Putnam
floor function
probability
inequalities
college contests

A [i]triangulation[/i] $ \mathcal{T}$ of a polygon $ P$ is a finite collection of triangles whose union is $ P,$ and such that the intersection of any two triangles is either empty, or a shared vertex, or a shared side. Moreover, each side of $ P$ is a side of exactly one triangle in $ \mathcal{T}.$ Say that $ \mathcal{T}$ is [i]admissible[/i] if every internal vertex is shared by $ 6$ or more triangles. For example
[asy]
size(100);
dot(dir(-100)^^dir(230)^^dir(160)^^dir(100)^^dir(50)^^dir(5)^^dir(-55));
draw(dir(-100)--dir(230)--dir(160)--dir(100)--dir(50)--dir(5)--dir(-55)--cycle);
pair A = (0,-0.25);
dot(A);
draw(A--dir(-100)^^A--dir(230)^^A--dir(160)^^A--dir(100)^^A--dir(5)^^A--dir(-55)^^dir(5)--dir(100));
[/asy]
Prove that there is an integer $ M_n,$ depending only on $ n,$ such that any admissible triangulation of a polygon $ P$ with $ n$ sides has at most $ M_n$ triangles.

I didn't want to draw the picture. All the stuff in green is me describing it.

We can choose $ M_n$ by the recursive estimate $ M_n\equal{}M_{n\minus{}1}\plus{}2\lfloor\frac n6\rfloor\plus{}1$, $ M_3\equal{}1$; in explicit form, we get $ M_n\le \frac{n^2\plus{}2n\minus{}8}{6}$ (grubby details omitted; this is equal to the recursive form at $ n\equiv 2\mod 6$, and within 1 everywhere except $ n\equiv 5\mod 6$).
The actual maximum is close to this estimate, since we can build a regular hexagon out of triangles for an admissible triangulation of a $ 6k$-gon using $ 6k^2$ triangles.

We prove that this $ M_n$ works by showing we can decompose any admissible triangulation of an $ n$-gon ($ n>3$) into an admissible triangulation of a $ \le (n\minus{}1)$-gon and a triangulation of a smaller polygon (at most $ 2\lfloor \frac n6\rfloor\plus{}3$ vertices) with no interior vertices.

Start with an admissible triangulation $ T$ of an $ n$-gon. To each vertex $ v$ of the $ n$-gon, we assign an integer $ D(v)$ equal to $ 4$ minus its valence. We have $ D(v)\le 2$, since there are two edges of the original polygon at $ v$.
Lemma: Let the vertices of the original polygon be $ v_1,v_2,\dots,v_n.$
$ \sum_{k\equal{}1}^n D(v_k)\ge 6.$
Proof: We use Euler's formula $ V\plus{}F\minus{}E\equal{}2$ for planar graphs. Let $ I$ be the number of internal vertices, so $ V\equal{}n\plus{}I$. Each face is a triangle except the outside and each edge lies on two faces, so $ E\equal{}\frac{3(F\minus{}1)\plus{}n}{2}$ (1). Finally, counting edges by their vertex connections, $ E\ge \frac12\left(6I\plus{}\sum_{k\equal{}1}^n 4\minus{}D(v_k)\right)\equal{}3I\plus{}2n\minus{}\frac12\sum_{k\equal{}1}^n D(v_k)$. By (1), $ F\equal{}\frac13(2E\minus{}n\plus{}3)$, so $ V\plus{}F\minus{}E\minus{}2\equal{}n\plus{}I\plus{}\frac23E\minus{}\frac13n\plus{}1\minus{}E\minus{}2\equal{}\frac23n\minus{}\frac13E\plus{}I\minus{}1\equal{}0$, and $ 2n\plus{}3I\minus{}3\equal{}E\ge 3I\plus{}2n\minus{}\frac12\sum_{k\equal{}1}^n D(v_k)$. This simplifies to the result $ \sum_{k\equal{}1}^n D(v_k)\ge 6.$
Corollary: $ M_3\equal{}1$.
Proof: With only three vertices, the only way to get $ \sum D(v_k)\equal{}6$ is to have each $ D(v)\equal{}2$. In this case, there are no interior vertices. With no interior vertices, there is only one triangle.

Now, we construct the promised decomposition. There are two cases:
If $ D(v)\equal{}2$ for some vertex $ v$, that vertex and its two neighbors are one of the triangles of $ T$. Cutting that triangle off, there are at most $ M_{n\minus{}1}\plus{}1$ triangles in $ T$. This is always the case for $ n\le 5$.
For the remainder, we suppose that $ n\ge 6$ and $ D(v)\le 1$ for all $ v$. We also assume that no edge of the triangulation is a diagonal of the polygon; that case gives us a much looser bound $ M_{k\plus{}2}\plus{}M_{n\minus{}k}$ for the number of triangles. Write the vertices in (counterclockwise) cyclic order $ v_1,v_2,\dots,v_n\equal{}v_0$, and consider the sums $ S_k\equal{}D(v_k)\plus{}D(v_{k\plus{}1})\plus{}\cdots\plus{}D(v_{k\plus{}m})$ (indices interpreted mod $ n$), where $ m\equal{}\lfloor \frac n6\rfloor$. The sum of the $ S_k$ is at least $ (\lfloor \frac n6\rfloor\plus{}1)\cdot \sum_{k\equal{}1}^n D(v_k)>n$, so some $ S_k$ is at least 2.
Given this $ k$, we construct a path of length $ m\plus{}1$ between the vertices $ v_{k\minus{}1}$ and $ v_{k\plus{}m\plus{}1}$ through the interior vertices; cutting along this path splits $ T$ into a triangulation of a $ (2m\plus{}3)$-gon into $ 2m\plus{}1$ triangles and an admissible triangulation of an $ (n\minus{}1)$-gon into at most $ M_n$ triangles.
The path follows all of the interior vertices $ w$ which are directly connected to $ v_{k\plus{}j}$ for some $ 0\le j\le m$ (the first also connects to $ v_{k\minus{}1}$ and the last connects to $ v_{k\plus{}m\plus{}1}$), and the triangles cut off are exactly the triangles with at least one vertex among $ v_k,v_{k\plus{}1},\dots,v_{k\plus{}m}$. We start with the vertex $ w_1$ connected to $ v_{k\minus{}1}$ and $ v_k$ and build along; there are $ 2\minus{}D(v_{k\plus{}j})$ internal vertices connected to $ v_{k\plus{}j}$, one of which was connected to $ v_{k\plus{}j\minus{}1}$. If this is more than one vertex, they are connected by a sequence of edges, closing the triangles with $ v_{k\plus{}j}$ as a vertex. For $ 0\le j\le m$, this process gives us a total of $ 1\plus{}\sum_{j\equal{}0}^m 1\minus{}D(v_{k\plus{}j})\equal{}m\plus{}2\minus{}\sum_{j\equal{}0}^m D(v_{k\plus{}j})\le m$ interior vertices on the path.
These $ \le m$ interior vertices together with $ v_{k\minus{}1}$ and $ v_{k\plus{}m\plus{}1}$ form a path of length $ \le m\plus{}1$, which cuts off the exterior path of length $ m\plus{}2$. The outside part is thus a triangulation of a polygon with no more than $ n\minus{}1$ sides, which is admissible since all of its interior vertices were interior vertices of the original polygon. The part we cut off is a triangulation of a polygon with no more than $ 2m\plus{}3$ sides using no interior vertices- this takes at most $ 2m\plus{}1$ triangles.

This would also benefit from a picture of the cutoff I'm describing.

Here's an approach I thought of that only uses graph theoretic ideas. There is no induction. (It's not a complete solution, but I think it shouldn't be that hard to finish.)

Let $ G \equal{} (V, E)$ be a graph with vertices $ V$ and edges $ E$. Let $ \text{Cr}(G)$ be the minimal number of crossings in a planar embedding of $ G$.

First we will show that $ \text{Cr}(G) \ge |E| \minus{} 3|V| \plus{} 6$. Euler's formula tells us that $ |V| \minus{} |E| \plus{} |F| \equal{} 2$ if the graph is planar. But we know that each 'face' has at least three edges bordering it, and each edge is counted exactly twice, so we have $ |E| \ge \frac{3}{2}|S|$. So if the graph is planar, then $ |V| \minus{} |E| \plus{} 2/3|E| \ge 2$, or $ |E| \minus{} 3|V| \plus{} 6 \le 0$. Now if we have an embedding of $ G$ with the minimal number of crossings, we can delete one edge in order to reduce the number of crossings by 1. If $ |E| \minus{} 3|V| \plus{} 6 \equal{} k$, we could delete an edge and remove at least one crossing, and since the expression is now $ k\minus{}1$ there are still crossings left. The result follows.

Now let's pick a random subgraph $ G' \subset G$. Each vertex is in $ G'$ with probability $ p$ (to be specified later), and the vertices are chosen independently. An edge is in $ G'$ if both of its endpoints were chosen. It's easy to see that
$ \mathbb{E}[|V'|] \equal{} p|V|$
$ \mathbb{E}[|E'|] \equal{} p^2|E|$
$ \mathbb{E}[\text{Cr}(G')] \equal{} p^4\text{Cr}(G)$
since an edge requires 2 vertices to be chosen and a crossing requires 4. Applying our inequality to $ G'$, we have
$ p^4\text{Cr}(G) \ge p^2|E| \minus{} 3p|V| \plus{} 6$.

Now, let's consider our particular case. We want to show that if there are too many interior vertices, then the graph actually couldn't be planar, which means it wasn't actually a legit admissible triangulation. Let $ |E| \equal{} n \plus{} k$, where the $ n$ are the edges on the outside of the polygon and $ k$ are the interior edges. There are $ (n \plus{} 2k)/3$ triangles, since each interior edge is involved in two triangles and each exterior one is involved in one triangle, and each triangle has three edges. Now each triangle has 3 vertices, so there are $ n \plus{} 2k$ vertices counting multiplicity. Suppose that exterior vertices are involved in $ \alpha$ triangles on average, and interior ones are involved in $ \beta$ on average. Then $ \alpha n \plus{} \beta m \equal{} n \plus{} 2k$, where $ |V| \equal{} n \plus{} m$. We have $ m \equal{} ((1\minus{}\alpha)n \plus{} 2k)/\beta$.

$ p^4\text{Cr}(G) \ge p^2(n \plus{} k) \minus{} 3p(n \plus{} m) \plus{} 6$
$ p^4\text{Cr}(G) \ge (p^2 \minus{} 3p \plus{} 3p(\alpha\minus{}1)/\beta)n \plus{} (p^2 \minus{} 6p/\beta)k \plus{} 6$

We know $ \alpha \ge 1, \beta \ge 6$. Suppose that $ \beta > 6$. Then we can choose $ p > 6/\beta$, and the RHS of the above expression will become positive if $ k$ is big compared to $ n$. (An explicit inequality can be written.)

So we have reduced to the case $ \beta \equal{} 6$. In this case the coefficient of $ k$ is nonpositive in our inequality, so this approach is ineffective. I don't know what to do with this, but I haven't really thought about it (I didn't think of this approach during the exam).

The interesting case is of course exactly $ \beta\equal{}6$. See the example I mentioned of a triangular lattice.
I don't think your approach does anything- the only thing we get there is my lemma, counting edges between exterior and interior vertices.


Solve in integers the equation $ xy(x^2\plus{}y^2)\equal{}2z^4$

Multiply by 8 to have $ 8xy(x^2\plus{}y^2)\equal{}(2z)^4$ or equivalently $ (x\plus{}y)^4\minus{}(x\minus{}y)^4\equal{}(2z)^4$

or $ (2z)^4\plus{}(x\minus{}y)^4\equal{}(x\plus{}y)^4$  which has no trivial solution due to the last theorem of Fermat  :)

Have a problem with your solution.
$ (x,y)\equal{}(t,t)$ is solution.
Use the equation
$ x^4\plus{}y^4\equal{}2z^4$
It has solution $ x\equal{}y\equal{}z$

But Fermat's equation has solutions only if $ x\minus{}y\equal{}0$ or $ 2z\equal{}0$.
Case 1: If $ x\minus{}y\equal{}0$, then $ x\equal{}y$ and $ z\equal{}\plus{}\minus{}x$ So the solutions are $ x\equal{}y\equal{}m$ and $ z\equal{}\plus{}\minus{}m$
Case 2: If $ z\equal{}0$ then $ (x\minus{}y)^4\equal{}(x\plus{}y)^4$ and so $ x\equal{}0$ or $ y\equal{}0$.
So the solutions are $ x\equal{}z\equal{}0$ and $ y\equal{}m$
$ x\equal{}m$ and $ y\equal{}z\equal{}0$

[quote="silouan"]  [b]which has no trivial solution[/b][/quote]

I have mentioned it  :wink:   . Non trivial means non zero(all)

[i]Sorry! Didnt want to upset you :lol: [/i]


Yeah, good luck and have fun everybody!  :lol:

Good Luck! :lol:
Can someone post the problems here after the deadline passes?

Sure, will do.
The deadline is next Monday, so you can probably expect the questions to be posted by Tuesday night.

I'm assuming you're talking about the Repechage.  Where do you get the problems?  Someone recently posted the problems (trying to get help), and I'd like to make sure it doesn't happen again.

What is the usual number of questions/the mark you need to get in CMOR order to qualify for the CMO?
At least 9 questions? (45/50)
I've heard that the 40-50 range people usually are almost all accepted.. is this true?

Well, I heard that last year's cutoff was around 42...

phew.. finished writing my solutions for 8 of the questions.. trying to solve the other 2 now.. :S
6:10 am here...
if there's anyone here that took the CMOR last year.. is this year's harder than last year's?

[quote="un_o"]
if there's anyone here that took the CMOR last year.. is this year's harder than last year's?[/quote]

Well, I took it last year, and got invited, and I have looked at the problems from this year, and they are almost the same difficulty, but this year's problems seem to require less previous knowledge than last years, i.e. more ingenuity...
 :lol:

I finally sent the solutions!!!! :D
SO, ... do we wait till tomorrow to discuss?
or is there no discussion at all?
btw... my solutions for 8 and 10 are crappy...-_-

I sent my solutions, but I only did 7 questions because I had too much homework.:( And maybe also because I only learned LaTex last week, so everything took insanely long to type.

can we discuss the questions from tomorrow?

I was able to solve #1 - #7 in about 2 hours

8 - 10  are a bit harder.....I solved them eventually and I handed it in to Mr.Ian VanderBurgh personally :P

The questions were generally harder than last year's, though they were still pretty easy. Since the due date has past, I am assuming I can discuss my solution for CMOR....

So, for #10, I just had to prove that the 'state' (or configuration of the balls) were arranged as a DAG. Since every state had a unique 'child' state (from the given), all I have to prove is that every state has 'a' (not necessarily one) parent state. Since proving this would mean that every state MUST have only ONE parent state (by means of the pigeonhole theory).

This means that the the initial state is part of some loop of states, and therefore it will be eventually returned to (since it has only ONE child and ONE parent state)

As for #9, it is formulating a formula using recursion, and then finding the general form of it.

#8 was just sub in powers of 3, and find that the exponents must be 6 mod 7, and then find an infinite family of powers of 3 that work.

#1 - #7 were pretty trivial, so I won't bother explaning my solutions as I am expecting that most people completed at least 1 - 7.

Wrong topic. See [url=http://www.mathlinks.ro/viewtopic.php?t=251132]this one[/url].


It's kind of funny when I found out that this series book was made specifically for my college: Rock Valley College.

What's sad, is that I can't find the complete solutions book.
This is very annoying.

The math book I'm working from is called:

[u]Beginning & Intermediate Algebra
An Integrated Approach[/u]
By: Gustafson and Frisk

I think this book is first edition. It has a copyright of 1996.

Now I have the book to work off. That's fine with me. But what isn't fine is not having the complete solution set. I can't even find it on the internet. I look for about an hour now on half.com, ebay, google, the publisher's website, other math websites, and I just can't find this one book that is suppose to have the answers. It seems as if the book doesn't exist.

The Book is suppose to be something like an Ancillary for the Instructor. Well, I'm teaching myself so I consider myself to be the instructor.

The book i'm looking for is called:

[u]Complete Solutions Manual
[/u] by: George Grisham

I'm hoping this would the first edition for the book I have with me. All I keep finding is the second edition of things. I need the first. Can someone help me?

Hi Windex, Everyone,

Publishers generally make it difficult to obtain the complete solution manuals to textbooks.  This is because if copies are made of a solution manual, this impacts their ability to make money off a textbook.  To get a copy of the solutions manual, you’ll most likely have to contact the textbook adoption salesperson at the publisher directly.  Then you’ll have to convince this person to sell you the full solutions manual.  I anticipate that this will be quite difficult.  You may need to get a Rock Valley College instructor to vouch to the textbook adoption salesperson on your behalf.

Good luck.

Yours truly,

M.E.B.


geometry
3D geometry
pyramid
tetrahedron
analytic geometry
trigonometry
vector

Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$  The vertices of its midpoint triangle are the midpoints of its sides.  A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle.  What is the volume of this pyramid?

Denote A' = (0, 0), B' = (34, 0), C' = (16, 24). The sides of the triangle $\triangle A'B'C'$ are $c' = A'B' = 34,$ $b' = C'A' = \sqrt{256 + 576} = 8 \sqrt{13},$ $a' = B'C' = \sqrt{324 + 576} = 30$. Let A, B, C be the midpoints of the sides B'C', C'A', A'B' forming the medial triangle $\triangle ABC$ with the sides $BC = a = \frac{a'}{2} = 15,$ $CA = b = \frac{b'}{2} = 4 \sqrt{13},$ $AB = c = \frac{c'}{2} = 17$. Denote $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ the internal angles of this triangle. Let D be the fold point of A', B', C', completing the tetrahedron ABCD with the opposite pairs of sides equal to a, b, c. Ignore the original coordinates of the points A', B', C' and choose the new coordinate system with origin at the point A, the positive x-axis along the ray (AB, and WLOG, assume that the z-coordinate of the point D is positive. Denote $\vec c = \vec{AB},\ \vec b = \vec{AC},\ \vec a = \vec{AD}$. The tetrahedron volume V is equal to

$V = \frac 1 6\ \vec a \cdot (\vec c \times \vec b) = \frac{abc}{6}\ \sin \alpha \cos \phi$

where $\phi$ is the angle between the vector $\vec a$ and the normal to the plane of the triangle $\triangle ABC$. We have to find this angle. Let $\vec n_c,\ \vec n_b,\ \vec n_a$ be unit vectors in the directions $\vec c,\ \vec b,\ \vec a$. Obviously, $\vec n_c = (1, 0, 0)$ and $\vec n_b = (\cos \alpha,\ \sin \alpha,\ 0)$. Let $\vec n_a = (x, y, z)$. Since $\widehat{(\vec n_c, \vec n_a)} = \beta,$ $\widehat{(\vec n_b, \vec n_a)} = \gamma,$ and since the endpoints of all 3 unit vectors are on a unit sphere centered at the origin A, we can write the following 3 equations for x, y, z:

$|\vec n_a|^2 = x^2 + y^2 + z^2 = 1$

$|\vec n_a - \vec n_c|^2 = (x - 1)^2 + y^2 + z^2 = 4 \sin^2 \frac \beta 2$

$|\vec n_a - \vec n_b|^2 = (x - \cos \alpha)^2 + (y - \sin \alpha)^2 + z^2 = 4 \sin^2 \frac \gamma 2$

Substituting $z^2 = 1 - x^2 - y^2$ to the 2nd equation, we get

$-2x + 1 + 1 = 4 \sin^2 \frac \beta 2,\ \ x = 1 - (1 - \cos \beta) = \cos \beta$

Substituting for $z^2$ and x into the 3rd equation yields

$1 - 2x \cos \alpha + \cos^2 \alpha - 2y \sin \alpha + \sin^2 \alpha = 4 \sin^2 \frac \gamma 2$

$y \sin \alpha = 1 - (1 - \cos \gamma) - x \cos \alpha = \cos \gamma - \cos \alpha \cos \beta$

$y = \frac{\cos \gamma - \cos \alpha \cos \beta}{\sin \alpha}$

Since $z = \cos \phi,$ we obtain

$\cos^2 \phi = z^2 = 1 - x^2 - y^2 = 1 - \cos^2 \beta - \frac{(\cos \gamma - \cos \alpha \cos \beta)^2}{\sin^2 \alpha}$

$\sin^2 \alpha \cos^2 \phi = \sin^2 \alpha - \cos^2 \beta \sin^2 \alpha - (\cos^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma + \cos^2 \alpha \cos^2 \beta) =$

$= 1 - \cos^2 \alpha - \cos^2 \beta (1 - \cos^2 \alpha) - \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma - \cos^2 \alpha \cos^2 \beta =$

$= 1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma$

Since $\alpha + \beta + \gamma = 180^\circ$,

$\cos^2 \gamma = \cos^2 (\alpha + \beta) =$

$= \cos^2 \alpha \cos^2 \beta - 2 \cos \alpha \cos \beta \sin \alpha \sin \beta + \sin^2 \alpha \sin^2 \beta =$

$\cos^2 \alpha \cos^2 \beta - 2 \cos \alpha \cos \beta \sin \alpha \sin \beta + (1 - \cos^2 \alpha)(1 - \cos^2 \beta) =$

$= 1 - \cos^2 \alpha - \cos^2 \beta + 2 \cos \alpha \cos \beta (\cos \alpha \cos \beta - \sin \alpha \sin \beta) =$

$= 1 - \cos^2 \alpha - \cos^2 \beta + 2 \cos \alpha \cos \beta (\cos(\alpha + \beta) =$

$= 1 - \cos^2 \alpha - \cos^2 \beta - 2 \cos \alpha \cos \beta \cos \gamma$

and consequently,

$\sin^2 \alpha \cos^2 \phi = 4 \cos \alpha \cos \beta \cos \gamma$

Using the cosine theorem formulas for $\cos \alpha,\ \cos \beta,\ \cos \gamma$, we get square of the tetrahedron volume in terms of the triangle sides a, b, c:

$V^2 = \frac{a^2b^2c^2}{36} \cdot \frac{(b^2 + c^2 - a^2) (c^2 + a^2 - b^2) (a^2 + b^2 - c^2)}{2a^2b^2c^2}) =$

$= \frac{1}{72}\ (b^2 + c^2 - a^2) (c^2 + a^2 - b^2) (a^2 + b^2 - c^2)$

Using the [url=http://mathworld.wolfram.com/Cayley-MengerDeterminant.html]Cayley-Menger determinant[/url] for a tetrahedron with the edges $d_{AB} = d_{CD} = c,\ d_{AC} = d_{BD} = b,\ d_{AD} = d_{BC} = a$ would yield the same result. Plugging in the numerical values:

$a^2 = 225,\ b^2 = 208,\ c^2 = 289$

$a^2 + b^2 - c^2 = 433 - 289 = 144$

$b^2 + c^2 - a^2 = 497 - 225 = 272$

$c^2 + a^2 - b^2 = 514 - 208 = 306$

$V^2 = \frac{144 \cdot 272 \cdot 306}{72} = 2 \cdot 272 \cdot 306 = 2^6 \cdot 3^2 \cdot 17^2$

$V = 8 \cdot 3 \cdot 17 = 408$

[hide="Perhaps a more accessible solution..."]suppose we have $A=(0,0)$, $B=(34,0)$, and $C=(16,24)$
$AB=34$, $BC=30$, and $AC=8\sqrt{13}$

Let the midpoints of $AB$,$BC$,and $CA$ be $D$,$E$, and $F$, respectively. Using the midline theorem, we get $DE=4\sqrt{13}$, $EF=17$, $FD=15$

Notice how the triangle has all half sides, so if we dialate the original coordinate system by $\frac{1}{2}$, we will have a triangle with half lengths.

This gives us a second cooridnate system with the new cooridinates: $D=(8,12)$, $E=(0,0)$, and $F=(17,0)$. Make sure you see that I did not just dialate the old points into the new ones, I made a triangle with half side lengths, and assigned the points $D$, $E$, and $F$ so they were the correct points.

Suppose our folded pyramid is $TDEF$ where $T$ is the vertex where $A$, $B$ and $C$ coincide. Along with $DE=4\sqrt{13}$, $EF=17$, $FD=15$, we have $TD=17$, $TE=15$, and $TF=4\sqrt{13}$

Draw height from $T$ to $\triangle DEF$, let this be $P$.

By the pythaogrean theorem, we have $h^2=17^2-PD^2=15^2-PE^2=(4\sqrt{13})^2-PF^2$

Let $P: (x,y)$ on our second cooridinate system. Using the distance formula, we get:

$h^2=17^2-\left[ (x-8)^2+(y-12)^2 \right]=15^2-\left[ x^2+y^2 \right]=(4\sqrt{13})^2-\left[ (x-17)^2+y^2 \right]$
using the 2nd and 4th part of the equality, we get $x=9$, and plugging that in to the second  and third parts, we get $y=0$, therefore we get $h^2=15^2-9^2-0^2=12^2 \implies h=12$

The area of $\triangle DEF=\frac{1}{2}17*12=102$

Using Volume$=\frac{1}{3}{b}{h}$, we get $V=\frac{1}{3}*102*12=\boxed{408}$[/hide]

A perhaps faster solution would be to notice that the projection of the top vertex of the tetrahedron on the plane formed by the base is the orthocenter of the biggest triangle with the given vertices; its coordinates can be found without much calculations (just find the intersection of any two altitudes). It turns out that the orthocenter is on one of the sides of the triangle, and the height of the tetrahedron follows.

Perhaps a simpler solution:

It is easy to see that the apex of the resulting pyramid will be directly on top of the orthocenter of the original triangle. First we find the coordinates of H. H is the intersection of $AH_A$ and $BH_B$ (where $H_A$ and $H_B$ are the foot of the altitudes from $B$ and $A$ and $A,B,C$ are the vertices of the original triangle.) The equation for $AH_A$ is $x=16.$ The slope for line $AC$ is $(24-0)/(16-34) = -4/3.$ Thus, the equation for $BH_B$ is of the form $y=(3/4)x + b,$ but since it passes through (0,0) b=0, so $y=(3/4)x.$

Thus, the coordinates for $H$ is $(16,12).$ So H lies on $B'C'$ where $A',B',C'$ are the midpoints of $BC,AC,AB$ respectively. $AH=12$ clearly, as it is $1/2$ the corresponding altitude of $ABC,$ which is 24. This is also the height of the pyramid. Now, $B'C' =17$ clearly, and since the altitude to $B'C"$ of triangle $A'B'C'$ is of length 12, its area is $17(12)/2 = 102.$ So, the volume of the pyramid is $1/3(102)(12) = 4(102) = 408.$

yetti with your amazing solutions.  :wink:

[quote="fire123"]Sorry to revive, but why is it obvious that the orthocenter is the apex of the pyramid?[/quote]

Bump. I would like to know also. I assumed the apex would be above the centroid.

Reviving to answer a question: Let ABC be the triangle, and D,E,F be the midpoints opposite to A,B,C.

Then when you fold A along EF, its projection to plane ABC is perpendicular to EF (and BC by extension). So it lies on the height from A to BC, and by similar logic it's on the orthocenter.


Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ be an increasing function. There are positive constants $ c_1,c_2$ such that $ f(x)\plus{}f(y)\minus{}c_1\leq f(x\plus{}y)\leq f(x)\plus{}f(y)\plus{}c_2$ for all $ x,y\in \mathbb{R}$.

Must there exist a real number $ k$ such that $ f(x)\minus{}kx$ is a bounded function?


[quote="GameBot"]Two golf carts start toward each other from opposite ends of a 1.8 mile road. The cart going 30 mph meets the other cart in 2 minutes. How many miles per hour was the other cart going?[/quote]

An important formula for problems like this is distance = rate x time. We know that the collective distance the two carts traveled is 1.8 miles, so Cart A's rate x time + Cart B's rate x time = 1.8. They met in 2 minutes, so for both of them, time = $ \frac{1}{30}$(2 minutes is 1/30 of an hour.) Cart A's rate was 30 mph, so Cart A's rate = 30. We can then plug in our values and solve for Cart B's rate, $ r$.

\[ 30 \times \frac{1}{30} \plus{} r \times \frac{1}{30} \equal{} 1.8 \implies r \equal{} \boxed{24} \]


[quote="frank44"]So how did everyone do?  I found it easier this year, but I ended up with a lot wrong while trying to go faster :(  I got a 60.[/quote]

i thought that it was harder than previous years...luckily i didnt make too many silly mistakes so i got a 79

[quote="vishalarul"]I got 57, but it's recorded as a 55 on the website... Lucky for me, it didn't affect my rank anyway (the person above me was supposed to get 58).[/quote]

they gave me a 78...what a bunch of losers :(

My score went up two points for some reason  :), but I still ended up only 24th in the 11th graders (does anyone know why they chose to put me at the bottom of the people who got 53s?

I think if your scores tie, it's alphabetized by your school's name


An integer $ n$ is randomly chosen from 1 to 100 inclusive.  Given that $ n$ is divisible by 4, what is the probability that $ n$ is divisible by 11?

There are 25 numbers divisible by 4 between 1 and 100 inclusive. Of these, only 2, 44 and 88, are divisible by 11 also. Thus, our answer is $ \boxed{\frac{2}{25}}$.

No, the 22 or 2 thing is just a typo. So the numbers are 44 and 88, and there are 25 total, so akhil's final answer $ 2/25$ is correct.

He meant that "only 2 numbers, 44 and 88,..."


Welcome to MathLinks forum participants from an AoPS forum moderator. It's good to see the combined forces board bringing people together from around the world. I enjoy the international perspective here because because I have lived outside the United States (mostly in Taiwan) for six years of my life.

I will try to round up more EASY problems for this Getting Started Forum. Now there are more high-level forums than ever before on this AoPS board, with the new combined forum list, so I will try to make this forum welcoming to very new beginners in problem-solving: people at the typical elementary (primary) school level in east Asia or eastern Europe, which is the junior high (middle school) level in the United States. People get better so fast when they get deeply into problem-solving that they soon forget how to pose EASY problems, so if you have some really easy problems for beginners, please post them here, with the title "Challenge Problem:" and encourage the least experienced participants to solve them.

I just moved a couple of threads called "Challenge Problems" to the Intermediate forum.

Students unfamiliar with this categorization of problems should know that the Getting Started forum should not get harder than easier AMC-10 problems.  Harder AMC-10 and AMC-12 problems, most AIME, most ARML, and similar problems should reside in the Intermediate forum.

The Advanced forum and the Mathlinks forums should house Olympiad level material.


Ar putea cineva sa posteze barajul de la Iasi? :)

[hide="prea tarziu"]1. Sa se rezolve in $Z$ ecuatia: $x^{3}-y^{3}=2xy+7$.
2.Fie $a\geq 2$ si sirul $x_{n}=1+a+a^{2}+...+a^{2n}$. Sa se arate ca exista o infinitate de numere prime care nu divid nici un termen al sirului.
3.nu mai tin minte exact enuntul
4.$n\geq 2$. Sa se arate ca oricum am colora cu $2$ culori o multime cu $\frac{n^{3}+5n}{6}$ numere consecutive, exista o submultime monocolora cu $n$ elemente astfel incat
$1\leq a_{2}-a_{1}\leq a_{3}-a_{2}\leq ... \leq a_{n}-a_{n-1}$.[/hide]

impresionanti astia de a 10-a de iau premii !

Padurariu Tudor a luat prima mentiune. voiai premiu I nu ? eh, asta e , sa nu te sinucizi din cauza asta !

1. Obs. ca daca $x^{3}-y^{3}= 2xy+7$, atunci $(-y)^{3}-(-x)^{3}= 2(-x)(-y)+7$, deci WLOG $x > 0$ ($x=0$ oricum nu merge si daca $x<0, y>0$ putem folosi inegalitatea mediilor).
Daca $x \leq y$, atunci nu exista solutie. Sa presupunem ca $x > y$. Rezulta ca $x^{3}-y^{3}\geq x^{3}-(x-1)^{3}$ si ca $2xy+7 \leq 2x(x-1)+7$. Asadar, $2x(x-1)+7 \geq x^{3}-(x-1)^{3}$, i.e. $x^{2}-x-6 \leq 0$. Singurii "icşi" care merg sunt $x=1,2,3$. Acuma introducem inapoi in ecuatie si obtinem singura solutie $(x,y) = (3,2)$. "Opusul" ei e $(x,y) = (-2,-3)$.

2. Fie $p > a$, astfel incat $a \not\equiv 1 \left( \bmod \, p \right)$. Observam ca $x_{n}\not\equiv 0 \left( \bmod \, p \right)$ daca si numai daca $(a-1) x_{n}\not\equiv 0 \left( \bmod \, p \right)$ daca si numai daca $a^{2n+1}\not\equiv 1 \left( \bmod \, p \right)$. Altfel zis, noi vrem ca $\text{ord}_{p}(a)$ sa fie par. De asemenea, stim ca daca $a^{k}\equiv 1 \left( \bmod \, p \right)$, atunci $\text{ord}_{p}(a) | k$. Daca vrem sa-l obligam pe $\text{ord}_{p}(a)$ sa fie par, nu exista prea multe variante pentru alegerea lui $k$. Zis mai de-a dreptul, luam $k = 2^{n}$. Se vede cu ochiul liber ca $m \leq n \, \Longleftrightarrow \, \, a^{2^{m}}-1 | a^{2^{n}}-1$, si de aici nu e deloc greu de dedus ca exista o infinitate de prime care divid numere de forma $a^{2^{n}}-1$ (altfel divizorii numerelor de forma $a^{2^{n}}+1$ ar fi si ei in numar finit => cand se face trecerea de la $n$ la $n+1$ pentru $n$ mare, singurul divizor prim al lui $a^{2^{n}}+1$ poate sa fie $2$ => $a^{2^{n}}+1 = 2^{l}$ => $a^{2^{n+1}}+2 a^{2^{n}}+1 = 2^{l+1}$, dar $a^{2^{n+1}}+1 = 2^{j}$ => $a^{2^{n}}= 2^{l}-2^{j-1}$ => $j=1$, contradictie).

PS: Imi cer scuze pentru lungimea parantezei, dar la inceput am crezut ca iese mult mai usor ultima parte, dupa care am descoperit o greseala...

exact asa au rezolvat`o majoritatea (pe problema 2)

eu am avut o solutie oarecum diferita.
ordinul lui $a$ e par daca si numai daca $|<a>|\subseteq \mathbb{Z}_{p}$ e par, si de aici rezulta ca $-1 \in <a>$.
pentru numere de forma $a^{s}+1$, daca iau $p|a^{s}+1$, $p>2$, atunci rezulta imediat ca $p$ este unul "bun", pt ca din $a^{s}\equiv-1$ si $a^{2n+1}\equiv 1$, rezulta $-1 \equiv (a^{s})^{2n+1}\equiv 1$, contradictie.
acum, sirul $a^{s}+1$ are o infinitate de divizori primi, pentru ca daca ar avea un numar finit $p_{1},p_{2},\ldots,p_{m}$, iau $q=\prod (p_{i}-1)$, si atunci numarul $a^{q}+1$ este congruent cu $2$ modulo toate $p_{i}$`urile. rezulta ca e de forma $2^{k}$ si ma uit si la $a^{2q}+1$ si analog asta e un $2^{l}$. insa chestia asta nu merge modulo $4$.

sau poate ca pur si simplu ai demonstrat gresit si nu iti dai tu seama unde. sunt destui cei care zic ca au facut tot dar corectorii sunt de alta parere

Sunt multe solutii frumoase pt problema 3. In afara de aia de pe site, ia $M$ mijlocul lui $AC$. Vezi ce iti iese :P

probabil ca aia de pe site e si asta care o postez io acu. insa asta am dat`o si in concurs.
[hide="solutie 3"]
fie $K$ intersectia dintre cercul circumscris triunghiului $\triangle ABE$ si dreapta $AC$. atunci $K$ este ortocentrul triunghiului $\triangle BDE$.
asta era partea "dificila" ca sa zic asa (ghicirea punctului). de aici, iese doar cu unghiuri (se foloseste faptu ca $ABKE$ si $BCDK$ sunt inscriptibile).
[/hide]

imi scapa mie ceva sau subiectele (de pe site) au extensia .exe ?

extensia este .exe dar ele nu sunt aplicatii . Trebuie doar sa schimbi extensia din .exe in .rar si sa le dezarhivezi normal apoi . ( Merge ... eu asa am facut )

:D aham. mersi. am incercat si eu .pdf si .doc dar n`a mers

imi place ca toti pe site din Bucuresti sunt din Vianu  :D
in realitate decat maky, pohoatza, cred ca sunt, dar vianu on top:D


I performed an expeiment in which i have to find themoles of water of crystallization in washing soda . So i got .30 g of washing soda, dissolved it in 30cmcubes of water and titrated this with 0.1M HCl (using methyul orange as indicator).  The solution changed colour when i added 25.80 cmcubes of  HCl.

I need help in the calculations after this, i have no idea what to do  :(

The titration reaction is

$ Na_2CO_3(aq) \plus{} 2HCl(aq) \longrightarrow 2NaCl(aq) \plus{} H_2O(l) \plus{} CO_2(g)$

and so in the equivalence point we have $ n(HCl) \equal{} 2 \cdot n(Na_2CO_3) \equal{}> n(Na_2CO_3) \equal{} \frac{[HCl] \cdot V(HCl)}{2} \equal{}$

$ \equal{} \frac{0.1\,mol/L \cdot 25.80 \times 10^{\minus{}3}\,L}{2} \equal{} 1.29 \times 10^{\minus{}3}\, mol$.

But:  $ n(Na_2CO_3) \equal{} \frac{m(Na_2CO_3)}{M(Na_2CO_3)} \equal{}> m(Na_2CO_3) \equal{} 1.29 \times 10^{\minus{}3}\, mol \, \cdot 105.99\, g/mol \equal{}$
$ \equal{} 0.137\,g$.

Hence, the mass of water of crystalization is m = 0.30 - 0.137 = 0.163 g, which corresponds to about 0.0102 moles of water.


A three-digit number $ \text{``}abc\text{''}$ has the property that the product of $ a$ and $ b$ is equal to $ c$. What is the greatest possible value of the three-digit number?

For the number to be as big as possible, $ a$ should be 9. Then since $ c<10$, $ b$ must equal $ 1$ and $ \text{``}abc\text{''}\equal{}919$.

Although the problem is not very clear, the digits do not have to be different. Since a is the hundreds place, it has to be the largest unit number, which is 9, the largest one digit. (sry) Then what is the largest number that 9 multiplies by that is only a one digit number. Well, 9 times 1 equals 9. 9 times 2 equals 18. Too large. Therefore, "abc" must equal 919.

am i understanding this wrong, because a*c=81 but b=1


trigonometry
complex analysis
calculus
calculus computations

Prove that
\[ \frac{1}{\cosh (\pi /2)}\minus{}\frac{1}{3\cosh (3\pi /2)}\plus{}\frac{1}{5\cosh (5\pi /2)}\minus{}\dots\equal{}\frac{\pi}{8}\]

With residue computation of $ f(z) \equal{} \frac {\cot z\, \coth z}{z^{4n \minus{} 1}}$

we can show that $ \sum_{k \equal{} 1}^\infty \frac {\coth(k\pi)}{(k\pi)^{4n \minus{} 1}} \equal{} \sum_{k \equal{} 0}^{2n} ( \minus{} 1)^{k \minus{} 1}\, \frac {\zeta(2k)}{\pi^{2k}}\, \frac {\zeta(4n \minus{} 2k)}{\pi^{4n \minus{} 2k}}$ .

We can do the same with

$ f(z) \equal{} \frac {\csc z\, \text{csch}\, z}{z^{4n \minus{} 1}} \; : \; \sum_{k \equal{} 1}^\infty \frac {( \minus{} 1)^k\,\text{csch}(k\pi)}{(k\pi)^{4n \minus{} 1}} \equal{} \sum_{k \equal{} 0}^{2n} ( \minus{} 1)^{k \minus{} 1}\, \frac {\eta(2k)}{\pi^{2k}}\, \frac {\eta(4n \minus{} 2k)}{\pi^{4n \minus{} 2k}}$

$ f(z) \equal{} \frac {\tan z\, \tanh z}{z^{4n \minus{} 1}} \; : \; \sum_{k \equal{} 0}^\infty \frac {\tanh\left((2k \plus{} 1)\frac {\pi}{2}\right)}{\left((2k \plus{} 1)\frac {\pi}{2}\right)^{4n \minus{} 1}} \equal{} \sum_{k \equal{} 0}^{2n} ( \minus{} 1)^{k \minus{} 1}\, \frac {\lambda(2k)}{\left(\frac {\pi}{2}\right)^{2k}}\, \frac {\lambda(4n \minus{} 2k)}{\left(\frac {\pi}{2}\right)^{4n \minus{} 2k}}$

$ f(z) \equal{} \frac {\sec z\, \text{sech}\, z}{z^{4n \plus{} 1}} \; : \; \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k\, \text{sech}\left((2k \plus{} 1)\frac {\pi}{2}\right)}{\left((2k \plus{} 1)\frac {\pi}{2}\right)^{4n \plus{} 1}}$$ \equal{} \sum_{k \equal{} 0}^{2n} ( \minus{} 1)^k\, \frac {\beta(2k \plus{} 1)}{\left(\frac {\pi}{2}\right)^{2k \plus{} 1}}\, \frac {\beta(4n \minus{} 2k \plus{} 1)}{\left(\frac {\pi}{2}\right)^{4n \minus{} 2k \plus{} 1}}$

where $ \eta(s) \equal{} \sum_{n \equal{} 1}^\infty \frac {( \minus{} 1)^{n \minus{} 1}}{n^s} \equal{} (1 \minus{} 2^{1 \minus{} s})\zeta(s)$ , $ \lambda(s) \equal{} \sum_{n \equal{} 0}^\infty \frac {1}{(2n \plus{} 1)^s} \equal{} (1 \minus{} 2^{ \minus{} s})\zeta(s)$ and $ \beta(s) \equal{} \sum_{n \equal{} 0}^\infty \frac {( \minus{} 1)^n}{(2n \plus{} 1)^s}$ .

In particular $ \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k\, \text{sech}\left((2k \plus{} 1)\frac {\pi}{2}\right)}{(2k \plus{} 1)\frac {\pi}{2}} \equal{} \frac {\beta(1)\,\beta(1)}{\frac {\pi}{2}\,\frac {\pi}{2}} \equal{} \frac14$ , so $ \sum_{k \equal{} 0}^\infty \frac {(\minus{}1)^k}{(2k \plus{} 1)\cosh\left((2k \plus{} 1)\frac {\pi}{2}\right)} \equal{} \frac {\pi}{8}$ .

Can you prove the following formulas?

$ \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k\, \text{sech}\left(\sqrt {3}\, (2k \plus{} 1)\frac {\pi}{2}\right)}{(2k \plus{} 1)\frac {\pi}{2}} \equal{} \frac {1}{12}$

$ \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k\, \text{sech}\left(\frac {1}{\sqrt {3}}\, (2k \plus{} 1)\frac {\pi}{2}\right)}{(2k \plus{} 1)\frac {\pi}{2}} \equal{} \frac {5}{12}$

$ \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k\, \text{sech}\left(\sqrt {3}\, (2k \plus{} 1)\frac {\pi}{2}\right)}{\left((2k \plus{} 1)\frac {\pi}{2}\right)^7} \equal{} \frac {1}{180}$

$ \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k\, \text{sech}\left(\frac {1}{\sqrt {3}}\, (2k \plus{} 1)\frac {\pi}{2}\right)}{\left((2k \plus{} 1)\frac {\pi}{2}\right)^7} \equal{} \frac {143}{27\cdot 180}$

I solved this problem using $ \text{sech} \left(\frac {\pi}{2}x\right) \equal{} \frac {4}{\pi}\sum_{n \equal{} 0}^{\infty}{( \minus{} 1)^n\frac {2n \plus{} 1}{x^2 \plus{} (2n \plus{} 1)^2}}$, which can be proved throw the Fourier series for $ \sinh (x)$. Throw it i can show that:

$ \sum_{m \equal{} 0}^{\infty}( \minus{} 1)^m\frac {1}{2m \plus{} 1}\left\{\text{sech} \left(\frac {\pi}{2}(2m \plus{} 1)\cdot\theta\right) \plus{} \text{sech} \left(\frac {\pi}{2}(2m \plus{} 1)\cdot\frac {1}{\theta}\right) \right\} \equal{} \frac {\pi}{4}$

and

$ \sum_{m \equal{} 0}^{\infty}( \minus{} 1)^m\frac {1}{(2m \plus{} 1)^7} \text{sech} \left(\frac {\pi}{2}(2m \plus{} 1)\cdot\theta\right) \equal{} \sum_{m \equal{} 0}^{3}( \minus{} 1)^m\theta^{2m}\beta(m \plus{} 1)\beta(7 \minus{} 2m) \plus{}$ $ \theta^6\sum_{m \equal{} 0}^{\infty}( \minus{} 1)^m\frac {1}{(2m \plus{} 1)^4} \text{sech} \left(\frac {\pi}{2}(2m \plus{} 1)\cdot\frac {1}{\theta}\right)$

$ \sum_{m \equal{} 0}^{\infty}( \minus{} 1)^m\frac {1}{(2m \plus{} 1)^4} \text{sech} \left(\frac {\pi}{2}(2m \plus{} 1)\cdot\frac {1}{\theta}\right) \equal{}$ $ \sum_{m \equal{} 0}^{1}( \minus{} 1)^m\frac {1}{\theta^{2m}}\beta(m \plus{} 1)\beta(4 \minus{} 2m) \plus{}$ $ \frac {1}{\theta^2}\sum_{m \equal{} 0}^{\infty}\sum_{n \equal{} 0}^{\infty}\frac {1}{(2n \plus{} 1)^2}\cdot\frac {1}{(2m \plus{} 1)^2 \plus{} \theta^2(2n \plus{} 1)^2}$

$ \sum_{m \equal{} 0}^{\infty}\sum_{n \equal{} 0}^{\infty}\frac {1}{(2n \plus{} 1)^2}\cdot\frac {1}{(2m \plus{} 1)^2 \plus{} \theta^2(2n \plus{} 1)^2} \equal{} \frac {\pi}{4\theta}\sum_{n \equal{} 0}^{\infty}\frac {\tanh\left(\frac {\pi\theta}{2}(2n \plus{} 1)\right)}{(2n \plus{} 1)^3}$

So

$ \sum_{m \equal{} 0}^{\infty}( \minus{} 1)^m\frac {1}{(2m \plus{} 1)^7} \text{sech} \left(\frac {\pi}{2}(2m \plus{} 1)\cdot\theta\right) \equal{} \sum_{m \equal{} 0}^{2}( \minus{} 1)^m\theta^{2m}\beta(m \plus{} 1)\beta(7 \minus{} 2m) \minus{}$ $ \theta^{4}\beta(2)^2 \plus{}$ $ \frac {\pi\theta^3}{4}\sum_{n \equal{} 0}^{\infty}\frac {\tanh\left(\frac {\pi\theta}{2}(2n \plus{} 1)\right)}{(2n \plus{} 1)^3}$

I should learn complex analysis first to prove your results :(, i had one year math at university :read:


LaTeX
combinatorics unsolved
combinatorics

We call a cross the shape obtained by removing the four corners from a  grid. How many non-overlapping crosses can be placed on a  chessboard, if each cross must cover exactly  squares?

This problem was proposed in [url=http://www.mathlinks.ro/Forum/topic-18483.html]Chinese Girls[/url]

[quote="kueh"]you missed the latex bits - 3x3 grid, 10x11 chess board, 5 squares.[/quote]

Yes. You are right.  :blush:

Wrong Message. I don't know how to delete this message.


function
real analysis
real analysis unsolved

How can I find the infinite sum of this:

(-1)^N * pi^(2N+1) divided by 4^(2N+1) * (2N+1)!

I know it's a series of some sort but I have no clue as to how to solve it. If I wanted to use Lagrange remainders I would have to find a function whose series is shown above.

The solution is to recognize the series. This is $ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{(2n\plus{}1)!}x^{2n\plus{}1}$, evaluated at $ x\equal{}\frac{\pi}{4}$. That is a power series you should have memorized.

Thank you. It's a sinus, therefore I can solve the whole thing more easily now!


calculus
integration
function
trigonometry
calculus computations

Thank you for your reply,,liyi.

That's true. :) Here is my idea.

$\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)'=\frac{4}{(e^x+e^{-x})^2}$.

kunny

Note that we can rewrite $\frac1{(e^x+e^{-x})^2}$ as $\frac14\mathrm{sech}^2x.$

And of course, $\int\mathrm{sech}^2x\,dx=\tanh x+C.$

(Which is exactly what kunny said.)

hmm.
i don't like reciting the integration formulae of hyperbolic functions  :( ... and i never do that  ;)
I always use the integration techniques to compute them...

On the one hand, defining the hyperbolic functions and inverse hyperbolic functions does not enlarge the list of "elementary function"; it just gives you alternate names for functions that you already had names for. But they have strong analogies (with occasional sign changes) with the trigonometric functions, and therein lies their power. Of course you don't have memorized what the integral of $\mathrm{sech}^2x$ is - but you do know that the integral of $\sec^2x$ is  $\tan x.$ With that, we have a plausible - and easily checked - guess about the integral of $\mathrm{sech}^2x.$ It's not a matter of memorizing more; it's a matter of being willing to use analogies like that.

There are still some slight differences.


Let $p>5$ is a prime number.
Let $A$is product of all odd  positive integer number in $[p+1;3p-1]$
Let $B$is product of all even  positive integer number in $[p+1;3p-1]$
Prove that    $1)$   $A \equiv 2^{p-1}.(p-1)!$  $(modp^3)$
                   $2)$   $2^{p-2}.B \equiv p!$  $(modp^4)$


1. $(x-m)^2+(y-n)^2=r^2$

2. $C(0,m\sqrt{3})$

3. Check [url=http://mathforum.org/dr.math/faq/formulas/faq.ag2.html#twotriangles]this page[/url].

[quote="sen"]hey guys,

sorry im just blank at the moment

and i need help on coordinate geometry

1) I was wondering firstly, how to graph a cirlce..

Let us say a circle with a radius of r when $(0, 0)$ is the centre, how about when $(m, n)$ is the centre?

If r is the radius and $(0, 0)$ is the centre, then we would have $x^2 + y^2 = r^2$ i think..

2) How would you plot an equilateral tiangle $\triangle ABC$ if $A(-m, 0) B(m, 0)$ what would the coordinates of C be?

3) Any particular formula to determine the coordinates of the incentre of circumcentre of a triangle?

thank you everybody[/quote]
[hide="2"]
well since it is equilateral, the length between A and B is 2m and the vertical distance to C is $m\sqrt{3}$ since the half triangle is 30-60-90 and the length opposite to the 30 degree angle is m. So $C(0, m\sqrt{3})$ or you could plot it below the x axis which is $C(0, -m\sqrt{3})$[/hide]

1)
For a circle centered at (k,l) with radius r the equation is:
\[ (x - k)^2 + (y - l)^2 = r^2  \]

2)
\[ C = (0,\sqrt{3m} )  \]
(Use that height of equilateral triangle is\[ \frac{{\sqrt 3 }}{2}  \]times the base.

3) For a triangle with vertices \[ \begin{array}{l} A = (x_1 ,y_1 ) \\ B = (x_2 ,y_2 ) \\ C = (x_3 ,y_3 ) \\ \end{array}  \]
The incentre is \[ I = (\frac{{ax_1 + bx_2 + cx_3 }}{{a + b + c}},\frac{{ay_1 + by_2 + cy_3 }}{{a + b + c}})  \]
where a is length of side BC, b is length of side AC and c is length of side AB.

I cannot remember the formula for the circumcentre.I tell you afterwards.

EDIT:OOPS others got there before me.


Two fair six-sided dice are rolled. One of them rolls an odd number, and the other rolls an even number. What is the probability that their sum is a prime number?

Well we only have 3 ways possible; assuming that the numbers on each die are 1, 2, 3, 4, 5 and 6.
So one person can have either
1 3 5
The other can have
2 4 6

The possible ways they can be summed are
$ 1\plus{}2$
$ 1\plus{}4$
$ 1\plus{}6$
$ 3\plus{}2$
$ 3\plus{}4$
$ 3\plus{}6$
$ 5\plus{}2$
$ 5\plus{}4$
$ 5\plus{}6$
In which the sums are
3
4
7
5
7
9
7
9
11
There are 9 sums; we include the ones that are the same since this will effect the roll (I'm not sure) and only 8 are prime... so our chance is $ \frac{8}{9}$
[b]OR IF IM WRONG IN THIS CASE[/b]
There are 9 sums but there are only 6 ways and out of those ways, 5 are prime so the chance is $ \frac{5}{6}$

I'm not sure which one is the right answer... I'm stupid...

[hide="..."]there are $ 6$ primes, out of the $ 9$ possible numbers, so the probability is $ \frac{6}{9}\equal{}\boxed{\frac{2}{3}}$. So $ 4$ and $ 9$ are definitely prime right?[/hide]

Augh! I failz...
in that case it would be $ \frac{6}{9}\equal{}\frac{2}{3}$
So in that case, the sums that are the same matter?

$ 1\plus{}4\equal{}4$?!?!?!?!?!? What is this madness you speak of?

[quote="joyofpi"]
The possible ways they can be summed are
$ 1\plus{}2$
$ 1\plus{}4$
$ 1\plus{}6$
$ 3\plus{}2$
$ 3\plus{}4$
$ 3\plus{}6$
$ 5\plus{}2$
$ 5\plus{}4$
$ 5\plus{}6$
In which the sums are
3
4
7
5
7
9
7
9
11
There are 9 sums; we include the ones that are the same since this will effect the roll (I'm not sure) and only 8 are prime... so our chance is $ \frac{8}{9}$
[/quote]
No...you have a sum wrong, $1+4\neq{4}$, it equals 5...and there are not $8$ prime sums even in the sums you listed, in fact, in your list, there are $6$ prime sums.  Let me list them again:
                     [b]Prime?[/b]
$ 1\plus{}2=3$ Yes
$ 1\plus{}4=5$ Yes
$ 1\plus{}6=7$ Yes
$ 3\plus{}2=5$ Yes
$ 3\plus{}4=7$ Yes
$ 3\plus{}6=9$ No
$ 5\plus{}2=7$ Yes
$ 5\plus{}4=9$ No
$ 5\plus{}6=11$ Yes

Now we can see there is $7$ prime sums out of $9$ prime sums.  Since all of those situations have equal probability, we need not worry about conditional probability.

Therefore, the probability of their sum is a prime number is $\fbox{7/9}$.


inequalities
function
logarithms
calculus
calculus computations

Prove that $ \ln (\ln q) \minus{} \ln (\ln p) < \frac {q \minus{} p}{e}$ for $ e\leq p < q$.


Person A has three brothers: persons B, C and D, and has no more brothers nor sisters. All of them live in the same street, and neither two of them live in the same house. Person A lives in the house which has 3 windows and 2 doors. Person B's house has as many doors as person's C house has windows, and person's B house has as many windows as person C's house has doors. Person D's house has as many windows as is the sum of the numbers of his brothers houses' windows, and has as many doors as is the sum of the numbers of his brothers houses' doors, and we know that person D's house has the same number of doors and windows. What is the probability that person A has a wife?

If you know some other paradoxical problems like this one write them here.

[quote="salmononpi"][hide] Person A is a girl. Therefor the answer is zero  :P  [/hide][/quote] What if she lives in Massachusetts? :?

[quote="djimenez"][quote="salmononpi"][hide] Person A is a girl. Therefor the answer is zero  :P  [/hide][/quote]

[hide]
Depending on where Perosn A lives, around an 8%!, but 0 in most of the places!
[/hide][/quote]
In America its 2%

[quote="mcalderbank"]wow, I was totally thrown off with all the talk about doors and windows. :D[/quote]

haha same, is there some trick to this...?

what about Australia? aren't civil unions legal there too?

Nobody seems interested in explaining why Person A is female in the first place...so I guess I will.
[hide]If Person B has d doors and w windows, Person C has w doors and d windows. So Person D has at least d+w doors and d+w windows. But d+w+2 is never equal to d+w+3, so the last condition is never satisfied if Person A is Person D's brother. So Person A must be Person D's sister.[/hide]
You may now continue with your discussion of...whatever you were discussing...


Find all integers $ x,y$ that satisfy  the equation $ x(x\minus{}y)\equal{}8y\minus{}7$

Rearranging, we have

$ y\equal{}\frac{x^2\plus{}7}{8\plus{}x}$.

Two integer solutions are made by setting the denominator equal to $ 1$ and $ \minus{}1$.

$ 8\plus{}x\equal{}1 \implies x\equal{}\minus{}7$, which says that $ y\equal{}56$, which gives the ordered pair $ (\minus{}7,56)$.

Doing a similar thing with $ 8\plus{}x\equal{}\minus{}1$, another ordered pair is $ (\minus{}9,\minus{}88)$.

To show that no other integer value of $ x$ produces an integer value for $ y$, consider $ x\equiv 0,1,2,3 \pmod4$, remembering that $ x^2\equiv 0,1 \pmod4$. Clearly, the fraction does not yield integer results for $ x\equiv 0,1,3 \pmod4$. Now consider $ x\equiv2 \pmod4$. This implies that $ x$ is even. However, if $ x$ is even, then the numerator is odd, and the denominator is even, which clearly does not yield integer results.

Therefore, our two ordered pairs are $ (\minus{}7,56)$, $ (\minus{}9,\minus{}88)$.

That's almost right, but you're missing a few.

[hide="Solution"]$ \frac{x^2\plus{}7}{8\plus{}x}\equal{}x\minus{}8\plus{}\frac{71}{x\plus{}8}$ after synthetic division, so $ x\plus{}8|71\implies x\plus{}8\equal{}\pm 1, \pm 71$

Then $ x\equal{}\minus{}9,\minus{}7,63,79$, so the solutions are: $ (x,y)\equal{}(\minus{}9,\minus{}88),(\minus{}7,56),(63,56),(79,\minus{}88)$[/hide]


number theory
Diophantine equation
continued fraction
number theory theorems

I saw a nice applet by [url=http://www.alpertron.com.ar/QUAD.HTM]Alpertron[/url] to solve Diopnatine equations. Sometimes it solves Diophantine problems by continued fraction. I tried to read the method but I did not fully understood the idea. How can we find the integer points of a given second degree Diophantine equation if we know the periodic part of the continued fraction explained in the same page?

Given the Diophantine equation

$x^{2}-dy^{2}= 1$

We consider the continued fraction of $\sqrt{d}$.  Specifically, calculate its convergents $\frac{p_{n}}{q_{n}}$.  In a period of the continued fraction, at least (exactly?) one of the convergents satisfies

$p_{n}^{2}-dq_{n}^{2}= 1$


We have 2 fixed points A and B.

$L_k$ is the locus of the point M in the plane such that $MA\cdot MB = k$. Prove that for a large enough k any line in the plane intersects $L_k$ in at most 2 points.

[quote="Myth"]:(
Any human solution?[/quote]

Any solution?

With this it is rather obvious:

http://mathworld.wolfram.com/CassiniOvals.html


algebra
polynomial
number theory unsolved
number theory

The polynomial $P(X)$ of degree n >= 5 with integer coefficients has n integer distinct roots $a_1, a_2, ..., a_n$, where $a_1=0$. Find all integer roots of polynomial $P(P(X))$.

Obviously, $a_1, ..., a_n$ are roots of $P(P(X))$ too. Now assume there is another integer root $a$ of $P(P(X))$. Since $a_1, ..., a_n$ are all roots of $P(X)$ we have $P(a) = a_m$ for some $m \in \{1; ...; n\}$. Furthermore, $a - a_i | P(a) - P(a_i)$ so $a_m - a_i | a_m \Leftrightarrow a_m - a_i | a_i$ for all $i \in \{1; ...; m-1; m+1; ... ; n\}$.
………


Can the reunion of a finite number of concave quadrilaterals form a convex polygon?

[hide="If you mean union..."]Yes.
Consider triangle $ ABC$, with $ D$ and $ E$ in the interior of the triangle, $ D$ closer to $ BC$ and $ E$ closer to $ AB$.  Quadrilaterals $ ABDC$ and $ AEBC$ are concave, but their union is convex triangle $ ABC$.[/hide]
[hide="If you mean intersection..."]Yes.
Consider triangle $ ABC$, with $ D$ on $ AB$, $ E$ on $ BC$, and points $ G$ and $ F$ outside the triangle, with $ F$ closer to $ AB$ and $ G$ closer to $ BC$.  Quadrilaterals $ ACFD$ and $ CAGE$ are concave, but their intersection is convex triangle $ ABC$.[/hide]


algebra
polynomial
induction
topology
advanced fields
advanced fields unsolved

Prove that in topological space $ X$, if $ V\subset X,V\ne X$ is an affine variety, $ \operatorname{int}V \equal{} \emptyset$ i.e. it has no interior points.

you can assume $ V$ is a hypersurface, ie its the vanishing set of a single polynomial, since in any case the interior of $ V(f)$ contains the interior of $ V(f_1,\ldots,f_r)$.  Then you want to show that in any neighborhood of a point in $ V$ you have points where $ f$ doesn't vanish.  With a little technical stuff, you can show that as long as $ f$ isn't zero, you can fix all but one variable so that the resulting polynomial in one variable isn't zero.  You probably do this by induction or something.  Then you just have to show that on some line segment, a nonzero polynomial in one variable doesn't always vanish, which is ok.  I'm assuming when you said topological space you meant some zariski topology over a field?


WELL ...... this goes to everybody....

POST about your KVPY interviews here.
So, post quick and make this thread alive.....

P.S:- Seniors are also required to provide expert views :P

 :)


calculus
integration
function
calculus computations

What is integral [size=150][b] (cos x)^(2n+1) dx .[/b][/size]

Do you know about Beta function ?[url]http://en.wikipedia.org/wiki/Beta_function[/url]

and use substitution method to go back to $ sinx$ and $ cosx$

Hi :)

Let $ u_n \equal{} \int cos^{2n\plus{}1}(x) dx$

than by integration by part we find that:

$ u_n \equal{} \frac{1}{2n\plus{}1} sin(x)cos^{2n}(x) \plus{} \frac{2n}{2n\plus{}1}u_{n\minus{}1}$ and $ u_0 \equal{} sin(x)$

 .....

by recurrecne we can fin a result that we want

PS if you're not find a finaly result , i will  complet the answer

thakns :)

even i get this reduction formulae but cant we get an ans without this u(n-1) ?etc....


$ ABCDEF$ is a conves hexagon such that $ < B$ $ \plus{}$ $ < D$ $ \plus{}$ $ < F$ $ \equal{}$ $ 360$ and $ \frac {AB}{BC}$ $ \times$  $ \frac {CD}{DE}$ $ \times$ $ \frac {EF}{FA}$ $ \equal{}$ $ 1$.prove that:

$ \frac {BC}{CA}$ $ \times$ $ \frac {AE}{EF}$ $ \times$ $ \frac {FD}{DB}$ $ \equal{}$ $ 1$


Just cross multiply to get $2(3x-2y)=4x+5y$ so then bring all the like terms together and you get $2x=9y$, so then $x=9$ adn $y=2$ works out so then plug those into the other expression and you get $31/26$.

This is exactly as you said Chess, however, couldn't $x=18$ and $y=4$, there are many many more solutions to this, you know...

[quote="Iversonfan2005"]Just cross multiply to get $2(3x-2y)=4x+5y$ so then bring all the like terms together and you get $2x=9y$, so then $x=9$ adn $y=2$ works out so then plug those into the other expression and you get $31/26$.....[/quote]

Which is exactly what I said... ;)  Why would there be more solutions?

Oh no, you're right, it will still simplify to that...so I guess it doesn't matter.

[quote="4everwise"]There can't be more solutions. These are one degree polynomials.[/quote]
thats not quite correct, system of equations with one equation of a polynomial of degree one with one unknown DOES have finite number of solutions
but, this is a system of equations with one equation of a polynomial of degree one on each variable with two unknowns
there are infinite number of real numbers $x,y$ satisfying the first equation because $\frac{x}{y}$ is only a proportion

Well, you are given one equation with two variables, so there would be an infinite number of solutions. However, the problem is stated so that all ordered pairs would give the same answer, since all answers are in the form $(9a,2a)$, as chess64 showed earlier.

[quote="chess64"][hide]
$6x-4y=4x+5y$
$2x=9y$
$(9,2)$ works, so
$\frac{27+4}{36-10}=\boxed {\frac{31}{26}}.$

[/hide][/quote]

Your solution isn't strict mathematically.

Strictly speaking we should answer as follows.

From $2x=9y\Longleftrightarrow x:y=9:2$, so we can set $x=9k,y=2k\ (k\neq 0)$.

Therefore $\frac{3x+2y}{4x-5y}=\frac{31k}{26k}=\frac{31}{26}\ (\because{k\neq 0)}$

kunny

What's that "because" dot thingy? $\text {I eat cheese} \because{\text {I like cheese.}}$ cool...

Anyways, I just used the easiest approach :) . That's what you would do in a real competition, because it's the fastest way. If this was some competition where you had to show your work, then I would use the ratios.


Let $P$ a point inside a triangle $\triangle ABC, A'=AP \cap BC, B'=AC \cap BP, C'=AB \cap CP$. If
\[ x = \frac{{AP}}{{PA'}}\ \ \ , \ \ \ {\rm{   }}y = \frac{{BP}}{{PB'}} \ \ \ ,\ \ \ {\rm{   }}z = \frac{{CP}}{{PC'}}$ \]
 show that
\[xyz = x + y + z + 2\]

[hide="Click here for the solution"]Since $x=\frac{AP}{PA^{\prime}}$, we have

$\frac{AA^{\prime}}{PA^{\prime}}=\frac{AP+PA^{\prime}}{PA^{\prime}}=\frac{AP}{PA^{\prime}}+1=x+1$,

and thus $\frac{PA^{\prime}}{AA^{\prime}}=\frac{1}{x+1}$. Similarly, $\frac{PB^{\prime}}{BB^{\prime}}=\frac{1}{y+1}$ and $\frac{PC^{\prime}}{CC^{\prime}}=\frac{1}{z+1}$. But [url=http://cut-the-knot.com/triangle/Gergonne.shtml]the Gergonne theorem[/url] states that

$\frac{PA^{\prime}}{AA^{\prime}}+\frac{PB^{\prime}}{BB^{\prime}}+\frac{PC^{\prime}}{CC^{\prime}}=1$;

thus,

$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$.

In other words,

$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}-1=0$.

But some calculations yield

$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}-1=\frac{x+y+z+2-xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}$;

thus, we must have x + y + z + 2 - xyz = 0, so that x + y + z + 2 = xyz. This solves the problem.[/hide]

  Darij

Here it follows a different proof (I use only Thales)  respect to that proposed [url=http://cut-the-knot.com/triangle/Gergonne.shtml]here[/url] for the Gergonne identity.

Let a, b and c be the three parallels by P to BC, CA and AB respectively. Let a^AB= Cb, a^AC=Bc, b^BA=Ca, b^BC=Ac, c^CA=Ba and c^CB=Ab.

The claim to prove is easily seen to be equivalent to

BaAb/AB + CbBc/BC + AcCa/CA = 2     (*).

To prove the latter note that BaAb/AB = CBa/CA and CbBc/BC = ABc/AC then the (*) is equivalent to (CBa + ABc + AcCa)/CA = 2 which is easily verified true.


algorithm
binomial coefficients
combinatorics open
combinatorics

Given n labeled nodes, how many ways are there to create a forest of non-rooted trees with at most k trees in the forest?

For example, if n=3 and k=2, the answer is 6:

a-b, c
a-c, b
b-c, a
a-b-c
a-c-b
b-a-c

if n=4 and k=2, the answer is 31 (I think...):

16 with one tree containing all 4 nodes (Cayley's formula)
12 with one tree containing 3 and the other containing 1 node (4*3--one node becomes the lone node, and one becomes the node between the other two in the 3-tree)
3 with two trees containing two nodes each (there are 3 choices for which node gets paired with node 'a').

I'm not expecting anything approaching a closed-form--this algorithm will be run on a computer.  However, an exponential-time algorithm is too slow--I'll need to run this for n around 50 and k around 20.  I have a solution for a forest on n nodes containing k *rooted* trees.  It uses [url=http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem]Kirchhoff's theorem[/url].  It seems like Kirchhoff's theorem should be useful for this, too, but I haven't yet found a solution.

I have posted it in [url=http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=118750]a German forum[/url], and I'll just repeat it here:

[quote="Bela Bollobas, [i]Modern Graph Theory[/i], VIII.5, problem 64:"]Prove that there are
$ \frac {1}{m!}\sum_{j \equal{} 0}^m\left( \minus{} \frac12\right)^j\binom{m}{j}\binom{n \minus{} 1}{m \plus{} j \minus{} 1}n^{n \minus{} m \minus{} j}\left(m \plus{} j\right)!$
acyclic graphs on $ n$ labelled vertices having $ n \minus{} m$ edges. Deduce that there are $ \frac12 \left(n \minus{} 1\right)\left(n \plus{} 6\right)n^{n \minus{} 4}$ forests with two components.[/quote]

Of course, $ n \minus{} m$ edges means $ m$ connected components for a forest. I hope this is what you mean when you say "$ m$ trees" (that is, I hope you do consider the graph with one vertex as a tree).

You asked for "at most $ m$ trees" rather than exactly $ m$ trees, so you need one more sum sign.

For $ m\equal{}n$ (that is, for the number of all acyclic graphs) there is a simpler formula:
http://www.emis.de/journals/JIS/VOL6/Callan/callan10.pdf

  darij

[quote="darij grinberg"]
$ \frac {1}{m!}\sum_{j \equal{} 0}^m\left( \minus{} \frac12\right)^j\binom{m}{j}\binom{n \minus{} 1}{m \plus{} j \minus{} 1}n^{n \minus{} m \minus{} j}\left(m \plus{} j\right)!$
[/quote]

Thanks for the help, Darij...but one question.  One term of this sum is going to have a $ \binom{n \minus{} 1}{m \plus{} m \minus{} 1}$ in it.  Say $ m > \frac {n}{2}$.  Is the binomial coefficient just zero in this case?  That seems to be commonly done, but not universally--the apache commons MathUtils package (the package I'm using to code this) throws an error if you tell it to compute $ \binom{n}{k}$ where $ k > n$.  Just want to make sure I'm not mis-reading it or something.

[quote="leptogenesis"]One term of this sum is going to have a $ \binom{n \minus{} 1}{m \plus{} m \minus{} 1}$ in it.  Say $ m > \frac {n}{2}$.  Is the binomial coefficient just zero in this case?[/quote]

Yes, $ \binom{n}{m}$ is always zero if $ n\geq 0$ and $ m > n$. At least, according to the following definition of binomial coefficients:
$ \binom{n}{m} \equal{} \left\{\begin{array}{c}\frac {1}{m!}\prod_{k \equal{} 0}^{m \minus{} 1}\left(n \minus{} k\right),\text{ if }m\geq 0; \\
0,\text{ if }m < 0\end{array}\right.$
for any $ n\in\mathbb{R}$ and $ m\in\mathbb{Z}$.
I am pretty sure the above formula uses this definition. At least, the "Deduce that there are $ \frac12 \left(n \minus{} 1\right)\left(n \plus{} 6\right)n^{n \minus{} 4}$ forests with two components." part would make it plausible.

  darij


geometry
circumcircle
trigonometry
ARML
Pythagorean Theorem
trig identities
Law of Sines

Any ideas on these?

1) In triangle ABC, AC = 18, and D is a point on segment AC for which AD=5. Perpendiculars drawn from D to segments AB and CB have lengths of 4 and 5 respectively. What is the area of triangle ABC?

Here, I drew BD and let the drawn perpendiculars to AB and BC be DP and DQ, respectively. I then let BP = n and BQ = m. From the pythagorean theorem, BD^2 = n^2 + 16 = m^2 + 25. Also, because triangles ABD and CBD have the same height, CD/CA = Area(CBD) / Area(ABD) = 5(m+12) / 4(n+3). However, solving these two simultaneous equations is pretty messy, and I think I screwed up on the arithmetic somewhere. IS there any easier way to do this problem?

2) From a point interior to a regular hexagon, perpendiculars drawn to the sides of the hexagon have lengths, in size order, of 3, 6, 15/2, 27/2, 15, and 18. How long is a side of this hexagon?

No idea where to start on this one.

I did #1 in a slightly simpler way. If E and F are the feet of the perpendiculars from D to AB and BC, then we have 2 right triangles ADE and BCF, from which we have sin A = 4/5 and sin C = 5/13. Since B = 180-(A+B), sin B=sin(A+C), in which we use the formula to get sin B = 63/65. Furthermore circumradius R by law of sines = AC/2*sinC = 65/7. Finally, [ABC]=sinA*sinB*AC*R = 4/5 *5/13 *65/7 *18=360/7. Ultimately, I dont know if this is easier than your way, but I tried. Is this form the ARML problem set on Nebraska's AMRL page?

[quote]Ultimately, I dont know if this is easier than your way, but I tried.[/quote]

Yup, lot less computation. :D

[quote]Is this form the ARML problem set on Nebraska's AMRL page[/quote]

Yeah, on that link I posted on some other thread.

Hmm, now that I've actually read your solution in depth, I am a bit confused here. How did you manage to bust out this formula:

[quote][ABC]=sinA*sinB*AC*R [/quote]

I mean, it can be derived from A = 1/2 * absin(C) with a few substitutions, but did you actually have that above formula memorized?

[color=cyan]That second problem isn't too bad at all.  It relies on the following pretty theorem, which I leave unproven, since if I proved it that would take all your fun (work) away:

The sum of the distances from any point to the edges of an equilateral convex polygon is constant.

This is true for normal distances for any point inside the polygon and for signed distances (I think) for any point in the plane of the polygon.  Prove it and you are most of the way done.  Or, just prove it for the case of a regular hexagon, which will do just as will.

With regards to your question of tetrahedr0n, he almost certianly did not have that formula memorized.  But why should he need to have -- if you know the Extended Law of Sines and any of the area formulae (except perhaps K = rs) you can get it with simple substitutions.[/color]

For the second problem, note that 3+18 = 6+15 = 15/2+27/2. Also note that all the segments from the point to the sides are perpendicular to the sides of the polygon. Why do you think it is necessary for all those sums to be equal? See if you can solve it now.

Thanks for the responses.

[quote]The sum of the distances from any point to the edges of an equilateral convex polygon is constant. [/quote]

But how can this help me, if the sum cannot be expressed as an expression involving the length of a side?

[quote]Why do you think it is necessary for all those sums to be equal?[/quote]

Well, since it is an equilateral hexagon, the distance from any side to its opposite side is constant (21 in this case). Let us call the hexagon ABCDEF, and draw segment AC, having M as its midpoint. AC = 21, so AM = 21/2. Drawing BM, ABM is clearly a 30-60-90 triangle. Thus AB, a side of the hexagon, if equal to 2 * 21/(2*Sqrt(3)) = 21/Sqrt(3) = 7*Sqrt(3). Right?

[color=cyan]You can use my theorem by picking the given point, P, with its given distances, and any other point you want.  So you could pick, say, a vertex of the hexagon and compute the distances to the other sides.  Or, even more prettily, pick the center of the hexagon.  Make sense?[/color]

[quote]You can use my theorem by picking the given point, P, with its given distances, and any other point you want. So you could pick, say, a vertex of the hexagon and compute the distances to the other sides. Or, even more prettily, pick the center of the hexagon. Make sense?[/quote]

Oh, I see. So picking a convenient point would allow me to manually compute the sum of the distances using the side length of the hexagon as a variable, and I could set that expression equal to the sum of the numbers that are given in the problem. Thanks for the help.

[quote]I mean, it can be derived from A = 1/2 * absin(C) with a few substitutions, but did you actually have that above formula memorized?
[/quote]

No, when I saw that I had one side and the three angles of the triabgle i thought i needed a formula for 1 side 2 angles. I quickly made the substitution into ab/2sinc to get what i needed.

[quote]No, when I saw that I had one side and the three angles of the triabgle i thought i needed a formula for 1 side 2 angles. I quickly made the substitution into ab/2sinc to get what i needed.[/quote]

Cool. Thanks.

[quote="Fierytycoon"]1) In triangle $ABC,$ $AC = 18$ and $D$ is a point on segment $AC$ for which $AD=5$. Perpendiculars drawn from $D$ to segments $AB$ and $CB$ have lengths of $4$ and $5$ respectively. What is the area of triangle $ABC$?[/quote]
[hide="Solution to 1"]Let the perpendicular from $D$ to $AB$ intersect $AB$ at $E,$ and let the perpendicular from $D$ to $BC$ meet $BC$ at $F.$ Draw the altitude from $B$ to meet $AC$ at $G.$ Note that $\triangle ADE \sim \triangle AGB$ and $\triangle CDF\sim \triangle CBG.$

Now, let $BG=x.$ It follows that $AG=\frac{3x}{4}$ and $GC=\frac{12x}{5}.$ Thus, $AC=AG+GC=\frac{3x}{4}+\frac{12x}{5}=18\implies x=\frac{40}{7}.$ It follows that\[[ABC]=\frac{1}{2}(BG)(AC)=\frac{360}{7}.\]
[/hide]

[quote="Fierytycoon"]2) From a point interior to a regular hexagon, perpendiculars drawn to the sides of the hexagon have lengths, in size order, of $3$, $6$, $15/2$, $27/2$, $15$, and $18.$ How long is a side of this hexagon?[/quote]
[hide="Solution to 2"]Let the side of the hexagon be $s.$ Note that the hexagon can be partitioned into six triangles with base $s$ and heights equal to the given perpendiculars. Setting two expressions for the area equal,\[\frac{1}{2}s(3+6+15/2+27/2+15+18)=\frac{6s^{2}\sqrt{3}}{4}.\]Solving, $s=7\sqrt{3}.$[/hide]


trigonometry
geometry
trapezoid
Pythagorean Theorem
geometry open

There are two chords AB,CD that are perpendicular to each other in a circle that has a radius of R.
Prove AC^+CB^+BD^+DA^ is constant.

p.s. I want to know what kind of process I have to think to find a strategy to solve this problem.

You can draw a diameter $ AE$ then prove that $ CBED$ is a trapezoid. Prove that $ AC^2\plus{}BD^2\equal{}AC^2\plus{}CE^2\equal{}4R^2$ ...

[quote="livetolove212"]You can draw a diameter $ AE$ then prove that $ CBED$ is a trapezoid. Prove that $ AC^2 \plus{} BD^2 \equal{} AC^2 \plus{} CE^2 \equal{} 4R^2$ ...[/quote]

I'm not quite sure how to do that. I think I have to use law of cosine but don't know how. I am thinking of making a relationship between the radius and each of the side of the quadrilateral.

Let the chords $AB,CD$ meet at $P.$ We shall show that $PA^2+PB^2+PC^2+PD^2$ is independent of $P.$ $ O$ is the center of the given circle and $ Q$ the second intersection of $ DO$ with $ (O).$ By Pythagorean theorem we get

$ PA^2 \plus{} PB^2 \plus{} PC^2 \plus{} PD^2 \equal{} \frac {_1}{^2}(DA^2 \plus{} AC^2 \plus{} CB^2 \plus{} BD^2)$

But since rays $ DQ,DP$ are isogonal WRT $ \angle ADB,$ it follows that $ ACQB$ is an isosceles trapezoid with $ CQ \parallel AB$ $ \Longrightarrow$ $ AC \equal{} BQ$ and $ CB \equal{} AQ.$  Thus, keeping in mind that $ DQ$ is a diameter, again by Pythagorean theorem we have:

$ PA^2 \plus{} PB^2 \plus{} PC^2 \plus{} PD^2 \equal{} \frac {_1}{^2}(DA^2 \plus{} AQ^2 \plus{} BD^2 \plus{} BQ^2)$

$PA^2 \plus{} PB^2 \plus{} PC^2 \plus{} PD^2 \equal{}  \frac {_1}{^2}(DQ^2 \plus{} DQ^2) \equal{} \frac {_1}{^2}(8R^2) \equal{} 4R^2.$


algebra
polynomial
complex numbers
algebra proposed

Prove that equation $z^{4}-2cz^{3}+2\bar{c}z-1=0$ -[i]which $c$ is a complex number[/i]- has a root $z$ with $|z|>1$, iff $|Re(c)^{\frac13}+Im(c)^{\frac13}i|>1$.

[quote="box of rocks"]What is the distinction with lower case c and lower case c with the bar over it that you both have?[/quote]

Do you not know what a complex conjugate is?  If $z=a+bi$ ($a,b\in\mathbb{R}$) then $\overline{z}=a-bi$.


32-nd Austrian Mathematical Olympiad 2001
Final Round
May 30–31, 2001
First Day :blush:

oke ^^ but I haven't found these problems.. so does somebody knows ??


function
algebra
polynomial
calculus
derivative
partial fractions
linear algebra

i have a problem with understanding the D operator. well this can be easily understood that it's distributive property & other linear properties like multiplying with constants etc. But how does it imply that
1. it can be factorised? ie (D^2+3D+2)y=(D+1)(D+2)y????????
2. 1/(F(D)) can be expanded binomially where F(D) is a function of D(polinomial)????
3. 1/F(D) can be partially factorized?????

That particular polynomials factor in particular ways is just a consequence of a few basic axioms (commutativity, distributivity, associativity, and the like) -- since the $ D$ operator satisfies these axioms, it behaves just like anything else that satisfies those axioms.  In particular, a factorization that's valid for a formal variable $ x$ satisfying those axioms (like, say, $ x^2 \plus{} 3x \plus{} 2 \equal{} (x \plus{} 2)(x \plus{} 1)$) must also be true in the particular case when we substitute $ x \equal{} D$.  Similarly, any other formal identity stays formally true when we substitute $ D$ in place of some formal variables.  As long as everything converges, formal identities can be replaced by actual equalities.

Another way of saying this: what possible obstruction is there to any of these things?

Well what is the relation between having associative, distributive properties to expanding negative binomials. again D is linear operatior how can u say that 1/f(D) too is a linear operator so that it can be operated as partial fractions? i got the statements in u gave in my books not the proof of the statement. pls reply.

If $ V$ is your space of functions (formal power series, $ C^\infty$ functions, something nice), there's a homomorphism $ \mathbb{R}[D] \to \mathrm{End}(V)$
sending $ D$ to the differentiation operator. So, you can do anything you'd do with polynomials on it. Since we can solve ordinary linear differential equations, it
could extend to a homomorphism $ \mathbb{R}(D) \to \mathrm{End}(V)$, which is why things like $ \frac{1}{1\plus{}D}$ make sense.

Actually, I suppose you would have to be a little more careful than that, since differential equations don't have unique solutions. For example, $ \frac{1}{D}$ would give you a map $ V \to V / \ker D$ (although this can be repaired in a not-natural way).

what is this R(D) to End (V). C to the power alpha. pls explain what u said.

$ {\mathbb{R}}[D]$ is the set of polynomials in the formal variable $ D$ with real coefficients.  $ \text{End}(V)$ is the set of endomorphisms of $ V$, i.e., linear maps from $ V$ to $ V$.  $ C^\infty$ are infinitely differentiable functions.  $ \mathbb{R}(D)$ is the set of rational functions with real coefficients in the formal variable $ D$.

what has it got to do with D operator? i relly didnt understand.

I have no idea what you're asking.  Please try to formulate (ideally using well-punctuated, grammatically correct sentences) a precise question about whatever it is that you don't understand.  It would probably also help you to reread carefully whatever text it is that you were reading when your questions arose.

i think i have stated what i wanted to know in the first thread quite clearly. i didnt disagree with D being linear operator but dont know why the 3 properties are implied with its being a linear operator.
1. partial factorisation of F(D).
2.factorisation of F(D).
3.negative binomial expansion of F(D).

And this has already been explained in two different ways.  Evidently, you do not understand some part of one or both of these explanations, but it is totally unclear which parts you do and do not understand.

[quote="gauss202"]It's probably the part about rings, homomorphisms, and endomorphisms. If someone hasn't taken a class in linear algebra or abstract algebra, they probably won't understand an explanation using those terms.[/quote]

Unfortunately that's probably the best language for expressing the relationships :\. Araditha, what context are you seeing this material in?

yes i dont understand any of those rings, homomorphisms. i searched those in the net though what i got there i just didnt understand that how can those be related to differentiation?
i am not asking those regarding any context. i am a student of 1st year physics as honours course & in our class of mathematical physics our prof used those properties while solving nonhomogeneous equations wid constant coeffficients. i dont like using properties what i dont understand why to use? can i get the proof anywhere? i asked it many people but none yet has given me any answer that i think solves at all my problems.
regarding those previous answers i got i really didnt understand how to use these rings, spaces, homomorphism things to soth real like D where we actually take a limit. pls reply.

And what part of my explanation (which does not include any of the words "ring," "endomorphism," etc.) is problematic?

If it helps, when I speak of "formal variables" I just mean that we aren't worried about what the "meaning" of $ D$ is -- it's enough to know that $ D$ satisfies the basic rules of arithmetic (e.g., $ D \cdot 2 \equal{} 2 \cdot D$, meaning that for any function $ y$ we have $ D(2y) \equal{} 2\cdot D(y)$, a property that you probably learned early in your first calculus class).

The identity $ \frac{1}{1 \minus{} D} \equal{} 1 \plus{} D \plus{} D^2 \plus{} \ldots$ is also a "formal identity," i.e., it is true in general if we're not interested in issues of convergence.  I work with "formal power series" a lot -- these are just like regular power series, but we don't care about analytic properties at all, and in this context the above (as well as the generalized binomial theorem) is a formal identity.  Roughly speaking, any "analytic identity" (like $ \frac{1}{1 \minus{} D} \equal{} 1 \plus{} D \plus{} D^2 \plus{} \ldots$) gives rise to a formal identity, and any formal identity gives rise to an analytic identity provided everything converges.  Another way to put this is just that the [i]definition[/i] of $ \frac{1}{1 \minus{} D}$ is the infinite sum $ 1 \plus{} D \plus{} D^2 \plus{} \ldots$ (whenever we're applying it to a function for which it converges), and the fact that we're dealing with a formal identity here means that this obeys the laws of arithmetic and you can do the sorts of natural things with it you would expect.

we are using all the properties of nos wid an operator like D. i dont understand how it really holds?
suppose we have an eqn D2y+Dy-20y=f(x)
now we write (D2+D-20)y=f(x)
or, (D+5)(D-4)y=f(x)
so, y=1/(D+5)(D-4).f(x)

now as D is only an operator how can we say that 1/(D+5)(D-4) ( which is also another operator) is same as 1/(D-4)9-1/(D+5)9.

again how can we say that it is same operator as 1/(D+5).1/(D-4)?????????

Okay.  Now, go re-read the post of mine that immediately precedes your own and which answers your question.

Dear Aradhita,

It seems to me I see what you don't understand in the subject:
some of my students have the same problem.

First of all, what is the meaning of $ DA$, where $ D$ and $ A$ are operators? By definition $ DA$ is the composition of $ D$ and $ A$, i.e. it is the operator which acts by the formula $ (DA)(x)\equal{} D(A(x))$. Check that if $ D$ and $ A$ are linear operators, then $ DA$ is also a linear operator! Remark, that for some operators $ DA$ and $ AD$ are not the same, so one has to be carefull with the order of multipliers!

$ D^2$ is defined as $ D \times D$, $ D^3$ as $ D \times D \times D$ etc.

Identity operator $ I$ which acts as follows: $ Ix \equal{} x$ for every $ x$ behaves like number 1: if one multiplies $ D$ by $ I$, one gets $ D$, i.e. $ ID\equal{}DI\equal{}D$. Check this! Sometimes instead of $ I$ one write $ 1$. I don't like this notation (it confuses numbers and operators), but it is popular.

Now, what is $ 1/D$? It is the inverse operator to $ D$ (another notation is  $ D^{\minus{}1}$). By definition if a linear operator $ A$ satisfies $ AD \equal{} DA \equal{} I$, then $ A$ is called inverse to $ D$, ond one writes $ A \equal{} 1/D$. Remark, that not all operators have inverse. Those that have inverse are called invertible. Check, that an operator is invertible if and only if it is bijective, and that the inverse to $ D$ is unique!

From this definition you can check that if $ U$, $ D$ are invertible, then $ D^{\minus{}1}U^{\minus{}1}$ is the inverse to $ UD$. In other words, $ \frac{1}{UD} \equal{} \frac{1}{D}\frac{1}{U}$.

Now concerning the formula  $ \frac{1}{I \minus{} D} \equal{} I \plus{} D \plus{} D^2 \plus{} \ldots$. It is valide if the series in the right hand side converges. To see this according to definition you just have to check two equalities:

\[ (I \plus{} D \plus{} D^2 \plus{} \ldots)(I\minus{}D )\equal{} I\]
and
\[ (I\minus{}D)(I \plus{} D \plus{} D^2 \plus{} \ldots)\equal{} I.\]

To do this just you just must open the brackets and to symplify.


function
Ring Theory
superior algebra
superior algebra unsolved

Suppose that $\phi$ is homomorphism from $C[0,1]$ (continous functions from $[0,1]$ to $\mathbb R$) to $\mathbb R$. Prove that there exist $0\leq\alpha\leq 1$ that $\forall f\in C[0,1]\ \ \ \ \ \phi(f)=f(\alpha)$

Is $C([0,1])$ a ring here? I mean, are the operations pointwise addition and multiplication?

Yes I meant the ring of continous function from $[0,1]$ to $\mathbb R$

I'm assuming you were referring to homomorphisms which are not identically zero.

Consider the restriction of $\varphi$ to the set of constant functions on $I=[0,1]$. If we identify every constant function defined by ${\bf a}(x)=a,\ \forall x\in I$ with $a\in\mathbb R$, the restriction of $\varphi$ induces a non-trivial ring homomorphism from $\mathbb R$ to itself, and the only such homomorphism is the identity (a very well-known problem; one must prove here that if $f: \mathbb R\to\mathbb R$ and $f(x+y)=f(x)+f(y),\ f(xy)=f(x)f(y),\ \forall x,y\in\mathbb R$, then $f$ is the identity on $\mathbb R$). In particular, this means that $\varphi$ is onto.

Now consider the map $\tilde\varphi: C(I,\mathbb R)/M\to\mathbb R$ that $\varphi$ induces on the quotient ring $C(I,\mathbb R)/M$, where $M$ is the ideal $\varphi^{-1}(0)$. This homomorphism is one-to-one by the definition of $M$, and it is onto because it has the same image as $\varphi$, namely $\mathbb R$. This means that $\tilde\varphi$ is an isomorphism, which, in turn, means that $C(I,\mathbb R)/M$ is a field. In other words, $M=\varphi^{-1}(0)$ is a maximal ideal in $C(I,\mathbb R)$. Suppose we prove that all maximal ideals $M$ of $C(I,\mathbb R)$ are of the form $M_x=\{f\in C(I,\mathbb R)\ |\ f(x)=0\}$. Then the desired result would follow: take $\alpha\in I$ such that $M=M_\alpha$. All $f\in C(I,\mathbb R)$ are of the form $f={\bf a}+m,\ m\in M$, which means that $\varphi(f)=\varphi({\bf a}+m)=\varphi({\bf a})=a$ (as discussed in the previous paragraph), which is equal to $f(\alpha)$.

All that's left now is to prove that result on the maximal ideals of $C(I,\mathbb R)$, namely that they are of the form $M_x=\{f\in C(I,\mathbb R)\ |\ f(x)=0\}$. Let $M=\{f_u\}_{u\in U}$ be a maximal ideal of $C(I,\mathbb R)$, and for each $u$ in the index set $U$, let $K_u=f_u^{-1}(0)$. The $K_u$'s are closed subsets of $I$. If $u_1,\ldots,u_t\in U$, then the function $f^2_{u_1}+\ldots+f^2_{u_t}$ has as its zero set precisely $\bigcap_{i=1}^tK_{u_i}$, which must thus be non-empty, for otherwise $M$ would contain the invertible element $f^2_{u_1}+\ldots+f^2_{u_t}$. This means that the collection $(K_u)_u$ of closed sets satisfies the finite intersection condition (every finite family has a non-empty intersection), and since this takes place in the compact space $I$, they must all have non-empty intersection. Let $x\in\bigcap_{u\in U}K_u$. We have $M\subseteq M_x$, so, since $M$ is maximal, we have $M=M_x$, which is what we wanted to prove in the first place.


number theory
least common multiple
number theory unsolved

given a sequence of natural numbers $ a_0 , a_1 ,....,a_n , ....$ such that $ a_0<a_1<..<a_n<..$

Prove that $ \frac{1}{[a_i,a_{i\plus{}1}]} \leq 1\minus{}\frac{1}{2^n}$

where [a,b] is defined as the LCM of a and b

anyone ?
I think this is an interesting one .

[quote="HTA"]given a sequence of natural numbers $ a_0 , a_1 ,....,a_n , ....$ such that $ a_0 < a_1 < .. < a_n < ..$

Prove that $ \frac {1}{[a_i,a_{i \plus{} 1}]} \leq 1 \minus{} \frac {1}{2^n}$

where [a,b] is defined as the LCM of a and b[/quote]
What is n? :huh:
if it is any number then we can always find contradiction. :huh: :wink:
I think you missed some kind of series of whose general term is $ \frac {1}{[a_i,a_{i \plus{} 1}]}$

I'm sorry  :blush:
the problem should be (with same given conditions)
$ \sum^{n \minus{} 1}_{i \equal{} 0}\frac {1}{[a_i,a_{i \plus{} 1}]} \leq 1 \minus{} \frac {1}{2^n}$

(it seems like i can not edit my first post )


pigeonhole principle
number theory open
number theory

Hi,

It is not difficult (pigeon-hole argument) to prove that for any given positive integers $ n,k,f$ and $ n>k$ there exist an infinite number of positive integers $ A$ which can be expressed as a sum of $ n$ distinct $ k^{th}$-powers in at least $ f$ different ways, that is $ A \equal{} a_1^k \plus{} \cdots \plus{} a_n^k$ where $ a_1 < ... < a_n$.

Is something known about the least possible value of $ A$?

More generally, for given $ n,k$ as above, is something known about the number of decompositions of a given integer $ A$ as a sum of $ n$ distinct $ k^{th}$-powers?

Any reference is welcome.
Thanks in advance.

Pierre.

[quote="pbornsztein"]Is something known about the least possible value of $ A$?[/quote]
Do you have any reason to suspect that the trivial upper bound obtained by the same pigeonhole argument is far away from the optimal one?

What are you really after, Pierre? (if not a secret, of course :) ).

There is no secret at all.
I'm working for the solution of this problem, proposed in a math journal that I'm supervising.
The proposer's answer is just the pigeon-hole bound, and I was wondering if it could be improved by some deeper stuffs (I've no idea about it).
Since it looks like a Waring's problem, I hoped there are some results...

Pierre.


geometry
incenter
circumcircle
inradius
trapezoid
AMC 10
similar triangles

There was one problem on AMC-10 that I didn't know what to do. Now, I know how to solve that one but is there an easy to find the distance of incenter and circumcenter of a triangle if all you're given is the length of the sides?

Thank you.

[Edit - ah, I didn't understand what you meant by [i]distance of circumcenter and incenter[/i].]

Do you mean circumradius and inradius?

If so, then yes there is a formula. Often, though, there would be some trick such as similar triangles or right angles etc etc. But if you want a formula:

Inradius:

\[r = \frac{1}{2}\sqrt{\frac{(b+c-a)(c+a-b)(a+b-c)}{a+b+c}}\]

Of course you won't remember that, so an easier way to remember:

\[r = \frac{\Delta}{s}\]

where $\Delta$ is the area (use Herons formula) and s is the semiperimeter.

Circumradius:

\[R=\frac{abc}{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}\]

Again, you won't remember that, so

\[R=\frac{abc}{4rs}\]

where r is the inradius calculated above and s is the semiperimeter.

Thanks for the formula but I was referring to exactly what I said --- > DISTANCE from incenter to the circumcenter.

I got this from one of mock AMC's.

It said something like:

What is the distance from the incenter to the circumcenter (could be reverse) to the triangle with sides 13,14, and 15 if the distance can be expressed as $\frac {\sqrt a}{b}$. Find a+b.

I had no idea to how to do that one. :-/

Well, in that case:
\[d=\sqrt{R(R-2r)}\]
and you can calculate R and r from the above formulae if you want it explicitly in terms of a, b, c.

(the [i]from[/i] and [i]to[/i] were the important words :))

Also, for that question, you can do it (if you don't know the formula) by forming a trapezoid by dropping perpendiculars from I and O to the side of lenght 14.  It'll take you longer, of course, but it'll work after a while.


could anyone make available the solution to ml 1editon 7 round
the following problem
For a set $ S$, let $ |S|$ denote the number of elements in $ S$. Let $ A$ be
a set of positive integers with $ |A| = 2001$. Prove that there exists a
set $ B$ such that

(i) $ B \subseteq A$;
(ii) $ \dsp |B| \ge 668$;
(iii) for any $ u, v \in B$ (not necessarily distinct), $ u + v \not\in B$.

[quote="pohoatza"]It is a particular case of a known theorem of Erdos: Let $ A \subset Z$ be a set of $ N$ nonzero integers. Then there is a sum-free subset $ B$ of $ A$ with $ |B| > |N|/3$. The most common proof is probabilistic. You can search for it on Google.[/quote]
thanks you gave reason to my work for last 15 days
 :lol:
:wink:
the google is not providing any good resources
could you suggest some

See, for example, here: [url]http://www.math.rutgers.edu/~kcostell/Math582/week4.pdf[/url].

arent there any elementary proofs
i cant understand some parts of it
i dont think that theory is expected at the usa tst or the ml contest
heartfelt thanks for your great help
altleast i now have a solution

does anyone have the official solution to this problem??? :)

I found some time ago an elementary proof of this (I think i was searching something about a Sankt Petersburg's problem). I also posted a nice article about this subject (an Erdos result related to Littlewood- Offord theorem). I hope you'll enjoy this :D.

@mychrom
could you tell me what is E[] and P[] these remain undefined functions in the paper
or could you tell me how does the statement in the other paper imply the problem

E and P are expectation and probability.

There is a proof in the opening chapter of Terence Tao's Additive Combinatorics textbook, which is part of the sample on Amazon and on his own website, if you want to google it.

This is NOT elementary enough :(

anyone understand this solution?

What!? It's perfectly elementary.

The expectation of $ f(x)$ if $ x$ is chosen from some set $ X$ is just
\[ E(f) \equal{} \frac {\sum\limits_{x\in X} f(x)}{|X|}
\]
It's just an easy shorthand for taking sums in some situations. I find 'probability' methods very useful for combinatorics problems.


I'm not really sure what level this is but:

$ \sum_{n\equal{}1}^\infty{{1 \over 2^n \minus{} 1}} \equal{}$ ?

I don't think there's any reason to suppose that it has a nice closed form.

[hide="Discussion"] This is the [url=http://mathworld.wolfram.com/LambertSeries.html]Lambert series[/url]

$ \sum_{n\equal{}1}^{\infty} \frac{x^n}{1 \minus{} x^n} \equal{} \sum_{n\equal{}1}^{\infty} \sigma_0(n) x^n$

where $ \sigma_0(n)$ is the divisor function, evaluated at $ x \equal{} \frac{1}{2}$.  I also don't think there isn't a particularly nice closed form, but you can calculate its binary digits at least. [/hide]


geometry
incenter
inequalities
analytic geometry
geometry unsolved

Given are $\triangle ABC$ inscribed $(O)$, $I$ inside triangle $IA\cap BC=\{A'\}$ and $IA\cap (O)=\{A''\}$ similar define $B',B'',C',C''$ find min values of $\frac{AA'}{AA''}+\frac{BB'}{BB''}+\frac{CC'}{CC''}$  :)

Result is $I$=incenter of triangle,but I can't give a proof can any body help me,please ?  :)

Read the extended problem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=143947#143947]http://www.mathlinks.ro/Forum/viewtopic.php?t=16697 post #2[/url]. If you set D = A', E = B', F = C' in the extended problem, then D' = A" and thus $w_{a}=\frac{AD}{AD^{\prime}}=\frac{AA^{\prime}}{AA^{\prime\prime}}$, and similarly $w_{b}=\frac{BB^{\prime}}{BB^{\prime\prime}}$ and $w_{c}=\frac{CC^{\prime}}{CC^{\prime\prime}}$). Hence, your problem becomes the question to find the minimum of $w_{a}+w_{b}+w_{c}$. But as the solution of the extended problem shows, $w_{a}\geq l_{a}$, $w_{b}\geq l_{b}$, $w_{c}\geq l_{c}$, so that $w_{a}+w_{b}+w_{c}\geq l_{a}+l_{b}+l_{c}$. As you can easily see from the proof, this inequality becomes an equality if the point I is the incenter of triangle ABC. Thus, $w_{a}+w_{b}+w_{c}$ is minimal if the point I is the incenter of triangle ABC.

  Darij

Ok Darij ! Your result is really a strong result! I found another proof 5 day ago:
If I have barycentric coordinate $(x,y,z)$ then
$\sum\frac{AA'}{AA''}=3-\sum\frac{yza^{2}}{(y+z)(yc^{2}+zb^{2})}$ we can see this problem is conseqeunce of this equal, but I hope a nice inequality
http://www.mathlinks.ro/Forum/viewtopic.php?p=582723
with this equal I want to find a ineq like Garfunkel's ineq for any point inside triangle but there isn't it(as you saw in above topic), Can you help me find the condition of $(x,y,z)$ for $\sum\frac{AA'}{AA''}\le \frac{9}{4}\leftrightarrow \sum\frac{yza^{2}}{(y+z)(yc^{2}+zb^{2})}\ge\frac{3}{4}$ true (with $x,y,z>0$ and $a,b,c$ is three side of triangle ) :P


Problem:
Let be given $167$ sets $A_1, A_2, ..., A_{167}$ satisfying the following conditions:
1. $\sum\limits_{i=1}^{167}|A_i| = 2004$.
2. $|A_j|=|A_i||A_i \cap A_j|$ for all $i,j \in {1,2,...,167}$ and $i \neq j$.
Calculate $| \bigcup \limits_{i=1}^{167}A_i |$, where $|A|$ denotes the number of elements of the set A.


modular arithmetic
quadratics
number theory unsolved
number theory

$ p$ is a prime $ \equiv 3 \, \text{mod} \, 4$ . And $ 3$ is a primitive root $ \text{mod} \, p$ .

Can $ 3^3$ also be a primitive root $ \text{mod} \, p$ ?

No. Since $ p \equiv 3 \pmod{4}$, $ \left(\frac{3}{p}\right) \equal{} \minus{} \left(\frac{p}{3}\right)$. Thus If $ p \equiv 2 \pmod{3}$ then $ 3$ is a quadratic residue modulo $ p$ so it can't possibly be a primitive root. Therefore $ p \equiv 1 \pmod{3}$ so if $ 3$ is a primitive root (has order $ p\minus{}1$) then $ 3^3$ has order $ (p\minus{}1)/3$, and it's not a primitive root.