geometry
3D geometry
tetrahedron
sphere
geometric transformation
homothety
ratio

There is a regular tetrahedron ABCD with side t.
V is a point inside ABCD such that VA = a, VB = b, VC = c and VD = d. How to determine t in terms of a, b, c and d?
Please help.

Let $A'B'C'D'$ be an arbitrary regular tetrahedron with the side $t' = A'B' = ...$. Let $(A'), (B'), (C'), (D')$ be spheres with radii $a, b, c, d$ centered at the vertices $A', B', C', D'$, respectively. The external and internal homothety centers $G_1, H_1$ of the spheres $(A'), (B')$ lie on the line $A'B'$ and they can be easily found. Similarly, the external and internal homothety centers $G_2, H_2$ of the spheres $(A'), (C')$ lie on the line $A'C'$ and the external and internal homothety centers $G_3, H_3$ of the spheres $(A'), (D')$ lie on the line $A'D'$.

Generalization of a similar theorem for circles: The locus of points in 3D space with a constant ratio of distances $\frac a b$ from 2 given points $A', B'$ is a sphere $(M_1)$ passing through the the homothety centers $G_1, H_1$ of the spheres $(A'), (B')$ and centered on the line $A'B'$, i.e., at the midpoint $M_1$ of the segment $G_1H_1$. Similarly, the locus of points with a constant ratio $\frac a c$ of distances from 2 given points $A', C'$ is a sphere $(M_2)$ passing through the the homothety centers $G_2, H_2$ of the spheres $(A'), (C')$ and centered on the line $A'C'$ at the midpoint $M_2$ of the segment $G_2H_2$. Finally, the locus of points with a constant ratio $\frac a d$ of distances from 2 given points $A', D'$ is a sphere $(M_3)$ passing through the the homothety centers $G_3, H_3$ of the spheres $(A'), (D')$ and centered on the line $A'D'$ at the midpoint $M_3$ of the segment $G_3H_3$.

The spheres $(M_1), (M_2)$ either intersect in a circle, touch at a point (circle of zero radius), or do not intersect at all. If the 2 spheres intersect in a circle, the sphere $(M_3)$ either intersects this circle at 2 points $V_1', V_2'$, touches this circle at a single point $V'$, or does not intersect it at all. If the spheres $(M_1), (M_2)$ touch at a point, the sphere $(M_3)$ either passes through this point $V'$ or it does not. Hence, we can have two solutions $V_1', V_2'$, one solution $V'$, or no solutiuon of the problem.

The side $t = AB$ of the desired tetrahedron $ABCD$ is then obtained by a dilation of the resulting figure by a factor $\frac{a}{a'} = \frac{VA}{V'A'}$:

$t = AB = A'B'\ \frac{VA}{V'A'}$

WLOG, asume that $a \le b \le c \le d$. Let the arbitrary regular tetrahedron $A'B'C'D'$ have the side $t' = A'B' =
A'C' = A'D' = ... = a + b$, to simplify the calculation. Put the coordinate origin at the point $A'$ and the x-axis along the line $A'B'$, the x-coordinate of the point $B'$ and the z-coordinate of the point $D'$ being negative. Let $g_1, h_1$ be the x-coordinates of the homothety centers $G_1, H_1$. Then

$g_1 = -a$

$\frac{h_1 + a + b}{h_1} = \frac b a$

$h_1 = \frac{a(a + b)}{b - a}$

Let $r_1$ be the radius of the sphere $(M_1)$ and $m_1$ the x-coordinate of its center $M_1$. Then

$r_1 = \frac{h_1 - g_1}{2} = \frac 1 2 [\frac{a(a + b)}{b - a} + a] = \frac{ab}{b - a} = \frac{ab(a + b)}{b^2 - a^2}$

$m_1 = g_1 + r_1 = -a + \frac{ab}{b - a} = \frac{a^2}{b - a} = \frac{a^2(a + b)}{b^2 - a^2}$

If the side of the equilateral tetrahedron was $a + c$, we would get similar expressions for the radius $r_2 = \frac{ac}{c - a}$ of the sphere $(M_2)$ and for the distance $m_2 = \frac{a^2}{c - a}$ of its center $M_2$ from the origin $A'$. But sice the side of the tetrahedron $A'B'C'D'$ is $a + b$, we have to scale these expressions by the factor $\frac{a + b}{a + c}$:

$r_2 = \frac{ac(a + b)}{c^2 - a^2}$

$m_2 = \frac{a^2(a + b)}{c^2 - a^2}$

Similarly, the radius $r_3$ of the sphere $(M_3)$ and the distance $m_3$ of its center from the origin $A'$ are

$r_3 = \frac{ad(a + b)}{d^2 - a^2}$

$m_3 = \frac{a^2(a + b)}{d^2 - a^2}$

The line $A'M_2 \equiv A'C'$ passes through the origin $A'$ and has the direction cosines

$(\cos{60^o}, \cos{30^o}, \cos{90^o}) = (\frac 1 2, \frac{\sqrt 3}{2}, 0)$

Consequently, the coordinates of the point $M_2$ are

$M_2 = m_2 (\frac 1 2, \frac{\sqrt 3}{2}, 0)$

The line $A'M_3 \equiv A'D'$ also passes through the origin, the coordinates of the point $D'$ are

$(a + b)\ (-\frac 1 2, -\frac{\sqrt 3}{6}, -\sqrt{\frac 2 3})$

the direction cosines of the line $A'D'$ are

$(\frac 1 2, \frac{\sqrt 3}{6}, \sqrt{\frac 2 3})$

and the coordinates of the point $M_3$ are

$M_3 = m_3 (\frac 1 2, \frac{\sqrt 3}{6}, \sqrt{\frac 2 3})$

The equations of the spheres $(M_1), M_2), (M_3)$ are then

$(x - m_1)^2 + y^2 + z^2 = r_1^2$

$(x - \frac{m_2}{2})^2 + (y - m_2\ \frac{\sqrt 3}{2})^2 + z^2 = r_2^2$

$(x - \frac{m_3}{2})^2 + (y - m_3 \frac{\sqrt 3}{6})^2 + (z - m_3 \sqrt{\frac 2 3})^2 = r_3^2$

Solving these 3 equations for $x, y, z$, we could get the coordinates of the intersections of the spheres $(M_1), (M_2), (M_3)$, i.e., of the points $V_1', V_2'$. But what we actually need is the distance $q = \sqrt{x^2 + y^2 + z^2}$ of the point $V'$ from the origin:

$q^2 - 2m_1x + m_1^2 = r_1^2$

$q^2 - m_2 (x + y \sqrt 3) + m_2^2 = r_2^2$

$q^2 - m_3 (x + y \frac{\sqrt 3}{3} + 2z \sqrt{\frac 2 3}) + m_3^2 = r_3^2$

Calculating $x, y, z$ from the equations of $(M_1), (M_2), (M_3)$:

$x = \frac{q^2 - (r_1^2 - m_1^2)}{2m_1}$

$y = \frac{\frac{q^2 - (r_2^2 - m_2^2)}{m_2} - x}{\sqrt 3} = \frac{q^2 - (r_2^2 - m_2^2) - m_2x}{m_2\sqrt 3}$

$z = \frac{\frac{q^2 - (r_3^2 - m_3^2)}{m_3} - x - y \frac{\sqrt 3}{3}}{2 \sqrt{\frac 2 3}} = \frac{q^2 - (r_3^2 - m_3^2) - m_3 (x + y \frac{\sqrt 3}{3})}{2m_3 \sqrt{\frac 2 3}}$

Substitute $x$ (the 1st equation) to the 2nd and 3rd equations for $y, z$. Then substitute $y$ (the 2nd equation with $x$ substituted) to the 3rd equations for $z$. Now, if we square $x, y, z$ and add the results, in a finite numer of minutes, we get the following quadratic equation for $q^2$:

$8m_1^2m_2^2m_3^2 = q^4 [3(m_2^2m_3^2 + m_1^2m_3^2 + m_1^2m_2^2) - 2m_1m_2m_3(m_1 + m_2 + m_3) ]\ -$

$- q^2\ 2[(r_1^2 - m_1^2)(3m_2m_3 - m_1m_3 - m_1m_2)m_2m_3\ +$
[color=white]. . . .[/color] $+ (r_2^2 - m_2^2)(3m_1m_3 - m_2m_3 - m_1m_2)m_1m_3\ +$
[color=white]. . . .[/color] $+ (r_3^2 - m_3^2)(3m_1m_2 - m_2m_3 - m_1m_3)m_1m_2]\ +$

$+ 3 [m_2^2m_3^2(r_1^2 - m_1^2)^2 + m_1^2m_3^2 (r_2^2 - m_2^2)^2 + m_1^2m_2^2(r_3^2 - m_3^2)^2 ]\ -$
$- 2m_1m_2m_3 [m_3(r_1^2 - m_1^2)(r_2^2 - m_2^2) + m_1(r_2^2 - m_2^2)(r_3^2 - m_3^2) + m_2(r_3^2 - m_3^2)(r_1^2 - m_1^2)]$

symmetrical with respect to the cyclic exchange of the indices 1, 2, 3, as expected. Substituting the expressions for $r_1, m_1, r_2, m_2, r_3, m_3$ in terms of $a, b, c, d$ and simplifying leads to the quadratic equation

$q^4[3(a^4 + b^4 + c^4 + d^4) - 2(a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + c^2d^2 + b^2d^2)]\ -$

$- q^2a^2(a + b)^2 2(a^2 + b^2 + c^2 + d^2) + 3a^4(a + b)^4 = 0$

The side $t = AB$ of the desired equilateral tetrahedron $ABCD$ is

$t = AB = A'B'\ \frac{VA}{V'A'} = (a + b) \frac a q$

$q = \frac{a(a + b)}{t}$

$\frac{1}{t^4}\ [3(a^4 + b^4 + c^4 + d^4) - 2(a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 +c^2d^2)]\ -$

$- \frac{1}{t^2}\ 2(a^2 + b^2 + c^2 + d^2) + 3 = 0$

$3t^4 - 2(a^2 + b^2 + c^2 + d^2)t^2\ +$

$+ 3(a^4 + b^4 + c^4 + d^4) - 2(a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 +c^2d^2) = 0$

or

$3t^4 - 2(a^2 + b^2 + c^2 + d^2)t^2 + 4(a^4 + b^4 + c^4 + d^4) - (a^2 + b^2 + c^2 + d^2)^2 = 0$

This quadratic equation is symmetrical to the cyclic exchange of $a, b, c, d$, as required. Putting

$u^2 = a^2 + b^2 + c^2 + d^2$

$v^4 = a^4 + b^4 + c^4 + d^4$

we get

$3t^4 - 2u^2t^2 + 4v^4 - u^4 = 0$

$t^2 = \frac{2u^2 \pm \sqrt{4u^4 - 12(4v^4 - u^4)}}{6} = \frac{u^2 \pm 2 \sqrt{u^4 - 3v^4}}{3}$

The solution exists if $u^4 - 3v^4 = (a^2 + b^2 + c^2 + d^2)^2 - 3(a^4 + b^4 + c^4 + d^4) \ge 0$. If 4 integers $a, b, c, d$ are found, such that this expression is a full square, the tetrahedron side would be rational and perhaps even integer.

Maybe this can help: It is known that every regular tetrahedron can be inscribed in a cube. [i]If you have a cube $ABCDA_1B_1C_1D_1$ then $ACD_1B_1$ is a regular tetrahedron.[/i] However, I don't see how this may be used at the moment.

[quote="luimichael"]Dear yetti,
I tried to verify the result of your formula with a special case as follows:
Regular tetrahedron ABCD with each side equals 2 units.
Consider the special point H which is the orthogonal projection of D on the face ABC.
With some calculations, a = b = c = 2/ :sqrt: 3 and d =  :sqrt: (8/3).
Substitute all these values of your equation, I obtained
3t^4 - 40/3 t 2 + 560/9 = 0.
But this equation does not give rise to a root of t = 2 by direct checking.
There must be something wrong anyway.
Yours, Michael.[/quote]

It is a nice check and I should have done it myself. But there is nothing wrong with the solution, you just made some arithmetic error.

$a = b = c = \frac{2}{\sqrt 3},\ \ d = \sqrt{\frac 8 3}$

$a^2 + b^2 + c^2 + d^2 = \frac 4 3 + \frac 4 3 + \frac 4 3 + \frac 8 3 = \frac{20}{3}$

$a^4 + b^4 + c^4 + d^4 = \frac{16}{9} + \frac{16}{9} + \frac{16}{9} + \frac{64}{9} = \frac{112}{9}$

$3t^4 - 2(a^2 + b^2 + c^2 + d^2)t^2 + 4(a^4 + b^4 + c^4 + d^4) - (a^2 + b^2 + c^2 + d^2)^2 =$

$= 3t^4 - \frac{2 \cdot 20}{3}\ t^2 + \frac{4 \cdot 112}{9} - \frac{400}{9} =$

$= 3t^4 - \frac{40}{3}\ t^2 + \frac{48}{9} = 3t^4 - \frac{40}{3}\ t^2 + \frac{16}{3} = 0$

$9t^4 - 40t^2 + 16 = 0$

$t^2 = \frac{40 \pm \sqrt{1600 - 4 \cdot 144}}{18} = \frac{40 \pm \sqrt{1024}}{18} = \frac{40 \pm 32}{18} =$

$ = \left(\frac{8}{18}\ \text{or}\ \frac{72}{18}\right) = \left(\frac 4 9\ \text{or}\ 4\right)$

But this is $t^2$. Then $t = \frac 2 3\ \text{or}\ 2$ as required. The first root corresponds to a point $V$ outside of the regular tetrahedron and the second one to a point $V$ on the face $\triangle ABC$ at its center.

:blush: Sorry for making such a simple mistake and thanks a lot for providing such a beautiful formula. I enjoy to see the symmetry of the coefficients and also the simplicity of the formula for this difficult problem to me.
 :)


$ x,y,z$ natural number . Solve:

$ 5(xyz\plus{}x\plus{}z)\equal{}222(yz\plus{}1)$

[hide="Hidden by mod"]It's easy if we use brute-force!
Note that the equation is equivalent to: (5x-222)(yz+1)=-5z
Now if z=0 we easily prove that the equation doesn;t have any solutions
So z#0. Now the RHS<=-5 which means LHS<=-5<0, WHICH MEANS X=<43
Just consider the case when x=0, we brute force totally 44 cases to get the result of x.The rest of the work is factorization into: (k-z)(l-y)=m with k,l,m are integers :)[/hide]

why don't we try another way . Rewrite it as :
$ x\plus{}\frac{1}{y\plus{}\frac{1}{z}}\equal{}44\plus{}\frac{1}{1\plus{}\frac{1}{\frac{2}{3}}}$
so we got $ x\equal{}44 ; y\equal{}1 ; z\equal{}\frac{2}{3}$

Very nice solution,HTA :) .But you will have to prove a lemma because it's not really obvious to match those terms! :P

Nice.

\rightarrow $ \rightarrow$.
\longrightarrow $ \longrightarrow$.

[quote="HTA"]why don't we try another way . Rewrite it as :
$ x \plus{} \frac {1}{y \plus{} \frac {1}{z}} \equal{} 44 \plus{} \frac {1}{1 \plus{} \frac {1}{\frac {2}{3}}}$
so we got $ x \equal{} 44 ; y \equal{} 1 ; z \equal{} \frac {2}{3}$[/quote]

But z should be a natural number.

I made a mistake , actually, it was :
$ x \plus{} \frac {1}{y \plus{} \frac {1}{z}} \equal{} 44 \plus{} \frac {1}{2 \plus{} \frac {1}{2}}$
after then , we obtain x=44;y=2;z=2
 :)


Let $ n$ be an odd integer, $ n > 3$. Prove that there exists an integer $ a$ satisfying $ n!$ divides $ a^n \plus{} 1$, but not divide $ a \plus{} 1$

It is not true for n=p. Let $ 5!|a^p\plus{}1$, then a odd and $ 8|a\plus{}1$, because $ a^4\minus{}a^3\plus{}a^2\minus{}a\plus{}1$ is odd.
$ 3|a^5\plus{}1\to 3|a\plus{}1$, $ 5|a^5\plus{}1\to a\not equiv 0\mod 5\to a\plus{}1\equiv a^5\plus{}1\mod 5$, therefore $ 120|a^5\plus{}1\to 120|a\plus{}1$.
It is not true for n - prime and true for odd composite n.

OK, n is an odd composite. You should give a proof for this problem if you know this. Thanks for noting.

This is practically an ISL problem (N4 of 2005).

[quote="thaithuan_GC"]Thanks for nohing.[/quote]
I hope this is meant to be "noting" and not "nothing"...

Let $ n$ is odd prime and $ n!|a^n\plus{}1$, then for prime $ p<n$ we get $ p^{ord_p(n!)}|a^n\plus{}1$. Because $ (p\minus{}1,n)\equal{}1$ must be $ p|a\plus{}1$ and because $ p\not |n$ must be $ ord_p(a^n\plus{}1)\equal{}ord_p(a\plus{}1)$. For p=n $ p|a\plus{}1$ and $ ord_p(n!)\equal{}1$. It give $ n!|a^n\plus{}1\to n!|a^n\plus{}1$.
Let $ n$ is odd composite, let $ p|a\plus{}1$, we can chose a, suth that $ a\equal{}\minus{}1 mod q^{ord_q(n!)}, q\not \equal{}p,q\le n$ and $ a\equal{}\minus{}1\mod p^{ord_p(n!)\minus{}1}$ for $ p|n,p<n$, suth that $ p^{ord_p(n!)}\not |a\plus{}1$.


linear algebra
matrix
algebra
system of equations
algebra unsolved

Solve equations

\[\left\{\begin{matrix}x^4+y^2-xy^3-\tfrac{9}{8}x=0 & \\

 y^4+x^2-yx^3-\tfrac{9}{8}y=0
& \end{matrix}\right.
\]


invariant
trigonometry
real analysis
real analysis unsolved

the limit is $\sqrt{a_{0}b_{0}}.$ :wink:

Definitely not.  $a_{0}> b_{0}$.  If $a_{n}> b_{n}$ then $a_{n}> a_{n+1}> b_{n}$ so $a_{n+1}> b_{n+1}> b_{n}$ and thus both sequences are monotone and bounded, so convergent, and every $a_{n}$ is an upper bound for the limits of both sequences.  But $a_{1}= \frac{2\cdot2}{2+1}< \sqrt 2$.

The limit, to the first few decimal places, is 1.209199576156145, but that's not a number I recognize.

[Removed due to JBL's post]
Convergence is easy: $a_{n}>a_{n+1}>b_{n+1}>b_{n}$, and since $a_{n+1}<\frac12(a_{n}+b_{n})$, $a_{n+1}-b_{n+1}<\frac12(a_{n}-b_{n})$. When the difference between an increasing sequence and a decreasing sequence goes to zero, both converge to the same limit.

Actually finding that limit is hard. We need an invariant, and I don't see an obvious one.
This is a pretty extreme question for a high school contest.

Plouffe's Inverter recognizes the number as $2\pi/\sqrt{27}$.

[quote="N.T.TUAN"]Define the sequences $a_{0}, a_{1}, a_{2}, ...$ and $b_{0}, b_{1}, b_{2}, ...$ by $a_{0}= 2, b_{0}= 1, a_{n+1}= 2a_{n}b_{n}/(a_{n}+b_{n}), b_{n+1}= \sqrt{a_{n+1}b_{n}}$. Show that the two sequences converge to the same limit, and find the limit.[/quote]
Denote $c_{n}=\frac{b_{n}}{a_{n}}(n=0,1,2,...)$, then $c_{0}=\frac{1}{2}$ and $c_{n+1}=\sqrt{\frac{1+c_{n}}{2}}(n=0,1,2,...)$. Now, I think we can know $c_{n}$. Do you agree?  :maybe:

Yes, I agree, and this is another way to show that the sequences converge to the same limit, but it doesn't give us the value of that limit.

JBL,

You not understand me, I said : We can find a formular give $c_{n}$  :wink:

On the other hand, the form of that (half-angle) and the number trichoplax posted strongly suggest trigonometry.

I'll come back later- I think I see the way now.

The limit is $\frac{\pi/3}{\sin(\pi/3)}=\frac{2\pi}{3\sqrt{3}}$. In general, $b_{n}=\frac{2^{n}\sin(2^{-n}\frac{\pi}3)}{\sin\frac{\pi}{3}}$ and $a_{n}=\frac{2^{n}\sin(2^{-n}\frac{\pi}3)}{\sin\frac{\pi}{3}\cos(2^{-n}\frac{\pi}{3})}$; these can be proved inductively, and the limit is easily evaluated.

If we start with an arbitrary $a_{0}>b_{0}$, the limit is $\frac{a_{0}b_{0}\arccos(\frac{b_{0}}{a_{0}})}{\sqrt{a_{0}^{2}-b_{0}^{2}}}$. There are similar formulas for the $a_{n}$ and $b_{n}$.


Let $ \Delta ABC$ be an arbitrary triangle . The incircle $ (I)$ touches $ AB,AC,BC$ at $ D,E,F$. The straightlines $ FE$ and $ FD$
cut the external bisector of $ \angle BAC$ in $ M,N$. Prove that $ \Delta IMN$ is always acute.

[geogebra]0ceaaeb3e0243f6f9d561ff43707f7e29a11107b[/geogebra]


Simplify $ \sqrt{50} \times \sqrt{6} \times \sqrt{27}$.

I think you can just multiply the numbers. It isn't that hard because all you have to do is \[ \sqrt{3\cdot27\times100}\equal{}9\cdot10\equal{}\boxed{90}.\]


geometry
calculus
analytic geometry
graphing lines
slope
quadratics
vector

Find the equation of a line perpendicular to 4x-3y=12 that passes through (2,3).

What is the area of the triangle whose sides lie on the x-axis, the y-axis, and the perpendicular line obtained in the previous problem? (You can use calculus for this one!)

Given the line 4x-3y=12, what is the distance between it and the origin? (this is an important formula, the distance from a generic point to a line, that i don't think many people know)

Give the generic equation of a line that divides the following circle into two equal parts: x:^2:+4x+y:^2:+6y=10.

Solve for x in terms of c: x(x-c)=1-c.

Have fun! Discuss, generalize, etc.

[hide]4x-3y=12  3y=4x-12  y=(4/3)x-4

Thus the slope of the line is 4/3. In order to find the slope of lines perpendicular to it we have to find the additive-multiplicative inverse of the slope (-1/s), which is -3/4. Since the line go through (2,3) point-slope would do, so y-3=(-3/4)(x-2)

Finding the interception of the line and the x/y-axis: For the y-intercept just plug in 0 for the previous equation and solve for x. y-3=(-3/4)(0-2)  y=4+1/2

x-intercept: 0-3=(-3/4)(x-2)  3/4x=4+1/2  x=6

Formula for area of a triangle: bh/2

Plug it in and get 4.5*3/2=13.5

The distance from a line to a point must be measured on a line perpendicular to the line and which goes through the point (First-Minimum Theorem), so just plug in 0 from problem 1 and get y=(-3/4)x. I believe the formula is y=mx and if n was the slope of the original line then m=-1/n

x(x-c)=1-c  x2-cx+(c-1) Using the quadratic formula you get x=(c:pm::sqrt:(c2-4*1*(c-1)))/2*1=(c:pm::sqrt:(c2-4c+4))2=(c:pm:c-2)/2=c-1, 1[/hide]

note: since, in the third problem, we are measuring the distance to the origin, we can indeed use Tare's method, wethinks (i haven't checked his answer yet). however, it might be useful to have the general formula. someone who knows it can post it with a proof. if no-one knows it, i'll post it at the end of this thread.

Oh whoops, I'll start over and solve for the generic equation Ax+By=C: [hide]So ax+by=c  by=c-ax  y=(-a/b)x+(c/b)

slope of the second line= -1/(-a/b)=b/a

Intersection of y=(b/a)x and y=(-a/b)x+c/b: y-y=(b/a)x-((-a/b)x+(c/b))  0=(b/a)x+(a/b)x-(c/b)  c/b=((b/a)+(a/b))x  c/b=((b2/ab)+(a2/ab))x  c/b=((a2+b2)/ab)x  x=cab/b(a2+b2)  x=ca/(a2+b2)

y=cb/(a2+b2)

By the Pythagorean Theorem the distance between the origin and the intersection is :sqrt:((ca/(a2+b2))2+(cb/(a2+b2))2=:sqrt:((c2a2/(a2+b2)2)+(c2b2/(a2+b2)2)=:sqrt:((c2a2+c2b2)/(a2+b2)2)=:sqrt:((c2(a2+b2))/(a2+b2)2)=:sqrt:(c2/(a2+b2))=c/:sqrt:(a2+b2)[/hide]

I'd love it if everyone saw what I wrote by quoting me, just to see how long this took for me...just in case I screwed up a calculation somewhere

Answer for the original question:[hide]plugging in 4 for a, -3 for b, and 12 for c into the formula you get 12/:sqrt:(42+(-3)2)=12/:sqrt:25=12/5(=2.4)[/hide]

good job. further questions: generalize to the distance from (m,n) to Ax+By+C=0.

generalize to higher dimensions, if you really want to. (think this is easiest with vectors, but can't remember).

I'm out (I have a presentation due tomorrow and i haven't done a thing about it...) *yawn*

Ok. This method is messy and unelegant. Plus it's in slope-int form, since I hate standard. Maybe it can be cleaned up....Messy derivation [hide]

Since the shortest distance will be along a perpendicular line,

we have the original

y=mx+b

and the new

y=-1/m*x+c c being the new y-intercept derived from point-slope form. Now use the fact that the equations are equal to each other to solve:

mx+b=-1/m*x+c

m+b=-1/m+c

m+1/m+b=c (This gives, by the way, an easier way to find c than point-slope form)

Now take the fact that c=m+1/m +b and substitute for c. Simplify. You get:

mx= -1/m*x+m+1/m

Sub in m and solve for x. Sub in x into y=mx+b and find y.[/hide]

Or kind of clean answer:[hide]

Take your equation y=mx +b.

Sub m into the equation mx=-1/m*x+m+1/m.

Solve for x.

Sub x into y=mx+b.

Solve for y.[/hide]

Is there a mistake in here?

Pointless green guy :mrgreen:

sigh i had wanted some of you guys to work on this. but considering the length of time, this thread will be open to tomorrow (remind me if i forget) and then i will clean it up perhaps and also post solutions. then i will continue posting problems (probably alg II, though maybe some geometry and later on some more alg I) (maybe even some precal) (don't worry, i won't go into some calculus problems unless there's a very strong audience desire for it)

Well...topper I can't think of a clearer way to do this without any specified formulas, so I guess you need to go through all these steps...good luck, you'll need it :twisted: </meanness+jk>

gasp you guys let me forget! grr...

well, prepare for gobs of problems. first, answers. i want SOLUTION for 4 before posting some more probs. yeah.

1.      -0.75x+4.5

2.      6*(9/2)/2=13.5

3.      2.4 [hide] general formula: if the line is Ax+By+C, and we want the distance to (m,n), it is given by |Am+Bn+C|/:sqrt:(A:^2:+B:^2:)[/hide]

4.      [hide]y=mx+2m-3[/hide]

5.      x=c-1,1

cool, huh?

tare's solutions to problems 1, 2, 3, and 5 were correct.


induction
Pythagorean Theorem
geometry
number theory unsolved
number theory

Given the Pythagorean theorem, $ a^2\plus{}b^2\equal{}c^2$ with $ c$ being the hypotenuse,
one example of two primitive pythagorean triples with the same value of $ c$ would be $ (13,84,85), (36,77,85)$

Is there an example of three or more primitive pythagorean triples with the same value of $ c$?

For any $ n$, there exists $ c$ such that there are at least $ 2^n$ different pairs of coprime integers $ a,b$ with $ a^2 \plus{} b^2 \equal{} c^2$.
For this to archieve, take $ n$ different primes $ p_1, ... , p_n \equiv 1 \mod 4$ and take $ c$ to be their product.

Let $ p$ be ine of these primes.
Now $ p \equal{} a^2 \plus{} b^2$ for some integers $ a, b$ (Fermat's 2-squares-theorem).
Then $ p^2 \equal{} (a^2 \minus{} b^2)^2 \plus{} (2ab)^2$ and this gives some coprime $ c,d$ with $ p^2 \equal{} c^2 \plus{} d^2$.

Assume by induction that you got $ 2^{n \minus{} 1}$ different pairs of coprime integers $ s,t$ with $ s^2 \plus{} t^2 \equal{} m$ where $ m$ is coprime to $ p$. For each such pair $ (s,t)$, consider the pairs $ (x,y) \equal{} (cs \minus{} dt,ds \plus{} ct)$ and $ (cs \plus{} dt,ds \minus{} ct)$. Now some calculation gives that all these pairs fulfill $ x^2 \plus{} y^2 \equal{} p^2m$ and all pairs are different and their entries are coprime. So we got $ 2^n$ pairs (and never used that $ m$ is a perfect square).

More conveniently, one would consider Gaussian integers, especially their unique factorisation. It gives immediatly the result, without any unnecessary calculation involved.

A completely unconstructive proof can be archieved by using http://www.mathlinks.ro/Forum/viewtopic.php?t=150965 and that "almost all" integers are not a sum of two squares.


[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given and let $ (K)$ be, the circle with diameter its side-segment $ BC,$ which intersects $ AB,\ AC,$ at points $ D,\ E,$ respectively. Through the midpoints $ M,\ N,$ of $ BD,\ CE$ respectively, we draw the lines perpendicular to $ BL,\ CL,$ where $ L$ is the midpoint of $ DE$ and let $ P$ be, their point of intersection. Prove that $ PB \equal{} PC.$[/color][/size][/b]

Kostas Vittas.

Dear Vittasko, your nice problem can be generalized:

Let $ ABC$ be a triangle and $ \left(K\right)$ is the circle with diameter BC. $ \left\{B,D\right\}\equal{}\left(K\right)\cap AB$, $ \left\{C,E\right\}\equal{}\left(K\right)\cap AC$ and $ L,M,N$ are
the midpoints of $ DE,BD,CE$, respectively. $ Q\in AL$ and the perpendicular to the line $ BQ$ through the point $ M$ intersects
the perpendicular to the line $ CQ$ through the point $ N$ at the point $ P$. Prove that $ PB\equal{}PC$.

$ \left\{V\right\}\equal{}AQ\cap BC$, $ \left\{X\right\}\equal{}CQ\cap AB$ and $ \left\{Y\right\}\equal{}BQ\cap AC$.
The perpendicular to the line $ BQ$ through the point $ M$ intersects the perpendicular to the line $ BC$ through the point $ K$ at the point $ P'$.
The perpendicular to the line $ CQ$ through the point $ N$ intersects the perpendicular to the line $ BC$ through the point $ K$ at the point $ P"$.
Let's prove that $ P'\equal{}P"\equal{}P$.

$ CD\bot AB$, $ BE\bot AC$ and $ L$ is the midpoint of $ DE$ $ \Rightarrow AV$ is a symmedian in triangle $ ABC$ $ \Rightarrow\frac{VB}{VC}\equal{}\frac{AB^{2}}{AC^{2}}$ (1)

It's easy to prove $ KM\bot AB$ and $ KN\bot AC$.

$ \frac{XA}{XB}\equal{}\frac{CA\cdot sin\left(\angle XCA\right)}{CB\cdot sin\left(\angle XCB\right)}\equal{}\frac{CA}{CB}\cdot\frac{sin\left(\angle KNP"\right)}{sin\left(\angle KP"N\right)}\equal{}\frac{CA}{CB}\cdot\frac{KP"}{KN}\Rightarrow$
$ \Rightarrow\frac{XA}{XB}\equal{}\frac{CA}{CB}\cdot\frac{KP"}{KN}$ (2)

$ \frac{YC}{YA}\equal{}\frac{BC\cdot sin\left(\angle YBC\right)}{BA\cdot sin\left(\angle YBA\right)}\equal{}\frac{BC}{BA}\cdot\frac{sin\left(\angle KP'M\right)}{sin\left(\angle KMP'\right)}\equal{}\frac{BC}{BA}\cdot\frac{KM}{KP'}\Rightarrow$
$ \Rightarrow\frac{YC}{YA}\equal{}\frac{BC}{BA}\cdot\frac{KM}{KP'}$ (3)

The lines $ AV,CX,BY$ are concurrent at $ Q\Rightarrow\frac{VB}{VC}\cdot\frac{XA}{XB}\cdot\frac{YC}{YA}\equal{}1$ (4)

(1),(2),(3),(4) $ \Rightarrow\frac{AB^{2}}{AC^{2}}\cdot\frac{CA}{CB}\cdot\frac{KP"}{KN}\cdot\frac{BC}{BA}\cdot\frac{KM}{KP'}\equal{}1\Rightarrow\frac{KP'}{KP"}\equal{}\frac{AB}{AC}\cdot\frac{KM}{KN}$ (5)

[...]-the area of the triangle ...
$ K$ is the midpoint of $ BC$ $ \Rightarrow\left[AKB\right]\equal{}\left[AKC\right]\Rightarrow KM\cdot AB\equal{}KN\cdot AC\Rightarrow\frac{KM}{KN}\equal{}\frac{AC}{AB}$ (6)

(5),(6) $ \Rightarrow\frac{KP'}{KP"}\equal{}1\Rightarrow P'\equal{}P"\equal{}P$.
$ K$ is the midpoint of $ BC$ and the point $ P$ lies on the perpendicular to the line $ BC$ through the point $ K$ $ \Rightarrow PB\equal{}PC$.

Best regards, Petrisor Neagoe :)

another proof for the generalization given by PETRY,
we obviously have :
$ MB^2\minus{}MQ^2\equal{}PB^2\minus{}PQ^2,NC^2\minus{}NQ^2\equal{}PC^2\minus{}PQ^2 \implies PB^2\minus{}PC^2\equal{}MB^2\minus{}MQ^2\minus{}(NC^2\minus{}NQ^2)$.
so we must prove that:$ MB^2\minus{}MQ^2\equal{}NC^2\minus{}NQ^2 \leftrightarrow BM^2\minus{}NC^2\equal{}MQ^2\minus{}NQ^2$,but we have $ BM^2\minus{}NC^2\equal{}MA^2\minus{}NA^2$.
which means that we must prove the perpendicularity of $ AQ$ and $ MN$.or in other words the perpendicularity of $ AL,MN$.
so we must prove $ ML^2\minus{}MA^2\equal{}NL^2\minus{}NA^2 \leftrightarrow \frac{BE^2}{4}\minus{}\frac{CD^2}{4}\equal{}MA^2\minus{}NA^2$,but we have $ \frac{BE^2}{4}\minus{}\frac{CD^2}{4}\equal{}\frac{BC^2\minus{}CE^2}{4}\minus{}\frac{BC^2\minus{}BD^2}{4}\equal{}BM^2\minus{}NC^2$.so we must prove that $ BM^2\minus{}NC^2\equal{}MA^2\minus{}NA^2$.which is easy!DONE!

Thank you a lot dear friends [b][size=100]Petrisor[/size][/b] and [b][size=100]shoki[/size][/b]. I didn't think the generalization given by [b][size=100]Petrisor[/size][/b] but, the solution I have in mind, may be also used there.

$ \bullet$ Because of $ ML\perp AC$ ( from $ ML\parallel BE$ and $ BE\perp AC$ ) and similarly, $ NL\perp AB,$ we have that the point $ L$ is the orthocenter of the triangle $ \bigtriangleup AMN$ and so, we have $ AL\perp MN.$

We see now that the lines through the vertices $ L,\ B,\ C$ of the triangle $ \bigtriangleup LBC$ and perpendicular to the sidelines $ MN,\ KM,\ KN$ respectively, of the triangle $ \bigtriangleup KMN,$ are concurrent at point $ A.$

That is, the triangles $ \bigtriangleup LBC,\ \bigtriangleup KMN,$ are [b][size=100]Orthologic[/size][/b] each other.

We conclude now, that the lines through vertices $ K,\ M,\ N$ of the triangle $ \bigtriangleup KMN$ and perpendicular to the sidelines $ BC,\ LB,\ LC$ respectively, of the triangle $ \bigtriangleup LBC,$ are also concurrent at one point, here the point $ P.$

Hence, we conclude that $ PK\perp BC$ $ \Longrightarrow$ $ PB \equal{} PC$ and the proof is copmpleted.

$ \bullet$ [b][size=100]NOTE.[/size][/b] - This problem has been used as [b][size=100]Lemma 2[/size][/b], for the solution of another problem appeared in a book of [b][size=100]Titeica[/size][/b], which posted [url=http://www.mathematica.gr/forum/viewtopic.php?f=50&t=2930][color=red]Here[/color][/url], by [b][size=100]Babis Stergiu[/size][/b].

Best regards, Kostas Vittas.


geometry
geometric transformation
inequalities
inequalities unsolved

The inequality is obviously wrong because it's inhomogeneous. Is there some factor missing?

  Darij

Well, this is the corect form, I mean this is the way it is written in the book. Perhaps it is false. Do you have a counter-example?

cheers! :D :D :cool:

Well, for instance take n = 3 and a1 = a2 = a3 = 2 and you get quite a nonsense... so I guess either you or the authors of the book have omitted some boundary condition, like a1a2...an = 1 or similar.

  Darij


geometry
parallelogram
circumcircle
geometric transformation
reflection
perpendicular bisector
angle bisector

The other common point is the symmetric of $A$ across the bisector of angle $\angle C$.

yes i agree. to see this take a special case when AB=2BC  hence we can see that if M  is the midpoint of AB  then we have that CM  is the perpendicular bisector of PQ for any choice of these points and hence we can see that in this case B  is our common point because the intersection of the circle with center M radius MB with sides BC and CD   are points that satisfy the property needed hence B IS OUR  common point. this suggests that in general our common point is the simetric of  A with respect to the internal bisector of C. I cant seem to prove this because the figure is almost impossible to draw correctly.

[quote="grobber"]Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$.[/quote]

Hmm, isn't the problem trivial? I will prove that the circumcircle of triangle APQ always passes through the reflection of the point A in the angle bisector of the angle BCD (of course, this point is a fixed point, so this will solve the problem). In fact, since CP = CQ, the triangle PCQ is isosceles, and thus, the angle bisector of the angle PCQ is the perpendicular bisector of the segment PQ. In other words, the angle bisector of the angle BCD is the perpendicular bisector of the segment PQ. But the center of the circumcircle of triangle APQ must lie on the perpendicular bisector of the segment PQ (since the center of a circle lies on the perpendicular bisector of any chord). Hence, the center of the circumcircle of triangle APQ lies on the angle bisector of the angle BCD. Thus, the circumcircle of triangle APQ is symmetric with respect to the angle bisector of the angle BCD (in fact, if the center of a circle lies on some line, then the circle is symmetric with respect to this line). Hence, as the circumcircle of triangle APQ passes through the point A, it must also pass the reflection of the point A in the angle bisector of the angle BCD. And we are done.

  Darij

[quote="darij grinberg"]

Hmm, isn't the problem trivial?
  Darij[/quote]
I agree, it is.

So with ABCD, we don't need parallelogram hypothesis ?

The fixed point is the reflection of $C$ over the angle bisector of $\angle A$. Because the center $O$ of $(PQC)$ lies on this line, this point must lie on $(PQC)$ as well.


Determine a positive constant $ c$ such that the equation

$ xy^{2} \minus{}y^{2} \minus{}x\plus{}y\equal{}c$

has precisely three solutions $ (x, y)$ in positive integers.

$ (y\minus{}1)(x(y\plus{}1)\minus{}y)\equal{}c$ and $ c$ must be integer because $ (x,y)$ positive integers. So if $ c$ is prime there are maximum 2 solutions $ (2,\frac{c\plus{}2}{3})$,$ (c\plus{}1,1)$ and if $ c\equal{}pq$ with $ p,q$ primes there are another two $ (p\plus{}1,\frac{p\plus{}q\plus{}1}{p\plus{}2})$,$ (q\plus{}1,\frac{p\plus{}q\plus{}1}{q\plus{}2})$. With can easily see that $ c\equal{}7*19$ hold.

Thanks Blackout.1.01. You seem to have written your ordered pairs back to front.

I get $ (x, y) \equal{} (\frac{c\plus{}2}{3}, 2), (1, c\plus{}1), (\frac{p\plus{}q\plus{}1}{p\plus{}2})$ or $ (\frac{p\plus{}q\plus{}1}{q\plus{}2}, q\plus{}1)$.

This works with your $ c\equal{}7 \times 19$, although I don't know how you got that; I got the smaller solution $ c\equal{} 2 \times 5$ which gives

$ (4, 2): 16\minus{}4\minus{}4\plus{}2\equal{}10$

$ (1, 11): 121 \minus{} 121 \minus{}1 \plus{} 11\equal{} 10$

and

$ (2, 3): 18\minus{}9\minus{}2\plus{}3 \equal{} 10$.

I got these, using your solution; I don't have a method of solution of my own for this.

[quote="AndrewTom"]Thanks Blackout.1.01. You seem to have written your ordered pairs back to front. [/quote]

yes sorry  :blush: .

[quote] This works with your $ c \equal{} 7 \times 19$although I don't know how you got that .[/quote]

$ (\frac {p \plus{} q \plus{} 1}{p \plus{} 2})$ or $ (\frac {p \plus{} q \plus{} 1}{q \plus{} 2}, q \plus{} 1)$ can't be both solutions because $ p\plus{}q\plus{}1\not\equal{}(p\plus{}2)(q\plus{}2)k$,        $ c\equiv 1 (mod3)$ so you can try $ p\equiv 1 (mod3)$, $ q\equiv1 (mod3)$ or $ \equiv\minus{}1$


inequalities
function
inequalities proposed

Prove that, for any positive real numbers $a,b$ and $c$,
\[ \frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}+\frac{a^2+b^2}{c^2+ab}\ge 2+\frac{a^2+b^2+c^2}{ab+bc+ca} \]

[quote="arqady"][quote="Tiks"]
And SOS,with$a\geq b\geq c$,$S_b\geq 0$,$S_a+S_b\geq 0$    .[/quote]
Why it is true, Tiks? Thank you.[/quote]
Dear Arqady the first one is easy to prove
We want to prove that $S_b\geq 0$
We have that
\[ \frac{1}{b^2+ac}-\frac{1}{ab+bc+ac}\geq 0 \](1)
and
\[ \frac{1}{b^2+ac}+\frac{b(a+c)(b^2-ac)}{(b^2+ac)(a^2+bc)(c^2+ab)}\geq 0 \](2)
<=>
\[ (a^2+bc)(c^2+ab)\geq (ab+bc)(b^2-ac) \]
but $a^2+bc\geq ab+bc$ and $c^2+ab\geq ab\geq b^2>b^2-ac$.
So $S_b=(1)+(2)\geq 0$
For $S_b+S_a\geq 0$
It is a little difficult work   :blush:
I have shown it with geting the sum $S_b+S_a$ to the comon denominator :huh: ,I would like you to find another solution,but by the way I would like to say that when $c=0$,$S_b+S_a\geq 0$ <=>
$(\frac{1}{a}-\frac{1}{b})^2\geq 0$.
I think that we can solve that ineq. with functional methods,
we can let $S_b+S_a$ to be a function from $c$,where $c\in{[0;b]}$.
Mybe from this we will have a shorter solution :? . ;)

[quote="kunny"]I think the sign of the given inequality is in reverse.

If my opinion is correct, I have solved.[/quote]
The sigin can't be in reverse,take $c=0$,then the given ineq is equivalent to $(\frac{a}{b}-\frac{b}{a})^2\geq 0$ ;) .

[quote="kunny"]Tiks, isn't $c$ postive? ;)[/quote]
Yes,but I think we can just take $c\to 0$,don't you think so? :?

[quote]And SOS, with $a\ge b\ge c$, $S_b\ge 0$, $S_a+S_b\ge 0$.[/quote]

Dear Tiks,
I'm afraid that $S_a+S_b\ge 0$ is not true since, unless I made mistake in calculation, I found a counter-example with $a=5$, $b=c=1$. It seems that it is negative for many values of $a\ge 5$ (with $b=c=1$).

[quote="libra_gold"]Expand This,

$((b^2+c^2)(b^2+ca)(c^2+ab)+(c^2+a^2)(c^2+ab)(a^2+bc)+(a^2+b^2)(a^2+bc)(b^2+ca))(ab+bc+ca)-(a+b+c)^2(a^2+bc)(b^2+ca)(c^2+ab)$

We will have $((6,1,1)+2(6,2,0))-((3,3,2)+2(4,4,0))$

which Muirhead can work![/quote]

Since I have no idea how SOS works (I think I will have to read through some papers until IMO  :? ), I have, obviously like libra_gold, not seen anything else than using brute force. But before expanding, you can take $a^2+b^2+c^2=1/2$ and use Jensen on $f(x)=1/x$. Then i'ts a bunch, you will come up to
\[ 3(5,1,0)+(3,3,0) + (3,2,1)\ge 4(4,2,0)+(2,2,2)  \]
where $(p,q,r)=\sum_{sym}a^pb^qc^r$.
We have $(5,1,0)+(3,3,0)\ge 2(4,2,0)$ by Am-Gm, $2(5,1,0)\ge (4,2,0)$ by Muirhead and obviously $(3,2,1)\ge (2,2,2)$.

Yimin

[quote="libra_gold"]
We will have $((6,1,1)+2(6,2,0))-((3,3,2)+2(4,4,0))$

which Muirhead can work![/quote]
Yes,
$\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}+\frac{a^2+b^2}{c^2+ab}\ge 2+\frac{a^2+b^2+c^2}{ab+bc+ca} \Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^6b^2+a^6c^2+a^6bc-a^3b^3c^2-2a^4b^4)\geq0.$
By the way, the best $k$, such that
 $\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}+\frac{a^2+b^2}{c^2+ab}\ge 3-k+\frac{k(a^2+b^2+c^2)}{ab+bc+ca}$ is true
for all positive $a,$ $b$ and $c$ is $k=1.$ ;)

Dear Ductrung, is your solution without full expansion?

Thank you very much.

Yes, Manlio. My solution does not need full expansion. Below is a hint
\[ \sum_{cyc}\frac{b^{2}\plus{}c^{2}}{a^{2}\plus{}bc}\equal{}\sum_{cyc}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{a^{2}\plus{}bc}\minus{}\sum_{cyc}\frac{a^{2}}{a^{2}\plus{}bc}\]
Using [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114737[/url], it suffices to show that
\[ \frac{a^{2}}{a^{2}\plus{}bc}\plus{}\frac{b^{2}}{b^{2}\plus{}ca}\plus{}\frac{c^{2}}{c^{2}\plus{}ab}\le\frac{(a\plus{}b\plus{}c)^{2}}{2(ab\plus{}bc\plus{}ca)}\]

Note that

$ \sum_{cycl}\frac{(b\plus{}c)^{2}}{a^{2}\plus{}bc}\equal{}\sum_{cycl}\frac{b^{2}\plus{}c^{2}}{a^{2}\plus{}bc}\plus{}\sum_{cycl}\frac{2bc}{a^{2}\plus{}bc}$
$ \equal{}\sum_{cycl}\frac{b^{2}\plus{}c^{2}}{a^{2}\plus{}bc}\plus{}2\sum_{cycl}\left(1\minus{}\frac{a^{2}}{a^{2}\plus{}bc}\right)$
$ \equal{}6\plus{}\sum_{cycl}\frac{b^{2}\plus{}c^{2}}{a^{2}\plus{}bc}\minus{}2\sum_{cycl}\frac{a^{2}}{a^{2}\plus{}bc}$

so that we can connect to (and provide an indirect but nice alternative solution to) a nice inequality by D. Grinberg and P. Scholze: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=60128[/url]


I have a suggestion. Let us say we have a post made by user A. Then we have many users replying to this including A. When multiple people reply there are different trains of thought, it becomes difficult to collate or refer to a certain point.

I found myself in situation where I had two messages in a post. As a reply to message/s of someother user, I wanted to give a reference to my second post, lets say. Only way  this can be done is by using quote, but this too becomes inefficent of there are multiple replies and its not always convinient to use 'search' on the page.

Suggestion: Index every message of post in the time order of posting. If a user wants to refer to something, he/she can say 'such and such' a thing was told in message #. It would be better if hyperlinks can be provided to each message to each post at the top and bottom of the post.

Every post has its own url -- in the aops skin, you can find this url by clicking the little page icon directly to the left of the "Posted: [day] at [time]" bit at the bottom of the post.  I have found very few threads here where the ability to easily quote has been insufficient to keep track of what's going on.

The same button is found in the Mathlinks layout. It's an anchor.


Two non-zero real numbers, $a$ and $b$, satisfy $ab=a-b$. Which of the following is a possible value of $\frac ab+\frac ba-ab$?

$\text{(A)}\ -2\qquad\text{(B)}\ -\frac12\qquad\text{(C)}\ \frac13\qquad\text{(D)}\ \frac12\qquad\text{(E)}\ 2$

[hide]Writing the expression under a common denominator, we get $\frac{a^2+b^2-(ab)^2}{ab}$. Substituting $ab=a-b$, we get $\frac{a^2+b^2-(a^2+b^2-2ab)}{ab}=\frac{2ab}{ab}=2$. [/hide]

[hide="useless comment"]
When I did this problem during a timed practice test, I realized to substitute ab=a-b about 35 seconds before time was up. [/hide]

[hide="Solution"]Note that $ab=a-b$ implies both $\tfrac ab = a+1$ and $\tfrac ba = 1-b$ (upon dividing by $b$ and $a$ respectively), so \[\frac ab + \frac ba - ab = 2 + (a-b-ab) =\boxed{2\textbf{ (E)}}.\][/hide]

Plug in a=-2, b=2 to get a/b+b/a-ab=-1-1+4=2 which is an answer! wow!

[hide=Solution]Note that $ab=a-b$ means $ab-a+b=0$, so $(a+1)(b-1)=-1$. Then $a=-2, b=2$ works, and plugging in gives $2 \implies \boxed{\text{(E)}}$.[/hide]


Triangle ABC is obtuse

Prove that -1 < CosACosB +CosBCosC + CosCCosA <= 0

First we have this lemma
For -1<a,b,c<1 we have ab+bc+ca+1 >0
This one comes from (1+a)(1+b)(1+c) + (1-a)(1-b)(1-c) >0
Apply for cosA,cosB,cosC we have for all triangle ABC
cosAcosB + cosBcosC +cosCcosA > -1

I think P=cosAcosB + cosBcosC + cosCcosA <= 0 cannot be true. Let A -> 90+
B=C -> 45-, P -> 1/2.

And I have another inequality
For all obstue triangle ABC we have
 P=cosAcosB + cosBcosC + cosCcosA < 1/2


z1,z2,z3 are complex numbers

a1,a2,a3 are from M={-1,1}

Prove that    |a1z1+a2z2+a3z3|^2>= |z1|^2+|z2|^2+|z3|^2


analytic geometry
graphing lines
slope
direct proportion

Hmm.. I don't really understand this problem.

If $y$ is directly proportional to $x,$ which of the following could be the graph that shows the relationship between $y$ and $x$?

[hide="Solution"]

The answer booklet says it's D. My question is, why isn't A or C the answer? I put A (I knew it had to A,C, or D) because I thought directly proportional meant that if $y$ increases, $x$ will increase as well. Or am I wrong? [/hide]

Well..the general form for a direct proportion equation is $y=kx$ where k is the constant of variation. If you compare to slope-intercept form, you see there is no $b$, and thus no y-intercept EDIT: y-intercept equals zero, actually.. Thus, it always has to go through the origin as in D.
(Wow...I actually learned something in school)

[quote="Silverfalcon"]I thought directly proportional meant that if $y$ increases, $x$ will increase as well. Or am I wrong? [/hide][/quote]

Yes, but they increase in the same proportion

[quote="tz"]Well I randomly guessed and found that the slopes are either 1 or -1 and some of the Y intercepts are the same, the only unique one is D, which the line goes through the orgin.[/quote]

The slope is not restrict to -1 and 1.  In fact, it can be anything $\in \mathbb{R}$ (it can be negative as well as positive).

Additionally, ALL the y-intercepts should be equal (to zero, that is).

ragnarok23 has it right.

[quote="ShoeFactory"]It is obviously D.

Slope does not matter because it is proportional

But it must go through origin bececause y = kx.[/quote]

Oh..

Haha.. I actually learned something from SAT!  :D

Something related...

For example if they said y is directly proportional to 1/x it means y = k/x

If y is inversely proportional to $x^2$ it means $y = \frac{k}{x^2}$

Yes it would be like that.

And it is D because y=kx, and so it must pass through the origin.


calculus
integration
limit
logarithms
calculus computations

Integral over what set? Without that information, the question is meaningless.

if you use the comparison integral of dx/x^2 it means that it is bounded from above ? right?

[quote="Kent Merryfield"]Then they both converge absolutely by limit comparison to $\int_{c}^{\infty}\frac{dx}{x^{2}}$ for some convenient $c>0.$[/quote]

for number 2 how can you use this comparison? you will then bound this integral from below and if the integral from below converges it does not tell anything about the integral

The [b]Limit[/b] Comparison test is different from the Comparison test that you're thinking of. You can find it [url=http://planetmath.org/encyclopedia/LimitComparisonTest.html]here[/url]. (Part 1 is all you need; Parts 2 and 3 are often not included in a calculus course).

[quote="EquinoX"]Decide whether the these integral converges or diverges?
1. the integral of 1/(4+z^2) dz
2. the integral of 1/(z^2-4) dz[/quote]
[quote="EquinoX"]for number 1. is from 2 to infinity
for number 2. is from 10 to infinity[/quote]
$\int_{2}^{+\infty}\frac{1}{4+z^{2}}dz=\frac{\pi}{8}=0.39269908169872415480783\cdots$;

$\int_{10}^{+\infty}\frac{1}{z^{2}-4}dz=\frac{\ln{3}-\ln{2}}{4}=0.10136627702704109549450\cdots$.

[quote="Ji Chen"]$\int_{2}^{+\infty}\frac{1}{2+z^{2}}dz=\frac{\pi}{8}=0.39269908169872415480783\cdots$[/quote]

First of all, this is wrong.  Second of all, your habit of posting answers without any solution or commentary is (1) annoying and (2) totally useless.  Cut it out.


vector
linear algebra
matrix
linear algebra unsolved

Prove that if $ M$ and $ N$ are $ n\times n$ matrices over a field $ F$ such that $ M^n\equal{}N^n\equal{}0$ but $ M^{n\minus{}1} \not\equal{} 0\not\equal{}N^{n\minus{}1}$, then $ M$ and $ N$ are similar.

[quote="The Captain"]Definitely discussed before. Since $ M^{n \minus{} 1}\neq 0$, there exists $ x$ such that $ M^{n \minus{} 1}x\neq 0$. Prove that $ x, Mx, \dots, M^{n \minus{} 1}x$ is a linearly independent sequence of vectors and find the representation of the the linear operator corresponding to $ M$ in the new basis obtained from the sequence above. Repeat this method for $ N$ to conclude.
[keywords:nilpotent, similar][/quote]

Thank you for your hint [b]The Captain[/b]. Here is the solution.

Let $ T_1: F^n\rightarrow  F^n$ be a linear operator such that $ T_1(A)\equal{}MA$.
Then $ [T_1]_B\equal{}M$ where $ B$ is the standart basis of $ F^n$.
$ c_1x\plus{}c_2Mx\plus{} \dots\plus{} c_nM^{n \minus{} 1}x\equal{}0 \Longrightarrow c_1M^{n\minus{}1}x\plus{}c_2M^nx\plus{} \dots\plus{} c_nM^{2n \minus{} 1}x\equal{}0$
$ \Longrightarrow c_1M^{n\minus{}1}x\equal{}0 \Longrightarrow c_1\equal{}0$ .
With the same process we get $ c_i\equal{}0$. So the set $ B_1\equal{}\{x, Mx, \dots, M^{n \minus{} 1}x\}$ are independent and form a basis for $ F^n$
$ T_1(x)\equal{}Mx \Longrightarrow [T_1(x)]_{B_1}\equal{}(0,1,0,...,0)^t$
$ T_1(Mx)\equal{}M^2x \Longrightarrow [T_1(Mx)]_{B_1}\equal{}(0,0,1,...,0)^t$
$ ..........$
$ T_1(M^{n\minus{}1}x)\equal{}M^nx \Longrightarrow [T_1(M^{n\minus{}1}x)]_{B_1}\equal{}(0,0,...,0)^t$

So $ [T_1]_{B_1}\equal{}\begin{pmatrix}
0 & 0 & ...& 0 &0 \\
1 & 0 &...& 0 & 0 \\
...&...&...&...&...\\
0 & 0 & ...& 1 & 0
\end{pmatrix}$

Now define $ T_2: F^n\rightarrow  F^n$ such that $ T_2(A)\equal{}NA$. By the similar way $ B_2\equal{}\{x, Nx, \dots, N^{n \minus{} 1}x\}$ form a basis for $ F^n$ and $ [T_2]_{B_2}\equal{}\begin{pmatrix}
0 & 0 & ...& 0 &0 \\
1 & 0 &...& 0 & 0 \\
...&...&...&...&...\\
0 & 0 & ...& 1 & 0
\end{pmatrix}$

So $ [T_1]_{B_1}\equal{}[T_2]_{B_2}$
$ \Longrightarrow P^{\minus{}1}MP\equal{}Q^{\minus{}1}NQ$
$ \Longrightarrow M\equal{}PQ^{\minus{}1}NQP^{\minus{}1}\equal{}(QP^{\minus{}1})^{\minus{}1}N(QP^{\minus{}1})$
So $ M$ and $ N$ are similar.

Hope this is a correct solution.


For a positive integer $a$, $a'$ is the integer obtained by the following method: the decimal writing of $a'$ is the inverse of the decimal writing of $a$ (the decimal writing of $a'$ can begin by zeros, but not the one of $a$); for instance if $a=2370$, $a'=0732$, that is $732$.

Let $a_{1}$ be a positive integer, and $(a_{n})_{n \geq 1}$ the sequence defined by $a_{1}$ and the following formula for $n \geq 1$:
\[a_{n+1}=a_{n}+a'_{n}. \]
Can $a_{7}$ be prime?

[hide="Answer."]Nope.[/hide]

[hide="(big?) Hint."]$11|a_{7}.$[/hide]

Obviosly, that $3|a_{n}\to 3|a_{n}'$, therefore, if $3|a_{1}$, then $3|a_{n}\ \forall n$ (exactly $a_{n}=a_{1}2^{n-1}(mod \ 9)).$
If $11|a_{n}$, then the sum of digites in even positions equal to sum of digits in odd positions by mod 11, therefore $11|a_{n}'$ too.
It give $11|a_{n}\to 11|a_{n+1}$. If $10^{k-1}\le a_{n}<10^{k}$ had odd number of digits (k is odd ), then $a_{n+r}, r\le 5$ had k+1 digits. It is easy to chek, that $11|a_{n+r+1}$, therefore $11|a_{7}$ for any $a_{1}$.

[quote="Rust"] If $ 10^{k \minus{} 1}\le a_{n} < 10^{k}$ had odd number of digits (k is odd ), then $ a_{n \plus{} r}, r\le 5$ had k+1 digits. It is easy to chek, that $ 11|a_{n \plus{} r \plus{} 1}$, therefore $ 11|a_{7}$ for any $ a_{1}$.[/quote]

I didn't seem to understand this step

Can s.o explain it please?

If $ a$ had even number of digits $ a\equal{}a_0\plus{}10*a_1\plus{}...\plus{}a_{m}10^{m}$ (m is odd), then $ a\equal{}a_0\minus{}a_1\plus{}a_2\minus{}...\minus{}a_m\mod 11$,
$ a'\equal{}a_m\minus{}a_{m\minus{}1}\plus{}...\plus{}a_1\minus{}a_0\mod 11 \equiv \minus{}a\mod 11$. Therefore $ 11(a\plus{}a')$ or if $ a_n$ had odd number of digits, then $ 11|a_{n\plus{}1}$.

excuse me, how do you know $ a_6$ has an odd number of digits, I think it's not necessary.

Can anyone help me to prove $ 11 / a_7$ I don't see how.

Answer: No.

We will prove first that one of $a_1,a_2,\ldots,a_6$ has an even number of digits. Suppose the contrary. Let $a_1$ has $k$ digits, where $k$ is odd. If $a_n$ has $k$ digits, then $a_{n+1}$ has at most $k+1$ digits, which has different parity with $k$. The numbers of digits in $a_1,a_2,\ldots,a_6$ are even, hence each of $a_1,a_2,\ldots,a_6$ has exactly $k$ digits. Let $p$ and $q$ be the first and the last digits of $a_1$. So $a_2$ starts and ends with $p+q$. Similarly, we get that $a_3,a_4,a_5,a_6$ start and end with $2p+2q,4p+4q,8p+8q,16p+16q$ respectively. But $16p+16q>10$, which implies that $a_6$ has more than $k$ digits, a contradiction. Therefore our claim holds, let $a_i$ ($i\le6$) be the integer with an even number of digits. The digits in odd positions of $a_i$ become digits in even positions of $a_i'$, and vice versa, therefore the sums of digits in even positions and odd positions of $a_i+a_i'$ are equal. Thus $11|a_{i+1}$. Note that if $11|a_k$ then $11|a_k'$ and hence $11|a_{k+1}$, which further implies $11|a_n$ for all $n\ge k$. Since $11|a_{i+1}$ where $i\le6$, then $11|a_7$. If $a_7$ is prime, then $a_7=11$. If $a_6$ has only one digit, then $a_6=a_6'$ and $11=a_7=2a_6$, which is impossible. Hence $a_6$ has two digits, $a_6=10$, then $a_5=5$, which leaves no possible value for $a_4$, a contradiction.


In the obtuse triangle $ABC$ with$\angle C > 90, AM = MB, MD \perp BC, EC \perp BC$ (D is on BC and E is on AB and M is on EB). If the area of $\triangle ABC$ is $24$, then the area of$\triangle BED$ is

A. 9
B. 12
C. 15
D. 18
E. Not uniquely determined

My answer is [b]E[/b].
This is so, because when I solved the question with the help of Analytical geometry I got the answer in terms of variables.


Im new here so Im not sure whether to post my questions. So I posted this in this section as well as the intermediate section.

I'm having difficulty with some number theory questions, can someone please help me out? Thanks.

1) Prove that square root of 3 cannot be expressed as (a/b) for any two integers a,b.

2) Let p be an odd prime. Then it can be proved that
[ (p-1)!/2 ]^2 is congruent to ((-1)^[(p+1)/2]) mod p
Use the above to show that if p is congruent to 1 (mod 4), then (-1/p) = 1

Please try not to post your questions in more than one forum in the future.  It gets frustrating.

1)

[hide] Suppose $\sqrt3 = a/b$ for integers a/b. Moreover, let this be the simplest form of the fraction. (so gcd(a,b)=1

Then $\frac{a^2}{b^2}=3$
and $a^2=3b^2$
Then $a^2$ is divisible by 3, and therefore so is a. Then we can write a=3k for some integer a
so $(3k)^2=3b^2$

carry on like this to get a contradiction[/hide]

2) see http://www.mathlinks.ro/Forum/viewtopic.php?t=46300 , post 5, lemma 1.


it's a great honor that africans have now their own community !! there are still many things to discover in and about africa including mathematical achievements , so if one has got any idea , it's welcome on this topic :)

Umm I don't know if I remember correctly but I think the guy who held the record for the largest number ever used was from South Africa - skewes number before Ronald Graham broke the record - I'll check

Emanuel Derman won financial engineer of the year was head of Quant strategies and firmwide risk at Goldman Sachs  and is a top researcher in Quantitative Finance was born and did his undergrad in South Africa.


When can we expect the official results? Are they mailed before the B testing date or do they wait to grade the B tests to send out with the results of the A test? My school's winter break starts at the end of next week, and I was wondering if I'd have to wait for school to start again in order to get the official results.

Not all answer forms have been received from schools yet.  Answer forms will be coming in to our office all this week.

We have begun scanning answer forms, but we will not complete the scanning process for the A-date contests for at least a week, maybe 10 days.  After that, we must do extensive editing of the answers scanned in, because the circles were not completely filled in, leaving gaps in names and school identifiers.  We also do some consistency checks and preliminary score analysis.

After that, we must do some preliminary statistical analysis to estimate award levels.

Due to all that, I expect that the first e-mailed reports can be no sooner than two weeks from today.  The first emailed reports will probably be in about 3 weeks.

Steve Dunbar
Director, American Mathematics Competitions

is it possible to send unofficial reports? as in our score without the consistency checks, etc? that shouldn't change for most people.

and for some reason, i remember that last year, my teacher had a list of scores available before the AMC B version.. so the people who already qualified for AIME / got a perfect score didn't need to take it again.

 :)  :)

[quote="probability1.01"]Err... if it takes 3 weeks, then wouldn't the people who qualified for AIME through AMC B be screwed? Because that barely leaves them any time to prepare for the AIME...[/quote]

They can prepare beforehand if needed.  Besides, one can always check ones scores upon receiving the answer booklet, right?  I always received it right after taking the thing, so i could check right away.

[quote="chess64"]And anyway, if you're doubting whether you'll even make it to AIME, you probably won't do well.[/quote]

Hmmm that's not true.

That is obviously not true.  People always say that losing is the first step to winning.  If person $x$ does horrible on AMC, let's say 100-110, then he/she will obviosuly try to study really hard for AIME (study as in practice).  I did horrible on COMC, due to some stupid mistakes of not understanding the english of the problems, so I'm practicing a lot.  I'm currently still a loser working my way up to a moderate winner. :P

[quote="dimak11214"]Its impossible to get a 100 on any AMC. You can get a 100.5 though. :P[/quote]

Impossible??? Apparently 162 people achieved the unachieveable last year on the AMC 10/12:
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05scoredist-1012AB.html

Btw, to get a 100 you need to get 15 right and leave 4 blank (and miss the rest).

[quote="lowest_reynolds"][quote="chess64"]When was the last time somebody got a 100 on AMC 12 and qualified for USAMO?[/quote]

Some eighth grader from Georgia got close to a 100, and did.  Brute forced the heck out of that AIME too. ;)[/quote]

Well its not like its a common thing or anything, like I said, [quote]...you [b]probably[/b] won't do well.[/quote]

I understand what you mean, but we shouldn't discourage those who didn't do that well on AMC from studying for AIME.

[quote="Fermatprime"]I qualified for USAMO back in eighth grade with almost no preparation, but then again, it was by the skin of my teeth. Actually, not even. It was by the skin OF the skin of my teeth. I got 101.5 on the AMC 12 and 7 (minimum qualifying score that year) on AIME. So... preparation is probably a good thing. Of course, all this was probably said three years ago.[/quote]

My point is that even if your AMC score is not that great, you still have a shot at USAMO so don't give up and study for AIME.

[quote="chess64"]When was the last time somebody got a 100 on AMC 12 and qualified for USAMO?[/quote]

When they were from Wyoming.

Just to comment, the correlation between AMC12 score and AIME score is pretty loose sometimes. The testing conditions are VERY different and that can make or break USAMO qualification. Of course given that, the chance of someone qualifying for USAMO with a 100 AMC12 does seem quite low to me, as they would need somewhere in the range of an 11-13 AIME score, which is by no means easy to get. That isn't to say they can't get a respectable score, or even an extremely good one, but statistics-wise is all I'm talking about.

[quote="Calculus"]On practice AMC 12 tests I can barely break 100, and on past AIME tests I can get any where from 7-10 depending on how many careless mistakes I make. (Thank god I get the benefit of the floor score.) I don't know why there is a huge difference between my scores. I guess I am not good at quickly solving problems. And also the only preparation I did was for the AIME. So it is definitely possible to qualify for the USAMO with a really low AMC 12 score.[/quote]

Yeah, for underclassmen that's definitely true, since AMC score is completely irrelevent.

[quote="beta"]
My point is that even if your AMC score is not that great, you still have a shot at USAMO so don't give up and study for AIME.[/quote]

And even if your AIME score isn't that great (but you still qualify), don't give up and do study for USAMO. If there's only a little correlation between AMC and AIME scores, there's probably even less between AIME and USAMO. (I can't say for sure, since they don't make any statistics on that available, but I ended up with a 10 on the USAMO that year. Then again, I'm probably the exception rather than the rule. Had I been a year older, I would have gotten into MOP easily. But they don't like to take 14-year-olds unless they have to.  :( )

Wait what about 14-year olds? I thought it was based on grade? I'm in 9th grade but I'm going to be 15 next september...does that mean I won't be able to make MOP? :?

I think he was talking about his grade. Im really sure they don't consider your age.


Prove that

$ a^3(\frac {b}{b^2\plus{}ca} \plus{}\frac {c}{c^2\plus{}ab}) \plus{} b^3(\frac {c}{c^2\plus{}ab} \plus{}\frac {a}{a^2\plus{}bc}) \geq a^2 \plus{} b^2$

for  $ a \geq b \geq c$

Can anyone help me to solve this one, please? I have spent a lot of days and nights on it  :( , but I cannot.

Here is my very ugly solution
$ a^3(c^2b\plus{}b^2a\plus{}c^2a\plus{}b^2c)(a^2\plus{}bc)\plus{}b^3(a^2c\plus{}c^2b\plus{}c^2a\plus{}a^2b)(b^2\plus{}ac) \geq$
$ (a^2 \plus{} b^2)(a^2\plus{}bc)(b^2\plus{}ac)(c^2 \plus{} ab)$
After opening the brackets

$ LHS\equal{}a^5c^2b\plus{}b^5c^2a\plus{}a^6b^2\plus{}b^6a^2\plus{}a^6c^2\plus{}b^6c^2\plus{}a^5b^2c\plus{}b^5a^2c$
$ \plus{}a^4c^3b\plus{}b^4c^3a\plus{}2a^3b^3c^2$

$ RHS\equal{}2a^4b^2c^2\plus{}2b^4a^2c^2\plus{}a^5b^3\plus{}b^5a^3\plus{}a^5c^3\plus{}b^5c^3\plus{}a^6bc\plus{}b^6ac\plus{}c^4a^3b\plus{}c^4b^3a$

$ \Rightarrow (a^6b^2\plus{}b^6a^2\minus{}a^5b^3\minus{}b^5a^3)\plus{}(a^6c^2\plus{}b^6c^2\minus{}a^4b^2c^2\minus{}b^4a^2c^2)\plus{}(a^5c^2b\plus{}b^5c^2a\minus{}a^4b^2c^2\minus{}b^4a^2c^2)\minus{}(a^5c^3\plus{}b^5c^3\minus{}a^4c^3b\minus{}b^4c^3a)\minus{}(a^6bc\plus{}b^6ac\minus{}a^5b^2c\minus{}b^5a^2c)\plus{}(2a^3b^3c^2\minus{}c^4a^3b\minus{}c^4b^3a)\geq 0$

İt is obvious that $ 2a^3b^3c^2\minus{}c^4a^3b\minus{}c^4b^3a\geq 0$ so let's prove that the sum of the others are also positive

$ a^2b^2(a\minus{}b)^2(a^2\plus{}ab\plus{}b^2)\plus{}c^2(a\minus{}b)^2(a\plus{}b)^2(a^2\plus{}b^2)\plus{}$
$ c^2ab(a\minus{}b)^2(a^2\plus{}ab\plus{}b^2)\minus{}c^3(a\minus{}b)^2(a\plus{}b)(a^2\plus{}b^2)\minus{}cab(a\minus{}b)^2(a\plus{}b)(a^2\plus{}b^2)\geq 0$

$ (a\minus{}b)^2ab(c^2\plus{}ab)(a^2\plus{}ab\plus{}b^2)\plus{}(a\minus{}b)^2(a\plus{}b)(a^2\plus{}b^2)(c^2a\plus{}c^2b\minus{}c^3\minus{}cab)\geq 0$

$ (a\minus{}b)^2(a^4b^2\plus{}b^4a^2\plus{}a^4c^2\plus{}b^4c^2\minus{}a^4bc\minus{}b^4ac\minus{}a^3b^2c\minus{}b^3a^2c\plus{}3a^2b^2c^2\plus{}a^3b^3\minus{}a^3c^3\minus{}b^3c^3\minus{}c^3a^2b\minus{}c^3b^2a\plus{}a^3c^2b\plus{}b^3c^2a)\geq 0$

İt is true because

$ a^4b^2\plus{}b^4a^2\plus{}a^4c^2\plus{}b^4c^2 \geq 2a^4bc\plus{}2b^4ac \geq a^4bc\plus{}b^4ac\plus{}a^3b^2c\plus{}b^3a^2c$
And
$ 3a^2b^2c^2\plus{}a^3b^3\minus{}a^3c^3\minus{}b^3c^3\minus{}c^3a^2b\minus{}c^3b^2a\equal{}(a^3b^3\minus{}a^3c^3)\plus{}(3a^2b^2c^2\minus{}b^3c^3\minus{}c^3a^2b\minus{}c^3b^2a) \geq 0$

Thank you very much, Cefer.  :)

Your solution works and it is not so ugly :wink:

When and where did I post this Lemma, Manlio? I don't remember it.

You used it to prove an inequality of Ductrung, but you used the mehod

$ (A \plus{} B \plus{} C) \minus{} C \equal{} B \plus{} C \geq 0$ with $ A \plus{} B \plus{} C \geq 0$ and $ C \leq 0$ to prove that $ A \plus{} B \geq 0$ which is your lemma

[quote="Vasc"]Write the inequality as
$ a^{n \minus{} 3}A \plus{} b^{n \minus{} 3}B \plus{} c^{n \minus{} 3}C \geq 0$,
where $ A \equal{} \frac {a^3b}{b^2 \plus{} ca} \plus{} \frac {a^3c}{c^2 \plus{} ab} \minus{} a^2$ etc.
Then, for $ a \geq b \geq c$ we have
$ a^{n \minus{} 3}A \plus{} b^{n \minus{} 3}B \plus{} c^{n \minus{} 3}C \geq b^{n \minus{} 3}(A \plus{} B) \plus{} c^{n \minus{} 3}C \geq c^{n \minus{} 3}(A \plus{} B \plus{} C) \geq 0$.[/quote]

You are right, Manlio. I found now this inequality of Ductrung in my new edition of Algebraic Inequality. :D


For any three positive reals a, b, c, we have

$\frac{a^3+3abc}{b+c}+\frac{b^3+3abc}{c+a}+\frac{c^3+3abc}{a+b} \ge 2(ab+bc+ca)$

inequality is equivalent to
\[
\sum {\frac{{a\left( {a^2  - b^2  - c^2  + bc} \right)}}
{{b + c}}}  \geqslant 0
\]

http://www.mathlinks.ro/Forum/viewtopic.php?t=32447

Lets rewrite or inequality as follows :
$\sum (\frac{a^3+3abc}{b+c}-a(b+c))\geq 0$ which is equivalent to :
$\sum \frac{a(a-b)(a+b-c)}{b+c}\geq 0$ or
$\sum \frac{a(a-b)(a-c)}{b+c}+\sum\frac{ab(a-b)}{b+c}\geq 0$. The first part is true from Vornicu Schur so we are left to prove that :
$\sum\frac{ab(a-b)}{b+c}\geq 0$ Now we have to prove that :
$\sum ab(a-b)(a+b)(a+c)\geq 0$ but this is equivalent to :
$\sum a^4b+a^3bc-b^3a^2-b^3ac\geq 0$ or simply :
$\sum a^4b \geq \sum b^3a^2$ but this is true by Muirheads.

[quote="RaduSorici"]
$\sum a^4b \geq \sum b^3a^2$ but this is true by Muirheads.[/quote]
Do not follow from Muirhead, since Muirhead can be only used for [b]symmetric inequalities[/b], and the inequality are not symmetric.

Well, actually the inequality $\sum a^4b \geq \sum b^3a^2$ is not true either. Moreover, I am sorry, RaduSorici, but already the very first step of your solution contains a calculation mistake.  :blush:

  darij

[quote="RaduSorici"]Lets rewrite or inequality as follows :
$\sum (\frac{a^3+3abc}{b+c}-a(b+c))\geq 0$ which is equivalent to :
$\sum \frac{a(a-b)(a+b-c)}{b+c}\geq 0$.[/quote]
The last ineq. is wrong.
The required ineq follows by Darij way, adding the inequalities

$\sum \frac{a(a-b)(a-c)}{b+c}\geq 0$
and
$\sum a^2+2abc\sum\frac{1}{b+c} \ge 2(ab+bc+ca)$ (Cauchy and Schur).

I will never ever use Muirheads again in my life :D
Sorry for the mistake! When I tried solving it again I got the same result as Vasc which gave a very pretty solution. I think he deserves to be named the Inequlity Top Solver(not really solving anything but giving hints and proposing new problems) in ML  :D

[quote="xtar"]For any three positive reals $ a, b, c,$ we have

$ \frac {a^3 \plus{} 3abc}{b \plus{} c} \plus{} \frac {b^3 \plus{} 3abc}{c \plus{} a} \plus{} \frac {c^3 \plus{} 3abc}{a \plus{} b} \ge 2( bc \plus{} ca\plus{}ab).$[/quote]$ \sum{\frac {a^3 \plus{} 3abc}{b \plus{} c}}\minus{}2\sum{bc}\equal{}\sum\frac{(b\minus{}c)^2[3bc\plus{}(b\plus{}c\minus{}a)^2]}{2(c\plus{}a)(a\plus{}b)}\geq0.$

Ji Chen got an incredible solution!
Here is my ugly solution!
We can easily make the simple expressions like the following:
$ \sum_{cyc}{\frac{a^3+3abc}{b+c}}=
\sum_{cyc}{\frac{a(a-b)(a-c)}{b+c}  + \frac{ab(a+c)}{b+c}+\frac{ac(a+b)}{b+c}}$
Likewise, we have the other expressions with the same expansion like above.
Hence, we need to use Vornicu- Schur and Cauchy Schwart for
$ \frac{ab(a+c)}{b+c}+\frac{ab(b+c)}{a+c} \geq 2ab$
Now, we get the result! :)


Putnam
algebra
polynomial
number theory
least common multiple
geometry
function

Questions from the 64th Putnam exam, Dec 6 2003.

A1
Let  n  be a fixed positive integer. How many ways are there to write  n
as a sum off positive integers,  n = a_1 + a_2 + ... + a_k,  with  k  an
arbitrary positive integer and  a_1 \le a_2 \l3 .. a_k \le a_1  + 1?
For example, with n=4 there are four ways: 4, 2+2, 1+1+2, 1+1+1+1.

A2
Let a1, a2, ..., a_n  and  b1, b2, ..., b_n  be nonnegative real numbers.
Show that
      (a1 a2 ... a_n)^{1/n} + (b1 b2 ... b_n)^{1/n} <=
                       ( (a1+b1) (a2+b2) ... (a_n + b_n) )^{1/n}

A3
Find the minimum value of
  | sin x + cos x + tan x + cot x + sec x + csc x |
for real numbers  x.

A4
Suppose that a,b,c,A,B<C  are real numbers, a\ne 0 and A \ne 0, such that
          | a x^2 + b x + c | <= | A x^2 + B x + C |
for all real numbers  x. Show that
              | b^2 - 4 a c | <= | B^2 - 4 A C |   .

A5
A Dyck n-path is a lattice path of  n  upsteps (1,1) and  n  downsteps (1,-1)
that starts at the origin  O  and never dips below the  x-axis.
A return is a maximal sequence of contiguous downsteps that terminates
on the  x-axis. For example, the Dyck 5-path illustrated has two returns,
of length  3  and  1  respectively.
              *
             / \
        *   *   *
       / \ /     \
      *   *       *   *
     /             \ / \
  O *---------------*---*

Show that there is a one-to-one correspondenc between the Dyck  n-paths
with no return of even length and the Dyck (n-1) paths.

A6
For a set  S  of positive integers, let  r_S(n)  denother the number of
ordered pairs  (s_1, s_2)  such that  s_1 \in S, s_2 \in S, s_1 \ne s2,
and  s_1 + s_2 = n.  Is it possible to partition the nonnegative
integers into two sets  A  and  B  in such a way that  r_A(n) = r_B(n)
for all  n ?

                ================================

B1
Do there exist polynomials  a(x), b(x), c(y), d(y)  such that
       1 + x y + x^2 y^2 = a(x) c(y) + b(x) d(y)
holds identically?

B2
Let  n  be a positive integer. Starting with the sequence
1, 1/2, 1/3, ..., 1/n  form a new sequence of  n-1  entries
3/4, 5/12, ..., (2n-1)/{2n(n-1)}  ny taking the averages of
two consecutive entries in the first sequence. Repeat the
averaging of neighbors on the second sequence to obtain a third
sequence of  n-2  entries and continue the final sequence produced
consists of a single number  x_n.  Show that  x_n < {2/n} .

B3
Show that for each positive integer  n,
    n ! = \Prod_{i=1}^n  lcm{1, 2, ..., [n/i] } .
(Here lcm denotes the least common multiple, and  [x]  denotes the
greatest integer \le  x .)

B4
Let  f(z) = a z^4 + b z^3 + c z^2 + d z + e = a(z-r_1)(z-r_2)(z-r_3)(z-r_4)
where  a,b,c,d,e  are integers, a \ne 0.  Show that if  r_1 + r_2  is a
rational number, and if  r_1 + r_2 \ne r_3 + r_4  then  r_1 r_2  is a
rational number.

B5
Let  A,B, and C  be equidistant points on the circumference of a circle
of unit radius centered at  O,  and let  P  be any point in the circle's
interior.  Let  a, b, c  be the distance from  P  to A, B, C  respectively.
Show that there is a triangle with side lengths  a, b, c  and that the
area of this triangle depends only on the distanct from  P  to  O.

B6
Let  f(x)  be  a continuous real-valued function defined on the interval
[0,1]. Show that
   \int_0^1 \int_0^1 | f(x) + f(y) | dx dy >= \int_0^1 |f(x)| dx.
---------------------------------------------------------------------
Since last Saturday a lot of answers, have a look

http://groups.google.fr/groups?dq=&hl=fr&lr=&ie=UTF-8&oe=UTF-8&threadm=co29tvcvqk4c6q6lf60fs95m2bbah01osv%404ax.com&prev=/groups%3Fhl%3Dfr%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26group%3Dsci.math

Mr.Omarjee

You can get the solutions from here.
"Thanks" to liyi for spoiling them.
http://www.unl.edu/amc/a-activities/a7-problems/putnam/


Help me with this problem:
  Determine all triples $(x, y, n)$ positive integers such that $gcd(x, n+1)=1$ and \[ x^n + 1 = y^{n+1} \]


calculus derivative trigonometry integration

great place to learn how to pick up girls/boys
shall we start writing??

Just try to be carefull with words here, as there are many kids lurking around. We don't want to sound to "adult" :)

sure we'll keep it on the pg level  ;)
for those of u who realllllly want to get serious..go to...haha jk

My advice for guys: The only case in which the words "pick-up line" should come into play is if there's a queue for your carpool.  To be genuine is to show you care.  If you consider romance a system to be beaten, you have no earthly business even considering a relationship.  You date because you enjoy the company of the potential-significant-other, not because your fingers and lips are getting itchy.

My advice for gals: If your crush is old-fashioned, break out the handkerchiefs and favors.  An ideal blend of being coy and friendly, if achieved, can be astonishingly effective.  If the traditional method is neither your nor your object of affection's style, it would smooth the way (not to mention, make our lives as guys simpler) to take the initiative.  Regardless of his opinion of you, he WILL be flattered.  Disclaimer: Said flattery is likely to vanish if you do not use discretion in your approach.  Use caution, and good luck.

Most likely; whether she can keep him and maintain her self-respect is another thing.  Guys can be as fickle as anyone else.  Many of us guys may be desperate, but we're not entirely without dignity.

Well, some of us, anyway.

[quote="Valentin Vornicu"] I don't think gals need advices. They almost always can get a guy if they really want to :) [/quote]

unless he's out of her league...*sigh*

Oh, yes, that reminds me:

Avoid anything that can remotely be construed as "stalking."  Girls will be annoyed (if not utterly creeped out), and will almost certainly NOT interpret your actions as "devotion."

Just trust me on this one.  I learned this the hard way.

[quote="Magnara"]Well, as this is the mathematician's guide:

Avoid integrating her over R.  I've found girls don't like that.  Wheh, the stories I could tell you.[/quote]

1/2 her^2?  huh?  I don't get it.

[quote="MathFiend"]why are these all tips for guyz to get gurlz? where's the gurlz for the guyz? :)[/quote]As I said almost all girls I personally know, including math loving girls, don't have any real problems when wanting to hook up a guy. Keeping him ... well that's another story ;)

Well, now that the forum's been restored, what's changed in the boy-girl dynamic?

:(
still hopelessly vying for a girl's attention....

I agree as well, but if you really want to know, check your PM's... ;)

[quote="#1287"][i]Moderator Note: The shirt is indeed quite inappropriate. However, no one without knowledge of Calculus I would be able to obtain any relevant meaning from it. Parents should be cautioned, but not very many middle schools and below know Calc.[/i][/quote]

Read a little bit more and you'll see that the website tells you how to calculate it...

I'm sure most middle schoolers can do the two calculations that are left...

[i]Moderator Note: You've seen high schoolers on this thread struggling with those calculations. So, perhaps not all middle schoolers would be able to do them. The description is not complete and someone who has no knowledge of integrals would probably be quite confused. However, you are right in that the site does basically calculate the answer for you[/i][/i]

Oh, I figured out the number, and I don't even know calculus  :D .  It's easy to figure out if you the slang.

I can't connect to the site.
Could anyone PM me about the number?

[hide="the t-shirt"]
[img]http://www.mathlinks.ro/Forum/files/t005.jpg[/img]
[/hide]

i don't know, i think this stuff is a bit tooo geeky to have any attractive/funny value to it

lol I'm a nerd, I find them amusing.  And proud of it too   :D

I love the shirt. But I go to a magnet school, where most people would be able to calculate it easily, so I would get in a lot of trouble for wearing it.

I'm going to the valentines dance, and I'm planning to ask my special other out to it. I like her and she likes me and we both know it but thats about it. We're good friends and I'm requesting advice on how to ask her out.  Quotes welcome

I rarely post on this forum but just suggestion:

Just be yourself. For what I know, it always work with me for girls when you're who you really are. Like pretending to be someone else would make you feel like you're great for beginning but then you wouldn't be able to keep it up since it's not [i]real you[/i].

And there's a small problem here. No offense but just this is little bit hard. I know you said you like her and she likes you but if everything works out in a way that this life goes, no one would have a problem, do they? :lol: But for reality, having relationships from the friendships aren't always the easiest thing in the world. You just have to like cool your that up feeling when you see her and try to be normal. And take time carefully where you and she are having most comfortable time.

I'm only telling you to do this because you posted "she likes me and we both know it." If she didn't know it but might just have a small interest, there are chances that things might not work out. I know that the breaking friendship is espeically painful if you lose by your heart....

Life goes on... Yet, pain would stay in your heart...

From a girl's point of view now.  Exactly what Silverfalcon said, just be you.

Like you said, you're good friends already.  That means she knows who you are, and she'll see a phony attempt at being something different.  Trying to be different might also make things awkard, and in some cases asking out a friend will ruin the friendship.  Just use your judgement.

You can go two routes, the serious, or the more playful.  It depends on what you think she'd prefer, and you too.  Besides, it sounds like there's not much a chance of her refusing, so what have you got to lose?  Just go for it!

But in the end, you're good friends.  Whatever you do, she'll be fine with it if you're really a good friend, as long as you stay the friend you are.

Good luck!

[quote="sran"]1. Could someone please explain the joke on the shirt?
2. What significance does integrating an individual over R have?[/quote]

1.  I'll give you a hint.  It evaluates to 69.

2.  That was in reference to my first comment, right?  It doesn't.  I was just amusing myself.  There was absolutely no hidden meaning there.

Magnara:
Yes, it was a reference to your comment.

On the matter of the integral: I evaluated it all right, but don't get the sig. And if its as bad as people are making it out to be, then I'm definitely not googling it, because my parents might cook me. Well.

Does anyone know if there is a dictionary of slang? I might be able to get away with that...

Thanks

sran - Don't bother. You will not find  [b]anything[/b] humorous   or anyting of value  in that than, say  a vulgar  slang not even worth paying attention.

For those, who find it is cute, or funnny   show that t-shirt to your parents  or anyone else whom you  want to impress     but  please stop putting messages in this math related form.

Mods and other postes  please cleanup/edit your comments

Regards.

P.S. And Valentin Vornicu : As an admin,  Please remove the link which is not appropiate here in this form.  Thanks.

The t-shirt can't hurt anyone that doesn't know what integrals are and how to compute them and if they do they are already old enough to find it funny once they compute it.

Therefore the link will remain there, but the topic will be locked so no further discussion about it is made. :)


[quote="Kouichi Nakagawa"]$ 3\left(\frac{1}{3}6^2\pi\minus{}\frac{1}{2}6^2\sin\left(\frac{2\pi}{3}\right)\right)\equal{}36\pi\minus{}27\sqrt{3}\approx66.33196$.
Thus, $ \boxed{66}$.

[b]Another solution[/b]
$ 6^2\pi\minus{}\frac{\sqrt{3}}{4}\left(2\left(\frac{6\times(3/2)}{\sqrt{3}}\right)\right)^2\equal{}36\pi\minus{}27\sqrt{3}$.[/quote]

Can you please explain in a easier way?

Sorry, but I don't really understand this. Can someone please show this with an explanation.

[quote="GameBot"]An equilateral triangle is inscribed in a circle. The radius of the circle is 6 in. What is the number of square inches in the area inside the circle but outside the triangle? Express your answer to the nearest whole number.[/quote]

[hide="Outline"]
Note that the altitudes in an equilateral triangle intersect each other in lengths with ratio $2:1$ (why?).
Also, the radius of the circle is basically the 2 part of the altitude's length (draw a triangle and draw altitudes-itll be clear).
Then, we have the length of the altitude to be:  $\dfrac{6}{2}\cdot 3 = 9$.
This means the side length of the triangle is:  $\dfrac{9\sqrt{3}}{3}$ (from 30-60-90).
The area of the circle is $36\pi$ and the area of the triangle is $\dfrac{81\sqrt{3}}{12} = \dfrac{27\sqrt{3}}{4}$.
Subtracting them gives the answer.[/hide]

[hide="Alternate"]

Draw from the centre of the circle radii which touch the tip of each vertex of the triangle.  Then look at a sector, and notice the area is just 1/3 the area of the circle - area of the segment of the circle.  Multiply by 3 to get the answer.
[/hide]

[hide="Alternate 2"]
Use $A = \dfrac{1}{2}ab\sin C$
[/hide]

Centroid Theorem states that the centroid (intersection of the medians of a triangle) is 2/3 the way to a vertex than to the median.

Might want to draw a picture to help see this... but the height is 6 * 3/2 = 9

Then, use 30-60-90 triangles to find the area of the triangle.

The area of the circle is simply 6 squared pi, or 36pi


real analysis
function
real analysis unsolved

Prove the existence of a set $A$ of zero Lebesgue measure such that any Riemann integrable function $f: [0,1]\to R$ is continuous in at least one point of $A$.

The set where a Riemann integrable function is not continuous is an $F_\sigma$ nullset, so it suffices to prove the existence of a nullset which is not contained in any $F_\sigma$ nullset. It is well-known that a nullset of the second category exists (the partition of the real line into a nullset and a set of the first category has been discussed before on the forum; the nullset appearing in this partition works, for instance). Such a set cannot be contained in an $F_\sigma$ nullset, because an $F_\sigma$ nullset is clearly of the first category.


For all $ a,b,c,d$ a positifs nombers prove that :

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{d\plus{}a}\plus{}\frac{d}{b\plus{}a}\geq2$

\begin{eqnarray*} \sum \frac {a}{b + c} & = & \sum \frac {a^2}{a(b + c)} \\
& \geq & \frac {(a + b + c + d)^2}{(a + b)(c + d) + (a + d)(c + b)} \\
\end{eqnarray*}
As $ (a + b + c + d)^2 - 2((a + b)(c + d) + (a + d)(c + b)) = (a - c)^2 + (b - d)^2\geq 0$, the result follows (alternately we could have used AM-GM on each of the terms in the denominator above, which is essentially the same thing)

hi
this is my prove :

this inequality help you...

$ \frac{U}{x}\plus{}\frac{V}{y} \geq \frac{4(Uy\plus{}Vx)}{(x\plus{}y)^2}$

$ \frac{a}{b\plus{}c}\plus{}\frac{c}{d\plus{}a}$  aply

$ \frac{b}{c\plus{}d}\plus{}\frac{c}{d\plus{}a}$ aply

[quote="mehdi cherif"]For all $ a,b,c,d$ a positifs nombers prove that :

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {d}{b \plus{} a}\geq2$[/quote]

Dear Mehdi Cherif,

the following identity is true for all $ a,b,c,d$:

\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} a} \plus{} \frac {d}{b \plus{} a} \minus{} 2 \equal{}
\]

\[ \equal{} \frac {1}{(a \plus{} b \plus{} c \plus{} d)^2}\left(2\,(a \minus{} c)^2 \plus{} 2\,(b \minus{} d)^2 \plus{} ( \frac {a}{b \plus{} c} \plus{} \frac {c}{a \plus{} d})\,(a \minus{} b \minus{} c \plus{} d)^2 \plus{} ( \frac {b}{c \plus{} d} \plus{} \frac {d}{a \plus{} b})\,(a \plus{} b \minus{} c \minus{} d)^2\right)
\]
Thank you for the nice inequality!

Sergey Markelov
markelov@mccme.ru

i'm sure
there is a mistake in the inequality

$ \frac{b}{c\plus{}d}$


Take a point $A$in the circle with radius 1 and consider the chord passing through $A.$ When $A,\ P,\ Q$ move, find the least value of $ \frac{1}{AP}+\frac{1}{AQ}.$

$\frac{1}{AP}+\frac{1}{AQ}=\frac{AP+AQ}{AP \cdot AQ}=\frac{PQ}{1-OA^{2}}$. $PQ$ takes it's least value when $PQ \perp OA$, that is when $PQ=2\sqrt{1-OA^{2}}$. So the least value of $\frac{1}{AP}+\frac{1}{AQ}$ is $\frac{2}{\sqrt{1-OA^{2}}}$.

After all, what's your answer, shobber. :maybe:

I misread your problem... I thought that point $A$ was fixed...... :blush:

So $\frac{1}{AP}+\frac{1}{AQ}\geq \frac{2}{\sqrt{1-OA^{2}}}\geq 2$. Equality holds when $A=O$.


quadratics trigonometry function HCSSiM MATHCOUNTS

:lol:  Ok the last one was easy, i got it in the first try, i have another one, check it out.

In the Art Gallery is a new conceptual art piece. The artist created a huge pile different paintings; periodically he picks up one of the paintings and hangs it on the wall, and adjusts it.

At 11:00, the painting titled "Number 30" was hung.

At 4:00, the painting titled "Number 240" was hung.

At 7:30, the painting titled "Number 315" was hung.

What painting will the artist hang on the wall at 9:20?

[hide]i didnt see any correlation between the time and the painting.. maybe ill see one later, but by numbers i would get 245.. not sure really ill get back to it..[/hide]

This problem is bad. There is no relationship at all between the times and numbers. (Linear, Quadratic, Cubic, Log, Sine, nothing!)

i can't find any relationship between the number of the painting and the time :?  :?

[quote="jimli"]Oh.  I might see correlation.  0:30,2:40,and 3:15 are army times.

This is probably wrong but:
of course there's a relationship.  Three points, so there must be quadratic one.  Just not an obvious quadratic.[/quote]

$f(x)=-\frac{288x^2}{119}+\frac{6438x}{119}+30$. That's as close as you'll get.  ;)

[quote="rcv"][quote="chess64"]This problem is bad.[/quote]
I think this is a nice problem.[/quote]

How exactly is this a nice problem? Does it have a cool answer/solution?

You advanced students are making this problem harder than necessary.

jimli, chess:  Of course, there will be a quadratic solution since there are three points.  But that is not the *simplest* solution, so it is not correct.  (Also, a quadratic function of that complexity would not be a "Basics" problem.)

Think "Basics".

is this a riddle or a math problem? if its the former, i dont want to spend any more time  :|

Oh how stupid can I be???  :mad:
Well, I probably wouldn't have thought of that anyway  :P

[quote="GoBraves"][hide="Answer with Solution"]The answer is $200$.

Solution: The painting number is just the angle between the clock hands, measured from the smaller hand to bigger hand in clockwise direction.[/hide][/quote]
i never would have thought of that

lol thats not math thats just fun and games  :P

i thought it would have to do with math more

New Hampshire State MATHCOUNTS Coach, frost13, has often posted the expression $30h-5.5m$ to the Middle School AoPS Forums.  (This formula applies to a variety of clock problems.  This formula is also published in John Cocharo's "Handbook of Math Solutions" book.  [Mr. Coacharo is a former Texas State MATHCOUNTS Coach, and a former member of the MATHCOUNTS question-writing committee.]  If you do MATHCOUNTS, or if you plan to do MATHCOUNTS, you should know this formula.)

Let's see how this formula might apply to this problem:

$\begin{array}{rcccc}
\hbox{11:00}&\rightarrow& 30\times 11-5.5\times \phantom{0}0=330 & \rightarrow&360-330=\phantom{0}30 \\
\hbox{4:00}&\rightarrow& 30\times \phantom{0}4-5.5\times \phantom{0}0=120  & \rightarrow&360-120=240 \\
\hbox{7:30}&\rightarrow& 30\times \phantom{0}7-5.5\times 30=\phantom{0}45 & \rightarrow&360-\phantom{0}45=315 \\
\hbox{9:20}&\rightarrow& 30\times \phantom{0}9-5.5\times 20=160  & \rightarrow&360-160=200
\end{array}$

I do not think this was a "trick question" (or even a very "tricky" question).

The first two times given were exactly 7 hours apart.  The picture numbers were 210 apart.  That provides is a nice linear relationship of 30 units per hour.  Even if you did not know the above formula, I think this provided a pretty good clue.

Please don't say this has nothing to do with math.  A large part of math is seeing relationships.  Great math students learn to look "outside the box" for these relationships.

:( THis is a good puzzle ,I rather would not waist time trying to figure something like  this in quadratic form


$\mathbb{N}=\left\{0,1,...............\right\}$
Lets $f: \mathbb{N}\to\mathbb{N}$ a bijective function.
Prove that there exists $(a,b,c)\in\mathbb{N}^3$ such that:
$a<b<c$ and $f(a)+f(c)=2f(b)$

It's sufficient to suppose $f$ surjective.
Let $B$ be the subset $B=\{t\in N || f(0) < f(t)\}$ , let $b$ be its minimum $b=min(B)$ and let $c$ be such that $f(c)= 2f(b)-f(0)$
$(0,b,c)$ satifies the required  condition
:cool:

I think we may have three $a,b,c$such that $f(a)+f(b)=2f(c)$ and $a+b=2c$ ,based on the condition $f$is bijective
but not sure about this supposition,i will try to prove it.


topology
advanced fields
advanced fields unsolved

Given the usual metric topology T on the real numbers R, show using Zorn's lemma that there exists a topology T' that contains T which is maximal under set inclusion with respect to the property that Q(the rationals) is dense.  Finally, show that T' is not the largest topology, only maximal.

Infact, for any topology $K$ s.t $\mathbb{Q}$ is dense, there exists a maximal topology $K'$ containing $K$ s.t $\mathbb{Q}$ is dense.
 Let $U$ ( resp. $V$ ) be the set consisted of an irrational $\alpha$ and all the numbers smaller (resp. larger ) than $\alpha$. If we denote $T*$ the topology generated by $U$ and the topology $T$ then it is easy to see that the rationals are dense in $T*$
 Let $T_u$ be a maximal topology containing $T*$ such that the rationals are dense in this topology as above. Similarly, we have $T_v$. Both of them are maximals and are different because if not, then $U$ and $V$ belong to a same topology $T_u$, thus $\{\alpha\} = U\cap V$ does not contain any rational, which contradicts the density of the rationals in $T_u$. That means there isn't any largest topology containing $T$ s.t $\mathbb{Q}$ is dense.


[size=75][color=darkblue]Geometry problem of a brazilian selection test to the Cone-Sul Olympiad.[/color]
[/size]
[color=darkred]EDIT[/color]
 :arrow: Consider a circle $ \alpha$ tangent to two paralell lines $ l_1$ and $ l_2$ at $ A$ and $ B$, respectively. Call $ C$ any point of $ l_1$, since it's not $ A$. Let $ D$ and $ E$ be two points in the circunference, such that they are not coincident and are at the  same side of $ AB$ as $ C$. Let $ CD\cap \alpha \equal{} \{F,D\}$ and $ CE\cap \alpha \equal{} \{G,E\}$. Let $ AD\cap l_2 \equal{} H$ and $ AE\cap l_2 \equal{} I$.Let $ AF\cap \l_2 \equal{} J$ and $ AG\cap \l_2 \equal{} K$. Show that $ JK \equal{} HI$.

Something in the enunciation must be incorrect. For instance, consider the situation where the chords $DE$ and $ AB$ are parallel and very close to each other. Then the projection of segment $\overline{DE}$ from $ A$ onto $ \ell_2$ tends to $ {\infty},$ while projection of $\overline{DE}$ from $ C$ onto $\ell_2$ tends to $ 0.$ So $ FG \equal{} HI$ makes no sense whatsoever.

I'm terribly sorry by a mistake in the exercise. The problem was edited.

Assume that $ JK \equal{} HI.$ Since $ (J,K)$ and $ (H,I)$ are the inverse of $ (F,G)$ and $ (D,E)$ under the inversion with center $A$ and power $AB^2,$ it follows that

$ \frac {JK}{FG} \equal{} \frac {AB^2}{AF \cdot AG} \ , \ \frac {HI}{ED} \equal{} \frac {AB^2}{AE \cdot ED} \ \Longrightarrow \frac {FG}{ED} \equal{} \frac {AF \cdot AG}{AE \cdot AD}$

On the other hand, projecting the quadrilateral $FGED$ on $\ell_2$ gives

$ \frac {AF \cdot AG}{AE \cdot AD} \equal{}\frac{FG}{ED}= \sqrt {\frac {CF \cdot CG}{CD \cdot DE}}$

Since $ (D,E)$ and $ (F,G)$ are inverse under the inversion with center $C$ and power $CA^2$

$\Longrightarrow \ \frac {FG}{ED} \equal{} \frac {CA^2}{CD \cdot CE}$

$\Longrightarrow \ \frac {CA^2}{CD \cdot CE} \equal{} \sqrt {\frac {CF \cdot CG}{CD \cdot DE}}$

Since $ CA^2 \equal{} CD \cdot CF\equal{}CE \cdot CG \Longrightarrow CD \cdot CE \equal{} CD \cdot CE,$ which is true.

I didn't know a solution like yours. Very good strategy. I would like to see more ideas and solutions to this problem, and if somebody want I will post my solution, that uses great tools of projective geometry.


Let triangle $ABC$ and $(O)$ is the circumcenter of triangle $ABC$ . Let $A',B',C'$ in $(O)$ such that $AA' , BB', CC'$ are parallel. Let $(l_{1})$ is the line through $A'$ and perpendicular to $BC$. Let $(l_{2})$ is the line through $B'$ and perpendicular to $CA$.Let $(l_{3})$ is the line through $C'$ and perpendicular to $AB$.Prove that $l_{1},l_{2},l_{3}$ is concurrent

Use following lemma:

Given are points $X,Y,A,B$. Then $XY \perp AB \iff XA^{2}-XB^{2}=YA^{2}-YB^{2}$.

and parallelity ($X'Y=XY'$ for $X,Y \in \{A,B,C \}$)

Let $X$ be point of intersection of $l_{1}$ and $l_{2}$. Try to prove $C'X \perp AB$ by using above lemma and parallelity as written above.


Consider the sequence $(f_{n}(x))$ ($n=0,1,2....$) of the functions defined on $[0,1]$ such that:
$f_{0}(x)=0$
and $f_{n+1}_(x)=f_{n}(x) + \frac{1}{2}(x-(f_{n}(x))^2}$
for every $n=0,1,2....$
Prove that  $\frac{nx}{2+n\sqrt{x}} \leq f_{n}(x) \leq \sqrt{x}$ for all $0 \leq x \leq 1$

If it's the first, then it can be proved that $f(x)\leq x^2$ which is stronger than $f(x)\leq\sqrt{x}$ for $x\in[0,1]$.

[quote="Rust"]$f_n(x) \ increase \ and \ \lim_{n}f_n(x)=x.$ $Were \ you \ find \ \sqrt x \ or \ x^2?$[/quote]
$\sqrt{x}$ is given in the original question and I may be wrong with $x^2$.

Look like a problem in MYM, If it's true, I think It shouldn't be discussed here.

I am sorry , I have a mistake in typing.
the right problem :
$f_{n+1}(x)=f_{n}(x) +\frac{1}{2}(x- (f_{n}(x))^2)$
{ This problem is from my teacher , not from MYM . Could someone tell me what "MYM" means? }

This problem is from "Problem in this issue" section of Vietnamese "Mathematics and Youth Magazines"-the newest version!!! you can see it here:(and all you need is a person who can translate it into E  :D ).. What are u doing,thonglao?find the solution of the problem by yourself...

and MYM mean Mathematics and Youth Magazine

and you can find it here [url=http://www.nxbgd.com.vn/toanhoctuoitre/?p=7&id=6&ReportID=45&ph=5]click here[/url]

$Ok! \ Consider \ function \ f(x,y)=y-y^2/2+x/2, \ when \ y<\sqrt x.$
$We \ have \ \frac{df(x,y)}{dy}>0, \ (y<\sqrt x<1).$
$Therefore \ f_n(x) \ increase, \ and \ f_n(x)<\sqrt x.$
$Let \ t_n=\sqrt x -f_n(x), \ we \ have \ t_{n+1}=t_n(1+\frac{t_n}{2}-\sqrt x)<t_n(1-\frac{\sqrt x}{2}).$
$Therefore \ t_n=\sqrt x -f_n(x)<\sqrt x (1-\frac{\sqrt x}{2})^n.$
It is better your approcsimation.


Let $ a,b,c>0$. Prove that:

$ \frac{(a\minus{}b\minus{}c)^2}{2a^2\plus{}(b\plus{}c)^2} \plus{} \frac{(b\minus{}c\minus{}a)^2}{2b^2\plus{}(c\plus{}a)^2} \plus{} \frac{(c\minus{}a\minus{}b)^2}{2c^2\plus{}(a\plus{}b)^2} \ge \frac{1}{2}$

[quote="thanhnam2902"]Let $ a,b,c > 0$. Prove that:

$ \frac {(a \minus{} b \minus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(b \minus{} c \minus{} a)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(c \minus{} a \minus{} b)^2}{2c^2 \plus{} (a \plus{} b)^2} \ge \frac {1}{2}$[/quote]
It's true for all reals $ a,$ $ b$ and $ c$ such that $ \prod_{cyc}\left(2a^2 \plus{} (b \plus{} c)^2\right)\neq0.$ :wink:

[quote="thanhnam2902"]Let $ a,b,c > 0$. Prove that:

$ \frac {(a \minus{} b \minus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(b \minus{} c \minus{} a)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(c \minus{} a \minus{} b)^2}{2c^2 \plus{} (a \plus{} b)^2} \ge \frac {1}{2}$[/quote]

$ \frac {(a \minus{} b \minus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(b \minus{} c \minus{} a)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(c \minus{} a \minus{} b)^2}{2c^2 \plus{} (a \plus{} b)^2} \ge \frac {1}{2}$
\[ \Leftrightarrow \sum {\frac {{4a^2 \plus{} ab \plus{} 4b^2 \plus{} 3ca \plus{} 3bc \minus{} c^2 }}{{2\left( {2a^2 \plus{} \left( {b \plus{} c} \right)^2 } \right)\left( {2b^2 \plus{} \left( {c \plus{} a} \right)^2 } \right)}}\left( {a \minus{} b} \right)^2 } \ge 0\]
let $ S_c \equal{} \frac {{4a^2 \plus{} ab \plus{} 4b^2 \plus{} 3ac \plus{} 3bc \minus{} c^2 }}{{2\left( {2a^2 \plus{} \left( {b \plus{} c} \right)^2 } \right)\left( {2b^2 \plus{} \left( {c \plus{} a} \right)^2 } \right)}}$, and $ S_a,S_b$ are similar.
without loss of generality, assume that $ a\ge b\ge c$, easy to prove that
$ S_b\ge0,S_c\ge0$
and
$ S_b \plus{} S_a$
$ \equal{} \frac {{4c^2 \plus{} ca \plus{} 4a^2 \plus{} 3bc \plus{} 3ab \minus{} b^2 }}{{2\left( {2c^2 \plus{} \left( {a \plus{} b} \right)^2 } \right)\left( {2a^2 \plus{} \left( {b \plus{} c} \right)^2 } \right)}} \plus{} \frac {{4b^2 \plus{} bc \plus{} 4c^2 \plus{} 3ab \plus{} 3ca \minus{} a^2 }}{{2\left( {2b^2 \plus{} \left( {c \plus{} a} \right)^2 } \right)\left( {2c^2 \plus{} \left( {b \plus{} a} \right)^2 } \right)}}$
$ \ge \frac {{4c^2 \plus{} ca \plus{} 4a^2 \plus{} 3bc \plus{} 3ab \minus{} b^2 }}{{2\left( {2c^2 \plus{} \left( {a \plus{} b} \right)^2 } \right)\left( {2a^2 \plus{} \left( {b \plus{} c} \right)^2 } \right)}} \plus{} \frac {{4b^2 \plus{} bc \plus{} 4c^2 \plus{} 3ab \plus{} 3ca \minus{} a^2 }}{{2\left( {2a^2 \plus{} \left( {b \plus{} c} \right)^2 } \right)\left( {2c^2 \plus{} \left( {b \plus{} a} \right)^2 } \right)}}$
$ \equal{} \frac {{3\left( {a \plus{} b} \right)^2 \plus{} 4c\left( {a \plus{} b \plus{} 2c} \right)}}{{2\left( {2c^2 \plus{} \left( {a \plus{} b} \right)^2 } \right)\left( {2a^2 \plus{} \left( {b \plus{} c} \right)^2 } \right)}}$
$ > 0.$
so, by SOS-theorem, ineq holds...


limit
real analysis
real analysis unsolved

[url=http://bildr.no/view/509255][img]http://bildr.no/thumb/509255.jpeg[/img][/url]
[url=http://bildr.no/view/509256][img]http://bildr.no/thumb/509256.jpeg[/img][/url]

I can not understand the sentence denoted  by red line .

Thanks

You are perfectly right, the paragraph you underlined in red is utter nonsense (I think many, or at least some of the materials from Bildr.no are written with undue haste).

The reasoning is in fact very simple. Since $ \ell_1 > \frac {f(y)} {y}$ for all $ y>0$, and $ \lim_{n \to \infty} \frac {f(y_n)} {y_n} \equal{} \ell_2$, it follows $ \ell_1 \geq \ell_2$. The exact same reasoning works for the indices reversed, so $ \ell_2 \geq \ell_1$. So $ \ell_1 \equal{} \ell_2$; in fact this common value is $ \sup\left \{ \frac {f(x)} {x} \ \mid \ x>0\right \}$. It is an example of the squeeze method.

[quote="mavropnevma"]You are perfectly right, the paragraph you underlined in red is utter nonsense (I think many, or at least some of the materials from Bildr.no are written with undue haste).

The reasoning is in fact very simple. Since $ \ell_1 > \frac {f(y)} {y}$ for all $ y > 0$, and $ \lim_{n \to \infty} \frac {f(y_n)} {y_n} \equal{} \ell_2$, it follows $ \ell_1 \geq \ell_2$. The exact same reasoning works for the indices reversed, so $ \ell_2 \geq \ell_1$. So $ \ell_1 \equal{} \ell_2$; in fact this common value is $ \sup\left \{\frac {f(x)} {x} \ \mid \ x > 0\right \}$. It is an example of the squeeze method.[/quote]

Thanks very much.


probability
combinatorial geometry
probability and stats

Seven points are selected randomly from $ S^1\subset\mathbb C$. What is the probability that origin is not contained in convex hull of these points?

wlog the $ z\equal{}1$ on the circle with six other points $ w_j\equal{}e^{iu_j}, j\equal{}1,..6$

then the origin is not in their convex hull  as long as $ \exists s\in (0,\pi), u_j \in [0, \pi\minus{}s)\cup ( 2\pi\minus{}s, 2\pi), \forall j$

rotating this circle by s radians would give us $ u_j\plus{}s\mod 2\pi\equal{}u_j'\in(0,\pi) \forall j$

so $ P(u_j'<\pi)\equal{}\frac{1}{2} \to P(\max u_j'<\pi)\equal{}\frac{1}{2^6}$

i dont know if i believe it, but i keep getting it as my answer =/

$ \frac{1}{2^6}$ is the probability that the last $ 6$ points lie on the semicircle with the middle at the first point. This is a proper subevent of the event in question ;).

if we reorder increasingly the $ Arg u_i$ ( from 0 to $ 2\pi$), the probability of the event is
$ P(Arg u_7 \minus{} Arg u_1 <\pi\}$
Of course to compute this we need to compute the law of the reordered points which were originally independent r.v. following a uniform law in $ [0;2\pi]$
which I think is quite classical.

[quote="TheBridgeTwo"]if we reorder increasingly the $ Arg u_i$ ( from 0 to $ 2\pi$), the probability of the event is
$ P(Arg u_7 \minus{} Arg u_1 < \pi\}$
[/quote]
I am not sure about this one ! seems a little bit more involved .. I think we've discussed about a very similar problem quite recently in this probability sub-forum, isn't it ?

You are right alekk to get the center in the convex hull we should have at least three points among the seven having the sum of their difference of their arguments superior to $ \pi$.

Well, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5715]this[/url] takes care of it (and of all obvious generalizations).

Can you check am I right or wrong?

WLOG, we can claim that (1.0) is selected and Lets note the angle x_1, x_2, ... , x_7 clockwise.
Then x_1+x_2+...+x_6 is less than 2*pi (x_1+...+x_7=2*pi)
If convex hull of 7 points do not contain (0, 0), for some i, x_i is more than pi

if x_i > pi, let y_i = x_i-pi
Then x_1+...+y_i+..x_6 < pi and the volume of the graph is 1/2^6 of x_1+...+x_6 <2*pi

So, the answer is 1 - 7/64 = 57/64...
So generalization, the answers is 1 - n/2^n-1

p.s. Thanks for finding typo @johnkwon0328

Can you check am I right or wrong?

WLOG, we can claim that $(1,0)$ is selected and Lets note the angle $x_1, x_2, ... , x_7$ clockwise.
Then $x_1+x_2+...+x_6$ is less than $2\times \pi$ since $x_1+...+x_7=2\times \pi$
If convex hull of $7$ points do not contain $(0, 0)$, for some $i$, $x_i$ is more than $pi$

if $x_i > \pi$, let $y_i = x_i-\pi$
Then $x_1+...+y_i+..x_6 < \pi$ and the volume of the graph is $1/2^6$ of $x_1+...+x_6 <2\times \pi$

So, the answer is $1 - \frac{7}{64} = \frac{57}{64}...$
So generalization, the answers is $1 - \frac{n}{2^{n-1}}$

Corrected an error, and made it easier to read...

Well, I think it is right. :)
I always get $ \frac{57}{64} $ .


hm...

I actually never learned how to prove but,

I just tested few number

41 * 11111 = 45551

and I just scrabbled the number as it was told

14555

and it was divisible by 41 and it is 355

Yetti -

 [hide]Loginc is correct.

You could simplify a littel by letting 'abcd ' (decimal digit value) ax X
and write  M = 10x+ e
       and   N = 10000 e + x
You get the same thing but you  have only x (one variable)  to deal with and not   4.
Some times that makes things a little easier. hope that  helps.[/hide]

Can you generalize? What other numbers have this property? Are there numbers that if they divide "abcedf", they would divide "fabcde" also? So forth.

[quote="ssupermath"]Let  abcde 5-digits number divisible 41.Prove  that  eabcd 5-digits number divisible  41.[/quote]
[quote]Since its divisible by 4, multiplying eabcd  by 10 doesnt matter. so 100000e-e is divisible by 41 if the above statement is true. 99999 is divisible by 41.[/quote]