function
algebra
polynomial
logarithms
functional equation
system of equations

(1) How many of the following sets of functions have the property that, given any two elements $ f(x)$ and $ g(x)$ of the set, the composition $ f(g(x))$ is in the set?

[list]1. functions of the form $ ax\plus{}b$
2, functions of the form $ ax^2\plus{}bx\plus{}c$
3. polynomial functions
4. polynomial functions with $ 12$ as a root[/list]

(2) Find all solutions to the functional equation $ 21f(x)\minus{}7f\big(\frac{1}{x}\big)\equal{}12x$.

Thanks a lot!

I'm not good with functional equations, but I can try number two.

[hide]$ x \ne 0$,
$ 21f(x)\minus{}7f(\frac{1}{x})\equal{}12x$.
If $ x\equal{}\frac{1}{x}$, the equation is $ 21f(\frac{1}{x})\minus{}7f(x)\equal{}\frac{12}{x}$.  Multiply original by three and combine to obtain $ 56f(x)\equal{}\frac{36x^2\plus{}12x}{x}\rightarrow f(x)\equal{} \frac{3\cdot(3x^2\plus{}1)}{14x}$.

We can also start with $ x\equal{}1 \rightarrow f(1)\equal{}\frac{6}{7}$, which checks.  But I don't know how to prove that this is the only solution, or if this is the only solution.[/hide]

Actually, the answers are

[hide="(1)"]$ f(x)\equal{}\log_{a}x\minus{}1$[/hide]

[hide="(2)"]$ f(x)\equal{}(9x\plus{}3/x)/14$[/hide]

Could someone please help me?  Thanks!

[hide="Question 1"]
1. Yes: the degree of f and g is 1, so the degree of f(g(x)) is 1*1=1, and it is linear.

2. No: the degree of f and g is 2, so the degree of f(g(x)) is 2*2=4, and it is not a quadratic.

3. Yes: a polynomial function raised to an integer power yields a finite polynomial, by the binomial theorem. Perform this for each term of f

4. No. Consider the polynomials f(x)=g(x) = x-12. Then f(g(x)) = x-24.
Another way of saying this is that if g(12)=0, then we require f(g(12))=f(0)=0, which was not originally asserted.[/hide]

[hide="Question 2"]$ 21f(x) - 7f\left(\frac {1}{x}\right) = 12x$

Plug in $ 1/x$:

$ 21\left(\frac {1}{x}\right) - 7f(x) = \frac {12}{x}$

Adding:

$ 14 \left( f(x) + f \left( \frac {1}{x} \right) \right) = 12 \left( x + \frac {1}{x} \right)$

Let $ f(x) = g(x) + \frac {6}{7} x$. Then, plugging in and canceling, we have:

$ g(x) + g \left( \frac {1}{x} \right) = 0$.

So all we need is an arbitrary function $ p(x)$ over $ x \in ( - \infty, - 1) \cup (1,\infty)$.

Then, $ g(x) = \left \{\begin{array}{cc} p(x) & |x| > 1 & - p(1/x) & |x| < 1 & 6/7 & |x| = 1 \end{array} \right \;$[/hide]

[quote="The Zuton Force"]
[hide="Question 2"]$ 21f(x) - 7f\left(\frac {1}{x}\right) = 12x$

Plug in $ 1/x$:

$ 21\left(\frac {1}{x}\right) - 7f(x) = \frac {12}{x}$

Adding:

$ 14 \left( f(x) + f \left( \frac {1}{x} \right) \right) = 12 \left( x + \frac {1}{x} \right)$

Let $ f(x) = g(x) + \frac {6}{7} x$. Then, plugging in and canceling, we have:

$ g(x) + g \left( \frac {1}{x} \right) = 0$.

So all we need is an arbitrary function $ p(x)$ over $ x \in ( - \infty, - 1) \cup (1,\infty)$.

Then, $ g(x) = \left \{\begin{array}{cc} p(x) & |x| > 1 & - p(1/x) & |x| < 1 & 6/7 & |x| = 1 \end{array} \right \;$[/hide][/quote]

Just because $ 14 \left( f(x) + f \left( \frac {1}{x} \right) \right) = 12 \left( x + \frac {1}{x} \right)$, doesn't mean that any solution satisfying that satisfies the original functional equation.

Suppose we had a system of equations: $ x+y=12$ and $ 2x+3y=29$. Adding yields $ 3x+4y=41$. Any pair $ (x,y)$ that satisfies the original system satisfies $ 3x+4y=41$, however, not every $ (x,y)$ that satisfies $ 3x+4y=41$ satisfies the original system.

How would one prove that $ \frac{3\cdot(3x^2\plus{}1)}{14x}$ is the ONLY solution?

[quote="Xantos C. Guin"]Just because $ 14 \left( f(x) \plus{} f \left( \frac {1}{x} \right) \right) \equal{} 12 \left( x \plus{} \frac {1}{x} \right)$, doesn't mean that any solution satisfying that satisfies the original functional equation.

Suppose we had a system of equations: $ x \plus{} y \equal{} 12$ and $ 2x \plus{} 3y \equal{} 29$. Adding yields $ 3x \plus{} 4y \equal{} 41$. Any pair $ (x,y)$ that satisfies the original system satisfies $ 3x \plus{} 4y \equal{} 41$, however, not every $ (x,y)$ that satisfies $ 3x \plus{} 4y \equal{} 41$ satisfies the original system.[/quote]

Mr. Guin, that is a lucid, intelligent, well thought-out objection.

On the note of my partial solution, however, consider what happens if you then subtract those two equations, thus creating another combination equation. This very strongly hints towards the real solution if you use the similar technique of subtracting out a linear.

[quote="mathwizarddude"]How would one prove that $ \frac {3\cdot(3x^2 \plus{} 1)}{14x}$ is the ONLY solution?[/quote]
azn's method works fine; I don't know why he doubted it. His steps show that $ f(x) \equal{} \frac {3\cdot(3x^2 \plus{} 1)}{14x}$ is a [b]necessary[/b] condition for $ f$ to be a solution of the equation; there can't be others. That it is sufficient is simple routine checking (in a contest solution you can just say "it is trivial to see that this works").


[quote="nttu"]Vasc gave a solution in "Old and New Inequalities" but it's hard to think . I also have a solution uses Schur's ineq[/quote]
   Vasc's solution :
  $x^{2}+xy+y^{2}= (x+y+z)^{2}-(xy+yz+xz)-(x+y+z)z$ :  :o
  So $\frac{(x+y+z)^{2}}{x^{2}+xy+y^{2}}= \frac{1}{1-(ab+ac+bc)-c}$ , where $a=\frac{x}{x+y+z}, b = .., c=..$
 The ineq can be written as $\sum \frac{1}{1-d-a}\geq 9$ , where $d=ab+ac+bc$
  After some calculations the ineq becomes :
  $d(3d-1)^{2}+(1-4d+9abc) \geq 0$ : obviously true
  What a strange solution !

OK, i rewrite here completely Vasc's solutions in this book
Considering the relation:
$x^{2}+xy+y^{2}=(x+y+z)^{2}-(xy+yz+zx)-(x+y+z)z$
we get:
$\frac{(x+y+z)^{2}}{x^{2}+xy+y^{2}}=\frac{1}{1-\frac{xy+yz+zx}{(x+y+z)^{2}}-\frac{z}{x+y+z}}$
Or $\frac{(x+y+z)^{2}}{x^{2}+xy+y^{2}}=\frac{1}{1-(ab+bc+ca)-c}$ where $a=\frac{x}{x+y+z},b=\frac{y}{x+y+z},c=\frac{z}{x+y+z}$
The inequality can be rewritten as:
$\frac{1}{1-d-a}+\frac{1}{1-d-b}+\frac{1}{1-d-c}\geq 9$
where $a,b,c$ are positive reals with $a+b+c=1$ and $d=ab+bc+ca$.
After making some computations the inequality becomes:
$9d^{3}-6d^{2}-3d+1+9abc\geq 0$ or $d(3d-1)^{2}+(1-4d+9abc)\geq 0$ which is [b]Schur's inequality[/b]

[quote="nttu"]I think this below ineq is stronger than the original ineq .
 $(ab+ac+bc)^{4}\geq 9a^{2}b^{2}c^{2}(2(a^{2}+b^{2}+c^{2})+ab+ac+bc)$ , where $a,b,c >0$[/quote]

Just in order to have it mentioned, this one is wrong for $a=b=1$ and $c=10$.

  darij


logarithms
calculus
calculus computations

find t such that :
$ lgP \equal{} \frac { \minus{} \delta{H}}{RT} \plus{} \frac {\delta{C_p}}{R} . lgT$

what is $ lg$ ?  is it $ \ln$ ?
also what is $ C_{p}, R,T,H,P$ ?
is it true that $ T\equal{}t$?
also what is $ \delta$ ?

[quote="Extremal"]what is $ lg$ ?  is it $ \ln$ ?
also what is $ C_{p}, R,T,H,P$ ?
is it true that $ T \equal{} t$?
also what is $ \delta$ ?[/quote]
 all 're  const!!!  and  t is T

Hmm... as I don't know which base are you thinking of when writing $ lg$ (it can be $ 2$, $ e$, or $ 10$, depending on agreement), I can't write an explicit solution, but I really can't see the problem here:

$ lgP \equal{} \frac { \minus{} \delta{H}}{RT} \plus{} \frac {\delta{C_p}}{R} \cdot lgT$

$ lgP \minus{} \frac {\delta{C_p}}{R} \cdot lgT \equal{} \frac { \minus{} \delta{H}}{RT}$

$ lg\frac {P}{T^{\frac {\delta{C_p}}{R}}} \equal{} \frac { \minus{} \delta{H}}{RT}$

$ \frac {P}{T^{\frac {\delta{C_p}}{R}}} \equal{} b^\frac { \minus{} \delta{H}}{RT}$

$ T\equal{}\left(\frac {P}{b^\frac { \minus{} \delta{H}}{RT}}\right)^{\frac {R}{\delta C_p}}$

where $ b$ is the base of $ lg$.

[quote="hsiljak"]Hmm... as I don't know which base are you thinking of when writing $ lg$ (it can be $ 2$, $ e$, or $ 10$, depending on agreement), I can't write an explicit solution, but I really can't see the problem here:

$ lgP \equal{} \frac { \minus{} \delta{H}}{RT} \plus{} \frac {\delta{C_p}}{R} \cdot lgT$

$ lgP \minus{} \frac {\delta{C_p}}{R} \cdot lgT \equal{} \frac { \minus{} \delta{H}}{RT}$

$ lg\frac {P}{T^{\frac {\delta{C_p}}{R}}} \equal{} \frac { \minus{} \delta{H}}{RT}$

$ \frac {P}{T^{\frac {\delta{C_p}}{R}}} \equal{} b^\frac { \minus{} \delta{H}}{RT}$

$ T \equal{} \left(\frac {P}{b^\frac { \minus{} \delta{H}}{RT}}\right)^{\frac {R}{\delta C_p}}$

where $ b$ is the base of $ lg$.[/quote]
$ (\frac {P}{b^\frac { \minus{} \delta{H}}{RT}})^{\frac {R}{\delta C_p}}$  expression has $ T$

This looks like a Thermodynamics problem. If yes, what is the context?

[quote="Hong Quy"] expression has $ T$[/quote]

 :blush: sorry... I hoped it's not another transcendental equation... back to numerical methods, then...


geometry
geometric transformation
rotation

Every time the pedals go through a 360° rotation on a certain bicycle, the  tires rotate three times. If the tires are 24 inches in diameter, what is the  minimum number of complete rotations of the pedals needed for the bicycle  to travel at least 1 mile?

(1) 12              (3) 561
(2) 281            (4) 5,280

[hide]24 inch diameter = 2 feet, so the circumference of the tire is $2\pi$ feet.  For the bicycle to travel 1 mile, or 5280 feet, the tires will need to spin $\frac{5280}{2\pi}= \frac{2640}{\pi}$ times.  If for every one pedal rotation, the tires rotate three times, the pedals need to rotate $\frac{2640}{3\pi}= \frac{880}{\pi}$ times.

Without a calculator, I approximated pi to 3 and ended up with 290.  Since pi is a little more than 3, the answer should be a little less than 290, and I got choice 2.[/hide]

Tell me, the 2 feet diameter concluded that the circumference of the tire is 2pi feet?

Explain that part again.

Thanks!

WOW YOU DO MAKE A LOT OF THREADS
lol
When 2 equals the diameter and the formula is $D*\pi$
so it's 2 pi, because of the circumference formula

Okay...much better.

Thanks for answering my question about the diameter being 2 feet.

I got it now.

Thanks!


calculus
function
derivative
calculus computations

[quote="trichoplax"][quote="Dvagner"]is it possible to compute
$\frac{d}{dt}(2 \uparrow\uparrow t)$[/quote]

No. The difficulty is that $2 \uparrow\uparrow t$ is only
defined when $t$ is a positive integer. (One could extend it to a differentiable function, but there is no standard way to do this.)[/quote]

Here's a way to extend the function to real numbers, I found it on the so-called [url=http://tetration.itgo.com/ident.html]Home page of Tetration[/url], look under the heading "A Simple Extension to Real Numbers".

how would one take the derivative of this real form?


superior algebra
superior algebra unsolved

Is there any difference between $ \mathbb{Z}/(p)$ and $ \mathbb{Z}/<p>$?

[quote="Kalle"]No? Except maybe that the first looks more like a quotient ring and the second a quotient group[/quote]

The set is the same but the first one is a field (if $ p$ is prime) the second a group. In the second one the multiplication isn't defined.

The notation often means the same. I've seen $ (p)$ and $ <p>$ for the ideal generated by $ p$.


geometry
parallelogram
area of a triangle
geometry solved

For some convex quadrilateral ABCD, find the locus of points P such that [PAB] + [PCD] = [PAD] + [PBC].

(Note, [X] means the area of polygon X.)

It is a quite classical problem.
Obviously [PAB]+[PCD]=1/2 [ABCD], which is fixed. The geometrical locus of P for which that sum is fixed is a segment, i.e. the intersetion between a certain line and the interior of $ABCD$. Take E the intersection between lines $AB$ and $CD$. Take the points X and Y on AB and DC such that $EX=AB$ and $EY=CD$. Then our sum of areas is [PEX]+[PEY]=[EXY]+[PXY], and [EXY] is a constant so P runs over all points on a line parallel to $XY$, which are also fixed points.
Now make the same trick for AD and BC and you will get that the locus is a) no point if these 2 segment do not intersect b) one point if they intersect at one point c) a infinity of points if the segments are the same.
More interesting would be to determine $ABCD$'s properties on each of these cases.
Oh.. If there is no intersection between AB and CD above, it's clear which is the locus.
The parallelogram is one of the case c) examples but I am afraid there are more quadrilaterals satisfying c).

[quote="ikap"]It is a quite classical problem.
Obviously [PAB]+[PCD]=1/2 [ABCD], which is fixed. The geometrical locus of P for which that sum is fixed is a segment, i.e. the intersetion between a certain line and the interior of $ABCD$. Take E the intersection between lines $AB$ and $CD$. Take the points X and Y on AB and DC such that $EX=AB$ and $EY=CD$. Then our sum of areas is [PEX]+[PEY]=[EXY]+[PXY], and [EXY] is a constant so P runs over all points on a line parallel to $XY$, which are also fixed points.
Now make the same trick for AD and BC and you will get that the locus is a) no point if these 2 segment do not intersect b) one point if they intersect at one point c) a infinity of points if the segments are the same.[/quote]

This is wrong. Actually, you get one segment as the locus of all points P such that [PAB] + [PCD] = 1/2 [ABCD], and one segment as the locus of all points P such that [PAD] + [PBC] = 1/2 [ABCD], but these two segments coincide, since the conditions [PAB] + [PCD] = 1/2 [ABCD] and [PAD] + [PBC] = 1/2 [ABCD] are equivalent!

In fact, the problem is really classic. I will solve it in the following modified variant:

[b]Problem.[/b] For some quadrilateral ABCD, find the locus of points P such that [PAB] + [PCD] = [PBC] + [PDA], where [XYZ] means the [i]directed[/i] area of a triangle XYZ.

I'm using directed areas in order not to get into distinguishing a dozen of cases.

So here is my [i]solution[/i]:

At first, assume that the quadrilateral ABCD is not a parallelogram. Then, the quadrilateral ABCD has (at least!) two non-parallel opposite sides. WLOG we assume that the two opposite sides AB and CD are not parallel. Hence, the lines AB and CD intersect at some point. Let E be this point. Furthermore, let X and Y be two points on the lines AB and CD such that EX = BA and EY = CD (with directed segments!). Then, the triangles PAB and PXE have equal areas (since they have equal bases and equal altitudes to these bases). So we have found [PAB] = [PXE]. Similarly, [PCD] = [PEY]. Thus, [PAB] + [PCD] = [PXE] + [PEY] = [PXEY] = [XEY] + [YPX].

Now, since ([PAB] + [PCD]) + ([PBC] + [PDA]) = [ABCD], the equation [PAB] + [PCD] = [PBC] + [PDA] is equivalent to [PAB] + [PCD] = [ABCD] / 2. From the identity [PAB] + [PCD] = [XEY] + [YPX] proved above, we see that the condition [PAB] + [PCD] = [ABCD] / 2 is equivalent to [XEY] + [YPX] = [ABCD] / 2, and therefore also to [YPX] = [ABCD] / 2 - [XEY].

But since the area of a triangle equals one half of the base times the corresponding altitude, the area [YPX] of the triangle YPX must equal one half of the base XY times the corresponding altitude, i. e. the distance from the point P to the line XY. In other words, $ \left[ YPX\right] = \frac12 \cdot XY \cdot d\left(P\right)$, where d(P) is the distance from the point P to the line XY. Therefore, the point P satisfies the condition [PAB] + [PCD] = [ABCD] / 2 if and only if the equation $ \frac12 \cdot XY \cdot d\left(P\right) = \left[ ABCD\right] / 2-\left[ XEY\right]$ holds. This equation can be rewritten as $ d\left(P\right) = 2 \cdot \frac{ \left[ ABCD\right] / 2-\left[ XEY\right] }{XY}$.

Hence, the equation [PAB] + [PCD] = [ABCD] / 2 is equivalent to the equation $ d\left(P\right) = 2 \cdot \frac{ \left[ ABCD\right] / 2-\left[ XEY\right] }{XY}$. But in the beginning of the solution, we saw that the equation [PAB] + [PCD] = [PBC] + [PDA] is equivalent to the equation [PAB] + [PCD] = [ABCD] / 2. Hence, the equation [PAB] + [PCD] = [PBC] + [PDA] is equivalent to the equation $ d\left(P\right) = 2 \cdot \frac{ \left[ ABCD\right] / 2-\left[ XEY\right] }{XY}$.

But the right hand side of this last equation is not dependent on the position of the point P; it is only dependent on the quadrilateral ABCD. Hence, the equation $ d\left(P\right) = 2 \cdot \frac{ \left[ ABCD\right] / 2-\left[ XEY\right] }{XY}$ gives a constant value for the distance d(P) from the point P to the line XY. Therefore, the locus of all points P fulfilling this equation is a line parallel to the line XY.

Thus, the locus of all points P fulfilling the condition [PAB] + [PCD] = [PBC] + [PDA] is a line parallel to the line XY.

Actually we can characterize this line more precisely by showing that this line passes through the midpoints of the diagonals AC and BD of the quadrilateral ABCD. In fact, if P is the midpoint of the diagonal AC of the quadrilateral ABCD, then the triangles PAB and PBC have equal areas (because of equal bases and equal altitudes), so that [PAB] = [PBC], and similarly [PCD] = [PDA], and thus the condition [PAB] + [PCD] = [PBC] + [PDA] is obviously fulfilled. This means that the point P lies on the locus. Similarly, if P is the midpoint of the diagonal BD, then P also lies on the locus. Hence, the locus (i. e., our line) must pass through the midpoints of the diagonals AC and BD.

This solves the problem for all quadrilaterals ABCD that are not parallelograms.

Now turn to the case when the quadrilateral ABCD is a parallelogram. In fact, in this case, the locus of all points P such that [PAB] + [PCD] = [PBC] + [PDA] is the whole plane.

[i]Proof.[/i] We will show that for every point P in the plane of the quadrilateral ABCD, the equation [PAB] + [PCD] = [PBC] + [PDA] holds. In fact, let a = AB = CD be the length of the equal sides AB and CD of the parallelogram ABCD, and let h be the distance between these two sides. Then, since the area of a parallelogram equals one of its sides times the corresponding altitude, the area of the parallelogram ABCD must equal $a \cdot h$. In other words, $\left[ ABCD\right] = ah$.

Let x and z be the distances from the point P to the sidelines AB and CD of the parallelogram ABCD. In other words, x and z are the lengths of the altitudes issuing from P in the triangles PAB and PCD. Now, since the area of a triangle equals one half times base times corresponding altitude, the area of the triangle PAB must equal its base AB times the corresponding altitude, i. e. we must have $ \left[ PAB\right] = \frac12 \cdot AB \cdot x$. In other words, $ \left[ PAB\right] = \frac{ax}{2}$. Similarly, $ \left[ PCD\right] = \frac{az}{2}$.

But since AB || CD, the perpendicular from the point P to the line AB and the perpendicular from the point P to the line CD are one and the same line. Thus, x + z = h. Hence, from $ \left[ PAB\right] = \frac{ax}{2}$ and $ \left[ PCD\right] = \frac{az}{2}$, we get

  $ \left[ PAB\right]+\left[ PCD\right] = \frac{ax}{2}+\frac{az}{2}= \frac{a\left(x+z\right)}{2}= \frac{ah}{2}= \frac{\left[ ABCD\right] }{2}$.

By analogous reasoning,

  $ \left[ PBC\right]+\left[ PDA\right] = \frac{\left[ ABCD\right] }{2}$.

Thus, [PAB] + [PCD] = [PBC] + [PDA] holds. This completes the proof that for a parallelogram ABCD, the locus of all points P such that [PAB] + [PCD] = [PBC] + [PDA] is the whole plane.

  Darij

Well, sorry to dissapoint you darij but my solution(modified) is correct and shorter than yours:). Of course I don't need to do the same thing again for $AD, BC$, cause they are equivalent so I get only that segment as the intersection of the line about which you are speaking and the interior (the problem was about the interior if I'm not wrong) Oh.. sorry I am wrong, I thought it was only about the interior, but it works the same way for the exterior.  Take Y and X on one side or on the other side of E as we want to determine the locus in a certain region of the plane(and the areas should be directed). I'm afraid to dissapoint you more darij, but this proofs also that the real locus is composed of 2 lines which intersect somewhere. You could have read my solution more carefully before saying that it's wrong, i mean the idea is not wrong only that i was proving for the interior.

oops, sorry.
It seems 2 must also be replaced by 4 in the number of solutions, cause we have 1 line on each side of XY at a fixed distance to XY.

[quote="ikap"]Well, sorry to dissapoint you darij but my solution(modified) is correct and shorter than yours:). Of course I don't need to do the same thing again for $AD, BC$, cause they are equivalent so I get only that segment as the intersection of the line about which you are speaking and the interior (the problem was about the interior if I'm not wrong)[/quote]

Hmm, actually your error was not that you didn't consider the exterior of the quadrilateral, but that you constructed two segments on which the point P must lie, and claimed that therefore the locus is "one point if they [the two segments - DG] intersect at one point", without noting that the two segments always coincide and hence the locus consists of infinitely many points. But I see you already noted this error.

What concerns my error, I actually gave a disclaimer. I said I solved a slightly [i]modified[/i] problem with [i]directed areas[/i]. And the locus for this modified problem is a line (or the whole plane in the case of a parallelogram). I do admit that I didn't consider the case of non-directed areas, which actually gives four segments. I didn't consider this case since I really hate distinguishing cases in geometry.

  Darij


[quote="Harry Potter"]Find maximum of:
$x_{1}^{3}.x_{2}^{2}+x_{2}^{3}.x_{3}^{2}+...x_{n}^{3}.x_{1}^{2}+n^{2(n-1)}.x_{1}^{3}.x_{2}^{3}...x_{n}^{3}$
With $x_{1}+x_{2}+x_{3}+...+x_{n}=1$ and n>2 ; n number >0[/quote]
This is my problem in Maths and Youth magazine, please don't discuss it anymore!

@To mod: could you please block this topic!

To toanhocmuonmau : I don't know it is the problem in Maths and Youth magazine . That is a problem my friend give me on  16th  May . And i solved it ; i post it on mathlinks because it is nice . I'm sorry if this is an ongoings contest.

[quote="vietnam2007"][quote="toanhocmuonmau"]Oh, I have sent it to Maths and Youth magazine in March! And now it has been publishing![/quote]

can you write the link please  :?:[/quote]
[url=http://www.nxbgd.com.vn/toanhoctuoitre/?p=7&id=6&ReportID=231&ph=21]See here[/url]
:)


Euler
integration
logarithms
function
calculus
calculus computations

show that Euler-gamma constant $\gamma$ is equal
\[\int_{0}^{1}\left( \frac{1}{1-x}+\frac{1}{\log x}\right) dx\]

[quote="boria"]show that Euler-gamma constant $\gamma$ is equal
\[\int_{0}^{1}\left( \frac{1}{1-x}+\frac{1}{\log x}\right) dx \]
[/quote]

We can show this for example by using digamma funcion:
$\psi (x) = \frac{\Gamma'(x)}{\Gamma(x)}= \int_{0}^{\infty}\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}dt$ and fact that $\gamma =-\Gamma'(1) =-\psi(1)$. If set $e^{-t}= u$ we have your integral.

thanks, nice idea ;0

could you give a sketch of proof of this equality :
\[-\gamma=\int_{0}^{\infty}\exp(-x)\log(x)dx\]

[quote="boria"]show that Euler-gamma constant $\gamma$ is equal
\[\int_{0}^{1}\left( \frac{1}{1-x}+\frac{1}{\log x}\right) dx \]
[/quote]
It's wellknow. In my book is picture like this
$\int_{0}^{1}\left( \frac{1}{1-x}+\frac{1}{\log x}\right) dx=\gamma$

When someone asks for something like this, they usually know that.  They just want to see how it's derived...


Let $ p$ be an odd prime.

Prove theres no primitive root $ mod \hspace{1 mm} 3p$ .

It shouldnt be too hard but I cant figure out how to do it.

Thanks.

[quote="t0rajir0u"]What is the maximum order of an element $ \bmod 3$?  $ \bmod p$?  If $ x^n \equiv 1 \bmod 3$ and $ x^n \equiv 1 \bmod p$ then...?[/quote]

Then:

$ x^n \equal{} 1 \bmod 3p$ .

I started with $ a^2 \equal{} 1 \bmod 3$ and $ a^{p \minus{} 1} \equal{} 1 \bmod p$ and didnt know how to continue.

even.

So according to the chinese remainder theorem theres $ x$ such that :

$ x \equal{} a^2 \equal{} 1 \bmod 3$ and
$ x \equal{} a^{p \minus{} 1} \equal{} 1 \bmod p$

so $ x^{p \minus{} 1} \equal{} 1 \bmod 3p$.

Is that good use of the CRT?

if the powers were coprime, lets say $ 5$ and $ 3$, then the resulting $ x$ from the chinese remainder theorem would fulfill $ x^{15}\equal{}1 \bmod 3p$ ?

Thx

I asked a more generalized question in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266807]this[/url] thread.  Here, you'll see a proof that any number with two distinct odd prime factors can't have a primitive root.


For x,y,z positive real numbers prove that

 x/y +(y/z)^1/2 + (z/x)^1/3 >3/2


Say I wanted to come up with a formula (in terms of n) for Sum(C,(n,i)), 4|i?  How would I go about doing that and what would such a formula be?

Well, you want to sum every fourth coefficient of

$(x+1)^{n}$

Right?  This is given by

$\frac{ (i+1)^{n}+(-1+1)^{n}+(-i+1)^{n}+(1+1)^{n}}{4}$

By the logic I gave you in the other thread.

Right, but would that give me the first term, the fifth term, the ninth term (x^n, x^n-4, x^n-8, etc.)?

Never mind, you're right.  I completely misinterpreted your response.  Many thanks!


geometry
incenter
ratio
trigonometry
geometry solved

G5. Let P , Q be points on BC in traingle ABC. Prove that if incircle radius of
ABP , ACQ are equal then the incircle radius of ABQ and ACP are also equal.

i think i have a computational proof of this one,after paying homage to the wacky world of trigonometry. :D .it's pretty long (as is the case with trig proofs).so i wont be posting it now (after my xams).and i'm sure sum1 here will come up with something better
say,amir cud u post iran 2003 on the forum
and btw ,is there a synthetic proof of this problem u posted?
cheers mate

Dear
 Try to prove this:
If the external center of similarity of the incircles ABP and ACQ
is T(the outer center of similarity of two circles: the intersection of their
external common tangents),then the external center of similarity of incircles of ABQ and ACP is also T.
(use this: The similarity centers of 3 similar shapes are collinear)

I also have a computational proof but the computations aren?t hard. Let circles C1, C2, C3, C4 be the inscribed circles of ABP, APC, ABQ, AQC respectively. Let I1, I2, I3, I4 be their centers, r1, r2, r3, r4 their radi and I the incenter of ABC and r its radius. In order to prove that the common external tangents of the circles (C1, C4); (C2, C3) and BC are concurrent it?s enough to prove that the lines I1I4, I2I3 and BC are concurrent. Since I1, I3, I are collinear and so are I2, I4, I we can apply Menelaus twice in the triangle IBC with transversals I1I4 and I2I3 respectively and put the condition that the intersection of  I1I4 and BC is the same of the intersection of  I2I3 and BC. After messing with the ratios a bit we are left with the following thing to prove:

If r1 and r2 are the inradi of triangles ABP and APC respectively for a pt P on (BC) then (r-r1)(r-r2)/r1r2 is the same no matter what P we choose. This isn?t hard to prove and I won?t prove it here. If smb wants me to, I?ll post the solution. We just write r1/(r-r1)=BI1/II1 and  r2/(r-r2)=BI2/II2, we project I, I1, I2 on BC and it follows easily from there.


find all$x \in R$ such that : sequence converge :
$\sum_{n=1}^{\infty}\frac{x^{n}}{1-x^{n}}$


geometry
rotation
inequalities
probability
geometry proposed

Let $ABC$ be a equilateral triangle and $M$ a point in its interior such that $\angle{BMC}=120^{\circ}$ and
  $\angle{CMA}=150^{\circ}$. Prove that $[AM]$, $[BM]$ and $[CM]$ can be the sides of a right-angled triangle.

See Theorem 1 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=297242#p297242]http://www.mathlinks.ro/Forum/viewtopic.php?t=23676 post #4[/url] for a more general version of the problem. Honestly, I have seen better problems in older issues of GM...

  Darij

We rotate the triangle $ABC$ about vertex $A$ until vertex $B$ coincides with vertex $C$.
[img]http://i19.photobucket.com/albums/b200/mariusdamian/problema.gif[/img]
We have $BM=CM'$ and triangle $AMM'$ is equilateral, so $AM=MM'$.
$\Delta AMM'$ is a right triangle and his sides are equal with $MA, MB, MC.$

[u]Darij is right ![/u] This problem is a particular case of a wellknown and old problem.

Let $\delta$ be the triangle with the lengths $MA$, $MB$, $MC$ of the its sides.
From the nice Damian's solution and from the his well performed figure result:

1. If we note $x=m(\widehat {BMC})$, $y=m(\widehat {CMA})$, $z=m(\widehat {AMB})$, then
the measures of the angles of the triangle $\delta$ are $x-60$, $y-60$, $z-60$.
2. For any point of the interior of the triangle $ABC$ there is the inequality
$MA+MB+MC\ge a\sqrt 3$, where $AB=BC=CA=a.$

With this opportunity, I shall allow to propose two problems:

3. What is the geometrical locus of the point $M$ for which the triangle $\delta$ is right ?.
4. What is the probability that the triangle $\delta$ be non-obtuse-angled ?


Does anyone know how to do this?

Suppose a plane is flying at a speed and direction which, in still air, would be 550 mph, 25 degrees north of west. Also, suppose there is a wind of 80 mph, 10 degrees south of east.

What is the resulting ground speed and direction of the plane in this wind condition?

This is a simple vector problem.  Although you can do this algebraically, I'm not since I don't want to go through the calculations.  So here's how you can do it with a ruler and protractor:

Set up a scale.  Maybe 1cm=100 mph.  Then set up a coordinate system.  Measure off an angle $ 25^{\circ}$ north of west (or $ 155^{\circ}$ from the other side).  Extend the segment forming the angle to 5.5cm (or whatever your scale is).  That is your plane velocity vector.  Now from the end of this vector, set up a new coordinate system with this point as the origin.  Form an angle $ 10^{\circ}$ south of east and extend it .8cm (or whatever your scale is).  This is your wind velocity vector.  Now you should have one vector eminating from the other.  From tail of the plane vector, connect it to the head of the wind vector to form a new vector, the resultant (it should form a triangle if you did it correctly).  This resultant is the vector you want.  Measure it's length (let's just say 3.5cm, I don't know what it actually is, at an angle of $ 35^{\circ}$ north of east, again just a random number).  Then you have that the plane will travel 350mph at $ 35^{\circ}$ north of east.


Find the subset $\large A \subset$ { $1,2,...,29$ } with the greatest number of elements that satisfies the following property : for any integer $k$ and two element $a,b$ of $A$ (not necessarily distinct), the number $a+b+30k$ is not the product of two consecutive integers.

My English is too poor!

see http://www.mathlinks.ro/Forum/viewtopic.php?p=346548#p346548  :wink:


geometry
3D geometry
MATHCOUNTS
algorithm

[quote="Aznwithabrain"]How do you find square roots of numbers manually by hand and get the exact square root?

Like the square root of 40320

Show All Work

Please[/quote]
Well, it's impossible to get an exact square root of a non-perfect square.

[quote="Aznwithabrain"]How do you find square roots of numbers manually by hand and get the exact square root?

Like the square root of 40320

Show All Work

Please[/quote]

[url]http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit_by_digit_calculation[/url]

Might be helpful  :|

[quote="Aznwithabrain"]How do you find square roots of numbers manually by hand and get the exact square root?[/quote]
The wikipedia link in the previous post includes some rather esoteric and high-level methods.  Here's another link that describes a couple of methods that are at a (high) Middle School level:
[url]http://www.mathpath.org/Algor/squareroot/algor.square.root.htm[/url]

If you choose to learn the classic (long-division) method, follow the pointer (near the bottom of the page) to an online square-root finder.  You provide the problem.  It provides the answer, and it even shows its work.

For 40320 specifically, we know it is $8!$ so we can easily factor that.  Others, we can either guess+check or factor.

After reading the book with webpage [url=http://web.mat.bham.ac.uk/C.J.Sangwin/euler/index.html]here[/url] (its from 1822)
I realized that math was in a way so much more before the use of calculators. In the book they develop a method that resembles division to compute both square roots and cube roots and shows how to generalize to nth roots (kind of). It's interesting but tough to explain and there's a bit of a learning curve.

EDIT: to find sqrt 40320, normally I would just notice that $200^{2}<40320$ but not by much then I would just compute $201^{2}$ and show that $200<\sqrt{40320}<201$ thus it isn't an integer.

But here's Euler's method or something like that

     1444 | 38
30|900
60|544
8|
68|544  (crucial step: we can stop since 8*68=544)
      0

oops sorry should have read the links before posting, one of them describes this I think

[quote="mwpl11"][quote="jhredsox"]isn't obvious that it's like 200.8^2?

jorian[/quote]

how is it obvious? can u explain it to me?[/quote]
It's more than $200^{2}$ and less than $201^{2}$
Also, it's closer to $201^{2}$ than it is to $200^{2}$

So, we can estimate it to be somewhere between $200.5^{2}$ and $201^{2}$

Well using any method to find square roots of numbers like that would take a long time, that's why it's never particularly useful. If you really want to know, just use Mathematica.

Leet, didn't you say you took a Mathcounts sprint (nationals) from a while ago and #1 was sqrt47 to the nearest hundreth or something?

jorian

It was in the 80s, and it was $\sqrt{something}$ to the nearest tenth, if I'm thinking of the same one.

It was $\sqrt{91}$ I think

That was quite easy, since you can see that the difference:

[code]
81<91<100

81 - - - - - - - - - - 91 - - - - - - - - 100

           10                         9

10/19=0.52...

[/code]

$0.52\approx 0.5$. So, $\sqrt{91}$ is approximately half way between those two square roots, so the answer is 9.5

PS It was 1985

I could have sworn it was sqrt 47

jorian

That was I think #3 on the 1984. There were 2: 1 in 1984, 1 in 1985

Edit; it was $\sqrt{55}$
Edit2; Wait. the state 1984 had $\sqrt{47}$

yes, that was it.

#3.....can you give us the question?

and your solution?

jorian

leet said it was trivial

jorian

It is trivial. Its 6.9, right :maybe: What's hard about it :maybe:

My solution:

[code]
36 - - - - - - - - - - 47 - 49

-          10              -   1

10/11 is approximately 0.9.

[/code]

So, its about 0.9 past $6^{2}$, so the answer is $\boxed{6.9}$

It was nearest TENTH!!!

Ahh I don't get The Long-Division-Type Square Root Algorithm

the instructions on [url]http://www.mathpath.org/Algor/squareroot/algor.square.root.htm[/url] are confusing...any help?

edit: reading the instructions for 10 mintues i figured it out but i am wondering if there is an easier set of instructions?

and how can you find the value of $a_{n-1}$ without a calcuator?

[quote="7h3.D3m0n.117"]I am wondering if there is an easier set of instructions?

How can you find the value of $a_{n-1}$ without a calculator?[/quote]
I don't know of any significantly easier instructions for the classic method.  It's mainly through practice that it gets easier.  [Remember when you first learned long-division?]  But, [url=http://mathforum.org/library/drmath/view/52610.html]here's another explanation[/url], and yet [url=http://www.du.edu/~jcalvert/math/sqrt.htm]another[/url].

If this is still too hard, and you still want to extract square roots by hand, you might check your local library for "The Trachtenberg Speed System of Basic Mathematics".  Trachtenberg's method for taking square roots is somewhat different.  Some people find Trachtenberg's method a little simpler than the classic method.  Personally, I think it's a toss-up.

As far as finding $a_{n-1}$...  This is just like getting the next digit of the quotient in a long division.  You have to make an estimate.  [This is the most frustrating part of learning long-division for many students.]  If your estimate is too big or too small, the error becomes apparent before too long, and you have to get out your eraser.

[quote="rcv"]As far as finding $a_{n-1}$...  This is just like getting the next digit of the quotient in a long division.  You have to make an estimate.  [This is the most frustrating part of learning long-division for many students.]  If your estimate is too big or too small, the error becomes apparent before too long, and you have to get out your eraser.[/quote]

Haha true. What I hate is that I have to solve $(10(a_{n})+a_{n-1})a_{n-1})$ as an estimate...


We do not need another poll, so this topic is locked.


Does anyone know of any books that contain polynomial problems like: "Let $ a_1,a_2,....,a_n$ and $ b_1,b_2,...,b_n$ be sets of distinct natural numbers such that the set of numbers $ a_i\plus{}a_j$ is a permutation of the numbers $ b_i\plus{}b_j$ prove n is a power of 2." ie problems that do not directly involve polynomials but in which they are helpful?

I think you might be interested in something like [url=http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf]generatingfunctionology[/url].

Any other suggestions? I was hoping for something not so advanced in theory and more focused on problem solving. Also I have already heard of this book.

I'm not sure if polynomials barbeau is what you're looking for.

a very good approach is the following: Polynomials, Victor Prasolov, Springer


abstract algebra
group theory
superior algebra
superior algebra unsolved

Any hint to prove that

if f \in Aut(G,G) has order p (prime) and G has order p+1 then G is abelian

actually every x has order 2.

If $ p$ is odd then $ G$ has some element of order 2.  What can you say about $ f(x)$?  What are the orbits of $ f$?

grr so easy now

Well since x has order 2, then f(x) has order 2 then apply f again an get p different elements of order 2
since f has order p

right?

but why is p prime? doesnt work for all odd orders?

Thx!

The solution requires that $ p$ be prime, since you need the fact that all non-identity elements belong to the same orbit.  If $ p$ were composite, $ f$ could have more than two orbits.

Im sorry could you tell me which is the orbit of f, I dont get it.

I know the definition of orbit, but just dont know what is the action here. Thanks

O(x) = {gx  | g \in G} and G acts on X

$ f$ turns $ G$ into a directed graph quite naturally, with each $ x$ pointing to $ f(x)$.  $ e$ points to itself, and the rest of the elements form a cycle... assuming that $ p$ is prime.  In other words, if you take any non-identity element of $ G$, and keep applying $ f$, you get every other non-identity element of $ G$.  That's only true when $ p$ is prime.  If $ p \equal{} 77$, for example, you could have many cycles of lengths $ 7$ and $ 11$, instead of one big cycle of length $ p$.

Draw a picture if that doesn't make sense.

If you want to get technical and reconcile the above paragraph with your definition of orbit, here's how: $ Aut(G,G)$ is a group under composition, and acts on $ G$ in the obvious way.  When I say "the orbits of $ f$", I'm talking about the orbits of the subgroup of $ Aut(G,G)$ generated by $ f$, acting on $ G$.  But this isn't what I was originally thinking of, I was just picturing little dots and arrows.

EDIT: What you may be missing here is that $ f$ can have order $ p$ without necessarily having order $ p$ when applied to any particular element.  When $ p$ is prime it's a different story, of course.


Hello everyone, I'm on a search for the worlds funniest youtube video...here are the rules.

1)A person may post up to 3 youtube videos that they deem funny

2)If a video has already been posted, please do not post it again

3)If you wish to join simply post the video's you wish to enter

4)I shall have 5 total "tournaments" of videos

5)The judges vote after 12-15 video's have been posted for that tournament

6)You may compete in all the tournaments as long as you have never won one

7) After the 5 tournaments we shall have a champions tournament!

8)Everyone who has posted at least 2 video's may vote in the champion's tournament (however if you have a video competing in the tournament you may not vote regardless)

9)Winning video is deemed worlds funnies Youtube video

10)If in any tournament, excluding champions, there is a tie. Winner shall be determined by the posters of the losing videos for that one tournament.

11) Judges will be changed for every tournament

12)If your video loses the tournament but you think it has potential you are allowed to post it again for the next tournament, but if it loses again you are not allowed to post it again

13)Obviously video's may not include X-Rated material,extreme amounts of swearing (I shall allow minimal amounts as I do understand most funny video's do have some swearing the words allowed are "minor swear words" so no use of the "F" word, and be reasonable, if you think it wont be allowed don't post it. Finally no offensive material. If it does video does have those things, poster may NOT post another video and that video is disqualified.

I will need 3 judges per tournament, please post in this thread if you want to be a judge for the first tournament. I will ask for more judges as the tournament goes on. Judges will be required to view all the videos for that tournament. However this is the honor system so therefore i will have to trust you. I will compete in the first tournament to set an example but am not allowed to win. I know thats a lot of rules but they are all designed to keep the game fair. Obviously do not create or obey rules created during the game.

[youtube]ubm5w0pWBGI[/youtube]

lol I wonder what happened after that...

[youtube]qd_syuD-N_k[/youtube]

I'm evil, so I will post something in spanish. BWAHAHAHAHAHAHAHAHAHAHAAAA!!! :diablo::evilgrin:

[youtube]b89CnP0Iq30[/youtube]

Then, I am still more evil. BWAHAHAHAHAHAHAHAHAAHH!!!!! :diablo: :evilgrin: :spider:

I'll pwn you with my self defense:

[youtube]2REG3-Wb5gM[/youtube]

EDIT: my third movie: [youtube]XBqY6cJD3CE[/youtube]

[youtube]IL7zdYFB7ns[/youtube]

I like this.   :rotfl:  :rotfl:

EDIT:  ok, ok, but it was funny

I have an idea: LET'S READ THE RUELS! :D

[quote]so no use of the "F" word, [/quote]

2 so far

[youtube]_fSpI4oZoDc[/youtube]

Will post the 3rd one.

EDIT: The second part MAY contain a "f" word. Not sure so plz check  :oops:

Re-edit: I deleted the second one so you can't watch it. It has f word, confirmed by 1=2

[youtube]PKnloiM-0Ns[/youtube]
My second on then.

It does.  :mad:  :mad:  :mad:  grr grr grr

I am so mad.

This one is epic

[youtube]in6RZzdGki8[/youtube]

I have counted more than 10 usages of the word in there. EDIT: 50

[size=200]NO SERIOUS CURSING! THE ABOVE VIDEO IS AN EXAMPLE OF WHAT NOT TO DO!!!!![/size]

how about whoever submits a video votes....as i am getting no judges here lol...

ok, i cant figure out how to post the video here, so i will say the name of it and you will find it on youtube.

1)Americans are NOT stupid.
2)Stupid Mario Bros Episode 1

Very funny :
http://img185.imageshack.us/img185/6585/howtobn2.png
EDIT: Is their supposed to be a real video on youtube with that title?

new rule: It actually has to be a movie.

err...that was pretty funny, lol...how to confuse an idiot. I was trying to get it to play for like 5 minutes. But yea..has to be a movie, would have thought that was a given....

gary zx's movie wont play. at all. How do you get it to play? [hide="."] YES i get the joke[/hide]


linear algebra
matrix
vector
induction
linear algebra solved

Let $A$ be a sub-algebra of $M_n(C)$ such that every matrix
in $A$ is nilpotent. Prove that there exists a basis $B=(e_1,..,e_n)$
such that every matrix in $A$ is triangular in the basis $B$.

I believe it is also true with the weaker assumption that whenever M,N are in the subspace A of a vector space End(V), we have
[M,N]=MN-NM is also in A (and each element of A is nilpotent). (I think those A are called Lie subalgebras of End(V)).
Here are the outlines:
It is enough to prove that each elements of A has a common eigenvetor with the eigenvalue 0.
We shall proceed by induction on the dimension of A when viewed as a vector space. Then consider A' to be the maximal proper subspace of A with the same properties as A. Use the induction hypothesis on the subspace $ad_{A'}$ of End((A/A')) to prove that A' is then an ideal of A, that is [M,N] $\in$ A' whenever M $\in$ A' and N $\in$ A. Then use the induction hypothesis again to get the conclusion.
If you need the details then just let me know.


There is only one pair of whole numbers whose product is 320,000 where neither factor contains the digit “0”.   What is the sum of these two numbers?

A.1137	B.1200	C.1444	D.1635	E.10,032

[hide="Solution"]$ 320000 \equal{} 2^5 \times 10^4 \equal{} 2^5 \times 2^4 \times 5^4 \equal{} 2^9 \times 5^4 \equal{} 512 \times 625$.

$ 512 \plus{} 625 \equal{} \boxed{1137}$. The answer is $ \boxed{A}$.

[/hide]

The answer is [hide]A. Sorry no time to explain right now[/hide]


Given a polynomial $f(x)$, supposed that $f(x)-f(a)$ can be  always divided by $x^3-a^3$ for all real numbers $a$.
Prove that we can write $f(x)=g(x^3)$ in terms of a certain polynomial $g(x)$.

We have $f(x)=m(x^3)+xn(x^3)+x^2p(x^3),m,n,p\in\mathbb R[X]$
Now $x^3-a^3|f(x)-f(a)\iff x^3-a^3|(x-a)n(a^3)+(x^2-a^2)p(a^3) \forall a\in\mathbb R$
Then we have $n(a^3)=p(a^3)=0\forall a\in\mathbb R$ so $n(x)=p(x)=0$.


in triangle ABC, AD is perpendicular to BC. Prove that for any point P on AD,

         BP^2 - PC^2 = BD^2 - DC^2

and conversely, if P satisfies the above then P lies on AD.

i have used your idea and solved the problem.

the way i solved the converse one is that let P doesn't lie on AD.
now,draw a perpendicular on AD ;let it be PM then join P to A and to B.

we have ultimately obtained the equation that
         BD^2 -CD^2 = BM^2 -CM^2   -----(1)
       = (BD+BM)(BD-BM)=(CD+CM)(CD-CM)
 but BD-BM = CD-CM
thus BD+BM = CD+CM
but it is impossible
hence the only way so that (1) is possible when M lies on D i.e. P on AD.


function
trigonometry
real analysis
real analysis unsolved

Prove that:
If all functions in a function sequence have the Darboux property, and the sequence is [b]uniformly convergent[/b], then the limit doesn't necessarily have the Darboux property. (Give an example). Let's say these funcions are R -> R.

[quote="pleurestique"]let $f_n$ be defined by

$f_n(x) = x$ on $(-\infty, -1]$,
$f_n(x) = -1+(1+\frac{1}{n})(x+1)$ on $(-1, 0)$,
$f_n(0) = 2$,
$f_n(x) = -\frac{1}{n}+(1+\frac{1}{n})x$ on $(0, 1)$,
$f_n(x) = x$ on $[1, \infty)$.

every $f_n$ has the Darboux property, while the sequence $f_n$ converges uniformly to $f$, where $f(x) = x$ on $\mathbb R - \{0\}$, $f(0) =2$, which doesn't have the Darboux property since $-1 = f(-1) < 0 < f(1) = 1$, but there is no $x$ for which $f(x) = 0$.[/quote]

Nope buddy. None of your $f_n$ has the Darboux property. If you take $x_n = - \frac{1}{2n}$ and $y_n = + \frac{1}{2n}$ then $\forall(n) (f_n(x_n)>0) \wedge (f_n(y_n)<0)$ and note that $\forall(x \in (-\frac{1}{2n},\frac{1}{2n})) f_n(x) \neq 0$.
[b]Attempt failed, the problem still open[/b]  :D

so modify it to this...

$f_n(x) = x$ on $(-\infty, \frac{1}{\pi/2-2n\pi})$,
$f_n(x) = x \sin \frac{1}{x}$ on $[\frac{1}{\pi/2-2n\pi}, \frac{1}{\pi/2+2n\pi}] - \{0\}$,
$f_n(0) = 2$,
$f_n(x) = x$ on $(\frac{1}{\pi/2+2n\pi}, \infty)$

these converge to the same $f$ as above

Nope fella: $x_n = 0$ and $y_n=\frac{1}{2^n}$.
$f_n(x_n)=0$
$f_n(y_n)>0$
and on $(0,\frac{1}{2^n})$ $f_n(x)$ is STRICTLY monotous and rising and there are NO values between 0 and $f_n(y_n)$.

In other words, please [b]THINK[/b] thrice before posting another trivial combination of some functions continous on the whole R excluding $\aleph_0$ points of discontinuity, because this problem is not so simple and you will either get non Darboux functions or the sequence will not be convergent uniformly.

all right, so what if i'm wrong again?? you wanna behead me or something?

[quote="demigod"]In other words, please [b]THINK[/b] thrice before posting another trivial combination of some functions continous on the whole R excluding $\aleph_0$ points of discontinuity, because this problem is not so simple and you will either get non Darboux functions or the sequence will not be convergent uniformly.[/quote]

Come on, be a little more polite.

Sorry, I was just irritated by having to think on where your example fails ;) I've thought on the problem for some time and a function sequence with discontinuity in 1 or n points doesn't work. I think I can easily proove it. One should look for the example among functions which are discontinous everywhere. Maybe I am wrong, and if you find an example among standard functions, that's gonna be :first: . But please think twice before posting something and leaving the thinking part to other people, this problem is quite non-trivial.  :D

Let $f$ be a function which takes every non-zero real value on every interval (and never vanishes), but such that $f(q_n)=\frac 1n,\ \forall n\ge 1$, where $(q_n)_{n\ge 1}$ is an enumeration of the rationals (such examples have been constructed before on the forum). Now take $f_n$ to be equal to $f$ everywhere except at $q_{n+1},q_{n+2},\ldots$, where it vanishes.

powerful technique, this enumeration of the rationals. thank you very much.


geometry
inequalities
perimeter
geometry unsolved

Let $ ABCD.A_1B_1C_1D_1$ be a cube. On its three skew edges, choose three points M, N, P. Determine the positions of the points M, N, P such that the triangle MNP has

1: The smallest permeter possible

2: The smallest area possible  :D

I am not sure about the meaning of "skew edges" (my English is not that good and the Google translator didn't help), but I will guess you are talking about edges wich are pairwise not coplanar.

The first part follows from the triangular inequality (i.e., $ \sqrt {a^2 \plus{} b^2 \plus{} c^2} \plus{} \sqrt {d^2 \plus{} e^2 \plus{} f^2} \ge \sqrt {(a \plus{} d)^2 \plus{} (b \plus{} e)^2 \plus{} (c \plus{} f)^2}$). Indeed, you can take $ M \equal{} (0, 1, m)$, $ N \equal{} (1, n, 0)$ and $ P \equal{} (p, 0, 1)$. The triangle's perimeter is

$ \sqrt {m^2 \plus{} (1 \minus{} n)^2 \plus{} 1} \plus{} \sqrt {1 \plus{} n^2 \plus{} (1 \minus{} p)^2} \plus{} \sqrt {(1 \minus{} m)^2 \plus{} 1^2 \plus{} p^2} \ge \sqrt {2^2 \plus{} 2^2 \plus{} 2^2}$

The equality occurs when $ m \equal{} n \equal{} p \equal{} 1/2$.

As for the second part, using Heron formula and opening (or computing the volume of the solid $ OMNP$, where $ O$ is the origin, and dividing this value by the distance from $ O$ to the plane $ MNP$), we see the area of our triangle is $ \frac {1}{2}\sqrt {\sum_{cyc} (1 \minus{} n \plus{} mn)^2}$. Now I don't know. But [url=http://www.mathlinks.ro/viewtopic.php?p=1525324#1525324]they[/url] know.

[quote="feliz"]I am not sure about the meaning of "skew edges" (my English is not that good and the Google translator didn't help), but I will guess you are talking about edges wich are pairwise not coplanar.

The first part follows from the triangular inequality (i.e., $ \sqrt {a^2 \plus{} b^2 \plus{} c^2} \plus{} \sqrt {d^2 \plus{} e^2 \plus{} f^2} \ge \sqrt {(a \plus{} d)^2 \plus{} (b \plus{} e)^2 \plus{} (c \plus{} f)^2}$). Indeed, you can take $ M \equal{} (0, 1, m)$, $ N \equal{} (1, n, 0)$ and $ P \equal{} (p, 0, 1)$. The triangle's perimeter is

$ \sqrt {m^2 \plus{} (1 \minus{} n)^2 \plus{} 1} \plus{} \sqrt {1 \plus{} n^2 \plus{} (1 \minus{} p)^2} \plus{} \sqrt {(1 \minus{} m)^2 \plus{} 1^2 \plus{} p^2} \ge \sqrt {2^2 \plus{} 2^2 \plus{} 2^2}$

The equality occurs when $ m \equal{} n \equal{} p \equal{} 1/2$.

As for the second part, using Heron formula and opening (or computing the volume of the solid $ OMNP$, where $ O$ is the origin, and dividing this value by the distance from $ O$ to the plane $ MNP$), we see the area of our triangle is $ \frac {1}{2}\sqrt {\sum_{cyc} (1 \minus{} n \plus{} mn)^2} \ (1)$. Now I don't know. But [url=http://www.mathlinks.ro/viewtopic.php?p=1525324#1525324]they[/url] know.[/quote]

Thank you very much . My english is very bad, too  :blush:  :blush:   :lol:  :lol:
(1) realy a inequality very hard. I think it is 1.5 when m = 0.5 ; n = 0 ; p = 1


This can of course be done with brute force algebra, but I like this way of organizing the work:

The given conditions can be restated as
$\left( \begin{array}{ccc}  1 & 1 & 1 \\  a & b & c \\ a^2 & b^2 & c^2 \end{array}\right) \cdot \left( \begin{array}{ccc} x \\ y \\  z \end{array}\right)  =  \left( \begin{array}{ccc} 1 \\ 1 \\  1 \end{array}\right) $.

We have $(\begin{array}{ccc} a^3 & b^3 & c^3 \end{array}) =(\begin{array}{ccc} abc & -ab-bc-ca & a+b+c \end{array}) \cdot \left( \begin{array}{ccc}  1 & 1 & 1 \\  a & b & c \\ a^2 & b^2 & c^2 \end{array}\right) $, so $a^3x+b^3y+c^3z = (\begin{array}{ccc} a^3 & b^3 & c^3 \end{array})\cdot \left( \begin{array}{ccc} x \\ y \\  z \end{array}\right) =$
$ (\begin{array}{ccc} abc & -ab-bc-ca & a+b+c \end{array}) \cdot \left( \begin{array}{ccc}  1 & 1 & 1 \\  a & b & c \\ a^2 & b^2 & c^2 \end{array}\right) \cdot \left( \begin{array}{ccc} x \\ y \\  z \end{array}\right) = $
$(\begin{array}{ccc} abc & -ab-bc-ca & a+b+c \end{array}) \cdot \left( \begin{array}{ccc} 1 \\ 1 \\  1 \end{array}\right) = $
$abc-ab-ac-bc+a+b+c=(a-1)(b-1)(c-1)+1 $.

Just follow [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15352]this[/url] generalization of the Newton sums (check out Harazi's post on that topic). Let $P_1=a+b+c,P_2=ab+bc+ca,P_3=abc$ and $S_i=\sum a^ix$. We get $S_3=P_1S_2-P_2S_1+P_3S_0$, and since $S_2=S_1=S_0=1$, we find $S_3=P_1-P_2+P_3=a+b+c-(ab+bc+ca)+abc=1-(1-a)(1-b)(1-c)$, Q.E.D.


Find all n such that $n^{4}+6n^{3}+11n^{2}+3n+31$ is a perfect square.

So I think it's safe to assume that $n$ is an integer

(otherwise, the expression is equal to each perfect square $\geq 36$ for one positive and one negative $n$ value)

[hide="Solution"]Let that be $f(n)$

$(n^{2}+3n+1)^{2}-f(n)=3n+30$, positive for $n>-10$
$f(n)-(n^{2}+3n)^{2}=2n^{2}+3n+31$, always positive.

$n$ cannot be greater than $-10$, or else $f(n)$ lies between two perfect squares.

$(n^{2}+3n+2)^{2}-f(n)=2n^{2}+9n-27$ is positive for all $n \leq-10$, so for $n \leq-10$, $f(n)$ is not a perfect square unless $f(n)=(n^{2}+3n+1)^{2}\implies (n^{2}+3n+1)^{2}-f(n) = 0$
$\implies 3n+30 = 0 \implies n=-10$

Thus, $\boxed{n=-10}$ is the only integer solution. Indeed, $f(-10)=71^{2}$[/hide]

You made a mistake.
$(n^{2}+3n+1)^{2}-f(n) = 3n-30$
Though it seems right other than that. Thanks.

And n is integer.


The sum of two numbers is 6. The difference of their squares is 12. What is the positive difference of the two numbers?

Just set up a system of equation

$ x \plus{} y \equal{} 6$
$ x^2 \equal{} y^2 \plus{} 12$

Then solve

$ x \equal{} 6 \minus{} y$
$ ( 6 \minus{} y )^2 \equal{} y^2 \plus{} 12$
$ 36 \minus{} 12y \plus{} y^2 \equal{} y^2 \plus{} 12$
$ 12y \equal{} 24$
$ y \equal{} 2$

Then solve for x.

$ y\equal{}2 x\equal{}4 x\minus{}y\equal{}2$

A more direct approach:

$ x \plus{} y \equal{} 6$

$ x^2 \minus{} y^2 \equal{} 12 \equal{} (x \plus{} y) (x \minus{} y)$

Substituting the quantity in the first equation into the second, we obtain:

$ 12 \equal{} 6 (x \minus{} y)$

$ x \minus{} y \equal{} 2$


Putnam
function
integration
calculus
calculus computations

Given that there is a unique function $f(x)$ such that $f(0)=1$, and $f'(x) = f(x)+\int_{0}^{1}f(t) \ dt$, find that function.

[hide="My Solution:"]Solving the differential equation $f'(x)-f(x) = \int_{0}^{1}f(t) \ dt$, gives:

$f(x) = Ce^{x}-\int_{0}^{1}f(t) \ dt\Rightarrow f(0)= C-\int_{0}^{1}f(t) \,dt \Rightarrow C=1+\int_{0}^{1}f(t) \ dt$.

Now, we have $f(x) = e^{x}+(e^{x}-1)\int_{0}^{1}f(t) \ dt$.

We also have $\int_{0}^{1}f(t) \ dt = \left(f'(x)-f(x)\right) \Rightarrow \int_{0}^{1}f(t) \ dt = \int_{0}^{1}e^{t}+(e^{t}-1)\left(f'(x)-f(x)\right) \ dt$

Let $I=\int_{0}^{1}f(t) \ dt$, then we have $I=e-1+(e-2)I\Rightarrow I=\frac{1-e}{e-3}$, and thus

$\boxed{f(x) = e^{x}+(e^{x}-1)\left(\frac{1-e}{e-3}\right)}$ [/hide]

$f(x) = f'(x)+k \Longrightarrow f'(x) = f''(x) \Longrightarrow f'(x) =C e^{x}$

[quote="Jimmy"]
We also have $\int_{0}^{1}f(t) \ dt = \left(f'(x)-f(x)\right) \Rightarrow \int_{0}^{1}f(t) \ dt = \int_{0}^{1}e^{t}+(e^{t}-1)\left(f'(x)-f(x)\right) \ dt$
[/hide][/quote]

I understand the first part but could you please explain how you got

$\Rightarrow \int_{0}^{1}f(t) \ dt = \int_{0}^{1}e^{t}+(e^{t}-1)\left(f'(x)-f(x)\right) \ dt$

Substitute $t$ into $f(x) : f(t) = e^{t}+(e^{t}-1)\int_{0}^{1}{f(t)}\,dt$
Then subst $\int_{0}^{1}{f(t)}\,dt = f'(x)-f(x)$ (the given diff equation), then you'll get it  :D


Hey I know this seems like an odd question, but can anyone define factoring for me and some of its basics?

Thanks.

That is odd. Factoring is defined as : [b]the resolution of an integer or polynomial into factors such that when multiplied together they give the integer or polynomial[/b]

Stated simply:

Factoring is just using the distributive law ($a(c+d)=ac+ad$) in reverse.


How much did money play a role in where you applied and decided to go to college?

A lot.  As it turned out, the school I liked the most was also one of the two schools that I could afford, so cost didn't actually affect where I decided on.  But it did eliminate almost every school I applied to, including most of my favorites.


geometry
incenter
circumcircle
parallelogram
trigonometry
geometry proposed

Let $ABC$ be an acute triangle with $AB \neq AC$. Let $D$ be the foot of the altitude from $A$ and $\omega$ the circumcircle of the triangle. Let $\omega_1$ be the circle tangent to $AD$, $BD$ and $\omega$. Let $\omega_2$ be the circle tangent to $AD$, $CD$ and $\omega$. Let $\ell$ be the interior common tangent to both $\omega_1$ and $\omega_2$, different from $AD$.

Prove that $\ell$ passes through the midpoint of $BC$ if and only if $2BC = AB + AC$.

Denote by $O_1$ the center of $\omega_1$ and by $O_2$ the center of $\omega_2$.
From [i]Thebault's Theorem[/i] we obtain that $O_1,I,O_2$ are collinear. Consequently, $I$ is the midpoint of $O_1O_2$.
Finally,(since $O_1DMO_2$ is cyclic) we get that $ID=IM$ ($M$ -the midpoint of $BC$), thus $2a=b+c$.

Can you detail more your soltution? It seems to be a great solution. My solutions uses Cayley th twice and a lot of calculations... Your solution seems to avoid the calculations.
The thebault theorem I found sais: On the sides of parallelogram ABCD erect squares -- all either on the outside or the inside of the parallelogram. Their centers then form another square. I don't see where is the link with this problem.

Actually, [i]Thebault's Theorem[/i] states that points $O_1,I,O_2$ are collinear.
A lemma used in its proof is the following fact:
If $P$ is the midpoint of $M_1M_2$ (these are the tangency points of the two circles with $BC$) then $IP||AD$.
Thus, in this case, we get that $I$ is the midpoint of $O_1O_2$.
Finally, it is enough to notice that:
\[ \angle{O_1DO_2}=\angle{O_1MO_2}=90^{\cdot} \]
and the conclusion follows.

[color=darkred][b][u]Lemma 1.[/u][/b] let $w_1$ be a circle which is interior tangent in the point $A$ to the circle $w$. For a point $M\in w_1$, $M\ne A$ denote the intersections $X,Y$ between the circle $w$ and the tangent-line in the point $M$ to the circle $w_1$. Then the ray $[AM$ is the bisector of the angle $\widehat {XAY}$.
[b][u]Lemma 2.[/u][/b] For the incenter $I$ and the centroid $G$ of the triangle $ABC$ we have
$\blacktriangleright IG\parallel BC\Longleftrightarrow b+c=2a\ .$
$\blacktriangleright IG\perp BC\Longleftrightarrow b=c\ \ \vee\ \ b+c=3a\ .$[/color]

[color=darkblue][b][u]Indication for proof the proposed problem.[/u][/b] The circles $w_1,w_2$ are interior tangent to the circle $w$. You can apply the above remarkable lemma for each of the circles $w_1,w_2$ and for each of the lines $BC,AD,l$.[/color]

Every one has used this theoerm < Thebault's Theorem> can some give link or its proof
Abdurashid

[quote="Sailor"]If $P$ is the midpoint of $M_{1}M_{2}$ (these are the tangency points of the two circles with $BC$), then $IP \| AD$.[/quote]
Could you (or anyone else) give a proof of this fact? The rest is quite clear, but I can't figure this part.

I didn't know the statement Sailor said, but I was pretty sure that the general statement of Thebault theorem says that: $ O_{1}, I, O_{2}$ are collinear [b]and[/b] $ \frac {O_{1}I}{O_{2}I} \equal{} \tan^{2}{\frac {\alpha}{2}}$, where $ \angle{ADB} \equal{} \alpha$, where $ AD$ is the "Thebault cevian".

In this case, $ \alpha \equal{} 90$, so the conclusion follows.

Here you can see, the statement of this thing, followed by another nice ones. [url]http://links.jstor.org/sici?sici=0002-9890%28200204%29109%3A4%3C362%3ATAOTGP%3E2.0.CO%3B2-A&size=SMALL&origin=JSTOR-reducePage[/url]

PS: sorry for the long link, i googled it.

Here is a full solution to this problem using Sawayama's Theorem and some length chasing.

[asy]
size(250);
defaultpen(linewidth(0.7)+fontsize(11pt));
pair A=dir(120),B=dir(210),C=dir(330),D=foot(A,B,C),I=incenter(A,B,C),Ap=(3*A+2*D)/5, Dp=1.4*D-0.4*A;
path alt = Ap--Dp;
real r1=0.222, r2=0.708;
pair O1=(D.x-r1,D.y+r1), O2=(D.x+r2,D.y+r2);
path c1=circle(O1,r1),c2=circle(O2,r2);
draw(A--B--C--A--D^^unitcircle^^c1^^c2);
pair M=(foot(O1,B,C)+foot(O2,B,C))/2, Mp=foot(M,O1,O2);
path ellp = reflect(M,Mp)*alt;
pair F=foot(O1,reflect(M,Mp)*A,reflect(M,Mp)*D),P1=intersectionpoint(O1--F,c1);
path ell=shift(P1-F)*ellp;
pair P=intersectionpoint(ell,B--C);
draw(incircle(A,B,C)^^foot(O1,B,C)--I^^foot(O2,B,C)--foot(O2,A,D),linetype("4 4"));
draw(rightanglemark(A,D,C,2));
dot(I^^foot(I,B,C)^^O1^^O2);
draw(ell);
label("$A$",A,dir(origin--A));
label("$B$",B,dir(origin--B));
label("$C$",C,dir(origin--C));
label("$D$",D,S);
label("$P$",P,S);
label("$F_1$",foot(O1,B,C),S);
label("$E_1$",foot(O1,A,D),dir(160));
label("$F_2$",foot(O2,B,C),S);
label("$E_2$",foot(O2,A,D),dir(20));
label("$I$",I,N);
label("$T$",foot(I,B,C),S);
label("$O_1$",O1,dir(95));
label("$O_2$",O2,N);
[/asy]
WLOG let $AB<AC$.  We first define several auxiliary points.  Let $O_1$ and $O_2$ denote the centers of $\omega_1$ and $\omega_2$ respectively.  Let the points of tangency of $\omega_1$ with $AD$ and $BD$ are $E_1$ and $F_1$ respectively; define $E_2$ and $F_2$ in a similar manner.  Denote by $I$ the incenter of $\triangle ABC$, and let $T$ be the point where its incircle is tangent to $BC$.  Furthermore, let $P=\ell\cap BC$.  Finally, let $P_1$ and $P_2$ be the points of tangency of $\ell$ with $\omega_1$ and $\omega_2$ respectively.

By Sawayama's Theorem, $E_1F_1$ passes through the incenter $I$ of $\triangle ABC$.  Since $DE_1=DF_1$, $\angle IF_1F_2=45^\circ$.  Similarly, $\angle IF_2F_1=45^\circ$.  Therefore $\triangle IF_1F_2$ is an isosceles right triangle, implying that $T$ is the midpoint of $\overline{F_1F_2}$.

Now we introduce a lemma:

[b]LEMMA: [/b] $P$ is the reflection of $D$ over $T$.

[i]Proof[/i].  First examine $\omega_1$.  Since $O_1F_1\perp BC$ and $O_1P_1\perp\ell$, quadrilateral $O_1F_1PP_1$ is cyclic, and combining this with $O_1F_1=O_1P_1$ gives that $PO_1$ bisects $\angle F_1PP_1$.  Similarly, $PO_2$ bisects $\angle F_2PP_2$.  As a result, $O_1P\perp O_2P$, which implies $\angle O_1PF_1$ and $\angle O_2PF_2$ are complementary $\implies \triangle O_1PF_1\sim\triangle PO_2F_2$.

Now let $r_1$ and $r_2$ be the radii of $\omega_1$ and $\omega_2$ respectively.  Remark that $PF_1+PF_2=F_1F_2=r_1+r_2$ from the fact that $AD\perp BC$ and that $PF_1\cdot PF_2=r_1r_2$ from the above similarity.  Thus, we may conclude that $\{PF_1,PF_2\}=\{r_1,r_2\}$.  However, since $AC>AB$, $r_2>r_1$, so the line $\ell$ will slope in such a way that $PF_1>PF_2$.  Thus $PF_1=r_2$ and $PF_2=r_1$.  Finally, note that $DF_1=r_1$ and $DF_2=r_2$; this reversal of lengths implies that $P$ and $D$ are reflections about the midpoint of $\overline{F_1F_2}$, which is $T$.  $\blacksquare$

This allows us to completely get rid of $\omega_1$ and $\omega_2$ and turn this problem entirely into a length-chasing one.  Assume general triangle notation, so that $a=BC$, etc.  We now prove the "only if" condition of the problem statement and then show that all our steps are reversible, thus proving the "if" condition as well.

[b]ONLY IF:[/b] Suppose that $P$ is indeed the midpoint of $\overline{BC}$.  Then $BP=CP=\tfrac a2$.  Furthermore, it is well-known that $BT=s-b$ and $CT=s-c$.  Therefore \[BD=2\cdot BT-BP=2(s-b)-\dfrac a2=a+b+c-2b-\dfrac a2=\dfrac a2+c-b.\]  Similarly, $CD=\tfrac a2+b-c$.  Now from the Pythagorean Theorem $AB^2-BD^2=AD^2=AC^2-CD^2$.  Plugging in everything we know gives \begin{align*}c^2-\left(\dfrac a2+c-b\right)^2&=b^2-\left(\dfrac a2-b+c\right)^2\\4c^2-(a+2c-2b)^2&=4b^2-(a+2b-2c)^2\\4(c^2-b^2)&=(a+2c-2b)^2-(a+2b-2c)^2\\&=[a+2c-2b-(a+2b-2c)][a+2c-2b+(a+2b-2c)]\\&=8a(c-b)\\\implies b+c&=2a\end{align*} as desired.

[b]IF:[/b] Now that all the algebra above has been hacked out, the "if" clause is easy.  Start with $b+c=2a$, and multiply both sides by $4(c-b)$.  Reversing all our steps above leads to the crucial equality \[c^2-\left(\dfrac a2+c-b\right)^2=b^2-\left(\dfrac a2-b+c\right)^2.\]  Remark that the converse of our use of the Pythagorean Theorem is true as well; in other words, if $AB^2-BD^2=AC^2-CD^2$, then $AD\perp BC$.  In this case, let $D'$ be the point for which $BD'=\tfrac a2+c-b$ and $CD'=\tfrac a2-b+c$.  By the perpendicularity criterion, $AD'\perp BC$.  Furthermore, since \[\left(\dfrac a2+c-b\right)+\left(\dfrac a2-b+c\right)=2\left(\dfrac a2\right)=a,\] points $B$, $D'$, and $C$ are collinear.  As a result, $D'\equiv D$.  Finally, we can reverse our original algebra using our Lemma to get $BP=CP=\tfrac a2$, proving this direction as well.

Now we have proven both directions, so we are done. $\blacksquare$

My solution:

Let $ I $ be the Incenter of $ \triangle ABC $ and $ T= \odot (I) \cap BC $ .
Let $ I_1 $ be the center of $ \omega_1 $ and $ T_1=\omega_1 \cap BC, S_1=\omega_1 \cap AD $ .
Let $ I_2 $ be the center of $ \omega_2 $ and $ T_2=\omega_2 \cap BC, S_2=\omega_2 \cap AD $ .
Let $ O $ be the circumcenter of $ \triangle ABC $ and $ M=AI \cap \odot (ABC), K= \ell \cap BC $ .

From [url=http://artofproblemsolving.com/community/c6h604601]here[/url] we get $ 2BC=AB+AC \Longleftrightarrow AI=IM $ . ... $ (1) $

From Sawayama's Theorem we get $ I \in S_1T_1, I \in S_2T_2 $ ,
so $ \angle IT_1T_2=\angle IT_2T_1=45^{\circ}  \Longrightarrow T $ is the midpoint of $ T_1T_2 $ ,
hence combine with $ T_1D=KT_2 $ (well-known) we get $ T $ is the midpoint of $ DK $ . ... $ (2) $

From $ (1), (2) $ we get $ K $ is the midpoint of $ BC \Longleftrightarrow 2BC=AB+AC $ .

Q.E.D

[quote=sailor]Finally, it is enough to notice that:
\[ \angle{O_1DO_2}=\angle{O_1MO_2}=90^{\cdot} \][/quote]

Why is $\angle{O_1MO_2}=90$? Thanks.

Does anyone know why? Thanks in advance.

[quote]Why is $\angle{O_1MO_2}=90$? Thanks.[/quote]
$P$ is of course a projection of $I$ on $AB$. Now since $I$ is a midpoint of $O_1O_2$, we get that $P$ is a midpoint of  $M_1M_2$. Besides due to fact that $OI \perp AI$, we have $PD=PM$, so if $IM_1=IM_2$, then $ID=IM$ q.e.d.

Dear Mathlinkers,
a link

Ayme J.-L., Sawayama and Thébault's theorem, Forum Geometricorum (2003) ; http://forumgeom.fau.edu/

Sincerely
Jean-Louis

[quote=Pinionrzek][quote] Besides due to fact that $OI \perp AI$.[/quote] [/quote]

Can you explain this in more detail?
Because that would mean that ,if $X$ is the midpoint of arc $BC$ not containing $A$,then $AI$=$IX$,and since $IX=BX = \frac{BC}{2\cos \frac{A}{2}}$ and $AI$=$\frac{\frac{b+c-a}{2}}{\cos \frac{A}{2}}$ we would directly get the conclusion.

Here's the easy explanation for why $\angle O_1MO_2 = 90$. We are assuming that the internal tangent other than $AD$ intersects $BC$ at $M$. Let this other internal tangent touch $\omega_1$ at $X_1$ and touch $\omega_2$ at $X_2$. Then, $O_1M$ bisects $\angle X_1MB$ and $O_2M$ bisects $\angle X_2MC$, so then $\angle O_1MO_2$ must be $90$.


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171804&start=1400
Well...since the old topic for this just..died? abruptly I decided to restart it. So just comment on the user above/below you

uhhh I suppose he/she likes the simpsons...

[u][b][color=darkred][size=200]TPAM IS THE COOLEST PERSON I KNOW!!!!!!!![/size][/color][/b][/u]

BELIEVE IT!

[size=200]AMOEBA/DUCK (TPAM) IS AMAZING TOO![/size]

TPAM is a 2-year old baby that i drowned  :P  :P  :P

TPAM is the cute quacker that let me die...

thank God for underwater breathing techniques :D

wait this isn't making any sense...

and mz we're the only ones replying ahahaha

Aww TPAM so modest...

TPAM IS A CUTIE!!! :O

...

TPAM can design a super-cool awesome website[color=white][size=9]*cough* no life *cough*[/size][/color] :lol:

[size=200]TPAM created the AoPS family tree!!! [/size]
[size=59](i think)[/size]

TPAM is stupid b/c he cannot figure that TPATPAM created the AOPS family tree

TPAM is still my auntie!
and TPATPATPAM is right

WRONG!!!! *ding ding ding*
TPAM used two  :P 's in his last post on this thread.
TPBM speaks English.

@the titan of doom: urusai baka means "shut up, idiot".

and no i do not speak english. i only know how to TYPE it. :P

TPAM was assuming that someone who can type english can also speak english...

TPBM has no social life. (if this is tru, contact me. we can be buddies! :D )

Weeeell...I have a social life, but not much of one...

TBAM is not not not a boy.

@pianogirl: at last! someone DIDNT mistake me for a boy! thank you!

and yes i think i am awesome. go me!

TPBM watches anime or reads manga at least a little...

TPBM or TPBTPBM or TPBTPBTPBM or etc. has taken the SAT (and if they're nice they'll pm me and tell me what its like... :oops: im worried, this year i'll be taking it... :(

Yes I have. I am feeling nice today but not nice enough to PM you. I'll write it here. Depending on whether or not this is the TIP program and your only in 7th grade or if this is "the real deal" the aspect is different. If its for the TIP program don't sweat it, it's not a very big deal and you'll do fine. If it's the real deal then study and I would recommend buying one of the SAT books. I hope this helps.

TPBM is not too awesome

ME IS AWESOME

TPAM was not feeling nice enough to pm TPATPAM.

TPBM likes pie?

Yes.

TPAM is a psychic.

TPBM has at least one letter in their username.

TPAM is correct, not that that is any guarantee of psychic abilities, because there are very few usernames without at least one letter in them.

TPBM has over 100 posts (I don't know...)

omg now that u say so, ur right! i amaze myself... :blush:

TPAM is a new member (as of august 22, 2008)

TPBM is not a new member (as of august 22, 2008)

Eh, i have over 1000 posts.  What am I?

TPAM has proven [b]her[/b]self as a psychic.

TPBM has proven him/herself as a psycho.

Wow your good, that is if crazy=phyco.

TPBM likes to lick lolipops.

actually, im terrified of them...loli loli loli POP! the mere thought of them gives me chills... :oops:  :D

TPAM is not terrified of the extremely terrifying race of aliens known as "lolipops".

TPBM has enough sense as to run at least a mile away when confronted by a "lolipop".

actually no. i'm the candy shop person in angie's family and i chuck yes. u guessed it lollipops at ppls.

TPAM is scared of me now
TPBM likes to eat candy (if not wow i'm scared)

TPAM is my granny! (and psychic!)
TPBM likes ice cream.


Let $ (C_1)$ and $ (C_2)$ are concentric circles, with center is point $ O$. Know that area of the ring (of the region) is $ 25\pi$. Let $ AB$ is a chord of $ (C_2)$ and $ AB$ contact with $ (C_1)$. Find leng of $ AB$.

$ A\equal{} \left ( R^2 \minus{} r^2 \right ) \pi$ and $ AB \equal{} 2\sqrt{R^2\minus{}r^2}$ so $ AB \equal{} 2\sqrt{\frac{A}{\pi}}$ in our case $ AB \equal{} 10$

[quote="¬[ƒ(Gabriel)³²¹º]¼"]$ A \equal{} \left ( R^2 \minus{} r^2 \right ) \pi$ and $ AB \equal{} 2\sqrt {R^2 \minus{} r^2}$ so $ AB \equal{} 2\sqrt {\frac {A}{\pi}}$ in our case $ AB \equal{} 10$[/quote]
[color=darkblue]Thank you very muck.[/color]


function
advanced fields
advanced fields unsolved

Hello,

The problem is to give an explicit solution to the number of ternary sequences of length n with a positive even number of 1s and a positive even number of 0s.  We can solve this problem with any method we want, but using recursion seems like the best bet.  I can get what seems to be a correct (maybe) recursive formula, but I'm having a difficult time getting an explicit solution for it.  If anyone could give me some hints to get started (perhaps recursion isn't the best way?)  it would be much appreciated.

Thank you

I just realized I was neglecting my good friend, the generating function.  I will post another reply after some experimentation with exponential GFs.

Exponential generating functions make the problem quite easy.

[hide]$\frac{3^n+(-1)^n + 6}{4} - 2^n$[/hide]


function
analytic geometry
graphing lines
slope
trigonometry

Won't do number 1 right now, but just find the slope and y intercept.

For number 2:
The period of $ y \equal{} a \sin bx$ is $ \frac {2\pi}{b}$
The amplitude is $ a$.

Number 3:
Multiply equation 1 by 3. You get 6x-12y=21. Adding the 2 equations together gives 0=42. This is false, so there is no solution.

[quote="bachelorgirl"]Am having real problems with the following question in a maths problems from my maths book. I'm trying to revise for a test ( and improve of course!) and am unsure about these:

1. Find the equation of the following straight lines. Give your answer in the form of y=mx+b
a) Through ( 3,-1) with slope of 1/3
b) Through (4,5) and (-3, 2)
c) Through (3,4) parallel to 3x-4y =5
d) Through (3,-2) parallel to the x-axis

2. What is the period and amplitude of the function y= 5sin3x ?

3. Simultaneous Equation:
2x- 4y=7
-6x +12y =21

Thank you very much![/quote]
[hide="1"]a) $ \minus{}1\equal{}1\plus{}b$, so $ b\equal{}\minus{}2$, so the line is $ y\equal{}\frac{1}{3}x\minus{}2$
b) Slope is $ \frac{3}{7}$. $ 2\equal{}\minus{}3(\frac{3}{7})\plus{}b$, so $ b\equal{}\frac{23}{7}$, so equation is $ y\equal{}\frac{3}{7}x\plus{}\frac{23}{7}$.
c) Since parallel, slope is $ \frac{3}{4}$. $ 4\equal{}\frac{9}{4}\plus{}b$. $ b\equal{}\frac{7}{4}$. Equation is $ y\equal{}\frac{3}{4}x\plus{}\frac{7}{4}$
d) Being parallel to the x-axis means that the slope is 0. Thus the equation is just $ y\equal{}\minus{}2$. [/hide]

[hide="2"]Period is $ \frac{2\pi}{B}$ in an equation of $ Asin(Bx\plus{}C)$. The amplitude is $ |A|$. So the period is $ \frac{2\pi}{3}$, and the amplitude is just $ 5$. [/hide]
[hide="3"]
Multiplying the first equation by 3 gets us $ 6x\minus{}12y\equal{}21$. Adding both equations results in $ 0\equal{}42$. Since this is not true, there is no solution set.[/hide]

Idk about the last two, but for 1, see this thread, and my post. Of course, you'll have to find the slope in different ways.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191175


analytic geometry
geometry
rotation
3D geometry

What is the resulting volume of the figure when the line segment $y=8x+4$, $-2<x<3$ is rotated about the x-axis?

The $<$ means less than or equal to. I'm still learning to use latex. :D

The line segment stretches between the points $(-2,-12)$ and $(3,28)$.  Rotating it around the x-axis will generate two cones, with the vertices at $(-.5,0)$.  One will have a radius of $12$ and a height of $1.5$, and the other will have a radius of $28$ and a height of $3.5$.  The formula for the area of a cone is $\frac13 \pi r^{2}h$.  For the first cone, this equals $72\pi$, and for the second it equals $\frac{2744\pi}{3}$.  The sum of these is, assuming I didn't make any stupid errors, $\frac{2960\pi}{3}$.

That was one of the ones I missed. lol

I skipped it basically.


inequalities AMC USA(J)MO USAMO function

Sorry if I'm wrong:
Maybe It's is similar to an old USAMO question..., I don't remember exactly, try to find it, arqady!!!

[quote="Lovasz"]Sorry if I wrong
[/quote]
No, I don't forgive you!  :)
And your back is white! :D

i hope i won't make any stupid mistake as i ain't that good in inequalities:
we have $n[(1-a_1)...(1-a_n)]^{1/n}\leq (n-\sum_{i=0}^na_i)$
thus $(n/{n-1})^{n-1}[(1-a_1)...(1-a_n)]\leq (n-\sum_{i=0}^na_i)^n/{n(n-1)^{n-1}}$
so the next step is to check if the inequality below is verified by the $a_i$:
$(n-\sum_{i=0}^na_i)^n/{n(n-1)^{n-1}}\leq (n-\sum_{i=0}^na_i)/{n\sum_{i=0}^na_i}$
which is equivalent to this easier inequality ;):
$(\sum_{i=0}^na_i)(n-\sum_{i=0}^na_i)^{n-1}\leq (n-1)^{n-1}$
for that inequality you just have to check the monotony of the function :$f: t\to t(n-t)^{n-1}$ for $t \in ]0,n]$ :).


A space shuttle is spotted flying at an angle of elevation of 17° when sighted from Cape Canaveral. At the moment it is sighted, it is directly overhead a ship located 12 mi from the Cape. The height of the shuttle at this moment is approximately?

This should go in HSB.

General strategy for this type of problem: Set up a right triangle, figure out which side you want to get, and set the sin/cos/tan of the known angle equal to the ratio of sides. Solve for the unknown side and you're done.


find all $ t \in \mathbb{R}$
such that $ \sqrt {7 \minus{} t} \plus{} \sqrt {7 \plus{} t} \in \mathbb{N}$

[hide="hint"]The only integers in the range are $ 4$ and $ 5$.  Just solve for both cases.[/hide]


lets say we have a m by n grid which is devided into mn one-by-one little squares.
what is the total number of rectangle that can be found in this grid?

[hide="not much of a hint"]
if m=n (i.e you have a n by n square), then the total number of rectangles is $(\frac{n(n+1)}{2})^2$[/hide]

There are $n+1$ vertical segments and $m+1$ horizontal segments. Since choosing exactly two unique vertical segments, and two unique horizontal segments uniquely determine a rectangle (the 4 intersections give the vertices of the rectangle), we can simply count how many ways there are to do this. Thus there are $\binom{n+1}{2}\binom{m+1}{2}$ ways to choose the bars, or equivalently, rectangles.

Applying this formula to the special case where $m=n$ gives the hint.

How could you do this the sides of the rectangle did not need to be alligned with the grid, i.e, if you could look that rotated rectangles with vertices on the grid?

It can also be written $\frac{m(m+1)n(n+1)}{4}$ or $T_{n}T_{m}$... :D


Sequence $ \{G_{n}\}$ is defined by: $ G_{0}\equal{}0, G_{1}\equal{}1$ and $ G_{n}\equal{}G_{n\minus{}1}\plus{}G_{n\minus{}2}\plus{}1$ for every $ n\geq 2$. Prove that for every positive integer $ m$ there exists two consecutive terms in the sequence that are both divisible by $ m$.

$ G_n\plus{}1\equal{}(G_{n\minus{}1}\plus{}1)\plus{}(G_{n\minus{}2}\plus{}1)$, Therefore $ G_n\equal{}F_{n\plus{}1}\minus{}1$.
If $ m|G_{n\minus{}1},G_n$, then $ F_{n\plus{}1}\equal{}1\mod m, m|F_{n\minus{}1}$. Let k is minimal number. suth that $ m|F_k$, then $ m|F_{2k},F_{2k\plus{}1}\equal{}1\mod m \to F_{2k\plus{}2}\equal{}1\mod m$, therefore work $ n\equal{}2k\plus{}1$.


Segundo nivel, problema 3:

Ocho niños de diferentes estaturas deben formar una fila ordenada de menor a mayor. Diremos que la fila tiene exactamente un error si hay un niño que está inmediatamente detrás de otro más alto que él, y todos los demás (salvo el primero de la fila) están inmediatamente después de uno más bajo. ¿De cuántas maneras pueden formar los ocho niños una fila con exactamente un error?

si fuesen  $n$ los  niños que se tiene que ordenar   con un error ,
primero se tiene que elegir el niño que no estara en su lugar.
 para el primer niño , hay $n-1$ niños mas altos que el y  $n-1$
lugares  donde  se estaria cometiendo un error  .
se nota que  para el niños del lugar $k$ hay $n-k$ mas altos que el
y por lo tanto  $n-k$  lugares  donde se puede cometer un error
entonces   la cantidad de formas para  ordenar a los niños   de tal manera
 que haya un error es la sumatoria de $1$ a $n-1$
para $n$ igual $8$ es  $28$

la respuesta correcta es 247. los 28 casos q has considerado son en los q solamente un niño esta fuera de lugar para causar el error, pero existen mas casos!


limit
linear algebra
matrix
real analysis
real analysis unsolved

Can anyone solve the following problem? I have been thinking for a while buy couldn't come up with an idea.

Let $\{ a_{n}\}$ be a bounded sequence. Suppose that for every bounded sequence $\{b_{n}\}$
we have
\[\limsup_{n\to\infty}(a_{n}+b_{n})=\limsup_{n\to\infty}a_{n}+\limsup_{n\to\infty}b_{n}. \]
Prove that $\{a_{n}\}$ is convergent.

Take $\varepsilon=\limsup a_{n}-\liminf a_{n}$. Let $\lim_{k\to\infty}a_{n_{k}}=\liminf a_{n}$. Take
\[b_{n}=\left\{\begin{matrix}\varepsilon,\ n\in\{n_{k}\}\\0,\ otherwise\end{matrix}\right.\]
Then $\limsup(a_{n}+b_{n})=\limsup a_{n}$ and $\limsup b_{n}=\varepsilon$. Hence $\varepsilon=0$

To RIP,

Thank you for your insight. But there is one step in your proof which I don't understand.

Namely, \[\limsup (a_{n}+b_{n}) = \limsup a_{n}\]

Could you explain that?

Obviously $\limsup(a_{n}+b_{n})\ge\limsup a_{n}$. Take any $\epsilon>0$. Then for all big n we have:1)$a_{n}<\limsup a_{n}+\epsilon$; 2) if $n\in\{n_{k}\}$ then $a_{n}<\liminf a_{n}+\epsilon.$ Hence $a_{n}+b_{n}<\limsup a_{n}+\epsilon$. Hence $\limsup(a_{n}+b_{n})\le\limsup a_{n}$

Or, let $b_{n}=-2a_{n}$, then
\[\limsup (-a_{n})=\limsup a_{n}+\limsup(-2a_{n}),\] hence
\[-\limsup (-a_{n})=\limsup a_{n},\] which is the same as
\[\liminf a_{n}=\limsup a_{n}\]


Oh ! if your problem is true then it's easy.it's old. :P  :D

[quote="silouan"]If $a,b,c>0$ and $a+b+c=3$  prove that

$\sqrt{\frac{a+b}{c+1}}+\sqrt{\frac{b+c}{a+1}}+\sqrt{\frac{c+a}{b+1}}\geq 3$[/quote]
$\sum_{cyc}\sqrt{\frac{a+b}{c+1}}\geq3\Leftrightarrow\sum_{cyc}(\sqrt{\frac{3-a}{1+a}}+\frac{1}{2}(a-1)-1)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-1)^2(3-a)}{\sqrt{\frac{3-a}{1+a}}+\frac{1}{2}(3-a)}\geq0.$ :)

My equivalent solution:

${\sqrt{\frac{a+b}{c+1}}} \ge \frac{a+b}{2}$.


Who was Fryer? I have heard of Galois and Hypatia but not of Fryer.

[quote="kool_dudy"]This was taken place today...

I did Fryer this morning.  I was wondering if anyone else took any of these three contests?[/quote]

I didn't, but what level of competition are they?  (High school?  Equivalent to AMC, AIME, or USAMO?)

They're Canadian. From what I heard, they're [b]easy[/b] for their grade levels. (Hypatia = grade 11; I'm not sure about the other two, having never written any of them.)

Fryer = 9
Galois = 10

They're easy contests if you were to comapre them with AMC.

THey're a bit like easy Olympiad questions where you have to prove and also explain your answers.

This is the last part of the last problem of Hypathia (gr 11), which shows exactly how "hard" the contest is:

4. (c)
3 green, 4 yellow and 5 red balls are placed in a bag. Two balls of different colours are selected at random, removed, and replaced with two balls of the third colour. Show that it is impossible for all of the remaining balls to be the same colour, no matter how many times this process is repeated.

Invitational math challenge (which these contests replaced) was so much tougher.


This trick is commonly used in school math clubs

You could simply use paper and multiply it out, but I guess that would defeat the purpose of trying to learn new tricks...  :huh:

Its also slower to use pencil and paper.

Although I more comomonly gaussintraining's method, there is another method that is still relative fast.

$ 46 \times 54 \equal{} (46 \times 50) \plus{} (46 \times 4) \equal{} \boxed{2484}$

Thanks.  :wink:


Prove that $ \sqrt{2} \plus{} \sqrt{3}$ is irrational and algebraic. Do the same for $ 2^{1/3} \plus{} 2^{2/3}$.

[hide]If $ \sqrt {2} \plus{} \sqrt {3}$ was rational, then it's square, $ 5 \plus{} 2 \sqrt {6}$ would be rational, contradiction. Let $ \sqrt {2} \plus{} \sqrt {3} \equal{} x$. Then $ x^2 \equal{} 5 \plus{} 2 \sqrt {6}$, so $ (x^2 \minus{} 5)^2 \equal{} 24$, so $ \sqrt {2} \plus{} \sqrt {3}$ is algebraic.
If $ 2^{2/3} \plus{} 2^{1/3}$ was rational, then $ 2^{2/3} \plus{} 2^{1/3} \plus{} 1$ would also be, so $ (2^{1/3} \minus{} 1)(2^{2/3} \plus{} 2^{1/3} \plus{} 1) \equal{} 1$ should be irrational, a contradiction. Anyways, if  $ 2^{2/3} \plus{} 2^{1/3} \equal{} x$ then $ (x \plus{} 1)(2^{1/3} \minus{} 1) \equal{} 1$, so $ 2 \equal{} (\frac {1}{x \plus{} 1} \plus{} 1)^3$, so $ 2^{2/3} \plus{} 2^{1/3}$ is algebraic.[/hide]

[quote="dgreenb801"]
If $ 2^{2/3} \plus{} 2^{1/3}$ was rational, then $ 2^{2/3} \plus{} 2^{1/3} \plus{} 1$ would also be, so $ (2^{1/3} \minus{} 1)(2^{2/3} \plus{} 2^{1/3} \plus{} 1) \equal{} 1$ should be irrational, a contradiction.[/quote]

Why should $ (2^{1/3} \minus{} 1)(2^{2/3} \plus{} 2^{1/3} \plus{} 1) \equal{} 1$ be irrational? But I get the first part there, thanks

[quote="aidan"][quote="dgreenb801"]
If $ 2^{2/3} \plus{} 2^{1/3}$ was rational, then $ 2^{2/3} \plus{} 2^{1/3} \plus{} 1$ would also be, so $ (2^{1/3} \minus{} 1)(2^{2/3} \plus{} 2^{1/3} \plus{} 1) \equal{} 1$ should be irrational, a contradiction.[/quote]

Why should $ (2^{1/3} \minus{} 1)(2^{2/3} \plus{} 2^{1/3} \plus{} 1) \equal{} 1$ be irrational? [/quote]
Because, an irrational times a rational is an irrational, and since $ 2^{1/3}$ is irrational, so is $ 2^{1/3}\minus{}1$.

Oh right - but do we have to prove that $ 2^{1/3}$ is irrational first?

Yes. We can prove it the same way we show $ \sqrt {2}$ is irrational. Suppose $ 2^{1/3} \equal{} \frac {m}{n}$, where the fraction $ \frac {m}{n}$ is in lowest terms, $ m$ and $ n$ are integers. Then $ m^3 \equal{} 2n^3$, so m must be even, so $ m \equal{} 2k$, so $ 4k^3 \equal{} n^3$, so $ n$ is even. But since $ m$ and $ n$ are both even, this contradicts the fact that $ \frac {m}{n}$ is in lowest terms, because we can take a 2 out of the numerator and denominator.


geometry
incenter
geometric transformation
vector
circumcircle
trigonometry
geometry solved

Let $\triangle ABC$ be a acute triangle and the points $M\in BC,N\in CA,P\in AB$ such that $MB=MP,MC=MN$. Prove that the points $N,P$, the incenter of the triangle $\triangle MNP$ and the orthocenter of the triangle $\triangle ABC$ are concyclically.

This one was an op02 problem, and I submitted a solution; here is just an extended writeup of this solution.

[color=blue][b]Problem Ukraine 6 / geo problem 192.[/b] Let $A_1$, $B_1$, $C_1$ be points on the sides BC, CA, AB, respectively, of an acute-angled triangle ABC such that $A_1B = A_1C_1$ and $A_1C = A_1B_1$. Let $I_1$ be the incenter of triangle $A_1B_1C_1$, and let H be the orthocenter of triangle ABC. Prove that the points $B_1$, $C_1$, $I_1$ and H lie on the same circle.[/color]

[i]Solution.[/i] Since $A_1B = A_1C_1$, the triangle $BA_1C_1$ is isosceles; thus, the angle at its apex is $\measuredangle BA_1C_1=180^{\circ}-2\cdot\measuredangle C_1BA_1=180^{\circ}-2B$. Similarly, $\measuredangle CA_1B_1=180^{\circ}-2C$. Hence,

$\measuredangle C_1A_1B_1=180^{\circ}-\measuredangle BA_1C_1-\measuredangle CA_1B_1=180^{\circ}-\left(180^{\circ}-2B\right)-\left(180^{\circ}-2C\right)$
$=2\left(B+C\right)-180^{\circ}=2\left(180^{\circ}-A\right)-180^{\circ}$          (since B + C = 180° - A by the sum of angles in triangle ABC)
$=180^{\circ}-2A$,

so that the sum of angles in triangle $A_1B_1C_1$ yields

$\measuredangle A_1C_1B_1+\measuredangle A_1B_1C_1=180^{\circ}-\measuredangle C_1A_1B_1=180^{\circ}-\left(180^{\circ}-2A\right)=2A$.

But since the point $I_1$ is the incenter of the triangle $A_1B_1C_1$, it lies on the angle bisectors of the angles $A_1C_1B_1$ and $A_1B_1C_1$ of this triangle, and thus $\measuredangle I_1C_1B_1=\frac{\measuredangle A_1C_1B_1}{2}$ and $\measuredangle I_1B_1C_1=\frac{\measuredangle A_1B_1C_1}{2}$. Hence, the sum of angles in triangle $B_1I_1C_1$ yields

$\measuredangle C_1I_1B_1=180^{\circ}-\measuredangle I_1C_1B_1-\measuredangle I_1B_1C_1=180^{\circ}-\frac{\measuredangle A_1C_1B_1}{2}-\frac{\measuredangle A_1B_1C_1}{2}$
$=180^{\circ}-\frac{\measuredangle A_1C_1B_1+\measuredangle A_1B_1C_1}{2}=180^{\circ}-\frac{2A}{2}=180^{\circ}-A=180^{\circ}-\measuredangle C_1AB_1$.

Therefore, the points A, $B_1$, $C_1$ and $I_1$ lie on one circle. If we succeed to show that the point H also lies on this circle, then we are done.

Well, let the perpendiculars to the lines AB and CA at the points $C_1$ and $B_1$ meet the line BC at the points X and Y, respectively. Also, let D be the midpoint of the side BC of triangle ABC. The foot of the altitude of triangle ABC issuing from the vertex B will be called B'. We will use directed segments.

If we denote by E and F the midpoints of the segments $CB_1$ and CB', then $CE=\frac{CB_1}{2}$ and $CF=\frac{CB^{\prime}}{2}$, so that $EF=CF-CE=\frac{CB^{\prime}}{2}-\frac{CB_1}{2}=\frac{CB^{\prime}-CB_1}{2}=\frac{B_1B^{\prime}}{2}$,

so that $\frac{B_1B^{\prime}}{EF}=2$. Now, the line BB' is perpendicular to the line CA (since it is the altitude of triangle ABC issuing from the vertex B). Since the points F and D are the midpoints of the sides CB' and BC of the triangle BCB', the line FD is parallel to the line BB', and hence also perpendicular to the line CA. The line $B_1Y$ is trivially perpendicular to the line CA. Finally, the line $EA_1$ is perpendicular to the line CA (since the triangle $CA_1B_1$ is isosceles and the point E is the midpoint of its base $CB_1$, so the line $EA_1$ joins the apex $A_1$ of this isosceles triangle with the midpoint of its base $CB_1$ and is thus perpendicular to its base $CB_1$, i. e. perpendicular to the line CA). Hence, the lines BB', FD, $B_1Y$ and $EA_1$ are parallel to each other, since they are all perpendicular to the line CA. Hence, by Thales, $\frac{YB}{A_1D}=\frac{B_1B^{\prime}}{EF}$. Since $\frac{B_1B^{\prime}}{EF}=2$, this becomes $\frac{YB}{A_1D}=2$, and thus $YB=2\cdot A_1D$. Similarly, $XC=2\cdot A_1D$. Hence, YB = XC; in other words, BY = CX. Moreover, since the segments are directed and the points B, C, X and Y all lie on the line BC, this becomes $\overrightarrow{BY}=\overrightarrow{CX}$.

Now, the translation with translation vector $\overrightarrow{BY}=\overrightarrow{CX}$ maps the point B to the point Y and the point C to the point X. Let this translation map the point H to some point Z; then, ZY || HB and ZX || HC (since translations map lines to parallel lines) and ZH || BC (since the translation vector $\overrightarrow{BY}=\overrightarrow{CX}$ lies on the line BC). Now, since ZY || HB, but on the other hand $HB\perp CA$ (since the point H is the orthocenter of triangle ABC), we have $ZY\perp CA$; on the other hand, $B_1Y\perp CA$. Hence, the lines ZY and $B_1Y$ coincide, i. e. the points Z, Y and $B_1$ all lie on one line perpendicular to the line CA. Thus, $\measuredangle ZB_1A=90^{\circ}$. Similarly, $\measuredangle ZC_1A=90^{\circ}$. Finally, ZH || BC, combined with $HA\perp BC$ (since the point H is the orthocenter of triangle ABC) entails $ZH\perp HA$, and thus < ZHA = 90°.

Since $\measuredangle ZB_1A=90^{\circ}$, $\measuredangle ZC_1A=90^{\circ}$ and < ZHA = 90°, the points $B_1$, $C_1$ and H must lie on the circle with diameter ZA. As a consequence, the points A, $B_1$, $C_1$ and H lie on one circle, i. e. the circle through the points A, $B_1$, $C_1$ and $I_1$ also passes through the point H.

As we saw above, this completes the solution.

  Darij

Let $A_1\in (BC),\ B_1\in (CA),\ C_1\in (AB)$ such that $A_1B_1=A_1C,\ A_1C_1=A_1B$.

I note $x=BA_1,\ y=CA_1\ (x+y=a)$, the projections $D,E,F$ of the orthocenter $H$

on the lines $BC,CA,AB$ respectively and the circumcircle $w$ of the triangle $B_1AC_1$.

$1^{\circ}.$ (like to [u]Darij's solution[/u]).

$m(\widehat {B_1I_1C_1})=90^{\circ}+\frac 12m(\widehat {C_1A_1B_1})=$

$90^{\circ}+\frac 12[180^{\circ}-m(\widehat {BA_1C_1})-m(\widehat {CA_1B_1})]=$

${180^{\circ}-\frac 12 [(180^{\circ}-2B)+(180^{\circ}-2C}]=B+C$

$\Longrightarrow m(\widehat {B_1I_1C_1})=$ $B+C\Longrightarrow I_1\in w.$

$2^{\circ}$ ([u]own[/u]). $\ C_1F=|AC_1-AF|=|c-2x\cos B-b\cos A|=$

$|a\cos B-2x\cos B|=|a-2x|\cos B=|x-y|\cos B\Longrightarrow$

$\tan (\widehat {C_1HF})=\frac{FC_1}{FH}=\frac{|x-y|\cos B}{AH\cos B}=\frac{|x-y|}{HA}$.

Analogously, $\tan (\widehat {B_1HE})=\frac{|x-y|}{HA}\Longrightarrow$.

Therefore, $\widehat {C_1HF}\equiv \widehat {B_1HE}\Longrightarrow \widehat {C_1HB_1}\equiv \widehat {CHB}\Longrightarrow$

$m(\widehat {C_1HB_1})=180^{\circ}-A\Longrightarrow H\in w.$


calculus
derivative
integration
complex analysis

Let $ u(x,y) \equal{} xe^{\minus{}x^{2}\minus{}y^{2}}$ for $ (x,y)$ in the plane.

My question is: how do I compute the Fourier transform of $ u$ in $ x$.

I feel like I have tried everything from :

frequency differentiation: $ \Im[t^{n}f(t)](w) \equal{} i^{n}\frac{d^{n}}{dw^{n}}\hat{f}(w)$

to

fourier transform of a derivative: $ \Im[f^{n}(t)](w) \equal{} (iw)^{n}\hat{f}(w)$

and I m not getting anywhere. I feel like God hates me. Well not really, but the level of frustration is pretty high. If anyone has any help to offer, that would be much appreciated.

Hi Mstrip.  You know, I'd first get the answer via Mathematica:

[code]In[5]:= FourierTransform[x*Exp[-(x^2 + y^2)], y, t]

Out[5]= (E^(-(t^2/4) - x^2)*x)/Sqrt[2][/code]

then work out how to get there:

\begin{align*}
\textbf{F}_y\left\{xe^{-(x^2+y^2)}\right\}&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} xe^{-(x^2+y^2)}e^{-izy}dy;\;z\in\mathbb{C} \\
&=\frac{1}{\sqrt{2\pi}}xe^{-x^2}\int_{-\infty}^{\infty}\exp\left\{-(y+iz/2)^2+(iz/2)^2\right\}dy \\
&=\frac{1}{\sqrt{2 \pi}}xe^{-x^2}e^{-z^2/4}\int_{-\infty}^{\infty}e^{-(y+iz/2)^2}dy \\
&=\frac{1}{\sqrt{2 \pi}}xe^{-x^2}e^{-z^2/4}\sqrt{\pi} \\
&=\frac{1}{\sqrt{2}}xe^{-x^2}e^{-z^2/4}
\end{align*}

oh haha. I guess I should have tried that, I wish I had a home version, since it does cost like 800 bucks right?

but thanks.

I should have first suggested just looking in a table.  In a table I find:

\[{ \textbf{F}_t\{e^{-at^2}}\}=
\sqrt{\frac{1}{2a}}\exp\{-\frac{z^2}{4a}\}\]

which is the central calculation above.  Also, if you're interested, you can get a student version of Mathematica for a 100 or so dollars.  Just log onto their site for info.


can you explain more about Ferrari way to solve equation order four ? I need it instanly .


calculus
MIT
college
geometry
linear algebra

Is there a university nearby? Many colleges have programs that let a high school student take a few classes.
Multivariable calculus and linear algebra are still a good idea- they're pretty much universal requirements.

I'm in a similar position, having taken BC Calculus last year as a sophomore. I'm using MIT's OpenCourseWare, where MIT has put resources - lecture notes, assignments, calendars, exams - for all of its courses online.
This year, I'm working on 18.024 Calculus with Theory II, which covers linear algebra and multivariable calculus in an introductory but rigorous fashion, which you might want to check out. OCW also has various other courses, which you can check out at [url]http://ocw.mit.edu/OcwWeb/Mathematics/index.htm[/url].

However, if you can find any good college courses in your area, I would also look into that

Another possibility is to stop taking math for a year at school and let that slot be a different class, in which case you can get credits in electives whose categories require credits. Although math classes help with learning math, they aren't the only places/oppertunities to learn math.

Hm, calc rulz, d'you think I'd be able to go straight to 18.024 after BC Calc? It does say that one year of calculus is expected for just 18.014 Calculus with Theory I, even though it seems like I'm covering almost all of what 18.014 is about.

If you spend 1-2 months reading through Apostol Volume I and working on 18.014, then you should be ready for 18.024. What you need is a rigorous background in calculus.

Hmm, sounds like I could do that next summer. Thanks.

[quote="davidyko"]Hey. So I'm a sophomore in BC Calc right now, which is the highest class my school currently offers. I've been thinking about doing multivariable calculus or linear algebra etc. at EPGY but I've heard that it's (EPGY) pretty easy, and I really want to challenge myself. What are some alternatives to taking classes at EPGY?[/quote]

Me too!! :o

...except that I'm at a small international school that doesn't offer AP courses, so I have to take Calculus with EPGY.

One thing that confuses me about EGPY is the Calc A-B-C thing (M040, M041, M042). I'm ready to take AP Calc BC; would that be enough for M041? The website lists M040 "or equivalent" as a prerequisite.

I asked on the official website but it'll probably be a few days before I get an answer from them.

Hope this question makes at least a little sense... I'm really tired :stretcher:


geometry
circumcircle
power of a point
radical axis
geometry unsolved

Let $ ABC$ be a non-equilateral triangle, and $ 0<t<\frac{1}{2}$ a real number. Let points $ A_1,A_2,B_1,B_2,C_1,C_2$ be on $ AB,AB,BC,BC,CA,CA$ respectively, such that

$ \frac{AA_1}{AB}\equal{}\frac{A_2B}{AB}\equal{}\frac{BB_1}{BC}\equal{}\frac{B_2C}{BC}\equal{}\frac{CC_1}{CA}\equal{}\frac{C_2A}{CA}\equal{}t$.

Prove that the radical axis of the circle passing through $ A_1,B_1,C_1$ and that passing through $ A_2,B_2,C_2$ is independent of $ t$.

Does somebody know how to solve this problem?

For now I could only prove that the centroid of $ ABC$ certainly lies on the radical axis of the two circles.  You only need this: the power of the centroid $ G$ about the circumcircle of the triangle $ ABC$ is $ \minus{} \frac {1}{9}\left(BC^2 \plus{} CA^2 \plus{} AB^2\right)$.

Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=91305 .

Strong numerical evidence suggests that
- this radical axis is the line joining the centroid and the symmedian point of triangle ABC;
- the midpoint between the symmedian points of triangles $ A_1B_1C_1$ and $ A_2B_2C_2$ also seems to lie on the line joining the centroid and the symmedian point of triangle ABC.

Maybe this can help... As for now, I have given up geometry and particularly this problem for lack of time.

  darij

Maybe you could use this: The power of the symmedian $ K$ of $ ABC$ with side lengths $ a,b,c$ about the circumcenter of $ ABC$ is
\[ \minus{} \frac {3a^2b^2c^2}{\left(a^2 \plus{} b^2 \plus{} c^2\right)^2}\,.
\]
We know that the values of $ a^2 \plus{} b^2 \plus{} c^2$ (from the last post I made) for the triangles $ A_1B_1C_1$ and $ A_2B_2C_2$ are equal.  So, only need to check that $ a^2b^2c^2$ of both triangles are also the same.  This is a heavy computation like darij said.

PS: This argument no longer holds, as darij pointed out (see his comment below).

Please don't forget that the symmedian points of $ \triangle A_1B_1C_1$ and $ \triangle A_2B_2C_2$ are not the one of $ \triangle ABC$.

And thanks for the nice proof of the two-families theorem!

  darij