quadratics
symmetry
Galois Theory
number theory
number theory unsolved

Prove that $ \{ \sqrt{p}$ : $ p$ is a prime$ \}$ is a linearly independent set over $ \mathbb{Q}$, the field of rational numbers.

P.S. : Any elementary proof will be greatly appreciated  :)

I gave two proofs [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275529]here[/url].  It's an interesting question which proof is more "elementary":  the first argument relies on both quadratic reciprocity and Dirichlet's theorem, which are highly nontrivial statements with highly nontrivial proofs, but the second argument uses the language of Galois theory.  I think the proof of the fundamental theorem of Galois theory is much easier than the proof of either QR or Dirichlet's theorem, and in principle it could be "unwound" into a much easier proof that doesn't make explicit use of the language of Galois theory, whereas you would be hard-pressed to do the same for the first proof.

On the other hand, for any particular finite subset you don't need either QR or Dirichlet's theorem to guarantee the existence of $ q$: you just have to find it.

[b]Edit:[/b]  The fundamental concept behind Galois theory is symmetry, so if you're not comfortable with the second proof as it's written, just think of it as exploiting conjugation $ \sqrt {a} \mapsto \minus{} \sqrt {a}$ a [i]lot.[/i]  One way to convince yourself that this works is that, without analytic structure, square roots are "anonymous":  you only know how a square root behaves based on what it squares to, and so you can't tell whether you're working with the positive or negative square root.  Thus if, for example,

$ \sqrt {2} \plus{} \sqrt {3} \equal{} r$

is true for some nonzero rational $ r$, it's true "abstractly" - without reference to the embedding of $ \sqrt {2}, \sqrt {3}$ into $ \mathbb{R}$ - and you've broken the symmetry:  symmetry implies that the statements

$ \sqrt {2} \minus{} \sqrt {3} \equal{} r$
$ \minus{} \sqrt {2} \plus{} \sqrt {3} \equal{} r$
$ \minus{} \sqrt {2} \minus{} \sqrt {3} \equal{} r$

must all also be true, which is absurd.  The first proof also takes advantage of this fact, but in a different way:  if the above is true "abstractly," it's also true $ \bmod p$ for any $ p$ not dividing the denominator of $ r$ (and since $ r$ is an algebraic integer by some standard results it must be an integer anyway), and this also leads to a contradiction (here, $ \bmod 7$).

By "complexity" (looking at the sum of all knowledge needed to understand the statements and their proofs): QR < Galois < Kummer (essentially Hilbert 90) < Dirichlet.
In fact, the idea behind the two middle proofs are indeed symmetry/invariants, whereas the two outer ones represent more fundamental structures of arithmetic /primes (and therefor are harder to generalize).
There were other proofs posted, one of them elementarizing the Galois one, another one doing the same with the Kummer theory approach. I prefer the latter because it easily generalizes to arbitrary roots, not only square roots.

And all together, this is a good example of the strong intervenings between Galois theory, Number Theory, Analysis and (but subtle) cohomology (as in Hilbert 90).

Edit (comment to t0r's edit):
What lies behind this are the different valuations on the rationals: The archimedean one, knowing "positiveness" and "modulus", and the non-archimedean ones, knowing "$ \mod p$" and "prime-like structures".


let $ A,B \in \mathcal{M}_n(\mathcal{R})$. $ det(A\plus{}B)$ and $ det(A\minus{}B)$ distinct 0.
let $ M\equal{}\left[\begin{array}{cc} A & B\\B & A \end{array}\right]$
prove that $ detM$ distinct 0

What about $ \det(M)\equal{}\det(A\plus{}B)\det(A\minus{}B)$?


(1)
Rewrite \[
E(a,b,c) = \sum {(a + b - c)(a - b)^2 }
\]
Then the given inequality is equivalent to
\[
\sum {\left( {a^2  + b^2  - c^2 } \right)(a - b)^2 }  \geqslant 0
\]
WLOG assume $ a \ge b \ge c$, if $b^2+c^2 \ge a^2$ it's done, else $a-c \ge b-c$, so
\[
\begin{gathered}
  \sum {\left( {a^2  + b^2  - c^2 } \right)(a - b)^2 }  \hfill \\
   = \left( {a^2  + b^2  - c^2 } \right)(a - b)^2  + \left( {b^2  + c^2  - a^2 } \right)(b - c)^2  + \left( {c^2  + a^2  - b^2 } \right)(c - a)^2  \hfill \\
   \geqslant \left( {a^2  + b^2  - c^2 } \right)(a - b)^2  + \left( {b^2  + c^2  - a^2  + c^2  + a^2  - b^2 } \right)(b - c)^2  \hfill \\
   = \left( {a^2  + b^2  - c^2 } \right)(a - b)^2  + 2c^2 (b - c)^2  \geqslant 0 \hfill \\
\end{gathered}
\]

(2) Given inequality is equivalent to
\[
\sum {\left( {a^2 b + a^2 c + b^2 a + b^2 c - c^2 a - c^2 b} \right)(a - b)^2 }  \geqslant 0
\]
WLOG assume $a \ge b \ge c$, if \[
b^2 c + b^2 a + c^2 a + c^2 b - a^2 b - a^2 c \geqslant 0
\]
then we are done, else
\[
\begin{gathered}
  \sum {\left( {a^2 b + a^2 c + b^2 a + b^2 c - c^2 a - c^2 b} \right)(a - b)^2 }  \hfill \\
   = \left( {a^2 b + a^2 c + b^2 a + b^2 c - c^2 a - c^2 b} \right)(a - b)^2  + \left( {b^2 c + b^2 a + c^2 a + c^2 b - a^2 b - a^2 c} \right)(b - c)^2  \hfill \\
   + \left( {a^2 b + a^2 c + c^2 a + c^2 b - b^2 a - b^2 c} \right)(a - c)^2  \hfill \\
   \geqslant \left( {a^2 b + a^2 c + b^2 a + b^2 c - c^2 a - c^2 b} \right)(a - b)^2  +  \hfill \\
  \left( {b^2 c + b^2 a + c^2 a + c^2 b - a^2 b - a^2 c + a^2 b + a^2 c + c^2 a + c^2 b - b^2 a - b^2 c} \right)(a - c)^2  \hfill \\
   = \left( {a^2 b + a^2 c + b^2 a + b^2 c - c^2 a - c^2 b} \right)(a - b)^2  + 2c^2 (a + b)(a - c)^2  \hfill \\
   \geqslant 0 \hfill \\
   \hfill \\
  b^2 c + b^2 a + c^2 a + c^2 b - a^2 b - a^2 c \geqslant 0 \hfill \\
\end{gathered}
\]

[quote="Vasc"]Let $a,b,c \ge 0$ and $E(a,b,c)=\sum a(a-b)(a-c)$. Prove that

(I)   $(a+b+c)E(a,b,c) \ge ab(a-b)^2+bc(b-c)^2+ca(c-a)^2$;

(II) $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})E(a,b,c) \ge a^2+b^2+c^2 -ab-bc-ca$.[/quote]

Some alternative proofs:

[color=blue][b]To inequality (I)[/b]:[/color]

Actually, inequality [b](I)[/b] holds for arbitrary reals a, b, c (not necessarily positive).

In fact, a simple calculation shows that

$\left(a+b+c\right)E\left(a,b,c\right)-\left(bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2\right)$
$=a^2\left(a-b\right)\left(a-c\right)+b^2\left(b-c\right)\left(b-a\right)+c^2\left(c-a\right)\left(c-b\right)$,

but http://www.mathlinks.ro/Forum/viewtopic.php?t=41284 post #4 Theorem 3 [b](b)[/b], applied to p = 2, yields $a^2\left(a-b\right)\left(a-c\right)+b^2\left(b-c\right)\left(b-a\right)+c^2\left(c-a\right)\left(c-b\right)\geq 0$. Thus, we have $\left(a+b+c\right)E\left(a,b,c\right)-\left(bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2\right)\geq 0$, so that

$\left(a+b+c\right)E\left(a,b,c\right)\geq bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2$,

and inequality [b](I)[/b] is proven. $\blacksquare$

[color=blue][b]To inequality (II)[/b]:[/color]

Inequality [b](II)[/b] indeed requires the nonnegativity of a, b, c.

We will derive [b](II)[/b] from [b](I)[/b]. In fact, we will show the stronger inequality

[b](III)[/b] $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2\right)\geq\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)$.

Using this inequality, from [b](I)[/b] it will follow that

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)E\left(a,b,c\right)\geq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2\right)$
$\geq\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)$;

now, division by a + b + c will transform this into

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)E\left(a,b,c\right)\geq a^2+b^2+c^2-bc-ca-ab$,

and inequality [b](II)[/b] will be proven.

Remains to establish [b](III)[/b]. This can be done in a rather tricky way by defining $x=\frac{b-c}{a}$, $y=\frac{c-a}{b}$ and $z=\frac{a-b}{c}$. Then,

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2\right)-\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)$
$=abc\left(x^2+y^2+z^2-yz-zx-xy\right)$.

Using the well-known inequality $x^2+y^2+z^2\geq yz+zx+xy$ for any three reals x, y, z, we get $x^2+y^2+z^2-yz-zx-xy\geq 0$, and thus

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(bc\left(b-c\right)^2+ca\left(c-a\right)^2+ab\left(a-b\right)^2\right)-\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)$
$\geq 0$,

so that the inequality [b](III)[/b] is proven. $\blacksquare$

  darij


Let $ G$ is a wheel graph with $ n$ vertex,how many spaning tree is there for $ G$?
thanks in advance

[hide][b]Hint[/b]:the answer is $ L_{2n} \minus{}2$ where $ L_n$ is the $ n$th locas number.[/hide]

Does anyone have idea to calculate the number of spanning trees in wheel with $n$ vertices?


geometry
circumcircle
geometric transformation
angle bisector
perpendicular bisector
geometry proposed

It is given a triangle and m, h, l are the median, height and angle bisector from a one and the same vetrex of the triangle (Let say C).
Express R by m, h and l. (R is circumradii of the triangle).

The construction of the triangle ABC given these 3 segments is the object of problem number 235 of the book of Luis Lopes "Manuel de construction de triangles"(in french), which is possible download for free from the russian site
http://ilib.mirror1.mccme.ru

Although I guess the question is a closed formula for R in terms of these lenghts, maybe giving the construction can help to find the closed form:
Triangle ABC will be constructed if we were able of construct triangle AH_aS_a and the point O (H_a, foot of altitude; S_a, foot of the bisector of angle A).It is well know that O belong to the perpendicular bisector of BC and to the line AE, being E the point diametrally opposed to A in the circumcircle.
i) Draw the triangle AH_aS_a, where angle AH_aS_a is 90º, AH_a= h, AS_a=w (the bisector of angle A). Let be r the line which pass trhough the points H_a and S_a. On the line r, draw the point M_a (will be the foot of median, mid point of BC) such that AM_a = m (the median). Draw the perpendicular( s, say) by M_a to the line r.
ii) Let  alpha = angle H_aAS_a. Draw the line t in a such way that if E belong to t, then angle EAS_a = alpha. The intersection of the lines s and t is the circumcenter O. Knowing A and O, draw the circumcircle k. Then B and C are the intersections of r and k.
(This is the translation of the construction given in the book of Luis Lopes I quoted before).

The question is to find the formula for R expressed by l, h, m.

With the help of construction is not hard to fin the closed formula.
Attached in pdf my solution

Excellent solution - just to know it is not mine problem and it is old one ... there is at least one more solution.

Thanks. I would like see another solution, if possible.

I'm very busy now, but when I've more time I'll post it.


geometry
trigonometry
inequalities
triangle inequality

In a triangle, if one side is 10 inches and the other is 11 inches, how long
is the other side of the triangle? The area of the triangle is 210.

Just plug into heron's formula

or you could use 1/2absinC to find the angle in between then use the law of cosines

To use Heron's formula, don't I need to know all three sides?

There is a missing side.

How do I find it?

you have the area
just plug it in
just write s in terms of the 3 sides and plug it in

Just because i dont want to make a new topic...

1. Use herons,

[hide]HERONS:
$A=\sqrt{(S)(S-a)(S-b)(S-c)}$, when $S=\frac{a+b+c}{2}$

Now plug in...
$a=10=\frac{20}{2}$
$b=11=\frac{22}{2}$
$c=x=\frac{2x}{2}$
So $S=\frac{21+x}{2}$

Then plug in and whatnot...

im definatley not doing it.
[/hide]

2. My question:

A triangle has one side of length 10, and another side of length 7.  What is the maximum possible area of the triangle?

[quote="Walk Around The River"]
2. My question:

A triangle has one side of length 10, and another side of length 7.  What is the maximum possible area of the triangle?[/quote]

Are you sure that it's not integer side lengths? If it is, you use triangle inequality, and then you find the answer by Heron's formula.

[quote="DiscreetFourierTransform"]Question:
Well, we have that the area is equal to $0.5(10)(7)\sin\theta$, which is maximized when $\sin\theta$ is. (right?)
Then, we know that $\sin$ has a maximum at 1. (we need to show this is achievable)
This is when the angle is 90 degrees, or we have a right triangle with two legs, one of length 7, and one of length 10?
So we have the hypotenuse = $\sqrt{149}$?
The area is then 35?

Is that how we'd do that one?[/quote]

Yes, that is correct, you dont really need to find the hypotenuse.  The trick is knowing that you want to maximize $\sin\theta$, which occurs when $\theta=90$.

The idea was to show that a right triangle could be formed with those lengths (satisfying the triangle ineq.) which means that the Maximum of sin is achievable, so I think I did need to show that the hypotenuse satisfies the triangle ineq.


induction
LaTeX
combinatorics proposed
combinatorics

There is a row of $ n\geq 2$ lightbulbs, and each of them has one of the two states "on" and "off". Every second, all lightbulbs simultaneously change their state in the following way:

- If a lightbulb $ X$ and its neighbours (the first and the last lightbulb have only one neighbour) are in the same state, then $ X$ turns off (if it is already turned off, then it remains turned off);
- in every other case, $ X$ turns on (if it is already turned on, it remains turned on).

Initially, all lightbulbs are turned off except for the first (i. e. leftmost) one, which is turned on.

[b]a)[/b] Prove that there are infinitely many integers $ n$ such that there is a time when all the lightbulbs are turned off.
[b]b)[/b] Are there odd integers $ n$ such that there is a time when all the lightbulbs are turned off?
[b]c)[/b] If $ n=1040$, is there a time when all the lightbulbs are turned off?

Seems to me that 2n bulbs all turn off after 2n iterations and odd numbers hit some sort of oscialltory state.Will post a proof soon :D

[quote]Seems to me that 2n bulbs all turn off after 2n iterations[/quote]

Oops,that was wrong,6 does not work.

I solved part a,2^n bulbs always become all off:

Describe the state at any second with a binary word with 1 at position k meaning the k th bulb is on and 0 otherwise.

From the rules specified:111 and 000 result in #0# {# depends on another triple) and any other sequence produces #1# in the next state.

Note that 111...111 in the next second becomes 000...000

Start by defining
State 1 :100...000(2^n -1) zeroes.
Claim:For the above state 1,
        state 2^n is :111...111(2^n times)

Lemma 1:The bulb at position k turns on for the first time exactly when state k is attained,
                 {This is an easy induction.}

I proceed by induction on n,the exponent of 2.
First note that the case n=1 is true (2 bulbs do result in the state: 11 in 2 steps)

Suppose case n were true,consider case n+1.
State 1:1000...000 (2^(n+1) zeroes).

Now note that the value of the first (2^n)-1 characters at
the state (2^n)-1  is determined  only by the first 2^n characters of the word by Lemma 1 as all the characters from postion 2^n to 2^(n+1) are zero.
Also note that after (2^n)-1 steps ,(2^n)-1 th position is 1(from Lemma 1).So the string looks like
this: .......10(2^n th position)00.......Thus  a ....1000..... results in a .... #10.....
So we are assured that the 2^n th place is 1 after 2^n steps immaterial of the trailing zeroes.
This combined with with our inductive assumption gives us

State 2^n: 111....11  (2^n ones)000...000(2^n zeroes).
Hence 2^n + 1:  000000...1 || 1...000000  ( (2^n)-1 zeroes on each side)

Note that this consists of 2 strings both corresponding to state 1 for case n,joined together with the first string reversed,ie 000000....01 and 100000...000 ( (2^n)-1 zeroes in each).
We now only have to enumerate cases to show each half of the state:(2^n)+1 reaches 11...11||11...11.
to show this we only have to show that the value at 2^n and (2^n)+1 th positions do not affect each other:
Consider: 01||10 and just  ||10 ,
                11||11 and just ||11,
                00||00 and ||00,
                finally 10||01 and ||01.
It is clear that position (2^n)+1 is unaffected by its same value occuring at position 2^n and vice versa.
Hence,each half of state (2^n)+1 of the case n+1 behave exactly as state 1 of case n.
Thus state 2^(n+1) is :11111...1111(2^(n+1) times).
Q.E.D

Hope that makes some sense.
Cheers
Bhargav

Ps:I still haven't become good with Latex so forgive me for that[/img]

[quote="nkht-tk14"]I think only $n=2^{k}$ such that exist second with n bulbs all turn off[/quote]

Proof, please  :)

If n even then exist k ;p with p be odd number
Write [tex]n=2^{k}.p[tex]
Prove by induction folowing n
My English is bad .
I wil post my solution in next time

Well, can anybody solve part (b) and (c)?

Can anyone found all n for which there exist a second when all the lightbulbs are turned off?

Our team leaders weren't able to do it  :P

[quote="mitchan88"]
Our team leaders weren't able to do it  :P[/quote]

Oh well, then nobody is likely to do it  :rotfl:

Okay, now for (b).

First it's easy to see that the step before all become $0$, it must be that all are $1$.

Now we can show easily by induction that each step from step 2 onward, each string of $0$ (or $1$, respectively) will consist of an even number of $0$ (or $1$, respectively), except if a string of $0$ is leftmost (or rightmost), then it might consist of either even or odd number.

So, of course it will never become all $1$ (because it is an odd string of $1$)

So we're now left with (c), and what's so special with the number $1040$...  :wink:


[hide="Solution"]
Note that if $ A\equal{} \{ a_1, \; a_2, \; \ldots a_m \}$ is a subset of $ S$, then $ A' \equal{} \{2009 \minus{} a_1, \; 2009 \minus{} a_2, \; \ldots 2009 \minus{} a_m \}$ is also a subset of $ S$. Let $ a_j \equal{} m(A)$. Then, $ 2009 \minus{} a_j \equal{} m(A')$, and the average of $ m(A)$ and $ m(A')$ is $ \frac{a_j \plus{} 2009 \minus{} a_j}{2} \equal{} \frac{2009}{2}$. Performing this average for each pair of subsets gives the global average median to be $ \frac{2009}{2}$ as well.
[/hide]

[quote="worthawholebean"]([b]3[/b]) Let $ S \equal{} \{1,2,\ldots,2008\}$. For any nonempty subset $ A\in S$, define $ m(A)$ to be the median of $ A$ (when $ A$ has an even number of elements, $ m(A)$ is the average of the middle two elements). Determine the average of $ m(A)$, when $ A$ is taken over all nonempty subsets of $ S$.[/quote]

[hide="alternatively?"]
I'm guessing you could do something REALLY smart involving symmetry to get $ \frac{2009}{2}$ fast

[/hide]

[quote="azjps"][quote="CircleSquared"]
[hide="alternatively?"]
I'm guessing you could do something REALLY smart involving symmetry to get $ \frac {2009}{2}$ fast

[/hide][/quote]

That's what The Zuton Force did, except he also showed it rigorously [is that a word? I'm not sure any more..].[/quote]

hmm, didn't notice.  i must see that.
and yes rigorously is a word

actually... he didn't consider all the cases where A=A'. Of course, in that case the median of A must be 2009/2, so the answer remains the same, but the proof is unrigorous.

what did it mean by the median of m(A)
if the set is (1,2,3) then the median is 2
and if it is (2,3,1) then 3 so if a set consists os odd number of elements then the sum + those who contain even number of medians??? how can i solve this??/ can anyone help plz  :o

Sets here are ordered sets. The median is not the number in the middle position, but the number with the median value. If there are two such numbers, take the arithmetic mean of the two.

Both of the sets you presented, the median value is 2.

Of course, TZF meant "unordered."  The two sets you wrote are the same set.


algebra
polynomial
real analysis
real analysis unsolved

Find two independent series solutions about $ x \equal{} 0$ of
\[ y'' \minus{} 2xy' \plus{} \lambda y \equal{} 0,\]for a constant $ \lambda$. Show that for $ \lambda \equal{} 2n$, with $ n$ a positive integer, the solutions are polynomials of degree $ n$ (These are the Hermite polynomials relevant for the solution of the simple harmonic oscillator in quantum mechanics).

[b]My progress:[/b]

Edit: $ x \equal{} 0$ is ordinary.

So we try $ y \equal{} \sum_{n \equal{} 0}^{\infty} a_nx^n$ and we get $ \sum_{n \equal{} 0}^{\infty} a_n(n(n \minus{} 1)x^{n \minus{} 2} \minus{} 2nx^n \plus{} \lambda x^n) \equal{} 0$. Considering the coefficient of $ x^{n \minus{} 2}$, we have $ n(n \minus{} 1)a_n \plus{} [\lambda \minus{} 2(n \minus{} 2)]a_{n \minus{} 2} \equal{} 0$.

This implies that $ n(n \minus{} 1)a_n \equal{} (2n \minus{} 4 \minus{} \lambda)a_{n \minus{} 2}$.

For $ n \equal{} 0$, we get $ 0 \cdot a_0 \equal{} 0 \Rightarrow a_0$ is arbitrary.

For $ n \equal{} 1$, we get $ 0 \cdot a_1 \equal{} 0 \Rightarrow a_1$ is arbitrary.

For $ n > 1$, $ a_n \equal{} \frac {2n \minus{} 4 \minus{} \lambda}{n(n \minus{} 1)}a_{n \minus{} 2}$.

From here, how do we find the solutions? In a similar problem the solution established that $ a_n \equal{} 0$ for odd $ n$ greater than $ 1$, while the even $ a_n$ formed a recurrence relation. Is this true here?

For some reason I can't edit the post, so there's a correction - instead of "the solutions are polynomials of degree n" it should be "one of the solutions is a polynomial of degree n"

Note: $ x\equal{}0$ is an ordinary point, not a singular point. This differential equation has no singular points anywhere.

I don't know what you think the definition of singular point is.

Just edited that, the first sentence was actually for another post. The rest of my solution (up to where I got stuck) is still valid, right?

When the question says "for $ \lambda \equal{} 2n$", it is best not to choose "$ n$" also as your index of summation.  Too confusing.

Thanks, we can change that to "show for $ \lambda \equal{} 2k$, with $ k$ a positive integer, one of the solutions is a polynomial of degree $ k$ - my working so far hasn't failed due to that oversight.

More importantly, I want to know how do we proceed from the recurrence relation? We have $ a_{n}\equal{}\frac{2n\minus{}4\minus{}\lambda}{n(n\minus{}1)}a_{n\minus{}2}$


function
number theory proposed
number theory

Find the smallest natural number $ n$ such that $ 3\tau\left(343n\right)\minus{}2\tau\left(41n\right)\equal{}3\tau\left(n\right)$

[b]No such $ n$ exists[/b].

[i]Proof:[/i] First, note $ 343\equal{}7^3$ and $ 7,41$ are prime. Set $ n\equal{}41^a\cdot 7^b\cdot m$ with $ a,b\in\mathbb{N}_0$ and $ m\in\mathbb{N}$ such that $ (m,7\cdot 41)\equal{}1$ holds. The equation transforms after division by $ \tau(n)\not \equal{}0$ and substitution of $ \tau(p^t)\equal{}\frac{p^{t\plus{}1}\minus{}1}{p\minus{}1}$ for primes $ p$ into $ 3\cdot \frac{7^{4\plus{}b}\minus{}1}{7^{1\plus{}b}\minus{}1}\minus{}2\cdot\frac{41^{2\plus{}a}\minus{}1}{41^{1\plus{}a}\minus{}1}\equal{}3$. This yields the contradiction $ 3\cdot 7^3<3\cdot \frac{7^{4\plus{}b}\minus{}1}{7^{1\plus{}b}\minus{}1}\equal{}3\plus{}2\cdot \frac{41^{2\plus{}a}\minus{}1}{41^{1\plus{}a}\minus{}1}<3\plus{}2\cdot 43$

[quote="SAPOSTO"]Hmm. There has been misunderstooding. You've used $ \tau n$ for sum of divisors. In Bulgaria $ \tau n$ denotes the number of divisors of n and $ \sigma n$ denotes the sum of it's divisors. As I can see from your Wikipedia it's conversely. So assume $ \tau n$ denotes the number of it's divisors and then solve the problem.[/quote]
Where does wikipedia state that $ \tau$ should be the sum of divisors? I've never seen (before hjbrasch's post) a use of $ \tau$ as the sum of divisors :) I've always used $ \tau$ as the number of divisors and $ \sigma$ as the sum of divisors. (It's also more intuitive, since $ \sigma$ usually has something to do with sums)

Okay, using your $ \tau$-definition and making the same kind of substitutions and transformations we get $ 3\cdot\frac{4\plus{}b}{1\plus{}b}\minus{}2\cdot\frac{2\plus{}a}{1\plus{}a}\equal{}3$, which is equivalent to $ 9\equal{}(7\minus{}2b)(a\plus{}2)$. This yields $ a\equal{}1,b\equal{}2$ or $ a\equal{}7,b\equal{}3$, i.e. all solutions of the equation are given by the set $ \{41\cdot 7^2\cdot m,41^7\cdot 7^3\cdot m|m\in\mathbb{N},(m,7\cdot 41)\equal{}1\}$. Its smallest element is obviously $ 2009\equal{}41\cdot 7^2$

@[b]Mathias_DK[/b]I haven't read it correctly. See [url=http://en.wikipedia.org/wiki/Divisor_function]here[/url]. $ \tau\left(n\right)\equal{}\sigma_{0}\left(n\right)$
@[b]hjbrasch[/b]That's the right answer [b]hjbrasch[/b].


A geometric figure has four vertices, A, B, C, and D, and the following distances between the vertices: |AB|=1, |AC|=1, ||AD|=11/20, |BC|=1, |BD|=11/20, and |CD|=11/20  This figure is a.....

If you know the answer please don't post it.

Quote:If you know the answer please don't post it.

I assume you mean if we have previously encountered the problem?

If thats what you mean, then I would say that it is a [hide]kite[/hide].

Right right... Thats what I mean, but you're wrong!

Osiris wrote:Wild guess: [hide]It's a tetrahedron with an equilateral base.[/hide] Seems to work though.

Oh yeah, forgot about [hide]3-d .. I was thinking of an equilateral triangle with the fourth vertex somewhere in the middle, which would be the view from above of a tetrahedrom  [/hide][/hide]

Im going to go with [hide]It isnt anything[/hide].

Ha! We need moer excplanations.  There are many wrong answers.

[hide]ABC is equilateral. Since D is equal distance from A,B,C, we would have to form a tetrahedron with D being on the line perpendicular to ABC through its centroid. But the minimum distance (achieved when D is the centroid) is actually greater than 11/20. So this thing cannot exist.[/hide]

How do you know its not a tetrahedron tetrahedr0n? :-D

This is completely random, but have you ever noticed that [tex]\frac{1}{\sqrt 3}\approx\gamma[/tex]?

Osiris wrote:Ah, of course...

[hide]Construct an equilateral triangle of side length 1. Then the distance between any vertex and the centre of the figure is

[tex]{{2}\over{3}}\sqrt{1^2 - \left({{1}\over{2}}\right)^2} \approx 0.577 > {{11}\over{20}}~.[/tex][/hide]

Heh, I should have followed up on my original thought.

What if it was 3d?  Then the lengths to the center could be any height!  it could be 1000!

But it cannot be less than the length from a vertex to its centre.

Um, they're saying that the distances to the center are too [b]small[/b] to reach together, even in 3D.

Oh... hmm maybe I didn't read that carefully enough. :-D


Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?

$ \text{(A)}\ 2\pi/3 \qquad
\text{(B)}\ 2\pi \qquad
\text{(C)}\ 5\pi/2 \qquad
\text{(D)}\ 8\pi/3 \qquad
\text{(E)}\ 3\pi$

he can reach a 360-120=240 degree sector of a 1 meter radius circle. thats 2/3 of the area of the entire circle so its $ 1^2\pi*2/3\equal{}2\pi/3$

The dog has a reach of 240 degrees, which is two thirds of a circle, around one vertex of the hexagon, and the radius of the circle is 2. So he has an area of $ 4\pi*2/3 \equal{} 8\pi/3$. Then he can reach around the other vertex with another radius 1 on both sides because the side length was only 1, but the radius is 2. Those two smaller circles are 60 degree sectors, so $ 2(1^2\pi * 1/6) \equal{} \pi/3$. Add the areas and you get $ 8\pi/3 \plus{} \pi/3 \equal{} 9\pi/3 \equal{} 3\pi$. $ E$

I agree with boarder3922， E is the right answer since 240/360*2^2pi+2*60/360*pi=3pi, which is E

huh wait @above doesn't that mean PMWC copied off of maa?

[quote=sugar_rush]What???????!!!!!!

The [url=https://artofproblemsolving.com/wiki/index.php/2005_PMWC_Problems/Problem_T7]2005 PMWC Problem T7[/url] is the [b]exact same[/b] problem as this one??!![/quote]

is one metre now one yard yay


probability
combinatorics open
combinatorics

In the eximination there are 30 questions written on the peace of а paper. One studen doesn't know answer for the two of the questions and another for the three. For which of the students it is better to get the paper with the questions first?

Note: papers all on the desk, and students can't see the side on which there is written the problems.

The order of students does not matter. The probability for each student to get a "bad" question is the same and independent on the order in which they get their questions. For the first student this probability is 2/30, for the second - 3/30.


geometry
IMO Shortlist
right angle
circles

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Let $G=NE\cap S_1,\ X=NE\cap MK,\ Y=KF\cap MN$. The quadrilateral $XNKY$ is ccylic because $\angle MXN=\angle MYK=\frac \pi 2$, so $\angle MNE=\angle FKM\ (*)$. If we show that the triangles $MNE,FKM$ are similar, then we would have $\angle NMK=\pi-\angle YMK=\frac \pi 2+\angle MKY=\frac \pi 2+\angle NME+\angle KMF$, which is exactly what we want.

We thus need to show that $\frac{MN}{NE}=\frac{FK}{KM}\ (**)$, which, together with $(*)$, proves the similarity of $MNE,FKM$.

$(**)\iff NE\cdot KF=BN\cdot BK\ (***)$, because $BN=MK,\ BK=MN$. Since all triangles $BCD$ are similar, no matter which line $CD$ we choose, there is a spiral similarity $\mathcal S$ cantered at $B$ which maps $D\to C$. $\mathcal S$ maps $BD\to BC$, so if $K'=\mathcal S(K)$, then $K'\in BC$ and $\frac{BK'}{K'C}=\frac{BK}{KC}=\frac{CN}{NB}$, meaning that $K'$ is the symmetric of $N$ wrt the perpendicular bisector of $BC$.

We then have $NE\cdot NG=BN\cdot NC=BN\cdot BK'$, and this proves $(***)$ because $\frac{K'F'}{KF}=\frac{BK'}{BK}$ and $K'F'=NG$, so $\frac{NG}{KF}=\frac{BK'}{BK}$.

[hide="Alternate Solution"]
We consider the point $ E'$ so that $ \angle FAE' \equal{} 90$.  Let the foot of the perpendicular from $ E'$ to $ BC$ be $ K'$ and the parallel from $ K'$ to $ BD$ meet $ CD$ at $ M'$.  Now, we have that\[ \angle NDF \equal{} \angle BAF \equal{} 90 \minus{} \angle BAE' \equal{} 90 \minus{} \angle E'CK' \equal{} \angle K'E'C\]and $ \angle E'K'C \equal{} 90 \equal{} \angle FND$.  Thus, $ \triangle DNF\sim \triangle E'K'C$, so $ \frac {DN}{E'K'} \equal{} \frac {NF}{K'C}\implies K'C\cdot ND \equal{} E'K'\cdot NF$.  Similarly, we may note that $ \triangle K'BE'\sim \triangle NFB$, meaning that $ \frac {K'E'}{NB} \equal{} \frac {K'B}{NF}$, so we may say that $ K'C\cdot ND \equal{} K'E'\cdot NF \equal{} NB\cdot K'B$.  It follows that $ \frac {M'C}{M'D} \equal{} \frac {K'C}{K'B} \equal{} \frac {NB}{ND}$, implying that $ M'N\parallel BC$, so $ M' \equal{} M$, meaning that $ K' \equal{} K$ and $ E \equal{} E'$.  Thus, $ \angle EAF \equal{} 90$.  From the above ratios, we have that $ EK\cdot NF \equal{} BN\cdot BK$, so $ EK\cdot NF \equal{} MN\cdot MK$ from parallelogram $ MKBN$.  Thus, $ \frac {EK}{MK} \equal{} \frac {MN}{NF}$ and \[ \angle EKM \equal{} 90 \plus{} \angle MKB \equal{} 90 \plus{} \angle MNB \equal{} \angle MNF\] implying that $ \triangle MKE\sim \triangle FNM$.  Thus,\[ \angle EMF \equal{} \angle KMN \minus{} (\angle KME \plus{} \angle NMF) \equal{} \angle KBN \minus{} (\angle EMK \plus{} \angle KEM)\]Noting that\[ \angle EMK \plus{} \angle KEM \equal{} 180 \minus{} \angle EKM \equal{} 180 \minus{} (90 \plus{} 180 \minus{} \angle CBD) \equal{} \angle KBN \minus{} 90\]we have that $ \angle EMF \equal{} 90$, as desired. [/hide]

[quote="77ant"]Strangely, $ \angle EMF\neq 90^{\circ}$ in my figure.
Why does it not hold in my figure? please help me  :([/quote]
I think that $ E$ has to be on the major arc $ CB$.

Dear Mathlinkers,
this problem appears also many time on Mathlinks.
I research a synthetic proof without calculation or transformations...
Is there one?
Sincerely
Jean-Louis

Hmm.... the problem seems to require $A$ to lie in between $C$ and $D$; I think [b]77ant[/b]'s diagram satisfies all the conditions.

Here's a more straightforward way to establish (***) from [b]grobber[/b]'s solution...

The relation is equivalent to $\triangle{ENB}\sim\triangle{BKF}$, or $\angle{EBN}+\angle{FBK}=90^\circ$. By the Law of Sines on $\triangle{BDF}$, we have
\begin{align*}
\frac{BD}{\sin\angle{BAD}}=\frac{BF}{\sin\angle{BDF}}
&=\frac{BF}{\sin(\angle{BAD}-\angle{FBK})}\\
&=\frac{BK}{\cos\angle{FBK}\sin(\angle{BAD}-\angle{FBK})}.
\end{align*}By symmetry,
\begin{align*}
\frac{BC}{\sin\angle{BAD}}=\frac{BC}{\sin\angle{BAC}}
&=\frac{BN}{\cos\angle{EBN}\sin(\angle{BAC}-\angle{EBN})}\\
&=\frac{BN}{\cos\angle{EBN}\sin(\angle{BAD}+\angle{EBN})}.
\end{align*}Thus $NM\|BD$ gives us
\[1=\frac{CN}{CB}+\frac{BN}{BC}=\frac{NM}{BD}+\frac{BN}{BC}=\frac{BK}{BD}+\frac{BN}{BC},\]whence plugging in and simplifying yields
\[\cos(\angle{FBK}+\angle{EBN})\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=0.\]But
\[0=\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=\sin(180^\circ-\angle{BCE}-\angle{KBF})\]is impossible since $0<\angle{BCE},\angle{KBF}<90^\circ$. Thus $\angle{FBK}+\angle{EBN}=90^\circ$, as desired.

All lengths and angles in this proof will be directed.

[b]Lemma 1[/b]: $\triangle CEN \sim \triangle FDK$.
[i]Proof:[/i] Let $FK$ intersect $S_2$ at $F'$. Note that $F'K \perp BD$, $\frac{DK}{KB} = \frac{DM}{MC} = \frac{BN}{NC}$, and $\angle BF'D = \angle BAD = \angle BEC$, so $\triangle ECB \sim \triangle F'BD$. Hence, $\angle ECN = \angle ECB = \angle F'BD = \angle F'FD = \angle KFD$, and $\angle FKD = \angle CNE = 90^{\circ}$, so $\triangle CEN \sim \triangle FDK$.

[b]Lemma 2[/b]: $\frac{DM}{MC} \cdot \frac{CE}{EB} \cdot \frac{BF}{FD} = 1$.
[i]Proof:[/i] Let $\theta = \angle CAB$ and $\alpha = \angle NEC$. It is easy to see that $\angle ECB = 90^{\circ} - \alpha$, $\angle BEN = 180^{\circ} - \theta - \alpha$, $\angle CBE = \theta + \alpha - 90^{\circ}$, $\angle BDF = \alpha$ (by lemma 1), and $\angle FBD = 180^{\circ} - \theta - \alpha$. Hence,
\begin{align*}
\frac{DM}{MC} \cdot \frac{CE}{EB} \cdot \frac{BD}{FD}
&=  \frac{DM}{MC} \cdot \frac{\sin(\theta + \alpha - 90^{\circ})}{\sin(90^{\circ} - \alpha)} \cdot \frac{\sin \alpha}{\sin(180^{\circ} - \theta - \alpha)}  \\
&= \frac{DM}{MC} \cdot \frac{- \cos(\theta + \alpha) \sin \alpha }{ \sin(\theta + \alpha) \cos \alpha }\\
&= \frac{DM}{MC} \cdot \frac{-\cot(\alpha + \theta)}{\cot \alpha} \\
&= \frac{DM}{MC} \cdot \frac{ \cot(180^{\circ} - \theta - \alpha) }{\cot \alpha} \\
&= \frac{DM}{MC} \cdot \frac{ \cot \angle BEN }{ \cot \angle NEB } \\
&= \frac{DM}{MC} \cdot \frac{CN}{NB} \\
&= 1.
\end{align*}

----------------------------------------------------------------------------------------------------------------------------

We have $\angle EAF = \angle EAB + \angle BAF = \angle ECB + \angle BDF = \angle ECN + \angle KDF = 90^{\circ}$, since $\triangle CEN \sim \triangle FDK$. Hence, it is sufficient to show that $AMFE$ is cyclic.

Consider an inversion about $A$ with radius 1. For any point $X$, let $X'$ denote the image of $X$ under this inversion. We wish to show that $E'$, $F'$, and $M'$ are collinear. $E'$ lies on $B'C'$ and $F'$ lies on $B'D'$, so by Menelaus's theorem, it is sufficient to show
\[ \frac{C'E'}{E'B'} \cdot \frac{B'F'}{F'D'} \cdot \frac{D'M'}{M'C'} = -1. \]
Note that
\begin{align*} \frac{C'E'}{E'B'} = \frac{CE \cdot C'A \cdot E'A }{EB \cdot E'A \cdot B'A} = \frac{CE}{EB} \cdot \frac{BA}{CA} \tag{1} \end{align*}
and
\begin{align*} \frac{B'F'}{F'D'} = \frac{BF \cdot B'A \cdot F'A }{FD \cdot F'A \cdot D'A} = \frac{BF}{FD} \cdot \frac{DA}{BA} \tag{2} \end{align*}
Let $CD$ intersect the line at infinity at $\infty$. We have $(M, \infty; C, D) = \frac{MC}{MD}$. Because cross-ratios are preserved under inversion, we have
\[ \frac{MC}{MD} = (M', A; C', D') = \frac{\frac{M'C'}{M'D'}}{\frac{AC'}{AD'}}, \]
or
\begin{align*} \frac{D'M'}{M'C'} = \frac{DM}{MC} \cdot \frac{AC}{AD}. \tag{3} \end{align*}
Combining (1), (2), (3), and lemma 2, we have
\begin{align*}
\frac{C'E'}{E'B'} \cdot \frac{B'F'}{F'D'} \cdot \frac{D'M'}{M'C'}
&= \left( \frac{CE}{EB} \cdot \frac{BA}{BC} \right) \cdot \left(\frac{BF}{FD} \cdot \frac{DA}{BA} \right) \cdot \left( \frac{DM}{MC} \cdot \frac{AC}{AD} \right) \\
&= \left( \frac{CE}{EB} \cdot \frac{BF}{FD} \cdot \frac{DM}{MC} \right) \cdot \left( \frac{BA}{BC} \cdot \frac{DA}{BA} \cdot \frac{AC}{AD} \right) \\
&= -1,
\end{align*}
as desired.

[hide]We can reconstruct the whole diagram from five points $B,C,D,M,E$. So let's fixed these points.

The existence of point $A$ ensures that $\angle CEB+\angle BFD=\angle BAD+\angle BAC=180^{\circ}$.

So point $F$ is the unique point satisfying: (1) $FK\perp BD$; (2) line $BD$ separates $F$ and $M$; (3) and $\angle CEB+\angle BFD=180^{\circ}$.

Now we assume that $F$ is the point satisfying (1), (2), and $\angle EMF=90^{\circ}$. We will prove that $\angle CEB+\angle BFD=180^{\circ}$.

We have \[\angle EMF=\angle EMN+\angle NMK+\angle KMF=\angle EMN+\angle MNC+\angle KMF=90^{\circ}-\angle MEN+\angle KMF\] so $\angle MEN=\angle KMF$. Similarly we get $\angle NME=\angle KFM$. So $\triangle MEN\sim\triangle FMK$.

Now we have $\frac{CN}{NE}=\frac{CN}{NM}\cdot\frac{NM}{NE}=\frac{MK}{KD}\cdot\frac{KF}{KM}=\frac{KD}{KF}$, so $\triangle CNE\sim\triangle FKD$. We also have $\frac{EN}{NB}=\frac{EN}{NM}\cdot\frac{NM}{NB}=\frac{MK}{KF}\cdot\frac{BK}{KM}=\frac{BK}{KF}$. Hence $\triangle EBN\sim\triangle BFK$. Therefore we conclude that \[\angle ECB=\angle CEN+\angle BEN=\angle FDK+\angle FBK=180^{\circ}-\angle BFD\] as desired.[/hide]

[b]Solution[/b]

Let $X$ be the point such that $E$ and $X$ are on opposite sides of $BC$ and $\triangle{CXB} \sim \triangle{DFB}$. First note that since $C$, $A$ and $D$ are collinear, $\angle{CXB}=\angle{DFB}=\angle{CAB}=180^\circ - \angle{CEB}$. Therefore $XBEC$ is cyclic. Let $X'$ be the point on $S_1$ such that $XX' \| BC$ and let $Y$ and $Y'$ be the projections of $X$ and $X'$, respectively, onto $BC$.

Now note that since the sides of $\triangle{MNC}$ and $\triangle{DKM}$ are parallel, it follows that

\[\frac{CN}{NB}=\frac{BK}{KD}=\frac{BY}{YC}=\frac{CY'}{BY'}\]
since $BXX'C$ is an isosceles trapezoid and $DFBK$, $CXBY$ and $BX'CY'$ are similar. The above equality implies that $Y'=N$ and therefore that $X'$ lies on $EN$. Since $BXX'C$ is an isosceles trapezoid, it follows that $BX=X'C$ and that $EX$ is the reflection of $EX'$ around the bisector of $\angle{CEB}$. Since the orthocenter of $\triangle{CEB}$, which lies on $EX'$ since $EX' \perp BC$, and the circumcenter of $\triangle{CEB}$ are isogonal conjugates, $EX$ is a diameter of $S_1$. This implies that

\[\angle{DBF}=\angle{CBX}=90^\circ - \angle{CBE}=\angle{NEB}\]
Therefore $\triangle{NEB} \sim \triangle{KBF}$ which implies that

\[\frac{EN}{KB} = \frac{BK}{KF} \quad \Rightarrow \quad \frac{EN}{NM}=\frac{MK}{KF}\]
since $MNBK$ is a parallelogram. This also implies that $\angle{MNE}=\angle{MKF}=90^\circ +\angle{CBD}$. Hence $\triangle{MNE} \sim \triangle{FKM}$ which implies that

\[\angle{EMF}=\angle{NMK}+\angle{EMN}+\angle{MFK}=\angle{CBD}+\angle{EMN}+\angle{MEN}=90^\circ\]
Hence $\angle{EMF}=90^\circ$ as desired. $\blacksquare$

Here is my proof. I hope I didn't make mistakes in using directed angles.

[hide="My Solution"]We'll use directed angles modulo $180$.

The key idea is that a spiral similarity $\delta$ about $B$ takes $S_2$ to $S_1$ and segment $BD$ to $BC$. This follows because $\angle BAC= \angle BAD$ implies arcs $BEC$ and $DAB$ have the same measure.

Let the image of $K$ and $F$ under $\delta$ be $K'$ and $F'$. We know $K'$ lies on $BC$, $F'$ lies on $S_1$, and $K'F' \perp BC$. Since
\[
\frac{K'C}{K'B} = \frac{DK}{KB} = \frac{AD}{AC} = \frac{NB}{NC},
\]
we also know $CK' = NB$.

We will show $\triangle MKF\sim \triangle ENM$. First, $\angle MKF = \angle MKB + \angle BKF = \angle BNM + \angle ENB = \angle ENM$. Second, because $\triangle CK'F' \sim \triangle CNE$ (because $CK' = NB$), we have
\begin{align*}
\frac{MK}{NE} &= \frac{NB}{NE} \\
&= \frac{CK'}{NE} \\
&= \frac{K'F'}{CN} \\
&= \frac{KF}{NM},
\end{align*}
so $\triangle MKF\sim \triangle ENM$. The last equality follows because $\delta$ has ratio $\frac{CB}{BD} = \frac{CN}{NM}$.

Now
\begin{align*}
\angle EMF &= \angle EMN + \angle NMK + \angle KMF \\
&= \angle EMN + \angle NMK + \angle NEM \\
&= \angle ENM + \angle NMK \\
&= 90 + \angle BNM + \angle NMK \\
&= 90,
\end{align*}
as desired.

[i]Remark.[/i] With the above understanding it's easy to show $\angle EAF$ as well. We have
\begin{align*}
\angle EAF &= \angle EAB + \angle BAF \\
&= \angle ECB + \angle BDF \\
&= \angle ECB + \angle BCF' \\
&= \angle CF'K' + \angle K'CF' \\
&= 90,
\end{align*}
where the second-to-last equality follows because $\triangle CNE \sim \triangle F'K'C$. [/hide]

[hide="Solution"]If $ABC$ and $DEF$ are distinct line segments in that order such that $AB/BC=DE/DF$ and if we let points $G$ and $H$ be on the perpendiculars from $B$ to $AC$ and $E$ to $DF$ such that $\angle AGC+\angle DHF=180$, then $\triangle BGC\sim\triangle DHE$ and $\triangle AGB\sim\triangle EHF$.

Proof: Consider the point $H'$ on the perpendicular from $E$ to $DF$ such that $\triangle BGC\sim\triangle EAH'$. Then $EH'/BC=DE/BG$, implying $EH'/AB=BC/AB*DE/BG=EF/BG$ as $BC/AB=EF/DE$ from definition. Then $EH'/AB=EF/BG$, and by SAS similarity $\triangle ABG\sim\triangle H'EF$.

This implies that $\angle DH'F+\angle AGC=\angle BAG+\angle BCG+90-\angle BAG+90-\angle BCG=180$ from the similar triangles equal angles, thus $H'=H$ as there is a unique $H$. $\blacksquare$
------

As $MKBN$ is a parallelogram, we have $CN/CB=NM/BD=BK/BD$, implying $CN/NB=BK/BD$ with some ratio manipulating. Also, $\angle BFD+\angle BAD=180,\angle CEB+\angle CNB=180$, and $\angle CAB+\angle BAD=180$, thus $\angle CEB+\angle BFD=180$.

We know apply our lemma to find using the similar triangles $EN/BK=NB/KF$, or using the parallelogram's congruent opposite sides,  $EN/MK=NM/KF$ and as $\angle ENM=90+\angle BNM=90+\angle MKB=\angle MKF$, thus by SAS similarity $\triangle NEM\sim\triangle KMF$ and

\[\angle EMF=180-\angle BNM-\angle EMN-\angle FMK=180-\angle BNM-\angle EMN-\angle MEN=180-\angle BNM-(180-90-\angle BNM)=90\][/hide]

We first prove that $\angle EBC + \angle FBD = 90^{\circ}$, and $\angle ECB + \angle FDB = 90^{\circ}$.
[hide="Proof of the previous assertion"]To do this, note that we can create a $\triangle A'B'C'$, so that there is a point $D'$ on $B'C'$ so that $\triangle A'D'B' \sim \triangle CEB$, and $\triangle A'D'C' \sim \triangle BFD$. So, we just need to prove that if $E'$ is foot of perpendicular from $D'$ to $A'B'$ and $F'$ is foot of perpendicular from $D'$ to $A'C'$, then if $E'F' || B'C'$ is given, then we have $\angle D'A'B' + \angle D'C'A' = 90^{\circ}$.
But we have $90^{\circ} = \angle D'F'A' = \angle D'F'E' + \angle E'F'A' = \angle D'C'A' + \angle D'A'B'$.
[/hide]

Now, this directly implies that $\angle EAF = \angle EAB + \angle BAF = \angle ECB + \angle FDB = 90^{\circ}$. So, we only need to prove that $E,M,A,F$ are concyclic.

Now, let $X = MK \cap EB$. Then $\angle XMA = \angle ACB = \angle AEB = \angle AEX$, hence $M,A,X,E$ are concyclic.

Now, using the fact that $\triangle EBN \sim \triangle BFK$, we get \[\frac{EB}{BF} = \frac{BN}{FK} = \frac{MK}{KF}\] Combining with $\angle EBF = \angle MKF$, we get $\triangle MKF \sim \triangle EBF$.

Thus, $\angle XEF = \angle XMF$, which gives $X,A,M,E,F$ concyclic, and we're done.

Let $E_1$, $F_1$ be the antipodes of $E,F$ with respect to their corresponding circles. It's not hard to see that the spiral similarity centered about $B$ sending $BC\cup S_1$ to $BD\cup S_2$ also sends $E_1$ to $F$ and $E$ to $F_1$. Therefore $A,E_1,F$ are collinear and $\angle EAF=90^\circ$. Now note that
\[
\frac{BN}{CN} = \frac{\cot\angle EBN}{\cot \angle ECN} = \frac{\sin \angle ECN}{\sin \angle EBN}\cdot \frac{\sin \angle CBE_1}{\sin \angle BCE_1} = (E,E_1;C,B)=(AE,AF;CD,AB)
\]
Now $\frac{BN}{CN}=\frac{DM}{CM}$. Consider an inversion about $A$, which maps $E,F$ to points $E',F'$ on lines $B'C'$ and $B'D'$. We want to show that $E',F',$ and $M'$ are collinear. But
\[
(C',D';A,M')=\frac{DM}{CM}=(AE,AF;CD,AB)=(AE',AF';C'D',AB')=(E',F';AB' \cap E'F', C'D' \cap E'F') = (C',D';A,C'D' \cap E'F')
\]
so $C'D'$ meets $E'F'$ at $M'$, as desired.

Probably the easiest G8

Let $EN$ intersect the circumcircle of $BCE$ again at $E'$, $EF'$ be a diameter of the circumcircle of $BCE$, and let $K'$ be the foot of the altitude from $F'$ to $BC$. Note that $BN=CK'$ and $E'N=F'K'$. Note that $\angle BF'C=\angle BAC=180-\angle BAD=\angle BFD$. Since $\frac{BK'}{K'C}=\frac{CN}{NB}=\frac{CM}{MD}=\frac{BK}{KD}$, $BF'CK'$ and $BFDK$ are similar. Now, note that $\frac{EN}{NC}=\frac{BN}{NE'}=\frac{CK'}{K'F'}=\frac{DK}{KF}$, and multiplying both sides by $\frac{BC}{BD}$ we have that $\frac{EN}{NM}=\frac{MK}{KF}$. Since $\angle ENM=\angle MKF=90+\angle CBD$, triangles $NEM$ and $KMF$ are similar. But since $MK\parallel BC\perp EN$ and $MN\parallel BC\perp FK$, $ME\perp MF$ as desired.

Let $EN$ intersect $S_1$ again at $E'$. Note that $\angle CE'B=\angle CAB=\angle DFB$, and $\dfrac{DK}{KB}=\dfrac{BN}{NC}$. This implies that $\triangle DFB\sim \triangle BE'C$, which you can see by fixing $B$ and $C$ on a circle and varying $A$ along an arc; clearly $\dfrac{BH}{HC}$ determines the location of $A$ where $H$ is the foot of the altitude from $A$. Then, $\angle FBD=\angle E'CB=90-\angle CBE\implies \dfrac{FK}{KB}=\dfrac{BN}{NE}\implies \dfrac{FK}{MN}=\dfrac{KM}{NE}$. Since $\angle FKM=90+\angle DKM=90+\angle MNC=\angle MNE$, so $\triangle ENM\sim \triangle MKE$. Then $\angle FME=\angle FMK+\angle KMN+\angle NME=\angle FMK+\angle DKM+\angle NFK=180-\angle FND=90$.

The problem itself is not hard. However, we must point out that the claim is true only when $A$ lies on segment $CD$. Otherwise, either $E$ or $F$ should lie on the arc containing $A$ to keep the claim true.

[hide=Solution][quote=ISL 2002 G8]Let two circles $S_{1}$ and $S_{2}$ meet at the points $A$ and $B$. A line through $A$ meets $S_{1}$ again at $C$ and $S_{2}$ again at $D$. Let $M$, $N$, $K$ be three points on the line segments $CD$, $BC$, $BD$ respectively, with $MN$ parallel to $BD$ and $MK$ parallel to $BC$. Let $E$ and $F$ be points on those arcs $BC$ of $S_{1}$ and $BD$ of $S_{2}$ respectively that do not contain $A$. Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$ prove that $\angle EMF=90^{\circ}$.[/quote]
[b][color=#000]Solution:[/color][/b] Extend $EA$ to intersect $S_2$ at $Y$ and $AF$ to intersect $S_1$ at $X$. Let $L$ be foot from $Y$ to $BD$. Note: $\Delta BEC$ $\sim$ $\Delta BYD$, $$\frac{BN}{NC}=\frac{BL}{LD}=\frac{MD}{CM}=\frac{KD}{BK}$$
Hence, $BK=LD$. This implies $F$ is $Y-$antipode in $\Delta YBD$ $\implies$ $\angle EAF=90^{\circ}$. Further, we also have, $\Delta BEN$ $\sim$ $\Delta BYL$ $\sim$ $\Delta FBK$ $\implies$ $ \tfrac{BN}{EN}$ $=$ $\tfrac{FK}{BK}$ $\implies$ $\tfrac{FK}{MK}$ $=$ $\tfrac{MN}{EN}$ $\implies$ $\Delta MNE$ $\sim$ $\Delta FKM$ \begin{align*} \angle CBD=\angle NMK=\angle EMF+\angle NME+\angle KMF=\angle EMF+180^{\circ}-\angle ENM =\angle EMF+\angle CBD-90^{\circ} \qquad \blacksquare \end{align*}

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2001 G8 was easier

[img width=50]https://media.discordapp.net/attachments/855230423172907008/954945615563989062/Screen_Shot_2022-03-19_at_8.32.51_PM.png[/img]
Note that $\frac{DK}{KM}=\frac{MN}{NC}$ so $\frac{DK}{KB}=\frac{BN}{NC}.$ Extend $FK$ to intersect $S_2$ at $H.$ Note that $\angle BHD=\angle BAD=\angle CEB.$ Let the transformation $S$ be from $C,B$ to $B,D$ respectively consisting of one reflection, preserving all angles and, by extension, length ratios. $N$ goes to $K$ and so the line $EN$ goes to the line $HK.$ Since $\angle CEB=\angle BHD,$ $E$ goes to $H.$

Thus, $\angle NBE=\angle HDB=\angle KFB.$ Therefore, $\angle NBE+\angle KBF=90^\circ.$ This implies that $\triangle NBE\sim\triangle KFB$ so $\frac{NE}{MK}=\frac{MK}{KF}.$ Additionally, by angle chasing $\angle MNE=\angle FKM$ so $\triangle MNE\sim\triangle FKM.$ Now, $\angle NME+\angle KMF=90^\circ-\angle MKB=\angle NMK-90^\circ,$ so we are done.

Let $FK$ meet $S_2$ again at $G,G'$ is reflection of $G$ across perpendicular bisector of $BD.$
Obviously $\measuredangle MNE=\measuredangle FKM$ and $|BN|:|NC|=|DM|:|MC|=|DK|:|KB|,$ so by spiral similarity $$BEC\stackrel{+}{\sim} BG'D\stackrel{-}{\sim} BGD\implies BNE\stackrel{+}{\sim} FKB\implies \frac{|MN|}{|FK|}=\frac{|BK|}{|FK|}=\frac{|EN|}{|BN|}=\frac{|EN|}{|KM|}\implies$$ $$\implies MNE\stackrel{+}{\sim} FKM\implies \measuredangle EMF=\angle (EN,MK)=\angle (EN,BC)=90^\circ \text{ } \blacksquare$$

nice problem
[hide = solution]
Consider the spiral sim $\tau$ sending $CB \mapsto BD$. Since \[\frac{BN}{NC} = \frac{DK}{KB} = \frac{DM}{MC}\] it follows that $N$ is mapped to point $K$ under $\tau$. Let $E'$ denote the map of $E$. We note that now $K, E', F$ collinear.

Consider triangle $\triangle FBD$ and note that since $E'$ lies on the $F$-altitude with \[\angle BFD + \angle BE'D = \angle BAC + \angle BEC = 180^{\circ}\] implying that $B, D, F, E'$ form an orthocentric system. Thus we have \[\angle EBN = \angle E'DK = \angle BFK\]
This implies that triangles $EBN$ and $BFK$ are similar. We therefore obtain \[\frac{EN}{BN} = \frac{BK}{KF}\]
Since $MNBK$ is a parallelogram, we have $MN = BK$ along with $MK = BN$. Thus we have \[\frac{EN}{NM} = \frac{EN}{BK} = \frac{BN}{KF} = \frac{MK}{KF}\]
Since \[\angle ENM = 90^{\circ} + \angle CBD = \angle MKF\] we have $\triangle ENM \sim \triangle MKF$. Since $EN$ is perpendicular to $BC$ which is in turn parallel to $MK$ it follows that the corresponding sides of these triangles are at $90^{\circ}$ with each other.

In particular, $EM \perp MF$, as desired.
[/hide]
[hide = remark]
Actually from simple angle chase we can also deduce $\angle EAF = 90^{\circ}$, so we in fact have points $E, F, M, A$ concylic.  :o
[/hide]


What is the smallest multiple of 6 greater than 115?

We could also quickly note that $ 120$ is a multiple of $ 6$, and since it is within $ 6$ greater than $ 115$, it is the smallest one.


Jo has given Aiden at least one toy car every birthday. The number of cars has corresponded to Aiden's age (one car for his first birthday, two cars for his second birthday, and so on). After receiving the cars for his 12th birthday, how many total toy cars will Aiden have received from Jo?

Here, we obviously do $ 1\plus{}2\plus{}3...\plus{}12$, and get $ \boxed {78}$.  A quick way to add any set of numbers (which form an arithmatic sequence) is to add the smallest and largest terms, and multiply it by half the number of terms.  In this situation, it would be $ (1\plus{}12) \times \frac {12}{2}$, which gives us $ \boxed {78}$ again.

The number of cars he receives is $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9 \plus{} 10 \plus{} 11 \plus{} 12$. Using the formula for sum of the first $ n$ numbers (which is $ \frac {n(n \plus{} 1)}{2}$), we get $ \frac {12(12 \plus{} 1)}{2} \equal{} \frac {12(13)}{2} \equal{} \frac {156}{2} \equal{} \boxed{78\text{ cars.}}$.

EDIT: hmm beaten by 2 min


[quote="warut_suk"]There is a positive integer in each square of a rectangular table. In each move, you may double each number in a row or subtract 1 from each number of a column. Prove that you can reach a table of zeros by a sequence of these permitted moves.[/quote]

It is sufficient to prove that you can put all numbers of a column to zero cause after that you proceed to the next column and so on. Take any column, put the smallest of the integers in it to 1 by substracting, then put it to two by doubling, substract again, etc, until all numbers in the column are 1, ergo 0. Go to next column.

Daniel


Find the Fibonnachi (correct my spelling if I am wrong :fool: ) sequence in Pascal's triangle, and prove your answer. Pictures would help.

P.S. How do you put pictures in the forum? :maybe:

Can you prove that the sum of the diagonals really does give the Fibonacchi Sequence?

[size=200]FIBONACCI[/size] is the correct spelling (no h)


No. I was thinking more of

[img]http://i111.photobucket.com/albums/n141/Felix_LaPubelle/Giant_Hamburger.jpg[/img]

Dang. How did you find a picture of my brother?

Also like... where the hell do you find all these random pictures?

[quote="serialk11r"][quote="moogra"]So I take it that you weight 100000 pounds?[/quote]
Dude just the section of road there probably is much more than that.[/quote]

Huh? What does that mean?

Like, you know how the picture has like, a section of a ROAD in it? Concrete is quite heavy.


geometry
circumcircle
probability
geometry proposed

W is a circle and A,B,D are points in your perimeter.
A is a point in arc BD not contains A.
O¹ is a center of excircle of ABC touching BC.
O² is a center of excircle of ACD touching CD.
Prove that when C movies the circuncenter of CO¹O²  intersect W in a fixed point.

I think one of the main difficulties is the fact that the points are not named properly :). Could you please correct it? For instance, you say "A is a point on arc BD not contains A" what does that mean? You have two $A$'s there :?

If I correctly understand Lopes' problem, it should be as follows:

[color=blue]Let W be a circle and A, B, D three points on its perimeter.
Let C be a point on the arc BD of this circle not containing A.
Let $O_1$ and $O_2$ be the excenters of the triangles ABC and ACD (both times the excenters opposite to the vertex A).
Prove that when the point C moves on the arc BD of the circle W, the circumcircle of triangle $CO_1O_2$ intersects the circle W at a fixed point.[/color]

Anyway, if this is what Lopes meant, then I can say that this problem is an analogue of the problem posted in the old thread http://www.mathlinks.ro/Forum/viewtopic.php?t=555 , but with excircles instead of incircles. With a great probability, this problem can be solved analogously to the one in that thread...

  darij


Heather compares the price of a new computer at two different stores. Store A offers $ 15\%$ off the sticker price followed by a $ \$90$ rebate, and store B offers $ 25\%$ off the same sticker price with no rebate. Heather saves $ \$15$ by buying the computer at store A instead of store B. What is the sticker price of the computer, in dollars?

$ \textbf{(A)}\ 750 \qquad \textbf{(B)}\ 900 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1050 \qquad \textbf{(E)}\ 1500$

[hide]
The equation is $ .85p\minus{}90\equal{}.75p\minus{}15\Rightarrow .1p\equal{}75$, so $ p\equal{}750$.  Answer is $ A$.
 [/hide]

wait, maybe i'm not getting this but it seems like 1050 also works: 1050 times .75= 787.5 and 1050 times .85 -90= 802.5 which is a difference of 15. 750 works too as he explained, i'm confused.

Yes, but it is a difference of $ 15$ the wrong way.  That would be if she saved by shopping at B instead of A.

forgot that little detail  :oops:  thanks

Suppose that $ x$ is the sticker price. At store A, Heather spends \[ (1\minus{}.15)\cdot x\minus{}90\] dollars. At store B, Heater spends \[ (1\minus{}.25)\cdot x\] dollars. The price at store A is $ 15$ dollars less than the price at store B. Hence \[ .85\cdot x\minus{}90\plus{}15\equal{}.75\cdot x\implies .10\cdot x\equal{}75\implies x\equal{}750\] Hence the answer is $ \boxed{A}$.

[hide=solution]$x=\text{sticker price}$
Then Store A gives $\cfrac{17}{20}x-90$ dollars and Store B gives $\cfrac{3}{4}x$ dollars.
$\cfrac{17}{20}x-90=\cfrac{3}{4}x-15\rightarrow\cfrac{1}{10}x=75\rightarrow x=\boxed{\textbf{(A)} ~750}$[/hide]

that is literally what post #6 said except you used fractions instead of decimals and skipped some intermediate steps


[hide]i used guess and check and i got that x=1, y=2 and z=3[/hide]

[quote="bryantacademy"]If x+y=z, and x, y, and z are consecutive digits, what are the variables for x, y, and z?[/quote]

didnt someone post a probelm very similar to this one not long ago?

[hide]answers are 1, 2 and 3  x, y , and z respectively the answer just popped into my head..[spelling corrected by solafidefarms][/hide]

[hide]
say that the 1st number is x the second would be x+1 or y and the third is x+2 or z

Then x+x+1=x+2
2x+1=x+2
x=1
y=2
z=3

no guess and check needed
[/hide]

yes, nonie, that was me. I posted that problem to see if I could stump the people in the Fun Forum. It was fun because I don't think that anyone had a clue what I was talking about!!!

well this isn't a particularily hard question
[hide]
x+y=z
x+2=y+1=z
x+y=y+1
x=1
y=x+1=2
z=x+2=3[/hide]

[quote="jensen"][quote="boxedexe"]Well, I am a high school student, so I am not answering this but giving a hint that is useful for any problem of this sort.

[hide="hint"]
Always see what they have in common, in this case, $x+y=z \rightarrow n+(n+1)=n+2$, since $x$,$y$, and $z$ are consecutive.  Then you just have to solve the equality for $n$.  $n$ will then be your first number, which is, in this case $x$.  For $y$ and $z$, you just have to add $1$ to $y$ and $2$ to $z$.
[/hide]
I hope that helps.[/quote]
i think this problem is so easy for you,you should be in olympiad section not here. :!:[/quote]

I go everywhere. :P

[quote="10000th User"][quote="surge"]$x+(x+1)=(x+2)$, $(x+1)+(x)=(x+2)$, $x+(x-1)=(x-2)...$

$x={1,2,-1,-2}$
$y={1,2,-1,-2}$
$z={3, -3}$[/quote]I'm sorry but I have no idea what you are trying to solve for...
Besides, -1 -2 are not digits, they are negative integers...[/quote]

tru dat, my mistake.

but wouldn't both x and y be either 1 or 2? Why else would he word the question oddly? (it doesn't say that x,y,z are consecutive digits in that order).

hello folks, new here
x+y=z
y=x+1
z=x+2
therefore
2x+1=x+2
x=1
y=2
z=3

Please stop bumping up 8 month old topics that have already been succesfully answered and provide solutions when you answer problems...

Just a thought...wouldn't you have to state that $x<y<z$?


Lets say that u have a boat in an ocean and you have a large rock on board. If you throw the rock overboard, what happens to the water level? More importantly why??

when u throw the rock, then the bouyant force decreases as the weight of the boat decreases, so, the boat gets raised in comparison to the earlier level. there is no raise in water level of the sea due to the rock as whatever the increase may be it will transmitted all over the ocean and the net result would be almost nil raise in level.

I transfer the problem from "a boat and a river" to "an ice cube floating in a glass full of water upto its edge".....
Consider that initially the ice-cube contained a coin on it.....and later it was gently dropped in the water....
What will happen now???? And, of course, WHY? :wink:

i think the water level will drop.

[hide]the ice cube is occupying the space of the weight of the water that is equal to its weight with coin.
so, when coin drops, the cube rises a little as its weight gets reduced.

secondly, the volume of water which was displaced by cube by coin's weight is more than the actual volume of coin as density of coin is much higher .
so, as change in volume is negative, so, water level drops.[/hide]

addition to my last post as i cant edit
[hide]if the cube does not float maybe due to higher average density than water then water level will still drop as density of ice is less than water and so it will occupy less space on malting.[/hide]
100th post! :showoff:
but is my right correct?

i haven't thought on it but from wat i remember there wud be no change in water level!!!

There should be  .. go through the link i gave we had the discussion like a year ago on this


[quote="SUPERMAN2"]Given $ a,b,c$≥$ 0$,prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{abc} \plus{} \frac {54abc}{(a \plus{} b \plus{} c)^3}$≥$ 5$[/quote]

It's very easy.It's only Am-Gm
$ LHS \ge \frac {(a \plus{} b \plus{} c)^3}{9abc} \plus{} \frac {81abc}{(a \plus{} b \plus{} c)^3} \minus{} \frac {27abc}{(a \plus{} b \plus{} c)^3} \ge 6 \minus{} 1 \equal{} 5$  :P


geometry
3D geometry
trigonometry
complex numbers

[hide="hint"]
Substracting the two equations we have
$ (x-y)(x^{2}+4xy+y^{2})=0\Rightarrow x-y=0$ or $ x^{2}+4xy+y^{2}=0$
[/hide]

[quote="BanishedTraitor"]A full solution would be nice. :)[/quote]

red_dog's hint basically solves the problem, but if you really want a solution, I'll post it here.

[hide="But I am not too sure about this solution"]red_dog said
$ (x-y)(x^{2}+4xy+y^{2})=0\Rightarrow x-y=0 \text{ or }x^{2}+4xy+y^{2}=0$

He/she obtained that by subtracting the two equations.
On the left,
$ x=y$
Notice that the right is:
$ x^{2}+4xy+y^{2}=0 \iff$
$ x^{2}+4xy+4y^{2}=3y^{2}\iff$
$ (x+2y)^{2}=3y^{2}\iff$
$ x+2y=\pm y\sqrt{3}\iff$
$ x=y(-2\pm \sqrt{3})$

So all solutions are of the form
$ (n,n)$
or
$ \left(n,\frac{n}{-2+\sqrt{3}}\right)$
or
$ \left(n,\frac{n}{-2-\sqrt{3}}\right)$

First form:
$ -2n^{3}= 1$
$ n = \sqrt[3]{-\frac{1}{2}}$
$ (x,y) = (\sqrt[3]{-\frac{1}{2}}, \sqrt[3]{-\frac{1}{2}})$

Second form and third form
The computations are heavy, so I'll just post what deej21 said:
$ (x^{3},y^{3}= \left( \frac{5 \mp 3 \sqrt 3}8 , \frac{5 \pm 3 \sqrt 3}8 \right)$[/hide]

EDIT: Wow, I didn't plug back in... I feel so stupid... I'm not sure if deej has the right answer, either.

vishalarul, we need solutions that apply to both simultaneous equations.  You just solved for the two equations red_dog got.

[hide="my solution"]
Note that neither x nor y can be 0, as then one of the equations would become 0 = 1.

Then, let $ z = \frac{x}{y}$

We can rewrite the given equations as the following:
$ z^{3}-3z = \frac 1{y^{3}}$
$ \frac 1{z^{3}}-3 \frac{1}{z}= \frac 1{x^{3}}$

By subtracting and factoring, we get $ (x-y)(x^{2}+4xy+y^{2}) = 0$
Thus, one of the two must be true:
$ x-y = 0 \implies x = y \implies z = 1$
$ x^{2}+4xy+y^{2}= 0 \implies \frac{x^{2}}{y^{2}}+4\frac{x}{y}+1 = 0$
$ \implies z^{2}+4z+1= 0 \implies z =-2 \pm \sqrt 3$

Case 1: $ z = 1$
Plus this into the rewritten equations above and get:
$ \frac 1{x^{3}}= \frac 1{y^{3}}=-2$
$ x^{3}= y^{3}= \frac{-1}2$

Case 2: $ z =-2 \pm \sqrt 3$
$ z^{3}=-26 \pm 15 \sqrt 3$
Note that the reciprocals of $ z$ and thus $ z^{3}$ are equal to their conjugates.
Plugging back into the rewritted equations above yields:
$ -20 \pm 12 \sqrt 3 = \frac 1{y^{3}}$
$ -20 \mp 12 \sqrt 3 = \frac 1{x^{3}}$
$ y^{3}= \frac{5 \pm 3 \sqrt 3}8$
$ x^{3}= \frac{5 \mp 3 \sqrt 3}8$

Thus, the solutions are:
$ (x^{3}, y^{3}) = \left( \frac{-1}2 , \frac{-1}2 \right), \left( \frac{5 \mp 3 \sqrt 3}8 , \frac{5 \pm 3 \sqrt 3}8 \right)$

edit: turns out the second solutions are fairly nice cubes, so the solutions can be simplified to:
$ \frac{(x, y)}{\sqrt[3] 4}= \left( \frac{-1}2 , \frac{-1}2 \right), \left( \frac{1 \mp \sqrt 3}4 , \frac{1 \pm \sqrt 3}4 \right)$
Not sure if that notation with dividing out the cube root of 4 is exactly proper, but it makes the rest of it look a lot nicer.  Nice enough to notice some very interesting connections to trig... I bet there is a very elegant solution there.
[/hide]

I plugged them into a calculator, so they *should* be right.  I also edited my post with a more simplified form, and with hope that someone finds a trig solution.  I ventured into complex numbers, for instance...
$ z =-2 \pm i \sqrt 3$
$ z = \sqrt 7 e^{\theta i}$ where $ \theta = \pi \pm \arctan \frac{ \sqrt 3 }2$
Although, $ z^{3}$ is going to be different from the cube of the original $ z$, in terms of the real and imaginary parts... so I don't know.

[quote="deej21"]I plugged them into a calculator, so they *should* be right.  I also edited my post with a more simplified form, and with hope that someone finds a trig solution.  I ventured into complex numbers, for instance...
$ z =-2 \pm i \sqrt 3$
$ z = \sqrt 7 e^{\theta i}$ where $ \theta = \pi \pm \arctan \frac{ \sqrt 3 }2$
I'm working on seeing where plugging it back into the equations I got in $ z$ leads... but I'm not optimistic.[/quote]

Hey! That's a good idea!

[hide]Let $ x=r\cos\theta$ $ y=r\sin\theta$. Then the equations simplify to
$ r^{3}\cos 3\theta = 1$
$ r^{3}\sin 3\theta =-1$

Square both, add:
$ r^{6}= 2$
$ r=\pm \sqrt[6]{2}$

This implies
$ 3\theta = 7\pi/4, 15\pi/4, 23\pi/4$
$ \theta = 7\pi/12, 5\pi/4, 23\pi/12$.

I guess that this practically finishes the solution, since $ x=r\cos\theta$, $ y=r\sin\theta$.[/hide]

Nice!  I had JUST thought of the triple angle formula and had a fun little eureka moment.  Those angles are pretty nice actually.

Nothing like staying up an extra two hours doing a math problem.


vector
topology
abstract algebra
group theory
absolute value
superior algebra
superior algebra unsolved

Show that $(\mathbb{R}, +)$ is isomorphic to $(\mathbb{C},+)$.

All we need to do is choose two bases of $\mathbb R$ and $\mathbb C$ as vector spades over $\mathbb Q$. The two bases have the same cardinality (the continuum cardinality), so the two vector spaces are both isomorphic to $\mathbb Q^c$, where $c$ is the continuum cardinality.

Here's a nicer one:

Are $(\mathbb C^*,\cdot)$ and $(\mathbb S^1,\cdot)$ isomorphic? (here, $(\mathbb S^1,\cdot)$ is the multiplicative group of complex number with absolute value $1$)

Enjoy :)

hmm no. However, one can construct an isomorphism between

$(\mathbb{C}^{*}/ \mathbb{R}^{*}_{+},\cdot)$ and $(S^{1},\cdot)$

by the isomorphism theorem and using the map $f(z) = \frac{z}{|z|}$.

How did you show that the two I asked about are not isomorphic?

well the problem is that you cant find a nice smooth function. the two dimensionalities are different, one is on a 2-manifold, the other's on a 1-manifold, so one cannot find a diffeomorphism between them. you can map the circle into $\mathbb{C}^{*}$ but it will be really jagged, and wont preserve any structure.

I don't want any smoothness properties for the isomorphism  ;). The question is a purely algebraic one: is there a group isomorphism between the two or not? :)

It seems that I found the solution, but it is rather long and boring.
May be I am missing something easy, but nevertheless here is it:

 Suppose that in general we have an abelian groups $ G', G'' $ with a subgroups
$ G' \supset N' $, $ G'' \supset N'' $, such that the following conditions
are satisfied :

* $ N',N''$ are isomorphic

* For every $ x' \in G' $ there exists $ n \in \mathbb{N} $ such that $ nx' \in N' $.
  For every $ x'' \in G'' $ there exists $ n \in \mathbb{N} $ such that $ nx'' \in N'' $.

* For every $ y' \in G' $ and $ n \in \mathbb{N} $ there exist exactly $ n $ elements
$ x' \in G' $, such that $ nx' = y' $.
  For every $ y'' \in G'' $ and $ n \in \mathbb{N} $ there exist exactly $ n $ elements
$ x'' \in G'' $, such that $ nx'' = y'' $.

Then $ G',G'' $ are isomorphic.
Let us prove this.
Suppose we have an isomorphism $ f: N' \rightarrow N'' $, then we will show that
it can be extended to an isomorphism $ G' \rightarrow G'' $.
Denote for a prime $ p $, by $ N'_{p} $ the set of all $ x' \in G' $ such that
$ p^{k}x' = 0 $ for large $ k $. Also denote by $ N''_{p} $ the set of all
$ x'' \in G' $ such that $ p^{k}x'' = 0 $ for large $ k $.
Then we have that every element $ x \in G' $ can be written as a sum
$ x = x_{1} + \ldots + x_{m} $ , where $ x_{i} \in N'_{p_{i}} $ for some
collection of primes $ p_{1},p_{2}, \ldots , p_{m} $.
Indeed, take some $ n \in \mathbb{N} $, such that $ nx \in N' $,
then if we take a prime decomposition $ n = p_{1}^{k_{1}}p_{2}^{k_{2}} \ldots p_{m}^{k_{m}} $
and denote $ n_{j} = \frac{n}{p_{j}^{k_{j}}} $, then $ gcd(n_{1},n_{2}, \ldots ,n_{k}) = 1 $,
so we can find $ a_{1},a_{2}, \ldots , a_{m} \in \mathbb{Z} $, such that
$ n_{1}a_{1} + n_{2}a_{2} + \ldots + n_{k} a_{k} =  1 $, then if we define
$ x_{j} = a_{j}n_{j}x $ then of course $ x = x_{1} + \ldots + x_{m} $ and
$ x_{j} \in N_{p_{j}}' $. Also note, that such representation is not unique, but if we have
$ x = x_{1} + x_{2} + \ldots + x_{m} = y_{1} + y_{2} + \ldots + y_{m} $, when
$ x_{i},y_{i} \in N'_{p_{i}} $, then $ x_{i} - y_{i} \in N' $ for every $ i $.
All this is true also about $ G'' $.
Take now some prime $ p $, and let's try to extend the isomorphism
$ f : N' \rightarrow N'' $ to homomorphism $ N_{p}' \rightarrow N_{p}'' $.

Denote by $ L_{p}' $ the group of all
$ x' \in G' $ such that $ px' = 0 $ and by $ L_{p}'' $ the group of all
$ x'' \in G'' $ such that $ px'' = 0 $. Then if we have $ L_{p}' \subset N' $ then
also $ L_{p}'' \subset N''$, and $ f(L_{p}') = L_{p}'' $.
Now, if $ L_{p}' \nsubseteq N' $, then $ L_{p}' \cap N' = \emptyset $ and
$ L_{p}'' \cap N'' = \emptyset $, so $ N' + L_{p}' \cong N' \oplus L_{p}' $
and $ N'' + L_{p}'' \cong N'' \oplus L_{p}'' $, so because $ L_{p}' $, $ L_{p}'' $
are isomorphic, we can extend $ f $ to an isomorphism
$ f: N' + L_{p}' \rightarrow N'' + L_{p}'' $.

Suppose we have already extended $ f $ to a monomorphism $ f: K' \rightarrow N_{p}'' $
for some $ N' + L_{p}' \subset K' \subset N_{p}' $. Denote $ K'' := f(K') \subset N_{p}'' $.
Now suppose that $ K' \subsetneq N_{p}' $. Then there exists $ x' \in
N_{p}' $, $ x' \notin K' $, such that $ y':= px' \in K' $. Denote $ y'' = f(y') \in K'' $.
Take some $ x'' \in N_{p}'' $, such that $ px'' = y'' $. Then $ x'' \notin K'' $, because
otherwise there exists $ z \in K' $ such that $ f(z) = x'' $, then
$ f(pz) = y'' = f(y') = f(px') $, so $ f(pz-px') = 0 $ so $ pz-px' = 0 $ so
$ z-x' \in L_{p}' $ and $ z \in K' $, so $ x' \in K' + L_{p}' = K' $ and we get a
contradiction. Therefore $ x' \notin K' $ ,  $ px' \in K' $
and $ x'' \notin K'' $, $ px'' \in K'' $. Therefore for every $ k \in \mathbb{Z} $,
not divisible by $ p $, we have $ kx' \notin K' $ , $ kx'' \notin K'' $.
Now if we define $ \overline{K}' $ to be the subgroup of $ N_{p}' $ which is all
sums $ kx' + z' $, where $ k \in \mathbb{Z} $ and $ z' \in K' $ and
$ \overline{K}'' $ to be the subgroup of $ N_{p}'' $ which is all
sums $ kx'' + z'' $, where $ k \in \mathbb{Z} $ and $ z'' \in K'' $.
Now for $ y \in  \overline{K}' $ if we represent it by $ y = kx' + z' $,
where $ k \in \mathbb{Z} $ and $ z' \in K' $, then define
$ \widetilde{f} (y) := kx'' + f(z') $. It is easy to see that $ \widetilde{f} $
is well defined, and it is injective, because if we have
$ f(y) = 0 $ then $ kx'' + f(z) = 0 $, so $ kx'' \in K' $, so $ k $ is divisible by $ p $,
therefore $ kx' \in K' $, so $ y \in K' $, and because $ f $ is injective, we get $ y = 0 $.

So we see that if we have a monomorphism $ f : K' \rightarrow N_{p}'' $, and
$ K' \subsetneq N_{p}' $, then it can be always extended to a monomorphism, defined
on a bigger subgroup of $ N_{p}' $. Now applying the Zorn lemma, we get that
$ f : N' + L_{p}' \rightarrow N'' + L_{p}'' $ can be extended to a monomorphism
$ f_{p} : N_{p}' \rightarrow N_{p}'' $. Then it must be also onto, because otherwise
we have an isomorphism $ f_{p}^{-1} : Im(f_{p}) \subsetneq N_{p}''
\rightarrow N_{p}' $, which cannot be extended to a monomorphism, defined on a
bigger subgroup.

Finally we see that for every prime $ p $, $ f : N' \rightarrow N'' $ can be extended
to an isomorphism $ f_{p} : N_{p}' \rightarrow N_{p}'' $.
Now, define $ F : G' \rightarrow G'' $ in the following way:
if we take $ x \in G' $ and a representation $ x = x_{1} + \ldots + x_{m} $ ,
where $ x_{i} \in N'_{p_{i}} $ for some collection of primes
$ p_{1},p_{2}, \ldots , p_{m} $, then define $ F(x) := f_{p_{1}}(x_{1}) +
f_{p_{2}}(x_{2}) + \ldots + f_{p_{m}}(x_{m}) $. Then $ F $ is well defined,
because if we have two representations
$ x = x_{1} + x_{2} + \ldots + x_{m} = y_{1} + y_{2} + \ldots + y_{m} $, when
$ x_{i},y_{i} \in N'_{p_{i}} $, then $ x_{i} - y_{i} \in N' $ for every $ i $,
so $ f_{p_{i}}(x_{i}-y_{i}) = f(x_{i} - y_{i}) $, so
$ ((f_{p_{1}}(x_{1}) + f_{p_{2}}(x_{2}) + \ldots + f_{p_{m}}(x_{m})) -
(f_{p_{1}}(y_{1}) + f_{p_{2}}(y_{2}) + \ldots + f_{p_{m}}(y_{m}))
 = f_{p_{1}}(x_{1} - y_{1}) + f_{p_{2}}(x_{2}-y_{2}) + \ldots + f_{p_{m}}(x_{m}-y_{m}) =
f(x_{1} - y_{1}) + f_{p}(x_{2}-y_{2}) + \ldots + f_{p}(x_{m}-y_{m}) =
f( x_{1} - y_{1} + \ldots + x_{m} - y_{m} ) = f(0) = 0 $. Also it is clear that
$ F $ is a homomorphism, and it is easy to see that it is one-to-one and onto.
Therefore $ F $ is an isomorphism.

 Let us now prove that  $ G' = ( \mathbb{S}^{1} , \cdot ) $ and
$ G'' = ( \mathbb{C}^{*} , \cdot ) $
are isomorphic. Take a Hamel basis for the $ \mathbb{R} $ over $ \mathbb{Q} $,
which contains $ 1 $ and denote by $ \Lambda $ the set of all element of this basis,
except $ 1 $. Then $ \Lambda $ has continuum cardinality and any non-trivial combination
with integer coefficients of elements of $ \Lambda $ cannot be an integer number.
Now elements in $ ( \mathbb{S}^{1} , \cdot ) $, of the form $ e^{ 2\pi i \lambda } $,
when $ \lambda \in \Lambda $. They are independent. Take now maximal set in
$ ( \mathbb{S}^{1} , \cdot ) $, which contains all those elements, and all elements
of it are independent. Denote by $ N' $ the subgroup of $ G' = ( \mathbb{S}^{1} , \cdot ) $,
generated by elements of this set. Then $ N' $ is a free abelian group with
continuum number of generators, and it is easy to see that the pair $ (G',N') $
satisfies all the conditions above.
Now for $ G'' = ( \mathbb{C}^{*} , \cdot ) $ do the same thing :
take all elements of the form $ e^{ 2\pi i \lambda } $, when $ \lambda \in \Lambda $,
then complete this set to a maximal set which contains independent elements,
and denote by $ N'' $ the subgroup of $ G'' = ( \mathbb{C}^{*} , \cdot ) $,
generated by this set. Then again, $ N'' $ is a free abelian group with
continuum number of generators, and the pair $ (G'',N'') $ satisfies all conditions
above. Also, $ N' $, $ N'' $ are free abelian groups with continuum number of generators,
so they are isomorphic. Therefore applying what we have shown, we get that
$ G' , G'' $ are isomorphic.

I did not (and will not ;)) read it, but I do believe we did about the same thing (when it comes to this problem; I see you have proven a very nice generalization :)).

haha hmm.. i'll try to digest that one after exams.


quadratics
modular arithmetic
number theory unsolved
number theory

For any prime $p$, prove that there exists integer $x_0$ such that $p | (x^2_0 - x_0 + 3)$ $\Leftrightarrow$ there exists integer $y_0$ such that $p | (y^2_0 - y_0 + 25).$

I think there might have been some other condition. Otherwise, you could choose $ (x,y) = (12,11) $. Then $ x^2 - x + 3 = y^2 - y + 25 = 135 $

You're reading it wrong.  It says that $y_{0}$ exists if and only if $x_{0}$ exists

The first one holds iff there is an $x_0$ s.t. $p|(2x_0+1)^2+11\ (*)$, while the second one holds iff there is an $y_0$ s.t. $p|(2y_0+1)^2+99\ (**)$. $(*)$ is equivalent to $-11$ being a quadratic residue $\pmod p$, while $(**)$ is equivalent to $-99=9\cdot(-11)$ being a quadratic residue, which is equivalent to $-11$ being a quadratic residue, and we're done.

The first assertion $\exists x\in\mathbb{Z}$ s.t. $p\mid(x^2-x+3)$ is equivalent to
\begin{align*}
    \exists x,d_1\in\mathbb{Z}\text{ s.t. }x^2-x+3=(x-\tfrac{1}{2})^2+\tfrac{11}{4}=d_1p&\\
    \iff(2x-1)^2=4d_1p-11&.
\end{align*}
We claim that the above is in turn equivalent to $-11$ being a QR $\pmod{p}$, i.e., $(\tfrac{-11}{p})=1$; indeed, necessity is apparent by taking the equation $\pmod{p}$. As for sufficiency, if $a,p-a$ are the "square roots" of $-11$, we can set $2x-1$ equal to whichever of $a,p-a$ is odd, such that $x$ is an integer. Then $(2x-1)^2+11=4(x^2-x+3)$ is divisible by $4p$, meaning $d_1=\tfrac{(2x-11)^2+11}{4p}$ is an integer and renders the above equation true; so we have the existence of the desired $x,d_1\in\mathbb{Z}$.

Similarly, the second assertion $\exists y\in\mathbb{Z}$ s.t. $p\mid(y^2-y+3)$ is equivalent to
\begin{align*}
    \exists y,d_2\in\mathbb{Z}\text{ s.t. }y^2-y+25=(y-\tfrac{1}{2})^2+\tfrac{99}{4}=d_2p&\\
    \iff(2y-1)^2=4d_2p-99&,
\end{align*}
which we analogously determine is equivalent to having $(-\tfrac{99}{p})=1$. Hence, it remains to show that $(-\tfrac{11}{p})=1\iff(-\tfrac{99}{p})=1$, which is clear since $(-\tfrac{99}{p})=(\tfrac{9}{p})(\tfrac{-11}{p})=(\tfrac{-11}{p})$.

Above, with $(\tfrac{-11}{p})$'s and $(\tfrac{9}{p})$'s thrown around, we implicitly require that $p>11$; fortunately, $p=2,3,5,7,11$ are easily verifiable by hand:
[list]
[*] $p=2$ renders both assertions false, because $x(x-1)$ and $y(y-1)$ are always even, and so $x^2-x+3$ and $y^2-y+25$ are always odd.
[*] $p=3$ renders both assertions true with $(x,y)=(0,2)$.
[*] $p=5$ renders both assertions true with $(x,y)=(2,0)$.
[*] $p=7$ renders both assertions false, because the residues of $n(n-1)\pmod{7}$ are $0,2,5,6$, so the residues of $x^2-x+3$ and $y^2-y+25$ are $3,5,1,2$ and $4,6,2,3$.
[*] $p=11$ renders both assertions true with $(x,y)=(6,6)$.
[/list]
All cases have been exhausted; the proof is complete. $\blacksquare$


Daca intr-un triunghi dublul unei laturi este egal cu diferenta dintre perimetru si arie (2p-S) atunci 16R+4r+p^2(r-2)=8.

Nu stiu cum ti se pare problema asta, dar dupa umila mea parere ea nu spune absolut nimic. :?
 Anyway, ce zici de asta?

         \[ (a+b+c)^{2}\geq 12\sqrt{3}(b+c-a). \] :)

Hmmm.... E si ea acolo o relatie! Nu spun ca arata bine :) , in schimb imi place ideea din care am obtinut-o. Referitor la aceasta inegalitate, ce sunt a,b,c? Laturi cu o anumita proprietate sau numere oarecare?

Tu ce crezi? :)  Sunt laturile unui triunghi cu proprietatea care ai dat-o tu. ;)


[quote="SUPERMAN2"]ABC is a acute triangle.Prove that
$ 2(a^2b \plus{} b^2c \plus{} c^2a)^2$≥$ (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)(a^2 \plus{} b^2 \plus{} c^2 \plus{} 4S\sqrt3)$[/quote]
Anyone has a solution?I think this is a very nice ineq.


Yesterday I was enrolled by USTC(University of Science and Technology of China), which was a pretty good one in China. Meanwhile, as I took a first prize it is not hard for me to enter the NK-SDB(试点班). But it is difficult for me to make a choice.

Certainly I will major in Maths in my college years, but which one is better?
Someone said NK-SDB is very good, perhaps better than any other one except PKU in China.
Someone said USTC is good as it is easy for going abroad for study there, but some famouse people in USTC are not always there.

What do you think about it? Please give me some recomments, thanks a million!

都很好
但是如果你确定你能be top的话，那就去USTC.USTC去顶级校的可能性大一些。
USTC的第一名基本上拿到Berkeley,Caltech,Stanford没什么问题。
但是在南开，顶多进Berkeley.南开原本有Berkeley传统，每年第一名都能进。但自从陈省身大牛仙逝后，情况有所改变。
前年试点班第一名被Berkeley拒，差点去了Wisconsin-Madison,还好后来从Wharton的waitinglist中转正。前年试点班第二名是去了Stanford，但做的是运筹学。
去年试点班第一名也没进Berkeley.普通班的一个人进了。

不过还要看发展，应该认识到的是，南开科大由于地理位置原因都在走下坡路。很多牛人都被挖走。
你是学竞赛的，应该知道余红兵吧？原科大教授，后来被苏州大学挖走了。现在南开的数学院长张伟平院士也有谣传去复旦。
不过南开换了饶子和作校长，饶子和很厉害，在南开做了一系列牛B的举措。特别是有南开百人计划，由学校推荐派人公费留学到国外著名大学然后回南开任教（只要教好像5年，而且可以交违约金违约）。去年刚和牛津剑桥等达成协议。
国家去年实施的公派留学计划，南开也获得了和清华北大同样的300名额，而且南开是全国唯一一个把名额分配给本科生的学校。这些都是饶子和的英明举措。而随着留学热潮的趋势，科大的出国优势会越来越小的。

如果你特别强的话，建议你还是去USTC，那里的氛围基本上是全民留学，很变态的。大一开始就一堆人上课不听课在底下背红宝书。我在南开待过一段儿，那里的出国氛围要淡很多。而且要注意的是，我听科大的人说，到后期数学系里面最强的基本都是高考上来的。

你没去尝试一下北大吗？在国内北大还是毫无疑问的NO。1
科大主要是物理化学，和北大平起平坐，比数学变态很多


there are 3 tables in a room, and each table has 4 seats.
12 people want to stay in the room for 5 days.
how should the people vary the seating arrangement so that after 5 days each person shares the table at lunch with every other person?

Yes, it is funny and not too difficult. I will not post the solution, but I observe that this problem has a quite unexpected connection with another one in this Forum.
[hide][url=http://www.mathlinks.ro/Forum/topic-82269.html]This one[/url], case 9x12. That solution is basically the same of 9 people/3 tables/3 seats each/4 days.[/hide]

i don't understand.
can you plz post complete solution?

As I said, my post is not a solution, but a remark. Anyway...
[hide]
In the thread I have linked, there is a board 9x12 painted with 3 colours. If you look carefully, you see that the columns are grouped in 4 groups of 3, in which colours are rotated. Each person in the table problem corresponds to a different row, each day is one of the 4 groups of columns, and each table is a different colour.
[/hide]


Not to be cocky, but I will win at states. Delaware is just sooooo easy.

[hide="problem 1"]
Total, Clock B gains 45+30=75 seconds=1.25 a day.
It has to catch up 15 minutes.
$\frac{15}{\frac 54}=12$[/hide]
[hide="problem 2"]
This was the one about Great Aunt Minnie and Great Uncle Paul right?

The easiest way is to guess and check.
Try $85^2=7225$.
Then for the first equation, you get that
$x=83$
Plug back into the second and it works!
So $(83,85)$[/hide]

[quote="tz"]Clock A loses 30 seconds a day. Clock B gains 45 seconds a day. If clock A is 10 minutes ahead of clock B, how many days will it take clock B to be 5 minutes ahead of clock A??

I got 60 but I'm not sure that's right.[/quote]

I think that you got $60$ because you made $45-30=15$ and then $10+5=15$ and $15$minutes = $900$seconds
and finally $900: 15=60$...is that your calculation ?

I think we must [b]add[/b] and not substract in the beginning !

So my solution is $45+30=75$ then $10+5=15$ and $15$minutes=$900$seconds.
Finally $900: 75=12$. The answer would be $12$ days. :roll:

[quote="biffanddoc"][hide="problem 2"]
This was the one about Great Aunt Minnie and Great Uncle Paul right?

The easiest way is to guess and check.
Try $85^2=7225$.
Then for the first equation, you get that
$x=83$
Plug back into the second and it works!
So $(83,85)$[/hide][/quote]

I think the question asked for the sum.

Are you making the poll just becuase your interested, or does it help with the problem in a way that I am not undertanding....I won't make it to nationals...... :(

At my chapter, I won the countdown round, but I am now stuck in the CA state, and there are a lot of kids in CA :(

[hide]I use algebra for first one.

600-30x=45x-300

This is easy to solve... Right?[/hide]


function
limit
real analysis
real analysis unsolved

Let $ f: (0,\infty) \rightarrow \mathbb{R}$ be a twice differentiable function. Suppose there exists $ K>0$ such that $ |f''(x)|<K$ for all $ x \in (0,\infty)$ and  that $ \lim_{x \to \plus{}\infty} f(x)\equal{}0$. Show that $ \lim_{x \to \plus{}\infty} f'(x)\equal{}0$

[hide]only i got was that $ f'$ is lipschtizian with constant $ K$[/hide]

We say that $ f'$ is simply [b]Lipschitz[/b], not Lipschitzian.

thank you JRav, it's called that way in portuguese so i wondered wrong :blush:


The NFL season will be starting soon, get your predictions in and we shall see in the end of the season...

Also, anyone up for an AoPS yahoo fantasy football league?

Carolina Panthers are going all the way this year for sure ;)

eagles  vs. steelers
           or
falcons vs colts

nfc: tampa bay, atlanta, or minnesota
afc: indianapolis or jaxsonville (gotta have a darkhorse)

[quote="noneoftheabove"]if your gonna have a dark horse how bout the steelers the jaguars?minnesota?[/quote]
well, you never know how it will end up. that's why they play the games. besides, i got dante culpepper and fred taylor on my fantasy football team.  :D

im intrested in a fantasy league are we going to start it know or later

[quote="A+MATH"][quote="math92"]i think the eagles will win it all.  My second pick is the Vikings and possibly the bears.[/quote]

What?! The Vikings!? Oh wait that would be sarcasm right?[/quote]
well they have a better defense than last year, and their offense is decent.
They got Sam Cowart as a linebacker and Burleson should be a good receiver.
Plus their running backs are awesome.  I don't expect them to win the Super
Bowl but they might get there or get to the NFC championships.

Patriots win folks.

Shame that none of you have been paying attention for the past 4 years.

who is going to start fantasy and i highly doubt that the patriots will win without weiss and crennel and brutsky and law

[quote="furious"]yes the patriots will be good, but does anyone really want to see them win every year?[/quote]

I do.

Yahoo draft times were pretty akward for before the season starting, so i started a NFL.com league, which is basically the same, but it also adds in yards allowed for defenses, which is better IMO.
1) Go to NFL.com
2) click on the button to sign up for fantasy football
3) register for an account
4) create a team name
5) search for the league "AoPS"
6) click on it
7) join it
8) type in password "mathrocks"
9) pre-rank your players if you wish
The draft is on Saturday at 7 PM eastern, which is 4 PM pacific

I got a fantisy team that was randomly generated :)

i dont get it what does ranking the players do

in case you can't make the live draft, then the computer automatically selects the players based on your list other than their own rankings

can u estimate how long the draft will be

from about half an hour to 45 minutes, usually towards the 30 minutes mark

I had an 8 person draft. It lasted an hour, everybody came and for the most part people were quick.

I'd give an hour and a half a good estimate for the draft.

JOIN PEOPLE!!!!!!!, WE ONLY HAVE 3 TEAMS SO FAR AND IF WE DON'T GET TO 12 THE LEAGUE WON'T EVEN START.

do u want to postpone the date to next week or something we dont have enough people? is the league over because when i went to check how many members we had there was no league


In order to compute $ 5^{100} \pmod{1001}$, you compute $ 5^{100}\pmod{7}$, $ 5^{100}\pmod{11}$ and $ 5^{100}\pmod{13}$. Then we use the chinese remainder theorem to take the pieces and put them back together.

For $ 5^{100}\pmod{7}$, write: $ 5^{100}\pmod{7}\equiv 5^96\cdot5^4 \pmod{7}\equiv 1\cdot 5^4\equiv( - 2)^4\equiv16\equiv2\pmod{7}$

For $ 5^{100}\pmod{11}$, write: $ 5^{100}\pmod{11}\equiv 1\pmod{11}$

For $ 5^{100}\pmod{13}$, write: $ 5^{100}\pmod{13} \equiv (5^{12})^8 \cdot 5^4\equiv (25)^2\equiv 1 \pmod{13}$.

Then write the system of congruences

$ x\equiv 2\pmod{7}$
$ x\equiv 1\pmod{11}$
$ x\equiv 1\pmod{13}$

Then we can write $ 5^{100} \equiv 2x_1 + 1x_2 + 1 + x_3$, where $ x_1,x_2$ and $ x_3$  found by solving the system in the usual way. Only I don't get where this system comes from! Is this system correct?

We also compute $ 5^{100}\equiv x_1 + 1 \equiv - 285.$ Is this correct? Why $ x_1+1$?

oh i see now; it is quite clear that the fact that $ 5^{100}\equiv 2x_1 \plus{} 1x_2 \plus{} x_3$ follows from the chinese remainder theorem; and one can then find $ x_1,x_2,x_3$ by finding solutions to the system

$ x\equiv 2 \pmod{7}$
$ x\equiv 1 \pmod{11}$
$ x\equiv 1 \pmod{13}$;

But it is not clear why $ 5^{100}\equiv x_1 \plus{} 1$...it has something to do with the fact that $ \sum_{i\equal{}1}^3 x_i \equal{}1$, but what exactly?


This is a nice inequality that I found in a book. If we have a sequence of positive real nos. $ a_{1},a_{2},....,a_{n}$ then prove that:
$ (a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n})^2 < {\frac{{\pi}^2}{6}}(a_{1}^2\plus{}2^2a_{2}^2\plus{}...\plus{}n^2a_{n}^2)$
 :)

[hide="solution"]Since $ \frac {\pi^2}{6} > \sum_{k \equal{} 1}^n \frac {1}{k^2}$ it's enough to prove that
\[ \left( \sum_{k \equal{} 1}^n a_k\right)^2 < \left(\sum_{k \equal{} 1}^n \frac {1}{k^2}\right)\left(\sum_{k \equal{} 1}^n k^2a_k^2\right)
\]
which is just Cauchy.[/hide]

It's well-known, that $ \sum_{k \equal{} 1}^\infty \frac {1}{k^2}\equal{}\frac {\pi^2}{6}$, so $ \frac {\pi^2}{6} > \sum_{k \equal{} 1}^n \frac {1}{k^2}$

[quote="Sunkern_sunflora"][quote="Svejk"]Since $ \frac {\pi^2}{6} > \sum_{k \equal{} 1}^n \frac {1}{k^2}$[/quote]
Plz prove it!  :P[/quote]

I think it's a well-known result.You can find at least a proof here: http://en.wikipedia.org/wiki/Basel_problem.

According to this [url]http://www.worldscibooks.com/etextbook/6063/6063_chap1.pdf[/url] it is Carlson inequality.

Can you explain me please how you can this [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=347945&p=1865866&hilit=Carlson#p1865866[/url] from it?

Thanks


There are n>=9 points in a plane. You know that if you choose any 9 of them you can find 2 circles, so that each of that 9 points lies on at least one of those circles. Prove that you can find 2 circles that each of that n points lies on at least one of those circles.

Also you may look at http://www.mathlinks.ro/Forum/viewtopic.php?t=838


Is it me or are there changes in the board? Like the new mathematics, why is it like this now?

rrusczyk can explain this better than I can, but this was being discussed in the moderators' forum. The problem is that far, far too much of the material being posted on the "lower" forums (middle school forums, high school basics, intermediate topics) has been too advanced for the forum in question, with the result that a beginner has no place to start with this site without becoming frustrated by difficulty or sophistication of what he's seeing.

Oh I see, I still need some getting used to the new forum...

So did MathCounts and MOEMS combine? Also, where did the Middle School Tournaments and the Other Problem Solving topics go?

ya mathcounts and moems combined. i thought mixing everything up makes it all confusing though...

[b]middle school[/b] tournaments is under [b]high school[/b] basics.  :rotfl:  :rotfl:

Not for me  :huh: ? I don't see any subforums in the High School Basics one.

It went in the category below "Mathematics": http://www.mathlinks.ro/index.php?f=215

oops! sorry powerofpi. i meant "[b]Middle School[/b] Tournaments" is in "Other US contests & programs" which is under "[b]High School[/b] Contests & Programs"

I found it, thanks to AIME15's signature.


Let $a_{1},a_{2}, \ldots ,a_{n}>0$ and $k>p \geq 0$ two natural numbers. Prove that:

$ \frac{a_{1}^k+ \ldots +a_{n}^k}{a_{1}^p+ \ldots +a_{n}^p} \geq ( \frac{a_{1}+ \ldots +a_{n}}{n})^{k-p} $.


inequalities
trigonometry
inequalities proposed

Let be $ ABC$ a triangle . Prove that :

$ \frac{\sin^2\ A\sin^2 B\sin^2\ C}{\sin^2\ A\sin^2\ B\plus{}\sin^2\ B\sin^2\ C\plus{}\sin^2\ C\sin^2\ A}$

$ \plus{} \frac{\cos^2\ A\cos^2 B\cos^2\ C}{\cos^2\ A\cos^2\ B\plus{}\cos^2\ B\cos^2\ C\plus{}\cos^2\ C\cos^2\ A}\le \frac{1}{3}$

[quote="arqady"][quote="alex2008"]Let be $ ABC$ a triangle . Prove that :

$ \frac {\sin^2\ A\sin^2 B\sin^2\ C}{\sin^2\ A\sin^2\ B \plus{} \sin^2\ B\sin^2\ C \plus{} \sin^2\ C\sin^2\ A}$

$ \plus{} \frac {\cos^2\ A\cos^2 B\cos^2\ C}{\cos^2\ A\cos^2\ B \plus{} \cos^2\ B\cos^2\ C \plus{} \cos^2\ C\cos^2\ A}\le \frac {1}{3}$[/quote]
Are you sure that it's true?  :wink:[/quote]
Counter example? :maybe:  :huh:

It's true and easy to be proved ^^ .
$ LHS \le \frac{1}{9}[ \sum sin^2A\plus{}cos^2A] \equal{}\frac{1}{3}$ .

[quote="arqady"][quote="alex2008"]Let be $ ABC$ a triangle . Prove that :

$ \frac {\sin^2\ A\sin^2 B\sin^2\ C}{\sin^2\ A\sin^2\ B \plus{} \sin^2\ B\sin^2\ C \plus{} \sin^2\ C\sin^2\ A}$

$ \plus{} \frac {\cos^2\ A\cos^2 B\cos^2\ C}{\cos^2\ A\cos^2\ B \plus{} \cos^2\ B\cos^2\ C \plus{} \cos^2\ C\cos^2\ A}\le \frac {1}{3}$[/quote]
Are you sure that it's true?  :wink:[/quote]
This inequality 's true :wink:

arqady , i'm sure this inequality is true .

Mitdac123 gave the solution

More detailed :

Let $ \sin^2\ A\equal{}a\ ,\ \sin^2 B\equal{}b\ ,\ sin^2\ C\equal{}c$

The inequality is equivalent with :

$ \frac{abc}{ab\plus{}bc\plus{}ca}\plus{}\frac{(1\minus{}a)(1\minus{}b)(1\minus{}c)}{(1\minus{}a)(1\minus{}b)\plus{}(1\minus{}b)(1\minus{}c)\plus{}(1\minus{}c)(1\minus{}a)}\le \frac{1}{3}$

From $ MA\minus{}MG$ we have that :

$ LHS\le \frac{\sqrt[3]{abc}}{3}\plus{}\frac{\sqrt[3]{(1\minus{}a)(1\minus{}b)(1\minus{}c)}}{9}\le \frac{a\plus{}b\plus{}c}{9}\plus{}\frac{3\minus{}(a\plus{}b\plus{}c)}{9}\equal{}\frac{1}{3}$


rotation
analytic geometry
geometry
incenter

[quote="El Chamo"]Let a plane shape (regular or irregular) perpendicular to the earth (we are on the earth!) whit the hole P on this, far from the mass center (qualitativ).
whit a pin in P, the shape rotates and then find the equilibrium. whit a pen i draw the vertical linie [i]l[/i] throught P on the shape.
I make the same whit another hole Q not on [i]l[/i] and far from the mass center, i obtain the linie [i]m[/i]
[i]l[/i] cut [i]m[/i] in G.
is G the mass center of the shape? when i repeat the process whit another hole, pass the neu linie again through G?[/quote]

If I understand what you are doing, yes, In practice.

In reality, you will not be able to be accurate enough for it to go through the same point, but it should be close enough for you to balance it there on something of reasonable size.

I think I managed a irregular object about 2' x 2' on the head of a nail (not the pointy bit)

[quote="SimonM"]

If I understand what you are doing, yes, In practice.

In reality, you will not be able to be accurate enough for it to go through the same point, but it should be close enough for you to balance it there on something of reasonable size.

I think I managed a irregular object about 2' x 2' on the head of a nail (not the pointy bit)[/quote]

In that case, i have a question  :?: :

Let $ A,B,C$ an equilateral triangle, $ P$ a point on $ AB$ such that: $ \frac{AP}{BP}\equal{}2$ and $ G$ the incenter.
Whit the pin in $ P$ i level the triangle and draw the linie [i]y[/i]. This linie pass throught $ G$, then is parallel to $ BC$, and divide $ ABC$ in two figures, say $ APQ$ (equilateral triangle) and $ PQCB$ (trapezoid).
The linie [i]x[/i] is the perpendicular to [i]y[/i] in G, the median of $ ABC$ respect $ A$. We can take [i]x,y[/i] as a coordinate system.
The incenters of $ APQ$ and $ PQCB$ are $ G_1$ and $ G_2$, where $ G_1\equal{}\minus{}G_2$ (superpositions property, $ G_1\plus{}G_2\equal{}G$)
then, the torque in $ G_1$ is the same as the torque in $ G_2$, respect $ P$ (really opposite), that is the mass of $ APQ$ = the mass of $ PQCB$, because $ PG_1G_2$ is isosceles. If the triangle has uniform density, the areas $ APQ$ and $ PQCB$ are the same, what is evident no true. Where is the error? please!

[quote="SimonM"][quote="El Chamo"]
Whit the pin in $ P$ i level the triangle and draw the linie [i]y[/i]. This linie pass throught $ G$, then is parallel to $ BC$[/quote]

I'm not sure, I haven't spent that much time reading what you've said. But I don't understand why something going through something else makes it parallel[/quote]

That´s geometry, $ GA$ is twice the distance $ GM$, where $ M$ is the midpoint of $ BC$, and $ AP$ is twice $ PB$