number theory
prime numbers
number theory unsolved

a)Find all prime numbers $p,q$ such that  $p,q\le2005$ and
$p^2+8$ is divisible by $q$,and $q^2+8$ is divisible by $p$.
b)the same thing for $p^2+4$ is div. by $q$,and $q^2+4$ is div.by $p$.

when p>q>3,

we can use this

p| q^2 +8
q| p^2 +8

pq| p^2 + q^2 +8

p^2 - kpq + q^2 +8 = 0

(q^2 + 8)/p is a solution which is smaller than q

and so on, we can find the beginning solution and make through them.

I guess that the next problem's solution is same process.


ceiling function
floor function
function
number theory
superior algebra
superior algebra unsolved

$n$ an integer, $\xi=e^{\frac{2i\pi}{n}}$.
Find a closed form for  $|\sum_{k=1}^{n}{\xi}^{k^3}|$

Let $e(x):=\exp(2\pi i x)$.  If $n=pq$ is composite, then by Chinese remainder theorem
\[\sum_{k=0}^{n-1} \xi^{k^3}=\sum_{k_1=0}^{p-1}\sum_{k_2=0}^{q-1} e\left(\frac{qk_1+pk_2}{pq}\right)=\sum_{k_1=0}^{p-1} e\left(\frac{k_1}{p}\right)\sum_{k_2=0}^{q-1} e\left(\frac{k_2}{q}\right).\]
Therefore it is sufficient to solve the problem when $n$ is a prime power. If $n=p^{2d}$ where $p$ is a prime, then we can transform the sum by writing $n=r+sp^d$ where $r,s\in\mathbb{Z}_{p^d}$ and summing over $r$ and $s$ independently as follows:
\[\begin{aligned}\sum_{k\in \mathbb{Z}_{p^{2d}}} e(k^3p^{-2d})&=\sum_{r,s\in \mathbb{Z}_{p^d}} e((r+sp^d)^3p^{-2d})\\&=\sum_{r,s\in \mathbb{Z}_{p^d}} e(((r^3+3r^2sp^d)p^{-2d})e(3rs^2+s^3p^d)\\
&=\sum_{r,s\in \mathbb{Z}_{p^d}} e(((r^3+3r^2sp^d)p^{-2d})\\
&=\sum_{r\in \mathbb{Z}_{p^d}} e((r^3p^{-2d})\sum_{s\in \mathbb{Z}_{p^d}}e(3r^2sp^{-d})
\\
&=\sum_{r\in \mathbb{Z}_{p^d}}\delta_{3r^2,0}p^d e((r^3p^{-2d}).
\end{aligned}\]
Let $d'=d$ if $p\neq 3$ and $d'=d-1$ if $p=3$. Then $p^d\mid 3r^2 \iff p^{\lceil d'/2 \rceil} \mid r$. Then the sum becomes
\[
\sum_{k\in \mathbb{Z}_{p^{2d}}} e(k^3p^{-2d})=p^d\sum_{r'\in \mathbb{Z}_{p^{\lfloor d'/2 \rfloor}}} e((r'p^{\lceil d'/2 \rceil})^3 p^{-2d})\]
which is nothing other than the original sum with $n=p^{2d-3\lceil d'/2 \rceil}$ times the factor of $p^d$.

Similar computations can be carried out when $n=p^{2d+1}$ for $d\geq 1$.

So, the problem reduces to computing the sum in the case $n$ is a prime number. About that I know close to nothing. Basically, as far as I understand, the Riemann hypothesis for function fields (which is now a theorem) [i]somehow[/i] implies an upper bound of the form $c\sqrt{n}$ on the sum, when $n$ is prime. I do not understand the `somehow' part myself, so I might well be wrong. I do not know whether the behavior of the argument of the sum is known, but I doubt it since the analogous question for Kloosterman sums is wide open (Sato-Tate conjecture).

So, my answer to your question is that the closed form for the sum is very unlikely to exist.

Boris

[quote="bbukh"]I do not know whether the behavior of the argument of the sum is known, but I doubt it since the analogous question for Kloosterman sums is wide open (Sato-Tate conjecture).

Boris[/quote]

Do you have references ( articles on the web) about Kloosterman sums and
What is Sato-Tate conjecture ?
Thank you for your reply Boris.

Googling yields hundreds of links, but I doubt whether any one them are readable. My reference is the book "Analytic number theory" by Henryk Iwaniec and Emmanuel Kowalski, AMS 2004. A Kloosterman sum is
\[S(a,b;c)=\sum_{x\in\mathbb{Z}_c^*}e\left(\frac{ax+bx^{-1}}{c}\right)\] where $x^{-1}$ is to be interpreted as inverse of $x$ in $\mathbb{Z}_c$.

Forget all the nonsense I said about the conjecture.  First of all, I misremembered it. Second of all, the sum is trivially real (conjugate and see what happens).

Boris


Square ABCD has its length in unit length (meaning its 1). If point E is set along line BC, and divide the triangle into two regions that one area is three times the other.  Smaller area is a right triangle.  FInd the length of line AE in triangle ABE.

You definitely need to rephrase that.  It doesn't make any sense the way it's written.

Never mind on this question
Just look at the question in difficult marathon except it's similar but I posted answer..

well, the answer is  :sqrt: 1.25, or  :rt5: /3, but do you still want us to do this?

I just posted that for how well I can solve the advance questions in the Mathcounts.. Basically, for fun... And major reason is I don't really know hard questions.. That's all... Nobody has to do it...

my bad. erm....... my finger slipped on the keyboard.... yeah.... :(


Prove that the sum of the interior angles in a triangle is $\pi$ (assuming Euclid's 5th postulate :D).

This is probably the most fundamental fact about triangles, but I was wondering how you would go about proving it. I came up with a way to prove it but I want to see if there's a simpler or more elegant method.

Umm... because $\pi$ radians are equal to $180$ degrees? jk, jk  :lol:

Hold on, I'll try.

[quote="Hamster1800"]I'd do it this way:

Draw any triangle and circumscribe a circle about it. each of the interior angles is half of the arc that it intercepts. so by the distributive property, the sum of the interior angles is half of the combined arc that they intercept. Since they intercept a whole circle, we get:

$a, b, c \text{ are measures of the interior angles}$

$a+b+c = \frac{2\pi}2 = \pi$[/quote]That works, but how do you prove "each of the interior angles is half of the arc that it intercepts"? I'm not challenging anything you said, I'd just like to see.

You draw two secants that intersect on the circle and the radii that intercept the same arc. Then draw the radius to the vertex of the inscribed angle. Given the fact that the exterior angle of a triange is the sum of the two non-adjacent interior angles, label the measures of the base angles of one triangle a and of the other triangle b. This gives the measure of the inscribed angle as a+b and the measure of the central angle, and thus the arc, 2a+2b.

unfortunately, as I wrote this i realised that to prove all of this you need that the sum of the interior angles of a triangle are $\pi$. But if you "fold" the three vertices of a triangle to a point on the triangle, you get a straight line, which is a straight angle which by definition is $\pi$ radians.

[quote="Hamster1800"]the fact that the exterior angle of a triange is the sum of the two non-adjacent interior angles[/quote]..which as you said is what we're trying to prove  :(

[quote="WindSlicer"]Can we use stuff we know about parallel lines to prove it? i.e. opposite interior angles are congruent?

I somehow fear this will turn circular.[/quote]I'm not quite sure, but I'm guessing that stuff comes from the 5th postulate, which I said you can assume. The proof that I came up with relies on parallel lines and the angles involved. I'll draw a picture and type it up tomorrow, but now I have to write an essay for english that I've been putting off  :D.

See diagram, I don't really feel like explaining it right now.  ;)

i think that, depending on what you consider trivial, the first part is all thats necessary

consider the following picture.
$MN\parallel BC\Rightarrow A_1=B_1, A_3=C_1$
$A_2+B_1+C_1=A_1+A_2+A_3=\pi$
done

[quote="polymorphic"]One way is to break up every triangle into two right triangles.  Another is to show that the sum of the exterior angles of any convex polygon is 360 degrees.  So in particular for a triangle, we have the sum of the exterior angles x + y + z = 360.  Sum of the interior angles is (180 - x) + (180 - y) + (180 - z) = 540 - 360 = 180.[/quote]

The exterior angles of any convex polygon is not 360 degrees.

(i.e. You need to prove it before ya use it :))

[quote="WindSlicer"]The exterior angles of any convex polygon is not 360 degrees.

(i.e. You need to prove it before ya use it :))[/quote]Especially since the sum of the interior angles directly follows from this, and vice versa.

[quote="shobber"]I wonder how people discovered it and proved it for the first time. Did they know paralel lines by that time??[/quote]The sum of the triangle's angles relies on Euclid's 5th postulate, which deals with parallel lines, so they've been known for a while.

I'd just draw a triangle, cut it out, rip off the corners, and put them together.  They should form a straight line.  Visual approaches are always simple. ;)

But that only proves it for that certain triangle, not all triangles in general.

Yep you could do that for 1,000,000,000 triangles and you still haven't "proven" the general case.

I've read sth about it... The book said that we can use examples to prove some of the geometry theorems.

Simple you draw a line parallel to one given side of the triangle through the other point on the triangle not on the line parallel to the line you just drew. Then by interior-exterior angle theorems we have drawn a 180 degree arc containing angles congruent to all the angles of the triangle.

ASSCI Diagram
______________
******/\******
*****/||\******
****[u]/||||\[/u]*****

Isn't it a theorem that an inscribed angle is 1/2 the arc it intersects, and if a triangle is inscribed in a circle, each of the angles are inscribed in the circle?

By the way, what is the book sth?


A rectangular grazing area is to be fenced off on three sides using part of a $ 100$ meter rock wall as the fourth side.  Fence posts are to be placed every $ 12$ meters along the fence including the two posts where the fence meets the rock wall.  What is the fewest number of posts required to fence an area $ 36$ m by $ 60$ m?

\[ \textbf{(A)}\ 11 \qquad
\textbf{(B)}\ 12 \qquad
\textbf{(C)}\ 13 \qquad
\textbf{(D)}\ 14 \qquad
\textbf{(E)}\ 16
\]

[hide="Click for solution"]
The fewest number of posts is used if the wall serves as the longer side of the rectangular grazing area.  Thus there are $ 6$ posts on the $ 60$ meter side (including the corners) and $ 3$ more posts on each $ 36$ meter side for a total of $ 12$ posts.
[/hide]


conics
hyperbola
geometry proposed
geometry

Let $A_1,A_2,A_3,B_1,B_2,B_3$ be six points in the plane such that for each $i,j$ we have $A_iB_j = i+j$.
What can be said about these six points?

Pierre.

I guess they must be on a line in the order $A_3A_2A_1B_1B_2B_3$ with $A_3A_2=A_2A_1=B_1B_2=B_2B_3=1,\ A_1B_1=2$. Is  it correct? I think I can prove it, but I don't like the proof one bit: it uses the intersections of some hyperbola branches and so on, so I would like to know if the result is Ok before I start typing it :). There can always be some mistake, of course.

As far as I remember, you are right Grobber.
Post your solution, I will look for mine in my archives....

Pierre.

It's rather messy and I'm not at all certain it works, but it seemed to work an hour ago so I'll post what I did :).

Construct circles of radii $(2,3,4),\ (3,4,5)$ centered at $A_1,\ A_2$ respectively. For each of the two points color the circle closest to it black (b), the next one red (r) and the next one green (g) (it will be easier to refer to the circles in this way; besides, that's how I drew the picture, so at least it's easier for me :)). Now color the intersections of the two circles of a certain color in that color. The two b points are the possible choices for $B_1$, the two r points are the two choices for $B_2$, and the two g points are the two choices for $B_3$. Now construct a b circle of radius $4$ around each b point, an r circle of radius $5$ around each r point, and, finally, a g circle of radius $6$ around each g point. From now on we deal only with these $6$ circles ($2$ of each color).

The necessary and sufficient condition for the existence of $A_3$ is the fact that there is a set of $3$ concurrent circles, one of each color, among the $6$ we're dealing with (and the intersection point would be $A_3$). We call both b points $B_1$, both  r points $B_2$ and both g points $B_3$ because there's no difference  among all the possible situations regarding the rest of the solution. Now pick a $B_i$ for each $i\in\overline{1,3}$ and let's work only with these three points and the three circles centered at the points. We assume they're concurrent. Then the branches of the hyperbolas having $(B_1,B_2),(B_2,B_3)$ as foci and passing through $A_1,A_2$ must intersect at another point $\ne A_1,A_2$, which is impossible unless the hyperbolas are degenerate (because two hyperbola branches can intersect in at most two points, and these are $A_1,A_2$), which gives us $A_1,A_2,A_3$ collinear.

Think it works?

I have not followed this MathLinks thread, but I think the same problem was discussed at Hyacinthos. See Hyacinthos messages #7462, #7476, #7481, #7482, #7486, #7521, #7541, which are all accessible through

  http://groups.yahoo.com/group/Hyacinthos/messages/7462?expand=1
  http://groups.yahoo.com/group/Hyacinthos/messages/7492?expand=1
  http://groups.yahoo.com/group/Hyacinthos/messages/7524?expand=1

  Darij


geometry
trigonometry
rotation
geometric transformation
number theory
relatively prime
trig identities

The two squares shown share the same center $O$ and have sides of length 1.  The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=431[/img]

Calling each of the unknown sides of the triangles in the regular octagon $x$, we use law of cosines to get $x^2=\frac{43^2}{99^2(2-\sqrt{2})}$, then using the sin-area formula an applying it by $8$ times for all the triangles we have $8(\frac{43^2}{99^2 \cdot 2-\sqrt{2}} \cdot \frac{\sqrt{2}}{2})$, but I really don't think this is the best way to solve it..... :( Never mind

The squares have length 1, thus the height of each traingle out of 8 is $\frac{1}{2}$.
As a result, the area of the octagon is $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}\cdot 8=\frac{86}{99}$
Thus $m+n=99+86=185$

Masoud Zargar

I got the same answer, but I don't know how you just assumed the height of the triangles was $\ \frac{1}{2}$.  Heres how I did it:

[hide]The rotated square divides the sides of the initial square into 3 segments.  One was length $\ \frac{43}{99}$  The other two we can call $\ x$ and $\ y$.  So $\ x+y=1- \frac{43}{99}$.  From now on I will call $\ \frac{43}{99}=a$.  So $\ x+y=1-a$.  Also these two segments and one side of the octagon form a right triangle, so $\ x^2 + y^2 =a^2$.  So $\ (x+y)^2 - (x^2 + y^2)=2xy=1-2a$.  Now the region in question is made up of a unit square is the area of 4 triangles of area $\ \frac{xy}{2}$.  So the total area of these triangles is $\ 2xy=1-2a$.  So the area of the desired region is equal to $\ 1-(1-2a)=2a= \frac{86}{99} = \frac{m}{n}$ so $\ m+n=185$[/hide]

Yeah I don't get the $\frac 12$ either.  I know the drawings aren't to scale, but the height looks nowhere near $\frac 12$.  The rotation of the square will change the height also, and I don't think it would be something elegant as $\frac 12$ when it intersects at an ugly length like $\frac{43}{99}$.  Although there is definetly something there since you got the right answer...

It probably has something to do with an invariance in the area of the 8 triangles overall.

I agree with Elemennop. Perhaps the reason boxedexe's method works is that the squares share the same center.

Yes, you can show that all those triangles are congruent.

If it's regular, than it's trivial, because even if it each one isn't one half, then let the height of one be $x$, and the other be $1-x$, and you get opposite triangles to be $\frac{1}{2}[\frac{43}{99}\cdot x+\frac{43}{99}(1-x)]=\frac{43}{198}$, and just do this four times for each pair of opposite triangles and you get the desired answer.

The height to which boxedexe refers is the perpendicular from the center of the squares to each of the bases AB, BC, CD, etc. This is why he is able to house area=base*height/2 with AB=43/99.

[quote="mna851"]I got the same answer, but I don't know how you just assumed the height of the triangles was $\ \frac{1}{2}$.[/quote]Are you guys talking about the same triangles?  :huh:

I believe he's splitting the octagon into eight triangles. Each triangle consists of a different side of the octagon, and its third vertex is at the center of the squares. Hence, the height of each of those eight triangles is half the length of the square's side, or 1/2.

And then, I guess it's intuitive that the octagon is equilateral. :)


Find all positive integers $n$ for which there exist distinct positive integers $a_{1},a_{2},\ldots,a_{n}$ such that \[\frac 1{a_{1}}+\frac 2{a_{2}}+\cdots+\frac n{ a_{n}}= \frac{ a_{1}+a_{2}+\cdots+a_{n}}n.\]

Does anyone has a solution?Difficult eh?

as my memory is still good
i think it's already somewhere in the forum
i m sure it is!
sorry if i have no solution
but thanks u remember me this problem i will think of it tonight :)

Someone in the Forum before days said this
  first of all I invented solution for all $n$ satisfying the Pell's equation $x^2-5(n+1)^2=-4$ :). After that, the great idea come to me and I improved my construction, so I think it covers all sufficiently large $n$ (maybe $n>20$ or something like that.

Could anyone explain with details this to me because I don't understand this :oops:

Could anyone help me with this problem. It is unsolved for a year .

I haven't  have solution yet;

But I can prove that there are infinitely many n satisfy.
I will still think of it tomorrow.

Silouan, I want to know that this problem have any application ????

WELLCOME
I Know less english , sorry if I write (F) . Thankyou !!!!

[quote="silouan"]Someone in the Forum before days said this
  first of all I invented solution for all $n$ satisfying the Pell's equation $x^{2}-5(n+1)^{2}=-4$ :). After that, the great idea come to me and I improved my construction, so I think it covers all sufficiently large $n$ (maybe $n>20$ or something like that.

Could anyone explain with details this to me because I don't understand this :oops:[/quote]
It is not true.
Let $x_{i}=x \forall i$, then $8x^{2}=4n(n+1)$. It is equavalent to Pell's equation:
$(2n+1)^{2}-2y^{2}=1$.
 Because y must be even solution is $2n+1+y\sqrt 2 =(3+2\sqrt 2)^{k}$ or
$n=(\frac{(\sqrt 2+1)^{k}-(\sqrt 2-1)^{k}}{2})^{2}$.
It can be write as 2 series \[n=(\sum_{i=0}^{m}C_{2m+1}^{2i}2^{i})^{2}, \ (k=2m+1)\] and \[n=2(\sum_{i=1}^{m}C_{2m}^{2i-1}2^{i-1})^{2}, \ (k=2m).\]

[quote="silouan"]Yes it is on the Forum but there ,there is no clear solution and I want another from this. :)[/quote]
Really, really, please don't do this, the forum is already mess enough!

[quote="Rust"]I prove, that $n=2k$  works for all k>1 and $n=4k+1$ for all k works.
Case $n=4k+3$ had solution, when n is not prime.[/quote]
Could you post your solution ?

BTW, ZetaX sorry for this one but I made a year before .

[quote="perfect_radio"]See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=junior&t=5454]this topic[/url].[/quote]

Thank you but as we can see there is no complete solution . :(

Read the problem again... Your solution is wrong.

Yes, I forget about distinct. But it had distinct solution for positive densyty numbers of n (I check density at least $\frac{1}{8}$).
For example $n=3$ $a_{1}=1, a_{2}=1, a_{3}=6$. $5*4^{2}-4=76$ is not square.
I think it had solution for all n>2.

Solution exist for all n>2 and can be defined in this way:
Let M is set of exlusive numbers (subset $\{i_{1}\le n\}$ or $\{i_{1},i_{2}|i_{j}\le n\}$ or $\{i_{1},i_{2},i_{3}|i_{j}\le n\}$).

(1) $a_{i}=i$ if i odd and $i\not \in M$, $a_{i}=2i$ if i even and $i\not \in M$.

For example $n=8k+4$ give one exlucive number $M=\{4k+2\}$ and $a_{4k+2}=4k+2$ $a_{i}$ defined by (1) for not exlusive numbers.


I've been doing many Statewide and local contest past tests. Usually, the only problems I miss are from not seeing that similar triangles are involved and not knowing which one it is. How can you improve at those problems and recognizing them?

I'm worried.

[quote="nutz_for2.718281828"]Just ask yourself "Are there any similar triangles?" It worked in the harder years in Mathcounts.[/quote]
how will that help?

notice equal angles and ratios... those are indicators of similar triangles

If you don't know what to do on a geometry problem, just label what you know on a diagram and label what you can find from these primary labels and the answer should come to you.

ya that is probably the best approach..(i'm no expert)..but so if you see that two triangles have the same angles or two sides and an angle..you know they are similar..the easiest example is when u have a large triangle and a smaller triangle inside of it such that they share a vertex and one side is parallel to the side of the larger triangle..you can clearly see that they are similar..

if there are any parrallel sides, most likely there is going to be a similar triangle involved.  just draw a good diagram and see if anything looks special.

also try working bakwards from what you are asked to find. "If I knew that, then I would know the answer, if I knew this I would know that, if I knew that ratio I would know this..." and so on. If it comes down to a ratio that you need to know, then similar triangles are probably in play.


logarithms
trigonometry
function
LaTeX
inequalities unsolved
inequalities

notice $\ln \sin x$ is concave on $[0,\pi]$
then:\[ \sin A\sin^2 \frac{B}{2}\sin^3 \frac{C}{3} \leq (\sin \frac{A+B+C}{6})^6=(1/2)^6 \]

[quote="SuperMath"]Let ABC be a triangle. Prove that:
$\large SinAsin^2 \frac{B}{2}sin^3 \frac{C}{3} \leq (\frac{1}{2})^6$[/quote]

Dear, SuperMath, For some functions in LATEX, please use \sin, \cos, \tan, \cot, \ln, \log, \exp for a better look.
\[ \sin,\;\; \cos,\;\; \tan, \;\;\cot,\;\; \ln,\;\; \log,\;\; \exp  \]
I hope you will be a SuperLATEX.


Can the repeating decimal .9999.... be written as a fraction?

Some people argue that .9 repeating = 1

So that the fraction is 1/1. Personally, I do not agree with this because 1 does not yield .9 repeating, it yields 1.

In retrospect, there is no fraction that when you divide the numerator by the denominator, you get .9 repeating.

$0.\overline{9}=1$ for several reasons. For instance, you should notice that:

$\frac{1}{9}=0.\overline{1}$
$\frac{2}{9}=0.\overline{2}$
$\frac{3}{9}=\frac{1}{3}=0.\overline{3}$

etc. so that:

$\frac{9}{9}=0.\overline{9}$

but $\frac{9}{9}=1$ so $0.\overline{9}=1$.

If you think about this, this actually makes sense, and there are several ways that people prove this. I'll post the link to a thread about this in a sec.

Edit: Here's the thread :D. http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=0.999&t=25371

[quote="furious"]i don't care how you "prove" that .99999...=1, i still don't consider them equal (maybe i'm stubborn). They are completely different numbers with different names.  they might be almost the same, but they are not exactly the same. i say they are different and i will not change my mind (i've had a math teacher try to convince me that the two are equal)[/quote]

What's the sum of the geometric series
\[ \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots ?\]

The problem with $0.\overline{9}$ is not that it's some number different than 1.  The problem is that it's a way to write 1 which makes no sense.  The fact that we would never write 1 this way to do anything other than display a concept makes our intuition go haywire at first.


quadratics
arithmetic sequence
geometric sequence
geometric series

A sequence of three real numbers forms an arithmetic progression with a first term of $ 9$. If $ 2$ is added to the second term and $ 20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 36\qquad
\textbf{(D)}\ 49\qquad
\textbf{(E)}\ 81$

[hide] Say the common difference is x.  We have that the arithmetic sequence is 9, 9+x, and 9+2x.  Now, the geometric sequence is 9, 11+x, 29+2x.  Since it is a geometric sequence, we have that $\frac{9}{11+x}=\frac{11+x}{29+2x}\Rightarrow x^{2}+22x+121=18x+261\Rightarrow x^{2}+4x-140=0\Rightarrow (x-10)(x+14)=0\Rightarrow x=10,-14$.  Now, to make the third term as small as possible, we use x=-14.  This means that 29+2x=29-28=1, which is the third term.  Therefore, the answer is $A$. [/hide]

Sorry for revive this, but I have a question. The above solution is very clever and I think it is correct. However, I got something different, and I don't see what my error is.

[hide="My Solution"]
Let the arithmetic sequence be $9, \ x, \ y$. Then, since the difference between consecutive terms remains constant, we have: \[ y - x = x - 9 \implies 2x = y + 9 \implies y = 2x - 9 \] The geometric sequence is: \[ \dfrac {y + 20}{x + 2} = \dfrac {x + 2}{9} \implies \dfrac {2x + 11}{x+2} = \dfrac {x + 2}{9} \] Cross-multiply to obtain a quadratic and solve: \[ 18x + 99 = x^2 + 4x + 4 \implies x^2 - 14x - 95 = 0 \implies (x+5)(x-19) = 0 \] Thus, $x = -5 \ \text{or} \ 19$. The third term of the geometric series is $29+2x$. So, the desired minimum is $29+2(-5)=19$, which is none of the choices.
[/hide]

Any help would be greatly appreciated!
Thanks in advance,
~ Ahaan. :)

The third term of the geometric progression in your solution is $2x+11$  NOT $29+2x$ so you get $2(-5)+11=1$.

Hi! I was wondering if I could have some help with this, please...
[hide=Incorrect work]9, 9 + x, 9 + 2x
9, 11 + x, 29 + 2x

9y = 11 + x
11y + xy = 29 + 2x

9y^2 = 2x + 29
9y^2 = 18y +7
9y^2 - 18y + 7 = 0[/hide]

That results in irrational values for y, which don't make any sense.

My solution:
[hide=Solution]
Let the common difference of the arithmetic sequence be $x$.
This means that the three terms of the sequence is: $9, 9+x, 9+2x$
The geometric sequence of these three terms would be: $9, 11+x, 29+2x$
We create an equation:
$\frac{29+2x}{11+x}=\frac{11+x}{9}$
We crossmultiply:
$9(29+2x)=(11+x)(11+x)$
Expanding, we have:
$261+18x=x^2+22x+121$
Moving $261$ and $18x$ to the right side, we have:
$x^2+4x-140=0$
We factor:
$(x+14)(x-10)$
Therefore, $x=10$ or $x=-14$
To make this as small as possible, we choose $-14$ as the common difference, so we subtract $28$ from $9$, to get $-19$. We add, $-19+20=1$.
Our solution: $\boxed{1}$.

[hide=@vsar0406]I think the word ratio could tip you off to use division instead of multiplying by a common ratio. You want to find the common ratio in terms of $d$ so the problem is a one-variable equation. So instead of multiplying, you could have tried dividing.

[hide=My Sol]So let the arithematic sequence be \[9, 9 +d, 9 + 2d\] where $d$ is the common difference. So following the problem statement, we find that the geometric sequence is \[9, 11 + d, 29 + 2d.\] Note that these terms have a common ratio, so we know that \[\frac{11+d}{9} = \frac{29+2d}{11 + d}.\] Solving for $d$ we get $-14, 10$. Since we want the smallest we choose $d = -14$. So the third term in the geometric sequence is $29 + 2(-14) = 29 - 28 = 1$ which is A. [/hide][/hide]


wILL there also be a contest this spring?

there will be one, as soon as I find the time to correct the last two rounds of the 4th edition ;) ... :D


function calculus integration geometry probability

(If reading this thread before this afternoon) [b]Are you ready?[/b]

(If reading this thread later) [b]How'd you do?[/b]

(I must remind you however that we cannot discuss any specific questions on the exam)

I was ready, except I forgot a calculator. I got that worked out though, and had one for the test.

Did fine - it was stats, of all things. 5

[quote="Kamior"]Took it, fairly sure I didn't miss anything. It was absurdly easy compared to previous years.[/quote]

I agree completely. My teacher made us do 2002 and 2003 tests for homework, and this year's test was much easier and more straightforward (althought 02 and 03 weren't difficult either :)).

If I wanted to self study AP Statistics, how much time would I have to invest?  Do you think I should do it?

If you were pretty serious about learning it, I'd say you could do it between 2 weeks and 2 months; it depends on how fast you move through it. But, you have to take into account the boredom factor: stat is [b]so[/b] different from normal math and [b]so[/b] boring that you may need a break in the middle. And by a break I mean burning your stat book and giving up all hope at completing the study. ;)

I self-studied stat in two days. I think I got most of the problems, but not necessarily a 5.

My opinion on stat: it is so NOT math, I refuse to remember anything I learned in those two days.

So Stat AP takes 2 days? lolol

BTW... What book should I use?

Sadly enough, most stats books are terrible.  If you're stuck using a book, I don't think it matters too much which one.  What I despise most about the stats AP is that they like to take points off for small mistakes/inaccuracies sometimes.  Most of it is memorizing terms and other random annoying stuff.  You can't understand most of the formulas w/o lots of calc and other higher level math.  And yes, stats is definitely NOT math... hehe...

When I was taking APstats, I always integrated the normal distribution function in order to get the density functions because I didn't feel like learning how to use the calculator functions. You can certainly do worse than to get a proper textbook on statistics (read: one that is filled with integrals and hopefully measures) and learn from that. If you use your knowledge on the AP exam, you'll probably confuse the graders, but if you solve a problem correctly by using approaches the graders don't know, I doubt they can justify taking off points (although now that I think about it, maybe they take off points for things like that on chemistry) for solving problems with calculus and measure theory.

And, just for fun, here is a math joke about statisticians that came up in my differential geometry class last week:

Statisticians are probably mathematicians.

[quote="Ragingg"]So basic calculus is required for AP Statistics?[/quote]
No. Basic (read: b-a-s-i-c) algebra is required for about three AP stats problems, which appear in the same form every time, just with different numbers. As for the rest of the problems, the knowledge expected to come out of your head ends at vocabulary and perhaps multiplication. Multiplication if you're lucky, that is. Usually just addition and subtraction.

It's just that it [i]can[/i] be done with Calculus, instead of looking things up in tables and using the calculator, which is the way it is taught.

I took it, it [i]did[/i] seem easier than practice free responses I had done... I've never taken any AP exam before though so I have no idea about the curve.

Ragingg, AP statistics is very simple, lots of people say it's the easiest AP, definitely don't waste time taking it at school, just self-teach...

The hardest thing on the exam that has to do with math is probability, which, if you're on this forum, probably isn't hard for you. You [i]can[/i] use calculus, but it's not necessary at all, they give you tables to calculate things.

I just used the Barron's review book (the course is that easy) and the free AP Free Response questions from the college board's website and I think I turned out okay... (watch out though, Barron's books have lots of mistakes, if something is weird, it's probably wrong) I took from December to May but I was lazy, a month or two is all you really need...

I did Barrons' as well and took a Stats course at the local community college. Shouldn't have done the latter; Barrons' is more than enough. In fact, some of the Barrons' questions are harder than those on the actual AP. -.-

I agree, Barrons is the best.  If you are taking the exam without taking the class, don't bother using a textbook, just get Barrons.  If you do get a textbook, get An Introductory to Statistics by Yates, Moore, and McCabe.  It is supposed to be the best, I think about 80% of schools use it.


A given convex polyhedron has no vertex which belongs to exactly 3 edges. Prove that the number of faces of the polyhedron that are triangles, is at least 8.

By the given condition $2E=\Sigma \deg v_i\ge 4V$ or $2V\le E$. Denote by $F_\Delta$ the number of triangular faces. Then $2E=\Sigma \deg f_i=3F_\Delta+k(F-F_\Delta)$. By the Euler's formula $4=2V-2E+2F\le 2F-E$ and
$$8\le 4F-2E=4F-3F_\Delta-k(F-F_\Delta)\le 4F-3F_\Delta-4F+4F_\Delta=F_\Delta.$$


In triangle ABC perpendiculars from C, A and B meet AB, BC and AC at D, E and F respectively. Prove that $2.CD^2=2.AD.BD+CF.AC+CE.BC$

Let the orthocentre be $H$.

Since $\triangle{CEH} \sim \triangle{CDB}$ so $CE \cdot BC = CH \cdot CD$. Also, $\triangle{AHD} \sim \triangle{CBD}$, so $AD \cdot BD= CD \cdot HD$. Hence:
\[AD \cdot BD+CE \cdot BC= CD\cdot (CH+HD)=CD^2\]
Since $\triangle{CFB} \sim \triangle{CEA}$, so $CE \cdot CB = CF \cdot CA$. Hence:
\[2CD^2=2AD \cdot BD + CF \cdot CA + CE \cdot CB\]

Another problem: (I'll change the notation.) Let $AD \perp BC$, $BE \perp AC$, $CF \perp AB$. And $H$ is the orthocentre. Prove:
\[AH \cdot AD + BH \cdot BE + CH \cdot CF = \frac{a^2+b^2+c^2}{2}\]

[hide="Solution"]$\triangle{AFH}\sim\triangle{ADB}$  so  $AH\cdot{AD}=AF\cdot{AB}$
Also, $\triangle{BFH}\sim\triangle{BEA}$  so  $BH\cdot{BE}=AB\cdot{BF}$
Thus $AH\cdot{AD}+BH\cdot{BE}=AB(AF+FB)=c^2$
$\triangle{CDH}\sim\triangle{CFB}$  so  $CH\cdot{CF}=a\cdot{CD}$

Now we have
$c^2+a\cdot{CD}=\frac{a^2+b^2+c^2}{2}$
$c^2=a^2+b^2-2a\cdot{CD}$
$c^2=a^2+b^2-2ab\cdot\frac{CD}{b}$
$c^2=a^2+b^2-2ab\cdot{cos C}$, which is true.[/hide]


The question is:
$\boxed{\text{Find the number of ways to split } 10^{15} \text{ into the sum of two perfect squares.}}$

I will wait for someone to post a solution, I have the nice trick first step or two I will post soon...

I'm not quite sure I see the bearing behind Altheman's hint...

Shyong nailed it with his post, those identities were what I based this problem on when writing it.

[hide="Here is my tentative solution for the maximum value:"]
Using the factorization Shyong used, we see that $(1^2+1^2)^{15}$ has exactly one way to split up into two squares:
Since $c=1=d$, the two ways to factor $(a^2+b^2)(c^2+d^2)$ -- $(ac+bd)^2+(ad-bc)^2$ or $(ac-bd)^2+(ad+bc)^2$ -- are the same. Thus if we perform $2^{15}$ in 14 steps, we see that for each step there is a unique value for a,b,c, and d.
Now we see that there are at most $2^{14}$ ways to split up $5^{15}$, therefore giving a maximum of $2^{14}$ for the answer to the problem.
Note: this is the brief sketchy halfbaked partly correct solution, lacking any proof at all, by counterexample I have already found there is definitely less than $2^{14}$ solutions. So NO ONE criticize this...
[/hide]

[quote="Altheman"][hide="hint"]mod 4[/hide][/quote]

Well if the question said $10^{15}+3$, then it would definitely help... :P

$a^2+b^2=2^{15}*5^{15}$, looking in mod 4, we see that the numbers must be 0 mod 4, and we repeat that to get

$a^2+b^2=2*5^{15}$, then you can repeat this looking mod 5, and the perfect square residues are -1,0,1,

either we have that the remainders are 0,0 or -1,1 mod 5

maybe some case work?

this helps some...but we still have a lot of case work that we would need to do, so there is still some insight that needs to happen, if we could get the LHS to be a perfect square, we could use the primitive pythagorean triples thing

[hide="I worked out the answer:"][size=150]3[/size][/hide]
Now wait for the proof and explanation...

it's interesting how after the first several terms, the two new possibilites produces by each of the three cancel out with another random repeat. I advise you to work out the first several terms; it is interesting to look at the subtle patterns.


[quote="MoreSubtleThan1+1"]Hello; I'm new at this forum!  Well, here's the problem...

Five girls (Alexandra, Betsy, Catherine, Deyola and Emily) travel with one boy (Frank) to a math contest.  They have four hotel rooms, numbered 1 through 4.  Each room can hold up to two people, and the boy has to have a room to himself.  HOw many different ways are there to assign the students to the rooms, including the way shown here?

Room 1   Room 2   Room 3   Room 4
  A,D        C,E          F           B

[hide]Answer is 360.[/hide]

I don't know how to solve this...
I used multiple different methods but still couldn't arrive at the answer.[/quote]

Seeing as there has to be 2 rooms with two girls...
The pairs would be $ \binom{5}{2}$
So there 10 possible pairs.
So in one of the rooms, there is one pair, meaning ten choices for the pair, and four choices for the room. So 40.
In the other room, you need to choose another pair, out of the three remaining girls, and there are three rooms to choose from, so 9.
In another room, you need to have a girl.
In another room, you need to have the boy.
Wait...
I get 720...
Argh.
I'll rethink this tomorrow.

In the first room, you can stick any of the 6 people.  In the second, you can stick any of 5.  In the third, you can stick any 3, and in the fourth, you can stick any 3.  $ 6\cdot5\cdot4\cdot3\equal{}360$.  I have no clue if this is right (probably not) or why it works.  I never like combinatorics.

[quote="JRav"]In the first room, you can stick any of the 6 people.  In the second, you can stick any of 5.  In the third, you can stick any 3, and in the fourth, you can stick any 3.  $ 6\cdot5\cdot4\cdot3 \equal{} 360$.  I have no clue if this is right (probably not) or why it works.  I never like combinatorics.[/quote]
It just works because fortune made it so.
But yeah...
Not really.
Because you aren't following the rules of the room things.

Wow, the Fundamental Counting Principle yet again...

Thank you.  If there are any other solutions (I'm talking to whoever is out there reading this), please tell me.  My target/team rounds on the school level are tomorrow, and I always had problems with counting (which I heavily studied recently), but sometimes I don't know which of the many formulas for counting to use and in what order.  I believe that the more ways I know to solve a problem, the greater chance of me finding one of those ways on the actual test!

EDIT:
But wait, when you did the 5P2 thing, don't you do 5!/3! since it's n!/(n-r)! rather than 5!/2!?  That seems to throw things out of balance.  Can you explain?

I kinda have an alternate solution

[hide]
We already know that the boy must be lonely, so all we need to find is the ways to put 5 girls in three rooms given the above.  Experimentation shows that there must be 2-2-1 or any permutation of that, so we have $ \binom{5}{2}\binom{3}{2}\binom{1}{1}\equal{}30$.  Then, we have 3 permutations of 2-2-1, so there are $ 90$ ways.  However, the boy could have been in any of 4 rooms to begin with, so we multiply by 4 to get $ \boxed{360}$. (note the use of symmetry)
[/hide]

[quote="MoreSubtleThan1+1"]Wow, the Fundamental Counting Principle yet again...

Thank you.  If there are any other solutions (I'm talking to whoever is out there reading this), please tell me.  My target/team rounds on the school level are tomorrow, and I always had problems with counting (which I heavily studied recently), but sometimes I don't know which of the many formulas for counting to use and in what order.  I believe that the more ways I know to solve a problem, the greater chance of me finding one of those ways on the actual test!
[/quote]
Oh wait, my solution is messed up.
Rewriting.

you have to know what the question is saying before you can jump to that conclusion

Wait Alan, post your solution, I just found out that my solution makes no sense.
 :|

Yeah I used the same method as Alan. It's really important to note that the information about the boy bassically means that he can be in any room, all we have to do is figure out how we can put 5 girls into 3 rooms (with max 2 into each room), then multiply by 4.


vector
combinatorics theorems
combinatorics

1.  Determine the number of vectors $ (x_1,...,x_n)$ such that each $ x_i$ is either 0 or 1, and $ \displaystyle\sum_{i\equal{}1}^n x_i >\equal{}k$

2.  How many vectors $ (x_1,...,x_k)$ are there for which each $ x_i$ is a positive integer such that $ 1<\equal{}x_i<\equal{}n$ and $ x_1<x_2<...<x_k$ ?

[hide]
1.  For each $ k \le j \le n$, there are $ \binom{n}{j}$ ways to choose the $ x_i$'s such that $ x_i \equal{} 1$, and the others are all zero.  Their sum is at least $ k$ by construction, and by summing over all such $ j$, we obtain

$ \sum_{j\equal{}k}^n \binom{n}{j}$.

No closed form expression for this sum is known for general $ n$ and $ k$.

2.  Suppose we have $ n$ balls, numbered from 1 to n inclusive.  Then we obtain an admissible vector $ (x_1, x_2, \ldots, x_k)$ by choosing exactly $ k$ of the $ n$ balls.  Conversely, every admissible ordered k-tuple corresponds to a choice of $ k$ balls, because no value may be repeated.  Therefore, the answer is simply

$ \binom{n}{k}$.
[/hide]


In round 4 question 1:   If we get answers for $(a,b)$ or $(c,d)$ --> $(x,y)$, do we need to write $(y,x)$ as a seperate answer?

We cannot answer questions about the round while it is in progress.


inequalities
trigonometry
function
inequalities open

[quote="borislav_mirchev"]Prove that: $ \xi_1 \xi_2 \sin^2 \varphi_3 \plus{} \xi_2 \xi_3 \sin^2 \varphi_1 \plus{} \xi_3 \xi_1 \sin^2 \varphi_2 \le \frac {1}{4}$
if $ \xi_1 \plus{} \xi_2 \plus{} \xi_3 \equal{} 1$ and
$ \xi_i$ - positive reals, $ \varphi_i$ - internal triangle angles.[/quote]
It's equivalent to $ \left(\xi_1 \plus{} \xi_2\cos2\varphi_3 \plus{} \xi_3\cos2\varphi_2\right)^2 \plus{} \left(\xi_2\sin2\varphi_3 \minus{} \xi_3\sin2\varphi_2\right)^2\geq0.$

what are the indexes in the second square? It is excellent idea, but how did you "guess" it?

[quote="borislav_mirchev"]what are the indexes in the second square?[/quote]
I have fixed it. Thank you!

the inequality is equivalent to $ x^2\plus{}y^2\plus{}z^2\minus{}2(xy\cos \alpha\plus{}yz\cos \beta\plus{}zx\cos \gamma)\geq 0$, where $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$.

if you consider the discriminant of this function it equals $ \minus{}4(y\sin \beta\minus{}z\sin \gamma)^2$, and we're done.

btw, $ x,y,z$ and $ \alpha,\beta,\gamma$ don't need to be positive..

[quote="borislav_mirchev"]Prove that: $ \xi_1 \xi_2 \sin^2 \varphi_3 \plus{} \xi_2 \xi_3 \sin^2 \varphi_1 \plus{} \xi_3 \xi_1 \sin^2 \varphi_2 \le \frac {1}{4}$
if $ \xi_1 \plus{} \xi_2 \plus{} \xi_3 \equal{} 1$ and
$ \xi_i$ - positive reals, $ \varphi_i$ - internal triangle angles.[/quote]

Have a look at Proposition 1.2.1 in the book: http://ultrametric.googlepages.com/tin2007.pdf


geometry
circumcircle
IMO Shortlist
pentagon

Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$.  Prove that the line $AP$ bisects the side $CD$.

[i]Proposed by Zuming Feng, USA[/i]

if the quadrilaterals ABCP and AEDP are cyclic then ADB and AEC are isoceles. am  i  missing something???

Hasan, I don't think that the triangles you mention are isosceles.
Please have a look at my picture.

Megus, I really like your proof. It is very nice!  :lol:

I found two proofs of this nice and easy problem.
Look at the picture for names of points.

[u][b]Proof 1:[/b] A very short, straightforward one:[/u]

Clearly, there exists a spiral similarity with center $A$, angle $BAC$ and factor $\frac{AC}{AB}$ that maps $ABCD$ onto $ACDE$.
Because $F$ and $G$ are the intersection points of the diagonals of two quadrilaterals that are similar, we know that
\[\frac{\overline{FA}}{\overline{FC}}=\frac{\overline{GA}}{\overline{GC}}\]
So if we call $\left\{S\right\}=AP \cap CD$, we know thanks to Ceva's theorem:
\[-1=\frac{\overline{FA}}{\overline{FC}}.\frac{\overline{SC}}{\overline{SD}}.\frac{\overline{GD}}{\overline{GA}}=\frac{\overline{SC}}{\overline{SD}}\]
So clearly $S$ is the midpoint of $CD$

[u][b]Proof 2:[/b] My second proof, which I think is more beautiful:[/u]

Clearly, $\Delta ABC, \Delta ACD, \Delta ADE$ are similar. Therefor, we can use the alternating segment theorem to prove that both $BC$ and $DE$ are tangent to the circumcircle of $\Delta ACD$ in points $C$ and $D$ respectively. If we call $Q$ the intersection of $BC$ and $DE$, we know that $AQ$ is a symmedian of $\Delta ACD$.

Now, because $\angle ACB = \angle AED$, we know that $ACQE$ is cyclic and $\angle CAQ = \angle CEQ$ (1)

Because $\angle ADB = \angle AEC$, we know that $AEDP$ is cyclic and $\angle PAD = \angle PED$ (2)

Combining (1) and (2), we get that $AP$ is the isogonal conjugate of $AQ$, and therefor, it is the median. The conclusion follows.

I hope you liked it  :)

[img]http://img262.imageshack.us/img262/8164/mathlinkssymmedianxk5.jpg[/img]

Thank you dear [b][size=100]jan[/size][/b], for the enjoyment from the beautiful ''proof 2'', of the [b][size=100]mattilgale's[/size][/b] problem.

Best regards, Kostas vittas.

[quote="Megus"]we have (from similarity of ABC, ACD,ADE):

$\frac{AB}{BC}=\frac{AC}{DC}=\frac{AD}{ED}$ and
$\frac{BC}{AC}=\frac{CD}{AD}=\frac{ED}{AE}$ from which we obtain:

$AB \cdot AD =AC^{2}$ and
$AC \cdot AE = AD^{2}$ and hence

$AB \cdot AE = AC \cdot AD$ or otherwise
$\frac{AE}{AC}=\frac{AD}{AB}$  hence triangles $ABD$ and $AEC$ are similar. Now it's just angle chasing. Use this similarity to have some equal angles and see that $ABCP$ and $AEDP$ are cyclic and tangent to line $CD$ respectively at $C$ and $D$. Hence if $M$ is intersection of AP and CD then $MD^{2}=MP \cdot MA = MC^{2}$ so $M$ is midpoint of $CD$ as wanted. QED[/quote]

As a remark to the [b][size=100]Megus’s[/size][/b] nice solution, we can say that the quadrilaterals $ABCD,$ $ACDE$ are similar, because of their angles, are equal one per one and so, we conclude that their homologous segment lines, form also equal angles.

Hence, as another approach, we have $\angle ADB = \angle AEC$ $,(1)$ and $\angle BDC = \angle CED$ $,(2)$

From $(1)$ we conclude that the quadrilateral $APDE$ is cyclic and so, $\angle DPE = \angle DAE = \angle DAC$ $,(3)$

From $(3)$ we have that the quadrilateral $AFPG,$ is also cyclic.

Now, it is easy to show that $\angle GFP = \angle PAD = \angle CED$ $,(4)$

From $(2),$ $(4)$ $\Longrightarrow$ $\angle GFP = \angle BDC$ $\Longrightarrow$ $FG\parallel CD$ $,(5)$

From trapezium $CDGF,$ we conclude that the segment line $AP$ bisects the segment $CD$ and the proof is completed.

Kostas Vittas.

Dear  Megus ,

 can you  please post  your solution with a little  more detail .  especially the angle chasing part,  ive tried to  finish it but cant find how.  thank  you  very  much

[quote="vittasko"]
As a remark to the [b][size=100]Megus’s[/size][/b] nice solution, we can say that the quadrilaterals $ABCD,$ $ACDE$ are similar, because of their angles, are equal one per one and so, we conclude that their homologous segment lines, form also equal angles.
[/quote]

Also we can say that their homologous points, divide the correspondent segments in to the same ratio.

That is from the similarity of  $ABCD,$ $ACDE,$ as a direct conclusion we have that $\frac{FA}{FC}= \frac{GA}{GD}$ and so, $FG\parallel CD.$

Kostas Vittas.

Well, with simple angle chasing one can see that $ABCP$ and $APDE$ are cyclic and their radical axis is $AP$.
It is also easy to see that $CD$ is a common tangent of circumcircles $\triangle ABC$ and $\triangle DEA$.
Therefore $AP$ passes through a point M in CD such that $MC^{2}=MD^{2}$.

As [b][size=100]Darij Grinberg[/size][/b] said at http://www.mathlinks.ro/Forum/viewtopic.php?t=154396, we can have a solution as a neat application of [b][size=100]Jacobi theorem[/size][/b].

Through vertices $C,$ $D$ of the given pentagon $ABCDE,$ we draw two lines parallel to $AD,$ $AC$ respectively and we denote as $F,$ their intersection point.

So, it is easy to show that the segment lines $DE,$ $DF,$ are isogonal conjugates with respect to the angle $\angle ADC$ and similarly, the segment lines $CB,$ $CF,$ are isogonal conjugates wrt $\angle ACD.$

Hence, because of $\angle DAE = \angle CAB,$ based on the [b][size=100]Jacobi theorem[/size][/b], we conclude that $P\equiv AF\cap CE\cap DB.$

Because of now the parallelogram $ADFC,$ we have that the segment line $AP,$ passes through the midpoint of the side segment $CD$ and the proof is completed.

Many thanks to [b][size=100]Darij,[/size][/b] Kostas Vittas.

I hope it is a nice solution, too

Let's call the interception of the diagonals F and G and M the middle of CD
Since ABC is similiar to ACD and ADE it is obvious that ABCD is similiar to ACDE
Therefore AF/FC = AG/GD
Ceva:
AF/FC*CM/MD*DG/GA=1, which is obviously true

Call the intersection of (ABC) and (ADE) is X. easy to prove that X = P. We can prove that CD is common targent of (ABC) and (ADE). Hence, AP is axial radical of (ABC) and (ADE) so AP passes through the midpoint of CD

Since triangle $ABC$ is similar to triangle $ADE$, $A$ is the spiral centre which takes $B$ to $C$ and $D$ to $E$, hence by a well known lemma, the point of intersection of $CE$ and $BD$ is concyclic with $ABC$ and $ADE$. Further, by easy angle chasing and the alternate segment theorem, $CD$ is tangent to both of these circles, hence since the power of the point of intersection of $AP$ and $CD$, $M$, is equal to $MP.MA$ for both circles, it is immediate that $MC$=$MD$.

[hide="Solution"]
From the angle equalities, we deduce that $\triangle{ADE}\sim\triangle{ACD}\sim\triangle{ABC}$ by AA similarity. Then a spiral similarity centered at $A$ takes $E$ to $C$ and $D$ to $B$. Therefore, $\triangle{AEC}\sim\triangle{ADB}$, so $\angle{AEC}=\angle{ADB}$ and $\angle{ACE}=\angle{ABD}$. It follows that $APDE$ and $APCB$ are cyclic, and line $AP$ is the radical axis of $(ADE)$ and $(ABC)$.

Furthermore, $\triangle{ADE}\sim\triangle{ACD}$ implies $\angle{AED}=\angle{ADC}$, so line $CD$ is tangent  to $(ADE)$. From $\angle{ABC}=\angle{ACD}$, we see that the line $CD$ is tangent to $(ABC)$. Hence, $CD$ is tangent to both circles.

Let line $AP$ intersect $CD$ at $M$. Note that $M$ lies on the radical axis, $AP$, of the two circles, so $MD^2=MC^2\implies MD=MC$. As a result, the line $AP$ passes through the midpoint of $CD$. // [/hide]

Let $R=AC \cap BD, Q=EC\cap AD, K=AE\cap CD, L=AD\cap BC$ and $M=AP \cap CD$
Ceva's theorem gives \[ \frac{MD}{MC}=\frac{QD}{QA}\frac{RA}{RC}\]
Menelaus' theorem in $ACL$ with line $D,R,B$ gives \[\frac{RA}{RC}=\frac{BC.DL}{BL.AD}\]
Menelaus' theorem in $ADK$ with line $E,Q,C$ gives \[\frac{QD}{QA}=\frac{CK.AE}{CD.KE}\]
$DLC$ and $EKD$ are similar so \[\frac{DL}{KE}=\frac{CD}{DE}\]
$ABC$ and $ADE$ are similar so \[\frac{BC}{DE}=\frac{AC}{AE}\]
$ABL$ and $ACK$ are similar, $AC$ and $AD$ are respectively their angles bisector  from $\hat{A}$ so \[\frac{AC}{BL}=\frac{AD}{CK}\]
so \[\frac{MD}{MC}=\frac{RA}{RC}\frac{QD}{QA}=\frac{BC.DL}{BL.AD}\frac{CK.AE}{CD.KE}=...=\frac{AC.CK}{BL.AD}=1\]
then $MC=MD$ as required ;)

[quote=newinolympiadmath][quote=metricpaper]Let $M$ be the intersection of ray $AP$ with line $DC$. Note that $$\triangle ABC\sim \triangle ACD \sim \triangle ADE.$$ Then quadrilaterals $AEDC$ and $ADCB$ are similar, so $\triangle DYC\sim \triangle EXD$ and $\triangle EXA\sim \triangle DYA$. From this we get $$\frac{CY}{XD}=\frac{DY}{EX}=\frac{YA}{AX},$$,so $\tfrac{CY}{YA}\cdot \tfrac{AX}{XD}=1$. By Ceva's, we have $\tfrac{DM}{MC}\cdot \tfrac{CY}{YA}\cdot \tfrac{AX}{XD}=1$, so $DM=MC$, i.e. $AP$ bisects $CD$.[/quote]
Then quadrilaterals $AEDC$ and $ADCB$ are similar, so $\triangle DYC\sim \triangle EXD$ and $\triangle EXA\sim \triangle DYA$

Can you please explain how exactly this implifies similarities?[/quote]

I'm sorry, I forgot to write that $X$ is the intersection of $EC$ with $AD$, and $Y$ is the intersection of $BD$ with $AC$.

What I meant by quadrilaterals $AEDC$ and $ADCB$ being similar is that one is a dilation and rotation of the other, with all angles being preserved. So the corresponding triangles inside them must be similar as well.

Note that the condition is equivalent to
\[
    \triangle ABC \sim \triangle ACD \sim \triangle ADE
\]

As such, by SAS congruence it follows that $\triangle ABD \sim \triangle ACE$
and that $\triangle BCD \sim \triangle CDE$.

Note that $\measuredangle AEP = \measuredangle ADP$ implies that $AEDP$ is cyclic.
Similarily, $ABCP$ is cyclic.

Then, since $\measuredangle PDC = \measuredangle PED$ and $\measuredangle PCD = \measuredangle PBC$
it follows that $(AEPD)$ and $(ABCP)$ are both tangent to $CD$.

As such, $CD$ is the common tangent of the two circles so the radical axis $AP$ bisects the tangent.

Why was this a g3 skull its basic sim triangles

Let F and G denote the intersection of diagonals AC with BD and AD with CE.
ABC ∼ ACD ∼ ADE. Consequently (by easy angle chasing, for example I did ABC=ACD, DAB=EAC, BCD=CDE) quadrilaterals ABCD and ACDE are similar. Then the ratios of the way AC is intersected by BD is preserved in the other shape (think of this as a dilation of the diagonals, dilations don't change ratios). Then AF/FC=AG/GD. Let AP intersect BC at M. Ceva’s theorem applied to triangle ACD implies AF/FC*CM/MD*DG/GA=1, and substituting earlier we get CM/MD=1. $\blacksquare$

Alternatively:
ABCP and APDE are cyclic by properties of spiral sims with P intersection of BD and CE. By the given angle equalities we get that CD is a tangent to both circles, hence AP the radical axis bisects the tangent $\blacksquare$

Let $X$ be where $AC$ intersects with $BD$, $Y$ be the point where $AD$ intersects with $CE$, and $Z$ be the point where $AP$ intersects with $CD$
Because of $AA$ similarity, $ABC$, $ACD$, and $ADE$ are similar, so $ABCD$ and $ACDE$ are similar and $\frac{AX}{XC}=\frac{AY}{YD}$
In triangle $ACD$, by ceva, $\frac{CX}{XA} * \frac{AY}{YD} * \frac{DZ}{ZC} = \frac{YD}{AY} * \frac{AY}{YD} * \frac{DZ}{ZC} = \frac{DZ}{ZC} = 1$, so $DZ=ZC$ and $AP$ bisects $CD$

Notice that $\angle ABC = \angle ACD$ and $\angle ADC = \angle AED$ by using the similarities given by the angle condition. So $\overline{CD}$ is tangent to $(ABC)$ and $(ADE)$. The result follows by radical axis.

[asy]
        /*
            Converted from GeoGebra by User:Azjps using Evan's magic cleaner
            https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
        */
        /*
            A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings.
        */
        pair A = (-24.48633,93.16852);
        pair B = (-90.79323,17.89501);
        pair C = (-20.84376,-46.81005);
        pair D = (112.16666,-46.58445);
        pair E = (248.03535,79.95556);
        pair Q = (45.76614,-108.42595);
        pair P = (32.98903,-21.43006);

        import graph;
        size(14.9cm);
        pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);

        pen cqcqcq = rgb(0.75294,0.75294,0.75294);
        draw(arc(A,25.75813,-131.37617,-88.50936)--A--cycle, linewidth(0.75) + blue);
        draw(arc(B,17.17209,-42.76962,48.62382)--B--cycle, linewidth(0.75) + red);
        draw(arc(B,19.17209,-42.76962,48.62382)--B--cycle, linewidth(0.75) + red);
        draw(arc(A,28.62015,-88.50936,-45.64256)--A--cycle, linewidth(0.75) + blue);
        draw(arc(C,20.03410,0.09717,91.49063)--C--cycle, linewidth(0.75) + red);
        draw(arc(C,22.03410,0.09717,91.49063)--C--cycle, linewidth(0.75) + red);
        draw(arc(A,31.48217,-45.64256,-2.77575)--A--cycle, linewidth(0.75) + blue);
        draw(arc(D,22.89612,42.96398,134.35743)--D--cycle, linewidth(0.75) + red);
        draw(arc(D,24.89612,42.96398,134.35743)--D--cycle, linewidth(0.75) + red);

        draw(A--B, linewidth(0.5));
        draw(B--C, linewidth(0.5));
        draw(C--A, linewidth(0.5));
        draw(C--D, linewidth(0.5));
        draw(D--A, linewidth(0.5));
        draw(D--E, linewidth(0.5));
        draw(E--A, linewidth(0.5));
        draw(C--E, linewidth(0.5));
        draw(D--B, linewidth(0.5));
        draw(C--Q, linewidth(0.5));
        draw(Q--D, linewidth(0.5));
        draw(A--Q, linewidth(0.5) + blue);
        draw(P--A, linewidth(0.5) + blue);

        dot("$A$", A, N);
        dot("$B$", B, NW);
        dot("$C$", C, SW);
        dot("$D$", D, SE);
        dot("$E$", E, NE);
        dot("$Q$", Q, dir(270));
        dot("$P$", P, dir(270));
[/asy]

Let $Q=BC\cap DE$.

Now we have that,
\begin{align*}
    \measuredangle QCD&=\measuredangle BCD\\
    &=\measuredangle BCA+\measuredangle ACD\\
    &=\measuredangle BCA+\measuredangle ABC\\
    &=\measuredangle BAC\\
    &=\measuredangle CAD
,\end{align*}
which gives us that $QC$ is tangent to $\odot(ACD)$. Similarly, we also get that $QD$ is tangent to $\odot(ACD)$ which further means that $AQ$ is the $A$-symmedian in $\triangle ACD$.

Now to finish, note that we clearly have that $\angle BAE$ and $\angle CAD$ share the common angle-bisector. So now by Isogonal Line Lemma on $\left\{AB,AE\right\}$ and $\left\{AC,AD\right\}$, we get that $AP$ where $P=BD\cap CE$ is isogonal to $AQ$ where $Q=BC\cap DE$. This means that $AP$ is the $A$-median in $\triangle ACD$ as we already had that $AQ$ is the $A$-symmedian which finishes.
-----

[b]Remark:[/b] We actually get a ton of other cyclicities like $\left\{\odot(ABCP),\odot(APDE),\odot(ABQD),\odot(ACQE)\right\}$ as $A$ is simply the center of spiral similarity taking $\overline{BC}\mapsto\overline{DE}$, that also maps $\overline{BD}\mapsto\overline{CE}$.

I just wanted to post a different solution using [b]isogonal lines lemma[/b]. It states that for a triangle ABC with pairwise isogonal lines AD-AG and AE-AF, then DE with FG and DF with EG intersecting at H and I yields that AH and AI are isogonal lines.
[i]Sketch.[/i] For J,K the intersections of AH with DF and EG, x=<PAS=<QAT, y=<SAX and z=<TAY, we have that sinKAP/sinKAY : sinTAP/sinTAY=(P,Y;K,T)=(S,Y;L,Q)=sinLAS/sinLAY : sinQAS/QAY after a flurry of simplifications reduces to sin(x+y)sinz=sin(x+z)siny, which reduces to tany=tanz or y=z. $\square$

Now, in the problem, AC-AD and AB-AE are isogonal line pairs, so for BD intersecting EG at F and DE intersecting BC at G, we get that AF-AG is an isogonal line pair as well (all wrt <CAD). Then, <CDG=180-ADC-ADE=180-ADC-ACD=CAD, so DG and analagously CG are tangent lines to (ACD). Now, it's clear that AG is a symmedian of ACD, whence AF is the median isogonal to it. $\blacksquare$

Took me some time to figure it out... I started by thinking that maybe Trigonometric form of Ceva's Theorem works... But simple stuff worked out.

[i]Solution:[/i] One can easily see that $ABCD \sim ACDE$. This would give us the ratio that
    \[\frac{AY}{YC} = \frac{AX}{XC}.\]
One application of Ceva's Theorem in $\triangle ADC$ at point $P$ would now solve the problem. $\blacksquare$

[hide=Lol] Let $F=AC\cap BD$ and $G=AD\cap CE$. Clearly $ABCDF\sim ACDEG$; specifically $\tfrac{AF}{CF}=\tfrac{AG}{DG}$, whence the result is trivial by Ceva's on $\triangle ACD$. [/hide]

Let $BD\cap AC=X$ and let $CE\cap AD=Y$. Notice that $ABCDX\sim ACDEY$ and so $\frac{AX}{XC}=\frac{AY}{YD}$. Finish by Ceva's.

Draw initially the points $A, B, C$ counter-clockwise, defining the rest, clearly, counter-clockwise too.
Notice that $CD$ is tangent to $(ABC)$ and $(AED)$, because
[hide]$\angle ACD= \angle ABC$ and $\angle AED= \angle ADC$ (well, it's given that $\angle CAD= \angle DAE$ and $\angle ACD= \angle ADE$, so you can notice by similar triangles or just that these equalities implies for sure that $\angle AED= \angle ADC$, as wanted).[/hide]
Now, let's redefine $P$. Let $P\neq A$ the second intersection of $(ABC)$ and $(ADE)$.
[hide=some formality, but you can skip this]
By the conditions that we put in our pentagon in the first line, we can ensure that $P$ lies in the arc $AC$ that doesn't contain $B$ and lies in the arc $AD$ that doesn't contain $E$.[/hide]
Finally, we can easily prove that $C, P, E$ are collinear (similar for $D,P, B$), this happens because $\angle APC= \angle BAC+\angle ACB$ and $\angle EPA= \angle ADE$ and look: $\angle BAC+\angle ACB+\angle ADE= \angle BAC+\angle ACB+\angle ABC= 180$.

We can finish from here, let $M= AP\cap CD$, so $M$ lies in the radical axes of $(ABC)= \omega_1$ and $(ADE)=\omega_2$

$Pow_{\omega_1}M= MC^2= MP\cdot MA$
$Pow_{\omega_2}M= MD^2=MP\cdot MA$
$\implies MC^2=MD^2\implies MC=MD$, as desired.

Let $\overline{BD}$ intersect $\overline{AC}$ and $F$, let $\overline{CE}$ intersect $\overline{AD}$ at $G$. Since quadrilateral $ABCD$ is similar to quadrilateral $ACDE$, it follows that $\frac{AF}{AC} = \frac{AG}{AD}$, hence by Ceva's theorem, $\overline{AP}$ bisects $\overline{CD}$.

Let $Q$ and $R$ be the intersections of $AC$ with $BD$ and $AD$ with $CE$ respectively. As quadrilaterals $ABCD$ and $ACDE$ are similar, $\frac{AQ}{QC} = \frac{AR}{RD}$. The result follows as a direct consequence of Ceva's theorem.

$\color{magenta}\boxed{\textbf{SOLUTION G3}}$

$\triangle ABC$ ∼ $\triangle ADE$ $\implies A$ is the center of the spiral similarity $S : BC \to DE.$  Hence $BD \cap CE \equiv P \implies A$ lies on the circumcircles of $\triangle PBC$ and $\triangle PDE$ that is $ABCP$ and $APDE$ are cyclic.
$\angle ABC = \angle ACD \implies  CD$ is tangent to $(ABC)$ Similarly, $CD$ is tangent to $(ADE)$
Let $M\equiv AP \cap CD$
By $\textbf{POP,}$ $$MD^2=MP.MA=MC^2 \implies MD=MC \blacksquare$$


Hello can someone help me simplify the following please:

1/x + 1/(x-a) + 1/(x-2a)....1/(x-na)

0 < x < 1

a is small and n is finite.

For my specific problem x is .98, n is 90 and a is .00001

Thanks for any pointers!

Jonathan.

Thanks for that Rust.  However surely multiplying by 'n/a' would make the last term very large, should that be '1/a' instead?

Cheers.


Assume in a democracy there is a $\frac{1}{3}$ chance of there being at least one revolution in a given year. What is the probability that over 10 years 3 of those years contained at least one revolution?

[quote="turboturtle"][quote="b-flat"][hide]There are $\binom{10}3$ ways to choose which 10 years contain a revolution.  The probability that 7 years do NOT have a revolution is $\left(\frac23\right)^{7}$ and the probability of 3 years having a revolution is $\left(\frac13\right)^{3}$.

$\therefore\frac{\binom{10}32^{7}}{3^{10}}$

$\frac{40\times128}{3^{9}}$

$\frac{5120}{19,683}$[/hide][/quote]
[hide] wait...wouldn't there be no need for the ten choose 3? because the problem never asks for which 10 years...it simply says what are the chances that we're going to have a revolution in these 10 years...so wouldn't the solution simply be $\frac{2^{7}}{3^{10}}$?  There are $\frac{1}{3}$ chance of there having a revolution in a year, and $\frac{2}{3}$ chance of there not being a revolution in a year. so isn't it $\frac{1^{3}}{3^{3}}$ chance of there being a revolution in 3 years, and then just $\frac{2^{7}}{3^{7}}$ possibility of there NOT being a revolution in the other 7 years...multiply together to get:
$\boxed{\frac{2^{7}}{3^{10}}}$
correct me if i'm wrong...I generally am...  :oops:
[/hide][/quote]
[hide]There is a $2^{7}/3^{10}$ for each arrangement such that 3 yrs have a revolution. We must account for all such arrangements, which there are ${10\choose3}$ of.[/hide]

eh heh heh...oops. Thanks, Pianoforte... :oops:


Given that

$ \cos\alpha \plus{}\cos\beta \minus{} \cos(\alpha\plus{}\beta)\equal{}\frac 3 2$

Find the acute angle $ \alpha,\beta$.

hello, i think in any triangle we have $ \cos(\alpha)\plus{}\cos(\beta)\plus{}\cos(\gamma)\le \frac{3}{2}$.
Sonnhard.

We have $ \cos\alpha \plus{}\cos\beta \minus{} \cos(\alpha\plus{}\beta)\equal{}\cos\alpha \plus{}\cos\beta \plus{} \cos(\pi\minus{}\alpha\minus{}\beta)\le \frac{3}{2}$ since $ \alpha\plus{} \beta\plus{} \pi\minus{}\alpha\minus{}\beta\equal{}\pi$
Therefore $ \alpha\equal{}\beta\equal{}\frac{\pi}{3}$


với mọi n tự nhiên lớn hơn 1 thì khoảng (n^2-n,n^2) tồn tại ít nhất 1 số nguyên tố.Không biết có cm chưa??? :?:

first , in this site , you have to post in English  :)
your problem can be translated into :
prove that , with any interger number n >1 , there exist at least 1 prime number in $[n^{2}-n , n^{2}]$
[hide]i think it's follows Bertrand's lemma
there exist at least 1 prime between n and 2n with n >1[/hide]

[quote="HTA"]first , in this site , you have to post in English  :)
[/quote]
ở đây  thì post tiếng Việt cũng được :P

[quote]i think it's follows Bertrand's lemma
there exist at least 1 prime between n and 2n with n >1[/quote]
Tôi không nghĩ vậy, hoặc nếu có , nó không đơn giản.

[quote="N.T.TUAN"][quote="HTA"]first , in this site , you have to post in English  :)
[/quote]
ở đây  thì post tiếng Việt cũng được :P[/quote]
cái hum post bài đó nó còn trong mục pre-olp :P

bài toán này đúng là ko đơn giản  :blush:

Mình đọc bài này trong câu chuyện toán học, đây là một bài tập cho vui vẻ nhưng lại khó


geometry
analytic geometry
conics
parabola
hyperbola
ellipse

I am in an Algebra 1 class, and I want to go to Algebra 2. But I must complete Geometry. Therefore I must take a Geometry test. Can you guys help me on some concepts of Geometry? I know some, like Symetry, corrosponding, Ajacent, complementry, sin cosin, tangent, cottangent, secant, csc, polar and retangular coordinates, graphing linear equations, parabola, hyperbola, Logrithams, etc. But I need to know the things that like a textbook would teach. So please tell me all you geniuses out there!! :D

Thank You very much. Please help :)  :)  :)  :)  :)  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :)  :lol:

Nothin but proofs

All you have to do is memorize all the postulates, thereoms etc about geometry and your set  :D

dont' forget your correlaries (i think that's how spell it) either. don't worry i already took my final exam, and i'm sure it's not too different from yours, and from what you've told us, i think you know more than you need to. :lol:

Why go to Algebra II? Go to AP Calc!!! (like me :P )

I believe that the way to best complete geometry is to think [i]abstractly[/i] and logically, because more words (proofs) are used in Geometry than in any other classroom math course (other than the stupid proofs you have to write for AP Calc).

And the constructions/Loci. The constructions were annoying.

My geometry class didn't have to many constructions, but when we did they were annoying

what about hyperbolas and parabolas? those were confusing.

[quote="gr8sk8r"]i hated constructions!! :bomb:[/quote]

I like constructions! they're not too hard  :D


Let $X= \{ 1, 2, 3, \ldots, 1000 \}$. For any non-empty subset $M$ of $X$, let $S$ be the sum of the largest and smallest numbers in $M$. Find the arithmetic mean of $S$ of all $M$.

1001. I think,  that if M had only one term z, then S(M)=2z.
Let x(M) is minimal term M, y(M) is maximal. Let $min_i$ -number of subsets, suth that x(M)=i, and $max_i$ - number of subsets suth that y(M)=1001-i. Let $f: X \to X$ defined f(z)=1001-z. It prove that $min_i=max_i.$  Let $N_0=2^{1000}-1$. Then average is
$\frac{1}{N_0} \sum_M [x(M)+y(M)]=\frac{1}{N_0} ( \sum_i imin_i +\sum_i (1001-i)max_i)=1001.$


1) How many integers between 1000 and 9999 have exactly one pair of equal digits, such as 4049 or 9902, but not 4449 or 4040? [hide]3888[/hide]

2) What is the sum of the smallest two positive real numbers whose digits are all 1s when written in decimal form and whose product and sum are the same?  Express your answer as a decimal to the nearest tenth. [hide]12.1[/hide]

[quote="andrewliu"]Can someone tell me what I'm doing wrong? for the 1st problem
casework:
#s 1000-2000
since the first digit is 1, you have [color=red]3[/color] choices left for another digit to be 1. then since another digit can't be 1, you have [color=red]9[/color] choices for your 2nd digit that you choose. since the last digit you choose can't be 1 or the 2nd digit that you chose, you have [color=red]8[/color] left.
so that's 3x9x8=216 for the ones 1000-2000. there are 9 "thousands" left so its 9x216=[color=green]1944[/color]
*what I find interesting about my answer is that it is exactly half of the correct one
messed up counting hurts[/quote]
Close... You forgot to consider that the remaining 2 numbers can be fliped.(i.e 4045 and 4054
So multiply your answer by 2 to get $ \boxed{3888}$

Since only 2 numbers matter in this question, and x can be in any three places, just do this:
$ 2(3\times9\times8)\times9\equal{}3888$

[hide="2"]we just need to find 2 #s that both add and multiply to equal the same number that use only ones. well, you have to be as small as possible, and 11 is a good choice to start off with. the reason being if we use all decimals, the sum is going to be higher than the normal ones while the product is going to be much lower. ex. 0.1+0.1=0.2, but 0.1*0.1=0.01
if we use 1; the product is going to stay the same, and under no circustances is x+1=1x. well, we see that 11*11=121 and 11+1.1=12.1, so we move the decimal place one over to get 11&1.1; whose sum and product is both 12.1

note: guess and check is a good way to tackle this problem once you find out the numbers have to be greater than one and one of the #s is 11(which is the best place to start[/hide]

[quote="spiders_hate_chestnut"]i dont get the second question

also
real numbers are just plain numbers, right?[/quote]

real numbers are made up of rational numbers and irrational numbers

rational numbers are the ones that can be expressed in the form p/q, where p and q are integers and $ \q\neq 0$

integers are made up of the positive integers, negative integers, and 0.

positive integers are 1,2,3,4,5,...

so real numbers are built up from the positive integers.

Another way to define a real number is basically all numbers that can be represented on the number line

So, $ \mathbb{N}\in$whole numbers$ \in\mathbb{Z}\in\mathbb{R}$.
In other words, the natural numbers are part of the whole numbers, which are part of the integers, which are part of the reals.

[hide="Algebraic Solution for #2"]Let us say the two numbers are $ x$ and $ y$.
Then $ x\plus{}y\equal{}xy$
Solving yields $ x\equal{}\frac {y} {y\minus{}1}$
Thus, we can see that the smallest possible $ y$ is $ 1.1$, yielding $ x\equal{}11$.
Or we can let $ y\equal{}11$, yielding $ x\equal{}1.1$.
Adding $ x$ and $ y$ gives us the desired result, $ 12.1$[/hide]

[quote="DonkeyKong"][quote="andrewliu"]Can someone tell me what I'm doing wrong? for the 1st problem
casework:
#s 1000-2000
since the first digit is 1, you have [color=red]3[/color] choices left for another digit to be 1. then since another digit can't be 1, you have [color=red]9[/color] choices for your 2nd digit that you choose. since the last digit you choose can't be 1 or the 2nd digit that you chose, you have [color=red]8[/color] left.
so that's 3x9x8=216 for the ones 1000-2000. there are 9 "thousands" left so its 9x216=[color=green]1944[/color]
*what I find interesting about my answer is that it is exactly half of the correct one
messed up counting hurts[/quote]
Close... You forgot to consider that the remaining 2 numbers can be fliped.(i.e 4045 and 4054
So multiply your answer by 2 to get $ \boxed{3888}$

Since only 2 numbers matter in this question, and x can be in any three places, just do this:
$ 2(3\times9\times8)\times9 \equal{} 3888$[/quote]
I don't think you switched the two numbers in your example (4045 and 4054) wouldn't it be 4540 if the first one was 4045?
you could think about it this way:
# of possilibilities for number value (# of possibilities of number place), of course, that would mean it would be exclusive... i think
so you have 9 choices as your thousands digit
- another digit would have to be 6 (which is one possibility), and there are 3 ways to insert it in the number.....thats 3 ways right now
-the other couldn't be 6, so that leaves 9 possibilities of  the next one, and there are 2 places to insert.... so thats 18 ways for this one
- the last one has 8 left, and there's only one more place to put it. so thats 8
8*18*3 for the #'s 6000-7000. THe question asks for 1000-9999
so that's 8*18*3*9=3888....... I am doing this right,right?

[quote="andrewliu"]I don't think you switched the two numbers in your example (4045 and 4054) wouldn't it be 4540 if the first one was 4045?
you could think about it this way:
# of possilibilities for number value (# of possibilities of number place), of course, that would mean it would be exclusive... i think
so you have 9 choices as your thousands digit
- another digit would have to be 6 (which is one possibility), and there are 3 ways to insert it in the number.....thats 3 ways right now
-the other couldn't be 6, so that leaves 9 possibilities of  the next one, and there are 2 places to insert.... so thats 18 ways for this one
- the last one has 8 left, and there's only one more place to put it. so thats 8
8*18*3 for the #'s 6000-7000. THe question asks for 1000-9999
so that's 8*18*3*9=3888....... I am doing this right,right?[/quote]
I don't think that's right actually... Because you might have two $ 0$s, and I don't think that's included in those cases. Also, you can permute the number in $ 12$ different ways, provided there is no $ 0$.
So I think the best way to approach this problem is case work, with one zero, two zeros, and zero zeros.


induction
number theory unsolved
number theory

No.

Just prove by induction that $a_{n}= 2^{n}$ for all $n$.

Pierre.

i have a solution.(and i want to post it :lol: )
I prove that  if $a_{n}=2^{a}+2^{b}$ then $a_{n+1}=2^{a}+2^{b+1}$ or $2_{a+1}+2^{b}$


inequalities
function
inequalities unsolved

\[ \left(1\right): \: a,\ b,\ c>0,\: a\plus{}b\plus{}c\equal{}1\Longrightarrow\frac{a}{\sqrt{b\plus{}c}}\plus{}\frac{b}{\sqrt{c\plus{}a}}\plus{}\frac{c}{\sqrt{a\plus{}b}}\ge\sqrt{\frac{3}{2}}\]

\[ \left(2\right): \: a,\ b,\ c>0,\: a\plus{}b\plus{}c\equal{}1\Longrightarrow\frac{a}{\sqrt{b\plus{}c}}\plus{}\frac{b}{\sqrt{c\plus{}a}}\plus{}\frac{c}{\sqrt{a\plus{}b}}\ge\frac{2}{3}\left(\frac{1}{\sqrt{1\plus{}a}}\plus{}\frac{1}{\sqrt{1\plus{}b}}\plus{}\frac{1}{\sqrt{1\plus{}c}}\right)\minus{}\sqrt{\frac{3}{2}}\]

[quote="Mîndrilă Claudiu"]
\[ \left(1\right): \: a,\ b,\ c > 0,\: a \plus{} b \plus{} c \equal{} 1\Longrightarrow\frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\ge\sqrt {\frac {3}{2}}
\]

\[ \left(2\right): \: a,\ b,\ c > 0,\: a \plus{} b \plus{} c \equal{} 1\Longrightarrow\frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\ge\frac {2}{3}\left(\frac {1}{\sqrt {1 \plus{} a}} \plus{} \frac {1}{\sqrt {1 \plus{} b}} \plus{} \frac {1}{\sqrt {1 \plus{} c}}\right) \minus{} \sqrt {\frac {3}{2}}
\]
[/quote]
[hide="hint for the first"]Consider the function $ f(x)\equal{}\frac{x}{\sqrt{1\minus{}x}}$[/hide]

[hide="Solution to (1)."]
The inequality can be written equivalently with help of the cyclic sums as
$ \small{\sum_{cyc}\left (\frac {a}{\sqrt {1 \minus{} a}} \minus{} \frac {1}{\sqrt {6}} \minus{} \frac {15}{4\sqrt {6}}\left (a \minus{} \frac {1}{3}\right )\right ) \equal{} \sum_{cyc}\frac {\sqrt {6}(5\sqrt {1 \minus{} a} \plus{} 2\sqrt {6})(3\sqrt {1 \minus{} a} \minus{} \sqrt {6})^2}{72\sqrt {1 \minus{} a}}\ge 0}$
and the equality holds iff $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$.
[/hide]

[hide="Solution to (2)."]
Here can be the inequality proved with help of cyclic sums again. Note that the inequality can be transformed to the equivalent inequality as
$ \sum_{cyc}\frac {a}{\sqrt {1 \minus{} a}}\ge\sum_{cyc}\left (\frac {\frac {2}{3}}{\sqrt {1 \plus{} a}} \minus{} \frac {1}{\sqrt {6}} \plus{} a \minus{} \frac {1}{3}\right ) \\
\Leftrightarrow \\
\sum_{cyc}\left (\frac {a}{\sqrt {1 \minus{} a}} \minus{} \frac {\frac {2}{3}}{\sqrt {1 \plus{} a}} \plus{} \frac {1}{\sqrt {6}} \minus{} a \plus{} \frac {1}{3}\right )\ge 0.$
But
$ \small{\left (\frac {x}{\sqrt {1 \minus{} x}} \minus{} \frac {\frac {2}{3}}{\sqrt {1 \plus{} x}} \plus{} \frac {1}{\sqrt {6}} \minus{} x \plus{} \frac {1}{3}\right )^{\prime} \equal{} \frac {1}{\sqrt {1 \minus{} x}} \plus{} \frac {x}{2\sqrt {(1 \minus{} x)^3}} \plus{} \frac {1}{3\sqrt {(1 \plus{} x)^3}} \minus{} 1 > 0}.$
Hence,
$ \min_{0\le x\le 1}\left (\frac {x}{\sqrt {1 \minus{} x}} \minus{} \frac {\frac {2}{3}}{\sqrt {1 \plus{} x}} \plus{} \frac {1}{\sqrt {6}} \minus{} x \plus{} \frac {1}{3}\right ) \equal{} \minus{} \frac {1}{3} \plus{} \frac {\sqrt {6}}{6} > 0.$
Thus, your inequality holds.
[/hide]

[hide="1"]Let $ f(x) \equal{} \frac {x}{\sqrt {1 \minus{} x}}$
$ f''(x) \equal{} \frac {4 \minus{} x}{4(\sqrt {1 \minus{} x})^5} > 0$ when $ x \in [0, 1)$
So $ f(x)$ is convex.
Then by Jensen
$ LHS \ge 3\frac {\frac {1}{3}}{\sqrt {1 \minus{} \frac {a \plus{} b \plus{} c}{3}}} \equal{} \sqrt {\frac {3}{2}}$[/hide]

1 is also holder, call the sum on the left side S. Then by Holder,
$ S^2 \cdot (a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b)) \ge (a \plus{} b \plus{} c)^3 \equal{} 1$
So we have to show
$ \sqrt {\frac {1}{2(ab \plus{} bc \plus{} ca)}} \ge \sqrt {\frac {3}{2}}$
Or $ ab \plus{} bc \plus{} ca \le \frac {1}{3}$ which is obvious since $ a \plus{} b \plus{} c \equal{} 1$

It`s very easy to proof that the function \[ f: \left[0,\ 1\right)\longrightarrow\mathbb{R},\: f\left(x\right)\equal{}\frac{1}{\sqrt{1\minus{}x}}\] is convex(derivates aren`t necesary ). Thus, according to Jensen inequality, we obtain: \[ ff\left(a\right)\plus{}f\left(b\right)\plus{}f\left(c\right)\ge3f\left(\frac{1}{3}\right)\equal{}3\sqrt{\frac{3}{2}}\Longleftrightarrow\sum\frac{1}{\sqrt{b\plus{}c}}\ge3\sqrt{\frac{3}{2}}\]
If \[ a\le b\le c\Longrightarrow\frac{1}{\sqrt{b\plus{}c}}\le\frac{1}{\sqrt{c\plus{}a}}\le\frac{1}{\sqrt{a\plus{}b}}\] and from Chebyshev`s inequality we obtain: \[ \sum\frac{a}{\sqrt{b\plus{}c}}\ge\frac{1}{3}\left(\sum a\right)\left(\sum\frac{1}{\sqrt{b\plus{}c}}\right)\ge\sqrt{\frac{3}{2}}\]

Problem (2) reduces at \[ \frac{f\left(a\right)\plus{}f\left(b\right)\plus{}f\left(c\right)}{3}\plus{}f\left(\frac{a\plus{}b\plus{}c}{3}\right)\ge\frac{2}{3}\left[f\left(\frac{a\plus{}b}{2}\right)\plus{}f\left(\frac{b\plus{}c}{2}\right)\plus{}f\left(\frac{c\plus{}a}{2}\right)\right]\] which is Popoviciu`s inequality applied for the convex functon  \[ f: \left[0,\ 1\right)\longrightarrow\mathbb{R},\: f\left(x\right)\equal{}\frac{1}{\sqrt{1\minus{}x}}\]

[quote="Mîndrilă Claudiu"]It`s very easy to proof that the function
\[ f: \left[0,\ 1\right)\longrightarrow\mathbb{R},\: f\left(x\right) \equal{} \frac {1}{\sqrt {1 \minus{} x}}
\]
is convex(derivates aren`t necesary ).[/quote]

Could you provide the proof?

We will prove that for any \[ x,\ y\in\left[0,\ 1\right)\Longrightarrow f\left(x\right)\plus{}f\left(y\right)\ge2f\left(\frac{x\plus{}y}{2}\right)\]

Indeed,

\[ f\left(x\right)\plus{}f\left(y\right)\ge2f\left(\frac{x\plus{}y}{2}\right)\Longleftrightarrow\frac{1}{\sqrt{1\minus{}x}}\plus{}\frac{1}{\sqrt{1\minus{}y}}\ge\frac{2}{\sqrt{1\minus{}\frac{x\plus{}y}{2}}}\equal{}\frac{2\sqrt{2}}{\sqrt{1\minus{}x\plus{}1\minus{}y}}\ (1)\]

Now denote $ \alpha^{2}\equal{}1\minus{}x,\ \beta^{2}\equal{}1\minus{}y$, and see that \[ \left(1\right)\Longleftrightarrow\frac{1}{\alpha}\plus{}\frac{1}{\beta}\ge\frac{2\sqrt{2}}{\sqrt{\alpha^{2}\plus{}\beta^{2}}}\Longleftrightarrow\frac{2}{\frac{1}{\alpha}\plus{}\frac{1}{\beta}}\le\sqrt{\frac{\alpha^{2}\plus{}\beta^{2}}{2}}\]

which is true.

Thus, the function \[ f: \left[0,\ 1\right)\longrightarrow\mathbb{R},\: f\left(x\right) \equal{}\frac{1}{\sqrt{1\minus{}x}}\] is convex.

[quote="Mîndrilă Claudiu"]We will prove that for any
\[ x,\ y\in\left[0,\ 1\right)\Longrightarrow f\left(x\right) \plus{} f\left(y\right)\ge2f\left(\frac {x \plus{} y}{2}\right)
\]
Indeed,
\[ f\left(x\right) \plus{} f\left(y\right)\ge2f\left(\frac {x \plus{} y}{2}\right)\Longleftrightarrow\frac {1}{\sqrt {1 \minus{} x}} \plus{} \frac {1}{\sqrt {1 \minus{} y}}\ge\frac {2}{\sqrt {1 \minus{} \frac {x \plus{} y}{2}}} \equal{} \frac {2\sqrt {2}}{\sqrt {1 \minus{} x \plus{} 1 \minus{} y}}\ (1)
\]
Now denote $ \alpha^{2} \equal{} 1 \minus{} x,\ \beta^{2} \equal{} 1 \minus{} y$, and see that
\[ \left(1\right)\Longleftrightarrow\frac {1}{\alpha} \plus{} \frac {1}{\beta}\ge\frac {2\sqrt {2}}{\sqrt {\alpha^{2} \plus{} \beta^{2}}}\Longleftrightarrow\frac {2}{\frac {1}{\alpha} \plus{} \frac {1}{\beta}}\le\sqrt {\frac {\alpha^{2} \plus{} \beta^{2}}{2}}
\]
which is true.

Thus, the function
\[ f: \left[0,\ 1\right)\longrightarrow\mathbb{R},\: f\left(x\right) \equal{} \frac {1}{\sqrt {1 \minus{} x}}
\]
is convex.[/quote]
The convexity can also be proved by considering the second derivative..


inequalities
function
inequalities solved

Given 0 \leq a \leq x1 \leq x2 \leq ... \leq xn \leq b prove that

(x1+x2+...+xn)(1/x1+1/x2+...+1/xn) \leq (1/4ab)((a+b)n)^2

We will prove that if $a_1,a_2,\dots,a_n\in [a,b]$ $(0<a<b)$ then
\[(a_1+a_2+\dots+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n})\leq
\frac{(a+b)^2}{4ab}n^2\]

P=(a_1+a_2+\dots+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n})=
(\frac{a_1}{c}+\frac{a_2}{c}+\dots+\frac{a_n}{c})(\frac{c}{a_1}+\frac{c}{a_2}+\dots+\frac{c}{a_n})\leq
\frac{1}{4}(\frac{a_1}{c}+\frac{c}{a_1}+\frac{a_2}{c}+\frac{c}{a_2}+\dots+\frac{a_n}{c}+\frac{c}{a_n})^2

Function f(t)=c/t+t/c have its maximum on [a,b] in a or b. We will choose
c such that f(a)=f(b). c=\sqrt{ab}. Then f(t)\leq \sqrt{a/b}+\sqrt{b/a}. Then

P\leq n^2(\sqrt{a/b}+\sqrt{b/a})^2 /4= n^2(a+b)^2/4ab.


[quote="kstan013"]At about 3:00 Central Time, the AoPS server was down and asked for something about the deletion of some file. Is everything fixed now?[/quote]Yes, it was an update we've done.

Thanks for reporting though!  :)


Prove that

$\displaystyle \frac{a_1^3}{a_2+a_3+a_4}+...+\frac{a_5^3}{a_1+a_2+a_3} \ge \frac{\sqrt{5}}{3} $

for all $a_i (i=1,..,5)$ positive real numbers  such that

$\sum_{i=1,n} a_i^2 \geq 1$

[quote="siuhochung"]wt if a_1=a_2=a_3=a_4=a_5 and  sum a_i^2 = 1?
it seems that minimum value is not sqrt 5/3?[/quote]

Hmm. That makes $a_1=a_2=a_3=a_4=a_5=\frac{\sqrt{5}}{5}$, and so each term is $\frac{\frac{\sqrt{5}}{25}}{\frac{3\sqrt{5}}{5}} = \frac{1}{15}$, and so the cyclic sum is $\frac{1}{3}$. Ah yes, unless its supposed to be squared on the numerator..

[quote="siuhochung"]no, the cyclic sum is 1/3.[/quote]
Oh thanks, :blush::blush::blush::blush:


An antacid tablet is in the shape of a right circular cylinder. The diameter of the base is $ \frac{3}{4}''$ and the tablet is $ \frac{3}{16}''$ thick. How many cubic inches are in the volume of the tablet? Express your answer as a common fraction in terms of $ \pi$.

Whenever this problem shows up, report it. You can go to the "Review Questions"(next to "Countdown" and "New Game") and go to that problem. On the bottom, next to "Ask on Forum" is "Report Error". If you click that, there will be a momentary pause, before a little "box" pops up that says something like "We will investigate the error" or something along those lines.

I reported every time I'd seen it, izzy. The problem is that this Q might not show up exactly as it'd appeared to us, for example, this one. I clicked both Report Error and Ask On Forum and as it looks, it's totally fine.

My point is that levans or his colleagues may mistake and consider this as a "false" report. :(

Yes, I see. I've reported several problems as well, with no result... :noo: ...this may be why...

[quote="GameBot"]An antacid tablet is in the shape of a right circular cylinder. The diameter of the base is $ \frac {3}{4}''$ and the tablet is $ \frac {3}{16}''$ thick. How many cubic inches are in the volume of the tablet? Express your answer as a common fraction in terms of $ \pi$.[/quote]

The radius is 3/8 so its 3/8 ^2 x 3/16 pi=27pi/1024

mewtwo55555, it's a cylinder, not a cone.

Doink, mewto55555 did it the right way. This is the right way to solve for a cylinder, not a cone.

And FYI, the problem on FTW! looks fine now.


[hide]this is an acute triangle, since the last angle has to be
70 degrees since 50+60+70=180[/hide]

[hide]
180-50-60=70

all angles are less than 90 degrees. It's an acute triangle
[/hide]

[hide]$50+60=110\\\\180-110=70$
since none of the angles are greater than 90 this is an acute triangle[/hide]

[quote="anirudh"][quote="236factorial"]Two angles of a triangle are 50 and 60. Is this an acute, obtuse, or right triangle?[/quote]

acute[/quote]

Show work next time  ;)

[hide]Solution:

Every triangle has 180 degrees total:

50+60=110

180-110=70 Degrees

Acute.[/hide]

It's an acute triangle because the sum of the 3 angles in a triangle is 180. Therefore, 180-50-60=70. ;)

Acute, but by now people will probably know it.

OK next problem!!

[quote="math92"]ok if the lengths of a triangle are 2, 3, and 6 is the triangle right, acute, obtuse, or not a triangle?[/quote] :roll:
I am confused because 2*3=6 and I can define a "product triangle" of lenghts $a$, $b$ and $c$ if  $a*b=c$. :oops:
How many "product triangles" are there ?

uhh...i don't know what product triangles are....do you mean that if the lengths of two sides multiplied together equals the third side?

If a product triangle is a triangle with one side being the product of the other two, the only product triangles are when one side is 1 (correct me if I'm not mistaken).

[quote="pianoforte"]unless they don't have to be integers (ex. 1/2,1/2,1/4). :?

anyways, 2-3-6 does not follow triangle inequality so it is not a triangle.[/quote]

Oh yeah, good point.  I was thinking of integers, though.

Um no it's not a triangle.
2 and 3 don't add up to 6. The sum of 2 of the smaller legs have to be bigger than the third leg for it to be a triangle.

yeah, you guys are right...nice job, it was on my math test so i put it on here.

[hide]not a triangle because $2+3\leq 6$[/hide]

[quote="mstoenescu"]I am confused because 2*3=6 and I can define a "product triangle" of lenghts $a$, $b$ and $c$ if  $a*b=c$. :oops:
How many "product triangles" are there ?[/quote]

There is a mistake in the problem.

If the unit of measure is $cm$, then in the first part of the equation $ab=c$ we have $cm^2$
but in the second part we have $cm$.

So, any triangle is a product triangle if we choose the proper unit of measure

In other words, given a triangle $KLM$ we can find another triangle $ABC$ which is similar to $KLM$ and $ab=c$

In fact, we can find two more triangles (if $KLM$ is not isosceles) with the properties $a=bc,b=ac$

For example, the triangle with sides $3-4-5$ can be transformed to $\triangle ABC$ with sides

$a = \frac{25}{12}, b = \frac{20}{12}, c = \frac{15}{12}$

$bc = \frac{300}{144} = \frac{25}{12} = a$

There are infinite procuct triangles:
1,1,1
1,2,2
1,3,3
1,4,4...

Of course un triangle with sides 2,3,6 does not exist but it was only a joke
I'm very happy because a lot of you had make of this "a very important question "
Thanks to everybody ! :ninja:


quadratics
invariant
number theory proposed
number theory

suppose that $p$ is odd number.
suppose that $A$ is a the set of reminders of the numbers $2^ka$ (in mod $p$)where $a$ is a fix natural number,and $k$ is a natural number,too..(remember that a set does not include equal members).
we define $f(x)$ in this way:if the remainder of $x$ when devided by $p$ is $r$ then if $r$ is greater than$\frac{p-1}{2}$ then $f(x)=r-p$ and otherwise $f(x)=r$.
prove that always $\sum f(x)=0$ where the sum is taken over all members of $A$.

i,ve arised to this brilliant problem while solving one problem.

that you said is absoloutly wrong!!
take $p=7$.
i think the problem is not so trivial!

[quote="gopherhole112"][hide]I'm fairly certain that $p\equiv 7 (mod\ 8)$ implies that 2 is a primitive root modulo $p$. This means since $p$ has no factors other than itself and 1, $a$ is not a factor of $p$. So $2^ka$ goes through all of the values from 1 to $p-1$. Half of these are greater than $\frac{p-1}{2}$, and half are less, so the sum is equal to zero.[/hide][/quote]
Your condition implies that $2$ is a quadratic residue, thus never a primitive root. Remark that there is no simple condition in being a primitive root.

maybe I'm missing something..
take p=7 and a=3
then $A=\{3,6,5\}$
f(3)=3
f(6)=-6
f(5)=-5
$\sum\neq 0$
I think f should be of the form: $f(x)=x$ if $x(mod p)\le(p-1)/2$ and $f(x)=x-p$ if $x(modp)> (p-1)/2$...
in this case with p=7 and a=3 we get that
f(3)=3
f(6)=-1
f(5)=-2
so $\sum = 0$...

you are right kuba i've edited it's mistake. :D

I think I have a proof for $p\not\equiv_8\pm1$.

Let $d$ be the order of $2$ mod $p$. Then if $d$ is odd we have that $(2^{(d+1)/2})^2=2$ so $2$ is a square modulo $p$, so $p\equiv_8\pm1$.
We have then that $d$ is even.

Now consider $A=\left\{2^ia : i\in\overline{0,d-1}\right\}$, where the values from $A$ are taken mod $p$. The ord of $2$ mod $p$ is $d$

which is even so $2^{d/2}\equiv_p-1$

Denote by $A_1$ those elements of $A$ which are greater than $(p-1)/2$. The problem can be reformulated like this

$\displaystyle\sum_{x\in A}x=p|A_1|$.

I will prove that  $x\in A-A_1$ iff $p-x$ in $A_1$. (only the implication $x\in A-A_1$ results $p-x$ in $A_1$, the other one is similar).

It is clear that $x\equiv_p 2^ia$ for some $i$.

$(p-1)/2<p-x\equiv_p -x\equiv 2^{d/2}x\equiv 2^{d/2+i}a\in A$.

It is obvious now that $|A_1|=|A-A_1|$. So $\displaystyle\sum_{x\in A}x=\sum_{x\in A_1}(x+p-x)=p|A_1|$.

So $\displaystyle\sum_{x\in A}f(x)=0$. q.e.d.

I will think for the other cases...

[quote="Rust"]It is lejander symbol[/quote]
It is called Legendre-symbol, with that writing!
[quote="Rust"]It is lejander symbol (2/p)=1, and period is odd, Therefore $2 \sum {2^ka}=\sum {2^k(p-a)}+\sum {2^ka}=0(mod p).$[/quote]
I don't know exactly what this sentence is intended to mean
 :?
(is it: the sum is $\equiv 0 \mod p$¿)

[quote="Kuba"]no, the sum from the problem is not modulo p, the sum means EQUAL with 0.[/quote]
My question was to Rust since I don't understand what he says; I know what the problem asks for. Or was yours to Rust too¿

hey ,the problem is valid for all odd $n$'s.

here's a solution (due to a friend of mine)

denote by $A_{<} =\{x\in A | x\le(p-1)/2\}$ and by $A_{>}=A-A_<$.
$A_e=\{x\in A | x \textmd{ is even }\}$. $A_o=A-A_e$.
The problem can be restated:
$S=\sum_{x\in A}x=p|A_>|$.

Now $2S=\sum_{x\in A}2x=\sum_{x\in A_e}x+\sum_{x\in A_o}(x+p)=\sum_{x\in A}x+p|A_o|$.
So it is enough to show that $|A_o|=|A_>|$.
Let $g: A_>\to A_o$ defined by $g(x)=2x-p$. We will prove that this is a bijection.

Let's show it is wel defined. Let $x\in A_>$ so $p>x>(p-1)/2$ so $2p>2x>p\Rightarrow p>2x-p>0$ and $2x$(mod $p$) is in $A$ and because $2x-p$ is odd it follows that it is in $A_o$.

It is easy to see that $g$ is injective.

We want to show that $h(x)=(x+p)/2, h: A_o\to A_>$ is also injective. $x\in A_o$ so $p>(x+p)/2>(p-1)/2$. We need to show that it is well defined: $(x+p)/2$ is in $A$ and is odd.
But $x\equiv 2^ia$ for some $i$. Then $(x+p)/2\equiv_p 2^{i-1}a$  and because $2^{i-1}a$ has a representant in $A$ it follows from $p>(x+p)/2>(p-1)/2$ that $(x+p)/2\in A_>$. We have thus that $g$ is a bijection, so $|A_>|=|A_o|$ and this is what we needed.

Let n is odd number, and T is subset $(\pm 1,\pm2, \dots, \pm m), m=\frac{n-1}{2}).$  Define $f(T)=(f(x)| x \in T),$ were
$f(x)=2x, \ if \ |2x|<n/2, \ f(x)=2x-nsign(x), \ if \ |2x|>m.$ If T is invariant subset (f(T)=T), then $S(T)=\sum_{x \in T} x=0,$ because 2S(T)=S(T), $2S(T)=S(T)+N_1n-N_2n=S(T)=0,$ since $N_1$ -number of terms change sign x from - to + equal to $N_2$ -number of terms change sign from + to -.

[quote="Rust"]$N_1$ -number of terms change sign x from - to + equal to $N_2$ -number of terms change sign from + to -.[/quote]
why¿


Find the maximum value of the expression $ \sum\frac{\left(ab\right)^n}{1\minus{}ab}$, where $ n\geq 2$ is a natural number and $ a$, $ b$, $ c$ are nonnegative reals such that $ a\plus{}b\plus{}c\equal{}1$.

Hereby, the $ \sum$ sign stands for cyclic summation.

This is a very nice problem but impossible to solve for me.
Have you a solution, Harazi?

Thank you very much.

Of course I do. And it is quite simple, though the problem is difficult-as far as I'm concerned.
   It is clear that the maximum is not attained when all variables are equal. We will show that it is obtained when a=b=1/2 and c=0 for example. Let us suppose that a>=b>=c. Of course, ab<=1/4, bc<=1/4 and ca<=1/4, so sum (ab)^n/(1-ab)<=4/3*sum (ab)^n. Also, we have of course 1/4^n>=a^n(b+c)^n=a^n(b^n+c^n+nb^n-1c+...)>=(ab)^n+(ac)^n+na^n*b^n-1*c>=(ab)^n+(bc)^n+(ca)^n. So, the expression is allways at most 4/3*1/4^n. For a=b=1/2 and c=0 it is 4/3*1/4^n.

Genial solution for its simplicity.

Thank you very much,Harazi.

Considering the number of times we use the "harazi inequality", perhaps this could be moved into "Theorems and Formulas", even if it's not a real theorem :D

Otherwise new members will probably be wondering what we are referring to every time we say that it reduces to this!

what's "harazi ineq" then?
i really don't know that ..

It says that if a,b,c>=0 have sum 1, then (ab)^n+(bc)^n+(ca)^n<=1/4^n. And it's not definitely a teorem, it's just a result I found playing with some inequalities. And I really doubt I firstly discovered it.

Maybe someone could give the topic with the solution of "Harazi inequality" ?


[color=green]I really really need help with this question: My homework is due tomorrow!!![/color]

[color=blue]Suppose you have a computer that takes 4.3ns to access main memory and 0.6 ns to access cache. Suppose you have a program which access memmory 200 million times. Finally, suppose you will hit the cache (the L1 and L2 combined) 99% of the time. Now, compute the ratio of the time the program would spend accessing memory id the computer does not have a cache, versus the time the program would spend accessing memory if the cmputer does have a cache. (Don't do the division, the answer can be left in the form (number)/(number) ). [/color]

[color=green]Thanks![/color]

If computer do not have cache at all, so it will require to access memory 200 million times, so 860 million ns or 860 milliseconds. If it has cache so it will access cache for (198 million times) 118.8 milliseconds and  main memory 2 million times or 8.6 miliseconds.
So ratio could be roughtly 430/64 (or 860/127.4). (no cache/with cache)


abstract algebra
vector
Ring Theory
superior algebra
superior algebra unsolved

Let A be a Noetherian ring of dimension n, P is a projective module over A, rk(P)=n and
f: P -> A
is a module homomorphism. Let J:=im(f)=/=A, is it right that ht(J)=n?

What restriction do you have on $ P$? What if $ f \equal{} 0$?

Edit: I actually meant to say the restriction on $ f$.

I don't have a restriction on P, but morphism f should be "generic". For example, let P=A^n then J=(x_1, ..., x_n), where x_i is image of vector (0,..,1,..,0), where 1 is on i-th place, in this case generic simply means that J can't be generated by k elements of A and k<n.

In the same time, I don't understand how to formulate a condition that f generic in general case of a projective P.


geometry
3D geometry
analytic geometry
algorithm
pyramid
combinatorics unsolved
combinatorics

We have a $ 2n\times2n\times2n$ cube divided into $ 8n^3$ unit cubes. Some are infected. If a non-infected cube has 3 neighbors which are infected (neighbor=shares a side), then it becomes infected. What is the minimum number of infected cubes at the starting point which is enough for the entire cube to be infected?

Give an example, and prove that a smaller number of cubes possible

I've got a nice example for $ 2n(3n \minus{} 1)$, and I proved that the number of cubes can't be less than $ 4n^2$, but what to do with the gap inside?

Would coloring just the squares with $ x\plus{}y\plus{}z$ divisible by $ 2n$ be enough?  I've (rather hurriedly) tried to work through the $ n\equal{}2$ example below.  The $ *$'ed squares are initially infected, and each square hopefully borders three squares which are either $ *$ or lower in number.

[code]
321*
21*5
1*36
*567

21*5
1*14
*123
543*

1*36
*123
321*
63*1

*567
543*
63*1
7*12
[/code]

After trying out kevin's idea, I think the min is $4n^2$.

The big cube consists of 2n layers. Denote them $L_1,L_2,..L_{2n}$.
Represent each cube with a lattice point. If the big cube is put into a 3d coordinate space,
infect each cube whose sum of coordinates $x+y+z \equiv k$ (mod 2n) for some $k$.
The infected cubes form 2 diagonal slices thru' the big cube.

Denote $L_i=(j,k)$ as the $i$th layer having infected squares of size $j$ and size $k$ at the
top left and bottom right corners respectively.
____________________________________________________________
Eg:
So $L_3$=(4.2,2) means the 3rd layer looks like:

****
****
***
**

space     **
space     **

where 4.2 indicates an incomplete square of size 4 filled up to 2-1 units away from its diagonal.
_____________________________________________________________
But the general algorithm is hard to write out, so I shall illustrate for n=3.

Then
$L_1=(6.1,0)$
$L_2=(5.2,1)$
..
$L_5=(2,4.2)$
$L_6=(1,5.1)$

I wish to show the region between the 2 pyramids will be infected.

This algorithm has n steps, each step having 2 substeps.

Step 1a: $L_4 \rightarrow L_3$

$L_3$=(4,3)
$L_4$=(3,3)

Step 1b: $L_3 \rightarrow L_5$

$L_2$=(5.2,1)
$L_3$=(4,3)
$L_4$=(4,3)
$L_5$=(4,4)

Step 2a: $L_5 \rightarrow L_2$

$L_2$=(5,4)
$L_3$=(4,4)
..
$L_6$=(1.5,1)

Step 2b: $L_2 \rightarrow L_6$

$L_1$=(6.1,0)
$L_2$=(5,4)
..
$L_6$=(5,5)

I must still explain $L_i \rightarrow L_j$..


function
integration
calculus
real analysis
real analysis unsolved

Let a continuous at $ [0,2]$ function $ f$  for which we have that $ \int_{0}^{2} (t\minus{}2)f(t)dt\equal{}0$

Prove that
a)there exist a root $ a$ of the function $ H(x)\equal{}\int_{0}^{x} tf(t)dt$ at the interval $ (0,2)$

b)there exist a $ k$ at the interval $ (0,a)$ such that $ a\int_{0}^{k} tf(t)dt\equal{}k^2\int_{0}^a f(t)dt$

Language notes:
- "at the interval" should be "in the interval", and "continuous at $ [0,2]$" should be "continuous on $ [0,2]$".
- "there exist" should be "there exists". The (third person) singular form of the verb is the one with the "s".
- That first sentence is missing a verb, and generally reads awkwardly. For a simple fix, make it "Let there exist..."; I would write it as "Let $ f$ be a continuous function on $ [0,2]$ such that $ \int_0^2 (t\minus{}2)f(t)\,dt\equal{}0$."

Content:
By the same principle as my posts [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275192]here[/url], we write $ \int_0^2 (2\minus{}t)f(t)\,dt\equal{}\int_0^2 \frac{2}{x^2}H(x)\,dx.$ This uses that $ \int_t^2 \frac{2}{x^2}\,dx\equal{}\frac{2\minus{}t}{t}.$
Since its integral is zero and $ \frac{2}{x^2}H(x)$ is continuous, it must be zero somewhere in the interval. That's part (a).

For part (b), we want to show that there is some $ k$ with $ \frac1{k^2}\int_0^k tf(t)\,dt\equal{}\frac1a\int_0^a f(t)\,dt$ given that $ \int_0^a tf(t)\,dt\equal{}0$. The original setting is irrelevant.
By essentially the same calculations as in (a), $ \int_0^a \frac1{x^2}\int_0^x tf(t)\,dt\,dx\equal{}\int_0^a (1\minus{}\frac{t}{a})f(t)\,dt\equal{}\int_0^a f(t)\,dt$. Dividing by $ a$, the average value of $ G(x)\equal{}\frac1{x^2}\int_0^x tf(t)\,dt$ is $ \frac1a\int_0^a f(t)\,dt.$ $ G(x)$ must take its average value somewhere, and we're done.