find the area of the triangle formed by the points (7,8) (12,5) (-4,3)

Or you can apply the distance formula, and see that the lengths of the triangle are $ 5$, $ 2\sqrt {65}$, and $ \sqrt {146}$. Now applying huron's formula, we get Area of triangle = $ \sqrt {6679}/4$

See, I have this calculator program that I made, that if I just enter the legs of a triangle, it'll tell me the area. I was kinda lazy.. :blush:  :blush:

Ahhh so technically speaking u coudn't have done this without using a calc.

I would be possible, however I do believe that it would be possible. However, it would be a lot more time consuming.

Um shoelace gives a rational result.

Where'd those supposed side lengths come from?

Those side lengths came from the distance between points formula. However, I might have messed up.  :blush:

Yea, I failed. $ 5^2 \plus{} 3^2\neq5^2$!!!!
DOPE!!!! :oops:

The correct side lengths should be $ \sqrt{34},2\sqrt{65}, and \sqrt{146}.$

Thus the correct area = $ 29$.

I definately failed......

well... i looked up the supposed "shoelace theorem" but i dont get it. it means...???

The shoelace theorem is a formula for finding the area of a polygon given the coordinates of its vertices, like the problem that you posted....

i realize that. but how do i apply the points in the formula??

The formula is at http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem. The coordinates are $ (a_1, b_1), (a_2, b_2), ... , (a_n, b_n)$ in clockwise order. Thus for example you have the coordinate (1,2), thus $ a_1\equal{}1$ and $ (b_1)\equal{}2$.

In Kouichi Nakagawa's post, let $ A(\minus{}4,\ 3),\ B(12,\ 5), C(7,\ 8), D(7,\ 3),\ E(12,\ 3)$, the desired area is $ [ADC]\plus{}[CDEB]\minus{}[AEB]$.

Just find distances using distance formula and then use heron's formula... ($ s\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$)


geometry
circumcircle
power of a point
radical axis
cyclic quadrilateral
geometry unsolved

Let $ABCD$ be an acute triangle with orhocenter $H$.Let $AG$ be the  altitude through $A$ and let $D$ be the midpoint of $BC$.The circles,whose diameters are $BC$ and $AD$ reapectively ,meet at $E$ and $F$.Prove that:
1)The circumciecles of $H;E;G$  and of $H;F;G$ are both tangent to the circle whose diameter is the side $B$
2)$H;E;F$ lie on the same line.

2)
Consider $BMNC$ where $M, N$ are the feet of altitudes of $AB, AC$. Clearly, $\angle AED = \angle AFD = 90^{\circ}$. So $EF$ is the polar of $A$ wrt circle with diameter $BC$. On the other hand the polar of $A$ passes through $BN \cap CM = H$. Hence $E, F, H$ are collinear.

2) It is well-known (and straight-forward to check) that $AH\cdot HG = BH\cdot HM$ ($M$ is the foot of the altitude $BH$). Hence $H$ lies on the radical axis of the two circles as desired.

1) Side $B$ should be replaced by side $BC$ in your post. $AEGF$ is a cyclic quadrilateral whereas $AEF$ is isosceles. Hence by simple angle chasing (applying the angle between chord and tangent lemma) we verify ourselves that, i.e. $AE$, is a mutual tangent of the circumcircle of $HGE$ and the circle with diameter $BC$.


group theory
superior algebra
superior algebra theorems

Please forgive the double post.

I am going to do an independent study on Group Theory next semester, and I've been looking for a decent text. I've had a semester of Number Theory. Any suggestions?

Dummit and Foote?  You won't outgrow it, although I don't know how it is as an introductory text.  There are certainly simpler treatments to be found (Fraleigh, Herstein, et al.)

I wouldn't buy Fraleigh... it's too "simple", the book gives you the feeling that the author is treating you as a baby or something. I love Crown, Fenrick & Valenza but that book is not so well-known and difficult to find.

I find e-book "Abstract Algebra: The basic graduate year" by Robert B. Ash (it can be found on author's homepage I guess - try looking for it) extremely nice - fancy why it is "graduate" while it starts from the beginning.


Given $ a,b,c > 0$ and$ a \plus{} b \plus{} c \plus{} abc \equal{} 1$
Find the minimum of the value:
$ Q \equal{} a^3 \plus{} b^3 \plus{} c^3$

hello, with the method of Lagrange Multipliers i have got
$ a^3\plus{}b^3\plus{}c^3\geq 3\minus{}9\left(\frac{2}{\sqrt{5}\minus{}1}\right)^{\frac{1}{3}}\plus{}9\left(\frac{1}{2}(\sqrt{5}\minus{}1)\right)^{\frac{1}{3}}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, with the method of Lagrange Multipliers i have got
$ a^3 \plus{} b^3 \plus{} c^3\geq 3 \minus{} 9\left(\frac {2}{\sqrt {5} \minus{} 1}\right)^{\frac {1}{3}} \plus{} 9\left(\frac {1}{2}(\sqrt {5} \minus{} 1)\right)^{\frac {1}{3}}$
Sonnhard.[/quote]
very  nice  :)

My proof is only by AM-GM :)  :lol:  hah hah

Let $ \alpha$ be the positive root of the equation $ x^3 \plus{} 3x \minus{} 1 \equal{} 0$  (1)
$ a^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3a\alpha^2$
$ b^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3b\alpha^2$
$ c^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3c\alpha^2$
$ \alpha^2(a^3 \plus{} b^3 \plus{} c^3)\ge 3abc\alpha^2$
$ \implies a^3 \plus{} b^3 \plus{} c^3\ge \boxed{\frac {3\alpha^2 \minus{} 6\alpha^3}{1 \plus{} \alpha^2}}$

[quote="Kyoshiro"]Let \alpha be the positive root of the equation $ x^3 \plus{} 3x \minus{} 1 \equal{} 0$  (1)
$ a^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3a\alpha^2$
$ b^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3b\alpha^2$
$ c^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3c\alpha^2$
$ \alpha^2(a^3 \plus{} b^3 \plus{} c^3)\ge 3abc\alpha^2$
$ \implies a^3 \plus{} b^3 \plus{} c^3\ge \boxed{\frac {3\alpha^2 \minus{} 6\alpha^3}{1 \plus{} \alpha^2}}$[/quote]
you can't solve equations:$ x^3\plus{}3x\minus{}1\equal{}0$ :P  :P  :P

[quote="Kyoshiro"][quote="quykhtn-qa1"][quote="Kyoshiro"]Let \alpha be the positive root of the equation $ x^3 \plus{} 3x \minus{} 1 \equal{} 0$  (1)
$ a^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3a\alpha^2$
$ b^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3b\alpha^2$
$ c^3 \plus{} \alpha^3 \plus{} \alpha^3\ge 3c\alpha^2$
$ \alpha^2(a^3 \plus{} b^3 \plus{} c^3)\ge 3abc\alpha^2$
$ \implies a^3 \plus{} b^3 \plus{} c^3\ge \boxed{\frac {3\alpha^2 \minus{} 6\alpha^3}{1 \plus{} \alpha^2}}$[/quote]
you can't solve equations:$ x^3 \plus{} 3x \minus{} 1 \equal{} 0$ :P  :P  :P[/quote]
There exist $ u$ such that $ x \equal{} u \minus{} \frac {1}{u}$ because the equation $ u^2 \minus{} ux \minus{} 1 \equal{} 0$ has $ \Delta \equal{} x^2 \plus{} 4 > 0$
Since then, let $ x \equal{} u \minus{} \frac {1}{u}$. We have:
$ \left(u \minus{} \frac {1}{u}\right)^3 \plus{} 3\left(u \minus{} \frac {1}{u}\right) \equal{} 1$
Simplifying, we obtain $ u^6 \minus{} u^3 \minus{} 1 \equal{} 0$ and $ u \equal{} \sqrt [3]{\frac {1 \plus{} \sqrt5}{2}}$
Therefore $ \alpha \equal{} \boxed{\sqrt [3]{\frac {1 \plus{} \sqrt5}{2}} \minus{} \sqrt [3]{\frac {2}{1 \plus{} \sqrt5}}}$[/quote]

IT VERY LONG :rotfl:  :rotfl:  :rotfl:

[quote="quykhtn-qa1"][/quote][quote="Kyoshiro"][quote="quykhtn-qa1"][quote="Kyoshiro"]Let \alpha be the positive root of the equation $ x^3 + 3x - 1 = 0$  (1)
$ a^3 + \alpha^3 + \alpha^3\ge 3a\alpha^2$
$ b^3 + \alpha^3 + \alpha^3\ge 3b\alpha^2$
$ c^3 + \alpha^3 + \alpha^3\ge 3c\alpha^2$
$ \alpha^2(a^3 + b^3 + c^3)\ge 3abc\alpha^2$
$ \implies a^3 + b^3 + c^3\ge \boxed{\frac {3\alpha^2 - 6\alpha^3}{1 + \alpha^2}}$[/quote]
you can't solve equations:$ x^3 + 3x - 1 = 0$ :P  :P  :P[/quote]
There exist $ u$ such that $ x = u - \frac {1}{u}$ because the equation $ u^2 - ux - 1 = 0$ has $ \Delta = x^2 + 4 > 0$
Since then, let $ x = u - \frac {1}{u}$. We have:
$ \left(u - \frac {1}{u}\right)^3 + 3\left(u - \frac {1}{u}\right) = 1$
Simplifying, we obtain $ u^6 - u^3 - 1 = 0$ and $ u = \sqrt [3]{\frac {1 + \sqrt5}{2}}$
Therefore $ \alpha = \boxed{\sqrt [3]{\frac {1 + \sqrt5}{2}} - \sqrt [3]{\frac {2}{1 + \sqrt5}}}$[/quote][quote="quykhtn-qa1"]

IT VERY LONG :rotfl:  :rotfl:  :rotfl:[/quote]
Can you post yours?

[quote="Kyoshiro"][/quote][quote="quykhtn-qa1"][/quote][quote="Kyoshiro"][quote="quykhtn-qa1"][quote="Kyoshiro"]Let \alpha be the positive root of the equation $ x^3 + 3x - 1 = 0$  (1)
$ a^3 + \alpha^3 + \alpha^3\ge 3a\alpha^2$
$ b^3 + \alpha^3 + \alpha^3\ge 3b\alpha^2$
$ c^3 + \alpha^3 + \alpha^3\ge 3c\alpha^2$
$ \alpha^2(a^3 + b^3 + c^3)\ge 3abc\alpha^2$
$ \implies a^3 + b^3 + c^3\ge \boxed{\frac {3\alpha^2 - 6\alpha^3}{1 + \alpha^2}}$[/quote]
you can't solve equations:$ x^3 + 3x - 1 = 0$ :P  :P  :P[/quote]
There exist $ u$ such that $ x = u - \frac {1}{u}$ because the equation $ u^2 - ux - 1 = 0$ has $ \Delta = x^2 + 4 > 0$
Since then, let $ x = u - \frac {1}{u}$. We have:
$ \left(u - \frac {1}{u}\right)^3 + 3\left(u - \frac {1}{u}\right) = 1$
Simplifying, we obtain $ u^6 - u^3 - 1 = 0$ and $ u = \sqrt [3]{\frac {1 + \sqrt5}{2}}$
Therefore $ \alpha = \boxed{\sqrt [3]{\frac {1 + \sqrt5}{2}} - \sqrt [3]{\frac {2}{1 + \sqrt5}}}$[/quote][quote="quykhtn-qa1"]

IT VERY LONG :rotfl:  :rotfl:  :rotfl:[/quote][quote="Kyoshiro"]
Can you post yours?[/quote]
OK.let $ x=\sqrt[3]{a+b}+\sqrt[3]{a-b}$
        =>$ x^3-3\sqrt[3]{a^2-b^2}x-2a=0$
       choose$ a=\frac{1}{2},b=\frac{\sqrt{5}}{2}$:P  :P  :P


geometry
circumcircle
search
incenter
analytic geometry
ratio
trigonometry

In an acute-angled triangle $ABC$, we consider the feet $H_a$ and $H_b$ of the altitudes from $A$ and $B$, and the intersections $W_a$ and $W_b$ of the angle bisectors from $A$ and $B$ with the opposite sides $BC$ and $CA$ respectively. Show that the centre of the incircle $I$ of triangle $ABC$ lies on the segment $H_aH_b$ if and only if the centre of the circumcircle $O$ of triangle $ABC$ lies on the segment $W_aW_b$.

This has been posted before quite a few times. Search for "BWM geometry" or something. I remember that this can be very easily done by trilinears, but that it is really tough if you don't know trilinears.

I can't find the problem anywhere on MathLinks, except of in the file with the proposed solutions of the IMO Shortlist 1997 (it's actually problem 16 from that shortlist) and in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=5697 where no actual solution is given. However, the problem is Germany 15 in the op02 project. Here are two solutions I submitted:

[hide="Problem"][color=blue][b]Problem.[/b] In an acute triangle $ ABC$, let $ H_a$ and $ H_b$ be the feet of the altitudes from the vertices $ A$ and $ B$, respectively, and let the angle bisectors of the angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ W_a$ and $ W_b$, respectively. Prove that the incenter $ I$ of triangle $ ABC$ lies on the segment $ H_aH_b$ if and only if the circumcenter $ O$ of triangle $ ABC$.[/color][/hide]

[hide="First solution using trilinear coordinates"]The incenter I of triangle ABC has the homogeneous trilinear coordinates

  I ( 1 : 1 : 1 );

the line HaHb has the homogeneous trilinear line coordinates

  HaHb ( cos A : cos B : - cos C );

hence, I lies on HaHb if and only if

  1 * cos A + 1 * cos B + 1 * (- cos C) = 0

holds, i. e. if cos C = cos A + cos B holds.

The circumcenter U has the homogeneous trilinear coordinates

  U ( cos A : cos B : cos C );

the line WaWb has the homogeneous trilinear line coordinates

  WaWb ( 1 : 1 : -1 );

thus, U lies on WaWb if and only if

  cos A * 1 + cos B * 1 + cos C * (-1) = 0

holds, i. e. if cos C = cos A + cos B holds.

This makes clear that I lies on HaHb if and only if U lies on WaWb, and this is equivalent to cos C = cos A + cos B. Actually, another condition equivalent to cos C = cos A + cos B is that the excircle opposite to C is congruent to the circumcircle of triangle ABC.[/hide]

[hide="First solution in a more detailed form"]We will use the notion of trilinear coordinates. If a triangle ABC is given, and P is a point in its plane, then we define the trilinear coordinates of P with respect to triangle ABC as follows:

Let $ x_1$ be the directed distance from P to the line BC. ["Directed" means: positive if P and A lie in the same halfplane with respect to BC, and negative if P and A lie in different halfplanes with respect to BC.] Similarly, let $ y_1$ and $ z_1$ be the directed distances from P to the lines CA and AB, respectively. Then, the triple $ \left(x_1;\;y_1;\;z_1\right)$ will be called the triple of "actual trilinear coordinates" of the point P with respect to triangle ABC. Instead of "trilinear coordinates", we will often just say "trilinears".

Now, of course, three actual trilinear cordinates for a point in a given plane are "too much", i. e. they cannot be independent of each other. In fact, for every point P, we have $ a x_1 \plus{} b y_1 \plus{} c z_1 \equal{} 2S$, where a, b, c are the sides and S is the area of triangle ABC. (Prove this for yourself!) Hence, we define the notion of homogeneous trilinear coordinates: If x, y and z are three real numbers (not all of them being 0) such that $ x : y : z \equal{} x_1 : y_1 : z_1$, then the triple ( x ; y ; z ) is called a triple of "homogeneous trilinear coordinates" of the point P with respect to triangle ABC. In other words, the homogeneous trilinears are three arbitrary real numbers proportional to the actual trilinears of P. [By the way, the actual trilinears $ \left(x_1;\;y_1;\;z_1\right)$ of P are homogeneous trilinears of P, too - here the factor of proportionality is 1.] Hence, the homogeneous trilinears are defined up to proportionality only. To indicate this, we will use the notion ( x : y : z ) rather than ( x ; y ; z ) for the homogeneous trilinears of a point P. Also, we will write P ( x : y : z ) to show that ( x : y : z ) are the homogeneous trilinears of the point P. [Strictly speaking, rather than "are the homogeneous trilinears", it is better to say "are homogeneous trilinears", for every point P has an infinity of triples ( x : y : z ) serving as its homogeneous trilinears. In fact, as I've already said, only the ratios x : y : z are important.]

It is not difficult to show that for any three real numbers x, y, z such that the sum ax + by + cz doesn't equal 0, there exists one and only one point P in the plane such that ( x : y : z ) are the homogeneous trilinears of P. If the sum ax + by + cz is 0, then there does not exist any point P in the Euclidean plane with homogeneous trilinears ( x : y : z ). Now, if not all three of the numbers x, y, z are 0, but ax + by + cz = 0, then we can interpret the point with homogeneous trilinears ( x : y : z ) as an infinite point. But this is not significant for the following.

A crucial fact about trilinear coordinates is that for any line g, there are three numbers k, l, m such that a point P with homogeneous trilinears ( x : y : z ) lies on the line g if and only if

  kx + ly + mz = 0.

In other words, any line g has an equation of the form kx + ly + mz = 0 in trilinear coordinates. Then, the triple ( k ; l ; m ) is called the "homogeneous line trilinear coordinates", or just "line trilinears" of the line g with respect to triangle ABC. Of course, these numbers k, l, m are defined up to proportionality again; hence, we will write ( k : l : m ) instead of ( k ; l ; m ).

[Why does any line g have an equation of the form kx + ly + mz = 0 in trilinear coordinates? Well, this can be shown, for instance, if you go back to the actual trilinears: since $ x : y : z \equal{} x_1 : y_1 : z_1$ for every point P, the equation kx + ly + mz = 0 is equivalent to $ kx_1 \plus{} ly_1 \plus{} mz_1 \equal{} 0$. Now, if you set a Cartesian coordinate system somewhere in the plane, you see that this equation is linear in the coordinates of P, so that it represents a line. There is, of course, a degenerate case when the terms containing the coordinates of P cancel each other and hence the equation is not linear anymore, so that it represents either the whole plane or the empty set. The whole plane is impossible (except of the forbidden case k = l = m = 0), while the empty set occurs if and only if k : l : m = a : b : c, as can be shown by considering the points with homogeneous trilinears ( 0 : - m : l ), ( m : 0 : -k ) and ( -l : k : 0 ), which always satisfy the equation kx + ly + mz = 0, and among which there is always at least one finite point unless k : l : m = a : b : c. We leave the details to the reader.]

If we have two points P and P' with their homogeneous trilinears P ( x : y : z ) and P' ( x' : y' : z' ), then the homogeneous line trilinears of their joining line PP' are

  PP' ( yz' - y'z : zx' - z'x : xy' - x'y ).

[This can be proved by solving a system of two homogeneous linear equation for three variables.]

Now, it is not difficult to establish the following homogeneous trilinears (think of the geometric significations of the points!):

[b]1.[/b] The incenter I of triangle ABC has the homogeneous trilinears

  I ( 1 : 1 : 1 ).

[b]2.[/b] The feet $ H_a$ and $ H_b$ of the A- and B-altitudes of triangle ABC have the homogeneous trilinears

  $ H_a\left(0: \cos C: \cos B\right)$;
  $ H_b\left(\cos C: 0: \cos A\right)$.

[b]3.[/b] According to the above formula for the line PP', the line $ H_aH_b$ has the line trilinears

  $ H_aH_b \left( \cos A : \cos B : \minus{} \cos C \right)$.

[b]4.[/b] The circumcenter U of triangle ABC has the homogeneous trilinear coordinates

  U ( cos A : cos B : cos C ).

[b]5.[/b] The "feet" $ W_a$ and $ W_b$ of the angle bisectors from A and B have the homogeneous trilinears

  $ W_a \left( 0 : 1 : 1 \right)$;
  $ W_b \left( 1 : 0 : 1 \right)$.

[b]6.[/b] According to the above formula for the line PP', the line $ W_aW_b$ has the line trilinears

  $ W_aW_b \left( 1 : 1 : \minus{} 1 \right)$.

Now, remember that a point with homogeneous trilinears ( x : y : z ) lies on a line with line trilinears ( k : l : m ) if and only if kx + ly + mz = 0. Hence, I lies on $ H_aH_b$ if and only if

  $ \cos A\cdot 1 \plus{} \cos B\cdot 1 \plus{} \left( \minus{} \cos C\right)\cdot 1 \equal{} 0$

holds, i. e. if cos C = cos A + cos B holds.

On the other hand, U lies on WaWb if and only if

  $ 1\cdot\cos A \plus{} 1\cdot\cos B \plus{} \left( \minus{} 1\right)\cdot\cos C \equal{} 0$

holds, i. e. if cos C = cos A + cos B holds.

This makes clear that I lies on $ H_aH_b$ if and only if U lies on $ W_aW_b$, and this is equivalent to cos C = cos A + cos B. Actually, another condition equivalent to cos C = cos A + cos B is that the excircle opposite to C is congruent to the circumcircle of triangle ABC (an easy exercise to the reader).[/hide]

[hide="Second solution without trilinear coordinates but still not what people call beautiful"]This problem, one of the hardest classic geometry problems being given at the BWM, has a simple solution using trilinear coordinates. No completely synthetic solution is known to me. Here is an elementary solution which is not free of calculations, but still rather manageable.

I will use directed segments throughout the solution. The sidelines $ BC$, $ CA$, $ AB$ of triangle $ ABC$ will be directed in such a way that $ BC > 0$, $ CA > 0$ and $ AB > 0$ (and consequently, $ CB < 0$, $ AC < 0$ and $ BA < 0$). We denote $ a \equal{} BC$, $ b \equal{} CA$ and $ c \equal{} AB$.

Since the triangle $ ABC$ is acute, both its incenter $ I$ and its circumcenter $ O$ lie inside the triangle, so that, instead of proving that the incenter $ I$ lies on the [i]segment[/i] $ H_{a}H_{b}$ if and only if the circumcenter $ O$ lies on the [i]segment[/i] $ W_{a}W_{b}$, it will be enough to show that the incenter $ I$ lies on the [i]line[/i] $ H_{a}H_{b}$ if and only if the circumcenter $ O$ lies on the [i]line[/i] $ W_{a}W_{b}$.

Now, we recapitulate two facts from the Ceva-Menelaos group. If they are not already known to you, you can try proving them on your own (it is quite simple; both times one can use the area method):

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46502][b]Van Aubel theorem:[/b][/url] Let $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$ be three points on the sidelines of a triangle $ ABC$. If the lines $ AA^{\prime}$, $ BB^{\prime}$, $ CC^{\prime}$ concur at a point $ P$, then

$ \frac {CP}{PC^{\prime}} \equal{} \frac {CA^{\prime}}{A^{\prime}B} \plus{} \frac {CB^{\prime}}{B^{\prime}A}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20289][b]Transversal theorem:[/b][/url] Let $ C^{\prime}$ be a point on the sideline $ AB$ of a triangle $ ABC$. Let $ A^{\prime}$ and $ B^{\prime}$ be two points on the sidelines $ BC$ and $ CA$, and let $ P$ be a point on the line $ CC^{\prime}$. Then, the point $ P$ lies on the line $ A^{\prime}B^{\prime}$ if and only if

$ AB\cdot\frac {C^{\prime}P}{PC} \equal{} AC^{\prime}\cdot\frac {BA^{\prime}}{A^{\prime}C} \plus{} C^{\prime}B\cdot\frac {AB^{\prime}}{B^{\prime}C}$.

Of course, in both theorems, we used directed segments.

Now, to solve our problem, we first consider not only the points $ W_{a}$ and $ W_{b}$ where the angle bisectors of the angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$, but also the point $ W_{c}$ where the angle bisector of the angle $ BCA$ meets the side $ AB$. Then, the lines $ AW_{a}$, $ BW_{b}$, $ CW_{c}$ are the angle bisectors of the three angles of triangle $ ABC$ and concur at its incenter $ I$. By the van Aubel theorem, we thus have

$ \frac {CI}{IW_{c}} \equal{} \frac {CW_{a}}{W_{a}B} \plus{} \frac {CW_{b}}{W_{b}A}$.

But since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have $ \frac {CW_{a}}{W_{a}B} \equal{} \frac {b}{c}$ and $ \frac {CW_{b}}{W_{b}A} \equal{} \frac {a}{c}$, so that

$ \frac {CI}{IW_{c}} \equal{} \frac {b}{c} \plus{} \frac {a}{c} \equal{} \frac {a \plus{} b}{c}$.

Hence,

$ \frac {W_{c}I}{IC} \equal{} \frac {IW_{c}}{CI} \equal{} 1: \frac {CI}{IW_{c}} \equal{} 1: \frac {a \plus{} b}{c} \equal{} \frac {c}{a \plus{} b}$.

On the other hand, for the circumcenter $ O$ of triangle $ ABC$, let the lines $ AO$, $ BO$, $ CO$ meet the sides $ BC$, $ CA$, $ AB$ of triangle $ ABC$ at the points $ O_{a}$, $ O_{b}$, $ O_{c}$, respectively. Then, the lines $ AO_{a}$, $ BO_{b}$, $ CO_{c}$ concur at the point $ O$, and the van Aubel theorem yields

$ \frac {CO}{OO_{c}} \equal{} \frac {CO_{a}}{O_{a}B} \plus{} \frac {CO_{b}}{O_{b}A}$.

Now, one can easily calculate these ratios using the Sine Law: For instance, the Sine Law in triangles $ O_{b}BC$ and $ O_{b}BA$ yields

$ \frac {CO_{b}}{O_{b}A} \equal{} \frac {BC\cdot\frac {\sin\measuredangle CBO_{b}}{\sin\measuredangle BO_{b}C}}{AB\cdot\frac {\sin\measuredangle ABO_{b}}{\sin\measuredangle BO_{b}A}}$.

Since $ \measuredangle BO_{b}C \plus{} \measuredangle BO_{b}A \equal{} 180^{\circ}$, we have $ \sin\measuredangle BO_{b}C \equal{} \sin\measuredangle BO_{b}A$, so that our formula simplifies to

$ \frac {CO_{b}}{O_{b}A} \equal{} \frac {BC\cdot\sin\measuredangle CBO_{b}}{AB\cdot\sin\measuredangle ABO_{b}} \equal{} \frac {a\cdot\sin\measuredangle CBO}{c\cdot\sin\measuredangle ABO}$.

Now, since the point $ O$ is the circumcenter of triangle $ ABC$, the central angle theorem yields $ \measuredangle BOC \equal{} 2\cdot\measuredangle BAC \equal{} 2A$, but since $ OB \equal{} OC$, the triangle $ BOC$ is isosceles, so that its base angle $ \measuredangle CBO$ equals

$ \measuredangle CBO \equal{} \frac {180^{\circ} \minus{} \measuredangle BOC}{2} \equal{} \frac {180^{\circ} \minus{} 2A}{2} \equal{} 90^{\circ} \minus{} A$.

Thus, $ \sin\measuredangle CBO \equal{} \cos A$. Similarly, $ \sin\measuredangle ABO \equal{} \cos C$. Hence,

$ \frac {CO_{b}}{O_{b}A} \equal{} \frac {a\cdot\sin\measuredangle CBO}{c\cdot\sin\measuredangle ABO} \equal{} \frac {a\cos A}{c\cos C}$.

Similarly,

$ \frac {CO_{a}}{O_{a}B} \equal{} \frac {b\cos B}{c\cos C}$.

Consequently,

$ \frac {CO}{OO_{c}} \equal{} \frac {CO_{a}}{O_{a}B} \plus{} \frac {CO_{b}}{O_{b}A} \equal{} \frac {b\cos B}{c\cos C} \plus{} \frac {a\cos A}{c\cos C} \equal{} \frac {a\cos A \plus{} b\cos B}{c\cos C}$.

Hence,

$ \frac {O_{c}O}{OC} \equal{} \frac {OO_{c}}{CO} \equal{} 1: \frac {CO}{OO_{c}} \equal{} 1: \frac {a\cos A \plus{} b\cos B}{c\cos C} \equal{} \frac {c\cos C}{a\cos A \plus{} b\cos B}$.

For the foot $ H_{b}$ of the altitude of triangle $ ABC$ from the vertex $ B$, the right-angled triangles $ AH_{b}B$ and $ CH_{b}B$ yield $ AH_{b} \equal{} AB\cdot\cos\measuredangle BAH_{b} \equal{} c\cos A$ and $ H_{b}C \equal{} BC\cdot\cos\measuredangle BCH_{b} \equal{} a\cos C$, so that $ \frac {AH_{b}}{H_{b}C} \equal{} \frac {c\cos A}{a\cos C}$. Similarly $ \frac {BH_{a}}{H_{a}C} \equal{} \frac {c\cos B}{b\cos C}$.

Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have $ \frac {BW_{a}}{W_{a}C} \equal{} \frac {c}{b}$, $ \frac {AW_{b}}{W_{b}C} \equal{} \frac {c}{a}$ and $ \frac {AW_{c}}{W_{c}B} \equal{} \frac {b}{a}$. The latter equation, together with $ AW_{c} \plus{} W_{c}B \equal{} AB \equal{} c$, leads to

$ \frac {AW_{c}}{AB} \equal{} \frac {b}{a \plus{} b}$ and $ \frac {W_{c}B}{AB} \equal{} \frac {a}{a \plus{} b}$.

Similarly to our proof of $ \frac {CO_{b}}{O_{b}A} \equal{} \frac {a\cos A}{c\cos C}$, we can show $ \frac {AO_{c}}{O_{c}B} \equal{} \frac {b\cos B}{a\cos A}$, and this, combined with $ AO_{c} \plus{} O_{c}B \equal{} AB \equal{} c$, yields

$ \frac {AO_{c}}{AB} \equal{} \frac {b\cos B}{a\cos A \plus{} b\cos B}$ and $ \frac {O_{c}B}{AB} \equal{} \frac {a\cos A}{a\cos A \plus{} b\cos B}$.

Remember that we have to show that the point $ I$ lies on the line $ H_{a}H_{b}$ if and only if the point $ O$ lies on the line $ W_{a}W_{b}$.

Apply the Transversal theorem to the points $ C^{\prime} \equal{} W_{c}$, $ A^{\prime} \equal{} H_{a}$, $ B^{\prime} \equal{} H_{b}$ and $ P \equal{} I$; what comes out is that the point $ I$ lies on the line $ H_{a}H_{b}$ if and only if

$ AB\cdot\frac {W_{c}I}{IC} \equal{} AW_{c}\cdot\frac {BH_{a}}{H_{a}C} \plus{} W_{c}B\cdot\frac {AH_{b}}{H_{b}C}$.

Upon division by $ AB$, this becomes:

$ \frac {W_{c}I}{IC} \equal{} \frac {AW_{c}}{AB}\cdot\frac {BH_{a}}{H_{a}C} \plus{} \frac {W_{c}B}{AB}\cdot\frac {AH_{b}}{H_{b}C}$.

Applying the formulas we found before, we see that this is equivalent to

$ \frac {c}{a \plus{} b} \equal{} \frac {b}{a \plus{} b}\cdot\frac {c\cos B}{b\cos C} \plus{} \frac {a}{a \plus{} b}\cdot\frac {c\cos A}{a\cos C}$.

This quickly simplifies to $ \cos C \equal{} \cos A \plus{} \cos B$. Thus, the point $ I$ lies on the line $ H_{a}H_{b}$ if and only if $ \cos C \equal{} \cos A \plus{} \cos B$.

On the other hand, we can also apply the Transversal theorem to the points $ C^{\prime} \equal{} O_{c}$, $ A^{\prime} \equal{} W_{a}$, $ B^{\prime} \equal{} W_{b}$ and $ P \equal{} O$; then, we obtain that the point $ O$ lies on the line $ W_{a}W_{b}$ if and only if

$ AB\cdot\frac {O_{c}O}{OC} \equal{} AO_{c}\cdot\frac {BW_{a}}{W_{a}C} \plus{} O_{c}B\cdot\frac {AW_{b}}{W_{b}C}$.

Division by $ AB$ transforms this into

$ \frac {O_{c}O}{OC} \equal{} \frac {AO_{c}}{AB}\cdot\frac {BW_{a}}{W_{a}C} \plus{} \frac {O_{c}B}{AB}\cdot\frac {AW_{b}}{W_{b}C}$.

Now we rewrite this using the formulas established before:

$ \frac {c\cos C}{a\cos A \plus{} b\cos B} \equal{} \frac {b\cos B}{a\cos A \plus{} b\cos B}\cdot\frac {c}{b} \plus{} \frac {a\cos A}{a\cos A \plus{} b\cos B}\cdot\frac {c}{a}$.

Again, this simplifies to $ \cos C \equal{} \cos A \plus{} \cos B$. Hence, the point $ O$ lies on the line $ W_{a}W_{b}$ if and only if $ \cos C \equal{} \cos A \plus{} \cos B$.

Thus, the condition for the point $ I$ to lie on the line $ H_{a}H_{b}$ and the condition for the point $ O$ to lie on the line $ W_{a}W_{b}$ are the same: the equation $ \cos C \equal{} \cos A \plus{} \cos B$ must hold. This proves that the point $ I$ lies on the line $ H_{a}H_{b}$ if and only if the point $ O$ lies on the line $ W_{a}W_{b}$. And the problem is solved.

[i]Note.[/i] The problem admits a generalization using the notion of isogonal conjugates. What is also notable is that all of the following conditions are equivalent:

- The incenter $ I$ of triangle $ ABC$ lies on the line $ H_{a}H_{b}$.
- The circumcenter $ O$ of triangle $ ABC$ lies on the line $ W_{a}W_{b}$.
- We have $ \cos C \equal{} \cos A \plus{} \cos B$.
- We have $ BH_{a} \plus{} H_{b}A \equal{} H_{a}H_{b}$.
- The circumcircle of triangle $ ABC$ and the $ C$-excircle have equal radius.
- The point $ O$ is the orthocenter of the triangle formed by the points where the $ C$-excircle of triangle $ ABC$ touches the lines $ BC$, $ CA$, $ AB$.

Thanks to Alexey A. Zaslavsky for finding the two latter conditions.[/hide]

[i]Remark.[/i] See Crux Mathematicorum problem #1939 by Christopher J. Bradley (Crux 1995/03 for a solution) for a closely related problem.

  darij

Thank you a lot Darij ! I didn't know the trilinears!

hello Darij,

  It was very strange to view your solution to this problem. You, who's name

  is linked with the magic word 'Pascal', used everything, but  'kitchen sink' and

  your favorite  tool.  The problem just begs to be solved by applying it after some

  minor lines producing.

   Thank you.

   M.T.

Darij offered a like solution (with the barycentric coordinates) in detail. The my solution differs from Darij's by the simplification of the calculations. The following chain of the equivalencies it is evidently:

\begin{eqnarray*} \blacksquare\ (0)\ \cos A+\cos B=\cos C &\Longleftrightarrow & a\cdot \tan B\cdot \tan C+b\cdot \tan C\cdot \tan A=c\cdot \tan B\cdot \tan A \ \ (1)\\ &\Longleftrightarrow & bc\cdot \sin 2A+ca\cdot \sin 2B=ab\cdot \sin 2C\ \ (2)\end{eqnarray*}

$O\ (\sin 2A,\ \sin 2B,\ \sin 2C)\ ;\ H\ (\tan A,\ \tan B,\ \tan C)\ ;\ I\ (a,\ b,\ c)\ .$

$H_a\ (0,\ \tan B,\ \tan C)\ ;\ H_b\ (\tan A,\ 0,\ \tan C)\ ;\ W_a\ (0,\ b,\ c)\ ;\ W_b\ (a,\ 0,\ c)\ .$

$I\in H_aH_b\Longleftrightarrow \left| \begin{array}{ccc} a&b&c\\ 0&\tan B&\tan C\\ \tan A&0&\tan C\end{array}\right| =0\Longleftrightarrow \ (1).$

$O\in W_aW_b\Longleftrightarrow \left| \begin{array}{ccc} \sin 2A & \sin 2B & \sin 2C \\ 0&b&c \\ a&0&c \end{array}\right|=0\Longleftrightarrow \ (2).$

This is a great problem. Here is my quite purely synthetic solution.
  We will show that $I$ lies on $H_aH_b \Longleftrightarrow R= r_c \Longleftrightarrow O$ lies on $W_aW_b$
  We first show that $I$ lies on $H_aH_b \Longleftrightarrow R= r_c$
  [hide="Proof"]
     Suppose $I$ lies on $H_aH_b$. Then $I$ must lie on the segment $H_aH_b$ and $\measuredangle{ACB}$ must be acute.
     We have
  $1 = \frac{H_aI}{H_aH_b} + \frac{IH_b}{H_aH_b} = \frac{r}{h_a cosC} + \frac{r}{h_b cosC}$
      This means $cosC = r(\frac{1}{h_a} + \frac{1}{h_b}) = r(\frac{1}{r} - \frac{1}{h_c}) = 1 - \frac{sin{\frac{C}{2}}}{2cos{\frac{A}{2}}cos{\frac{B}{2}}}$
      Hence, $4sin{\frac{C}{2}}cos{\frac{A}{2}}cos{\frac{B}{2}} = 1$ or $r_c = R$.
     Now we will show the converse part. Suppose $R=r_c$, then $4sin{\frac{C}{2}}cos{\frac{A}{2}}cos{\frac{B}{2}} = 1$. From this, we imply that $\measuredangle{BCA}$ is acute. Otherwise, we will have $1 = 4sin{\frac{C}{2}}cos{\frac{A}{2}}cos{\frac{B}{2}} \geq \sqrt{2} > 1$ (contradiction).
     Therefore the ray $CI$ is between the rays $CH_a$ and $CH_b$. If $I'$ is the intersection of the ray $CI$ and $H_aH_b$, then we can easily obtain that the distance from $I'$ to $CA$ is $r$. Thus $I'$ is the incenter of $ABC$. Therefore, $I$ lies on $H_aH_b$. And the theorem is proved.
      [/hide]
   We will now show that $O$ lies on $W_aW_b \Longleftrightarrow R= r_c$
   [hide="Proof"]
       Call $I_c$ the excenter of $ABC$ wrt to the side $AB$. Let the excircle $(I_c)$ touch the sides $BC, CA, AB$ at $A', B', C'$. Suppose $M, N$ are the midpoints of $B'C', A'C'$ respectively. Then $M, N$ are respective the intersections of $AI_c$ and $B'C'$, $BI_c$ and $A'C'$.
       Notice that $A'M$ is the polar of $W_a$ wrt $(I_c)$ and $B'N$ is the polar of $W_c$ wrt $(I_c)$. Hence if we call $G$ the intersection of $C'M$ and $B'N$ which is also the centroid of triangle $A'B'C'$, then $W_aW_b$ is the polar of $G$ wrt $(I_c)$.
       We will show that $O$ lies on $I_cG$ and $\frac{I_cG}{I_cO} = \frac{r_c}{3R}$.

      [hide="Proof"]
   Consider the inversion center $I_c$ with constant $r_c^2$. Under this inversion, triangle $ABC$ is mapped to the medial triangle of $A'B'C'$.  Call $O_c$ the circumcenter of the medial triangle of $A'B'C'$, then $\frac{I_cO_c}{I_cO} = \frac{r_c}{2R}$
       Notice that $O_c$ is also the nine-point center of $A'B'C'$. Thus $\frac{I_cG}{I_cO_c} = \frac{2}{3}$.
       Hence we obtain that $\frac{I_cG}{I_cO} = \frac{r_c}{3R}$.
       [/hide]
    Now, $O$ lies on $W_aW_b \Longleftrightarrow O$ is the inverse image of $G$ wrt $(I_c) \Longleftrightarrow I_cG.I_cO = r_c^2 \Longleftrightarrow \frac{r_c}{3R}.I_cO^2 = {r_c}^2 \Longleftrightarrow R^2 + 2Rr_c = 3Rr_c \Longleftrightarrow R=r_c$.
          [/hide]

The Treegoner's solution is a very nice metrical solution and it isn't a pure syntetic solution. The problem of finding a pure synthetic solution remains openly.

Hi guys I remember this problem,it is from IMO short list 1997 ,you can see it this
[url=http://www.kalva.demon.co.uk/short/sh97.html]www.kalva.demon.co.uk/short/sh97.html[/url]problem 16,I solved it two years ago :P  :D

A synthetic proof of a generalization can be found here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=402242&p=2241207#p2241207

[color=darkred][b][u]Generalization[/u].[/b] Let $ABC$ be a triangle. Let $M$ , $N$ and its isogonals $P$ , $R$ respectively. For $X$ define $\left\{\begin{array}{c}
X_a\in AX\cap BC\\\\
X_b\in BX\cap AC\\\\
X_c\in CX\cap AB\end{array}\right\|$ ,

i.e. $\triangle X_aX_bX_c$ is the cevian triangle of $X$ w.r.t. $\triangle ABC$ . Prove that $\boxed {\ N\in M_bM_c\ \iff\ P\in R_bR_c\ \iff\ \frac ux=\frac vy+\frac wz\ }$ ,

where the points $M$ , $N$ have the  barycentrical coordinates $(x,y,z)$ and $(u,v,w)$ respectively  w.r.t.  $\triangle ABC$ .[/color]

[color=darkblue][b][u]Proof[/u].[/b] Observe that $N\in M_bM_c\iff \left|\begin{array}{ccc}
u & v & w\\\\
x & y & 0\\\\
x & 0 & z\end{array}\right|=0\iff$ $\frac ux=\frac vy+\frac wz$ and

$P\in R_bR_c\iff$ $\left|\begin{array}{ccc}
\frac {a^2}{x} & \frac {b^2}{y} & \frac {c^2}{z}\\\\
\frac {a^2}{u} & \frac {b^2}{v} & 0\\\\
\frac {a^2}{u} & 0 & \frac {c^2}{w}\end{array}\right|=0\iff$ $\left|\begin{array}{ccc}
\frac {1}{x} & \frac {1}{y} & \frac {1}{z}\\\\
\frac {1}{u} & \frac {1}{v} & 0\\\\
\frac {1}{u} & 0 & \frac {1}{w}\end{array}\right|=0\iff$ $\frac ux=\frac vy+\frac wz$ .[/color]

Dear Mathlinkers,
for a generalization and the synthetic proof

http://jl.ayme.pagesperso-orange.fr/Docs/Deux%20couples%20de%20points%20isogonaux.pdf

Sincerely
Jean-Louis


Prove that for all positive integers $n$, $n>1$ the number $\sqrt{ \overline{ 11\ldots 44 \ldots 4 }}$, where 1 appears $n$ times, and 4 appears $2n$ times, is irrational.

I think, that we should consider this number $\pmod{7}$, but I don't know when a number is divisible by $7$, could someone answer :!:

u don't need that at last u just have to prove that $\overline {11...1}$ where $1$ appears $n>1$ times isn't a perfect square which is sort of easy

A number is divisible by 7 when

$\frac{n-u}{10} - 2u$  where $n$ is the number we're testing for divisibility and $u$ is the units digit is divisible by 7

The number can be written as

$N=\frac{10^n-1}{9}\cdot 10^{2n}+4\cdot\frac{10^{2n}-1}{9}$, or after simplification

$N=\frac{10^{3n}+3\cdot 10^{2n}-4}{9}$

Denote $a=10^n.$ Then

$N=\frac{a^3+3a^2-4}{9}=\frac{a^3-a^2+4a^2-4}{9}=\frac{(a-1)(a^2+4a+4)}{9}=\\=(a-1)\left(\frac{a+2}{3}\right)^2$, or

$N=(10^n-1)\left(\frac{10^n+2}{3}\right)^2$

The fraction in the parentheses is integer, since the sum of the digits of $10^n+2$ is three, but the number $10^n-1$ can never be a square for $n>1$ because it's congruent to $3$ modulo $4$. Therefore $\sqrt{N}$ is irrational.

k=1111.....11 olsun. burada 111 ... 11 lerin sayısına n dersek kolaylıkla cozum gelir.


linear algebra
matrix
linear algebra theorems

:blush:   please show me how to find the invers of a matrix $M_{3}$


a,b are natural nos. with b>2.show that$ 2^b\minus{}1$never divides $ 2^a\minus{}1$

But for $ a\equal{}bc$ it follows $ 2^b \minus{} 1 \mid 2^a \minus{} 1 \equal{} (2^b)^c \minus{} 1$.

I'm pretty sure it should be that $ 2^b \minus{} 1$ does not divide $ 2^a \plus{} 1$.

Lemma, for odd b:
assume for the sake of contradiction that $ 2^{b}\minus{}1$ divides $ 2^{a}\plus{}1$. Let k divide $ 2^{b}\minus{}1$. Then k also divides $ 2^{a}\minus{}1$. So $ 0 \equiv 2^{b}\minus{}1\equiv 2^{a}\plus{}1\mod{k}$. Then
$ 2^{b}\equiv 1\mod{k}$
$ 2^{ab}\equiv 1\mod{k}$
$ 2^{a}\equiv \minus{}1\mod{k}$
$ 2^{ab}\equiv \minus{}1\mod{k}$
;contradiction.
For even b:
(unrigorous)Divide b by 2 until it becomes odd, then call this integer c.(unrigorous) Since c|b, $ 2^c\minus{}1|2^b\minus{}1$. Assume for the sake of contradiction that $ 2^{b}\minus{}1$ divides $ 2^{a}\plus{}1$. Then $ 2^{c}\minus{}1$ divides $ 2^{a}\plus{}1$. But this is not true by our lemma, and we are done.

what i dont understand is why this is true even if if b divides a

i solved Zhero's problem, which is different. Your problem is not always true because if b|a, $ 2^{b}\minus{}1 | 2^{a}\minus{}1$

[hide="hint"] Write $ a\equal{}bq\plus{}r$, as clearly $ a>b$.[/hide]

[hide="solution"] From the hint, we write $ a\equal{}bq\plus{}r$ and so $ \frac{2^a\plus{}1}{2^b\minus{}1}\equal{}2^{a\minus{}b}\plus{}\frac{2^{a\minus{}b}\plus{}1}{2^b\minus{}1}\equal{}2^{a\minus{}b}\plus{}2^{a\minus{}2b}\plus{}\frac{2^{a\minus{}2b}\plus{}1}{2^b\minus{}1}$ and so on until you get $ \frac{2^a\plus{}1}{2^b\minus{}1}\equal{}2^{a\minus{}b}\plus{}2^{a\minus{}2b}\plus{}...\plus{}\frac{2^r\plus{}1}{2^b\minus{}1}$. As $ r<b$ and $ b>2$, $ \frac{2^r\plus{}1}{2^b\minus{}1}$ is positive and less than $ 1$,so $ 2^b\minus{}1\not{|}{2^a\plus{}1}$.[/hide]


Find $ax^5+by^5$ if the real numbers a, b, x, y, satisfy the equations
\begin{eqnarray*} ax + by &=& 3, \\ ax^2 + by^2 &=& 7, \\ ax^3 + by^3 &=& 16, \\ ax^4 + by^4 &=& 42. \end{eqnarray*}

Just use the fact that $(ax^n+by^n)(x+y)-(ax^{n-1}+by^{n-1})xy=ax^{n+1}+by^{n+1}.$ This probelm has been posted at least a couple of times.

Masoud Zargar

another (cool) solution...

[hide="..."]consider the sequence: $a_n=ax^n+by^n$

it is a common result of linear recursions that $a_n$ satisfies:

$(p)a_n+(q)a_{n+1}=a_{n+2}$ where $p$, $q$ are constants

so, $a_1=3$, $a_2=7$, $a_3=16$, $a_4=42$, and we want $a_5$, from that point, it is a trivial system of equations[/hide]

Wow, this is pretty easy (considering that the pbm. was #15 in AIME)


Let $D:M_n(K) \to M_n(K)$ be linear map, not identically zero, with property $D(AB)=D(A)D(B)$.Prove that $D(X)=C^{-1}XC$ for all $X \in M_n(K)$ and some $C \in M_n(K)$

This is equivalent to finding all authomorphisms of $L(C^n)$. I posted some time ago the much more difficult problem of proving that for arbitrary vectorial space $V$ the authomorphisms of the algebra $L(V)$ are interiour. Has anyone any idea for that problem?


AMC
USA(J)MO
USAMO
inequalities
inequalities proposed

It's equivalent to $\displaystyle \sum\frac{a^2}{2a^2+(b+c)^2}\geq \frac{1}{2}$.We have $(b+c)^2\leq 2(b^2+c^2)$,so the problem is solved. :)

This is the weakest inequality from

$\frac{(qa+b+c)^2}{2a^2+(b+c)^2} + \frac{(qb+c+a)^2}{2b^2+(c+a)^2} + \frac{(qc+a+b)^2}{2c^2+(a+b)^2} \le \frac{(q+2)^2}{2}$,

where $0 \leq q \leq 2+\sqrt 6$. I think the inequalities for $q=4$ and $q=\frac{(q+2)^2}{2}$ are more interesting.  :(


easy  :)  :)  :)
to whom of you who like it it comes from Journal  Parabola from thi site http://problemcorner.org/

[quote="Rzeszut"]Indeed easy.
[hide]Assume that they are. Then $\frac{\sqrt{2}+\sqrt{5}}{2}=\sqrt{3}$ which gives after squaring $7+2\sqrt{10}=12$. It can be simplified into the form $\sqrt{10}=\frac52$, which is obviously false. [/hide][/quote]

The problem says just "terms", not "consecutive terms".

[hide="Correct solution"]If there existed such an arithmetic progression, then for some integer $m,n$ and some real $d$ we would have $\sqrt{3}-\sqrt{2}=md,\sqrt{5}-\sqrt{3}=nd$, giving $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}={m\over n}\in\mathbb{Q}$. But it's quite easy to prove that LHS is irrational.[/hide]

[quote="Farenhajt"]

... giving $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}={m\over n}\in\mathbb{Q}$. But it's quite easy to prove that LHS is irrational.[/quote]

How would you prove this?

[quote="gauss202"][quote="Farenhajt"]

... giving $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}={m\over n}\in\mathbb{Q}$. But it's quite easy to prove that LHS is irrational.[/quote]

How could you prove this?[/quote]
[hide]Assume that $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}={m\over n}\in\mathbb{Q}$. It is equivalent to $n\left(\sqrt3-\sqrt2\right)= m\left(\sqrt5-\sqrt3\right)$, which squared gives $3n^{2}+2n^{2}-3n\sqrt6= 5m^{2}+3m^{2}-2m\sqrt{15}$. Thus $a\sqrt6+b\sqrt{15}=c$ for some $a,b,c,\in\mathbb{Q}$. But it is impossible, since after squaring one more time it gives that $\sqrt{90}$ is rational, which is clearly false. [/hide]

#2:

[hide="answer"]
Putting $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}$ in the form of a single fraction gives you $\frac{3n^{2}+6n+2}{n(n+1)(n+2)}$. The denominator is obviously divisible by 3 while the top obviously isn't ($3n^{2}+6n+2 = 3(n^{2}+2n)+2$), meaning the denominator cannot consist of only powers of 2 and 5 when in lowest terms - it must include a 3. Therefore $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}$ must be nonterminating.
[/hide]


For a positive integer $n$, let $r(n)$ denote the sum of the remainders when $n$ is divided by $1,2,...,n$ respectively.
Prove that $r(k)=r(k-1)$ for infinitely many positive integers $k$

Let k be any power of 2 :) and try to prove that the sums of the remainders are the same.


Let $n>3$ be a positive integer. Consider $n$ sets, each having two elements, such that the intersection of any two of them is a set with one element. Prove that the intersection of all sets is non-empty.

[i]Sever Moldoveanu[/i]

Let the sets be $A_1, A_2, \dots , A_n$.  Let $A_1 = \{a,b\}$ and suppose $A_2 = \{a,c\}$ where $a,b,c$ are distinct elements. Now, any other $A_i$ must contain either $a$ or $b$ (and not both) or $a$ or $c$ (and not both).  Thus, $A_i = \{b,c\}$ or $\{a,x\}$ for some element $x$.  Because $n > 3$, no $A_i$ can be $\{b,c\}$ or else some other $A_j = \{a,x\}$ with $x \neq b,c$ and has no common element with $A_i = \{b,c\}$.  Thus, all the sets contain $a$... qed


Compute the exact value of:

${3 + \cfrac{1}{4 + \cfrac{1}{3 + \cfrac{1}{4+\cdots}}}} $

Set the fraction equal to $x$. Then $x=3+\frac{1}{4+\frac{1}{x}}$. This reduces to $4x^2-12x-3=0$ Since the value must be positive, the answer is $\frac{12+\sqrt192}{8}$

Your answer is unacceptable if this was real contest problem (which it was).

Please simplify your answer. ALWAYS. :)

[quote="Bob"]$\frac{3}{2}+\sqrt3$ (I'm assuming my solution is correct)[/quote]

 :yup:  :10:


Could  anyone  explain  me  about   Fenchel-Legendre  trasformations......................please

        I suppose this is  something  like  that  $f*(x) = \sup_{a<s<b}(xs-f(s) )$

       but  I  would  be very obliged if you  let  me  know  some  more  interesting  theorems  about  these  transformations............... :lol:


number theory
relatively prime
algebra proposed
algebra

I found this nice problem in Russian Journal "Mathematics in School" (suggested by A.B.Spivak)

A sequence $\{a_n\}$ is defined as following: $a_1 = 1$, $a_2 = 2$, $a_n$ ($n\geq 3$) is the least positive integer different from $a_1$, $a_2$, ..., $a_{n-1}$ and all numbers which relatively prime with $a_{n-1}$ (Particularly $a_3 = 4$, $a_4 = 6$, $a_5 = 3$, $a_6=9$, $a_7=12$, $a_8=8$, ...) Show that this sequence contains all positive integers.

Namdung

[i]Edited by Myth[/i]

This problem appeared on the Tournament of the towns 2002 I think.
And indeed, the solution is very nice and elegant. :)

I don't know nice solution. Maybe someone will tell us?
I will write my approach.

Suppose the contrary and $n$ is the minimal number, which is not in $\{a_k\}$.

1. Let $k$ be such number that $\{1,...,n-1\}\in\{a_1,...,a_k\}$, $M=\max_{i=1..k}a_k$.
Then $2sn\not\in \{a_k\}$ for all $s\geq M$. [b](1)[/b]

2. Let's prove that $\{a_k\}$ contains infinitely many even numbers. [b](2)[/b]
Suppose the contrary. It will mean that the set of prime divisors of $\{a_k\}$ is finite! Indeed, if some $a_m$ contains a new prime factor $p$, then $a_{m+1}=p$ and $a_{m+2}=2p$.
Let $p_1$, ..., $p_l$ are all prime divisors of $\{a_k\}$. Then we necessarily have $2p_1...p_lMn\in\{a_k\}$. Contradiction with [b](1)[/b].

3. From [b](1)[/b] and [b](2)[/b] by considering $2Mn$ we immediately obtain contradiction.

I would also like to see a nice solution to this problem if it is different from Myth's.

[quote="Myth"]Let $p_1$, ..., $p_l$ are all prime divisors of $\{a_k\}$. Then we necessarily have $2p_1...p_lMn\in\{a_k\}$. Contradiction with [b](1)[/b].

3. From [b](1)[/b] and [b](2)[/b] by considering $2Mn$ we immediately obtain contradiction.[/quote]
Sorry for my ignorance but I will never understand such delicate and short arguments by myself  :(.Maybe I'm the only one that misunderstand Myth's proof,but can I bother you to help?


Βρείτε το ελάχιστο (min) της παράστασης
Α=[(α/β)^(β/γ)]+[(β/γ)^(γ/α)]+[(γ/α)^(α/β)]+[(α^3)+(β^3)+(γ^3)]/αβγ

Ξέχασα τα α,β,γ>0

Βρείτε το ελλάχιστιο της παράστασης:

$ A \equal{} (\frac {a}{b})^{\frac {b}{c}} \plus{} (\frac {b}{c})^{\frac {c}{a}} \plus{} (\frac {c}{a})^{\frac {a}{b}} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{abc}$ αν $ a,b,c>0$

[hide]Θέστε οπου $ \dfrac{a}{b}\equal{}x$ , $ \dfrac{b}{c}\equal{}y$ και $ \dfrac{c}{a}\equal{}z$ . Παρατηρείστε πως γραφετε ολη η παρασταση ως προς x,y,z[/hide]

Αν ήταν  $ \left( \frac {a}{b}\right)^{\frac {a}{b}} \plus{} \left( \frac {b}{c}\right)^{\frac {b}{c}} \plus{} \left( \frac {c}{a}\right)^{\frac {c}{a}}$

θα ήταν πιο εύκολη αλλά τώρα με παιδεύει ακόμα :(

* η αντικατάσταση σου kritikos  σίγουρα βοηθάει ... και εγώ έτσι την προσπαθώ

θα δούμε τελικά τι θα γίνει ...

Η παράσταση γράφεται ισοδύναμα $ x^y \plus{} y^z \plus{} z^x \plus{} x/z \plus{} y/x \plus{} z/y$
με $ xyz \equal{} 1$
Από AM-GM η ελάχιστη τιμή θα είναι $ 6$ αν $ x^y*y^z*z^x > \equal{} 1$
Τώρα θεωρούμε ότι ο $ z$ είναι ο μικρότερος λόγω κυκλικότητας άρα μικρότερος της μονάδας και διαιρούμε με τις δυνάμεις του $ z$. Μετά είναι προφανές. :wink:

Γράψε λίγο πιο αναλυτικά σε παρακαλώ Κώστα γιατί έχω κάτι αμφιβολίες

Γράφω αγγλικά για να μην αλλάζω γλώσσα συνέχεια. :D

Einai $ a/b \equal{} x$ $ b/c \equal{} y$ kai $ c/a \equal{} z$. Profanos $ xyz \equal{} 1$.

Me aytes tis antikatastaseis i parastasi grafetai isodynama:

$ A \equal{} x^y \plus{} y^z \plus{} z^x \plus{} x/y \plus{} y/z \plus{} z/x$

Apo AM-GM pairnoyme $ A > \equal{} 6*sqrt6(x^y*y^z*z^x)$

Tha deixoyme oti elaxisti timi einai to $ A \equal{} 6$ to opoio epitygxanetai otan $ x \equal{} y \equal{} z \equal{} 1$

Arkei gi ayto na einai $ x^y*y^z*z^x > \equal{} 1$

Esto $ z$ o mikroteros ek ton trion.

Einai $ x^y*y^z*z^x \equal{} (xyz)^z*x^[y \minus{} z]*z^[x \minus{} z] \equal{} x^[y \minus{} z]*z^[x \minus{} z]$

Omos ayti i parastasi einai megaliteri tis monadas giati $ y \minus{} z > \equal{} 0  kai  x \minus{} z > \equal{} 0$

Ζητώ συγγνώμη που κάνω λάθος σε θέματα που αφορούν γενική παιδεία Β λυκείου... :oops:  Παρακαλώ κύριε Στεργίου να διαγραφεί το προηγούμενο ποστ. :o

Η ανισότητα είναι αρκετά sharp. Λύνεται με ανάλυση? Δεν μπορώ να βρω κάτι που να προχωράει την προηγούμενη απόπειρα... :?:

Και εγώ σ΄ αυτό είχα σταματήσει !

Να δείξω δηλαδή ότι  αν xyz = 1 τότε $ x^y  y^z  z^x \ge 1$

Αν δείξουμε αυτό ( αν ισχύει ) ,η άσκηση βγαίνει αμέσως.

Δυστυχώς δεν ισχύει νομίζω... πάρε παράδειγμα $ (x,y,z)\equal{}(1/2,1/2,4)$  :wink:

Φαίνεται η ανισότητα δεν είναι και τόσο χαλαρή...

Νομίζω πως αρκεί να βρούμε το min της:

$ A \equal{} y\ln x \plus{} x\ln z \plus{} z\ln y$

(που έχω την υποψία πως είναι το $ 0$, για $ x\equal{}y\equal{}z\equal{}1$ αλλά θέλει ψάξιμο!)

Όταν $ xyz \equal{} 1$

(προφανώς $ x,y,z > 0$)

Δεν έχω χρόνο να τη κοιτάξω καθόλου. Πάντως η ιδέα μου είναι να τη γράψω ως εκθετική και να 'χω κοινές βάσεις. Και επίσης από ό,τι φαίνεται το min πάει για 6, άλλα είναι αρκετά σφιχτή!

Στέλιος

Για την ιστορία να πω ότι η ακόλουθη άσκηση που ανήκει episis στον xp έμεινε άλυτη τόσο στο περιοδικό Crux όσο και στο αγγλικό φόρουμ (αν και η παραπάνω είναι αρκετά ευκολότερη) . :wink:

Έστω  $ x,y,z > 0$ με  $ xyz \equal{} 1$ Νδο

$ 2\sqrt {3(x^{y} \plus{} y^{z} \plus{} z^{x})} \plus{} x \plus{} y \plus{} z\geq 9$

Για αυτή έχω μια αρκετά περίπλοκη λύση με χρήση βαριών θεωρημάτων και αρκετής ανάλυσης

Κάνουμε χρήση των $ x^y\equal{}e^{y\ln x}$ , $ e^x\geq x\plus{}1$ , $ \ ln x\geq\frac{x\minus{}1}{x}$ και AM-GM....


logarithms
integration
complex numbers
algebra
partial fractions
calculus
calculus computations

Prove that for any positive integer $p$, $\sum_{k=0}^{\infty}\frac{(-1)^{k}}{pk+1}=\frac1{p}\sum_{k=0}^{p-1}a_{k}\left[\log (-a_{k})-\log (1-a_{k})\right]$ where $a_{k}=e^{i(2k+1)\frac{\pi}{p}}$.

Can anyone simplify the sum on the right? I have also that it $=\frac1{p}\sum_{k=0}^{p-1}a_{k}\log\left( \frac{a_{k}}{a_{k}-1}\right)$, but that's seems worse... the complex numbers make it difficult for me to simplify in the general case. The first few values for $p=1,2,3,4$ are $\log 2,\frac{\pi}{4},\frac1{3}\log 2+\frac{\pi}{3\sqrt{3}},\frac{\sqrt{2}}{8}\left[\pi+\log (3+2\sqrt{2})\right]$, repectively.

It's probably hopeless in general. The sum from $-\infty$ to $\infty$, on the other hand, is nice for almost all complex values of $p$. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=21316]here[/url].

For your formula, we have $\sum_{k=0}^\infty \frac{(-1)^{k}}{1+pk}=\sum_{k=0}^{\infty}\int_{0}^{1}(-1)^{k}x^{pk}\,dx$
$=\int_{0}^{1}\sum_{k=0}^\infty (-1)^{k}x^{pk}\,dx= \int_{0}^{1}\frac1{1+x^{p}}\,dx$.

Partial fractions gives your formula. There's no reason to expect simplification.

I don't think it is a good idea to start with the RHS.  Here's my suggestion:

If we set $f(x)=\sum^\infty_{k=0}\frac{(-1)^{k}x^{pk+1}}{pk+1}, |x|<1$
Then $f'(x)=\frac{1}{x^{p}+1}=\frac{1}{\Pi^{p-1}_{k=0}(1+xe^{i(2k+1\pi/p)})}$

Factoring out the last expression should help.  Though f(x) is generally not defined on |x|=1, but it converges at x=1 so you can still use f(x) to approach f(1).  But I may be wrong.

[quote="Victor96"]I don't think it is a good idea to start with the RHS.  Here's my suggestion:

If we set $f(x)=\sum^\infty_{k=0}\frac{(-1)^{k}x^{pk+1}}{pk+1}, |x|<1$
Then $f'(x)=\frac{1}{x^{p}+1}=\frac{1}{\Pi^{p-1}_{k=0}(1+xe^{i(2k+1)\pi/p})}$

Factoring out the last expression should help.  Though f(x) is generally not defined on |x|=1, but it converges at x=1 so you can still use f(x) to approach f(1).  But I may be wrong.[/quote]

I used exactly the approach jmerry posted,  followed by yours and justified it by Abel's theorem. Using partial fractions I integrated over [0,x) and taking the limit as $x\to 1^{-}$ gave me the right-hand sum in my OP. I like the way you guys think  :lol: ! It just seemed like a sum over the $p^{th}$ roots of unity ought to simplify nicely despite the logs.


Just wondering.

I have received help from all of these, but I would say the most helpful was my state coach.

What about other. For me it was old nats test I got on my owwn.

Yeah. You should add an option for self-teaching.

NSFY, you have a math tutor? What does he teach you?

[quote="chess64"][quote="jb05"]Lucky! I wish I had a math tutor. As of now (and I just finished precalc), I have had three real classes since 1st grade. Those were 4th grade math (at the time I was in 3rd grade), algebra (7th grade), and precalc this year. I could use a tutor.[/quote]

Me too. What do you mean by "real classes"?[/quote]

Maybe he means classes he took at school, and that he learned the rest on his own...

For nationals specifically, my state coach. However, for MathCounts in general, definitely my school coach; he was amazing. I think that he would be the best coach for the state team, but our teams were never good enough for him to be able to, except for one year, I think.


1)$n^5-5n^3+60n^2-56n$
If n is =ve and >1,prove that the above is a multiple of 120.

2)How many numbers can be formed with digits 1,2,3,4,3,2,1 so that the odd digits occupy odd places?

3)Can someone tell me why the following is true:-(I'm new to combinatorics)

(n$C$r)*r!=n$P$r
We use this to find what n$C$r gives.

2)the combinations in which the odd digits cand be placed on odd positions are $1133,1313,1331,3113,3131,3311$ so we remain to place the followng combinations $224,242,422$ from the first one in each set we get the numer $1212343$ thus we have $6*3=18$ solutions.

[hide="#1"]\begin{eqnarray*}n^5-5n^3+60n^2-56n &=& n(n^4-5n^2+60n-56)\\ &=& n(n^4-n^2-4n^2+4n+56n-56)\\ &=& n[n^2(n^2-1)-4n(n-1)+56(n-1)]\\ &=& n[n^2(n+1)(n-1)-4n(n-1)+56(n-1)]\\ &=& n(n-1)[n^2(n+1)-4n+56]\\ &=& n(n-1)[n^2(n+1)-4n-4+60]\\ &=& n(n-1)[n^2(n+1)-4(n+1)+60]\\ &=& n(n-1)[(n+1)(n^2-4)+60]\\ &=& n(n-1)[(n+1)(n-2)(n+2)+60]\\ &=& (n-2)(n-1)n(n+1)(n+2)+60n(n-1)\end{eqnarray*}

The first term is the product of $5$ consecutive integers. Among $5$ consecutive integers, at least one is divisible by $4$ and [u]another one[/u] is divisible by $2$, at least one is divisible by $3$ and exactly one is divisible by $5$. Hence that term is divisible by $4\cdot 2\cdot 3\cdot 5=120$

The second term is $60$ times the product of two consecutive integers. Among two consecutive integers, exactly one is even, hence that term is divisible by $60\cdot 2=120$.

As both terms are divisible by $120$, the whole expression is divisible by $120$.[/hide]

[quote="shadysaysurspammed"][quote="Farenhajt"][hide="#1"]$\begin{eqnarray*}n^5-5n^3+60n^2-56n &=& n(n^4-5n^3+60n-56)$.[/hide][/quote]

Just a tiny typo there,otherwise everything's perfect. :)[/quote]

Thanks, I missed it :) Edited now.

[quote="spix"][quote="shadysaysurspammed"][quote="spix"]$3) C_r^n=\frac{n!}{r!(n-r)!},P_r=r!$[/quote]
No that's not what i have doubts about,In the book,this is the method to $Derive$ the thing for $C_r^n$.So why to multiply $C_r^n$ by $r!$ to get $P_r^n$?

And your ans for the 2nd ojne is wrong.If anyone gives the answer for the 2nd one,please give the method as well.[/quote]

yes I see... I edited my post...  maybe this will help :) $C_n^k=\frac{A_n^k}{P_n}$[/quote]
Well I don't really get it :( What's $A_n^k$?Any typo with $P$?Anyway please can you or anyone explain this to me?

$A_n^k$ represents arrangements of $n$ taking only $k$ of them $A_n^k=\frac{n!}{(n-k)!}$ sorry about the typo it should be$C_n^k=\frac{A_n^k}{P_k}$ maybe I didn`t understand your question coul you be more explicit please?

[quote="mathmanman"]Actually, your $A$ is his $P$. :)[/quote]
I still don't get it :(  :(  :( .I'll think more on it....

P.S-DOesn't explicit mean cheap?Like on album covers its written explicit content,or i'm just dumb enougfh to think that.?

it means be more clear because it is not very clear what you mean by question 3

[quote="shadysaysurspammed"]

2)How many numbers can be formed with digits 1,2,3,4,3,2,1 so that the odd digits occupy odd places?

[/quote]

[hide]Let $abcdefg$ be the number with $a, c, e, g$ as odd places and $b, d, f$ as even places.
There are $6$ different dispositions for $a, c, e, g$ and $3$ for $b, d, f$. So there are $18$ numbers with that property.[/hide]

Spix, what he's asking is :
"Why does $C_n^r \cdot r!$ = $A_n^r$.
In other words, what is an arrangement, explain why is this "r!" here etc." ;)

[hide="hint for 1"]$n^5-5n^3+60n^2-56n$
We know $n^5 \equiv n \bmod 5$ and $n^3 \equiv n \bmod 3$ so:
Above mod 3 is equiv to $n^5+n^3+n \equiv n + n + n \equiv 0$
Above  mod 5 is equiv  to $n^5-n \equiv n-n \equiv 0$
Above mod 8 is  equiv to $n^5-5n^3+4n^2 \equiv n^2(n-1)(n-4)$
if $n$ is even above is obviously $0 \bmod 8$
if $n$ is odd, $n^2 \equiv 1 \bmod 8$ so above is $n + 3n + 4 + 0 \equiv 4(n+1) \equiv 0 \bmod 8$ (remember $n+1$ is even) [/hide]

[quote="123s"]N choose R = N!/R!(N-R)!
N permute R = N!/(N-R!)

To get permutations from choses, you just multiply by R! This is because, in choosing, the order of the objects do not matter. This means that by permutating, you have overcounted each group by R! (as each group can be ordered in that many different ways) times. For example, you would count AB and BA as separate groups when permutating, but they would be equivilant when choosing.

I agree with Andreas and Spix's solutions for number 2. Can you tell us what is wrong with them?[/quote]
No,the solutionms are right,the answer in the book must be wrong.

Using a simple example,can someone eplain me the 3rd one?
Eg:-I have 3 books labelled 1,2,3.In how many ways can i choose 2 books?

Yes, but what shadysaysurspammed wanted is (I guess) : in $C_3^2$ ways, which indeed equals 3, and these ways are shown by Spix! :)

Well, first we must cover ordinary permutations.

A permutation is an arrangement where ALL the elements of the given set appear exactly once. Hence the number of permutations is $n\cdot (n-1)\cdot(n-2)\dots 2\cdot 1$, because on the first place you can put any of the $n$ elements, then on the second place you can put any of the remaining $n-1$, on the third any of the remaining $n-2$ etc. As you know, that product is denoted $n!$ and read "n factorial".

Now the arrangements. These are defined as [i]ordered[/i] subsets of a cardinality $r$, chosen from a given set of $n$ elements, and their number is denoted $nAr$. For the first place you can choose any of $n$ elements, for the second place any of the remaining $n-1$ elements, for the third any of the remaining $n-2$ elements, and so on, and for $r$-th place you can choose any of the remaining $n-r+1$ elements. Multiplying all that, you get $nAr=n(n-1)(n-2)\dots(n-r+1).$ That can be written as $nAr=\frac{n!}{(n-r)!}.$

And now combinations. These are defined as [i]unordered[/i] subsets of a cardinality $r$, chosen from a given set of $n$ elements, and their number is denoted $nCr$. Let's begin with the arrangements, whose number is $nAr.$ If any number of arrangements consist of the same elements, but possibly in the different order, they are counted as only ONE combination, since for the combinations the order is irellevant. Hence, we must count all the arrangements consisting of the same elements. Obviously, $r$ elements can be ordered in $r!$ ways (see Permutations), therefore every (unordered) combination can be represented by $r!$ (ordered) arrangements, which means that there's $r!$ times less combinations than arrangements. Hence $nCr=\frac{nAr}{r!}=\frac{n!}{r!(n-r)!}$. This number is most commonly denoted ${n\choose r}$ (read "n choose r") instead of $nCr$.

Here's another way to deal with 2) using encoding.

We can deal with the even places and the odd places independently. So, we must distribute 1,3,3,1 in 4 places and 2,4,2 in 3 places. We can "encode" 1331 as a word and find all permutations of the word with $\frac{4!}{2!2!}=6$ and we can encode 242 as a word and find $\frac{3!}{2!}=3$ Therefore the answer is $18$.

The way this encoding method works is that we first pretend that all numbers/letters are distinct and then we divide it out to account for indistinctness. For example, you have $xxyyzz$; We first pretend they are instinct - so $6!$ and then we divide out for the indistinct parts - i.e. that both xs are same, both ys are same, both zs are same - $2!$, $2!$, $2!$. so :$6!/(2!2!2!)$


conics hyperbola Asymptote geometry circumcircle

[hide]B., because the asymptotes are $y=\pm\frac{b}{a}$.[/hide]

The asymptotes are -b/a x so its B.  Also notice that the graphs are up and down (the locus is on the y axis) so it has to be either A or B.

[hide="Answer"]

Since the asymptotes are $y=\pm \frac{2}{3}x$, we can set the equation of the desired hyperbola as $(3x-2y)(3x+2y)=k\ (k\neq 0)\ \cdots [E]$. Let $F(0,\ 2\sqrt{13})$ be the focus of the curve, $\overline{OF}$ is the circumradius of a rectangular, we can find a right triangle with $2: 3: \sqrt{13}$, from this $[E]$ will pass through the point $(0,\ 6)$, yielding $k=-4\cdot 6^{2}$. Thus the desired equation is $9x^{2}-4y^{2}=-4\cdot 6^{2}\Longleftrightarrow-\frac{x^{2}}{16}+\frac{y^{2}}{36}=1$, the answer is $A$. [/hide]


On a digital clock showing hours and minutes, how many different readings between noon and midnight contain at least two 3's?

In all the hours except from 3 to 4, there is 1: __ : 33.
From 3 to 4, 2 threes means that the minute must have at least 1 three. There are 15 threes. Thus, there is a total of 11+15=26 threes from noon to midnight.


induction
number theory solved
number theory

I'll post some problems about the sum of digits (Arne suggested some1 should do this). In all the problems let S(n) be the sum of digits of n.

Prove that for any positive integer n S(8n)/S(n) >=1/8.

This one is again cute..

we have that S(n)=S(1000n)=S(125*8n)<=S(8n)*S(125)=8S(8n).

cheers! :D :D

The property S(ab) <= S(a) S(b) seems to be well - known ?
It feels natural but how do you prove this ?

This is also something that may sometimes come in hand!

S(a+b)<=S(a)+S(b). (try to prove it) --> also Arne's question!

cheers! :D :D

Seems that nobody posted an answer to this one! :?

I'll take a shot! :D :D
We have four important propeties:
[b]a)[/b] s(x+y)<=s(x)+s(y) for any pozitive x,y
[b]b)[/b] s(x1+..+xn)<=s(x1)+..+s(xn) for any pozitive numbers.
[b]c)[/b] s(nx)<=ns(x) for any n,x natural.
[b]d)[/b] s(xy)<=s(x)s(y) for any x,y natural.

[b]Proofs:[/b]
for [b]a)[/b] we have that if the sum of the digits is of the same order and is striclty <10, then in s(x+y) the contribution of these digits is the same and if the sume is >=10 then their sum is strictly bigger than the contribution that they have for s(x+y).
[b]b)[/b] pure induction
[b]c)[/b] we get it from [b]b)[/b] if x1=..=xn=x.
[b]d)[/b] results from the fact that if we consider that the numbers x and y are multiplied written one under the other for every digit of y we apply [b]c)[/b] and so we get [b]d)[/b].

that's all!!!!
cheers! :D :D

I think everybody understood the simplicity of S(ab)>S(a)S(b) that is why nobody answered your question.

One more problem about S(n).

Is there 2003-digit number n with nonzero digits only such that S(n)|n ?

1222221122111...111222...222

There are 1561 "1"s for the 111...111 section (including those already shown) and 432 "2"s for the 222...222 section.

Sum of digits = 2442 = 2 * 11 * 111 and it is obvious that 1221 | 12222211221 (the numbers at the beginning).

I apologise for giving such a lousy construction, as my brain is currently stuck in biology mode (exams in 2 days time).

You impressed me, Valiowk!

How about common case of n-digit number?


a,b,c are positive real numbers.

prove that

sqrt(a/b+c) + sqrt(b/c+a) + sqrt(c/a+b)  >=2 .

which is in Macedonia 1995.

sorry i'm poor at text letters.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=46428 , http://www.mathlinks.ro/Forum/viewtopic.php?t=27481 , http://www.mathlinks.ro/Forum/viewtopic.php?t=63801 , http://www.mathlinks.ro/Forum/viewtopic.php?t=67223 and http://www.mathlinks.ro/Forum/viewtopic.php?t=48825 .

  dg


DEAR ALL,
              CAN ANY ONE GIVE ME THE DETAILS OF RAMANUJAN MATHEMATICAL OLYMPIAD AND HOW TO ENROLL FOR IT.I AM UNAWARE OF THE SOURCE OF THE EXAM AND LONG TO WRITE IT.

THANK YOU

I assume that you are from gujarat.If you are from Gujarat,you can contact the Regional Coordinator for mathematical Olympiads in Gujarat,Mr UDAYAN PRAJAPATI,Department of Mathematics,Gujarat University.
Ramanujan Ganit Spardha is meant for classes 8,9,10.If you are in 11,12th you can give RMO.
You can give an email to:      ganit_spardha@yahoo.co.in.
Mr Prajapati will send you the application form for Ramanujan Ganit Spardha.

If u can give me your mail address, i can also fwd the application form to you.He has sent the form to all of his contacts.

do not delay,the Ramanujan Ganit Spardha will take place on 4th october 2009.You have to register before 1st october

is rgs only for residents of Gujarat? :huh:

all i know is that its held in gujarat.......i have no idea about the the other states...because it is organised by the suganitam trust in gujarat...


A man buys a house for $10000$ dollars and rents it. He puts $ 12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325$ dollars a year taxes and realizes $ 5\frac{1}{2}\%$ on his investment. The monthly rent is:

$\textbf{(A)}\ 64.82\text{ dollars} \qquad
\textbf{(B)}\ 83.33\text{ dollars} \qquad
\textbf{(C)}\ 72.08\text{ dollars} \qquad
\textbf{(D)}\ 45.83\text{ dollars} \qquad
\textbf{(E)}\ 177.08\text{ dollars}$

[hide]Let $ x$ be the rent amount. Now, since the man puts aside $ 12.5\%$ of the rent, that means he keeps $ 87.5\%$ of it. Assuming that the total time is one year, then we know that an increase of $ 5.5\%$ on his investment is simply $ (.055)(10000)\equal{}550$. There are $ 12$ months in a year, so now we can simple put this in an equation.

$ (12)(.875)x\minus{}325\equal{}550$
$ 12x\equal{}10000$
$ x\equal{}83.33$
Therefore the answer is $ \boxed{B}$[/hide]

[quote="nick42"]yep, he's right. you can also guess & check if you don't know any other way.
 :coolspeak:[/quote]

that would be like poking pins in your eyes with the numbers in the choices.


trigonometry
limit
trig identities
complex numbers

I know all the basic trig identities: pythagorean, sum <-> product, half/double/triple... but i was looking for cooler ones. Does anyone know any?

For example,

$\displaystyle\frac{\sin x+\sin3x+\sin5x+\ldots+\sin(2n+1)x}{\cos x+\cos3x+\cos5x+\ldots+\cos(2n+1)x}=\tan(n+1)x$

Anything you know, interesting or not, would be much appreciated.

Hmm, lets see.

Recursively, we can reduce something of the form $\sin (nx)$ and $\cos (nx)$ as well as $\tan (nx)$ to:

$\sin (nx) = 2\sin((n-1)x)\cos{x}-\sin((n-2)x)$

$\cos (nx) = 2\cos((n-1)x)\cos{x}-\cos((n-2)x)$

$\tan (nx) = \frac{\tan((n-1)x)+\tan{x}}{1-\tan((n-1)x)\tan{x}}$

Try with n=2 and compare with the well-known double angle formulas.  I'm pretty sure that this holds true only for positive integers, but I haven't tried with decimal vaules of n.

I know that it works, Its not to bad of a proof with induction. I haven't tried it, but it would probably also work using:

$sin\theta=\displaystyle\frac{e^{i\theta}-e^{-i\theta}}{2i}$

and such formulas. But I'm looking for other cool identities that you wouldn't learn in your average trig class.

there was a cool thing i say for $tan(n\theta)$ and the binomial theroem but i dont remember it. ill look for it and post it if i find it.

Cool trig identities, you say?

$\alpha + \beta + \gamma = n\pi \implies \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma$
$\sin x = x \left( 1 - \frac{x^2}{\pi^2} \right) \left( 1 - \frac{x^2}{4\pi^2} \right) \left( ... \right)$
$\sin x + \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + ... = \frac{x}{2}$
$\frac{\sin x}{x} = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} ...$

[quote]Too bad you can't simplify $e^{i \theta}$ without cis, which makes it a kind of trivial identity.[/quote]

Not so!  Using this identity we get $\sinh x = i \sin ix$, thus extending $\sin$ easily to the complex numbers, and it also provides a way to extend $\sin^{-1}$ to the complex numbers as well :)

Are these identities the sort of thing you are looking for? If so, I can post  more.At least five of my posts concerning trigonometry on this forum are still unanswered.If you enjoy trigonometric questions you might like to to try them.

TanX + 2Tan2X + 4Cot4X = CotX

cosec(pi/2) + cosec(pi/4) +cosec(pi/8) = cot(pi/16)

If A+B+C =90 degrees then

CosecA.CosecB.CosecC- cotB.tanC- cotC.tanB- cotC.tanA- cotA.tanC- cotA.tanB- cotBtanA = 2.

Tan20.Tan40.Tan60.Tan80 =3.

(3+Cos4x)/(1-cos4x) = [(CotX)^2  + (TanX) ^2 ]/2

[hide="Here is the first method."] Denote $\omega =\cos x+i\sin x$. It is well-known $\overline {\omega}=\cos x-\sin x=\omega ^{-1}$ and for any $k\in Z$ we have $\omega ^k=\cos kx+i\sin kx$. Denote $C_n=\sum\limits_{k=0}^n \cos (2k+1)x$ and $S_n=\sum\limits_{k=0}^n \sin (2k+1)x$.
$C_n+iS_n=\sum \limits_{k=0}^n [\cos (2k+1)x+i\sin (2k+1)x]=\sum\limits_{k=0}^n \omega ^{2k+1}=$
$\omega \cdot \frac{\omega ^{2(k+1)-1}}{\omega ^2 -1}$ $=\frac{\omega ^{2(n+1)}-1}{\omega -\overline {\omega}}=\frac{-1+\cos 2(n+1)x+i\sin 2(n+1)x}{2i\sin x}=$
$\frac{-2\sin ^2 (n+1)x+2i\sin (n+1)x\cos (n+1)x}{2i\sin x}=$
$\frac{\sin (n+1)x\cos (n+1)x+i\sin ^2 (n+1)x}{\sin x}\Longrightarrow$
$\boxed {\ S_n=\frac{\sin ^2 (n+1)x}{\sin x}\ ,\ C_n=\frac{\sin (n+1)x\cos (n+1)x}{\sin x}\ }\ \Longrightarrow$$\frac{S_n}{C_n}=\tan (n+1)x\ .$[/hide] [hide="Here is the second method."] $2S_n\sin x=\sum\limits_{k=0}^n 2\sin x\sin (2k+1)x =\sum\limits_{k=0}^n [\cos 2kx-\cos (2k+2)x]=$
$1-\cos 2(n+1)x=2\sin^2 (n+1)x\Longrightarrow \boxed {\ S_n=\frac{\sin^2 (n+1)x}{\sin x}\ }\ \ \ (1).$
$2C_n\sin x=\sum\limits_{k=0}^n 2\sin x\cos (2k+1)x=\sum\limits_{k=0}^n \sin 2(k+1)x-\sin 2kx=$
$\sin 2(n+1)x=2\sin (n+1)x\cos (n+1)x\Longrightarrow \boxed {\ C_n=\frac{\sin (n+1)x\cos (n+1)x}{\sin x}\ }\ \ \ (2).$[/hide] [hide="Here is a remark : the method of the mathematical induction."] The relations $(1)$ and $(2)$ from the second method can be proved easily with the method of the mathematical induction. Denote $T_k=\sin (2k+1)x$, $k\in \overline {0,n}$ and in this case $S=\sum\limits_{k=0}^n T_k$.
Must prove $T_0=S_0\ (=\sin x)$ (what is truly) and for any $n\in N$, $T_{n+1}=S_{n+1}-S_n$.

$T_{n+1}=S_{n+1}-S_n\Longleftrightarrow \sin (2n+3)x=\frac{\sin ^2(n+2)x-\sin^2 2(n+1)x}{\sin x}\Longleftrightarrow$

$2\sin (2n+3)x\sin x=2\sin^2 (n+2)x-2\sin^2 (n+1)x\Longleftrightarrow$

$\cos 2(n+1)x-\cos 2(n+2)x=[1-\cos 2(n+2)x]-[1-\cos 2(n+1)x]$,

what is truly. Therefore, for any $n\in N$, $S_n=\frac{\sin ^2(n+1)x}{\sin x}\ .$[/hide] [hide="Here is other exercises from the same family."]
[b]Exercises.[/b] Ascertain the expressions $A$, $B$, $C$, $D$ and $L$, where :
$\blacktriangleright\ 1^{\circ}.\ A=\sum\limits_{k=1}^n k\sin k$ and $B=\sum\limits_{k=1}^n k\cos k\ .$
$\blacktriangleright\ 2^{\circ}.\ C=\sum\limits_{k=1}^n k^2 x^k$ and $D=\sum\limits_{k=1}^n k^3\binom {n}{k}\ .$
$\blacktriangleright\ 3^{\circ}.\ L_1=\lim_{n\to \infty} \sum\limits_{k=1}^n \frac{k}{2^k}$ and $L_2=\lim_{n\to \infty} \sum\limits_{k=1}^n \frac{k^2}{2^k}\ .$[/hide]


function
integration
real analysis
real analysis unsolved

Is there a distance on the set of integrable functions from $[0,1]$ in $R$?

Doesn't $d(f,g)=\int_0^1 |f(x)-g(x)|dx$ work?

btw, do you mean Riemann-integrable or Lebesgue-integrable?

The space is complete in Lebesgue's sense; but incomplete in Riemann's sense.

I don't think your "distance" is really a distance liyi : $d(f,g) = 0$ doesn't imply always that $f = g$ for non continuous functions

There is always a "stupid" distance on every set :

$d(x,x) = 0$ and $d(x,y) = 1$ if $x \neq y$


These are acid base reactions.

In the first reaction, the conjugate acid of $\text{OCl}^{-}$ and the conjugate base of $\text{H}_{2}\text{O}$ are produced.

In the second reaction, $\text{HOCl}$ dissociates into a hydrogen ion and its conjugate base.


S17

What is the number of integer solutions of the equation \[ (x^{2}-3x+1)^{x+1}= 1\]

See this topic. :wink:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159833[/url]

This equation has 4 integer solutions : -1 , 0 , 1 , 3 .

Haha ok.. but anyway there are 4 solutions, i missed out the one that many others are likely to miss. $ x^{2}-3x+1$ could be equal to $ -1$ and the power positive. We get another solution from here, giving 4 solutions in all.  :)


[quote="chien than"]Let $ a;b$ be real numbers satisfying $ 3(a\plus{}b)\geq 2|ab\plus{}1|$
Prove that $ 9(a^{3}\plus{}b^{3})\geq |a^{3}b^{3}\plus{}1|$[/quote]
See  here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=608853
the sqing's solution.


trigonometry
function
inequalities
inequalities proposed
n-variable inequality

Let $b_i=\tan(a_i)$.  We are given $\sum_{i=0}^n \frac{b_i-1}{b_i+1} \geq {n-1}$.  Rearranging this gives $\sum_{i=0}^n \frac{1}{b_i+1} \leq 1$

Let $c_i=\frac{1}{b_i+1}$, so $0<c_i<1$.  We have $\sum \frac{1}{b_i} = \sum \frac{c_i}{1-c_i}=\sum (1-\frac{1}{1-c_i})$.  This last function is concave on $(0,1)$, so it follows that for a fixed $\sum c_i$ that $\sum \frac{1}{b_i}$ is maximized when all the $c_i$ are equal.  Since increasing a $c_i$ decreases the corresponding $b_i$ and thus increases $1/{b_i}$, it follows that $\sum \frac{1}{b_i}$ is maximized when all the $c_i$ are equal to $\frac{1}{n+1}$, i.e.
\[ \sum_{i=0}^n \frac{1}{b_i} \leq \frac{n+1}{n} \]
This means the harmonic mean of the $b_i$ is at least $n$, so the geometric mean must also be at least $n$.

[quote="kevinatcausa"]$\sum \frac{1}{b_i} = \sum \frac{c_i}{1-c_i}=\sum (1-\frac{1}{1-c_i})$.  This last function is concave on $(0,1)$...[/quote]

$\frac{c_i}{1-c_i}=\frac{1}{1-c_i} -1$, not sure whether your argument still works.

[hide="solution"]

Let $b_k=\tan{{a_k-\frac{\pi}{4})}, k=0,1,2,...,n}$  It follows from the hypothesis that for each $k, -1<b_k<1$ and
$1+b_k\geq \sum_{0\le l \not = k\le n} (1-b_l) (1)$
applying AM-GM to the positive numbers $1-b_l, l=0,1,2,...,k-1,k+1,...,n$  we obtain:
$\sum_{0\le l \not = k\le n} (1-b_l)\geq n(\prod_{0\le l \not = k\le n} (1-b_l))^{1/n} (2)$

from (1) and (2) it follows that
$\prod{k=0}^{n} (1+b_k)\geq n^{n+1}(\prod_{l=0}^{n}(1-b_l)^n)^{1/n}$
thus:
$\prod_{k=0}^{n}\frac{1+b_k}{1-b_k}\geq n^{n+1}$
becuase:
$\frac{1+b_k}{1-b_k}=\frac{1+\tan{(a_k-\frac{\pi}{4})}}{1+\tan{(a_k-\frac{\pi}{4})}}=\tan{ (a_k-\frac{\pi}{4}+\frac{\pi}{4})}=\tan{a_k}$
as desired :) [/hide]

Solution with smoothing!

Rearrange in the obvious way, then let $ b_{i} \equal{} \tan{a_{i}}$.

Lemma: When $ b_{i},b \minus{} {j}\ge1$, we have $ \dfrac{2}{1 \plus{} \sqrt {b_{i}b_{j}}}\le\dfrac{1}{1 \plus{} b_{i}} \plus{} \dfrac{1}{1 \plus{} b_{j}}$.
Proof: Expand, it becomes $ (b_{i} \minus{} b_{j})^{2}(b_{i}b_{j} \minus{} 1)\ge0$.

What this means is, if we have $ b_{i}b_{j}\ge1$, we can change them into $ \sqrt {b_{i}b_{j}},\sqrt {b_{i}b_{j}}$ so that the product of the $ b_{i}$'s remains the same and the $ b_{i}$ still satisfy the $ \displaystyle\sum_{i\equal{}0}^{n}\dfrac{1}{1\plus{}b_{i}}\le1$ condition, and furthermore as long as we can find two terms whose product is at least 1, we can make everything equal, hence reducing the problem to the equality case.

Now we can prove that we can always find two terms whose product is at least 1. If all of the $ b_{i}$ are at least 1 to start with, we're done; note that if $ b_{i}b_{j}\ge1$, the square root is also at least 1. Now, if more than two of the $ b_{i}$ are less than 1, we find that the sum must be more than 1 (because their corresponding terms will be more than 1/2), contradiction. So only one of them can be 1. Say $ b_{0} < 1$. We claim that there's another term $ b_{i}$ with $ b_{0}b_{i}\ge1$. Assume the opposite, so $ b_{i} < 1/b_{0}$ for all $ 1\le i\le n$. Then,

$ \displaystyle\sum_{i \equal{} 0}^{n}\dfrac{1}{1 \plus{} b_{i}} \equal{} \dfrac{1}{1 \plus{} b_{0}} \plus{} \displaystyle\sum_{i \equal{} 0}^{n}\dfrac{1}{1 \plus{} b_{i}} > \dfrac{1}{1 \plus{} b_{0}} \plus{} n\cdot\dfrac{b_{0}}{1 \plus{} b_{0}}\ge1$,

contradiction. Thus, we can operate on $ b_{0}$ and $ b_{i}$, making every term greater than 1, and proceed as before.

Now it's left to consider $ b_{0} \equal{} b_{1} \equal{} \cdots \equal{} b_{n}$, but this is obvious.  :)

It was brought to my attention that there's a hole in my above proof - we have to prove that we can indeed keep operating until all the $ b_{i}$ are equal. This actually isn't true, but we can prove, intuitively, that they get closer and closer together. Here's an outline (I haven't actually worked out all the details).

Define Let $ g$ be the Geometric Mean of all the terms, and define $ c_{i}$ to be $ g/b_{i}$ when this is at least 1, or the reciprocal when it's not, so $ c_{i} > 1$. Note that if the terms are not all equal, we can find two terms, on on each side of $ g$. It's not too hard to show that after operating on these, the product of the $ c_{i}$ goes down. The limit as the number of operations goes to infinity should be 1 (which I don't have a rigorous proof for yet, I'll post one if/when I find it). This means that $ \text{max}b_{i}/\text{min}b_{i}$ must to 1 as well.

Using the $ \displaystyle\sum\dfrac{1}{1 \plus{} b_{i}}$ condition, we can get that $ \text{max}b_{i}\ge n$. Assume the product of the $ b_{i}$'s is less than $ n^{n \plus{} 1}$, letting it be $ n^{n \plus{} 1} \minus{} x$, say. But note that the product is at least $ (\text{min}b_{i})^{n \plus{} 1}$, which, since max/min goes to 1, will surpass $ n^{n \plus{} 1} \minus{} x$ after a sufficiently large number of operations.

Again, I still have to show that the limit is 1...hopefully I'll get back to this.

OK I think this works. The limit clearly exists because the product is strictly decreasing and always stays above 1. Let the limit be L. Now, we can show that the product always decreases by a factor of $ c_{i}$ or $ c_{j}$ depending on which side of $ g$ $ b_{i}b_{j}$ is on. Thus, we have $ (L\plus{}\epsilon)c_{i}>L$, and we find that $ L/(L\plus{}\epsilon)<c_{i}$. Thus, as $ \epsilon$ goes to 0, $ c_{i}$ goes to 1, and thus the product of the $ c_{i}$'s goes to 1.

[quote="kevinatcausa"]Let $b_i=\tan(a_i)$.  We are given $\sum_{i=0}^n \frac{b_i-1}{b_i+1} \geq {n-1}$.  Rearranging this gives $\sum_{i=0}^n \frac{1}{b_i+1} \leq 1$

Let $c_i=\frac{1}{b_i+1}$, so $0<c_i<1$.[/quote]

Adopting this notation, another way to finish this by smoothing is to see that if we have $\sum{c_i}=k\le1$, then we wish to minimize $P=\prod{b_i}=\prod\left(\frac{1}{c_i}-1\right)$. But we can show that if some $u,v$ exist with $c_u<k/(n+1)<c_v$, then changing $(c_u,c_v)$ to $(k/(n+1),c_u+c_v-k/(n+1))$ will not change the sum of the $c_i$'s but will decrease the desired product, since
\begin{align*}
\left(\frac{1}{c_u}-1\right)\left(\frac{1}{c_v}-1\right) &> \left(\frac{1}{\frac{k}{n+1}}-1\right)\left(\frac{1}{c_u+c_v-\frac{k}{n+1}}-1\right)\\
&= \left(\frac{n+1}{k}-1\right)\left(\frac{n+1}{(n+1)(c_u+c_v)-k}-1\right)\\
\Longleftrightarrow\frac{1-c_u-c_v}{c_uc_v} &> \frac{(n+1)^2}{k[(n+1)(c_u+c_v)-k]}-\frac{n+1}{k}-\frac{n+1}{(n+1)(c_u+c_v)-k)}\\
&=\frac{(n+1)^2}{k[(n+1)(c_u+c_v)-k]}-\frac{(n+1)^2(c_u+c_v)}{k[(n+1)(c_u+c_v)-k]}\\
&=\frac{(n+1)^2(1-c_u-c_v)}{k[(n+1)(c_u+c_v)-k]}.
\end{align*}But $c_u+c_v<\sum{c_i}\le1$, so this is equivalent to
\[\frac{1}{c_uc_v} > \frac{(n+1)^2}{k[(n+1)(c_u+c_v)-k]},\]and because $(n+1)(c_u+c_v)>(n+1)c_v>k$, we can expand and rearrange to get\[0>k^2-k(n+1)(c_u+c_v)+c_uc_v(n+1)^2=(k-c_u(n+1))(k-c_v(n+1)),\]which is true because $c_u<k/(n+1)<c_v$ by assumption. Since this smoothing process brings more of the $c_i$ to $k/(n+1)$ each time, it clearly terminates and minimizes $P$ when all $c_i=k/(n+1)$, so noting that $k\le1$,
\[P=\prod\left(\frac{1}{c_i}-1\right)\ge\left(\frac{n+1-k}{k}\right)^{n+1}\ge n^{n+1},\]as desired.

[b]Edit:[/b] Darn... so evidently
\[P=\prod{b_i}=\prod\left(\frac{1}{c_i}-1\right)\ge\prod\left(\frac{c_0+\cdots+c_{i-1}+c_{i+1}+\cdots+c_n}{c_i}\right)\ge n^{n+1}\]by AM-GM.

Another way to smooth is to note that if not all $b_i$ are equal, then we can find $b_m$ greater and $b_n$ less than the geometric mean of all $b_i$. Then, we prove that if $a<b$ and $e>0$, then $\frac{1}{a+1} + \frac{1}{b+1} > \frac{1}{ae+1} + \frac{1}{\frac{a}{e}+1}$ by expanding, collecting terms, blahblahblah. Then, at each step, we can choose $e$ so that either $b_me$ or $\frac{b_n}{e}$ is equal to the geometric mean of all $b_i$, so we eventually get to the equality case.

Although the "infinite" smoothing by replacing two terms with their geometric mean is intuitive, it requires some limit stuff to rigorize, which may be frowned upon...

Sorry for reviving an old topic, using Jensen's inequality on the function $ln(\frac{1+x}{1-x})$, but it is concave on $(-1,0]$ and convex on $[0,1)$. How does smoothing solve this problem so we can use Jensen's inequality here?

Thank you! :)

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=266622

Solution from [i]Twitch Solves ISL[/i]:

Let $x_i = \tan(a_i - \frac{\pi}{4})$. Then we have that \[ \tan a_i = \tan(a_i - 45^{\circ} + 45^{\circ}) = \frac{x_i + 1}{1 - x_i}. \] If we further substitute $y_i = \frac{1 - x_i}{2} \in (0,1)$, then we have to prove that the following statement:
[b][color=red]Claim:[/color][/b]  	If $\sum_0^n y_i \le 1$ and $y_i \ge 0$, we have 	\[ \prod_{i=1}^n \left( \frac{1}{y_i}-1 \right) \ge n^{n+1}. \]
[i]Proof.[/i]  	Homogenizing, we have to prove that 	\[ \prod_{i=1}^n \left( \frac{y_0+y_1+y_2+\dots+y_n}{y_i}-1 \right) \ge n^{n+1}. \] 	By AM-GM, we have 	\[ \frac{y_1+y_2+y_3+\dots+y_n}{y_0} 		\ge n \sqrt[n]{\frac{y_1y_2y_3\dots y_n}{y_1}}. \] 		Cyclic product works. $\blacksquare$

[b][color=red]Remark:[/color][/b]  	Alternatively, the function $x \mapsto \log(1/x-1)$ 	is a convex function on $(0,1)$ so Jensen inequality should also work.

\begin{align*}
& \tan(a_i - \dfrac{\pi}{4})\\
&= \dfrac{\tan a_i - \tan \dfrac{\pi}{4}}{1+ \tan a_i \tan \dfrac{\pi}{4}}\\
&= \dfrac{\tan a_i - 1}{\tan a_i + 1}
\end{align*}

Let $x_i = \tan a_i \forall i$,
So
$$\tan(a_i - \dfrac{\pi}{4}) = \dfrac{x_i - 1}{x_i + 1}$$

Since $a_i \in (0,\dfrac{\pi}{4}) \forall i$, we must have $x_i > 0 \forall i$.
Then we have $-1 < \tan(a_i - \dfrac{\pi}{4}) < 1 \forall i$.

We now apply a local adjustment.
We want to minimise $xy$ given a fixed $\dfrac{x-1}{x+1} + \dfrac{y-1}{y+1} = c$.
Let us denote $x' = \dfrac{x-1}{x+1}$,
Thus $x= \dfrac{1+x'}{1-x'}$, $x'+y'=c$.

\begin{align*}
&xy\\
&= \dfrac{(1+x')(1+c-x')}{(1-x')(1-c+x')}\\
&= \dfrac{-x'^2 + cx' + c + 1}{-x'^2 + cx' - c + 1}
\end{align*}
Which reaches minimum at $x'= \dfrac{c}{2}$,
Where $x'=y'$.

Consider the following operation.
Each time we take the 2 $x_i'$s with the greatest difference,
And set them to be equal to their arithmetic mean.
This decreases the product of all $x_i$s as shown.
Each time this is done,
The maximum difference between 2 $x_i'$s decrease
So eventually we have
$$x_i' = x_j' = \dfrac{n-1}{n+1} \forall i,j \in \mathbb{Z_0^+}, i,j \geq 0$$

We then evaluate the required product of all $x_i$s under this worst case.
For any $0 \leq i \leq n$,
\begin{align*}
&x_i\\
&= \dfrac{1+ x_i'}{1- x_i'}\\
&= \dfrac{1+ \dfrac{n-1}{n+1}}{1- \dfrac{n-1}{n+1}}\\
&= \dfrac{n+1+n-1}{n+1-n+1}\\
&=n
\end{align*}

Thus we conclude
$$ x_0x_1 \dots x_n \geq n^{n+1} \forall n$$
$\square$

[quote=Potla][quote="MithsApprentice"]Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that \[ \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1.  \] Prove that \[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.  \][/quote]
What about AM-GM? ;)
Let $x_i=\tan a_i$ and rewrite the given condition into $\frac 1{x_0+1}+\frac1{x_1+1}+\cdots+\frac 1{x_n+1}\leq 1.$ We have to show that $x_0x_1x_2\cdots x_n\geq n^{n+1}.$
Note that using the given condition, we have
\[\frac{x_0}{x_0+1}\geq \frac{1}{x_1+1}+\frac 1{x_2+1}+\cdots+\frac 1{x_n+1}\geq n\sqrt[n]{\frac 1{(x_1+1)(x_2+1)\cdots (x_n+1)}}.\]
Getting its analogues and multiplying all the relations together, we get
\[x_0x_1x_2\cdots x_n\geq n^{n+1}.\]
Equality occurs iff $x_i=n\forall i=0,1,\cdots n.\Box$
:)[/quote]

This. Was. Beautiful.

Let $x_i = \tan(a_i - \pi/4)$ for each $i$. Then it suffices to show that $$\prod \left(\frac{x_i+1}{1-x_i}\right) \geq n^{n+1} \iff \sum \log\left(\frac{x_i+1}{1-x_i}\right) \geq (n+1) \log n.$$ For $y_i = \frac 12(1-x_i)$, the inequality becomes $$\sum \log\left(\frac 1{y_i} - 1\right) \geq (n+1)\log n,$$ which is immediate by Jensen.

[color=#9a00ff][b]Remark.[/b][/color] Interestingly enough, direct Jensen on the first display doesn't quite work, which motivates the substitution.

Define $t_i = \tan(a_i) \ge 0$.
Then \[ \sum_{i=0}^n \frac{t_i - 1}{t_i + 1} \ge n - 1 \]
\[ 1 \ge \sum_{i=0}^n \frac{1}{t_i + 1} \]
By GM-HM it follows that \[ \sqrt[n+1]{(t_0 + 1)\dots (t_n + 1)} \ge \frac{n + 1}{\frac{1}{t_0 + 1} + \dots + \frac{1}{t_n + 1}} \] and thus \[ (t_0 + 1)(t_1 + 1)\dots (t_n + 1) \ge (n+1)^{n+1}. \]
Smoothing together $t_i$ and $t_j$ while preserving $(t_i + 1)(t_j + 1)$ decreases $t_it_j$, thus $t_0 \dots t_n$ is minimized when $t_0 = t_1 = \dots = t_n = n$.


Find the adherence of the set of complex nxn matrices M for which there exists $k\geq 1$ such that $M^k=I$.

I think this is the set B of matrices whose eigenvalues have
modulus equal to 1.

Let E={M s.t there exists some k such that M^k=I}
Then it is clear that the set A of diagonalizable matrices whose
eigenvalues have modulus equal to 1 is in the adherence of E. But it
is not hard to see that the adherence of A is B. Let M a mtrice in B:
then there exists a basis where M is triangular with (d1,d2,..,dn) on
the diagonal. But M is the limit point of the sequence of
matrices M_n where:
M_n(i,j)=M(i,j) for i<>j and M_n(i,i)=di*(Tn)^i and Tn=exp(2*i*Pi/n)
There exist a sufficiently large N0 such that for n>N0, M_n is in A.
So B=Adh(A). is it right ?

could you please explain what does it mean the word "adherence", i can't understand clear meaning of this word because of my poor english either math stupidness

the adherence of X is the intersection of all closed sets that contain X ...


Is any 2004 Iberoamerican Olympiad contestant here? Does anyone have today's test?

Carlos

HY Carlos I am pascual from colombia and i participated in IMO but i didnt take this OIM because i want to go next year because its gonna be in colombia....Anyway, some friends who took the test told me they will email it to me as soon as they could, so ill post it here...

BTW do you think Brasil is going to win again? I think thAt if you dont win we ill do it! We were close in IMO!

Hi Pascual, thanks very much in advance.

Well, the four students in this Ibero were silver medal in IMO (Fabio won silver in Japan 2003; Alex won silver in USA 2001; Rafael Hirama and Gabriel won silver this year in Greece), so I'm confident Brazil will do very well in this Ibero.

But Colombia, Mexico, Peru and Argentina (though Yue is not there) are not out of the running, so I think it's gonna be a great competition among these countries...

Anyway, good luck and good job to all the students taking this test!

Regards, Carlos

The 2004 OIM's problems are posted in http://www.teorema.mat.br/phpBB2/viewforum.php?f=4
But they are in Portuguese language... :?

I read it in portuguese and "traslate" it to english.....Here are the problems, from the first day:

1.[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=117424#117424[/url]
2.[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=117425#117425[/url]
3.[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=117427#117427[/url]

Why didnt Fabio Won silver this year? I thought he would get a gold!

Brasil going to win again? I thougth we won last year! Maybe my memory is playing with me!
 Anyway, please post here anything you know about the contest, not only the problem but when you know, the medalls  (in fact someone told me there are no medals this year).
  How Brasil did? And Colombia?

Sorry, You are right Carlos, You won last year because you had a super
-team, but Brasil Won at iberoamerican level this year in the IMO.

Alex: 42 points
Fabio: 40 points
Gabriel: 37 points
Rafael: 39 points

But I don't know the scores for getting gold, silver or bronze medals yet.

It seems you were right, Brazil is winning the olympiad again and again and again.

As soon as you know more results or medalls post it please.
 Thank you.

I have posted all the problems. here are the links. BTW the problems were easy and problem 6 is a well known.

4.[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=117721#117721[/url]
5.[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=17064[/url]
6.[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=117721#117721[/url]

The results are in [url]http://www.iberolimp.uji.es/htmdocs/medal.htm[/url].

Regards, Carlos.

Well, looks like Brasil won again, 4 gold medals, but colombia also did it great, Santiago Saavedra scored perfect and esteban Gonzales scored 41. Well congratulations to Brasil and colombia, and to all the other teams.

We ill have to wait until 2005 to see what happens there!

Pascual: Really great for colombia! Send congratulations to Santiago and jonathan (creo que a los demas no los conozco)....
  And congratulations Brasil too....

Why the 4th and 6th problem are same from the link above?


This graph shows the linear relationship between the time in seconds, $ x$, for Caroline's walk and the distance in meters, $ y$, Caroline is from her starting point. The graph passes through the point $ (20,30)$. According to the graph, how many meters will Caroline walk in exactly one hour?
[asy]pair A,B,C,D,E,F,G;

A=(-20,0);
B=(120,0);
C=(0,-20);
D=(0,120);
E=(-6.2,-8);
F=(60,90);
G=(70,105);

dot(F);

draw(A--B,Arrows);
draw(C--D,Arrows);
draw(E--G,Arrows);

label("(20,30)",G,SE);
label("$y=1.5x$",G,N);[/asy]

There are $ 60 \cdot 60 \equal{} 3600$ seconds in an hour.  Thus, we plug in $ 3600$ for $ x$ in $ y \equal{} 1.5x$, and we get $ y \equal{} 1.5 \cdot 3600 \equal{} \boxed{5400}$ meters.


At first two different pipes running together will fill a tank in 20/3 minutes. The rate that water runs through each of the pipes is then adjusted. If one pipe, running alone, takes 1 minutes less to fill the tank at its new rate, and the other pipe, running alone, takes 2 minutes more to fill the tank at its new rate, then the two running together will fill the tank in 7 minutes. Find in what time the tank will be filled by each pipe running alone at the new rates?
Help me plz. I can't solve it.

Suppose the first pipe could originally fill the tank in $x$ minutes,
and the second in $y$ minutes.

Then,

 $\frac{1}{x}+\frac{1}{y}= \frac{3}{20}$

After the adjustment,
the first pipe can fill the tank in $x-1$ minutes, while
the second takes $y+2$ minutes.

Then,

 $\frac{1}{x-1}+\frac{1}{y+2}= \frac{1}{7}$

 Using the first equation, write $y$ in terms of $x$ ,
 substitute this in the second equation to obtain the required values.

of course it will be a mess, but i s'pose that's because the values are not nice.

[hide="Complete Solution"]Let $a_{1}$ and $b_{1}$ be the original times it took for the pipes to fill the tank, and $a_{2}$ and $b_{2}$ be the new times.

$\frac1{a_{1}}+\frac1{b_{1}}=\frac3{20}$

$\frac1{a_{2}}+\frac1{b_{2}}=\frac17$

$\frac1{a_{1}-1}=\frac1{a_{2}}$, or $\frac1{a_{1}}=\frac1{a_{2}+1}$

$\frac1{b_{1}+2}=\frac1{b_{2}}$, or $\frac1{b_{1}}=\frac1{b_{2}-2}$

Substituting the last two equations into the first gives $\frac1{a_{2}+1}+\frac1{b_{2}-2}=\frac3{20}$.

Now the two equations that have just the new times in them are multiplied out:

$7a_{2}+7b_{2}=a_{2}b_{2}$ and $20\left(a_{2}+1+b_{2}-2\right)=3\left(a_{2}+1\right)\left(b_{2}-2\right)$

The second equation can be simplified to $20a_{2}+20b_{2}-20=3a_{2}b_{2}+3b_{2}-6a_{2}-6$, and even further to $26a_{2}+17b_{2}-14=3a_{2}b_{2}$.

Now, the first equation is multiplied by $3$, and is subtracted from the second (removing the $a_{2}b_{2}$).

When simplified, the new equation is $5a_{2}-4b_{2}=14$.

The first equation can be factored into $\left(a_{2}-7\right)\left(b_{2}-7\right)=49$ by SFFT. From here, $b_{2}$ is isolated: $b_{2}=\frac{49}{a_{2}-7}+7$.

Plugging this into the other equation, $5a_{2}-4\left(\frac{49}{a_{2}-7}+7\right)=14$.

After some algebra, it is found that $a_{2}=14$, and from their, $b_{2}=14$.

After the adjustment, both pipes will fill the tank is $\boxed{14\text{ minutes}}$ alone.[/hide]

[quote="thientran1210"]thank you very much. But i want to know how can you get 7a2+7b2=a2b2--> (a2-7)(b2-7)=49, i mean how can you think about it, but not the other way. Sorry about my equation.[/code][/quote]

What do you mean by "not the other way"?

However, the use of SFFT is very unnecessary because it is easier to just arrange it in  the form $kb=m$ like: $b_{2}(a_{2}-7)=7a_{2}$ to see that $b_{2}=\frac{7a_{2}}{a_{2}-7}$.

[hide="my solution"]Let's say it takes $m$ and $n$ minutes for the first and second pipe to fill in the tank after the adjustment. Then,
\[\frac{1}{m+1}+\frac{1}{n-2}= \frac{3}{20}. \]
And also, from the new rates,
\[\frac{1}{m}+\frac{1}{n}= \frac{1}{7}. \]
It is solving the system...
\[20(m+1+n-2)=3(m+1)(n-2)\\ 7(m+n)=mn \]
Expand the first equation: $20(m+n)-20=3mn-6m+3n-6$
Since $mn=7(m+n)$, $20(m+n)-20=21(m+n)-6m+3n-6$
Then $m+n=6m-3n-14$ and $5m=4n+14$. Substituting,
$5mn=7(5m+5n) \rightarrow (4n+14)n=63n+98 \rightarrow 4n^{2}-49n-98=0$.
Solving, $(4n+7)(n-14)=0$, so $n=14$ and $m=70/5=14$.[/hide]


Many journalists, in time pressure or maybe just plain laziness, resort to stock photos when writing articles that are to be accompanied with a photo.

Sometimes the use of photo stock are blatently obvious.

In this game, you play the journalist. Go on a news site (or wikipedia, whatever), cut a paragraph or two from a news article, and then post your photo to go with it! (Google Image Search, basically. To post an image in a thread, when doing a google image search, click "view full size image", copy the URL from it, and then post
[code][img]http://URL OF IMAGE[/img][/code]

[quote]
WASHINGTON -- Senator Barack Obama decisively swept the Democratic presidential primaries today in Maryland, Virginia, and the District of Columbia to [b]seize the momentum[/b] over Senator Hillary Clinton, who has lost eight straight primaries and caucuses since Super Tuesday.
[/quote]
http://www.boston.com/news/politics/politicalintelligence/2008/02/obama_mccain_sw.html

[img]http://www.antonine-education.co.uk/Physics_AS/Module_2/Topic_7/Momentum_4.gif[/img]

[quote]Prime Minister Kevin Rudd opened a new chapter in Australia’s tortured relations with its indigenous peoples with a comprehensive and moving apology for past wrongs.[/quote]

[hide]
[img]http://www.squeep.com/~fek/misc/sepia-backgrounds/writing-2.png[/img]
[/hide]

Um done right?

Edit: Oh, sorry...  :oops:

Request: Please place all images under a hide tag.

http://www.associatedcontent.com/article/853687/new_milk_containers_i_like_my_new_milk.html?cat=3
bumped cuz i found it humorously relevant


The set of rational numbers is the same size as the set of integers. There is a proof on p273 in Art of Problem Solving, Volume 2. However, I think I found a flaw in that proof. 1/2 corresponds to 2, 2/4 corresponds to 12, but 1/2 = 2/4, so they're really the same number. So we don't have a 1-1 correspondence anymore.

Well that shows that you can at least map each rational number to an integer, and to show that you can map each integer to a rational number, then just map it onto itself. I'm not sure if this logic is correct or not, but it shows that you can go strictly from one to the other, and if you can do that both ways, then shouldn't they be the same size?

I'm not sure if you understood what I said. I meant to say that 1/2 is the same rational number as 2/4, but they correspond to different integers. It's a problem with unreduced fractions.

That's easy to overcome: If during the mapping you stumble upon a rational already covered, just skip it. You'll still map exactly one rational to exactly one natural.

Is this the proof with the grid with all integers on each axis?
Why not put all integers on one axis and all prime positive integers on the other axis, and then put two integers on each resulting square?

It's correct if you only use fully reduced fractions.


If I select any day at random, what is the probability it willl be July 7th?

[hide]In 4 years there are 4*365+1=1461 days.

Thus the probability is 4/1461 if were not using the fact that 1900 wasn't a leap year :P[/hide]

Well, we are using a fact that 1900 wasn't a leap year...
a nasty fraction,
calculators allowed  :P

Umm... I don't think I quite understand this question. What's the time period?

1900 was a leap year... but 2000 was not. That's what's confusing me. How big of a time period is this using?

You got it the wrong way. 1900 wasn't a leap year.

Leap years are every four years unless its a multiple of 100 unless its a multiple of 400.  [hide]In 400 years there are 400*365+97=146097 days, 400 are July 7, so 400/146097[/hide]

i was looking at the question and i said "well, isn't it 1/365?"
but then i thought about leap years and nat mc's solution cleared it up for me. :)

The exact fraction depends on the time period, and the year it starts and ends. As everyone said, you don't know how many leap years are in a time period if you don't know what years it encompasses. For example, there is one leap year in the year 2004, but no leap years in the year 2001. Also even with the 400-year rule, there are still difficulties, because of Caesar's funny 400s-day year in the 400s B.C., the Julian vs. Gregorian calendar, and the ancient calendars that did not use the same names of the months.

But for all practical purposes, it's a bit under $\frac{1}{365.25}$

However , in peroid of 4000 years , the Gregorian calender is again advanced by one more day  :?

Really? I didn't even know that. I guess this problem wasn't such a good problem....
It is nearly impossible to just follow one pattern for long, because soon the day differences will add up.