analytic geometry
graphing lines
slope
number theory
relatively prime

Below is an example of a square grid of dots with 3 dots along each side. In a similar square grid of dots with 8 dots along each side, how many distinct lines pass through the dot at the bottom left corner and at least one other dot of the grid?
[asy]dot((0,0)); dot((0,1)); dot((0,2));
dot((1,0)); dot((1,1)); dot((1,2));
dot((2,0)); dot((2,1)); dot((2,2));[/asy]

I shall explain what he did.

[hide="Solution"]Two lines with the given properties are different if and only if they have different slopes. Therefore we are counting the number of different slopes we can get. Now the slope of a line is the rise over the run, so if the line passes through the coordinate (n,m), then it has a slope of m/n. The coordinate that the line passes through has the following two restrictions placed on the coordinates:

1. mn=0 (at least one of m or n is zero)
or
2. $ 1\leq m\leq 7$ and $ 1\leq n\leq 7$.

If 1 holds, then the slope of the line is either 0 or undefined, and there are two lines that have this property.

If 2 holds, then we will have to count distinct slopes. Now we only have to count the pairs (m,n) such that m and n are relatively prime, because the slopes to all of these points are each different, and we will have counted them all (because if GCD(m,n) is greater than 1, then we can cancel out something from the numerator and denominator of m/n to get that some point (y,x) such that (0,0), (y,x), and (n,m) are collinear, and GCD(y,x)=1, and we would have already counted this line.

Now the n such that m=1 and GCD(m,n)=1 are 1, 2, 3, 4, 5, 6, and 7. This gives 7 lines.
The n such that m=2 and GCD(m,n)=1 are 1, 3, 5, and 7. This gives 4 more lines.
If n=3, then m is either 1, 2, 4, 5, or 7. This gives 5 lines.
If n=4, then m is either 1, 3, 5, or 7. This gives 4 more lines.
If n=5, then m is either 1, 2, 3, 4, 6, or 7. This gives 6 more lines.
If n=6, then m is either 1, 5, or 7. This gives 3 more lines
If n=7, then m is either 1, 2, 3, 4, 5, or 6, and we hae 6 more lines.
We therefore have 7+4+5+4+6+3+6=35 lines. Add the two lines we found in case 1 to get [b]37[/b] total lines.[/hide]

The slopes of these lines are in the form x/y where x and y are integers from 1-7. Using this, we 49 different lines, but we have to subtract ones we counted twice. The slope 1 was counted 7 times (1/1, 2/2 ...) so we subtract 6 and get 43. The slope 2 was counted 3 times (2/1, 4/2, 6/3) and 1/2 is the same, because it is just the inverse. So, we subtract 4 to get 39. Lastly, 3 and 1/3 were counted twice (3/1, 6/2), so we subtract 2 to get [b]37[/b].


[hide]
First think what is the formula for an infinite geometric series?  :)
[/hide]

[hide="Hint"]$ \sum_{i\equal{}0}^\infty x^i \equal{} \frac{1}{1\minus{}x}$[/hide]


For those addicted to neverending stories- the same rules apply as in Never Ending Story- try to keep all content at at a PG level.

Word limit- 5-25

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

One sunny day, hero-wannabe Randy started out on his quest to...

find the biggest porkchop in the land of Udoria. He goes to

Because the other one's really pointless, and this one will be soon too.

Does that make any sense? I didn't think so either..

[quote="hwenterprise"]A moderator should put this Neverending Story as a place where posts don't count, so people can't spam to earn posts.

The story begins:

It all started with a dark and windy night in the land of America, Samuel Gates was looking out the back window at the clouds racing above...

(Sorry, 27 words...:blush:)[/quote]

and suddenly lightning struck a tree that was right next to his house.

c'mon guys, one never ending story is bad enough, now two?  :huh:  This is not good.....

.. which shut out all the power in the city...

The city was plummeted into darkness, and then, the terrible rath of the storm knocked down a pillar supporting the main water system...

...Samuel Gates watched as the storm became fiercer and fiercer, the sky darker and darker, the rain heavier and heavier, to the point when...

the water all flowed into the city called Nwe Olreans and...

.....cardwacko9999 lost his AoPs account in the violent tides of the admin, for creating an additional pointless topic.......

except the admin relented because he was cool and he realized spammers are basically really fun people at heart

No guys. Just no. One of these things is plenty. -.-

Locked.


algebra
polynomial
number theory unsolved
number theory

:)  Hey People!!!  :)

This is a problem I don't have any idea of solving it:

Let p(x) be the polynomial x^3 + 14*x^2 - 2*x + 1.
Let p(n)(x) denote p(p(n-1)(x)), p(1)(x)=p(x)
Show that there is an integer N such that p(N)(x) - x is divisible by 101 for all integers x.

(PS.: Do exist infinitely many N?)

Already on this site. Go to :
http://www.mathlinks.ro/viewtopic.php?p=14148&highlight=#14148
..........


Is there an easy way to use generating functions to find the number of positive integer solutions to $ 3x\plus{}4y\plus{}5z\equal{}100$?

How about this ... Let
$ g_3(u) \equal{} u^3 \plus{} u^6 \plus{} u^9 \plus{} \dots$, exponents multiples of $ 3$, and similarly
$ g_4(u) \equal{} u^4 \plus{} u^8 \plus{} u^{12} \plus{} \dots$ ,
$ g_5(u) \equal{} u^5 \plus{} u^{10} \plus{} u^{15} \plus{} \dots$ .
Now in the product $ g_3(u) g_4(u) g_5(u)$ isn't the coefficient of $ u^a$ equal to the number of solutions of $ 3x\plus{}4y\plus{}5z\equal{}a$ ?

Yea, I got stuck on finding that coefficient without casework.

[hide="solution"]
The generating function would be \[ (x^3\plus{}x^6\plus{}x^9\plus{}\dots)(x^4\plus{}x^\plus{}x^{12}\plus{}\dots)(x^5\plus{}x^{10}\plus{}x^{15}\plus{}\dots)\] and the coefficient we want is the coefficient of $ x^{100}.$

Factor $ x^3$ out of the first, $ x^4$ out of the second, and $ x^5$ out of the third. It would be \[ x^{12}(1\plus{}x\plus{}x^2\plus{}x^3\plus{}\dots)^3\equal{}x^{12}(1\plus{}3x\plus{}6x^2\plus{}10x^3\plus{}\dots).\] We just need to find the coefficient of $ x^{88}$ in the second term. In general, the coefficient of $ x^{k}$ in the power series \[ 1\plus{}3x\plus{}6x^2\plus{}10x^3\plus{}\dots\] is $ \frac{(k\plus{}1)(k\plus{}2)}{2}.$ Just substitute $ k\equal{}88$ to get \[ \frac{89\cdot90}{2}x^{100}\equal{}\boxed{4005}x^{100}.\] Therefore, there are $ 4005$ solutions to the equation \[ 3x\plus{}4y\plus{}5z\equal{}100.\][/hide]

When you factor $ x^3$ out, it should be $ (1\plus{}x^3\plus{}x^6\plus{}...)$.

Thanks! That solution was pretty good. :D

Sure, except that it's also wrong.  The correct generating function is $ \frac {x^{12}}{(1 \minus{} x^3)(1 \minus{} x^4)(1 \minus{} x^5)}$, and computing the coefficient of $ x^{100}$ from this generating function doesn't significantly simplify the computation; it's easier to divide into cases, for example based on the value of $ z$.  Computing an exact formula from this generating function is possible but highly tedious and won't help much unless you're either interested in asymptotics or in exact values for very large $ n$.

If you're interested, the asymptotic is $ \frac{(n\minus{}12)^2}{2 \cdot 3 \cdot 4 \cdot 5}$, so there should be approximately $ 64$ solutions.  The problem is that the terms of lower order are periodic in a nontrivial way.

How did you get the asymptotic from the generating function?  :)

The slick way is to compute the volume of the region $ 4y \plus{} 5z \le 88$ and divide by $ 3$.  The more general way is to look at the smallest pole on the positive real axis, which is $ x \equal{} 1$, and look at what happens to the function "locally" there.  It's not hard to see that near $ x \equal{} 1$,

$ \frac{1}{(1 \minus{} x)^3(1 \plus{} x \plus{} x^2)(1 \plus{} x \plus{} x^2 \plus{} x^3)(1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)} \approx \frac{1}{(1 \minus{} x)^3 3 \cdot 4 \cdot 5}$

and the Taylor series expansion of $ \frac{1}{(1 \minus{} x)^3}$ is precisely $ {n\plus{}2 \choose 2}$ by the generalized binomial theorem.  Usually one only takes the leading term because periodicity in the other terms makes those terms less useful, but in this case the denominator has no other double roots so the linear term is also trustworthy, which I should've included.

The method of looking at the nearest singularity ... there is a basic explanation in the book "generatingfunctionology" by Wilf


geometry
trapezoid
circumcircle
trigonometry
cyclic quadrilateral
trig identities
Law of Sines

[quote="Altheman"](Maybe the problem is a bit easy, but it is nice. :) )

Let $ A_1A_2B_2B_1$ be a cyclic quadrilateral (circumcircle $ \mathcal{O}$) for which $ A_1A_2//B_1B_2$. Let the tangent to $ \mathcal{O}$ at $ A_1$ intersect $ B_1B_2$ at $ Y$ and let the tangent to $ \mathcal{O}$ at $ B_1$ intersect $ A_1A_2$ at $ X$. Let $ M_1$ and $ M_2$ be the midpoints of $ A_1X$ and $ A_2X$, respectively. Let $ N_1$ and $ N_2$ be the midpoints of $ B_1Y$ and $ B_2Y$, respectively. Prove that $ M_1M_2$, $ N_1N_2$, and $ XY$ are concurrent.[/quote]
Sorry, but isn't the line $ M_1M_2$ equal to $ A_2X$ and the line $ N_2N_1$ equal to $ B_2Y$ and thus $ XY$, $ M_1M_2$, and $ N_1N_2$ are not concurrent?  Is the problem $ M_1N_1$, $ XY$, and $ M_2N_2$?  Or did I misinterpret the problem?

Assuming we want to prove $ M_1N_1$, $ M_2, N_2$ and $ XY$ are concurrent, it suffices to show that $ \frac {M_1X}{N_1Y} = \frac {M_2X}{N_2Y}$.  Rearranging things a bit, it suffices to show $ \frac {A_1X}{A_2X} = \frac {B_1Y}{B_2Y}$.  Because $ B_1X$ is the geometric mean of $ A_1X$ and $ A_2X$ and $ A_1Y$ is the geometric mean of$ B_1Y$ and $ B_2Y$, it suffices to show that $ \frac {A_2X}{B_1X} = \frac {B_2Y}{A_1Y}$.  This follows by the law of Sines, because $ m\angle{B_2A_1Y}=m\angle{A_2B_1X}$ and $ \sin{\angle{B_1A_2X}}=\sin{\angle{A_1B_2Y}}$.

Alternatively, we can show that $ \frac {YB_1}{XA_1} \equal{} \frac {B_1B_2}{A_1A_2}$ as follows.
$ \angle YA_1B_1 \equal{} \angle A_1A_2B_1$, $ \angle A_1B_1Y \equal{} \angle B_1A_1A_2$, $ \angle XB_1A_1 \equal{} \angle B_1B_2A_1$, and $ \angle XA_1B_1 \equal{} \angle A_1B_1B_2$.  Now,
\[ \triangle YB_1A_1\sim \triangle B_2A_2A_1\implies \frac {YB_1}{B_2A_2} \equal{} \frac {B_1A_1}{A_2A_1}
\]
\[ YB_1\times A_2A_1 \equal{} B_2A_2\times B_1A_1 \equal{} (A_1B_1)^2
\]
Also, we notice that
\[ \triangle XA_1B_1\sim \triangle A_1B_1B_2\implies \frac {XA_1}{A_1B_1} \equal{} \frac {A_1B_1}{B_1B_2}\implies XA_1\times B_1B_2 \equal{} (A_1B_1)^2
\]
Thus, $ XA_1\times B_1B_2 \equal{} YB_1\times A_1A_2\implies \frac {YB_1}{XA_1} \equal{} \frac {B_1B_2}{A_1A_2}$.


ratio
LaTeX
geometric series
absolute value

1. A set of consecutive positive integers beginning with 1 is written. Then one number is deleted. The average of the remaining number is 35 7/17. What number was deleted.

2.S1 is the sum of an infinite geometric series with first term a and common ratio r.
S2 is the sum of an infinite geometric series with first term a^2 and common ratio r^2. This is true for all Sn  and absolute value of r is less than 1 and absolute value of a is greater than 1. then determine the sum of the infinite series
1/S1+1/S2+1/S3+1/S4+...

1. Since the average is 35 5/17 we know that highest number would be in the 70 ish place. We know that the number of numbers is 69 since it is close to 70 and 69-1 is divisible by 17. Then we find out that the missing numbers has to be 15 since (69*70/2)-(35 5/17*68)=15.

[hide="2"]
S1=$ \frac{a}{1-r}$

S2=$ \frac{a^2}{1-r^2}$

etc

so we are looking for

$ S=\frac{1-r}{a}+\frac{1-r^2}{a^2}....=\left(\frac{1}{a}+\frac{1}{a^2}+...\right) - \left(\frac{r}{a}+\frac{r^2}{a^2}+...\right)$

so we have two different infinite geometric series.

$ S=\frac{\frac{1}{a}}{1-\frac{1}{a}}-\frac{\frac{r}{a}}{1-\frac{r}{a}}=\frac{1}{a-1}-\frac{r}{a-r}=\frac{a-r-r(a-1)}{(a-1)(a-r)=\frac{a-ar}{(a-1)(a-r)}$[/hide]

[quote="Mewto55555"][hide="2"]
S1=$ \frac {a}{1 - r}$

S2=$ \frac {a^2}{1 - r^2}$

etc

so we are looking for

$ S = \frac {1 - r}{a} + \frac {1 - r^2}{a^2}.... = \left(\frac {1}{a} + \frac {1}{a^2} + ...\right) - \left(\frac {r}{a} + \frac {r^2}{a^2} + ...\right)$

so we have two different infinite geometric series.

$ S = \frac {\frac {1}{a}}{1 - \frac {1}{a}} - \frac {\frac {r}{a}}{1 - \frac {r}{a}} = \frac {1}{a - 1} - \frac {r}{a - r} = \frac {a - r - r(a - 1)}{(a - 1)(a - r)} = \frac {a - ar}{(a - 1)(a - r)}$[/hide][/quote]

Fixed your $ \text{\LaTeX}$; the preview button's there for a reason ;)


logarithms
inequalities
inequalities proposed

Let $ a, b, c$ be real numbers with $ abc\equal{}1$.

Let $ p(x)\equal{}2x^3\plus{}3x^2\minus{}6x\plus{}1$.

Prove that $ \sum \frac{p(a)}{a^2} \ge 0$.

[hide="solution"] The sum is $ \frac {p(a)}{a^2} \plus{} \frac {p(b)}{b^2} \plus{} a^2b^2p\left( \frac {1}{ab}\right)$.  Expanding everything out, after some tedious algebra you find it can be factored as
\[ \frac {(a \plus{} b \minus{} 3ab \plus{} a^2b^2)^2}{a^2b^2}
\]
which is non-negative. [/hide]

[quote="djuro"]Let $ a, b, c$ be real numbers with $ abc \equal{} 1$.

Let $ p(x) \equal{} 2x^3 \plus{} 3x^2 \minus{} 6x \plus{} 1$.

Prove that $ \sum \frac {p(a)}{a^2} \ge 0$.[/quote]
$ \sum_{cyc}\frac {p(a)}{a^2}\geq0\Leftrightarrow\sum_{cyc}\left(\frac {p(a)}{a^2} \minus{} 6\ln a\right)\geq0,$ which is true because
$ \left(\frac {p(a)}{a^2} \minus{} 6\ln a\right)' \equal{} \frac {2(a \minus{} 1)^3}{a^3}.$

I have a little nicer solution. Keep trying!   :wink:

What do you think is this inequality sharp or not?

[quote="JoeBlow"][quote="arqady"]
$ \sum_{cyc}\frac {p(a)}{a^2}\geq0\Leftrightarrow\sum_{cyc}\left(\frac {p(a)}{a^2} \minus{} 6\ln a\right)\geq0$[/quote]

That'd be a slick proof for $ a,b,c > 0$, but they need not be positive, so the right hand sum is complex.[/quote]
Oh yes, you are right. Thank you!
I have read "abc=1, cyclic sum p(a)/a^2 is positive" : " $ a,b,c$ positive. :(

[quote="djuro"]Let $ a, b, c$ be real numbers with $ abc \equal{} 1$.

Let $ p(x) \equal{} 2x^3 \plus{} 3x^2 \minus{} 6x \plus{} 1$.

Prove that $ \sum \frac {p(a)}{a^2} \ge 0$.[/quote]
It's equivalent to $ (ab \plus{} ac \plus{} bc \minus{} 3)^2\geq0.$

[quote="arqady"]It's equivalent to $ (ab \plus{} ac \plus{} bc \minus{} 3)^2\geq0.$[/quote]

Yeah, that's exactly the way I have created it!   :D


Let XYZT be a rectangle. Let's take A the midpoint of XY and B the midpoint of XT. Let C be the intersection point of TA with ZB. Find AX/XY,BX/XT,CB/CZ, CT/CA.

[hide="Solution"]For convenience, I will solve the problem stated as follows:
[b]Let $ ABCD$ be a rectangle. Denote by $ M$ and $ N$ the midpoints of $ AB$ and $ AD$, respectively. The lines $ DM$ and $ CN$ meet in $ P$. Find the ratios $ \frac{AM}{AB}$, $ \frac{AN}{AD}$, $ \frac{PN}{PC}$, $ \frac{PD}{PM}$. [/b]
It is trivial that $ \frac{AM}{AB}\equal{}\frac{AN}{AD}\equal{}\frac{1}{2}$.
To compute the last two ratios, I will use areas. We get $ \frac{PN}{PC}\equal{}\frac{[DNM]}{[DCM]}\equal{}\frac{[DAM]}{2\cdot [DCB]}\equal{}\frac{[DAB]}{4\cdot [DCB]}\equal{}\frac{1}{4}$. At last, $ \frac{PD}{PM}\equal{}$ $ \frac{[NDC]}{[NMC]}\equal{}$ $ \frac{[ADC]}{2\cdot ([ABCD]\minus{}[DNC]\minus{}[NAM]\minus{}[MBC])}\equal{}$ $ \frac{[ADC]}{4\cdot [ABC]\minus{}[DAC]\minus{}[DAB]/2\minus{}[ABC]}\equal{}\frac{2}{3}$.
[b]Remark[/b]. The result is the same when $ ABCD$ is a parallelogram. However, this is an easy example - areas can be successfully used on problems which are hardly attackable by other elementary means.
[/hide]


group theory
abstract algebra
topology
invariant
linear algebra
matrix
absolute value

Here's a nice one:
  There exists a neighbourhood $V$ of $I_n$ in $GL_n(C)$ that does not contain any non-trivial (with more than one element) subgroup of $GL_n(C)$.

We may assume that we're working with the norm $\|M\|=\sqrt{\sum_{i,j=1}^n|a_{ij}|^2}$, because it induces the usual topology on $\mathcal M_n(\mathbb C)$.

Since our norm is invariant to unitary similarities, if we assume that our neighbourhood contains a non-trivial subgroup which thus has at least one non-trivial element $A$, then it also contains $UAU^*$ for any unitary matrix $U$. We can choose $U$ s.t. $UAU^*$ is upper triangular, so we may assume WLOG that $A$ is upper triangular, for simplicity. If we choose the neighbourhood small enough s.t. the eigenvalues (which must clearly have absolute value $1$) all have positive real part (i.e. the distance between such an eigenvalue and $1$ is $<\sqrt 2$), we're done, because given any $|z|=1,z\ne 1$, there is an integer $n$ s.t. the real part of $z^n$ is negative.

I hope it's correct.

I don't know what I was thinking :?. All I have shown here is that we can find a small enough neighbourhood $\mathcal V$ around $I_n$ s.t. if $A$ is an invertible matrix s.t. $A^n\in\mathcal V,\ \forall n\in\mathbb Z$, then the eigenvalues of $A$ are all equal to $1$. This doesn't mean that the matrix itself is $I_n$. :?

I'll give it more thought.

Edit: Actually, that's easy to take care of, since we can now assume we're working with upper-triangular matrices $A$ with $1$ on the main diagonal. By looking at the entries $1,2$ in $A^n$, we can show that $a_{1,2}=0$. Then we go on to $a_{1,3}$, and so on. We will find all the entries of $A$ which are not on the main diagonal to be $0$.

i can't understand why if $A$ is in the group then $UAU*$ is in the group  :?
what am I missing ?

[quote="alekk"]if $A$ is in the group then $UAU*$ is in the group  :? [/quote]

That is not what I said. This is what I meant:

Assume you have a ball $\mathcal B$ of a certain radius $r$ (under the mentioned norm) around $I_n$ which contains a group $<A>$, where $A\ne I_n$. Then $U\mathcal B U^*$ is the same neighbourhood around $I_n$ (this transformation preserves the norms of the matrices) and it contains $<UAU^*>$. From here we continue as above.

I've already received two PM's with questions about this problem, so you people are beginning to make me doubt it. Now, is it correct or not? :?

Grobber, probably people wouldn't doubt about it if you wrote it a little bit clearer.  ;)

Is at least the proof that there is a small enough neighbourhood of $I$ s.t. any matrices $A$ with the property that the group $<A>$ belongs to this neighbourhood must have all eigenvalues equal to $1$ clear now?

Yes, grobber, but this was quite clear.  :P  ;)

Ok, then, let's take it step by step: is the part where I said that we could assume $A$ to be upper-triangular clear and correct as well? :)

Grobber, I don't need to understand like a baby, you know? I just said that the fact some peoples which are not quite idiot did not understand means that you weren't clear.


first of all question says $ 4^x \plus{} 3^x \equal{} 2^x \plus{} 5^x$
Let f(x) = $ 4^x \plus{} 3^x \minus{} 2^x \minus{} 5^x$
f(1) = 0 so clearly 1 is a soln
f(0) = 0 so clearly 0 is a soln
f(2) < 0 and $ 5^x > 4^x \plus{} 3^x$ for x > 2
implies no soln for x > 2
also for x < -2 clearly there is no soln as $ 4^x \plus{} 3^x < 2^x$
Lemme think after this  :)

This is just an application of Mean Value theorem isnt it?
One can use it for the function f(t) = t^{x} to the t  in intervals [2,3] and [4,5] by writin the problemas 3 (to the x) - 2(to the x) = 5 (to the x)- 4 (to the x).indeed one must have c1 in (2,3) and c2 in(4,5) so that xc1(to the x-1) =3 (to the x) - 2(to the x) and similarly for the other thing. Now since c1 and c2 are in disjoint intervals, c1 and c2 are different forcing x=1 as the only solution apart from the trivial x=0.  :D

perfect da this was the solution...though it looks easy the method is so ingenious :)  :lol:
ok ur suppose dto post it as a different thread...i will move it :)

Write it as $ 5^x \minus{} 4^x \equal{} 3^x \minus{} 2^x$

The function $ x^a \minus{} (x \minus{} 1)^a$ is strictly increasing when $ x \geq2$ if a is not 0 or 1. So, if $ 5^x \minus{} 4^x \equal{} 3^x \minus{} 2^x$, then, f'(x)=0 for all x. i.e. a=0 or a=1.

One more method occurred to me later

Again write as $ 5^x$ - $ 4^x$ = $ 3^x$ - $ 2^x$

Now consider the function f(x) = $ x^a$

f'(x) = $ ax^{a\minus{}1}$

f"(x) = $ a(a\minus{}1)x^{a\minus{}1}$

Now, f"(x) is analogous to acceleration, it is rate at which f'(x) is increasing and here f"(x) is non-zero when a is not 0 or 1.

But here rate of increase is zero. Hence f"(x) = 0 i.e. a=0 or a=1

i don't get ur last method can u explain  that a bit?

I meant this

f'(x) > 0 tells us $ 5^x > 4^x$ and $ 3^x > 2^x$

But if f"(x)>0 then $ \frac {5^x \minus{} 4^x} {5 \minus{} 4} > \frac {3^x \minus{} 2^x} {3 \minus{} 2}$

And the opposite inequality if f"(x)<0

But here $ 5^x \minus{} 4^x \equal{} 3^x \minus{} 2^x$

This means f"(x) = 0


Problem1
Let ABC be a triangle and construct squares ABED,BCGF,ACHI externally
on the sides of ABC.Show that the points D,E,F,G,H,I are concyclic
if and only if ABC is equilateral or isoceles right.

Try complex numbers.. I think.. :?

And I also think that vecten's theorem should be of some help...I think..

:D

The center of this circle must be O, because the middle perpendiculars to DE and AB are coinside.

AB=2R sin C

OE^2=R^2+EB^2-2*R*EB cos EBO = R^2+4R^2 sin^2 C - 4R^2 sinC cos C=R^2(3+2cos(2C)+2sin(2C)).

So this six points concyclic iff cos 2C+sin2C=cos 2A+sin2A=cos 2B+sin2B
but this can be only in one case (very easy).


algebra
polynomial
search
algebra unsolved

$ P(x),Q(x),R(x),S(x)$ are polynomial such that
$ P(x^5)\plus{}Q(x^5)\plus{}x^2R(x^5)\equal{}(x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1)S(x)$
Prove that $ x\minus{}1|P(x)$.

I think you have a slight typo.

If I'm right, then http://www.mathlinks.ro/Forum/viewtopic.php?search_id=509040721&t=89664


It is Harmonic series. As I know, There is no answer of closed form.

if you can give a simple forme to this sum ..then am gonna give you Fields Medal  :rotfl:


Given the side length of a regular n-gon, find the area.

is it to something to do with a sequence?

I'm pretty sure there's some trig way...

EDIT: http://mathworld.wolfram.com/RegularPolygon.html

EDIT 2: This problem is also in the intermediate forum, so this topic is locked.


Find the exact value of $ \frac {\frac {2 \sin (140) \sec (280)}{\sec (220)} \plus{} \frac {\sec (340)}{\csc (20)}}{\frac {\cot (200) \minus{} \tan(280)}{\cot (200)}}$

It's a lot less cluttered now.

Let $ \theta \equal{} \pi/9$ (radians).  Then

$ \frac{2\frac{\sin 140 \, \sec 280}{\sec 220} \plus{} \frac{\sec 340}{\csc 20}}{\frac{\cot 200 \minus{} \tan 280}{\cot 200}} \equal{} \frac{2\frac{\sin 7\theta \, \sec 14\theta}{\sec 11\theta} \plus{} \frac{\sec 17\theta}{\csc \theta}}{\frac{\cot 10\theta \minus{} \tan 14\theta}{\cot 10\theta}} \equal{} \frac{2\frac{\sin 2\theta \, \sec (\minus{}4\theta)}{\minus{}\sec 2\theta} \plus{} \frac{\sec \theta}{\csc \theta}}{1 \minus{} \tan (\minus{}4\theta) \, \tan \theta}$

$ \equal{} \frac{\minus{}2 \sin 2\theta \, \sec 4\theta \, \cos 2\theta \plus{} \frac{\sin \theta}{\cos \theta}}{1 \plus{} \tan 4\theta \, \tan \theta} \equal{} \frac{\minus{}\frac{\sin 4\theta}{\cos 4\theta} \plus{} \tan \theta}{1 \plus{} \tan \theta \, \tan 4\theta}$

$ \equal{} \frac{\tan\theta \minus{} \tan 4\theta}{1 \plus{} \tan\theta \, \tan 4\theta} \equal{} \tan(\theta \minus{} 4\theta) \equal{} \minus{}\tan 3\theta$

$ \equal{} \minus{}\tan \frac{\pi}{3} \equal{} \minus{}\sqrt{3}$.


trigonometry
logarithms
integration
calculus
calculus computations

i think nobody can show that. because there is a error on it. the rigth formulae is

${\sum_{n=0}^{\infty }\frac{\frac{1}{3 n+2}+\frac{3}{3 n+1}}{27^{n}}=3 \log \left(\frac{13}{4}\right)}$

just put n=0 and see that the sumation is greater than 1


Let $a$ and $b$ be positive integers such that $\frac{2}{3} < \frac{a}{b} < \frac{5}{7}.$.Find the value of $a + b$ when $b$ has the minimum value....
 :? A thorough explanation would be appreciated....


I know this is a VERY silly question but I don't really see how to do this right now:

I want to show that an inner product (defined on a real vector space) is a non degenerate bilinear form, so basically I need to show that the matrix $A$ of the bilinear form is invertible (where $\langle v, w \rangle = x^TAy$, I guess this is standard notation). But how? I know that $x^TAx = 0$ implies $x = 0$ for all $x$, but then I'm stuck.

If A is a diagonalyzing matrice then you can write $y^tDy=0$ where D is the diagonal matrice with eigenvalues onto the diagonal. Write? If $d_i=0$ for some i then we cam take a y with y_i not 0 then we show that there exist an y not identically zero, which contradicts that y=0. So d_i isn´t zero for all i and thus A is invertible. Is this write?
   And if A isn´t a diagonayizing matrice, how to procede?

if it were not invertible, there would be a vector $0 \neq x \in \ker A$, contradiction...

Right, how could I not see that? Thanks.

I don't think you even have to use matrices

definition of an inner product f :

$f(v,w)=f(w,v)$
f is linear in its first component (or the second, that is convention)
$f(v,v)\geq 0$ with equality only if $v=0$

definition of a nondegenerate (symmetric) bilinear form
there is no nontrivial vector v such that for all w in the space $f(v,w)=0$

Well if f is an innerproduct, and v is a vector, let us suppose $f(v,w)=0$ for all w in the space
then also it holds for $w=v$ thus $f(v,v)=0$
the last part of the definition of inner product implies that $v=0$, thus f is a nondegenerate symmetric bilinear form

does this answer your question?

Yes, but the point is that we got that exercise in class before we learnt the theorem that there is no nontrivial vector v such that for all w in the space f(v,w)=0 iff f is non-degenerate. So I was looking for a way to do it without using the theorem. Anyways, I was really tired last night :D

Strange, then what is the definition of nondegenerate?  Because in our class that was the definition?

Our definition:

Let $\langle \cdot, \cdot \rangle$ be a bilinear form on a finite dimensional vector space $V$ with matrix $A$ (with respect to some basis of $V$). The rank of $\langle \cdot, \cdot \rangle$ is defined as the rank of $A$. Then $\langle \cdot, \cdot \rangle$ is non degenerate if and only if the rank of $\langle \cdot, \cdot \rangle$ is equal to $\dim V$.

Aah i see
well tastes differ of course, but I'd go for the other way around, which is more geometrical
A little subtly is now that one might wonder if the rank is not dependent upon the basis, and that requires proof
(not difficult however, as the matrices are congruent regarding a different basis)

Yes, that was another exercise we got in class: verify that the rank of a bilinear form is independent of the choice of a basis. By the way, I don't like the matrix approach either :(


khane haye yek matrix $ n \times n$ ra ba addadhaye namanfi por mikonim be shartike jame adade har satr va sotun $ 1$ shavad sabet konid mitavan $ n$ khane az matrix entekhab kard ke mosbat bashand va hich kodam ham satr va ham sotun nabashand.

ye ghazie ghavitar has,avval ye tarif,badam ghazie:

[b]Tarif:[/b] matrise $ P$ ra yek matrise tartibi namim,hargah derayehaye an $ 0,1$ bashand va dar har satro sotoone an daghighan yek $ 1$ be kar rafe bashad.

[b]Ghazie:[/b] [i]farz konid $ Q$ matrisi $ n\times n$ bashad ke derayehaye an aadadi haghighi va namanfi bashand,hamchenin jame derayehaye har satr va har sotoon barabare $ \ell$ bashad.sabet konid chenin matrisi ra mitavan be soorate $ Q \equal{} c_1P_1 \plus{} c_2P_2 \plus{} \ldots \plus{} c_mP_m$ nevesht ke dar an $ c_1,c_2,\ldots ,c_m$ aadadi haghighio namanfiand ke majmooeshan barabare $ \ell$ ast va $ P_1,P_2,\ldots ,P_m$ matrishai $ n\times n$ va tartibi hastand.[/i]

masale kheily rahat ba phillip hall dar miad.

dar morede ghazie ke BaBaK Ghalebi goft ,esbatesh kheily rahat ba hamin masale dar miad.mishe ro tedad arraye haye nasefre matrix esteghra zad. tebghe masale yek matrix be shekle $ aB$ ke $ B$ tartibie va $ a\leq 1$ vojode darad(a adad minimum dar n addade entekhab shodast) .agar matrixe avaliaro $ A$ begirim un mughe $ A\minus{}aB$ tedae arraye nasefre kamtari darad(albate $ A\minus{}aB$ sharte masala ro darad)
bra ghazie esbate digari ham darin?

oon ghaziei ke goftam ye tamrine maroofe ghazie Halle,hamoontori ke khodetam gofti...


If anybody has information and/or application material for the 2006 ARML team, could you please post it here?  Thanks.


geometry
geometric transformation
reflection
projective geometry
geometry unsolved

A line meets the lines containing sides $ BC,CA,AB$ of a triangle $ ABC$ at $ A_1,B_1,C_1,$ respectively. Points $ A_2,B_2,C_2$ are symmetric to $ A_1,B_1,C_1$ with respect to the midpoints of $ BC,CA,AB,$ respectively. Prove that $ A_2,B_2,$ and $ C_2$ are collinear.

If $ {A_1}'$ is the harmonic conjugate of $ A_1$ WRT $ BC,$ then the harmonic conjugate $ {A_2}'$ of its reflection $ A_2$  about the midpoint $ M_a$ is clearly the reflection of $ {A_1}'$ about $ M_a.$ By similar reasoning, $ A{A_2}',B{B_2}',C{C_2}'$ concur at the isotomic conjugate $ U'$ of the tripole $ U$ of the transversal $\overline{A_1B_1C_1}.$ Then, $ A_2,B_2,C_2$ are collinear on the trilinear polar of $ U'.$

Let $ M,N,P$ respectively be the midpoint of $ BC,CA,AB$. Since $ A_1,B_1,C_1$ are collinear, we obtain:

$ \frac {A_1B}{A_1C}.\frac {B_1C}{B_1A}.\frac {C_1A}{C_1B} \equal{} 1$

Since $ A_2,B_2,C_2$ respectively are the symmentric points of $ A_1,B_1,C_1$ wrt $ M,N,P$, therefore, we obtain:

$ \frac {A_2B}{A_2C}.\frac {B_2C}{B_2A}.\frac {C_2A}{C_2B} \equal{} 1$, which follows that $ A_2,B_2,C_2$ are collinear. $ \square$


Let a and b be integers not divisible by the prime number p. If a^p = b^p (mod p), prove that a^p = b^p (mod p^2).
Please help me!

Have you posted it in the right forum? Anyway use Fermats little theorem to show that $a-b$ is congruent to zero mod $p$ then use the fact that $a^{p}-b^{p}=(a-b)(a^{p-1}+\dots+b^{p-1})$ to finish your proof.

[hide="Solution"] Let $c \equiv \frac{a}{b}$.  We are given $c^{p}\equiv c \equiv 1 \bmod p$.  Let $c = 1+pk$.  Then

$(1+pk)^{p}= 1+p(pk)+\frac{p(p-1)}{2}(pk)^{2}+... \equiv 1 \bmod p^{2}$ [/hide]


Def: A compact space X is maximal compact iff every strictly larger (finer) topology on X is noncompact.

Problem:
1) A compact space X is maximal compact iff every compact subset is closed.

For one direction of (1), I can argue as follows:

Proof: Suppose that every compact subset of X is closed with respect to $ \tau_{1}$. Suppose on the contrary there exists a $ \tau_{2}$ that is strictly finer than $ \tau_{1}$ such that $ (X, \tau_{2})$ is compact. Let C be a closed subset of $ (X, \tau_{2})$. Then since every closed subset of a compact space is compact, C must be compact in $ (X, \tau_{2})$. But then C is compact in $ (X, \tau_{1})$ because $ \tau_{1} \subset \tau_{2}$ and since every compact subset of $ ( X, \tau_{1})$ is closed by the hypothesis,
C is closed in $ (X, \tau_{1})$, which means that $ \tau_{2} \subset \tau_{1}$, a contradiction. Hence if every compact subset of X is closed then X is maximal compact.

The other direction is the one I would need help on.

For that other direction, use the contrapositive. Suppose $ K$ is a non-closed compact set. Define a new finer topology $ \tau'$ on $ X$, generated by the open sets of the original topology $ \tau$ and $ P=X\setminus K$. Every open set in $ \tau'$ is of the form $ (U\cap P)\cup V$ for some open sets $ U$ and $ V$ in $ \tau$.
Now, take an open (in $ \tau'$) cover of $ X$ $ \{(U_i\cap P)\cup V_i\}$. Split it into $ \{U_i\cap P\}\cup \{V_i\}$; the subdivided open cover has a finite subcover if and only if the original did. Now, $ \{V_i\}$ is an open cover of $ K$, which has a finite subcover $ \{G_j\}$. Also, $ \{U_i\}\cup \{V_i\}$ is an open cover of $ X$ in $ \tau$, which has a finite subcover ${ \{H_k\}\cup\{G_l}$ (splitting among the two parts). The combination $ \{H_k\cap P\}\cup \{G_j\}\cup \{G_l\}$ is a finite open cover of $ X$ with respect to $ \tau'$, and is a subcover of our original (subdivided) cover.
This could be done for any open cover, so the finer topology $ \tau'$ is compact. Hence, $ X$ is not maximal compact.


I noticed the Resources page has lots of problems from previous years of math contests. I was just wondering, is there a page with answers to these problems?

I'm not sure what you're referring to.  Our resources page from the left sidebar?  Where are the problems?

You mean the Contests section, right?  We are still working on those. We're more focused on gettting the problems up right now.  There are still solutions linked to some of them. For example, http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2002 ...the S's on the right side of the page link to the solutions.  The process of putting all of the problems and solutions into the database is a long and slow one.  Bear with us  :)

The resources pages will be upgraded at some point to provide easier means to input solutions. For now you can click on the problem's number to discuss it on the forum.


geometry
rectangle
calculus
calculus computations

Find the solutions $u(x,y)$ to
\[ u\frac{\partial^2 u}{\partial x\partial y} = \frac{\partial u}{\partial x}\frac{\partial u}{\partial y} \]

Very nice indeed.

[hide="Idea"]

$u\frac{\partial^2 u}{\partial x\partial y} = \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}$
$\Rightarrow \frac1u\left(u\frac{\partial^2 u}{\partial x\partial y} - \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}\right)=0$
$\Rightarrow \frac{\partial }{\partial x}\left(\frac1u\frac{\partial u}{\partial y}\right)=0$ or $\frac{\partial }{\partial y}\left(\frac1u\frac{\partial u}{\partial x}\right)=0$

[/hide]

You can complete from here!

Another idea, which turns this nonlinear equation into a truly simple linear PDE.

[hide="hint"]Let $u=e^v.$ (Deal with the possibility that $u$ changes signs later)[/hide]

[hide="answer"]$u(x,y)=f(x)g(y)$[/hide]

[quote="jmerry"] answer $u(x,y)=f(x)g(y)$[/quote]
That is indeed true if $u$ doesnt vanish anywhere. But you may have terribly complicated solutions looking like countably many different islands of this kind supported by rectangles floating in the ocean where $u=0$...


Carolina Panthers

I hope patriots go 0-16.

philly!!! come back and repeat as NFC champs, but this time, win the Super Bowl :D

As long as Belicheck is still coach, New England Patriots are the team to beat.

this will not happen, BUT GO COWBOYS!!!!

[quote="GoBraves"]I hope patriots go 0-16.[/quote]
I hope the Patsies play the Steelers in New England for the AFC Championship, the Steelers are supposed to lose by 20, and the Patriots are up by 6 with 10 seconds to play, Brady fumbles it, and the Steelers run it back for a touchdown!(and win)

[quote="plokoon51"]philly!!! come back and repeat as NFC champs, but this time, win the Super Bowl [/quote]
I live near Philly, so I want them to win the NFC, but not the Superbowl. That spot is reserved for the STEELERS :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

[quote="ln(dx/dy)"][quote="GoBraves"]I hope patriots go 0-16.[/quote]
I hope the Patsies play the Steelers in New England for the AFC Championship, the Steelers are supposed to lose by 20, and the Patriots are up by 6 with 10 seconds to play, Brady fumbles it, and the Steelers run it back for a touchdown!(and win)[/quote]

Lol, very entertaining story.

Too bad there's a such thing as a "quarterback kneel".

[quote="KrNboy10"]the raiders are gonna be killin it next year.... :lol: anyways.... now they got randy moss, and trade for micheal vick.... tom brady ain't nothin!

by the way, hope the patriots suck next year!!! 0-16  :D  :D  :D[/quote]

LOL, RAIDERS??? They still don't have any defense at all!

mark my words, the $Minnesota Vikings - Randy Moss = Vikings^{2}$

This year $Vikings \geq Any Past Superbowl Winning Team$

[quote="shobber"]NFL, is that sth about the SUPER BOWL?[/quote]

what is 'sth'?  I think that what has been posted pertains to next year: aka Year of the Dynasty.

[quote="Jhyon8149"]Who do you think the 49ers will draft with their #1 pick? I like them but there is consensus on the best player this year.... A bunch of good players but no superstar....[/quote]

they (ESPN) say they'll go after a QB like Alex Smith of Utah or Aaron Rodgers of Cal. but i thought they still had Ken Dorsey, who should be a pretty able (future) QB...

then again, the 49ers aren't really my all-time favorite team in the world

Patriots! Yeah! They're gonna win again! 4 in 5 years!

...erm, I could swear I remember someone saying something similar before  :lol:

[color=red][size=200]BUCCANEERS!!!!!!!![/size][/color]

they happen to be the team that won the super bowl between all the patriot's wins. nobody seems to realize that. and they'll do it again this year.

[quote="miraculouspostmaster"]Go Kansas City Royals!!!,
er,  Chiefs!! (whichever one plays football)[/quote]

LOL. i can tell you are a real sports fan.

[quote="ln(dx/dy)"]Who do you think will be really good next year in the NFL?[/quote]

All Philadelphia Super Bowl would be amazing. If not, Steelers making to the Super Bowl, which is WAY overdue.

I agree with GoBraves....0-16 for the Patriots would be pretty cool.  :P

[quote="Dcadu9lvr"][quote="plokoon51"][quote="ankur87"][b][size=200][color=green]Eagles!![/color][/size][/b][/quote]

ankur87, i like your thinking :D[/quote]

As do I! :D[/quote]

And I admire your guys' ability to recognize good thinking.

[quote="ankur87"][/quote][quote="Dcadu9lvr"][quote="plokoon51"][quote="ankur87"][b][size=200][color=green]Eagles!![/color][/size][/b][/quote]

ankur87, i like your thinking :D[/quote]

As do I! :D[/quote][quote="ankur87"]

And I admire your guys' ability to recognize good thinking.[/quote]

and i scorn ur inability to recognize good thinking....
[b]COWBOYS!!![/b]

Up until the late 90s, your "team" was consisted mainly of jailbirds and problem-players.

And Michael Irvin is a terrible TV personality  :P


hello, we want to solve the equation with respect to $ x$. The range of definition is $ p\neq0,q\neq0,x\neq0$ and $ p\plus{}q\neq\minus{}x$. Further we have
$ \frac{1}{p\plus{}q\plus{}x}\equal{}\frac{x(p\plus{}q)\plus{}pq}{pqx} \Leftrightarrow pqx\equal{}(p\plus{}q\plus{}x)(pq\plus{}x(p\plus{}q))$. Simplifying this we get the equation
$ 0\equal{}(x^2\plus{}x(p\plus{}q)\plus{}pq)(p\plus{}q)$. If $ p\plus{}q\equal{}0$ then solves $ x \in \mathbb{R}, x\neq0$ the equation. Assuming $ p\plus{}q\neq0$ and solving the quadratic equation we get $ x_1\equal{}\minus{}q$ or $ x_2\equal{}\minus{}p$.
Sonnhard.


Given a triangle $ABC$ let  \begin{eqnarray*} x &=& \tan\left(\dfrac{B-C}{2}\right) \tan \left(\dfrac{A}{2}\right) \\ y &=& \tan\left(\dfrac{C-A}{2}\right) \tan \left(\dfrac{B}{2}\right) \\ z &=& \tan\left(\dfrac{A-B}{2}\right) \tan \left(\dfrac{C}{2}\right). \end{eqnarray*} Prove that $x+ y + z + xyz = 0$.

We now by trigonometry \[ \displaystyle tan(p)tan(q)=\dfrac{cos{p-q}-cos(p+q)}{cos(p-q)+cos(p+q)}  \] So $x=\dfrac{cos(B)-cos(C)}{cos(B)+cos(C)}$ and so on. If we replace these values in the equality it is equivalent to$\Sigma{(cos(B)-cos(C))(cos(A)+cos(B))(cos(A)+cos(C))}=0$. The sum of these three is:\[ \Sigma{cos^2(A)cos(B)-\Sigma{cos(A)cos^2(B)}=0 }\]. The first side is \[ (cos(A)-cos(B))(cos(B)-cos(C))(cos(C)-cos(A)) \]. I don't know if I am wrong or your problem is not correct.

[b]Lemma[/b]: $\tan{\frac{(C-A)}{2}}\tan{\frac{(B-C)}{2}}\tan{\frac{(A-B)}{2}}=\tan{\frac{(C-A)}{2}}+\tan{\frac{(B-C)}{2}}+\tan{\frac{(A-B)}{2}}$.
[u]Proof[/u]: The statement of the lemma is equivalent to $\tan{\frac{(C-A)}{2}}=-\frac{\tan{\frac{(B-C)}{2}}+\tan{\frac{(A-B)}{2}}}{1-\tan{\frac{(B-C)}{2}}\tan{\frac{(A-B)}{2}}}=-\tan{\frac{(A-C)}{2}}$. But this equality is obvious since $\tan{x}=-\tan{-x}$.

[color=red]Proof[/color]: Put the given expressions of $x,y,z$ into the identity we need to prove to obtain $\tan{\frac{B-C}{2}}\tan{\frac{A}{2}}+\tan{\frac{C-A}{2}}\tan{\frac{B}{2}}+\tan{\frac{A-B}{2}}\tan{\frac{C}{2}}+\tan{\frac{A}{2}}\tan{\frac{B}{2}}\tan{\frac{C}{2}}\tan{\frac{(C-A)}{2}}\tan{\frac{(B-C)}{2}}\tan{\frac{(A-B)}{2}}=0$

Using the statement of the lemma this identity is equivalent to \[ \tan{\frac{B-C}{2}}\tan{\frac{A}{2}}+\tan{\frac{C-A}{2}}\tan{\frac{B}{2}}+\tan{\frac{A-B}{2}}\tan{\frac{C}{2}}+\tan{\frac{A}{2}}\tan{\frac{B}{2}}\tan{\frac{C}{2}}(\tan{\frac{(C-A)}{2}}+\tan{\frac{(B-C)}{2}}+\tan{\frac{(A-B)}{2}})=0 \]
$\Longleftrightarrow\tan{\frac{B-C}{2}}\tan{\frac{A}{2}}(1+\tan{\frac{B}{2}}\tan{\frac{C}{2}})+\tan{\frac{C-A}{2}}\tan{\frac{B}{2}}(1+\tan{\frac{A}{2}}\tan{\frac{C}{2}})+\tan{\frac{A-B}{2}}\tan{\frac{C}{2}}(1+\tan{\frac{A}{2}}\tan{\frac{B}{2}})=0$
$\Longleftrightarrow\sum_{cyc}\tan{\frac{B-C}{2}}\tan{\frac{A}{2}}(1+\frac{\cos{\frac{B-C}{2}}}{\cos{\frac{B}{2}}\cos{\frac{C}{2}}})=0$
$\Longleftrightarrow\sum_{cyc}\sin{\frac{B-C}{2}}\sin{\frac{A}{2}}=0$
$\Longleftrightarrow\sum_{cyc}\frac{1}{2}(\cos{\frac{B-C-A}{2}}-\cos{\frac{A+B-C}{2}})=0$
But the last identity is trivial since $\cos{-x}=\cos{x}$.

it's well known that in a triangle (a+b)/(a-b)=tan(A/2+B/2) / tan(A/2-B/2)

since tan (A/2)= tan(90-B/2-C/2)=ctg(B/2+C/2)=1/tan(B/2+C/2)

then we have those three eqautions

x=(b-c)/(b+c)
y=(c-a)/(c+a)
z=(a-b)/(a+b)

we can put now
e=a+b
f= b+c
g=c+a

and a-b= g-f, b-c= e-g and c-a= f-e

and then the eqution is pretty easy to prove..

[hide="Here is a similar proof with the nice Felixx-s proof."][u]Theorem of the tangent[/u] ([b]Mollweide[/b]) $x\equiv \tan \frac{B-C}{2}\cdot \tan \frac A2=\frac{b-c}{b+c}$ a.s.o. $\Longrightarrow$

$\{ \begin{array}{cc} 0\cdot a+(x-1)\cdot b+(x+1)\cdot c=0\\ (y+1)\cdot a+0\cdot b+(y-1)\cdot c=0\\ (z-1)\cdot a+(z+1)\cdot b+0\cdot c=0\end{array}\Longrightarrow | \begin{array}{ccc}0 & x-1 & x+1\\ y+1 & 0 & y-1\\ z-1 & z+1 & 0\ \end{array}|=0\Longrightarrow$

$(1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)\Longrightarrow x+y+z+xyz=0\ .$[/hide]

If we notice carefully , we dont even need the assumption that $A,B,C$ is angles in a triangle . :) The result remains true for any real $A,B,C$  ;)

Simply use Napier's Analogy, which gives x = (b-c)/(b+c), etc. So, x + y + z  = -(a-b)(b-c )(c-a )/[(a+b)(b+c)(c+a)] = -xyz. Hence proved.

A similar question from KVS-JMO 2001 paper:

Simplify:
(a-b )/(a+b)  +  (b-c )/(b+c)  +  (c-a )/(c+a)  +     [(a-b )(b-c )(c-a )]/[(a+b)(b+c)(c+a)]

[hide=a problem of worth!] so we have $x=\tan{(\frac{B-C}{2})}\cdot \tan{\frac{A}{2}}$\\
$A=90^{\circ}-(\frac{B+C}{2})$\\
$x=\frac{\tan\left(\dfrac{B-C}{2}\right)}{\tan\left(\dfrac{B+C}{2}\right)}$ $\implies x=\frac{\sin\left(\dfrac{B-C}{2}\right)\cdot \cos\left(\dfrac{B+C}{2}\right)}{\sin\left(\dfrac{B+C}{2}\right)\cdot \cos\left(\dfrac{B-C}{2}\right)} $\\
we get $x=\frac{\sin{B}-\sin{C}}{\sin{B}+\sin{C}}\implies \frac{1-x}{1+x}=\frac{\sin{C}}{\sin{B}}$\\
similarly compute $\frac{1-y}{1+y}$ and $\frac{1-z}{1+z}$ to get \\
$\frac{1-x}{1+x}\cdot \frac{1-y}{1+y}\cdot \frac{1-z}{1+z}=1\implies (1-x)(1-y)(1-z)=(1+x)(1+y)(1+z)$\\
consider a cubic equation in $"t"$ with roots $x,y,z$ s.t.\\
$P(t)=t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz$\\
we have $P(1)=-P(-1) \implies \boxed{x+y+z+xyz=0} \blacksquare$[/hide]


Six circles are tangent to the seventh one, such that each of them is tangent to its two neighbours.  Prove, that three lines joing opposite points of tangency are concurrent in a point.

It is also on [url=http://mathworld.wolfram.com/SevenCirclesTheorem.html]mathworld.[/url]

Dear Mathlinkers,
 a nice proof can be find on the website of Stanley Rabinowitz
http://www.mathpropress.com/stan/
Sincerely
Jean-Louis


calculus
integration
superior algebra
superior algebra unsolved

need to find orthogonal base  of {1 ,sinx ,cosx}  x (o or 2p) iner product <f,g>= 0 integral  2p (f(x)g(x))dx     [/img]

Exactly where do you get stuck?

Also, this belongs in the linear algebra forum.

Please be more specific. I'll give you one eigenvector which I found by inspection, $ (1,1,1)$.


group theory
abstract algebra
superior algebra
superior algebra solved

show An is only subgroup of index 2 of Sn.

It was posted before that $A_{n}$ is the only nontrivial normal subgroup if $n>4$ (can be seen by intersecting with the simple group $A_{n}$).
But every subgroup of index $2$ is normal -> done.

Another argument: the problem is equivalent to determining a surjective homomorphism onto $\mathbb{Z}/(2)$. But that requires us to send at least one transposition to $1 \mod 2$ (they generate $S_{n}$), thus they require us to send all transpositions there (conjugate them).
Thus the kernel will always be $A_{n}$.

will you please give details of An is only nontrivial normal subgroup of Sn where n>4


[quote="vardanch"]a,b,c are positive real numbers, such that you can't construct with them a triangle.
$Prove$ $that$
$\sum {\frac{a}{b} + \sum {\frac{b}{a} \ge 7} }$.[/quote]
Let $a\geq b+c.$
Since $\frac{b}{c}+\frac{c}{b}\geq2$ remain to prove that
$a\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{b+c}{a}\geq5\Leftrightarrow$
$\Leftrightarrow(b+c)a^2-5bca+(b+c)bc\geq0.$
Let $f(x)=(b+c)x^2-5bcx+(b+c)bc.$
But  $\frac{5bc}{2(b+c)}\leq b+c$ and $f(b+c)\geq0.$
Hence, $f(a)\geq f(b+c)\geq0.$ :)

$PROBLEM$
[quote="vardanch"]a,b,c are positive real numbers, such that you can't construct with them a triangle.
$Prove$ $that$
$\sum {\frac{a}{b} + \sum {\frac{b}{a} \ge 7} }$.[/quote]
$SOLUTION$

Let $a\leq b<c$ $\Rightarrow$ $a+b<c$ $\Rightarrow$ $5(a+b+c)\geq 8(a+b)+2c$
$\Rightarrow$ $(a+b+c)\geq 2/5(4(a+b)+c)$ $\Rightarrow$ by $(A(m)\geq G(m))$
$(a+b+c)(1/a+1/b+1/c)\geq (a+b+c)(4/(a+b)+1/c)\geq 2/5(4(a+b)+c)(4/(a+b)+1/c)\geq 10$
Q-E-D.

[quote="vardanch"]If $a,b,c$ are positive real numbers, such that you can't construct with them a triangle. Prove that:
$\sum {\frac{a}{b} + \sum {\frac{b}{a} \ge 7} }$.[/quote]
Here's mine.
Suppose $a \ge b \ge c$, normalizing by $a+b+c=2$, hence $a \ge 1$.
Let denote the sum is $f(a,b,c)$. Consider:
\begin{eqnarray*}f(a,b,c)-f\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)&=&\frac{a}{bc(b+c)}(b-c)^2+\frac{(b-c)^2}{bc} \ge 0 \\ \implies f(a,b,c) &\ge& f\left(a,\frac{b+c}{2},\frac{b+c}{2}\right) \\ f(a,b,c)&\ge& \frac{4a}{2-a}+\frac{2-a}{a}+2 \\ &\ge& 7\ \ ( \text{since }\ a \ge 1).\end{eqnarray*}


calculus
integration
calculus computations

Evaluate $ \int_0^2 \frac{1}{\sqrt {1 \plus{} x^3}}\ dx$.

Really?
[url=http://www.wolframalpha.com/input/?i=integral+1%2Fsqrt%281%2Bpower%28x%2C3%29%29+dx+from+x%3D0+to+2]integral 1/sqrt(1+power(x,3)) dx from x=0 to 2[/url] - Wolfram|Alpha

Kunny, it's not like you to post integrals that do not have elementary closed form solutions.

We can obtain a series expression, however.


geometry
3D geometry
octahedron
symmetry
geometric transformation
rotation
prism

Olá!
I'm having difficulties with the following problem:

[code]A polyhedron is made of two regular octahedra, glued together that they share a common face. Find the elements of symmetry (rotation axes, mirror planes, centres of inversion, inversion axes and mirror axes) for that polyhedron and determine the point group.[/code]

What I think I know:
- one mirror plane (the glued face)
- one three-fold rotation axis (through the centroid of the triangles that are parallel to the glued face)
- no centre of inversion

Is there any another axis or plane that I'm not seeing?

To be honest, it is a chemical problem actually, but I think you guys from the maths department can tackle it better than chemists.  :D

the plane perpendicolar to the common face and true one side of the common equlateral triangle and that pass true the opposit vertex of that side, it is a mirror plane . So 3 other mirrors planes, one for each side.
the line that join one vertex of the common equilateral triangle to the midpoint of the opposit side it's a rotation axes. So other 3 rotation axes.
This 3 line are also inversion axes.
There isn't a centres of inversion cause also a Triangular Rectangular Prism don't have it.


Hi - I have a couple of questions regarding multivariable problems..

The first is, given three points P(-2,4,0), Q(1,2,-1), R(-1,1,2), how can you prove that the triangle with these vertices is an equilateral triangle. I was thinking of using the Pythagorean thm to find the distances between the lines (from P to Q, Q to R and R to P) and to see if all the distances matched up. Is this the correct way to solve this problem?

Secondly, given a point A(5,1,3) how do you know if this lies on a line or not? This one I have no idea how to solve..

Finding the three distances is probably the conceptually simplest way of determining whether the triangle is equilateral - so your instincts are correct. There are other methods, including finding two distances and one angle. (Find vectors, say $ \vec{PQ}$ and $ \vec{PR},$ and, via the dot product, the angle between those vectors.)

As for finding whether a point is on a line: what information about that line do you have? Do you have points? Equations?

That question makes no sense. You're not giving us all of your information. Of course, every point lies on [b]a[/b] line - infinitely many different lines.

$ P,\ Q,\ R\in x\plus{}y\plus{}z\equal{}2$.

[quote="prime_lover"]all I know is that I have to determine if that point lies on a line in $ R^{3}$[/quote]

So, the answer is YES.


How can I prove the following inequality [b]properties[/b]:

$1. \quad a \ge b \Longleftrightarrow a+c \ge b+c, \ \ \ \forall c \in \mathbb{R}$

$2. \quad a \ge b \implies ac \ge bc, \ \ \ \forall c \in \mathbb{R}^+$

and this:

$\left ( {\frac{a+b}{2}} \right )^2 > ab, \quad a \ne b$

[quote="Ramanujan"]Axiomatic proof ????  :|[/quote]

HOW ??

$1)$  $\boxed{a\geq b \Longrightarrow a+c \geq b+c}$ is an axiom at an ordered field like $\mathbb{R}$.
Now since $a+c \geq b+c$ adding $(-c)$ we take $a+c \geq b+c \Longrightarrow a+c +(-c) \geq b+c +(-c) \Longrightarrow a \geq b$

$2)$ $a \geq b \Longrightarrow a + (-b) \geq b+ (-b) \Longrightarrow a-b \geq 0$
Now another axiom say that  $\boxed{a \geq 0 \, \wedge b \geq 0 \Longrightarrow a \cdot b \geq 0}$.
We have $a-b \geq 0$ and $c \geq 0$ so $c \cdot (a-b) \geq 0 \Longrightarrow ca-cb\geq 0 \Longrightarrow ca-cb +cb \geq cb \Longrightarrow ca \geq cb$.

$3)$  For this i am not sure that it needs a proof. Maybe it comes from $x^2 \geq 0 , \, \forall x \in \mathbb{R}$.
Here is an idea: If $x \geq 0$ then $x^2 \geq 0$ using the second axiom. If $x \leq 0$ then $- x \geq 0$ and again $(- x)^2 \geq 0$ or  $x^2 \geq 0$.
Now take $x= \frac{a-b}{2}$ and bring it to the form you want using the axioms . I think that is.  :)

I am not sure why  $(- x)^2 = x^2$ . Anyone help ??  :D

Thank you for help. I think I found the 3rd one:

$a \ne b \Longleftrightarrow a>b \vee b>a$

$i) a > b \Longleftrightarrow a-b>0 \Longleftrightarrow (a-b)^2>0\Longleftrightarrow a^2 +b^2 -2ab>0\Longleftrightarrow a^2+b^2+2ab>4ab \Longleftrightarrow \boxed{\left (\frac{a+b}{2}\right)^2>ab}$

$ii) b>a \Longleftrightarrow b-a>0 \Longleftrightarrow (b-a)^2>0\Longleftrightarrow b^2+a^2-2ab>0 \Longleftrightarrow b^2+a^2+2ab>4ab \Longleftrightarrow \boxed{\left (\frac{a+b}{2}\right)^2>ab}$

$\therefore \boxed{\left (\frac{a+b}{2}\right)^2>ab, \quad a\ne b}$


I am applying to some colleges for Fall 2008 and wanted to know whether I must send 2 high school teacher recommendations apart from the counselor recommendation. Can't I send 1 high school recommendation and 1 from a university mathematics professor? I think that a weak recommendation from school would further weaken my already weak chances of getting in. Besides, I live in India and people in my high school know almost nothing about writing recommendations. I am applying only to need-blind colleges like Northwestern, Boston College, Cornell, Stanford. Could anyone guide me in this? I am posting this here because those colleges don't reply until weeks after I send them any e-mail.

I myself am using a recommendation from a teacher at a university and another from a retired teacher.  It's my understanding that as long as the teachers taught classes that you took while in high school, this is valid.

The important thing is to find a teacher who knows you well enough to mention specific information about you and above to find a teacher who likes you and will give you a strong recommendation. If your most recent math teacher teaches at a college rather than at a high school, that should be fine.

Thanks for all the info. My low Gpa in 10th grade and lack of international contests like IMO make it necessary for me to have a strong recommendation, which is why I want to send one from him. The people at the high school I go to do not know me well enough to give a strong recommendation.

Are you applying to the institutions that are need-blind for international students?  They are as follows:

   1.  MIT, Massachusetts Institute of Technology in Massachusetts
   2. Harvard University in Massachusetts
   3. Princeton University in New Jersey
   4. Yale University in Connecticut
   5. Williams College in Massachusetts
   6. Middlebury College in Vermont

Actually, I am not an international student since I am a US citizen. However, I have been living outside USA for 16 years. Hence, all states will treat me as a domestic but out-state applicant. That is why I cannot apply to state colleges which might be easier to get into.


[quote="Interval"]Can someone explain the concept of infinity?

QUESTIONS:

(1) What is infinity divided by infinity? Explain.

(2) Why is any number over x = 0?

SAMPLES:

(1/x) = 0

(2/x) = 0

(3/x) = 0, etc.....[/quote]
Please don't start this debate again.

$\frac{\infty}{\infty}$ is what is called an indeterminate form:

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

Now for your second question, think of the limit $\lim_{x\to\infty}\frac{a}{x},\ a\in\mathbb{R}_{+}$.

Well one way to think about it is that you don't know the relative size of the infities.  Consider $\lim_{x\to\infty}\frac{e^{x}}{x}$ this gives you $\frac{\infty}{\infty}$, but the limit is $\infty$ because the infinity in the numerator is greater.

You really have to analyze the situation from which the infities arose.

[quote="jli"][quote="Interval"]Can someone explain the concept of infinity?

QUESTIONS:

(1) What is infinity divided by infinity? Explain.

(2) Why is any number over x = 0?

SAMPLES:

(1/x) = 0

(2/x) = 0

(3/x) = 0, etc.....[/quote]

there are different infinities.  the lim x-> infinity of x/x is 1 while the lim x-> infinity of x^2/x is infinity.

also, nobody said that any number over x is 0 :maybe:  :huh:[/quote]

I think he meant for the x to represent infinity... in which case, Interval, use the latex code [code]\infty[/code] to use the symbol for infinity.

Anyways, imagine we have the fraction $\frac1x$. Then, say that $x$ gets really big. Not just big; HUGE. That is to say, it approaches infinity. Then, think about what happens when the denominator gets large. The fraction gets smaller. One hundredth is smaller than one tenth; one trillionth is alot smaller than one hundredth. But no, infinity is alot bigger than a trillion. Thus, it will continue to shrink, and shrink, and shrink, until it becomes so infinitesimally small that it approaches zero. Even if we multiply it by some number, ANY constant number $k$, it will still shrink until it approaches zero.

Translated to math, this means that for $k \in \mathbb{R}$ we have that $\lim_{x \rightarrow \infty^{+}}\left( \frac{k}x \right) =0$.

[quote="Interval"](1) What is infinity divided by infinity? [/quote]

I think a better question to ask would be

[quote="Interval"](0) [b]What is infinity?[/b][/quote]

Good question!

First of all, it's very important to realize that [b]infinity is not a number in the normal sense[/b].  It's just a concept.  Operations involving infinity are not necessarily well-defined.

A rigorous treatment of the infinite begins in calculus (real analysis), where we consider [b]limits.[/b]  Rather than naively "plugging in infinity," we say that

$\lim_{x \to \infty}f(x) = L$

If $| f(x)-L |$ can be made arbitrarily small.  This is, effectively, $f(\infty)$, but keep in mind that [b]sometimes limits don't exist.[/b]

[quote="moogra"][quote="Interval"]You said:

infinity/infinity is called an indeterminate form.

Why is it INDETERMINATE?

What about the other question?[/quote]

So (roughly speaking) 0/0 might mean 1, or it might mean 0 (and in fact, by appropriate examples, it can be made to mean anything); so it is indeterminate.[/quote]

Yes; here's something of a demonstration.

Let's say that $\frac{0}{0}= a$.  What does this mean?  Well, by the definition of division it means

$0 = 0 \cdot a$

But this is true for [b]any finite a.[/b]  Hence $\frac{0}{0}$ is [b]not well-defined.[/b]


inequalities
algorithm
function
floor function
algebra
polynomial

Does there exist any general algorithm with which one can find the number of solutions to an inequality of the form $ \sum^{n}_{k\equal{}1}c_{k}x_{k} \leq c_{n\plus{}1}$
where all the $ c_{i}$ are given constants and all the $ x_{i}$ are positive integers?

Use generating functions:
$ \displaystyle{(1 \plus{} a^{c_1} \plus{} a^{c_1^2} \plus{} \cdots a^{c_1^{\lfloor log_{c_1}{c_{n \plus{} 1}} \rfloor}})(1 \plus{} a^{c_2} \plus{} a^{c_2^2} \plus{} \cdots a^{c_2^{\lfloor log_{c_2}{c_{n \plus{} 1}} \rfloor}})\cdots(1 \plus{} a^{c_n} \plus{} a^{c_n^2} \plus{} \cdots a^{c_n^{\lfloor log_{c_n}{c_{n \plus{} 1}} \rfloor}})}$
The answer will be the sum of the coefficients of $ a^{z} : 0 \le z \le c_{n\plus{}1}$.

Sorry for butchering the latex. :oops:

Generalizing Pick's theorem to higher dimensions and then considering the right simplex might be the best way, though all the $ \gcd$'s involved to compute boundary points will make things hairy. [url=http://mathworld.wolfram.com/EhrhartPolynomial.html]Ehrhart polynomials[/url] are this generalization; that link shows that even doing this for a 3-term sum on the left side is not pretty.


MATHCOUNTS
search
AwesomeMath
summer program
FTW

The winners from this year's national Mathcounts competition were recognized at the White House on Tuesday.  The top four kids and their coaches along with the 1st place team were given an hour and a half behind the scene tour of the White House.  They got a guided tour by a secret service agent who told them all sorts of interesting stories about things in every room that they went into.  They got to go into the library and the china room and the reception room that is used to welcome visiting heads of state.  They also  went into the green room, the blue room and the main ball room.

After the tour they were taken to the West Wing entrance and were escorted into the Roosevelt room where they waited for the President.  Several got to sit in the Presidential chair at the conference table.  The President spent around 20-30 minutes with the entire group in the Oval Office where he told them about the history of the Presidential desk (Old Resolute) and how it was presented to the United States fby the Queen of England for their efforts in saving an English ship that had become trapped in the ice in the search for the Northwest Passage.

Afterwards he presented every member of the group with either a stick pin for the ladies or a tie clip to the gentlemen.  They all had the United States seal and the President's signature on them.  They also received a silver round page marker that had the Presidential seal  on it.  They also had pictures taken  by the White House photographer that will be mailed to them in the next few weeks.  The entire group then had their pictures taken by the national press corp for the local papers.  If you go to the White House web site their is a picture of the entire group with the President.

It was a really cool afternoon.  :lol:  :lol:  :lol:

I looked at the website and the picture; why wasn't Bobby there? :(

Bobby was missing because he was representing the USA at the World Primary Math Contest in Hong Kong.  He could not get back in time to join us at the White House.  I was able to get a set of the presents for him from the President.

The world Primary Math Contest is an intermational math contest that usually has around 40 countries involved.  The problems are a lot like Mathcounts problems but much harder.  He was on the team last year and he had a perfect score and his team won the award for the best non-Asian team at the competition.  I have not talked to him yet to see how he and the team did this year.  I think he gets back today.

Wait, doesn't Kevin Tian also do that? Or did he just miss that contest and go to the White House? Or did Kevin not make the PMWC?

[quote="JBoyd"]From what I understood, Kevin qualified to be on the team but declined for some reason.  He is at Awesome Math camp right now.[/quote]Oh, so I guess he went to AwesomeMath immediately after the White House ceremony?

Oh, I see. He's beast :D

And why isn't Bobby going to AwesomeMath? :(

wow.... meeting the president. I guess that is another good reason for preparing for mathcounts. But then again, Bush will be out of office so we never know how Obama and Mccain think about math. Hopefully they love it as much as we do.  :lol:

[quote="JBoyd"]Yes, all  three of the other Texas kids went to camp right after the Presidential trip.  Kevin did compete in the World Primary Math contest last year when they also won the award for the top non-Asian team in the competition.[/quote]

Hahaha, they award awards for the top non-asian team? Sounds like a beastly competition.

And I supposed the teams from china, japan, indonesia, etc. pwn everybody?

Sure. The Asians probably do, so that's why there exists an award for the top non asian team

[quote="RussianRocket"]In Asia math is probably so much more publicized and accepted then in the US. It sourta sucks here...[/quote] [b] Asia ftw![/b]

[quote="gauss1181"][quote="RussianRocket"]In Asia math is probably so much more publicized and accepted then in the US. It sourta sucks here...[/quote]
Yes I agree. The average fifth grader in China has the mathematical mind of an average 7TH GRADER IN THE USA![/quote]

Nah... more like the average high-schooler...

Only the best AoPS like Bobby Shen compare to the Chinease people.

Maybe it is also so much more accepted in China because there are so many people and not that many job slots.
I think if you are in the top 1% of smart people then you are accepted into college for free.

That means you got a scholarship if you are that smart...

[quote="isabella2296"]And that's just the [b]average[/b] fifth grader...

Heh, in my school kids think I'm a "genius" because I skipped two grades of math, but in China I'd just be average.[/quote]

i skipped two grades too...and we wouldn't be average we'd be way below...

in 1st grade they do long multiplication and division :(

its scary

I did multiplication and long division in the summer before 1st grade....

:( i learned it in 1st grade, but they didnt teach it in school till late 3rd and 4th.

Yes, that's true. My brother taught me all that stuff.

i learned by myself with my brothers textbooks o_o
this is going off-topic.

They have textbooks on multiplication/long division?!?

Well in the 4th grade glencoe books...yes..theres stuff about it

[quote="AIME15"]multiplication in 3rd, division in early 4th[/quote]

I think I learned that in first grade.

Anyways, this is getting out-of-hand.

I spam.  A lot.

i did too. but in our school system thats when you learn it :P


The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different
sets of three points of this grid can be the three
vertices of an isosceles triangle?

[asy]size(50);
dot((0,0));
dot((10,0));
dot((20,0));
dot((0,10));
dot((10,10));
dot((20,10));
dot((0,20));
dot((10,20));
dot((20,20));[/asy]

If we count the different possibilities, we find there are $ \boxed{4}$.

Err...  I count at least 24.

If you consider the four corners of a square, you can group any three to make a 45-45-90 isosceles triangle.  In this way we may form 4 triangles for each small square, and 4 for the large square.

Beyond that, you may select any two adjacent corners of the big square and connect them with the midpoint of the side opposite, and that accounts for 4 more triangles.

That's how I got 24, lemme know if I forgot any.

EDIT:  Argh, yes, there are a lot more! :P See Kiseki's note below.... you get 4 more by pairing any two corners with the center, 4 more by grouping any set of three of the edge midpoints, and 4 more by grouping two adjacent midpoints with the far corner.

Do the triangles need to be distinct or can they be congruent? That makes a difference.

EDIT: Then I am incorrect - I believe 24 is the right answer.

they can be congruent
all the question says is to choose three points for the triangle
so as long as the triangle is not in the exact same place, it counts

ummm...the answer wuz 36...but i counted 24 too...

1-1-sqrt2 * 16
2-2-2sqrt2 * 4
2-sqrt2-sqrt2 * 8
2-sqrt5-sqrt5 * 4
sqrt2-sqrt5-sqrt5 * 4

Hey, thats 36!

For small right triangles in a 1*1 square, there are 16
For right triangles in a 1*2 square, there are 16 (running count 32)
For right triangles in the large square there are 4 (running count 36)
For non-right triangles in a 1*2 rectangle, there are 8 (running count 44)
for non right triangles in the large square, there are 8 (running count 52)

Did i miss any? I think there are 52 triangles, much more than 36.

I think right triangles in 1*2 square are not "isoceles".  So the answer is really 36... thank you very much :)


Let $ x;y$ be positive integer numbers satisfying $ xy|x^2\plus{}y^2\plus{}2$
Caculate $ A\equal{}\frac{x^2\plus{}y^2\plus{}2}{3}$

Given $ xy|x^2\plus{}y^2\plus{}2$, therefore $ \frac{x^2\plus{}y^2\plus{}2}{xy}$ is and integer. Given that 3 can only be formed of positive integers by $ 3.1$. Without loss of generality, let $ x\equal{}3,y\equal{}1$

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

[quote="SimonM"]Given $ xy|x^2 \plus{} y^2 \plus{} 2$, therefore $ \frac {x^2 \plus{} y^2 \plus{} 2}{xy}$ is and integer. Given that 3 can only be formed of positive integers by $ 3.1$. Without loss of generality, let $ x \equal{} 3,y \equal{} 1$

$ A \equal{} \frac {x^2 \plus{} y^2 \plus{} 2}{xy}$
$ A \equal{} \frac {3^2 \plus{} 1^2 \plus{} 2}{3}$
$ A \equal{} 4$[/quote]
No one is telling you that xy=3, you have no guarantee that A is an integer.
[hide="idea"]Root flipping[/hide]

[quote="SimonM"]Given $ xy|x^2 \plus{} y^2 \plus{} 2$, therefore $ \frac {x^2 \plus{} y^2 \plus{} 2}{xy}$ is and integer. Given that 3 can only be formed of positive integers by $ 3.1$. Without loss of generality, let $ x \equal{} 3,y \equal{} 1$

$ A \equal{} \frac {x^2 \plus{} y^2 \plus{} 2}{xy}$
$ A \equal{} \frac {3^2 \plus{} 1^2 \plus{} 2}{3}$
$ A \equal{} 4$[/quote]

good solution.

Let $ \frac{x^2\plus{}y^2\plus{}2}{xy}\equal{}k$,hence
$ x^2\minus{}x(ky)\plus{}y^2\plus{}2\equal{}0$     $ (1)$.
Let us consider 2 cases:
1 case:
$ x\equal{}y\Rightarrow x^2|2$,thus $ x\equal{}y\equal{}1$. So $ A\equal{}\frac{4}{3}$.
2 case:$ x>y>1$.
Let $ x_0,y$,be a solution of the $ (1)$,such that $ x_0\plus{}y$ is less than any other $ x\plus{}y$,where $ (x,y)$ is a solution of $ (1)$ too.
Also we know that there exist one more solution of $ (1)$,we call it $ x_1$.
Then from Vieta:
$ x_0\cdot x_1\equal{}y^2\plus{}2$.Since $ y^2\plus{}2,x_0>0$,hence $ x_1>0$.And
$ x_1\equal{}\frac{y^2\plus{}2}{x_0}\leq x_0$.Equality doesn't occurs,because $ x_0^2\equal{}y^2\plus{}2$,has no solutions in positive integers.
Hence we have a contradiction to the min. of $ x_0\plus{}y$.Thus $ y\equal{}1$.
$ (1)\equal{}x\plus{}\frac{3}{x}\in\mathbb{N}$.Hence $ x\equal{}1$ or $ x\equal{}3$.
If $ x\equal{}1$ then as we've already known $ A\equal{}\frac{4}{3}$.
If $ x\equal{}3$ then $ A\equal{}4$.

[quote="Altheman"]thats the general idea of these problems...comment:

i) there is not necessarily a unique pair x,y so x+y is min

what if (4,3), (5,2) were solutions...[/quote]
OK,i think we must suppose that $ x$ is a minimal solution to this equation,or not?


function
real analysis
real analysis unsolved

[quote="mli"]Is it true that every distribution in the linear transformation sense ( http://en.wikipedia.org/wiki/Distribution_(mathematics) )
is also a distribution in the probabilistic sense, and, if so, do the notions of weak convergence match? [/quote]

No, these are two unrelated uses of the word "distribution".

A particular example on $ \mathbb{R}:$

$ \delta',$ defined so that its action on a test function $ \varphi$ produces the number $ \minus{} \varphi'(0),$ is a perfectly reasonable distribution in the Schwartz or generalized function sense.

I don't see any way that $ \delta'$ makes any probabilistic sense at all.

I think Schwartz made an infelicitous choice of terminology when he chose the word "distribution." We might be better off if "generalized function" or something like that had become the more widely accepted term. What is the usual term in Russian?

[Spelling note: yes, there is a "t" in this Schwartz. I'm referring to L. Schwartz (20th century, French), not H.A. Schwarz (19th century, German), whose name you see in a number of other places.]

I'm sorry...I misphrased that question.

Is it true that every distribution can be interpreted as a signed measure on Lebesgue-measurable sets?

err...never mind.  The above is still a good example.

A signed measure is a linear functional acting on a space of continuous functions. A distribution is a linear functional acting on a space of $ C^{\infty}$ functions. Hence you can always differentiate a distribution, whereas you can't count on being able to differentiate a signed measure.

Distributions were built to be differentiated - they're associated with the study of differential equations.


geometry
3D geometry
algebra unsolved
algebra

a is positive integer. For all positive integer n the number 4(a^n+1) =b^3 for any positive integer b. Prove that a=1

Hint:
$ (a\plus{}1)(a^2\plus{}1)(a^3\plus{}1)$ would have to be a cube  :wink:

If $ a>1$ then the above expression can be placed strictly between two consecutive cubes.


The circles S and S' meet at A and B. A line through A meets S, S' again at C, D respectively. M is a point on CD. The line through M parallel to BC meets BD at K, and the line through M parallel to BD meets BC at N. The perpendicular to BC at N meets S at the point E on the opposite side of BC to A. The perpendicular to BD at K meets S' at F on the opposite side of BD to A. Show that ∠EMF = 90o.  ;)

See:

http://www.mathlinks.ro/Forum/viewtopic.php?t=22165
http://www.mathlinks.ro/Forum/viewtopic.php?t=22201
http://www.mathlinks.ro/Forum/viewtopic.php?t=17322
http://www.kalva.demon.co.uk/short/soln/sh02g8.html
http://www.mathlinks.ro/Forum/viewtopic.php?t=15588 post #2 file IMO_2002_shortlist.pdf page 21 .

  Darij


Prove or disprove:
Every [b]directed[/b] graph with no leaves[i](vertex of degree 1)[/i] has two vertices of equal degree.

This holds only for finite graphs.
Moreover, I think you have to be more precise about the degrees in your statement. Are you talking two times about indegrees? Outdegrees? the sum of them?

Pierre.

I mean by the degree of a vertex that $ deg(v)\equal{} indeg(v) \minus{} outdeg(v)$
and also for finite directed graphs.
How con I prove that?

[quote="fesghelbahal"]I mean by the degree of a vertex that $ deg(v) \equal{} indeg(v) \minus{} outdeg(v)$
[/quote]

Then it is not true.
Let $ A,B,D$ be the vertices with directed egdes $ A \minus{} > B,A \minus{} > D,B \minus{} > D,$  :wink:

Pierre.

Oops, You are right. :oops:
I forgot an assumption: suppose that the degree of each vertex is nonzero!

Still not true...
Consider vertices $ A,B,C,D,E$ and oriented edges $ A \to E,B \to E,C \to E,D \to E,A \to D,B \to D,C \to D,A \to C$.
The respective degrees of $ A,B,C,D,E$ are $ \minus{} 3, \minus{} 2, \minus{} 1,2,4$.

Pierre.

Pierre, \to makes a much nicer-looking arrow in LaTeX.

My fault.
I think It's better to write the whole question here again:

Prove or disprove:
Every finite [b]directed[/b] graph with $ | \delta | \geq2$ has two vertices of equal degree.
Where $ deg(v)\equal{}indeg(v)\minus{}outdeg(v)$

[hide]PS: the assumption $ | \delta | \geq2$ is equal to that the graph has no leaves and no vertex of zero degree![/hide]

Seems still untrue...
Consider graph with vertices $ A_1,A_2,A_3,B_1,B_2,B_3$ and oriented edges $ B_j \to A_i$ for $ i,j \equal{} 1,2,3$, and $ A_{i \plus{} 1} \to A_i$ for $ i \equal{} 1,2$, and $ B_3 \to B_2$.
The respective degrees of $ A_1,A_2,A_3,B_1,B_2,B_3$ are $ 4,3,2, \minus{} 3, \minus{} 2, \minus{} 4$.  :wink:

Pierre.


If \[x \equal{} \frac { \minus{} 1 \plus{} i\sqrt3}{2}\qquad\text{and}\qquad y \equal{} \frac { \minus{} 1 \minus{} i\sqrt3}{2},\] where $ i^2 \equal{} \minus{} 1$, then which of the following is [i]not[/i] correct?

$ \textbf{(A)}\ x^5 \plus{} y^5 \equal{} \minus{} 1 \qquad \textbf{(B)}\ x^7 \plus{} y^7 \equal{} \minus{} 1 \qquad \textbf{(C)}\ x^9 \plus{} y^9 \equal{} \minus{} 1$
$ \textbf{(D)}\ x^{11} \plus{} y^{11} \equal{} \minus{} 1 \qquad \textbf{(E)}\ x^{13} \plus{} y^{13} \equal{} \minus{} 1$

We have $ x\equal{}e^{i \frac{2 \pi}{3}}$, $ y\equal{}e^{\minus{}i\frac{2 \pi}{3}}$

Now $ x^n\plus{}y^n\equal{}e^{i \frac{2n \pi}{3}}\plus{}e^{\minus{}i\frac{2n \pi}{3}}\equal{}2 Re(e^{i \frac{2n \pi}{3}})\equal{}2 cos (\frac{2n \pi}{3})\equal{}1$
So $ cos (\frac{2n \pi}{3})\equal{}\frac{1}{2}$ , checking $ n(mod \;\; 3)$ we know $ n\equal{}1, 2(mod \;\; 3)$ is the solution. So the answer is $ C$

Did you used Muavre's theorem for power of complex number  ?
 I'd like to know more for this theorem

He used Euler's formula more than DeMoivre.

$ e^{i\theta}\equal{}\cos\theta\plus{}i\sin\theta$

Anyway, DeMoivre's theorem is simple: $ (\cos\theta\plus{}i\sin\theta)^n\equal{}\cos n\theta\plus{}i\sin n\theta$, which follows from Euler's formula.

[quote="stephencheng"]We have $ x \equal{} e^{i \frac {2 \pi}{3}}$, $ y \equal{} e^{ \minus{} i\frac {2 \pi}{3}}$

Now $ x^n \plus{} y^n \equal{} e^{i \frac {2n \pi}{3}} \plus{} e^{ \minus{} i\frac {2n \pi}{3}} \equal{} 2 Re(e^{i \frac {2n \pi}{3}}) \equal{} 2 cos (\frac {2n \pi}{3}) \equal{} 1$
So $ cos (\frac {2n \pi}{3}) \equal{} \frac {1}{2}$ , checking $ n(mod \;\; 3)$ we know $ n \equal{} 1, 2(mod \;\; 3)$ is the solution. So the answer is $ C$[/quote]
I think it should be $ x^n \plus{} y^n \equal{} 2\cos(2n\pi/3) \equal{} \minus{} 1$ or $ \cos[2\pi/3 \plus{} 2(n \minus{} 1)\pi/3] \equal{} \minus{} 1/2\implies n\equiv \pm 1\mod 3$.

EDIT: fixed the $ \pm$ sign...

[quote="enndb0x"]So DeMoivre's theorem is just a case of Euler's theorem  ?[/quote]

Sort of, yes. Euler's formula was discovered after DeMoivre's theorem, so the earlier proof of DeMoivre's theorem involves binomial expansion or something... but now DeMoivre's theorem is in a sense generalized by Euler's formula


factorial
number theory
prime numbers
number theory proposed

Hi everybody,

I have found a way to write factorial (p-1)! where p is prime

(p-1)! = (a(1)p-1)*(a(2)p-1)*(a(3)p-1)*....*(a(k)p-1)

a(i) is an integer >0

I want to build some table of the coefficients a(i) to understand how it works.
If we can predict the values of a(i) it will be very helpful.
Any idea will be helpful.

Example :

12!= (1*13-1)*(2*13-1)*(5*13-1)*(6*13-1)*(25*13-1)

a(1)=1
a(2)=2
a(3)=5
a(4)=6
a(5)=25

Thank you for any comment

[quote="fedja"]Have you ever looked at the proof of Wilson's theorem?[/quote]

Yes I did.

Here is an algo to find the a(i)'s

1. For p create a list L={p-1,p-2,p-3,.....,p-(p-2),p-(p-1)}

Example : p=23

List L={22,21,20,19,.....,4,3,2,1}

2. From left to the right and for each number of the list solve ((p-i)*x) mod p = p-1
x is the first number solving the equation

Example :

22x mod 23 = 22 x=1 then store a(1)=(22*1 + 1)/23=1 remove 22 from the list L
21x mod 23 = 22 x=12 then store a(2)=(21*12 + 1)/23=11 remove 21 and [u]12[/u] from the list L (edited because of mistake)
20x mod 23 = 22 x=8 then store a(3)=(20*12 + 1)/23=7 remove 20 and [u]8[/u] from the list L (edited because of mistake)
19x mod 23 = 22 x=6 then store a(4)=(19*6 + 1)/23=5 remove 19 and 6 from the list L

and so on until the last couple of the list L

3. Sort the a(i) (ascending order) and print the new list of a(i) ordered with new indexes a(j)

My goal is not to compute (p-1)!
My goal is to see if there is some formula to write directly the a(i)'s
Something like a Pascal triangle.

I forgot something.
If someone can program it I will thank him a lot if he can give me the list of the a(i) for the prime numbers between 100 and 120.

Thank you for any help


modular arithmetic
number theory proposed
number theory

Let $ a,b,c,d\in\mathbb{N}$ such that $ \frac{a}{c}\equal{}\frac{b}{d}\equal{}\frac{ab\plus{}1}{cd\plus{}1}$
Prove that $ a\equal{}c$

We have:
\[ \frac{ab}{cd}\equal{}\left(\frac{ab\plus{}1}{cd\plus{}1}\right)^2\Leftrightarrow \frac{(ab\plus{}1)^2}{ab}\equal{}\frac{(cd\plus{}1)^2}{cd}\Leftrightarrow ab\plus{}2\plus{}\frac{1}{ab}\equal{}cd\plus{}2\plus{}\frac{1}{cd}\Leftrightarrow ab\plus{}\frac{1}{ab}\equal{}cd\plus{}\frac{1}{cd}\]
Then either $ ab\equal{}cd$ or $ ab\equal{}\frac{1}{cd}$. But $ ab$, $ cd$ - integer numbers, so in the second case $ ab\equal{}cd\equal{}1$, and we have $ ab\equal{}cd$ anyway. Then
\[ \frac{a}{c}\equal{}\frac{ab\plus{}1}{cd\plus{}1}\equal{}1\]

Let $ \frac{x}{y}\equal{}\frac{a}{c}$ where $ (x,y)\equal{}1$.  Then $ x|a,x|b$ implies that $ ab\plus{}1\equiv1\pmod{x}$.  Since we must have $ x|ab\plus{}1$ also, $ x\equal{}1$.  Similarly, we find $ y\equal{}1$ and so $ a\equal{}c$.


algebra
polynomial
function
quadratics
complex numbers
algebra proposed

Find all real variables $ a$, for which the sum of all real roots of the equation $ x^4\minus{}5x\plus{}a\equal{}0$ is $ a$.

[hide]If there are no real roots (if this case is allowed), we must have an empty sum for $ a\equal{}0$, but this has a real root of course. Similarly, if there are four real roots, the sum of the real roots is the sum of all roots, also 0. But the polynomial $ x^4\minus{}5x$ only has two real roots.

So we reduce to the case of two real roots and a complex conjugate pair $ z,\overline{z}$ such that $ z\plus{}\overline{z}\equal{}\minus{}a,z\minus{}\overline{z}\neq 0$. Subtract the equations for $ z,\overline{z}$ to get
\[ 0\equal{}z^4\minus{}\overline{z}^4\minus{}5(z\minus{}\overline{z})\equal{}(z\minus{}\overline{z})((z\plus{}\overline{z})(z^2\plus{}\overline{z}^2)\minus{}5)\]
This gives $ a\neq 0$, $ z^2\plus{}\overline{z}^2\equal{}\minus{}\frac{5}{a}$, and $ |z|^2\equal{}\frac{1}{2}a^2\minus{}\frac{5}{2a}$.

Now add the equations for $ z,\overline{z}$ to get
\[ 0\equal{}2a\minus{}5(\minus{}a)\plus{}z^4\plus{}\overline{z}^4\equal{}7a\plus{}(z^2\plus{}\overline{z}^2)^2\minus{}2|z|^4\]
\[ \equal{}7a\plus{}\frac{25}{a^2}\minus{}\frac{1}{2}\left( a^2\minus{}\frac{5}{a}\right)^2\]
\[ \Rightarrow 0\equal{}14a^3\plus{}50\minus{}(a^3\minus{}5)^2\equal{}25\plus{}4a^3\minus{}a^6\]
\[ \Rightarrow a\equal{}\boxed{\sqrt[3]{2\pm \sqrt{29}}}\][/hide]

[quote="scorpius119"]
\[ a \equal{} \boxed{\sqrt [3]{2\pm \sqrt {29}}}
\]
[/quote]

The sum of the roots is always $ > 0$. It is easy to see from the graphic of the function $ f(x) \equal{} x^4 \minus{} 5x$. So $ a \equal{} \sqrt [3]{2 \plus{} \sqrt {29}}$.

Yes, I should have checked $ |z|^2\geq 0$ to get $ a^3\geq 5$. At least for $ a\equal{}\sqrt[3]{5\plus{}\sqrt{29}}$, we can actually find $ z,\overline{z}$ with sum $ \minus{}\sqrt[3]{5\plus{}\sqrt{29}}$ and product $ \frac{\sqrt{29}}{2\sqrt[3]{5\plus{}\sqrt{29}}}$ because the discriminant of the resulting quadratic is negative.


If a positive integer is even, divide it by two. It it is odd, subtract one from it. Repeat this until the number is $ 0$. Let $ s(n)$ be the number of subtractions when input number is $ n$. How to compute $ \sum_{i \equal{} 1}^ns(i)$ effectively, in the cases $ 1\leq n\leq 10^{20}$?

[hide="Hint"]Use base $ 2$[/hide]

[hide="Formula for s(n)"]Consider $ n$ in base $ 2$ then $ s(n)$ is the number of ones on that base[/hide]

[hide="Formula for the sum for special n"]
Consider $ n\equal{}2^k$, consider all the $ i$ numbers on base $ 2$ with $ k\plus{}1$ digits there are in total $ 2^k$ numbers and $ (k\plus{}1)\cdot 2^k$ digits in that base and half of them are $ 1$ then that sum $ \equal{}(k\plus{}1)\cdot 2^{k\minus{}1}$[/hide]

Is it true?

This is what I found too. But if $ 2^k < n < 2^{k \plus{} 1}$ for some $ k\in\mathbb{Z}^\plus{}$, how to evaluate the sum $ \sum_{i \equal{} 2^k \plus{} 1}^n s(i)$?

[quote="MeKnowsNothing"]This is what I found too. But if $ 2^k < n < 2^{k \plus{} 1}$ for some $ k\in\mathbb{Z}^ \plus{}$, how to evaluate the sum $ \sum_{i \equal{} 2^k \plus{} 1}^n s(i)$?[/quote]

[hide="Hint"]Find a recursive formula, notice that for all $ i$ they always have the same amount of digits in base $ 2$ and start with a $ 1$[/hide]
[hide="Hint2"]$ n \equal{} 2^k \plus{} m$ with $ 0 < m < 2^k$[/hide]
[hide="Hint3"]Apply several times that recursive formula as the amount of $ 1$ on $ n$'s base $ 2$ expresion[/hide]


trigonometry
geometry
ratio
geometry unsolved

[quote="canada"]Can anybody draw picture to this problem

Let a convex quadrilateral $ APQC$ have its sides $ AP$, $ PQ$ and $ QC$ tangent to a minor circular arc $ ABC$ at the points $ A$,$ B$ and $ C$, respectively.
Let $ E$ be the projection of $ B$ onto $ AC$. Let semicircle with $ PQ$ as diameter cut $ AC$ at $ H$ and $ K$, with $ H$ between $ A$ and $ K$.

Now problem comes:
Without using triginometry, prove that $ BE$ bisects $ \angle PEQ$ and that $ PH$ bisects $ \angle APB$.[/quote]

Thank you very much man

May you take gold this year at IMO!!! :first:

[quote="canada"][color=darkred]Let a convex quadrilateral $ APQC$ have its sides $ AP$, $ PQ$ and $ QC$ tangent to a minor circular arc $ ABC$ at the points $ A$,$ B$ and $ C$, respectively.
Let $ E$ be the projection of $ B$ onto $ AC$. Let semicircle with $ PQ$ as diameter cut $ AC$ at $ H$ and $ K$, with $ H$ between $ A$ and $ K$.
Prove that $ BE$ bisects $ \angle PEQ$ and that $ PH$ bisects $ \angle APB$.[/color][/quote]
[color=darkblue][b]I don't like this bushy enunciation ![/b] I"ll present an equivalent enunciation.[/color]
[quote="Virgil Nicula"][color=darkred]Let $ w=\mathcal C (O)$ be the circumcircle of the triangle $ ABC$. Denote the intersections $ \{\begin{array}{c}E\in AC\ ,\ BE\perp AC\\\ J\in AC\cap BB\\\ P\in AA\cap BB\ ,\ Q\in CC\cap BB\end{array}$.
The circle with the diameter $ [PQ]$ cuts the line $ AC$ in the points $ H$, $ K$, where $ H\in (AK)$. Denote $ R\in PK\cap QH$.
Prove that the ray $ [EB$ is the bisector of the angle $ \widehat{PEQ}$ and $ O\in PH\cap QK\cap BR$.[/color][/quote]
[color=darkblue] Upon a time I posted the following very usual lemmas :[/color]

[color=darkred][b]Lemma I.[/b] Let $ ABCD$ be a convex quadrilateral for which $ \{\begin{array}{c}BA=BC\\\ \widehat{ADB}\equiv\widehat{CDB}\end{array}$. Then $ DA=DC$ or the given quadrilateral is cyclically.

[b]Lemma II.[/b] Let $ A,C,B,D$ be four collinear points in thie mentioned order and let $ P$ be a point which doesn't belong to the line $ AB$.
Prove that if two from the following sentencies are truly, then and another sentence is truly : $ \{\begin{array}{cc}1\blacktriangleright & \frac{CA}{CB}=\frac{DA}{DB}\\\ 2\blacktriangleright & PA\perp PB\\\ 3\blacktriangleright & \widehat{CPB}\equiv\widehat{DPB}\end{array}$.
[b]Remark.[/b] The first sentence states that the division $ (A,B,C,D)$ is harmonically, i.e. the points $ C$, $ D$ are conjugate w.r.t. the points $ A$, $ B$.[/color]

[color=darkblue][b]Proof.[/b] Denote the intersection $ D\in AA\cap CC$. Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline{ACJ}$ and the triangle $ PDQ$ :

$ \frac{JP}{JQ}\cdot\frac{CQ}{CD}\cdot\frac{AD}{AP}=1$ $ \implies\over{(CD=AD)}$ $ \frac{JP}{JQ}=\frac{AP}{CQ}$ $ \implies\over{\|\begin{array}{c}PA=PB\\\ QB=QC\end{array}\|}$ $ \frac{JP}{JQ}=\frac{BP}{BQ}$, i.e. the point $ B$ is the conjugate of the point $ J$ w.r.t. the points $ P$, $ Q$.
Apply the [b]lemma II[/b] : $ \{\begin{array}{ccc}HP\perp HQ & \Longrightarrow & \widehat{AHP}\equiv\widehat{BHP}\\\\ EB\perp EJ & \Longrightarrow & \boxed{\widehat{PEB}\equiv\widehat{QEB}}\\\\ KP\perp KQ & \Longrightarrow & \widehat{BKQ}\equiv\widehat{JKQ}\end{array}$.  If denote the point $ R\in PK\cap QH$, then $ BR\perp PQ$ because $ \{\begin{array}{c}HQ\perp HP\\\ KP\perp KQ\end{array}$
and the division $ (P,Q,B,J)$ is harmonically (the point $ R$ is the orthocenter of the triangle $ PSQ$, where $ S\in PH\cap QK$). Thus,  $ \boxed{O\in BR}$.
Apply the [b]lemma I[/b] to the convex quadrilaterals $ APBH$ and $ CQBK$ :
$ 1\blacktriangleright \{\begin{array}{c}PA=PB\\\ \widehat{AHP}\equiv\widehat{BHP}\end{array}\|$ $ \implies$ $ HA=HP$ because contrary, the quadrilateral $ APBH$ is inscribed in the circle with the diameter $ [PR]$, what is falsely.

$ 2\blacktriangleright \{\begin{array}{c}QC=QB\\\ \widehat{BKQ}\equiv \widehat{CKQ}\end{array}\|$ $ \implies$ $ KC=KB$ because contrary, the quadrilateral $ CQBK$ is inscribed in the circle with the diameter $ [BR]$, what is falsely.
In conclusion, $ \{\begin{array}{c}HA=HB\\\ KC=KB\end{array}\|$ $ \implies$ $ \boxed{O\in PH\cap QK}$. Accurately, $ O\in PH\cap QK\cap BR$.

[b]Remark.[/b] Here is a nice related problem about which I think that it is the [b]"spring"[/b] of the proposed problem.[/color]

[quote="Virgil Nicula"][color=darkred]

$ \mathrm{IN}\blacktriangleright$ The incircle $ w=\mathcal C(I)$ of the triangle $ ABC$ touches it in the points $ \{\begin{array}{c}M\in BC\\\ N\in CA\\\ P\in AB\end{array}$.
Denote $ \{\begin{array}{c}U\in IB\cap NP\\\ V\in IC\cap NP\\\ R\in BV\cap CU\end{array}$. Prove that the points $ U,V$ belong to the circle with the diameter $ [BC]$ and $ R\in IM$.
$ \mathrm{EX}\blacktriangleright$ The $ A$- exincircle $ w=\mathcal C(I_{a})$ of the triangle $ ABC$ touches it in the points $ \{\begin{array}{c}M\in BC\\\ N\in CA\\\ P\in AB\end{array}$.
Denote $ \{\begin{array}{c}U\in I_{a}B\cap NP\\\ V\in I_{a}C\cap NP\\\ R\in BV\cap CU\end{array}$. Prove that the points $ U,V$ belong to the circle with the diameter $ [BC]$ and $ R\in I_{a}M$.[/color][/quote]

I have solved first part of problem with similarity and some angle chasing
I think that it is a nice and simple solution and I put it as attachment.

Does anybody have solution to second part of problem with using only similarity, Menelaus ( I mean without this ratios, isogonal conjugate points etc. becous it is a little bit difficult to understand it)

[quote="April"]Denote: $ D=AP\cap CQ,\,J=AC\cap PQ$
We have $ DB,\,PC,\,QA$ are concurrent, so $ (J,\,B,\,P,\,Q)=-1\implies H(J,\,B,\,P,\,Q)=-1$
On the other hand, $ HB\perp HJ,$ hence: $ \angle PHB=\angle QHB$[/quote]

[b] Figure[/b]

[b]Stergiu[/b] what did you want to say with above post.


limit
real analysis
real analysis unsolved

$\forall n\in \mathbb{N}$ Define $x_{n+1}=x_{n}(1-x_{n})$ which $0<x_{i}<1$ prove that:
$\lim_{n\to\infty}nx_{n}=1$

$\lim_{n\to\infty}nx_{n}=\lim_{n\to\infty}\frac{1/x_{n}}{n}=\lim_{n\to\infty}(\frac{1}{x_{n+1}}-\frac{1}{x_{n}})$ by Stolz-Cesàro theorem. Now note that $\lim_{n\to\infty}x_{n}=0$.

this is the classical trick that appears while studying sequences of the form
$x_{k+1}=x_{k}-a x_{k}^{n}+...$
I'm sure that the sequences $x_{k+1}=\sin(x_{k})$, $x_{k+1}=\ln(1+x_{k})$, etc..  appeared on the forum. Take a look at it  :)

This problem is part of a list at
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=151928[/url]


Two integers are chosen randomly between 0 and 80 inclusive. The sum of these integers is divided by 9. What is the probability that the remainder is 3?

Well since the sum of any 2 integers can go up to 159 then we get that the sum could be 12,21,30,39,48,57,66,75,84,93,102,111,120,129,138,147, and finally 156

With the sums known we just list all the cases for the sums
12 Cases: 1,11 2,10 3,9 4,8 5,7
21: 1,20 2,19 3,18 4,17 5,16 6,15 7,14 8,13 9,12 10,11
30: 1,29 2,28 3,27 4,26 5,25 6,24 7,23 8,22...........

We continue to list all the cases until they are found then we divide the total cases by 80!/(78!*2!)
Remember that in the cases 1,11 is the same as 11,1 because we used the combination formula to find the total cases so order doesn't matter

MMPC win.

It can be noted that there are an equal number of numbers that have the same remainder when divided by 9. That is:
9 numbers =0mod 9
9 numbers=1 mod 9
9 numbers=2 mod 9
......
9 numbers=8 mod 9.

Now there are 9 possible numbers you can pick (n=0mod9, n=1mod9, n=2mod9, etc), and for any one you pick, there will be only one other number you can pick to get 3mod9 (eg. if you pick n=1mod9, you NEED to pick n=2mod9) So the probability is (9x1)/(9x9)=1/9.

It is important to note in the problem whether or not they are [b] distinct [/b] integers.


probability
articles
modular arithmetic
trigonometry

1) A Word of 6 letters is formed from a set of 16 different letters of the english alphabet(with replacement). Find out the probability that exactly 2 letters are repeated??

My solution = (16 x 15 x 14 x 13 x 12 x 5)/16^6

But the given solution is (6C2 x 16 x 15 x 14 x 13 x 12)/16^6

Whats the flaw???

2) A College employs 8 professors on their staff. their respective probability of remaining in employment for the next 10 years is 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. The Probability that after 10 years at least 6 of them will still work in the college is?

How to go about solvng this one??? any simple approaches??

3) Mony inc uses a quality check scheme on produced items before they are sent into the market. the plan is as follows: A set of 20 articles is readied and 4 of them are chosen at random. if any one of them  is found to be defective then the whole set is put under 100% screening again. If no defectives are found, the whole set is sent to the market. What is the probability that a box containing 4 defective articles will be sent into the market?

4) if the integers m & n are chosen at random from 1 to 100, then the probability that a number of the form 7^m + 7^n is divisible by 5 is?

This one seems daunting...any simple approaches to this one? Can we use geometric probability for this one??? if so how???

3:
There are $%Error. "comb" is a bad command.
{16}{4}$ ways to choose 4 nondefective objects, and $%Error. "comb" is a bad command.
{20}{4}$ total; therefore, the probability is $\frac{16\times15\times14\times13}{20\times19\times18\times17}= \frac{364}{969}$.

4:
[hide]Well, the powers of 7 mod 5 are 2, 4, 3, 1, 2, 4, 3, 2, etc.

In order to have $7^{n}+7^{m}|5$, then there are 4 possibilities: $7^{n}\equiv1\pmod{5}$ and $7^{m}\equiv4\pmod{5}$, or vice versa; and $7^{n}\equiv2\pmod{5}$ and $7^{m}\equiv3\pmod{5}$, and vice versa.

There are $25\times25$ ways of choosing each of these 4 possibilities; therefore, the probability is $\frac{2500}{10000}= \frac{1}{4}$.
[/hide]

hey not_trig

Can you explain the logic behind 3) and is there a way of solving 4) without mod???(i'm a bit uncomfortable with mod! :( )

[quote="fizsle"]1) A Word of 6 letters is formed from a set of 16 different letters of the english alphabet(with replacement). Find out the probability that exactly 2 letters are repeated??

My solution = (16 x 15 x 14 x 13 x 12 x 5)/16^6

But the given solution is (6C2 x 16 x 15 x 14 x 13 x 12)/16^6

Whats the flaw???[/quote]

I'm not used to thinking with permutations, but here goes...

The first four letters, non-repeated, are chosen in a certain arrangement: A B C D = 16*15*14*13
There are 12 possibilities for the repeated letter 'E', and there are 6C2 ways to place the two 'E's into A B C D. (Look up balls and urns formula)
So you end up with 16*15*14*13*12*6C2 for the numerator.

In your solution you aren't counting the arrangements of the repeated letter.

#4 -- I'm guessing any other way of doing it will still use mod rationality just without saying the word 'mod'.

$7^{a}\equiv n \pmod{5}$ means that $7^{a}$ leaves remainder $n$ when divided by 5. In other words, $7^{a}= 5k+n$ where $k$ is some integer.

If you have $7^{a}\equiv n \pmod{5}$ and $7^{b}\equiv m \pmod{5}$ and $n+m = 5$, then $7^{a}+7^{b}= 5k+n+5j+m = 5(k+j+1)$; thus, divisible by 5.

[quote="fizsle"]
2) A College employs 8 professors on their staff. their respective probability of remaining in employment for the next 10 years is 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. The Probability that after 10 years at least 6 of them will still work in the college is?

How to go about solvng this one??? any simple approaches??
[/quote]

Possibilities: All 8, only 7 (8 possbilities), only 6(56 possiblities).

Find probability for each of these 65 cases and then add them up.

(.2x.3x.4x.5x.6x.7x.8x.9)^10 + (.3x.4x.5x.6x.7x.8x.9)^10 + (.2x.4x.5x.6x.7x.8x.9)^10+...+(.2x.3x.4x.5x.6x.7x.8)^10+ (.4x.5x.6x.7x.8x.9)^10 +(.2x.5x.6x.7x.8x.9)^10+...(.2x.3x.4x.5x.6x.7)^10

In total, there are 1+8+56 = 65 cases.

I'm thinking of a shorter way to do this now...

#2...there must be another way.....probability of 65 cases is too long to solve in under 2 mins!..#1...yes now i see the flaw!

For #2, only 6 isn't 56 possibilities. It's $\binom{8}{2}=28$ possibilities. Even so... that's quite a few. (Order doesn't matter.)