inside a hollow conducting sphere at equilibrium

[color=blue] Yes, that's an example[/color]

no its not true and stop these stupid question and tell me how does entropy of a galaxy exceeds its pulsating dilation effect?
 :D

If the electrostatic field is null everywhere in a region (not only in one point), means that the electric potential is constant: null or not depends on the reference.


modular arithmetic
number theory proposed
number theory

Define a sequence $ \{a_n\}\in\mathbb{N}$ such that $ a_1\equal{}2,\ a_{n\plus{}1}\equal{}2a_n^2\minus{}a_n\plus{}2\ (n\equal{}1,\ 2,\ \cdots).$
Find the great common divisor of $ a_{n\plus{}2}$ and $ a_n.$

write $ a_{n \plus{} 2} \equal{} 2(2a_n^2 \minus{} a_n \plus{} 2)^2 \minus{} (2a_n^2 \minus{} a_n \plus{} 2) \plus{} 2$, which will expand to give you a multiple of $ a_n$, plus 8. thus $ \gcd(a_n,a_{n \plus{} 2})|8$. now the sequence $ \pmod{8}$ goes $ 2, 0, 2, 0, 2, ...$. so

$ \gcd(a_n,a_{n \plus{} 2}) \equal{} 2$ if $ n$ is odd,

$ \gcd(a_n,a_{n \plus{} 2}) \equal{} 8$ if $ n$ is even.


Hi all,
I have one question on this theorem: What is ''period-$m$ point'' ?  :maybe:
Thanks!

I think it's a point such that $f^{(m)}(x)=x$, where $f^{(m)}=f\circ f\circ\dots\circ f$ is the $m$th iterate of $f$.


analytic geometry
geometry
rotation
MATHCOUNTS

[hide="Solution"]Consider this, for each pair of points less than four away, there are two circles that satisfy this. There are ten of these which makes [b]$20$[/b] circles.[/hide]

[quote="perfect628"][hide="correction"]You're right, except that 2 sets of those points are exactly 2 away. So, those have only 1 passing through, so the answer is 20-2=18.[/hide][/quote]
Actually, they have to be exactly $4$ away, becaus the diameter is $4$.

Oops, well I actually meant that. There are 2 sets [b]4[/b] units away.

[quote="perfect628"]Oops, well I actually meant that. There are $2$ sets $4$ units away.[/quote]
Oh, I didn't include those in my counting. [hide="Real Answer"]$22$[/hide]

Err, I don't get it. Is it $\binom62$ for the pairs? Real solution please >__>
Yes, the answer is indeed [hide]18[/hide]

[quote="sonny"]The graph shows six labeled points. How many distinct circles of radius 2-units are in the coordinate plane and pass through exactly two of the labeled points on this graph?[/quote]
[hide]
I draw a 2-inch radius circle around each of the original six points in the drawing.  Wherever exactly 2 of those circles intersect, I have placed a red dot.  [I've placed 18 red dots, so I think that is the answer to the problem.]
[asy]
size(200,200);
pair p1,p2,p3,p4,p5,p6;
p1=(-1,-1); p2=(1,-1); p3=(3,-1);
p4=(-1,1); p5=(1,1); p6=(3,1);
filldraw(Circle(p1,.1));
filldraw(Circle(p2,.1));
filldraw(Circle(p3,.1));
filldraw(Circle(p4,.1));
filldraw(Circle(p5,.1));
filldraw(Circle(p6,.1));
draw(Circle(p1,2),dotted);
draw(Circle(p2,2),dotted);
draw(Circle(p3,2),dotted);
draw(Circle(p4,2),dotted);
draw(Circle(p5,2),dotted);
draw(Circle(p6,2),dotted);
pair c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18;
c1=p1+rotate(60)*(p2-p1);
c2=p1+rotate(-60)*(p2-p1);
c3=p2+rotate(60)*(p3-p2);
c4=p2+rotate(-60)*(p3-p2);
c5=p4+rotate(60)*(p5-p4);
c6=p4+rotate(-60)*(p5-p4);
c7=p5+rotate(60)*(p6-p5);
c8=p5+rotate(-60)*(p6-p5);
c9=p1+rotate(60)*(p4-p1);
c10=p1+rotate(-60)*(p4-p1);
c11=p2+rotate(60)*(p5-p2);
c12=p2+rotate(-60)*(p5-p2);
c13=p3+rotate(60)*(p6-p3);
c14=p3+rotate(-60)*(p6-p3);
c15=p1;
c16=p3;
c17=p4;
c18=p6;
dot(c1,red);
dot(c2,red);
dot(c3,red);
dot(c4,red);
dot(c5,red);
dot(c6,red);
dot(c7,red);
dot(c8,red);
dot(c9,red);
dot(c10,red);
dot(c11,red);
dot(c12,red);
dot(c13,red);
dot(c14,red);
dot(c15,red);
dot(c16,red);
dot(c17,red);
dot(c18,red);
[/asy]
If you draw a 2-inch radius red circle centered at each red dot, that circle will reach back to exactly two of the original points.  I've only drawn a couple of red circles, here, because 18 red circles makes a real mess!
[asy]
size(200,200);
pair p1,p2,p3,p4,p5,p6;
p1=(-1,-1); p2=(1,-1); p3=(3,-1);
p4=(-1,1); p5=(1,1); p6=(3,1);
filldraw(Circle(p1,.1));
filldraw(Circle(p2,.1));
filldraw(Circle(p3,.1));
filldraw(Circle(p4,.1));
filldraw(Circle(p5,.1));
filldraw(Circle(p6,.1));
draw(Circle(p1,2),dotted);
draw(Circle(p2,2),dotted);
draw(Circle(p3,2),dotted);
draw(Circle(p4,2),dotted);
draw(Circle(p5,2),dotted);
draw(Circle(p6,2),dotted);
pair c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18;
c1=p1+rotate(60)*(p2-p1);
c2=p1+rotate(-60)*(p2-p1);
c3=p2+rotate(60)*(p3-p2);
c4=p2+rotate(-60)*(p3-p2);
c5=p4+rotate(60)*(p5-p4);
c6=p4+rotate(-60)*(p5-p4);
c7=p5+rotate(60)*(p6-p5);
c8=p5+rotate(-60)*(p6-p5);
c9=p1+rotate(60)*(p4-p1);
c10=p1+rotate(-60)*(p4-p1);
c11=p2+rotate(60)*(p5-p2);
c12=p2+rotate(-60)*(p5-p2);
c13=p3+rotate(60)*(p6-p3);
c14=p3+rotate(-60)*(p6-p3);
c15=p1;
c16=p3;
c17=p4;
c18=p6;
if (false)
{
dot(c1,red);
dot(c2,red);
dot(c3,red);
dot(c4,red);
dot(c5,red);
dot(c6,red);
dot(c7,red);
dot(c8,red);
dot(c9,red);
dot(c10,red);
dot(c11,red);
dot(c12,red);
dot(c13,red);
dot(c14,red);
dot(c15,red);
dot(c16,red);
dot(c17,red);
dot(c18,red);
}
dot(c1,red);
draw(Circle(c1,2),red);
// draw(Circle(c2,2),red);
// draw(Circle(c3,2),red);
// draw(Circle(c4,2),red);
// draw(Circle(c5,2),red);
// draw(Circle(c6,2),red);
// draw(Circle(c7,2),red);
// draw(Circle(c8,2),red);
// draw(Circle(c9,2),red);
// draw(Circle(c10,2),red);
// draw(Circle(c11,2),red);
// draw(Circle(c12,2),red);
// draw(Circle(c13,2),red);
// draw(Circle(c14,2),red);
// draw(Circle(c15,2),red);
dot(c16,red);
draw(Circle(c16,2),red);
// draw(Circle(c17,2),red);
// draw(Circle(c18,2),red);
[/asy]
@Aequilipse:  ${6\choose 2}=15$ is the number of pairs of dotted circles.  Each pair can meet at 0, 1, or 2 points.  You can do some casework.  2 of these 15 pairs don't intersect at all.  2 of these 15 pairs intersect at 1 point.  And 11 of these 15 pairs intersect at 2 points.  All together, this is $2\times 0+2\times 1+11\times 2=24$ intersections to consider.  However, if you look at the drawings, there are 2 places where three dotted circles meet, so we lose 6 of our possible intersection points.  Net result is $24-6=\boxed{18}$ distinct circles that meet the conditions of the problem.
[/hide]

I think the answer is not 18

[hide]combination of C(6,2)=15
But we need to exclude 2 combinations, which is
(-1,1) and (3,-1)
(-1,-1) and (3,1)
That give us 13 combinations.
But most combination can produce 2 distinct circle. So result is 13*2=26 circles.

There are two sets of points only produce 1 distinct circle
(-1,1) and (3,1)
(-1,-1) and (3,-1)

Minus there are two set of circles that duplicates...
When the circle has a center point of (1,1) or (1,-1), it will past through 3 dots, so we will need to remove 2 more circles.

So we have 26-2-2= 22 distinct circles for these pionts.

[/hide]

...

This is a MathCounts State problem, and the answer is 18.

@rhvg, this is the [b]sprint[/b] round, not the target. And not everybody has a Ti-89 Titanium.

yes, right, the answer is 18, but it just can't be counted even using case work. You can only count it one by one and get the answer.  I found out now, my answer 22 is wrong because the circle contain {(-1,-1),(1,1),(3,-1)} and the circle contain {(-1,1),(1,-1),(3,1)} was triple counted... You can only count it one by one or just draw circles on these 6 points to get the right answer. :(

Hmm ... I wouldn't say that casework is impossible. I've given a relatively quick casework solution below.

[hide="Solution"]
[u]Case 1: Center of Circle is On Given Points[/u]
We note that since the center needs to be $2$ units away from exactly two other given points, only the four points on the left and right of the given set will work. The center two points are $2$ away from three other points, so they cannot be the centers of a circle. Hence, in this case, there are $4$ centers, and thus $\boxed{4}$ circles that work.

[u]Case 2: Center is Outside the Given Points[/u]
We see that the center must be $2$ units away from exactly two other points. This means that if we choose a center from outside the given set, any two adjacent points on the same "edge" will work. The center will form an isosceles triangle of leg length $2$ with those two given adjacent points. Note, however, that any two non-adjacent points will not work, as we cannot construct an isosceles triangle of leg length $2$ without creating a "flat" triangle. Also note that any two adjacent points that are adjacent through the inside of the set will not work, as we cannot construct an isosceles triangle of leg length $2$ outside the set.  Hence, we note that there are $6$ pairs of adjacent points, and thus $\boxed{6}$ possible circle centers.

[u]Case 3: Center is Inside the Given Points[/u]
Again, we note that we can construct a circle through exactly two points if the two points are adjacent to each other. However, this time, we can include points that are adjacent "internally" to each other, since the center doesn't have to be outside the set. Hence, we see that for "vertical" sets of adjacent points, there are $\boxed{4}$ possible centers (the center vertical pair has $2$, while the two edge pairs each have $1$), and for "horizontal" sets of adjacent points, there are also $\boxed{4}$ possible centers (one for each horizontal pair). Hence, we have a total of $\boxed{8}$ circle centers in this case.

In total, we have:
\[4+6+8=\boxed{18\text{ circle centers}}\]
and thus $18$ circles.

And we are done. [/hide]

Apologies for the rather ambiguous solution. If you have any questions, feel free to ask. :)


induction
real analysis
real analysis solved

1) Prove that the sequence (an)n>=1 with a1=1 and a2=0 and:
5^(a_(n+1))=3^a_n+2^(a_(n-1))+(3n-8)(a_n-a_(n-1)) converges.
(Ploiesti Olympiad)
2) Let the sequence (x_n)n>=0 defined by x0=1 and x1=2  and:
12^x_n=2*9^(x_(n-1))+10*3^((x_(n-2)). Study the convergence of x_n and if it is convergent compute its limit.
(Arges Olympiad - 2003)

cheers! :D :D

for the second problem the answer is simple: let us prove by induction that x_n \in (2,3) forall n, and that xn+1 > xn. Suppose this holds for all values k < n+1.

 Then we have  2.93 +10.33 =123 >  12xn+1  >  2.81+90 > 144

Also it's obvious that x2 = log12 191 >2 = x1 > x 0  , then

 12xn+1   = 2.9 xn +10.3xn-1 >  2.9 xn-1 +10.3xn-2  =  12xn  .

Therefore xn is convergent, and it is easy to see from the recurrence relationship that its limit can only be 3.

The 1st one is also easy!

if we compute a3 and a4 we get that a3=1 and a4=1 so by induction we prove that an=1 for n>=3 and then we get that it is convergent...
too easy! :D :D :D :D

cheers!!!


Let $ z\in \mathbb{C}$, $ f(z)\equal{}az^2\plus{}bz\plus{}c$, $ a,b,c\in \mathbb{C}$,

if $ |z|\leq1$, then $ |f(z)|\leq 1$

Then, when $ |z|\leq 1$, find the range of value: $ |az\plus{}b|$


inequalities
number theory unsolved
number theory

\[ {S_n} \equal{} \left\{ {k \in N /1 \le k \le n \minus{} 1/\gcd (k,n) \equal{} 1} \right\}\]
Prove or disprove:
\[ \sum\limits_{k \in {R_n}} k  \le {{n\varphi (n)} \over 2}\]

\[ \sum\limits_{k \in {R_n}} k \equiv {{n\varphi (n)} \over 2}
\], because $ (n,k)\equal{}1$ if and only if $ (n,n\minus{}k)\equal{}1$.


Using the generally accepted notations for triangles ($h_a$ is the height from $A$), prove the following inequality:

\[a^2  + 4h_a ^2  \le \left( {b + c} \right)^2 \]

Your inequality $a^2+4h_a^2\leq \left(b+c\right)^2$ is clearly equivalent to $4h_a^2\leq\left(b+c\right)^2-a^2$. But

$\left(b+c\right)^2-a^2=\left(\left(b+c\right)+a\right)\left(\left(b+c\right)-a\right)=\left(a+b+c\right)\left(-a+b+c\right)$.

If we denote by $s=\frac{a+b+c}{2}$ the semiperimeter of triangle ABC, we have a + b + c = 2s and - a + b + c = (a + b + c) - 2a = 2s - 2a = 2 (s - a). Thus,

$\left(b+c\right)^2-a^2=\left(a+b+c\right)\left(-a+b+c\right)=2s\cdot 2\left(s-a\right)=4s\left(s-a\right)$.

Hence, the inequality in question, $4h_a^2\leq\left(b+c\right)^2-a^2$, rewrites as $4h_a^2\leq 4s\left(s-a\right)$. Upon multiplication with $\frac{1}{16}a^2$, this becomes $\frac14 a^2h_a^2\leq \frac14 a^2s\left(s-a\right)$.

But if S is the area of triangle ABC, then, on the one hand, $S=\frac12 ah_a$, while, on the other hand, the Heron formula yields $S=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. Hence, $\frac12 ah_a=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. Squaring this, we get $\left(\frac12 ah_a\right)^2=s\left(s-a\right)\left(s-b\right)\left(s-c\right)$; in other words,

$\frac14 a^2h_a^2 = s\left(s-a\right)\left(s-b\right)\left(s-c\right)$.

Hence, the inequality we have to prove, $\frac14 a^2h_a^2\leq \frac14 a^2s\left(s-a\right)$, is equivalent to $s\left(s-a\right)\left(s-b\right)\left(s-c\right)\leq \frac14 a^2s\left(s-a\right)$. This immediately simplifies to

$\left(s-b\right)\left(s-c\right)\leq \frac14 a^2$.

But this inequality is almost trivial: By AM-GM, applied to the numbers s - b and s - c, we have

$\left(s-b\right)\left(s-c\right)\leq \left(\frac{\left(s-b\right)+\left(s-c\right)}{2}\right)^2$.

But (s - b) + (s - c) = 2s - (b + c) = (a + b + c) - (b + c) = a, and thus this becomes

$\left(s-b\right)\left(s-c\right)\leq \left(\frac{a}{2}\right)^2=\frac14 a^2$,

so that our inequality is proven. $\blacksquare$

[b]EDIT:[/b] I just noted that I de-facto posted a different solution of this problem at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187381#p187381]http://www.mathlinks.ro/Forum/viewtopic.php?t=28185 post #4[/url]. In fact, there I proved that $b+c\geq 2\cdot BX$ and $2\cdot BX=\sqrt{a^2+4h_a^2}$, so that $b+c\geq\sqrt{a^2+4h_a^2}$ and $\left(b+c\right)^2\geq a^2+4h_a^2$.

  darij

Just  use : [tex] h_a\leq w_a=\frac{2bc}{b+c}\sqrt{\frac{s(s-a)}{bc}}\leq\sqrt{s(s-a)}[/tex]  and  find  [tex] h_a\leq\sqrt{s(s-a)} [/tex] . Square this  inequality :)
Also  note  that  [tex]  w_a [/tex]  is bisector  of  angle  A   and  s is  semiperimeter  of the  triangle.


Consider 2 positive numbers $ a, b$ such that $ a\neq b$ and there exist 3 positive numbers $ x, y, z$ satisfy
$ x\plus{}y\plus{}z\equal{}xyz\equal{}a; xy\plus{}yz\plus{}zx\equal{}\frac{a}{b}$.
Find the smallest value of
$ P\equal{}\frac{2a^3b\minus{}3a^2\plus{}5ab}{a\minus{}b}$

[quote="nmtvn"]Consider 2 positive numbers $ a, b$ such that $ a\neq b$ and there exist 3 positive numbers $ x, y, z$ satisfy
$ x \plus{} y \plus{} z \equal{} xyz \equal{} a; xy \plus{} yz \plus{} zx \equal{} \frac {a}{b}$.
Find the smallest value of
$ P \equal{} \frac {2a^3b \minus{} 3a^2 \plus{} 5ab}{a \minus{} b}$[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 10, and we can't discuss it now


Find general solution of the equation y'' -4y' +3y =cosh x

[img]http://img23.imageshack.us/img23/5714/dsc00309kko.jpg[/img]

Am i correct?? please, appreciate it, thank you.

hello, i have found
$ y(x)\equal{}c_1 e^x\plus{}c_2 e^{3 x}\minus{}\frac{1}{16} e^{\minus{}x} \left(4 e^{2 x} x\plus{}2 e^{2 x}\minus{}1\right)$
Sonnhard.

can you show me the steps and what is your ( let h(x))? i cant do it, i tried already, get back what i post, don't how to get your answer


[color=brown][quote="balan razvan"]In the triangle $ ABC$ denote $ M,N,P$ the middles of $ AB,BC,CA$  and let $ U\in$[$ MB$), $ V\in$ [$ NC$), $ W\in$ [$ PA$).Prove that $ [UVW]\geq [MNP]$ (Laurentiu Panaitopol)[/quote][/color]

It is known that $ [MNP]=\frac{1}{4}[ABC]$ therefore it remains to prove that $ [UVW]\geq\frac{1}{4}[ABC]$ or $ \boxed{\frac{[ABC]-[UVW]}{[ABC]}\leq \frac{3}{4}}$ $ \clubsuit$
Let $ \|\begin{array}{cccc} AU=x\cdot AB\\
\\
BV=y\cdot BC\\
\\
CW=z\cdot CA \end{array}$. By how wy have chosen the points $ U,V,W$ we have $ x,y,z\in$ [$ \frac{1}{2},1)$
$ \frac{[AUW]}{[ABC]}=\frac{AV\cdot AW}{AB\cdot AC}=x(1-z)$  and then $ \clubsuit$ becomes $ x(1-z)+y(1-x)+z(1-y)\leq \frac{3}{4}$ or $ \boxed{x+y+z-(xy+yz+zx)\leq \frac{3}{4}}$ $ \spadesuit$
we have $ \sum(x-\frac{1}{2})(y-\frac{1}{2})\geq0$ or $ x+y+z-(xy+yz+zx)\leq \frac{3}{4}$ who is $ \spadesuit$ therefore the conclusion


geometry
incenter
angle bisector
geometry proposed

Triangle ABC has a right angle at A. We have a circle having its center on line AB and which is tangent to BC. Let T be the point where the tabgebt from C (not BC) touches the circle. D is the midpt of BC. AD \cap CT = {M}. Prove that MA=MT.

I found this in an old issue of GM. It was proposed by Romania for the 11'th IMO.

Let C' be the rflection of C with center A. CT \cap C'B=Q.

TM=MA  <=>  TM+MD=BC/2.

Now the statement is follows from some formulas for triangle BQC (we observe that the circle is incircle for this triangle):
MD=BQ/2. TM=MQ-TQ=CQ/2-(BQ+QC-BC)/2.

Beautifull and simple problem!!!

Take $E$ – centre of the circle, $F$- tangency with $BC$, $G$ – midpoint of $CE$ & centre of circle $\odot CTAEF$. $MT = MA \Longleftrightarrow G$ on the angle bisector of $\angle CMD$. But $\triangle ADC$ is isosceles, $DG\parallel AB$, hence $DG\perp AC$, i.e. $DG$ angle bisector of $\angle ADC$ and $G$ – incenter of $\triangle MDC$, done.

Best regards,
sunken rock


In a perfectly smooth straight groove on the surface of a horizontal table there lie n perfecltly smooth ,perfectly elastic and identical particles none of which touches the other.
The particles are now given velocities along the groove.
Show that at most n(n-1)/2 collisions can take place.

I don't think this works, but it's worth a try.

After a collision between two particles, each particle continues where the other would have gone, and with the sames speed as the other one initally had. Therefore, we can assume that the particles just go "past" each other. And then it's obvious...

Any comments?

Actually, I think your solution is all right Arne. And a nice one too. :coolspeak:

So it does work? Hehe. That's cool :) It's a nice question.


geometry incenter circumcircle Euler conics hyperbola parabola

Let $ D$ be a point on side $ BC$ of triangle $ ABC$. $ E, F$ is respectively the incenter of triangle $ ABD$ and $ ACD$. Prove that if $ B,E,F,C$ is on 1 cirle then:

$ \large \frac {AD \plus{} BD}{AD \plus{} CD} \equal{} \frac {AB}{AC}$

Lemma: Let $ ABC$ be a triangle and $ O$ its circumcenter. Let $ X,Y$ be a point on $ AB,AC$ such $ B,X,Y,C$ are cyclic. Then $ AO$ is perependicular to $ XY$.

Let $ I \equal{} BE\cap CF$. Then $ I$ is incenter of $ ABC$. Let $ K$ be a circumcenter of $ BIC$. Since $ K$ is intersection of $ AI$ and circumcircle of $ ABC$, $ K$ lie on $ AI$. By lemma,  $ IK$ is perependicular to $ EF$. Therefore, $ AI$ is perpendicular to $ EF$. Let $ X \equal{} DE\cap AB,Y \equal{} DF\cap CA$. $ \angle EAF \equal{} \angle EAD \plus{} \angle DAF \equal{} \frac {1}{2}\angle A \equal{} \angle XAI$. So $ \angle IAF \equal{} \angle XAE \equal{} \angle EAD$. So $ AD,AI$ are isogonal line wrt $ AEF$. Since $ AI$ is perpendicular to $ EF$, circumcenter of $ AEF$ lie on $ AD$. Let $ O$ be circumcenter of $ AEF$.Since $ \angle OEA \equal{} \angle EAO \equal{} \angle XAE$, $ AX$ is parallel to $ OE$. Similarly, $ AY$ is parallel to $ OF$. Then $ \frac {AX}{OE} \equal{} \frac {AD}{OD} \equal{} \frac {AY}{OF}$. So $ AX \equal{} AY$. $ \frac {AD \plus{} BD}{AD \plus{} CD} \equal{} \frac {1 \plus{} \frac {BD}{AD}}{1 \plus{} \frac {CD}{AD}} \equal{} \frac {1 \plus{} \frac {XB}{AX}}{1 \plus{} \frac {YC}{AY}} \equal{} \frac {AY\cdot AB}{AX\cdot CA} \equal{} \frac {AB}{CA}$

Very nice your solution, dear [b]Leonhard Euler[/b]! I have another approach to prove that $ AX\equal{}AY$. Here it goes:

Denote points $ X\equal{}DE\cap AB$, $ Y\equal{}DF\cap AC$, $ P$, $ Q$ be the intersections of $ EF$ with $ AB$, $ AC$, respectively. We have $ \angle APQ\equal{}\angle PBE\plus{}\angle PEB\equal{}\frac{B}{2}\plus{}\angle FCB\equal{}\frac{B\plus{}C}{2}$. Similarly, we also have $ \angle AQP\equal{}\frac{B\plus{}C}{2}$, it implies $ AP\equal{}AQ$.
Now, draw a line passes through $ A$ and parallel to $ PQ$, intersects $ DX$, $ DY$ at $ S$, $ T$, respectively. We have $ \frac{EP}{EI}\equal{}\frac{AP}{AI}\equal{}\frac{AQ}{AI}\equal{}\frac{FQ}{FI}$, it follows $ \frac{IE}{IF}\equal{}\frac{EP}{FQ}$.

On the other hand, by the [b]Thales' theorem[/b] $ \frac{IE}{IF}\equal{}\frac{AS}{AT}$. Hence, we get $ \frac{EP}{EQ}\equal{}\frac{AS}{AT}$, i.e. $ \frac{EP}{AS}\equal{}\frac{EQ}{AT}$, which implies $ \frac{XP}{XA}\equal{}\frac{YQ}{YA}$, i.e. $ XY\parallel PQ$. It means $ AX\equal{}AY$.

How to construct this cevian AD ?

  T hank you.
  M.T.

Is the construction of point D really difficult or  so elementary  that noone wants
  to bother? If it is indeed too easy, then please give me a hint, for I do not see it.

  Thank you.
  M.T.

x = AD, y = BD, z = CD. 3 equations for 3 unknowns:

$ y \plus{} z \equal{} a,\ \ \frac{x \plus{} y}{x \plus{} z} \equal{} \frac{c}{b},\ \ a(x^2 \plus{} yz) \equal{} yb^2 \plus{} zc^2$

Solving for x:

$ \frac{x^2}{s(s \minus{} a)} \plus{} \frac{(b \minus{} c)^2}{abc}\ x \equal{} 1$

T is the tangency point of the $ \triangle ABC$ incircle (I) with BC. J, K are incenters of $ \triangle ABD, \triangle ACD.$ The common internal tangent of the incircles (J), (K) other than AD goes through T. AD cuts the center line JK at Z. ABTZ, ACTZ are tangential => JK tangent to a hyperbola $ \mathcal H$ passing through B, C with foci A, T. JK bisects the angle $ \angle AZT$ between the two common internal tangents $ DA \equiv ZA$ and $ ZT.$ The intersection $ Z \equiv AD \cap JK$ is the tangency point of JK with the hyperbola $ \mathcal H.$ Its pedal circle is centered at the midpoint M of AT and has diameter $ AB \minus{} TB \equal{} AC \minus{} TC \equal{} s \minus{} a.$ If U, V are tangency points of the incircle (I) with CA, AB and $ X \equiv TU \cap CI,\ Y \equiv TV \cap BI$ midpoints of TU, TV, then the isosceles $ \triangle MXY \sim \triangle AUV$ are centrally  similar with similarity center T and coefficient 1/2, they have parallel sides, and $ MX \equal{} MY \equal{} \frac {AU}{2} \equal{} \frac {AV}{2} \equal{} \frac {s \minus{} a}{2}.$ Thus (M) with radius MX = MY is the hyperbola pedal circle.  For BCKJ to be cyclic, bisector AI of the $ \angle A$ is perpendicular to JK, which is tangent to $ \mathcal H.$ AI meets (M) at P on the opposite side of I than A and a normal to AI at P meets the bisectors BI, CI of the $ \angle B,\ \angle C$ at J, K. The $ \angle JDK \equal{} \angle JTK \equal{} 90^\circ$ are right, a circle with diameter JK meets BC at T, D, and AD is the desired segment.

It turned out to be a very nice construction both ways.
   It looks like a parabola with focus A and BC as directrix might also
   allow a solution tangent to it for the strike three.

   Thank you.

   M.T.


Here's a nice math problem we had the other day in class:

If the sum of $ \frac{1}{3}$ of an even integer and $ \frac{3}{4}$ of the next consecutive even integer is equal to $ 21$, what is the odd integer that falls between these two even integers?

[hide="Solution"]
$ a\in Z$
$ b\in Z$
$ a\neq b$
$ a\equiv 0\pmod{2}$
$ b\equal{}a\plus{}2$

Therefore, $ \displaystyle\frac{a}{3}\plus{}\displaystyle\frac{3b}{4}\equal{}21$

Substituting, $ \displaystyle\frac{a}{3}\plus{}\displaystyle\frac{3(a\plus{}2)}{4}\equal{}21$

Simplifying, $ \displaystyle\frac{a}{3}\plus{}\displaystyle\frac{3a\plus{}6}{4}\equal{}21$

To simplify further, $ 12(\displaystyle\frac{a}{3}\plus{}\displaystyle\frac{3a\plus{}6}{4}\equal{}21)$

After multiplying and simplifying, $ 13a\equal{}234$

Therefore, $ a\equal{}18$ and $ b\equal{}20$

So, $ 19$ is the odd number.

[/hide]

[hide="Really?"]Let $ a$ be the smaller even integer.

$ a \plus{} \frac{3(a\plus{}2)}{4} \equal{} 21$
$ \frac{4a\plus{}3(a\plus{}2)}{4} \equal{} 21$
$ 4a\plus{}3a\plus{}6\equal{}84$
$ 7a\equal{}78$

$ a\equal{}\frac{78}{7}$, which isn't an integer.[/hide]

RisingFlame, your solution is wrong. Notice, that $ a$ should be over $ 3$.

Hmm...for some odd reason, the 1/3 didn't show up on my screen before.


geometry
rectangle
LaTeX
analytic geometry

this isan easy question,but the 2 most intelligent boys in my class could not get it.(includes riddler AND EMINEM)
there is a circle inscribed in a  square.if a rectangle of length and breadth 12 and 6 is inscribed in the corner of the square and one of its vertex touches the circle,find the radius of the cicle.
I DONT KNOW HOW TO DRAW THINGS HERE,NOR CAN I DO THE SQUARE ROOT THING.WILL SOME ONE TELL ME HOW TO SO THAT THE QUESTION BECOMES MORE CLEAR!
RIDDLER CANT REPLY ON THIS CAUSE HE KNOWS THE SOL.

You can type square root as \sqrt{x} but with a dollar sign before and after the code. Try it.
You get this: $\sqrt{x}$

To add a picture to your message, just click on 'add an attachment'
You can find the button when you're posting a message, it's next to 'options'.

I don't understand the question (the part where a rectangle is inscribed in a corner). You shold add a picture.

I think he means something like this (see attachment):

[quote="10000th User"][quote="Iversonfan2005"]When you mean that the rectangle is inscribed in the "corner of the square", you mean, in one of the spaces not occupied by the circle? If so, I don't know how it could be solved ;) Are you sure there's no more additional information? Perhaps it's $18\sqrt{2}$ because there is a $45,45,90$ triangle beside the rectangle, which has a hypotenuse of $12$ which is the length of the rectangle. Thus, each leg of the triaangle is $6\sqrt{2}$, from there, I thought that this was $\frac{1}{6}$ of the length of the square, so I multiplied it by $3$ to obtain the radius, thus getting $18\sqrt{2}$.[/quote]
I think your answer is wrong. Radius is [color=white]30[/color][/quote]

Oh, I know, because Behemoth's drawing explains it. I was thinking that the rectangle was placed diagonally.

If you draw it, 10000th User's answer seems to be right, but I havent found the solution.

EDIT:

Nevermind, here is the solution. Use the formula of a circle "(x-a)^2 + (y-b)^2 = r^2", assume it is at the origin, and plug in the w and l of the rectangle: "(r-12)^2 + (r-6)^2 = r^2". This simplifies to "0 = r^2 - 36r + 180". Solving for r yields [color=white]30 and 6[/color].

(sorry I dont know Latex)

If you draw the circle, both answers give sense; if the circle is radius 6:

Another way you can solve it is if you can set up coordinates, and make the center of the circle as the origin ;)

That's what I did, and it ended up being Pythagorean as in 10000th User's drawing.


"Prove that if the entries in each row of an nxn matrix add up to zero then the determinant of the matrix is zero"

If A has a row of zeros, then det(A) =0

proof:

each term of det(A) includes one factor that contain each row, hence each term has a zero factor.  Thte sum of zeros is zero.

Does that sound like a good enough proof?

TIA!

[quote="Kent Merryfield"]No, I wouldn't accept that proof. We didn't say our matrix had a row of zeros, just that each row added up to zero. And I do not understand this sentence at all: "each term of det(A) includes one factor that contain each row, hence each term has a zero factor." That makes no sense to me.

Here's another route: let $u$ be the column vector in which each entry is $1.$

What is $Au$? And what does that tell you about whether $A$ is invertible?

Or for another approach, do column elimination on $A$ (or, if you don't like that, do row elimination on $A^T.)$[/quote]

I guess he misunderstood the question and he used the formula $det A = \sum_{\pi \in S_n} A_{\pi(1)1}...A_{\pi(n)n}$.

[quote="Kent Merryfield"]Or for another approach, do column elimination on $A$ (or, if you don't like that, do row elimination on $A^T.)$[/quote]

This seems most natural to me.  We can reduce it into upper-triangular form by adding multiples of rows to other rows, which means that every row in the upper triangular form still sums to zero, hence the row with (at most) a single nonzero entry in it is in fact a zero row.  But this means that the determinant is...

We can add the second, . . . , nth column to the first, and the value of the determinant will not change. Then the elements of the first column will be the sums of each row which are zero. The new determinant has a zero column so it is zero.


[size=150]Let $ x,y,z \geq 0$,$ t > 0$,and $ x \plus{} y \plus{} z \equal{} txyz$,prove that:
$ xy \plus{} yz \plus{} zx \plus{} x \plus{} y \plus{} z \geq \frac {3(3 \plus{} \sqrt {3t})}{t}.$
BQ[/size]

the inequality is equivalent to:
$ (x\plus{}y\plus{}z)^{2}\plus{}(xy\plus{}yz\plus{}zx)(x\plus{}y\plus{}z)\geq9xyz\plus{}3\sqrt{3xyz(x\plus{}y\plus{}z)}$
we have : $ xy\plus{}yz\plus{}zx\geq \sqrt{3xyz(x\plus{}y\plus{}z)}$
then let's prove that
$ (x\plus{}y\plus{}z)^{2}\plus{}(xy\plus{}yz\plus{}zx)(x\plus{}y\plus{}z)\geq9xyz\plus{}3(xy\plus{}yz\plus{}zx)$
but:$ (xy\plus{}yz\plus{}zx)(x\plus{}y\plus{}z)\geq9xyz$ and $ (x\plus{}y\plus{}z)^{2}\geq3(xy\plus{}yz\plus{}zx)$.....


A light flashes every 1 minute and 15 seconds. Another flashes every 1 minute and 40 seconds. Suppose they flash together at a certain time. What is the shortest amount of time that will elapse before both lights will again flash together?

[hide="solution"]The first light flashes every 75 seconds and the second every 100 seconds. The least common multiple of the two is 300, so 5 minutes will elapse until they again flash together. [/hide]

so convert both times to seconds.
you get one  light flashes every 75 seconds. The other flashes every 100 seconds.

Find the Least Common Multiple of 75 and 100. It is 300. So, the next time the lights will flash together is in 300 seconds or 5 minutes.


inequalities
trigonometry
geometry unsolved
geometry

$ A_1$, $ B_1$ and $ C_1$ are the projections of the vertices $ A$, $ B$ and $ C$ of triangle $ ABC$ on the respective sides. If $ AB \equal{} c$, $ AC \equal{} b$, $ BC \equal{} a$ and $ AC_1 \equal{} 2t AB$, $ BA_1 \equal{} 2rBC$, $ CB_1 \equal{} 2 \mu AC$. Prove that:

\[ \frac {a^2}{b^2} \cdot \left( \frac {t}{1 \minus{} 2t} \right)^2 \plus{} \frac {b^2}{c^2} \cdot \left( \frac {r}{1 \minus{} 2r} \right)^2 \plus{} \frac {c^2}{a^2} \cdot \left( \frac {\mu}{1 \minus{} 2\mu} \right)^2 \plus{} 16tr \mu \geq 1
\]

We have \[ \frac{AC_1}{AB}=2t=\frac{b \cos A}{c}=\frac{b^2+c^2-a^2}{2c^2} \] \[ \implies t=\frac{b^2+c^2-a^2}{4c^2} \]
Substituting, \[ \left( \frac{t}{1-2t}\right)=\frac{\frac{b^2+c^2-a^2}{4c^2}}{\frac{2c^2-(b^2+c^2-a^2)}{2c^2}}=\frac{1}{2} \left(\frac{b^2+c^2-a^2}{c^2+a^2-b^2} \right)=\frac{1}{2} \left( \frac{b \cos A}{a \cos B} \right) \] \[ \implies \frac{a^2}{b^2} \left( \frac{t}{1-2t}\right)^2= \frac{1}{4} \left( \frac{\cos A}{\cos B} \right)^2  \] Analogously, we have \[ \frac{b^2}{c^2} \left( \frac{r}{1-2r}\right)^2=\frac{1}{4} \left( \frac{\cos B}{\cos C} \right)^2 \] \[ \frac{c^2}{a^2} \left( \frac{\mu}{1-2 \mu}\right)^2=\frac{1}{4} \left( \frac{\cos C}{\cos A} \right)^2 \] On the other hand, we get \[ 16tr \mu=\frac{1}{4} \left(\frac{(b^2+c^2-a^2)(c^2+a^2-b^2)(a^2+b^2-c^2)}{a^2b^2c^2} \right)=2 \cos A \cos B \cos C \] Therefore, the inequality reduces to \[ \left(\frac{\cos A}{\cos B} \right)^2 + \left(\frac{\cos B}{\cos C} \right)^2 + \left(\frac{\cos C}{\cos A} \right)^2 + 8 \cos A \cos B \cos C \ge 4 \] which is well known but I can't find a link, so a proof.

Lemma: For reals $ x,y,z>0, xyz \le 1, $ we have $ \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge x+y+z $.
Applying the lemma to $ 4 \cos^2A, 4 \cos^2 B, 4 \cos^2 C $ (note that $ \prod \cos A \le \frac{1}{8} $), we get \[ \left(\frac{\cos A}{\cos B} \right)^2 + \left(\frac{\cos B}{\cos C} \right)^2 + \left(\frac{\cos C}{\cos A} \right )^2 \ge 4 \cos^2A+4 \cos^2 B+4 \cos^2 C \] therefore, \[ \left(\frac{\cos A}{\cos B} \right)^2 + \left(\frac{\cos B}{\cos C} \right)^2 + \left(\frac{\cos C}{\cos A} \right)^2 + 8 \cos A \cos B \cos C \] \[ \ge 4 \cos^2A+4 \cos^2 B+4 \cos^2 C + 8 \cos A \cos B \cos C = 4 \] because $ \cos^2 A+ \cos^2 B+ \cos^2 C+2 \cos A \cos B \cos C =1 $. $ \blacksquare $


geometry
invariant
parallelogram
geometry unsolved

In the quadrangle $ A_{1}A_{2}A_{3}A_{4}$ ,Suppose $ B_{i}$ is midpoint of side $ A_{i}A_{i\plus{}1}$,Prove the area of quadrangle $ A_{1}A_{2}A_{3}A_{4}$ is five times the area of quadrangle $ C_{1}C_{2}C_{3}C_{4}$.

who can slove it?please help me,thank you.

could you show more details?i dont understand  what you mean

Among the properties of the affine transformations there is the following two: Any quadrilateral can be transformed in a square by means of an affine transformation; and "The quotient of areas is preserved in one affine transformation". If we transforms the two quadrilaterals of the problem in squares, in such way that the points Bi be the mid points of AiAi+1, then the quotient of the areas of A1A2A3A4 and C1C2C3C4 is the same. Therefore it suffices solve the problem in the case of a square A1A2A3A4 and the square obtained joining each vertex with the mid point Bi (in the way of the original picture). And in this case is almost inmediat that this quotient must be 5.
You can find a nice account of the properties of the affine transformations in the book of I.M.Yaglom Geometric transformations III. I am sure that it is also possible find out properties in Internet.

Unfortunately, that won't do it.  An affine transformation takes parallel lines to parallel lines, so the affine image of any square is always a parallelogram.

[quote="28121941"]The area is an affin invariant; therefore it suffices to prove the result in the case in which A1A2A3A4 is a square, and Bi the mid points of the sides. The result is then classic and easy. :D[/quote]

Who can solve it by using pure geometrical?Thanks!

After checking some cases, it turns out that the result is not true (in general).

It is easy to show that $ ($ [b][size=100]Virgil Nicula's[/size][/b] wording $ ),$

$ (XYZT) \equal{} (AQT) \plus{} (BMX) \plus{} (CNY) \plus{} (DPZ) \equal{} S_{1}$ $ ,(1)$ and $ (BNYX) \plus{} (DQTZ) \equal{} (AMXT) \plus{} (CPZY) \equal{} S_{2}$ $ ,(2)$

From $ (1),$ $ (2),$ it is enough to prove that $ S_{2} \equal{} 1,5\cdot S_{1}$ $ ,(3)$

But, although we can check, by the drawing, that the equalities $ (1),$ $ (2)$ are true, however the equality $ (3),$ it doesn't hold.

So, in my opinion also, the problem is not true in general.

Kostas Vittas.

[quote="wxd686"] [color=darkred]Let $ ABCD$ be a convex quadrilateral. Denote the midpoints $ M$ , $ N$ , $ P$ , $ Q$ of the sides $ [AB]$ , $ [BC]$ , $ [CD]$ , $ [DA]$

and $ X\in MC\cap BQ$ , $ Y\in MC\cap ND$ , $ Z\in PA\cap ND$ , $ T\in PA\cap BQ$ . Prove that $ [ABCD] \equal{} [XYZT]$ .[/color] [/quote]

Denote $ [ABC] \equal{} d$ , $ [BCD] \equal{} a$ , $ [CDA] \equal{} b$ , $ [DAB] \equal{} c$ and $ [ABCD] \equal{} S$ . Observe that $ a \plus{} c \equal{} b \plus{} d$ .

$ I.\blacktriangleright\ \left\|\begin{array}{ccc} \frac {[BAT]}{[BAQ]} \equal{} \frac {BT}{BQ} & \implies & [BAT] \equal{} \frac c2\cdot\frac {BT}{BQ} \\
 \\
\frac {[BMX]}{[BAQ]} \equal{} \equal{} \frac {BM}{BA}\cdot\frac {BX}{BQ} & \implies & [BMX] \equal{} \frac c4\cdot\frac {BX}{BQ}\end{array}\right\|$ $ \implies$ $ [MATX] \equal{} [BAT] \minus{} [BMX]$ $ \implies$ $ \boxed {\ [MATX] \equal{} \frac c2\cdot\left(\frac {BT}{BQ} \minus{} \frac {BX}{2\cdot BQ}\right)\ }\ (1)$ .

$ \blacktriangleright\ \frac {TB}{TQ} \equal{} \frac {[ABP]}{[AQP]} \equal{}$ $ \frac {[ABC] \plus{} [ACP] \minus{} [BCP]}{\frac 14\cdot [ACD]} \equal{}$ $ \frac {d \plus{} \frac b2 \minus{} \frac a2}{\frac b4} \equal{}$ $ \frac {4d \plus{} 2b \minus{} 2a}{b}$ $ \implies$ $ \frac {TB}{BQ} \equal{} \frac {4d \plus{} 2b \minus{} 2a}{4d \plus{} 3b \minus{} 2a}$ $ \implies$ $ \boxed {\ \frac {BT}{BQ} \equal{} \frac {2(c \plus{} d)}{a \plus{} d \plus{} 3c}\ }\ (2)$ .

$ \blacktriangleright\ \frac {XB}{XQ} \equal{} \frac {[CBM]}{[CQM]} \equal{}$ $ \frac {\frac 12\cdot [ABC]}{[AMC] \plus{} [AQC] \minus{} [AMQ]} \equal{}$ $ \frac {\frac d2}{\frac d2 \plus{} \frac b2 \minus{} \frac c4} \equal{}$ $ \frac {2d}{2b \plus{} 2d \minus{} c}$ $ \implies$  $ \frac {BX}{BQ} \equal{} \frac {2d}{2b \plus{} 4d \minus{} c}$ $ \implies$ $ \boxed {\ \frac {BX}{BQ} \equal{} \frac {2d}{a \plus{} b \plus{} 3d}\ }\ (3)$ .

$ \blacktriangleright$ Using $ (1)$ , $ (2)$ , $ (3)$ obtain $ [MAXT] \equal{} \frac c2\cdot\left[\frac {2(c \plus{} d)}{a \plus{} d \plus{} 3c} \minus{} \frac 12\cdot\frac {2d}{a \plus{} b \plus{} 3d}\right]$ , i.e. $ \boxed {\ [MAXT] \equal{} \frac {c(c \plus{} d)}{a \plus{} d \plus{} 3c} \minus{} \frac {cd}{2(a \plus{} b \plus{} 3d)}\ }$ .

$ II.\blacktriangleright\ \left\|\begin{array}{ccc} \frac {[DYC]}{[DNC]} \equal{} \frac {DY}{DN} & \implies & [CDY] \equal{} \frac a2\cdot\frac {DY}{DN} \\
 \\
\frac {[DZP]}{[DNC]} \equal{} \equal{} \frac {DP}{DC}\cdot\frac {DZ}{DN} & \implies & [DZP] \equal{} \frac a4\cdot\frac {DZ}{DN}\end{array}\right\|$ $ \implies$ $ [PCYZ] \equal{} [CDY] \minus{} [DZP]$ $ \implies$ $ \boxed {\ [PCYZ] \equal{} \frac a2\cdot\left(\frac {DY}{DN} \minus{} \frac {DZ}{2\cdot DN}\right)\ }\ (4)$ .

$ \blacktriangleright\ \frac {YD}{YN} \equal{} \frac {[DMC]}{[NMC]} \equal{}$ $ \frac {[AMC] \plus{} [ACD] \minus{} [AMD]}{\frac 14\cdot [ABC]} \equal{}$ $ \frac {\frac d2 \plus{} b \minus{} \frac c2}{\frac d4} \equal{}$ $ \frac {2d \plus{} 4b \minus{} 2c}{d}$ $ \implies$ $ \frac {DY}{DN} \equal{} \frac {2d \plus{} 4b \minus{} 2c}{3d \plus{} 4b \minus{} 2c}$ $ \implies$ $ \boxed {\ \frac {DY}{DN} \equal{} \frac {2(a \plus{} b)}{b \plus{} c \plus{} 3a}\ }\ (5)$ .

$ \blacktriangleright\ \frac {ZN}{ZD} \equal{} \frac {[ANP]}{[ADP]} \equal{}$ $ \frac {[CNA] \plus{} [PAC] \minus{} [CNP]}{\frac 12\cdot [ACD]} \equal{}$ $ \frac {\frac d2 \plus{} \frac b2 \minus{} \frac a4}{\frac b2} \equal{}$ $ \frac {2d \plus{} 2b \minus{} a}{2b}$ $ \implies$  $ \frac {DZ}{DN} \equal{} \frac {2b}{2d \plus{} 4b \minus{} a} \equal{}$ $ \implies$ $ \boxed {\ \frac {DZ}{DN} \equal{} \frac {2b}{c \plus{} d \plus{} 3b}\ }\ (6)$ .

$ \blacktriangleright$ Using $ (4)$ , $ (5)$ , $ (6)$ obtain $ [PCYZ] \equal{} \frac a2\cdot\left[\frac {2(a \plus{} b)}{b \plus{} c \plus{} 3a} \minus{} \frac 12\cdot\frac {2b}{c \plus{} d \plus{} 3b}\right]$ , i.e. $ \boxed {\ [PCYZ] \equal{} \frac {a(a \plus{} b)}{b \plus{} c \plus{} 3a} \minus{} \frac {ab}{2(c \plus{} d \plus{} 3b)}\ }$ .

$ \blacktriangleright\ [XYZT] \equal{} S \minus{} [BCM] \minus{} [ADP] \minus{} [MAXT] \minus{} [PCYZ] \equal{}$ $ S \minus{} \frac d2 \minus{} \frac b2 \minus{} [MAXT] \minus{} [PCYZ]$ . In conclusion,

$ \boxed {\ [XYZT] \equal{} \frac S2 \minus{} \frac {c[2(c \plus{} d)(a \plus{} b \plus{} 3d) \minus{} d(a \plus{} b \plus{} 3c)]}{2(a \plus{} b \plus{} 3d)(a \plus{} d \plus{} 3c)} \minus{} \frac {a[2(a \plus{} b)(c \plus{} d \plus{} 3b) \minus{} b(b \plus{} c \plus{} 3a)}{2(c \plus{} d \plus{} 3b)b \plus{} c \plus{} 3a)}\ }$ , where $ S \equal{} a \plus{} c \equal{} b \plus{} d$ .

[b][u]Remark.[/u][/b] More precisely, here is the expressions for all areas which appears in the quadrilateral $ ABCD$ :

$ [AQT] \equal{} \frac {bc}{2(d \plus{} a \plus{} 3c)}\ \ ;\ \ [AMXT] \equal{} \frac {c[2(c \plus{} d)(a \plus{} b \plus{} 3d) \minus{} d(d \plus{} a \plus{} 3c)]}{2(a \plus{} b \plus{} 3d)(d \plus{} a \plus{} 3c)}$ .

$ [BMX] \equal{} \frac {cd}{2(a \plus{} b \plus{} 3d)}\ \ ;\ \ [BNYX] \equal{} \frac {d[2(d \plus{} a)(b \plus{} c \plus{} 3a) \minus{} a(a \plus{} b \plus{} 3d)]}{2(b \plus{} c \plus{} 3a)(a \plus{} b \plus{} 3d)}$ .

$ [CNY] \equal{} \frac {da}{2(b \plus{} c \plus{} 3a)}\ \ ;\ \ [CPZY] \equal{} \frac {a[2(a \plus{} b)(c \plus{} d \plus{} 3b) \minus{} b(b \plus{} c \plus{} 3a)]}{2(c \plus{} d \plus{} 3b)(b \plus{} c \plus{} 3a)}$ .

$ [DPZ] \equal{} \frac {ab}{2(c \plus{} d \plus{} 3b)}\ \ ;\ \ [DQTZ] \equal{} \frac {b[2(b \plus{} c)(d \plus{} a \plus{} 3c) \minus{} c(c \plus{} d \plus{} 3b)]}{2(d \plus{} a \plus{} 3c)(c \plus{} d \plus{} 3b)}$ .

You can now verify if the [b]Vittasko[/b]'s relations is truly, i.e.

$ [XYZT]\equal{}[AQT]\plus{}[BMX]\plus{}[CNY]\plus{}[DPZ]$ and $ [BNYX]\plus{}[DQTZ]\equal{}[AMXT]\plus{}[CPZY]$ .

I am very sorry for my bad interpretation of the affine properties, and thanks to Naoki Sato for have pointed out my mistake.
I think I have located the special case of the problem when ABCD is a parallelogram, in the russian book of Prasolov, Zadachi po planimetrii, probl. 4.61, page 90, with solution in page 102. But i am not good translating russian...

[quote="vittasko"]It is easy to show that $ ($ [b][size=100]Virgil Nicula's[/size][/b] wording $ ),$

$ (XYZT) \equal{} (AQT) \plus{} (BMX) \plus{} (CNY) \plus{} (DPZ) \equal{} S_{1}$ $ ,(1)$ and $ (BNYX) \plus{} (DQTZ) \equal{} (AMXT) \plus{} (CPZY) \equal{} S_{2}$ $ ,(2)$[/quote]
$ \bullet$ Because of $ N,\ Q,$ are the midpoints of the side-segments $ BC,\ AD$ respectively, we have $ (BQA) \equal{} (BQD)$ $ ,(1)$ and $ (DNB) \equal{} (DNC)$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ (BQD) \plus{} (DNB) \equal{} \frac {S}{2}$ $ ,(3)$ and $ (BQA) \plus{} (DNC) \equal{} \frac {S}{2}$ $ ,(4)$ where $ S \equal{} (ABCD).$

From $ (3)$ $ \Longrightarrow$ $ (BNYX) \plus{} (XYZT) \plus{} (DQTZ) \equal{} \frac {S}{2}$ $ ,(5)$

$ \bullet$ Because of $ M,\ P,$ are the midpoints of the side-segments $ AB,\ CD$ respectively, we have $ (CMA) \equal{} (CMB)$ $ ,(6)$ and $ (APC) \equal{} (APD)$ $ ,(7)$

From $ (6),$ $ (7)$ $ \Longrightarrow$ $ (CMA) \plus{} (APC) \equal{} \frac {S}{2}$ $ ,(8)$ and $ (CMB) \plus{} (APD) \equal{} \frac {S}{2}$ $ ,(9)$

From $ (8)$ $ \Longrightarrow$ $ (AMXT) \plus{} (XYZT) \plus{} (CPZT) \equal{} \frac {S}{2}$ $ ,(10)$

From $ (5),$ $ (10)$ $ \Longrightarrow$ $ \boxed{(BNYX) \plus{} (DQTZ) \equal{} (AMXT) \plus{} (CPZT)}$ $ ,(11)$

$ \bullet$ From $ (4)$ $ (BMX) \plus{} (AMXT) \plus{} (AQT) \plus{} (CNY) \plus{} (CPZY) \plus{} (DPZ) \equal{} \frac {S}{2}$ $ ,(12)$

From $ (10),$ $ (12)$ $ \Longrightarrow$ $ \boxed{(XYZT) \equal{} (BMX) \plus{} (AQT) \plus{} (CNY) \plus{} (DPZ)}$ $ ,(13)$

Kostas Vittas.

If some of the mathlinkers had the electronic edition of the romanian journal Gazeta matematica B, there is published in the number 10, year 1991, pp.362-366, an interesting note about a problem of areas in a convex quadrilateral, which can illustrate a bit more the carefully analysys of Virgil Nucula and Vittas Kostas.


search
inequalities unsolved
inequalities

dont read  it , it has been desccused hundered times before , i tried to delete it but i just couldnt

harazi is gonna kill me

for $x,y,z >0$ prove :
$(xy+yz+zx)(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2})\geq \frac{9}{4}$

very very old question...........
I think you can search in the forum.


Find all $ x,y,z,\in R$ such that  $ 1 \plus{} x^4\le 2(y \minus{} z)^2,1 \plus{} y^4\le 2(z \minus{} x)^2,1 \plus{} z^4\le 2(x \minus{} y)^2$

hello, i have found
$ (x\equal{}\minus{}1\land ((y\equal{}0\land z\equal{}1)\lor (y\equal{}1\land z\equal{}0)))\lor (x\equal{}0\land
((y\equal{}\minus{}1\land z\equal{}1)\lor (y\equal{}1\land z\equal{}\minus{}1)))\lor (x\equal{}1\land ((y\equal{}\minus{}1\land
z\equal{}0)\lor (y\equal{}0\land z\equal{}\minus{}1)))$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have found
$ (x \equal{} \minus{} 1\land ((y \equal{} 0\land z \equal{} 1)\lor (y \equal{} 1\land z \equal{} 0)))\lor (x \equal{} 0\land ((y \equal{} \minus{} 1\land z \equal{} 1)\lor (y \equal{} 1\land z \equal{} \minus{} 1)))\lor (x \equal{} 1\land ((y \equal{} \minus{} 1\land z \equal{} 0)\lor (y \equal{} 0\land z \equal{} \minus{} 1)))$
Sonnhard.[/quote]
 you're right.
We have:$ (1\minus{}x^2)^2\plus{}2x^2 \leq 2(y\minus{}z)^2$ then $ x^2 \leq (y\minus{}z)^2$
Similar, $ y^2 \leq (x\minus{}z)^2$ nad$ z^2 \leq (x\minus{}y)^2$
suppose that, $ x \leq y \leq z$ we have:
$ |x|\plus{}|z| \leq z\minus{}y\plus{}y\minus{}x\equal{}z\minus{}x \leq |x|\plus{}|z|$
Thus $ x<0$,$ z>0$ and $ \minus{}(1\minus{}x^2)^2\equal{}0, (1\minus{}z^2)^2\equal{}0$ or $ x\equal{}\minus{}1,z\equal{}1,y\equal{}0$
 Roots of the problem:$ (\minus{}1,0,1)$ and permutable


Massaru has 3 pencils in his bookbag.  Two pencils are blue and one is red.  If he randomly selects two pencils, what is the probability that they are the same color?  Express your answer as a common fraction.

[hide]$\frac 23 * \frac 12 = \frac 13$[/hide]

Is there any even easier way to solve this one?

[hide]label them 1,2, and 3. 1 and 2 being blue and 3 red. pairs are 12,13,23. only12 works. so 1/3. it may not be the best method but it is one.[/hide]

Only two are of the same colors so there is only one way to get a pair of same color pencils. And there are $3$ ways of choosing two pencils so the answer is 1/3.

[hide]You have to get the blue pencils (There are 2 blue pencils and only one red).  So the probability for geting blue the first time is $\frac{2}{3}$.  The second time, It's $\frac{1}{2}$ because there are only 2 pencils left.  Multiply them together to get $\frac{1}{3}$

$\frac{2}{3} * \frac{1}{2}=\frac{1}{3}$[/hide]

my answer is [hide]$\frac13$[/hide]
and i have a question about probability: when should we add and when should we multiply the probabilities of different times??

nice name *thumbs up* man, i truly want that name :[

[hide]For a problem with only 3 choices, we can just write out the possibliities, but i wont od that.

3 choose 2 is $\frac{3!}{(2!)(1!)}$. This is equivalent to 3. Now we have to subtract 2 from this number 3 because there are two possibliities that dont work: B1, R and B2, R, so we have 1. Sicne we're finding a probabliity, we put this one over 3 (because there are 3 pencils). In the end we have [b]$\frac{1}{3}$. There's your answer.[/b][/hide]

to answer your question...
you add the probabliities if the events are unrelated
you multiply the probabilities if they are related.

2/3 for the first,
1/2 for the second,

1/3 same color?[/hide]

[hide] 3!/ (2!)= 3, There is only one possible wasy of getting two of teh same colour, and thats if you pick both blues.  Thus teh probability is, [b]1/3[/b][/hide]

[hide]2/3 is the probability he gets blue on the first draw and 1/2 is the probability he gets another blue on the second draw.

Therefore, 1.2 (2/3) = 1/3[/hide]


If $f(x)$ is a polynomial, what are all solutions to $f(2x)=f(x)$? What if $f(x)$ could be any function?

assuming you were talking of poynomials from $\mathbb R \rightarrow \mathbb R$
if $f$ is a constant polynomial we are done...
let $deg(f)=n>0$
then if $a$ is a root so is $2a$,$4a$....so practically if this has one non-zero real root this has many such non-equal so it  can't have non zero real roots ...
so if
$f(x)=cx^{n}$ thsi gives $f(x)=0$ a constant function ...
now for the very same reason it can't have complex rots also

As $x$ increases without bound, any polynomial in $x$ is eventually strictly increasing or strictly decreasing; hence a polynomial function cannot satisfy this equality.

But there is the generic solution:
[hide]
$f(x)=g( \{ log_{2}|x| \})$ where $\{ \; \}$ denote the fractional part and $g$ is any function defined over at least $[0,1)$.

Note that $f(2x)=g( \{ log_{2}|2x| \})=g( \{ 1+log_{2}|x| \}) = g( \{ log_{2}|x| \}) = f(\pm x)$

It is continuous for $x \not = 0$ if $g(0)=g(1)$, and is smooth if $g(0)=g(1)$ [u]and[/u] $g'(0)=g'(1)$ (verify the last statement though, I'm tired).

[b]Important:[/b] f(x) will be defined for all real numbers except 0 under this generic model even if g(x) is a constant function (you still can't take the log of 0). Although, if g(x) is a constant function, you can just do f(x)=g(x).
[/hide]


can anyone tell me whatz brute force.?
i am new here? :lol:

Don't make so many topics
Brute force is when you do the problem without actually using what you're supposed to.
A few methods:
Calculus
Listing out all possibilities
Guess and Check


Just use De Mouivre's Theorem. The rest is just trig stuff...

Thanks, why the question want us to do -40-42i before doing 40+2i?

Don't bother with trig- the roots are Gaussian integers. There really isn't much difference between finding the square roots of $-40-42i$ and $40+42i$.

[quote="jack1234"]Find the square roots fo the complex number -40-42i.
Hence
(i) Find the square roots of the complex number 40+42i,
(ii) solve the equation (z+1)^2 + 160 + 168i = 0 for all complex roots.[/quote]

For (i), note that $20^{2}+21^{2}= 29^{2}$.  :)


Given: triangle WZY is congruent to triangle WXY, and triangles WZY and XZY are isoceles.
Prove: WXYZ is a rhombus

okay, draw a typical-loooking parallelogram, and the draw the diagonals. the intersection of the diagonals is P, the top left corner of the parallelogram (p.) is W, the top right of the p. is X, the bottom left of the p. is Z, and the bottom right of the p. is Y.

sorry about the bad drawing...

one more..... and p.s the first correct answer is chosen as best

given: triangle LGK is congruent to triangle MJK and GHJK is a parallelogram
prove: GHJK is a rhombus

draw another typical parallegram, and then label the top left K. top right J, bottom left G, bottom right H, and intersection of the diagonals M. then draw a triangle with KG as its base. the tip of the triangle is labeled L.

that is about right... except l and k should be connected.

okay, now i need 2 2 column proofs. And please help me....!


inequalities symmetry induction logarithms calculus derivative

[hide]
You could be totally lame and use Lagrange multipliers... it comes out to deducing $a = b$ from $\frac{bcd}{(a+1)^3} = \frac{acd}{(b+1)^3}$, or

$(a-b)[(a^2+b^2)(a+b)+3(a^2+ab+b^2)+3(a+b)+1] = 0$,

and since the right term is always positive we must have $a = b$. Similarly, we conclude $a = b = c = d = 1$ at the minimum, which gives us the desired result.[/hide]

[quote="blahblahblah"]Hard problem

Try proving

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac{1}{1+ab}$[/quote]
With condition $ab \ge 1$.

[quote="MrWise"]Let $a,b,c,d > 0$ and $abcd=1$
Prove that
$\frac{1}{(a+1)^2}+\frac{1}{(b+1)^2}+\frac{1}{(c+1)^2}+\frac{1}{(d+1)^{2}} \geq 1$[/quote]

[b]An elementary proof[/b]

Let $a=\frac x y , b = \frac y z , c = \frac z t, d = \frac t x$ where $x, y , z , t > 0$.
The problem becomes showing $\frac{y^2}{(x+y)^2}+\frac{z^2}{(y+z)^2}+\frac{t^2}{(z+t)^2}+\frac{x^2}{(t+x)^2}\geq 1$.

Note that
$1=(\frac x{x+y} + \frac y {x+y})^2=\frac{x^2}{(x+y)^2} + \frac{y^2}{(x+y)^2} +\frac{2xy}{(x+y)^2}$---(1)
$1=(\frac y{y+z} + \frac z {y+z})^2=\frac{y^2}{(y+z)^2} + \frac{z^2}{(y+z)^2} +\frac{2yz}{(y+z)^2}$---(2)
$1=(\frac z{z+t} + \frac t {z+t})^2=\frac{z^2}{(z+t)^2} + \frac{t^2}{(z+t)^2} +\frac{2zt}{(z+t)^2}$---(3)
$1=(\frac t{t+x} + \frac x {t+x})^2=\frac{t^2}{(t+x)^2} + \frac{x^2}{(t+x)^2} +\frac{2tx}{(t+x)^2}$---(4)

Adding these four equations and then re-distribute terms we obtain:
$4 = U + V + \frac{2xy}{(x+y)^2}+\frac{2yz}{(y+z)^2}+\frac{2zt}{(z+t)^2}+\frac{2tx}{(t+x)^2}$---(*),
 where $U=\frac{x^2}{(x+y)^2}+\frac{y^2}{(y+z)^2}+\frac{z^2}{(z+t)^2}+\frac{t^2}{(t+x)^2}$ and $V=\frac{y^2}{(x+y)^2}+\frac{z^2}{(y+z)^2}+\frac{t^2}{(z+t)^2}+\frac{x^2}{(t+x)^2}$.

Now, observe that $\frac{2xy}{(x+y)^2}\leq 2\cdot \frac{ (\frac{x+y}2)^2}{(x+y)^2}= \frac 1 2$ by AM-GM inequality.
Simliary, $\frac{2yz}{(y+z)^2}\leq \frac 12$, $\frac{2zt}{(z+t)^2}\leq \frac 12$ and also $\frac{2tx}{(t+x)^2}\leq \frac 1 2$.

Combine these results with (*), we have $U+V\geq 4-4\cdot \frac 1 2 = 2$.

Case 1  When $V\geq U$
Then $2V\geq U+V\geq 2 \implies V \geq 1$, which is exactly what we want.

Case 2 When $U\geq V$
Then $2U\geq U+V\geq 2 \implies U \geq 1$. This result is true for all $x, y, z , t > 0$
We can make use of symmetry, apply the permutation: x exchanges with y and z exchanges with t, then the requried inequality is recovered.
Q.E.D.

[b]Remark[/b]
1.  We can extend the result to n variables:
Given $a_1, a_2, ....,a_n>0$ and $a_1\ a_2 ... a_n =1$, then $\frac 1 {(1+a_1)^2}+ \frac 1{(1+a_2)^2}+... +\frac 1{(1+a_n)^2}\geq \frac n 4$.
2. I am seeking a proof of the general case by mathematical induction. It will be much appreciated if anyone can help me with that. :)

I remember darij gives a nice prove using only cauchy

$(a+b)(ab+1)\geq b(a+1)^2$
$(b+a)(ab+1)\geq a(b+1)^2$

and the rest should be easy  :)

Nice solution, luimichael.
There's about 3 solution for this inequality. I remember that iandrei2 divided a proof by contracdition, which is also good.

Hi! Lovasz, thanks for your compliment. :)

In terms of simplicity, the method mentioned by shyong is better.

Well, if the general case for any $n$ is true, then why not take a trial by induction ?

It may become the fourth method!  :idea:

[quote="luimichael"]
Case 1  When $V\geq U$
Then $2V\geq U+V\geq 2 \implies V \geq 1$, which is exactly what we want.

Case 2 When $U\geq V$
Then $2U\geq U+V\geq 2 \implies U \geq 1$. This result is true for all $x, y, z , t > 0$
We can make use of symmetry, apply the permutation: x exchanges with y and z exchanges with t, then the requried inequality is recovered.
Q.E.D.
[/quote]

I don't think this proof works. In case 2, $U\ge 1$ isn't true for all $x, y, z , t > 0$. It's only true under the assumption that $U\geq V$. This means you can't exchange variables freely.

I am sorry, my brain is probably blocked at this moment. I hope someone else can make a proper judgement.

[quote="MrWise"]Let $a,b,c,d > 0$ and $abcd=1$
Prove that
$\frac{1}{(a+1)^2}+\frac{1}{(b+1)^2}+\frac{1}{(c+1)^2}+\frac{1}{(d+1)^{2}} \geq 1$[/quote]

Jensen's on $f(x) = \frac{1}{ (e^x + 1)^2 }$.

$x = \ln a, y = \ln b, z = \ln c, w = \ln d$
$x + y + z + w = 0$
$\sum_{cyc} \frac{1}{ (e^x + 1)^2 } \ge \frac{4}{(e^{0} + 1)^2} = 1$

:)

Edit:  The general case is, of course, obvious from this proof.  :D

[quote="t0rajir0u"]Jensen's on $f(x) = \frac{1}{ (e^x + 1)^2 }$.
[/quote]

Are you sure $f(x)$ is convex on the entire interval $(-\infty,+\infty)$?

To present a solution here, you probably should calculate the 2nd derivative to verify the convexity.

Oh.  $f(x)$ is not, in fact, convex $\forall x$.   :|

Edit: I have misread the statement stated by fuzzylogic.
And I know what you mean now, fuzzylogic.

Let us denote $F(x,y,z,t) = \frac{x^2}{(x+y)^2}+\frac{y^2}{(y+z)^2}+\frac{z^2}{(z+t)^2}+\frac{t^2}{(t+x)^2}$.

Then we have $F(x,y,z,t)+F(y,x,t,z) \geq  2$---(*)
Let $S = inf \{F(x,y,z,t)| x, y , z, t >0 \}$, whose existence is guaranteed because $F$ is bounded below by 0.
By symmetry of the role of x, y, z, and t , $S = inf\{F(x,y,z,t)| x, y , z, t >0\}= inf \{F(y,x,t,z)| x, y , z, t >0 \}$.
Therefore, by (*), we have $2S\geq 2 \implies S\geq 1$.
Finally, $F(x,y,z,t)\geq S\geq 1$.

I hope this can remedy the mistake in my proof.

[quote="luimichael"]Then we have $F(x,y,z,t)+F(y,x,t,z) \geq 2$---(*)

Therefore, by (*), we have $2S\geq 2 \implies S\geq 1$.
[/quote]

I am not sure how (*) implies that $2S\ge2$.

But [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30724]here[/url], iandrei2 showed us a very nice one.

Oh... fuzzylogic misunderstood luimichael, and luimichael made a typo in his solution. It should be rewrite kile this:
\[ a,b>0,\ a+b \ge 2 \implies 2a \ge a+b \ge 2 \implies a\ge 1 \ (\text{suppose}\ a \ge b)  \]

[quote="silouan"]I think the best solution is the following.....

[quote="blahblahblah"]Hard problem

Try proving

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac{1}{1+ab}$[/quote][/quote]
So nice!
\[ \geq\frac{1}{1+ab}+\frac{1}{1+cd} \]
\[ =\frac{1+1+ab+cd}{1+abcd+ab+cd}=1 \]

[quote="blahblahblah"]Hard problem

Try proving

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac{1}{1+ab}$[/quote]

And how do we prove this?

[quote="blahblahblah"]If you expand it it reduces to

$ab(a^2+b^2)+1\geq a^2b^2+2ab$, but

$ab(a^2+b^2)+1\geq 2a^2b^2+1=a^2b^2+(1+a^2b^2)\geq a^2b^2+2ab$[/quote]

Really nice :!:  I tried expanding, but didn't success


In a right-angle triangle $ ABC$, $ AB$ is the hypotenuse and $ CF$ an altitude. The circle through $ F$ centered at $ B$ and the circle of the same radius centered at $ A$ have a common point on the side $ BC$. Determine the ratio $ FB: BC$.


Let $ A_{1}, A_{2},...., A_{m}$ be subsets of $ {1,2,3,...,n}$. If there exist a prime p and a positive integer k such that $ |A_{i}|$ (the number of elements) is not divisible with $ p^{k}$ for any i in 1,n but $ |A_{i}\cap A_{j}|$ is divisible by $ p^{k}$  for any i,j (different) in 1,n than $ m\leq n$.


Ngày 20 tháng 10 năm 2007, Sở GD HN kết hợp với Hội Toán học Hà Nội đã tổ chức kỳ thi Chọn Đội tuyển Việt Nam dự Olympiad Khoa học Quốc tế (IJSO 2007). Sau đây là kết quả chọn đội tuyển

1) Hoàng Tuấn Anh
2) Đào Tùng Anh
3) Nguyễn Nhật Chi
4) Mai Quang Huy
5) Nguyễn Anh Quân
6) Nguyễn Bá Kiều Oanh

Các bạn 1,2,3,4 là học sinh trường Hanoi Ams, bạn 5 từ THCS Trưng Vương, bạn 6 là học sinh trường THCS Nguyễn Trường Tộ. Kỳ thi sẽ diễn ra từ ngày 2 đến 11 tháng 12 năm 2007 tại Đài Loan.

Chúc mừng các em.

em nghe nói còn có bạn LÊ Vũ Huy khá giỏi , nhưng lần này lại trượt ....
với lại em cũng băn khoăn là Sở đã có thông tin gì về lịch học bồi dưỡng , hay tương tự như thế chưa ạ ?

ai có kết quả thì post lên cho mọi người xem với xem với!

Năm nay, đoàn VN dành được 4 HCĐ; tất cả đều thuộc về các bạn trai.

Số lượng các nước tham gia kỳ thi này mỗi năm lại tăng lên. Năm nay chứng kiến sự tham gia của nhiều nước phát triển nữa. Thành tích đội mình năm nay không quá tệ. Ngay cả UK cũng chỉ có một HCĐ (không có HCV, HCB).


Does anyone have the answer key to 2009 Mathcounts nationals competition? If you do, do you mind sending me a copy? I need one badly to check my answers. Thanks. :lol:

PM me your answers and I'll tell you what you missed.

or maybe draops wants to know what the answers are.

I can send you a copy (don't worry, for free)

thanks, i would just like to know the answers :lol:  e-mail- ericdraops@gmail.com

Sorry, I don't really have the test either. I wrote all over the test that my friend made a copy of for me, and I currently don't have a blank copy right now.....

[quote="mathgenius1"]This is kinda funny...

I have the answers but not the test!

I can give you the answers if you can give me the test! :P[/quote]
Do you have anything else to trade? If not, just PM me and I will give you the tests.

Also, this might belong in the "MATHCOUNTS Resources HERE!" forum.


invariant
advanced fields
advanced fields unsolved

Let $ L_{p}$ be the standard proof system - 3 axioms and Modus Ponens:
$ (A1) A \implies (B \implies A)$
$ (A2) (A \implies (B \implies C)) \implies ((A \implies B) \implies (A \implies C))$
$ (A3) (\minus{}B \implies \minus{}A) \implies ((\minus{}B \implies A) \implies B))$
(minus means negation)

It is known that $ L_{p}$ is complete. Let's replace (A3) with a new rule of inference: if $ \minus{}B \implies A$ and $ \minus{}B \implies \minus{}A$ can be proved (with $ (A1),(A2)$ and $ MP$), then we can also prove $ B$. Is this new system, $ L'_{p}$, complete?

It seems to me that these two systems are trivially isomorphic.  In particular, we can prove (A3) by substituting $ \neg A$ for $ A$.

EDIT: Nevermind, it's not as trivial as I thought.  I suppose the entire question is whether (A3) is indeed provable using your rule of inference.

Yep... And I thought that the deduction theory shows exactly that ($ B$ is provable from $ \sum \cup A$ iff $ A \implies B$ is provable from $ \sum$), but It doesn't, because we also need to prove that if $ A \implies ( \minus{} B \implies C)$ and $ A \implies ( \minus{} B \implies \minus{} C)$ can be proved,  $ A \implies B$ can also be proved. Intuitively it seems as if a proof cannot be constructed but I can't point exactly why. So I will keep trying...

Maybe we can find some sort of invariant that would be violated by such a proof?


Just how many "harmonious prime series"are they in any given musical notes? Does the "ocean of prime numbers" really have a strong connection to every musical composition?
Just curious.

                                                  Respectfully,
                                                 A.Eddington

What do you mean?

Curious.

Your fellow curious person,
V.M.Funcia

err ok i was faking the sig but what did you mean?

There are a infinite amount of prime harminious series...

 :spider:

[quote="pizza1512"]There are a infinite amount of prime harminious series...
[/quote]

We have a finite range of hearing, so we can't have an infinite number of harmonies, and we dont have limitless distinction b/w notes, you may be using infinite in as a literary term, but just because infinity is commonly misunderstood i wanted to clarify

I don't think there are any in a technical sense.  I was looking into harmonies and major chord and minor chord periods and here are some things I figured out

you can define frequency based on note:
$f(n) = c 2^{\frac{n}{12}}$
what this means is every 12 half steps (or semitones) the frequency doubles. (there are 12 half steps in an octave)
A has a frequency of 440 hz which you can use to define c (you can use other frequences but 440 is the std frequency.  You can look at different harmonies like a perfect fifth (7 half steps) but the constant of proportionality b/w the 2 notes is irrational, same for all harmonies.  You may have seen different ratios for the differernt harmonies like [url=http://thinkzone.wlonk.com/Music/12Tone.htm]here[/url] but they're approximations. On the site the harmonies have some a lot of primes so it depends how technical you want to get, the diatonic scales, most common ones have 7 notes.


How does one make an arc symbol in latex?

The package [b]yhmath[/b] gives the command \wideparen, so write \usepackage{yhmath} in the preamble in order to use this command.
If you want other sorts if arcs, parentheses,.... cf. [url=http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]The Comprehensive LaTeX Symbol List[/url]

[quote="stevem"]Do you mean $\stackrel{\frown}{AB}$ which is given by \stackrel{\frown}{AB}?[/quote]
I think that this one is smaller.

No había encontrado la solucion a este problema. Revise muchos manuales y nada.

Me has salvado de caer demente al no encontrar la respuesta. Gracias.

[i]Please post in English on this forum[/i]
Google translator gives
[quote]He had not found the solution to this problem. Check many manuals and nothing.

I've saved it from falling insane not to find the answer. Thank you.[/quote]
Answered in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url] 10

The question here has been asked many times before, and can easily be found by searching. A summary: there's no really good way to do it on the forum, and several packages that contain solutions for your own documents.


integration
logarithms
calculus
derivative
calculus computations

given just that

$ \Gamma\,(y)\;\equal{}\;\int_{0}^{\infty}\;x^{y\minus{}1}\;e^{\minus{}x}\;\textbf dx$

and

$ \psi\,(y)\;\equal{}\;\frac{d\,^{2}}{dy^{2}}\;\left(\;\ln\,(\Gamma(y))\;\right)$

show :D

$ \psi\left(\frac{3}{2}\right)\;\equal{}\;\boxed{\frac{\pi^{2}}{2}\minus{}4}$

$ \psi(z)\plus{}\gamma\equal{}\int_{0}^{1}\frac{1\minus{}t^{z\minus{}1}}{1\minus{}t}dt$ is after differentation $ \psi'(z)\equal{}\int_{0}^{1}\frac{\minus{}t^{z\minus{}1}\ln t}{1\minus{}t}dt$

$ \psi'(n)\equal{}\int_{0}^{1}\sum_{k\equal{}0}^{\infty}\minus{}t^{n\minus{}1\plus{}k}\ln t\, dt\equal{}\sum_{k\equal{}0}^{\infty}\frac{1}{(n\plus{}k)^{2}}\equal{}\sum_{k\equal{}n}^{\infty}\frac{1}{k^{2}}\equal{}\frac{\pi^{2}}{6}\minus{}\sum_{k\equal{}1}^{n\minus{}1}\frac{1}{k^{2}}$

Hence $ \psi'(1)\equal{}\frac{\pi^{2}}{6}$ and $ \psi'(2)\equal{}\frac{\pi^{2}}{6}\minus{}1$ .

Logarithmate the duplication formula $ \Gamma(z)\,\Gamma\left(z\plus{}\frac{1}2\right)\equal{}2^{1\minus{}2z}\sqrt{\pi}\,\Gamma(2z)$

$ \ln\Gamma (z)\plus{}\ln\Gamma\left(z\plus{}\frac{1}2\right)\equal{}(1\minus{}2z)\ln 2\plus{}\ln\sqrt{\pi}\plus{}\ln\Gamma(2z)$ and differentiate

$ \psi(z)\plus{}\psi\left(z\plus{}\frac{1}2\right)\equal{}\minus{}2\ln 2\plus{}\psi(2z)\cdot 2\quad .$ After that differentiate again

$ \psi'(z)\plus{}\psi'\left(z\plus{}\frac{1}2\right)\equal{}\psi'(2z)\cdot 4$ and set $ z\equal{}1$ .

It follows $ \psi'\left(\frac{3}2\right)\equal{}4\psi'(2)\minus{}\psi'(1)\equal{}\frac{\pi^{2}}{2}\minus{}4$


calculus
integration
trigonometry
real analysis
real analysis unsolved

For $ p\in \mathcal N^*$ ascertain the improper definite integral $ \int_0^{\infty}\frac {1}{(1 \plus{} x^2)(1 \plus{} x^p)}\ \mathrm {dx}\ .$

[hide="Big hint"]Substitution $ t \equal{} \frac1x$ helps a lot  :wink:[/hide]

[hide="Answer"]The value of this integral is independent of p. To see this make the substitution $ x \equal{} \tan t$ to get

$ I \equal{} \int_0^{\pi/2}\frac{1}{1\plus{} \tan^pt}\,dt \equal{} \int_0^{\pi/2}\frac{\cos^pt}{\cos^pt \plus{} \sin^pt}\,dt$.

If we now make $ t \equal{} \frac{\pi}{2} \minus{} u$ we get

$ I \equal{} \int_0^{\pi/2}\frac{\sin^pu}{\sin^pu \plus{} \cos^pu}\,du$

and so

$ I \plus{} I \equal{} 2I \equal{} \int_0^{\pi/2}\frac{\cos^px}{\cos^px \plus{} \sin^px}\,dx \plus{} \int_0^{\pi/2}\frac{\sin^px}{\sin^px \plus{} \cos^px}\,dx$

$ 2I \equal{} \int_0^{\pi/2}\frac{\sin^px \plus{} \cos^px}{\cos^px \plus{} \sin^px}\,dx \equal{} \int_0^{\pi/2}dx \equal{} \frac{\pi}{2}$.

Therefore, $ I \equal{} \frac{\pi}{4}$ for every p in N.[/hide]


Let $p$ be the smallest prime divisor among all the prime divisors of the numbers $n_i$. Assume $p|n_2$. We then have $p|(2^{n_1}-1,2^{p-1}-1)=2^{(n_1,p-1)}-1=1$, and we're done.


function
group theory
linear algebra
linear algebra unsolved

[probably well known, but nice for sure  ;) ]

Find all functions $f : S_n \to  M_n(R)$ such that $f(x o y) = f(x) + f (y), \forall x, y \in S_n$
($S_n$ is the set of permutations of n degree)

So $f$ is a group homomorphism from a finite group ($(S_n,\circ)$) to a torsion free group ($(\mathcal M_n(\mathbb R),+)$). Obviously, every such homomorphism must be trivial, i.e. it must send every element of its range to the identity of the second group. In our case, $f(\pi)=0,\ \forall\pi\in S_n$.


i need to figure out how to prove that a vertex of a triangle and two excenters are concurrent or essentially that excentral triangal exists. Please help, anything you say is helpful. Thank you.

Do you know what is meant by an "external angle bisector"? You might like to consider whether one, or both of the relevant excentres lies on the external angle bisector of a particular angle!


ratio
geometry
circumcircle
inequalities
trigonometry
vector
area of a triangle

1. The ratio of the circumcircle radius to incircle radius of a 4,5,6 triangle is?

2. In triangle ABC $a=5, b=4$ and $\cos(A-B)=\frac{31}{32}$ Find c.

3. If a^2+b^2+c^2=1 find maximum value of ab+bc+ca.

[hide="3"]
By Cauchy-Schwarz inequality, we get \[ (ab+bc+ca)\le (a^2+b^2+c^2)(b^2+c^2+a^2)=1^2=\boxed {1}.  \] That's my first time ever using this inequality, so I could be wrong.
[/hide]

edit:

[hide="1"]
$s=\frac{15}{2}$

$\Rightarrow K=\sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)}$

$=\frac{15\sqrt{7}}{4}$

$R=\frac{4(5)(6)}{4\left(\frac{15\sqrt{7}}{4}\right)}=\frac{8}{\sqrt{7}}$

$r=\frac{\frac{15\sqrt{7}}{4}}{\frac{15}{2}}=\frac{\sqrt{7}}{8}$

$\therefore \frac{R}{r}=\boxed {\frac{64}{7}}$

Could be wrong.
[/hide]

#2
[hide]$c=6$.
$\cos(A)\cos(B)+\sin(A)\sin(B)=\frac{31}{32}$. Using the sine law, $\sin(A)=\frac{5\sin(B)}{4}$. Sub in this to above, we get: $\frac{5(\sin(B))^2}{4}+\cos(A)\cos(B)=\frac{31}{32}$. $(\sin(B))^2=1-(\cos(B))^2$. Now apply the cosine law to $A$ and $B$, then we get $\cos(A)=\frac{c^2-9}{8c}$ and $\cos(B)=\frac{c^2+9}{10c}$. Sub in these values and u get $c=6$ [/hide] :)

What's the Cauchy-Schwarz Inequality?  I just thinking that the greatest value would be if they were all equal, in which case $ab+bc+ac=a^2+b^2+c^2=\boxed{1}$

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

The acutal inequality is written in $\sum$ form.

[quote="236factorial"][hide="1-so annoying like the girls on my swim team, yet an excellent problem"]The area, $A$, using Heron's formula, is $\frac{15\sqrt7}{4}$.

$R=\frac{abc}{4A}=\frac{120}{15\sqrt7}=\frac{8\sqrt7}{7}$.

$r=\frac{A}{s}=\frac{\frac{15\sqrt7}{4}}{\frac{15}{2}}=\frac{\sqrt7}{2}$.

So $\frac{R}{r}=\frac{\frac{8\sqrt7}{7}}{\frac{\sqrt7}{2}}=\frac{16}{7}$. [/hide][/quote]

Oops, I had an 8 in my denominator for $r$ instead of 2. :blush:

[hide="more on 3"]I believe the application of Cauchy Schwarz would be
\[ 3(a^2+b^2+c^2) \ge (a+b+c)^2 \implies a^2+b^2+c^2 \ge ab+bc+ca \]

An even simpler Cauchy application: $ab+bc+ca=(a, b, c) \cdot (b, c, a)=(a^2+b^2+c^2)\cos{\theta}$ where $\theta$ is the angle between the two vectors...[/hide]

[quote="K81o7"]An even simpler Cauchy application: $ab+bc+ca=(a, b, c) \cdot (b, c, a)=(a^2+b^2+c^2)\cos{\theta}$ where $\theta$ is the angle between the two vectors...[/quote]

That's what I did, but not in vector form. ;)

But where did you get $ab+bc+ca \le (a^2+b^2+c^2)(b^2+c^2+a^2)$?

yea i think it should be:

$(a^2+b^2+c^2)(b^2+c^2+a^2) \ge (ab + bc + ac)^2 \implies 1 \ge (ab+bc+ac)^2 \implies 1 \ge (ab+bc+ac)$

O.K a dumb extension on the third one I thought of....

After find $ab+bc+ca=1$,prove that $a=b=c$(And if you want it to be a bit  challenging,do it without seeing my method or even better,  [hide="without"]inequalities  [/hide]   :D )
:rotfl: Stupid easy extension.

[hide="My method for #3"]$a^2+b^2 \geq 2ab$

$b^2+c^2 \geq 2bc$

$c^2+a^2 \geq 2ac$

Add all three

$a^2+b^2+c^2 \geq ab+bc+ca$

And the rest is easy... [/hide],if a,b,c are positive reals


AMC
AIME
algebra
polynomial
binomial theorem

um ok so i understand part of this quesiton but i would like a more thourough explanation of the end.  Here is the problem

$ P_0(x) \equal{} x^3 \plus{} 313x^2 \minus{} 77x \minus{} 8$ for integers $ n\ge1$ define $ P_n(x) \equal{} P_{n \minus{} 1}(x \minus{} n)$ what is the coefficient of $ x$ in $ P_{20}(x)$?

Ok so i understand how to do up to the point where you plug $ (x \minus{} 210)$ into the formula!  But then i dont understand how to find the coefficient on $ x$ if someone could explain in a little more detail than in the solution provided below i would really appreciate it!

[url=http://www.artofproblemsolving.com/Wiki/index.php/1993_AIME_Problems/Problem_5]Solution i dont understand!!![/url]

The solution simply means to substitute all instances of $ x$ in the original polynomial $ P_0(x)$ with $ (x\minus{}210)$.

and then do what??? thats what i dont understand!

Oh. It's just a really badly worded answer. The "cubic" and "square" just refer to the first and second terms. (And then the binomial theorem is used on them.)

[quote="Brut3Forc3"]Oh. It's just a really badly worded answer.[/quote]

Hey  :P .

I've fixed a bit, my writing skills were really bad (and still are).

oh ok thank you very much, i understand!!!YAY(im terrible at AIME 3-4 average! :o )


Se considera numerele reale pozitive $\ a,b,c$ astfel incat $\ 0<(a,b,c)<1$.Demonstrati ca avem inegalitatea
 $\ a^3b+a^2c+ab+b^2c+b^2c^3<abc+(abc)^2+(abc)^3+3$.

edit : nu mai conteaza ce am scris in postu anterior, initial nu era $3$`ul in membrul drept in enuntul problemei

Da, asa e.Am corectat eu.Problema poate fi facuta in cel putin 3 moduri.

[quote="maky"]inegalitatea se scrie ca
$(1-a^{3}b)(1-b^{2}c^{3})+(1-a^{2}c)(1-b^{2}c)+(1-ab)+abc>0$.
deoarece $a,b,c<1$ este evident.
putem chiar sa renuntam la $abc$ din membrul drept[/quote]
    Eu am spart inegalitatea in
$\ (1-ab)(c-1)<0,(1-b^2c)(a^2c-1)<0,(1-b^2c^3)(a^3b-1)<0$.

[quote="pohoatza"]Monotonia functiilor pt clasa a 7a ;))[/quote]
    Era doar o idee.Eu asa ma gandisem prima data...Daca vrei, pot sa postez solutia completa folosind monotonia functiilor(pentru pureblack:iti spun eu ca nu mai e rezolvare de un rand ;) )

Zii,...stii sa o faci cu monotonia functiilor??E destul de usor.


Let n be a positive integer and let P1,P2.....,Pn be distinct 2 element subsets of the set
S={a1,a2.....,an}
such that if Pi intersection Pj not equal to null set then {ai,aj} is one of the P's . Prove that each of the a's appears in exactly two of the P's.

Does anyone have an answer to this sum?  :)

why not?. I think this is what he means  :)
Let n be a positive integer and let $ P_{1},P_{2}.....,P_{n}$ be distinct 2 element subsets of the set
 S = {$ a_{1},a_{2}.....,a_{n}$}
such that if$ P_{i} \cap P_{j}$ not equal to null set then {$ a_{i},a_{j}$} is one of the P's .
[b]Prove that each of the a's appears in exactly two of the P's.[/b]


floor function
inequalities
number theory proposed
number theory

Let $a,b$ be positive integers and $a|b^2,b^2|a^3,a^3|b^4,b^4|a^5,...$ prove that $a=b$.

Take any prime $p$ and let $v_p(n)$ be the highest exponent such that $p^{v_p(n)}$ divides$n$.
By the given conditions we have $(2n-1) v_p(a) \leq 2n v_p(b)$ and also $2n v_p(b) \leq (2n+1) v_p(a)$ for all $n$, so $(2n-1) v_p(a) \leq 2n v_p(b) \leq (2n+1) v_p(a)$ which gives $\frac{2n-1}{2n} v_p(a) \leq v_p(b) \leq \frac{2n+1}{2n} v_p(a)$. But taking limits gives instantly $v_p(a)=v_p(b)$, so $a=b$.

Sorry, but I wasn't able to understand your solution. Or at least I think...Let me see if I got it right. Take $p$ a prime number such that $p|a,b$. And consider $v_p(a),v_p(b)$ as you defined. Knowing that

$a^n|b^{n+1}\; \mbox{and}\; b^{n+1}|a^{n+2}$

we'd have

$(n+1)v_p(b)\geq nv_p(a)\; \mbox{and}\; (n+2)v_p(a) \geq (n+1)v_p(b)$

putting them together, we'd get

$(n+2)v_p(a)\geq (n+1)v_p(b)\geq nv_p(a), n\rightarrow \infty$

and then follows immediatelly the conclusion $v_p(a)=v_p(b)$

but why do we have $a=b$

and correct me if I got the solution in a wrong way...
thanks

You got it the right way :)
To come from $v_p(a)=v_p(b)$ to $a=b$, you have to consider not only one prime $p$, but all primes $p$ (the conclusion $v_p(a)=v_p(b)$ works for any prime).
So all prime factors occure in $a$ and $b$ the same number of times, so $a$ and $b$ have exactly the same prime factors with the same exponents $v_p$, thus they are identical.

Thank for your solution,now what do you think about it:
Can you introduce a function(f) from N to N that for numbers a,b both smaller than m,from this assertion:
$a|b^2,b^2|a^3,a^3|b^4,...,a^{f(m)}|b^{f(m)+1}$we can get the result a=b.

$f(n): = \left\lfloor log_2 (m) \right\rfloor$ should do (and is optimal) when $a,b \leq m$:
We will only use the last two conditions: $a^{f(m)-1} | b^{f(m)}$ and $b^{f(m)} | a^{f(m)+1}$.
These give us
$(1-\frac{1}{f(m)}) v_p(a) \leq v_p(b) \leq (1+\frac{1}{f(m)}) v_p(a) \iff -\frac{1}{f(m)} v_p(a) \leq v_p(b) - v_p(a) \leq \frac{1}{f(m)} v_p(a)$.
Now we have just to avoid $|v_p(a) - v_p(b)| \geq 1$, which is (by $\frac{1}{f(m)} v_p(a) \geq |v_p(a) - v_p(b)|$) excluded iff $\frac{1}{f(m)} v_p(a)<1$.
Now trivially $v_p(a) \leq log_p(a)$, so we want $\frac{1}{f(m)} log_p(a) <1 \iff log_p(a) < f(m)$.
Since $2$ is the smallest prime, we just need $log_2(a) \leq f(m)$ and the condition for the other primes will follow.
When we look back now, we see that all these inequalities can be sharp, so this $f$ is optimal.
To show directly that $f$ is optimal:
Take $m=a=2^n, b=2^{n-1}$. When we have $f(n) = \left\lfloor log_2 (m)\right\rfloor -1=n-1$ (or less), the conditions are fulfilled ($a^{n-2}=2^{n^2-2n} | 2^{n^2-n}=b^{n-1}, a^{n-1}=2^{n^2-n} | 2^{n^2-n}=b^n$), but they are not equal.

$a>b+k$.For n is large enough then$a^m>b^n$ with$m=n-1$.


This is a thread for the disscussion of chess, ratings,  puzzles and other things.

To start it off, I am rated somewhere between 1100 and 1400, depending on the day I play, who I play, and if I am black or white. For white, I open with e4, and with the max lange attack if I get symetrical e pawns. Otherwise I just wing it, though I have looked into the Kings Indian attack. For black, I'm trying to learn the king's indian and the sicilian, though I play much better with white.

I also enjoy doing chess tactics drills and puzzles. I can do the tactics things much better than the real puzzles, like the ones by Sam Loyd, like [url=http://en.wikipedia.org/wiki/Excelsior_%28chess_problem%29]this one.[/url] I found a great database of tactics drills [url=http://www.uni-klu.ac.at/~gossimit/c/tactic.htm]here[/url]. Over 10,000 drills, plus a link the easier training ones.

One great resource of chess knowledge, especially on people and openings, opening traps especially, is wikipedia.

So, um, tell about yourself and post some chess puzzles?

To start the puzzles , try the mate in three, number 3W0001, [url=http://www.entertainmentjourney.com/3w.htm#3W0001]here.[/url]

My rating was as high as 1630, but its dropped now to 1570ish. I havnt played any USCF games for about the last 6 months...i seem to have lost interest in the game.

[quote="Gyan"]I would expect a few in this crowd to have 2000+ ...[/quote]
Who?
And how do you get that high!? Do you need to be a prodigy?

i have 1198. :blush:
i only have played once online and i lost.

So, what are everyone's openings? Training methods? Favorite tactics?

I like attacking f7 when playing white. In the Max Lange, after bishop to c4, I sneak in a Queen to f3. Qxf7 is mate if not blocked.

This is a great way to win. I love sneaking it in while in the middle game. They almost never see the mate coming.

I quit over a year ago but I was about 1300.


If m<n, find minimum n : (m,n)=1 and the decimal representation of m/n contains the digits 2, 5, 1 consecutively. Help!

First, we'd like to place those 2, 5, 1 somewhere.  If m/n = .blahblah251blahblah..., we can multiply by a large power of 10 and subtract out an integer to make it look like .251blahblahblah.

Now, we need the smallest n such that there exists an m such that .251 :le: m/n < .252

See what you can do from there.

I figured this one out but looking back on it
251/1000 <= x/y < 252/1000

easily simplifies to finding min(x)|Floor[1000x/251]-Floor[1000x/252]>=1.

Yet is there any way to do this other than trial and error? (Knowing that the gap in the interval (1000x/252,1000x/251] increases as x increases).

Here was how I did it -- we know it looks like .251blah > 1/4, but not by much.  Thus, we know our number has the form (n + 1)/(4n + k), where k = 0, 1, 2, or 3.  Then, you solve the equations for that being less than 252/1000 but greater than 251/1000.  It actually isn't hard to see that n will be min when k is max (although, of course, since there are only 4 cases for k you don't really need to see that) and it works out not badly at all.

Yeah I found another way to do it:

0.251 = [0,3,1,61,1,3] = [0,3,1,61.75]
0.252 = [0,3,1,30,2] = [0,3,1,30.5].

[0,3,1,61.75]<=x<[0,3,1,30.5]

Yet [0,3,1,x]=(x+1)/(4x+3). Let x=a/b, then [0,3,1,x]=(a+b)/(4a+3b). 4(a+b)-(4a+3b)=b and (4a+3b)-3(a+b)=a, so (4a+3b,a+b)|a and (4a+3b,a+b)|b. Yet WLOG (a,b)=1 since otherwise, divide through numerator and denominator of (a+b)/(4a+3b) by (a,b) yielding strictly smaller denominator. So (4a+3b,a+b)=1. Clearly, the smallest denominator is then obtained when 4a+3b is minimized, or when b=1, and a=31. So 4a+3b=127.

All hail continued fractions... which I learned today!


inequalities
combinatorics solved
combinatorics

At a tennis tournament there were $2n$ boys and $n$ girls participating. Every player played every other player. The boys won $\frac 75$ times as many matches as the girls. It is knowns that there were no draws. Find $n$.

[i]Serbia[/i]

So there are boys (otherwise $n =0$ is a trivial answer :D)??? what is known about their number?

Pierre.

in fact there are 2n boys and n girls playing at a tenning tournament. the problem is extremely simple: the boys won all the games they played between them namely n(2n-1) and the also won x games with the girls. the girls won the games between then, namely $\displaystyle \frac {n(n-1)}2 + 2n^2-x$, and all the games between them and the boys which the boys didn't win (obviously).

we thus obtain $5n(2n-1) + 5x = \displaystyle 7\frac {n(n-1)}2 + 14n^2-7x$, and we get $x$ out of this and we obtain that $x$ is an integer iff $n$ gives residues 5 or 3 modulo 8 (or something like that).

I see no combinatorics in this problem ... do you?

Valentin:Oh,no ,Miss "Can't understand it" is still bothering us with her infinite unsolved problems! :D
Iris:Yeah, you're right,but i'm getting prepared for the 8th JBMO,so...I need some help because most of the problems are very difficult for me :? (Valentin:Really??!!I hadn't noticed that ):D  :P
Irisi:I dont understand smth in the solution of the "extremly simple problem"(i'm usind the inverted commas for myself,becayse i know that it's very simple for you ;) )
1-What does it mean "the boys won all the matches beetwen them"?Does that mean every boy won every match he played with the other boys?That's impossible,because if A won with B than B lost with A(obviously),where A and B are boys.
2-Suppose i understand that when somebody will explain it,but how do you prove that?(this is may be absolutely easy but... :maybe: )
3-after working with the expresion you get x=n(5n-1)/8.How do you find what residues n should have,exept the case when 8 is a divisor of n?(the dump can't understan this too :D )
4-Suppose we found the residues of n,then will have more than one solution,an infinity .But the problem said :"find n",not all the values of n,although by knowing that n is the number of girls in a tennis tornament,we can't have very high values of it.
That's all.
Valentin:Urraaaaaah!!!!

Iris  to Mr Know it all(that'sd not irony,you seem to know everything about everything in maths,lucky you  :)

It's quite a shame for Serbia to propose at JBMO 2000 a problem that was given at Moldova NMO 1982, grade 10...

It's really easy.
There are (9n^2-3n)/2 matches played and therefore (15n^2-5n)/8 matches won by girls. But those matches played by girls only must be won by girls. So (15n^2-5n)/8 ≥n(n-1)/2 , simplify and we get 11≥n . And the rest has posted by the others.
May be this is too obvious so the solutions before did not state it .

@ stephencheng
your inequality yields $ n\geq \frac{1}{11}$

and how about the rest ?

[quote="Iris Aliaj"]
Irisi:I dont understand smth in the solution of the "extremly simple problem"(i'm usind the inverted commas for myself,becayse i know that it's very simple for you ;) )
1-What does it mean "the boys won all the matches beetwen them"?Does that mean every boy won every match he played with the other boys?That's impossible,because if A won with B than B lost with A(obviously),where A and B are boys.

[/quote]

I think this means that : In all games that two players were boys,no girl won the game,this is obviously since the game will win one of the boys, or draw.(but there is no draw).So you should answer the questions :

1.In how many games both players was boys
2.In how many games both players was girls
3.In how many games a player was boy,and another girl.

sorry for bad english !

Solution by [b]Burhan Can Karaca[/b]:

The match which was played a Boy vs. a Boy, has a Boy winner. It's same about Girl vs. Girl. So,

Boy vs. Boy --> $\binom{2n}{2}=2n^2-n$

Girl vs. Girl --> $\binom{n}{2}=\dfrac{n^2-n}{2}$

Now let's assume that Boys won $x$ of the matches which was played a Boy vs. a Girl. So Girls won $2n^2 - x$ of them.

$B$=Amount of the matches Boys won $=2n^2-n+x$

$G$=Amount of the matches Girls won $=\dfrac{n^2-n}{2} + 2n^2 - x$

$\dfrac{7}{5}=\dfrac{B}{G}=\dfrac{2n^2-n+x}{\dfrac{n^2-n}{2}+2n^2-x}$

From this, we have $8x=5n^2+n \Longrightarrow 8x=n(5n+1)$

Hence, $n$ must be equivalent $0$ or $3$ in $\mod8$


In trapezoid TRAP, TR || AP, TA ⊥ PR, TR = 6, TA = 19, and PR = 180. Compute AP.

Are you sure it is $180$???? because the answer will come in fractions or else I might be wrong.


Prove that if all $x_i$ are posititve rational numbers and $\sum_{i=1}^n \sqrt{x_i}$ is rational then all $\sqrt {x_i}$ are rational.

[admin edit: made it $\LaTeX$ compliant]

Let a_i =  \sqrt (x_i). Thus (a_i) 2  is a rational number for each i, and S = a_1 + ... + a_n is a rational number. We will prove that a_i is a rational number for each i.

Suppose, for a contradiction, that it is not the case. Wlog, we may assume that a_1 is not a rational number. Note that a_1 =/= 0.

Let P(t) be the polynomial defined by
P(t) = prod( t - a_1  \pm  a_2  \pm .... \pm  a_n)  (1)
where the product is over all the 2 n-1  choices of signs of the a_i's for i > 1.
Note that, if i > 1, then changing a_i into -a_i does not affect P(t), thus expanding P(t), each of these a_i's will appear only with even powers.

So, expanding, we group the terms according to the parity of the exponent of a_1, and we obtain P(t) = P(t, a_1, a_2,...,a_n) = N(t, (a_1) 2 ,..., (a_n) 2 ) - a_1*D(t, (a_1) 2 ,..., (a_n) 2 ).  (2)
Note that N and D are polynomials with rational coefficients.

Moreover, from (1), we have P(S, a_1,..., a_n) = 0.
Now, since a_1 is not a rational number, from (2) with t = S, we deduce that :
D(S, (a_1) 2 ,..., (a_n) 2 ) = 0.
In another hand, from (2), we have :
-2(a_1)*D(S, (a_1) 2 ,..., (a_n) 2 ) = P(S, a_1,..., a_n) - P(S, -a_1, a_2,...,a_n) = - P(S, -a_1, a_2,...,a_n)
from which we deduce that
0 = D(S, (a_1) 2 ,..., (a_n) 2 )
= P(S, -a_1, a_2, ..., a_n) / (2a_1)
= prod (2( a_1 + sum{i in I} a_i))
where the sum is over all the non-empty subsets of {2,...,n}
This leads to a contradiction, since each factors of the product is a positive number.
The conclusion follows....

Pierre.


i think this contracdition is $ a,b,c$ are 3 bases of a triangle and the prof is not difficult
@1 you can use $ a \plus{} b > c$........
@2 you see $ a \equal{} 2RsinA$ so....

[quote="ghostchungho"]$ a$,$ b$,$ c$ are the lenght of a triangle .
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{(a \plus{} b \plus{} c)^2} < \frac {1}{2}$
prove that we can't find a real number $ < \frac {1}{2}$ and satisfy this inequality.[/quote]

For all $ a,b,c$ the following identity is true:
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{(a \plus{} b \plus{} c)^2} \equal{} \frac {1}{2} \minus{} {\frac {\left( a \plus{} c \minus{} b \right) \left( a \plus{} b \minus{} c \right) \plus{} \left( b \plus{} c \minus{} a \right) \left( a \plus{} b \minus{} c \right) \plus{} \left( b \plus{} c \minus{} a \right) \left( a \plus{} c \minus{} b \right) }{2 \left( a \plus{} b \plus{} c \right) ^{2}}}
\]
This clearly proves that the left part is less then $ \frac 1 2$ if $ a,b,c$ are lengths of sides of triangle.

Sergey Markelov

[quote="Allnames"]So how about if All angels of $ ABC$ are less than $ 90$ degree[/quote]

No difference, maximum is the same: $ \frac 1 2$. Try the triangle with sides $ 1,1$ and $ \epsilon$, where $ \epsilon$ is very small.

Sergey Markelov

Thank Markelov,how about it
$ a^2 \plus{} b^2 \plus{} c^2\geq k(a \plus{} b \plus{} c)^2$ find the best $ K$
where angel $ A$ is greater than 90 degree

$ \frac{a^2\plus{}b^2\plus{}c^2}{(a\plus{}b\plus{}c)^2} \ge \frac{(a\plus{}b\plus{}c)^2}{3(a\plus{}b\plus{}c)^2} \equal{}\frac{1}{3}$

Because $ 3(a^2\plus{}b^2\plus{}c^2)\ge (a\plus{}b\plus{}c)^2$ by Holder..


[quote="FieryHydra"][quote="kimby_102"][quote="1234567890"]Let $p$ be an odd prime. Prove that $2^{p}-1$ is not divisible by $p$.[/quote]
[hide]
by Femat's last theorem, $2^{p}\equiv 2 (mod p)$.
so $2^{p}-1 \equiv 1 (mod p)$.[/hide][/quote]
I think you meant Fermat's Little Theorem. His Last Theorem would be too big a theorem to use here!  :wink:[/quote]
oops... yes i meant fermat's little theorem..


calculus
derivative
real analysis
real analysis unsolved

Show that if $ f: \Omega \to \mathbb{R}$ where $ \Omega$ is an open subset of $ \mathbb{R}^n$ and if each partial derivative $ D_jf$ exists at every point of $ \Omega$ and is bounded, then $ f$ is continuous in $ \Omega.$

[hide]Let $ M > 0$ be a bound for $ D_j f \ (j \equal{} 1, \cdots, n)$. Suppose $ B_{\delta}(\mathrm{x}) \subset \Omega$ and $ \mathrm{y} \in B_{\delta}(\mathrm{x})$.

If we let $ \mathrm{y}_0 \equal{} \mathrm{x}$ and $ \mathrm{y}_j \equal{} \mathrm{y}_{j\minus{}1} \plus{} (y_j \minus{} x_j)\mathrm{e}_j$ for $ j \equal{} 1, \cdots, n$, mean value theore applied for $ f(\mathrm{y}_{j\minus{}1} \plus{} t\mathrm{e}_j)$ shows that $ | f(\mathrm{y}_j) \minus{} f(\mathrm{y}_{j\minus{}1}) | \leq M |y_j \minus{} x_j|$. Thus

$ | f(\mathrm{y}) \minus{} f(\mathrm{x}) | \leq \sum_{j\equal{}1}^{n} | f(\mathrm{y}_j) \minus{} f(\mathrm{y}_{j\minus{}1}) | \leq \sum_{j\equal{}1}^{n} M |y_j \minus{} x_j| \leq nM \| \mathrm{y} \minus{} \mathrm{x} \|$

and $ f$ is continuous at $ \mathrm{x}$. So $ f$ is continuous.[/hide]


I was wondering how to do this problem without just listing all of the combos.

1. The 600 students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different  time. A computer randomly assigns each student to of of three lunch groups. What is the probability that three friends , Al, Bob and Carol?

 If anyone has a clever way to do this, please let me know.
Thanks

THe question is that what is the probability that they will get into the same lunch group.

[hide="Click for solution"]
Al has $ \frac{1}{1}$ chance of being assigned in a lunch group.  Bob has $ \frac{1}{3}$ chance of being in the same group, and Carol has $ \frac{1}{3}$ chance also.  Multiplying, we get $ \boxed{\frac{1}{9}}$.
[/hide]

Yes, the key idea here is that the lunch groups are assigned [b]randomly[/b], and are not necessarily of equal size. Therefore, each person would have a $ \frac 13$ chance of being assigned to any given lunch group.

If each lunch group was designed beforehand have 200 people, then the answer would be
[hide="Really complicated fraction"]
Let Al be assigned to a group. Then:
There are 199 seats left out of 599, so Bob has a $ \frac {199}{599}$ chance of being in the same lunch group as Al, and:
There are now 198 seats left out of 598, so Carol has a $ \frac {198}{598}$ chance of being in the same lunch group as Al and Bob, so multiply them together to get:
\[ \frac {19701}{179101} \approx 0.109999
\]
Probability that Al, Bob, and Carol are in the same lunch group.

[b]Off-topic:[/b] Notice that the above fraction is VERY close to $ \frac{11}{100}$... why?
[/hide]