A triangle has vertices $ A(\minus{}4, \minus{}1)$, $ B (2, \minus{}1)$ and $ C(1, 3)$. What is the area, in square units, of triangle $ ABC$?

(-4, -1) and (2, -1) lie on a horizontal line, so we can use that as the triangle's base, which has length 2+4=6. (1, 3) is 1+1=2 units above the base, so the altitude of the triangle is 2. Thus the area is (6)(2)/2=6 square units is our final answer.

Or one can shoelace it.

[color=blue]For more on shoelace, please see the entry titled "The Shoelace Theorem" in my blog. You can click the "coordinates" or "geometry" tag to get to it. -Izzy[/color]

EDIT:My edit this time: Aaaaah she can control my posts for advertising purposes!

[quote="ernie"]Like I have said in a previous post, shoelace is not the wisest decision if you have either $ (x_1,y_1)$ and $ (x_1,y_2)$ or $ (x_1,y_1)$ and $ (x_2,y_1)$.[/quote]
True, it would take longer than gauss's approach so it would be more favorable under a time limit.

I mean if you don't notice it immediately you can immediately start bashing Shoelace. And if you're arithmetic is really fast, you might be able to beat the time on that approach. But then again, gauss's approach is nicer.

Just bash shoelace and you get [hide]12[/hide]


Show that a necessary and sufficient condition for three points a, b, and c in the complex plane to form an equilateral triangle is that $a^2+b^2+c^2=bc+ca+ab$.

This is straightforward, of course, but I find it very nice and it can be helpful when one looks for equilateral triangles whose orientation is not known.

  Darij

We have :
$a^2+b^2+c^2 - ab - bc - ca = (a+jb+j^2c)(a+j^2b + jc)$,
where $j = e^{2i \frac {\pi} 3}$.

Hence result.

Pierre.

Hmmm, this was just covered here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=16380[/url]


Source: British Math Olympiad, Round 1 (Wednesday, January 14, 1998)

Divide a 5x5 square into 25 unit squares.  One of the numbers 1, 2, 3, 4, and 5 is inserted into each unit square so that each row, each column, and each of the two diagonals contains each of the five numbers once and only once.  The sum of the numbers in the squares immediately below the diagonal from top left to bottom right is called the score.

a) Show that it is impossible for the score to be 20.
b) What is the highest possible score?

If the score were 20, each of the score boxes would be 5. The diagonal has numbers from 1-5. The box in the diagonal which has the 5 will be in the same row/column as one of the score boxes. Oopsies.

I meant "oopsies"  as a polite way of saying "CONTRADICTION!"

I don't get it. Can you tell me what contradicted what? (I know, I'm stupid.)

You're not stupid; I'm just inexplicable. anways, you see that if we have all 5's in the 4 score boxes, and there must be a 5 in the main diagonal from upper left to bottom right, wherever the 5 is in that diagonal, it will be in the same row or column as a 5 in one of the score boxes. That is a contradiction, because you can't have the same number twice in a given row or column.

Oh, now I see.

By the way, is [hide]19[/hide] the answer for b?

I don't think so; I've tried it. Are you sure you've got all the digits in your main diagonal?

I don't think 18 works either. I then decided just going down and testing each number was not the best way to solve the problem, so I gave up temporarily.

Tare wrote:This reminds me of the queens puzzle where you have to fit n number of queen onto a n by n board where no two queens are in the same row, column, or diagonal...

OK, my point is that[hide] as the puzzle shows, there can be only one number in the score. For example, there can't be two 5s. So I'm guessing the answer is 14.[/hide]

differ away....however I don't understand your reasoning for the part in spoiler. Why can there not be two 5's in the score?

[hide]I don't think two of the same numbers can be written in the same diagonal either, if you want don't want two same numbers in a column or a row. Thus only one of each number can be in the score. (Well you could actually try it, and you should come up with 10 different placements in which the above statement is true.)[/hide]

I think you could have two being the same.. well I don't see any instant reason why not.

[color=cyan]I think Tare is treating the "minor diagonal" (I'm making that term up) which is the score as if it were one of the diagonals which, as the problem states, must have each number occuring once.  That is, no repeats.  Which gives you a maximum of 14.  But that isn't what the statement of the problem means.  (Although you could redefine diagonals, either to what Tare means or to include the five squares that would make up a diagonal if the whole thing were a torus.  That is, the score plus the corner opposite.  Someone else explain what I mean.)[/color]

Oh, I meant diagonal lines made up of 3, 4, or 5 squares. (It includes the usual diagonals, but it excludes the 3 squares at the corners, just to make this simpler.)
A diagram:
ABCXY
DABCX
EDABC
ZEDAB
QZEDA
or
YXABC
XABCD
ABCDE
BCDEZ
CDEZQ

Where as the A,B,C,D,E,X,Y,Z, and the Qs are the "diagonals" I'm reffering to.
In this case D in the upper diagram would be the "score."
Note that not all lines have a number in them. 4 of the diagonls would be empty.
and Joel, I think this would also stand in the question's case, since this only require that the question says "no two same numbers can be on the same row or column"

[color=cyan]But Tare, reread the question.  It says:
[quote="mathfanatic"]One of the numbers 1, 2, 3, 4, and 5 is inserted into each unit square so that each row, each column, and each of [i][b]the two diagonals [/b][/i]contains each of [b][i]the five numbers once and only once[/i][/b][/quote]
(emphasis added).[/color]


what is "jension  en quality ".
how to use it in proving simple  en equality .
please help...........

http://mathworld.wolfram.com/JensensInequality.html
http://en.wikipedia.org/wiki/Jensen's_inequality
http://planetmath.org/encyclopedia/JensensInequality.html

For example, letting $ f(x) \equal{} \log x$ provides a proof of AM-GM.

or http://www.artofproblemsolving.com/Wiki/index.php/Jensen%27s_Inequality

AoPSWiki is your friend.


ceiling function
logarithms
algebra proposed
algebra

Let $n$ be a fixed natural number.
[b]a)[/b] Find all solutions to the following equation :
\[ \sum_{k=1}^n [\frac x{2^k}]=x-1  \]
[b]b)[/b] Find the number of solutions to the following equation ($m$ is a fixed natural) :
\[ \sum_{k=1}^n [\frac x{2^k}]=x-m  \]

The answer for b is $\sum_{i=0}^{m}\binom{n}{i}$

rust,the number $ x $ must be an integer,not necessarily a positive one  :wink:

In this solution sketch, $m$ is a nonnegative integer.  We know that there are $\sum_{i=0}^m \binom{n}{i}$ nonnegative solutions from earlier replies.  If we include negative integers, it seems really messy.  Note that $m<n$ must happen for a negative solution to exist.  (Thus, if $m \geq n$, there are $\sum_{i=0}^m\binom{n}{i}=\sum_{i=0}^n \binom{n}{i} = 2^n$ solutions.)

If $m=n-1$, $x=-1$ is the only negative solution, whence there are in total \[\sum_{i=0}^{n-1}\binom{n}{i}+1=2^n\] solutions in this case.  From now on, we assume that $m \leq n-2$.

There are \[\sum_{\ell=0}^{n-1} \binom{\ell}{n-2-m}=\binom{n}{n-1-m}=\binom{n}{m+1}\] negative solutions $x\geq -2^n$.  The number of solutions $x<-2^n$ is more complicated and given by this monstrous summation \[{\sum_{\ell=1}^{\left\lceil{\log_2(n-m)}\right\rceil} \sum_{j=2^{\ell-1}+m+1}^{2^{\ell}+m}}\binom{n}{j}=\sum_{i=m+2}^{2^{\left\lceil{\log_2(n-m)}\right\rceil}+m}\binom{n}{i}\,.\]  Therefore, the total number of solutions for $0 \leq m \leq n-2$ is \[\sum_{i=0}^{2^{\left\lceil{\log_2(n-m)}\right\rceil}+m} \binom{n}{i} = \sum_{i=0}^n \binom{n}{i}=2^n\,.\]
The same answer holds if we take $m$ to be an arbitrary nonnegative real number.  Judging from how simple the final answer looks, I think there might be a simple counting argument and it probably has something to do with expressing each nonzero integer as $\sum_{j=0}^N a_j 2^j$ where $a_j \in \{-1,+1\}$).

Can you explain how do you get the answer on b)?

We will be using the well known fact that
$$ \sum_{j \ge 1} \left\lfloor \frac{x}{2^j}  \right\rfloor = x - \text{sum of digits of } x \text{ when written in base } 2  \qquad \qquad (\star)$$
It directly follows that all solutions to [b](a)[/b] are
$$ \boxed{2^0,2^1,\ldots,2^n} $$
We solve part [b](b)[/b] now. Write
$$ x = \underbrace{\sum_{i=0}^{n-1} 2^ia_i}_{=y \text{ (say)}} + \underbrace{2^n \cdot l}_{=z \text{ (say)}}$$
where each $a_i \in \{0,1\}$ and $l \in \mathbb Z_{\ge 0}$ (note as $x=0$ is clearly never a solution, so we don't need to worry about all $a_i$ and $l$ being equal to $0$). Using $(\star)$ we get
$$ \sum_{k=1}^n \left\lfloor \frac{y}{2^k} \right\rfloor = \sum_{j \ge 1}  \left\lfloor \frac{y}{2^j} \right\rfloor = y - \sum_{i=0}^{n-1} a_i $$
On the other hand it is easy to compute
$$ \sum_{k=1}^n \left\lfloor \frac{z}{2^k} \right\rfloor = z \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) = z \left( 1- \frac{1}{2^n} \right) = z - l $$
Now observe
$$  \sum_{k=1}^n \left\lfloor \frac{x}{2^k} \right\rfloor = \sum_{k=1}^n \left\lfloor \frac{y}{2^k} \right\rfloor + \sum_{k=1}^n \left\lfloor \frac{z}{2^k} \right\rfloor = \left(y - \sum_{i=0}^{n-1} a_i \right) + (z - l) = x - \left( l + \sum_{i=0}^{n-1} a_i \right)$$
Hence the given equation is equivalent to
$$ l + \sum_{i=0}^{n-1} a_i = m $$
Note for a fixed choice of $l,m$, there are
$$ \binom{n}{m-l} $$
solutions to the above equation, as we have to choose some $m-l$ of the $n$ $a_i$'s to equal $1$. Hence total number of solutions to [b](b)[/b] is
$$ \sum_{l \ge  0} \binom{n}{m-l} = \boxed{ \sum_{i=0}^m \binom{n}{i} } $$
This completes the proof. $\blacksquare$

[b][color=#f00]Remark: [/color][/b] In this proof we have assumed $x \ge 0$. For $x < 0$, weird things start happening like for $n=2$, $x=-1$ is a solution to [b](a)[/b]. The Key reason is that $(\star)$ is no longer true for negative $x$. The reason is the following:

 Basically for $x \ge 0$, if we write
$$ x = \sum_{i \ge 0} a_i2^i $$
where $a_i \in \{0,1\}$, then we have
$$ \sum_{j \ge 1} \left\lfloor \frac{x}{2^j} \right\rfloor \textcolor{red}{=} \sum_{i \ge 0} \left( \sum_{j=0}^i \left\lfloor \frac{a_i2^i}{2^j} \right\rfloor \right) = \sum_{i \ge 0} \left(a_i2^i - a_i \right) = x - \sum_{i \ge 0} a_i $$
while if we try to do the same thing with $x < 0$ by taking $a_i \in \{0,-1\}$, then the red equality isn't true as some $-1$'s also pop, like for example
$$ \left\lfloor \frac{-1 - 2}{2} \right\rfloor = \left\lfloor \frac{-1}{2} \right\rfloor + \left\lfloor \frac{-2}{2} \right\rfloor \textcolor{blue}{-1} $$
where that extra $\textcolor{blue}{-1}$ is not there if $-1,-2$ are replaced by $1,2$.

I got a somewhat complicated form which seems right (and also much easier to calculate)
Assume $x \geq 0$ first of all and now let's introduces some notations ,
$v_p(a)$ is the highest power of $p$ in $a$ and $s_p(a)$ is the sum of digits in the base $p$ representation of $a$ where $p$ is a prime.

a) Suppose $x \leq 2^{n+1} - 1$ then ${LHS} = v_2(x!) = x - s_2(x) = x - 1 \implies s_2(x) = 1 \implies x \in \{ 1 , 2 , \cdots , 2^n\}$
Now suppose $x \geq 2^{n+1} \implies {LHS} = v_2(x!) - \sum_{k \geq n+1} \left\lfloor \frac{x}{2^{n+1}} \right\rfloor = x - s_2(x) - \sum_{k \geq n+1} \left\lfloor \frac{x}{2^{n+1}} \right\rfloor = x - 1 \implies s_2(x) + \sum_{k \geq n+1} \left\lfloor \frac{x}{2^{n+1}} \right\rfloor = 1$ but note that
$s_2(x) + \sum_{k \geq n+1} \left\lfloor \frac{x}{2^{n+1}} \right\rfloor \geq 1 + 1 = 2$ , Thus no solution here .

Therefore all the solutions are $\{ 1 , 2 , \cdots , 2^n\}$.

b)Let $\binom{a}{b} = 0$ if $b > a$ for $a,b$ positive integers.
Now we begin with our solution ,
Suppose $x \leq 2^{n+1} - 1 \implies \text{The length of x as a binary string is atmost n}$
Then ${LHS} = x - s_2(x) = x - m \implies s_2(x) = m$ , Thus it suffices to count the number of binary strings of length $\geq m$ such that it has got $m$ $1$'s in it and that's exactly
$$ \binom{m}{m} + \left(\binom{m+1}{m} - \binom{m}{m} \right) + \left(\binom{m+2}{m} - \binom{m+1}{m} + \binom{m}{m} \right) + \cdots = \binom{n+1}{m}$$

Now suppose $$x \geq 2^{n+1} \implies {LHS} = x - s_2(x) - \sum_{k \geq n+1} \left\lfloor \frac{x}{2^{n+1}} \right\rfloor = x - m \implies s_2(x) + \sum_{k \geq n+1} \left\lfloor \frac{x}{2^{n+1}} \right\rfloor = m$$
Note that for $2^{n+i+1} > x \geq 2^{n+i}$ forall $i \geq 1 \implies \sum_{k \geq n+1} \left\lfloor \frac{x}{2^{n+1}} \right\rfloor = \sum_{k \geq n+1} \left\lfloor \frac{2^{n+i}}{2^{n+1}} \right\rfloor = 2^i - 1$ and also that the length of $x$ as a binary string is $= n+i+1$ with a leading $1$.
Therefore for $2^{n+i+1} > x \geq 2^{n+i}$,
$$ s_2(x) + 2^i - 1 = m \implies s_2(x) = m + 1 - 2^i$$
for $i \leq \left\lfloor log_2(m) \right\rfloor$ and the number of solution to this is $\binom{n+i}{m - 2^i}$ , Therefore in total forall $i$ the number of solution is $\sum_{i=1}^{\left\lfloor log_2(m) \right\rfloor} \binom{n+i}{m-2^i}$

Thus the total number of solution to this is $\left(\sum_{i=1}^{\left\lfloor log_2(m) \right\rfloor} \binom{n+i}{m-2^i}\right) + \binom{n+1}{m}$.


[hide]The "resting points" are the plateaus in the lines. Anne first rested at approx. 3800 ft, and Ben first rested at approx. 4000 ft.

Anne leaves the resting point at 1 hour, and Ben arrives there at 2 hours. Therefore, the difference is $\boxed{1\text{ hour}}$.[/hide]

:!:  :rotfl:  :blush:  Whoops, I posted the wrong problem.  (Using the same graph) Ben and Anne hike to the top of a mountain and back by the same route.  At one time Anne is changing her elevation twice as fast as Ben.  At another time in their hike Ben is changing his elevation three times as fast as Anne.  How long is it from the beginning of one of these times to the beginning of another.

I can't find the interval where Ben's slope is three times Anne's slope  :blush:

look at the middle of the descent. are you sure this doesn't say, round to the nearest hour?

I have to agree that the anwer key is wrong. :P

[quote="Phoenix257"]No rounding involved.  Do you mean at 4.75 hours?

[hide]Anne is going twice as fast as Ben at 3.75 hours.  The answer says that the time when Anne is twice as fast as Ben and the time when Ben is 3 times as fast as Anne are 5.25 hours apart.  I'm pretty sure the answer key is wrong  [/hide][/quote]
Actually Anne also went twice as fast as Ben in the begining. Anne got to the rest point in less than 1 hour (about 0.8 hours). ben took more than 1.5 hours (about 1.6 hour) to get there.
therefore Anne traveled twice as fast as Ben in the begining.
Ben went three times as fast as Anne in the last stage...
i think that makes it about 5.25 hours.
ben started going 3 times as fast around 5.75 hours.
Anne started going twice as fast around 0.5 hour.
that's a 5.25 hour difference


limit
symmetry
topology
real analysis
real analysis unsolved

Let $ A\subseteq M$ and define $ d(x,A)\equal{}\inf\{d(x,a): a\in A\}$.  Prove that $ \left|d(x,A)\minus{}d(y,A)\right|\leq d(x,y)$.

Well we went through a proof, which I will supply, but I think the TA might have miswrote something because one of his conclusions doesn't seem to follow.
================================================================================================================
Define $ \lim_{n_{i}\rightarrow\infty}d(x,y_{n_{i}})\equal{}d(x,A)$ and $ \lim_{n_{i}\rightarrow\infty}d(y,y_{n_{i}})\equal{}d(y,A)$, where $ (x_{n})$ and $ (y_{n})$ are sequences, and $ (x_{n_{i}})$ and $ (y_{n_{i}})$ are subsequences.

First assume $ \limsup_{n\rightarrow\infty}\left|d(x,y_{n})\minus{}d(x,x_{n})\right|\equal{}0$.  Then we can choose subsequences such that $ \lim_{n_{i}\rightarrow\infty}\left|d(x,y_{n_{i}})\minus{}d(x,x_{n_{i}})\right|\equal{}0\Rightarrow \left|d(x,A)\minus{}d(y,A)\right|\equal{}0\leq d(x,y)$.
Now assume $ \limsup_{n\rightarrow\infty}\left|d(x,y_{n})\minus{}d(x,x_{n})\right>0$.  Then $ \exists x_{n_{i}},y_{n_{i}}$ such that $ \left|d(x,x_{n_{i}})\minus{}d(y,y_{n_{i}})\right|\leq d(x,y_{n_{i}})\minus{}d(y,y_{n_{i}})\leq d(x,y)$.
================================================================================================================
I think the problem may be in the definition but I'm not totally sure.  Thanks.

The proof neglects to mention what properties the sequences and subsequences satisfy, which would probably clear up whatever questions you have about the argument (for example, for the argument to make any sense, $ \{x_{n_i}\}$ and $ \{y_{n_i}\}$ ought to be approaching some very specific points in $ A$).

They approach the $ a$ such that $ d(x,A)\equal{}\inf\{d(x,a)\}$ and likewise for $ y$.

A different approach, elementary in my opinion; for $ z\in A$, $ d(x,A)\leq d(x,z)\leq d(x,y) \plus{} d(y,z)$, now take infimum over $ z\in A$, we have $ d(x,A) \minus{} d(y,A)\leq d(x,y)$. For symmetry ( we could have started with $ d(y,A)$ ), we have your claim.

Edit: who, or what is TA  :huh: ?

Nice argument-it definitely makes sense.  Is it rigorous enough?

TA is teaching assistant-usually a grad student that assists the professor by grading papers, going over homework problems, teaching a little bit,...  They typically have their own class time (usually called a section or conference or something to that effect) to do this stuff and the TA provided that solution during his section.

Mr Doe shows the correct approach. Now you can start learning about the famous Hausdorff metric on closed subsets of a metric space.

I don't see it in my book.  Maybe I'm looking in the wrong place or maybe it's not there so, what is it?


induction
LaTeX
combinatorics unsolved
combinatorics

Answer: n=2^k-1
I solve this problem by using complex induction principle.
I don't know how to use LATEX, so I can't write my solution.

[quote="anonymouslonely"]I don't understand... how can this permutation which has $ n+1 $ terms become a permutation with $ n $ terms?[/quote]

Good point. I fixed the problem.

does somebody have a solution? it looks so strange!
frejer,how do you solve it more exactly?

This is (probably) the official solution.

Let $P=(1,n,n-1,\ldots,3,2,0)$ denote the starting permutation, and let $R$ denote the desired finishing permutation $(1,2,\ldots,n,0)$. We observe that if $n=1$ or $2$, then $P$ is trivially regular. Now we suppose that $n>2$ and consider the following three cases:

[b]Case 1:[/b] $n>2$ is even. If $n$ is even, say $n=2k$ and $k>1$, then we obtain from $P$ after $k-1$ legal transportations the following permutation:
$$(1,n,0,n-2,n-1,n-4,n-3,\ldots,4,5,2,3).$$
No further legal transportations are possible, and we conclude that $P$ is not regular in this case.
[b]Case 2:[/b] $n$ is of the form $2^j-1$ for some $j\ge 1$. For $i=1,2,\ldots, j$ define the permutation $P_i$ as follows:
$$P_i=(1,2,\ldots,2^i-1,0; n-2^i+1,n-2^i+2,\ldots,n-1,n;n-2\cdot2^i+1,\ldots,n-2^i;\ldots; 2^i,\ldots,2\cdot 2^i-1).$$
The permutation $P_i$ consists of $(n+1)/2^i$ "blocks " of $2^i$ numbers, each one indicated by a semicolon. Note that the members within each block (except the first) are in increasing numerical order. We now note that $P_i$ is obtained from $P$ by precisely $\frac{n-1}{2}$ legal transportations, and that $P_{i+1}$ is obtained from $P_i$ by precisely $\frac{n+1}{2}$ legal transportations for $i=1,2,\ldots, j-1$. But $P_j=R$ and thus $R$ is obtained from $P$ by precisely $\frac{j(n-1)}{2}$ legal transportations. Hence, $P$ is regular whenever $n$ is of the form $2^j-1$.

[b]Case 3:[/b] $n$ is odd, but not of the form $2^j-1$. Then $n$ can be written uniquely as $n=(2k+1)2^j-1$, when $j,k\ge 1$. In this case we can define $P_1, P_2,\ldots, P_j$ as in Case 2; also as in case 2, $P_j$ is obtained by $P$ by precisely $\frac{j(n-1)}{2}-1$ legal transpositions. However, in this case $P_j\ne R$; but after exactly $k$ further legal transportations on $P_j$ we obtain the following permutation:
$$(1,2,\ldots,2^j;n-2^j+1,\ldots,n;0,n-2\cdot 2^j+2,\ldots,n-2^j;\ldots; 2\cdot2^j,\ldots,3\cdot 2^j-1; 3\cdot 2^j,\ldots,2\cdot 2^j-1).$$
No further legal transportations are possible, and we conclude that $P$ is not regular.
All numbers are now accounted for and we infer that $P$ is regular if $n=2$ or $n=2^j-1$ for $j=1,2,\ldots$.

[quote=shobber]Given a permutation ($a_0, a_1, \ldots, a_n$) of the sequence $0, 1,\ldots, n$. A transportation of $a_i$ with $a_j$ is called legal if $a_i=0$ for $i>0$, and $a_{i-1}+1=a_j$. The permutation ($a_0, a_1, \ldots, a_n$) is called regular if after a number of legal transportations it becomes ($1,2, \ldots, n,0$).
For which numbers $n$ is the permutation ($1, n, n-1, \ldots, 3, 2, 0$) regular?[/quote]
Does a_i=0 for i>0 mean only that a_0#0?


algebra
polynomial
superior algebra
superior algebra unsolved

If we assume AC then it is pretty easy to construct uncountable proper subfields of $ \mathbb{R}$. Can we do it without AC?

Apparently the von Neumann numbers provide such an example.  Found this in Wagon's book "The Banach-Tarski Paradox", on Google.  For any real $ t > 0$, define $ v_t \equal{} \sum_{n \equal{} 1}^\infty 2^{2^{[tn]} \minus{} 2^{n^2}}$.  The idea is that $ v_t$ is easier to approximate by rationals than any $ v_s$ with $ s < t$, so they are algebraically independent.  If $ K \equal{} \mathbb{Q}(\{v_t | t\in (0,1)\})$, this implies that $ K\neq \mathbb{R}$.

The citation for this is present in Wagon's book, but I can't see it–something written by von Neumann.

What a crazy example! Forgive the ignorance - but why does the fact that they are algebraically independent mean that K is a proper subfield of the reals? Why could the collection of $ v_{t}$ with t between 0 and 1 not be a transcendence basis for the reals over the rationals? (Also - L is awesome.)

Well, if we had $ v_1 \in K$, for example, then $ v_1$ would be a polynomial combination of smaller $ v_t$ so they would not be algebraically independent.

(yeah, this is pretty much my favorite avatar ever)


HMMT
AMC
AIME
geometry
ARML
algebra
partial fractions

are AIME problems harder than HMMT problems [b]based upon your experience[/b]?

AIME problems tend to be more computationally heavy; HMMT problems are more often "see the trick and you're done" type things. That said, time pressure is much greater at HMMT and I think there tends to be a greater rate of careless mistakes; (problems solved)/(correct answers) tends to be lower at HMMT, at least for me. These kinds of questions aren't really answerable; just do the problems.

The contests have completely different characters. It's very difficult to compare, as CatalystOfNostalgia said.

This question is like asking "Are the red sox better than the yankees?"
Although a few devout fans will say that the red sox are better, the numbers don't lie...
Also, AIME is far more prone to starting off with a couple of questions that are far easier than most of what hmmt has, like the first 5 on AIME would not fit onto any HMMT question beyond number 1, guaranteed, and even 6-10 are usually equal to like 2-5 on HMMT. HMMT is much harder, basically.

The HMMT is definitely harder, although not as hard to prepare for. My score went from about 15 to 35/100 in a week of doing mass numbers of HMMT tests. Probably this is because the problems tend to not change much over the years; always something on substituting for a variable, an infinite series and/or partial fractions, factorization, sum of roots, counting using a unique method, power of a point, etc. Still, the harder problems are generally original and will take some more creativity than on the AIME.

The time factor also often forces the competitor to use the first method that he/she thinks of, leaving little time to think of an elegant solution if it is not there initially. So you have to be quick in identifying relevant key steps.

The three hours on the AIME allows for plenty of looking in the sky and pondering about a better solution than brute force. However, it is much harder to prepare for it because it goes everywhere. Sure, your score will increase if you do problems, but most likely because you develop your problem-solving ability, not because you get used to the problems.

[quote="Smartguy"][quote="diophantient"]This question is like asking "Are the red sox better than the yankees?"[/quote]

Ehem, THEY ARE BETTER THAN THE YANKEES!!!

(look at where I am from if you didn't already)[/quote]
Ah, and your statement proves my point completely. I'm not a yankees fan myself, but no one can deny the statistics, and it's not uncommon for people that get 8, 9, 10, 11, 12 on AIME to get like... single digit scores at HMMT in one of their topics.

[quote="diophantient"]...and it's not uncommon for people that get 8, 9, 10, 11, 12 on AIME to get like... single digit scores at HMMT in one of their topics.[/quote]

Thanks for that, brother.

by the way Carl, what did you get on the geometry round?

You can go to extended division II (B? ugh I still don't know which one) and see :roll:

the problem with HMMT is that it's easy to get stuck on a problem and then you're like ARGH

[quote="not_trig"]the problem with HMMT is that it's easy to get stuck on a problem and then you're like ARGH[/quote]

Thats what happened to me with all the geometry problems except the first one.

I like the fact that the solution comes quickly after the trick. I find it more satisfying. Computation was always one of my weaknesses, and that's why I love my calculator. I wish there were more competitions that focused on the quick tricks.

just asking: is hmmt invitational contest or is it a contest u sign up for?  :)

HMMT problems are written  by a group of young people (and future mathematicians?) . AIME problems are written by a group of young people's fathers (experienced mathematicians).

Computational skill is basic of the basic math skills.

[quote="toadoncart"]is hmmt invitational contest or is it a contest u sign up for?  :)[/quote]

Why don't you go look at their website and find out?  ;)  http://web.mit.edu/hmmt/www/


Does there exist a bounded real function $f$ defined on the reals s.t. $f(1)=1$ and $f\left(x+\frac 1{x^2}\right)=f(x)+f^2\left(\frac 1x\right)$?

A cute but very easy one, I know :)

$f(x)=x$ ? Or I'm missing something ?

[edit] oh, I see now bounded in the name of the topic but I see no additional conditions in the text  :?

I forgot to mention that $f$ has to be bounded, although I mentioned that in the title of the thread :). I'll edit it right away.

$f$ doesn't reach is $\sup$ since if $\exists x_0, f(x_0)=\sup f, f(x_0+\frac{1}{x_0^2})=f(x_0)+f(\frac{1}{x_0})^2>\sup f$ absurd

Ok. So? Or was it simply meant to be an observation?

I'm still searching how can we use f(1)=1. Yes that was just an observation.

Suppose $M=\sup|f(x)|\Rightarrow -M\leq f(x)\leq M$.
Let $x=2 :f(2)=f(1)+f(1)^2=2\Rightarrow M\geq 2(1)$
We have:$f(\frac{1}{x}+x^2)-f(\frac{1}{x})=f(x)^2\leq 2M$,
$\Rightarrow |f(x)|\leq\sqrt{2M}$.
Thus,$\sqrt{2M}\geq M\Rightarrow M\leq 2(2)$
From (1),(2) we have :$M=2$ then $f(x)\leq 2,\forall x\in \mathbb{R},x\neq 0$
Let $x=\frac{1}{2}:f(\frac{1}{2}+4)=f(\frac{1}{2})+f(0)^2\Rightarrow 4=f(\frac{9}{2})-f(\frac{1}{2})$
Since $-2\leq f(\frac{9}{2}),f(2)\leq 2\Rightarrow f(\frac{9}{2})=2,f(2)=-2$
Let $x=2:f(2+\frac{1}{4})=f(2)+f(\frac{1}{2})^2=2+(-2)^2=6>2$,a contradiction.So we can conclude that there no exist a bounded real function $f$


limit
integration
induction
geometric series
calculus
calculus computations

1.  Let $a_{1}$=1 and define the sequence recursively by $a_{n+1}=\sqrt{a_{1}+...+a_{n}}$.  Find $\lim_{x\to\infty}\frac{a_{n}}{n}$.

2.  Find $S = \frac{1^{2}}{2^{1}}+\frac{2^{2}}{2^{2}}+\frac{3^{2}}{2^{3}}+\frac{4^{2}}{2^{4}}+\frac{5^{2}}{2^{5}}+...$

3.  Let $f$ be infinitely differentiable and suppose that \[f(1/n) = \frac{n^{2}}{n^{2}+1}\] for all positive integers n.  Find $f^{(4)}(0)$.

4.  Find the value of $\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$.

[hide="2."]
Consider $2S-S$ and combining terms with equal denominators. With the new expression for $S$, do the same thing. The result is a geometric series and a constant.
[/hide]

[hide="4."]
$\lim_{n \rightarrow \infty}\sum_{k=n+1}^{2n}\frac{1}{k}= \int_{n}^{2n}\frac{1}{x}dx$[/hide]

1. It's easy to prove by induction that $a_{n}\le\frac{n+1}2.$    (1)
Here's the rest of sol.
It's easy to prove by induction that :1)$a_{n}\ge n/4$, thus $a_{n}\to\infty$;
2) $a_{n+1}^{2}=a_{n}^{2}+a_{n},a_{n+1}=a_{n}\sqrt{1+1/{a_{n}}}$.
Let's take $\epsilon>0$. Then for all n sufficiently large
$a_{n+1}>a_{n}(1+\frac1{(2+\epsilon)a_{n}})=a_{n}+\frac1{2+\epsilon}$, hence
$a_{n}>\frac n{2+\epsilon}(1+o(1))$. Since $\epsilon$ is arbitrary, we have \[\liminf_{n\to\infty}\frac{a_{n}}n\ge1/2.\] Now look at (1).
[b]1/2 [/b]is the answer.

sorry RIP but could u please further explain your proof, i dont quite follow.

how you found the two different expressions for $f\left(\frac{1}{n}\right)$

The 1st one is because f is infinitely differentiable.
The 2nd one is because $f(1/n)=(1+\frac1{n^{2}})^{-1}.$


divide {$1,2,...,n$} into $A,B,C$,and $|A|,|B|,|C|>\frac{n}{4}$,show that exist $a$ in $A$,$b$ in $B$,$c$ in $C$,such that the sum of the two of $a,b,c$ is equal to the other.

shencaili,
At this time, I can't solve this problem. But I know similary problem, they are http://www.mathlinks.ro/Forum/viewtopic.php?p=167295#p167295
 or

 http://www.mathlinks.ro/Forum/viewtopic.php?highlight=contest&t=3781

N.T. Tuan, the second link you posted contains a full solution of this question.

Thank you.
Maybe it's really hard,much harder than I thought.


how do you put a repeat sign over multiple numbers..

e.g. 0.1234[b]1234[/b].....

How about \overline:

0.\overline{1234} =

$0.\overline{1234}$

In China, it is like this : $0.56718718718\cdots = 0.56\dot71\dot8$

In Romania it's $0,12(34) = 0.1234343434...$ (notice the comma instead of the dot to symbolize the decimal and integer part separation).

Wow, i didn't know there was variation in things like that.  Do they teach the $.\overline{78}$ notation in school?  Which way do you prefer?  Which way do most people prefer in your country?


Hello Everyone.

I seam to be in a bit of a predicament. I am currently finishing my 3rd last year at high school, ie from the start of next year i have that year and the next until i officially finish high school. Through acceleration and self teaching i have learnt all the mathematics in high school. Due to a flaw in my exam methodology, in my recent Olympiad i didnt do too well (eg i lost about 7 marks in silly calculation errors or failing to read the question exactly, as well i lost 5 points for not putting down my proof for the last question which was right... i have [b]no[/b] idea why i didn't...).

I really would like to make it to the IMO, and i think that it is an achievable goal whence i improve on my exam methodology. Further more there is a high chance i will make the 2011 IPhO. But i am between a rock and a hard place because their is also the possibility of me pursuing University study in pure mathematics for next year. What should i do??? Out of curiosity, just say i talked on a regular basis with a university professor whom was giving me work to do and helping me out, would that make me not able to participate in the IMO/IPhO? Should i continue with mathematics competitions or start university mathematics "officially"?

I would really appreciate all comments =)

[quote="kurt.math"]i talked on a regular basis with a university professor whom was giving me work to do and helping me out, would that make me not able to participate in the IMO/IPhO? [/quote]
As far as I know, neither talking to a university professor on a regular basis, nor doing mathematical research in high school disqualifies you from the participation in the IMO as long as you are officially a high schooler. Quite a few people managed to attend university lectures and even to write their first research paper(s) when in high school and to get gold on the IMO simultaneously. So these activities absolutely do not need to be mutually exclusive. As long as you like them both and have enough time, I do not see why you should sacrifice any of them for the sake of the other. After all, Terence Tao apparently didn't and he is not the worst role model you can think of :P.

Is it legal to go to two international olympiads in the same year?

Yes, if you can find and qualify for two that don't have time conflicts or other rules imposed by your country. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=203128]here[/url].

[quote="Kent Merryfield"]I am personally aware of U.S. students who were enrolled in university mathematics courses - in some cases, graduate level mathematics courses - in the same year in which they competed at the IMO. That's all perfectly legal, as long as they were still officially enrolled in high school, which they were.[/quote]

Could you define "enrolled"? Does this mean that they were "formally enrolled"? Did they get credit for doing the course?


Let $ X,Y$ be two independent random variables which are both distributed $ \mathrm{Pois}(\lambda)$. Find $ P(X>2Y)$.

I tried tackling this problem with different approaches, but I always end up with series whose value I don't know. Any ideas?

I think that there is not many different ways:
$ P(X > 2 Y) \equal{} \sum_{y\equal{}0}^{\infty} \sum_{x \equal{} 2y\plus{}1}^{\infty} P(X\equal{}x) P(Y\equal{}y)$
and you plug the right value of $ P(X\equal{}x)$ and $ P(Y\equal{}y)$. I do not really know how you would solve this exercice without summing series

That's what I did, but as said, the resulting series are difficult to solve. If anyone knows of some way of finding these sums, I'd be happy to hear.


I've heard that there was a russian mathematician who prove that there exist formula which generates prime numbers. Do you know something about this?

You can write down lots of formulas that produce only primes. But there is no usefull one (meaning: it gives an effective way to find arbitrary large prime numbers). Take a look at http://en.wikipedia.org/wiki/Formula_for_primes for some.


x,y,z是实数,且x(4-x)=y,y(4-y)=z,z(4-z)=x;x,y,z不全相等；
求x+y+z的值
 :pilot:

是的,在重庆.
你呢?
我看你汉语也不差,不太像个美国人呀!
还有___非常感谢你的解答.
 :)

不用谢. 我是在北京出生. 在那里上到小学五年级然后来美国的. 所以我还记得很多中文.

你看过那道不等式了吗？
你能再帮我解决一下它吗？
谢谢了


If a motorist increases her speed by 8 miles per hour, she will cover a distance of 120 miles in 30 fewer minutes. What is her original rate of speed in miles per hour?

(This is a #3 on a Target for some year...)

[quote]There's gota be some quick clever way of doing this problem.[/quote]
[hide]I think its somewhere between 30 and 40 but i can't get the number exactly.  Maybe there's some formula you could use for this?[/hide]

[quote="kidwithshirt"]good one... i dun know.....

I m checking some algebra stuff on polynomial..[/quote]

Lol!  Good luck with that!  :D hehe

[quote="kidwithshirt"]ahh! my equation doesnt really tell anything.... thats very pathetic[/quote]
maybe you just need a second equation to aide your first :D lol

[quote="kidwithshirt"]i dun know i just messed up somehow, maybe i m not right

a is the original speed and b is the original time

(a+8)(b-30)=120

........................................

wat next?

seems like there is no limit for the time that the motorist used....[/quote]

Actually the time is in hours (since you're driving in MPH), so it's 1/2, not 30....

Ok, so if your going 8 mph faster, your going to go a certain destance (120 mi) in 30 min faster than if u were going the original speed.  so, let x represent the original speed. so,........god this is way harder then it seems.  u no what, screw it i give up.  abslugger34 has it anyway, so ya.

[quote="abslugger34"]solution here[hide]
x=# of mph (original)
y=# of hours (original)

y=120/x
(x+8)(120/x-1/2)
120-x/2-4+960/x=120
960/x-x/2=4
(1920-x 2 )/(2x)=4
1920-x 2 =8x
x 2 +8x-1920=0
(x-40)(x+48)=0
x has to be positive, so
x=40
[b]40 mph [/b] answer[/hide]

complicated, but it works if you're good at algebra, like me  :lol:[/quote]

If he's right, then i was close!  I told you it had to be 30-40 mph! :P

I think he is absolutely right.... lol :D


analytic geometry
graphing lines
slope
function

[quote="bryantacademy"]Is there a formula to find the points on a Cartesian system that, when the points are found and substituted, I can get m in the formula y=mx + b? B is the y intercept.

Basically, I am asking if there is a formula to pinpoint the exact points on a graph that would be used with the slope-intercept method in finding m.

I can do this kind of problem, and after three or four tries of wrong points, get it right.[/quote]

Hmm, I'm not totally understanding your question.
If you have a line, any 2 points on that line can be used to find the slope...
$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$, I believe...

[quote="bryantacademy"]Basically, I am asking if there is a formula to pinpoint the exact points on a graph that would be used with the slope-intercept method in finding m. [/quote]
Well, as long as it's a linear function, you can choose any 2 points to find the slope. There are no "wrong" or "right" points to choose.

It doesn't matter what points you use to find the slope, however the most commonly used points and the easiest to find points are the $x$ and $y$ intercepts.

To find the $y$ intercept, set $x$ to 0.

To find the $x$ intercept, set $y$ to 0.

Hope this helps.  :)

if you're trying to find the x and y interecepts just like it was mentioned above set x=0 and/or y=0. The y-intercept formula is (y2-y1)/(x2-x1). Hope this helps

[quote="bryantacademy"]I know that I was talking smart, but really I don't understand you guys. What I need to know is if there is a formula to find the points to substitute into m in the equation y=mx+b. B is the y-intercept. But finding m is trickier. My textbook, Saxon Algebra I (2nd edition), says that the formula for finding m is

|Slope| = |Rise| over |Run|

or

|m| = |rise| over |run|

This is confusing to me because a: I do not know what the rise and the run are, and b: it seems that there are infinite number of possibilities for those two points to be set to to where they can be subbed to get m.

This seems like this should be easy, so I really want to conquer it.  :?:  :blush:  :?:  :?:[/quote]
Well, that "formula" (about slope = rise/run) isn't really a formula; it's more like a way to help you understand what slope means. To actually calculate the slope, you pick [i]any[/i] two points on the line. Divide the difference of their y-coordinates (which is the rise) by the difference of their x-coordinates (which is the run). So then if there are two points $(a,b)$ and $(c,d)$ on a line, the slope of the line is
$m=\frac{d-b}{c-a}$.
For example, for the line $y=x+1$, you can pick any two points on that line, say $(1,2)$ and $(3,4)$, and the slope is
$m=\frac{4-2}{3-1}=\frac{2}{2}=1$.
If you picked two other points on the line, say $(7,8)$ and $(5,6)$, you would find that the slope is
$m=\frac{6-8}{5-7}=\frac{-2}{-2}=1$.
In other words, no matter which two points on the line that you pick, the slope between them is always the same.

what they mean by rise over run is that the run is the x-value and the rise is the y-value. Hope this helps

[quote="bryantacademy"]b: it seems that there are infinite number of possibilities for those two points to be set to to where they can be subbed to get m.[/quote]

Absolutely correct. Rise over run isn't a formula though.
Just select any 2 points on the line you want, then you can find m.

If the points are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, then

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$


search
superior algebra
superior algebra unsolved

Let $R$ be a ring s.t $\forall x \in R$ $x^{4}=x$ prove that $R$ is commutative.
(Actually i know there is a general theorem which says it's true for all $n\in \mathbb{N}$ but i am looking for an easier proof for this one.)

This is the easiest way: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=52344333&t=57581  :ninja:


Hello,

Prove that :    $ \displaystyle 8 \prec \sum\limits_1^{48} {\frac{1}{{\sqrt {2n \plus{} 1} }}}  \prec 10$

Comparison of the area of a box for $ y\equal{}\frac{1}{\sqrt{2x\plus{}1}}$.

[quote="Bugi"][quote="drase"]Hello,

Prove that :    $ \displaystyle 8 \prec \sum\limits_1^{48} {\frac {1}{{\sqrt {2n \plus{} 1} }}} \prec 10$[/quote]

Wait, what does $ \prec$ mean? Should it be $ <$?[/quote]

yes $ <$


conics
parabola
geometry proposed
geometry

Given is a triangle $ \triangle DBC$ in which $ DB \equal{} DC$. Let $ A$ be variable point in the interior of triangle $ \triangle DBC$. The perpendiculars to $ BC$ trough the mid-points of $ AB$ and $ AC$ meet $ DB$ and $ DC$ at $ P$ and $ Q$, respectively.
Find the locus of $ A$ for which $ P$, $ A$ and $ Q$ are collinear.

It is a parabola passing through B,C and the midpoint of the altitude from D to BC, with the altitude as major axis.


logarithms
quadratics
number theory unsolved
number theory

$\alpha$ is the least primitive root of p(is prime) then $\alpha\geq c.logp$

This conflicts with Artin's conjecture (saying that every nonsqare is primitive root $\infty$ often). You mean $\leq$ ¿

I think problem is true?but I know it is very famous theorem. People , do you know?

Well, it's true for $c=0$ ;)

But the statement ($c>0$ assumed) is not true for all primes $p$ as said above (Artin's conjecture even says that $2$ is primitive root $\infty$ often, clearly contradicting your original statement).

The correct statement in fact is:

There are $\infty$ many primes $p$ such that the least primitive root $\alpha_{p}\mod p$ is $>c \cdot \log p$, $c>0$ some constant.

Claim: For all $n$ is integer then there exits $p$ is prime such that the least primitive root $mod$ $p > n$ and can you prove that your correct statement? : lol:
$Certain, formerly , your father is a thief , He steal stars in the sky and put them in your eyes$

Thats much easier: take a prime $p$ such that $2,3,...,n$ are quadratic residues $\mod n$ (for this it suffices that all primes $q\leq n$ are quadratic residues). For example, take $p \equiv 1 \mod 8n!$, or $\equiv 1 \mod 8P$, where $P$ is the product of the primes $\leq n$.


geometry
3D geometry
inequalities
function
rearrangement inequality
inequalities proposed

[quote="far-reaching"]x,y,z - positive.
$ x^5 \plus{} y^5 \plus{} z^5 \minus{} x^3z^2 \minus{} y^3x^2 \minus{} z^3y^2\ge 0$[/quote]
Yes, this inequality is true  :) .You can use AM-GM
$ x^5\plus{}x^5\plus{}x^5\plus{}z^5\plus{}z^5 \ge 5x^3z^2$
$ y^5\plus{}y^5\plus{}y^5\plus{}x^5\plus{}x^5 \ge 5y^3x^2$
$ z^5\plus{}z^5\plus{}z^5\plus{}y^5\plus{}y^5 \ge 5z^3y^2$
Add 3 inequlities we have Q.E.D :)

[quote="peine"]Why don't use rerangemnt inequality?[/quote]
What do you mean?

Then is this right $ x^{2n\plus{}1}\plus{}y^{2n\plus{}1}\plus{}z^{2n\plus{}1}\minus{}x^{n\plus{}1}y^n\minus{}y^{n\plus{}1}z^n\minus{}z^{n\plus{}1}x^n\ge 0$ for some $ n\in \mathbb{N}$ :blush:

[quote="Mathias_DK"][quote="far-reaching"]Then is this right $ x^{2n + 1} + y^{2n + 1} + z^{2n + 1} - x^{n + 1}y^n - y^{n + 1}z^n - z^{n + 1}x^n\ge 0$ for some $ n\in \mathbb{N}$ :blush:[/quote]
Yes it is :)
Consider the two sequences $ (x^{n + 1}, y^{n + 1}, z^{n + 1})$ and $ (x^n, y^n, z^n)$. They are obviosly similarly sorted when $ n \in \mathbb{N}$, so according to the rearrangement inequality we have the desired :)[/quote]

that was the same argument i gave in 2nd post.
[quote="far-reaching"]no , xyz are arbitrary numbers[/quote]
 why do you think its false.?

[quote="ith_power"]that was the same argument i gave in 2nd post.[/quote]
Almost. But you can't assume $ x \ge y \ge z$.... (That's not necessary to use the rearrangement inequality, but assuming $ x \ge y \ge z$ makes the solution invalid!) Noticing that the functions $ x \to x^n$ and $ x \to x^{n\plus{}1}$, both are increasing, is what kills the problem  :!:

[quote="geniusbliss"][quote="ith_power"][quote="far-reaching"]x,y,z - positive.
$ x^5 \plus{} y^5 \plus{} z^5 \minus{} x^3z^2 \minus{} y^3z^2 \minus{} z^3x^2\ge 0$[/quote]
I assume it is as you can write $ x \ge y \ge z$ then same for cubes and squares then...
$ \sum x^5 \ge \sum x^3z^2$[/quote]
well ith_power even if u do assume $ x \ge y \ge z$
$ z^5 \ge z^3x^2$ is not true![/quote]

Thats not the point. Its not term by term valid inequality. in general by rearrangement inequality [from the two series, one of squares and another of cubes], then [b]sum on lhs[/b] is greater than [b]sum on rhs.[/b]


What ordered pair of positive integers (m, n ) satisfies
 7m + 12n = 43?

You know that 43 -7m needs to be postive and a multiple of 12.
43-7m needs to be even ( odd -odd = even)
So you should only try m=5,3,1.
m=1 works

m=1, n=3


εστω  $ x,y,z\geq 0$ οι οποιοι ικανοποιουν την χαρακτηριστηκη ιδιοτητα της μοναδιαιας σφαιρας δηλ.
$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$
νδο: $ \frac {x}{1 \minus{} x^2} \plus{} \frac {y}{1 \minus{} y^2} \plus{} \frac {z}{1 \minus{} z^2}\geq \frac {3\sqrt {3}}{2}$

Μια αλλη πρόχειρη ιδέα να χρησιμοποιήσουμε τριγωνομετρία πχ x=tana όμως δεν το έχω κοιτάξει ακόμη

[quote="Durandal"][quote="DIMITRIS T"]εστω  $ x,y,z\geqq$ οι οποιοι ικανοποιουν την χαρακτηριστηκη ιδιοτητα της μοναδιαιας σφαιρας δηλ.
$ x^2 + y^2 + z^2 = 1$
νδο: $ \frac {x}{1 - x^2} + \frac {y}{1 - y^2} + \frac {z}{1 - z^2}\geq \frac {3\sqrt {3}}{2}$[/quote]

Ένα [hide="hint"]Από σφαιρική συμμετρία και εναλλαγή μεταβλητών (ή και Lagrange), το ακρότατο της $ \phi(x,y,z) = \frac {x}{1 - x^2} + \cdots$ πάνω στη σφαίρα εμφανίζεται όταν $ x = y = z$. Εύκολα μπορεί να δειχθεί ότι αυτό είναι ελάχιστο, οπότε προκύπτει το ζητούμενο.[/hide]
H άσκηση γενικεύεται για $ n$ μεταβλητές, αρκεί η RHS να γίνει $ \frac {n\sqrt {n}}{n - 1}$.

Cheerio,

Durandal 1707[/quote]
το ξεσκισες  :P  (καλα εκανες!)

παντως η ασκηση μπορει να λυθει με πολυ πιο elementary τροπο :wink:

[quote="DIMITRIS T"]εστω  $ x,y,z\geqq$ [u]μη αρνητικοι αριθμοι[/u] οι οποιοι ικανοποιουν την χαρακτηριστηκη ιδιοτητα της μοναδιαιας σφαιρας δηλ.
$ x^2 + y^2 + z^2 = 1$
νδο: $ \frac {x}{1 - x^2} + \frac {y}{1 - y^2} + \frac {z}{1 - z^2}\geq \frac {3\sqrt {3}}{2}$[/quote]

να μια απλη λυση:κατ αρχην πρεπει να δουλεψουμε στο καθε κλασμα του πρωτου μελους χωριστα και να εκμεταλλευτουμε την ισοτητα που μας δινει η ασκηση...
οποτε αρκει να δειξουμε οτι:

$ \frac {x}{1 - x^2} \geq \frac {3\sqrt {3}}{2}x^2$

πραγματι αυτη η ανισοτητα ισχυει...και ειναι ισοδυναμη με την
$ \frac {2}{3\sqrt {3}} \geq x - x^3 \Leftrightarrow \frac {2}{3\sqrt {3}} + x^3 \geq x$

που ισχυει διοτι $ LHS = x^3 + \frac {1}{3 \sqrt {3}} + \frac {1}{3 \sqrt {3}}\geq 3\sqrt[3]{x^3\frac {1}{3^2\sqrt {9}}} = x$ απο AM-GM
QED :lol:


A positive integer $n$ has exactly 12 positive divisors $\displaystyle 1=d_{1}<d_{2}<...<d_{12}=n$.
Let $m=d_{4}-1$. We have $\displaystyle d_{m}=(d_{1}+d_{2}+d_{4})d_{8}$.
Find $n$

Don't you just love russian problems? This one is so sweet ! It is clear that we must have $ d_1+d_2+d_4\leq d_5 $. But $ d_1+d_2+d_4|d_m|n $ and also $ d_1+d_2+d_4>d_4 $. Thus $ d_1+d_2+d_4=d_5 $ and also $ d_m=n=d_{12} $. Thus, $ d_4=13$. And so $ d_5=14+d_2 $. But $ d_2 $ is the smallest prime factor of n and since $ d_4=13 $ we have $ d_2=2,3,5,7,11 $. Also, n must have at most 3 prime factors (he has 12 divisors!). If $ d_2=2$ then $ d_5=16 $ and thus 4 and 8 are two divisors smaller than $ d_4 $ and greater than $ d_2$, false. The same discussion for the other cases. The most interesting is when $ d_2=3$ . In this case, since n has 12 divisors and since it is a multiple of $ 3\cdot 13\cdot 17$, we must have $ n=9\cdot 13\cdot 17$ or $ n=3\cdot 169\17$ or $ 3\cdot 13\cdot 289$. Only the first case is a solution, namely 1989.


how is this done? what equation do i use?

The heat capacities of aluminum and water are 0.902 J g-1 K-1 and 4.184 J g-1 K-1, respectively. What is the final temperature of the aluminum-water mixture if a 175.0-g sample of aluminum, initially at 85.0oC, is added to 50.0 g of water, initially at 20.0oC

Make MC*delta T equations for both water and aluminum, letting x or another variable be the temperature change; then set them equal to each other (because heat gained = heat lost), and solve for X

would it be like this

qwater= 50gH20*4.184Jg-1k-1* 20 degrees?

qaluminum= 175g*0.902Jg-1k-1*85 degrees?

or does change in Temperature = 65 degrees?
Please help?

qwater= 50gH20*4.184Jg-1k-1*(x- 20) degrees

qaluminum= 175g*0.902Jg-1k-1*(85-x) degrees

I explained it wrong the first time; x represents the final temperature,  :oops:


Prove that the polynomial    $P(x)=2x^8+3x^6-5x^3-4x+6$   has no real roots

[hide]$\sum_{cyc}a = 0$

$\sum_{cyc}ab = \frac{3}{2}$

$(\sum_{cyc}a)^2 = \sum_{cyc}a^2 + 2\sum_{cyc}ab = 0$

$\implies \sum_{cyc}a^2 = \frac{-3}$

this proves that they cannot all be real, but how do you prove that they are all not real?[/hide]

It suffices to show that $P(x) > 0$ for all real values of $x$.
$P(x)>0 \Leftrightarrow 2x^8+3x^2+6>5x^3+4x$---(*)
Clearly, the inequality is true for all $x\leq 0$.

For $0<x<1$,
$2x^8+3x^6>5(x^8\cdot x^6)^{1/5}=5x^{14/5}>5x^3$ and $6>4x$
Adding these two inequlaities, we have (*).

It remains to prove that (*) is true for $x\geq 1$
Consider $P'(x)=16x^7+18x^5-15x^2-4=x^7+15(x^7-x^2)+14x^5+4(x^5-1)>0$ for all $x>1$.
Therefore, $P(x)$ is strictly increasing for $x\geq 1$.
$P(x)\geq P(1)=2>0$ for all $x\geq 1$.

In conclusion, $P(x)>0$ for all real vaues of $x$.
The problem is solved.

[quote="silouan"][quote="luimichael"]
For $0<x<1$,
$2x^8+3x^6>5(x^8\cdot x^6)^{1/5}$[/quote]

You are wrong here. You don't use the AM-GM correctly.[/quote]

Thanks for pointing out the mistake.
How about this?
$2(x^8-x^3)+3(x^6-x)+6>3x^3+x$ for $0<x<1$.

[quote="luimichael"]
How about this?
$2(x^8-x^3)+3(x^6-x)+6>3x^3+x$ for $0<x<1$.[/quote]

How to prove this ?? :huh:

Theorectically, we can construct the Sturm sequence and apply Sturm's separation theorem to the interval (0,1).
But it shouldn't be the method for Pre-Olympic question, right?  :(


[quote="Agrippina"]Is there a way to search for all unread posts in a particular subforum (eg Olympiad Section)?  If not, could it be added?

I also wish there was a way to find all members from a particular country, and a page with statistics on member demographics including country of origin.[/quote]The second one is in progress (a new version of the site is due soon). I'm not sure if I'll implement the statistics though, or I might do it only for registred members anyway to save server power :)

As for the first, go to the portal page ( http://www.mathlinks.ro/Forum/portal.php ) and click on the [url=http://www.mathlinks.ro/Forum/search.php?search_id=newposts]View posts since last visit[/url]. However it won't show you only posts from Olympiad Section, but I might implement this in the future (thanks for the tip).

Ooh!  Being clever, I just worked out a way to do essentially the same thing:  use the search function for the string " ", limiting the forum to "Olympiad Section" and the time to "last 1 day".  But it would still be cool to have a way to filter out the ones you've read.

[quote="Agrippina"][quote="Valentin Vornicu"]
As for the first, go to the portal page ( http://www.mathlinks.ro/Forum/portal.php ) and click on the [url=http://www.mathlinks.ro/Forum/search.php?search_id=newposts]View posts since last visit[/url]. However it won't show you only posts from Olympiad Section, but I might implement this in the future (thanks for the tip).[/quote]
Hope you do!  Anyway though, you're doing a great job as it is.  :lol:[/quote]Thanks  :blush:  :P


inequalities
function
trigonometry
inequalities unsolved

A,B,C are angles of a triangle

Proof:

sinA/2 + sinB/2 + sinC/2 <= 3/2

(tanA/2)^2 + (tanB/2)^2 + (tanC/2)^2 >= 1

Please use inequality of A-G-M

hello, the function $ \sin(x/2)$ is in the interval $ [0,\pi]$ concave and by the Jensen inequality we get
$ \frac{\sin(\alpha/2)\plus{}\sin(\beta/2)\plus{}\sin(\gamma/2)}{3}\le \sin(\frac{\alpha\plus{}\beta\plus{}\gamma}{6})\equal{}1/2$
from here we get
$ \sin(\alpha/2)\plus{}\sin(\beta/2)\plus{}\sin(\gamma/2)\le3/2$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the function $ \sin(x/2)$ is in the interval $ [0,\pi]$ concave and by the Jensen inequality we get
$ \frac {\sin(\alpha/2) \plus{} \sin(\beta/2) \plus{} \sin(\gamma/2)}{3}\le \sin(\frac {\alpha \plus{} \beta \plus{} \gamma}{6}) \equal{} 1/2$
from here we get
$ \sin(\alpha/2) \plus{} \sin(\beta/2) \plus{} \sin(\gamma/2)\le3/2$
Sonnhard.[/quote]

Yes,thanks.But could you try an other way,by using A-G-M inequality.

frist one

let $ a \equal{} y \plus{} z,b \equal{} z \plus{} x,c \equal{} x \plus{} y,x,y,z>0$, that we have
$ \sin\frac {A}{2} \equal{} \sqrt {\frac {yz}{(z \plus{} x)(x \plus{} y)}}$, $ \sin\frac {B}{2} \equal{} \sqrt {\frac {zx}{(x \plus{} y)(y \plus{} z)}}$, $ \sin\frac {C}{2} \equal{} \sqrt {\frac {xy}{(y \plus{} z)(z \plus{} x)}}$.
so
\[ \sin\frac {A}{2} \plus{} \sin\frac {B}{2} \plus{} \sin\frac {C}{2} \le \frac {3}{2}\]\[ \Longleftrightarrow \sqrt {\frac {{xy}}{{\left( {y \plus{} z} \right)\left( {z \plus{} x} \right)}}} \plus{} \sqrt {\frac {{yz}}{{\left( {z \plus{} x} \right)\left( {x \plus{} y} \right)}}} \plus{} \sqrt {\frac {{zx}}{{\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)}}} \le \frac {3}{2},\]by AM-GM, we have\[ \sqrt {\frac {{xy}}{{\left( {y \plus{} z} \right)\left( {z \plus{} x} \right)}}} \plus{} \sqrt {\frac {{yz}}{{\left( {z \plus{} x} \right)\left( {x \plus{} y} \right)}}} \plus{} \sqrt {\frac {{zx}}{{\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)}}}\]\[ \le \frac {1}{2}\left( {\frac {y}{{y \plus{} z}} \plus{} \frac {x}{{z \plus{} x}}} \right) \plus{} \frac {1}{2}\left( {\frac {z}{{z \plus{} x}} \plus{} \frac {y}{{x \plus{} y}}} \right) \plus{} \frac {1}{2}\left( {\frac {x}{{x \plus{} y}} \plus{} \frac {z}{{y \plus{} z}}} \right)\]\[ \equal{} \frac {3}{2},\]
so...

second one
similarly, we have $ \tan^2\frac {A}{2} \equal{} \frac {yz}{x(x \plus{} y \plus{} z)}$ and so on, so
\[ \tan^2\frac {A}{2} \plus{} \tan^2\frac {B}{2} \plus{} \tan^2\frac {C}{2}\ge 1\]
\[ \Longleftrightarrow x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \ge xyz(x \plus{} y \plus{} z)\]
\[ \Longleftrightarrow \frac {1}{2}\sum{z^2(x \minus{} y)^2} \ge 0.\]
so...

BTW, the stronger inequality——Garfunkel-Bankoff inequality:
\[ \tan^2\frac{A}{2}\plus{}\tan^2\frac{B}{2}\plus{}\tan^2\frac{C}{2}\ge2\minus{}8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}.\]

First we prove in any triangle $ cosA \plus{}cosB \plus{}cosC$  $ \leq$ $ \frac{3}{2}$ by AM-GM inequality.

$ cosA \plus{}cosB \plus{}cosC$ =     1+ $ 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$=      $ 1\plus{} 4\frac{(s\minus{}a)(s\minus{}b)(s\minus{}c)}{abc}$ where 2s =a+b+c and $ \sin\frac{A}{2}$ =$ \sqrt\frac{(s\minus{}b)(s\minus{}c)}{bc}$ etc
By the AM-GM inequality we have the three inequalities:

a=(s-b)+(s-c) $ \geq$ $ 2\sqrt{(s\minus{}b)(s\minus{}c)}$ equality when b=c
b=(s-c) +(s-a) $ \geq$ $ 2\sqrt{(s\minus{}a)(s\minus{}c)}$  equality when c=a
c=(s-a) +(s-b) $ \geq$ $ 2\sqrt{(s\minus{}a)(s\minus{}b)}$ equality when a=b and multiplying them together we get

$ abc$ $ \geq$ $ 8(s\minus{}a)(s\minus{}b)(s\minus{}c)$ with equality when a=b=c. Therefore $ 4\frac{(s\minus{}a)(s\minus{}b)(s\minus{}c)}{abc}$ $ \leq$ $ \frac{1}{2}$ and  thus cosA +cosB +cosC $ \leq$ $ \frac{3}{2}$ with equality when A=B=C

Now consider a triangle PQR such that P=$ \frac{\pi\minus{}A}{2}$ Q=$ \frac{\pi\minus{}B}{2}$ R =$ \frac{\pi\minus{}C}{2}$

Then cosP +cosQ +cosR =$ sin\frac{A}{2} \plus{}sin\frac{B}{2} \plus{}sin\frac{C}{2}$ $ \leq$ $ \frac{3}{2}$

For the second question $ \tan^2\frac{A}{2}$ +$ \tan^2\frac{B}{2}$ $ \geq$ $ 2\tan\frac{A}{2}\tan\frac{B}{2}$ and two similar inequalities.
Adding them together we have $ \sum\tan^2\frac{A}{2}$ =$ \sum\tan\frac{A}{2}\tan\frac{B}{2}$ =1

:lol:  Very nice! Learning

But, is there any other way?

To show that $ \sin\frac{A}{2}$ +$ \sin\frac{B}{2}$ +$ \sin\frac{C}{2}$ $ \leq$ $ \frac{3}{2}$ is equivalent to showing that $ \cos\frac{B\plus{}C}{2}$ +$ \sin\frac{B}{2}$ +$ \sin\frac{C}{2}$ $ \leq$ $ \frac{3}{2}$

Now $ \cos\frac{B\plus{}C}{2}$=$ \cos\frac{B}{2}\cos\frac{C}{2}$ - $ \sin\frac{B}{2}\sin\frac{C}{2}$ $ \leq$ $ \frac{\cos^2\frac{B}{2} \plus{}\cos^2\frac{C}{2}}{2}$  -  $ \sin\frac{B}{2}\sin\frac{C}{2}$ =$ \frac{(1\minus{}\sin^2\frac{B}{2}) \plus{}(1\minus{}\sin^2\frac{C}{2})}{2}$ - $ \sin\frac{B}{2}\sin\frac{C}{2}$

 = 1 -$ \frac{ (\sin\frac{B}{2} \plus{} \sin\frac{C}{2})^2}{2}$

Therefore $ \cos\frac{B\plus{}C}{2}$ +$ \sin\frac{B}{2}$ +$ \sin\frac{C}{2}$ $ \leq$ 1- $ \frac{x^2}{2}$ +x where x=$ \sin\frac{B}{2}$ +$ \sin\frac{C}{2}$ and therefore the inequality $ \sum\sin\frac{A}{2}$ $ \leq$ $ \frac{3}{2}$ will be proved if we can show

1- $ \frac{x^2}{2}$ +$ x$ $ \leq$ $ \frac{3}{2}$. But this is equivalent to $ (x\minus{}1)^2$ $ \geq$ 0 which is clearly true.

Equality occurs when x=1 ,that is when B=C =$ \frac{\pi}{3}$


Putnam
function
linear algebra
matrix
geometry
3D geometry
geometric transformation

Let $ f$ be a real-valued function on the plane such that for every square $ ABCD$ in the plane, $ f(A)\plus{}f(B)\plus{}f(C)\plus{}f(D)\equal{}0.$ Does it follow that $ f(P)\equal{}0$ for all points $ P$ in the plane?

This one probably came from
http://www.mathlinks.ro/Forum/viewtopic.php?p=327816#327816

This is my first time doing the Putnam.

Any way, when I am doing this problem, I thought it's pretty trivial because it doesn't specify anything about the length so I assume length 0 is a possibility.

However, my prof said, after the exam, that a degenerate square is not considered a square. So I don't know.....

Really, I think the Putnam people should really make themselves clear (10 pts gone b/c of them not being clear  :mad:  could've spent the whole 3 hours rather than just 20 minutes assuming that the side length can be 0)

Are you serious? You should've known you weren't going to get any points for degenerate squares.

Ah well, I knew this question was trivial but I would have never thought of picking a point and then drawing the squares around it or however it was supposed to be solved.  I broke it into 3 cases, with a base case in which i drew the x-axis so the square was bi-sected, which would satisfy the assumption, and the other 2 cases involved shifting the square up or down which would change the sum to either greater than or less than 0.  Probably a reach, and since it was so simple im sure it will be graded much more strictly which doesn't bode well for me haha!

To Elemennop:

well, this is my first time doing Putnam and really, how I am supposed to know this?   This is a competition and I always think that as long as I don't contradict the assumptions made in the problem, I can solve it how ever I want.

For any points $ a,b,c$ where $ b$ is the midpoint of $ a$ and $ c$, consider auxillary points $ d,e,h,g$ constructed as follows:

[img]http://rpi.edu/~naamay2/a1.jpg[/img]

From the image, we can see that

$ 1. f(a) \plus{} f(d) \plus{} f(b) \plus{} f(h) \equal{} 0$
$ 2. f(e) \plus{} f(d) \plus{} f(g) \plus{} f(h) \equal{} 0$
$ 3. f(e) \plus{} f(b) \plus{} f(g) \plus{} f(c) \equal{} 0$
(1) and (2) give us
$ f(a) \plus{} f(b) \equal{} f(e) \plus{} f(g)$
and so, from (3),
$ f(b) \plus{} f(c) \plus{} f(a) \plus{} f(b) \equal{} 0$
meaning that
$ \minus{}2f(b) \equal{} f(a) \plus{} f(c)$

So for any $ x,y$,
$ f((x\minus{}1,y\minus{}1)) \plus{} f((x\plus{}1,y\plus{}1)) \equal{} \minus{}2f((x,y))$ and $ f((x\plus{}1,y\minus{}1)) \plus{} f((x\minus{}1,y\plus{}1)) \equal{} \minus{}2f((x,y))$

Note that $ (x\minus{}1,y\minus{}1),(x\plus{}1,y\minus{}1),(x\plus{}1,y\plus{}1),(x\minus{}1,y\plus{}1)$ form a square, so
$ f((x\minus{}1,y\minus{}1)) \plus{} f((x\plus{}1,y\minus{}1)) \plus{} f((x\plus{}1,y\plus{}1)) \plus{} f((x\minus{}1,y\plus{}1)) \equal{} 0$
Thus, $ 0 \equal{} \minus{}2f((x,y)) \plus{} \minus{}2f((x,y)) \equal{} \minus{}4f((x,y))$

So $ f((x,y)) \equal{} 0$ for any $ x,y$

Well, from hindsight, I also have a solution that works for length side > 0

consider a midpoint on the square... call it p, where we can draw some cross intersecting at p that divides the square into 4 quadrants.

suppose quadrant A (top left) = a + b + p + c = 0.

quadrant B (bottom left) = c + p + d + e =0.

quadrant C (top right) = b + f + g + p = 0.

quadrant D (bottom right) = p + g + h + d = 0.

add them up, you find that 4p = 0 or p = 0.

$ h$ is the better label - better to fix the diagram than the computations below it.

Even if the hypothesis had been restricted to squares with sides parallel to the axes, the result will still follow - and I assume that many people have proofs that only use such squares.

[quote="yonatan"][quote="WelcomeGhosts"]I think in your diagram you labeled the point "f" instead of "h"?[/quote]

That I did. I thought I uploaded the fixed diagram, but apparently I forgot that I need to enter my password at the prompt.

Fixed now. Thanks for pointing it out.[/quote]

No problem, that is a nice solution.  I guess what I don't understand is why it is necessary to draw more than one square?  I would have never thought to draw more than one square, but drawing in the x-axis such that it bisects a square ABCD made more sense to me...so without posting my entire proof, I'm wondering if pursuing the x-axis idea was worth it or just completely incorrect?

[asy]size(120); pathpen = linewidth(1); real r = 0.1, r2 = 2/3*r; path s = (0,0)--(1-r,0)--(1-r,1-r)--(0,1-r)--cycle;
D(shift(r/3,r/3)*s);D(shift(-1+r2,r/3)*s);D(shift(r/3,-1+r2)*s);D(shift(-1+r2,-1+r2)*s); D((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,red+linewidth(1));D((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,blue+linewidth(1));
for(int i = -1; i < 2; ++i) for(int j = -1; j < 2; ++j) D((i,j),linewidth(7)); MP("P",(0,0),(2,2));[/asy] Summing $ f$ over the vertices of the black squares, then subtracting $ f$ over the vertices of the red square, then subtracting twice of $ f$ over the vertices of the blue square, gives $ 4f(P) \equal{} 0$, and thus $ f(P)$ is identically zero for all points $ P$.

[quote="azjps"][asy]size(150); pathpen = linewidth(1); real r = 0.1; path s = (0,0)--(1-r,0)--(1-r,1-r)--(0,1-r)--cycle;
D(shift(r/2,r/2)*s);D(shift(-1+r/2,r/2)*s);D(shift(r/2,-1+r/2)*s);D(shift(-1+r/2,-1+r/2)*s); D((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,red+linewidth(1));D((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,blue+linewidth(1));
for(int i = -1; i < 2; ++i) for(int j = -1; j < 2; ++j) D((i,j),linewidth(8)); MP("O",(0,0),(2,2));[/asy]

Summing $ f$ over the vertices of the squares inscribed in the black squares and subtracting $ f$ over the vertices of the red square, and subtracting twice of $ f$ over the vertices of the blue square, gives $ 4f(O) \equal{} 0$.[/quote]

this was my solution.  Once you found this picture it was very straightforward.

[quote="azjps"][asy]size(150); pathpen = linewidth(1); real r = 0.1; path s = (0,0)--(1-r,0)--(1-r,1-r)--(0,1-r)--cycle;
D(shift(r/2,r/2)*s);D(shift(-1+r/2,r/2)*s);D(shift(r/2,-1+r/2)*s);D(shift(-1+r/2,-1+r/2)*s); D((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,red+linewidth(1));D((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,blue+linewidth(1));
for(int i = -1; i < 2; ++i) for(int j = -1; j < 2; ++j) D((i,j),linewidth(8)); MP("P",(0,0),(2,2));[/asy]

Summing $ f$ over the vertices of the black squares, then subtracting $ f$ over the vertices of the red square, then subtracting twice of $ f$ over the vertices of the blue square, gives $ 4f(P) \equal{} 0$, and thus $ f(P)$ is identically zero for all points $ P$.[/quote]

Yes!
Without any algebra...I don't why I couldn't think of this during the comp.

Here's an argument which uses only squares with sides parallel to the axes.

Start with square $ ABCD.$ Partition it into four smaller squares. Form a sum in which you add the values of the function at the four vertices of the lower left and upper right smaller squares, and subtract the values at the four vertices of the upper left and lower right smaller squares. The values at all of the internal points cancel out, leaving us with $ f(A)\minus{}f(B)\plus{}f(C)\minus{}f(D)\equal{}0.$ Combine that with $ f(A)\plus{}f(B)\plus{}f(C)\plus{}f(D)\equal{}0$ to get that $ f(A)\plus{}f(C)\equal{}0$ and $ f(B)\plus{}f(D)\equal{}0.$ So two points on the same $ 45^{\circ}$ line have values that are negatives.

Now take another square; call it $ ABCD$ again. Let $ E$ be the point in the center. By the previous argument $ f(A)\equal{}\minus{}f(E)\equal{}f(B)\equal{}f(C)\equal{}f(D).$ So $ 4f(A)\equal{}0$ and $ f(A)\equal{}0.$

As to why I would rank this as the easiest problem on the exam: it's because it's hard to find an approach that doesn't work. (Well, maybe trying to write giant matrix equations won't be too productive.) Many different starting ideas can be made to work in the end.

I gave a nonstandard solution like this:

Consider an n by n grid of points, which has n^2 points in it. Taking squares along the grid (not taking diagonals) the grid contains n^3 / 3 - n^2 / 2 + n / 6 many squares. Thus for sufficiently large n there are more squares than points. Each square corresponds to a linear equation and the whole grid corresponds to a linear system. Thus at some large enough n we have more equations than variables in our system. I argued at length from the construction of the grid that any such finite grid had all linearly INDEPENDENT equations - that each square was a different statement. Thus the system becomes overdetermined and has only the trivial solution.

Strange, and it doesn't actually work. Overdetermined systems can have nontrivial solutions if there are too many linear dependence relations.

Suppose some statement ABCD (meaning f(A) + f(B) + f(C) + f(D) = 0 ) given by the square ABCD can be a linear combination of other statements. Then we need to add A, B, C, and D from other equations. Squares share 0, 1, 2, or 4 points - never 3. 4 is trivial and 0 is irrelevant. If we use squares that share 1 point, we must add together the four corners of ABCD from four surrounding squares. Then we have 12 unneeded points which we must subtract. Yet this is conceptually the counterpart to the problem of getting four points at the corners - trying to add one is just like trying to subtract three. Thus our addition and subtraction of squares must continue outward ad infinitum. Alternatively, if we choose squares that share 2 points, we get ABCD from two opposing sides. Then we have 4 extra points to subtract out, and our task becomes the counterpart of adding 4 points at the sides. We add and subtract laterally outwards ad infinitum.

Therefore in any finite grid all the statements are linearly independent. The solution, if inelegant, is valid.

You can't have it both ways. In an overdetermined system, the equations are necessarily linearly dependent. So, there are dependence relations you're missing. Where do they come from? We have squares of many sizes, and can retrace the same area several times. Here's one sample dependence relation: Take a $ 4\times 4$ grid, covering the corners of a $ 3\time 3$ region. The four $ 1\times 1$ squares in the corners cover each point once with their vertices. The four $ 2\times 2$ squares in the corners do as well; take the difference, and that's a linear combination of eight squares which is zero.

[quote="Kent Merryfield"]$ f(A) \plus{} f(C) \equal{} 0$ and $ f(B) \plus{} f(D) \equal{} 0.$ So two points on the same $ 45^{\circ}$ line have values that are negatives.[/quote]
You can take this as a lemma; for [b]any[/b] distinct points $ P, Q$ we have $ f(P) \plus{} f(Q) \equal{} 0$.  Then take three distinct points $ P, Q, R$.

I thought of a 3D analog of this problem.

Let f(P): R^3 --> R have the property that if A, B, C, D, E, F, G, and H are vertices of a cube then
f(A)+f(B)+f(C)+f(D)+f(E)+f(F)+f(G)+f(H)=0.

Prove that f(P)=0 for all points P in R^3.

This is doable but quite tricky compared to putnam problem A1.

I didnt actually take the putnam this year, but have glanced over the problems and was shocked that I got this one so quickly. However, in trying to figure out how different people approach these kinda things, and answer the question:
   "How would I have thought of that?"

I just wanted to ask the following question:
   "What observations did you make that got you to the end."
Mine went something like:
1. Lets take a point in the plane, probably one thats not pathological (whatever that means at this stage), but probably a point that is in the first quadrant that isnt the origin would work.
2. Whats the most natural way to make a square with the point? At this point my gut instinct was to reflect the point across the y-axes, the x-axes, and both the x and y-axes.
3. Your then left with a square, and one relation on the vertices of the square. This means that your going to need to get at least 3 more relations to say something about f(p). With a little thought its easy to see that the least fussy way to do this is to split the square into four congruent subsquares each with vertex on the origin.
4. We then have five relations, and its not immediate what to do with these, but with a goal in mind you could notice that when you add the relations of the four subsquares you get the following sum
\[ f((\minus{}x,y))\plus{}f((x,y))\plus{}f((x,\minus{}y))\plus{}f(\minus{}x,\minus{}y)\]
plus twice some other vertices and four times the origin. The relation from the first square we made by reflection kills off this first sum, and if you visually remove this vertices from the diagram you notice you're left with a square on the vertices (0,y),(0,-y),(x,0),(-x,0). This gives you another relation which lets you kill off the vertices which arent the origin, leaving you with
\[ 4f((0,0))\equal{}0\].
5. Now we were trying to prove something about f(p), and we proved something about f((0,0))=0, but its not that hard to see that you can "flip" the argument and start on p, extend out four equal and perpendicular lines and make a square surrounding p.

The other question I had was about a generalization of the problem, namely if we have a n-ary relation
\[ f(a_1)\plus{}\ldots\plus{}f(a_n)\equal{}0\] for each (regular?) n-gon in the plane does it necessarily follow that $ f(p)\equal{}0$ for each point in the plane?

Jamie: look at the link in Albanian Eagle's post #2 in this topic.

If f were an analytic function in the complex plane, you could use [url=http://mathworld.wolfram.com/GausssMean-ValueTheorem.html]Gauss's mean value theorem[/url] to show that the center of any square is zero by placing the vertices of the square in a circle.

I don't think you can assume the function is analytic though...

[quote="t0rajir0u"]$ f$ doesn't even have to be continuous.[/quote]

$ f$ is identically zero, but doesn't have to be continuous!  That tricky old $ f$ has some life yet!

What I mean is that you can't assume that $ f$ has any properties beyond the functional equation until you prove them from the functional equation.

Consider the same problem for an equilateral triangle (ABC) instead of a square (ABCD) such that f(A)+f(B)+f(C)=0. Does it follow f(X)=0 for all points X in the plane? A geometric solution should pop in front of eyes! :-)

Yes, that doesn't take long. Adjoin $ \triangle BCD$ to $ \triangle ABC.$ Conclude that since $ f(A)\plus{}f(B)\plus{}f(C)\equal{}f(B)\plus{}f(C)\plus{}f(D),$ then $ f(D)\equal{}f(A).$ But $ A$ and $ D$ could have been any two distinct points, so $ f$ is constant. But the only constant $ f$ can be is zero.

I actually forget what my reasoning was; I did this competition prior to taking MVC.  However, I do believe that I arrived at the correct result.  Seeing the geometry on the first page of this thread truly helps me to understand the logic behind this proof.

does it work for other polygons?

by drawing a regular hexagon, i have shown that "square" can be replaced by "equilateral triangle"

[quote="sexynsmartjenny"]I just want to add that this problem does not only hold for squares, but also any n-point convex polygon for n greater than 3.  I don't think the polygon even has to be regular.

And I just want to add, that since the first time I took Putnam in 2005, this porblem has been the easiest I have ever seen.[/quote]

how do you prove that it works for all n greater than (or equal to) 3?

and, not restricting the polygon to be regular actually weakens the statement.

[quote="JBL"][quote="Andy Loo"]and, not restricting the polygon to be regular actually weakens the statement.[/quote]  Jenny's claim is, "For any (fixed) polygon $P$, if the function $f$ satisfies $\sum_{s \in S} f(s) = 0$ for any set $S$ of points similar to the vertices of $P$ then $f \equiv 0$."  This is substantially stronger than the problem statement; I have not tried to prove or disprove it.[/quote]

OMG, if Joel cannot prove or disprove it, I must certainly have made a blunder and a fool of myself.

I don't really remember what I was thinking, since it was so long ago.  But it must have been this

let $N$ be a point in $R^2$, then draw a convex polygon with this point.  Draw as many vertex as you want, as long as it is convex.  Let's say $N$ vertices.  Label them $1,2,3,...,N$
then according to myself condition, given a $f$ s.t. $f(1)+f(2)+....+f(N)=0$
Well, draw a line between $N_1$ and $1$ and you get a polygon of $N-1$ vertices, so according to my condition,
$f(1)+f(2)+...+f(N-1)=0$  Therefore, $f(N)=0$

OK, I think that was what I did.  Now I await your laughter.

BTW, What is "similar" and what is this $\equiv$?

[quote="JBL"]Two polygons are "similar" if one can be gotten from the other by a combination of scaling, rotating and translating.  The symbol "$\equiv$" is being used to denote "is identically 0" i.e., is 0 for all inputs.

Ok, the problem I thought you were asking is much harder than the problem Andy Loo thought you were asking, and it turns out that Andy Loo was right.  As Andy Loo noted in his second post, allowing all polygons makes the proof basically immediate (as you've shown) and is a substantial weakening, not a generalization, of the given problem.

Also, for anyone else who is interested, this discussion appears to extend here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=549&t=417806[/quote]

Yes, I know.  You have to fix $n$.  I wasn't sure which one I meant either.

Nice functional geo.

Let $X$ be an arbitrary point in the plane. Construct four squares around it with side length $1$ like so:
[asy]
pair A,B,C,D,P,Q,R,S,X;
A=(0,2);
B=(2,2);
C=(2,0);
D=(0,0);
P=(1,2);
Q=(2,1);
R=(1,0);
S=(0,1);
X=(1,1);
dot(A);
dot(B);
dot(C);
dot(D);
dot(P);
dot(Q);
dot(R);
dot(S);
dot(X);
label(A,"A",NW);
label(P,"P",N);
label(B,"B",NE);
label(Q,"Q",E);
label(C,"C",SE);
label(R,"R",N*(-1)); // for some reason S doesn't work here
label(D,"D",SW);
label(S,"S",W);
label(X,"X",SW);
draw(A--B--C--D--A);
draw(P--Q--R--S--Q);
draw(S--P--R);
[/asy]
From this, we get:
\begin{align*}
f(A)+f(P)+f(X)+f(S)&=0,\\
f(P)+f(B)+f(Q)+f(X)&=0,\\
f(X)+f(Q)+f(C)+f(R)&=0,\\
f(S)+f(X)+f(R)+f(D)&=0,\\
f(A)+f(B)+f(C)+f(D)&=0,\\
f(P)+f(Q)+f(R)+f(S)&=0.\end{align*}
Adding the first four equations and using the last two yields $\boxed{f(X)=0}$ for all points $X$.

I apologize for not having a picture.

Let $E,F,G,H$ be midpoints of segments $AB,BC,CD,DA$ respectively. It follows that:
$$0=f(A)+f(B)+f(C)+f(D),$$
$$0=f(E)+f(F)+f(G)+f(H),$$
$$0=f(A)+f(E)+f(P)+f(H),$$
$$0=f(B)+f(F)+f(P)+f(E),$$
$$0=f(C)+f(G)+f(P)+f(F),$$
$$0=f(D)+f(H)+f(P)+f(G).$$
Add the last four, subtract the first once and the second twice to get $0=4(P)$, and the claim follows. $\square$

very nice problem!\\

diagram for the problem:\\

[asy]
pair A,B,C,D,P,Q,M,N,O;
A=(0,2);
B=(2,2);
C=(2,0);
D=(0,0);
P=(1,2);
Q=(2,1);
M=(1,0);
N=(0,1);
O=(1,1);
dot(A);
dot(B);
dot(C);
dot(D);
dot(P);
dot(Q);
dot(M);
dot(N);
dot(O);
label(A,"A",NW);
label(P,"P",N);
label(B,"B",NE);
label(Q,"Q",E);
label(C,"C",SE);
label(M,"M",N*(-1));
label(D,"D",SW);
label(N,"N",W);
label(O,"O",SW);
draw(A--B--C--D--A);
draw(P--Q--M--N--Q);
draw(N--P--M);
[/asy]

So consider $M,N,P,Q$ to be midpoints of sides $CD,AD,AB,BC$ respectively  and  $O$ to be the center of square. now $MNPQ$ also forms a square so we get:\\

$f(A)+f(B)+f(C)+f(D)=0$ and $f(M)+f(N)+f(P)+f(Q)=0$\\

now also we have that $MOQC , OPBQ, OPAN, OMDN$ to be also square so we get that:\\

$f(N)+f(P)=f(C)+f(O)$\\

$f(Q)+f(P)=f(D)+f(O)$\\

$f(Q)+f(M)=f(A)+f(O)$\\

$f(M)+f(N)=f(B)+f(O)$\\

which gives us $2(f(M)+f(N)+f(P)+f(Q))=f(A)+f(B)+f(C)+f(D)+4f(O)$\\

which gives $4f(O)=0\implies f(O)=0$ , now we can shift this point any where in the place and hence we get that $f(P)=0$ $\forall$ points $P$ in the plane. $\blacksquare$

Sketch:

Fundamentally, we can contradict by saying suppose there exists some point $p$ such that $f(p) \neq 0$. We can now construct a non-degenerate square where $p$ serves as the center. Then we can make $4$ equations, and see that $f(p) = 0$.

As mentioned there, this is genralized to n-gon by Romania TST 1996 problem: https://artofproblemsolving.com/community/c6h51933p327816

[quote=aansc1729]As mentioned there, this is genralized to n-gon by Romania TST 1996 problem: [url=https://artofproblemsolving.com/community/c6h51933p327816] see [/url] [/quote]


ratio geometry rectangle LaTeX quadratics SFFT algebra

I was doing the 1994 Nationals and I couldn't figure these problems out:

Sprint, Number 8:
Given the equation $ \frac {1}{m} \plus{} \frac {1}{n} \equal{} \frac {2}{15}$, where $ m$ and $ n$ are positive integers, find the value of $ m \plus{} n$ such that $ m > n$ and $ m$ is not a multiple of $ n$.

[hide="Answer"]
32
[/hide]

Target, Number 4:
There are positive integers $ k,n,$ and $ m$ such that $ \frac {19}{20} < \frac {1}{k} \plus{} \frac {1}{n} \plus{} \frac {1}{m} < 1$. What is the smallest possible value of $ k \plus{} n \plus{} m$?

[hide="Answer"]
12
[/hide]

Sprint, Number 9:
Use the pattern below to express $ 100^2$ in the form $ a^2 \plus{} b^2 \minus{} c^2$.
Then find $ a \plus{} b \plus{} c$.
\[ 12^2 \equal{} 8^2 \plus{} 9^2 \minus{} 1^2\]

\[ 14^2 \equal{} 10^2 \plus{} 10^2 \minus{} 2^2\]

\[ 16^2 \equal{} 12^2 \plus{} 11^2 \minus{} 3^2\]

\[ 18^2 \equal{} 14^2 \plus{} 12^2 \minus{} 4^2\]
[hide="Answer"]
194
[/hide]

Sprint, Number 10:
The ratio of the length of a rectange to its width is the same as that of the diagonal to the length. If the width is $ 2$, how many units are in the length of the diagonal?

[hide="Answer"]
1+$ \sqrt{5}$
[/hide]

As I'm still not finished trying to figure out how I got some problems wrong, I'll edit this post later to include some more problems.

(There are probably going to be a lot more because my score was 19/46; very embarrasing)

[hide="Sprint number 8"]
What I got (probably wrong too):

$ \frac1m \plus{} \frac1n \equal{} \frac {m \plus{} n}{mn} \equal{} \frac {2}{15}$

Thus, $ m \plus{} n \equal{} 2$. Although the two numbers aren't obvious, they do exist.
[/hide]

That could be true but it could be that $ m \plus{} n \equal{} 4$ and $ mn \equal{} 30$.
Anyway
[hide="solution for target"]
We want low numbers. So lets start with 2 and 3. 19/20= .95
$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4}$ Doesn't work

$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{5}$ Doesn't work

$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{6}$ Doesn't work

$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{7}$ Works. It equals about .97

So the answer is $ 2 \plus{} 3 \plus{} 7 \equal{} \boxed{12}$
[/hide]
[hide="sprint number 8"]
We know the unit fraction rule:

$ \frac {1}m \plus{} \frac {1}n \equal{} \frac {m \plus{} n}{mn} \equal{} \frac {2}{15}$

So we just build up $ \frac {2}{15}$ into $ \frac{32}{240}$. This is $ \frac {1}{12} \plus{} \frac {1}{20}$
So the answer is $ 12 \plus{} 20 \equal{} \boxed{32}$

[/hide]

[quote="PowerOfPi"][hide="Sprint number 8"]
$ \frac1m \plus{} \frac1n \equal{} \frac {m \plus{} n}{mn} \equal{} \frac {2}{15}$

Thus, $ m \plus{} n \equal{} 2$. Although the two numbers aren't obvious, they do exist.
[/hide][/quote]

[hide="correction"]Uh... m+n can't be 2 if they're positive integers, and they multiply to 15, as you imply.
Rather, m+n would be a multiple of 2 (call it 2x) and mn would be a multiple of 15 (15x, and x is the same in both parts).[/hide]

edit: beaten, darn.

So the solutions i have so far are:
[hide="solution for target number 4"]
We want low numbers. So lets start with 2 and 3. 19/20= .95
$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4}$ Doesn't work

$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{5}$ Doesn't work

$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{6}$ Doesn't work

$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{7}$ Works. It equals about .97

So the answer is $ 2 \plus{} 3 \plus{} 7 \equal{} \boxed{12}$
[/hide]
[hide="sprint number 8"]
We know the unit fraction rule:

$ \frac {1}m \plus{} \frac {1}n \equal{} \frac {m \plus{} n}{mn} \equal{} \frac {2}{15}$

So we just build up $ \frac {2}{15}$ into $ \frac {32}{240}$. This is $ \frac {1}{12} \plus{} \frac {1}{20}$
So the answer is $ 12 \plus{} 20 \equal{} \boxed{32}$

[/hide]
[hide="sprint number 9"]
So you use the pattern to find out that a is 96 by noticing that a is always the number minus 4. 100-4 = 96. You find out b is 53 because b increases by 1 every two. So 100-12 = 88. 88/2 = 44. 44 + 8= 53. And using that same formula you find out c is 45.
Add them together and you get $ \boxed{194}$
[/hide]

sorry i dont know latex
[hide="10"]
so like, we know that L/W=D/L
we can express D as sqrt L^2+W^2
so L/W=sqrt(L^2+W^2)/L
And they give that W=2
so L/2=sqrt(L^2+4)/L
Cross multiplying we get
L^2=2sqrt(L^2+4)
which can be simplified into
L^4-4L^2-16=0
let L^2=x
x^2-4x-16=0
solve as a quadratic, find L^2 and plug into
D=sqrt(L^2+W^2).[/hide]

[hide="sprint 8"] Multiply by $ 15mn$ to get $ 15n\plus{}15m\equal{}2mn$. This rearranges and factors as $ (2m\minus{}15)(n\minus{}\frac{15}{2})\equal{}\frac{225}{2}\implies(2m\minus{}15)(2n\minus{}15)\equal{}225$, where we multiplied by 2 to get rid of the fractions.

Since you know $ m>n$, $ 2m\minus{}15>2n\minus{}15$ so just try setting $ 2m\minus{}15$ equal to factors of $ 225$ greater than 15. We find $ 2m\minus{}15\equal{}25$ works, as $ (m, n)\equal{}(20, 12)$, so $ m\plus{}n\equal{}32$.[/hide]

[hide="Target #4"] Let $ k\le{n}\le{m}$. If $ k\equal{}3$, then clearly not all three of $ k, n, m$ can be $ 3$ as the sum is equal to $ 1$. Thus, at least one must be greater than $ 3$, which gives a largest possible sum of $ \frac{1}{3}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\equal{}\frac{2}{3}\plus{}\frac{1}{4}\equal{}\frac{11}{12}<\frac{19}{20}$.

Thus, $ k\equal{}2$, so $ \frac{9}{20}<\frac{1}{n}\plus{}\frac{1}{m}<1$. Clearly, $ n\ne{2}$ If $ n\equal{}3$, then $ \frac{7}{60}<\frac{1}{m}<\frac{2}{3}$. From here, we can see that $ m\equal{}7$ is the smallest number that satisfies this.

Thus, the smallest possible $ k\plus{}m\plus{}n$ is $ 2\plus{}3\plus{}7\equal{}12$.[/hide]

[hide="Sprint #9"] We can guess that $ a\equal{}100\minus{}4\equal{}96$. We can also guess that $ c\equal{}\frac{a}{2}\minus{}3$, so $ c\equal{}45$. Now we can guess that $ b\equal{}a\minus{}c\plus{}2$, so $ b\equal{}53$ and $ a\plus{}b\plus{}c\equal{}96\plus{}53\plus{}45\equal{}194$.[/hide]

[hide="Sprint #10"] Let the length be $ a$ so the diagonal is $ \sqrt{a^2\plus{}4}$.

From the ratio given, we have $ \frac{a}{2}\equal{}\frac{\sqrt{a^2\plus{}4}}{a}$ so $ a^2\equal{}2\sqrt{a^2\plus{}4}$ Squaring both sides, we have $ a^4\equal{}4a^2\plus{}16$ which rearranges to $ a^4\minus{}4a^2\minus{}16\equal{}0$. We use the quadratic formula to get $ a^2\equal{}\frac{4\pm{\sqrt{80}}}{2}\equal{}\frac{4\pm{4\sqrt{5}}}{2}\equal{}2\pm{2\sqrt{5}}$. As $ \sqrt{a^2\plus{}4}$ is a positive real number, we must have $ a^2\equal{}2\plus{}\sqrt{5}$. Thus, the length of the diagonal is $ \sqrt{6\plus{}2\sqrt{5}}$ which is just $ \sqrt{(1\plus{}\sqrt{5})^2}\equal{}1\plus{}\sqrt{5}$.[/hide]

[quote="pytheagle"]I was doing the 1994 Nationals and I couldn't figure these problems out:

Sprint, Number 8:
Given the equation $ \frac {1}{m} \plus{} \frac {1}{n} \equal{} \frac {2}{15}$, where $ m$ and $ n$ are positive integers, find the value of $ m \plus{} n$ such that $ m > n$ and $ m$ is not a multiple of $ n$.

[hide="Answer"]
32
[/hide]

[color=blue]Multiply by $ 15mn$ to get $ 15m\plus{}15n\equal{}2mn$. Use SFFT to get $ (2m\minus{}15)(2n\minus{}15)\equal{}225$. Since m is not a multiple of n, the solution is $ m\equal{}20$, $ n\equal{}12$ and the sum is 32. [/color]

Target, Number 4:
There are positive integers $ k,n,$ and $ m$ such that $ \frac {19}{20} < \frac {1}{k} \plus{} \frac {1}{n} \plus{} \frac {1}{m} < 1$. What is the smallest possible value of $ k \plus{} n \plus{} m$?

[hide="Answer"]
12
[/hide]

[color=blue]If k is the smallest number and it is 3, there must be at least one number of k,n,m greater then 3. If there is only 1 4, the sum is $ \frac{11}{12}$, not greater then $ \frac{19}{20}$. So $ k\equal{}2$. So $ n\equal{}3$, $ m\equal{}7$, and the sum is 12.[/color]

Sprint, Number 9:
Use the pattern below to express $ 100^2$ in the form $ a^2 \plus{} b^2 \minus{} c^2$.
Then find $ a \plus{} b \plus{} c$.
\[ 12^2 \equal{} 8^2 \plus{} 9^2 \minus{} 1^2\]

\[ 14^2 \equal{} 10^2 \plus{} 10^2 \minus{} 2^2\]

\[ 16^2 \equal{} 12^2 \plus{} 11^2 \minus{} 3^2\]

\[ 18^2 \equal{} 14^2 \plus{} 12^2 \minus{} 4^2\]
[hide="Answer"]
194
[/hide]

[color=blue]The pattern is $ (2x)^2\equal{}(2x\minus{}4)^2\plus{}(x\plus{}3)^2\plus{}(x\minus{}5)^2$. Letting $ x\equal{}50$ gives $ a\plus{}b\plus{}c\equal{}96\plus{}53\plus{}45\equal{}194$.[/color]

Sprint, Number 10:
The ratio of the length of a rectange to its width is the same as that of the diagonal to the length. If the width is $ 2$, how many units are in the length of the diagonal?

[hide="Answer"]
1+$ \sqrt{5}$
[/hide]

[color=blue]$ \frac{l}{2}\equal{}\frac{\sqrt{l^2\plus{}4}}{l} \implies l^2\equal{}2\sqrt{l^2\plus{}4}$.

Let $ y\equal{}l^2$. Then we have $ y^2\equal{}4y\plus{}16 \implies y^2\minus{}4y\minus{}16\equal{}0 \implies y\equal{}2\plus{}2\sqrt{5}$.

This makes the diagonal $ \sqrt{6\plus{}2\sqrt{5}}\equal{}1\plus{}\sqrt{5}$.[/color]

As I'm still not finished trying to figure out how I got some problems wrong, I'll edit this post later to include some more problems.

(There are probably going to be a lot more because my score was 19/46; very embarrasing)[/quote]

On #8 sprint, can you explain how you used SFFT?