geometry
analytic geometry
circumcircle
incenter
trigonometry
ratio
conics

Given six real numbers $a$, $b$, $c$, $x$, $y$, $z$ such that $0 < b-c < a < b+c$ and $ax + by + cz = 0$.

What is the sign of the sum $ayz + bzx + cxy$ ?

Why is this a triangle geometry problem?

Cuz a,b,c are the sides of a triangle perhaps? :?

And the sum looks to be pretty negative, because it will always be:
(say g<h<i)

-ig -ih +gh times something
because it are triangle sides this can never get positive.

[quote="Peter VDD"]Why is this a triangle geometry problem?

Cuz a,b,c are the sides of a triangle perhaps? :?[/quote]

Okay, this is the starting point of a solution. What is really nice is that the solution uses triangle geometry and almost no calculation. Of course, there are more or less straightforward ways but the triangular one is most enlightening. One tip finally: It makes use of trilinear coordinates. If this says nothing to you, I can explain.

[quote="Peter VDD"]And the sum looks to be pretty negative, because it will always be:
(say g<h<i)

-ig -ih +gh times something
because it are triangle sides this can never get positive.[/quote]

I'm sorry, this is somehow unclear to me: What are g, h, i? Are they positive?

  All the best,
  Darij

I have been asked about the explanation of trilinear coordinates. Here are some fundamentals important for my solution of the problem.

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

[b]1.[/b]

If we try to prove triangle geometry theorems with the help of Cartesian coordinates, we observe that the calculations get tedious very quickly and the formulas are long and ugly. Cartesian coordinates are not really suitable for triangle geometry; an adequate coordinate system should have three rather than two coordinate axes, namely the sidelines of the triangle. Hence we try to find a good coordinate system analogous to the Cartesian one but having three coordinate axes.

Consider a triangle ABC and a point P in its plane. Let $x_{1}$, $y_{1}$ and $z_{1}$ denote the distances from P to the lines BC, CA, AB, where $x_{1}$ is considered positive if P and A lie in the same halfplane with respect to the line BC and negative if P and A lie in different halfplanes with respect to the line BC, and similarly $y_{1}$ and $z_{1}$ are signed.

The distances $x_{1}$, $y_{1}$, $z_{1}$ are called [b]actual[/b] or [b]normalized trilinear coordinates[/b] of the point P with respect to triangle ABC. Now, it is easy to see that for any point P, the coordinates $x_{1}$, $y_{1}$, $z_{1}$ satisfy the equation

(1)   $ax_{1}+by_{1}+cz_{1}=2S$,

where a, b, c are the sidelengths and S is the area of triangle ABC. Hence, the actual trilinear coordinates are not independent from each others. This is a disadvantage, and in order to eliminate this disadvantage, we slightly modify the definition of the coordinates:

Let x, y, z be any three real numbers such that $x=kx_{1}$, $y=ky_{1}$, $z=kz_{1}$, where k is some real number not equal 0.

Then, the numbers x, y, z are called [b]homogeneous trilinear coordinates[/b] of the point P with respect to triangle ABC. The homogeneous trilinear coordinates x, y, z of a point P are usually written in the form

  P ( x : y : z ).

Of course, now every point P has an infinity of triples (x, y, z) which can be considered as homogeneous trilinear coordinates of P. However, it can be easily seen that every triple of real numbers (x, y, z) defines one and only one point P having this triple as homogeneous trilinear coordinates, unless ax + by + cz = 0. If ax + by + cz = 0, there is no point P in the Euclidean plane with homogeneous trilinears ( x : y : z ). Note that x = y = z = 0 is forbidden (there is no point with homogeneous trilinear coordinates ( 0 : 0 : 0 )); to the case ax + by + cz = 0 we will return later.

"Trilinear coordinates" is usually abbreviated as "trilinears", hence we will speak of "actual trilinears" and "homogeneous trilinears".

[b]2.[/b]

The following homogeneous trilinears of some well-known points are easily established:

  A ( 1 : 0 : 0 ),
  B ( 0 : 1 : 0 ),
  C ( 0 : 0 : 1 ),
  circumcenter of triangle ABC ( cos A : cos B : cos C ),
  orthocenter of triangle ABC ( sec A : sec B : sec C ),
  centroid of triangle ABC $\left(\frac{1}{a}: \frac{1}{b}: \frac{1}{c}\right)$,
  incenter of triangle ABC ( 1 : 1 : 1 ),
  A-excenter of triangle ABC ( -1 : 1 : 1 ),
  B-excenter of triangle ABC ( 1 : -1 : 1 ),
  C-excenter of triangle ABC ( 1 : 1 : -1 ),
  symmedian point of triangle ABC ( a : b : c ).

For any point P with the homogeneous trilinears P ( x : y : z ), we have x = 0 if and only if P lies on the line BC, and similarly for y and z.

Generally, it can be shown that the homogeneous trilinears P ( x : y : z ) of a point P are given by the following equations:

  $\frac{y}{z}=\frac{CA^{\prime}}{A^{\prime}B}\cdot\frac{c}{b}=\frac{\sin\measuredangle CAP}{\sin\measuredangle PAB}$;
  $\frac{z}{x}=\frac{AB^{\prime}}{B^{\prime}C}\cdot\frac{a}{c}=\frac{\sin\measuredangle ABP}{\sin\measuredangle PBC}$;
  $\frac{x}{y}=\frac{BC^{\prime}}{C^{\prime}A}\cdot\frac{b}{a}=\frac{\sin\measuredangle BCP}{\sin\measuredangle PCA}$,

where A', B', C' are the points where the lines AP, BP, CP meet BC, CA, AB, respectively, and the angles are directed angles modulo 360°. This form is related to the Ceva theorem (in fact, the Ceva theorem in its affine and trigonometric forms is a direct corrolary of these equations).

[b]3.[/b]

If $x_{1}$, $y_{1}$, $z_{1}$ are the actual trilinear coordinates of a point $P_{1}$, and $x_{2}$, $y_{2}$, $z_{2}$ are the actual trilinear coordinates of a point $P_{2}$, then the actual trilinears $x_{3}$, $y_{3}$, $z_{3}$ of a point $P_{3}$ which lies on the line $P_{1}P_{2}$ and divides the segment $P_{1}P_{2}$ in the ratio $\frac{P_{1}P_{3}}{P_{3}P_{2}}=n$ are given by

  $x_{3}=\frac{x_{1}+nx_{2}}{1+n}$,
  $y_{3}=\frac{y_{1}+ny_{2}}{1+n}$,
  $z_{3}=\frac{z_{1}+nz_{2}}{1+n}$.

One of the remote corollaries of this fact is the following important theorem:

For any line g in the plane of triangle ABC, there exist three real numbers k, l, m such that a point P with its homogeneous trilinears P ( x : y : z ) lies on the line g if and only if kx + ly + mz = 0. This equation, kx + ly + mz = 0, is the equation of the line g in homogeneous trilinears.

I. e., the equation of a line in homogeneous trilinears is a linear equation without the free member.

Conversely, for any three reals k, l, m, there is a line g such that the equation of g is kx + ly + mz = 0, unless k / a = l / b = m / c. Again, we will return to the k / a = l / b = m / c case later.

[b]4.[/b]

As we have seen, a homogeneous equation of first degree in trilinears represents a line. Generally, a homogeneous equation of second degree gives a conic; however, the most important of all conics related to triangle ABC is the circumcircle of the triangle, having the equation ayz + bzx + cxy = 0, or, equivalently, a / x + b / y + c / z = 0 (of course, the latter making sense only for nonzero x, y, z). This is easy to verify synthetically.

[b]5.[/b]

Now, in the previous considerations, we have excluded some degenerate cases. At first, we have said three reals x, y, z define a point P unless ax + by + cz = 0; later, we have said three reals k, l, m define a line unless k / a = l / b = m / c. Now what is up with these "unless" cases?

Actually, if we define abstract "points" P by their homogeneous coordinates x, y, z with ax + by + cz = 0, and an abstract "line" i by its equation kx + ly + mz = 0 with k / a = l / b = m / c - or, equivalently, the equation of i is ax + by + cz = 0, and if we abstractly define the incidence relation by stipulating that a point is defined to "lie" on a line if the line equation is fulfilled for the homogenous trilinears of the point, then it is clear that all abstract points P with ax + by + cz = 0 lie on the abstract line i.

It can also be shown that any abstract point lies on an infinite set of non-abstract, i. e. real lines, and all of these real lines are parallel to each other. Actually, an abstract point may be identified with a class of parallel lines. Speaking in terms of projective geometry, the abstract points are (or can be identified with) the infinite points of the projective plane. The abstract line i is the line at infinity.

Hence, we see that homogeneous trilinear coordinates are a very suitable coordinate system for the projective plane. In contrast to actual trilinear and Cartesian coordinates, they naturally integrate infinite points (note that infinite points don't have actual trilinears!).

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

I am sorry for the length of this; now we come to the solution of the problem.

It is easy to see that a, b, c are the sides of a triangle (a, b, c are all positive, and the triangle inequalities are fulfilled).

We are going to show that for all x, y, z with ax + by + cz = 0 except of the trivial degenerate case x = y = z = 0, the expression ayz + bzx + cxy is negative.

In fact, we may consider the point P with homogeneous trilinear coordinates P ( x : y : z ); this point lies on the line at infinity, since ax + by + cz = 0. Now, in order to deal with a term uniquely defined from P, we will consider the expression $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}$ rather than simply ayz + bzx + cxy. We have to show that the latter expression, ayz + bzx + cxy, is negative; but we will show that the former one, $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}$, is negative, what is obviously equivalent, since the denominator, $x^{2}+y^{2}+z^{2}$, is always > 0.

Note that the expression $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}$ is always definite, since the denominator is strictly > 0 (we have precluded the case x = y = z = 0).

Regard a point P with homogeneous trilinear coordinates P ( x : y : z ) - so that ax + by + cz = 0 - continuously moving on the line at infinity. If the expression $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}$ would change sign, there would be some point P such that $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}=0$.

But this leads to ayz + bzx + cxy = 0, and hence the point P lies on the circumcircle of triangle ABC. But the line at infinity and the circumcircle have no common points (I admit this has to be proven algebraically, but the proof is not too hard); hence, this is impossible. Thus, the expression $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}$ cannot change sign for P on the line at infinity. It is therefore enough to find the sign of this expression for some concrete position of P (for instance, for the point P where the line at infinity meets BC : x = 0, y = -c, z = b) - this sign turns out to be negative, and then the sign is negative for all other positions of P.

Qed..

Of course, this is not the simplest solution; the solution I participated with was a rather elementary one. Generally, I dislike any solution using continuity or other analysis arguments, but this one is just elegant, don't you think so?

  Darij

Just looked up Kvant and found this here:

[b]Problem M1166, solved in Kvant 11/1989, page 26:[/b]

Prove the inequality $ a^2pq \plus{} b^2qr \plus{} c^2rp \leq 0$, where a, b, c are the sidelengths of a triangle, and p, q, r are three real numbers satisfying p + q + r = 0.

This is stronger than the above German TST problem (because sidelengths of a triangle are always the squares of the sidelengths of another triangle). But there is also a very simple algebraic way to solve this problem:

Since p + q + r = 0, we have

$ \left(b \plus{} c\right)^2pq \plus{} b^2qr \plus{} c^2rp \equal{} \left(b^2q \plus{} c^2p\right)\left(p \plus{} q \plus{} r\right) \minus{} \left(bq \minus{} cp\right)^2$
$ \equal{} 0 \minus{} \left(pq \minus{} cp\right)^2\leq 0$.

The numbers pq, qr, rp cannot all be negative, since this would force their product $ pq\cdot qr\cdot rp \equal{} \left(pqr\right)^2$ to be negative as well, what is absurd since this product is a square. Hence, at least one of the numbers pq, qr, rp is $ \geq 0$. We can WLOG assume that $ pq\geq 0$. Since a, b, c are the sidelengths of a triangle, the triangle inequality yields a < b + c; combining this with $ pq\geq 0$, we get $ a^2pq\leq\left(b \plus{} c\right)^2pq$. Hence,

$ a^2pq \plus{} b^2qr \plus{} c^2rp\leq\left(b \plus{} c\right)^2pq \plus{} b^2qr \plus{} c^2rp\leq 0$,

and the problem is solved.

  Darij

[quote="wojto111"]darij grinberg why you use expression $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}$ . In all your prove you don't use that you divided by $x^{2}+y^{2}+z^{2}$[/quote]

In fact, the expression $\frac{ayz+bzx+cxy}{x^{2}+y^{2}+z^{2}}$ is a function of the point P, while the expression ayz + bzx + cxy isn't - we are working in [b]homogeneous[/b] coordinates.

  darij

[quote="darij grinberg"]I'm sorry, this is somehow unclear to me: What are g, h, i? Are they positive?[/quote]Dunno what I meant 2 years ago, probably this. Your $0 < b-c < a < b+c$ is equivalent to the more handlable $b+c>a>b>c>0$. Take WLOG $x>0$ (we can safely assume $x,y,z$ nonzero).

If $y,z$ have different signs we find a negative sign immediately, so to have a positive sign we need $y,z<0$. For easyness call  $x=g,y=-h,z=-i$ (so $g,h,i>0$, so we get to prove from $ah=bh+ci$ that $ahi<bgi+cgh$. From $(b+c)^{2}>a^{2}$ it follows: \[a(bgi+cgh)=(bh+ci)(bi+ch)\ge(b+c)^{2}(hi)>a^{2}hi,\] division by $a$ gives the requested negative sign.


algebra
polynomial
number theory
greatest common divisor
algebra unsolved

Find all polynomials $P(x)$ with real coefficients which satisfy the relation
$(x^3+3x^2+3x+2)P(x-1) = (x^3-3x^2+3x-2)P(x)$

Oops, my confusion. Yesterday maybe I was sleepy and I was thinking about putting in $x$ values. Of course it is irrelevant.


Hi, I'm totally confused by this problem. I understand how the soluble ions will not form together, but I don't understand how one of these choices could even work:

The net ionic equation for formation of an aqueous solution of $Al(NO_3)_3$ via mixing solid $Al(OH)_3$ and aqueous nitric acid is _____.
A) $Al(OH)_3(s)+3HNO_3(aq) \rightarrow 3H_2O(l)+Al(NO_3)_3(aq)$
B) $Al(OH)_3(s)+3NO_3^-(aq) \rightarrow 3OH^-(aq)+Al(NO_3)_3(aq)$
C) $Al(OH)_3(s)+3NO_3^-(aq) \rightarrow 3OH^-(aq)+Al(NO_3)_3(s)$
D) $Al(OH)_3(s)+3H^+(aq) \rightarrow 3H_2O(l)+Al^{3+}(aq)$
E) $Al(OH)_3(s)+3HNO_3(aq) \rightarrow 3H_2O(l)+Al^{3+}(aq)+NO_3^-(aq)$

The correct equation is D.  The nitrate ions will be spectators, and the the hydrogen ion will combine with the hydroxide ion from aluminum oxide to form water.

Okay thanks, I get it now. I realize why I couldn't get it before. What I was confused about was why they even showed the aluminum being there, as it wasn't part of the reaction, but now I see that it's solid, so therefore wouldn't be pulled apart from the hydroxide. Ok:) thanks a lot.


In the quadratic function y=(-2x+7)(8x-4), the vertex of this function is at (2,-9) right? and the maxmium value of this function would be 2? I found this using the -b/2a thing.

Can somebody explain to be when in ax^2+bx+c, if a<0 the equation is -b/2a, but even when a>0 it is also -b/2a?

Also how would you solve the problem that goes
"let f(x) be a linear function such that for all x, f(5x+1)=2x+3. Find f(x)."

I don't know where to begin in this.

[quote="Farcus"]In the quadratic function y=(-2x+7)(8x-4), the vertex of this function is at (2,-9) right? and the maxmium value of this function would be 2? I found this using the -b/2a thing. [/quote]

The maximum is at (2,36). :D

[quote="Farcus"]Can somebody explain to be when in ax^2+bx+c, if a<0 the equation is -b/2a, but even when a>0 it is also -b/2a? [/quote]

That's because when we complete the square, we get something like
$y=a(x+b/2a)^{2}+stuff$ and to maximize this when a<0, $a(x+b/2a)^{2}$ has to be 0 so $x=-b/2a$. When a>0, we no longer want to maximize the value(because the max will be infinity), but we want to minimize it. $a(x+b/2a)^{2}$ has to be 0 to get the minimum, and $x=-b/2a$

[quote="Farcus"]"let f(x) be a linear function such that for all x, f(5x+1)=2x+3. Find f(x)." [/quote]

we can let $f(x)=ax+b$.
then $f(5x+1)=5ax+a+b=2x+3$
so $5a=2$ and $a+b=3$
now you can solve the equations to get the answer :D

hope that helps

so it doesn't matter if a<0 or a>0? I mean at the end its all -b/2a right?

Complete the square.

When you have $y=ax^{2}+bx+c$, where $a>0$, we have $y=a(x+\frac{b}{2a})^{2}-\frac{b^{2}}{4a}+c$. By the trivial inequality, the minimum occurs when $x+\frac{b}{2a}=0$, and $x=\frac{b}{2a}$. When $a<0$, this is the maximum occurs. Then what you're left with is the actual minimum (or maximum)


function
limit
logarithms
complex analysis
advanced fields
advanced fields unsolved

Hi!

I don't quite know where to start here:
Let $f(z) = \sum c_nz^n$ be an entire function of finite genus mu.
Prove that: $\lim_{n\to\infty} c_n(n!)^{\frac 1{\mu+1}}=0$.

Any ideas appreciated  :)

Thanks,
Amar

[i]Edited by Myth[/i]

Excuse me my ignorance. What is genus here?
I think it is the same as russian "порядок" ("order" in english), i.e. entire function $f$ has a finite order iff $|f(z)|=O(e^{|z|^A})$ as $|z|\to\infty$ for some $A$, and $\inf A$ is the order of $f$.

It immediately follows from the following general theorem about genus of entire functions.

[b]Theorem.[/b]
Function $ f(z) = \sum_{n=0}^{\infty} c_nz^n$ is a function of finite genus $\mu$ iff \[\liminf_{n\to\infty}\frac{\ln(1/|a_n|)}{n\ln n}=\frac{1}{\mu}.\]

Hi Myth,

thanks for your reply. In the meantime I tried to do this using Cauchy's Estimate,
but this seems not to work out.
Sorry, but I didn't know the theorem you stated.
How can you conclude "immediately", using this theorem, that the given limit
is 0?

Thanks again.

Acordingly to this theorem, we can state that $\frac{\ln(1/|a_n|)}{n\ln n}>\frac{1}{\mu+1}$ for all large $n$. It gives that $a_nn^{\frac{n}{\mu+1}}<1$. On the other hand, due to Stirling's formula, $a_n(n!)^{\frac{1}{\mu+1}}\sim a_n(n^ne^{-n}\sqrt{2\pi n})^{\frac{1}{\mu+1}}=\big(a_nn^{\frac{n}{\mu+1}}\big)\cdot (e^{-n}\sqrt{2\pi n})^{\frac{1}{\mu+1}}\to 0$, as the first factor is bounded and the second factor tends to zero.


For natural $ m,n$ prove:

$ |\sqrt{2}\minus{}\frac{m}{n}| > \frac{1}{3n^2}$

Because $ \dfrac{m}{n}$ is rational, and $ \sqrt {2}$ is not, $ \dfrac{m}{n} \neq \sqrt {2}$. Among these four statements one of them must be true: $ 0 < \dfrac{m}{n} < \sqrt {2}$,
$ \sqrt {2} < \dfrac{m}{n} < \dfrac{3}{2}$,
$ \dfrac{3}{2} \le \dfrac{m}{n} < \sqrt {2} \plus{} 1$, and $ \dfrac{m}{n} \ge \sqrt {2} \plus{} 1$.

If $ \dfrac{m}{n} < \sqrt {2}$, then $ m^2 < 2n^2$ and $ m^2 \le 2n^2 \minus{} 1$, so
$ 2n^2 \minus{} m^2 \ge 1$. [1]
Since $ m < n \sqrt {2}$, $ n\sqrt {2} \minus{} m > 0$, so
$ (n\sqrt {2} \minus{} m)^2 \equal{} 2n^2 \minus{} (2\sqrt {2})nm \plus{} m^2 > 0$. [2]
Adding inequalities [1] and [2] and dividing by $ (2\sqrt {2})n^2$ we get
$ |\sqrt {2} \minus{} \dfrac{m}{n}| \equal{} \sqrt {2} \minus{} \dfrac{m}{n} > \dfrac{1}{(2\sqrt {2})n^2} > \dfrac{1}{3n^2}$, as desired.

If $ \sqrt {2} < \dfrac{m}{n} < \dfrac{3}{2}$ then
$ m^2 > 2n^2$ and $ m^2 \ge 2n^2 \plus{} 1$, so
$ m^2 \minus{} 2n^2 \ge 1$. [3]
Since $ m > n\sqrt {2}$, $ m \minus{} n\sqrt {2} > 0$, so
$ (m \minus{} n\sqrt {2})^2 \equal{} m^2 \minus{} (2\sqrt {2})nm \plus{} 2n^2 > 0$. [4].
Adding inequalities [3] and [4] and dividing by $ 2mn$ we get
$ |\sqrt {2} \minus{} \dfrac{m}{n}| \equal{} \dfrac{m}{n} \minus{} \sqrt {2} > \dfrac{1}{2mn} \equal{} \dfrac{1}{m} \cdot \dfrac{1}{2n} > \dfrac{2}{3n} \cdot \dfrac{1}{2n} \equal{} \dfrac{1}{3n^2}$, as desired.

If $ \dfrac{3}{2} \le \dfrac{m}{n} < \sqrt {2} \plus{} 1$ then consider the value of $ n$.
If $ n \equal{} 1$ then $ \dfrac{3}{2} \le m < \sqrt {2} \plus{} 1 < 3$, so by elimination $ m \equal{} 2$.
$ |\sqrt {2} \minus{} \dfrac{m}{n}| \equal{} \dfrac{m}{n} \minus{} \sqrt {2} \equal{} 2 \minus{} \sqrt {2}$.
Since $ 2 \minus{} \sqrt {2} > 2 \minus{} \dfrac{3}{2} \equal{} \dfrac{1}{2} > \dfrac{1}{3} \equal{} \dfrac{1}{3n^2}$,
$ |\sqrt {2} \minus{} \dfrac{m}{n}| > \dfrac{1}{3n^2}$ and we are done.
If $ n \ge 2$ then $ n^2 \ge 4$ and $ 3n^2 \ge 12$, so
$ \dfrac{1}{3n^2} \le \dfrac{1}{12}$.
$ |\sqrt {2} \minus{} \dfrac{m}{n}| \equal{} \dfrac{m}{n} \minus{} \sqrt {2} \ge \dfrac{3}{2} \minus{} \sqrt {2}$. [5]
$ 288 < 289$, so $ 2 < \dfrac{289}{144}$, $ \sqrt {2} < \dfrac{17}{12}$, $ 12\sqrt {2} < 17$, $ 18 \minus{} 12\sqrt {2} > 1$, and
$ \dfrac{3}{2} \minus{} \sqrt {2} > \dfrac{1}{12} \ge \dfrac{1}{3n^2}$. [6]
Combining inequalities [5] and [6] we get
$ |\sqrt {2} \minus{} \dfrac{m}{n}| > \dfrac{1}{3n^2}$ and we are done.

If $ \dfrac{m}{n} \ge \sqrt {2} \plus{} 1$ then
$ |\sqrt {2} \minus{} \dfrac{m}{n}| \equal{} \dfrac{m}{n} \minus{} \sqrt {2}\ge (\sqrt {2} \plus{} 1) \minus{} \sqrt {2} \equal{} 1 > \dfrac{1}{3n^2}$, so we are done.


The areas of the two adjacent squares are 256 square inches and 16 inches, respectively, and their bases lie on the same line. What is the number of inches in the length of the segment that joins the centers of the two inscribed circles? Express your answer as a decimal to the nearest tenth.

[asy]draw((4,0)--(4,4)--(0,4)--(0,0)--(5,0)--(5,1)--(4,1));
draw(Circle((2,2),2));
draw(Circle((4.5,.5),.5));
draw((2,2)--(4.5,.5));[/asy]

Hello
Radius of bigger circle=$ 8$inches
Radius of smaller circle=$ 2$ inches
Length of segment=$ \sqrt{6^2\plus{}10^2}$
                         =$ 11.66$
Thank u.


AMC AIME calculus MATHCOUNTS USA(J)MO USAMO ARML

My son is in 7th grade.   He took AMC 12B and he just found out that he qualified for AIME with a score of 103.  He didn't get to take AIME,  probably because the  middle school just didn't participate in it (I don't know for sure, though).  I tried to look at statistics on the AMC web site, but I didn't find much for 7th graders.  I assume that not many middle school kids take in the first place.  Nevertheless, if any of you have an idea how this score looks compared to other 13 year olds, I would appreciate the info.  Thanks

it is no doubt a great score; the 12 tests were designed for juniors and seniors to try to advance to the AIME on.

there is still time to take the AIME II, although the chances are extremely small

which school did your son take the 12B at? why didn't he take the AIME with them?

if you contact the AMC offices in time, you [i]might[/i] be able to arrange something with them

you'll probably have to pay for 10 of the AIME II exams themselves and have them shipped on like, overnight shipping

note that the best method would've been to take the AIME I with the school that he took the AMC 12B at

its a good thing that he's only in 7th grade, he has a bright future ahead of him!

Unfortunately, the AMC hasn't compiled the statistics for this years AMC yet. But, you could always look at last years results, assuming similar difficulties

[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05scoredist-grade12A.html]www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05scoredist-grade12A.html[/url]

A 103 is an excellent score for a middle schooler. :)

Thanks for your replies.
I don't think the school participated in AIME, if they did, I can't imagine why the teacher would leave my son out.  It might participate in the alternate AIME.  I emailed the teacher about this, but we have spring break now, so I probably won't hear about it untill the 20th.

I have another question.  I plan to get the Art of Problem Solving books.  Should I get both Vol I and Vol II,  and are the solution manuals very useful?

defenately
for AMC: Vol 1 +solution manual
(the solution manual is actually very important, although the students might know most of the material in the Vol 1, but the problems at end of each chapter are very very very good practices for AMC)

for AIME/Olympiad: Vol 2+solution Manual
for this one you defenatly need the solutions (at least i did :) )

BTW, my friend, a 9th grader in AP Calculus, also got a 103 on AMC 12. So yeah, it's a pretty good score for a 7th grader! :yup:

[quote="justaparent"]I don't think the school participated in AIME, if they did, I can't imagine why the teacher would leave my son out.  It might participate in the alternate AIME.  I emailed the teacher about this, but we have spring break now, so I probably won't hear about it untill the 20th.[/quote] I would call the AMC on Monday to see whether the school is registered for the alternate AIME.  If not, maybe you could ask the AMC to send it to the school.  You might need to pay the registration fee (something like 25 dollars).  The alternate AIME is on March 22, so there's not a lot of time left.

Last year, only 46 7th graders qualified for taking AIME.  So, it is really a good accomplishment.

I took the AMC 12 as an 8th grader and qualified for AIME.  I took the AIME also! There's no reason that your son shouldn't.

Yes that is a good score.
Most 11th and 12th graders dont do that well.
Keep your child's interest in math alive.
You are lucky enough to hear about the AMC's in 7th grade.
I never knew they existed until after last year's test.

Lucky ... our school does not allow middle schoolers to take part in any high school contests whatsoever.

That stinks.
Last year our middle school was invited to a highschool math invitational since we were beating alot of the highschools that competed in the non-invitational competitions.
Unfortunately, State Mathcounts was on the same day.
I don't think anything like that will happen for a while.

Just a correction.  There were 42 7th graders who took the aime last year not 46.  You should definately try to take the AIME this year but if it doesn't work out you could just get a copy of the test because 7th graders almost never make it to the USAMO so don't be too sad.

You never know.  Twins on our (Washington's) ARML team qualified as 8th graders.  :)

You can get information about how this score ranks for a 7th grader compared to other grades at the Score Distribution by Grade page at:
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06-scoredist-grade12AB.html

That and other statistics just went up yesterday.

Steve Dunbar
Director, AMC

So it looks like a 103 is 17th out of all the 7th graders.

I haven't posted here in a while, but...

When I took the AMC12A in seventh grade last year, I got a 100.5.  No one in my school system had EVER made it to the AIME, so my parents and I really just needed to make sure my school knew, and we had lots of meetings about scheduling it and making sure I actually got to take the test.  Qualifying for the AIME but not taking it is something that just shouldn't happen.

I was mistaken earlier.  My son will take the alternate AIME next Wednesday.  That should be interesting, considering he is just taking Algebra I!  I think that AMC 12 has enough problems which don't require knowledge of math higher than that, but not AIME.  Well, it's a good start anyway.

103 in the 7th grade with only Alg 1 is really awesome.

Purchase AoPS volumes 1 and 2. 1 for now but im sure you will want 2 later, so just get them with solutions for reduced price.

Have him try the practice AMC's in his spare time in the contests section in AoPS.


integration
calculus
calculus computations

We can't understand your problem for 1).

You should learn Latex. :wink:

$ \int \frac1{x^{\frac13} \plus{} x^{\frac14}} \plus{} \frac{\ln(1 \plus{} x^{\frac16})}{x^{\frac13} \plus{} x^{\frac12}}\ dx$ I suppose.

[quote="Tyl"]$ \int \frac1{x^{\frac13} \plus{} x^{\frac14}} \plus{} \frac {\ln(1 \plus{} x^{\frac16})}{x^{\frac13} \plus{} x^{\frac12}}\ dx$ I suppose.[/quote]

Hi :) !!!

if that's the integral than i answer:

let $ x \equal{} t^{12}$ than the integral become :

$ 12 \int \left( \frac {t^{11}}{t^4 \plus{} t^3} \plus{} \frac {t^{11} ln(1 \plus{} t^2)}{t^4 \plus{} t^6 } \right) dt$

          $ \equal{} 12 \int \sum_{k \equal{} 0}^{7}( \minus{} 1)^{7 \minus{} k}t^k dt \minus{} 12 \int \frac {dt}{1 \plus{} t} \plus{} 12\int \frac {t^7ln(1 \plus{} t^2)}{1 \plus{} t^2}dt$

           $ \equal{} 12 \sum_{k \equal{} 1}^{8}\frac {( \minus{} 1)^{8 \minus{} k}t^k}{k} \ \minus{} \ 12ln(1 \plus{} t) \ \plus{} \ 12F(t) \ \plus{} \ Cte$

with:

$ F(t) \equal{} \int \frac {t^7ln(1 \plus{} t^2)}{1 \plus{} t^2}dt \\
\equal{} ln(1 \plus{} t^2) \left( \frac {t^2}{12} ( 2t^4 \minus{} 3t^2 \plus{} 6) \minus{} \frac {1}{4}ln(1 \plus{} t^2) \plus{} \frac {11}{12} \right) \minus{} \frac {t^6}{18} \plus{} \frac {5t^4}{24} \minus{} \frac {11t^2}{12}$

and thanks

Hi :) !!

for second integral i think that you want : $ \int \frac{\sqrt{cos(2x)}}{sin(x)}dx$

If YES than my answer is:

Let $ t\equal{}\sqrt{cos(2x)}$

than the integral become :

$ \int \frac{\sqrt{2}t^2}{(1\minus{}t^2)\sqrt{1\plus{}t^2}}dt \equal{} \sqrt{2} \int \frac{\minus{}dt}{\sqrt{1\plus{}t^2}} \plus{} \sqrt{2} \int \frac{dt}{(1\minus{}t^2)\sqrt{1\plus{}t^2}}$

that's Not difficult to calcul that ...

if you want the finaly result i'm here to give it ;)

and thanks :)

[quote="Zed_darkx"]Mathema*
Thanks ,  But can you give the step wise solution  or actual procedure where you used the Summation in the 1st  questition.
And I got the 2nd one , Thanks anyway :)[/quote]

Hi :)

Ok with pleaser :)

about summation isn't a big things because $ \int \frac {t^8}{1 \plus{} t}dt \equal{} \int \frac {t^8 \plus{} 1}{t \plus{} 1}dt \plus{} \int \frac{\minus{}dt}{1 \plus{} t}$

we know that $ (t^8 \plus{} 1) \equal{} (t \plus{} 1) \sum_{k\equal{}0}^7 ( \minus{} 1)^{n \minus{} k}t^k$ than the rest are easy with calculs....

and thx

PS: if there's other things i'm here ;)

[quote="mathema*"]Hi :) !!

for second integral i think that you want : $ \int \frac {\sqrt {cos(2x)}}{sin(x)}dx$

If YES than my answer is:

Let $ t \equal{} \sqrt {cos(2x)}$

than the integral become :

$ \int \frac {\sqrt {2}t^2}{(1 \minus{} t^2)\sqrt {1 \plus{} t^2}}dt \equal{} \sqrt {2} \int \frac { \minus{} dt}{\sqrt {1 \plus{} t^2}} \plus{} \sqrt {2} \int \frac {dt}{(1 \minus{} t^2)\sqrt {1 \plus{} t^2}}$

that's Not difficult to calcul that ...

if you want the finaly result i'm here to give it ;)

and thanks :)[/quote]

The Last part of the integral , i am not able to do.

Hi :) !!!

what integral ???

you mean that: $ \int \frac{dt}{(1\minus{}t^2)\sqrt{1\plus{}t^2}}$ ????

[quote="mathema*"]Hi :) !!!

what integral ???

you mean that: $ \int \frac {dt}{(1 \minus{} t^2)\sqrt {1 \plus{} t^2}}$ ????[/quote]
Yes , or maybe i am making a calculation mistake.

Hi :) !!!

you can see that : $ \frac{1}{(1-t^2) \sqrt{1+t^2}} = \frac{1}{2} \left( \frac{\sqrt{1+t^2}}{1-t^2} + \frac{1}{\sqrt{1+t^2}} \right)$

or you can let it as here fome after with a changing of varaible and a big calculs we find that:

$ \int \frac{dt}{(1-t^2) \sqrt{1+t^2}} = \frac{1}{\sqrt{2}} \left( ln \left( \sqrt{\frac{t+1}{t-1}} \right) + ln \left( \sqrt{ \frac{1 + \sqrt{2t^2 + 2 }+t}{1 + \sqrt{2t^2+2}-t} \right)} \right) +Cte$

and thx :)


How many solutions $(m,n)$ does

$m^3+n^3+99mn=33^3$
have.

What is the domain of $m$ and $n$?

[hide]From the factorization $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$, use $x=m,\ y=n,\ z=-33$.

The equation becomes $(m+n-33)(m^2+n^2+33^2+33m+33n-mn)=0$. The first factor gives $m+n=33$.

If $m+n\neq 33$, then $m^2+n^2+33^2+33m+33n-mn=0$. Multiplying by 2 and forming squares, $(m-n)^2+(m+33)^2+(n+33)^2=0$. The only real solution to this is $m=n=-33$ by trivial inequality, but there are other complex solutions.[/hide]

[hide="to get number of solutions"]
For $m+n=33$, either $m$ or $n$ must be positive because if $m,\ n\leq 0$, then $m+n\leq 0$. But if one were positive and the other were negative, then $mn<0$ which is outside domain. Then $m$ and $n$ are nonnegative. The equation $m+n=33$ has 34 solutions in nonnegative integers, given by (0,33) (1,32) (2,31) ... (33,0).

For the second factor, the only real (and therefore only integer) solution is (-33,-33). Since $(-33)^2>0$ this is in domain. So there are a total of $\boxed{35}$ solutions.
[/hide]

[quote="maokid7"]could you plese further clearify. i still do not understand.[/quote]
The key point here, (whch makes the problem interesting) is:
[quote]$m$ and $n$ are integers such that $mn\ge{0}$[/quote]
(may be maokid can  edit the  very first post, and include this  edit there)
Anyway,  As suggestd above, You want solutions of  $(m+n-33)(1/2)((m-n)^2+(m+33)^2+(n+33)^2)=0$
[hide][b]Assuming $m$ and $n$ are integers.[/b]
Case I: BOTH, m and n are NON-NEGATIVE.
    There are 34 solutions $(0,33),(1,32),(2,31) \dots (33,0)$  where $m+n=33$ [hide]
Because  $(m-n)^2+(m+33)^2+(n+33)^2$  will be greater than zero (remember  its sum of squares) [/hide]
Case II: BOTH, m and n  are negative;
Here m+n-33 can not be zero, but $(m-n)^2+(m+33)^2+(n+33)^2$ is zero when $m=n=-33$
Rest of the cases where one  number  is nonnegaive, and other  is negative will  give $mn<0$
Hope this helps. [/hide]

what i still do not get is why $m+n$ must equal $33$

ooh silly me..i see why now.
if $s=m+n$ then $s^3=m^3+n^3+3(m+n)(mn)$
and since $m+n=33$ u just substitue that in and u have your answer.

thanks i finally see it now.  :)


A nice problem:

Let every variable/constant that occurs here be an integer.

Let $A(a,b)$ mean that some property is true for the pair $(a,b)$. We know two rules about this property:
[b]a)[/b] $A(x,y) \implies A(mx,ny)$ for any $m,n$.
[b]b1)[/b] $A(x,z)$ and $A(y,z)$ together imply $A(x+y,z)$.
[b]b2)[/b] $A(z,x)$ and $A(z,y)$ together imply $A(z,x+y)$.

We know for some $k$ that $A(k-1 , k-1)$, $A(k,k)$ and $A(k+1 , k+1)$. Find the smallest positive $d$ (depending on $k$) for that we can prove $A(d,d)$ with the given rules.

in rule (a), do you mean $\Leftrightarrow$ ?

No, just one direction... The other one would make it trivial, wouldn't it¿  :D

Are you expecting something larger than $k - 1$?  Because the rules don't let you "climb down": once you have some data, you can only generate data with at-least-as-large values of x and y, unless you begin with $k\leq0$.

Can't $m,n$ be negative?

Either way, I probably need a slap in the face for this: [hide]since the second property doesn't change the second variable, it must be an integer multiple of $k-1$,$k$ or $k+1$, and the smallest is of course $k-1$ (unless k=1, then it's k, or k=0, then it's k+1, and for k negative we just do m=n=-1 to get it positive)

I don't think you'd post it if it were this easy though. :?[/hide]

[quote="Peter VDD"]Can't $m,n$ be negative?[/quote]

Even so, you can't climb down in the y coordinate (as you note), and the question seems silly if we're going to consider negative numbers (since, once we have one solution, we can then generate arbitrarily big-negative solutions).

Sorry, the rule b) exists also in the symmetric version  :blush:

But $m,n$ are any integers, they are not forced to be positive; but $d$ has to be positive.

[hide="I think the answer is..."]...that $d=\gcd(k^2-1,k^2-k,k^2+k)$. :)[/hide]

[hide="That should be..."]correct (but can be simplified)[/hide]

[hide="Well obviously..."]I'll leave something for the other people out there.
[hide="But if you really want to know..."]$d=1$ if k even, $d=2$ if k odd. (but I tried to avoid case study ;))[/hide][/hide]

[hide="You could..."]write it simpler without cases as $\gcd(k-1,k+1)$; but $d=1$ iff $k$ is even, not inverse ;)
[hide="But..."]all this hiding looks a bit funny :D [/hide][/hide]

[hide="Er..."]You're right, as always. ;)
[hide="And most of all..."]Hide tags r00l. ^^[/hide][/hide]


inequalities
AMC
AIME
USA(J)MO
USAMO
articles
AoPS Books

can you give me some examples of problems involving AM-GM. i know AIME II #3 was one.

There are a lot of them.  The AoPS books have plenty on them.  Check google and I'm sure you'll find some.

Should AM-GM really be in the High School Basics section? It's quite advanced, since we usually don't learn inequalities in high school. I didn't even know what it was until like after I found I made USAMO this year.

A few that I remember from Engel.

[b]Exercise 1:[/b]  Show that $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$.

[hide="Hint"] AM-GM tells us that $ a^2 \plus{} b^2 \ge 2ab$. [/hide]
[b]Exercise 2:[/b]  Let $ x_1, x_2, ... x_n$ be positive reals with product $ 1$.  Show that $ (1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_n) \ge 2^n$.

[hide="Hint (quite a giveaway!)"] AM-GM tells us that $ \frac {1 \plus{} x_k}{2} \ge \sqrt {x_k}$. [/hide]
Here's a fun inequality I managed to solve using AM-GM alone:

Let $ a, b, c \ge 0$.  Prove that $ a^5 \plus{} b^5 \plus{} c^5 \ge 5abc(b^2 \minus{} ac)$.

[quote="Carbon57"]Should AM-GM really be in the High School Basics section? It's quite advanced[/quote]

The basic principle of the two-variable case is strikingly simple.  The $ n$-variable case is a straightforward intuitive leap, especially if you understand the proof by Jensen's.  Certainly the first inequality I cite above is close to (if not quite) HSB material.

Take a look at 2004 USAMO #5: http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2004.pdf.

Also, the Wikipedia article on AM-GM gives one example of finding a minimum using AM-GM: http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Example_application.

I'll bet the AoPS books have tons more interesting examples, however.  :)

[hide="first one"]
by AM-GM,$ \frac{a\plus{}b} {2}\ge \sqrt{ab}$ so $ a^2\plus{}b^2\plus{}2ab \ge 4ab$
 so $ a^2\plus{}b^2 \ge 2ab$. similarly, $ b^2\plus{}c^2 \ge 2bc$ and $ a^2\plus{}c^2 \ge 2ac$. adding gets $ 2(a^2\plus{}b^2\plus{}c^2) \ge 2ab\plus{}2bc\plus{}2ac$, os $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}ac\plus{}bc$[/hide]

thanks to everyone who helped

[quote="abacadaea"]by AM-GM,$ \frac {a \plus{} b} {2}\ge \sqrt {ab}$ so $ a^2 \plus{} b^2 \plus{} 2ab \ge 4ab$
 so $ a^2 \plus{} b^2 \ge 2ab$. [/quote]

You can get this in one step by using AM-GM on $ a^2, b^2$ instead of $ a, b$.

I'm not sure if you even needed AM-GM for that one. We have
\[ (a - b)^2 \ge 0 \\
a^2 + b^2 \ge 2ab.
\] :)

That's exactly the (simplest) proof of AM-GM in two variables.  There's a similar proof for four variables.

For three variables:  $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} \frac {1}{2} \left( a \plus{} b \plus{} c \right) \left( (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \right) \ge 0$.

The generalization to five and more variables is difficult (along these lines); there is a great thread about it I can't find.

Yeah, it's basically,

$ \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n}\ge\sqrt [n]{a_1a_2...a_n}$

Let $ f(a,b,c) \equal{} a^5 \plus{} b^5 \plus{} c^5 \minus{} 5abc(b^2 \minus{} ac)$.

Then $ f(a,b,c) \minus{} f(t,b,t)$ where $ t \equal{} \sqrt {ac}$ is equal to $ a^5 \plus{} c^5 \minus{} 2t^5\ge 0$ by AM-GM on $ a^5$ and $ c^5$.

Thus it suffices to prove that
\[ 2t^5 \plus{} b^5 \minus{} 5t^2b(b^2 \minus{} t^2)\ge 0
\]
If $ t \equal{} 0$, then it is obvious, otherwise, normalize to $ t \equal{} 1$. Then we have a one variable inequality:
\[ b^5 \minus{} 5b^3 \plus{} 5b \plus{} 2\ge 0\iff (b \plus{} 2)(b^2 \minus{} b \minus{} 1)^2\ge 0
\]
of course, I could rephrase this into an AM-GM proof...

[url=http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_9]Here[/url] is a classic problem. Again, there are way too many.

[quote="Altheman"]of course, I could rephrase this into an AM-GM proof...[/quote]

Here's my proof using AM-GM alone:  set $ t \equal{} \frac{10 \sqrt{5} \minus{} 11}{2} \equal{} (\sqrt{5} \minus{} 2)^{ \frac{5}{3} }$.  Then $ 5 \sqrt[5]{t^3} \plus{} 2 \sqrt{5(1 \minus{} 3t)} \equal{} 5$ and

$ a^5 \plus{} tb^5 \plus{} tb^5 \plus{} tb^5 \plus{} c^5 \ge 5\sqrt[5]{t^3} ab^3 c$,
$ (1 \minus{} 3t)b^5 \plus{} 5a^2 bc^2 \ge 2 \sqrt{5(1 \minus{} 3t)} ab^3 c$.

Summing gives the desired result.

Oh man...let $ t\equal{}\sqrt{ac}$

$ a^5 \plus{} c^5\ge 2t^5$

$ (b \plus{} 2t)(b^4) \plus{} t^2(b \plus{} 2t)(b \plus{} t)^2\ge 2b^2(b \plus{} 2t)(b \plus{} t)t$

summing gives the result


How would you prove the theorem on Apollonius Circle in the following page

http://planetmath.org/encyclopedia/ApolloniusCircle.html

Could you explain how to do inversion? Thanks.

Inversion is not necessary.

Consider two fixed points $ B,C$ and a point $ A$ that belows to the locus, that is $\frac {_{AB}}{^{AC}} = k = \text{const}.$ Internal and external bisectors of $ \angle BAC$ meet the line $ BC$ at two fixed points $X,Y,$ since by angle bisector theorem, we have that  $ \frac {_{XB}}{^{XC}} = \frac {_{YB}}{^{YC}} = \frac {_{AB}}{^{AC}} = k.$ Since the internal and the external bisector of $ \angle A$ are perpendicular, then the locus of $ A$ is the circumference with diameter $\overline{XY}.$


[quote="mateivld"]This inequality is false.Take $ a\rightarrow 0$  :P[/quote]

Yes, because if $ b=c\to \infty$ then $ LHS$ will be $ 2$ and $ RHS\to \infty$

So, automatically my question about the ineq is wrong :)

[quote="BaBaK Ghalebi"]but the following one is true:
$ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq \frac{x+y+z}{\sqrt[3]{xyz}}$[/quote]
$ \sqrt[3]{xyz}\geq \frac{3xyz}{xy+yz+zx}$ by $ GM-HM$. Then, after making some computations, we need to prove

$ 2(x^{2}z+y^{2}x+z^{2}y)\geq x^{2}y+y^{2}z+z^{2}x+3xyz$ which is easier

[quote="BaBaK Ghalebi"]but the following one is true:
$ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq \frac{x+y+z}{\sqrt[3]{xyz}}$[/quote]

Yes i know this is true but i think this has no relation with our original problem. Because i tried to solve it using all i know, but with no luck. :(

[quote="Zamfirmihai"]
$ 2(x^{2}z+y^{2}x+z^{2}y)\geq x^{2}y+y^{2}z+z^{2}x+3xyz$ which is easier[/quote]

Wrong. Try $ x=3$, $ y=1$ and $ z \rightarrow 0$

For Babak's inequality use $ 2\frac{x}{y}+\frac{y}{z}\geq  \frac{x}{\sqrt[3]{xyz}}$, by AM-GM.

[quote="pohoatza"][quote="Zamfirmihai"]
$ 2(x^{2}z+y^{2}x+z^{2}y)\geq x^{2}y+y^{2}z+z^{2}x+3xyz$ which is easier[/quote][/quote]

 :blush: Thanks for correcting me, I had a mistake at the end..

[quote="sinajackson"][quote="nayel"]Let $ a,b,c>0$. Prove that
\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{a+b+c}{2\sqrt[3]{abc}}\]
[/quote]

Now if we put $ b+c=x$, $ c+a=y$ and $ a+b=z$ then:

$ a=\frac{z+y-x}{2},b=\frac{x+z-y}{2},c=\frac{x+y-z}{2}$

and nayel's inequality will be:

$ \frac{x+y-z}{z}+\frac{y+z-x}{x}+\frac{z+x-y}{y}\geq\frac{x+y+z}{\sqrt[3]{(x+y-z)(y+z-x)(z+x-y)}}$

i know it might have got a little more complicated now but maybe we can use Babak's inequality here to solve it....! :wink:[/quote]

No, we still can't solve it using that inequality! :P

there are two solutions for the problem that I gave:
[hide]
$ \boxed{\text{solution 1: }}$

we have:

$ RHS=\frac{x+y+z}{\sqrt[3]{xyz}}=\sqrt[3]{\frac{x^{3}}{xyz}}+\sqrt[3]{\frac{y^{3}}{xyz}}+\sqrt[3]{\frac{z^{3}}{xyz}}$

$ =\sqrt[3]{\frac{x^{2}}{yz}}+\sqrt[3]{\frac{y^{2}}{zx}}+\sqrt[3]{\frac{z^{2}}{xy}}$

now put $ a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ now we have to prove that:

$ a+b+c\geq \sqrt[3]{a^{2}b}+\sqrt[3]{b^{2}c}+\sqrt[3]{c^{2}a}$ which is true by rearangement inequality ($ m^{3}+n^{3}+t^{3}\geq m^{2}n+n^{2}t+t^{2}m$)...

$ \boxed{\text{solution 2: }}$

WLOG we can assume that $ xyz=1$ so we have to prove that:

$ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq x+y+z$

now by $ AM-GM$ we have that:

$ \frac{x}{y}+\frac{z}{x}+\frac{z}{x}\geq 3\sqrt[3]{\frac{z^{2}}{xy}}=3\sqrt[3]{\frac{z^{2}.z}{xy.z}}=3\sqrt[3]{z^{3}}=3z$

and similary we get that:

$ \frac{y}{z}+\frac{x}{y}+\frac{x}{y}\geq 3x$

$ \frac{z}{x}+\frac{y}{z}+\frac{y}{z}\geq 3y$

now adding the above three inequalities will give us the desired resault...

(I think the solution 2 is what [b]pohoatza[/b] meant...)[/hide]

can someone solve my problem? :maybe: if anyone finds a solution, please post it. i will be really greatful. :)

EDIT: thank you very much BaBaK.  :blush:

nayel as discussed above,your inequality is wrong,take $ a=\frac{1}{100},b=c=1$  :wink:


[quote="mkn19"]How many numbers less than 1387 (or if you can more generaly n) can be written as a power of three or a sum of some powers of three?[/quote]

If powers may be identical (and nothing in your problem forbids this) : $ n$ numbers (including $ 0$, so $ n\minus{}1$ excluding $ 0$).
For $ n\equal{}1387$, this gives $ 1387$ numbers (including $ 0$, so $ 1386$ excluding $ 0$).

If powers must be distinct : all numbers whose representation in base $ 3$ contains only digits $ 0$ and $ 1$

$ 1387_{10}\equal{}1220101_3$ so $ 2^7\equal{}128$ numbers (including $ 0$, so $ 127$ excluding $ 0$)


Al walks down to the bottom of an escalator that is moving up and he counts 150 steps.  His friend, Bob, walks up to the top of the escalator and counts 75 steps.  If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time?  (Assume that this value is constant.)

[hide="Solution"]We want to form two equations with the information given. If we let $m$ be the number of steps on the escalator and $n$ be the number of steps that are shuffled in the time Bob takes 25 steps, $m+2n=150$ and $m-3n=75$. Solving these linear equations yields $m=120$.[/hide]

If we let the number of steps be $ x$ and the speed of the escalator $ e$ and the speed of Bob $ b$, then the time it takes Bob to finish the trip would be $ x/(e \plus{} b)$. Since Bob moved through 75 steps and his speed for each step is $ b$ then the time is also equal to $ 75/b$. Similarly for Al we have $ x/(3b \minus{} e) \equal{} 150/(3b)$
$ x/(e \plus{} b) \equal{} 75/b\implies xb/(e \plus{} b) \equal{} 75$
$ x/(3b \minus{} e) \equal{} 150/3b\implies$
$ 3bx/(3b \minus{} e) \equal{} 150 \equal{} 2xb/(e \plus{} b) \equal{} (3bx \plus{} 2bx)/(3b \minus{} e \plus{} e \plus{} b) \equal{} 5bx/(4b) \equal{} 150$
$ x \equal{} 150*4/5 \equal{} 30*4 \equal{} 120$

Is this correct?
I checked wiki but it said the time for Bob to complete the trip should be $ 75/e$, but I disagree since it was Bob who walked through the steps and therefore we should account his walking speed for the 75 steps he walked (instead of escalator's speed)... so do you guys think there's a mistake in the wiki (or am I missing something)???
(wiki also got 120 as the ans btw)

EDIT: Here's the link [url]http://www.artofproblemsolving.com/Wiki/index.php/1987_AIME_Problems/Problem_10[/url]

/bump could someone answer this pls? i am still v confused thx

lcm(m, n) * gcd(m, n) = mn
think that should help idk


algebra
polynomial
function
algebra unsolved

Let $ P(x)$ be a fourth degree polynomial with four positive real roots. Prove that equation
\[ (1\minus{}4x) \frac{P(x)\minus{}P'(x)}{x^2} \plus{} P'(x)\minus{}P''(x)\equal{}0\]
has four positive real roots too.

Does the problem holds with condition, that roots of $ P$ and roots of equation are real?

[quote="micf"]Let $ P(x)$ be a fourth degree polynomial with four positive real roots. Prove that equation
\[ (1 \minus{} 4x) \frac {P(x) \minus{} P'(x)}{x^2} \plus{} P'(x) \minus{} P''(x) \equal{} 0\]
has four positive real roots too.[/quote]

Let $ f(x)\equal{}P(x)e^{\minus{}x}$ : this function has $ 4$ positive real zeroes and its limit is zero at $ \plus{}\infty$. So $ f'(x)\equal{}(P'(x)\minus{}P(x))e^{\minus{}x}$ has at least $ 4$ positive real roots.

So $ P(x)\minus{}P'(x)$ has at least $ 4$ positive real roots. Since this is a fourth degree polynomial, it has exactly $ 4$ positive real roots.

Let then $ g(x)\equal{}(P(x)\minus{}P'(x))\frac{e^{\minus{}\frac 1x}}{x^4}$ : this function has $ 4$ positive real zeroes and its limit is zero at $ \plus{}\infty$. So $ g'(x)\equal{}(1\minus{}4x)(P(x)\minus{}P'(x))\plus{}x^2(P'(x)\minus{}P''(x))\frac{e^{\minus{}\frac 1x}}{x^6}$ has at least $ 4$ positive real roots.

So $ (1\minus{}4x)(P(x)\minus{}P'(x))\plus{}x^2(P'(x)\minus{}P''(x))$ has at least $ 4$ positive real roots. Since this is a fourth degree polynomial, it has exactly $ 4$ positive real roots.

Hence the result.

[quote="micf"]Does the problem holds with condition, that roots of $ P$ and roots of equation are real?[/quote]

I dont undertstand at all what you mean.

[quote="micf"]I mean this:

[color=blue]Let $ P(x)$ be a fourth degree polynomial with four real roots. Prove that equation
\[ (1 \minus{} 4x) \frac {P(x) \minus{} P'(x)}{x^2} \plus{} P'(x) \minus{} P''(x) \equal{} 0\]
has four real roots too.[/color]

And obviously thanks for replying  :D[/quote]

Are you joking ?

I understood this phrase, since I answered (and, btw, repeating exactly the same text without any change would not help me to understand).

What I did not understand is the second question : "Does the problem holds with condition, that roots of $ P$ and roots of equation are real?"

I thought it was clear in my post.
Obviously I was wrong :(


At an awards ceremony 280 personalized certificates were given out in the categories and quantities shown here.  Julie earned certificates in both fine arts and mathematics.  The principal grabbed a certificate out of the stack at random after all of the science and social studies certificates were given out.  What is the probability that he pulled out one of Julie's certificates?  Express your answer as a common fraction.

\begin{tabular}{|c|c|}
  \hline
  \textbf{Type of Award} & \textbf{Number of Awards} \\
  \hline
  Mathematics &\end{tabular}35$ \\
  \hline
  Science &$38$ \\
  \hline
  Social Studies &$29$ \\
  \hline
  Business &$40$ \\
  \hline
  Physical Education &$23$ \\
  \hline
  English &$25$ \\
  \hline
  Fine Arts &$30$ \\
  \hline
  Foreign Language &$60$ \\
  \hline
  \end{tabular}$

[quote="GameBot"]At an awards ceremony 280 personalized certificates were given out in the categories and quantities shown here.  Julie earned certificates in both fine arts and mathematics.  The principal grabbed a certificate out of the stack at random after all of the science and social studies certificates were given out.  What is the probability that he pulled out one of Julie's certificates?  Express your answer as a common fraction.

[code]$\begin{tabular}{|c|c|} \hline \textbf{Type of Award} & \textbf{Number of Awards} \\
\hline Mathematics &$35$\\
\hline Science &$38$\\
\hline Social Studies &$29$\\
\hline Business &$40$\\
\hline Physical Education &$23$\\
\hline English &$25$\\
\hline Fine Arts &$30$\\
\hline Foreign Language &$60$\\
\hline \end{tabular}$[/code][/quote]

35 math
38 science
29 social studies
40 business
23 physical education
25 english
30 fine arts
60 foreign language

so 2/(35+40+23+25+30+60)=[b]2/113[/b].


induction
ceiling function
logarithms
algebra proposed
algebra

Given an integer $ m\geq 2$, and two real numbers $ a,b$ with $ a > 0$ and $ b\neq 0$. The sequence $ \{x_n\}$ is such that $ x_1 \equal{} b$ and $ x_{n \plus{} 1} \equal{} ax^{m}_{n} \plus{} b$, $ n \equal{} 1,2,...$. Prove that
(1)when $ b < 0$ and m is even, the sequence is bounded if and only if $ ab^{m \minus{} 1}\geq \minus{} 2$;
(2)when $ b < 0$ and m is odd, or when $ b > 0$ the sequence is bounded if and only if $ ab^{m \minus{} 1}\geq\frac {(m \minus{} 1)^{m \minus{} 1}}{m^m}$.

[b]For Part 1:[/b] let us divide the whole sequence by b, call it $ {y_n}$. Clearly $ {y_n}$ is bounded iff $ {x_n}$ is bounded. The first few terms of $ {y_n}$ are as follows:
$ 1$
$ ab^{m \minus{} 1} \plus{} 1$
$ ab^{m \minus{} 1}(ab^{m \minus{} 1} \plus{} 1)^m \plus{} 1$
...and so on. Figure out the recursive definition of $ {y_n}$ yourself.

[b]($ \Leftarrow$)[/b] Note since $ a > 0$ and $ b < 0$, it follows that $ ab^{m \minus{} 1} < 0$. Assuming $ ab^{m \minus{} 1}\geq \minus{} 2$, then $ \minus{} 1 < ab^{m \minus{} 1} \plus{} 1 < 0$.

Then we will prove by induction on $ n$ that every $ y_n$ lies in $ [0,1]$. Base case is $ y_1 \equal{} 1$.

Inductive step: Assume $ y_n \in [0,1]$. Since $ m \geq 2$, $ (y_n)^m \in [0,1]$ as well. Since $ \minus{} 1 < ab^{m \minus{} 1} < 0$, $ ab^{m \minus{} 1}(y_n)^m \in [ \minus{} 1,0]$. So $ y_{n \plus{} 1} \equal{} ab^{m \minus{} 1}(y_n)^m \plus{} 1 \in [0,1]$.

By mathematical induction it follows that $ {y_n}$ is bounded by $ [0,1]$, and we are done.

[b]($ \Rightarrow$)[/b] We will prove the contrapositive. Assume $ ab^{m \minus{} 1} < \minus{} 2$, then $ ab^{m \minus{} 1} \plus{} 1 < \minus{} 1$.

Let us prove by induction that $ y_n < \minus{} 1$ for all $ n > 1$. Base case ($ n \equal{} 2$) is already on the previous line.

Inductive step: Assume $ y_n < \minus{} 1$. Since $ m \geq 2$ and $ m$ is even, $ (y_n)^m > 1$.

It follows that $ y_{n \plus{} 1} \equal{} ab^{m \minus{} 1}(y_n)^m \plus{} 1 < ab^{m \minus{} 1} \cdot 1 \plus{} 1 < \minus{} 1$.

By mathematical induction it follows that $ y_n < \minus{} 1$ for all $ n > 1$.

Now let $ N$ be a negative real number. We will show that there exists $ n \in \mathbb{N}$ such that $ y_n < N$.

In fact, just let $ n \equal{} \left\lceil \dfrac{\log_{ab^{m \minus{} 1} \plus{} 1}(N)}{m} \right\rceil$. Easy to check that it works.

(This $ n$ is a little too big- it guarantees that $ (y_{n \minus{} 1})^m$ will be less than $ N$.)

Therefore $ {y_n}$ is unbounded if $ ab^{m \minus{} 1} < \minus{} 2$. We are finally done!
[i]
Notes:  Maybe we could log the whole sequence! This might give a neater solution but I can't see how to use the $ ab^{m\minus{}1} \geq 2$ condition. The above solution is mechanical but immediate, and I guess the latter matters more in a competition.[/i]

[b]Part 2:[/b] Next time perhaps.

[quote="limsk1"]I guess problem (2) is wrong.

try $ m\equal{}3, a\equal{}3, b\equal{}\minus{}1$.[/quote]
The Chinese version is"<=",not">="

Assuming that (2) means that the sequence is bounded iff \( ab^{m-1} \leq \frac{(m-1)^{m-1}}{m^m} \).

[b]Lemma[/b]: \( \{ x_n \} \) is bounded iff \( ar^m + b = r \) has a positive real root.
[i]Proof[/i]: \( (\Leftarrow) \) Let the root be \( \alpha \). Then we show via induction that \( \{ x_n \} \) is bounded above by \( \alpha \). Clearly by definition as \( b < 0 \) and \( \alpha > 0 \), \( x_1 < \alpha \). Suppose \( x_k < \alpha \). Then \( x_{k+1} = ax_k^m + b < a \alpha^m + b = \alpha \), as desired.
\(( \Rightarrow) \) Suppose otherwise. Then consider the minimum positive value that \( ar^m + b - r \) can take; call it \( \beta \) for convenience. However, remark then that the difference between two consecutive terms in our sequence, \( x_{k+1} - x_k \) is equal to \( ax_k^m + b - x_k \geq \beta > 0 \), which means that \( \{x_n \} \) is unbounded, contradiction.

By the lemma, let us rearrange our equality \( ar^m + b = r \) as \( ar^{m-1} + \frac{b}{r} = 1 \)
To finish off, we apply AM-GM (the form of which is motivated by the bound in question) to the aforementioned, namely
\( ar^{m-1} + \frac{b}{r} = ar^{m-1} +\underbrace{ \frac{b}{(m-1)r} + ... + \frac{b}{(m-1)r}}_{m - 1 \ \text{times}} \geq m \sqrt[m]{ \frac{ab^{m-1}}{(m-1)^{m-1}}} \), which rearranges to the desired.


AMC
USA(J)MO
USAMO
AIME
function
AMC 10
AMC 10 A

Here is what I have as the answers for the 2005 AIME1. (Unconfirmed)

Explanations: http://students.imsa.edu/~albertni/AIME%202005/AIME1%202005%20Solutions.pdf

1. 942
2. 012
3. 109
4. 294
5. 630
6. 045
7. 150
8. 113
9. 074
10. 047
11. 544
12. 025
13. 083
14. 936
15. 038

Any questions, comments, or CORRECTIONS would be greatly appreciated. I feel the solution to Problem 14 is incredibly clever (14 was the one problem on the test I simply could not figure out how to do).

EDIT: I have updated the solutions with an alternate solution for problem 14 (one that is probably more accessible). Additionally, I now consider all answers to be confirmed.

Looks like I got 8 right then. *hopes its enough for USAMO... but isnt overly sure*

Meh... now I see what I did wrong on 13. Horribly over counted, oh well. And I knew 15 wasnt as simple as I took it to be, I just miss read the quetion  :D

[quote="darktreb"][quote]How did youi brute force 9?[/quote]

How do you [i]not[/i] brute force 9?[/quote]

Maybe I've just got my definition of brute force wrong then. Well, I guess it's like brute forcing. Yeah...

Consider your answers confirmed. :-) (Sorry, I'd already posted a set on another thread.)

Darktreb, if you got all of those during the test, congrats on a 15. :-)

Well, I can't say I got all of those answers...but I can confirm a few of them. I wrote C++ programs that did #2,#3,and #12, and those are indeed correct, except for 3, on which the program got 2 for the answer... Also-get this-I guessed a random number from 0-999 on 6-and I got 45!!!! What kind of luck is that? :D Oh well, I guess a 6-7 is pretty good for a 13 year old...No. I must not end my post with a triple period.

[quote="xxreddevilzxx"]I brute forced #4, #9 and #12.... i was OFF BY 1 on 12... UGGH[/quote]
I was off by two...WAAAAAAAA, that might (cough, cough) be the difference between a trip to the USAMO (well, a trip to the town's high school) and staying at home. Oh well, there are still four years ahead of me.

[code]]
// AIME2005Inumber2.cpp : Defines the entry point for the console application.
// This is the program to solve #2 on the 2005 AIME I

#include "stdafx.h"
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
	int total=0;
	for(int k=1;k<2006;k++)
	{
		for(int n=1;n<2100;n=n+k)
		{
			if (n==2005)
			{
				cout << k << endl;
				total++;
			}
		}
	}
	cout << "Total: " << total << endl;
	int stop;
	cout << "Hit any key, then Return/Enter to end" << endl;
	cin >> stop;

	return 0;
}

[/code]
[code]
// Primefactorizer.cpp : Defines the entry point for the console application.
// This is the program to solve number 12 on the 2005 AIME I
#include <stdafx.h>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;

int base[100];
int power[100];
int nopfacs=0;
int facs=1;
int number;
int totalfacs=0;
int even=0;
int odd=0;

int primefinder(int prime);
void pfactor(int no);

int _tmain(int argc, _TCHAR* argv[])
{

	for(int g=1;g<=2005;g++)
	{
		facs=1;

	for (int n=0;n<=100;n++)
	{
		base[n]=1;
		power[n]=0;
	}
		pfactor(g);
		for(int t=1;t<=nopfacs;t++)
		{
			facs=facs*(power[t]+1);
		}
		totalfacs=totalfacs+facs;
		if(totalfacs%2==0) even++;
		else odd++;
	}

	cout << abs(even-odd);

	return 0;
}

int primefinder(int prime)
{
	int prm=1;
	float fprime=prime;
	for (int k=2;k<=sqrt(fprime);k++)
	{
		if(prime%k==0)
		{
			prm=0;
			break;
		}
	}

	return prm;

}

void pfactor(int no)
{
	int oldno=no;
for(int r=2;r<=(oldno);r++)
{
	if(!oldno/2<r<oldno)
	{
	if(primefinder(r)==1)
	{
		if (no%r==0)
		{
			//insert body of function here: what to do if number is divisible by second number
			nopfacs++;
			base[nopfacs]=r;
			power[nopfacs]=1;
			no=no/r;
				while(no%r==0)
				{
					no=no/r;
					power[nopfacs]++;
				}
		}

	}
}
}

}

[/code]

[code]
// Primefactorizer.cpp : Defines the entry point for the console application.
// This is the program to solve #3 on the 2005 AIME I
#include <stdafx.h>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;

int base[100];
int power[100];
int nopfacs=0;
int facs=1;
int number;

int primefinder(int prime);
void pfactor(int no);

int _tmain(int argc, _TCHAR* argv[])
{

	for(int g=1;g<=2500;g++)
	{
		facs=1;

	for (int n=0;n<=100;n++)
	{
		base[n]=1;
		power[n]=0;
	}

		pfactor(g);
		for(int t=1;t<=nopfacs;t++)
		{
			facs=facs*(power[t]+1);
		}
		if (facs==5) cout << g << endl;

	}

	return 0;
}

int primefinder(int prime)
{
	int prm=1;
	float fprime=prime;
	for (int k=2;k<=sqrt(fprime);k++)
	{
		if(prime%k==0)
		{
			prm=0;
			break;
		}
	}

	return prm;

}

void pfactor(int no)
{
	int oldno=no;
for(int r=2;r<=(oldno);r++)
{
	if(!oldno/2<r<oldno)
	{
	if(primefinder(r)==1)
	{
		if (no%r==0)
		{
			//insert body of function here: what to do if number is divisible by second number
			nopfacs++;
			base[nopfacs]=r;
			power[nopfacs]=1;
			no=no/r;
				while(no%r==0)
				{
					no=no/r;
					power[nopfacs]++;
				}
		}

	}
}
}

}

[/code]

#2: This is simply the number of factors of 2005-1 = 2004.  12.
#4: a^2 + 5 = b^2 + 7b
b^2 + 7b - a^2 - 5 = 0
a = (-7 +/- sqrt(49 + 4a^2 + 20))/2 = (-7 +/- sqrt(69 + (2a)^2))/2
So (2a)^2 + 69 must be a perfect square.  Since 69 is odd, we know that
1+3+5+... + 69 = 35^2 is a perfect square and thus 2a =34 and a=17.  17^2 + 5 = 294

And I have to go now, but I'll post some more later :)

I actually found a solution to #4. Which was really ridiculous. I went home and found out that I got #3 wrong. Damit

[quote="yif man12"][quote="hockeyadi23"][quote="yif man12"]I got the same answers as those for 1-12.

Hehe, I did 4 by brute forcing it. Luckily, it didn't take too long.[/quote]
Aren't you taking the alternates? And what did you get last year?[/quote]

I am taking the alternate. My friend gave me the problems and I did them for practice.

Last year I got a 105 on the AMC10A because of 4 dumb mistakes and didn't take the AIME. Got a 3 in 7th grade though.[/quote]
How did you take the AIMEs in 7th grade? I thought there was no way a sevy coult take the AMC 12/10s...

I scored high enough on the AMC8 and the school let me take the 10.

[quote="filletwho"]In that case, I think I hardly ever use brute force( anymore) unless i'm checking a solution and don't want to risk getting it wrong.[/quote]

That's a good thing, a very good thing. I still resort to brute force sometimes.

It's not always a bad thing to try to brute force a problem.  Sometimes you will realize along the way an elegant approach to the problem.  Just don't make brute force the first thing you turn to in a problem ;) .

[quote="joml88"]Just don't make brute force the first thing you turn to in a problem ;) .[/quote]

Unless of course the brute force is less brutal than the normal solution. :-) Too bad "brute force" is not an acceptable solution for #6 on the USAMO. Heck, for that matter it's a shame no points are awarded for ingenius statements like "for the remainder of this solution, please see <insert 2004 USAMO non-senior Winner>'s paper".

[quote]Heck, for that matter it's a shame no points are awarded for ingenius statements like "for the remainder of this solution, please see <insert 2004 USAMO non-senior Winner>'s paper".[/quote]

Too ingenious for the graders, sorry. Less ingenuity please.

[quote]Just don't make brute force the first thing you turn to in a problem[/quote]

I agree completely. I think in almost all the problems I do I may try to brute force something somewhere along the line. It's not always a bad thing, and has never greatly hindered my results on contests. It's just working on the fine balance of the timing that's important.

[quote="xxreddevilzxx"]i hate brute forcing... i t takes  too long.[/quote]

But on this AIME, many of the solutions required less time for you to brute force than to actually find the elegant solution to...(#13 anyone?)

[quote]I tried to find an elegant solution to 13 based on combinations.[/quote]

I haven't seen or heard about any sort of truly elegant solution to #13.

Well, some are definitely better than others. I'd say the one from the Math Jam at least requires a little bit of thought whereas just listing out 83 paths by cases (and yes, I do know of ppl that did that) probably didn't.

[quote="ASquare"]  How could a Speed Cuber miss that???[/quote]
Off-topic: there's no correlation between ability to [b]speed[/b]cube and to problem solve. :lol:

Yeah, but the cube was basically a Rubik's cube.

I was under the impression speed cubers were trained to match the colors to the sides...not calculate the probabilities of their work.

Can someone explain to me IN DETAIL how to to number 13?

Look at the math jam for solutions of all the problems.


function
real analysis
integration
calculus
LaTeX
algebra
partial fractions

If f(w) and h(w) are two Fourier Transforms, is there a formula for f(w).h(w)? (i.e. multiplication of F.T.'s) If not then how does one do that? (please assume both functions are complex enough that partial fractions would make things even worse)

Is there a formula? IS THERE A FORMULA?

What we've got here is, in my eyes, the single most important identity associated with the Fourier transform. In words: [b]convolution[/b] one one side becomes pointwise multiplication on the other.

Conventions about defining the Fourier transform differ - where to put the $2\pi,$ which sign to put in the exponent. Rather than trying to guess which version of notation you're using, I'll just put up my own.

Definition of the Fourier transform on $\mathbb{R}:$

For integrable $f,\,\hat{f}(s)=\int_{-\infty}^{\infty}f(x)e^{-2\pi isx}\,dx$

Under favorable circumstances, we can recover $f$ by the "synthesis equation:"

$f(x)=\int_{-\infty}^{\infty}\hat{f}(s)e^{2\pi isx}\,ds.$

Now we define the convolution of two functions $f$ and $g$ (presumed integrable):

$f*g(x)=\int_{-\infty}^{\infty}f(x-y)g(y)\,dy.$

Comments: convolution is commutative, associative, and distributes over addition.

The theorem:

$\widehat{(f*g)}(s)=\hat{f}(s)\hat{g}(s).$

Oh, and while we're at it, if $h(x)=f(x)g(x),$ then $\hat{h}(s)=\hat{f}*\hat{g}(s).$

Variations of this are also true for the Fourier transform on $\mathbb{T},\,\mathbb{Z},$ and $\mathbb{Z}/(N\mathbb{Z}).$

I appreciate your answer. The problem that I am dealing with requires to find a constant K for which $F[f(t).g(t)] = K(\hat{f}*\hat{g})(\omega)$ (please read $F[f(t).g(t)]$ as the Fourier Transform of $f(t).g(t)$). This is how far I could get (which isn't really much):

$F[f(t).g(t)] = \int_{-\infty}^{\infty}e^{-i\omega t}f(t)g(t)dt$
and
$K(\hat{f}*\hat{g})(\omega) = K/2\pi \int_{-\infty}^{\infty}e^{i\omega t}f(t)g(t)d\omega$

The terms $f(t)g(t)$ on the 2nd line are derived by taking the F.T. of $\hat{f}(\omega).\hat{g}(\omega)$ (which I am not sure of its correctness!)

Please advise. Thanks.

P.S> I hope the difference in notation is not too confusing...

It looks to me like you're using this notation:

$F[f(t)]=\hat{f}(\omega)=\int_{-\infty}^{\infty}e^{-i\omega t}\,dt$ (analysis equation)

$f(t)=\frac1{2\pi}\int_{-\infty}^{\infty}\hat{f}(\omega)e^{i\omega t}\,d\omega$ (synthesis equation)

(We're assuming that $f$ and $\hat{f}$ satisfy whatever regularity conditions might be needed to make the pairing of analysis and synthesis equations - Fourier transform and inverse Fourier transform - valid.)

Now define $(\hat{f}*\hat{g})(\omega)=\int_{-\infty}^{\infty}\hat{f}(\omega-\alpha)\hat{g} (\alpha)\,d\alpha.$

Put that into the synthesis equation above:

$\frac1{2\pi}\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}\hat{f}(\omega-\alpha)\hat{g} (\alpha)\,d\alpha\right]e^{i\omega t}\,d\omega$

(Of course we're going to interchange the order of integration; that's what Fourier analysts do.)

$=\frac1{2\pi}\int_{-\infty}^{\infty}\hat{g}(\alpha)\left[\int_{-\infty}^{\infty}\hat{f} (\omega-\alpha)e^{i\omega t}\,d\omega\right]\,d\alpha$

Make the substitution $\omega'=\omega-\alpha,\,\omega=\omega'+\alpha$ in the inner integral:

$=\frac1{2\pi}\left[\int_{-\infty}^{\infty}\hat{f}(\omega')e^{i\omega't}\,d\omega' \right] \left[\int_{-\infty}^{\infty}\hat{g}(\alpha)e^{i\alpha t}\,d\alpha\right]$

We'd need another $\frac1{2\pi}$ to make that $f(t)g(t).$ Hence we have

$\frac1{2\pi}\int_{-\infty}^{\infty}(\hat{f}*\hat{g})(\omega)e^{i\omega t}\,d\omega = 2\pi f(t)g(t).$

Hence, $F(f(t)g(t))=\frac1{2\pi}(\hat{f}*\hat{g})(\omega)$

(Unless, notationally, you want to include a $\frac1{2\pi}$ in the definition of convolution.)

---

The need to constantly handle $2\pi$ one way or another in this is one reason why I prefer the notation of my own earlier post, with $2\pi$ in the exponent. Think of it as expressing frequency in Hz rather than radians/sec.

[quote="Kent Merryfield"]The need to constantly handle $2\pi$ one way or another in this is one reason why I prefer the notation of my own earlier post, with $2\pi$ in the exponent. Think of it as expressing frequency in Hz rather than radians/sec.[/quote]

That is actually the engineer's convention for writing fourier transforms, but you probably know that.

Thanks for taking the time to answer my question Kent Merryfield. I appreciate it.

[quote="Kent Merryfield"][quote="blahblahblah"]That is actually the engineer's convention for writing fourier transforms, but you probably know that.[/quote]
The advanced books by Eli Stein and Guido Wiess (members of the "family of Zygmund") also put the $2\pi$ in the exponent; those are the sources from which I first learned much of this.[/quote]

I did a quick count of the texts I own; Lax, Rudin, and Strichartz divide out by $\sqrt{2\pi}$, Lieb and Loss, Stein, and Folland put it in the exponent.  But I think Folland was one of Stein's students, so that's not exactly surprising.

As a $\LaTeX$ note: I think $\widehat{f}$ (\widehat{f}) produces nicer results than $\hat{f}$ (\hat{f})


topology
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Let $(X,d)$ be a metric space with the property that every subspace of $X$ consisting of $n+3$ points can be embedded isometrically in the Euclidean space $\mathbb R^{n}$. Prove that $X$ can be embedded isometrically in $\mathbb R^{n}$. Also, show that $n+3$ cannot be replaced with $n+2$.

An example to show that $(n+2)$ is not enough: let $v_{0},\dots,v_{n}$ be the vertices of a regular $n$-simplex $\Delta_{n}$, and let $c$ be its center. Add an abstract point $c'$ by declaring that $d(c',x_{i})=d(c,x_{i})$ and $d(c,c')$ is twice the distance from $c$ to the boundary of $\Delta_{n}$. The space $X=\{v_{0},\dots,v_{n},c,c'\}$ cannot be embedded into $\mathbb R^{n}$ since $c'$ has nowhere to go. But if one of the vertices, say $v_{0}$, is removed, then $c'$ can be placed so that $c$ and $c'$ are symmetric about the affine span of $v_{1},\dots,v_{n}$.


Let $n$ be a positive integer of the form $4k+1$, $k\in \mathbb N$ and $A = \{ a^2 + nb^2 \mid a,b \in \mathbb Z\}$. Prove that there exist integers $x,y$ such that $x^n+y^n \in A$ and $x+y \notin A$.

n=1 positive integer of the form $4k+1$. Prove that $x+y\in A$ and $x+y\not \in A$. :P

I think that in Romania $0$ is not a positive integer so $k\neq0$  :?:

Already posted by me last year...
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Moldavian&t=46004

A little problem with solution given there.
[quote="maxal"]because $u^2 + n*v^2 \equiv 0\pmod p$ has no solutions[/quote]
case $u\equiv v\equiv0$ failes entire solution.


quadratics
function
number theory unsolved
number theory

Find all prime p such that the equation  x2+y2=2003+pz, has solutions in x,y,z element of Z.

Strange as it may be, I found that any prime is a solution. So, take p any prime. Suppose that the equation doesn't have a solution. Then it follows that whenever x is in {0,1,...,p-1}, 2003-x^2 is not a quadratic residue modulo p. Define the function f on {1,2,...,(p-1)/2} which assigns to x the class modulo p of 2003-x^2. This function is clearly injective and its image is included in the set of quadratic non-residues modulo p, which has (p-1)/2 elements. It follows that its image is exactly the set of non-residues modulo p. Since 2003 is also non-residue, it follows that we can find x such that f(x)=modulo p class of 2003. That is x^2 is divisible by p and x is in {1,2,...,(p-1)/2}, false.

I couldn't understand your solution.
By contradiction you assume that there is a prime p such that
2003-x2 =/= y2 (Mod p).
The you define f(x)=2003-x2 (Mod p)
and take x=1,x=2,...x=(p-1)/2.
After? Sorry, I'm bit stupid today... :D

Well, f is clearly injective. Thus, its image has at least (p-1)/2 elements. But we have seen that it's image is contained in the set of quadratic non-residues modulo p, since otherwise we would have the desired conclusion. Since this set of quadratic non-residues has (p-1)/2 elements, we deduce that it is in fact the image of f. But f(0) is also non-residue (otherwise we can find x such that x^2 is 2003 modulo p, and thus we can take y=0) and so it is f(u) for a certain u between 1 and (p-1)/2. Thus, u^2 is divisible ny p, false. Is it clear now?

Small unrelated note: This is the 2nd question I've seen from the Kazakhstan City Olympiad 11th grade which is the same as the 2003 Singapore MO questions...the other question is [url=http://www.mathlinks.ro/viewtopic.php?t=4870]this[/url].

Thanks, I got it now.
Valiowk, where did you see the Singapur questions? Do you have them?


geometry
trapezoid
rotation
geometric transformation
rectangle
symmetry

[quote="mihail911"][b]Prove any line that passes through the center of a square bisects its area[/b][/quote]
There are two possibilities. 1)The line is the diagnol 2)The line is not the diagnol. Suppose #1 then the proof is complete. Suppose #2 then you have 2 trapezoids, now use the formula for computing the area of a trapezoid.

You can also prove that the trapezoids are congruent through SASAS for quadrilaterals

[quote="t0rajir0u"]How about you just rotate one of the trapezoids onto the other?  :)[/quote]
i thought the exact same thing, but you can't really write that as a proof. unless:
"make the center of the square the center of rotations for two congruent trapezoids. Q.E.D"?

Pretty much.  It's perfectly legitimate; a $ 180^{\circ}$ rotation through the center of the square takes either of the trapezoids to the other.

ok, i dont know how to draw pictures so i'll describe the best i can. if you have a square ABCD with a line through the middle, you can mark the intersection points X on AB and Y on CD. Let W be on CD such that WX is perpendicular to CD and Z on AB such that YZ is perpendicular to AB. Draw in the line segments. Now you have two triangles which you can prove congruent using parallel line theorems, and the proof that the rectangles are congruent should be easy from there.

Consider square $ ABCD$. $ \ell$ intersects $ AB$ at $ E$ and $ CD$ at $ F$.

$ EA \equal{} FC\Leftrightarrow EB \equal{} FD\Longrightarrow FA \equal{} EC\Leftrightarrow$
$ \triangle CEB\cong\triangle AFD\Leftrightarrow \triangle FEA\cong\triangle EFC\Longrightarrow$
$ \triangle AFD \plus{} \triangle FEA \equal{} \triangle CEB \plus{} \triangle EFC\Leftrightarrow$
$ AEFD\cong CFEB\Leftrightarrow AEFD\equal{}\frac{ABCD}{2}\equal{}CFEB$

EDIT: typo   :oops:

I'm sure that this also work for all rectangles, circle, and regular n-gons. Can anyone prove this?

Actually, it fails for equilateral triangles.  :)  A sufficient requirement is $ 180^{\circ}$ rotational symmetry, which is satisfied by rectangles, circles, and regular $ 2n$-gons.

So, for regular 2n+1-gons, any line that passes through the center and one of the vertex must bisect the area. Is this correct? How to prove it?

Well, that's trivial; a line that passes through the center of a regular $ 2n\plus{}1$-gon and a vertex is a line of symmetry!


IMO Shortlist
geometry
3D geometry
polyhedron
Euler

Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it. Call a pair of points of the polyhedron [i]antipodal[/i] if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes. Let $A$ be the number of antipodal pairs of vertices, and let $B$ be the number of antipodal pairs of midpoint edges. Determine the difference $A-B$ in terms of the numbers of vertices, edges, and faces.

[i]Proposed by Kei Irei, Japan[/i]

Here is a rather ugly and messy solution based on continuity. The idea is somewhat similar to one of the solutions of IMO 2006/6.

We prove that $ A-B=V-1$ by induction on the number of vertices of the polygon $ P$. Firstly we note that slightly moving the vertices of the polyhedron does not change $ A$ and $ B$, so we may move the polyhedron such that no line connecting two vertices is parallel to a face. Pick up a vertex $ V$ of the polyhedron. Let's construct $ P'$ the unique convex polyhedron containing the vertices of the original polyhedron without $ V$. The faces not containing $ V$ of the original polyhedron will
still be faces of $ P'$, but a couple of new faces and edges may appear. Now let's move $ V$ continuously inwards until it meets a face of $ P'$. The idea is to prove that $ A-B$ does not change while we move it.

Let's analyze how $ A$ changes. Consider a pair of opposite vertices $ X,Y$. Let's seek the planes $ \pi$ such that $ X$ is the highest and $ Y$ is the lowest with respect to $ \pi$. To each plane $ \pi$ not parallel to $ XY$ we may assign the unique ray $ l$ perpendicular to it such that the projection of $ X$ is farther than the projection of $ Y$ on $ l$. Now let's assume that $ l$ is of form $ [OP$ where $ O$ is a fixed point in space and let's find the possible positions of $ P$. We must satisfy the
conditions that $ X$ is above all its neighbors and $ Y$ below all its neighbors. Generally if we want to ensure that $ U$ is above $ V$ we must have $ P$ in the above halfplane determined a plane throug $ O$ perpendicular to $ UV$. So we get a set of regions and $ P$ exists if and only if these regions have non-empty intersection. The intersection of these regions always has the property that if a point $ K$ belongs to them then the whole ray $ [OK$ does. Thus they are in correspondence with plane regions: pick up a plane not passing through $ O$ and replace each region by the intersection of this region with it. What plane is taken actually matters, but what we need now is that we can apply Helly's theorem: all the regions intersect if and only if every three of them do.

Calla plane $ \pi$ separating for $ S,T$ if all the other points are sandwitched between the planes parallel to $ \pi$ through $ S$ and $ T$.

Let's investigate a limit case, i.e when the mobile vertex $ V$ passes through some point $ V_{1}$ and $ A$ changes. It means some pair of opposite points disappears or conversely is created. Say $ S,T$ is modified. It means that if we draw the regions for $ S,T$ as above, for $ V=V_{1}$ some three of them will have a degenerate intersection i.e. a line, and when $ V$ passes through $ V_{1}$, at one side of $ V_{1}$ the intersection will be empty and at other non-empty. As three regions have a common line, it means that three edges through $ S,T$ are coplanar (we mean their directions). Therefore, $ A$ may change only when an edge becomes parallel to a face.

Let's turn to analyzing $ B$. The midpoints of edges $ KL$ and $ MN$ are opposite if and only if $ K,L$ and $ M,N$ are opposite with respect to the plane parallel to $ MN$ and $ KL$. If we pass to parallel lines, it means that corresponding ray $ l$ through $ O$ parallel to $ MN$ and $ KL$ is contained in the non-empty intersection of corresponding regions for $ K$ (or $ L$) and $ M$ (or $ N$). The limit case is when $ l$ falls on one of the planes $ \pi$ bounding the region. It means $ KL, MN$ and the edge perpendicular to that plane are coplanar. Say $ KL,KK',MN$ are coplanar. Again $ B$ changes if and only if an edge becomes parallel to a face.

So we need to analyse the change in $ A$ and $ B$ when an edge $ ST$ becomes parallel to a face $ F=A_{1}A_{2}\ldots A_{n}$. It follows from the reasonings above that $ A$ and $ B$ change only when the plane parallel to the face $ A_{1}A_{2}\ldots A_{n}$ is separating for $ S$(or $ T$) and $ A_{i}$(at the limit point). By continuity it means that in the neighborhood of the limit point a plane close enogh to $ (A_{1}A_{2}\ldots A_{n})$ sandwiches all points except $ S,T,A_{1}, \ldots A_{n}$ between $ S,T$ and $ A_{i}$. So only the relationship between these points matters.

We need to analyse two cases: when $ V$ comes close to the limit point $ V_{1}$ and when it passes through it. Let's analyse each of them.

In the first case, assume that $ S$ was below $ T$ (we take distances from points to $ F$). Then it is clear for any $ A_{i}$ we can take a plane separating $ A_{i}$ and $ T$. However there are problems with $ S$. If we project the vector pointing from $ S$ to $ T$ on $ F$ then there exists a plane separating $ A_{i}$ and $ S$ if and only if the vector drawn from $ A_{i}$ does not point inwards the polygon. So if we have $ k$ consecutive points from where the vector points inwards the polygon and $ l$ onsecutive points from where the vector points outwards, then there are exactly $ l$ opposite pairs of form $ S,A_{i}$. Now if we pass to $ B$, we see that the pair $ ST, A_{i}A_{i+1}$ is separating if and only the same vectors points outwards drawn from both $ A_{i},A_{i+1}$. There are exactly $ l-1$ such edges. So the difference between $ A$ and $ B$ is $ n-1+C$ (where $ C$ is the difference between pairs of opposite vertices and edges except the ones counted here, as we know $ C$ is unchanged).

It is rather clear by analogy that after passing through the limit point the difference will stiil be $ n-1+C$. (we could still have $ S$ below $ T$ and have exactly the same pairs or have $ T$ below $ S$ and now $ k$ would exchange with $ l$).

We have thus established that moving $ V$ does not affect $ A-B$. It thus remains to see what happens when $ V$ gets close enough to a face $ F$. If we take $ U$ the vertex farthest from $ F$, clearly that $ U,V$ will be the only pair of opposite
vertices containing $ V$.

Now let's remove $ V$. We claim no pair of opposite vertices  appears. Indeed let $ IJ$ be a "new" pair. Then there is a plane $ \pi$ such that all other vertices are between $ I$ and $ J$ according to the plane. Then taking $ V$ close enough to  the face $ F$ will ensure that $ V$ is also between $ I$ and $ J$ so $ IJ$ is actually an "old" pair, contradiction. Also no pair of opposite vertices except $ UV$ disappears. Indeed if $ I,J$ was an "old" pair then all vertices were sandwiched between $ I,J$ and this condition holds by removing $ F$. Hence $ A$ decreases by $ 1$. We now show that $ B$ remains unchanged.

If a pair of opposite edges disappears, it must be a pair $ VI, KL$. But this is impossible as $ V$ tends to a point strictly inside the face $ F$ so $ V$ must be below at least some vertex of the face. If a pair of opposite edges appears, then it must be a pair of form $ MN, KL$. Then all the vertices of the polygon must be between $ MN$ and $ KL$ wrt the plane parallel to $ MN$ and $ KL$, except $ V$. However $ V$ tends to be on the face $ F$, which implies that the plane parallel to $ MN$ and $ KL$ is parallel to the face $ F$. Therefore $ M,N,K,L$ are on the same face $ F$. But this is clearly impossible.

So $ A-B$ has decreased by 1 for the new polyhedron $ P'$. If has one vertex less, so applying the induction hypothesis ensures

the claim. (Note that for a tetrahedron the condition holds).

Denote the number of faces, edges and vertices of our polyhedron $ P$ by $ F,E,V$.

For any face $ \Phi_{i}$ ($ i=1,2,\dots,F$) of $ P$ consider the point $ A_{i}$ on a unit sphere such that $ OA_{i}$ is a vector of outer unit normal to the face $ \Phi_{i}$. For any edge $ e$ of $ P$, which is common to faces $ F_{i}$ and $ F_{j}$, draw an arc on a sphere, which joins points $ A_{i}$ and $ A_{j}$. This arc is the set of such points $ X$ on a sphere, that a plane with "outer" normal $ OX$, which passes through the edge $ e$, does not contain inner points of $ F$. We get a graph $ G$ on a sphere, which is dual to the graph of initial polyhedron $ P$. So, $ G$ has $ V$ faces, $ E$ edges and $ F$ vertices.

Draw the graph $ -G$ on a sphere, which is symmetric to $ G$ with respect the origin. The union of edges of graphs $ G$ and $ -G$ is some graph on a sphere, we add the points, in which edges of $ G$ and $ -G$ intersect, to the set of vertices of this graph (call it $ G'$). Note that the number of added vertices equals to $ 2B$. The number of edges of $ G'$ equals $ 2E+4B$, the number of vertices equals $ 2F+2B$, the number of faces equals $ 2A$ (any face of $ G'$ is an intersection of faces of $ G$ and $ -G$. It corresponds to an antipodal pair of vertices). So, by Euler formula,

$ 2A-2E-4B+2F+2B=2$, $ A-B=1+E-F=V-1$.

[hide]
Let, $E$ be the number of edges, $V$ the number of vertices, $F$ the number of faces.

Some informal notation: let $\#(V,V)$ be the number of antipodal ordered pairs of vertices, $\#(E, V)$ the number of antipodal ordered pairs of an edge and a vertex (i.e. some plane containing the edge and the parallel plane through the vertex contain the polyhedron between them), etc.

Algebraically, my solution boils down to:

$2A = \#(V,V) = V + \#(V,E) - \#(V,F)$ (1)

$\#(V,E) = E + \#(E,E) = E + 2B$ (2)

So,

$2A - 2B = E+V-F = 2V-2$

by Euler's formula, and therefore $A-B = V-1$.

To prove (1), count the number of pairs $\#(V,V)$ as follows. Fix a vertex $v$. There must be some vertex $u$ antipodal to it (consider the one farthest away, say). Now, look at all such $u$; they form a planar graph where edges are edges of the polyhedron. The total number of such $u$ is thus $1$ plus the number of edges in the graph(= the number of edges antipodal to $v$) minus the number of cycles(= the number of faces antipodal to v). Summing over all $v$, we get (1).

To prove (2), count the number of pairs $\#(V,E)$ as follows. Fix an edge $e$. There must be some vertex $u$ antipodal to it (consider the one farthest away, say). Now, look at all such $u$; they form a planar graph where edges are edges of the polyhedron. The total number of such $u$ is thus $1$ plus the number of edges in the graph(= the number of edges antipodal to $e$) minus the number of cycles (=0 since no face is parallel to $e$). Summing over all $e$, we get (2).

Note that $\#(V, F) = F$ since there is a unique vertex $v$ antipodal to each face, namely the one farthest away.
[/hide]

Thinking in 3D was hardest part for me! Rename A,B as X,Y

for a vertex V I construct a planar graph where vertices are the vertices of the polygon that form antipodal pair with V and edges V1V2 are such that midpoint of V1V2 and V are antipodal. By continuity of the plane at V, is not hard to see that the faces of this graph are faces of the polyhedron such that their center is antipodal with V (except the exterior face of the graph). By Euler formula we get

# vertices - # edges + # faces = 1

where #X refers to the number of elements X that are antipodal with V. Summing over all V, we get

2X - (V,A) + (V,C) = V

where (V,A) and (V,C) refers to the number of pairs of vertices and edges, and vertices and faces, respectively, that are antipodal.

Now construct such a graph for edges instead of vertices, use Euler's formula in that graph and sum over all edges. we get

(V,A) - 2Y + 0 = A

where V,A is number of vertices and edges, and 0 is because no face is antipodal to an edge (they are not parallel).

Notice also (C,V) = C, number of faces in polyhedron, since for each face there is exactly one vertex antipodal to it.

Finally, add the two equations.

2X-2Y+C = A+V, so 2X-2Y=A+V-C

Using that V+C-2=A by Euler's formula on the whole graph, we get X-Y=V-1, which is the answer.

We claim that the answer is $V-1$.

First, choose an arbitrary vertex $v$, and draw a plane, $P$, which coincides with a face touching $v$. Now, there exists a unique vertex $u$ of the polyhedron such that $(u,v)$ form an antipodal pair and the corresponding plane can be parallel to $P$. Call the plane parallel to $P$ passing through $u$ $P'$.

Now, consider the smooth transformation by which $P$ goes from its current face to a new one by pivoting along an edge incident to $v$. Then, it switches pivots to the other edge incident to $v$, and it continues moving until $P$ returns to its initial orientation. During this cycle, we adjust $u$ every time $P'$ coincides with an edge. Note that, when this happens, we get an antipodal pair of midpoint edges. In this way, when $P$ traces a cycle, $u$ will form a polygon $f(v)=u_1u_2\ldots u_k$. We call the region bounded by $f(v)$ which does not include $u$ the "domain" of $u$. From construction, the edges incident to $v$ are part of $k$ antipodal pairs of midpoint edges.

Note that the domain of $u$ roughly corresponds to the reflection of the angle of space subtended by the faces incident to $v$. This means that, if we repeat this construction over all vertices of the polyhedron, all domains are disjoint (since the angles sweeped by all processes perfectly cover the sphere) and 2 domains share part of their border if and only if the vertices they correspond to have a common edge.

Now, $A$ can be computed by $\sum_v V(f(v))$ where $V(f(v))$ counts the number of vertices in a domain, and $B$ is $\sum_v E(f(v))$ where $E(f(v))$ counts the number of edges in the boundary of a domain. To compute $A-B$, consider the graph $G$ formed by edges of the boundaries of domains. Let the set of vertices, faces, and edges of $G$ be $V',F',E'$. Also, for convenience, we will assume pairs in $A,B$ are ordered for now.

When computing $A-B$ over vertices in $G$, we contribute $\sum_{f\in F'}V(f)=2|E'|$ to $A$ and $|E'|$ to $B$ for a net total of $|E'|$. However, we neglect to include domains which only have $1$ or $2$ boundary points. Each of these two edge cases contribute a net sum of $1$ to $A-B$, and recall that there are $V$ domains, so there are $V-|F'|$ such edge cases. Hence, $G$ contributes $|E'|+(V-|F'|)=V+|V'|-2$.

On the other hand, vertices not part of $G$ contribute nothing to $B$, so they contribute $V-|V'|$ to $A-B$. Adding these two cases and dividing by $2$ because each pair was counted twice, our final answer is $\frac{V+|V'|-2+V-|V'|}{2}=V-1$, as desired.

Let $\mathcal{P}$ denote the polyhedron. Let $V,E,F$ denote the number of vertices, edges, faces respectively of the polyhedron $\mathcal{P}$ and note that since it is convex, hence genus $0$, we have $V-E+F = 2$. Label the vertices of the polyhedron $\mathcal{P}$ with $1,2,\cdots, V$ and define a coloring on $\mathcal{S}^2$ as follows: A point $(a,b,c) \in \mathcal{S}^2$, which must satisfy $a^2+b^2+c^2=1$, will be colored with the color $i$ if the function $f(x,y,z) = ax+by+cz$ is maximized at vertex $i$. Note that points can be colored with multiple colors, but the set of such multi-colored points has measure zero on $\mathcal{S}^2$.

This coloring defines a simple graph $S$ on $\mathcal{S}^2$ such that the face number $i$ of $S$ corresponds to the $i$-monochromatic domain of $\mathcal{S}^2$ (the set of all points which are colored with $i$), the edges of $S$ correspond to boundaries between monochromatic domains, and the vertices of $S$ correspond to intersections of three or more monochromatic domains. Note that $S$ is isomorphic to the dual graph of the graph of $\mathcal{P}$. Furthermore, the edges of $S$ are segments of great circles. If we let $V',E',F'$ denote the number of vertices, edges, faces respectively of $S$, we have $V' = F$, $E' = E$, and $F' = V$.

Take the reflected graph $S'$ on $\mathcal{S}^2$, defined as reflection of $S$ about the origin $(0,0,0)$. Consider the new graph $T$ [i]topologically-induced[/i] by $S \sqcup S'$; i.e. $T$ has not only the vertices of $S$ and the vertices of $S'$, but also new vertices formed by intersections of edges from $S$ with edges from $S'$, etc. The problem condition implies that edges from $S$ and edges from $S'$ do not intersect except in their interiors; they do not intersect at their endpoints. (This also implies that no vertex of $S$ coincides with a vertex of $S'$.) Let $k$ denote the number of intersections of an edge from $S$ with an edge from $S'$.

Note that antipodal pairs of vertices of $\mathcal{P}$ correspond to antipodal faces of $T$, and antipodal midpoints of edges of $\mathcal{P}$ correspond to antipodal pairs of intersections of edges from $S$ with edges from $S'$. This means that $2B = k$. Furthermore, letting $V'', E'', F''$ denote the number of vertices, edges, and faces respectively of $T$, we see that $2A = F''$.

It's time to calculate $V'', E'', F''$ in terms of $V', E', F'$. Note that \[V'' = 2V' + k\] because $T$ contains the vertices of $S$, the vertices of $S'$, and new vertices formed by intersections of edges from $S$ with edges from $S'$. Furthermore, \[E'' = 2(E'+k)\] because each intersection increases $E''$ by 2 (it splits two edges in half). Because $V'' - E'' + F'' = 2$, we calculate \[F'' = 2F' + k - 2.\] Thus \[A - B = \frac{1}{2}(2A-2B) = \frac{1}{2}(F'' - k) = \frac{1}{2}(2F' - 2) = F' - 1 = \boxed{V - 1}.\] $\blacksquare$