In this PDF http://www.math.uiuc.edu/~mjunge/540-material.pdf
 on page 26 he claims that

f_n(x) + 2*epsilon/3 <= f_n(y) + epsilon near the end of the proof.

I don't see how he gets this because the delta(x)'s he chose only work for that particular function in the sequence, and aren't guaranteed for functions after that function in the sequence.

Is the fact that f_n(x) is increasing necessary?  Is there a counterexample when this is not so?

Thanks,
Beoulve

Edit: forgot to include link

The book I'm using is Elementary Classical Analysis by Mardsen.  The particular excercise this relates to is Excercise 8 at the end of Chapter 5.

Please post that proof here, because I haven't that book!  :oops:

Are you having trouble accessing the pdf?

I don't see any PDF files here ... thankfully. In general, a question stated in your own words is more likely to get a response than an attached file.

It appears that we are to discuss the following theorem: if $f_{n}\to f$ on $[a,b]$, $f$ is continuous, and $f_{n}(x)\le f_{n+1}(x)$ for all $n$ and all $x\in [a,b]$, then the convergence is uniform. If so, then your second question should answer the first one: $\delta(x)$ work for subsequent terms because the sequence is increasing.

A counterexample with $\{f_{n}\}$ not being an increasing sequence: $f_{n}(x)=\max(0,1-n|x-1/n|)$ converge to $f\equiv 0$ but not uniformly.

Ack! It seems I never attached the link.

http://www.math.uiuc.edu/~mjunge/540-material.pdf

Thanks for answering the question mlok!


probability
inequalities
geometry
function
3D geometry
tetrahedron
expected value

Imagine that I keep a running total of random numbers from $[0,1]$ until the sum exceeds 1. So I pick a number, write it down, pick another number, add it to the number on my paper... stoping when the sum exceeds one. What is the expected number of numbers I add up until I exceed 1? The numbers are selected with uniform probability. I think it's either $1/2$ or $e$

You mean you think it's either $2$ or $e$.

[hide="Possibility"] Consider the $n$-dimensional graph of the inequality $x_{1}+x_{2}+...+x_{n}\ge 1$ as a partition of the unit $n$-cube.

If you can characterize the area of this inequality as a function of $n$ then the expected number can be found.  Unfortunately, while it's not so bad for $n = 2$ I don't see an obvious way to generalize... but it's late. [/hide]

It can't be 2 because with 2 numbers, you have a $\frac{1}{2}$ chance of getting a sum of larger than 1, and with fewer than 2 numbers you have a 0 chance of breaking 1.  So the expected value is at least $2\cdot \frac{1}{2}+3\cdot \frac{1}{2}= \frac{5}{2}$.

In general, the probability that $n$ numbers randomly selected from $[0, 1]$ have sum greater than 1 appears to be $1-\frac{1}{n!}$.  Geometrically, the $\frac{1}{n!}$ is the volume of a rectangular $n$-simplex, the analog of a tetrahedron where all angles between edges at a given vertex are right, each of whose edges at that vertex has length 1.  Alternatively, it's what you get when you cut off the lower corner of the rectangular box $[0, 1]^{n}$, since the points $(0, 0, \ldots, 1), (0, 0, \ldots, 1, 0), \ldots, (0, 1, \ldots, 0), (1, 0, \ldots, 0)$ define the hyper-plane $x_{1}+x_{2}+\ldots+x_{n}= 1$.

Then the expected value is $\sum_{n = 2}^{\infty}n((1-\frac{1}{n!})-(1-\frac{1}{(n-1)!})) = \sum_{n = 2}^{\infty}\frac{1}{(n-2)!}= e$, as you conjectured.

[quote="JBL"]
In general, the probability that $n$ numbers randomly selected from $[0, 1]$ have sum greater than 1 appears to be $1-\frac{1}{n!}$.

[/quote]

why is this?

[quote="JBL"]That was what the next paragraph attempted to explain, although actually it explained the complementary result, that the probability that $x_{1}+\ldots+x_{n}\leq 1$ is $\frac{1}{n!}$.  For the geometric reasoning, it helps to examine first the cases $n = 2$ and $n = 3$ (the square and cube).[/quote]

but the whole: the volume of the n-simplex with blah blah right angles, etc

i don't think thats clear...

Altheman - what JBL was saying about the volume of an n-simplex is just some basic definitional stuff. If you want a more in-depth definition of an n-simplex and its properties, you should be able to find one on Wikipedia or somesuch.

To anyone - I understand the geometric solution, but can anyone offer an algebraic one? (I'm not that comfortable with geometry in 2-space, much less n-space. :P)

Oh, and by the way, Watson - I told you it was e! I told you! Ha!  :) (Okay, done gloating now.) Cya on friday!
-Katie C.

May be this one is not geometric...

Let's consider a similar problem. We will add natural numbers from the set $\{1,2,...,n\}$ and than take a limit.
Suppose we already collected $n$ we have to wait 0 turns, $e_{n}=0$.
Suppose we collected $n-1$, so $e_{n-1}=1$ (1 turn)
$e_{n-2}=\frac{1}{n}e_{n-1}+\frac{n-1}{n}e_{n}+1=(1+\frac{1}{n})$
$e_{n-3}=\frac{1}{n}e_{n-2}+\frac{1}{n}e_{n-1}+\frac{n-2}{n}e_{n}+1=(1+\frac{1}{n})^{2}$
...
$e_{0}=(1+\frac{1}{n})^{n-1}$
The limit of $e_{0}$ is $e$

er, nevermind what this post was before I edited it - I get it now. **sheepish** Thanks guys!


If $ r$ is a rational number given by $ r\equal{}\frac{m}{n}$, where $ m$ and $ n$ are integers and $ n>0$, let $ S_r(a)\equal{}\left\{x\in{R}: 0\leq{x^n}\leq{a^m}\right\}$, and define $ a^r\equal{}\sup{S_r(a)}$. ($ a>1$)
 Prove that $ z\equal{}a^r$ is the unique positive root of the equation $ z^n\equal{}a^m$.
(Hint: there is a constant $ K$ such that if $ 0<\varepsilon<1$, then $ (1\plus{}\varepsilon)^n<1\plus{}K\varepsilon$. Hence if $ x^n<a^m<y^n$, there exists an $ \varepsilon>0$ such that $ x^n(1\plus{}\varepsilon)^n<a^m<\frac{y^n}{(1\plus{}\varepsilon)^n}$.)

 I don't know how to use the hint. Please, help me. Thanks.

[quote="J.Y.Choi"]If $ r$ is a rational number given by $ r = \frac {m}{n}$, where $ m$ and $ n$ are integers and $ n > 0$, let $ S_r(a) = \left\{x\in{R}: 0\leq{x^n}\leq{a^m}\right\}$, and define $ a^r = \sup{S_r(a)}$. ($ a > 1$)
 Prove that $ z = a^r$ is the unique positive root of the equation $ z^n = a^m$.
(Hint: there is a constant $ K$ such that if $ 0 < \varepsilon < 1$, then $ (1 + \varepsilon)^n < 1 + K\varepsilon$. Hence if $ x^n < a^m < y^n$, there exists an $ \varepsilon > 0$ such that $ x^n(1 + \varepsilon)^n < a^m < \frac {y^n}{(1 + \varepsilon)^n}$.)

 I don't know how to use the hint. Please, help me. Thanks.[/quote]
Warning : you can use the normal exponent operations only with integer exponents.
We have $ a^m > 0, n \geq 1$ . It's easy to see that $ a^r > 0$
$ 1 \leq a^m \Longrightarrow 1\in S_r(a)  \Longrightarrow  a^r > 0$
$ 0<a^m < 1 \Longrightarrow (a^m)^{n-1}<1 \Longrightarrow (a^m)^n<a^m \Longrightarrow a^m \in S_r(a)  \Longrightarrow  a^r > 0$

If $ 0< \varepsilon < 1$ then $ (1 + \varepsilon)^n < 1 + 2^n\varepsilon$. Let's suppose that $ (a^r)^n\neq a^m$, so $ (a^r)^n< a^m$ or $ (a^r)^n> a^m$ .
If $ (a^r)^n< a^m$ , let's denote $ \varepsilon=max(\frac 12, \frac{\frac {a^m}{(a^r)^n}-1}{2^n})$ so $ (a^r)^n<(a^r)^n(1+\varepsilon)^n<(a^r)^n(1+2^n\varepsilon)\leq a^m\ \Longrightarrow\ a^r(1+\varepsilon)\in S_r(a)$
so $ a^r$ is not the superior bound of ${ S_r(a)}$ (contradiction)

If $ (a^r)^n> a^m$ , let's denote $ \varepsilon=max(\frac 12, \frac{\frac {(a^r)^n}{a^m}-1}{2^n})$ so $ a^m\leq \frac{(a^r)^n}{1+2^n\varepsilon}< \frac{(a^r)^n}{(1+\varepsilon)^n} < (a^r)^n$, so  ${ 0 \leq x^n \leq a^m \Longrightarrow x <\frac {a^r}{1+\varepsilon}}<a^r$,
so $ a^r$ is not the samallest superior bound of ${ S_r(a)}$ (contradiction).

Finally $ (a^r)^n=a^m$  :cool:

Why did you set $ \varepsilon\equal{}\max\left(\frac{1}{2},\frac{\frac{a^m}{(a^r)^n}\minus{}1}{2^n}\right)$? And how do we know $ 0<\frac{\frac{a^m}{(a^r)^n}\minus{}1}{2^n}<1$ to satisfy $ 0<\varepsilon<1$?

Diogene, where are you??  Or please, anybody help me. Thanks.

I am very sorry, I made a mistake  :blush: : put $ min$ (in place of $ max$) .
...
If $ (a^r)^n < a^m$ , let's denote $ \varepsilon = min(\frac 12, \frac {\frac {a^m}{(a^r)^n} - 1}{2^n})$ so $ (a^r)^n < (a^r)^n(1 + \varepsilon)^n < (a^r)^n(1 + 2^n\varepsilon)\leq a^m\ \Longrightarrow\ a^r(1 + \varepsilon)\in S_r(a)$
...
If $ (a^r)^n > a^m$ , let's denote $ \varepsilon = min(\frac 12, \frac {\frac {(a^r)^n}{a^m} - 1}{2^n})$ so $ \ a^m\leq \frac {(a^r)^n}{1 + 2^n\varepsilon} < \frac {(a^r)^n}{(1 + \varepsilon)^n} < (a^r)^n$, so ${ 0 \leq x^n \leq a^m \Longrightarrow x < \frac {a^r}{1 + \varepsilon}} < a^r$
...
 :cool:


I have come across this observation from one of my students.  It states that "If two n-sided polygons intersect each other, then the total number of line segments(diagonals) that can be drawn from the vertices of the ploygons to all the intesecting points is n(6n-11)"
He wanted this to be checked for its validity and earlier existence in the History.

I think it's not correct. Try $ n\equal{}3$.

Another case :

Assume that $ n$-gons intersect only in one point(let that point be a common vertex of both polygons).

two $ n$-gons can intersect in almost any number of points, so there is not a formula depending only on $ n$.

Suppose two $ n$-gons intersect in $ k$ points.  For each point of intersection (which we assume is not a vertex of either polygon), we draw in $ n \minus{} 2$ segments from the vertices of one of the polygons (disregard the endpoints of the side of the polygon that the point of intersection lies on).  There are $ 2n$ vertices between the two polygons, but we only use $ n\minus{}2$ from each polygon, giving $ 2(n \minus{} 2)^2$ segments to each point of intersection, for a total of $ 2k(n \minus{} 2)^2$ segments.


If we are using Latex, Are we allowed to winzip all of the files we're gonna use, and send them by email?

Yep.  Don't even have to zip 'em if you don't want to.


function
floor function
algebra unsolved
algebra

For any real number $ x$, let $ [x]$ denote the greatest integer which is less than or equal to $ x$. Define

$ q(n) \equal{} [\frac{n}{[\sqrt{n}]}]$ for $ n\equal{}1, 2, 3, ... .$

Determine all positive integers $ n$ for which $ q(n) > q(n\plus{}1)$.

between successive squares, $ \lfloor \sqrt{n} \rfloor$ is constant, and thus $ q(n)$ is non-decreasing. thus the only places where $ q$ can possibly decrease are in going from $ k^2\minus{}1$ to $ k^2$, for some integer $ k$. in fact, $ q(k^2\minus{}1)\equal{}q(k^2)\plus{}1$.

in conclusion, $ n$ satisfies the condition in question if and only if it is one less than a square.


The distance from the earth to the sun is 93,000,000 miles, and light travels at 186,000 miles per second.  How many seconds does light from the sun take to reach the earth?


limit
logarithms
inequalities
factorial
calculus
calculus computations

Thanks, Kent  :lol: . I was working for a long time with factorails, double factorials and Stirling's formula, but I did not get any conclusive thing


inequalities
trigonometry
inequalities proposed

Prove that in any acute tringle $ABC$ the following inequality holds:

                            \[ m_{a}\leq\frac{1}{2}\cdot\sqrt{(b+c)^{2} cos A+a^{2}}. \]

using the fact that
$4m_a^2=2(b^2+c^2)-a^2$, we see that the inequality to be proven is equivalent to
$2(b^2+c^2)-a^2\leq (b+c)^2\cos A+a^2 \\\Leftrightarrow 2(b^2+c^2-a^2)\leq (b+c)^2 \cos A \\\Leftrightarrow 4bc\cos A\leq (b+c)^2 \cos A \\\Leftrightarrow 0\leq (b-c)^2$
and we're done!

Nice, campos. My solution is the same. ;)


[quote="perfect_radio"]Find $\displaystyle n \in \mathbb N$, $\displaystyle n \geq 2$, and the digits $\displaystyle a_1,a_2,\ldots,a_n$ such that
\[ \displaystyle \sqrt{\overline{a_1 a_2 \ldots a_n}} - \sqrt{\overline{a_1 a_2 \ldots a_{n-1}}} = a_n .  \][/quote]

[hide]
We see that $\overline{a_1 a_2 \ldots a_{n-1}}$ has to be a square, so let $\overline{a_1 a_2 \ldots a_{n-1}} = x^2$ and $a_n = y$.

We want to solve

$\sqrt{10x^2+y}-x = y$

$9x^2-2xy+(y-y^2) = 0$.

Quadratic formula says

$x = \frac{y}{9} \pm \frac{\sqrt{10y^2-9y}}{9}$.

Then obviously $y = 9$ so $x = 4, -2$. Checking, we see that $x = 4$ is right, so

$\sqrt{169}-\sqrt{16} = 9$.[/hide]


modular arithmetic
combinatorics unsolved
combinatorics

A regular polygon of $ n$ sides ($ n\geq3$) has its vertex numbered from 1 to $ n$. One draws all the diagonals of the polygon. Show that if $ n$ is odd, it is possible to assign to each side and to each diagonal an integer number between 1 and $ n$, such that the next two conditions are simultaneously satisfied:

(a) The number assigned to each side or diagonal is different to the number assigned to any of the vertices that is endpoint of it.

(b) For each vertex, all the sides and diagonals that have it as an endpoint, have different number assigned.

Label each edge with endpoints $i,j\in\overline{1,n-1}$ with the number in $\overline{1,n}$ congruent to $i+j\pmod n$. Now for each $i\in\overline{1,n-1}$ all numbers in $\overline{1,n}$ appear on edges adjacent to $i$ except for one, which is different from $n$ (otherwise we'd have $2i\equiv 0\pmod n\Rightarrow i\equiv 0\pmod n$, contradiction; the implication is a consequence of the fact that $n$ is odd). Label the edge $in$ with this number. This is the labeling we're looking for.

This is a little hard to explain, so it might be a little long. (For the sake of explanation I present the case when $n=11$ while detailing the solution).

WLOG suppose all numbers are positioned clockwise ( if not then just relabel the numbers). Let us assign to every edge the number corresponding to the vertex opposite to it ( we can guarantee there is one since $n$ is odd) (see figure 1).
Now pick a vertex,  say 1, draw diagonals to the endpoints of the opposite edge, we will call these diagonals "main diagonals". As you can see, we end up with two symmetric shapes , each of which has $n$ numbers on it, specifically; $1, 2, 3, 4,..., n$ and there will be a "middle" or "pivot" number on each side, in our example 9 and 4 are pivot numbers(Remark: The pivot number could be either on an edge or a vertex), and that will be the assigned number to the main diagonal nearest to it. Now if you draw all diagonals from vertex 1, there will always be a pivot number for each diagonal (this because of the parity of $n$) and that's gonna be the one you'll assign to it (See figure 2). Finally, to see that every diagonal connected to 1 is different, notice by symmetry on our pick of the numbers of the edges that the endpoint (different from 1) of any diagonal is counted as an edge on the other shape formed by the other main diagonal i.e. the other symmetrical half, thus there can't be two diagonals with the same number.


induction
ceiling function
function
inequalities
calculus
integration
5th edition

Ok, let's gather our solnutions. I will post my solution from home. Anyhow, can any one give solution of 3? It seemed easiest problem at first but oh! no! it was the hardest one! :o

By the way, valentine why haven't u posted round 4 problem? I am a bit puzzled!

and one thing more, when will u give the result of edition 4? And when will u give result of edition 5? After IMO? :D

[quote="mahbub"]By the way, valentine why haven't u posted round 4 problem? I am a bit puzzled!

and one thing more, when will u give the result of edition 4? And when will u give result of edition 5? After IMO? :D[/quote]The problems have just been posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=239943#p239943]here[/url]. I had some troubles with my internet connection, thus the delay.

I hope to publish soon the results of the 4th edition. The results of the 5th edition should also be available before the IMO :)

So, it seemed trivial at first... To me too, when I found it, but then I realized the opposite. :P  :P

[quote="erdos"]I know this is a stupid question  :blush:  But I am not familiar with the notation $\underset{1\leq i<j\leq n}{\sum }x_{i}x_{j}\left| a_{i}-a_{j}\right| $  !!! Can anyone expand it please ??  :lol:  :lol:[/quote]

say $n=4$

the sum means: $x_{1} x_{2} | a_{1}-a_{2}| +x_{1} x_{3} | a_{1}-a_{3}| + x_{1} x_{4} | a_{1}-a_{4}|+x_{2} x_{3} | a_{2}-a_{3}|+x_{2} x_{4} | a_{2}-a_{4}|+ x_{3} x_{4} | a_{3}-a_{4}|$

i didn't manage to do anything with 3 from round 3! Anyone did better?

Thanks [b]perfect_radio[/b] Now it's clear  :lol:

plz some one post soln of 3. here are my solns for 1 and 2.

5+5=10pages solution. soln of 2 was very messy. specially the extra part.

the soln of 1 was easy!

ah.. so strange.
I remember I thought q1 for around 15 min.. and wrote the case when $x_1=1$, then i found that i can't find infinite 1. Am I missing something? (I don't really have time these two weeks and I used all the time i can spare for q3.. and i can't solve it, so i didn't hand in my solution  :( )

[quote="mahbub"]may be u lost some thing, but for x1=1, it takes too many terms about 15 terms maybe. but for 2,3,4,5 they takes comparatively less terms.[/quote]
do you mind listing the case for x_1=1 and x_1=2?

Oh! plz some one come up with the solution of 3.

hmmm... here are my solutions.. 3 not included :P i still haven't seen a soln to the one in round 2, so i guess we'll have to wait for this one also

the soln to the first one is extremely long :(, i'm not quite sure whether it's correct
(pengshi's solution looks nicer)

Here are my solutions to this round. I didn't solve problem 3 either.
{\bf Problem 1:} Let $\{ x_n \}_n$ be a sequence of positive rational numbers, such that $x_1$ is a positive integer, and for all positive integers $n$, $x_n= {2(n-1)\over n}x_{n-1}$ if $x_{n-1}\leq 1$, and $x_n = {(n-1)x_{n-1}-1 \over n}$, if $x_{n-1}>1$.

{\bf Solution:} We will show that $x_n =1 $ for infinitely many $n's$, which is clearly sufficient. Define an integer sequence $\{ y_n \}_n$ such that $y_n = nx_n$, then $y_n=2y_{n-1}$ if $y_{n-1}\leq n$ and $y_n=y_{n-1}-1$ if $y_{n-1}> n$.

	First, we show by induction that $y_n \geq n-1$ for all $n$. This is obviously true for $n=1$ and 2. Suppose that it is true for $n=k$. If $y_k \geq k+1$, then $y_{k+1} \geq k =(k+1) -1$. Suppose that $k \geq y_k \geq k-1$, then $y_{k+1} \geq 2(k-1) \geq (k+1)-1$, since $k\geq 2$. Now, everytime $y_n$ take the value of $n$ or $n-1$, we call it a "pass", and we call the two kinds of passes "class n passes" and "class n-1 passes" resepetively. For $n \geq 4$, it is easy to see that as $n$ increases, $y_n$ will decrease by one each time until it makes a pass, after which it doubles and decrease by one each time again (since for $n \geq 4$, $2(n-1) > n+1$). If the sequence makes a pass at index $n=k$, then let $t=y_k$. The sequence will make a pass again at $y_{k+1+ \left\lceil {2t-k-1 \over 2} \right\rceil}$, since $y_n-n$ decreases by 2 each time after $y_n$ doubles. It follows that the sequence will make infinitely many passes.

	We now show that the $y_n$ will make infinitely many class n passes. Suppose on the contrary that it dosn't, then there exists $M$ such that for all $n>M$, the $y_n$ never makes a class n pass. Since we've shown that the sequence will make infinitely many passes, all the passes after $n>M$ must be class n-1 passes. Let the sequence makes a pass at index $g$ where $g>M$. Let $y_g = g-1 = 2^mp$ where p is odd. Using the formula above, we see that the sequence will make a pass again at index $g'= g+1+\left\lceil {2(g-1)-g-1 \over 2} \right\rceil=g+1+ \left\lceil {g-3 \over 2} \right\rceil$, which has a value of $t_{g'}=2(g-1)- \left\lceil {g-3 \over 2} \right\rceil$. Since this must be a class n-1 pass, we have $g+1+ \left\lceil {g-3 \over 2}\right\rceil -1= 2(g-1)- \left\lceil {g-3 \over 2}\right\rceil$, which implies that $g$ must be even. Now, $g'=g+1+ \left\lceil {g-3 \over 2}\right\rceil=2^mp+2^{m-1}p= 2^{m-1}(3p)$, which divides a smaller power of 2 than $g$. Thus, we can repeat this argument $m$ times and get a $g''$ which is odd, which would be a contradiction since all the $g$'s must be even. Therefore, $y_n =n$ for infinitely many $n$'s and as a result, $x_n =1 $ for infinitely many $n's$.
{\bf Q.E.D}

{\bf Problem 2:} Let $0<a_1 <a_2 < \cdots <a_{16} <122$ be 16 integers. Prove that there exists integers $(p,q,r,s)$, with $1 \leq p < r \leq s <q \leq 16$, such that $a_p+a_q=a_r+a_s.$

{\bf Solution:} We will prove the problem with a bound of 123 instead of 122. Suppose on the contrary that integers $(p,q,r,s)$ as described in the problem don't exist, then for each $1 \leq i, j,i',j' \leq 16$, with $(i,j) \neq (i',j')$, we have $a_i-a_j \neq a_{i'}-a_{j'}$. (Or else we can easily construct $(p,q,r,s)$ by properly sorting $(i,j', i', j)$.) Let $b_i=a_{i+1}-a_i$ for $1 \leq i \leq 15$. Then by the above, all the $b_i$'s are distinct, along with $\sum_{i=u}^{v}{b_i}$ for all $1 \leq u \leq v \leq 15$. Furthermore, $\sum_{i=1}^{15}{b_i} = a_{16}-a_1 \leq (123-1)-(0+1)=121$. Since $1+2+3+\cdots +14 +15 =120$, we have the following two cases.

	Case 1: The $b_i$'s are a permutation of $\{1,2,3,\cdots,15\}$. Note that to avoid repeats, 1 can only be beside 15, so 1 must be either at the beginning or the end of the permutation. WLOG, $b_1=1$, then $b_2 = 15$. Now, 2 can not be adjcaent to anything less or equal to 13 since all the $\sum_{i=u}^{v}{b_i}$ are distinct. Moreover, 2 can not be beside 14 since 16=1+15 is taken. This means that 2 can only be next to 15, so $b_3 =2$. However, anything at $b_4$ would produce a contradiction.

	Case 2: The $b_i$'s are a permutation of $\{1,2,3,\cdots,13,14, 16\}$. We use the same type of reasoning as above. Note that 1 can only be next to 14 or 16. Suppose that 1 is next to 14, then 2 can only be next to 16. Thus, 2 is either at the very beginning or very end. WLOG, $b_1 =2$, $b_2 =16$. Continuing, 3 can only be next to 14 or 16. If 3 is at the very end, then, $b_{15}=3$, $b_{14}=14$, $b_{13}=1$, and anything at $b_{12}$ would produce a contradiction. If 3 is in the middle, then it must be adjacent to both 14 and 16 and so we have $b_3=3$, $b_4=14$, and $b_5=1$.  However, anything at $b_6$ would produce a contradiction. Thus, 1 is not adjacent to 14. Suppose that 1 is next to 16 and not 14, then 1 must be at the beginning or the end of the permutation. WLOG, $b_1=1$, $b_2=16$, then 2 can only be next to 13. This means that 3 can only be next to 16, and $b_3=3$. However, anything at $b_4$ would create a contradiction.

	This covers every case, so we are done.
{\bf Q.E.D}

It is not fair, may be no one solved p2 on r3. so they did not publish. But u said u solved, again u r not publishing it is unfair :mad:

Anyhow, grobber what about ur solution?

It is possible to get a better bound as follows.

Following pengshi, let $b_i=a_{i+1}-a_i$. Since $b_i+b_{i+1}=a_{i+2}-a_i$, the $b_i$'s and sums of consecutive pairs are all distinct. This gives 29 distinct numbers whose sum is less then $3\sum b_i$, and therefore $a_{16}-a_1 > 145$, and 122 can be replaced by 147.

By also using consecutive triplets, this can be further improved. I am not sure what is the best bound achievable by this method, and I would be surprised if it is anywhere near the true bound.

Defining $y_n=nx_n$, we show that $y_n=n$ infinitely often. We have $y_{n+1} = y_n-1$ if $y_n>n$ or $2y_n$ otherwise. Consider $y_n$ just before such doublings. Either $y_n=n$, or $y_n=n-1$. Suppose that from some time on only the latter occurs.

If a doubling occurs at an even time $n$, then the next doubling occurs at time $3n/2$. It follows that eventually a doubling must occur at an odd time $n$, but then $y_m=m$ for $m=3n-1$.

i am a bit surprised to see 147 as a possible bound. really surprised! :huh:

Anyhow, one thing not related to math. i think i saw once a spider emotion. but i can't see that in left list. why? and i think i saw champion emotion, can't see that also. were these elemenated?

[quote="mahbub"]i am a bit surprised to see 147 as a possible bound. really surprised! :huh: [/quote]I was kinda expecting this :P :)

following the method.
if we consider $b_i=b_i+b_{i+1}+b_{i+2}$
we can get $a_{16}-a_{1} \geq 151$
then we  can replaced by 152. :D

well zhaobin have u solved p3?

No one???

Valentin, can u give us some hint on that?

no :( ,sorry
I try to use largrange multping,but I faided.
the another problem is 2nd-p2 ;)

I apologise for not posting the solution of nb 3 problem, but I could not enter mathlinks this week. It seems it wasn't solved by many people. Ok, here it is my idea:
    Consider the convex function $ f(x)=\frac{1}{x+\frac{1}{n-1}}$. Use generalized Popoviciu inequality $\sum {f(x_i)}+n(n-2)\cdot f(\frac{x_1+...+x_n}{n})\geq (n-1)\sum {f(\frac{x_2+...+x_n}{n-1})}$. After that it reduces to a very simple inequality. Do you manage to conclude with this hint?

[quote="siuhochung"]btw, harazi, how can you think of popoviciu?! I only tried induction, mixing variables and your integration method...  :([/quote]He invented the problem, so that's how he thought of popoviciu :D

[quote="siuhochung"][quote="perfect_radio"][quote="harazi"]
Consider the convex function $ f(x)=\frac{1}{x+\frac{1}{n-1}}$. Use generalized Popoviciu inequality $\sum {f(x_i)}+n(n-2)\cdot f(\frac{x_1+...+x_n}{n})\geq (n-1)\sum {f(\frac{x_2+...+x_n}{n-1})}$. After that it reduces to a very simple inequality. Do you manage to conclude with this hint?[/quote]

did anyone manage to finish it off? i just get another inequality which looks as ugly as the first one :([/quote]
Using the hint,
$\sum f(x_i) - \sum f(\frac{S-x_i}{n-1}) \ge (n-2) (\sum f(\frac{S-x_i}{n-1}) - n f(\frac{S}{n})) \ge 0$. The last one is true by jensen.
btw, harazi, how can you think of popoviciu?! I only tried induction, mixing variables and your integration method...  :([/quote]

 :huh:
the second sum is $(n-1)\cdot$ RHS
the first sum is $(n-1)\cdot \frac{1}{(n-1)x_{i}+1}$

how does it imply the given ineq ?

sorry if i'm missing smth very obvious

Can someone please show me what to do after the Popoviciu? Since the inequality you get afterwards is not simple to me at all. Thanks.

Actually, I figured it out. After applying Generalized Popoviciu, we need to prove:
\[\sum{(n-1)^2 \over n-1+x_i} \geq \sum{1 \over x_i + {1 \over n-1}} + {n(n-2) \over {S \over n} + {1 \over n-1}}\]
Subtracting the first term on the right from the left, rearranging, the inequality becomes equivalent to:
\[\sum{1 \over x_i + {1 \over x_i} + n-1 + {1 \over n-1}} \geq {1 \over {S \over n} + {1 \over n-1}}\]
We now apply jensen to $f(y_i)={1 \over y_i + n-1 + {1 \over n-1}}$ where $y_i = x_i + {1 \over x_i}$, and the above follows from the fact that $S \geq n$, which can be found using cauchy. Nice problem, Harazi.

[quote="harazi"]Thanks!  ;) . This is why I thought of generalized Popoviciu, because I created it.[/quote]
Oh, valentin and harazi, that explains the source of this ingenious solution.  :P

I give a nice solution once  a year on the forum and then Valentin must say that in fact I give the solution to a problem created by me. It's not fair! This demoralizes me... :(  :(  :P  :P  :P  :P

OK next time I won't tell anyone that it's your problem :D

well i would be glad if some one help me!

If we let $f(x)=\frac{1}{x+\frac{1}{n-1}}$ (according to harazi)

then according to popoviciu's generalized inequality we will get:

$\sum {f(x_i)} + n(n-2)f(\frac{s}{n}) \geq (n-1)\sum{ f(\frac{s-x_i}{n-1})} = (n-1)^2*RHS$

So how to finish the problem now? :(

Another thing, $f(x_1)=\frac{n-1}{nx_1-x_1+1}$ not the one given in the problem. Again i have not used the fact: $\sum{x_i} = \sum{\frac{1}{x_i}}$ fact yet! :(

Is the solution I posted above not correct?
I think that it answers all your questions.

[quote="mahbub"]how r u bringing that?[/quote]mahbub, please no chatspeak, ok? :) (write "you" instead of "u", "are" instead of "r", etc).

[quote="mahbub"]Ok. One thing more. Why do you say not to use only capital letter? (actually my exam is running. so i shortened my writing as possible.)[/quote]

well, in general using capital letters for all alphabets is quite rude.

hey no one wanna cler me on that inequality?

[quote="mahbub"]Well, after subtracting i got:
$x_i+\frac{x_i^2}{n-1}+\frac{1}{n-1}+\frac{x_i}{(n-1)^2}$ at denominator

how r u bringing that?[/quote]

Multiply the top and bottom by $(n-1) / x_i$, then you get $n-1+{1\over n-1}+ (x_i+{1\over x_i})$ at the bottom. Now, you treat $x_i+{1\over x_i}$ as a whole and apply jensen, following my previous post.

I hope that this is clear now.


solve:
all triples $ (a,b,c)$ such that ,
$ a^2 \minus{} 2b^2 \equal{} 1 ,
 2b^2 \minus{} 3c^2 \equal{} 1,
 ab \plus{} bc \plus{} ca \equal{} 1$

Hello!!

Maybe $ (\sqrt{2},\frac{1}{\sqrt{2}},0)$ and $ (\minus{}\sqrt{2},\frac{\minus{}1}{\sqrt{2}},0)$ are the only solution .

yep your final answer is correct .
plz post ur solution

[hide="solution"]
$ ab\plus{}bc\plus{}ca\equal{}1 \Longrightarrow (b\plus{}c)(b\plus{}a)\equal{}1\plus{}b^2 \qquad(1)$

$ a^2\minus{}2b^2\equal{}1 \Longrightarrow (a\minus{}b)(a\plus{}b)\equal{}1\plus{}b^2 \qquad(2)$

$ a\plus{}b\not \equal{} 0$ otherwise $ c(a\plus{}b) \plus{} ab \equal{} 1 \Rightarrow ab\equal{}1$ which is impossible

So from (1) and (2), $ a\equal{}2b\plus{}c (*)$

Similarly
$ ab\plus{}bc\plus{}ca\equal{}1 \Longrightarrow (c\plus{}b)(c\plus{}a)\equal{}1\plus{}c^2 \qquad(3)$

$ 2b^2\minus{}3c^2\equal{}1 \Longrightarrow 2(b\minus{}c)(b\plus{}c)\equal{}1\plus{}c^2 \qquad(4)$

Therefore $ a\plus{}c\equal{}2(b\minus{}c) \Rightarrow a\equal{}2b\minus{}3c (**)$

From $ (*)$ and $ (**)$ clearly $ c\equal{}0$

Hence $ a\equal{}2b \Longrightarrow (2b)b \plus{} 0(b\plus{}c) \equal{} 1 \Longrightarrow b\equal{}\pm \frac{1}{\sqrt{2}}$

And $ a\equal{}\pm\sqrt{2}$

This gives $ (a,b,c) \equal{} (\sqrt{2},\frac{1}{\sqrt{2}},0)$ or $ (\minus{}\sqrt{2},\minus{}\frac{1}{\sqrt{2}},0)$ as the only solutions  :) [/hide]


geometry
parallelogram
geometry proposed

Given a ball $K.$ Find the locus of the vertices $A$ of all parallelograms $ABCD$ such that $AC\leq BD,$ and the diagonal $BD$ lies completely inside the ball $K.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


number theory
prime numbers
algebra proposed
algebra

Solve in prime numbers:

$  \{\begin{array}{c}\ \ 2a - b + 7c = 1826 \
3a + 5b + 7c = 2007 \end{array}$

Multiply the first equation by $ 5$ and add to the second. We get $ 13a\plus{}42c\equal{}11137$. Since $ 42c$ and $ 11137$ are both divisible by $ 7$ and $ a$ is prime, must be $ a\equal{}7$. Then easily follows $ c\equal{}263$ and $ b\equal{}29$.

It is funny I am author of this problem)

5x first eq + second gives 13a+42c=11137. RHS is 0 mod 7 so LHS mod 7 means 13a is divisible by 7, or a=7. Plug it into our equation and we get c=263 and b=29, which indeed satisfies the equation.


[quote="illcrowflu"]I must prove that if $p$ is an odd prime and $p \neq 5$, then $p^{2}+1$ or $p^{2}-1$ is divisible by $10$.

The way I am proving this is that $p$ must be a number such that the digit is the ones place is $1,3,7$ or $9$.  Squaring any number with a $1$ or a $9$ in the ones place results in a number with a $1$ in the ones place and then $p^{2}-1$ will be divisible by $10$.  Squaring any number with a $3$ or a $7$ in the ones place will result in a number with a $9$ in the ones place and then $p^{2}+1$ will be divisible by $10$.[/quote]
Yes, It is true!


inequalities
trigonometry
inequalities solved

Here is the problem posted at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=761

But I couldn't reply since the topic is locked.

What is "topic locked" mean? This is the first time I encounter the problem.

[quote]Let $x,y,z>0 $, then
\[ \frac{x}{x+ \sqrt {(x+y)(x+z)}}+ \frac{y}{y+ \sqrt{(y+z)(y+x)}} + \frac{z}{z+ \sqrt {(z+x)(z+y)}} \leq 1. \]
[/quote]

It looks impossible, actually it's very simple. Here we go:

By AM-GM, we have ${\displaystyle  \sqrt {(x+y)(x+z)}} \leq \frac{(x+y)+(x+z)}{2} = x + \frac{y+z}{2}$.

Hence:

(1) $\displaystyle \frac{x}{x+ \sqrt {(x+y)(x+z)}} = \frac{x( \sqrt {(x+y)(x+z)} - x) }{xy+yz+zx} \leq \frac {x(y+z)}{2(xy+yz+zx)}$

Similarly,

(2) $\displaystyle \frac{y}{y+ \sqrt{(y+z)(y+x)}} \leq \frac {y(z+x)}{2(xy+yz+zx)}$

(3) $\displaystyle \frac{z}{z+ \sqrt {(z+x)(z+y)}} \leq \frac {z(x+y)}{2(xy+yz+zx)}$

Adding (1), (2) and (3) together, we get the desired result.

If we let $a=\sqrt{y+z}$, $b=\sqrt{z+x}$ and $c=\sqrt{x+y}$, then $a,b,c$ are sides of an acute triangle. And, we have:

$x= (b^2+c^2-a^2)/2$, $y= (c^2+a^2-b^2)/2$ and $z= (a^2+b^2-c^2)/2$.

The inequality is equivalent to:

$\displaystyle \frac{\cos{A}}{1+\cos{A}} + \frac{\cos{B}}{1+\cos{B}} + \frac{\cos{C}}{1+\cos{C}} \leq 1$.

Or equivalent to:

$\displaystyle \frac{1}{1+\cos{A}} + \frac{1}{1+\cos{B}} + \frac{1}{1+\cos{C}} \geq 2$,

which is equivalent to:

$\displaystyle \sec^2 \frac{A}{2} + \sec^2 \frac{B}{2} + \sec^2 \frac{C}{2} \geq 4$.

The last one is one of the trig inequalities posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2873

we have (x+y)(x+z)=x^2+x(y+z)+yz>=2xsqrt(yz)+x(y+z)=x(sqrt(y)+sqrt(z))^2
so x+sqrt[(x+y)(x+z)]>=sqrtx(sqrtx+sqty+sqrtz)
and x|(x+sqrt[(x+y)(x+z)])<=sqrtx|sqrtx+sqrty+sqrtz
we have similar inequalities and adding them we get the result


geometry
parallelogram
vector
cyclic quadrilateral
geometric transformation

Let $ ABCD$ is a rectangle. Let $ N$ is midpoint of $ CD$, let $ H$ is project orthogonal of $ B$ on $ AC$, let $ M$ is midpoint of $ AH$. Prove that $ \angle BMN \equal{} 90^0$.

He meant [url=http://mathworld.wolfram.com/Projection.html]projection[/url].

I see that $ \triangle ADB\sim \triangle MNB$, but I don't see how I could prove this...

[hide="Solution"]
Since $ \angle BAH \equal{} \angle BDC$, we have $ \triangle BAH \sim \triangle BDC$. Since $ M$ is the midpoint of $ \overline{AH}$ and $ N$ is the midpoint of $ \overline{DC}$, it follows that $ \triangle BMH \sim \triangle BNC$, and so $ \angle BMC \equal{} \angle BNC$. Thus $ BMNC$ is a [[cyclic quadrilateral]], from which it follows that $ \angle BMN \equal{} 180 \minus{} \angle BCN \equal{} 90^{\circ}$.
[/hide]

Edit: Typo fix, thanks.

Might it be possible to solve this through a spiral similarity (centered at $ B$, which sends $ \triangle BHC$ to $ \triangle BAD$)?

[quote="azjps"]
Since $ \angle BAH \equal{} \angle BDA$, we have $ \triangle BAH \sim \triangle BDC$. [...][/quote]

I think you mean $ \angle BAH \equal{} \angle BDC$, but otherwise nice solution.

[quote="azjps"]
Might it be possible to solve this through a spiral similarity (centered at $ B$, which sends $ \triangle BHC$ to $ \triangle BAD$)?[/quote]

That's what I had in mind, except with $ \triangle BMN$.

Thank you very much. I will post my solution.

This is my solution:

$ \bullet$ Let $ P$ is midpoint of $ BH$ $ \Longrightarrow MP // AB$ and $ MP \equal{} \frac {1}{2} AB$ $ \Longrightarrow MP // NC$ and $ MP \equal{} NC$ $ \Longrightarrow MPCN$ is parallelogram $ \Longrightarrow MN // CP$, (1).

$ \bullet$ $ MP // AB$ $ \Longrightarrow MP \perp BC \Longrightarrow P$ is orthocentre point of $ \triangle BCM$ $ \Longrightarrow CP \perp BM$, (2).

$ \bullet$ Since (1) & (2) $ \Longrightarrow MN \perp BM \Longrightarrow \angle BMN \equal{} 90^0$.

This is a nice problem. Can you have other solution?

Solution 1: A spiral similarity maps $ AHB$ to $ DCB$ center $ B$. Since $ M$, $ N$ are midpoints, that spiral similarity maps $ MHB$ to $ NCB$, but that means that $ MNB$ is similar to $ HCB$, so we're done.

Solution 2: A spiral similarity center $ B$ maps $ CHB$ to $ DAB$, so the average of these triangles, $ NMB$ is similar to the original triangles, so we're done.

[quote="Altheman"]A spiral similarity center $ B$ maps $ CHB$ to $ DAB$, so the average of these triangles, $ NMB$ is similar to the original triangles, so we're done.[/quote]

Ok, so that argument does work.  :)

Edit: Yeah, I've seen mean geometry before, but not for a while, so I wasn't sure.

Realize that it is not an obvious result. I've heard of it called "MGT" or "Mean Geometry Theorem." I believe that on an olympiad, you would want to cite outline a proof...it is not particularly long or difficult, but it is not obvious.

Using the dot product of vectors, we will get the result easily! I'll let you guys try solving it by this method!(I;m also busy right now) :wink:

$ \Vec {BM}.\Vec {MN} \equal{} 0 \Longrightarrow \angle BMN \equal{}90^0$!! :rotfl:


linear algebra
matrix
induction
combinatorics proposed
combinatorics

Let $ n\geq 2$ an integer and consider a $ n*n$ matrix with all its elements 1 or -1. Each element is the product of its neighbours (with a common side). Prove that we can find an infinity of n for wich all the numbers must be 1.

NObody? :(  C'mon :P  at least give me some hint! I will appreciate a solution  using linear algebra.

well I didnt use linear algebra...  :maybe:
[hide="hint"]note that if a $ n*n$ table had the propertyy,then a $ (2n-1)*(2n-1)$ table has the property too,also proved by induction I think...[/hide]

Look at [b]iandrei[/b]'s solution:
http://www.mathlinks.ro/viewtopic.php?t=3819
As a resume, [b]iandrei[/b] proves that for $ n\geq2$, for a $ 2^{n}-1\times2^{n}-1$ board there is a single arrangement - when all numbers are 1.


linear algebra
matrix
algorithm
graph theory

The only number that can be adjacent to $ 8$ is the number $ 3^2 \minus{} 8 \equal{} 1$, because
$ 2^2 \minus{} 8 \equal{} \minus{} 4 < 0$,
$ 4^2 \minus{} 8 \equal{} 8$ and
$ 5^2 \minus{} 8 \equal{} 17 > 16$
So we cannot put the numbers in a circle, because then we would need two adjacent numbers to $ 8$
The same thing happens with $ 16$, which can have only $ 9$ as adjcacent

So, if we put the numbers in a line, we should put $ 8$ and $ 16$ at the edges

We make a table
[img]14363[/img]
In the row - [b]i[/b] we can see all the numbers that can be adjacent to the number [b]i[/b]. From the table we can see that the numbers $ 8,16$ have only one adjacent number each

The corresponding graph is the following:
[img]14369[/img]

From this graph we can see that there is only one solution. The order is $ 8 \minus{} 1 \minus{} 15 \minus{} 10 \minus{} 6 \minus{} 3 \minus{} 13 \minus{} \cdots \minus{} 16$ (or the inverse order)

[quote="Altheman"]Can we arrange the numbers $ 1$, $ 2$, ..., $ 16$ on a circle such that the sum of each two adjacent numbers is a perfect square? Same question for a line.[/quote]
This is a prime example of a question where reading carefully is helpful, since the way the question is phrased it's definite that the first is not possible (otherwise the second would be pointless)

Nice use of graph theory. I just use the same argument on 8 and 16 in words, and showed an example to prove the line part, basing it on the fact that 8 and 16 must be at the end.

Well done. My method was similar:

For each number $ 1$,$ 2$,...,$ 16$, associate one point with each number in a graph.

Let $ M$ be the matrix where $ m_{ij}\equal{}i\plus{}j$. Then let $ f: \mathbb{N}\rightarrow \{0,1\}$ such that if $ x$ is a perfect square, $ f(x)\equal{}1$, else $ f(x)\equal{}0$. Then $ f(M)\equal{}A$ where $ A$ is the adjacency matrix of the graph (of course $ A$ is symmetric since $ i\plus{}j\equal{}j\plus{}i$).

In graph theory terms, the first question asks us to determine if there are any Hamiltonian cycles. However $ 8$ and $ 16$ each have degree $ 1$ (in terms of undirected edges) so there are no cycles.

The next question asks us to find a Hamiltonian path. There isn't any good algorithm for this, but given the graph, it is pretty easy. We know $ 8$ and $ 16$ must be on the edges, then we follow the path starting from $ 16$.


For $ a,b > 1,$ prove that there exists infinite number of pairs $ (a,b)$ such that $ \displaystyle \frac {a}{b} \plus{} \frac {a \minus{} b}{\frac {b}{a}}$ is an integer.

[hide="The bestest solution ever"]Set $ a \equal{} b$.  :wink: [/hide]
[hide="Or..."]$ \frac {a}{b} \plus{} \frac {a \minus{} b}{\frac {b}{a}} \equal{} \frac {a}{b} \plus{} \frac {a(a \minus{} b)}{b} \equal{} \frac {a \plus{} a(a \minus{} b)}{b} \equal{} \frac {a(a \minus{} b \plus{} 1)}{b}$ So, even limiting yourself to $ a,b \in \mathbb{N}$, you can have $ b|a$ or $ b|a \minus{} b \plus{} 1 \Rightarrow b|a \plus{} 1$.[/hide]

Yes, that is well-done. I should've put restriction of $ a \neq b$ but I forgot.

It is equivalent to $ b|(a^2\plus{}a\minus{}ab)$ or $ b|(a^2\plus{}a)$ which is still trivial.

$ \frac {a}{b} \plus{} \frac {a \minus{} b}{\frac {a}{b}} \equal{} \frac {a}{b} \plus{} \frac {a(a \minus{} b)}{b}$???

isnt it $ \frac {a}{b} \plus{} \frac {a \minus{} b}{\frac {a}{b}} \equal{} \frac {a}{b} \plus{} \frac {b(a \minus{} b)}{a}$????

Brut3Forc3 wrote the denominator of the second fraction incorrectly.  Reread the original problem.


AMC
AIME
geometry
3D geometry
pyramid
USA(J)MO
USAMO

How hard was the AMC?  I'm not really in a position to judge.  I thought it was a little tougher than the ones I had seen, but I may be wrong.  I ended up with a 120, when I took the 2001 and got a 132 the night before.  Maybe it was just testing conditions that hurt me.

It was easier for people aiming to make the AIME, but harder for people aiming for a perfect. This was because most questions were easy, but a few problems (particularly 18, 24, and 25) were very easy to mess up (I missed 24 and left 25 blank).

I left a couple blank that I shouldn't have, like 16 and 17.  I wasn't quite sure if 1/3bh applied to non-right pyramids, so I didn't answer that.  I think I was way too conservative, because I made 4 dumb mistakes last year, getting a 98.

I got 132.5, and I got 125.5 on practice 2001 a few days earlier.  If I had done the 2001 like I had done the 2005, I would probably have gotten around 118.5 on 2001.

Anyone wanna guess what index will be needed to qualify for USAMO?  Of course, AIME hasn't occurred, but maybe considering an AIME of average difficulty?

It was easier than the Mock AMC's (on average) for sure.  :D

I can attest to the much easier than the MOCK ones, but for me, it was really easy. But again, easiness doesn't guarantee that I do well. I screwed this really easy one up alot.

I guessed well on 10.24.

[hide="my solution"]I figured that n is a square of a prime, since GPF= :sqrt: n, and thus we have  :sqrt: prime^2+48 as the GPF of n+48. Now we see that if n!=4(plug it in), n+48 is not  a square, since from n^2=>(n+1)^2 is n^2+n+(n+1) and 48 does not fit that formula, so  :sqrt: n+48 is not integral, so [/hide]

I think 1-20 are okay(like last year except there's no tricky question about the dollars and cents ),

21-23??? What's wrong? Why make them so easy? It really really upset me that I didn't attempt #21 and 23(i got both in 5 minutes after the test).

#24 and 25 are still pretty hard.

You did a lot of casework?  Perhaps I did it incorrectly then.

18) Call a number "prime-looking" if it is composite but not divisible by 2, 3, or 5.  The three smallest prime-looking numbers are 49, 77, and 91.  There are 168 prime numbers less than 1000.  How many prime-looking numbers are there less than 1000?

a)100
b)102
c)104
d)106
e)108

[hide]I just found multiples of 2, 3, and 5 of 1000, then add back the double counted ones (unless you never subtracted them to begin with), then subtract the number of primes.  The answer is (a), I believe.[/hide]

On that problem I didn't realize that 2, 3, and 5 were both prime and multiplies of (yes, this sounds idiotic now) 2, 3, and 5.  I also forgot about 1 being in its own group so I came up with 98.  I knew my answer could not have been too far off so I said 100...thank god they didn't put 98 as an answer..phew.

At first I got 101, then I realized 1 is neither prime nor composite. If they put 101 as a choice, I might've gotten it wrong

[quote="blahblahblah"]Like I said, ignore me.  It's straightforward inclusion-exclusion.[/quote]

ah, my pardons.  I tend to leave pages open in my browser and not refresh them.

[quote="frankthetank"]Yeah, but my friend actually wrote a program on his calculator to do that problem.  It took 36 minutes.[/quote]

 :)

That makes me smile. Yesterday before I went to bed, I programmed [i]my[/i] calculator ([url=http://education.ti.com/us/product/tech/v200/features/features.html]TI Voyage 200[/url]) and it took a couple of minutes. I'm not too familiar with the programming though, so I ended up making the program output a string with every number that was prime-like. Then I took the length of the string and after subtracting six (the three two digit prime-like number) and dividing by three, I got that in total there are 100 prime-like numbers

On the test I subtracted out the primes before I did the exclusion thing so I got some huge number like 400. Later I was informed that exclusion only works if you have an arithmatic sequence of numbers and that I was supposed to subtract out the primes first... Whoops...

I know this is a strange question, but what calculator + code did your friend use?

[quote="blahblahblah"]It's significantly easier than the 12B was last year.  I did Q19-24 in about 15 minutes combined (zabelman gave me the questions).  I did a lot of casework on 18 though; can anyone write out the question for me again?

..

Hahahaha, I just realized the shortcut to 18.  Ignore me, people![/quote]

When I looked #18, I stored noticed that it gave the number of primes, but then thought to myself "how willl this help" and forgot it was even there.  Then, when I realized I DID need the number of primes, I started listing them, and then stopped.  Didn't want to count them.  Boy, oh boy was I angry at myself!

JB

When I first read that problem I accidently skipped over the part where they told you 168 primes so i was like..."wtf, they expect us to know how many primes are less than 1000?" So then I reread the problem and calmed down once I saw 168  :D .

[quote="avolfson"][quote="frankthetank"]Yeah, but my friend actually wrote a program on his calculator to do that problem.  It took 36 minutes.[/quote]

I know this is a strange question, but what calculator + code did your friend use?[/quote]

I don't have his code, but he had a TI-83 plus, which is probably why it took 36 minutes.  Fortunately, I lent him an additional calculator before the test started.

[quote="zabelman"]Writing a program for this problem is very fun to do! So in the last 10 minutes I decided to write one. I came up with the following within a few mins (TI-83 Plus, Basic syntax):

[code]0->K
For(A,2,1000)
Disp A
If fPart(A/2) $\neq$ 0 and fPart(A/3) $\neq$ 0 and fPart(A/5) $\neq$ 0
1+k->k
End
Disp K[/code]

(I put in the line "Disp A" so I could watch it count up to 1000: I *hate* waiting for a program to finish running without knowing if it is going at all.) Anyway, it took about 2 minutes to run, and then it told me 265. So after subtracting off the 165 leftover primes, thats 100. Yay, it worked![/quote]

Hah, that's quite good.  In fact, it would have been faster to do this then inclusion-exlcusion by hand (work on another problem while the calc does a search).  I wouldn't have thought of using fpart; mod(a,x) came to mind when people said code.

[quote="frankthetank"]Yeah, but my friend actually wrote a program on his calculator to do that problem.  It took 36 minutes.[/quote]

Oh, I definately should have thought to do that.  I ended up with 96 because I forgot 1,2,3, and 5  :blush:.  I imagine I could make that program in under 5 minutes on an 89, though it would have taken a little while to run.

[quote="Ultimatemathgeek"]When would be the scores for AMC A and AMC B available?

I haven't heard from my teacher about my score.[/quote]

The answer key is already posted on the UNL website

[url]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05AMC12AB-answers.html[/url] so you can go check your answers

You probably won't get your official score until another two weeks

[quote="zabelman"]Writing a program for this problem is very fun to do! So in the last 10 minutes I decided to write one. I came up with the following within a few mins (TI-83 Plus, Basic syntax):

[code]0->K
For(A,2,1000)
Disp A
If fPart(A/2) $\neq$ 0 and fPart(A/3) $\neq$ 0 and fPart(A/5) $\neq$ 0
1+k->k
End
Disp K[/code]

(I put in the line "Disp A" so I could watch it count up to 1000: I *hate* waiting for a program to finish running without knowing if it is going at all.) Anyway, it took about 2 minutes to run, and then it told me 265. So after subtracting off the 165 leftover primes, thats 100. Yay, it worked![/quote]
Rock on, Zack.  I never thought to use such easy syntax.  Wish I could program better--but on to the AIME.  And my 89 from the night before the test (no, I didn't bother using it anyway) won't be useful there.

I still don't see why the last problem was such a stress.  It's not even maths--maybe that's the problem--hardly, just some counting applied to a picture tha makes some easy sums of 8s...  All in all, this test was easy, but I made simpler mistakes than I thought possible and crashed 20 points from a near-perfect, best-in-State score.  All well, I'll go try that program.


let f(x) = 6x^4 - 12x^3 + 3x^2 + 3x + 3

Find the value of x that maximizes the function

If I'm not mistaken, the function monotonically increases for $ \approx x\geq 3$ so the answer must be $ \boxed{\infty}$.

I forgot to include the interval is from 0 to 1.


calculus
integration
probability
function
real analysis
real analysis solved

Hey guys,
I stumbled upon a problem, and I can't find a way to relate it to the FTC Part I:

For what value of x does: integral (from 0 to x) e^(-t^2) dt = 0.6?

Any help would be appreciated.

 :cool:

This is going to have to be approximated one way or another. One approach is to use a table of the cumulative probability function of the standard normal distribution. Such tables are widely available. To do that, you would need to divide by $\sqrt{2\pi}.$ You would be looking on the table for the (approximate) value of $z$ such that $P(0<Z<z)=\frac{.6}{\sqrt{2\pi}}.$

There would be numerical methods that would be more accurate than that table - but it's a start.

I went ahead and worked a more accurate approximation.

Let $F(x)=\int_0^xe^{-t^2}dt=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{k!(2k+1)}.$

Evaluate $F(x)$ for each needed $x$ by adding up a partial sum of this power series. It suffices in the region we're in to take 15 terms.

Let $x_0$ be some initial guess - I used $x_0=1$ - and define a sequence recursively by

$x_{n+1}=x_n-\frac{F(x_n)-.6}{e^{-x_n^2}}.$

This is Newton's method. It converges rapidly to the value we're looking for - in fact, 5 iterations get us all the accuracy we're going to get. To the accuracy of a spreadsheet, the result is $x\approx .69888164835462.$


I found the answer to the question, it was a true and false.  The answer is false but why? because I got true and i can't figure it out.

In triangle PQR, measure of angle P = measure of angle R = 50.  If T lies on segment PR and measure of angle PQT = 42, then PT < TR

[hide="hint"]Draw a picture  :) [/hide]

[hide="solution"]Because $\triangle PQR$ is isoceles, $PT=TR$ if and only if $QT\perp PR$.  However, $42<90$, so you must move the side over for this to happen, so $PT>TR$.

You can also use the Hinge Theorem some here[/hide]

I get part of it until moving the side over. were you assuming temporarily that QT perpendicular PR? and if you use the hinge theorem, then wouldn't it make the triangle PQT congruent to triangle RQT?


calculus
integration
function
derivative
calculus computations

Hello

I am trying to solve for the constant $C$ in the following equation:

$\frac{dW}{dR}= \frac{-0.02W+0.00002RW}{0.08R-0.001RW}$

I want to show that $\frac{R^{0.02}W^{0.08}}{e^{0.00002R}e^{0.001W}}= C$

So far I have gotten to: $(0.08-0.001W)\frac{dW}{W}= (-0.02+0.00002R)\frac{dR}{R}$

What would I do next?

Any help is appreciated.

Thanks

Do you mean that $C$ is the constant of integration obtained when the given differential equation is solved?

If $C$ is a constant of integration, there are two points to make:
1. $C$ depends strongly on the method of solution. There are many different ways we can do this, and many different constants we can write down.
2. You need an initial condition to say anything.

So, here are two questions:
What's the initial condition?
What's the form of the solution?

There was no initial condition given. This is a model for rabbit and wolf populations. Also, it is impossible to solve for $W$ as an explicit function of $R$. Thats why I need to show that the constant obtained by solving the diff eq. is equaled to the above. But I cant show this without an initial condition?

Ah, that's it. You're not supposed to "solve for $C$" at all, and there's no need to solve the equation. You want to show that a certain quantity depending on $R$ and $W$ is constant in time. To do this, ignore everything you have so far toward solving the equation, and differentiate that quantity. If you can show its derivative is zero, you have what you want.


You have forgotten an important condition, namely that the numbers a, b, c and x, y, z are such that the number arrays (a; b; c) and (x; y; z) are equally sorted. With this condition, the inequality is proved on http://www.mathlinks.ro/Forum/viewtopic.php?t=5517 .

  darij

it's a problem from Romania in the 90's.


function
integration
calculus
calculus computations

Let $ (X, \mathcal{F})$ and $ (Y, \mathcal{G})$ be measurable spaces. Suppose that $ K: X\times \mathcal{G}\to [0,\infty]$ is such that for each $ x\in X$, the function $ B\to K(x,B)$ is a measure on $ (Y, \mathcal{G})$ and for each $ B\in \mathcal{G}$, the function $ x\to K(x,B)$ is $ \mathcal{F}$- measurable.

Given such a $ K$, a measure $ \mu$ on $ (X, \mathcal{F})$ and an $ \mathcal{G}$- measurable nonnegative function $ f$ on $ Y$, I [b]have shown[/b] that
1)   $ \nu(B)\equal{}\int K(x,B) \mu(dx)$ is a measure on $ (Y, \mathcal{G})$ and
2)   $ g(x)\equal{}\int f(y) K(x,dy)$ is an $ \mathcal{F}$- measurable function on $ X$.

Can anyone show that $ \int f d\nu \equal{}\int \left [\int f(y) K(x,dy)\right ] \mu(dx)$?

Thanks in advance.

This should immediately follow from what you've done for characteristic functions of sets. Now just use the linearity of the integral to extend it to all simple functions and the monotone convergence theorem to get the result for all non-negative functions.


Let there's a function $ f: \mathbb{R}\rightarrow\mathbb{R}$
Find all functions $ f$ that satisfies:
a) $ f(2u)\equal{}f(u\plus{}v)f(v\minus{}u)\plus{}f(u\minus{}v)f(\minus{}u\minus{}v)$
b) $ f(u)\geq0$

[quote="wangsacl"]Let there's a function $ f: \mathbb{R}\rightarrow\mathbb{R}$
Find all functions $ f$ that satisfies:
a) $ f(2u) \equal{} f(u \plus{} v)f(v \minus{} u) \plus{} f(u \minus{} v)f( \minus{} u \minus{} v)$
b) $ f(u)\geq0$[/quote]
Take $ u\to \minus{} u ,v\to \minus{} v$ then we have
$ f( \minus{} 2u) \equal{} f( \minus{} u \minus{} v)f(u \minus{} v) \plus{} f(v \minus{} u)f(u \plus{} v) \equal{} f(2u)$
Therefore $ f(x) \equal{} f( \minus{} x),\forall x\in R$
So $ f(u \plus{} v) \equal{} f( \minus{} u \minus{} v),f(u \minus{} v) \equal{} f(v \minus{} u)$ :
$ f(2u) \equal{} 2f(u \plus{} v)f(u \minus{} v)$
Let $ x \equal{} u \plus{} v,y \equal{} u \minus{} v$ then $ f(x \plus{} y) \equal{} 2f(x)f(y)$
Take $ x\to \minus{} x$ then $ f(y \minus{} x) \equal{} 2f( \minus{} x)f(y) \equal{} 2f(x)f(y)$
It gives :$ f(y \minus{} x) \equal{} f(x \plus{} y) \Rightarrow f(x) \equal{} 0$
Easy to check $ f(x)\equiv 0$ is solution of this equation .
I think you have a typo  :wink:

Ha?
No, I'm not...

Hm... $ f(x)\equiv0$?
What does it mean?

I have another solution, $ f(x)\equal{}\frac{1}{2}$

I'm sorry .
$ f(x)\equiv c$
Then $ c\equal{}2c^2\Leftrightarrow c\equal{}\in\{0,\frac{1}{2}\}$
But I think ,we don't need condition ii.

wangsacl, why don't you post the exact wording please?

@Johan : Because I'm not holding the copy of the problem. I just recall the problem from my memory. Sorry for my bad grammar.

Take $ u \equal{} v \equal{} 0$ and we have $ f(0)^2 \minus{} f(0) \equal{} 0$ so $ f(0)\in{\{0,\frac {1}{2}\}}$. If $ f(0) \equal{} 0$ we have $ f(x) \equal{} 0$. If $ f(0) \equal{} \frac {1}{2}$ and $ u \equal{} v$ we get
\[ f(2u) \equal{} \frac {f(2u)}{2} \plus{} \frac {f( \minus{} 2u)}{2}\Rightarrow f(u) \equal{} f( \minus{} u)
\]
Then we replace this in
\[ f(2u) \equal{} f(u \plus{} v)(u \minus{} v) \plus{} f(v \minus{} u)f( \minus{} u \minus{} v)\Rightarrow f(2u) \equal{} 2f(u \plus{} v)f(u \minus{} v)
\]
If $ v \equal{} 0$ we get $ f(2u) \equal{} 2f(u)f(u)\Rightarrow f(2u) \equal{} 2f(u)^2$ and i'm trying to look for something more  :D

Find all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$  so that
a) $ f(2u)=f(u+v)f(v-u)+f(u-v)f(-u-v)$ for all  $u,v \in \mathbb{R}$ and
b) $ f(u)\ge  0$ for all  $u \in \mathbb{R}$

[this is the exact wording]


quadratics
number theory proposed
number theory

Let $a_1,\ldots,a_{2004}$ be integers with the property that for all $n\geq 1$ the number $(a_1)^n+(a_2)^n+\cdots+(a_{2004})^n$ is the square of an integer. Prove that at least $21$ numbers are $0$.

Noone? And I thought that this problem is so beautiful. It seems I was wrong. So be it. I will let you another 5 days and after that I will post a solution.

Umm... why is it 21? Let k be the number of non-zero terms. For any prime p larger than all the a_i, k is a quadratic residue mod p (just consider n = p-1). I think it is a well-known result of QR that this implies k is a square.

probability1.01 not exacly :). We must take such $p$ that numbers $2004, 2004-1,...2004-21$ are all non quadratic residue $mod p$. Such p we pruduct using Chinese theory and Dirichlet theory.

EDIT : sorry I wrongly read.

I tell you why $k<2004-44^{2}$. In numbers $2004,2003,....$ its probably the situation such that we have 3 numbers form $a^{2}sl$, $b^{2}lm$, $c^{2}ms$. Where $s,l,m$ are prime numbers. This 3 numbers can be all at once non quadric residue.

Don't see what's incorrect with probability1.01's reasoning...

we're waiting for a slution !!!!!!!!:sleeping:  can any one post a solution ?

Your problem is easy. Let $p>N=max(|a_{i}|,2004)$ is prime. Let $n=p-1$, then $\sum_{i}a_{i}^{n}\equiv k(mod \ p)$, were k is number of terms, suth that $a_{i}\not =0$. From your condition we get $(\frac{k}{p})=1 \ \forall p>N$. Therefore k is perfect square.
It mean $k=l^{2}<2004\Longrightarrow l\le 44$ therefore at least $2004-1936=68$ terms are 0.
For example $a_{i}=m^{2},i=1,2,...,1936, a_{i}=0,i=1937,...,2004$, then $\sum a_{i}^{n}=(44m^{n})^{2}.$


induction
function
algebra proposed
algebra

[quote="TTsphn"]$ 0 < x_{1}< 1$
 $ x_{n+1}= x_{n}+{\frac{x_{n}}{n^{2}}}$
Prove that $ x_{n}$converge
 2)$ x_{1}= 29$
  $ x_{n+1}=\sqrt{3}+{\frac{x_{n}}{\sqrt{x_{n}^{2}-1}}}$
 Find a such that
     $ x_{2k}< a < x_{2k+1}$
 @to Rust:It is easy to know that a is limit of this sequence so can you post a full solution?[/quote]
1. $ x_{n+1}=x_{1}\prod_{k=1}^{n}(1+\frac{1}{k^{2}})<x_{\infty}=x_{1}\frac{sh(\pi)}{\pi}$.
2. Let $ f(x)=\sqrt 3+\frac{x}{\sqrt{x^{2}-1}}$. Obviosly $ f'(x)=-(x^{2}-1)^{3/2}<0,\ x>1$.
Let $ a$ is root $ a=f(a)$, Obviosly $ a>\sqrt 3+1$, therefore $ -1<f'(a)<0$. It give $ x_{2k}<a<x_{2k+1}$.

[quote="TTsphn"]a)On the first problem we can use
1)$ x_{n}$is monotone
2)$ x_{n}< 2$ for all n
It is means that $ x_{n}$is converge
b)I thinkl you are wrong because it easy to check that
$ x_{2k}$ is monotone and $ x_{2n\minus{}1}$ is decreasing it mean that you must solve system equation to check that 2 sequence converge in common point.[/quote]
First problem.
1. Obviosly.
2. $ x_{n}$ may be any real. If you say $ x_{n}< 2x_{1}$ you are wrong $ x_{2}\equal{} 2x_{1}, x_{3}\equal{}\frac{5}{2}x_{1}$ and $ x_{\infty}\equal{} x_{1}\frac{e^{\pi}\minus{}e^{\minus{}\pi}}{2\pi}$.
Second problem.
Because $ \minus{}1 < f'(x) < 0, x_{1}> a$ we get $ y_{n}\equal{} (x\minus{}a_{n})(\minus{}1)^{n\minus{}1}>0$ and decrease.

2)$ x_{n}$is bound prove by induction
3)You must concide function
$ g(x)\equal{}f(f(x))\minus{}x$ and prove it monotone or increasing to prove the system has uniqge solution.
More than you must prove that my question .


FInd the prime factors of  $ 9^8\plus{}7^6\plus{}5^4\plus{}3^2\plus{}1$. I need the detailed solution, not the answer. thanks

how to find this without any aids (calculators, math software) and only use simple formula and or logic?


Come on guys, comment on ur performance in FT 11 (the last test before the REAL TEST!!)

Start with yourself no? Dont worry i wont die of a stroke soon :D ive had quite a few number of shocks today so dont hesitate to even say if you got 380 odd :D

oops sorry!
i forgot to do that!
and don't worry, in such a test if i had got 380 odd marks, then it would have been me who would have a died of a stroke.

anyway, my score is 221

PHYSICS- 46 + 57 = 103
CHEMISTRY- 13 + 23 = 36 (?!?!?!)
MATHS - 23 + 49 = 72

my analysis would be- i could have scored more in physics, considering that i gave 1 1/2 hours in each paper to it. chemistry is always a problem, and maths is an improvement! most certainly my best performance in a FIITJEE Full Test.

btw in ur centre, did they take the answer sheet, or did they refuse saying that they won't evaluate it?

They refused :rotfl:

Well anyway let the bigheads post their marks, then the puny fella (thats me :D ) shalt reveal mine :)

Got 233.......I think

Physics - 78
Chemistry - 80
Maths - 75

Not much to boast about......... :| But yeah plenty of mistakes in solns to point out!

well probable errors and corrections for FIITJEE solns........!

paper 1:

given answer---------->correct answer
Physics:
bicycle ques:       B ------------->C
(wonder how they will provide torque for front wheel!) :rotfl:
comprehension:
q3(areal velocity):  B---->C        (confirmed by achu and chrono)

Chemistry:
(rubber)
multi correct        B,C---------->B,C,D
(if C is true so is D........exothermic rxn. delta H is -ve....diff gives positive! [b]simple math[/b])

Math:
compr. Q3
(explain how inverse will exist for many one function)  :maybe:

Hypothesis: Question paper setters are not the same people who provide solns...as far as FIITJEE FULL Tests are concerned this is always true! :rotfl:

Well chronoz got the right answer for the physics question as far as i know :)

And seeing that he is god of logic ( :P ), ( i think its imperative that u know that by now :D ), he ob did it by physics.

got it right ? according to FIITJEE or according logic and physics?

did you get the multi correct about NaF in paper 2 chem???????? :?:

can someone post the chem question plz. the bicycle thingie also if possible.
forget it if its too long

Paper 2:

Phy:
Assert -reasoning (Doppler)....C---->D
there is no relative motion b/w source and observer along the line joining the two!

Chem:
match the following: (thermo)
first are three fine but the last.......getting A->r,B->r,C->p, D->p.......... :maybe:(help needed)

NaF ques(multi correct).....B,C---->A,C :maybe: (help needed)

equilibrium ques(multi)..A,C,D----->C,D
pressure will not affect solid-liquid equilibrium!

Math:
sin^-1[sec x] ques
query: A is also an answer!

comprehension:
prob for periodic ques(Q2).....D--->B
expect many to falter here
f1,f2,f4.taken two at a time + one func f3+f5=[x]-x=-{x} :ninja:
hence 4 func(sample space), 3 with period two.......therefore, ans. (B)3/4

(multi correct)
Given natural no. n........A,B,C---->A,C
they mised out '-' sign in soln hence B is infact wrong.......try it for a value say .5

In case any mistakes are found in the above mentioned errors...feel free to point out (as if that was necessary to mention.......people here at Aops are would gladly do that)!

bicycle is perfect ans pramod.... if yku want an expl call me at 100. or i'll ask rohit to post proper expl.. but the third ans in gravi comp is wrong crap.

me 227
91 + 67 + 69

horrible sols. worse than usual pathetic... im annoyed.

can you guys achu pramod phoenix post the other commendable performances inur city. chaappli got 253 here.

i am not aware of stunning performances, coz many didn't give it. maybe pramod can tell u of someone (he knows a lotta guys!!)

and can u plz tell why there will be no friction in the front wheel??

i got 252
the twins from my place r getting 270 odd
anit is getting 340
how did he get that i m myself surprised
he doesnt deserve it :mad:

@ashwath

I rarely come to aops in academic pursuit...
its a place for me to chill out... and spam the place with my gibberish...

So I am perfectly comfortable with the private fights fought publicly unless of course it becomes way too personal...

[quote]I think that basically shows India in a poor light to all outsiders who read this forum. [/quote]

poor light...hmmmmmmm...
They have to check if their bulbs glow dimmer when they come to the Indian forum...
Its a problem with their [b]bulbs[/b] and not with us...(no pun intended)

poor light...hmmmmmmm...
They have to check if their bulbs glow dimmer when they come to the Indian forum...
Its a problem with their [b]bulbs[/b] and not with us...(no pun intended)[/quote]

i think they've switched off their bulbs for earth hour..

achu imo a score of 230 if ft11 was jee would have got us something in the 500s.

pramod ill ask madness to post the answer. i'm confident its right.

chronoz thala U back hmmmmm......... so madness lied about that typing problem ah??? :P

[quote="madness"]No he uses an on screen keyboard :D

And btw chronoz, i refuse to type it out unless i get a favour :D[/quote]

rohit havent u read that lesson i had written in our english book :P

[i][u][b][color=red][size=150]"NEVER HELP IN THE CUSTOMARY SENSE DEVELOP YOURSELF INTO A HARMONIUS INDIVIDUALS WHO WORKS TOWARDS THE BETTERMENT OF THE SOCIETY!!!![/size] [/color]:[/b][/u][/i]  :police:

[quote="madness"]Hmmm, well you are going wayyyyyyyy off topic mr. einstien :huuh: So if you want to continue further, kindly do so in the metathread.

Also i might kindly want to point out that it should be develop yourselves into harmonius individuals or into a harmonius individual and not what you wrote :D[/quote]
no i am sure i am rite as you follow what i wrote but my actual edicts have been copied to the golden guide!!!! and this general discussion da not IITJEE thread for one not to enjoy or gossip as i come to mathlinks just like what soumya comes for the same reason refreshment!!!! :D  :D  :spider:

[quote="chemrock"]Enough of jalras and NV's. And by the way, why did u not come mr.einstein? Scared/playing safe. :evilgrin:

I'm getting 266:
Physics-88
Maths-82
chem-96[/quote]

spidey :spidy:  is never scared of anything my friend i had fever thought could take some rest thats why!!!! :play_ball:  :spider:

well in paper 2 math que one of my claims is wrong
the multi :( -correct ques is perfectly alright!...my mistake :mad:

forgot to use chain rule :rotfl:

Honestly, i'd say that even if i get a rank, ill get only from 2000 - 5000 odd :(

[quote="pramod.best"]@madness,chronoz: wishing speedy recovery of whoever's faulty keyboard......

however, kindly enlighten me with your so-called 'correct' soln. of the bicycle question..... :starwars: .....achu and i are very keen to be proved wrong by u guys!!!!!!

[quote]speak up where do you see urself[/quote]

@achu: at the beginning of the journey i saw myself at AIR 1 .....thought i was born to be AIR 1........until i was acquainted with people like u and our fellow mates @ aops!!!......Now, I'm seriously considering my chances of making it big if not exactly vanquishing IIT-JEE 2008.......let me just say..lets wait ad watch!!!!!!! :weightlift:[/quote]

How can you pramod?? how dare you?
You better not even dream abt such stuff when I am around... :rotfl:  :P
 :P

@Soumya, thats why he stopped dreaming. He woke up, saw you and realized the futility of his dream and the complications of reality.

Madness, why being such a hypocrite especially in the sea of hypocrites.

@pramod: Just because you got it wrong, doesn't mean answer is wrong?? Even though you maybe the lord of the physics of the 31 dimensions.

Okay, its given that the cycle moves down with pure rolling. So at any instant, the mg force pulls you down and threatens to accelarate you, making you lose your Pure Rolling. So what do you do? You press the brakes, which clamps the back wheel's rolling thereby v>rw, so friction acts on the back wheel and is sufficient enough to counter mg sin. What about the front? Front is already in pure rolling and nobody has disturbed it. Moreover, friction on rear balances sin component of gravity. And so front wheel is not disturbed.

This was how I thought and wrote the answer.. and fiitjee agrees with me. So great minds think alike.

Or

[hide]Fools seldom differ[/hide]

@ chronoz:

i think u are right. actually, when i agreed with pramod, i totally forgot that the question involved applying brakes, and thought that it was discussing a simple case of a bicycle rolling down the plane. (however, i got the answer wrong in the exam, because i rather foolishly thought that the front and rear wheels were connected through the chain!! :oops: )
long time since i rode a bicycle!!!


If $n$ and $a$ are positive integers, prove that $n^{4}+a$ will be composite for an infinite number of positive integers $a$.

Define $a$ as $4m^{4}$, where $m>n, m\in Z$ (of which there are an infinite number).  Using the factorization Phelpedo posted, we just need to show that $n^{2}-2nm+2m^{2}> 1$ (since $n$ and $m$ are positive integers, other part of the factorization is obviously greater than 1).

We know $m>n$, therefore $2m^{2}>2nm$ and since all numbers are integral, $2m^{2}-2nm\geq 1$.  You then add on $n^{2}$, which is at least 1, and you get $2m^{2}-2nm+2n^{2}\geq 2$, meaning $n^{4}+a$ factors in a way other than 1 and itself, and is therefore composite.

Um, as stated it is trivial: if $n$ is odd, $n^{4}+a$ is composite for all odd $a$ and if $n$ is even, $n^{4}+a$ is composite for all even $a$. I think you mean [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=363654#p363654]this problem[/url].

Yes, sorry, that came out as a VERY trivial proof.  :P I was going by memory, and I didn't remember the problem...


$t_n$ defined by its first term $t_0$ positive integer $\geq 2$
$t_{n+1}=2t_n^2-1$

Compute $\sum_{n\geq 0}\frac{1}{2t_02t_1...2t_n}$

nice problem.
let $t_0=\frac{x+1/x}{2}$ ;)


prove that if 3 lines are concurrent then their poles are collinear.

it must have been posted before but anyway,
use the fact that if a point $ P$ is on the polar of $ A$ then the polar of $ P$ passes through $ A$ .
we also have : if 3 points are collinear then their polars are concurrent

[quote]if 3 points are collinear then[/quote]
then what?

[quote]it must have been posted before but anyway, [/quote]
can anybody post the link to this problem

is it too easy or what?
just post the link if so


calculus
derivative
analytic geometry
number theory
algebra
linear equation
number theory unsolved

Hello,

I found this problem in a book and I haven't been able to solve it. It is Putnam's style (I think). It requires from calculus just the idea of the implicit derivative (to find a tangent) and it is on the edge between Number Theory and Algebra for the rest (at least that is what I think) Therefore, I thought this was the best place to post it. [b]Moderators[/b], if you consider that it should be moved, go ahead!

This is the problem. Suppose that in the curve $y^2=x^3-n^2x$ one has a point $(x_0,y_0)$ where both coordinates are rational. Prove that the tangent line to this curve at $(x_0,y_0)$ intersect the curve in a point $(x_1,y_1)$ such that $x_1,\ x_1-n,\ x_1+n$ are squares of rational numbers.

Now, after almost a page of algebra, I get to the point where $x_1=(x_0^2+n^2)^2$. In this case it is evident that $x_1$ is the square of a rational number, but what about the other two. Now, I think my calculations are accurate, but may be I got something wrong, you may check if you want.

Best,

[quote="Fiachra"]If you have a linear equation for $x_1$ and $y_1$, you must have that $y_1$ is rational. Then we also know that $(x_1+n)(x_1-n)$ is a perfect square, which seems very close.[/quote]

It is celar that $y_1$ is a rational (basically after you compute the tangent line), and therefore there is no doubt that $(x_1+n)(x_1-n)$ is a perfect square, but how is that very close to the fact that each of the factor is a perfect square.

For example, take $n=5$ and $x_1=\frac{25}{4}$, then $y_1=\frac{75}{8}$, but neither $x_1-n$ nor $x_1+n$ is a square (but $(x_1+n)(x_1-n)$ is). Of course, according to the problem, there can be no rational point $(x_0,y_0)$ in the curve such that the tangent through it passes through $(x_1,y_1)$.

I suppose you avoid $x_0=-n,\, 0,\, n$ (with $\sqrt{2n}\not \in \mathbb{Z}_+$). In other case, if $(x_0,y_0)$ is a point of the curve $y^2=x^3-n^2x$ which are rationals, then
\[
y'_0=\frac{x_0^2}{y_0}+\frac{y_0}{x_0} \ \iff \ y-y_0=\left[\frac{x_0^2}{y_0}+\frac{y_0}{x_0}\right](x-x_0).
\]
Then, apply this equation in the original curve we get that
\[
y_0^2+(3x_0^2-n^2)(x-x_0)+\left(\left[\frac{x_0^2}{y_0}+\frac{y_0}{x_0}\right](x-x_0)\right)^2=x^3-n^2x
\]
then applying algebraic properties of the solution $(x_0,y_0)$ we get that
\[
x=\left[\frac{x_0^2}{y_0}-\frac{y_0}{x_0}\right]=\frac{n^4}{y_0^2}
\]
which is a square of a rational, and $y_1^2=x_1^3-n^2x_1$ then
\[
x_1^2-n^2=\frac{y_1^2}{x_1}
\]
then $(x_1-n)(x_1+n)$ is a square of a rational. And as djimenez told before we can find examples where $x_1-n$ is not a square of a rational.

Best regards.

RobertuX


inequalities
AMC
USA(J)MO
USAMO
inequalities solved

a=( m^(m+1) + n^(n+1) )/(m^m + n^n). If m,n,a are positive integers prove the inequality:
a^m + a^n >= m^m + n^n.

This one was already discussed! check the proposed problems topic!

cheers! :D:D

Unfortunately I couldn't find this problem in Proposed Problems Section. Are u sure it is from that Section?

Yes! :D

http://www.mathlinks.ro/phpBB/viewtopic.php?t=223

cheers! :D :D

I saw that you discassed it but there is no solution. Taking n>=m IMHO is a dead-end.  :(

I think this works.

Define $x = \frac{a}{m} - 1, y = \frac{a}{n} - 1$. We have $x = n^n \cdot \frac{\frac{n}{m} - 1}{m^m + n^n}$, $y = m^m \cdot \frac{\frac{m}{n} - 1}{m^m + n^n}$. Now, $(1+x)^m \ge 1 + mx \Rightarrow \frac{a^m}{m^m} \ge 1 + n^n \cdot \frac{n-m}{m^m+n^n} \Rightarrow a^m \ge m^m + m^mn^n\frac{n-m}{m^m+n^n}$. Adding the similar inequality given by $(1+y)^n \ge 1+ny$ gives the result.

It's from USAMO 1991: [url=http://www.kalva.demon.co.uk/usa/usoln/usol914.html]www.kalva.demon.co.uk/usa/usoln/usol914.html[/url]


what are moderators and how do you become one?

Moderators are people who warn, edit, and delete useless, bad, and/or stupid posts.
You can become one if you're really really good in the forums :)

Moderators are people responsible for certain sections of the message board.  The duty of the moderators is primarily to encourage useful discussion.  Moderators of the math sections of the board will throw out problems or ideas for discussion a few times a week.  Moderators will also have the ability to split threads, delete offensive posts, and move posts to other boards, and they can do each of these at their discretion.

We picked out the initial set of moderators based *very loosely* on the activity of folks on the site to date.  If you haven't been asked yet, please don't be offended.  As you can see, we haven't finished naming moderators.  We expect to have 1-3 per board (depending on the activity of the board).  If you are interested in being a moderator, pm me and let me know what areas you are interested in.  We may not have a place for you right away, but as time goes by, we'll likely have attrition among our current moderators and have other opportunities.


algebra
polynomial
linear algebra
linear algebra unsolved

Let $ A$ and $ B$ mtrices s.t $ A^HB\equal{}0$
Show that $ P_A\equal{}I\minus{}P_B$

[hide="Idea"]It seems that ranges of A and B are orthogonal, but I couldn't go further[/hide]

I'm not sure what your notation means.  Does $ A^H$ denote transpose and $ P_A$ denote characteristic polynomial?

$ A^H$ is hermitian of $ A$, $ P_A$ denotes the projection matrix.


hi folks,

because the other wish list topic is getting crowded with different sorts of stuff, I am creating this other wish list, which I will more closely manage.

any post without relevance will be erased without notice (ie wish for things that can already be set up in the options menu, etc.)

for now, on the To Do list there are the following item

1. Each user chooses which category should be displayed on the main index [in progress]

2. WAPgateway [released, still adding features]

3. Integrated nav bar [in progress]

4. User chooses style of unread / read posts. Normal mode, or keep unread posts. [on the waiting list]

Ah, great idea Valentin! :)

I still got one suggestions unanswered: [list][*] About the structure: since the newer people complain they don't want subforums in pre-olympiad and such, perhaps it may be a good idea to allow people to pick a subject when posting? Like a topic status to pick from either "Algebra", "Combinatorics", etc.
That doesn't require any hassle on different forums, but it does have the advantage that you can immediately have what you're looking for.[/list]

Just my 2 cents.

Greetz, Peter

In the mathlinks style, is there a "mark all posts read" button on every forum? If not, can you add one?

yes, there is one in ML style. :)

I'm not sure it works though with the flag keeping...

Do you think it would be possible to make it so that  is hidden when you put it in spoiler?  Right now, it is not.  For example, the following is in spoiler:

[hide]

[/hide]

For now, whenever I want to post a solution without ruining it for other people, I am stuck using fractions like 1/2 instead of  .

Some of us already had this idea. I think it is technically very hard, if not even impossible. But who knows...

  Darij

It is technically almost impossible to make this happen. Not in the way LaTeX is rendered on this site anyway.

It might be hard to make it come out in spoiler per se, but what happens if you include the color package so that people can post in white? Then again, if you tell us the colors of the backgrounds, we can make the $\text{\TeX}$ in exactly those colors.

It doesn't matter. It can't work because you need to change the image as people select it. Also there are 4 different backgrounds, and there are also users that don't use spoilers (selected the option not to appear).

This would mean something like rendering 5 images instead of 1, which would induce decrease of site speed and increase by 5 times the number of files in the /pictures/ directory, which is as is already huge!.

... perhaps you can help this people giving an option to hide all replies to a topic, and reveal the rest after a button click :maybe:

May sound stupid but not more stupid than the spoiler system as is... heck now they even got a good reason not to learn latex! :P

It's a good idea Peter. I think that this will indeed solve many of some people's problems. I will try to find if there is already a similar mod (I belive I saw such a thing on the VBulletin 3.0.3 board, where you could choose the style of viewing a thread - you can either use this style, or the old replies system like this one here http://lists.suse.com/archive/suse-linux-e/2001-Jun/0472.html) ;)

Perhaps even better: make [b]some parts[/b] of a reply revealable upon mouseclick (while the rest of the is automatically shown; in fact, not everything must be hidden). For instance:

[quote="darij grinberg"]Nice problem, thanks. There is a little typo: $a_i$ should be replaced by $a_j$.

[u]Click here to see the solution.[/u][/quote]

  Darij

It would be nice if, in the my classes section, the listing of transcripts had names, and not just the date. Clearly, a lot (well, the Algebra class) already knows what they're discussing, b/c they say in the forum. However, it would be nice to see "Synthetic Divison (9/30)" in the transcript list instead of "Class Transcript from 9/30" so that you could easliy use the the transcripts for references.

[quote="Valentin Vornicu"]Darij, that would mean using dhtml into posts, which can be extremelly messy. I don't work with dhtml, neither does Richard, so I guess we can't pull this off (properly). In plus it would not work on some browsers.

Hopefully something like this will be available in phpbb 2.2 :)

Meanwhile the collapsing forums mod is up and running. You can select what categories are open, and the computer remembers them. You can also save a specific state in the config, and when you want to use it, just push the restore button.[/quote]

Couldn't you just have two versions of the page, one where spoilers are displayed, and one where they aren't, and you would click a link to move between them. To avoid showing latex, you just wouldn't add the link to the image (ie, no <IMG> tag in the source). (I haven't actually used the forum much, and certainly haven't looked at the source much, but that doesn't sound too hard)

[quote="Valentin Vornicu"]I think I did this a long time ago :D[/quote]
Oh, i see...i did not pay attention to the [b]go to page[/b] thing...... :blush:
Thanks for all of your help! :D

Go to the Memberlist tab at the top.  It should take you [url=http://www.artofproblemsolving.com/Forum/memberlist.php]here[/url].  It should already be sorted by post (descending).

[quote="jhollenbeck"]how do we sort members by number of posts? i was going to make a post about that because i can't figure it out. :oops:[/quote]

To add to what joml88 said...there is a dropbox on the memberlist page that allows you to sort members in several ways, including post count.

What about an option for the author of a Thread to delete his thread :?:

Also sometimes a thread exists but a user does not notice it and starts a new one with the same topic.

So if the author can delete his own thread then things will be simplified.

If we let users delete their own threads, we would be handing the ability to censor directly to users.  There are other problems with such an option, so it's unlikely we would ever institute it.

i don't think they would let anyone be an admin becuase you didn't help build the site.

[quote="MCrawford"][quote="dogseatcheese"]How do you become an administrator? :D  ;)[/quote]

Administrators are AoPS employees.[/quote]
can i work for AoPS?

In light of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87291]these[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87476]two[/url] posts, I would like to officially request the creation of a Self Pity forum.  This forum will serve a vital purpose in the community, giving complainers of all stripes an ideal location to commiserate.  The AMC forum is likely to especially benefit from this in the aftermath of the administration of all AMC contests.

Anyhow, think about it.

One qualification for an admn. is to be really old.....

I wish the GFF posts are deleted soon and there is no country bias thing

[quote="Valentin Vornicu"][quote="shadysaysurspammed"]One qualification for an admn. is to be really old.....[/quote]There is no such criterion.[/quote]

No,by really old i meant about 20+  :D


ABCD is a rectangle with AB = 16 cm and BC = 12 cm. Points E and F lie on sides AB and CD so that AECF is a rhombus.
What is the length of EF?

hello, the length of $ \overline{EF}\equal{}15cm$.
Sonnhard.

$ AE\equal{}EC\equal{}CF\equal{}AF\equal{}x$
Consider $ \triangle BEC$
We have $ EC^2\equal{}BE^2\plus{}BC^2$
$ x^2\equal{}(16\minus{}x)^2\plus{}144$
$ 32x\equal{}400$
$ x\equal{}12.5$

And $ AC\equal{}20$

So, $ EF\equal{}2\sqrt{12.5^2\minus{}10^2}\equal{}15$


\[ \left\{ \begin{array}{l}
 4x \plus{} 3y \plus{} 1 \equal{} 3\sqrt {\left( {x \plus{} y} \right)\left( {2x \plus{} y \plus{} 1} \right)}  \\
 x^2  \plus{} y^2  \equal{} 2 \\
 \end{array} \right.
\]

hello, squaring the first equation and simplifying we get $ y\equal{}\minus{}\frac{2}{3}x\plus{}\frac{1}{3}$, substituting this
in the second equation we have $ x^2\plus{}(\minus{}\frac{2}{3}x\plus{}\frac{1}{3})^2\equal{}2$. Solving the last equation by $ x$ we have $ x_1\equal{}\frac{17}{13}$ and $ x_2\equal{}\minus{}1$, from here we get $ y_1\equal{}\minus{}\frac{7}{13}$ and $ y_2\equal{}1$, these are all solutions of our system.
Sonnhard.

Squaring and factoring the first equation gives (2x+3y-1)(x+1)=0, so x = -1 and y = 1 (as y=-1 doesn't work) and then substitution in the second equation gives the other solution: x = 17/13 and y = -7/13.

[quote="toancapba"]
\[ \left\{\begin{array}{l} 4x \plus{} 3y \plus{} 1 \equal{} 3\sqrt {\left( {x \plus{} y} \right)\left( {2x \plus{} y \plus{} 1} \right)} \\
x^2 \plus{} y^2 \equal{} 2 \\
\end{array} \right.
\]
[/quote]
Put
\[ \begin{array}{l} \left\{\begin{array}{l} u \equal{} x \plus{} y \\
v \equal{} 2x \plus{} y \plus{} 1 \\
\end{array} \right. \\
\Rightarrow 2u \plus{} v \equal{} 3\sqrt {uv} \Leftrightarrow 2\sqrt {\frac {u}{v}} \plus{} \sqrt {\frac {v}{u}} \equal{}3 \\
\end{array}
\]


[hide="Hint"]Count the number of paths that [i]do[/i] pass through the point, and subtract this from the total number of paths.[/hide]

i have tried it
but i dont know how to count the paths :oops: !!!
please could you show your solution
with details please
im really bad at combinatorics
i would be very happy if you tell me
where can i learn combinatorics
to do this kind of problems
well, good bye  :)

Well to get from $ (0, 0)$ to $ (i, j)$ we need to make $ i\plus{}j$ moves, $ i$ of which are rights and $ j$ of which are ups. Similarly to get from $ (i, j)$ to $ (m, n)$ we need to make $ m\plus{}n\minus{}i\minus{}j$ moves, $ m\minus{}i$ of which are rights and $ n\minus{}j$ of which are ups. From here it should be straightforward how to calculate the number of paths from $ (0, 0)$ to $ (m, n)$ going through $ (i, j)$.

you say that to get (2,2) we have 2+2=4 options, but
there are 6  :o
i'm trying to see something interesting, but
i feel i'm far to get it  :oops:

The number of [i]moves[/i] (not the number of [i]ways[/i]) to get to $ (m,n)$ from $ (i,j)$ is $ m\plus{}n\minus{}i\minus{}j$.

EDIT: I made a careless mistake.

[quote="pelao_malo"]We have a $ m\times n$ grid situated in the cartesian plane, with a corner in $ (0,0)$ and the opposite corner in $ (m,n)$. We kill a point $ (i,j)$ with $ 0\le i\le m$ and $ 0\le j\le n$, and you can't pass over it. In how many ways you can travel from $ (0,0)$ to $ (m,n)$ if you can only move one point in the positive direction (to the right or up)?

 :([/quote]

$ \binom{m \plus{} n}{m} \minus{} \binom{i \plus{} j}{i} \cdot \binom{m \plus{} n \minus{} i \minus{} j}{m \minus{} i}$ is what you're lookin for

$ \binom{m \plus{} n}{m}$ represents the number of ways to get from $ (0,0)$ to $ (m,n)$

$ \binom{i \plus{} j}{i}$ represents the number of ways to get from $ (0,0)$ to $ (i,j)$

$ \binom{m \plus{} n \minus{} i \minus{} j}{m \minus{} i}$ represents the number of ways to get from $ (i,j)$ to $ (m,n)$

i probably made a careless mistake or two in there, by the way


[quote="cezar lupu"][color=blue]
 Let $a, b, c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$
 Prove that
                                  \[ a^{2}+b^{2}+c^{2}+1 < 2abc(a+b+c).  \]

 [/color][/quote]

Let $x=\frac1{a}$, $y=\frac1{b}$ and $z=\frac1{c}$.

The inequality becomes $x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2<2(xy+yz+zx)$ with $x+y+z=1$

From $x+y+z=1$ we have that $2(xy+yz+zx)=1-(x^2+y^2+z^2)$, so it remains to prove that:

$x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2+x^2+y^2+z^2<1$

Could anyone finish the prove?

A more harder one is $\displaystyle \sum a^2+1<\frac{1}{2} abc(\sum a)$

[quote="shyong"][quote="Incumbuss"]$2a^{2}+2b^{2}+2c^{2}>a^{2}+b^{2}+c^{2}+1.$[/quote]

This is not true ....[/quote]
No no,that is true cause $a^2,b^2,c^2>1$


This is my first time to chapter. I need your help to understand some rules. I know how to get the overall scores for the written contest(sprint & target), but I am not sure how to calculate the scores for the countdown. Suppose I got 40 points and placed 10th from the written. If I beat the 9th and 8th in the countdown, is my overall placement for the chapter jump to 8th? If this is the case, the countdown weights a lot, right?
Thanks a lot.

Yes, to all of your questions.

If your countdown is official:

10th against 9th.  Loser is 10th.

Winner of previous match against 8th.  Loser is 9th.

Etc.

So the most you can drop is 1 spot but you can go all the way up to first place.  Of course, you need to make top 10 first :D

That is only if the countdown is official. If it isn't, it doesn't matter at all in your final placing.

Yes, they must hold your chapter countdown this way to make it official. If it is any other way, then it's unofficial and just for fun. For example, my final rankings are just based upon the written competition, and the countdown is a single elimination bracket with 16 people.

for a countdown to be official, it has to be in the format mentioned (but i like a bracket with 12 people much better :) )

Thanks a lot. But how do I know if my chapter is official or not?

Your chapter? Um, if you can qualify for state, then I guess it's official... Are you talking about the entire competition, or just countdown? I recommend you first try very hard to do well, of course. Then, if you make it, try your best at countdown to do well. At chapter, ask your coordinator if countdown is official at state. Make sure you know if the state countdown is official. or not. Chapter should be okay, because a lot more people make the next level.


Prove that there do not exist two consecutive perfect numbers (number is called perfect if it is equal to the sum of its proper divisors).

Doesn't that follow by the formula for generating perfect numbers with mersenne primes?

Basically, Euler's characterization is used in a proof but then you have to prove that neither $2^{p-1}(2^p-1) +1$ nor $2^{p-1}(2^p-1) -1$ can't be perfect.

Yes.Suppose n is perfect number ,we will prove that:n-1 and n+1 is not.Indeed:
$n=2^{p-1}(2^p-1)=4^{\frac{p-1}{2}}(2.4^{\frac{p-1}{2}}-1)=4.(2.4-1)=4(\mod 12)$
->n-1≡-1(mod 4)
σ(n-1)=n-1/d+d≡0(mod 4) but 2(n-1)≡-2(mod 4) .Contradiction.
n+1≡-1(mod 3).Hence:
σ(n+1)=n+1/d+d≡0(mod 3) but 2(n+1)≡-2(mod 3).It's also false.


ratio
trigonometry
geometry
trig identities
Law of Sines

How does one prove that equal angles in two triangles implies ratios of corresponding sides equal? I think it has something to do with the law of sines, but I'm not sure exactly how to proceed.

Say you have two angles t1 and t2 in two triangles. For one triangle, call [b]a[/b] the side opposite to the angle t1 and [b]b[/b] the side opposite to t1. For the other triangle, call [b]A[/b] the side opposite to the angle t1 and [b]B[/b] the side opposite to t2. By the sine law:

a/sin(t1) = b/sin(t2) and
A/sin(t1) = B/sin(t2)

So a/b = sin(t1)/sin(t2) = A/B

i.e. the ratios are the same.

There's a bit of a problem with using the law of sines here - using sine at all essentially assumes AA Similarity.  Try proving it without the law of sines.  (1337 - if you're still stuck in 3 days, ask Coach Frost to borrow his Intro Geometry book and work through the similarity chapter.  The proof is there.)

Ok...here is a pretty random proof (?):

Given $\triangle{ABC}$ and $\triangle{DEF}$, where $\angle{A}\cong\angle{D}$, $\angle{B}\cong\angle{E}$, and $\angle{C}\cong\angle{F}$.

Let E' and F' be points on sides AB and AC, respectively, such that AE'=DE and AF'=DF.

By SAS, $\triangle{AE'F'}\cong\triangle{DEF}\implies\angle{AE'F'}\cong\angle{E}$.

Since $\angle{B}\cong\angle{E}$, $\angle{B}\cong\angle{AE'F'}$.

If E'=B, then triangle ABC and AE'F' are the same triangle, and AB/DE = AC/DF because each fraction is equal to 1.

If E' and B are not equal, then lines E'F' and BC are parallel. It is not too hard to prove that the triangles must now be similar, by considering ratios of areas, same altitude, etc.

i have to say that the proof is pretty tricky...but cool...basically place the two triangles ontop of each other sides/vertex coincide, the other sides are parallel...the hard part is where to go from here...here is a hint: (as in try and figure it out now...) use area ratios and the parallel line shows two areas are equal...

EDIT: this is basically what you want to do based on the last two sentences of the previous post