calculus
integration
function
geometry
geometric transformation
reflection
calculus computations

Let $ f$ be a continuous function with $ f(0)\equal{}0,\,f(1)\equal{}1,\,f'(x)>0$ and $ \int_0^1f\equal{}\frac13.$ Find $ \int_0^1f^{\minus{}1}(x)\,dx.$

I can solve this by putting $ x\equal{}f(t)$ but my problem is how to prove that $ f$ is a surjection, since I know that is injective 'cause is a strictly increasing function.

It's surjective by the intermediate value theorem.

The statement also has a geometric interpretation (which is not necessarily the same as as proof): $ \int_0^{1}f^{ \minus{} 1}(x)\,dx$ is the area of a region that is congruent to the region under the graph of $ f$; the congruence being reflection about $ y \equal{} x$.

[quote="mlok"]It's surjective by the intermediate value theorem. [/quote]
Oh thanks, but how would you prove it?

Prove the intermediate value theorem?
http://en.wikipedia.org/wiki/Intermediate_value_theorem

No, no, I know what the IVT is, it's just that I don't know how to apply it to show the surjection.

What does "$ f$ is surjective" mean in this context?  What does the Intermediate Value Theorem say in this context?

[quote="mlok"]$ \int_0^{1}f^{ \minus{} 1}(x)\,dx$ is the area of a region that is congruent to the region under the graph of $ f$;[/quote]
Over (to the left of) the graph of $ f,$ not under the graph. Anyway, draw the picture for yourself. You'll see what he means.


hi!!

Let $ a,b,c$ the side lengths of a triangle,prove that:

$ \frac{a}{3a\minus{}b\plus{}c}\plus{}\frac{b}{3b\minus{}c\plus{}a}\plus{}\frac{c}{3c\minus{}a\plus{}b}\ge 1$.

thanx.

[quote="reason"]hi!!

Let $ a,b,c$ the side lengths of a triangle,prove that:

$ \frac {a}{3a \minus{} b \plus{} c} \plus{} \frac {b}{3b \minus{} c \plus{} a} \plus{} \frac {c}{3c \minus{} a \plus{} b}\ge 1$.

thanx.[/quote]

let $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x,x,y,z>0$, ineq becomes

$ {\frac {x\plus{}y}{4x\plus{}2y}}\plus{}{\frac {y\plus{}z}{4y\plus{}2z}}\plus{}{\frac {z\plus{}x}{4z\plus{}2x}} \ge 1,$

clearing of fractions, ineq becomes

$ xy^2\plus{}yz^2\plus{}zx^2 \ge 3xyz,$

obvious true by AM-GM.

All values of $ k$ for which the inequality
\[ \frac {a}{ka \minus{} b \plus{} c} \plus{} \frac {b}{kb \minus{} c \plus{} a} \plus{} \frac {c}{kc \minus{} a \plus{} b}\ge \frac{3}{k}\]
is true for all triangle are $ k\equal{}1$ and $ k\equal{}3.$

[quote="arqady"]All values of $ k$ for which the inequality
\[ \frac {a}{ka \minus{} b \plus{} c} \plus{} \frac {b}{kb \minus{} c \plus{} a} \plus{} \frac {c}{kc \minus{} a \plus{} b}\ge \frac {3}{k}\]
is true for all triangle are $ k \equal{} 1$ and $ k \equal{} 3.$[/quote]

$ k\equal{}2$ is also true, and I think it holds for $ 1\le k \le 3$.

[quote="reason"]hi!!

Let $ a,b,c$ the side lengths of a triangle,prove that:

$ \frac {a}{3a \minus{} b \plus{} c} \plus{} \frac {b}{3b \minus{} c \plus{} a} \plus{} \frac {c}{3c \minus{} a \plus{} b}\ge 1$.

thanx.[/quote]
we have:
$ 4\sum \frac {a}{3a \minus{} b \plus{} c} \equal{} \sum \frac {4a}{3a \minus{} b \plus{} c} \equal{} 3 \plus{} \sum \frac {a \plus{} b \minus{} c}{3a \minus{} b \plus{} c} \equal{} 3 \plus{} \sum \frac {{\left(a \plus{} b \minus{} c \right)}^{2}}{\left(a \plus{} b \minus{} c \right)\left( 3a \minus{} b \plus{} c \right)}$
$ \geq 3 \plus{} \frac {{\left( a \plus{} b \plus{} c \right)}^{2}}{\sum\left(a \plus{} b \minus{} c \right)\left( 3a \minus{} b \plus{} c \right)} \equal{} 4$
$ \Rightarrow \sum \frac {a}{3a \minus{} b \plus{} c}\geq 1$
Equality holds for $ a \equal{} b \equal{} c$

[hide=*][quote=arqady][quote="kuing"][quote="arqady"]All values of $ k$ for which the inequality
\[ \frac {a}{ka - b + c} + \frac {b}{kb - c + a} + \frac {c}{kc - a + b}\ge \frac {3}{k}\]
is true for all triangle are $ k = 1$ and $ k = 3.$[/quote]

$ k = 2$ is also true, and I think it holds for $ 1\le k \le 3$.[/quote]
Yes, you are right. I don't know to solve the following system:
\[ \{\begin{array}{l}k > 0 \\
k^2 + 2k - 3\geq0 \\
- k^2 + 2k + 3\geq0 \end{array}\]
 :D  :([/quote][/hide]
Let $ a,b,c$ the side lengths of a triangle,prove that:
\[ \frac {a}{ka - b + c} + \frac {b}{kb - c + a} + \frac {c}{kc - a + b}\ge \frac {3}{k}\]
Where $ 1\le k \le 3.$
[b]Klamkin[/b] (1982):
Let $ a,b,c$ the side lengths of a triangle,prove that:
\[ \frac {a}{k(b + c)-a} + \frac {b}{k(c + a)-b} + \frac {c}{k(a + b)-c}\ge \frac {3}{2k-1}\]
Where $  k \ge 1.$
[url=https://artofproblemsolving.com/community/c6h501238p2816500]here[/url]


probability
function
algebra
polynomial
number theory
relatively prime
geometric series

Ten identical crates each of dimensions $ 3$ ft $ \times$ $ 4$ ft $ \times$ $ 6$ ft.  The first crate is placed flat on the floor.  Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.  Let $ \frac{m}{n}$ be the probability that the stack of crates is exactly $ 41$ ft tall, where $ m$ and $ n$ are relatively prime positive integers.  Find $ m$.

[hide="Haha, we posted this problem at the same time so I changed the other one to #12."] Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:

$ 3a \plus{} 4b \plus{} 6c \equal{} 41$;
$ a \plus{} b \plus{} c \equal{} 10$.

Subtracting 3 times the third from the first gives $ b \plus{} 3c \equal{} 11$, or $ (b,c) \equal{} (2,3),(5,2),(8,1),(11,0)$. The last doesn't work, obviously. This gives the three solutions $ (a,b,c) \equal{} (5,2,3),(3,5,2),(1,8,1)$. In terms of choosing which goes where, the first two solutions are analogous.

For $ (5,2,3),(3,5,2)$, we see that there are $ 2\cdot\dfrac{10!}{5!2!3!} \equal{} 10\cdot9\cdot8\cdot7$ ways to stack the crates. For $ (1,8,1)$, there are $ 2\dbinom{10}{2} \equal{} 90$. Also, there are $ 3^{10}$ total ways to stack the crates to any height.

Thus, our probability is $ \dfrac{10\cdot9\cdot8\cdot7 \plus{} 90}{3^{10}} \equal{} \dfrac{10\cdot8\cdot7 \plus{} 10}{3^{8}} \equal{} \dfrac{190}{3^{7}}$, which clearly is irreducible. Thus, our answer is the numerator, $ \boxed{190}$.[/hide]

[hide]
Since each crate will add at least $ 3$ ft to the height, we can subtract it off.  If we let $ a$ be the number of $ 4$ ft crates and $ b$ be the number of $ 6$ ft crates, we have $ a\plus{}3b\equal{}11$ and $ a\plus{}b\leq 10$.  The possibilities are $ (a,b)\equal{}(2,3)$, $ (5,2)$, or $ (8,1)$.  Thus the number of legal towers is

$ \binom{10}{2}\binom{8}{3}\plus{}\binom{10}{5}\binom{5}{2}\plus{}\binom{10}{8}\binom{2}{1}\equal{}5130$.

The total number of possible towers is $ 3^{10}$.  So the probability is

$ \frac{5130}{3^{10}}\equal{}\frac{190}{3^7}$

And $ m\equal{}190$.
[/hide]

EDIT: We keep posting the same thing at the same time!  :roll:

Heh, someone I know did the same thing (although he answered $ 190\plus{}729\equal{}919$, which I presume is what you meant).

I thought 10 canceled out with the 3s... ya.

Dude, I got $ \frac{570}{3^{8}}$
Need to learn how to reduce.
:mad:

can you use balls and urns to find the answer?

I added an extra case, so 5130+1260=6390, divide by 9 and I got 710/3^8 :(

I also panicked because I assumed it wanted $ m\plus{}n$.  After working through the solution twice I finally saw that it just asked for $ m$.   :roll:

[quote="calc rulz"]I thought of it in terms of generating functions. You want to find the coefficient of $ x^{41}$ in
\[ \left(\frac {x^3 \plus{} x^4 \plus{} x^6}{3}\right)^{10}
\]
[/quote]Would you find the coefficient using the same method as the generic solution?

Yes, essentially, although the solution by generating functions makes it clear that this is just an application of the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]multinomial theorem[/url], just like many counting problems can be solved using just the binomial theorem.

Let us consider a generating function approach. Let the exponent of $ x$ denote the height of the stack. $ f\equal{}\frac{1}{3}\cdot (x^3\plus{}x^4\plus{}x^6)$ represents on block. The coefficient of $ x^k$ represents the probability of a height of $ k$ occuring. If we take $ f$ to the tenth power, we obtain all combinations of $ 10$ blocks in the expansion of that polynomial, because the exponents add and the coefficients multiply like they should in this case. Let us consider
\[ (3f)^{10}\equal{}(x^3\plus{}x^4\plus{}x^6)^{10}\equal{}...\plus{}a_{41}x^{41}\plus{}...\]
We can equivalently factor out $ (x^3)^{10}$ from both sides and we want
\[ (1\plus{}x\plus{}x^3)^{10}\equal{}...\plus{}k*x^{11}\plus{}...\] Now lets apply the multinomial theorem. We can expand the left where $ a_i\ge 0$ represents the number of times that we choose the $ i$th factor: \[ \sum_{a_1\plus{}a_2\plus{}a_3\equal{}10}\frac{10!}{a_1!a_2!a_3!}(1)^{a_1}(x)^{a_2}(x^3)^{a_3}\equal{}...\plus{}k*x^{11}\plus{}...\]
Since we want the terms with an exponent of eleven on $ x$, we need to have $ a_2\plus{}3a_3\equal{}11$, which has solutions $ (a_1,a_2,a_3)\equal{}(2t\minus{}1,11\minus{}3t,t)$ where $ t\equal{}1,2,3$. Then we must evaluate the coefficient which is \[ \sum_{k\equal{}1}^3\frac{10!}{(2t\minus{}1)!(11\minus{}3t)!(t)!}\equal{}90\plus{}2520\plus{}2520\equal{}9\cdot (10\plus{}280\plus{}280)\]

The final probability is then $ \frac{9(570)}{3^{10}}\equal{}\frac{570}{3^8}\equal{}\frac{190}{3^7}$. It follows that the answer is $ \boxed{190}$.

I just generated the number possibilites based upon the number of crates stacked with height 6.
It can be arranged in only 3 ways:
6,6,6,4,4,3,3,3,3,3
6,6,4,4,4,4,4,3,3,3
or 6,4,3,3,3,3,3,3,3,3

The number of ways to arrange the first two are 10C5*5C3, which is 2520.
The number of ways to arrange the last one is 10C8*2, which is 90.
(2520+2520+90)/3^10 simplifies to 190/3^7, 190 is the answer.

For those that used the generating function/counting coefficient method: (IRT #12 and #15)
Why did you divide by $ 3$?
(in $ f \equal{} \frac {1}{3}\cdot(x^3 \plus{} x^4 \plus{} x^6)$ and $ \left(\frac {x^3 \plus{} x^4 \plus{} x^6}{3}\right)^{10}$

The coefficient of $ x^k$ is the probability that $ k$ occurs. For example, $ 4$ occurs with a probability of $ \frac{1}{3}$, so we multiply. Since all the probabilities are equal, we can just factor the one-third out.

[quote="Altheman"]The coefficient of $ x^k$ is the probability that $ k$ occurs. For example, $ 4$ occurs with a probability of $ \frac {1}{3}$, so we multiply. Since all the probabilities are equal, we can just factor the one-third out.[/quote]
OH I see why everything works now...
That's an interesting way to look at the problem.

[quote="cognos599"]that's what i did. i got 190/3^7 and im like oh shoot 3^7 >1000 so i panicked and just assumed it was 190/3^6 and said 190 + 729 = 919.[/quote]

Hey that's exactly what happened to me! I should've done Diophantine equations because it's much easier. Also thought it was m+n.  :(

[quote="davidyko"]Exactly what I did.
Dumb anecdote: One of my friends, having simplified it to $ \frac {190}{3^7}$, assume that the answer had to be in the form $ m \plus{} n$, and therefore concluded that he must have been off by a factor of 3 in the denominator, and answered $ 190 \plus{} 729 \equal{} 919$.  :P

EDIT: Yes, yes, typing fail.[/quote]

[quote="cognos599"]that's what i did. i got 190/3^7 and im like oh shoot 3^7 >1000 so i panicked and just assumed it was 190/3^6 and said 190 + 729 = 919.[/quote]

[quote="E^(pi*i)=-1"]Heh, someone I know did the same thing (although he answered , which I presume is what you meant).[/quote]

Yup, yup, I did the same.  :ewpu:

I almost started out with a denominator of 3^9, and assumed that when it said that the first box was laid "flat" on the floor, it meant that it had minimum height...but rethought it out and said it was more safe to assume all 10 boxes had a variable height.

[quote="LawOfSigns"]I tried solving it that way at first, but when I saw 3^10, I thought that was too large a number and tried to find the probability without considering the order of the boxes.  I got the wrong answer, but I'm not sure where I went wrong.

I found the 3 correct arrangments, and had 3 as my numerator.  I then used a dividers and slots method to count the denominator, and got 12 choose 2 (arranging 10 boxes and 2 dividers), or 66.  I reduced 3/66 to 1/22, and bubbled in 001.  Why didn't this approach work out?[/quote]

not all posibilities are equally likely, I believe

[quote="LawOfSigns"]I tried solving it that way at first, but when I saw 3^10, I thought that was too large a number and tried to find the probability without considering the order of the boxes.  I got the wrong answer, but I'm not sure where I went wrong.

I found the 3 correct arrangments, and had 3 as my numerator.  I then used a dividers and slots method to count the denominator, and got 12 choose 2 (arranging 10 boxes and 2 dividers), or 66.  I reduced 3/66 to 1/22, and bubbled in 001.  Why didn't this approach work out?[/quote]

There are more possible possibilities with 2 of your arrangements than the other.  There were 2520 possibilities each with those 2 arrangements, and 90 possibilities for the last arrangement, so you had to add up those 3 numbers and simplify.

[quote="LawOfSigns"]I tried solving it that way at first, but when I saw 3^10, I thought that was too large a number and tried to find the probability without considering the order of the boxes.  I got the wrong answer, but I'm not sure where I went wrong.

I found the 3 correct arrangments, and had 3 as my numerator.  I then used a dividers and slots method to count the denominator, and got 12 choose 2 (arranging 10 boxes and 2 dividers), or 66.  I reduced 3/66 to 1/22, and bubbled in 001.  Why didn't this approach work out?[/quote]

Wow, I did the exact same thing. I'm pretty sure it's because one of the three valid arrangements has less ways to be created than the other two (whereas if there were an equal number of ways to create all three of them, this method likely would have worked.)

I really messed this problem up:  spending 40 minutes multiplying out generating functions to get the wrong answer.    :(

EDIT:  Was there an easy way to find the coefficient of $ x^{41}$ ($ x^{11}$ after factoring out $ x^{30}$)?
I couldn't figure out any method of expressing it as a geometric series either.
EDIT (AGAIN):  Nevermind, there's an explanation on the other page.   :blush:

Ugh! Stupid mistakes are killing me. Instead of $ (3 ft, 4 ft, 6 ft) \equal{} (1,8,1), (3,5,2), and (5, 3, 2)$ I failed and did $ 3\plus{}2\equal{}4$ to get $ (3 ft, 4 ft, 6 ft) \equal{} (1,8,1), (3,5,2),$and$ ($[b][size=150]4[/size][/b]$ , 3, 2)$  :mad:

[hide=Very Nice Problem]
Call the $10$ heights: $h_1,h_2$ and so on, till $h_{10}.$
Notice that the sum of these terms is $41.$
But note that the minimum height is $3.$
For simplification purposes, rewrite the equation as:
$h_1+h_2+. . . + h_{10}=11,$ where all the terms on the LHS are non-negative integers.

We can divide this into cases.
Also, be careful: the $h$'s can now be $0, 1,$ or $3.$

Case 1: Only one $3$
$\binom{10}{1}*9=90$

Case 2: Two $3$'s
$\binom{10}{1}\binom{8}{5}*8*\binom{7}{2}$

Case 3: Three $3$'s
$\binom{10}{3}\binom{7}{2}$

These sum to $5130,$ and the simplifications aren't too hard.
[/hide]


Hello!

I'm having a problem on my TI-89. When I turn it on, the left part of the calculator is dark, yet if you look hard enough, you can still see past the "darkness." As time passes after it turns on, some of that black stuff fades away, but not completely.

What is wrong with it, and is there a way to fix it?

Thanks in advance!

Thanks. Even though it doesn't solve the problem completely, but fading the screen a bit makes it manageable.

Thanks, again!


Let x,y,z,t>0 and xy+xz+xt+yz+yt+zt=6.Prove that
      sum \sqrt ((x 4 +1)/2) \leq x 2 +y 2 +z 2 +t 2

Please check this solution I might be wrong.....:D
From the given condition using AM-GM it can proved that
 \sum x2  \geq 4.......(1)

From the inequality  \sqrt (( \sum x 2 )/n)  \geq  (\sum x)/n .......(Root Mean Squares  \geq AM) We have
( \sum  \sqrt( ( x4 + 1 )/2))/4  \leq  \sqrt (( \sum x 4 +4)/8) .......(*)

Applying (RMS  \geq AM) again we have
 \sqrt (( \sum x 4 )/4)  \geq   (\sum x 2  )/4

From which we have that  \sum x 4 \geq 4

Substituting in * we have
 \sum  \sqrt( ( x4 + 1 )/2)  \leq 4...(2)

From (1) & (2)...We have the required result.

It is true that I didn't understand very well those radicals (that is why I use sqrt), but I do not think that we have sum sqrt((x^4+1)/2)<=4. I think its the inverse, which doesn't help us very much.
    Anyway, I think I've found a solution:
  Let a=x/sqrt(2) and so on. WE have sum ab=3 and we must prove that sum sqrt((1+4a^4)/2)<=2 sum a^2. But using Cauchy and a nice identity we have sum sqrt((1+4a^4)/2)=sum sqrt((2a^2-2a+1)/2)*sqrt((2a^2+2a+1))<=sqrt( 2sum a^2+4-2 sum a))*sqrt[(2sum a^2+4+2sum a)/2]. So, we must prove that 8(sum a^2)^2>=4(sum a^2)^2+16sum a^2+16-4(sum a)^2. This reduces to (sum a^2-2)^2+(sum a)^2>=8. So it suffices to prove that sum a>=2sqrt(2), or sum x>=4, which comes from Mc-Laurin.

I think so.We have sum \sqrt ((x 4 +1)/2) \geq 4
This is my solution
     \sqrt ((x 4 +1)/2)+x  \leq x 2 +1
  so sum \sqrt ((x 4 +1)/2) +sumx \leq sumx 2 +4
  and sumx \geq 4 by Mc-Laurin
qed

Hi Hazari and Khotram if my interpretation of the radicals is correct...then the inequality (*) holds good.....with equality occuring when x=y=z=t....

This would invariably result in the maximum value of the LHS in (*).... giving  \sum  \sqrt ((x 4 +1)/2)  \leq 4

I would like to know your suggestions on this....

Kalyan

Very nice, Khoa. I also created a problem based on this fact (sqrt(x)+sqrt((x^2+1)/2)<=x+1), but I prefered the more interesting solution with Cauchy.
   Dear Kalyan, suppose you are right, so sum sqrt((x^4+1)/2)<=4. Since x^4+1>=2x^2, it follows that sum x<=4, which is false.

I spent a lot of time on this problem and could never think the sol is so simple.
 :(  :(  :(
I have another solution
we have
(x 4 +1)=[(x 2 +1)/ \sqrt 2+x][(x 2 +1)/ \sqrt 2-x]
Using B.C.S we have
  \sum  \sqrt (x 4 +1)<=(x 2 +y 2 +z 2 +t 2 +4) 2 -(x+y+z+t) 2
 now we must prove that
(x 2 +y 2 +z 2 +t 2 +4) 2 -(x+y+z+t) 2 <=(x 2 +y 2 +z 2 +t 2 ) 2
<=> (x 2 +y 2 +z 2 +t 2 -2)(x 2 +y 2 +z 2 +t 2 -4)>=0 (True for x 2 +y 2 +z 2 +t 2 >=4)


Hey, were doing review for our exams and i ran into an interesting question. It says calculat the sum of the first 5 terms of this geometric series: 2, 6i, -18.... I know that a =2, r= 3i but i cant take it from there. Any help would be great.

[quote="math92"]use the forumula

$\frac{a}{1-r}$
i think thats the formula.

Try not to click the submit button twice in the future.[/quote]

Actually that is a formula for the infinite geometric series.

As for eh-jay you can just add up the terms, not that difficult or use the formula for the partial geometric series.

[quote="mathgeniuse^ln(x)"][quote="math92"]use the forumula

$\frac{a}{1-r}$
i think thats the formula.

Try not to click the submit button twice in the future.[/quote]

Actually that is a formula for the infinite geometric series.

[/quote]
Sorry about that. :blush:

[quote="kunny"]For $S=a_1+a_2+a_3+a_4+a_5,$
$rS=a_2+a_3+a_4+a_5+a_6.$
Thus substracting both sides, we have $(1-r)S=.....$[/quote]
thats what kunny posted in the other topic.

[quote="eh-jay"]i can get the awnser just by adding them but for full marks ill ned to use the formula. i think that is Sn=a(r^n-1)
                                            -----------
                                                r-1[/quote]The forum doesn't take multiple spaces. Best way is to $\LaTeX$ but I don't know if you learned it yet. I'm pretty sure you meant $S_n=a+ar+ar^2+\dots+ar^{n-1}=a\frac{r^n-1}{r-1}$

Is there something like finite differences that you can use for a geometric sequence/series? I know you cant use finite differences but I just want to know if you can use something similar to it.

You mean like:
\[ a-ar-ar^2-ar^3-...-ar^n \]?

[hide]The sum of a finite geometric series is calculated by the formula:

$S_n=\frac{a_1(1-r^n)}{1-r}$

$a_1=2$
$r=3i$
$n=5$

$S_5=\frac{2(1-(3i)^5)}{1-3i}$[/hide]

for this example sinc eyou have 5 terms it is obvious your series is finite therefore you know you have to use (1-r^n/1-r^n) multiplied by initial terms. r=3i and a1=2

If you have the formula Sn= a(1-r^n)/1-r then you just substitute the corresponding values in tis case 2(1-3i^5)/1-3i. can't you just take out and exponent five from the numerator hence you get 2(0)^5 or simply 2 as your sum. Please someone respond if my algebra is failing.


Δεν ξέρω ρε παιδιά για εσάς αλλά εμένα μου αρέσουν υπερβολικά πολύ οι ασκήσεις από τη jbmo.... :)
Να λοιπόν μία:
Let $ a$  be positive real number such that $ a^3\equal{}6(a\plus{}1)$ . Prove that the equation  $ x^2\plus{}ax\plus{}a^2\minus{}6\equal{}0$  has no real solution.

Έχω βρει μία πολύ όμορφη λύση αλλά θέλω να δω και τις δικές σας.....

Πολυ μου αρεσουν τα pure αλγεβρικα προβληματα.
Η εξισωση εχει διακρινουσα $ D \equal{} 3(8 \minus{} a^2)$
Θα δειξουμε οτι $ a^2 > 8$
Αν $ a\geq3$ τοτε $ a^3 \minus{} 6a \equal{} a(a^2 \minus{} 6)\geq3(9 \minus{} 6) > 6$ ατοπο
Αρα $ a < 3$.Τοτε
$ a^2 \equal{} 6(1 \plus{} \frac {1}{a}) > 6(1 \plus{} \frac {1}{3}) \equal{} 8$

[quote="wallace"]και στην χειρότερη περίπτωση, υπάρχει και ο Bolzano......[/quote]

Εντάξει, αλλά Bolzano σε [b]Junior[/b] BMO;  :o

You can find it here: http://www.mathlinks.ro/viewtopic.php?p=874747#p874747

All JBMO problems are in the resources. BTW, JBMO'09 is in a few days!

H λύση μου:
$ a^2 \equal{} 6\left(1 \plus{} \frac {1}{a}\right) \equal{} 6 \plus{} \frac {6}{a}$
Επομένως αρκεί νδο:
$ x^2 \plus{} ax \plus{} \frac {6}{a} \equal{} 0$
δεν έχει λύση.
Πρέπει $ D < 0 \Longleftrightarrow a^2 \minus{} \frac {24}{a} < 0 \Longleftrightarrow a^3 < 24 \Longleftrightarrow a < 3$
Απομέμως ακεί νδο $ a < 3$
$ a^3 \minus{} 6a \minus{} 6 \equal{} 0 \Longleftrightarrow (a \minus{} 3)(a^2 \plus{} 3a \plus{} 3) \equal{} \minus{} 3$
Aλλα $ a^2 \plus{} 3a \plus{} 3 > 0$ επομένως αναγκαστικά $ a < 3$  :)

Kαλή επιτυχία και για τους juniors που θα διαγωνιστουν σε λίγες μέρες στην jbmo :!:


inequalities
calculus
derivative
function
topology
real analysis
calculus computations

Suppose $(X,d)$ be metric space and $X$ is a compact set.The map $T: X\to X$ satisfying: $d(T(x),T(y))<d(x,y)$ for all $x,y\in X$.
Prove that T has unique fixed point.
I know this problem is well-known but didn't solve.Can anyone give me some information about it.Thanks.

doing the lim x-->y, we see that $d(T(x),T(y)<\epsilon$ so T is a uniformly continuous application.
Now, it´s easy to visualize the situation of the problem.  Using your inequality, we see that the derivative of two distinct points is less then one, so, it´s very simple to see that our function is crossing the identity function (non parallel lines) one time .
I admit that there are some dark points in my solution, but i think that this is right for the case that X is ilimited. I am not so sure about the case when X is limited. Hope that i am not wrong in my partial solution!

$T(x)$ is continuous hence function $f(x)=d(x,T(x))$ is continuous. $X$ is compact hence there's $\min_{X}f(x)=f(x_{0})$. If $f(x_{0})>0$ then $f(T(x_{0}))<f(x_{0})$. Contradiction. Hence $T(x_{0})=x_{0}$. Obviously there's only one fixed point


Find all the function $f: R \to R$ satisfying:
   $f(x^2+f(y))=y+f^2(x)$

Sorry. There was a trouble in posting message. :blush:
Mod, please delete one of two same topic.
Thanks.


There are 2 exercises in Anton's text which I am not sure whether I did it correctly or not. Especially the later half of the proof of problem 2, I don't know what I wrote about. Please check them for me if you would like to.

[b]1.[/b] Let $ S$ be a basis for an $ n$-dimensional vector space $ V$. Show that if $ \mathbf v_1,\mathbf v_2,\cdots,\mathbf v_r$ form a linearly independent set of vectors in $ V$, then the coordinate vectors $ (\mathbf v_1)_S,(\mathbf v_2)_S,\cdots,(\mathbf v_r)_S$ form a linearly independent set in $ \mathbb R^n$, and conversely.
[hide="Proof."] Let $ S=\{\mathbf u_1,\cdots,\mathbf u_n\}$. Since $ S$ is a basis for $ V$, we have

$ \mathbf v_j=\sum_{i=1}^n k_{ij}\mathbf u_i$ for $ j=1,2,\cdots,r$.

Thus $ (\mathbf v_j)_S=\sum_{i=1}^n k_{ij}\mathbf e_i$.

Assume that $ (\mathbf v_1)_S,(\mathbf v_2)_S,\cdots,(\mathbf v_r)_S$ are linearly dependent. Then the equation

$ \sum_{j=1}^r x_j(\mathbf v_j)_S=\mathbf 0$

has a nontrivial solution $ x_j=\alpha_j, j=1,2,\cdots,r$. Besides,

$ \sum_{j=1}^r \alpha_j(\mathbf v_j)_S=\sum_{j=1}^r \alpha_j\left(\sum_{i=1}^n k_{ij}\mathbf e_i\right)=\sum_{i=1}^n\left(\sum_{j=1}^r k_{ij}\alpha_j\right)\mathbf e_i=\mathbf0$.

Since $ \mathbf e_1,\cdots,\mathbf e_n$ are linearly independent,

$ \sum_{j=1}^r k_{ij}\alpha_j=0$ for $ i=1,2,\cdots,n$.

Hence $ \sum_{i=1}^n\left(\sum_{j=1}^r k_{ij}\alpha_j\right)\mathbf u_i=\sum_{j=1}^r \alpha_j\left(\sum_{i=1}^n k_{ij}\mathbf u_i\right)=\sum_{j=1}^r \alpha_j\mathbf v_j=\mathbf0$,

As $ \alpha_j$ are not all zero, it contradicts the fact that $ \mathbf v_1,\cdots,\mathbf v_r$ are linearly independent. Thus $ (\mathbf v_1)_S,\cdots,(\mathbf v_r)_S$ are linearly independent. The proof for the coverse is similar. $ \Box$[/hide]

[b]2.[/b] Using the notation from above exercise, show that if $ \mathbf v_1,\mathbf v_2,\cdots,\mathbf v_r$ span $ V$, then the coordinate vectors $ (\mathbf v_1)_S,(\mathbf v_2)_S, \cdots,(\mathbf v_r)_S$ span $ \mathbb R^n$, and conversely.
[hide="Proof."]Since $ \mathbf v_1,\mathbf v_2,\cdots,\mathbf v_r$ span $ V$,

$ \forall k_j\in\mathbb R: \sum_{j=1}^r k_j\mathbf v_j\in V;\,\forall\mathbf w\in V\,\exists!k_j\in\mathbb R: \mathbf w=\sum_{j=1}^r k_j\mathbf v_j$.

Let $ S=\{\mathbf u_1,\cdots,\mathbf u_n\}$ be a basis for $ V$. Then

$ \forall m_i\in\mathbb R: \sum_{i=1}^n m_i\mathbf u_i\in V,\,\sum_{i=1}^n m_i\mathbf e_i\in\mathbb R^n$;

$ \forall\mathbf w\in V$ and $ \forall\mathbf x\in\mathbb R^n, \exists!m_i\in\mathbb R: \mathbf w=\sum_{i=1}^n m_i\mathbf u_i,\, \mathbf x=\sum_{i=1}^n m_i\mathbf e_i$

since $ \{\mathbf e_1,\cdots,\mathbf e_n\}$ is a basis for $ \mathbb R^n$. Thus

$ \mathbf v_j=\sum_{i=1}^n l_{ij}\mathbf u_i$ for $ j=1,2,\cdots,r$,

and $ (\mathbf v_j)_S=\sum_{i=1}^n l_{ij}\mathbf e_i$.

Hence for any $ \mathbf w\in V$, we have

$ \mathbf w=\sum_{i=1}^n m_i\mathbf u_i=\sum_{j=1}^r k_j\mathbf v_j=\sum_{j=1}^r k_j\left(\sum_{i=1}^n l_{ij}\mathbf u_i\right)=\sum_{i=1}^n\left(\sum_{j=1}^r k_jl_{ij}\right)\mathbf u_i.$

So $ \sum_{i=1}^n m_i\mathbf e_i=\sum_{i=1}^n\left(\sum_{j=1}^r k_jl_{ij}\right)\mathbf e_i =\sum_{j=1}^r k_j\left(\sum_{i=1}^n l_{ij}\mathbf e_i\right)=\sum_{j=1}^r k_j(\mathbf v_j)_S$

since $ \mathbf u_1,\cdots,\mathbf u_n$ are linearly independent. But for any $ m_i\in\mathbb R$, we have $ \mathbf w\in V$, thus

$ \forall m_i\in\mathbb R\, \exists k_j\in\mathbb R: m_i=\sum_{j=1}^r k_jl_{ij}$.

And for any $ \mathbf x\in\mathbb R^n$, we have $ \mathbf x=\sum_{i=1}^n m_i\mathbf e_i$. Consequently,

$ \forall\mathbf x\in\mathbb R^n: \mathbf x=\sum_{i=1}^n m_i\mathbf e_i=\sum_{j=1}^r k_j(\mathbf v_j)_S$.

Thus $ \mathbb R^n\subseteq\text{span}\,\{(\mathbf v_1)_S,\cdots,(\mathbf v_r)_S\}$. But $ (\mathbf v_1)_S,\cdots,(\mathbf v_r)_S\in\mathbb R^n$, so $ \text{span}\,\{(\mathbf v_1)_S,\cdots,(\mathbf v_r)_S\}\subseteq\mathbb R^n$. Therefore, $ (\mathbf v_1)_S,(\mathbf v_2)_S, \cdots,(\mathbf v_r)_S$ span $ \mathbb R^n$. The proof for the converse is similar. $ \Box$[/hide]

Yes, this looks correct. There would probably be other ways to phrase this, as well.

They seem not hard, so I think there would be a better way to write the proof.


Divide and express your answer in scientific notation:
\[ \frac{2\cdot 10^{13}}{4\cdot 10^{10}}.\]

I will report this problem. Scientific notation problems should not be in the FTW problem set.

That being said, the solution just for the sake of increasing mathematical knowledge:

$ \frac{2}{4}=\frac{1}{2}=0.5$
||
||
\/
$ 0.5\cdot10^3=\boxed{5\cdot10^2}$


geometry
rectangle
Coloring
combinatorics
counting
IMO Shortlist

A square $ (n \minus{} 1) \times (n \minus{} 1)$ is divided into $ (n \minus{} 1)^2$ unit squares in the usual manner. Each of the $ n^2$ vertices of these squares is to be coloured red or blue. Find the number of different colourings such that each unit square has exactly two red vertices. (Two colouring schemse are regarded as different if at least one vertex is coloured differently in the two schemes.)

I think it's 2^(n+1) - 2. Here is my solution. Let the vertices of the top row be assigned a random coloring. Then suppose that two vertices that are next to each other have the same color. Then obviously the coloring of the rest of the vertices is fixed. There are 2^n - 2 colorings of the top row so that there are at least two vertices next to each other with the same color.(total=2^n and there are two that alternate the colorings)If the vertices of the top row are colored alternately we have 2 ways so in total we have 2^n ways. So the total number is 2^n - 2 + 2^n = 2^(n+1)

[quote=JBL]Color the top row of points arbitrarily ($ 2^{n+1}$ ways).

If it alternates red-blue-red-blue for its entire length (2 possibilities), proceed as follows: coloring the leftmost dot of the second row (2 ways) determines the entire second row, and it is also alternating.  Then coloring the leftmost dot of the third row (2 ways) determines the entire third row, and it is alternating.  And so on.  Thus, we have from these possibilities a total of $ 2^{n+1}$ colorings.

If the top row does not alternate red-blue-red-blue for its entire length ($ 2^{n+1}-2$ ways), there are two consecutive dots of the same color.  Note that these dots determine the color of the two dots directly below them, and as a result the first row determines the second row.  Moreover, the second row now also has two consecutive dots of the same color, so the third row and all future rows are determined.  Thus, from these options we have in total $ 2^{n+1}-2$ ways.

All together, this gives us $ 2^{n+2}-2$ total valid colorings.  (In an $ m \times n$ rectangle, this answer would be $ 2^{m+1}+2^{n+1}-2$ colorings.)

A follow-up question: how many ways are there to color the vertices with 4 colors so that each unit square has one vertex of each color?[/quote]

Just to not cause confusion to the readers, here n was taken as n+1.

Nice bijective action!

[hide=motivation]
After fiddling around a little, notice that it is quite hard to count directly because of all the conditions. Then, realize that there may be an easier route to consider - a bijection involving the colors.
[/hide]

[hide=solution]
Note that if there were any two adjacent squares having the same color (say that it is in the first row), the vertices directly underneath would be predetermined, and thus, the other adjacent square(s) would have 3 of their 4 vertices colored, resulting in one way to color the rest of the vertices. Hence we separate the problem into two cases.

[b]Case 1:[/b] There are two adjacent squares with the same color
There are $2^n-2$ ways to do this. There is only one way to color the rest.

[b]Case 2:[/b] There are no two adjacent squares with the same color
There are 2 ways to do this. Now for each succeeding row, there are two ways to color (switching between $GBGB...$ or $BGBG...$). Hence, there are $2*2^{n-1}=2^n$ ways for this case.

Summing up, we obtain $\boxed{2^{n+1}-2}$ as our final answer.
[/hide]


What is the sum of all integers $ x$ that satisfy $ 1 < (\pi \minus{}1 )x < 10$?

We know that $ \pi \minus{} 1$ is around $ 3.14$ so $ 0.46 < x < 4.67$. Therefore, the sum is $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \equal{} \boxed{10}$

We see that $ 1$ is the lowest satisfying integer, and $ 4$ is the highest satisfying integer. All integers in between also satisfy the expression. $ 1\plus{}2\plus{}3\plus{}4\equal{}\boxed{10}$.


So, how did the early round of admissions go for WA people? Or your friends, if you aren't a senior.

[quote="drunner2007"]well i got into MIT and Caltech.[/quote]

Are you from WA? :huh:

best friend got into Stanford SCEA, another friend got deferred from MIT :(  As for me, I'm doing RD to Stanford and to some other schools.

I'm curious about Adam Hesterberg, Jeffrey Wang, and Young.  Those guys can probably get in anywhere - I wonder what school they will choose to go to

[quote="Phoenix257"][quote="paladin8"][quote="Phoenix257"]Jeffrey Wang[/quote]

Yes, I'm very curious as well.[/quote]

 :blush: Oops, I didn't realize that you were Jeffrey Wang :blush:  :blush: So, are you choosing Stanford as your school?[/quote]

Lol, it's ok. I haven't decided for sure yet, but chances are good :).

[quote=jack74]My cousin got a scholarship into UC Stanford, but declined and went to UC San Diego because it is closer apparently[/quote]

WHY DO YOU HAVE THAT LOGO NOW??? and WHY DID YOU REVIVE A 10 YEAR OLD THREAD???

[quote=jack74]Logo? What logo?[/quote]

avatar/logo

Yeah, forgot to lock my computer and my brother pranked me. I don't know of any better avatars at the moment.

[quote=hwl0304][quote=jack74]My cousin got a scholarship into UC Stanford, but declined and went to UC San Diego because it is closer apparently[/quote]

Because UC Stanford is a thing.

See, this is why we need moderation...[/quote]

Also, if you don't know where California is, then maybe you should get your eyes checked. Thank you.

[quote=B-mode]He is like the main troller of wa forum, along with his other 2 multis[/quote]

For the last time those aren't my multis.

[quote=jack74]My cousin got a scholarship into UC Stanford, but declined and went to UC San Diego because it is closer apparently[/quote]

Ya, mine goes there too.

My friend is smart and went to Princeton :o


Find the values maximum and minimum of $ \left|\frac{z\minus{}i}{z\plus{}i}\right|$ when $ |z|\equal{}3.$

Let $ z \equal{} a \plus{} bi$. Then the task is to optimize the norm of $ \frac {a \plus{} (b \minus{} 1)i}{a \plus{} (b \plus{} 1)i} \equal{} \sqrt {\frac {a^2 \plus{} (b \minus{} 1)^2}{a^2 \plus{} (b \plus{} 1)^2}}$ subject to $ a^2 \plus{} b^2 \equal{} 9$.

Substituting $ a^2 \equal{} 9 \minus{} b^2$ we get (neglect the square root since optimizing the original expression is dual to optimizing the expression inside the sqrt)

$ \frac {10 \minus{} 2b}{10 \plus{} 2b} \equal{} f(b)$

Since $ f'(b) \equal{} \frac { \minus{} 2(10 \plus{} 2b) \minus{} 2(10 \minus{} 2b)}{(10 \plus{} 2b)^2} \equal{} \frac { \minus{} 40}{(10 \plus{} 2b)^2}\neq 0$
we just have to check the boundary condition $ |b|\le3$

$ f(3) \equal{} 1/4$ and $ f( \minus{} 3) \equal{} 4$.

Now we take the square root to get $ 2$ and $ 1/2$.

I would like to see a geometric approach without calculus involved.

A geometric approach might involve inversion in the complex plane... Let $ z\equal{}3e^{i\theta}$ and then try to describe the curve in the complex plane I suppose (I'm having difficulty visualizing this)

What is "inversion"? I'm not familiar with this term...

[url]http://en.wikipedia.org/wiki/Inversive_geometry[/url]

It turns the plane inside out, so points $ z$ in the complex plane map to $ \frac{1}{z}$ (I think).

I don't think inversion is necessary to solve this problem.

$ |z\minus{}a|,\ a\in \mathbb{C}$ means the distance between $ z$ and $ a$. If you use this, solution follows directly.

[hide]The condition $ |z|\equal{}3$ describes $ x^2\plus{}y^2\equal{}9$ in the $ xy$ plane, which is a circle with radius $ 3$ centered at $ (0,0)$. Also, let $ \minus{}i$ lie on $ (\minus{}1,0)$ and $ i$ lie on $ (1,0)$.
So if we draw two segments from $ \minus{}i$ and $ i$ to the circle, the problem is to minimize/maximize the ratio (length of segment from $ \minus{}i$)/(length of segment from $ i$).

To minimize it, we want the length of segment from $ \minus{}i$ to be the least and the length of segment from $ i$ to be the greatest. If you just look at the graph, it becomes obvious that this occurs when both segments are drawn to the point $ (\minus{}3,0)$ on the circle. So, the minimum is $ \frac{3\minus{}1}{3\plus{}1} \equal{} \frac{1}{2}$.

Similarly, we get that the max is $ \frac{3\plus{}1}{3\minus{}1}\equal{}2$.[/hide]


Solve the equation:
$cos^{2}x+cos^{2}(2x)+cos^{3}(3x)=1$

hello, your equation is equivalent to
$\left( \cos \left( x \right)  \right) ^{2} \left( \cos \left( x
 \right) +1 \right)  \left( 4\, \left( \cos \left( x \right)  \right)
^{2}-3 \right)  \left( 16\, \left( \cos \left( x \right)  \right) ^{4}
-16\, \left( \cos \left( x \right)  \right) ^{3}-8\, \left( \cos
 \left( x \right)  \right) ^{2}+8\,\cos \left( x \right) +1 \right)
=0$
Sonnhard.

I would start by asking the proposer about the last term in the left hand side. Is it really cubed?


integration
LaTeX
calculus
calculus computations

t (seconds) : 0, 10, 20, 30, 40, 50, 60, 70, 80
v(t) (feet per second) : 5, 14, 22, 29, 35, 40, 44, 47, 49

Rocket A has positive velocity  v(t) after being launched upward from an initial height of $ 0$ feet at time $ t = 0$ seconds. The velocity of the rocket is recorded for selected values of t over the interval $ 0 < = t < = 80$ seconds, as shown in the table above.

a) Find the average acceleration of rocket A over the time interval $ 0 < = t < = 80$ seconds. Indicate units of measure.

b) Using correct units, explain the meaning of $ \int_{10}^{70}v(t)dt$ in terms of the rocket's flight. Use a midpoint Riemann's sum with 3 subintervals of equal length to approximate $ \int_{10}^{70}v(t)dt$.

c) Rocket B is launched upward with an acceleration of $ a(t) = \frac {3}{(t + 1)^\(1/2)}$ feet per second per second. At time $ t = 0$ seconds, the initial height of the rocket is 0 feet, and the initial velocity is 2 feet per second. Which of the two rockets is traveling faster at time $ t = 80$ seconds? explain your answer.

my god, please help me asap. :(
and please excuse my LaTeX

On the $ \text{\LaTeX}$- you don't want \v, as that's an upside-down hat.

How are position, velocity and acceleration related? What can you do with the Fundamental Theorem?

[hide="solution a"]
$ a_{\text{avg}}\equal{}\frac{v_f\minus{}v_i}{t_f\minus{}t_i}$ so $ a_{\text{avg}}\equal{}\frac{(49\minus{}5)\text{ft/sec}}{(80\minus{}0)\text{sec}}\equal{} \frac{44\text{ft/sec}}{80\text{sec}}\equal{}\boxed{.55\text{ft/sec/sec}}$
[/hide]

[hide="solution b"]
$ \int_{10}^{70}[v(t)]dt$ gives the distance the rocket traveled, in feet, during the time interval $ 10\leq t\geq70$.
Your three subdivisions are: $ [10,30],[30,50],[50,70]$. Their midpoints are $ 20,40,60$, their widths are each 20, and their heights are $ 22,35,44$.
This makes your approximation (need approx equals code) $ \int_{10}^{70}[v(t)]dt\equal{}22*20\plus{}35*20\plus{}44*20\equal{}440\plus{}700\plus{}880\equal{}\boxed{2020}$
[/hide]

I have no time for c, sorry.


function
floor function
algebra proposed
algebra

Find all reals $x$ satisfying $0 \le x \le 5$ and
$\lfloor x^2-2x \rfloor = \lfloor x \rfloor ^2 - 2 \lfloor x \rfloor$.


Let $x,$ $y$ and $z$ are non-negative real numbers. Prove that
$5(x^4y^2+y^4z^2+z^4x^2)+42x^2y^2z^2\geq19(x^3y^2z+y^3z^2x+z^3x^2y).$

Note: the proof uses the author's technique.Clearly the given ineq is true if x or y or z = 0. So we can assume x,y, and z >0. Dividing both sides by (xyz)^2 we got a new inequality that needs to be proved namely:

New Ineq: 5(x^2 + y^2 + z^2) + 42 >= 19(x + y + z)  with the new condition: xyz = 1 and x,y, and z > 0.
Proof of New Ineq: At this point, one may use the problem's author own strategy. That is: let x = min{x,y,z} then
yz >= 1 and let f(x,y,z) = 5(x^2 + y^2 + z^2) + 42 - 19(x + y + z). One can verify that:
f(x,y,z) - f(x,(yz)^(1/2), (yz)^(1/2)) = (y^(1/2) - z^(1/2))^2.[5(y^(1/2) + z^(1/2))^2 - 19] >= 0 since the first factor is clearly non-negative while the second one is thru AM-GM ineq >= 5(4) - 19 = 1 > 0 because yz >= 1.
So we complete the proof by showing the last ineq: f(x,y,y) >= 0 (*) with the condition that x.y^2 = 1 and y >= 1.
Using x = 1/y^2 and y >= 1 one can check that:
(*) <=> (y - 1)^2.(10y^4 - 18y^3 - 4y^2 + 10y +5) >= 0 for y>= 1. Thus the problem is considered solved once we verify that:
(**) 10y^4 - 18y^3 - 4y^2 + 10y + 5 >= 0 for y >= 1.
But (**)'s LHS = (y^2 - 2)^2 + 9y^4 - 18y^3 + 10y + 1 = (y^2 - 2)^2 + 9y(y^3 - 2y^2 + 1) + y + 1 >=
(y^2 -2)^2 + 9y(y^2 - 2y + 1) + y + 1 = (y^2 -2)^2 + 9y(y - 1)^2 + y + 1 > 0 for y >= 2.
So we only need to lastly verify that (**)'s LHS is true for 1 <= y <= 2.  Put y = u + 1 we reduce it to:
10u^4 + 22u^3 + 2u^2 - 12u + 3 > 0 (***) for 0 <= u <= 1. At this point plot a graph of
v = g(u) = (***)'s LHS and check that the graph stays above the x-axis in [0 , 1] by using a TI-83.
It seems we got a proof.

You're using a graphing calculator to do a proof?!?


geometry
angle bisector
Pythagorean Theorem

this is a local math contest question.

Anyway section of a tile floor is shown.

Line segment MX is the angle bisector of <WMT

A. justify that point X is the midpoint of WT

later it asked a situation
consider it b i guess so yeah b
B. Length of MW is 10 inches. length of WT is 14 inches and MX is perpindicular to WT.
What is length of MX to nearest hundreth of an inch.
(show your explination and reasoning)
p.s picture is the attachment

perhaps i am missing something but if not, this isnt intermediate level (see the guidelines on the top of the forum, or look at other posts for sample difficulty level)

First of all, the diagram shows that
MW=MT
and we know
angle WMX = angle XMT, MX=MX so XMT and WMX are congruent by SAS. Thus WX=XT, which means X is the midpoint of WT
b) pythagorean theorem gives sqrt(100-49=sqrt(51) which is about 7.14

[quote="ISTormrage"]perhaps i am missing something but if not, this isnt intermediate level [/quote]

Sure...


conics
hyperbola
function
geometry
blogs
set theory
projective geometry

As mentioned above, what is the meaning of "infinity?"
People always use this word in daily life, but I am not sure if they really understand the meaning of infinity.
 :idea:

[size=150][color=green]Infinity=Without End[/color][/size]

its Symbol was introduced by John Wallis!!!

$ \infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty$

Are you asking about mathematical conceptions of infinity?  (Its use in casual conversation is largely an issue of hyperbole and does not concern itself much, one way or another, with the mathematical concept of infinity.)

I suppose one could talk about the [b]set-theoretic[/b] concept of infinity, which one can define as any set $ S$ that does not biject to a set of the form $ \{ 1, 2, ... n\}$ for some natural number $ n$.  Of course, it is well-known that there are many different cardinalities of infinite sets, and indeed this is an interesting and unexpected characterization of the infinite.

Secondly, we could talk about the [b]analytic[/b] concept of infinity, which is just a quantity without a bound.  For example, we can let a function increase without bound, or we say that a function is unbounded as it approaches a point, and we express both of these concepts using infinity.

[size=150][i]From Webster's Dictionary:[/i][/size]
[b]Pronunciation: in-'fi-n&-tE
Function: noun
Inflected Form(s): plural -ties
1 a : the quality of being infinite b : unlimited extent of time, space, or quantity : BOUNDLESSNESS
2 : an indefinitely great number or amount <an infinity of stars>
3 a : the limit of the value of a function or variable when it tends to become numerically larger than any preassigned finite number b : a part of a geometric magnitude that lies beyond any part whose distance from a given reference position is finite <do parallel lines ever meet if they extend to infinity> c : a transfinite number (as aleph-null)
4 : a distance so great that the rays of light from a point source at that distance may be regarded as parallel[/b]

[size=200][i]From Wikipedia:[/i][/size]
[b]
The word infinity comes from the Latin infinitas or "unboundedness." It refers to several distinct concepts (usually linked to the idea of "without end") which arise in philosophy, mathematics, and theology.

In mathematics, "infinity" is often used in contexts where it is treated as if it were a number (i.e., it counts or measures things: "an infinite number of terms") but it is a different type of "number" than the real numbers. Infinity is related to limits, aleph numbers, classes in set theory, Dedekind-infinite sets, large cardinals,[1] Russell's paradox, non-standard arithmetic, hyperreal numbers, projective geometry, extended real numbers and the absolute Infinite.
[/b]

t0rajir0u gave a very good description. To add. The word "infinite" means "not finite". And in mathematics things aren't finite tend to have some interesting properties behind them. But as t0rajir0u implicitly states "infinity" has different meaning in different areas of mathematics. In [u]set theory[/u] we work with mostly infinite sets. In [u]analysis[/u] we work with limits and similar things. But no where in mathematics is the word "infinity" treated as a philosophical topic. I have found from seeing many forums that "infinity" creates unusually long threads, which aren't mathematical at all rather philosophical. I can state right now that "infinity" is a rigorously and perfectly defined concept in mathematics which therefore leaves no room for philosophy. Finally "infinity" unlike in other uses, such as English vocabulary, [b]doesnt[/b] mean "without end", yes its true that is the essence behind "infinity", but rather "infinity" is defined in totally different ways rather than "without bound".

By the way, using Webster's for explaining mathematical concepts isn't the best source to use.

I have posted two sets of explanations as above because I think that many people do not really understand the meaning of infinity.
(just like what I have posted in my BLOG, most people do not understand fractions.)

For students in Mathematics like me, the concept of "infinity" should be clear enough,but honestly, most people do not know what they are talking about in daily life.... sigh~

Yet, another example, I remember that once a while there was an advertisement in Hong Kong saying that "Our toothpaste contains Sodium Chloride!", and I bet many people found that such a brand of toothpaste was very good BECAUSE [b]IT CONTAINS SODIUM CHLORIDE! lol~lol~lol~lol~lol~~~~~[/b]  :D  :D  :D  :D  :D
**************************************************
[size=150][color=indigo][i]MASTER ASKED(with a soft tone): Boy, what do you see in your eyes?
Student answered(concretely): Nothing, but INFINITY! [/i][/color][/size]


Can someone give me an idea how to approach this problem ??
http://students.iiit.ac.in/~abhilashi/ques.pdf

Thanks in advance.

Uhm, well, the only thing that comes to my mind is just trying all the possible regular expressions and checking if they satisfy your problem. The first one that does is the answer.

You could try to improve this though. One idea I have is to divide the regular epressions that don't solve your problem into two piles: the ones that could be possibly extended to give the correct answer and the ones that are completely worthless and should be thrown away:

A = ccca, cccb, cccc

B = caab

R1 = ab{c}* - throw away
R2 = {c}* - could be extended (to {c}*{abc} for example)

To decide whether the expression can be extended or not, you only need to find out if it matches some prefix of each word in A.

When some regular expression is thrown in the "worthless" pile, you won't have to check any other regexps that have this one as a prefix.


function
algebra
polynomial
combinatorics unsolved
combinatorics

Let $ f(x_{1},x_{2},...,x_{n})$ is a polynomial for n variables.

And let S=sum of coefficient of terms whose all exponent is even.

(example: $ f(x) \equal{} (1 \plus{} x)^{k} \minus{} > S \equal{}$ $ {k} \choose {0}$+$ {k} \choose {2}$+... $ \equal{} 2^{k \minus{} 1}$)

problem: find S

[hide]Prove that S=$ \frac {f(1,1,...,1) \plus{} f(1,1,..., \minus{} 1) \plus{} .. \plus{} f( \minus{} 1, \minus{} 1,..., \minus{} 1)} { 2^{n}}$[/hide]

And how can we generalize it  even (multiple of 2) to multiple of n??

Is it too easy??

Sorry for my poor English;;[/hide]

[quote="dmath328"]And how can we generalize it  even (multiple of 2) to multiple of n??[/quote]

I've seen this described on the forum as a "roots of unity filter."  The key observation is that if $ z$ is an $ n$-th root of unity then $ \sum_{i \equal{} 0}^{n \minus{} 1}z^i \equal{} \begin{cases} n, & z \equal{} 1 \\
0, & \text{otherwise.}\end{cases}$


[quote="Extremal"]Sure . as $ n \to \infty$ then serie is less than some constant $ C$[/quote]

More details please.

I think you can estimate it as :
$ ... \leq \frac {c_{1}}{n \minus{} 1} \plus{} \sum _{i \equal{} 2}^{n \minus{} 2} \frac {c_{i}}{(n \minus{} 2)^{2}} \plus{} \frac {c_{n \minus{} 1}}{n \minus{} 1}\to 0$ as $ n \to 0$

[quote="bookworm_vn"]Does
\[ \sum\limits_{k \equal{} 1}^{n \minus{} 1} {x^k \left( {1 \minus{} x} \right)^{n \minus{} k} }
\]
converge uniformly to $ 0$ in $ [0,1]$?[/quote]

i  don't think so.  :maybe:  at least not for $ x\in[0,1]$

$ \textbf f(x)\; \equal{} \; \sum\limits_{k \equal{} 1}^{n \minus{} 1} {x^k \left( {1 \minus{} x} \right)^{n \minus{} k} }\; \equal{} \; \minus{} x^n \minus{} (1 \minus{} x)^n \plus{} \sum\limits_{k \equal{} 0}^{n} {x^k \left( {1 \minus{} x} \right)^{n \minus{} k} }$

$ \equal{} \; \minus{} x^n \minus{} (1 \minus{} x)^n \plus{} \left(\dfrac {x^{n \plus{} 1} \minus{} (1 \minus{} x)^{n \plus{} 1} } { 2x \minus{} 1 } \right)$

 $ \left(\quad \because \quad x^{n \plus{} 1} \minus{} y^{n \plus{} 1} \equal{} (x \minus{} y) \left( \sum_{k \equal{} 0}^{n}x^ky^{n \minus{} k} \right)\quad \right)$

$ \textbf f(x)$ doesn't appear to be monotone decreasing for $ x\in[0,1]$ (which is a criteria for uniform convergence)

but for either one of $ x\in\left[0,\dfrac{1}{2}\right]$ or  $ x\in\left[\dfrac{1}{2},1\right]$, i would say it does.  :|

[quote="misan"][quote="bookworm_vn"]Does
\[ \sum\limits_{k \equal{} 1}^{n \minus{} 1} {x^k \left( {1 \minus{} x} \right)^{n \minus{} k} }
\]
converge uniformly to $ 0$ in $ [0,1]$?[/quote]

i  don't think so.  :maybe:  at least not for $ x\in[0,1]$[/quote]

Try this \[ 1 \geqslant 1 \minus{} x^n  \minus{} \left( {1 \minus{} x} \right)^n  \equal{} \sum\limits_{k \equal{} 1}^{n \minus{} 1} {\left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right)x^k \left( {1 \minus{} x} \right)^k }  \geqslant n\sum\limits_{k \equal{} 1}^{n \minus{} 1} {x^k \left( {1 \minus{} x} \right)^k } .
\]


function
abstract algebra
calculus
integration
superior algebra
superior algebra unsolved

I am working on some exercise and I would like to know if it is true or not (and how to prove it) that if $ p$ is prime and $ p(x)$ is irreducible in $ Z[x]$, then

\[ Z[x]/(p,p(x)) \equal{} (Z/pZ)[x] / (p(x))\]

where in the right hand, $ p(x)$ is taken mod $ Z/pZ$

danken!

I think the identity holds although I'm not sure (my solution isn't rigorous and is mainly a "feeling").

Consider the function $ \phi : \mathbb{Z}[x]/(p,p(x)) \rightarrow \mathbb{Z}_p[x]/(p(x))$ sending an equivalence class $ f(x) \plus{} (p,p(x))$ to $ f_p(x)\plus{}p_p(x)$ where $ f_p$ and $ p_p$ are the images of $ f$ and $ p$ through the canonical map. I think this is an isomorphism (its surjective, and to show the kernel is trivial, use the fact that $ \mathbb{Z}_p[x]/(p(x))$ is an integral domain). But again, I'm not sure about this.

[quote="morova"]I am working on some exercise and I would like to know if it is true or not (and how to prove it) that if $ p$ is prime and $ p(x)$ is irreducible in $ Z[x]$, then
\[ Z[x]/(p,p(x)) \equal{} (Z/pZ)[x] / (p(x))
\]
where in the right hand, $ p(x)$ is taken mod $ Z/pZ$

danken![/quote]

HINT Use the universal property of factor rings to factor $ R \rightarrow R/(I \plus{} J)$ through $ R/I$.
This is often called the Second Isomorphism Theorem for rings.


Let $ x_1 \equal{} 3,x_{n\plus{}1} \equal{} x_n ^2 \minus{} 2x_n \plus{} 2$
Prove that $ \sum {\frac {1}{{x_n }}} \le 0.6$
sorry I have edited

hello, do you mean $ x_{n\plus{}1}\equal{}x_n^2\minus{}2x_n\plus{}2$ with $ x_1\equal{}3$?
Sonnhard.

[quote="ocha"]Isn't this identical to the question you posted in the HCM city TST thread[/quote]
It is not really simillar. Can you write your solution in detail?
Another problem for you
$ x_{1}\equal{}2$
$ x_{n\plus{}1}\equal{}x_{n}^{3}\minus{}3x_{n}\plus{}2$
Prove that $ \sum{\frac{1}{{x_{n}}}}\ > 0.75$ :)

This isn't a very nice solution:

Note the first few vaules $ 3,5,17,257$

$ \frac{1}{x_{n\plus{}1}} \equal{} \frac{1}{x_n\minus{}2}\minus{}\frac{2}{x_{n\plus{}1}\minus{}2}$

So $ \sum_{k\equal{}1}^N \frac{1}{x_k} \equal{} \frac{1}{x_1\minus{}2} \minus{} \left(\sum_{k\equal{}2}^N \frac{1}{x_k\minus{}2}\right) \minus{} \frac{1}{x_N\minus{}2}$

And the inequality is equivalent to

$ \left(\sum_{k\equal{}2}^N \frac{1}{x_k\minus{}2}\right) \plus{} \frac{1}{x_N\minus{}2} \ge \frac{2}{5} (*)$

We can check this inequality holds for $ N\equal{}2,3,4$

By induction $ x_k > 2 \Rightarrow x_k^2\minus{}2k \plus{} 2 > 2 \Rightarrow x_{k\plus{}1} > 2$

Since $ x_1\equal{}3$ we know all the terms are positive, and for $ N > 4$

$ \left(\sum_{k\equal{}2}^N \frac{1}{x_k\minus{}2}\right) \plus{} \frac{1}{x_N\minus{}2} > \sum_{k\equal{}2}^4 \frac{1}{x_k\minus{}2} \equal{} \frac{1}{3}\plus{}\frac{1}{15} \plus{} \frac{1}{255} > \frac{2}{5}$

So the inequality is true.
[/hide]

[hide="Hint"]Complete the square, let $ x_n \equal{} y_n \plus{} 1$.[/hide]

[hide="The relevant sum"]$ \sum_{n\equal{}0}^{\infty} \frac{1}{2^{2^n}\plus{}1} \approx 0.596063 \ldots$[/hide]

[b]@ ocha:[/b] Your first equation should be $ \frac{1}{\mathbf{x_{n}}}\equal{}\frac{1}{x_{n}\minus{}2}\minus{}\frac{2}{x_{n\plus{}1}\minus{}2}$, where the index in the first term is different from what you wrote. I haven't checked to see if this affects your answer or not.

Gah, it stopped letting me edit my post while I was in the middle of previewing it a minute ago. Here's what I meant to write:

[hide="Solution"]$ S \equal{} \sum_{n \equal{} 0}^{\infty} \frac {1}{2^{2^n} \plus{} 1}$

The first four terms are $ \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{17} \plus{} \frac {1}{257}$

The next term is $ \frac {1}{256^2 \plus{} 1} < \frac {1}{256^2 \minus{} 1} \equal{} \frac {1}{255} \cdot \frac {1}{257}$.

Thus, we come up with the geometric-series upper bound \[ S < \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{17} \plus{} \frac {1}{257} \left(1 \plus{} \frac {1}{255} \plus{} \frac {1}{255^2} \plus{} \ldots \right)\] \[ \equal{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{17} \plus{} \frac {1}{257} \cdot \frac {256}{255}\] \[ < \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{17} \plus{} \frac {1}{250}\] \[ < \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {3}{50} \plus{} \frac {1}{250}\] \[ \equal{} \frac {224}{375}\] \[ < \frac {225}{375}\] \[ \equal{} \frac {3}{5}\]
[/hide]


tamame adade avvale p ra biabid ke be mod har adade avvale digari,squarefree bashand.

P be mode har adade avale digar squarefree bashe,daghighan yani chi?

yeni inke tamame aadade avval mese $ p$ ro peida konid ke be ezae tamame aadade avvale $ q$ ke $ q\not\equal{} p$ dashte bashim $ p\equiv \textrm{ khali az morabba' }\pmod q$

[quote="BaBaK Ghalebi"]yeni inke tamame aadade avval mese $ p$ ro peida konid ke be ezae tamame aadade avvale $ q$ ke $ q\not \equal{} p$ dashte bashim $ p\equiv \textrm{ khali az morabba' }\pmod q$[/quote]
bazam nafahmidam :oops: P mode q adadaye mokhtalefi mitune beshe ke momkene baziashun khali az moraba bashan,va baziashun nabashan...

daghighan,hala in masale hame $ p$hai ro mikhad peida kone ke oon baghimoondeha hame khali az morabba bashan,yekhorde pichide be nazar miad,vali ziad masale sakhti nis... :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group, and let $ T$ be an automorphism of $ G$ such that $ T^3 \equal{} I$ (where $ I$ means the identity automorphism) and such that $ T\left(x\right) \equal{} x$ for $ x\in G$ implies $ x \equal{} e$.

Prove that, for every prime $ p$ dividing the order $ o\left(G\right)$ of $ G$, any Sylow $ p$-subgroup of $ G$ is normal in $ G$.

Observe that if $ \varphi$ is an automorphism of $ G$ of prime order, with only the identity as a fixed point (a "regular automorphism"), then $ G\rtimes \left<\varphi\right>$ is a Frobenius group, so by a theorem of Thompson, $ G$ must be nilpotent (and in fact, the nilpotence class of $ G$ is bounded by a constant depending only on the order of $ \varphi$).  But there are elementary proofs of the following:

If $ \varphi$ has order 2, then $ G$ must be abelian.

If $ \varphi$ has order 3, then $ G$ has nilpotence class no more than 2.

[quote="berti"][quote="Eigenray"]Observe that if $ \varphi$ is an automorphism of $ G$ of prime order, with only the identity as a fixed point (a "regular automorphism"), then $ G\rtimes \left < \varphi\right >$ is a Frobenius group[/quote]Why is this?[/quote]
One definition of Frobenius group is: $ G$ is Frobenius if there is a subgroup $ 1 < H < G$ such that $ H\cap H^g \equal{} 1$ unless $ g\in H$.

Here, to show that $ G\rtimes H$ is Frobenius, where $ H \equal{} \left < \varphi\right >$, we need to check that if $ (g^{ \minus{} 1},1)(1,\varphi)(g,1) \equal{} (1,\varphi^i)$ for some $ i$, then $ g \equal{} 1$.  But the left hand side is $ (g^{ \minus{} 1}\varphi(g),\varphi)$, and by assumption the only way this could lie in $ H$ is if $ g \equal{} 1$.

For the current problem the first step is to show that $ g\mapsto g^{ \minus{} 1}\varphi(g)$ is a bijection.  Then you can show that $ g\varphi(g)\varphi^2(g) \equal{} 1$ for all $ g$.  Then there is a series of tricks to show that $ [g,\varphi(g)] \equal{} 1$ for all $ g$, and then that $ [g,g^h] \equal{} 1$ for all $ g,h$.  Somehow this implies $ G$ must be nilpotent.

For details see:
B. H. Neumann, “Groups with automorphisms that leave only the neutral element fixed,” Arch. Math.,7, No. 1, 1–5 (1956).

For the final conclusion that $ G$ must be nilpotent, he cites:
F. W. Levi, "Groups in which the commutator operation satisfies certain algebraic conditions," J. Indian math. Soc.6, 87–97 (1942).

But I don't have online access to this one.


let a,b,c>0
prove that
 \sum a ^3 + 6(\sum (cyclic)a 2 b) \geq 3( \sum (cycle)ab 2 )+12abc.
try it  :D  :D (may be it's easy)

It is my solution
If a \geq b \geq c (1) then Arosisi's inequality <=>( \sum  a^3+3* \sum a 2 b) +3(a-b)(b-c)(a-c)-12abc \geq 0
By Cauchy we have  \sum a^3+3* \sum a 2 b  \geq 12abc  and because  (1) 3(a-b)(b-c)(c-a) \geq 0

If a \geq c \geq b.Easily we can rewrite the inequality
(a-b ) 2 (2b+2c-a)+(b-c) 2 (2c+2a-b)(c-a) 2 (2a+2b-c) \geq 0

You can sol it  easilly.Is it right???????

Today I had a nice solution
Arosisi's inequality <=>  \sum b(2a-b-c) 2  \geq 0

what a nice solution minhthang
i couldn't think about your second solution
my solution is the same as your first sol  :D
can you tell me how could you changed my ineq to that form ?


algebra
polynomial
inequalities
triangle inequality

i dont think i understand.  why $a$= what you said?  how does the conclusions follow?

Oops! :P you are right, two guys. I will edit.

kunny

i still dont understand how that shows $|q(x)|\leq2$.  because if $a=b=-1$, $c$=1 all the given conditions are true but $|a|+|b|+|c|$=3.


Heard of it and seen it done, but never knew how to do it myself. Decided now was the time to stop procrastinating and start learning the Completing Square method. Can you guys teach me?

This might be useful: http://www.youtube.com/watch?v=xGOQYTo9AKY

Take a quadratic:

$ ax^2 \plus{} bx \plus{} c \equal{} 0$

Get rid of the square coefficient first:

$ x^2 \plus{} dx \plus{} e \equal{} 0$

Where d = b/a and e = c/a.

Now as you probably know, $ (x \minus{} r)^2 \equal{} x^2 \minus{} 2rx \plus{} r^2$ and $ (x \plus{} r)^2 \equal{} x^2 \plus{} 2rx \plus{} r^2$. Note how the linear term always has an even coefficient. So that means you take d, and divide by 2, and square it. The ideal square is:

$ x^2 \plus{} dx \plus{} \left(\frac d2\right)^2$

Therefore, you must add $ \left(\frac d2\right)^2$ and subtract it back. You would get:

$ x^2 \plus{} dx \plus{} \left(\frac d2\right)^2 \minus{} \left(\frac d2\right)^2 \plus{} e$
$ \equal{} \left(x \plus{} \frac d2\right)^2 \plus{} \left(e \minus{} \left(\frac d2\right)^2\right)$


algebra
polynomial
binomial coefficients
real analysis
real analysis unsolved

Let $P$ the polynomial:$P(x)=a_0+a_1x+...+a_nx^n$.Then all its roots are real if and only if $\forall i,  a_i^2\geq a_{i-1}a_{i+1}$.I hope it's not also posted already.

Killer, are you really sure that the condition is sufficient (I say this because I think there were some other stronger formulae that were necessary).  I really have some doubts. As for necessity, yes it has been posted and proved.

Obvious counterexample to sufficiency: $1+x+x^2$ has no real zeros.

To eliminate this possibility, add in the binomial coefficients: $p(x)=\sum_k\binom nk\cdot a_kx^k$. Then $a_k^2\ge a_{k-1}a_{k+1}$, with equality when there is a root of multiplicity $n$. There's still more, of course; see $x^3-ax^2+ax-1$, where $a$ is tiny.


graph theory
combinatorics unsolved
combinatorics

Find the number of the induced subgraphs of a nonlabeled $ (p,q)$ graph. BTW Is this a solved problem?

$ (p, q)$ graph is a complete bipartite graph with $ p$ vertices on one side and $ q$ on the other?  (I think probably not.)  What then?

he probably meant a grpah with $ p$ vertices and $ q$ edges... :maybe:

Okay, but then obviously it is not well-defined: For a graph on 4 vertices with 2 disjoint edges, there are 5 induced nonempty subgraphs (one with one vertex, two with two vertices, one with three vertices and the entire graph), but for a graph on 4 vertices with 2 edges sharing an endpoint, there are 7 (one with one vertex, two with two vertices, three with three vertices and the entire graph).  Maybe you can find an answer for particular families, but in general there is no reason to expect a nice answer.

Thank you both. I just quoted the question from my teacher and I know that in that case it is not well defined as you said, but what he meant was to answer the question on a given nonlabeled graph. ie we are given $ G(V,E)$. I don't think that there still exists a nice answer to this one.(has anyone seen a paper or sth on this subject?)


inequalities
geometry
trigonometry
ratio
IMO Shortlist
angle bisector
area of a triangle

Find the largest constant $C$ such that for all $a,b,c$ positive real numbers such that $b<a$ and $c<a$:

$\frac{a}{a-b}\sqrt{\frac{b^2-ba+a^2}{b^2+bc+c^2}}+\frac{a}{a-c}\sqrt{\frac{c^2-ca+a^2}{c^2+cb+b^2}}\geq C$.

When does equality occur ?

Lets use "$K$" for the constant, instead of $C$

Let $A$, $B$, $C$ be points in the plane, so that $OA = a$, $OB = b$, $OC = c$,
$A$ is on the $x$ axis, and $B$ is in the first quadrant, $C$ is in the 4th quadrant, and $\angle BOA = \angle AOC = 60^{\circ}$.

Then we have $\frac{a}{a-b}\frac{AB}{BC} + \frac{a}{a-c}\frac{AC}{BC} \geq K$.

Wolog, suppose $b > c$, and let $a = c+y+z$, $b = c+y$, so that $y,z >0$.

Now $\frac{c+y+z}{z}\frac{AB}{BC} + \frac{c+y+z}{y+z}\frac{AC}{BC} \geq \frac{c+y+z}{y+z}\frac{AB+AC}{BC} \geq \frac{c+y+z}{y+z} \geq 1$.

Equality in the first when $y = 0 \Leftrightarrow b=c$; equality in the second when $AB + AC = BC$, equality in the third when $c$ approaches zero.

So your answer is $K=1$ ? If yes it is false ... (for instance if $b=c$, e.g $a=4$, $b=1$, $c=1$ gives $\frac{8\sqrt(39)}{9}>5$).

I don't understand how it is false.

Let the expression be F(a,b,c).  All you have shown is that F(4,1,1) > 5.  This does not preclude the existence of some a,b,c such that F(a,b,c) < 5.

What we are looking for is the largest K so that $F(a,b,c) \ge K$ for all a,b,c.  Atleast, this is the problem statement.

Am I wrong?

[quote="Singular"]equality in the second when $AB + AC = BC$, equality in the third when $c$ approaches zero.[/quote]
I think you are wrong because the equality of these two cannot attain at the same time. So your constant is weaker.

The answer is indeed bigger.
Singular: You have said that equality is obtained if and only if $b=c$, haven't you ?

[quote="Igor"]Nobody  :P  ?[/quote]I give "my tongue to the cat" (in French, this means to admit defeat :wallbash: ) Could you please tell us the answer and say how you guys came up with such a strange inequality ?

[quote="Igor"]Find the largest constant $C$ such that for all $a,b,c$ positive real numbers such that $b<a$ and $c<a$:

$\frac{a}{a-b}\sqrt{\frac{b^2-ba+a^2}{b^2+bc+c^2}}+\frac{a}{a-c}\sqrt{\frac{c^2-ca+a^2}{c^2+cb+b^2}}\geq C$.[/quote]

Well, there are not many ways to construct such a problem, and it is particularly difficult to hide all traces of the construction. So, a number of people will note that expressions like $\sqrt{b^2-ba+a^2}$ are closely related to 60° angles, and the fact that the best constant C equals 4 strongly reminds of [url=http://www.kalva.demon.co.uk/short/soln/sh02g2.html]the problem G2 from the IMO Shortlist 2002[/url]. And indeed, it's now clear that the problem arose from an attempt to algebraically kill that shortlist problem, but it turned out to be quite resistant to this line of attack and thus, instead of becoming trivial, gave rise to an algebraic inequality which is not easier. Did I get the history of origins of the problem right? ;)

Now here is a detailed solution of your problem. We will use directed angles modulo 360°.

Let A, B, C, D, E, F be six points in the plane such that FA = a, $FB=\frac{ab}{a-b}$, $FC=\frac{ac}{a-c}$, FE = b, FD = c, and such that < BFE = < EFA = < AFD = < DFC = 60°. Then,

< BFA = < BFE + < EFA = 60° + 60° = 120°;
< EFD = < EFA + < AFD = 60° + 60° = 120°;
< AFC = < AFD + < DFC = 60° + 60° = 120°;
< CFB = - < BFC = - (< BFA + < AFC) = - (120° + 120°) = - 240° = 120°.

Also, < BFD = < BFE + < EFA + < AFD = 60° + 60° + 60° = 180°, so that the point F lies on the segment BD; similarly, the point F lies on the segment CE.

Finally, note that sin 120° = sin (180° - 60°) = sin 60°.

Denote by $\left|P_1P_2P_3\right|$ the area of a triangle $P_1P_2P_3$. Then, using the fact that the area of a triangle equals $\frac12$ times the product of two sides times the sine of the angle between them, we have

$\left|BFE\right|=\frac12\cdot FB\cdot FE\cdot\sin\measuredangle BFE=\frac12\cdot\frac{ab}{a-b}\cdot b\cdot\sin 60^{\circ}$;
$\left|EFA\right|=\frac12\cdot FE\cdot FA\cdot\sin\measuredangle EFA=\frac12\cdot b\cdot a\cdot\sin 60^{\circ}$;
$\left|BFA\right|=\frac12\cdot FB\cdot FA\cdot\sin\measuredangle BFA=\frac12\cdot\frac{ab}{a-b}\cdot a\cdot\sin 120^{\circ}$
$=\frac12\cdot\left(\frac{ab}{a-b}\cdot b+b\cdot a\right)\cdot\sin 60^{\circ}=\frac12\cdot\frac{ab}{a-b}\cdot b\cdot\sin 60^{\circ}+\frac12\cdot b\cdot a\cdot\sin 60^{\circ}$
$=\left|BFE\right|+\left|EFA\right|$.

The last equation yields that the point E lies on the segment AB. Similarly, the point D lies on the segment AC. Thus, the point F, being the point of intersection of the segments BD and CE, must lie inside the triangle ABC. Also, since < BFA = < AFC = < CFB = 120°, the sides of triangle ABC subtend the same angles at the point F. Hence, by [url=http://www.kalva.demon.co.uk/short/soln/sh02g2.html]the IMO Shortlist 2002 problem G2[/url], we have $AB+AC\geq 4\cdot DE$.

By the cosine law, applied to the triangle AFD, we have

$AD=\sqrt{FA^2+FD^2-2\cdot FA\cdot FD\cdot\cos\measuredangle AFD}$
$=\sqrt{a^2+c^2-2\cdot a\cdot c\cdot\cos 60^{\circ}}=\sqrt{a^2+c^2-2ac\cdot\frac12}=\sqrt{a^2+c^2-ac}$.

Similarly, $AE=\sqrt{a^2+b^2-ab}$. By the cosine law, applied to the triangle EFD, we have

$DE=\sqrt{FE^2+FD^2-2\cdot FE\cdot FD\cdot\cos\measuredangle EFD}$
$=\sqrt{b^2+c^2-2\cdot b\cdot c\cdot\cos 120^{\circ}}=\sqrt{b^2+c^2-2bc\cdot\left(-\frac12\right)}=\sqrt{b^2+c^2+bc}$.

Since < BFE = < EFA = 60°, the line FE is the angle bisector of the angle AFB. Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we thus have $\frac{AE}{EB}=\frac{FA}{FB}$. In other words, $\frac{AE}{EB}=\frac{a}{\left(\frac{ab}{a-b}\right)}=\frac{a-b}{b}$. Thus, $\frac{EB}{AE}=\frac{b}{a-b}$, and $\frac{AB}{AE}=\frac{AE+EB}{AE}=1+\frac{EB}{AE}=1+\frac{b}{a-b}=\frac{a}{a-b}$, so that $AB=\frac{a}{a-b}\cdot AE=\frac{a}{a-b}\cdot\sqrt{a^2+b^2-ab}$. Similarly, $AC=\frac{a}{a-c}\cdot\sqrt{a^2+c^2-ac}$. Finally, we know that $DE=\sqrt{b^2+c^2+bc}$. Thus, the inequality $AB+AC\geq 4\cdot DE$ becomes

$\frac{a}{a-b}\cdot\sqrt{a^2+b^2-ab}+\frac{a}{a-c}\cdot\sqrt{a^2+c^2-ac}\geq 4\cdot\sqrt{b^2+c^2+bc}$.

Upon division by $\sqrt{b^2+c^2+bc}$, this becomes

$\frac{a}{a-b}\cdot\sqrt{\frac{a^2+b^2-ab}{b^2+c^2+bc}}+\frac{a}{a-c}\cdot\sqrt{\frac{a^2+c^2-ac}{b^2+c^2+bc}}\geq 4$.

This is exactly your inequality, with the constant on the right-hand side being 4. Equality holds for $b=c=\frac12a$, so the constant 4 cannot be improved. And the problem is solved.

  Darij

I haven't time to read you proof , but
[quote="darij grinberg"]
Well, there are not many ways to construct such a problem, and it is particularly difficult to hide all traces of the construction. So, a number of people will note that expressions like $\sqrt{b^2-ba+a^2}$ are closely related to 60° angles, and the fact that the best constant C equals 4 strongly reminds of [url=http://www.kalva.demon.co.uk/short/soln/sh02g2.html]the problem G2 from the IMO Shortlist 2002[/url]. And indeed, it's now clear that the problem arose from an attempt to algebraically kill that shortlist problem, but it turned out to be quite resistant to this line of attack and thus, instead of becoming trivial, gave rise to an algebraic inequality which is not easier. Did I get the history of origins of the problem right? ;)[/quote]
if perfectly correct  :lol:


hello,

solve in natural numbers the equation : $ \displaystyle {\rm{ x}}^2  \plus{} y^2  \equal{} 1997\left( {x \minus{} y} \right)$

I don't know if it's right but, I'll give it a try.

$ x^2 \plus{} y^2 \equal{} z^2$
$ z^2\equal{}1997(x\minus{}y)$
$ 1997$ is prime so for $ z^2$ to be a perfect square,

$ x\minus{}y \equiv 0 \mod1997$
$ \boxed{x \equiv y \mod1997}$

Let k=gcd(x,y) so x=km and y=kn, where gcd(m,n)=1. Then $ k(m^2 \plus{} n^2) \equal{} 1997(m \minus{} n)$. But the gcd of $ m^2 \plus{} n^2 \equal{} (m \minus{} n)^2 \plus{} 3mn$ and m-n is either 1 or 3, because any common factor would have to divide 3mn and m-n. So we either have m-n=k or m-n=3k. In the last case, we get a contradiction mod 4. So m-n=k, so $ m^2 \plus{} n^2 \equal{} 1997$. Brute forcing, we get $ 29^2 \plus{} 34^2 \equal{} 1997$. So m-n=5 and the solution is ($ 5 \cdot 34, 5\cdot 29$)= (170, 145).

[quote="dgreenb801"]Let k=gcd(x,y) so x=km and y=kn, where gcd(m,n)=1. Then $ k(m^2 \plus{} n^2) \equal{} 1997(m \minus{} n)$. But the gcd of $ m^2 \plus{} n^2 \equal{} (m \minus{} n)^2 \plus{} 3mn$ and m-n is either 1 or 3, because any common factor would have to divide 3mn and m-n. So we either have m-n=k or m-n=3k. In the last case, we get a contradiction mod 4. So m-n=k, so $ m^2 \plus{} n^2 \equal{} 1997$. Brute forcing, we get $ 29^2 \plus{} 34^2 \equal{} 1997$. So m-n=5 and the solution is ($ 5 \cdot 29, 5\cdot 34$)= (145,170).[/quote]

Would you explain how

$ \boxed{m^2 \plus{} n^2 \equal{} (m \minus{} n)^2 \plus{} 3mn}$ comes?

[quote="modularmarc101"]$ x^2 \plus{} y^2 \equal{} z^2$[/quote]
And where was it stated that $ x^2 \plus{} y^2$ had to be a square?   :|

Nowhere ..stupid solution doesnt make any sense :P

[quote="vinskman"]
$ \boxed{m^2 \plus{} n^2 \equal{} (m \minus{} n)^2 \plus{} 3mn}$ comes?[/quote]

Sorry, my bad, but the proof is still similar (Even though I can't do Algebra 1), $ m^2 \plus{} n^2 \equal{} (m \minus{} n)^2 \plus{} 2mn$, so the $ gcd(m^2 \plus{} n^2,m \minus{} n)$ is either 1 or 2, we've already covered if it's 1 so if $ m \minus{} n \equal{} 2k$, then $ m^2 \plus{} n^2 \equal{} 3994$, brute forcing (m,n)=(5,63) and k=(m-n)/2=29 so the solution is ($ 63 \cdot 29$, $ 5 \cdot 29$)=(1827,145).

I'm not sure if there is an easier way than brute forcing.


algebra
polynomial
invariant
vector
abstract algebra
linear algebra
linear algebra unsolved

I've heard someone saying that $A,B$ are similar if and only if $ rank P(A)= rank P(B)$ for any polynomial $P$. Is this true? I've tried to prove it, but no chance. Does anyone know similar result? It would be really a very nice result.

Probably you already heard about that if you have $A,B \in M_n(C)$ with the same
characteristic polynomial and the same minimal polynomial they are similar

This is true for $n=2, n=3$ I think, but it is not true for higher $n$. Come on, Moubi, I would have realised immediately the answer if it had been this question.   ;)

see Leichtman page ?? for the counter example (at "invariant de similitude")  :D

[quote="Moubinool"][quote="harazi"] it is not true for higher $n$[/quote]

you have a couterexample[/quote]

I think it's pretty easy to provide one:

$\left(\begin{array}{cccc}0&0&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)$ and $\left(\begin{array}{cccc}0&0&0&0\\1&0&0&0\\0&0&0&0\\0&0&1&0\end{array}\right)$.

In both cases, the characteristic polynomial is $x^4$ and the minimal polynomial is $x^2$.

When harazi said "it is not true for higher $n$", I thought he was referring to the problem he posted :), and I think this discussion is getting confusing :): alekk, when you say "counterexample", are you referring to harazi's problem, or to Moubinool's problem?

[quote]if you have $A,B \in M_n(C)$ with the same
characteristic polynomial and the same minimal polynomial they are similar[/quote]

Without loss of generality consider nilpotent $A,B$.  They define two filtrations on the vector space they operate upon.  The minimal polynomial just gives the length of the filtration and the characteristic polynomial (i.e. its degree) gives the dimension of the space.  But it takes more information to specify a filtration, unless the dimension is very small ($2$ or maybe $3$, as grobber indicated) so that there are no choices.

Do you mean that $A$ and $B$ are $n\times n$-matrices?

Clearly, if $A$ and $B$ are similar, then ranks of $P(A)$ and $P(B)$ are equal.

Let's reconstruct Jordan form of $A$ by information of ${\hbox{rank}}(P(A))$ for any polynomial $P$. We may reconstruct all eigenvalues (as such numbers that ${\hbox{rank}}(A-\lambda)$ is less then $n$. Further, ${\hbox{rank}}(A-\lambda)$ is the number of Jordan blocks, which correspond to $\lambda$. Then, we know dimensions of kernels of $(A-\lambda)^k$, and those sizes of all Jordan boxes (say, number of cells $1\times 1$ plus twice the number of larger cells equals $\dim \ker (A-\lambda)^2$ and so on).

Matrices with the same Jordan form are similar.

this proof works with the assumption that the field is algebraicly closed. In general, to prove the assertion of harazi we have to use cyclic spaces decomposition (I don't know the name in english  :blush: )

In general, we may consider $p_{\lambda}(A)$ instead $A-\lambda$, where $p_{\lambda}$ is a minimal polynomial of eigenvalue $\lambda$ (over initial field, $\lambda$ belongs to its algebraic closure).


:maybe: What exactly is a greedy algorithm :(


Prove that, in a triangle $ ABC$,

$ \cos A\cos B \cos C \leq \frac{1}{8}$.

It is an easy problem , but im a little lazy to write the full solution, i'll give u a hint  :)

[hide="hint"]use the identity $ cosA \cdot cosB \cdot cosC \equal{} \frac {1}{2}(1 \minus{} cos^2A \minus{} cos^2B \minus{} cos^2C)$

then Jensen kills the right side of the equality[/hide]

Thanks Tiger100.

I proved your identity which is equivalent to $ \sin^{2}A \plus{} \sin^{2}B\plus{}\sin^{2}C \equal{} 2 \plus{} 2\cos A \cos B \cos C$ and then used Jensen's inequality to get

$ \frac{3}{2}(\frac{1}{3} \minus{} (\frac{\cos^{2}A}{3} \plus{} \frac{\cos^{2}B}{3}\plus{}\frac{\cos^{2}C}{3}))$

$ \leq \frac{1}{2} \minus{}\frac{3}{2}\cdot \cos(\frac{A\plus{}B\plus{}C}{3})^{2}$

$ \frac{1}{2} \minus{} \frac{3}{2}\cdot \frac{1}{4} \equal{} \frac{1}{8}$.


inequalities
quadratics
inequalities proposed

Give positive real numbers $ a,b,c$.Prove that
$ \frac{\sqrt{3(a^3\plus{}b^3\plus{}c^3)}}{3abc}\geq\frac{\sqrt a\plus{} \sqrt b\plus{}\sqrt c}{ab\plus{}bc\plus{}ca}$

After clearing denominators and squaring both sides the inequality becomes:
$ (a^3\plus{}b^3\plus{}c^3)(ab\plus{}bc\plus{}ca)^2\ge 3a^2b^2c^2(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2$   (*)
Now, by AM-GM $ (ab\plus{}bc\plus{}ca)^2\ge (3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}})^2\equal{}9a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}$
By Muirhead, $ a^3\plus{}b^3\plus{}c^3\ge (a\plus{}b\plus{}c)a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}$
Then, $ (a^3\plus{}b^3\plus{}c^3)(ab\plus{}bc\plus{}ca)^2\ge 9a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}(a\plus{}b\plus{}c)a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}$
$ \equal{}9a^2b^2c^2(a\plus{}b\plus{}c)\equal{}(3(a^2\plus{}b^2\plus{}c^2))(3(a\plus{}b\plus{}c))\ge 3(a^2b^2c^2)(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2$ by Quadratic-Arithmetic Mean inequality. This is the same as (*) and the result follows.

[quote="rem"]
Then, $ (a^3 \plus{} b^3 \plus{} c^3)(ab \plus{} bc \plus{} ca)^2\ge 9a^{\frac {4}{3}}b^{\frac {4}{3}}c^{\frac {4}{3}}(a \plus{} b \plus{} c)a^{\frac {2}{3}}b^{\frac {2}{3}}c^{\frac {2}{3}}$
$ \equal{} 9a^2b^2c^2(a \plus{} b \plus{} c) \equal{} 3a^2b^2c^2(3(a \plus{} b \plus{} c))\ge 3(a^2b^2c^2)(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2$ by CBS inequality. This is the same as (*) and the result follows.[/quote]
 :D


If you increase the length of a metal bar, will its density increace, decrease, or stay the same?

(i should know this) :(

As long as you are using the same metals, its density will stay the same.

They will each increase at the same weight. Let m1 be the mass of the original pole and v1 be the volume of the original pole. You then add on a portion with mass m2 and volume v2. If it is the same material, m2/v2=m1/v1

the density remains the same. density simply refers the the amount of "stuff" in a certain space.

[quote="ln(dx/dy)"]But Density = Mass/Volume
What if its mass increased faster than it's volume or the opposite?[/quote]

That is not possible, at least no if you are using the same metal.

[quote]That is not possible, at least no if you are using the same metal.[/quote]
What???? all you have to do is to heat it. ... (or just pull a wire made out of  metal  the wire will be hotter and the density will change)

depends on how u increase the length: if it is heated (as it probably would be), it's density would decrease (i think) but it would slowly go back to the same density (assuming the addition is of the same metal)


Pascal\u0027s Triangle
binomial coefficients
number theory proposed
number theory

I would like to ask help for you with these problems!

1-Let $n$ be a positive intenger.Let $A(n)$ denote the number of factor 2 in $n!$, and let $B(n)$ denote the number of 1s in the binary expansion of $n$.Show that the number of odd entries in the n-row of Pascal's Triangle is $2^{B(n)}$, and that $A(n)+B(n) = n$.

2-For a positive intenger $n$ ,let $a(n)$ and $b(n)$ denote the number of odd coefficients in the nth-row of Pascal's Triangle that are congruent to 1 and 2 mod 3 respectively.Prove that $a(n)-b(n)$ is always a power of 2.

3-Let $a_{n}$ be the last non-zero digit in the decimal representation of  the number $n!$.Is the sequence $a_{1},a_{2},a_{3}, \dots$ periodic after a finite number of members?

Thanks!!


geometry
3D geometry
perimeter
quadratics
calculus
Pythagorean Theorem

Here are several problems just for fun. Enjoy!

$\text{Number Problems}$
$1.$ What fraction added to its reciprocal gives $\frac{13}{6}$?
$2.$ Find all sets of $(x;y)\in\mathbb{R}$ such that the differnce of $x$ and $y$ is $7$ and the difference of whose cubes is $1267$.
$3.$ Find two numbers whose difference is 10 and the sum of whose squares is 250.
$4.$ Solve for $x$: $\sqrt{x-9}+\sqrt{x}=\frac{45}{\sqrt{x}}$.

$\text{Geometry Problems}$
$1.$ A right triangle has one side equal to $25.3$ and a perimeter of $68.4$.  Find the length of the hypotenuse and the other leg.
$2.$ A rectangular sheet of brass is twice as long as it is wide.  Squares, $3\times 3$ are cut from each corner, and the ends are turned up to form an open box having a volume of $648$.  What were the dimensions of the original sheet of brass?
$3.$ A casting in the shape of a cube is seen to shrink $0.175$ on a side, with a reduction in volume of $2.48$.  Find the original dimensions of the cube.
Enjoy!

Oh, yes, sorry.  I forgot to mention that.

Yes, would you please show your work? :)

here's what i did ..... :)

1. x+1/x= 13/6, jus solve this equation
2. x-y=7, $x^3-y^3$ = 1267 then factorise x^3-y^3 as ( x-y)(X^-xy+y^2) and put put x-y =7 in this , and it'll come

3. almost same , x-y=10 , $x^2+y^2=250$ , replace x^2 by (10 + y)^2 and solve the quadratic

4.  just take $\sqrt{x}$ of the RHS to the LHS and multiply it . then square the equation

hope that's adequate ;)

Can you explain your work for number 2 more? I'm at the step you just described, and I think I'm going in circles.

Edit: Never mind, got it. That was ugly.

Geometry problem #1

[hide]

Let $x$, $y$, and $z$ be the sides of a triangle, where $z$ is the hypotenuse.

$z + y + z = P$
$25.3 + y + z = 68.4$
$y + z = 43.1$

By the Pythagorean theorem,

$x^2 + y^2 = z^2$
$25.3^2 + y^2 = z^2$
$z^2-y^2=640.09$

By substitution,

$(43.1-y)^2 - y^2 = 640.09$
$y = 14.1$ (rounded to one decimal place)

Solving for z,

$14.1 + z = 43.1$
$z = 29$

Checking the solutions,

$25.3 + 14.1 + 29 = 68.4$
$25.3^2 + 14.1^2 = 29^2$

Therefore, the lengths of the other sides are 14.1 and 29.
[/hide]

If there is a simpler way to do this, let me know please.

Geometry Problem #2

These problems are cute, my Calculus teacher liked to put an optimization twist on these.

[hide]

Let $x$ be the width of the brass sheet, then $2x$ is the length.

$V=lwh$
$648=3(2x-6)(x-6)$
$648=3(2x^2-12x-6x+36)$
$648=6x^2-54x+108$
$x^2-9x-96=0$

Solving this equation gives us $x=15.28, -6.28$. A dimension cannot be negative, so we're left with $x=15.28$.

$w=2x$
$w=30.56$
$l=x$
$l=15.28$

The original dimensions of our brass were 30.56 x 15.28.
[/hide]

Yeah, the volume isn't exactly 684 with these dimensions, but it's close enough. Round off error. That's how you do it.


Kevin has 4 red marbles and 8 blue marbles. He arranges these 12 marbles randomly, in a ring. Determine the probability that no two red marbles are adjacent?

Here's a hint:  How many ways are there to distribute 8 blue marbles among four boxes so that there is at least one marble in each box?


The side length of a square is 16 cm long. The midpoints of each side are joined to form a second square. The process of joining the midpoints of the sides of the innermost square is repeated. What is the number of centimeters in the perimeter of the fifth square?

Sides:
16
8sqrt2
8
4sqrt2
4
Perimeter: 4*4=16


I'm a soph and want to qualify for USAMO...

and I usually get 7-9 on AIME prac...

But I don't know what kind of book to use to prepare for USAMO...

Does any one have any recommendations??

Thanks,

Try this link: http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php

and look under the "ready for more" and "for the pros" section.  If you haven't already, you'll want to get Art and Craft of Problem Solving, since that's an excellent introduction to the kind of thinking you'll need for the USAMO, as well as a broad introduction to many sorts of problem-solving (with the notable exception of geometry).  Once you're ready for more in-depth prep, you can get books like Mathematical Olympiad Challenges, Engel's Problem-Solving Strategies, and topic books like Geometry Revisited.  I'm not sure what a 7-9 indicates on the AIME: if they're consistently the first 7-9, then you'll want to focus on Art and Craft-level problem-solving first and pick up Volume II of the AOPS series.  If you can hit the harder ones pretty consistently and just make mistakes or need breadth, then you can start stepping up to some of the harder books that I mentioned.


Find all positive integers $ (a,b)$ such that
\[ b^2(a^2\plus{}b^2\minus{}2ab\minus{}a\minus{}b)\equal{}(a\plus{}b)^2(a\minus{}b)\]

Denote $ d \equal{} \gcd(a,b)$, $ a \equal{} da'$, $ b\equal{}db'$, therefore $ \gcd(a',b') \equal{} 1$.
Then $ b'^2(d(a'^2\plus{}b'^2\minus{}2a'b')\minus{}a'\minus{}b')\equal{}(a'\plus{}b')^2(a'\minus{}b')$. If a prime $ p \mid b'$, then $ p \mid a'\plus{}b'$ or $ p \mid a'\minus{} b'$, whence $ p \mid a'$, absurd. Therefore $ b'\equal{}1$.

It follows $ d(a'\minus{}1)^2 \equal{} (a'\plus{}1)a'^2$, easily seen to only have solutions $ a'\equal{}2$, $ d\equal{}12$, and $ a'\equal{}3$, $ d\equal{}9$.

Therefore the only positive integer solutions are $ (a,b) \equal{} (24,12)$ and $ (27,9)$.


geometry
trigonometry
function
inequalities unsolved
inequalities

Let ABCD have M inside ABCD.Put ^MAB=x;^MBC=y;^MCD=z;^MDA=t;
Prove that min{x;y;z;t} <= pi/4

I believe this is a generalization of Brocard points; basically just plug and chug away with Law of Sine and Cosine in relation to all four angles assumed to be >45, and you'll get a contradiction.

Can you post your solution more detail????

Let $w$ the angle of Brocard .

Then we know that $cotw=cota+cotb+cotc\geq\sqrt{3}$ where a,b,c are the angles of the triangle using the fact that cot is decreasing function we have that $w\leq\frac{\pi}{6}$  what we want

I don't know the Brocard point ,some one can eplain to me???


Solve the system

x^2 + 5y^2 = z^2
5^x2 + y^2 = t^2

hello, do you mean
$ x^2\plus{}5y^2\equal{}z^2$
$ 5x^2\plus{}y^2\equal{}t^2$?
Sonnhard.

well... assuming that you mean integers
adding, we get that
$ 6(x^2\plus{}y^2)\equal{}z^2\plus{}t^2$
the LHS must be divisible by 3, so both $ z$ and $ t$ are divisible by 3 (this is casework)
Then we have $ 6(x^2\plus{}y^2)\equal{}9(z'^2\plus{}t'^2) \implies 2(x^2\plus{}y^2)\equal{}3(z'^2\plus{}t'^2)$
Then the RHS must be divisible by x, implying that x and y are divisible by 3
so any solution to $ 6(x^2\plus{}y^2)\equal{}z^2\plus{}t^2$
must have a solution with each variable 3 times smaller.
If we are working in integers, this means that each variable must be 'infinitely divisible by 3', or that it's 0.
Only $ (0,0,0,0)$ works


[quote="mysmartmouth"]You know those problems on MathCounts that have something like:

DOG
[u]+  DOG[/u]
CAT

Well, I never really got those problems fully until I read a book my little brother gave me.  Is anyone familiar with the Wayside School books?  They are children's books, (NOT MATH BOOKS) and I reccomend them for anyone looking for a light, funny read.  But the author also wrote a math book called [url=http://www.amazon.com/exec/obidos/tg/detail/-/0590457268/103-7526440-6663808?v=glance]Sideways Arithmetic From Wayside School.[/url]  It might be too simple for a lot of you, but it helped me with several math ideas that I see in MathCounts, and you should at least check it out from the library if you don't understand problems like the one above.  Except now I feel stupid, because my little brother had a book about something I didn't even fullly understand.  My only condolence was that he had no idea why ELF + ELF equaled FOOL. ;)[/quote]

I've read that book before, and its a pretty good book too.

[hide]ELF + ELF = FOOL

Well, F = 1

EL1 + EL1 = 1OOL

So L = 2.

E21 + E21 = 1OO2
O = 4

E21 + E21 = 1442

E = 7

721 + 721 = 1442

:o .

I liked that book too.[/hide]

That was fun.

In #2 he is fair by giving Q a problem dedicated to it.

[quote="Treething"] :o [/quote]

Whenever I see an "emoticon" like that one, I have to put my mouse over it to see what it means. :blush:

[quote="236factorial"]I like those wayside school books. They test your logic.
Can you really solve your dog question?

how about this?

ONE
[u]x 9[/u]
NINE

What are the two solutions to NINE?[/quote]

I don't feel like getting that book. Can anyone tell me how you're supposed to do these?

[quote="236factorial"]I like those wayside school books. They test your logic.
Can you really solve your dog question?

how about this?

ONE
[u]x 9[/u]
NINE

What are the two solutions to NINE?[/quote]

i got 325 * 9 = 2925
and  i dont know if each letter represents a different number
so 875 * 9 = 7875 ????


13. Let ♣(x) denote the sum of the digits of the positive integer x. For example, ♣(8) = 8 and ♣(123) = 1 + 2 + 3 = 6. For how many two-digit values of x is ♣(♣(x)) = 3?

$ (A) 3 \qquad (B) 4 \qquad (C) 6 \qquad (D) 9 \qquad (E) 10$

I found 10.  We first have those whose digits total 3, so then once we place 3 in the function, we get 3 back: 12, 21, and 30.  Now no two digits can sum up to 21 or 30, but there are 7 that sum up to 12: 39, 93, 48, 84, 57, 75, and 66.  Since they all sum to 12, when we put 12 back in the the function we get 3.  So there are $ \boxed{10}$.


A friend of mine I was talking with a few days ago said, "I don't know how anyone could willingly sign away their right to not kill other people."

[quote="peter"]If any one country had no army, they would be overrun, so the only way to get rid of armies is to have all countries have no armies, because then there are no threats.  Only problem is, if one country secretly makes an army, they can take over everywhere.[/quote]

Well, that is not necessarly true. On December 1st, 1949 the goverment of Costa Rica abolished the army, and there is no army since then. We have survived without an army, even with revolutions around all the time.  Not even once we have had attempts of overrun for insiders or outsiders!

Some people say: Costa Rica has no army but has police!, yes, as any country, we have a Police force, and they have some specialized SWAT kind units, but the police in my country is less than 0.13% of the population. On a quick serch on the web I couldn't find that data about USA, but I am sure it is higher (Just a guess).

djiminez, I think you should make a poll out of this issue. I do not like the fact the US army has become a 'police' force in the world, and that we need to spend an absurdly large portion of our budget on military equipment.

 For one thing, American teenagers (in general, not here of course) rank almost last in developed countries for math skills. I think money should be spent to rectify this problem.

Yet, I do respect a country's ability to have an army.

[quote="beta"]I definitely don't want to fight a war, I don't want get killed by other people that I don't know, or kill anyone that I don't know.[/quote]

I wouldn't want to kill anyone that I do know...why don't we just not kill.

as little as the general population is educated in this country, US still got all the smart people, technology, etc...

and for the original topic, having an army is "disgusting".. but not having an army while others do is stupid. What if you get attacked? The pride soldiers get is the pride of being a patriot and the ability to defend their country, i think..

and although having such a huge military, the US isnt really making a bad move politically. The US gets its way with everything, and although policing the world isn't a good thing, it is the best thing the US can do for itself.

[quote="kentliu"] having an army is "disgusting".. but not having an army while others do is stupid. [/quote]

Yes, it is that stupid that Costa Rican have lived more than a half a century without one, with lots of advantages and very few drawbacks. If you read, Costa Rica is by far the most prosperous Central American country, with the best living and education standards in the region. Do you have an idea why? Main reason: WE DON'T HAVE AN ARMY!, we don't expent most of our budget in paying soldiers and buying weapons. We expend more money in health and education!

Let's make just a few of math! I talk about the US, because here is where I currently live. I remember I hear in a documental not long ago (I can't find it in the web now, but may be someone can) that the US Army budget is officially a little more than $\$300 billion (add to it the black budget!). Now, we heard pretty often than in most of Africa most of the people are starving to death. The population on Africa I believe is less than 700 million people. Say that 15% of the people handle to have a fairly descent life (I tend to think that it should be more than that, but I don't know the data!) then we have around than 600 million people starving. That is, the money of the US is enough to give
$\$500 in food to each hungry person in Africa. Just to give you an idea how big is the spent money in an Army, at least by the US!

But any way, the original post does not ask if an army is or not necessary, is about what does people think about the army on their respective countries.

In my state (California), teachers have not received a raise and are about to go on strike. We are billions of dollars in debt and our governor, Schwarzenegger, is not doing a lot, accoridng to teachers, to help the problem. He has went back on many of his promises and our school and many others are in need of money for supplies.

I hate how America spends so damn much money on war. That money could be used to fix problems in places that need it. Our class sizes are getting out of hand and teachers barely get paid, and are underappreciated, so there is little incentive to go into that field.

I read an interesting quote once. "It'll be the day when schools have all the money they need and the military has to hold a bake sale to buy more weapons." - I can't remember from who....

There is a major difference between the US and Costa Rica.  As being one of the "supreme powers" in the world, it faces a lot of risk from different things. Just look back in history, and see what horror could have occured if the US had no military.

There are a lot of threats towards the US...and today they are higher than ever.  An army is necessary to a certain extent.  It all comes down to what you do with that army.

[quote="Farazy"]There is a major difference between the US and Costa Rica.[/quote]

A major difference, there are thousands of major differences, believe me, I have lived in both countries, and I know what I say!  :D  :D  :D

[quote="Farazy"]As being one of the "supreme powers" in the world, it faces a lot of risk from different things. Just look back in history, and see what horror could have occured if the US had no military.

There are a lot of threats towards the US...and today they are higher than ever.  An army is necessary to a certain extent.  It all comes down to what you do with that army.[/quote]

I am not claiming that there is no need to have an army. I said that it is something unnecesary in a [b]rational world[/b], unfortunatelly, most of the time this is not a rational world  :( .

On the other hand, even when an army may be necessary, I still think that $\$300 billion a year plus black budget is a little excesive for an army!

On the other hand, I don't doubt that being the country that handle more than half of the economic resourses in the world, the US needs to have some kind of protection, and that there are very real threats to the national security, but I still think that more than half (I would say, more than 80%) of the fears on the collective mind of the US population are no more than paranoia (just my personal opinion).

On the other hand, not to have an army does not imply not to have the willing of the population to defend itself in case it is necessary. I honestly admire the swiss option (well, at least as far as I have heard, directly from a few swiss people and reading a little) that every citizen is trained to defend itself and its country if necessary, but they are neutral in foreing conflicts, and as far as I know, there is no need for them to use their army to stop inside attempts of overrun the goverment. And well, I have heard that now is Germany, but for decades it was Switzerland the economically more powerful contry in Europe (I am not totally sure about that fact).

i don't mind countries having a resonable-sized army for defense purposes but when a country tries to force their ideas to other countries and keep people in other countries from being "sadamized" (but the war in iraq was not to gain control over a huge oil supply...that must be a bonus), then i have a problem. the war is very costly, like ankur87 mentioned, and the money could be spent on a lot better purposes.

I am proud to have military parents who care so much for our country that they'd give up their entire lives to dedicate themselves to serving our country.

[quote=peter_]If any one country had no army, they would be overrun, so the only way to get rid of armies is to have all countries have no armies, because then there are no threats.  Only problem is, if one country secretly makes an army, they can take over everywhere.[/quote]

Many countries, sovereign countries, have no armies.

[quote=peter_]If any one country had no army, they would be overrun, so the only way to get rid of armies is to have all countries have no armies, because then there are no threats.  Only problem is, if one country secretly makes an army, they can take over everywhere.[/quote]

It's sort of like playing split or steal, except 195 countries have the option and picking steal provides a greater net win

[quote=djimenez]Before saying something that may be misunderstood, let me explain something first. I am from Costa Rica, as far as I know, the only country in the world without an army, and we haven't have one for almost 60 years. I grew up being thought that an army is something unnecessary in a rational society, something gross. Therefore, in my opinion, to be on an army, to be a soldier, is something disgusting.

Nevertheless, when I came to the US, even when so many people would never join the army, there is a huge (and I mean a HUGE) number of people that consider the army a carrier, and to be a soldier a good profession, even encourage their sons and daughters to become one. Even, I have met a lot of people that are in the army (or navy or whatever other division of the forces of this country) and they use to be really proud to be a soldier, marine, etc.

Therefore, I would like to know on your respective countries, what is thought about having an army, and to be a soldier, and of course, what is your personal opinion.[/quote]

yo you're from costa Rica that is a beautiful place i love it there i went there for vacation once and visited a bunch of volcanos and hot springs it was really fun.
I don't think Costa Rica needs an army when they aren't really threatening people and or defending themselves since America is a country of it's size and has a decent amount of enemy's and also help other country's in aiding disasters.
i don't know much about costa Rica's history when it comes to the military so don't quote me on this


Triangle $ABC$ is right angled at $B$. The bisector of $\angle A$ meets $\overline{BC}$ at $D$. Let $G$ denote the centroid of $\triangle ABC$. Show that $GD\parallel AB$ if and only if $\angle C=30^{\circ}$.

[hide="hint"]
too ez :D
1.use the fact of centroid [2:1]
2.then use the theorem for the parallel lines on $GD$ and $AB$
3.angle bisector theorem
Hope that u have got it....
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