geometry
3D geometry
tetrahedron
geometric transformation
reflection
vector
parallelogram

Let $h_1$, $h_2$, $h_3$, $h_4$ be the altitudes of a tetrahedron and $d_1$, $d_2$, $d_3$ be the distances between opposites edges. Prove the equality \[\frac{1}{h_1^2}+\frac{1}{h_2^2}+\frac{1}{h_3^2}+\frac{1}{h_4^2}=\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}.\]

Myth, I am a newcomer to stereometry, hence it would be very nice if you could check the following solution. I will try to write it up as briefly as I can. ;)

In the following, the (non-directed) area of a polygon X will be denoted by [X], and the (non-directed) volume of a polyhedron Y will be denoted by [[Y]].

Also, in a parallelepiped, the distances between opposite faces will be called [i]altitudes[/i] of the parallelepiped (so that every parallelepiped has three altitudes).

Now, let ABCD be our tetrahedron, let V = [[ABCD]] be its volume, and let $S_A=\left[BCD\right]$, $S_B=\left[CDA\right]$, $S_C=\left[DAB\right]$, $S_D=\left[ABC\right]$ be the areas of its side-faces. WLOG assume that the altitudes $h_1$, $h_2$, $h_3$, $h_4$ of the tetrahedron ABCD are the altitudes to the side-faces BCD, CDA, DAB, ABC, respectively. Then, by the well-known formula for the volume of a tetrahedron, the volume V of the tetrahedron ABCD equals $V=\frac13h_1S_A=\frac13h_2S_B=\frac13h_3S_C=\frac13h_4S_D$. Thus, $h_1=\frac{3V}{S_A}$, $h_2=\frac{3V}{S_B}$, $h_3=\frac{3V}{S_C}$, $h_4=\frac{3V}{S_D}$.

Now, through any pair of opposite edges of the tetrahedron ABCD, draw two parallel planes. Altogehter, we get six planes which border a parallelepiped P. Four of the eight vertices of this parallelepiped P are the vertices A, B, C, D of our initial tetrahedron ABCD; the other four vertices will be denoted by A', B', C', D' in such a way that the points A', B', C', D' are the reflections of the points A, B, C, D with respect to the center of the parallelepiped P. The parallelepiped P has three pairs of parallel side-faces: the parallel side-faces AC'BD' and A'CB'D, the parallel side-faces AD'CB' and A'DC'B, and the parallel side-faces AB'DC' and A'BD'C.

Since the distance between two skew lines is the distance between the two parallel planes drawn through these two lines, the distances $d_1$, $d_2$, $d_3$ between pairs of opposite edges of the tetrahedron ABCD are the altitudes of the parallelepiped P. We can WLOG assume that the distance $d_1$ is the altitude perpendicular to the side-faces AC'BD' and A'CB'D, that the distance $d_2$ is the altitude perpendicular to the side-faces AD'CB' and A'DC'B, and that the distance $d_3$ is the altitude perpendicular to the side-faces AB'DC' and A'BD'C.

If we denote by W = [[P]] the volume of the parallelepiped P, and by $T_B=\left[AC^{\prime}BD^{\prime}\right]=\left[A^{\prime}CB^{\prime}D\right]$, $T_C=\left[AD^{\prime}CB^{\prime}\right]=\left[A^{\prime}DC^{\prime}B\right]$, $T_D=\left[AB^{\prime}DC^{\prime}\right]=\left[A^{\prime}BD^{\prime}C\right]$ the areas of its side-faces, then, since the volume of a parallelepiped equals the area of one of its side-faces multiplied with the altitude to this face, we have $W=T_B\cdot d_1=T_C\cdot d_2=T_D\cdot d_3$. Hence, $d_1=\frac{W}{T_B}$, $d_2=\frac{W}{T_C}$, $d_3=\frac{W}{T_D}$.

Now, I will show that W = 3V. The proof is put into a hider, so that those for whom this fact is old news can read on.

[hide="Proof of W equals 3V"]
[i]Proof of W = 3V.[/i] We will use vectors.

Let K be the centroid of triangle BCD. Then, $\overrightarrow{A^{\prime}K}=\frac{\overrightarrow{A^{\prime}B}+\overrightarrow{A^{\prime}C}+\overrightarrow{A^{\prime}D}}{3}$. On the other hand, since the vectors $\overrightarrow{A^{\prime}B}$, $\overrightarrow{A^{\prime}C}$ and $\overrightarrow{A^{\prime}D}$ span up the parallelepiped P and the point A is the vertex of this parallelepiped opposite to the vertex A', we have $\overrightarrow{A^{\prime}A}=\overrightarrow{A^{\prime}B}+\overrightarrow{A^{\prime}C}+\overrightarrow{A^{\prime}D}$. Thus, $\overrightarrow{A^{\prime}K}=\frac{\overrightarrow{A^{\prime}A}}{3}$. Hence, the point K lies on the line A'A and satisfies $A^{\prime}K=\frac{A^{\prime}A}{3}$. Thus, $3\cdot A^{\prime}K=A^{\prime}A$, and it follows that

$AK=A^{\prime}A-A^{\prime}K=3\cdot A^{\prime}K-A^{\prime}K=2\cdot A^{\prime}K$,

and thus $\frac{A^{\prime}K}{AK}=\frac12$.

Now, we know that $h_1$ is the altitude of the tetrahedron ABCD to the plane BCD. Let $h_1^{\prime}$ be the altitude of the tetrahedron A'BCD to the plane BCD. Then, by the area formula for a tetrahedron, we have $\left[\left[ABCD\right]\right]=\frac13h_1\left[BCD\right]$ and $\left[\left[A^{\prime}BCD\right]\right]=\frac13h_1^{\prime}\left[BCD\right]$. Thus, $\frac{\left[\left[A^{\prime}BCD\right]\right]}{\left[\left[ABCD\right]\right]}=\frac{h_1^{\prime}}{h_1}$.

Now, since $h_1$ and $h_1^{\prime}$ are the altitudes of the tetrahedra ABCD and A'BCD, respectively, to the plane BCD, i. e. the distances from the points A and A' to the plane BCD, while the point K lies on the plane BCD and on the line A'A, the Thales theorem yields $\frac{h_1^{\prime}}{h_1}=\frac{A^{\prime}K}{AK}$. Since $\frac{\left[\left[A^{\prime}BCD\right]\right]}{\left[\left[ABCD\right]\right]}=\frac{h_1^{\prime}}{h_1}$ and $\frac{A^{\prime}K}{AK}=\frac12$, this becomes $\frac{\left[\left[A^{\prime}BCD\right]\right]}{\left[\left[ABCD\right]\right]}=\frac12$. Equivalently, $\left[\left[A^{\prime}BCD\right]\right]=\frac12\cdot\left[\left[ABCD\right]\right]$. Similarly, $\left[\left[B^{\prime}CDA\right]\right]=\frac12\cdot\left[\left[ABCD\right]\right]$, $\left[\left[C^{\prime}DAB\right]\right]=\frac12\cdot\left[\left[ABCD\right]\right]$, $\left[\left[D^{\prime}ABC\right]\right]=\frac12\cdot\left[\left[ABCD\right]\right]$. Since the parallelepiped P is formed by the tetrahedra [ABCD], [A'BCD], [B'CDA], [C'DAB], [D'ABC], we have

$W=\left[\left[P\right]\right]=\left[\left[ABCD\right]\right]+\left[\left[A^{\prime}BCD\right]\right]+\left[\left[B^{\prime}CDA\right]\right]+\left[\left[C^{\prime}DAB\right]\right]+\left[\left[D^{\prime}ABC\right]\right]$
$=\left[\left[ABCD\right]\right]+\frac12\cdot\left[\left[ABCD\right]\right]+\frac12\cdot\left[\left[ABCD\right]\right]+\frac12\cdot\left[\left[ABCD\right]\right]+\frac12\cdot\left[\left[ABCD\right]\right]$
$=3\cdot\left[\left[ABCD\right]\right]=3V$,

and the equality W = 3V is proven.[/hide]

Now, having the formula W = 3V, we can transform the equalities $d_1=\frac{W}{T_B}$, $d_2=\frac{W}{T_C}$, $d_3=\frac{W}{T_D}$ into the form $d_1=\frac{3V}{T_B}$, $d_2=\frac{3V}{T_C}$, $d_3=\frac{3V}{T_D}$. Combining these with $h_1=\frac{3V}{S_A}$, $h_2=\frac{3V}{S_B}$, $h_3=\frac{3V}{S_C}$, $h_4=\frac{3V}{S_D}$, we see that the identity in question,

$\frac{1}{h_1^2}+\frac{1}{h_2^2}+\frac{1}{h_3^2}+\frac{1}{h_4^2}=\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}$,

is equivalent to

$\frac{1}{\left(\frac{3V}{S_A}\right)^2}+\frac{1}{\left(\frac{3V}{S_B}\right)^2}+\frac{1}{\left(\frac{3V}{S_C}\right)^2}+\frac{1}{\left(\frac{3V}{S_D}\right)^2}=\frac{1}{\left(\frac{3V}{T_B}\right)^2}+\frac{1}{\left(\frac{3V}{T_C}\right)^2}+\frac{1}{\left(\frac{3V}{T_D}\right)^2}$.

This simplifies to

$S_A^2+S_B^2+S_C^2+S_D^2=T_B^2+T_C^2+T_D^2$.

Now, how to prove this identity? Well, probably there is a nice geometric way, but I don't see it, so my proof uses vectors and their scalar and cross products:

We will denote by $\overrightarrow{v}\cdot\overrightarrow{w}$ the scalar product and by $\overrightarrow{v}\times\overrightarrow{w}$ the cross product of two vectors $\overrightarrow{v}$ and $\overrightarrow{w}$. Also, for any vector $\overrightarrow{v}$, the notation $\overrightarrow{v}^2$ will mean the scalar product of this vector with itself: $\overrightarrow{v}^2=\overrightarrow{v}\cdot\overrightarrow{v}=\left|\overrightarrow{v}\right|^2$.

Now, introduce the vectors $\mathbf{b}=\overrightarrow{A^{\prime }B}$, $\mathbf{c}=\overrightarrow{A^{\prime }C}$ and $\mathbf{d}=\overrightarrow{A^{\prime }D}$. Since the quadrilateral A'CB'D is a parallelogram, $\overrightarrow{A^{\prime }B^{\prime }}=\overrightarrow{A^{\prime }C}+\overrightarrow{A^{\prime }D}$; in other words, $\overrightarrow{A^{\prime }B^{\prime }}=\mathbf{c}+\mathbf{d}$. On the other hand, since the vertices A' and B' of the parallelepiped P are opposite to the vertices A and B, we have $\overrightarrow{A^{\prime }B^{\prime }}=\overrightarrow{BA}$. Thus, $\overrightarrow{BA}=\mathbf{c}+\mathbf{d}$.

The parallelogram A'CB'D is spanned up by the vectors $\overrightarrow{A^{\prime }C}=\mathbf{c}$ and $\overrightarrow{A^{\prime }D}=\mathbf{d}$. Thus, the area of this parallelogram equals $\left[A^{\prime}CB^{\prime}D\right]=\left|\mathbf{c}\times\mathbf{d}\right|$. In other words, $T_B=\left|\mathbf{c}\times\mathbf{d}\right|$. Similarly, $T_C=\left|\mathbf{d}\times\mathbf{b}\right|$ and $T_D=\left|\mathbf{b}\times\mathbf{c}\right|$.

The triangle BCD is spanned up by the vectors $\overrightarrow{BC}$ and $\overrightarrow{CD}$. Thus, the area of this triangle equals

$\left[BCD\right]=\left|\frac12\cdot\overrightarrow{BC}\times\overrightarrow{CD}\right|=\left|\frac12\cdot\left(\overrightarrow{A^{\prime}C}-\overrightarrow{A^{\prime}B}\right)\times\left(\overrightarrow{A^{\prime}D}-\overrightarrow{A^{\prime}C}\right)\right|$
$=\left|\frac12\cdot\left(\mathbf{c}-\mathbf{b}\right)\times\left(\mathbf{d}-\mathbf{c}\right)\right|=\left|\frac12\cdot\left(\mathbf{c}\times\mathbf{d}-\mathbf{c}\times\mathbf{c}-\mathbf{b}\times\mathbf{d}+\mathbf{b}\times\mathbf{c}\right)\right|$.

Since $\mathbf{c}\times\mathbf{c}=\overrightarrow{0}$ (in fact, the cross product of each vector with itself is 0) and $-\mathbf{b}\times\mathbf{d}=\mathbf{d}\times\mathbf{b}$, this becomes

$\left[BCD\right]=\left|\frac12\cdot\left(\mathbf{c}\times\mathbf{d}+\mathbf{d}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}\right)\right|$.

In other words,

$S_A=\left|\frac12\cdot\left(\mathbf{c}\times\mathbf{d}+\mathbf{d}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}\right)\right|$.

The triangle DAB is spanned up by the vectors $\overrightarrow{BD}$ and $\overrightarrow{BA}$, where we know that $\overrightarrow{BA}=\mathbf{c}+\mathbf{d}$. Thus, the area of this triangle equals

$\left[DAB\right]=\left|\frac12\cdot\overrightarrow{BD}\times\overrightarrow{BA}\right|=\left|\frac12\cdot\left(\overrightarrow{A^{\prime}D}-\overrightarrow{A^{\prime}B}\right)\times\left(\mathbf{c}+\mathbf{d}\right)\right|$
$=\left|\frac12\cdot\left(\mathbf{d}-\mathbf{b}\right)\times\left(\mathbf{c}+\mathbf{d}\right)\right|=\left|\frac12\cdot\left(\mathbf{d}\times\mathbf{c}+\mathbf{d}\times\mathbf{d}-\mathbf{b}\times\mathbf{c}-\mathbf{b}\times\mathbf{d}\right)\right|$.

Since $\mathbf{d}\times\mathbf{d}=\overrightarrow{0}$ (because the cross product of each vector with itself is 0) and $\mathbf{d}\times\mathbf{c}=-\mathbf{c}\times\mathbf{d}$ and $-\mathbf{b}\times\mathbf{d}=\mathbf{d}\times\mathbf{b}$, this becomes

$\left[DAB\right]=\left|\frac12\cdot\left(-\mathbf{c}\times\mathbf{d}-\mathbf{b}\times\mathbf{c}+\mathbf{d}\times\mathbf{b}\right)\right|=\left|\frac12\cdot\left(\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{d}-\mathbf{d}\times\mathbf{b}\right)\right|$.

In other words,

$S_C=\left|\frac12\cdot\left(\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{d}-\mathbf{d}\times\mathbf{b}\right)\right|$.

Similarly,

$S_D=\left|\frac12\cdot\left(\mathbf{c}\times\mathbf{d}+\mathbf{d}\times\mathbf{b}-\mathbf{b}\times\mathbf{c}\right)\right|$;
$S_B=\left|\frac12\cdot\left(\mathbf{d}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}-\mathbf{c}\times\mathbf{d}\right)\right|$.

Thus, using the expressions for $T_B$, $T_C$, $T_D$, $S_A$, $S_B$, $S_C$ and $S_D$ obtained so far, the equality in question,

$S_A^2+S_B^2+S_C^2+S_D^2=T_B^2+T_C^2+T_D^2$,

becomes

$\left|\frac12\cdot\left(\mathbf{c}\times\mathbf{d}+\mathbf{d}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}\right)\right|^2+\left|\frac12\cdot\left(\mathbf{d}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}-\mathbf{c}\times\mathbf{d}\right)\right|^2$
$+\left|\frac12\cdot\left(\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{d}-\mathbf{d}\times\mathbf{b}\right)\right|^2+\left|\frac12\cdot\left(\mathbf{c}\times\mathbf{d}+\mathbf{d}\times\mathbf{b}-\mathbf{b}\times\mathbf{c}\right)\right|^2$
$=\left|\mathbf{c}\times\mathbf{d}\right|^2+\left|\mathbf{d}\times\mathbf{b}\right|^2+\left|\mathbf{b}\times\mathbf{c}\right|^2$.

Since $\left|\overrightarrow{v}\right|^2=\overrightarrow{v}^2$ holds for every vector $\overrightarrow{v}$, this equality rewrites as

$\left(\frac12\cdot\left(\mathbf{c}\times\mathbf{d}+\mathbf{d}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}\right)\right)^2+\left(\frac12\cdot\left(\mathbf{d}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}-\mathbf{c}\times\mathbf{d}\right)\right)^2$
$+\left(\frac12\cdot\left(\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{d}-\mathbf{d}\times\mathbf{b}\right)\right)^2+\left(\frac12\cdot\left(\mathbf{c}\times\mathbf{d}+\mathbf{d}\times\mathbf{b}-\mathbf{b}\times\mathbf{c}\right)\right)^2$
$=\left(\mathbf{c}\times\mathbf{d}\right)^2+\left(\mathbf{d}\times\mathbf{b}\right)^2+\left(\mathbf{b}\times\mathbf{c}\right)^2$.

But this equality is a simple corollary of the algebraic identity

$\left(\frac12\cdot\left(x+y+z\right)\right)^2+\left(\frac12\cdot\left(y+z-x\right)\right)^2+\left(\frac12\cdot\left(z+x-y\right)\right)^2+\left(\frac12\cdot\left(x+y-z\right)\right)^2$
$=x^2+y^2+z^2$,

applied to the vectors $x=\mathbf{c}\times\mathbf{d}$, $y=\mathbf{d}\times\mathbf{b}$ and $z=\mathbf{b}\times\mathbf{c}$. This solves the problem.

  Darij

Choose an arbitrary point and denote vectors from this point to the vertices of the tetrahedron as $a,b,c,d$. Then \[\frac{1}{h_a^2}=\frac{[BC,BD]^2}{(BA,[BC,BD])^2}=\frac{[BC,BD]^2}{(3V)^2}=\frac{[b-c,b-d]^2}{9V^2}=\frac{([c,d]-[b,d]+[b,c])^2}{9V^2},\] \[\frac{1}{d_{AB,CD}^2}=\frac{[AB,CD]^2}{([AB,CD],AC)^2}=\frac{[AB,CD]^2}{([AB,AC],CD)^2}=\frac{[AB,CD]}{9V^2}=\frac{[a-b,c-d]^2}{9V^2}=\frac{([a,c]+[b,d]-[a,d]-[b,d])^2}{9V^2}.\]
Sum up.

Oh, thanks, that's very nice, although I had to derive the formulas you used for myself, since they were anything but known to me...

  Darij

A newbie question: in Myth's solution, what do the notations [a,b] and (a,b) denote?

Well, if $\overrightarrow{x}$ and $\overrightarrow{y}$ are two vectors, then $\left[\overrightarrow{x},\overrightarrow{y}\right]$ denotes their cross-product (I usually write $\overrightarrow{x}\times\overrightarrow{y}$ for it), and $\left(\overrightarrow{x},\overrightarrow{y}\right)$ denotes their scalar product (I denote it by $\overrightarrow{x}\cdot\overrightarrow{y}$).

The formulas used by Myth are:

[b](1)[/b] If P is a point in the space, and p is a plane passing through three points O, X and Y, then the distance from the point P to this plane p is

$\frac{\left(OP,\left[OX,OY\right]\right)}{\left|\left[OX,OY\right]\right|}$.

[b](2)[/b] If AB and CD are two skew lines in the space, then the distance between these lines is

$\frac{\left(\left[AB,CD\right],AC\right)}{\left|\left[AB,CD\right]\right|}$.

The proofs of these formulas are rather obvious consequences of the geometric interpretations of the cross and scalar products (the cross product of two vectors is a vector perpendicular to the plane spanned up by the two vectors, and the scalar product of two vectors equals the length of the projection of one of them on the other one, multiplied by the length of the other one).

  Darij


group theory
abstract algebra
superior algebra
superior algebra unsolved

If any p-subgroup Sylow of an abelian group $ G$ is cyclic prove that $ G$ is cylic.

if your group is finite, it follows from the structure theorem for finite abelian groups.

Again assuming that $ G$ is finite, if $ G$ is abelian, then $ G$ is nilpotent and hence a direct product of its Sylow subgroups. As the distinct Sylow subgroups have pairwise coprime order, they commute and hence if one chooses a generator from each Sylow, then the product of these generators will be a generator for $ G$.

You can also see it very directly.  For $ p^\alpha$ dividing $ |G|$, $ \alpha$ maximal, there exists some $ x_p$ of order $ p^\alpha$.  It follows that the product of all $ x_p$ generates $ G$.

The structure theorem seems to imply that this is still true if the group is infinite but finitely generated.  What about infinitely generated abelian groups?


The digits 1, 3, and 5 can be rearranged to form six distinct 3-digit numbers.  Let $Z$ be the product of these six numbers.  $(Z=622092199590225).$  What is the largest prime factor of $Z$?  Do not use a calculator.  Show your work.

[hide]622092199590225=3^15*5^2*7*13*17*19*59
I don't think there is more work to show. Just try it. It's over.[/hide]

[hide]Let's break up the numbers:
$513=3^3\cdot17$

$153=3^2\cdot17$

$315=3^2\cdot5\cdot7$

$351=3^3\cdot13$

$135=3^3\cdot5$

$531=3^2\cdot59$

(Note that each number has 3 as a factor, since 1+3+5=9, div. by 3).

Since we are multiplying all three numbers together, all the prime factors above are going to be in Z. The largest prime factor, therefore, is 59. [/hide]

[quote="Treething"]On a side note, how in the world did you calculate that number?[/quote]

Just use your computer's calculator... it can hold 15 digits.  :)

You can multiply it out by hand. I'd guess it would take maybe 5 minutes, max.

rcv, are non middle schoolers not even allowed to post in the MC forum, or what?If his solution helps someone, why not show it?

High-schoolers are supposed to only post a solution if the middle-schoolers can't get it and the mods, in their finite wisdom, don't decide to move it up.

[quote="JesusFreak197"]Is it allowable to post an alternate solution, or maybe something more in-depth, once a few middle schoolers have solved a problem already?[/quote]

Do you have a better solution? go ahead.  :)

Now that there is a solution from a Middle School student, I have restored Seaward's original response.  Please read the guidelines in the Announcements for this Forum.  If a problem is easy for you, you should not be the first to answer it.  Middle School students should always have the first crack at Challenge problems.  After a couple of days, more advanced students and teachers are invited to share their wisdom in providing alternative answers and explanations.

Back on topic...  236factorial's response is very good, and shows the main insight required to solve this problem without a calculator.  A slightly deeper insight might allow you to solve this problem even more quickly.  Any ideas from Middle Schoolers?  [More advanced students, please wait until tomorrow to provide your insights.]

[quote="236factorial"]Do you have a better solution? go ahead.  :)[/quote]

No, not in this case; I was asking it in regard to rcv's post, as a general rule. By his most recent post, I will assume that it is allowable once it has been left for a couple days and answered by middle schoolers already. :)

[quote="jhollenbeck"]$3^15*5^2*59*19*17*13*7=Z$ so 59 :idea: is the largest factor[/quote]

Do you mean $3^{15}$? The code is 3^{15}. Notice the brackets. For an exponent more than one digit in length, use brackets.

And you mean prime factor.

Is there any way to do the problem without having to prime-factor the huge number or the six numbers? Because speed is everything :).

thanks for reminding me mathnerd, i forgot to do that :blush: i have edited my post

[quote="mathnerd314"]Is there any way to do the problem without having to prime-factor the huge number or the six numbers? Because speed is everything :).[/quote]

[hide]You have 6 numbers, composed of 3, 1, 5... You know the largest prime factor cannot possibly be 5, so you don't have to factor 315, and 135. That leaves us with 513, 153, 531, 351... Each of these numbers is divisible by 9 since the sum of the digits is 9. Since we are looking for the largest prime factor, we divide 531 first, to find that it is 59 after it is divided by 9. Seeing as how the quotient obtained when dividing the other 3 numbers by 9 will obviously be smaller than 59, 59 must therefore be the largest prime factor.[/hide]

bob123 has got it!
[hide]Don't start with the 15-digit number.  Start with the six 3-digit factors, since they are [b]much[/b] easier to deal with.  Then notice the sum of the digits (1+3+5) is 9.  By the divisibility rule for 9, each of the six numbers is divisible by 9.  So the factors are:

\begin{eqnarray}
Z&=&622092199590225\\
&=&135\times 153\times 315\times 351\times 513\times 531 \\
&=&9\times\frac{135}9\times 9\times\frac{153}9\times 9\times\frac{315}9\times 9\times\frac{351}9\times 9\times\frac{513}9\times 9\times\frac{531}9
\end{eqnarray}

Remember we are just looking for the largest prime factor.  As bob123 suggests, let's start with the largest number in equation $(3)$.

$\frac{531}9=59$.  But $59$ is prime.  There is no need to perform the remaining divisions, since none of the other quotients can be larger than $59$.[/hide]

Wow! That seems so easy now I've seen the answer!

Great solutions. Wish I'd thought of them


please let me know how to find the length of the adhesive tape  roll
 :D
i want to know  how many meter of tape winded on the spool (paper core)

vinaya,
You need to tell me and other people more information for me to solve for the length (I think that radius of the spool, radius to outer tape, and thickness of tape should be enough).

But I can give you a formula for it:

Let r be the radius of the spool, R the radius to the outer tape, and T the thickness of the tape.
If you think about it, the tape and the spool are in the shape of a cylinder. The total volume of the spool and the tape is (pi)R^2*h, where h is the height of the spool (This will cancel out later).

The volume of the spool is (pi)r^2*h. Therefore the volume of the tape is h*pi(R^2-r^2). In order to find the cross-sectional area of the tape, we must divide by the height, h to get:
A=pi(R^2-r^2)
Divide that by the thickness of the tape to get the approximate length of the tape:
L=pi(R^2-r^2)/T

Somebody should check my work though, since I probably overlooked something.

The formula looks right to me..
but maybe you should specify r as the OUTER radius of the spool.


$ \overline{WA}$ divides rectangle $ WXYZ$ into two parts whose areas are in the ratio of 7 : 1. What is the ratio $ \frac{XA}{AY}$? Express your answer as a common fraction.

[asy]currentpen = fontsize(10pt);
pair W = (0,3), X = (8,3), Y = (8,0), Z = (0,0), A = (8,3/4*3);
draw(W--X--Y--Z--cycle); draw(W--A); label("$W$",W,NW); label("$X$",X,NE); label("$Y$",Y,SE); label("$Z$",Z,SW); label("$A$",A,E);[/asy]

$ \frac{XA}{AY}\equal{}\frac{[AXW]}{[AYZW]}$ Therefore $ \frac{XA}{XY}\equal{}\frac{1}{1\plus{}7}\equal{}\boxed{\frac{1}{8}}$

FTW says the answer is actually 1/3, although I also recieved an answer of 1/8.

[geogebra]90caee22d578df1b15bbabc6b838d5c676827f9b[/geogebra]

The ratio of 7 to 1 indicates that seven triangles congruent to AXW can fit inside WXYZ. This creates four "rows" (see diagram) in WXYZ. Therefore the ratio of the segments XA and AY are 1 to 3.

lol i just took a leap of faith and used my fingers to kinda estimate and i got it correct :)


[quote="kimmystar94"]Hi, I'm pretty new to Trig, and I am currently VERY stuck.
Could anyone help me into showing me how you figure out the sign of the sines and cosines and what the quadrants help to do?

Thank you!

(and I'm sorry if this is in the wrong board or if this is too general of a topic)[/quote]

I _think_ such a topic would fit better into high school basics.

For your question, just remember the definition; all its properties derive from that, so that approach can never be wrong. ;)

[quote="kimmystar94"]Hi, I'm pretty new to Trig, and I am currently VERY stuck.
Could anyone help me into showing me how you figure out the sign of the sines and cosines and what the quadrants help to do?

Thank you!

(and I'm sorry if this is in the wrong board or if this is too general of a topic)[/quote]
Remember a few rules of thumb for the sign:
$ \sin \minus{}\theta\equal{}\minus{}\sin \theta$
$ \cos \minus{}\theta\equal{}\cos \theta$
$ \sin \pi \minus{} \theta\equal{} \sin \theta$
$ \cos \pi \minus{} \theta \equal{} \minus{} \cos \theta$

okay..
i think that MIGHT have helped..
i'm not sure..
now back to my trig homework and finding out stuff...

any more suggestions?

The quadrants tell you whether an angle is positive or negative, you need to work out what the angle would be in the 1st quadrant, then put the sign in front

could you expand more of that?
i think that's what i'm having trouble with...
because some of the identity stuff we're doing has plus-minus signs and
i don't know which one to choose when...

remeber "All Students Take Calc"  to remeber which value is positive in each quadrint, I all, II Sin, III tan, IV cos.

I think this is what you are looking for, hopes it helps

CAST might be shorter  :P

Explanation: Starts from the fourth qudrant, going up counterclockwisely. [b]C[/b]osine [b]A[/b]ll [b]S[/b]ine [b]T[/b]angent.

THAT's what i was looking for!!!

THANKS SOOO MUCH!!!!
now i can do my homework without worries...
i've heard all students take calc, but i didn't know what significance it had..

Thanks!!!!


Number the cells in a $2n \times 2n$ grid from $1$ to $4n^2$, starting at the top left and moving left to right along each row, then continuing at the left of the next row down and so on. Colour half the cells black and half white, so that just half the cells in each row and half the cells in each column are black.

Let $a_{ij}$ be the entry in the $i$th row and $j$th column. Prove that for any coloring, $1\leq i,j\leq 2n$, $a_{ij}$ and $a_{ji}$ have the same color.  Note that this nearly immediately gives a proof of the original problem.

What is original problem?
Also your statement is incorrect:
bbww
wwbb
bwbw
wbwb

Am I missing something?

No, you're not (as usual) missing anything.  Thanks, I should have said give a counterexample if such a counterexample does exist.  I tried to prove the problem this way initially; clearly, it was a wrong conjecture.

In any event, there is a simple solution, and the wording of the problem can be found here:

http://www.kalva.demon.co.uk/putnam/putn01.html (it's B1)

I asked you what was original question.

[i]Show that however this is done the sum of the black squares equals the sum of the white squares.[/i]

[quote="Myth"]Actually, the problem is super obvious![/quote]

Out of curiosity, what was your solution?

Just write $a_{ij}=2n(i-1)+j$ and consider separately matrices with $a'_{ij}=2n(i-1)$ and $a''_{ij}=j$.

A little observation from Kedlaya's site, which should be fun to prove:

[quote]Namely, if $ A$ and $ B$ are $ \{ 0, 1 \}$ matrices with the same row and column sums, then there is a sequence of operations on $ 2 \times 2$ matrices of the form \[ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \] or vice versa, which transforms $ A$ into $ B$.[/quote]

(I didn't find anything via "search".)


LaTeX
algorithm
greatest common divisor
number theory
Euclidean algorithm
relatively prime
superior algebra

Sorry for bothering you all with such a mundane question, but I'm confused with one crucial part of a common proof of Wilson's Theorem. The proof relies on the fact that only 1 and p-1 are multiplicative inverses of themselves, while 2 to p-2 will be able to pair off such that ab=1 (mod p), where a and b are two different elements of Zp. Why is it that 2 to p-2 can pair off such that, when multiplied, they will all cancel out to 1 mod p? Apparently it comes from the fact that it is Zp is a group, and by definition of a group, each element of Zp has a multiplicative inverse. But WHY is this true? Thanks in advance for the help. If I'm unclear, I'll gladly clarify for you guys; just ask.

and of course, not that this needs to be stated, but Wilson's Theorem states that (p-1)! = -1 (mod p), where p is a prime number. On a sidenote, I'd like to use LaTeX, but I haven't yet taken the time out to learn how to use LaTeX (I have it...just haven't learned how to use it yet; I don't really have the time to learn it unfortunately).

I mean, I can see how for primes up to 13 that indeed 2 to p-2 pair off such that the remainder of the products is 1, and then you are left with (p-1)! = -1 (mod p), but again, I don't understand why Zp (sorry for no LaTeX again, but this is the set of integers modulo p) is a group (specifically, why every element of Zp has a multiplicative inverse). Any help would be much appreciated!

It's the Euclidean algorithm: for any $ m$ relatively prime to $ p$, we have by definition that $ (m, p) \equal{} 1$ and so by the Euclidean algorithm there exist integers $ a, b$ such that $ am \plus{} bp \equal{} 1$.  Taking this expression modulo $ p$ gives $ am \equiv 1$ and so (the equivalence class of) $ a$ is the inverse of (the equivalence class of) $ m$.

Hey JBL,

Thanks a lot, I really appreciate it. I just looked up the proof of the Euclidean algorithm, which of course leads to the extension of the Euclidean algorithm (which, like you said, implies that there exist integers a, b such that ax + by = gcf (x, y), in this case ax + bp = gcf (x, p) = 1). Makes perfect sense now, thanks again!


calculus
integration
function
trigonometry
calculus computations

Integrate $ 4 \sin^2 x$ with respect to $ x$.

[quote="Patterns_34"][hide="Huge Hint"]$ \cos 2x \equal{}$[hide]$ \cos^2 x \minus{} \sin^2x \equal{} 1 \minus{} 2\sin^2 x$[/hide][/hide][/quote]

[hide="solution"] So I need to integrate $ 2 \minus{} 2 \cos (2x)$ with respect to $ x$: $ \int (2 \minus{} 2 \cos (2x)) dx \equal{} 2x \minus{} \sin 2x \plus{} c$ for an arbitrary constant $ c$. [/hide]


Let f in AC[0,1] monotonic,Prove that if m(E)=0 then m(f(E))=0

More generally I think, http://www.mathlinks.ro/Forum/viewtopic.php?t=58742


I was thinking of going to ARML this year, and I had some questions.
What kind of math do you do on practices/real competition? How demanding arethe practices? Is it hard to get in?
If you attended ARML before, please talk about your experience, share you stats etc.

Thanks! :)

Regarding what kind of math in ARML, Silas has explained well in the following post.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190944[/url]

As far as I know, everyone can get in. We will be given ARML problems from the previous years during the practices. The results of the individual rounds will be recorded. After several practice sessions, the participants will be grouped into teams (each with 15 people, e.g. SFBA A team, SFBA B team, SFBA C team, etc.), then focus on team rounds and power rounds.

I have attended the last year's SFBA ARML teams for the first time. It's fun, largely because:
 - you have opportunity to meet many students with similar interest (i.e. math) in mind;
 - and there are other fun things such as rubik's cube, chess, puzzles, card games, frisbee, etc.
 - you will get an SFBA T-shirt, designed by a team member (that could be you).
 - you will get souvenirs (e.g. ARML mug, ARML tag, ARML certificate),
 - and if you are lucky, you may also get medals, books and even cash prices.
 - you will get college dorm experience (not as comfortable as a hotel  :D )
 - and more ... (please add)

I think I will also participate this year. I definitely hope SFBA teams can do much better this year!

I see.  Well unfortunately I've got standardized testing AND finals the week after the contest (it's odd, there's usually a two week gap)

So if I do apply it'll be a bit of a stretch.

Huh? Finals?
Finals week is the most awesome week the entire semester...

I've had finals the week after ARML for at least four of my six years.  Never stopped me.

Neither did a speculative date (cough cough).


Sandy plans to paint one wall in her bedroom. The wall is 9 feet high and 12 feet long. There is a 2-foot by 4-foot area on that wall that she will not have to paint due to the window. How many square feet will she need to paint?

The wall, including the window, is 9*12=108 square feet, but the window is 8 square feet, so she will have to paint $ 108\minus{}8\equal{}\boxed{100}$ square feet.


[hide="Hint"]How many weeks are there in a year?[/hide]

[quote="hyderman"]hello

i need help understanding this question ... please help

How many people are needed to guarantee that at least two were born on the same day of the week and in the same month (perhaps in different years)?[/quote]

[hide]the worst case is when every day of the week is taken in every month so that is 12*7=84

then add one to get 84[/hide]

[quote="hyderman"]hello

i need help understanding this question ... please help

How many people are needed to guarantee that at least two were born on the same day of the week and in the same month (perhaps in different years)?[/quote]
[hide]
there are 12 months in a year. there are seven days in a week
so there are 84 combinations in all
with 85 people, you can garentee that there will be at least one repeat.
[/hide]


Prove that :

$ \lfloor\sqrt{n\plus{}5}\plus{}\sqrt{n\plus{}6}\rfloor\equal{}\lfloor\sqrt{4n\plus{}22}\rfloor$ $ \ ,\ (\forall)n\in \mathbb{N}\cup \{\minus{}5,\minus{}4,\minus{}3,\minus{}2,\minus{}1\}$

[quote="alex2008"]Prove that :

$ \lfloor\sqrt {n \plus{} 5} \plus{} \sqrt {n \plus{} 6}\rfloor \equal{} \lfloor\sqrt {4n \plus{} 22}\rfloor$ $ \ ,\ (\forall)n\in \mathbb{N}\cup \{ \minus{} 5, \minus{} 4, \minus{} 3, \minus{} 2, \minus{} 1\}$[/quote]

$ \sqrt {n\plus{}5} \plus{}\sqrt {n\plus{}6} \le \sqrt{2[(n\plus{}5)\plus{}(n\plus{}6)]} \Rightarrow \sqrt{n\plus{}5}\plus{}\sqrt{n\plus{}6} \le \sqrt{4n\plus{}22}$
$ \sqrt {n\plus{}5} \plus{}\sqrt{n\plus{}6} \ge \sqrt{4n\plus{}21} \Leftrightarrow \sqrt{(n\plus{}5)(n\plus{}6)} \ge n\plus{}5$ after squaring both sides, and which is obviously true.

So $ \sqrt{4n\plus{}22} \ge \sqrt {n\plus{}5} \plus{}\sqrt {n\plus{}6} \ge \sqrt{4n\plus{}21}$. Also mod 4 we know $ 4n\plus{}22$ cannot be a square , done.


let $R$ and $r$ denote radius of circumcircle and incircle of triangle $ABC$, and $s$ be the semiperimeter of
triangle $ABC$. show that :
$s\leq 2R+\left(3\sqrt{3}-4\right)r$.

Sorry to interfere in this unknown field for me, but I wanted to ask Darij whether he knows where I posted a nice proof for this problem.

Finally I have located it: http://www.mathlinks.ro/Forum/viewtopic.php?t=28182 .

  darij


LaTeX
superior algebra
superior algebra theorems

How will you draw the lattice diagram of $Z_{2}$ X $Z_{4}$, under addition modulo 2 and addition modulo 4 respectively? Which group is cyclic?

I have tried drawing it but i don't know if my answer is correct.

Thanks.

By the way how can one draw a lattice in $LaTeX$?


geometry
3D geometry
probability
number theory
prime numbers

[quote="sharkman"]A fair number cube is labeled 1 to 6.  If the cube
is tossed 300 times, about how often would a prime number be expected
to show up on the top face?[/quote]

[hide="hmm..."]

the primes are 2, 3, 5, so one half of the cube is prime....therefore approximately 150 times

[/hide]

[quote="sharkman"]A fair number cube is labeled 1 to 6.  If the cube
is tossed 300 times, about how often would a prime number be expected
to show up on the top face?[/quote]
[hide]There are only 3 prime numbers 1-6. So its a 1/2 chance of a prime number coming up. Since its tossed 300 times it is $300\cdot\frac{1}{2}=150$.[/hide]

[hide]6 faces on a cube rite

3 of the numbers r prime, out of 6 faces

thats $\frac{3}{6}= \frac{1}{2}$

 $\frac{1}{2}\cdot 300 = 150$
[/hide]

[hide]Since half of the numbers on a die are prime, the likely hood of one being rolled is 1/2. So .5 of 300 is 150.[/hide]

[hide="Solution"]
 Probability that the number is prime is $\frac{1}{2}$. Therefore,
$E(\text{The Number is Prime})= 150$[/hide]
    Bharath.

[quote="nicholaskramer"]Just incase noone else explained well...

Probability for equally likely events is (ways to win or succeed)/(total ways). there are 3 prime numbers you can roll (2 3 and 5) and 6 total ways so the probability of rolling a prime is 3/6 or 1/2. this means for every 2 dice rolls you are expected to roll one prime number. 1/2 OF 300 = 150 so 150 expected prime rolls.[/quote]

...please do not revive old threads...

[quote="nicholaskramer"]Sorry!! I didn't notice the date on this thread. Ill make sure to pay attention to that from now on =)[/quote]

 :D  :lol: it happens

I would not consider 6 days to be too long (as a moderator); nicholaskramer was fine in his post.

It only really starts to matter after a week.  Classroom math isn't a very busy forum, so posts a week old can still be on the first page.


Prove that the polynomial $ P(x) \equal{} x^{999} \plus{} x^{888} \plus{} ... \plus{} x^{111} \plus{} 1$ is divisible by $ Q(x) \equal{} x^9 \plus{} x^8 \plus{} ... \plus{} x \plus{} 1$.

Let $ A \equal{} \sum_{m \equal{} 0}^9 x^m$.
So, $ \sum_{n \equal{} 0}^9 x^{111n} \equal{} \sum_{i \equal{} 1}^9 \left( \sum_{j \equal{} 1}^{11i} \left( x^{10j \minus{} 9 \plus{} i}A \minus{} x^{10j \minus{} 10 \plus{} i}A \right) \right) \plus{} A$.
Thus $ \sum_{m \equal{} 0}^9 x^m \left|\sum_{n \equal{} 0}^9 x^{111n} \right.$.

Interesting...

There goes a solution without crazy sums:

Treating $ Q(x)$ as a geometric progression, we get $ Q(x) \equal{} \frac {x^{10} \minus{} 1}{x \minus{} 1} \ , x \neq 1$

To find the roots of $ Q(x)$, we do $ x^{10} \minus{} 1 \equal{} 0$ while respecting the existence condition.

$ x^{10} \equal{} 1 \Rightarrow x \equal{} cis \left( \frac {2n\pi}{10} \right) \ , n \equal{} 1,2,...,9$
In a similar way, we find the roots of $ P(x)$: $ x \equal{} cis \left( \frac {2k\pi}{1110} \right), k \equal{} 1,2,...,1110$ excluding $ x \equal{} cis \left( \frac {2k\pi}{111} \right) , k \equal{} 1,2,...,111$ (the multiples of ten).

It's easy to see that, when $ k$ is a multiple of 111, $ x$ is a root of $ P(x)$ and $ Q(x)$. Thus, all roots of $ Q(x)$ are also roots of $ P(x)$ and the divisibility is proved.


algebra
polynomial
geometry
3D geometry
superior algebra
superior algebra unsolved

find minimal polynomial the element w+$ %Error. "display" is a bad command.
\sqrt[3]{2}$ on $ %Error. "display" is a bad command.
\mathbf{Q}$.we have w= $ %Error. "display" is a bad command.
1/2 + \surd 3  /2      i$.we know $ %Error. "display" is a bad command.
\ w^3 =1$.

Set $ x\equal{}w\plus{}\sqrt[3] 2$.
Then $ x\minus{}w \equal{} \sqrt[3] 2$, so we cube to get $ (x\minus{}w)^3 \equal{} 2$. Now try to get rid of $ w$, too.


integration
calculus
probability
LaTeX
function
derivative
special factorizations

If $ X \sim N(0,1)$, find the pdf of $ Y = |X|$.

So $ F_{Y}(y) = P(Y \leq y) = P(|X| \leq y) = P(X \leq |y|)$

So ${ F_{Y}(y) = \int_{0}^{|y|} \frac {1}{\sqrt {2 \pi}}e^{ - u^{2}/2}} \ du$

Then I would differentiate the result. Now I am thinking that this will involve the $ \text{erf}$ function. How would I integrate that?

Would the cdf of $ Y$ be: $ \frac {1}{2} \left(1 + \text{erf} \frac {u}{\sqrt {2}} \right)_{0}^{|y|} = \frac {1}{2} \left(1 + \text{erf} \frac {|y|}{\sqrt {2}} \right) - \frac {1}{2}$?

Then differentiate this to get the pdf?

No, you don't need to evaluate the integral to differentiate it (FTC!). Also, the limits are wrong.

... if $ y$ is positive. If $ y$ is negative, there's nothing there.

The pdf is just the right half of the standard normal distribution:

\[ f\left(y\right) \equal{} \begin{cases} 0 & y < 0 \\ \frac{1}{\sqrt{2\pi}}e^{\minus{}\frac{1}{2}y^2} & y\geq 0 \end{cases} \]

[quote="Kent Merryfield"]You have to multiply that density by $ 2.$ As you've written it, that's not a probability density because its integral is not $ 1.$[/quote] :oops_sign:

[quote="Dr. No"]The pdf is just the right half of the standard normal distribution:
\[ f\left(y\right) = \begin{cases} 0 & y < 0 \\
\frac {1}{\sqrt {2\pi}}e^{ - \frac {1}{2}y^2} & y\geq 0 \end{cases}
\]
[/quote]

as a $ \text{\LaTeX}$ aside, you don't need to do the \left( \right) thing for all parentheses; $ f(y)$ 'looks' better than $ f\left(y\right)$ :)

Just so the answer is posted somewhere in this thread: the density is
\[ f(y)\equal{}\begin{cases}0&y<0\\ \sqrt{\frac{2}{\pi}}\,e^{\minus{}\frac12y^2}&y>0\end{cases}\]
If you ask me what $ f(0)$ is, I'll say that I don't know and I don't care - it's a matter of no importance.

The next exercise would be to compute the mean and variance of this distribution - or perhaps go beyond that to compute the moment generating function (or characteristic function).

In light of Professor Merryfield's correction to the pdf, the moment generating function is:

\[ M(t) \equal{}\int_{\minus{}\infty}^{t} e^{y}f(y)\, dy \]

\[ M(t) \equal{}\int_{0}^{t} \sqrt{\frac{2}{\pi}}e^{y\minus{}\frac{y^2}{2}}\, dy \equal{} \sqrt{e}\left[Erf\left(\frac{\minus{}1 \plus{} t}{\sqrt{2}}\right) \minus{}Erf\left(\minus{}\frac{1}{\sqrt{2}}\right)\right] \]

That's not what I recognize as the moment generating function. That function, as I understand it, is
\[ M(t)\equal{}E(e^{tY})\equal{}\int_{\minus{}\infty}^{\infty}e^{ty}f(y)\,dy\equal{}
\sqrt{\frac{2}{\pi}}\int_0^{\infty}e^{ty\minus{}\frac12y^2}\,dy\]
After completing the square, we will wind up with a representation using Erf.

Perhaps more useful is the moment, directly. We can compute that

$ E(Y^k)\equal{}\frac{2^{k/2}}{\sqrt{\pi}}\,\Gamma\left(\frac{k\plus{}1}2\right).$

For even $ k$, this is exactly the same as the corresponding moment for the normal distribution; but for odd $ k,$ they differ. (The odd degree moments of the normal distribution are all zero.)

In particular, $ E(Y)\equal{}\sqrt{\frac{2}{\pi}},\ E(Y^2)\equal{}1,$ and $ \text{Var}(Y)\equal{}\frac{\pi\minus{}2}{\pi}.$

I was going to ask about that, but hadn't figured out what to ask yet. I knew something wasn't quite right about it. The integral is correct in a mechanical sense as the anti-derivative of the expression I gave actually is: $ \sqrt{e}Erf\left(\frac{\minus{}1 \plus{} y}{\sqrt{2}}\right)$, but when I differentiated it, twice in order to compute the variance, I obtained a negative value. In my own defense, when I screw-up like this it isn't because I'm being careless necessarily -- in my current circumstances I have almost no access to my library, my texts are in storage, and I'm writing most of my stuff from memory. It is kind of difficult to remember things that one hasn't used in more than a decade.

After much more plug - and - grind, using Professor Merryfield's helpful correction,
\[ M(t) \equal{} \int_{ \minus{} \infty}^{\infty} e^{ty}f(y)\,dy \equal{} \int_{0}^{\infty}\sqrt{\frac{2}{\pi}} e^{ty \minus{} \frac {y^2}{2}}\,dy
\]

\[ M(t) \equal{} e^{\frac {t^2}{2}}\left(1 \plus{} erf\left(\frac {t}{\sqrt {2}}\right)\right)
\]

\[ M'(t) \equal{} t e^{\frac {t^2}{2}}\left(1 \plus{} erf\left(\frac {t}{\sqrt {2}}\right)\right) \plus{} \sqrt {\frac {2}{\pi}}\implies E\left[Y\right] \equal{} \sqrt {\frac {2}{\pi}}
\]

\[ M^{\prime\prime}(t) \equal{} e^{\frac {t^2}{2}}\left(1 \plus{} t^2\right)\left(1 \plus{} erf\left(\frac {t}{\sqrt {2}}\right)\right) \plus{} \sqrt {\frac {2}{\pi}}t\implies E\left[Y^2\right] \equal{} 1
\]

\[ Var(Y) \equal{} 1 \minus{} \frac {2}{\pi} \equal{} \frac {\pi \minus{} 2}{\pi}
\]


logarithms
analytic geometry
graphing lines
slope
function
inequalities

Hello

Let $ d >c >b  >a$. Prove that $ a^b b^c c^d d^a \ge b^a c^b d^c a^d$

[hide="Solution"]
It is equivalent to
$ (\frac {c}{a})^{d \minus{} b} \ge (\frac {d}{b})^{c \minus{} a}$
Taking the $ \ln$ of both sides,
$ \frac {\ln c \minus{} \ln a}{c \minus{} a} \ge \frac {\ln d \minus{} \ln b}{d \minus{} b}$
But note that these are the slopes of the lines on $ f(x) \equal{} ln x$ between $ (a, \ln a)$ and $ (c, \ln c)$ and the line between $ (b, \ln b)$ and $ (d, \ln d)$. As $ f(x) \equal{} \ln x$ is a concave function and $ a < b < c < d$, this inequality must be true.[/hide]


continued fraction
number theory unsolved
number theory

I know of some methods of finding the continued fraction representation of certain numbers $ x\in\mathbb{R}$.  However, if we are only given a continued fraction, is there a general method for finding what it converges to?

You can calculate the rational approximations. This means only considering the first $ n$ entries.
More exactness is in general not possible, but you may be able to express it as some more common number sometimes.


[quote="Phelpedo"]Show that among any $n+2$ integers, either there are two whose difference is a multiply of $2n$ or there are two whose sum is divisible by $2n$.[/quote]

[hide]Well if any two of them have the same modulo class in 2n, then automatically the difference is a multiple of 2n, so assume they are all different.

So this means that the classes possible are {0,1,2,...,2n-1}

Notice that the sets that compose for sum of modulo 2n are

{1,2n-1}, {2, 2n-2}, {3,2n-3}...{n-1,n+1}, {0}, {n}.

And differences it would be impossible to get 0(mod 2n) since they are all different mods so no difference of mods can be 0.

So consider only sums.

By pigeonhole principle, there are n+1 holes and n+2 pigeons so two pigeons have to go into one 1, hence the set will have two elements from the same group and hence proof is complete.[/hide]


How did everyone do on the USAMTS overall?

Score Prize

96-100  "Gold Winner"
76-95    "Silver Winner"
60-75    "Bronze Winner"
40-59    "Honorable Mention"

What are the prizes?

I'm short of >10 points because the graders somehow skipped some of my answers in round 3+4 (#2 in both!) Does anyone have similar experiences?

[quote="zabelman"]100 :D[/quote]

Congratulations. What tips do you have for a first-time participant in USAMTS starting out next year?

(Anyone else is welcome to answer too; but I have to ask the guy that aced all four tests. I'm way beyond my coaching ability when the guys on my math team start doing essay-proof-based tests.)

[quote="fairyofwind"]100 here too :)[/quote]

Congratulations to you too, and the same question  :D  What would you recommend to first-time USAMTS participants with an inadequate coach (that is, my math team here)?

97.  Darn the triangle counting problem on round 2.

[quote="tokenadult"]What would you recommend to first-time USAMTS participants with an inadequate coach (that is, my math team here)?[/quote]

My math team also has an inadequate coach. It doesn't seem to matter. The kids I had who were interested in USAMTS just gave it a whirl on their own and did fine. The program is generous with partial credit, and they sort of learned as they went along. AOPS vol 2 is helpful to ease students into proofs.

whoa, they sent out results?  when did u guys get them?

And did the prizes (books/certificates) come with your results?

[quote]What would you recommend to first-time USAMTS participants with an inadequate coach (that is, my math team here)?[/quote]

I'll tell you how I got into writing USAMTS proofs in 7/8th grade. I first went after the problems that asked "find the answer" rather than "prove this result," because I treated these just like any other question and just found the answer. For the solution I (tried to) type up, I just explained my thought process as I worked my way to the answer. Of course, I got counted off because there were many mental "leaps" that needed to be more rigorously explained, but thats just how the mind works... 3 steps ahead of logic. Anyway, after doing these for a while, I got used to the whole concept of using stuff to prove other stuff (instead of saying something is true because it looks like it is), and once that concept kicks in you're golden! You just have to master the art of slowing your mind down enough to where you can actually put step-by-step explanation into words rather than leaps of intuition and hope.

If you're new to proofs, I would recomment this approach. If you understand what is necessary for a legit proof, then all I can recommend is DO MORE PROBLEMS! CONSTANTLY! Oh, and actually writing a proof to a problem so that someone else can comment on it is also a very good idea. (But I wouldn't do this with official USAMTS problems!)

I got my letter today, but unfortunately the awards haven't come yet.

Zabelman has good suggestions for getting started on proofs. Another suggestion I got from my neighbor, who was on a previous IMO team, is to just read and try to follow someone else's proof, then put it away for a couple of weeks. Then come back to the problem and try to write your own version of the proof.

That only works if you can do it. I would never be able to write my own solution to a problem once I have seen another once since I'll just remember it. But if you can, I imagine it would be a good technique.

[quote="ComplexZeta"] I would never be able to write my own solution to a problem once I have seen another once since I'll just remember it. [/quote]

I think the idea was that you would try to remember and recreate the same solution you had read, but after enough time had passed for you to forget the details. Remember, this was a suggestion for people with no proof-writing experience who do not know the thinking process yet to come up with their own solutions.

yup, got letter in mail today...total: 23, 25, 24, 14 = 86, Silver.

I will definitely participate in USAMTS next year. It seemed to be easy for me because there's no exam pressure and time pressure, which are my major weaknesses.

25+25+25+20 = 95= silver :(

I was too lazy to even attempt one of the problems on the last round because I was under the belief that 91-100 was gold. So I had a perfect on everything I bothered to do...

Got my R4 scores in the mail today.  Total: 24 + 23 + 24 + 25 = 96 =  Gold :)

That was a really close one.


what exactly does this symbol mean:   <<, less than less than?

Hm.. I don't know of any standard meaning in mathematics (though there is one in computing :)).. have you got an example of it being used in context?

for any fixed real number c > 0, the inequality d(n) << (log(n))^c as  n approaches infinity does not hold...sorry i cant figure out how to use latex in the posts...

I don't know the mathematical meaning,
but I think << means "less than ([u]by A LOT[/u])" in science.

are you sure they didn't make a typo? :?

My physics teachers also use that sign to express [b]less than [i]by a Lot[/i][/b]

Yeah, in my physics book it has << and says much less than.  It's probably just used mostly in science.  Mathematicians usually only care about the absolute relation, i.e. greater than, less than or equal but in physics it might be important that something is much greater or less than something else.  Then you could use an approximation of a formula without worrying about huge error.

When I see the "<<" operator, I think of the binary-shift...


AMC
AIME
USA(J)MO
USAMO
function
AIME I
geometric sequence

Ok, so I'm a sophomore who barely qualified for the AIME. And since I'm going to take the AIME, I definitely do not want a 0! (however, my math skills are not so great because I can only answer 15/25 of AMC questions on average without the stress of time limits).

So... I took a brief glimpse through some past AIME exams and it appears that with the 3 hour time limit, I could perhaps use pure brute force on a couple of the problems. For those who've taken the AIME - have any of you used brute force on a couple just to avoid getting a zero?

Yes.

Last year I knew I had no chance of USAMO so I didn't really try as much as I should have.  I didn't want a zero though.  I should have gotten a 4 or something (stupid mistakes) but it ends up that I got a 1.  The one that I got, I brute forced.

[quote="Simfish"]Ok, so I'm a sophomore who barely qualified for the AIME. And since I'm going to take the AIME, I definitely do not want a 0! (however, my math skills are not so great because I can only answer 15/25 of AMC questions on average without the stress of time limits).[/quote]
Sounds like me...
I used brute force on one question on the AMC 12B, actually, and got it right.
For the AIME, where you have a 3-hour luxury of time, brute force should prevent a 0.

Same situation here.

Brute force is plug n' chug. And of course many other ways are considered brute force too.

I'm going to do brute force if I don't have time to do past AIME tests. Gah.

DO NOT attempt to brute force a problem - even if you have a remote chance of getting it otherwise, you probably have a better chance that way. Last year, I saw a problem I brute forced - except I got it wrong (a counting problem, off by 10.) So instead of a 7 I got a 6. If I spent half the time I used for calculations on actually attempting the problem, I might have gotten it right,

Yes. Last year, as an eighth-grader, I brute-forced every problem I could (except for one), hoping to get a problem or two.

I got 7. So, yeah, brute-forcing (or hybrid brute force and elegance) is VERY valuable on the AIME.

tetrahedr0n- there is a difference between brute forcing when you are trying to qualify for USAMO and when you are just trying to prevent a 0.  When there is a possibility that you wont get any, I say do whatever it takes, within the rules, to rack up a point or two.

I know that I'm capable of solving the first few without brute force, but I just get carried away so easily...

I won't be surprised if I brute force all of them.

But it does work... however, depends on what you mean by brute force.

There's a difference between trying a few possibilities and tackling 100 possibilities...

[quote="1234567890"][quote="filletwho"]If you're really desperate, just guess something that has to do with the year 2005. I find that the randint() function on your calculater works well too.[/quote]

No calculators, am I correct?[/quote]

Sorry, I was talking about competitions in general. In fact I was thinking about my friend who isn't "math-oriented". On one of the middle school mml's, he made a dice out of paper for guessing. I think he got like a 30. So that's pretty good considering he guessed on each one.

Brute-force will usually get you the wrong answer anyway so brute-forcing is probably not a good idea for AIME

I remember when I first started practicing the AIME, I couldn't get any. There was one problem which I made a list. It took four pages of all the permutations. I didn't miss any permutations because I spent my three hours organizing the list. It was fun.

I kind of brute forced a combinatorics problem on one of the mock AIMEs. Off by 10 :blush:

for counting problems, brute force is NOT a good idea.  other than that, some of the easier problems are generally brute force-able.

brute forcing counting problems usually just means a lot more casework. it's more tedious, but it can either be more or less dangerous than a simpler solution

Brute force is usually not effective in permutations.......unless you can estimate that there are only a few possibilities. Otherwise, try to focus on get the answer without listing them! good luck in AIME, the first few questions are straight-forward if you had a good score in AMC 12.

I used brute force on every counting problem but one last year. I got an 8, but I wouldn't recommend using brute force on every problem. The official exam was the first time I sat down and took the 3 hour test, so I guess diligence and beginner's luck helped.

if u want to do all 15 problem ,then brute force isn't a great idea,but if u just want to get something like 6-10,which leave u a little extra time,then just go with whatever way u find,as long as it can lead u to the answer

i dont think plug and chug brute forcing is good for AIME unless you have no choice. Al though one contest that i know brute forcing is good is Mu Alpha Theta :D  i have personlaly brute forced a lot in that contest.

[quote="frankthetank"]On the 2003 AIME, I brute forced the problem about triangles and cubes.  I wrote down every freaking triangle.  I got it right, but I'm really ashamed of doing it because 2003 was an easy year and that was an easy problem.[/quote]

Don't feel bad. I did the same thing. And I also tried to list out all the binary number from 1-2003 on the binary problem. I gave up after 350 or so.

actully when ur math skills are limited and time is not a problem go ahead and brutal force one out...but be very careful and neat so u don't make a stoopid mistake which waste time and points...

i brutal through a test and got a 8 before...just make sure u can't do any of the others before u start or u might be wasting valueable time..

If you have taken practice tests and only attempted two or three, and you believe you have no hope on the other problems, I would brute force and try to find a nice solution to back it up.

[quote="Zratonx"]actully when ur math skills are limited and time is not a problem go ahead and brutal force one out...but be very careful and neat so u don't make a stoopid mistake which waste time and points...
[/quote]

That's almost impossible, I almost never brute-force, I often come up with a good solution and I still make stupid mistakes. So brute-forcing => you'll get it wrong almost for sure, since brute-force often require a ton of calculation(although i say this probably because if I find a good solution for a problem but like added wrong, and someone who brute-forced the problem got it right, I'd feel bad :lol: )

Maybe the only type of problem that can be brute-forced are something like this:

9. A sequence of positive integers with a_1 = 1 and a_9 + a_10 = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all n >= 1, the terms a_(2n-1), a_(2n), a_(2n+1) are in geometric progression, and the terms a_(2n), a_(2n+1), and a_(2n+2) are in arithmetic progression. Let a_n be the greatest term in this sequence that is less than 1000. Find n + a_n.

So you can just go listing numbers and find n( I didn't do this, just listing a few terms out would give you the pattern, but if you don't see it, for this one you can actually go list numbers).

That sounds like one heck of a problem, how would you solve that, brute force OR nice?

Write a list of all of the possible ways until you get to n_9 + n_10 = 646 :) .


I wasn't sure where to ask this, but I didn't want to revive the old thread...

Let $ a, b, c$ be real numbers. Suppose that for all real numbers $ x$ such that $ |x|$ $ \le$ 1, we
have $ |ax^2 \plus{} bx \plus{} c|$ $ \le$ $ 100$. Determine the maximum possible value of $ |a| \plus{} |b| \plus{} |c|$.

Anyways, I understand the solution, but I can't find the mistake in my reasoning so if you could let me know where I'm messing up that would be great!

I let $ x \equal{} \minus{}1$ $ \Rightarrow$ $ |a \minus{} b \plus{} c|$ $ \le 100$

Then why can't you let $ a$ equal any number, and then cancel it out with the same negative number $ b$?

For example why can't you use $ a \equal{} 10^{321}$ and b$ \equal{} \minus{}10^{321}$?

Having the inequality hold for $ x \equal{} \minus{} 1$ says nothing about whether it holds for [b]all[/b] $ |x| \le 1$.  If $ a$ and $ b$ are large and positive then the inequality won't hold at $ x \equal{} 1$.

Are you sure you understand the question?

I see now :)   As you suggested I didn't really understand the question - thanks for the help.


function
limit
number theory proposed
number theory

Prove that F(F(n))/(sqrt5) is asymptotic to F(F(n-1))F(F(n-2)) where F(n) is the nth fibonacci number

Let $\phi=\frac{1+\sqrt{5}}{2}$, then
$F(n)=\left[\frac{\phi^{n}}{\sqrt{5}}\right]$ ,[x] is nint function (well-known)

so,
$\lim_{n\rightarrow \infty}\frac{\frac{F(F(n))}{\sqrt{5}}}{F(F(n-1))F(F(n-2))}=\lim_{n\rightarrow \infty}\frac{\phi^{\frac{\phi^{n}}{\sqrt{5}}}}{\phi^{\frac{\phi^{n-1}}{\sqrt{5}}}\phi^{\frac{\phi^{n-2}}{\sqrt{5}}}}=\lim_{n \rightarrow \infty}\phi^{\frac{1}{\sqrt{5}}(\phi^{n}-\phi^{n-1}-\phi^{n-2})}=1$
QED


The U.S. produces about 5.5 million tons of apples each year. Of the total, $ 20\%$ is mixed with other products, and the remainder is split equally between apple juice and fresh apple sales. How many million tons of apples are used for apple juice? Express your answer as a decimal to the nearest tenth.

[quote="superllamel"]$ 5.5*0.4 \equal{}$
$ 2.2$ lbs.[/quote]

The reasoning is that half of the remaining 80% is 40%, and so

$ 5.5 \cdot .4 \equal{} 2.2$ [i]million tons[/i], not lbs.


AMC
AIME
quadratics
algebra
quadratic formula
AIME I

darkquantum wrote:The solutions to
 log_225 (x) + log_64(y) = 4, log_x(225) - log_y(64) = 1 are
 (x, y) = (x1, y1) and (x2, y2). Find log30(x1y1x2y2).

[hide]letting log_225 (x)=a and log_64(y)=b, we have a+b=4, and 1/a-1/b=1. Plugging in and solving, I get b-a=ab => (4-a)-a=(4-a)a => 4-2a=4a-a:^2: => a:^2:-6a+4. Solving with the quadratic equation, I get a=3 :pm: :rt: 5, and b=1 :pm: :rt: 5. We get our x and y values by taking 225 and 64 to a and b, respectively. Multiplying all together and taking the log base 30, as the problem asks, we get log_30(225^(3+:rt:5)225^(3-:rt:5)64^(1+:rt:5)64^(1-:rt:5)) = log_30(225^(6)64^(2)) = log_30(15^12*2^12)=log_30(30^12)=12[/hide]

Yes that is correct.

This is the way i did it, as well

but im wondering

is there a way to do this one without actually solving for x and y?

in the quadratic formula, we didn't really need to know what the plus/minus term was, since it cancelled out, but besides that, I don't think there were any simplifications.

i kno wat u mean
when i took this AIME, i did it the exact same way
i was just wondering if we didn't actually have to solve for x and y

what type of calculator does one get to use on the AIME?

which makes it that much more FUN!

you can find details on all these competitions at http://www.unl.edu/amc


geometry
calculus
integration
function
calculus computations

Find the area enclosed by the graph $4y^{2}=x^{2}-x^{4}$

First of all, notice that $x^{2}-x^{4}\geq 0$ if and only if $| x | \leq 1$, (or $-1 \leq x \leq 1$). Also note that the curve $4y^{2}= x^{2}-x^{4}$ is symmetric in both the x-axis and the y-axis. So we can just evaluate the area in the first quadrant (EDIT: and multiply it by 4 of course). Rewriting the equation then yields $y = \frac{1}{2}x\sqrt{1-x^{2}}$. (Of course, this only holds in the first quadrant.)

So the area equals
\[4 \int_{0}^{1}\frac{1}{2}x \sqrt{1-x^{2}}dx =-\int_{0}^{1}-2x \sqrt{1-x^{2}}dx \]
.

Substituting $u = 1-x^{2}$ then implies $du =-2x dx$, so our area equals
\[-\int_{1}^{0}\sqrt{u}du = \int_{0}^{1}\sqrt{u}du \]
,

and the latter expression equals $\frac{2}{3}$.

EDIT: Mistaken by a factor 2 while taking the square roots of both sides. I hope its all correct now

no

the area of this function is zero

because it is symetrical to x axis

and he did not make x a real variable. so we need to assume that x is complex and y too

Ehmm.. the area ENCLOSED by the curve $x^{2}+y^{2}= 1$ would be zero too according to your reasoning, and the area is clearly $\pi$... I think your problem is counting the area below the x-axis as negative, but negative areas don't exist in this particular use of the word "area". You may be confused with "normal" evaluations of integrals.

And the use of the words "area" and "enclosed by" already imply that the variables used are real (so the points (x,y)) would be in the plane.

didint see the enclosed

huauhuhau

i think is under


vector
calculus
derivative
rotation
calculus computations

if $ \frac{d\vec{a}}{dt} \equal{} \vec{w} \times \vec{a}$
and $ \frac{d\vec{b}}{dt} \equal{} \vec{w} \times \vec{b}$
$ \frac{d}{dt}\vec{a}\times\vec{b} \equal{} ?$

[hide="Hint"]In general, $ (\vec{a} \times \vec{b})^' = (\vec{a}^'\vec{b}) \times (\vec{b}^'\vec{a})$.

[/hide]

ok...
$ \frac {d}{dt}\vec{a}\times\vec{b} \equal{} (\vec{w} \times \vec{a} \times \vec{b}) \plus{} (\vec{a} \times \vec{w} \times \vec{b}) \implies [(\vec{w}\cdot\vec{b})\vec{a} \minus{} (\vec{w}\cdot\vec{a})\vec{b}] \plus{} [(\vec{a}\cdot\vec{b})\vec{w} \minus{} (\vec{a}\cdot\vec{w})\vec{b}] \implies (\vec{w}\cdot\vec{b})\vec{a} \plus{} (\vec{a}\cdot\vec{b})\vec{w} \minus{} 2(\vec{a}\cdot\vec{w})\vec{b}$
it is ok?

Both $ \vec{a} , \vec{b}$,are constant modulus vectors that rotates at the same angular velocity $ \vec{\omega}$, perpendicular to the plane defined by $ \vec{a} , \vec{b}$.
So the angle $ \vec{a} , \vec{b}$ is constant and so is the vectorial product $ \vec{a} \times \vec{b}$.

Therefor$ {d \over{dt}} \vec{a} \times \vec{b} \equal{} 0$.

We could also find it out using double vector product.

ah, interesting, but I didn't understand you =p  :ninja:
How do find it using the double vector product?
is $ (\vec{w}\cdot\vec{b})\vec{a}\plus{}(\vec{a}\cdot\vec{b})\vec{w}\minus{}2(\vec{a}\cdot\vec{w})\vec{b}$ equal to 0? why $ \vec{a},\vec{b}$ are constant?

I said (rather posted) that their moduli or norm are constant, not the vectors theirselves.
Only their directions rotates with angular velocity $ \vec{\omega}$ .
That's because their time derivative are perpendicular to theirselves.

ok, now I don't understand the angular velocity thing and why $ {d\over{dt}}\vec{a}\times\vec{b}\equal{} 0$

For the angular velocity, we could show it , for instance, like that:

Let $ \vec{a}$ be a vector with a constant norm $ a$, in x,y plane , whose direction makes angle $ \theta \equal{} \omega t$ with the x axis.

So $ \vec{a} \equal{} a [ cos (\omega t). \vec{i} \plus{} sin (\omega t). \vec{j}] \equal{} > {{d \vec{a}} \over {dt}} \equal{} \omega.a [ \minus{} sin (\omega t). \vec{i} \plus{} cos (\omega t ).\vec{j}] \equal{} \omega \vec{k} \times \vec{a}$, where  $ \omega \vec{k} \equal{} \vec{\omega}$, so \[ \frac{d\vec{a}}{dt} \equal{} \vec{w} \times \vec{a}.\]


can anyone define factorial in $ R$?
for example how compute $ (0.1)!$

Use gamma function $ x!\equal{}\gamma(x\plus{}1)$.

Gamma-function is not the same as factorial. It can be used to compute factorials and many calculators use that but factorial is defined only for non-negative integers. Therefore 0.1! is not defined. It is capital Gamma, not lowercase gamma.


geometry
power of a point
radical axis
geometry proposed

Four points A,B,C,D lie on a circle. Prove that three radical axes of three pairs of circles respectively with diameters (AB,CD); (BC,DA) and (AC,BD) are concurrent.

$ K, L, M, N$ are midpoints of $ AB, BC, CD, DA$ and $ U, V$ are midpoints of $ AC, BD.$ $ X \equiv AB \cap CD,$ $ Y \equiv BC \cap DA,$ $ Z \equiv AC \cap BD.$

$ \overline{XA} \cdot \overline{XB} \equal{} \overline{XC} \cdot \overline{XD}$ $ \Longrightarrow$ $ X$ is on radical axis of the circles $ (K), (M)$ with diameters $ AB, CD.$ The K- and M-altitudes of the $ \triangle KXM$ are intersecting at its orthocenter $ T$ are 2 maltitudes of the cyclic $ ABCD$ intersecting at its anticenter, hence $ T$ is the anticenter. Radical axis of $ (K), (M)$ perpendicular to the center line $ KM$ is the X-altitude of the $ \triangle KXM,$ passing through its orthocenter $ T.$

In exactly the same way, the radical axis of $ (L), (N)$ with diameters $ BC, DA$ and the radical axis of $ (U), (V)$ with diameters $ AC, BD$ pass through the anticenter $ T.$


Putnam
vector
analytic geometry
induction
graph theory
combinatorics unsolved
combinatorics

Prove that a list can be made of all the subsets of a finite set in such a way that

i. the empty set is first in the list
ii. each subset occurs exactly once, and
iii. each subset in the list is obtained either by adding one element to the preceding subset or by deleting one element of the preceding subset.

This is called Gray code, and it's equivalent to writing the $2^n$ binary strings of length $n$ in a sequence s.t. each string differs from the preceding string by one digit. It can thus be solved by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14995]this[/url].

Which Putnam does it come from, btw?

As I see, a solution was already posted, but let me just post my solution (hoping that it's correct; the only thing I'm sure about is that I can't write it down in a way it will be clear):

At first, two conventions:

- I will say [b]n-hypercube[/b] for [b]n-dimensional hypercube[/b].
- Two vertices of a graph are called [b]connected[/b] if there is an edge joining them.

Now to the solution of the problem:

If we call our finite set S and its elements $a_1$, $a_2$, ..., $a_n$, then we can attach to every subset T of S a vector $v\left(T\right) = \left( x_1;\;x_2;\;...;\;x_n\right)$, where the number $x_i$ is 0 if the element $a_i$ doesn't belong to the subset T and 1 if the element $a_i$ belongs to the subset T. Conversely, of course, every vector of the form $\left( x_1;\;x_2;\;...;\;x_n\right)$, where every of the coordinates $x_i$ is either 0 or 1, corresponds to a subset of S. Hence we have found a one-to-one correspondence between all $2^n$ subsets of the set S and all possible $2^n$ vectors $\left( x_1;\;x_2;\;...;\;x_n\right)$, where every of the coordinates $x_i$ is either 0 or 1.

The empty set is a subset of S and corresponds to the vector (0; 0; ...; 0).

Also note that for two subsets T and T' of S, we can rewrite the clumsy condition "The set T' is obtained either by adding one element of S to the set T or by deleting one element of T" in the following, much more appealing form: "The vectors corresponding to the sets T and T' have n - 1 coordinates in common". (This formulation means that if these vectors are $\left( x_1;\;x_2;\;...;\;x_n\right)$ and $\left( y_1;\;y_2;\;...;\;y_n\right)$, then there exists one i such that $x_i \neq y_i$, and for all $j \neq i$, we have $x_j = y_j$.)

Hence, we can translate the problem in the language of vectors: Prove that a list can be made of all vectors $\left( x_1;\;x_2;\;...;\;x_n\right)$, where every of the coordinates $x_i$ is either 0 or 1, in such a way that

i. The vector (0; 0; ...; 0) is first in the list;
ii. Each possible vector $\left( x_1;\;x_2;\;...;\;x_n\right)$, where every of the coordinates $x_i$ is either 0 or 1, occurs exactly once in the list;
iii. Each vector of the list has with the precedent vector n - 1 coordinates in common.

Now, consider the graph whose vertices correspond to all the vectors $\left( x_1;\;x_2;\;...;\;x_n\right)$, where every of the coordinates $x_i$ is either 0 or 1. We connect two vertices of this graph by an edge if and only if the corresponding vectors have n - 1 coordinates in common. If we find a Hamiltonian cycle in this graph, then we can start with the vertex (0; 0; ...; 0) and form a list of vectors satisfying the properties i., ii. and iii., what will immediately solve the problem. So it remains to find a Hamiltonian cycle in our graph.

But our graph is isomorphic to the graph formed by the vertices and the edges of an n-hypercube. Thus, it will be enough to prove the existence of a Hamiltonian cycle on the n-hypercube (for $n \geq 2$).

But this is simple by induction. For a 2-hypercube (i. e. a square), I leave the problem to the bored reader. Now assume we can find a Hamiltonian cycle on the n-hypercube. We will prove that we can find a Hamiltonian cycle on the (n + 1)-hypercube.

In fact, the (n + 1)-hypercube H consists of two n-hypercubes $H_a$ and $H_b$, each of them lying on a hyperplane of order n.

These two n-hypercubes $H_a$ and $H_b$ are connected with each other, i. e. each vertex of one of these two n-hypercubes is connected with the corresponding vertex of the other one. Now, by the assumption of our induction, we can find a Hamiltonian cycle on each of these two n-hypercubes. Let $A_1 A_2 ... A_k A_1$ be a Hamiltionian cycle on the n-hypercube $H_a$, with $A_1$, $A_2$, ..., $A_k$ being the vertices of $H_a$ (so that $k = 2^n$). Now, for every i, there is exactly one vertex $B_i$ of the n-hypercube $H_b$ connected with the vertex $A_i$ of the n-hypercube $H_a$. Then, obviously, the cycle $B_1 B_2 ... B_k B_1$ is a Hamiltonian cycle on the n-hypercube $H_b$. So we have found a Hamiltonian cycle $A_1 A_2 ... A_k A_1$ on the n-hypercube $H_a$ and a Hamiltonian cycle $B_1 B_2 ... B_k B_1$ on the n-hypercube $H_b$. Consider the union of these two cycles, i. e. the set of the edges $A_1 A_2$, $A_2 A_3$, ..., $A_k A_1$, $B_1 B_2$, $B_2 B_3$, ..., $B_k B_1$. This set is not yet a Hamiltonian cycle on the whole (n + 1)-hypercube H, but it can be easily modified to form a Hamiltonian cycle: Erase the edges $A_1 A_2$ and $B_1 B_2$ and add the edges $A_1 B_1$ and $A_2 B_2$ instead. The resulting set of edges is a Hamiltonian cycle on the (n + 1)-hypercube H.

Thus, the induction step is done, and the proof is complete.

  Darij


modular arithmetic
number theory
number theory proposed

Let p a prime number>=5 and $ (a,b)\in{Z}$
prove that:
$ ab^{p} \minus{} ba^{p}\equiv0[6]$
thanx.

The expression is clearly divisible by $ 2$ ($ x^2 \equiv x \pmod 2 \implies ab^p \equiv ab \equiv a^pb \pmod 2$)
If one of $ a,b$ is divisibly by $ 3$, we are also done.
Otherwise, $ ab^p\minus{}ba^p \equiv 0 \pmod 3 \leftrightarrow (ba^{\minus{}1})^{p\minus{}1} \equiv 1 \pmod 3$
And this is true since order of $ ba^{\minus{}1}$ divides $ 2$ (by Fermat's Little Theorem) and $ p\minus{}1$ is even.
As you see, it is enough that $ p \neq 2$.

[quote]The expression is clearly divisible by $ 2 (x^2 \equiv x \pmod 2 \implies ab^p \equiv ab \equiv a^pb \pmod 2)$
If one of a,b is divisibly by 3, we are also done.
Otherwise, $ ab^p\minus{}ba^p \equiv 0 \pmod 3 \leftrightarrow (ba^{\minus{}1})^{p\minus{}1} \equiv 1 \pmod 3$
[color=red]And this is true since order of $ ba^{\minus{}1}$ divides 2 (by Fermat's Little Theorem) and $ p\minus{}1$ is even[/color].
As you see, it is enough that $ p \neq 2$.

[/quote]

Can you explain?
thanx.

there is my solution:
by using the generalisation of little fermat's theorem we have:
$ a^{\phi(x)}\equiv1[n]$ with $ a\wedge n\equal{}1$ and $ \phi(x)$ is Euler's function.
if $ 3|(a,b)$ we see that's easy t o prove the result!
so we let $ 3\nmid(a,b)$ ,then $ b\wedge3\equal{}1$ and $ \phi(3)\equal{}2$ so:
$ a^{\phi(3)}\equiv1[3]$
$ b^{\phi(3)}\equiv1[3]$
==>$ a^{p\minus{}1}\equiv1[3]$ (p-1 is even)
$ b^{p\minus{}1}\equiv1[3]$
==>$ b^{p\minus{}1}\minus{}a^{p\minus{}1}\equal{}0[3]$
==>$ ab^{p}\minus{}a^{p}b\equal{}0[3]$
Done!

As $ p$ is odd,

$ ab^p\minus{}ba^p\equal{}ab\minus{}ba\equal{}0(mod 2)$ and

$ ab^p\minus{}ba^p\equal{}ab\minus{}ba\equal{}0(mod 3)$


[quote="ashrafmod"]A diagonal in a convex polygon is a line joining two non-adjacent vertices.  Take an n-sided polygon and draw all possible diagonals.  Suppose that no three diagonals ever intersect at a common point.  How many crossings are there in the interior of the polygon?[/quote]

[hide]You can choose 4 different points and there is only one way to connect the points by pairs so that they intersect at a point, so it's just $\binom n4$...?[/hide]

You know, diagonals can intersect at the vertices... so maybe the question ought to be reworded.

also don't forget that they must intersect inside the polygon

That shouldn't be a problem. The polygon is convex.

no, its still possible for them to intersect outside the polygon

The point is that four any four points you can draw the diagnols such that there is a crossing.  So for the four points you picked you obviously won't get a crossing if you draw it like that.  But if you draw it the other way you will.


If $ 4^{x\plus{}1} \equal{} 2$, express $ 4^{x\minus{}1}$ as a common fraction.

Notice that $ 4^{\frac{1}{2}} \equal{} 2$, so we can say $ x \plus{} 1 \equal{} \frac{1}{2} \implies x \equal{} \minus{} \frac{1}{2}$. Therefore, $ 4^{x \minus{} 1} \equal{} 4^{\minus{}\frac{1}{2} \minus{} 1} \equal{} 4^{\minus{}\frac{3}{2}} \equal{} \frac{1}{8}$.


Solve in positive integers the equation $\frac{19}{x^{2}}+\frac{79}{y^{2}}=\frac{z}{1979}.$

[hide="Comments"]
This is memorable as one of the first Olympiad number theory problems I have solved  :)
[/hide]

[hide="Solution"]

[b]Claim: The only solution in $(x, y, z)$ is $(1, 1, 98 \cdot 1979) = (1, 1, 193942).$[/b]

[b]Proof:[/b]
We manipulate the equation to get

$1979(79x^{2}+19y^{2}) = x^{2}y^{2}z$ ... (1)

Taking the equation $\bmod{1979},$ we have $x^{2}y^{2}z \equiv 0 \bmod{1979}.$ This yields three cases:

[b]Case 1: $x^{2}\equiv 0 \bmod{1979}$[/b]

Since $1979$ is prime, $x^{2}\equiv 0 \bmod{1979}\Leftrightarrow x \equiv 0 \bmod{1979}.$ Thus, for some $x_{1}\in \mathbb{Z}^{+},$ let $x = 1979x_{1}.$ Substitute into (1) to yield

$1979(79 \cdot 1979^{2}x_{1}^{2}+19y^{2}) = 1979^{2}x_{1}^{2}y^{2}z$
$\Leftrightarrow 79 \cdot 1979^{2}x_{1}^{2}+19y^{2}= 1979x_{1}^{2}y^{2}z$ ... (2)

But $\bmod{1979},$ $y^{2}\equiv 0 \bmod{1979}\Rightarrow y \equiv 0 \bmod{1979},$ so let $y = 1979y_{1}$ for some $y_{1}\in \mathbb{Z}^{+}.$ Substitute into (2) to get:

$79 \cdot 1979^{2}x_{1}^{2}+19 \cdot 1979^{2}y_{1}^{2}= 1979x_{1}^{2}\cdot 1979^{2}y_{1}^{2}\cdot z$
$\Leftrightarrow 79x_{1}^{2}+19y_{1}^{2}= 1979x_{1}^{2}y_{1}^{2}z$
$\Leftrightarrow \frac{19}{x_{1}^{2}}+\frac{79}{y_{1}^{2}}= 1979z.$

Since $x_{1}, y_{1}, z \in \mathbb{Z}^{+},$ we have $x_{1}^{2}| 19$ and $y_{1}^{2}| 19.$ But the only solution that may occur happens when $x_{1}= 1, y_{1}= 1,$ which contradicts the fact that $z \in Z^{+}.$

[b]Case 2: $y^{2}\equiv 0 \bmod{1979}$[/b]

This case is equivalent to the previous case.

[b]Case 3: $z \equiv 0 \bmod{1979}$[/b]

Let $z = 1979z_{1}$ for some $z_{1}\in \mathbb{Z}^{+}.$ Substitute into (1) to get

$1979(79x^{2}+19y^{2}) = x^{2}y^{2}\cdot 1979 z_{1}$
$\Leftrightarrow 79x^{2}+19y^{2}= x^{2}y^{2}z_{1}$
$\Leftrightarrow \frac{19}{x^{2}}+\frac{79}{y^{2}}= z_{1}$
$\Leftrightarrow x = 1, y = 1, z_{1}= 98.$

This is the only solution as explained towards the end of case 1. Thus, the only solution in $(x,y,z)$ is $(1, 1, 98\cdot 1979) = (1, 1, 193942).$[/hide]

i didn't see your solution. :wink:

rewrite the equation as $1979(19y^{2}+79x^{2})=x^{2}y^{2}z\Longrightarrow x^{2}\mid 1979\cdot 19y^{2}$.since 1979, 19 and 79 are primes, either $x=1$ or $x^{2}\mid y^{2}$ and for the same reason $y=1$ or $y^{2}\mid x^{2}$. if $(x,y)\neq (1,1)$, we get $x=y$ and $x^{2}z=1979\cdot 98$ thus $x=7,z=1979\cdot 2=3958$. so we have the solutions $(1,1,193942),(7,7,3958)$.


Demonstrate that there is no fonction f:R->R so that
|f(x)-f(y)|<2006
 for any x different from y,x,y belong to R.[/b][/hide][/list][/quote][/code]

Any function such that $\text{max}(f)-\text{min}(f) < 2006$.   :rotfl:

(Is $\mathbb{R}$ the codomain or the range?)


induction
number theory proposed
number theory

Let $d > 0$ be an arbitrary real number. Consider the set $S_{n}(d)=\{s=\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}|x_{i}\in\mathbb{N},s<d\}$. Prove that $S_{n}(d)$ has a maximum element.

With induction is simple: for $ n=1$ it's obvious. Denote $ \sup(S_{n}(d))=u$, then a number $ s>u-e$($ 0<e<u$ , $ e$ fixed) will have an element(suppose $ x_{n}$) $ d>\frac{1}{x_{n}}> \frac{u-e}{n}\rightarrow \left[\frac{1}{d}\right]<x_{n}\le \left[\frac{n}{u-e}\right]$, so it can take a finite number of values. By induction, for every $ x_{n}$ , $ S_{n-1}\left(d-\frac{1}{x_{n}}\right)$ will have a maximum element. So we will have a finite number of maximum elements, adding to each one it's $ \frac{1}{x_{n}}$ we can choose a maximum element for $ S_{n}(d)$.


In the topic posted by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15100]Dr Sonnhard Graubner[/url], darij proved the following inequality using Maple:

[quote="darij grinberg"]
[i]If a, b, c, x, y, z are six positive numbers such that $a \leq b \leq c$ and $x \leq y \leq z$, then

  $\displaystyle \left( bz-cy\right) ^{2}+\left( cx-az\right) ^{2}+\left( ay-bx\right) ^{2}<\left( \frac{\left( a+b+c\right) \left( x+y+z\right) }{2}\right) ^{2}$. [/i]
[/quote]

Adding $(ax+by+cz)^2$ to both sides, it's equivalent to:

$\displaystyle ( a^2+b^2+c^2)(x^2+y^2+z^2)<(ax+by+cz)^2+ \left( \frac{\left( a+b+c\right) \left( x+y+z\right) }{2}\right) ^{2}$.

Here is my conjecture:
[quote]Let $0< a_1\le a_2 \le \cdots \le a_n$ and $0 < b_1 \le b_2 \le \cdots \le b_n$ be two increasingly sorted sequences of positive reals, then

\[
\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right) < \left(\sum_{i=1}^n a_ib_i\right)^2 +
\frac{1}{4}\left(\sum_{i=1}^n a_i\right)^2\left(\sum_{i=1}^n b_i\right)^2.
\]
[/quote]

I have verified it for $n=1, 2, 3, 4$ using Mathematica with the help of a friend.

This is very similar to problem 299 (p.70-71) in the [url=http://www.amazon.com/exec/obidos/tg/detail/-/0486277097/qid=1093280415/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-9903304-2984762?v=glance&s=books&n=507846]The USSR Olympiad Problem Book[/url], we might be able to borrow some idea from its solution.

This is really a masterpiece! It is so nice and there is such a beautiful solution! Observe that the right-hand side is at least $ (\sum_{i=1}^n {a_i})\cdot (\sum_{i=1}^n {b_i})\cdot (\sum_{i=1}^n {a_i\cdot b_i}) $ by the AM-GM inequality. Next, it is clear that $\sum_{i=1}^n {(a_i)^2}<a_n\cdot\sum_{i=1}^n {a_i} $ and analogous for $ b_j $. Multiplying these yields the conclusion. We even see that we obtain an even stronger inequality: it suffices to have in the right-hand side instead of $(\sum_{i=1}^n {a_i\cdot b_i})^2 $ just $ a_n\cdot b_n\cdot\sum_{i=1}^n {a_i\cdot b_i} $. But your form is fmuch more elegant.

Thank you, Fuzzylogic, for the generalization. And thank you, Harazi - you proved the generalization of the generalization of the inequality Sonnhard originally posted!

Actually let me add some information about the original inequality (the one with the sides and the angles of the triangle). It was first posted in the [url=http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=23580]Matheplanet thread "Ungleichung für ein Dreieck"[/url]. According to what Buri told about it, this inequality was created by D. Blanusa in 1953 and subsequently proven by T. Leko, but the proof was very tedious.

  Darij

Thank you, harazi, for the wonderful proof! I am sure Dr Sonnhard Graubner will be very happy to see this.

hello,thank you very much for your generalization and this nice proof.
Regards,Sonnhard


If determinant is zero, can we conclude that any of the rows/columns is a linear combination of the other rows/columns?
I know that if any of the rows/columns is a linear combination of the other rows/columns then the determinant is zero.
Thanks for your help in advance

[quote="AngelaS"][quote="mtphaovibul"]the answer is yes, since the determinant equal to zero, so the matrix is not invertible, so it is implies that the column/row are linearly dependant.[/quote]

Thanks mtphaovibul![/quote]

I just found out that the answer is no! We can write [b]some[/b] coloumns/rows as a linear combination of the other columns/rows.

well that's true, I didn't read it carefully,


[quote="8parks11"]What is the probability that [b]you[/b] will roll a 3 before you roll both a 1 and a 2 if [b]we[/b] roll a fair 6-sided die repeatedly?[/quote]
do you mean roll a (3 before 1) and (3 before 2), or roll a 3 before (roll both 1 and 2) (parallelism error btw)
If you mean the first case, let $ P$ be the probability you want to find.
then you get
\[ P \equal{} \frac16 \plus{} \frac36*P\]
since the $ \frac16$ is the probability you get a 3 on the first roll, and $ \frac36$ is the probability you get a 4,5 or 6 on the first roll, and move on to the second.
multiplying by 6 on both sides, you get
\[ 6P \equal{} 1 \plus{} 3P\]

\[ 3P \equal{} 1\]

\[ P \equal{} \frac13\]

If the problem you solved had been intended, the wording would have been "roll a 3 before either a 1 or a 2."

We can solve the question that was asked using something like the argument you used, but there's a simpler way: [hide="main idea"]suppose we keep rolling the die indefinitely, then only go back to check at the end whether we won or lost.  After you have a long enough string of rolls so that it includes at least one of 1, 2 and 3, isolate the positions of the first 1, the first 2 and the first 3.  Under which relative orders of these three rolls do you win?  What are the probabilities of these relative orders?[/hide]  (This idea also works for the version of the problem that you solved.)

it's obvious that the problem can only be asking for the probability of (rolling a 3) before (rolling a 1 and a 2).

the number of sequences of length $ n$ all of whose terms come from $ \{1, 2, 3, 4, 5, 6\}$, whose last term is 3 and no term before it is 3, and such that 1 and 2 do not both occur in the sequence, is $ 2\cdot 4^{n\minus{}1}\minus{}3^{n\minus{}1}$. thus the probability is

\[ \sum_{n\equal{}1}^{\infty} \frac{2\cdot 4^{n\minus{}1}\minus{}3^{n\minus{}1}}{6^n}\equal{}\frac{1}{3}\sum_{n\equal{}1}^{\infty} \left( \frac{2}{3} \right)^{n\minus{}1}\minus{}\frac{1}{6} \sum_{n\equal{}1}^{\infty} \left( \frac{1}{2} \right)^{n\minus{}1} \equal{} 1\minus{}\frac{1}{3}\equal{}\frac{2}{3}\]

Well, the original poster has been extremely rude and removed the original text of the problem (rendering this discussion largely unintelligible) instead of just providing a link to the previous discussion when apologizing.  Here's the previous discussion (including the same misunderstanding of the problem together with a creative variety of wrong answers and a couple of correct solutions):
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=577379279&t=122067

[quote="AceOfDiamonds"][quote="JBL"]If the problem you solved had been intended, the wording would have been "roll a 3 before either a 1 or a 2."
[/quote]
well that could be interpreted as roll a 3 before a 1  or roll a 3 before a 2[/quote]  Not by someone who read carefully and understands how the English language works  :wink:  (I agree that this is an easy question to misunderstand, but it's also true that the question is not actually ambiguous.  Ditto for the alternate version as I phrased it.)


This question may be old, but can someone give me names of problem books or any other useful books or hints to help me for next year's USAMO? :huh:

in my opinion, AoPS volume 2 is a very good book for stepping toward the olympiad level...it have many proof problems which are ideal for the USAMO...if you have already finished that then read Paul Zeitz's (i think) The Art and Craft of Problems Solving....this is considered by many to be AoPS Volume 3...it's supposed to be good for people preparing for such higher competitions as the USAMO/IMO...

You're right. The question is old.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101391

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102652

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82958

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82043

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81296

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79077

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79014

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77740

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50068

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31888

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29802

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11418

Thanks a lot! I'll try AoPS2 and see what happens!  :)


Euler
Gauss
real analysis
geometry
calculus
topology

Who do you suppose are the greatest mathematicians of each initial?
My opinion(partly): A--Abel B--Bernoulli C--Cauchy D--Descartes E--Euler F--Fermat G--Gauss H--Hilbert...

Iverson, Jacobi, Kronecker, Lagrange, Maclaurin, Newton

Archimedes?

...

O seems to be particularly hard, then Poincare, Q ...., Riemann, Schwarz or Serre, Turing, U ...., von Neumann, Weierstrass or Weyl, X ...., Y ...., Zariski?

In no particular order....

P-ascal
N-ewton
J-acobi
L-eibniz
G-alois
C-antor

Euler may be hard to beat in the E-spot, but there is Erdos. Are they the two most prolific mathematicians ever?

L is pretty heavy on French mathematicians from the turn of the 19th century- a rather interesting time and place.

E-uclid                                                 .

Descartes doesn't seem too convincing in 'D', and I think maybe a pair of Fields medalists might belong there - Drinfeld or Deligne (I mean, really, there aren't many others; de la Vallee Pousin, de Moivre, Dirichlet?).  Also:

Fourier over Fermat,
Ito in I?
L is too hard to call, between Lagrange and Liouville and Leibniz and Lebesgue
Minkowski over Maclaurin
Turan over Turing (I think that was a typo on my part)
Yau of Calabi-Yau fame in the 'Y' spot (another Fields medalist)

anyone else have anything to add?

[quote="biffanddoc"]yeah but geometry is cooler than calculus :coool:  :cool:[/quote]

yeah, in what world?

T-Turing

Z - Zeb ;)

Seriously, are there ANY great mathematicians whose names start with Z :?:

Also...

G - Gauss
E - Euler (greatest mathematician of all time, right?)
F - Fermat
C - Conway
N - Newton

[quote="biffanddoc"]NOOOO i like Euclid
I mean, come on geometry is the best ever :coool: :coool:
Everyone learns geometry
Less than half the population learns calc.[/quote]

Which means only the cool people learn calc.

which means cool people like euclid :coool: :coool:

Euler>Euclid, but it's really hard to compare because they came from very different time periods.

Euclid was cool; he was modest as well and would allow Euler to take the spot, but...why can't E be for both Euler [i]and[/i] Euclid, i mean if schrodinger's cat can be two things at once...you get the idea here, this isn't the fourth grade...

Zariski, of Zariski topology fame, did very significant work in algebraic geometry.

[quote="zeb"]Darnit, I guess there are people with Z-names......[/quote]

it's not too late to die in a duel set for tomorrow and write down everything you know about math on a piece of paper...

Wouldn't that prove that I know very, very little? (Not to say that I don't.)

By the way, I've found some very beautiful proofs of the Goldbach and the twin prime conjectures which this post is too short to contain...

[quote="PenguinIntegral"][quote="biffanddoc"]for e:euclid, i mean he basically invented geometry, but close between him and euler[/quote]
And euler basically invented calculus.[/quote]

ummm,did newton do that too?
Newton
Archimedes(he's awesome bcos of the stuff he did and + he is prettty ancient,is he a mathmetician?)
Ramanujan
$Nilesh$

pythogorean helped create geometry
but Eculid all the way :!:  :!: !!!!!!!!!!!!!!!!!!!!!

yeah more people are siding with me!

take that euler

Euler>all other E's

How about young mr Terrence Tao for T, double T's too, hah

he's like, 29 and being considered for Fields Medal this year, "Mozart of mathematics without attitude problems"

[quote="zeb"]Wouldn't that prove that I know very, very little? (Not to say that I don't.)[/quote]

that's how Galois died, in a duel, and the night before he wrote down everything he knew about mathematics (or so the story goes)

Hmm. this sounds like fun.

A - Archimedes (Greece)
B- Jacob Bernoulli (Switzerland)
C - Cauchy (France)
D - Descartes (France)
E - Euler (Switzerland)
F - Fermat (France)
G - Gauss (German)
H - Hilbert (German)
I - Iwasawa (Japan)
J - Jacobi (German)
K - Felix Klein (German)
L - Lagrange (France)
M - Morse (USA)
N - Newton (England)
O -
P - Poincare (France)
Q - Quillen (USA)
R - Riemann (German)
S - J.P. Serre (France)
T - Turing (England)
U - Ulam (Poland)
V - Veblen (USA)
W - Weil (France)
X -
Y - S.T. Yau (China)
Z - Zorn (German)


Wow. All these forums have become packed with threads. There should be a monthly deletion or something.

Delete this while you are at it, but I just wanted to comment on the "crowdedness" of this place.

You don't need to delete old posts.. i mean, they get spread over more than one page anyway. Forums don't delete old posts like that.

Actually, you're right. I think we should delete the fun and games forum. The other ones have useful information/problems in them.

We already have auto-pruning and that's enough as it is. As for crowdedness I highly doubt your suggestion will go through, for a forum is supposed to have a vast archive unlike a chatroom or instant messages, and by deleting them that would mean destroying thousands upon thousands of precious math problems yet to be rediscovered. In fact if the boards aren't packed to the brim with threads, chances are people aren't interestend in that board, and that's a [b]bad[/b] thing.

P.S. You're avatar is broken

P.P.S. In the future please post non-MATHCOUNTS related comments in a non-MATHCOUNTS releted board, thank you.

See Tare! This is what I mean! Why did you move that thing here? It belongs in the MATHCOUNTS forum.

Why? Give me one good reason. (which you need to tell one of the admins because now it is out of my power to move it back)

And no, I have not moved it because the board was getting full, I did it becasue this thread is off-topic, it's not even about math, and therefore doesn't belong in the MATHCOUNTS board, do you understand?

Fun and Games is a nice magnet for posts unrelated to mathematics.  It seems to be the easiest way to keep the problem solving forums pure to their purpose.

[quote="Tare"][quote="white_horse_king88"]NO! You moved that weird complicated problem to the F and G! Why would anyone do that? Math belongs in Math , Fand G is F and G(although it should be deleted altogether(f and G)).

I misposted this thread. Lo siento.[/quote]
I ave no idea what you're talking about...can you be more specific?[/quote]
i thinnk what he means by that weird complicated math problem was (mathcounts challenge: complicated?) i remember it was in f and g at one time.

Oh yeah, I remember now :-x That was Neal repeatedly sending it to F&G to tick me off and say to me "This problem is so stupid it's not even worthy of calling it a math problem, *spits to problem*" without actually saying that. Don't worry, I moderate them both, so i have the upper hand :lol:

1) What is your proof that is was me?
2) You do not have the upper hand on Bob and me.

[quote="MCrawford"]Those who spend their time engaging real problems have the upper hand over those who spend their time becoming problems.[/quote]

'tis true, I didn't reallly mean Bob and I had an upperhand over Tare; it was mostly out of anger because Tare jumped to the conclusion that I moved it to annoy him, even though it was not me.

I'm sorry Neal, I did the same thing people did to me and I should've known better. Mmm now I know who the real culprit is...


Subiecte:

1)a)Pe axa numerelor reale de origine O se considera punctele A,B,C cu abscise numere intregi.Aratati ca cel putin o distanta dintre doua puncte ale multimii{O,A,B,C} este un multiplu de trei .
   b)Fie a,b doua nr reale a<b a.i. (a,b) :cap: Z=0 . Daca I=[a-2005,b+2005] , cate elemente are multimea I :cap: Z? Discutie in raport cu a si b.

[quote="rockymarcianocosmin99"]
   b)Fie a,b doua nr reale a<b a.i. (a,b) :cap: Z=0 . Daca I=[a-2005,b+2005] , cate elemente are multimea I :cap: Z? Discutie in raport cu a si b.[/quote]

As dori daca se poate o clarificare a notatiei, si anume: in enuntul de mai sus, prin 0 s-a notat cumva multimea vida? Sau intentia a fost de fapt (a,b) :cap: Z = {0} (multimea formata cu elementul 0)? Multumesc.

0 vrea sa fie multimea vida acolo la punctul b)


TripleM wrote:What is the side length of the biggest square that can fit inside a cube of side length n?

Solution: (?)

[hide]I think the answer is 3 :rt2: n/4.

My reasoning is that the optimal area should be the cross section of a plane slicing through the cube. Thus, I had the plane slicing through two adjacent edges, both x away from their shared vertex, and the same thing happened in the opposite region of the cube.

We can then get a side of the cube by either  :sqrt: (2x^2) or  :sqrt: (2(n-x)^2 + n^2). Solving, we get x = 3n/4, and so I believe that the answer is 3 :rt2: n/4, but I am not positive.

[/hide]

The problem is, you haven't proven that that is the maximum, you just calculated a possible value for it.

JBL, that is quite true. But I have a feeling that this is the correct answer...I'm sure some explanation involving moving this plane could verify it.


geometry
parallelogram
geometry unsolved

Let be given a triangle $ ABC$. Points $ P$ on side $ AC$ and $ Y$ on the production of $ CB$ beyond $ B$ are chosen so that $ Y$ subtends equal angles with $ AP$ and $ PC$. Similarly, $ Q$ on side $ BC$ and $ X$ on the production of $ AC$ beyond $ C$ are such that $ X$ subtends equal angles with $ BQ$ and $ QC$. Lines $ YP$ and $ XB$ meet at $ R$, $ XQ$ and $ YA$ meet at $ S$, and $ XB$ and $ YA$ meet at $ D$. Prove that $ PQRS$ is a parallelogram if and only if $ ACBD$ is a cyclic quadrilateral.

+If $ ABCD$ is a cyclic quadrilateral.
Since $ \frac{DR}{RB}\equal{}\frac{YD}{YB}\equal{}\frac{DC}{AB}\equal{}\frac{XD}{XA}\equal{}\frac{DS}{SA}$ we get $ RS//AB$ and $ \frac{RS}{AB}\equal{}\frac{DC}{DC\plus{}AB}$
Similarly, $ PQ//AB$ and $ \frac{PQ}{AB}\equal{}\frac{DC}{DC\plus{}AB}$
So $ PQRS$ is a parallelogram
+If $ PQRS$ is a parallelogram.
Denote $ L$ the intersection of $ SQ$ and $ RP$ we have $ LR\equal{}LP$
$ \Rightarrow$ triangle $ XRP$ is scalene
Therefore $ \angle XRP\equal{}\angle XPR$ or $ \angle RYD\plus{}\angle YDR\equal{}\angle RYB\plus{}\angle ACB$
Thus $ \angle YDB\equal{}\angle ACB \Rightarrow ABCD$ is cyclic.


6. In triangle ABC, let D, E be points on AB, AC, respectively, and let BE and CD meet at
point P. If the areas of triangles ADE, BPD, and CEP are 5, 8, and 3, respectively, find
the area of triangle ABC.

7. Bob has 11 socks in his drawer: 3 different matched pairs, and 5 socks that dont match with
any others. Suppose he pulls socks from the drawer one at a time until he manages to get a
matched pair. What is the probability he will need to draw exactly 9 socks?

An idea  for 7: [hide]  The only way he can pull out 9 socks without getting a pair is if he pulls out all 5 unmatched ones and one from each pair.  How many ways are there of doing this?  Then how many ways are there of choosing 9 socks?[/hide]

hint for 6:[hide]
In any triangle ABC with D on AB you have $\frac{|ACD|}{|BCD|} = \frac{AD}{BD}$ where $|ABC|$ denotes the are of ABC.

If you haven't seen this you should convince yourself of this fact before you carry on - its pretty easy to show.

Now let $|DPE| = x$ and $|CPB| = y$, and use the fact above to set up some equations for $x$ and $y$.

[/hide]


The positive reals $x$ and $y$ satisfy $x^2+y^3 \geq x^3+y^4$.
Prove that $x^3+y^3 \leq2$