AMC
AIME
ARML
geometry
number theory
AoPS Books

I am preparing for the upcoming AMC-12 (and AIME, I'm fairly sure), and I had some questions regarding the best way to study for. I have already signed up for the AMC-12 Problem Series class, and for the Introductory Number Theory class, because I never touched the stuff till ARML last year. In terms of preparation, what else is suggested? The options I'm considering include:

Introduction to geometry: The problem with this class is that I can solve all but 1 or 2 of the post-test problems within 2-3 minutes; the others took me 15-20 minutes ( :blush: ). Yet there definitely exist AMC geometry problems that would give me one heck of a lot of trouble. What should I do?

Introduction to Counting/Probability: I cannot get 2-3 of the post-test problems. Again, I feel weak on this.

AoPS books: How much depth do they cover? How strong are they?

Please bear in mind (Yes, I whine now) that I have less free time to work on math than I would like due to compromises with parents in areas of Literature, etc. I can only spend 3-6 hours every week (not counting class time), though that number should rise to 6-8 by February (an activity ends).

What are the recommendations?

Thanks

I would definitely recommend the AoPS books.  A full list of what the books cover can be found [url=http://www.artofproblemsolving.com/Books/AoPS_B_V1.php]here[/url] (for volume 1) and [url=http://www.artofproblemsolving.com/Books/AoPS_B_V2.php]here[/url] (for volume 2).

It sounds to me like you probably don't need the Intro to Geometry class.  Since you solved all of the post-test problems (and most of them pretty quickly) the class would probably mostly be a review.  Sadly, there is not (yet) and Intermediate Geometry class, although AoPS volume 1 has several chapters on geometry, if you still feel that you are weak in this area.

As for counting/probability, it also sounds to me that you may not want the Intro course.  However, they do offer an Intermediate class.  Have you looked at the pre/post tests for that one?  Perhaps that class might be more appropriate.


geometry circumcircle inradius inequalities

Find all right triangles for which its circumradius R and inradius r satisfy the equation R = 2r.

[hide]There are none.  The relation R = 2r holds iff the triangle is equilateral:

http://mathworld.wolfram.com/EulersInequality.html[/hide]

Mod edit--Please put your comments/solutions in spoiler! Thank you --JSRosen3

[hide]R = abc/(4[ABC])

r = [ABC]/s

If R = 2r, then abc/(4[ABC]) = 2[ABC]/s, i.e. s(abc) = 8[ABC]^2

But [ABC] = sqrt[(s)(s-a)(s-b)(s-c)], so 8[ABC]^2 = 8s(s-a)(s-b)(s-c).

Thus, the equality becomes s(abc) = 8s(s-a)(s-b)(s-c)

abc = 8(s-a)(s-b)(s-c)

abc = (-a+b+c)(a-b+c)(a+b-c)

...and now I'm stuck.  It seems like we want to establish some inequality involving the two sides.  AM-GM would be nice, since the equality condition for it is when a=b=c, which is the solution.  But I don't see anything.  Maybe expansion:

abc = -a^3 - b^3 - c^3 +ac^2 + ca^2 + ab^2 + ba^2 + bc^2 + cb^2 - 2abc

a^3 + b^3 + c^3 + 3abc = ac^2 + ca^2 + ab^2 + ba^2 + bc^2 + cb^2

(a+b+c)^3 = 3abc + 4(ac^2 + ca^2 + ... + cb^2)[/hide]

[hide]

for mathfanatic:

a^3 + b^3 + c^3 + 3abc = ac^2 + ca^2 + ab^2 + ba^2 + bc^2 + cb^2

a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)=0

but that is >=0 (i think that is schur's), so its only true in the equality condition a=b=c

[/hide]

[hide]I'm confused... the problem asks for right triangles... so none?[/hide]

[hide]There are none, but it doesn't seem fair to me to assume the inequality R  >= 2r in doing this problem. The solution I had in mind was writing R and r in terms of the sides of the triangle and working from there.[/hide]


Ice cream pops cost 75¢ each and ice cream cones cost 95¢ each. What is the cost of 4 pops and 2 cones?

A) $1.20
B) $3.00
C) $4.90
D) $6.00


Details are given at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=334066#334066]Ultimate guide to inserting Pictures in LaTeX[/url]


Miguel started with $ \$18$. After he bought a compact disc, his remaining money was $ \frac45$ of what he had spent for the compact disc. How many dollars did the compact disc cost?

Since his remaining money was $ \frac45$ of his CD, his remaining money was $ \frac{4}{4\plus{}5}$ of what it was before.

Thus, he had $ \$8$ left and the answer is $ \boxed{\$10}$.

A quick and efficient way to do this problem is to set up a linear equation.

18 - c = (4/5)(c)

Solve for c to get 10.
_-_-_-_-_-_-_-_-_-_-_-_

~pi man!!  :nhl:


geometry
inradius
circumcircle
incenter
Euler
trigonometry
radical axis

can anyone help me wif diz:
prove the radius of the apollonius circle (the apollonius circle that is externally tangent to the three excircles) is r^2 + s^2 / 4*r (r = inradius, s= semiperimeter as usual).
generally, how would you find the radius of a circle which is externally tangent to three known circles ?

[quote="vinoth_90_2004"]prove the radius of the apollonius circle (the apollonius circle that is externally tangent to the three excircles) is

$\frac{r^2+s^2}{4r}$

(r = inradius, s = semiperimeter as usual).[/quote]

This is a piece of triangle geometry far from being simple, hence please excuse me for the long depiction.

The nine-point circle of triangle ABC is the circumcircle of triangle A'B'C', where A', B', C' are the midpoints of the sides BC, CA, AB. The center N of the nine-point circle is the circumcenter of the triangle A'B'C'. The radius of the nine-point circle is R/2, where R is the circumradius of triangle ABC. The nine-point circle touches all three excircles externally.

The Spieker point S of triangle ABC is defined as the incenter of triangle A'B'C'. The incircle of triangle A'B'C' has center S and is called the "first Spieker circle" of triangle ABC. Let r be the inradius and s the semiperimeter of triangle ABC. The circle with center S and radius $\frac{1}{2}\sqrt{r^2+s^2}$ will be called the "second Spieker circle" of triangle ABC. Now, I state the following deeper result which I will prove later:

[b]Lemma 1.[/b] This "second Spieker circle" of triangle ABC is orthogonal to the three excircles of triangle ABC.

Hence, the inversion with respect to this second Spieker circle lets the excircles fixed. On the other hand, the nine-point circle, which touches the three excircles externally, is mapped by this inversion to a circle which touches the three excircles internally (i. e. which touches the three excircles and encompasses them). This is the Apollonius circle!

Hence, we can state the result that the Apollonius circle is the image of the nine-point circle in the inversion with respect to the second Spieker circle.

Now, it is easy to prove that the inversion with respect to a circle with center M and radius k maps a circle with center O and radius r to a circle with radius

$\frac{k^2 r}{\left| MO^2-r^2\right| }$.

In our case, the second Spieker circle has center S and radius $\frac{1}{2}\sqrt{r^2+s^2}$, while the nine-point circle has center N and radius R/2. Hence, the inverted image, i. e. the Apollonius circle, has radius

$\frac{\frac{1}{4}\left( r^2+s^2\right) \frac{R}{2}}{\left| SN^2-\left( \frac{R}{2}\right) ^2\right| }$.

Now, remember the Euler formula which states $IO^2=R^2-2Rr$ if I is the incenter and O is the circumcenter of a triangle ABC. Apply this to the triangle A'B'C', whose incenter is S, whose circumcenter is N, whose circumradius is R/2, and whose inradius is r/2. Then you get

$SN^2=\left( \frac{R}{2}\right) ^2-2\frac{R}{2}\frac{r}{2}$.

Thus, after a bit of calculation, the above expression for the radius of the Apollonius circle takes the form

$\frac{\frac{1}{4}\left( r^2+s^2\right) \frac{R}{2}}{\left| -2\frac{R}{2}\frac{r}{2}\right| }=\frac{\frac{1}{4}\left( r^2+s^2\right) \frac{R}{2}}{2\frac{R}{2}\frac{r}{2}}=\frac{r^2+s^2}{4r}$,

qed..

I conclude with the [i]proof of Lemma 1.[/i]

Let $I_a$, $I_b$, $I_c$ be the three excenters of triangle ABC. The lines $I_bI_c$, $I_cI_a$, $I_aI_b$ are the external angle bisectors of the angles CAB, ABC, BCA and pass through A, B, C, respectively.

Let $X_b$, $X_c$, $Y_c$, $Y_a$, $Z_a$, $Z_b$ be the orthogonal projections of the points B, C, C, A, A, B on the lines $I_bI_c$, $I_bI_c$, $I_cI_a$, $I_cI_a$, $I_aI_b$, $I_aI_b$, respectively. (You are reading right: Each point is projected onto two lines, and on each line we project two points!) These six orthogonal projections have bags of properties. For instance, some simple angle chase shows that the points $X_b$ and $Y_a$ lie on the line joining the two points where the C-excircle touches the lines CA and AB, while the points $Y_a$ and $Z_a$ lie on the line B'C' joining the midpoints B' and C' of the segments CA and AB. While the first of these properties won't be used in the following, the second one will be used, and therefore we will give a proof of it later. But actually, what are we aiming at? Our first goal is to show that the points $X_b$, $X_c$, $Y_c$, $Y_a$, $Z_a$, $Z_b$ lie on one circle, namely - did you expect it? - on the second Spieker circle of triangle ABC.

It is merely obvious that the point $Y_a$ lies on the circle with diameter AB. The center of this circle is the midpoint C' of AB. Hence,

$\measuredangle BC^{\prime }Y_a=2\cdot \measuredangle BAY_a=2\cdot \left( 90^{\circ }-\measuredangle ABY_a\right) =2\cdot \left( 90^{\circ }-\left( 90^{\circ }-\frac{B}{2}\right) \right)$
$=B=\measuredangle ABC=\measuredangle AC^{\prime }B^{\prime }$,

and thus, the point $Y_a$ lies on the line B'C'. Similarly, the point $Z_a$ lies on that line, but, strictly speaking, we won't need even this in the further.

The length $C^{\prime }Y_a$ equals the radius of the circle with diameter AB, i. e. it equals $C^{\prime }Y_a=\frac{1}{2}AB=\frac{1}{2}c$. Now, let X be the point where the incircle of triangle A'B'C' touches the side B'C'. Remember that the center of this incircle is the Spieker point S of triangle ABC. The triangle $SXY_a$ is right-angled, with one catet SX = r/2 (since r/2 is the inradius of triangle A'B'C'), and the other catet $Y_aX=C^{\prime }X+C^{\prime }Y_a=\frac{1}{2}\left( s-c\right) +\frac{1}{2}c=\frac{1}{2}s$. (Actually, we have $C^{\prime }X=\frac{1}{2}\left( s-c\right)$ because the corresponding segment in triangle ABC is s-c, and each linear value of triangle A'B'C' is one half of the corresponding linear value of triangle ABC). Hence, after the Pythagoras theorem, the squared hypotenuse is $SY_a\;^2=SX^2+Y_aX^2=\left( \frac{1}{2}r\right) ^2+\left( \frac{1}{2}s\right) ^2=\frac{1}{4}\left( r^2+s^2\right)$. Similarly, one gets the same value for $SY_c\;^2$, $SZ_a\;^2$, $SZ_b\;^2$, $SX_b\;^2$ and $SX_c\;^2$. Therefore, the six points $X_b$, $X_c$, $Y_c$, $Y_a$, $Z_a$, $Z_b$ lie on the circle with center S and radius $\frac{1}{2}\sqrt{r^2+s^2}$, i. e. on the "second Spieker circle" of triangle ABC. Now it only remains to show that this second Spieker circle is orthogonal to the three excircles!

This goes as follows. Let D be the foot of the perpendicular from the C-excenter $I_c$ to the line BC. We have

$\measuredangle I_cCX_c=\measuredangle I_cCA+\measuredangle ACX_c=\measuredangle I_cCA+\left( 90^{\circ }-\measuredangle CAX_c\right) =\frac{C}{2}+\left( 90^{\circ }-\left( 90^{\circ }-\frac{A}{2}\right) \right)$
$=\frac{C}{2}+\frac{A}{2}=90^{\circ }-\frac{B}{2}=\measuredangle I_cBD$

and $\measuredangle I_cX_cC=90^{\circ }=\measuredangle I_cDB$; hence, the triangles $I_cX_cC$ and $I_cDB$ are similar, and we obtain

$\frac{I_cX_c}{I_cD}=\frac{I_cC}{I_cB}$.

On the other hand,

$\measuredangle AI_cB=180^{\circ }-\measuredangle I_cAB-\measuredangle I_cBA=180^{\circ }-\left( 90^{\circ }-\frac{A}{2}\right) -\left( 90^{\circ }-\frac{B}{2}\right)$
$=\frac{A+B}{2}=90^{\circ }-\frac{C}{2}$

and thus

$\measuredangle I_cCD=\frac{C}{2}=90^{\circ }-\left( 90^{\circ }-\frac{C}{2}\right) =90^{\circ }-\measuredangle AI_cB$
$=90^{\circ }-\measuredangle X_bI_cB=\measuredangle I_cBX_b$.

Also, obviously $\measuredangle I_cDC=90^{\circ }=\measuredangle I_cX_bB$. Hence, the triangles $I_cCD$ and $I_cBX_b$ are similar, entailing

$\frac{I_cC}{I_cB}=\frac{I_cD}{I_cX_b}$.

Combining with the previous equality, we get

$\frac{I_cX_c}{I_cD}=\frac{I_cD}{I_cX_b}$,

or $I_cX_c\cdot I_cX_b=I_cD^2$. But $I_cD$ is just the radius of the C-excircle of triangle ABC. Hence, the two points $X_b$ and $X_c$ are mutually inverse with respect to the C-excircle of triangle ABC. Therefore, the second Spieker circle of triangle ABC, passing through these two points, must be orthogonal to the C-excircle! Similarly, the second Spieker circle is orthogonal to the A-excircle and the B-excircle, too. The proof of Lemma 1 is complete.

Note that Lemma 1 can be rewritten as follows:

The Spieker point S of triangle ABC is the radical center of the three excircles of triangle ABC, and its power with respect to any of these excircles is $\frac{1}{4}\left( r^2+s^2\right)$, where r is the inradius and s is the semiperimeter of triangle ABC.

There are some much easier ways to show that S is the radical center of the three excircles, but most of these ways don't allow to find directly the power of S. One can also bash Lemma 1 by some trigonometry (actually, doing this intelligently, one can keep the calculation at less than 3 pages...), but this is not the kind of proof I like to post at MathLinks. Another analytic proof was given by Peter Y. Woo at

http://www.woopy.woo.org/math/spieker.htm

Finally note another consequence of our proof of Lemma 1. If ABC is an acute-angled triangle, and AX, BY, CZ are its altitudes, then triangle XYZ is called the "orthic triangle" of triangle ABC, and it is easy to show that A, B, C are the excenters of triangle XYZ. Now, let $B_{x}$, $C_{x}$, $C_{y}$, $A_{y}$, $A_{z}$, $B_{z}$ be the orthogonal projections of the points Y, Z, Z, X, X, Y on the lines BC, BC, CA, CA, AB, AB. Then, after what we have proved above, these projections $B_{x}$, $C_{x}$, $C_{y}$, $A_{y}$, $A_{z}$, $B_{z}$ lie on one circle, namely the second Spieker circle of triangle XYZ. This circle is called the "Taylor circle" of triangle ABC. We have hereby not only proved the existence of the Taylor circle (for an acute-angled triangle only; however, for an obtuse-angled triangle, everything is analogous by replacing an excenter by the incenter), but also the fact that the Taylor circle of an acute-angled triangle is the second Spieker circle of its orthic triangle, and consequently, the center of the Taylor circle of an acute-angled triangle is the Spieker point of its orthic triangle.

[quote="vinoth_90_2004"]generally, how would you find the radius of a circle which is externally tangent to three known circles ?[/quote]

Generally, this is possible but it should be [i]very[/i] messy...

  Darij

[quote="N.T.TUAN"]Please check link : http://www.woopy.woo.org/math/spieker.htm .[/quote]

Indeed, I have no idea why the guy got offline, but traces remain:
http://web.archive.org/web/20051026093051/jacob.biola.edu/~woopy/math/spieker.htm
The graphic is missing, however...

  darij


trigonometry induction function conics parabola quadratics algebra

1.  Find $(sin 15^{\circ} + sin75^{\circ})^6$.

2.  Find least positive x for which $cos 5^{\circ} = sin 25^{\circ} + sin x^{\circ}$.

3.  Find least positive x for which $tan x = \frac{cos5^{\circ} cos 20^{\circ} + cos 35^{\circ} cos 50^{\circ} - sin 5^{\circ} sin 20^{\circ} - sin 35^{\circ} sin 50^{\circ}}{ sin5^{\circ} cos20^{\circ} - sin35^{\circ}cos50^{\circ} + cos5^{\circ}sin20^{\circ} - cos35^{\circ}sin50^{\circ}}$

4.  $(cos 22.5^{\circ})^2$ is a root of $x^2 + bx + c = 0$ where $b,c \in Q$.  Determine $(b,c)$.

5.  Determine $arcsin\frac{4}{5} +arccos\frac{12}{13} + arcsin\frac{16}{65}$.

6.  John types the number two into a calculator, then presses the buttons arctan, sin, arctan, sin, arctan, sin,... etc.  After the sin key has been pressed 13 times, the display contains the decimal $0.abababab$., where a,b are digits.  Find $(a,b)$.

7.  Find smallest integer p for which $sin x = \frac{x}{p}$ has exactly 1999 solutions.

8.  Determine $cos \frac{\pi}{7} - cos \frac{2\pi}{7} + cos \frac{3\pi}{7}$

9.  Let p,q,r be roots of $1993x^3 - ax^2 + bx + c = 0$, where $a,b,c \in R$.  Given $arctan (p) + arctan (q) + arctan (r) = \frac{\pi}{4}$, determine $a+b+c$.

10.  $\prod_{k=1}^{45} sin(2k-1) = \frac{1}{2^n}$.  Determine the rational number $n$.

Singular, putting \ before each trig name makes them look better.

1
[hide]
We rewrite statement in parentheses as $\sin(45 - 30) + \sin(45 + 30)$
Using $\sin(a+b)=\sin a \cos b + \cos a \sin b$, this simplifies to $\frac{\sqrt 6}2$
So we need $\left(\frac{\sqrt 6}2\right)^6 = \left(\frac 6 4 \right)^3 = \frac{27}{8}$
[/hide]

2
[hide]
Rewrite as $\sin x = \sin 85 - \sin 25$
Using $\sin a - \sin b = 2\sin\left(\frac{a - b}{2}\right)\cos\left(\frac{a + b}{2}\right)$, we get
$\sin x = \cos 55$
$x = 35$
[/hide]

[hide="4."]
$\cos\frac\pi8=\frac{\sqrt{2+\sqrt2}}2 \ \ \ \Rightarrow \cos\frac\pi8=\frac{2+\sqrt{2}}4$

Now let $\cos\frac\pi8=x$, then we have $4x-2=\sqrt2$
squaring both sides gives us $16x^2-16x+4=2$
$\therefore (b,c)=(-1,\frac18)$[/hide]

[color=blue][b](7)[/b][/color] The line $y = \frac x p$ cuts the curve $y = \sin x$ 2x in every period (until it stops cutting it), except in the 2 periods around x = 0, because there the root x = 0 is common to the periods $(-2 \pi, 0),\ (0, 2 \pi)$. For 1999 roots, we need 2000 periods, 1000 periods for x > 0 and 1000 periods for x < 0. If the line $y = \frac x p$ passes through the point $\left(1998 \pi + \frac \pi 2,\ 1\right)$, it will surely have 2 roots in the 1000th period $(1998 \pi, 2000 \pi)$, because $\frac{\text d}{\text dx} \left[\sin\left(2000 \pi + \frac \pi 2\right)\right] = 0$ and $\frac{\text d}{\text dx}\left(\frac x p\right) = \frac 1 p > 0$. It will obviously have no roots in the next 1001st period $(2000 \pi, 2002 \pi)$. Hence, we put $\frac{1998 \pi + \frac \pi 2}{p} = 1$, $p = 1998 \pi + \frac \pi 2 \doteq 6278.4729$ and take p = 6279. Since the line $y = \frac{x}{6279}$ passes slightly below the point $\left(1998 \pi + \frac \pi 2,\ 1\right)$, there might be a smaller integer p, such that the line $y = \frac x p$ still intersects the sine wave in the 2000th period at 2 points. Close to the point $\left(1998 \pi + \frac \pi 2, 1\right)$, we can approximate the sine function by an upside down parabola with the vertex at this point and quadratic coefficient $- \frac 1 2$:

$y = \sin x \doteq -\frac 1 2 \left(x - 1998 \pi - \frac \pi 2 \right)^2 + 1$

Now we can test if the following quadratic equation has any roots

$-\frac 1 2 \left(x - 1998 \pi - \frac \pi 2 \right)^2 + 1 = \frac x p$

$x^2 + \left(\frac 2 p - 3997 \pi \right) x + \frac{3997^2 \pi^2}{4} - 2 = 0$

The discriminant of this quadratic equation is

$D = b^2 - 4ac = \left(\frac 2 p - 3997 \pi \right)^2 -  3997^2 \pi^2 + 8 = \frac{4}{p^2} - \frac{4 \cdot 3997 \pi}{p} + 8$

When the discriminant is zero, the line $y = \frac x p$ just touches the parabola approximating the sine wave and since this parabola is very slightly below the sine wave, we will still have 2 roots. Solving the quadratic equation D = 0 for p

$D = \frac{4}{p^2} - \frac{4 \cdot 3997 \pi}{p} + 8 = 0$

$2p^2 - 3997 \pi p + 1 = 0,\ \ \ p = \frac{3997 \pi + \sqrt{3997^2 \pi^2 - 8}}{4} = 6278.4728$

Thus there is no integer smaller than 6279, such that the line $y = \frac x p$ still intersects the sine wave in the 2000th period at 2 points. Since the line  $y = \frac{x}{6279}$ passes slightly below the point $\left(1998 \pi + \frac \pi 2,\ 1\right)$, we have to verify that it does not intersect the sine wave in the next 2000st period very close to the point $\left(2000 \pi + \frac \pi 2,\ 1\right)$. In a similar way as above, we approximate the sine wave close to this point by

$y = \sin x \doteq -\frac 1 2 \left(x - 2000 \pi - \frac \pi 2 \right)^2 + 1$

and test if the following quadratic equation for p = 6279 has any roots

$-\frac 1 2 \left(x - 2000 \pi - \frac \pi 2 \right)^2 + 1 = \frac x p$

$x^2 + \left(\frac 2 p - 4001 \pi \right) x + \frac{4001^2 \pi^2}{4} - 2 = 0$

The discriminant of this quadratic equation is

$D = b^2 - 4ac = \left(\frac 2 p - 4001 \pi \right)^2 -  4001^2 \pi^2 + 8 = \frac{4}{p^2} - \frac{4 \cdot 4001 \pi}{p} + 8$

Substituting p = 6279,

$D \doteq 0.0000001015 - 8.0073337837 + 8 = -0.0073336823 < 0$

Thus the line $y = \frac{x}{6279}$ does not intersect the sine wave in the 2001st period and the integer p = 6279 therefore solves the problem (7).

[quote="deej21"]Singular, putting \ before each trig name makes them look better.
[/quote]

I know, but there was alot of sin and cos in there, so i was lazy.

[color=blue][b](8)[/b][/color]

$\cos \frac \pi 7 + \cos \frac{3\pi}{7} - \cos \frac{2\pi}{7} = \cos \frac \pi 7 + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} = \cos \frac{13\pi}{7} + \cos \frac{11\pi}{7} + \cos \frac{9\pi}{7}$

The angles $\frac{\pi}{7},\  \frac{3\pi}{7},\  \frac{5\pi}{7},\  \frac{7\pi}{7} = \pi,\  \frac{9\pi}{7},\  \frac{11\pi}{7},\  \frac{13\pi}{7}$ are the oriented angles of the vertices of a regular heptagon inscribed in a unit circle, with one vertex at the point (-1, 0). The vertices of this regular heptagon can be represented by the complex numbers

$z_i = \cos \frac{(2k - 1)\pi}{7} + i \sin \frac{(2k - 1)\pi}{7},\ \ k = 1, 2, 3, 4, 5, 6, 7$

These 7 complex numbers are the roots of the 7th degree polynomial

$z^7 + 1 = (z + 1)(z^6 - z^5 + z^4 - z^3 + z^2 - z + 1) = 0$

The roots of the 2nd factor are $z_1, z_2, z_3, z_5, z_6, z_7$, because the root $z_4 = \cos \pi + i \sin \pi = -1$ has been factored out. The degree of a polynomial with reciprocal coefficients is normally reduced by the substitution

$x = z + \frac 1 z = \frac{z^2 + 1}{z} = \frac{(z^2 + 1) \bar z}{|z|^2} = \frac{|z|^2 z + \bar z}{|z|^2}$

Since all our roots are on the unit circle, $|z_k|^2 = 1$ and

$x_k = z_k + \bar z_k = 2\text{Re}(z_k) = 2 \cos \frac{(2k - 1)\pi}{7},\ \ k = 1, 2, 3$

The cubic equation for x is

$x^3 - x^2 - 2x + 1 = 0$

which is easily verified by substituting $x = z + \frac 1 z$ and multiplying the equation by $z^3$:

$z^3 + 3z + \frac 3 z + \frac{1}{z^3} - z^2 - 2 - \frac{1}{z^2} - 2z - \frac 2 z + 1 = 0$

$z^3 - z^2 + z - 1 + \frac 1 z - \frac{1}{z^2} + \frac{1}{z^3} = 0$

As a result, $x_1 = 2 \cos \frac{\pi}{7},\ x_2 = 2\cos \frac{3\pi}{7},\ x_3 = 2\cos \frac{5\pi}{7}$ are the roots of the above cubic equation for x. The sum of the roots is equal to the coefficient at the power 2 taken with the opposite sign, i.e., to 1:

$x_1 + x_2 + x_3 = 2 \cos \frac{\pi}{7} + 2\cos \frac{3\pi}{7} + 2\cos \frac{5\pi}{7} = 1$

$\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} = \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} - \cos \frac{2\pi}{7} = \frac 1 2$

[color=red]Who in the blazes put this problem into the Pre-Olympiad section ?[/color]  :?

Yetti

Looks like yetti is on a solving rampage, so i'll see if i can do the last one...
I'm guessing it's in degrees, btw.
$\sin{1^{\circ}}\sin{3^{\circ}}\sin{5^{\circ}}...\sin{89^{\circ}}=\frac{1}{2^n}$
$\sin{1^{\circ}}\cos{1^{\circ}}\sin{3^{\circ}}\cos{3^{\circ}}...\sin{45^{\circ}}=\frac{1}{2^n}$
$\frac{\sin{2^{\circ}}\sin{6^{\circ}}\sin{10^{\circ}}...\sin{86^{\circ}}}{2^{\frac{45}{2}}}=\frac{1}{2^n}$
$\sin{2^{\circ}}\cos{4^{\circ}}\sin{6^{\circ}}\cos{8^{\circ}}...\sin{42^{\circ}}\cos{44^{\circ}}=2^{\frac{45}{2}-n}$

Have to think more...

(9)Where you at on this one, yetti? Too easy for ya?
Well anyways, by Viete's relations on $1993x^3 - ax^2 + bx + c = 0$ we get $p + q + r = -\frac{-a}{1993}$ and $pq + qr + pr = +\frac{b}{1993}$ and $pqr = -\frac{c}{1993}$. Since $\arctan p + \arctan q + \arctan r = \frac{\pi}{4}$, $\tan(\arctan p + \arctan q + \arctan r) = \tan(\frac{\pi}{4}) = 1$. For now, say: $\arctan p = A, \arctan q = B$ and $\arctan r = C$. So, by a well known an easily provable identity:

$\tan(A+B) = \displaystyle\frac{\tan A + \tan B}{1-\tan A \tan B}$

$\tan((A+B)+C) =\displaystyle\frac{\tan(A+B) + \tan C}{1-\tan(A+B) \tan C} = \displaystyle\frac{\frac{\tan A + \tan B}{1-\tan A \tan B} + \tan C}{1-\frac{\tan A + \tan B}{1-\tan A \tan B}\tan C}$

$= \displaystyle\frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan A \tan C - \tan B \tan C}$

Since $\arctan p = A, \arctan q = B$ and $\arctan r = C$, we know $\tan A = p, \tan B = q$ and $\tan C = r$

$\Rightarrow \tan(A+B+C) = \displaystyle\frac{p + q + r - pqr}{1 -pq - pr - qr} = \tan(\frac{\pi}{4}) = 1$

$\Rightarrow (p + q + r) + (- pqr) = 1 - (pq + pr + qr)$

$\Rightarrow -\frac{-a}{1993} + -(-\frac{c}{1993}) = 1 - (+\frac{b}{1993})$

$\Rightarrow a + b + c = 1993$.

[quote="Pork_Chop8"](9)Where you at on this one, yetti? Too easy for ya?[/quote]

I was in bed. I did the problems in the posted order, it was late by the time I typed in the solution of (8), and I had to take some sleep. The difficulty is increasing from (1) to (10), so (9) is not an easy one, I knew that after solving (8), but I did not give (9) a try.

As for the problem [color=blue][b](10)[/b][/color], K81o7 got the correct answer (it is easy to calculate in a spreadsheet, once you get the wind) and here are some more details. Assuming the angles are in degrees (otherwise the problem would not make sense), the solution of (10) is similar to the solution of (8), only instead of a regular heptagon inscribed in a unit circle, we have to deal with a regular 180-gon iscribed in a unit circle, symmetrical with respect to both coordinate axes, but with no vertex on the coordinate axes, because

$P = \prod_{k=1}^{45} \sin(2k-1) = \sqrt{\prod_{k=1}^{90} \sin(2k-1)}$

Then we have to figure out the absolute term of a 90-degree polynomial for 90 roots $\sin(2k-1),\ k = 1, 2, ..., 90$, which is equal to their product. First of all, the problem can be reduced to regular 90-gon with 2 vertices od one coordinate axis, which will decrease the polynomial order from 90 to $\frac{90}{2} - 1 = 44$, just the way K81o7 did it.

$P = \prod_{k=1}^{45} \sin(2k-1) = \frac{\sin(45^{\circ})}{2^{22}} \cdot \prod_{k=1}^{22} 2 \sin(2k-1) \cos(2k-1) =$

$= \frac{\sqrt 2}{2^{23}} \cdot \prod_{k=1}^{22} \sin(4k-2) = \frac{\sqrt 2}{2^{23}} \cdot \prod_{k=1}^{22} \cos(4k) = \frac{\sqrt 2}{2^{23}} \cdot \sqrt{\prod_{k=1}^{44} \cos(4k)} =$ [color=white].[/color]$\frac{\sqrt 2}{2^{45}} \cdot \sqrt{\prod_{k=1}^{44} 2 \cos(4k)}$

The last transition from sines to cosines corresponds to the rotation of the coordinate axes by $\pm 90^o$ to put 2 vertices of the regular 90-gon from the y-axis to the x-axis. It remains to be shown that

$(?)\ \ Q = \prod_{k=1}^{44} 2 \cos(4k) = 1$

The vertices of the regular 90-gon with 2 vertices at the points (1, 0), (-1, 0), respectively, can be represented by complex numbers

$z_k = \cos(4k) + i \sin(4k),\ \ k = 0, 1, 2, ..., 89$

which are the roots of the 90-degree polynomial

$z^{90} - 1 = (z^{45} - 1)(z^{45} + 1) = 0$

$(z - 1) (z + 1) (z^{44} + z^{43} + z^{42} + ... + z + 1)(z^{44} - z^{43} + z^{42} - ... - z + 1) = 0$

$\frac{(z - 1)(z + 1)}{z^{44}}(z^{22} + z^{21} + ... + \frac{1}{z^{21}} + \frac{1}{z^{22}})(z^{22} - z^{21} + ... - \frac{1}{z^{21}} + \frac{1}{z^{22}}) = 0$

In the above steps, we factored out the roots $z_0 = \cos(0^{\circ}) + i \sin(0^{\circ}) = 1$ and $z_{45} =  \cos(180^{\circ}) + i \sin(180^{\circ}) = -1$ and prepared the remaining 2 factors for the substitution

$x = z + \frac 1 z = z + \bar z = 2 \text{Re}(z) = 2 \cos \phi$

because z is on a unit circle and $|z|^2 = 1$ ( compare to the solution of (8)). The resulting 44-degree polynomial for x given as a product of two 22-degree factors has the roots

$x_k = 2 \cos{4k},\ \ k = 1, 2, ..., 44$

and we have to find the absolute term of this polynomial equal to the product of these 44 real roots multiplied by $(-1)^{44} = 1$, so we want to show that this absolute term is equal to 1. We can show by induction, that for an even 2n, the absolute term of both factors is equal to $(-1)^n$ and the absolute term of their product to $(-1)^n (-1)^n = 1$. For 2n = 2,

$z^2 + z + 1 + \frac 1 z + \frac{1}{z^2} = (z^2 + 2 + \frac{1}{z^2}) + (z + \frac 1 z) - 1 = x^2 + x - 1$

$z^2 - z + 1 + \frac 1 z  + \frac{1}{z^2} = (z^2 + 2 + \frac{1}{z^2}) - (z + \frac 1 z) - 1 = x^2 - x - 1$

and (-1)(-1) = 1 (this is the case of a regular pentagon, which is why the quadratic equations for the golden ratio poped up). Assume that the statement is correct for some 2n. Then

$\sum_{k=0}^{4(n + 1)}z^{2(n+1)-k} = z^{2(n+1)} + \frac{1}{z^{2(n+1)}} + \sum_{k=0}^{4n}z^{2n-k}$

By the induction assumption, the absolute term of the sum on the right side is $(-1)^n$, when expressed in terms of x, so we want to show that the absolute term of $z^{2(n+1)} + \frac{1}{z^{2(n+1)}}$ is $2 (-1)^{n+1}$, when expressed in terms of x, because then we would have $2 (-1)^{n+1} + (-1)^n = (-2 + 1)(-1)^n = (-1)^{n+1}$ and we would be done. Again, this claim can be shown by induction for even 2n. For 2n = 2,

$z^2 + \frac{1}{z^2} = (z^2 + 2 + \frac{1}{z^2}) - 2 = x^2 - 2$

and the absolute term is $2 (-1)$ as required. Assume that the claim is correct for some 2n. Then, using the binomial formula (where $C_q^p = \frac{p!}{q!(p - q)!}$ are the binomial coefficients),

$z^{2(n+1)} + \frac{1}{z^{2(n+1)}} =$

$= x^{2(n+1)} - C_1^{2(n+1)} (z^{2n} + \frac{1}{z^{2n}}) - C_2^{2(n+1)} (z^{2(n-1)} + \frac{1}{z^{2(n-1)}}) - ...\ -$

${- C_n^{2(n+1)} (z^2 + \frac{1}{z^2}}) - C_{n+1}^{2(n+1)}$

By the induction assumption, the absolute term of this polynomial, when expressed in terms of x, is equal to

$-C_1^{2(n+1)} \cdot 2(-1)^n - C_2^{2(n+1)} \cdot 2(-1)^{n-1} - ... - C_n^{2(n+1)} \cdot 2(-1) - C_{n+1}^{2(n+1)} =$

$= (-1)^{n+1} \sum_{k=1}^{2n+1}(-1)^{k-1} C_k^{2(n+1)}$

where we used the fact that $C_q^p = C_{p-q}^p$. We want to show that this is equal to $2(-1)^{n+1}$, because then we would be done. Hence, we need

$(?)\ \ \sum_{k=1}^{2n+1}(-1)^{k-1} C_k^{2(n+1)} = 2$ [color=white].[/color] or [color=white].[/color] $\sum_{m=0}^{2(n+1)}(-1)^m C_m^{2(n+1)} = 0$

But since this is the binomial formula for $0 = (1 - 1)^{2(n+1)}$, it is trivially true. Going back, we proved that

$Q = \prod_{k=1}^{44} 2 \cos(4k) = 1$

$P = \prod_{k=1}^{45} \sin(2k-1) = \frac{\sqrt 2}{2^{45}} \sqrt Q = \frac{\sqrt 2}{2^{45}} = \frac{1}{2^{\frac{89}{2}}}$ [color=white].[/color] (the exponent is $\frac{89}{2}$)

[color=red]Ditto for posting this problem in the Pre-Olympiad section.[/color]

Yetti

I have found a much nicer solution to number 8 and I know you'll love it yetti. I sort of double posted, but it's pertinent here.
(8) $\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} =  \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}$
Let's try to solve:
$\frac{1}{\sin x} = \frac{1}{\sin 2x} + \frac{1}{\sin 3x}$.

$\Rightarrow \frac{1}{\sin x} - \frac{1}{\sin 3x} = \frac{1}{\sin 2x}$

$\frac{1}{\sin x} - \frac{1}{\sin 3x} = \frac{\sin 3x - \sin x}{\sin x \sin 3x}$

$^{nota} \sin 3x - \sin x = \sin (2x+x) - \sin (2x-x) = (\sin 2x \cos x + \sin x \cos 2x) - (\sin 2x \cos x - \sin x \cos 2x) = 2\sin x \cos 2x ^{nota}$

$\Rightarrow \frac{\sin 3x - \sin x}{\sin x \sin 3x} = \frac{2\sin x \cos 2x}{\sin x \sin 3x}$

$\Rightarrow \frac{2\sin x \cos 2x}{\sin x \sin 3x} = \frac{1}{\sin 2x} \Rightarrow 2\sin 2x \cos 2x = \sin 3x$

$\Rightarrow \sin 4x = \sin 3x$.

Now, to solve that, we know $\sin a = \sin b$ if $a = b$ which means $4x = 3x$ or $x = 0$. Since $\sin 0 = 0$, by the equality given, a division by $0$ would occur, which makes $0$ an extraneous root for $x$.
Before we move on, we need to prove $\sin ((2k-1)\pi - a) = \sin a$ for integer $k$. This is sort of trivial but hey I'm doing it. We get $\sin ((2k-1)\pi - a) = \sin ((2\pi)k-\pi - a) = \sin (-\pi - a)$

$= \sin(-\pi) \cos(-a) + \sin(-a)\cos(-\pi) =0 \cdot \cos(a) + (-\sin(a))\cdot (-1) = \sin a$

So, from that, we can say $\sin a = \sin b$ if $a + b = (2k-1) \pi$ for any integer $k$. Since $a = 3x$ and $b = 4x$, we get $a + b = 7x$ and it quickly follows that $x = \frac{(2k-1)\pi}{7}$.

(2)First, we know from Demoive's that $e^{i \theta} = \cos \theta + i \sin \theta$.

$\Rightarrow (e^{i \theta})^2 = \cos^2 \theta + 2\cos \theta (i \sin \theta) + (i \sin \theta)^2 = (\cos^2 \theta - \sin^2 \theta) + i(2\cos \theta \sin \theta)$

$(e^{i \theta})^2 = e^{i 2\theta} = cos 2\theta + i \sin 2\theta$

$\Rightarrow \sin 2\theta = 2\cos \theta \sin \theta$

$(e^{i \theta})^3 = \cos^3 \theta + 3\cos^2 \theta (i \sin \theta) + 3\cos \theta (i \sin \theta)^2 + (i \sin \theta)^3$

$= (\cos^3 \theta - 3\cos \theta \sin^2 \theta) + i(3\cos^2 \theta \sin \theta - \sin^3 \theta)$

$(e^{i \theta})^3 = e^{i 3\theta} = cos 3\theta + i \sin 3\theta$

$\Rightarrow \sin 3\theta = 3\cos^2 \theta \sin \theta - \sin^3 \theta = \sin \theta(3\cos^2 \theta - \sin^2 \theta) = \sin \theta(4\cos^2 \theta - 1)$

Plugging these into our original equation, we get:

$\frac{1}{\sin x} = \frac{1}{2 \sin x \cos x} + \frac{1}{\sin x(4 \cos^2 x - 1)} \Rightarrow 1 = \frac{1}{2 \cos x} + \frac{1}{4 \cos^2 x - 1}$

$\Rightarrow (4 \cos^2 x - 1)(2 \cos x) = (4 \cos^2 x - 1) + (2 \cos x)$

$\Rightarrow  8 \cos^3 x - 4 \cos^2 x - 4 \cos x + 1 = 0$.

Even though we have $x = \frac{(2k-1)\pi}{7}$, it still doesn't hold for some values of $k$ as it previously appeared. When $k=4$, $x = \pi$ and $\sin x = 0$ which makes our equation impossible. So, after a little inspection, we see that $\frac{\pi}{7},\frac{3\pi}{7},\frac{5\pi}{7},\frac{9\pi}{7},\frac{11\pi}{7}$ and $\frac{13\pi}{7}$ are the only values of $x$ pertinent since everything else cycles mod $2\pi$. Additionally, we have

$\cos \frac{13\pi}{7} = \cos (2\pi - \frac{\pi}{7}) = \cos \frac{\pi}{7}$

$\cos \frac{11\pi}{7} = \cos (2\pi - \frac{3\pi}{7}) = \cos \frac{3\pi}{7}$

$\cos \frac{9\pi}{7} = \cos (2\pi - \frac{5\pi}{7}) = \cos \frac{5\pi}{7}$

So, after this we see that $\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}$ and $\cos \frac{5\pi}{7}$ are the three solutions to the polynomial. So, by Viete's theorem, we know the sum of the roots of the polynomial is $-\frac{-4}{8} = \frac{1}{2}$. So, since we know the roots, we have, simply:

$\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} = \frac{1}{2}$
QED

Very nice (everything checks). I knew that $\cos \frac \pi 7, \cos \frac{3\pi}{7}, \cos \frac{5\pi}{7}$ are roots of the cubic equation $8x^3 - 4x^2 - 4x + 1 = 0$, but I could not find out why by trigonometry without the regular heptagon. On the other hand, I realized that the heptagon solution can be simplified. Once the root $z_4 = -1$ is factored out from the 7th degree polynomial equation $z^7 + 1 = 0$, we get the 6th degree polynomial equation

$z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = 0$

with the roots $z_k = \cos \frac{(2k-1)\pi}{7} + i \sin \frac{(2k-1)\pi}{7},\ k = 1, 2, 3, 5, 6, 7$. Because of symmetry of the regular heptagon with respect to the real axis, they are 3 complex conjugate pairs, we can add them up right here and the imaginary parts cancel out. So

$z_7 = \bar z_1,\ z_6 = \bar z_2,\ z_5 = \bar z_3$

$z_1 + z_2 + z_3 + z_5 + z_6 + z_7 =  (z_1 + \bar z_1) + (z_2 + \bar z_2) + (z_3 + \bar z_3) =$

$= 2\ \text{Re}(z_1) + 2\ \text{Re}(z_2) + 2\ \text{Re}(z_3)) = 2 (\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}) = -(-1) = 1$

$\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} = \frac 1 2$

In fact, even factoring the 7th degre polynomial equation $x^7 + 1 = 0$ is not necessary, just from knowing the 7 complex roots $z_k,\ k = 1, 2, 3, 4, 5, 6, 7$, we can write right away

$z_1 + z_2 + z_2 + z_4 + z_5 + z_6 + z_7 = 0$

$(z_1 + \bar z_1) + (z_2 +  \bar z_2) + (z_3 +  \bar z_3) + z_4 = 0$

$2\ \text{Re}(z_1) + 2\ \text{Re}(z_2) + 2\ \text{Re}(z_3))  + z_4 = 0$

$2 (\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}) - 1 = 0$

etc.


Find the best value of $ k > 0$ to this ineq is true for all $ a,b,c\geq\ 0$
$ 1\plus{}(\frac {2b \plus{} c}{2a \plus{} c})^3 \plus{} ( \frac {2c \plus{} a}{2b \plus{} a})^3 \plus{} ( \frac {2a \plus{} b}{2c \plus{} b})^3\geq\ k((\frac {3a \plus{} 2b \plus{} c}{a \plus{} 2b \plus{} 3c})^3 \plus{} (\frac {2a \plus{} b \plus{} 3c}{2a \plus{} 3b \plus{} c})^3 \plus{} (\frac {a \plus{} 3b \plus{} 2c}{3a \plus{} b \plus{} 2c})^3)$
An other gift to can_hang 2007

Any proof or opinion? I dont think it is easy

[quote="Allnames"][quote="Allnames"]Find the best value of $ k > 0$ to this ineq is true for all $ a,b,c\geq\ 0$
$ 1 \plus{} (\frac {2b \plus{} c}{2a \plus{} c})^3 \plus{} ( \frac {2c \plus{} a}{2b \plus{} a})^3 \plus{} ( \frac {2a \plus{} b}{2c \plus{} b})^3\geq\ k((\frac {3a \plus{} 2b \plus{} c}{a \plus{} 2b \plus{} 3c})^3 \plus{} (\frac {2a \plus{} b \plus{} 3c}{2a \plus{} 3b \plus{} c})^3 \plus{} (\frac {a \plus{} 3b \plus{} 2c}{3a \plus{} b \plus{} 2c})^3)$
An other gift to can_hang 2007[/quote]
It quite hard,really
Where are you,arquady,vasc,and can_hang2007?[/quote]
I think it don't too hard . Can U tell me k = ? And When does equal occur? :lol:

[quote="Math pro"]my solution,all names! :blush:
we'll prove that k=4/3
 let a+b+c=1
-->LHS=1 +   [(1-a+b)/(1-b+a)]^3+...
    RHS=4/3.{[(2-a+b)/(2-b+a)]^3+...
set x=a-b,y=b-c,z=c-a -->x+y+z=0
 f(x)=[(1-x)/(1+x)]^3 - 4/3 [(2-x)/(2+x)]
then find f''(x) and use Jensen -->we have done![/quote]
I have said,i dont need some sentences like them,where is your full proof,
I can say,Your proof that you posted is wrong( and not cleary,Why can we use Jensen or Why is f(x)(yours) a convex,and how to used Jensen to prove my problem)
Detail it ,please


I have heard that Canadians are now officially allowed to participate in the USAMO..

Does that mean we are allowed to join USAMTS as well?

So, I can't say for sure, but I was under the impression that USAMTS was limitted to US students because it was US government funded, graded by the NSA and so on.  I also thought that there was an international competition with exactly the same questions.  This is all somewhat tenuous memory, though.  You should wait for someone who actually knows something to confirm or deny.

At this point, the USAMTS is still limited to US citizens and residents, as its primary sponsor is the National Security Agency (i.e. the US government).  We are attempting to raise funds to support an expansion to international students, and will certainly let you know via this website if we succeed.


Euler
algebra
polynomial
superior algebra
superior algebra unsolved

Let $ \varepsilon_n\equal{}\exp\left(\frac{2\pi i}{n}\right)$.
What (and why) is the dimension of $ \mathbb{Q}(\varepsilon_n)$ over $ \mathbb{Q}$ ?

The dimension is $ \varphi(n)$ where $ \varphi$ is the Euler-Totient function. It's not so easy to prove this though. Can you do the case when $ n$ is a prime number?

Thanks for your (very fast) answer :)

If $ n$ prime, different powers of $ \varepsilon_n$ will be linearly independent, right?
There are $ n \minus{} 1$ of them and all of them belong to the space.
So the answer for $ n$ prime is $ n\minus{}1\equal{}\varphi(n\minus{}1)$.

I assume you're working in the context of field extensions and you want to show that $ [\mathbb{Q}(\epsilon_n): \mathbb{Q}] \equal{} \varphi(n)$. If we can show that that the minimal polynomial of $ \epsilon_n$ has degree $ \varphi(n)$, we're done. The minimal polynomials are the [url=http://en.wikipedia.org/wiki/Root_of_unity#Cyclotomic_polynomials]cyclotomic polynomials[/url]. In particular, for $ p$ a prime number, the cyclotomic polynomial $ \Phi_p$ is given by $ \Phi_p(x) \equal{} x^{p \minus{} 1} \plus{} x^{p \minus{} 2} \plus{} \ldots \plus{} 1$. Can you show that it is irreducible? Can you conclude that $ [\mathbb{Q}(\epsilon_p): \mathbb{Q}] \equal{} p \minus{} 1$? Can you do the general case?

[quote="yannick"] Can you show that it is irreducible? [/quote]Eisenstein works

[quote="yannick"] Can you conclude that $ [\mathbb{Q}(\epsilon_p): \mathbb{Q}] \equal{} p \minus{} 1$? [/quote]If I know that the poly is irreducible, I can use known theorem about roots of irreducible polynomials.

[quote="yannick"] Can you do the general case?[/quote]I'm not sure... Could you write a sketch of proof?

this is not straight forward, but can be found in any introduction into cyclotomic fields.


In my independent study class, I am stuck between having to choose to take either differential equations or linear algebra. I have just finished multivariable calculus, and I do not know which is the next logical course to take. Which course is more suitable? Which is more important? Also, can anybody recommend any good textbooks for either course?

In my opinion, linear algebra is much more fundamental.  I think it should be taken before any calculus course.

In general, though, you should be aware that college mathematics doesn't proceed in a linear (pardon the pun; I should say well-ordered) fashion, with an obvious choice of next course to take at each step.

So do you suggest that I should take linear algebra next; i.e. it is more useful to take linear algebra first? It seems to me that linear algebra involves more memorization while differential equations seems to give more freedom. Is that true?

For the specific things you'll have to learn in linear algebra, that might be true, but talking about linear operators and vector spaces is so fundamental to the rest of mathematics that there's no excuse for not learning about them as soon as possible.

Here's a secret:  a lot of differential equations is really linear algebra.

OK, thanks. I'll take linear algebra next then.

I second the vote for linear algebra.  It's hard to overstate the importance of linear algebra.

Also, linear algebra is not memorization.  A typical linear algebra course is usually more conceptual and less memorization-oriented than a typical differential equations course.

Calculus and linear algebra are like two pillars of mathematics. Differential equations is balanced elegantly (but delicately) on top of both of these pillars. I generally advise people to expose themselves to linear algebra concepts before taking multivariable calculus, too, since the Jacobian is a mere consequence of linear algebra concepts.

If you take introductory linear algebra as a class, there's the danger of being "tricked" into believing, for example, that a list of numbers is an arrow, or oversimplifying the relationship between a matrix and a linear transformation, or such things, or being conditioned to think in terms of $ \mathbb{R}^n$, or believing there's some holy standard basis out there, or other things. The flavor of linear algebra is lost in such courses.

If you want to take it as an independent study, you have all the control! I would suggest taking a standard book in intro linear algebra, such as "Linear Algebra with Applications" by Otto Bretscher, and read it *concurrently* with "Linear Algebra Done Right" by Sheldon Axler. They have distinct sets of advantages, both of which manifest themselves immediately if you read them concurrently, I think (they also make up for each other's shortcomings). Don't use the intro book by David Lay, it's pretty useless, and will not mesh with the Axler.

I'm sure several other people will have suggestions on what will be the most fruitful use of your time, as well.


geometry
circumcircle
ratio
trapezoid
trigonometry
calculus
geometric transformation

Let $ABC$ be a triangle, and let the tangent to the circumcircle of the triangle $ABC$ at $A$ meet the line $BC$ at $D$. The perpendicular to $BC$ at $B$ meets the perpendicular bisector of $AB$ at $E$. The perpendicular to $BC$ at $C$ meets the perpendicular bisector of $AC$ at $F$. Prove that the points $D$, $E$ and $F$ are collinear.

[i]Valentin Vornicu[/i]

my solution can be found [url=http://www.mathlinks.ro/viewtopic.php?t=41]HERE[/url], I am expecting other solutions :D

[color=red][Moderator edit: Just for the sake of completeness, I post this solution here:

Because both triangles DBE and DCF are right-angled triangles, it is sufficient to prove that they are similar to obtain D,E,F collinear. All we must prove is that DB/DC=BE/FC (1)

Now let's compute each ratio: for the first one, we'll use the fact that AD is tangent to the circumcircle of ABC. Therefore <DAB=<ACB thus the triangles DAB and DCA are similar, which gives us:
DB/DC=AB/AC=AD/BC => DB/DC = c2/b2  (2)

Let's compute now BE and EF. Let M be the midpoint of [BC]. Then in the right-angled trapezoid [EBMO] because <EOM=<A+<B it follows that <BEO = <C. Using the latter fact we can compute easily the length of
BE=(a/2)cotB+RcosA
and by the same arguments FC=RcosA+(a/2)cotC . Now we have to prove using (2) that:    (RcosA+(a/2)cotB)b2 = (RcosA+(a/2)cosC)c2   which is obvious after some computations.
Thus the three points are collinear.

][/color]

The point of this problem is to compute DB/DC. But you can easily prove that DB/DC = (AB/AC)^2. Let D' be the intersection of EF and BC. By using Thales theorem you can proveit easily. Thanks. :D

I found it slightly strange that the [i]proposer's[/i] solution uses trigonometry when a relatively straightforward synthetic solution can be found :?

I think treegoner's hint is more than enough to solve the problem :) (I don't know what Thales' theorem is, and I didn't use the intersection of EF and BC) - my solution doesn't quite reach half a page.

The trigonometric solution is kind of short too: BE=AB/2sinB, CF=AC/2sinC, and we then use the fact that DB/DC=AB2/AC2 (this is well-known; the tangent in A to the circumcircle is the ex-symmedian of ABC corresponding to A). Then, as Treegoner said, Thales works fine.

[Moderator edit: Someone said there is another solution by inversion.]

[quote="A1lqdSchool"][size=150][b]"I don't know what Thales' theorem is"[/b][/size] =>
Are you sure?? It's very well-known Theorem.[/quote]

not only well-known, most people would use this intuitively, without knowing it has a name, so don't worry about it

it's just the rules of x/x = x/x in triangles of the same form :maybe:

[quote="Valiowk"]I found it slightly strange that the [i]proposer's[/i] solution uses trigonometry when a relatively straightforward synthetic solution can be found :? [/quote] not so strange. I didn't tought much about the problem when I proposed it, and one solution sufficed.

I had no idea it will make it to the actual contest :D:D.

I don't know what this following theroem is called in your country, but it is a basic one of geometry. I call it Thales 's theorem
    Let ABC be a triangle and D, E are in AB and AC. Then we have
        DE//BC iff AD/AB = AE/AC.
     What do you call it in your country ? :D

The interesting thing about the theorem is that Thales only proved it for rational ratios, an the prooff coul only be completed after calculus was invented, so the theorem was used intuitively (without a complete proof) for a long time.

What you call Thales theorem is known as Thales theorem in Russia, but in Germany it is called "Strahlensatz"; moreover, the German "Satz von Thales" is the fact that the diameter of a circle suspends a 90 angle.

However, it is really useless to discuss about such naming issues.

Of course, the problem is simple. I wonder why Valentin solves it by heavy trigonometry, whereas there are simple solutions either synthetic or with a quite little use of trigonometry (sine law).

  Friendly,
  Darij Grinberg

Let C1 and C2 be the circumferences of centre F and centre E that intersect at A. Clearly BC is a common tangent line to these two circles, we will show that D is their homothecy centre. Indeed, let A' be the second intersection point of AD with C1, we have that <BAD=<CAB=<CA'A. Hence CA' || AB, which implies the result.

Dear Mathlinkers,
let A*B*C* the tangentiel triangle of ABC and O the circumcenter of ABC
According to the Desargues's theorem, A*O is the perspectrix of the triangle BEC* and CFB*;
in consequence, these triangle are perspective
...
Sincerely
Jean-Louis

Thank you for nice solutions dear [b][size=100]arniszt[/size][/b] and [b][size=100]Jean-Louis[/size][/b].

Let us to present an alternative approach based on the [b][size=100]douple ratio[/size][/b] theory.

$ \bullet$ Let be the points $ B'\equiv AD\cap OF,\ C'\equiv AD\cap OE,\ E'\equiv BC\cap OE$ and $ F'\equiv BC\cap OF.$

It is clear that the line segments $ C'B,\ B'C,$ tangent to the circumcircle $ (O)$ of $ \bigtriangleup ABC,$ at points $ B,\ C,$ respectively.

In the configuration of the pencils $ B.COEC',\ C.BOFB',$ where $ O$ is the circumcenter of $ \bigtriangleup ABC,$ we see that   congruent angles are formed, by their homologous rays $ ($ $ \angle CBO \equal{} \angle BCO$ and $ \angle OBE \equal{} \angle OCF$ and $ \angle EBC' \equal{} \angle FCB'$ $ ).$

So, we conclude that these pencils have congruent [b][size=100]douple ratios[/size][/b] ( = [b][size=100]cross ratios[/size][/b] ).

That is we have $ (B.COEC') \equal{} (C.BOFB')$ $ ,(1)$

Because of now, the line segments $ OE,\ OF,$ intersect the pencils $ B.COEC',\ C.BOFB'$ respectively, we have that $ (B.COEC') \equal{} (E',O,E,C')$ $ ,(2)$ and $ (C.BOFB') \equal{} (F',O,F,B')$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ (E',O,E,C') \equal{} (F',O,F,B')$ $ ,(4)$

From $ (4),$ we conclude that the line segments $ E'F'\equiv BC,\ EF,\ B'C',$ are concurrent at one point, here the point $ D$ and the proof is completed.

[b][size=100]REMARK[/size][/b]. – We can say the same words, also when $ E,\ F,$ are points on the mid-perpendiculars of the side-segments $ AB,\ AC$ respectively, such that $ \angle EBC \equal{} \angle FCB$ and in this general case also hold, the proofs presented by [b][size=100]arniszt[/size][/b] and [b][size=100]Jean-Louis[/size][/b].

Kostas Vittas.

This problem also can be solve by Thales's Theorem. :)
In my point view,if using this theorem, your solution may be become clear and easy. :roll:

Friendly. :P

Dera now and never,
can you post your solution?
Thanks in advance
Sincerely
Jean-Louis

Here we have again that Greek invention called  PARABOLA

   with focus in A and directrix BC;    OE,OF  are tangents,

   BE, CF   are diameters.

   M.T.

I see no one posted my solution yet, so... :P

Let $ O$ be the circumcenter of $ ABC$, and $ R$ the circumradius.
If we prove that $ \triangle DAO \sim \triangle DBE$, then analogously $ \triangle DAO \sim \triangle DCF$, and we conclude that $ \triangle DBE \sim \triangle DCF$, and as Valentin noted, that implies that $ D,E,F$ are collinear.

Since $ \angle DAO \equal{}\frac{\pi}{2}\equal{}\angle DBE$, we just need to prove that:

$ \frac{AO}{BE}\equal{}\frac{AD}{DB}$

Using the sine law in $ AEB$ and $ ABC$, we obtain:

$ \frac{EB}{AB}\equal{}\frac{\sin(90\minus{}\beta)}{\sin(2\beta)} \Leftrightarrow \frac{EB}{2R \sin\gamma}\equal{}\frac{\cos\beta}{2\sin\beta \cos\beta} \Leftrightarrow \frac{AO}{BE}\equal{}\frac{\sin \beta}{\sin {\gamma}}$

On the other hand, using the sine law on $ ABD$, we get

$ \frac{AD}{DB}\equal{}\frac{\sin\beta}{\sin\gamma}$, since AD tangent to the circumcircle $ \Rightarrow$ $ \angle DAB \equal{} \pi\minus{}\gamma$

This way, $ \frac{AO}{BE}\equal{}\frac{\sin\beta}{\sin\gamma}\equal{}\frac{AD}{DB}$

$ QED$

[quote="Valentin Vornicu"] [color=darkred]Let $ ABC$ be a triangle, and let the tangent to the circumcircle of the triangle $ ABC$ at $ A$ meet the line $ BC$ at $ D$. The perpendicular to $ BC$ at $ B$ meets the perpendicular bisector of $ AB$ at $ E$. The perpendicular to $ BC$ at $ C$ meets the perpendicular bisector of $ AC$ at $ F$. Prove that the points $ D$, $ E$ and $ F$ are collinear ([i]Valentin Vornicu[/i]).[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b][/color] $ \left\|\ \begin{array}{c}
\frac {DB}{DC}\equal{}\frac {c^2}{b^2}\\\\
BE\equal{}\frac {c}{2\sin B}\\\\
CF\equal{}\frac {b}{2\sin C}\end{array}\ \right\|\ \Longrightarrow\ \frac {BE}{CF}\equal{}\frac {DB}{DC}\ \Longrightarrow\ D\in EF$ .

$ DB/DC\equal{}BE/FC$
How does this mean $ D,E,F$ are collinear points.
Please tell me even if it is so obvious. Pl dont ignore this doubt.

Thanks in advance.

That means DCF and DBE are similar triangles. Since they share a vertex and a side (and E,F lie on the same side of that side) the conclusion follows.

Thanks a lot. Sorry for this stupid doubt.

Also, this line is perpendicular to Euler line of $ABC$.

Let $XYZ$ be the orthic triangle. Note $\angle BAX = \angle EAB = \angle ZCB$ so if $AE \cap \odot ABC = Q$ then $Q \in CH$ where $H$ is the orthocentre of $ABC$ and define $S= AF \cap \odot ABC$, so $BH \cap \odot ABC = S$. Define $R, P$ to be the antipodes of $C, B$ respectively. Apply pascals theorem on hexagons $BPAAQC$ and $BRAASC$, giving $D, E, AP \cap CQ = K$ are collinear and $D, F, AR \cap BS = L$ are collinear. Its suffice to show $D, K, L$ are collinear.Consider the circles through $KZA$ and $LYA$, and note they are tangent to $AC, AB$ and have diametres $AK, AL$. Consider the circle with diametre $AD$, obviously passing through $X$. It suffice to show they are co-axial. Take an inversion with power $AB \cdot AZ = AC \cdot AX$ about $A$. $X$ goes to $H$, $BC$ go to $YZ$ and $D$ is on $\odot AYZ$ such that $AK \parallel YZ$. Let $TUV$ be the anti-complementary triangle of $ABC$. The intersection of $\odot KZA \cap \odot LYA$ goes to $T$, and obviously $X$ is the circumcentre of $TUV$ and $YZ$ is antiparallel to $UV$ so $TX \perp YZ$ but trivially $TD \perp YZ$ ($AX$ is diametre of $\odot AYZ$), so it follows $T, X, D$ are collinear.

Alternative:

Consider the circles with centres $E, F$ through $C, B$. Notice that if they intersect at $P$, by angle chasing it lies on $BHC$. Let $AH$ meet $BHC$ again at $Q$ and note $PBQC$ is a harmonic quadrilateral, so the tangents meet at $D$, so $DA = DP$ and the conclusion follows (Co-axial circles).

Note though that as my solution points out, this problem is equivalent to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=734440&sid=cd467778a69094898ff0094e7e39b2db#p734440]USA 2005 TST P6[/url]

It appears USA probably based their problem on this one...

Here is another (correct I hope) approach.

[i][b][u]Lemma:[/u][/b](attachment at the bottom)Let $DK$ be tangent to the circle,with $K\neq A$ and $Z\equiv KB\cap l_{AB}$ where $l_{AB}$ the perpendicular bisector of $AB$.

Then $DZ\parallel AB$.

The tangent-chord theorem gives $\hat{BCK}=\hat{BKD} \ \bf \color{red} (1)$.

Also $\hat{ACK}=\hat{ABZ}=\hat{BAZ}$.

The latter is equivalent to $\hat{ACB}+\hat{BCK}=\hat{BAD}+\hat{DAZ}$

The tangent-chord theorem gives $\hat{ACB}=\hat{BAD}$ thus $\hat{BCK}=\hat{DAZ}$.

Now $\bf \color{red} (1)$ gives $\hat{BKD}=\hat{DAZ}\Leftrightarrow \hat{ZKD}=\hat{DAZ}$.

Thus,the quadrilateral $DZAK$ is cyclic.This gives $\hat{DZK}=\hat{DAK}$.The tangent-chord theorem gives $\hat{DAK}=\hat{ACK}$.

We also have $\hat{ACK}=\hat{ABZ}$.Thus $\hat{ABZ}=\hat{DZK}\Leftrightarrow \boxed{DZ\parallel AB}$[/i]

Back to our problem(attachment at the middle),let $P,Q$ be the orthogonal projections of $B$ upon $AB,BC$ respectively.
Since $DZ\parallel AB$ the quadrilateral $BDZE$ is cyclic.

Thus,the orange angles are equal because of the cyclic quadrilateral we referred to and because $PK\parallel l_{AB}$.

But one of them,$\hat{PKB}$ equals $\hat{BQP}$.We finally get $\hat{PQD}=\hat{PQB}=\hat{EDB}\Leftrightarrow \boxed{PQ\parallel DE}$.

Let $R$ be the orthogonal projection of $K$ upon $AC$ and $H\equiv l_{AC}\cap CK$ (attachment on the top)

According to the lemma,$DH\parallel AC$.Since $l_{AC}\parallel RK$ we get that the quadrilaterals $DHKF,KQRC$ are cyclic.

Thus,the blue angles are equal (angle-chasing).

This however gives $\boxed{DF\parallel QR}$.Since $\overline{PQR}$ is the Simson line $l$ of $K$ with respect to $\triangle{ABC}$

we get that $DE\parallel l$ and $DF\parallel l$ thus $DE\parallel DF$ which proves the collineartiy.

Perform the Inversion with center $ D $ and factor $ {DA}^2 $ $ = $ $ DB $ $ \cdot $ $ DC $, then this inversion swap $ \odot (E, EB) $ and $ \odot (F, FC) $ $ \Longrightarrow $ $ D, $ $ E, $ $ F $ are collinear .

[b][i][size=150][color=#600]My Solution[/color][/size][/i][/b]
------------------------------
It is enough to show that \[\frac{FC}{EB}=\frac{CD}{BD}=\frac{CB+BD}{BD}=1+\frac{CB}{BD}\] Let's say that $\angle ACB=\angle BAD=\alpha$,$R=\text{circumradius of triangle ABC}$ and $\angle ABC=\beta$. Then we have $$\boxed{AB=2R\sin\alpha \ ,  \ AC=2R\sin\beta \ , \ \ BC=2R\sin(\alpha+\beta)}  \  \text{(By Sine Law)}$$ And again By Sine Law in triangle $ABD$  we have that \[\boxed{BD=\frac{2R\sin^2\alpha}{\sin{\beta-\alpha}}}\] And we know $BC$ and $BD$ it remain to find $EB$ and $CF$. By sine Law in triangles $ABE$ and $CFA$ we have that \[\boxed{EB=\frac{R\cdot \sin\alpha}{\sin \beta} \  , \  \  CF=\frac{R\cdot\sin\beta}{\sin\alpha}}\] And we must prove that \[\frac{FC}{EB}=1+\frac{CB}{BD}\to \frac{CB}{BD}=\frac{FC}{EB}-1\longrightarrow \frac{\sin(\alpha+\beta)\cdot\sin(\beta-\alpha)}{\sin^2\alpha}=\frac{\sin^2\beta-sin^2\alpha}{sin^2\alpha}\longrightarrow \sin(\alpha+\beta)\cdot\sin(\beta-\alpha)=\sin^2\beta-sin^2\alpha \] And The lat part is [url=http://artofproblemsolving.com/community/c4h1180692p5712586]here[/url]
----------------------------------

From the problem condition it follows that it's enough to show that $\frac{BE}{BD}=\frac{CF}{CD}.$ Let's mark $\angle OAB$ with $\alpha$ and $\angle OBC$ with $\beta.$
( $O$ is center of the circle)
Using $\triangle EBA \sim \triangle OCA,$ we have $\boxed{\frac{BE}{AB}=\frac{r}{CA}}$ (where $r$ is radius of the circle)    $(1)$
By $\text{sines'\ law:}$
In $\triangle ABD :$ $\boxed{AB=\text{sin}(90^\circ-2\alpha-\beta)\cdot \frac{BD}{\text{cos}\alpha}}$
In $\triangle OFC :$ $\boxed{r=\text{cos}{\alpha}\cdot \frac{CF}{\text{sin}(\alpha+\beta)}}$
In $\triangle CAD :$ $\boxed{CA=\text{sin}{(90^\circ-2\alpha-\beta)}\cdot \frac{CD}{\text{sin}(\alpha+\beta)}}$
Making the above substitutions in $\boxed{\frac{BE}{AB}=\frac{r}{CA}}$ $,$ we get $\frac{BE}{BD}=\frac{CF}{CD}$ $,$ as desired.
[color=#0f0]Note:[/color] Proving that $\frac{BE}{BD}=\frac{CF}{CD},$ we have $\angle CDF=\angle BDE,$ so the points $D,E,F$ are collinear.

Can anyone please post a solution using Inversion please

Imagine using circles of positive radius smh.

[hide=HaPpY NeW yEaR]$DB, DC$ are tangent to the circle $\omega$ with diameter $BC$ so since
 [list]
[*] $EB^2 = EA^2$ and $FC^2 = FA^2$
[*] $DB \cdot DC = DA^2$
[/list]
$\overline{DEF}$ is the radical axis of $\omega$ and the circle with center $A$ and radius $0$.[/hide]

WLOG $AB<AC$
Let $\omega_1$ and $\omega_2$ be the circles centered and $E,F$ and passing through $B,C$ respectively. Both $\omega_1$ and $\omega_2$ pass through $A$. Let $D'$ be their exsimilicenter. i.e. the intersection of $BC$ and $EF$. Also Let $G$ be the second intersection of $\omega_1$ and $AD'$. We have $BG\parallel CA$ because of the homothety at $D'$ swapping $\omega_1$ and $\omega_2$. $$\measuredangle D'AB=\measuredangle GAB=\measuredangle GBD=\measuredangle ACD=\measuredangle ACB$$ Hence $D'A$ is tangent to $(ABC)$ at $A$ and $D'=D$. Since $\overline{D'EF}$, we have $\overline{DEF}$

$AD\cap \odot (E,EB) = [A,L]$.
Obviously $E$ is on perp bisector of $BL$.
Also by Reim $BL\parallel AC$.
So $D$ - center of homothety $BLE\mapsto CAF$.

Complex bash cause why not
let the circle around $ABC$ be the unit circle / from AOꞱDA you get $\frac{d-a}{\overline{d}-\overline{a}}=-a^2\longrightarrow \overline{d}=\frac{2a-d}{a^2}$ from bc being a chord $\longrightarrow \overline{d}=\frac{c+b-d}{bc}\longrightarrow \frac{c+b-d}{bc}=\frac{2a-d}{a^2}\longrightarrow a^2c+a^2b-da^2=2abc-dbc\longrightarrow d=\frac{a^2(b+c)-2abc}{a^2-bc}$ from OEꞱAB you get $\frac{e-0}{\overline{e}-0}=ab$ from BEꞱCB you get $\frac{b-e}{\overline{b}-\overline{e}}=bc$ combining them you get $e=\frac{a(b-c)}{a-c}$ you get F by this way aswell because it's symmetric. Now we just have to prove collinearity we just need do calculate d-e and d-f for d-e you get $d-e=\frac{ac(a-b)(b+c-2a)}{(a^2-bc)(c-a)}$ for d-f you get the same but just switch b with c so plugging everything in you just need to prove that $\frac{\frac{ac(a-b)}{c-a}}{\frac{\overline{ac}(\overline{a-b})}{\overline{c-a}}}=\frac{\frac{ab(a-c)}{b-a}}{\frac{\overline{ab}(\overline{a-c})}{\overline{b-a}}}$ which isn't hard to prove :)


$ f,g : R \longrightarrow R$.prove that exist $ x,y \in [0,1]$ such that:

$ |xy\minus{}f(x)\minus{}g(y)| \geq \frac{1}{4}$


Find the values of $x$ such that the following inequality holds: $\sqrt[3]{x-a}+\sqrt[3]{x-b}+\sqrt[3]{x-c}>0$ whrere $a,b,c\in\mathbb{R}$. This may be Pre-Olympiad.

I've got a partial solution. Only for $x > a$ and $x > b$ and $x > c$ and a,b,c are unequal each other.

[hide=""Click here to show that""]
Let $\alpha = \sqrt[3]{x - a}$, $\beta = \sqrt[3]{x - b}$, $\gamma = \sqrt[3]{x - c}$

We have: $\alpha + \beta + \gamma > 0$ and $\alpha\beta + \alpha\gamma + \beta\gamma > 0$.

And so...

$(\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 + \alpha\beta + \alpha\gamma + \beta\gamma) > 0$
$\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma > 0$

Then...

$(x - a) + (x - b) + (x - c) > 3\sqrt[3]{(x - a)(x - b)(x - c)}$
$\therefore$
$\frac{(x - a) + (x - b) + (x - c)}{3} > \sqrt[3]{(x - a)(x - b)(x - c)}$

It holds for every positive $(x - a)$, $(x - b)$, $(x - c)$, unless $a = b= c$

Therefore the inequality $\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} > 0$ hols (not only, I guess) for $\forall x, x - a > 0$ and $x - b > 0$ and $x - c > 0$ and a,b,c unequal each other. In other words, x > m such that m is the larger value between a,b,c.
[/hide]

Apologize me if I've got mistakes...

I think I've got one estimate $ x \geq \frac{a+b+c}{3} $ by using the inequality..

$ {( \frac{ {\alpha}^3 + {\beta}^3 + {\gamma}^3}{3})}^{\frac{1}{3} } \geq \frac{{\alpha} + {\beta} + {\gamma}}{3} \\
\\ \alpha = \sqrt[3]{x-a} ..... $

Is this right???


Let $ P(x)\equal{}(x\minus{}a)^2\plus{}b^2$, where $ a,b$ are reals.

Find the minimal $ n\in N$, such that for any such $ P$, there exist $ n\plus{}1$ polynomials $ Q,R_1,R_2,...,R_n$ with real coefficients such that $ P(x)\equal{}Q(x)^2\plus{}x(1\minus{}x)(R_1(x)^2\plus{}R_2(x)^2\plus{}...\plus{}R_n(x)^2)$.

if $ P(x) \equal{} Q(x)^2 \plus{} x(1 \minus{} x)(R_1(x)^2 \plus{} R_2(x)^2 \plus{} ... \plus{} R_n(x)^2)$
we have $ \forall x\in\mathbb{R}: \ R_1(x)^2 \plus{} R_2(x)^2 \plus{} ... \plus{} R_n(x)^2\ge 0$
then $ \exists A,B\in\mathbb{R}[X]: \ R_1(x)^2 \plus{} R_2(x)^2 \plus{} ... \plus{} R_n(x)^2 \equal{} A(x)^2 \plus{} B(x)^2$
so $ P(x) \equal{} Q(x)^2 \plus{} x(1 \minus{} x)(A(x)^2 \plus{} B(x)^2)$
then so if $ n$ exists we will have $ n\le 2$

[quote="aviateurpilot"]if $ P(x) \equal{} Q(x)^2 \plus{} x(1 \minus{} x)(R_1(x)^2 \plus{} R_2(x)^2 \plus{} ... \plus{} R_n(x)^2)$
[/quote]

And how do you know the $ R_i$'s exist?

[quote="aviateurpilot"]
then $ \exists A,B\in\mathbb{R}[X]: \ R_1(x)^2 \plus{} R_2(x)^2 \plus{} ... \plus{} R_n(x)^2 \equal{} A(x)^2 \plus{} B(x)^2$
[/quote]

Why is that?

And is the answer $ n \equal{} 2$ or $ n \equal{} 1$?  :D

well, i think it's well-known lemma that. $ n\equal{}1$ is enough

$ (x\minus{}a)^2\plus{}b^2\equal{}(\alpha x\plus{}\beta)^2\plus{}x(1\minus{}x)\gamma^2$ for some $ \alpha ,\beta ,\gamma\in\mathbb{R}$.
You can find them yourself.

Hmm... I don't think it's that easy to find these

What's there to think?  :huh:

$ (x \minus{} a)^2 \plus{} b^2 \equal{} (\alpha x \plus{} \beta)^2 \plus{} x(1 \minus{} x)\gamma^2$ gives you $ 3$ equation's with $ 3$ unknown variables $ \alpha ,\beta ,\gamma$.
Solve it, which is easy, and you'll see.
(You just need $ \alpha,\beta,\gamma\in\mathbb{R}$, they don't need to be pretty)


MATHCOUNTS
geometry
perimeter
rectangle
angle bisector

This problem is open to all 8th graders and below:

Each of three congruent circles sit inside an equilateral triangle such that each of the circles is externally tangent to the other two and each circle is tangent to two sides of the triangle.  If each circle has radius 12, what is the perimeter of the triangle?

For an extra challenge: If you connect the three points where the circles are pairwise externally tangent you wind up with another equilateral triangle.  What is the perimeter and area of that smaller triangle?

If you draw from the center 2 radii of the circles to the points of tangent on the sides, and then draw a segment through the center of that circle to the closest vertex, you will have 2 right triangles. since by connecting a vertex of the triangle to the circle will create its angle bisector, the created triangles are 30 - 60 - 90s, so the longest leg is 12 root 3. then drawing a segment through the centers of 2 circles will create a rectangle with the dimensions of 12 x 24. thus the perimeter is:
3 ( 24 + 12 root 3) = 72 + 36 root 3

i thought this was an interesting question to bring up

[b]Extra[/b]:  The triangle has sides composed of 2*radius each.  This makes the perimeter 12*2*3=72.  The area of an equilateral triangle is  :rt3: s :^2: /4.  With s equal to 2r, the equation becomes  :rt3: r :^2: .  Substitute r for 12 and we get 12 :^2:  :rt3:  :arrow: 144 :rt3: .

Ergo, the perimeter is 72 and the area is 144 :rt3: .


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that:
\[ a^4 \plus{} b^4 \plus{} c^4 \minus{} 2(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2) \plus{} \frac{9a^2b^2c^2}{ab\plus{}ac\plus{}bc}\geq0\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ a^4 \plus{} b^4 \plus{} c^4 \minus{} 2(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2) \plus{} \frac {9a^2b^2c^2}{ab \plus{} ac \plus{} bc}\geq0\]
[/quote]
$ LHS \geq a^4 \plus{} b^4 \plus{} c^4 \minus{} 2(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2) \plus{} \frac {9a^2b^2c^2}{a^2\plus{}b^2\plus{}c^2} \geq 0$


Problem.Give $ A\in M_n(K)$. Find $ rank(adjA)$ depend $ r(A)$

Hello everybody, I'm new student. Can you give name of books about matrix?

We must consider three cases:

I. If $ r(A) \equal{} n$ then $ \det(A) \neq 0$, and $ A \cdot adj(A) \equal{} \det(A) I_n$ shows that $ r[adj(A)] \equal{} n$.

II. If $ r(A) < n \minus{} 1$, then all n-1 minors of A are zero and so all elements of $ adj(A)$ are zero. Therefore $ adj(A) \equal{} O_{n}$ and $ r[adj(A)] \equal{} 0$.

III. It remains to consider the case $ r(A) \equal{} n \minus{} 1$. We start with a lemma, the proof of which I will leave as exercise:

[b]Lemma.[/b] Let $ A \in M_{m \times n} (\mathbb{K})$ have rank r. If there is a matrix $ B \in M_{n \times p} (\mathbb{K})$ such that $ AB \equal{} O$ then $ r(B) \leq n \minus{} r$.

Now, $ r(A) \equal{} n \minus{} 1$ implies that $ A \cdot adj(A) \equal{} O$, and from the lemma we must have $ r[adj(A)] \leq n \minus{} (n \minus{} 1) \equal{} 1$. But $ r(A) \equal{} n \minus{} 1$ also implies that there is at least one n-1 minor of A different from zero: then, we have $ adj(A) \neq O$ which implies $ r[adj(A)] > 0$. Therefore, $ r[adj(A)] \equal{} 1$. $ \Diamond$


[quote="Bayasgalan"]Is it  true by Muirhead's inequality ?       $ [2,2,0]\succ[0,0,4]$   ????

that

2>0
2+2>0+0
4+0+0=0+0+4[/quote]
First of all we use $ ()$ instead of $ []$. And $ (2,2,0) \succ (0,0,4)$ is wrong! Both sides must be decreasing so $ (4,0,0) \succ (2,2,0)$ is the right result.


Prove that for any $n,k \in \mathbb N$ (such that $n-k\geq 1$) we have:

$L_n = \frac{F_{n+k}+F_{n-k}}{F_k},$ if $k$ is odd

$L_n = \frac{F_{n+k}-F_{n-k}}{F_k},$ if $k$ is even

where
$F_1=1, F_2=1$
$L_1=1, L_2=3$

$F_n = F_{n-1} + F_{n-2}, \ \ \forall n>2$
$L_n = L_{n-1} + L_{n-2}, \ \ \forall n>2$

This follows from known formulas
$F(i)=\frac{a^i-b^i} {\sqrt 5}$
$L(i)=a^i+b^i$
where $a=\frac{1+\sqrt 5} 2$
$b=\frac{1-\sqrt 5} 2$
So after rewriting the equality is
$a^{n+k}-b^{n+k}-a^{n-k}(-1)^k+b^{n-k}(-1)^k=a^{n+k}-b^{n+k}+a^nb^k+b^na^k$.
The result now folows from $ab=-1$, because $a^nb^k=a^{n-k}(-1)^k$ and
$b^na^k=b^{n-k}(-1)^k$.


i forgot the way you do this type of problem...
A master padlock has a dial with 28 numbers.  How many different 3 number combinations are there where no two consecutive numbers can be the same?

[hide="Solution"]To do this, multiply the number of ways to get the first number, number of ways to get the second, and number of ways to get the third.

There are $ 28$ possible numbers for the first number.

There are $ 27$ possible numbers for the second, since it cannot be the same as the first one.

There are $ 27$ possible numbers for the third, since it cannot be the same as the second one.

28 x 27 x 27 = 20,412 ways[/hide]


properties of big O and little o:

[url]http://planetmath.org/?op=getobj&from=objects&id=7048[/url]
[url]http://en.wikipedia.org/wiki/Big_O_notation#Properties[/url]

[color=red]What kind of textbook can find them ?[/color]

http://www.brpreiss.com/books/opus4/html/page60.html#theoremi

and

"Cormen"  algorithrm


Can zacacox or possibly zman answer this?

So with the AMY Packets that we got, is it a requirement that we write up the solutions and send them in or bring them to camp? Because I did alot of the problems, but didn't write up solutions, and I got the packet like 3 months ago so I don't have it anymore. If it is necessary that we write up the solutions, is there any way I can get another one?

i dont think you have to because they were due like amonth after they mailed them to you.  i dont think you have to write up the induction packet solution either

can somebody else back me up? :maybe:

I e-mailed Mr. Feng and he said there were no homework packets sent out...which is a bit confusing since I thought Titu specifically said that AMP students were supposed to do some of them in preparation for camp...

My best guess is that Titu printed some extra copies of an AMY (AwesomeMath Year Round) packet--since induction was the topic for AMY March (see [url=http://www.awesomemath.org/yearround.shtml]here[/url])--but that maybe he forgot to send it to everyone?  The whole "send your solutions in within 3 weeks" part is probably just an artifact of it being an AMY packet originally (since those people are supposed to send their stuff in to be graded).

Keone

Hey guys,

Sorry I've been out of touch and haven't answered this; I'm back in the States now.

The AMY packet sent out at the end of the school year to AMSP participants is [b]not[/b] mandatory. It's intended as a sample for people who haven't seen what an AMY segment looks like. Certainly, you ought to look through it (and work it, if you like), but we won't necessarily spend classtime on it. You may certainly discuss it and the problems it contains together.

Also, we haven't sent out homework problems, and we probably won't. It looks like we'll have plenty of time at camp to do the work that we asked students to do beforehand last year.

Sorry for the delay in responding to this!

-Zac

[quote="13375P34K43V312"]Oh, also a semi-related question, does our enrollment at AMP this summer give us AMY next year for free?[/quote]
According to the website, it gives you a discounted price.


If the roots of Lx^2-px+q=0 be in the ratio p:q, prove that √(p/q)+ √(q/p)+√(n/l)=0

some one please help

sorry i wrote the equation wrong its supposed to be lx^2+nx+n=0.
Sorry.

[hide="Solution"] We have the quadratic $x^{2}+\frac{n}{l}x+\frac{n}{l}= (x-r)(x-s) = x^{2}+(-r-s)x+rs$

Let $u = \sqrt{ \frac{r}{s}}+\sqrt{ \frac{s}{r}}$.  Then

$u^{2}= \frac{r}{s}+\frac{s}{r}+2 = \frac{r^{2}+s^{2}}{rs}+2 = \frac{(r+s)^{2}}{rs}= \frac{ \frac{n}{l}^{2}}{ \frac{n}{l}}= \frac{n}{l}$

So $u = \sqrt{ \frac{n}{l}}$.  I think this is the result you were after, since square roots are themselves non-negative. [/hide]

Because it's not true.  What you should actually have written is that $\sqrt{ \frac{p}{q}}+\sqrt{ \frac{q}{p}}= \sqrt{ \frac{n}{l}}$, or something along those lines, because [b]square roots are always non-negative.[/b]

Anyway, if you're trying to figure out why I haven't used $p, q$, that's because if the roots $r, s$ are in the ratio $p : q$ then $\frac{r}{s}= \frac{p}{q}$ and vice versa.

[quote="Fouad1013"]but that still doesn't prove that √(p/q)+ √(q/p)+ √(n/l)=0[/quote]

For that to be true, all three of p, q, and l must be 0. Plug that into the equation. :wink:


probability
combinatorics unsolved
combinatorics

Find the number of the permutation (a_1 ,.....a_n) of {1,2...,n} such that :
a_i  is not equal to i for all i =1,2,...,n

It is a standart exercise for an inclusion-exclusion formula.

Let $A_i$ be a sets of permutation preserving $i$ on its place. Then we are to calculate $n!-|A_1\cup A_2\cup ...\cup A_n|$.

We have \[|A_1\cup A_2\cup ...\cup A_n|=|A_1|+...+|A_n|-|A_1\cap A_2|-...-|A_{n-1}\cap A_n|+...+(-1)^{n+1}|A_1\cap A_2\cap ...\cap A_n|.\] It is clear that $|A_{i_1}\cap A_{i_2}\cap...\cap A_{i_k}|=(n-k)!$. Therefore \[|A_1\cup A_2\cup ...\cup A_n|=C_n^1\cdot (n-1)!-C_n^2\cdot (n-2)!+...+(-1)^{n+1}C_n^n(n-n)!=\] \[=n!\cdot\left(\frac{1}{1!}-\frac{1}{2!}+...+\frac{(-1)^{n+1}}{n!}\right).\]
And final answer is \[n!\cdot\left(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-...+\frac{(-1)^{n}}{n!}\right).\]

I hope it is correct.

yeah, I have found one, but we must see $D_n$ as the number of ways to put rooks on an $n*n$ chessboard without atacking between them and such that there is none in the diagonal, try to do the interpretation by yourself with this clue :)

[quote="Christian Hirsch"]Does anybody happen to know if there's a combinatoric interpretation for the recurrence relation
[tex]D_{n+1}=(n+1)D_n+(-1)^{n+1}[/tex]

Where [tex]D_n[/tex] is the number of permutations without fixed points.[/quote]

to the best of my knowledge, one can look up the paper by Remmel in the European Journal of Combinatorics,v4(1983),371-374.

[quote="Myth"]
And final answer is \[n!\cdot\left(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-...+\frac{(-1)^{n}}{n!}\right).\]

I hope it is correct.[/quote]

Sure it is  :) .
And for large $n$, it means that the probability to choose a permutation with no fixed point is about $\frac 1 e.$

Pierre.


Hey all.

I got a problem that I can't solve. I know I have to do a integral, but I just can't find the method for it. This is what the problem looks like:

Two veichles are moving in the same direction on a trac. Their speed is: A= 20 * e^0,03t and B= 12. t is the time counted from the beginning of the observation. How long will it take before A catches up with B, if A start 5000 meters behind B.

The two "main" tasks are:
1. Create an equation to decide the time
2. Decide a approximate value for the time in seconds.

Lets call the point $A$ start from $O$ and measure the displacement $A_{O}$ and $B_{O}$ from $O$.

At time $t$,
$A_{O}=\int 20 \cdot e^{0.03t}\textrm{d}t$ and
$B_{O}=12t+5000$

Integrating, $A_{O}=\frac{5}{3}e^{0.03t}+c$, and as $t=0, A_{O}=0$,

$A_{O}=-\frac{5}{3}$.

We want the time when $A_{O}=B_{O}$;

$\frac{5}{3}( e^{0.03t}-1)=12t+5000$
$\frac{5}{3}e^{0.03t}= 12t+\frac{15005}{3}$
$e^{0.03t}=\frac{36t}{5}+3001$

and then you can take logs and calculate... I hope I haven't made any silly errors here.

Thanks for your input.

I'm crying when I'm saying this but could you explain / simplify it abit?

I understand the steps done, but shouldn't both be done as integrals (and primitising them) and take them minus eachother.


hi! please ask here all the administrative questions about this edition of the contest.

I will not answer on PM or e-mail to these questions, but I will answer here, so that other users can see this, and stop asking the same questions over and over again :D

Okay, here are the first two questions:

1) Do the problems in different rounds of the Contest ascend in difficulty?
2) Problem 1: I am sorry, I may be confused with different notations but f : L --> P doesn't explicitely mean that every element of P is the image of an element of L ?
[I wrote: Is it necessary to participate in round 1 in order to take part in the next rounds? Now I see it... "The is no need to register for the contest - you can start in which ever round you want / feel apropriate."]

I am sorry for asking these: the answers are surely in some thread, but I am rather new here and can't finally figure out what is where in this forum.

  Darij

1) no, they do not ascent in difficulty, at least they haven't untill now. I have no special desire in doing that, and moreover the difficulty level is relative to each person :D

2) yes f:L->P means that all the elements of L are mapped into elements of P. However it does not imply that each element of P is a mapping of an element of L, in that particular case we would be facing an [i]onto[/i] or [i]surjective[/i] function.

3) glad you figured it out by yourself :D.

I'm really sorry if this is a stupid question, but where are the questions posted?  I've tried looking, but can't find them.

look at the "contest" link at the top of the page, between "Forum" and "downloads".

If you are coming to this site via aops, I don't think there is a link to the contest anyway. I can't find it anyway. Its on the mathlinks page though.

http://www.mathlinks.ro/contest.php

  dg

The contest is over since it's started on MAy 26th and there is only 2 weeks to finish it, right?

right the third edition is over :)
but the 4th edition will follow in fall this year.

good luck :)

MathLinks used be another message board untill they joined up with Art of Problem Solving.

[quote="takuanitromars36"]What exactly is MathLinks?[/quote] besides being another board, it's a concept that we are now trying to promote, that is Math Links, as in Math Connects (people) :) ;)

[quote="Peter VDD"]when will we know the results of the next ML contest :D[/quote]I am sorry for this HUGE delay in results. They are ALMOST finished, however I did not find time to finalize them until now.

They will surely be available before the next edition, which will start this fall.

[quote="ikap"]sorry v.v. but what does "this fall" mean?[/quote]fall means autumn in american english (probably because it's the season when the leaves fall :)).

September 22 $\le$ Fall $<$ December 21   :D

(For this year, at least!)

the contest usually begins around the first week(s) of october I think...

Can't wait for it to begin... Valentin, any exact dates known yet? :)


I am somewhat of a beginner at LaTeX(though I have been posting on Mathlinks.ro for 4 months!). I hope to learn more about LaTeX and recently dowloaded Miktex, Texniccentre etc... Though at the same time, at this moment, I do not plan to go 'big' regarding Math typesetting and will most probably restrict typesetting in LaTeX to just the odd research paper or problem/solution set etc. So I need a beginner's into to LaTeX while at the same time, I don't think I want to learn LaTeX to its root!!! Although I recently downloaded Knuth's "TeXbook" 's source, as you might know, reading a non-compilable LaTeX document doesn't exactly teach much.(If you know a way to compile ie, please tell!!!) I recently came across the recommended-for-beginners " LateX : A doucment prep. " br Leslie Lamport. Do you think it is good enough for me keeping in mind my context. I believe "LaTeX Companion" is probably too "big" and too "comprehensive" for my need.Which other book would you recommend???

Try this manual i attached. It's a free document - it says so inside  :D.
Hope it helps.

Yes, I know that one. It's an excellent guide to LaTeX for beginners!

Yes, I already have that and it is good. However, at least for me, nothing surpasses reading paper books!!!1

I also suggest [i]A Guide to LaTeX[/i] by Kopka and Daly.

[i]The LaTeX Companion[/i] is too advanced for beginners.

Isn't Leslies lamport's "$ \text{\LaTeX} $ goos enough(for me)!?!?


algebra
polynomial
inequalities unsolved
inequalities

Let polynomial $ax^{3}$   $+x^{2}$+bx+1 has 3 root $\in$ R+ Find min: P=$\frac{5a^{2}-3ab+2}{a^{2}(a-b)}$

bui_haiha write :
[quote="bui_haiha2000"]Let polynomial $ax^{3}+x^{2}+bx+1$has 3 root $\in R^{+}$.
Find min: $P={\frac{5a^{2}-3ab+2}{a^{2}(a-b)}}$[/quote]

The minimum of $P={\frac{5a^{2}-3ab+2}{a^{2}(a-b)}}$ is $12\sqrt{3}$, provided $ax^3+x^2+bx+1$ has 3 roots in $\mathbb{R}^{+}$. :roll:

Can you post your solution more detail,???

Why don't you try this way.
First let $x_1, x_2, x_3$ the three roots of the equation then $x_1,x_2,x_3>0$.
Using Vietè theorem we have:
$\{ \begin{array}{l} x_1+x_2+x_3=x_1x_2x_3=-\frac{1}{a} \\ x_1x_2+x_2x_3+x_3x_1=-\frac{b}{a} \end{array} $

Transform the expression into an expression with $x_1,x_2$ and $x_3$. Then we just have to find the minimum of a symetric expression. It 's not difficult. You can try

I try this way much time but I fail,some one help me????

I am not too good at such problems...
Using Viete, we get:
$\{ \begin{array}{l} x+y+z=xyz=-\frac{1}{a} \\ xy+yz+zx=\frac{b}{a} \end{array} $.
Then $a=\frac{-1}{xyz}$, $b=(xy+yz+xz)a=-(xy+yz+zx)\cdot\frac{1}{xyz}$.
Then the given expression becomes:
$\frac{ \frac{5}{(xyz)^{2}} - 3\frac{(xy+xz+zx)}{(xyz)^{2}} + 2}{ \frac{xy+yz+zx-1}{(xyz)^{3}} }$.
Let $p=xyz$.
The expression rewrites as:
$\frac{5p-3p(xy+yz+xz)+2p^{3}}{xy+yz+zx-1}$. Or,
$E=\frac{2p+2p^{3}-3p(xy+yz+zx-1)}{xy+yz+zx-1}$.
Since $x+y+z=xyz$, and $x,y,z>0$ then $xy+yz+zx\geq9$. So $xy+yz+zx-1$ is positive.
To minimize $E$, thus, we have to maximize $xy+yz+zx$.
Since, $xy+yz+zx\leq \frac{(x+y+z)^{2}}{3}=\frac{(xyz)^{2}}{3}=\frac{p^{2}}{3}$, then
$E\geq \frac{2p-3p(\frac{p^{2}}{3}-1)+2p^{3}}{\frac{p^{2}}{3}-1}=F=\frac{15p+3p^{3}}{p^{2}-3}$.
Suppose $F\geq k$.
Then $F=3p^{3}+15p+3k\geq kp^{2}$.
Since $F=p^{3}+p^{3}+p^{3}+15p+3k\geq5\sqrt[5]{p^{10}45k}=p^{2}\cdot 5\sqrt[5]{45k}$.
Set $k=5\sqrt[5]{45k}$, and follows $k=\sqrt[4]{5^{5}\cdot 45}$.
Check that $p=\sqrt{15}$, satisfies $p^{3}=15p=3k$.
Then minimum is $\sqrt[4]{5^{5}\cdot45}$.
Note: The condition doesn't say anything that the roots must be different. :)
Hope I didn't do any mistake.


algorithm
induction
function
Euler
topology
invariant
binomial coefficients

It is a well-known fact that binomial coefficients are integers, usually proved by floor function's characteristics.

However, Is there a way to prove only using some basic theorms such as Division Algorithm?

I'm very interested in second solution :lol:

Thanks

[quote="bjh7790"]It is a well-known fact that binomial coefficients are integers, usually proved by floor function's characteristics.[/quote]

THE standard proof that binomial coefficients are integers uses induction and the recurrence relation: $ \binom{n}{r}\equal{}\binom{n\minus{}1}{r\minus{}1}\plus{}\binom{n\minus{}1}{r}$. For $ n\geq 0$, you can also use the combinatorial interpretation of $ \binom{n}{r}$ as the number of $ r$-element subsets of $ \left\{1,2,...,n\right\}$ (unfortunately, this doesn't extend to negative $ n$).
Floor function is overkill (unfortunately, some harder facts about integrality need the floor function approach, but $ \binom{n}{r}\in\mathbb{Z}$ is not one of them!).

  darij

Oh... I thought the floor function one was the standard... :blush:

Thanks a lot!

Why isn't the standard solution the combinatorial interpretation?  When $ n \ge 0$, there are exactly $ k!$ ordered lists of $ k$-tuples from an $ n$-element set for every unordered list.  When $ n < 0$ you can use the identity

$ {\minus{}n \choose k} \equal{} (\minus{}1)^k {n\plus{}k\plus{}1 \choose k}$.

By far the most satisfying way to prove that a power series has integer coefficients is to give them a combinatorial interpretation.  Otherwise, it's not at all obvious that a power series like $ e^{e^x \minus{} 1}$, for example, has integer Taylor coefficients.

[quote="t0rajir0u"]Why isn't the standard solution the combinatorial interpretation?[/quote]

Alas, it is (wishful thinking on my side). I don't like it too much due to the additional work one has to do for $ n<0$.

  darij

There is actually a beautiful proof that encompasses both cases at once, but it requires the notion of [url=http://math.ucr.edu/home/baez/counting/]Euler measure[/url] to makes sense of negative cardinalities as a sort of non-homotopy-invariant Euler characteristic on certain classes of sets.


abstract algebra
category theory
group theory
vector
Ring Theory
calculus
integration

Hi,

tensor products are quite new to me, especially when I can't see them as spaces of multilinear forms.
In order to have more insight, some remarks on these questions could help me a lot

Let $R$  be a ring (with unity but need to be commutative), and $M$ a right R module, $N$ a left R module

$M\otimes N$ is then an abelian group known as the tensor product

now here are my questions :

1) if an element in the tensor product of form $x\otimes y$ is zero, what can I say about $x$ and $y$
one of them zero?  guess not  :oops:   But what changes when I assume $R$ has no zero divisors, or maybe even that $R$ is a field??

2) let $M$ and $N$ be right and left modules respectively again, and they are finite
what can you say about the number of elements in that tensor product?

All hints or extra info are welcome too! :roll:

Hi :-)

I hardly dare to answer ... non-commutative rings are so ugly ;-)

1) In general, we have the result
\[ \mathbb{Z}/n \otimes_\mathbb{Z} \mathbb{Z}/m \cong \mathbb{Z}/gcd(m,n)  \]
So, in case $m,n$ relatively prime, the tensor product  $\mathbb{Z}/n \otimes_\mathbb{Z} \mathbb{Z}/m$ is trivial. But of course, the "components" of the "elementary" tensors don't have to be trivial if $m,n > 1$. For example it holds $\overline{1} \otimes \overline{2} = 0$ in $\mathbb{Z}/2 \otimes_\mathbb{Z} \mathbb{Z}/3$.

When $R$ is a field, then you can show the following:
\[ x \otimes y = 0 \Rightarrow x = 0 \vee y = 0  \]
Just think of linear forms and extending basis.

2) I don't think there's a general formula for the number of elements. However, by the explicit construction of tensor products, we have
\[ |M \otimes_R N | \leq min(|R|,|\mathbb{Z}|)^{|M| * |N|}  \]
And when $R$ is commutative and $M,N$ are free $R$-Modules, then their tensor product is also free of rank rank(M)*rank(N), hence
\[ |M \otimes_R N | = |R|^{rank(M) * rank(N)}  \]

Non commutating rings do irk you right? :D
What irks me most is that even the most basic properties like $Hom_{R}(M,N)$ is an $R$-module are usually not true (it is just an abelian group)

Hm, I really do need to know this.
It's like my professor  handles this as something very easy, for instance she leaves this as just an exercise :

if $M$ is a left $R$-module, denote by $FM$ the underlying abeliang group :
this map $f: FM\rightarrow R\otimes M : y\rightarrow 1_{R}\otimes y$ is an isomorphism

[hide]This is part of proving that the covariant additive functor $R\otimes_{R}$ and the 'forget functor' from $R-mod$ to $Ab$ are naturally isomorphic[/hide]
well surjectivity is quite easy, but injectivity, well , there is my problem again  :huh: , and nothing is explicitly assumed about the ring (division ring, commutativity,...)

Do you have any ideas?

But it [i]is[/i] an easy exercise :-)

Construct an inverse homomorphism (of course with the universal property).

Okay, that is indeed easy, it is however nice that you say use universal property, that is THE way to define maps on them I assume (it seems in class we often just say $x\otimes y$ generates it, there is your map)

But what if for instance I would want to prove $\mathbb{Q}$ is flat $\mathbb{Z}$ module

in $Q\otimes M$ every element is of form $r\otimes m$
is a map $1_{\mathbb{Q}}\otimes f$ injective when $f$ is injective

well it maps$r \otimes f$ onto $r\otimes f(m)$

if I can prove the latter is only zero when the first is zero, I need to be able to answer my question about when the product is zero, right?
This is just another example of how important (in my opinion) this is? :o

Right, the universal property is THE way to define maps from tensor products, as they are completely determinded by this property. The same thing holds for other universal constructions, especially quotient spaces (which is hidden by explicit constructions!).

There are many approaches for the flatness of $\mathbb{Q}$, but your one is not correct ... $\mathbb{Q} \otimes M$ is just [i]generated[/i] by the $r \otimes m$.

The (imho) best proof uses the evident isomorphism of functors $R-\text{Mod} \to S-\text{Mod}$:
\[ S^{-1} R \otimes - \cong S^{-1} - \]
where $S$ is a multiplicative subset of the ring $R$. Since localization of modules is exact, this implies, that $S^{-1} R$ is a flat $R$-module. Applying this to $R = \mathbb{Z}, S=\mathbb{Z} - {0}$, we get the flatness of $\mathbb{Q}$.

Woops, that is complicated, localization? :?

I have never heard of that.

But what is wrong the claim that in $\mathbb{Q}\otimes M$ every element is of form $q\otimes n$?

I mean, I have the option of doing this trick finitely many times :

$(\frac{a}{b}\otimes n_1 ) + (\frac{c}{d}\otimes n_2 )=\frac{1}{b d } \otimes ( (a d) (n_1) + (b c) (n_2 ))$

Oh, I didn't know this trick :)

Localization is not complicated, just new to you :)

http://en.wikipedia.org/wiki/Localization_of_a_module

Other proofs I know would be far more complicated. If you wish, I can post them, but I don't recommend that ;).

Yes, but as cool or easy that trick is, it doesn't answer my question

if $f$ is injective
is $1\otimes f$ injective

well it maps every element (and every element has this form) $a\otimes n$ onto $a \otimes f(n)$

so what to do from here?  I mean, say well $f(n)=0\rightarrow n=0\rightarrow a\otimes n=0$

[quote="-oo-"]Right, the universal property is THE way to define maps from tensor products, as they are completely determinded by this property. The same thing holds for other universal constructions, especially quotient spaces (which is hidden by explicit constructions!).
[/quote]

I really like things simple   :roll:  , so I said, I'm gonna follow oo's guidance and try to prove ( in $\mathbb{Z}-mod$ to make it easy) this without any hard constructions , really nice and clean, 100% correct
$(M\otimes N)\otimes U\cong (M\otimes N)\otimes U)$

That's a little complicated though, I have to consider bilinear maps here.
So in order to get a map from the first to the latter by universal property, as you suggest, I woul have to make a bilinear map from $(M\otimes N)\times U$ to $(M\otimes N)\otimes U$ ?

Or doesn't this approach work well here in the end?

Sorry for all the questions, I am preparing for an algebra exam, but it is one of my favorite courses and I really wanna get to the bottom of everything.  I truly admire your knowledge. :|

What about the localization to prove the flatness of $\mathbb{Q}$? :)

You can immediately prove the associativity of the tensor product in $R$-Mod, it doesn't make thinks more complicated. By the way, I don't think that you really mean $(M \otimes N) \otimes U \cong (M \otimes N) \otimes U$. ;)

You're right, we have to find a bilinear map $(M \otimes N) \times U \to M \otimes (N \otimes U)$. But what can we do with an $x \in M \otimes N$ and an $u \in U$? Well, let $u \in U$ fixed, and define $\alpha_u : M \otimes N \to (M \otimes N) \otimes U$ by $m \otimes n \mapsto m \otimes (n \otimes u)$. This works because of the  universal property of $M \otimes N$. Now define $(M \otimes N) \times U \to M \otimes (N \otimes U)$  by $(x,u) \mapsto \alpha_u(x)$. By the universal property, this yields an homomorphism
\[ \beta : (M \otimes N) \otimes U \to M \otimes (N \otimes U) ,~ (m \otimes n) \otimes u \mapsto m \otimes (n \otimes u)  \]
Construct an explicit inverse homomorphism $\gamma$ (yes, by applying twice the universal property), and show $\beta \gamma = id$ and $\gamma \beta = id$ by checking these identities on the "elementary tensors" (which suffices, because they generate the tensor product). Note, that these isomorphisms are [i]natural[/i], i.e. isomorphisms of functors $R-\text{Mod}^3 \to R-\text{Mod}$, namely $(M,N,U) \to (M \otimes N) \otimes U$ and $(M,N,U) \to M \otimes (N \otimes U)$.

By the way, this stuff is quite well-known, you can find this in texts about (homological) algebra. So please don't admire "my" knowledge ... And of course, you don't have to be sorry for these questions!

\[ \text{Flatness of~} \mathbb{Q}  \]
[b]1. Lemma.[/b] Let $R$ be a commutative ring, $M,~N$ two $R$-modules, and consider an arbitrary element of $M \otimes_R N$, say $x = \sum m_i \otimes n_i$. If $x=0$ in $M \otimes_R N$, then there is a finitely-generated submodule $M'$ of $M$ containig the $m_i$ and a finitely-generated submodule $N'$ of $N$ containig the $n_i$ so that $x=0$ in $M' \otimes N'$.

Proof. Recall the explicit construction of the tensor product:
\[ M \otimes_R N = F(M \times N) / U  \]
where $F$ means the free $R$-module and $U$ is the submodule generated by elements of the form $(m+m',n)-(m,n)-(m',n)$, $(m,n+n')-(m,n)-(m,n')$, $(rm,n)-r(m,n)$, $(m,rn) - r (m,n)$. Now $x=0$ means that $\sum (m_i,n_i)$ lies in $U$, i.e. it is a linear combination of the mentionned elements. In these occur some elements of $M$ in the first components. Let $M'$ be the submodule of $M$, which is generated by these elements and the $m_i$. Define $N'$ similarly, and also $U'$. Then of course $\sum (m_i,n_i) \in U'$, i.e. $x = 0$ in $M' \otimes_R N'$. $\square$

[b]2. Lemma.[/b] Let $A$ be an abelian group, and $A \to A \otimes \mathbb{Q}$ the homomorphism $a \mapsto a \otimes 1$. Its kernel is $t(A)$, the torsion subgroup of $A$.

Proof. If $a \in t(A)$, then $na = 0$ for some $n>0$. It follows $a \otimes 1 = na \otimes \frac{1}{n} = 0$. Now let $a \in A$ s.t. $a \otimes 1 = 0$. By Lemma 1, there's a finitely-generated subgroup $B$ of $A$ which contains $a$, and s.t. $a \otimes 1 = 0$ in $B \otimes \mathbb{Q}$. Now by the structure theorem for finitely-generated abelian groups, we have
\[ B=\left\langle e_1 \right\rangle \oplus \cdots \oplus \left\langle e_m \right\rangle  \]
for some $e_i \in B$. We assume w.l.o.g., that $e_{n+1} , \cdots , e_m$ are torsion, and $e_1 , \cdots , e_n$ not. Then we have
\[ \left\langle e_i \right\rangle \otimes \mathbb{Q} \cong \left\{\begin{array}{cc} 0 & i>n \\ \mathbb{Q} & i \leq n \end{array}\right.  \]
For $i \leq n$, the isomorphism is given by $z e_i \otimes q \mapsto z q$. This yields an isomorphism
\[ B \otimes \mathbb{Q} \cong \left\langle e_1 \right\rangle \otimes \mathbb{Q} \oplus \cdots \oplus \left\langle e_n \right\rangle \otimes \mathbb{Q} \cong \mathbb{Q}^n  \]
which is given by $(z_1 e_1 + \cdots + z_m e_m) \otimes q \mapsto (z_1 q,\cdots,z_n q)$. So, if we write $a = z_1 e_1 + \cdots + z_m e_m$, the equation $a \otimes 1 = 0$ implies $z_i = 0$ for $i \leq n$, i.e. $a$ is torsion. $\square$

[b]3. Lemma.[/b] If $A$ is an abelian group, then every element of $\mathbb{Q} \otimes A$ has the form $\frac{1}{q} \otimes a$, which is zero if and only if $a$ is torsion.

Proof. That every element of $\mathbb{Q} \otimes A$ has the form $\frac{1}{q} \otimes a$, follows inductively from
\[ \frac{p}{q} \otimes a + \frac{r}{s} \otimes a' = \frac{1}{qs} ((ps) a + (rq) a')  \]
Since $\mathbb{Q} \otimes A$ is a vector space over $\mathbb{Q}$ with $q(s \otimes a) = (qs) \otimes a$, we have $\frac{1}{q} \otimes a = 0 \Leftrightarrow \frac{1}{q} (1 \otimes a) = 0 \Leftrightarrow 1 \otimes a = 0$. This is equivalent to $a$ torsion, by Lemma 2. $\square$

[b]4. Theorem.[/b] $\mathbb{Q}$ is flat.

Proof. Let $A,B$ be abelian groups and $f : A \to B$ a monomorphism. Then $1_\mathbb{Q} \otimes f$ maps $\frac{1}{q} \otimes a$, which is an arbitrary element by Lemma 3, to $\frac{1}{q} \otimes f(a)$. But by Lemma 3, $\frac{1}{q} \otimes f(a) = 0 \Leftrightarrow f(a)$ torsion $\Leftrightarrow a$ torsion $\Leftrightarrow \frac{1}{q} \otimes a = 0$. Thus, $1_\mathbb{Q} \otimes f$ is a monomorphism.

Here an alterniative proof. We have the following characterization of flat modules:

[b]Theorem 1. [/b] A $R$-module $M$ is flat if and only if the map $I \otimes M \to M , a \otimes m \mapsto am$ is injective for all ideals $I$ of $R$.

You can find the proof in the script http://www.math.ku.dk/ma/kurser/mat4ha/ha.pdf, page 97. $\square$

[b]Theorem 2. [/b] Over PIDs  flat modules coincide with torsionfree modules.

Proof. Let $R$ be a PID and $I$ an ideal of $R$, say $I=aR$, and $M$ a $R$-module. Assume $M$ is torsionfree. Every element of $I \otimes M$ has the form $a \otimes m$, since $(a r_1 \otimes m_1) + \cdots + (a r_n \otimes m_n) = a \otimes (r_1 m_1 + \cdots + r_n m_n)$. By $I \otimes M \to I$, it is mapped to $am$. For $a=0$, the map is clearly injective, and otherwise it follows from torsionfreeness. So by Theorem 1, $M$ is flat. Now assume $M$ is flat, and let $a \in R$ , $a \neq 0$. Multiplication with $a$ yields a map $R \to R$, which is injective because $R$ is an integral domain. Then $M \otimes R \to M \otimes R$ is also injective, thus $M \cong M \otimes R \to M \otimes R \cong M$, which is given by $m \mapsto am$. So, $am = 0$ implies $m$, and $M$ is torsion-free. $\square$

In particular, $\mathbb{Q}$ is flat.

Another one ... :-)

[b]Theorem.[/b] If every finitely-generated submodule of $M$ is flat, then $M$ is flat.

Proof. $M$ is the direct limit of its finitely-generated submodules, i.e. $M=Colim(F)$, when $F$ is the forgetful functor $P \to R$-Mod and $P$ is the preorder of the finitely-generated submodules of $M$. There's a natural isomorphism
\[ M \otimes N \cong Colim(F) \otimes N \cong Colim(F- \otimes N) = (Colim \circ G)(N)  \]
where $G$ denotes the functor $R-\text{Mod} \to (R-\text{Mod})^P , ~ N \mapsto (p \mapsto F(p) \otimes N)$. Now each projection $N \mapsto F(p) \otimes N$ is exact, because $F(p)$ is flat by assumption. So, $G$ is exact. But Colim is also an exact functor $(R-\text{Mod})^P \to R-\text{Mod}$, since $P$ is directed. Therefore, $M \otimes -$ is a composition of exact functors, hence also exact, i.e. $M$ is flat. $\square$

In particular, torsionfree modules over PIDs are flat (here, finitely-generated submodules are even free by the structure theorem).

You can also prove this theorem easily with the first lemma above:

Let $f : A \to B$ be an monomorphism, and let $x \in \ker(f \otimes M)$, say $x = \sum a_i \otimes m_i$ and $\sum f(a_i) \otimes m_i = 0$ in $B \otimes M$. By the lemma, there's a finitely generated submodule $N$ of $M$ s.t. $N$ contains every $m_i$ and $\sum f(a_i) \otimes m_i = 0$ holds in $B \otimes N$. Since $N$ is flat by assumption, $f \otimes N$ is injective, which implies $\sum a_i \otimes m_i = 0$ in $A \otimes N$ and hence in $A \otimes M$. $\square$

I think, this is the simplest proof for the flatness of $\mathbb{Q}$: Use the above theorem, and the fact, that $\mathbb{Q}$ is locally cyclic (which both can be proven without many previous knowledge). Then every finitely-generated subgroup of $\mathbb{Q}$ is either $\cong \mathbb{Z}$ or $\cong 0$, hence flat, so that $\mathbb{Q}$ is flat, too.