AMC
AMC 10
AMC 10 A
symmetry
blogs
number theory
least common multiple

Here's a topic for discussing problems on the 2005 AMC10A. I'll start off.

Embarrassingly, I spent 20 minutes on #24 by misreading the instructions and not noticing that n>1. Luckily, there was no choice for 2, because that's what I got first.  :blush:

#17 might have been the most interesting problem, since it can be solved with a symmetry approach like the ones discussed in [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=60]Mr. Rusczyk's blog[/url].  Well, sort of.  Since we have an arithmetic sequence,

AB + BC + CD + DE + EA = 5(middle segment)

But AB + BC + CD + DE + EA = 2(A+B+C+D+E) = 2(3+5+6+7+9) = 60

So middle segment = 12

[quote="keta"]Man, I was so disappointed after missing #22 about the sets of multiples of 4's and 6's. You had to look for the LCM of those numbers, and my mind ignorantly thought for a moment that LCM (4,6) = 24, so I picked 333, which I thought should have been 334. Right after the test, I realized my mistake.  :([/quote]

dude lcm of 4 and 6 is 12

Yeah, he said he realized his mistake right after the test.  Anybody else think #24 was a really nice problem?  The whole difference of squares...and number 8 was just the diagram used to prove the Pythagorean Theorem, which I'm sure most of us are familiar with.

Most of them were great. I agree that 17 and 24 were cool. Was there any way to do 25 besides $A=\frac 12 ab\sin\theta$? I would have gotten that one had I known the formula before taking the test. Instead I omitted it and our proctor explained the formula afterwards.

i personally like the [i]trefoi[/i]l problem
also the one where we had an arithmetic sequence.
hated octagon problem (forgot that the sqrt(2) was divided by 2... D'OH!) hence I got 7 intead of 7/2

What I didn't like about 25 was that my friend who isn't really good at math assumed the the triangles were right in 25, and got it right, but I checked, and found that they weren't, and didn't answer it.

I'm really annoyed at myself about 25 now. I knew the 1/2ab(sinC) formula but I only looked at it in the last 2 minutes because I thought: Wow, #25, must be too hard, let's check my answers first. :(

Kudos to you who got it :D

i screwed up on reading 21 and thought it was the other way around....
i can't believe I missed 12 for number 15
also, i marked number 23 wrong even though I had the right answer...
and I couldn't really get an "actual" solution to 19 w/o guessing

Intuition is often not good enough my friend  :P
There were 4 cases where B was 2nd and A won.
One of them consisted of 4 games (ABAA I think).
let p be the probability of a 5 game occurence with A winning and B winning the second game
so we can write 2p (for the ABAA game) + 3p = 1
so 5p = 1
probability of a 5 game win that fits the requirements is 1/5
(I put 1/4 on the test...)

hmm okay.. maybe i dont understand it well b/c i dont really know the prerequisities.... (i didnt take the AMC A, so this is based upon what other ppl told me)

5 games; Team B won game 2 and you want to find the probability that Team A wins the series?

no, team B wins game 2, team A wins the series.

Find probability that team B wins game 1 also.

[quote="yif man12"][quote="sky9073"]For number 25, you don't need the sine formula... just draw a couple of lines.
Connecting DC, the areas of triangles ADC and DBC have the ratio 19/25, (same height)
and similarly, connecting ED, the areas of triangles AED and EDC have the ratio 14/42.
I almost left 24 blank, but got it in the last 3 minutes... is trial and error the only way to do it?[/quote]

Well, we know that both n and n+48 have to be squares of prime numbers, and 48 can be expressed as 23+25 or 9+11+13+15 or 3+5+7+9+11+13, so there's only 3 possibilities for n and n+48 being squares. Check all 3 and you'll find that only n=121 and n+48=169 works.[/quote]

Alternatively, let n = k^2 and n+48 = l^2.  Then l^2-k^2 = 48, so (l+k,l-k) = (1,48), (2,24),... (48,1).  Of these, only the pairs where k = 1, 4, or 11 work, so n = 1, 16, or 121.  But n > 1, and 16 isn't the square of a prime, so only n=121 works.

[quote="pkothari13"]does anyone know why that soccer problem is 1/5?? my intuition would tell me it would be 1/2, but apparently not....[/quote]

As for your intuition, it is true that team B has a 1/2 chance of winning game 1.  However, that's not the question - the question is actually 'out of all the cases where team B wins game 2 but team A wins the series, what is the probability that team B will win the first game'.  Since the set of cases where team B wins game 2 but team A wins the series is a different set from the set of all games, it's no longer 1/2 - we have additional conditions that we must satisfy.

The four possible outcomes, and the probability that they occur, is
ABAA = (1/2)^4 = 1/16
ABABA = (1/2)^5 = 1/32
ABBAA = 1/32
BBAAA = 1/32

We are given that one of these occured (since B won game 2, A won series).  One of these occurs 1/16+1/32+1/32+1/32 = 5/32 of the time, so now we need to multiply all of the probabilities by 32/5 so that they add up to 1 (since we know that one of them must occur):

ABAA => 2/5
ABABA => 1/5
ABBAA => 1/5
BBAAA => 1/5

And there you have it.

Even though a "best of 5" series usually stops when one team wins 3 games, we can imagine that the series continues for the full 5 games.  For example, the event ABAA splits into two outcomes ABAAA and ABAAB.  That way, it is clear that there are 5 equally-likely outcomes, and we don't have to worry about some outcomes being weighted more than others.

Adding such fictitious games to "best of" series often makes such problems a lot easier.  I recall using that trick in the past, but I can't think of the examples now.


find integer $a$ such that:$x^2-x+a|x^{13}+x+90$


algebra
polynomial
LaTeX
superior algebra
superior algebra unsolved

Prove that sum and product of 2 algebric integers is an algebric integer.

Proof:Suppose a,b are algebric integers then a satisfies a minimal monic polynomial with integer coefficients i.e. let
$a^n +a_1a^{n-1}+...+a_n=0$.
 So, $a^n =-(a_1a^{n-1}+...+a_n)$ …………..(i)
Thus all the higher powers of a can be written in the form of linear combinations of 1,a,….a^n-1
So this set generates Z[a]. Similarly if b satisfies a minimal monic polynomial of degree m, then 1,b,…,b^m-1 generates Z[b].
Now we claim that Z[a,b] is generated by elements of the form aibj where I=0(1)n-1
J=0(1)m-1.
To see this take any element p belonging to Z[a,b]. then p is of the form
 $P= ååx_ij a_ib_j$ . Since there are only a finitely many terms in the expression, starting from the 1st term check the power of a. If it is less than n, we leave the term unchanged and go to the next term. If the power is greater than n we reduce it by the help of (i).After reducing the power of a from each term we then reduce the power of b by the same method. By these operations each term becomes a product of polynomials where the degree of a & b are atmost n-1 and m-1 respectively. Multiplying all the terms and rearranging them we can write p as a linear combination of aibj . Thus aibj generate Z[a,b] as claimed and as the no of such generators is finite we have Z[a,b] a finitely generated Z-module.

Is my proof correct?It certainly is easy for a double starred herstein problem

Please use $\LaTeX$ correctly (e.g. use a^{n-1} to get $a^{n-1}$, not as you do a^(n-1) ).

I am very sorry I dont know latex. :blush:
I have a lot of problems with the summation sign.I know its difficult for you to read but plz tell me if my soln is correct or not.

[quote="the game"]Prove that sum and product of $2$ algebric integers is an algebric integer.

Proof:Suppose $a,b$ are algebric integers then a satisfies a minimal monic polynomial with integer coefficients i.e. let
$a^n +a_1a^{n-1}+...+a_n=0$.
 So, $a^n =-(a_1a^{n-1}+...+a_n)$ …………..(i)
Thus all the higher powers of a can be written in the form of linear combinations of $1,a,....a^{n-1}$
So this set generates $\mathbb{Z}[a]$. Similarly if $b$ satisfies a minimal monic polynomial of degree $m$, then $1,b,...,b^{m-1}$ generates $\mathbb{Z}[b]$.
Now we claim that $\mathbb{Z}[a,b]$ is generated by elements of the form $a^i b^j$ where $i=0...n-1$
$J=0...m-1$.
To see this take any element $p$ belonging to $\mathbb{Z}[a,b]$. then $p$ is of the form
 $p= \sum_{i,j} x_{i,j} a^ib^j$ . Since there are only a finitely many terms in the expression, starting from the 1st term check the power of $a$. If it is less than $n$, we leave the term unchanged and go to the next term. If the power is greater than $n$ we reduce it by the help of (i).After reducing the power of $a$ from each term we then reduce the power of $b$ by the same method. By these operations each term becomes a product of polynomials where the degree of $a$ & $b$ are atmost $n-1$ and $m-1$ respectively. Multiplying all the terms and rearranging them we can write $p$ as a linear combination of $a^ib^j$ . Thus $a^ib^j$ generate $\mathbb{Z}[a,b]$ as claimed and as the no of such generators is finite we have $\mathbb{Z}[a,b]$ a finitely generated $\mathbb{Z}$-module.

Is my proof correct?It certainly is easy for a double starred herstein problem[/quote]
The above should be what you want :) (click on quote to see the plain text for the formulas)

To the prove:
The part "finetely generated $\implies$ algebraic integers" is missing.
The rest seems ok.


From what I've heard, you should only apply to RSI in 10th grade if you intend to graduate in 11th grade.


invariant
linear algebra
linear algebra unsolved

This is just beautiful: suppose that $A,B$ are invertible complex matrices which commute to their commutator. Then all eigenvalues of the commutator are roots of the unity.

I suppose the commutator here is $ABA^{-1}B^{-1}$ and not $AB-BA$.

Let $\lambda$ be an eigenvalue of $[A,B]=ABA^{-1}B^{-1}$. because $A,B$ commute with $[A,B]$, the $\lambda$-eigenspace of $[A,B]$ is invariant under both $A$ and $B$. By restricting attention to this eigenspace, we may as well assume that $[A,B]=\lambda$.

We now have $AB=\lambda BA$. Let $\sigma$ be an eigenvalue of $A$ and let $x$ be a $\sigma$-eigenvector of $A$. We then have $ABx=\lambda BAx=\lambda\sigma Bx$, i.e. $Bx$ is a $\lambda\sigma$-eigenvector of $A$. We repeat this procedure to conclude that for all positive integers $n,\ \lambda^{n}\sigma$ is an eigenvalue of $A$. Since $A$ has finitely many eigenvalues, this can only happen if $\lambda$ is a root of unity.


I'm hoping to get a summation symbol like this:

$ \displaystyle\sum_{k\equal{}0}^n$

in a fraction like this: $ \displaystyle\frac{\sum_{k\equal{}0}^n}{\sum_{j\equal{}0}^u}$.

But I can't figure out how to get the summation sign to render stylistically in a fraction.

Help appreciated. Thanks!

Use \sum\limits...
$ \frac{\sum\limits_{k\equal{}0}^{n}}{\sum\limits_{j\equal{}0}^{u}}$


calculus
integration
function
trigonometry
logarithms
complex numbers
calculus computations

Compute the "antiderivative" of the function $ln(sinx)$

There is no elementary antiderivative. The best we can do is to evaluate some definite integrals for that function.

Starting with the argument I just posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152690]here[/url] but taking the real part instead of the imaginary part, we get this result:

$\frac1n\ln|2\sin2\theta|=\ln|1-e^{i\theta}|=\sum_{n=1}^{\infty}\frac{\cos n\theta}{n}.$

Integrating, we get

$\int_{0}^{x}\frac12\ln|2\sin2\theta|\,d\theta=\sum_{n=1}^{\infty}\frac{\sin nx}{n^{2}}.$

One could, with sufficient patience, use this to calculate certain definite integrals, such as

$\int_{0}^{\pi}\ln\sin x\,dx.$ Carcul claimed as much.


function
real analysis
real analysis solved

Does there exists a real function f such that for all real x the set of solutions of the equation f(t)=x is in bijection with the reals?

Of course. f(0,abcdef...)=0,ace..., where 0,abcde.. is a decimal representation of a number.

What a simple example! Thank you. I start to ask some very easy questions, but I like them. I don't know why.
  Another (maybe stupid question): does there exist a continuous function with those properties?

Yes, it exists, There exists a continious function f which maps the segment to a square. If f(t)=(x(t),y(t)), then x(t) is such a function.


algorithm
analytic geometry
graphing lines
slope
trigonometry
ratio

I am looking for some justification as to why when you have a line in Ax+By+C=0 form you divide by  :pm:   :sqrt: A :^2: +B :^2:  to get it into normal form and then the coefficents represent cos :phi: x + sin :phi: y - p =0.

Does anyone know where I can find a link that would tell me this?

I have been trying to do this myself but cannot find justification for the C term.

Thanks!

What do you mean that you cannot find a justification for the C term?  You need to explain more what you're asking about for us (well, me at least) to understand and help.

Okay well I am trying to find a justification as to why one preforms a certain algorithim to get from general form of an equation of a line, Ax+By+C=0 to Normal form, xcos( :phi: )+ysin( :phi: )-p=0.

The algorithim is that you find sqrt(A^2+B^2), and give it a sign that is opposite of C's sign.  Then you divide all terms of Ax+By+C=0 by this quantity, and now the line is in normal form.

I have figured out why this is true for the A/ :pm:  :sqrt: A :^2: +B :^2: and B/:pm:  :sqrt: A :^2: +B :^2: but I can't figure out why it works w/the C term.  (FYI I did this by changing Ax+By+C=0 to y=(-A/B)x+(-C/B), since the slope of this line is -A/B I constructed a triangle w/ A*n, B*n, and n*sqrt(A^2+B^2) leg lengths, n is a constant, and formed trig ratios w/  90+ :phi: , to find out why the coefficents were in the normal form now.)

[quote="mrblueskies"]I have figured out why [b]this[/b] is true ...[/quote]
(emphasis mine)

What is the "this" you are refering to?  I understand everything you've written.  I understand the operation you're carrying out.  What I don't understand is what you think is happening to specific coefficients -- what does the "this" in that sentence refer to?  "Normal form" is a property of the equation of the line; it isn't a property of the individual coefficients.

"this" as in the algorithim, I found out why this algorithim transformes the line so the coefficents of x and y represent cos phi and sine phi.  I cannot however find out why the new C term after preforming the algorithim represents p.

Hey,

Well...if you have the line Ax + By + C = 0, dividing by  :pm:  :sqrt: (A^2 + B^2) will give you the hypotenuse of a right triangle with sides A and B. In our case, since the slope is -A/B, say n = arctanA/B. Dividing gives us that sin(n)x + cos(n)y - C/root(A^2 + B^2) = 0. Is that the p you are trying to "justify"?

I don't understand what you think p is -- it's just a number, which is the result of dividing C by :sqrt:(A2 + B2).  What significance are you attributing to it?  All it's doing is telling you that there is some constant left over.

p is the perpendicular distance of the line from the origin.

If the line is at an angle  :theta: to the x-axis (so that tan( :theta: )= -A/B) then the perpendicular distance p  of the line from the origin is given by

p = ycos( :theta: ) - xsin( :theta: )

if the line crosses the y-axis above the origin, or

p = xsin( :theta: ) - ycos( :theta: )

if the line crosses the y-axis below the origin, where (x,y) is any point on the line.

In the first case, set  :phi: =  :theta: + :pi: /2 and in the second case set :phi: =  :theta: - :pi: /2, and then in both cases you have

p = xcos( :phi: ) + ysin( :phi: )

and, also in both cases, tan( :phi: ) = -cot( :theta: ) = B/A


integration
number theory unsolved
number theory

:roll: What is the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot n$?

$\Gamma(n+1) = \int_{0}^{\infty} t^n e^{-t} \, dt$

Alternately:

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

For completness sake:
$n! = \Gamma ( n + 1 )$ for all positive integer $n$.

[quote="sagar1331"]:roll: What is the product 1*2*3*4*.........n?[/quote]
My friend says:
Evgeny: $1*2*...*n=n*(n-1)*...*1$.
Roman: It's very silly question. ;)

I agree with Roman..  :P

(Sorry for the useless post)

Oh, well, just to make it more valuable : http://mathworld.wolfram.com/Factorial.html

;)


inequalities
AMC
USA(J)MO
USAMO
inequalities solved

given positive reals a1,a2,... a5 satisfy  \sum 1/(1+a)=1
Prove that  \sum a/(4+a 2 ) \leq 1.

you joined CWMO?
I got a silver this yr..

The official solution used a substitution.
anyone wants me to post it?

[quote="liyi"]West Math Olympiad?[/quote]

yep, something like Western China Mathematical Olympiad

Very beautiful and quite challenging. Do you see anything similar with Usamo 2003 or Japan 1997?
   Let 1/(1+a_i)=x_i. Then a_1=(x_2+x_3+x_4+x_5)/x_1 and so on. The inequality becomes sum x_1*(x_2+x_3+x_4+x_5)/(4x_1^2+(x_2+x_3+x_4+x_5)^2)<=1. We prove that this is true for all x_i, not necessarily with sum 1. Since it is homogenuous, we can assume that sum x_i=5. It becomes sum x_i*(5-x_i)/(4+(x_i-1)^2)<=5. It is obvious that LHS is at most sum x_i(5-x_i)/4 and from now on everything is clear since sum x_i=5 and we can apply Cauchy ( rest in peace; beautiful inequality had he discovered).

For $x>0$, we have $\displaystyle \frac{x}{4+x^2} \leq \frac{1}{20}\left(1+\frac{15}{1+x}\right)$. In fact, this is equivalent to $(x-4)^2(x+4)\geq0$.

Adding, we get $\displaystyle  \sum_{i=1}^5\frac{a_i}{4+a_i^2} \leq \frac{1}{20}\sum_{i=1}^5 \left(1+\frac{15}{1+a_i}\right) = 1$.

[quote="fuzzylogic"]For $x>0$, we have $\displaystyle \frac{x}{4+x^2} \leq \frac{1}{20}\left(1+\frac{15}{1+x}\right)$. [/quote]

Somebody asked me how I came up with this inequality. I have to give credit to [b]hxtung[/b]. I learned this technique from his post  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180]Poland 1996 inequality[/url].  This technique can be applied to Japan 1997 and USAMO 2003 as well.

For this particular problem,  I expect that  $\displaystyle \frac{x}{4+x^2} \leq \frac{ax+b}{1+x}$ for some $a,b$. This is equivalent to
$f(x) = (ax+b)(4+x^2) - x (1+x) \geq 0$, where $f(x)$ is a cubic polynomial.

So how do we find $a,b$? I expect the equality holds when all $a_i$ are equal to 4 (from the given condition). So I expect $f(4)=0$. Further, I expect $(x-4)^2$ divides $f(x)$, otherwise $f(x)\geq0$ can't be true. So $f'(4)=0$ too. From $f(4)=f'(4)=0$, I found that $a=\frac{1}{20}$ and $b=\frac{4}{5}$. The rest is just verify that the inequality is true.

This problem can be extended to $n$ variables as follows:
[quote]Let $a_1,a_2,\ldots, a_n > 0$ satisfying $\displaystyle \sum_{i=1}^n \frac{1}{1+a_i}=1$. Prove that
\[
\sum_{i=1}^n \frac{a_i}{(n-1)+a_i^2} \leq 1.
\][/quote]

Cheers! :D  :D


ratio
combinatorics unsolved
combinatorics

Numbers $ 1,2,...,64$ are written on a $ 8\times 8$ board. For every two numbers $ a,b$ with $ a>b$ in the same row or column, the ratio $ \frac{a}{b}$ is calculated. The characteristic of the board is defined as the least of these ratios. Find the greatest possible value of a characteristic.

Is the answer 64/7 or am I being stupid?

It has to be at most $ \frac{64}{14}$ (there are only 13 numbers less than 14, so 14 appears in the denominator of some ratio).

Ah, sorry - somehow I completely missed "or column" when I read that!

64/14 is indeed right.

[hide="Ponderings"]Let the characteristic be $ r$. In the row containing 1, note that the next smallest entry is at least $ r$. The next after that is at least $ r^2$; otherwise, the ratio between the two would be less than $ r$, meaning that $ r$ would not be the characteristic. Then by similar reasoning, the largest entry in the row is at least $ r^7$. To maximize $ r$, you want the largest entry in this row to be as high as possible. Therefore, $ r<\sqrt[7]{64}$.

You can do better by considering the maximum among the least entries in each row. This maximum is at least 8, so $ 8r^7<64$.

Further note that since the entries are integers, you will always be considering the ceiling of the $ r$ powers at each step, so that deflates it even more. As an example, the second smallest entry in the row containing 1 is at least 2. The maximum among second smallest entries in each row is at least 16. Then we have $ 16r^6<64$.

I'm pretty sure this continues to the 7th smallest (or second largest) entry in the row, suggesting the maximum characteristic is less than $ \frac{64}{56}$.[/hide]

This method doesn't lend itself to determine whether characteristics can actually be achieved, though.


calculus
integration
algebra
function
domain
Ring Theory
algorithm

Let [b]F [/b]be a field. State under what circumstances, if any, [b]F[/b] is respectively an Euclidean Domain, Principal Ideal Domain, Uniqe Factorization Domain. Justify your answers

Thank you very much in advanced

What do you mean? If you have a certain field which satisfies the criteria for those domains, it'll be that.

For instance construct a norm over your field as follows
$ N(a) \equal{} 1$ for any $ a \not \equal{} 0$
$ N(0) \equal{} 0$
Then your field is an Euclidean Domain.

To be a PID, its really easy to see, since every element in a field is a unit. Hence with the Euclidean Algorithm there won't be any non-zero remainders...

There's no need to define a Euclidean valuation since a field is vacuously Euclidean: there are no nonzero remainders upon division. Since Euclidean => PID => UFD all else follows. But, of course, it's trivial to directly verify the other claims  too.


Roslyn has ten boxes. Six of the boxes contain pencils, three of
the boxes contain pens, and two of the boxes contain both pens
and pencils. How many boxes contain neither pens nor pencils?

Call the number of boxes with neither pens nor pencils $ x$.  Using [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]PIE[/url], we know that $ 6\plus{}3\minus{}2\plus{}x\equal{}10\implies x\equal{}\boxed{3}$.


combinatorics proposed
combinatorics
Bulgaria

Find the greatest positive integer $n$ such that we can choose $2007$ different positive integers from $[2\cdot 10^{n-1},10^{n})$ such that for each two $1\leq i<j\leq n$ there exists a positive integer $\overline{a_{1}a_{2}\ldots a_{n}}$ from the chosen integers for which $a_{j}\geq a_{i}+2$.

[i]A. Ivanov, E. Kolev[/i]

[hide="Does it help?"]$ 1)$ Notice that for all the $ 2007$ numbers $ a_1 \ge 2$
$ 2)$ If $ n\equal{}4k$ then a choice covering the maximum amount of pairs is $ 24682468\cdots2468$ covering $ 6 \cdot \frac{k(k\plus{}1)}{2}$ pairs
$ 3)$ There are $ \frac{n(n\plus{}1)}{2}$ pairs[/hide]

[quote="mszew"]
$ 2)$ If $ n \equal{} 4k$ then a choice covering the maximum amount of pairs is $ 24682468\cdots2468$ covering $ 6 \cdot \frac {k(k \plus{} 1)}{2}$ pairs
[/quote]

Umm... how can you prove this is maximum?

Moreover, the other $ n$s won't cover this many pairs...

[url=https://dgrozev.wordpress.com/2022/07/10/packing-balls-method/]A solution and some other stuff[/url] related to this problem.


rotation
combinatorics unsolved
combinatorics

An $m$x$n$ board is divided into unit squares. An L-triomino consists of three unit squares: a central square and two outer squares. An L-triomino is located in the upper left corner of the board, the central square covering the corner square of the board. In a move one may rotate the triomino around the midpoint of one of its outer squares by multiples of 90 degrees. For which $m$ and $n$ is it possible to move the triomino to the lower right corner of the board using a finite number of moves?


Find all integers triples (p,q,r) such that:
$ r|p^q\plus{}1 , q|r^p\plus{}1 , p|q^r\plus{}1$

An idea:I will not write the whole solution but an effective idea is to consider the multiplicative order (http://en.wikipedia.org/wiki/Multiplicative_order) of,p mod r, r mod q and q mod r. Then using elementary properties you will reach the solution
If you have questions about my post ask

Solution
[hide]The solution is (2,3,5) and permutations[/hide]

Daniel


I'm trying to come up with a formula to calculate the radius for the largest circle whose interior contains exactly n numbers of lattices.  n =0,1,2,3 through 9.
I can draw the circles with 0, 1, 2, 3 etc lattice points but I can't seem to come up with a formula whereby I can just enter the n value (0 through 9) and it computes the radius.
I figure there must be a formula perhaps one formula for even numbers for n and one for odd numbers.  The reason I say this is because for the even numbers for n the center point of the cicle is between two lattice points where the odd n's form triangles for the center point.
Mayber theres just some simple pattern that explains it.
PLEASE HELP AS THIS IS DRIVING ME CRAZY!

Can someone please help me on this problem.

i know that there must be a formula that will work for n=even and another for n=odd.  I can graph the various points but need to come up with a formula for computing the radius.

Thanks for any help offered,

What exactly are lattices inside a circle?

Well I know pretty much nothing about lattice points and such, but here's what seems to be a good website if you haven't seen it: http://mathworld.wolfram.com/PointLattice.html.

I do not think there will be a "nice" formula in the way that you hope.  For each number of lattice points, there is such a radius, but it is unlikely to have a nice expression in terms of n, I'm afraid.

Are these any particular points, like say the first n along the number line (ok its trivial but just to give an example) or just n random points?


how do you prove that the interior angles of a triangle are concurrent using ceva's theorum?

Try to look around at the forums to determine the most appropriate place to post questions like this.  I'll move this to a high school forum.

Please don't post the same thing twice :!:


Putnam
linear algebra
matrix
college contests

Alan and Barbara play a game in which they take turns filling entries of an initially empty $ 2008\times 2008$ array. Alan plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all entries are filled. Alan wins if the determinant of the resulting matrix is nonzero; Barbara wins if it is zero. Which player has a winning strategy?

Pick two columns.  If Alan plays in one column, Barbara places the same real number in the corresponding position int he other column.  If Alan plays in neither column, Barbara makes an arbitrary move in another column.  Since there are an even number of positions not in either column, when the game ends two columns are identical and Barbara wins.

Yay!  That's what I said!  (Except much wordilier.  "Wordilier" should be a word.)

(BONUS: The professor who runs the Putnam at my school is named Barbara.)

Barbara can even pair off all the columns, and force a rank of no more than 1004.

The natural question people down at my school were asking was this: what if the matrix was $ n \times n$ and $ n$ was odd? Picking two columns doesn't work because Alan can run Barbara out of the remaining spaces.

Well it was already given n = 2008, an even. However, whether n is even or odd doesn't matter.

Let i denote the ith row and let j denote the jth column of the matrix.

The solution I put was if Alan fills entry a_ij with a number where 0 < i, j <= 2008, then if Barbara fills out entry a_IJ where I = 2009 - i and J = j with the same number, the matrix will at least have two identical rows, and by the property of the determinants, the determinant of the matrix will be 0. Therefore, Barbara will always win with this strategy.

EDIT: Looking in my reference, it seems so long as the matrix has two identical rows or columns for any n x n matrix, the determinant will be 0.

EDIT 2: Will I still get full credit if I justified that "since at least a pair of rows are identical, the determinant of matrix is 0" by "property of determinants"? Or was I supposed to provide proof to this property as well?

I made up a first-player win proof... and forgot to check that who moves. This should work for any odd n (on even n a mirroring strategy works for a second-player win).

Since exchanging rows and columns does not affect the magnitude of the determinant I'll just order them by move first played in. Alan's first move is arbitrary, and Barbara can either play in the same row/column or in a new row, new column. Regardless, Alan plays a number so that the position

x1 x1
--- x2

is reached, where x1, x2 are real numbers. Barbara may fill in the -- slot or play in a new row, new column. If she plays in the spot and doesn't fill in x2, then the two rows are LI, so she should play x2 for optimality (we'll get back to this in the end). Regardless of which she does, Alan forces the following position:

x1 x1 ---
x2 x2 ---
--- --- x3

Now consider if Barbara plays in any of the four marked spots - Alan can immediately play a new number in the other spot to force LI of the rows/columns, for example

x1 x1 ---
x2 x2 ---
x4 x5 x3

And so Barbara is forced into yet a new row/column. Alan again copies her move to force:

x1 x1 --- ---
x2 x2 --- ---
--- --- x3 x3
--- --- --- x4

and her dilemma continues. Notice when the second x4 is placed, all 8 empty positions are forbidden to her - Alan can force all 4 rows/columns to be LI by picking suitable new numbers. Thus assuming she does not give up LD on rows/columns, eventually the position becomes a 2x2-block diagonal matrix with equal entries on each row, with one remaining row. Alan plays a new number in it (since 4 moves per block), and now Barbara must suicide until Alan forces every single row/column to be LI and wins.

Back to the possibility that Barbara gives up making two rows LD earlier, to wit

x1 x1 ---
x2 x3 ---
--- --- x4

with x4 being Alan's new move. Now rows 1/2 are LI as are columns 1/2, and suppose Barbara plays in this submatrix, case 1:

x1 x1 ---
x2 x3 ---
x5 --- x4

Alan can put a suitably huge number in the bottom row such that columns 1-3 are all LI, the other case is:

x1 x1 x5
x2 x3 ---
--- --- x4

whence Alan places another huge number in the rightmost column. Thus giving up rows cannot improve Barbara's condition and we (on no errors in the above) can conclude it's a first-player win.

PS: I wrote this on my A2, except I forgot that when the entire matrix was 2x2-block diagonal... Alan moves and suicides next.

Wow, I can't believe I missed this one! And, I thought A1 was way too easy!

[color=red]This proof is wrong

First:  any 2n*2n matrix can have its determinant defined by the difference in the sums of the left-wise and right-wise diagonal products.  Then, any left-wise and right-wise diagonal intersect in either 2 or 0 elements.

Let the first player place any number in the M_i,j position.

By definition, the left-wise and right-wise diagonals that intersect at M_i,j also intersect at M_p(i),p(j), where p(x) = (x+1004)mod2008.

If Barbara places the integer "0" at M_p(i),p(j), then she makes the diagonal's products equal 0 for both diagonals.

In addition, as long as she continues that pattern, every diagonal will eventually have a product of 0, so both sums will equal 0 and the difference of sums will equal 0.[/color]

[quote="Kent Merryfield"][quote="collegebookworm"]First:  any 2n*2n matrix can have its determinant defined by the difference in the sums of the left-wise and right-wise diagonal products. [/quote]
This is not true. The determinant of a $ 4\times 4$ matrix has $ 24$ terms, not $ 8$ terms.[/quote]
Frosh's mistake...folly...I guess I shouldn't take Calc 2 then.

What does that have to do with linear algebra? They're unrelated classes, and there's no shame in getting something wrong when you haven't seen it yet.

A fellow test-taker at my university told me what he wrote for his answer, which makes sense to me.
When Alan places a number in any spot in the array, Barbara places the same number in the column, but one row up or down.  That was she forces the rows to be linearly dependent, and the resulting determinate is zero.  I think that's the gist of what he said.

I spent a long time thinking about what would make the determinate easier to calculate, for what reason I don't know.  Also, if Alan was particularly dim, Barbara could fill a whole row up with zeros and win the game.

[quote="jaws1216"]I tried something a bit different.  I had Barbara responding with 0's based on where Alan played.  If Alan played in an m,n spot which wasn't on the boundary, Barbara would play in the m+1, n-1 spot.  Or if there was already a 0 on that diagonal, then play in the m+1, n+1 spot.  On the boundary the idea was similar, broken up into cases to make sure the move happened on the diagonal if possible.  Would this be enough to guarantee that the matrix' determinant was 0?[/quote]
My gut would say no. Regardless, if you don't know the answer to your question, it means you didn't sufficiently prove it on the exam, thus won't get points (1 if you're very lucky). So it doesn't really affect your score whether that works or not.

On another note, I have a question stemming from this problem:
How much attention does Barbara actually have to pay? Specifically, how many moves must she make non-arbitrarily? 2008 is easy--just do the two identical rows method. Currently I think that this is also the best we can achieve, but I'm not entirely sure how to prove it. Any thoughts?

[quote="Sly Si"]I don't really like this problem.  It reminds me of the kind of problem I write when I have writer's block: something that looks kind of cool, but reduces down to something quite trivial.  I'll take the 10 free points, though.[/quote]

same, what a POS problem.

I paired up all the rows by identifying row i with row 2009 - i, except I...accidentally wrote 2008 - i. I later stated that row 1 was paired with row 2008, so it was clearly a typo. How bad is this?

Here's another quick proof.

Let Alan place his number, say $ r$, at $ a_{ij}$. Have Barbara place the same number at $ a_{(2008\minus{}i\plus{}1)j}$. It is clear that Barbara can always place a number down. After this process, the matrix will be symmetrical vertically. Since $ 2008$ is even, then upon determinant expansion along the first column, the sign of the term with $ a_{i1}$ will be the opposite of $ a_{(2008\minus{}i\plus{}1)1}$. The determinants of the submatrices corresponding to these elements will be identical. Thus, the determinant will end up as zero.

What if 2008 was replaced by 2009?  Who has a winning strategy?

Try it with 3 before asking about 2009. And look at the 2002 Putnam - it's not exactly the same problem, but ...

We are discussing this on an internal mailing list.
According to the discussion, B can guarentee the rank of the matrix is no more than ceiling(n/2).

If $A$ places a real number in row 1, $B$ places $0$ in row 2 underneath $A$. If $A$ places a real number in row 2, $B$ places $0$ in row 1, above $A$. By generalized laplace, the determinant is $0$.

Why didn't I just make it so two rows are the same? I don't know   :|


trigonometry
algebra
polynomial
absolute value
linear algebra
linear algebra unsolved

$A \in \mathcal{M}_{n}(\mathbb{Z})$, $\parallel A\parallel _{\infty}=\max_{\parallel X\parallel =1}\parallel AX\parallel $, where $\parallel . \parallel $ is the euclidian norm.
Show that either $\parallel A\parallel _{\infty}$ could be written $2 \cos (\pi /k)$ with $k \in \mathbb{N}$, $k\geqslant 2$, or $\parallel A\parallel _{\infty}\geqslant 2$.

Hint: use Kronecker's theorem on polynomials whose roots have modulus 1.

Nice hint harazi, I hope that won't spoil the fun :lol:

Surely not, there is still some work to do after that!

Opps, I think I didn't correctly evaluate the difficulty. First, I thought it is a direct consequence of Kronecker's theorem, but it's not really like that. Ok, here's my proof. Suppose that the norm is smaller than 2 and introduce $B=(0,A,^{t}A, 0)$ notation by blocs, the first two elements being the first line.  It is not difficult to see that $C$ has the same norm as $B$ and thus smaller than 2. The idea is that hopefully, $C$ has all eigenvalues real numbers, because it is symetric. Moreover, any eigenvalue is surely in $(-2,2)$ because its norm is smaller than $2$. Because $C$ has integer coefficients, its characteristic polynomial is monic with integer coefficients and  all roots in $(-2,2)$. A direct application of Kronecker's theorem shows that all of its roots are of the form $2\cos (2\pi r)$ for some rational $r$. Because $C$ is diagonalisable in orthonormal basis, it is enough to prove that the maximal absolute value of a root of $P$ must be of the form $2\cos (\pi/k)$ for some integer $k$. Of course, using the irreducibility of the cyclotomic polynomial, we infer that if $2\cos (2\pi u/v)$ is a root of $P$, so is $2\cos(2\pi/v)$. A small case analysis (whether $v$ is odd or not) shows that indeed the maximal modulus of a root of $P$ has the form $2\cos(\pi/k)$ and this finishes the proof, except that as usual I forgot the case when $P$ is of the form $X^{2n}$, that is $C$ is nilpotent and symmetric, thus 0 and we find again the desired form. [/code]

Nice proof,
A detail
[quote="harazi"]... It is not difficult to see that $C$ has the same norm as $B$ and thus smaller than 2....[/quote]
you meant probably "It is not difficult to see that $B$ has the same norm as $A$ and thus smaller than 2", etc. because I don't see any $C$ matrix.

Yes, sorry, C is so close to B on french keyboard...


function
algebra
difference of squares
special factorizations

Refer to the picture.

I made this question up because of something my friend showed me.

I couldn't think of a way to describe this question so I had to draw it with MS Paint.  Below in the drawings are actually all squares although some might look like rectangles.  Heh...

Anyways, my question is...

In Square 1, there is 1 square that can be found.
In Square 2, there are 5 squares that can be found.
In Square 3, there are 14 squares that can be found.
In Square 4, there are 30 squares that can be found.

a) Make a formula that will be able to solve for Square N where N is any number from 1 to  :inf: .
b) Solve for Square 50.

Note: I'm not sure if this question can be solved however...

I think this is an easy question but I couldn't solve it ... as always.

Another friend of mine showed me the answer to the formula from a book...(highlight for spoilers)

[hide]n(n+1)(2n+1)/6[/hide]

Edit : Typos

Yep, that formula your friend showed you is correct.

Here's a question that will help you begin the path to figuring out why that is so:

What is (the number of squares in the 50th square) - (the number of squares in the 49th square)?
(Of course, 50 could be replaced there by any number, and the 49 by a number one less.)

Thx for your advice.

I thought about that as well.

The difference of squares between the consecutive large squares is always the square of a number.

For example, Square 1 has 1 square in total.  Square 2 has 5 squares in total therefore the difference is 4 which is the square of 2.  This pattern repeats.

But after that, I can't really figure out how the factors of the formula all get pieced together.

Okay, that's great -- so in other words, we could describe the number of squares in the [i]n[/i]th diagram as the sum of the first [i]n[/i] square numbers, right?

Now, one thing you can do is to check that the formula you have obeys that law -- that, starting with any value of n, if you increase n by 1, the function will increase in value by (n + 1)^2.  This is really just an implimentation of the mathematical method known as induction.  See if you can make it work in this case.

[quote="JBL"]
Now, one thing you can do is to check that the formula you have obeys that law -- that, starting with any value of n, if you increase n by 1, the function will increase in value by (n + 1)^2.  This is really just an implimentation of the mathematical method known as induction.  See if you can make it work in this case.[/quote]

When I did my work, I tried doing something just like (n+1)^2 and that's where I halted all my steps because I couldn't find any more solutions.


Suppose $ A$ has a right-inverse $ B$. Then $ AB \equal{} I$ leads to $ A^{T}AB \equal{} A^{T}$ or $ B \equal{} (A^{T}A)^{\minus{}1}A^{T}$. But that satisfies BA = I; it is a left inverse. What step is not justified?

Can someone point this one out, I can't seem to see any mistakes. Assuming A is m by n, it seems to work out.

THanks

$ A^T A$ is not necessarily invertible, is it?

Hi bythecliff,
Let $ A$ be a $ (m,n)$ matrix and $ B$ a $ (n,m)$ matrix.
if $ m\not\equal{}n$ you can't have $ AB\equal{}I_m$ and $ BA\equal{}I_n$ because $ rank(AB),rank(BA)\leq{inf(m,n)}$.
If $ m\equal{}n$ it's wellknown that $ AB\equal{}I$ implies $ BA\equal{}I$: indeed
$ AB\equal{}I$ implies $ det(A)\not\equal{}0$ implies $ A$ is invertible implies $ B\equal{}A^{\minus{}1}$.


geometry
circumcircle
search
geometry proposed

The incircle k of triangle ABC touches the sides at points A',B',C'.
For any point K on k, let d be the sum of the distances from K to the sides of the
triangle A'B'C'. Prove that KA + KB + KC > 2d.

I have an algebraic solution, but I am looking for a solution using Erdös Mordell

Let distances from $ K$ to the $ B^{'}C^{'},C^'A^',A^'B^',BC,CA,AB$ be $ x,y,z,a,b,c$. Then we have $ x^2 = bc,y^2 = ca,z^2 = ab$.(this can be proved as follow: Let $ P,Q,R$ be projection of $ K$ on $ AB,B^'C^',CA. \angle KQP = \angle KC^'P = \angle KB^'C^' = \angle KRQ$ and similarly, $ \angle KPQ = \angle KQR$. Hence triangle $ KQP$ and triangle $ KRQ$ are simialr, and we obtain $ x^2 = bc$). By Erdös Mordell theorem, $ KA + KB + KC\geq 2(a + b + c)\geq 2(\sqrt {bc} + \sqrt {ca} + \sqrt {ab}) = 2(x + y + z) = 2d$. Obviously, Equal isn't hold. Hence $ KA + KB + KC > 2d$.

[quote="m.candales"]Sorry but I can't contain my curiosity:

What are F1 and F2 in your avatar (I am supposing that O and N are the circuncircle and the Nagel point)  :oops:[/quote]
$ F_1,F_2$ is two Fermat point and $ N$ is nine-point center. My avatar is picture of Lester theorem( http://www.mathlinks.ro/viewtopic.php?search_id=298908544&t=16494)


$f: R\to R$
$fx)= x+m$, for x<=1
        $2mx-1$, for x>1
If the function is surjective, what is the $m$'s interval ?

$f$ automatically takes all values in $(-\infty,m+1]$. We then need that $2mx-1$ for $x>1$ take values in the rest of $\mathbb{R}$. So the minimum of $2mx-1$ must be $m+1$ or less. When $x\to1$, $2mx-1\to 2m-1$, so:

$2m-1\leq m+1$

yielding,

$m\leq2$

However, it's obvious that $2mx-1$ must go to $\infty$ for some $x$ also, which means that $m>0$. Hence,

$m\in(0,2]$


linear algebra
matrix
combinatorial geometry
linear algebra unsolved

Let $ A \in \mathcal{M}_n(\mathbb{C})$, $ H(A)\equal{}\{ X^*AX \quad | \quad  X \in \mathbb{C}^n, X^*X\equal{}1 \}$ ($ X^*$ denotes transpose conjugate).
For which $ n\in \mathbb{N}^*$, do we have the following statement:

If $ H(A)$ is the convex hull of the eigenvalues of $ A$, then $ A$ is normal.

Answer : For $ n\leq 4$
 This year it's really a nice problem at ENS , especially in compare with the other one ,but unfortunately I didn't succeed ... :(

I have a partial proof: with Facts 1-2 we can prove that the answer is YES if $ n\leq3$. Unfortunately we cannot conclude if $ n\equal{}4$.

There exists $ P$ unitary matrix s.t. $ P^{\minus{}1}AP$ is triangular. Thus we may assume that $ A\equal{}[a_{ij}]$ is upper triangular with $ a_{ii}\equal{}\lambda_i$. $ X^*AX\equal{}\sum_i{\lambda_i|x_i|^2}\plus{}\sum_{j>i}a_{ij}\bar{x_i}x_j$ where $ X\equal{}[x_i]$ satisfies $ \sum_i{|x_i|^2}\equal{}1$. Let $ \mathcal{C}$ be the convex hull of $ (\lambda_i)$ and $ \mathcal{B}$ its edge.
The question is: must $ A$ be a diagonal matrix ?
Fact 1: let $ i,j$ s.t. $ \lambda_i\not\equal{}\lambda_j$ and $ [\lambda_i,\lambda_j]\subset\mathcal{B}$.
Let $ X$ s.t. $ x_k\equal{}0$ except $ x_i\equal{}\alpha_i/\sqrt{2},x_j\equal{}\alpha_j/\sqrt{2}$ where $ |\alpha_i|\equal{}|\alpha_j|\equal{}1$. $ X^*AX\equal{}\lambda_i/2\plus{}\lambda_j/2\plus{}\bar{\alpha_i}\alpha_j{a_{ij}}/2$. $ X^*AX\in\mathcal{C}$ and $ \lambda_i/2\plus{}\lambda_j/2\in\mathcal{B}$. Then the complex number $ \bar{\alpha_i}\alpha_j{a_{ij}}$ is always in the same half plane for all $ \alpha_i,\alpha_j$. This fact implies that $ a_{ij}\equal{}0$.
Fact 2: let distinct $ i,j$ s.t. $ \lambda_i\equal{}\lambda_j$.
If $ \mathcal{C}\not\equal{}\{\lambda_i\}$ then $ \bar{\alpha_i}\alpha_j{a_{ij}}$ is in a fixed angular sector (eventually in a half plane) for all $ \alpha_i,\alpha_j$. This implies that $ a_{ij}\equal{}0$.
If $ \mathcal{C}\equal{}\{\lambda_i\}$ then $ \bar{\alpha_i}\alpha_j{a_{ij}}\equal{}0$ and $ a_{ij}\equal{}0$.


abstract algebra
function
algebra
polynomial
topology
Ring Theory
superior algebra

At the site http://www.math.uga.edu/~bensondj/ I found this remarkable remark:

[b]Foundations sometimes matter
The projective dimension of $\mathbb C(x,y,z)$ as a module over $\mathbb C[x,y,z]$ is:
two if the continuum hypothesis holds and three otherwise.[/b]

I assume C(x,y,z) is a function field over the complex, and C[x,y,z] is a polynomial ring, so I understand that the former is a module over its subring. What is the "projective dimension" here?

LiMa

Hi Lima ;-),

is the general definition unclear?

http://planetmath.org/encyclopedia/ProjectiveDimension.html

I guess that results of homological algebra may involve foundations by the recursive construction of homotopy equivalences of arbitrary projective resolutions of a module. But I can't imagine what does this matter in the case of finite resolutions  :maybe:. Well, perhaps CH admits the construction of curious modules, which may shorten the projective dimension. :-)

... after a short google-session ;-)

CH plays even often a role dealing with projective dimensions. For an example, see

mpcs.usip.edu/faculty/vas/homepage_sa_UMDja/Semisimple_abstract.pdf

But this seems to be quite difficult to understand :/


function
logarithms
real analysis
real analysis unsolved

Can anyone give an example of a function $ f \in L^p(\Bbb R)$ such that $ f \notin L^h(\Bbb R)$ for any $ h > p$? ($ p$ is finite)

Try $ f(x)\equal{}|x\ln^2x|^{\minus{}1/p}$ for $ x\in (0,1/2)$ and zero otherwise.


calculus
integration
trigonometry
function
absolute value
complex analysis

MAIN QUESTION:

Does      $ |(cos(R^2 cos(2 \theta)) \plus{} i sin(R^2 cos(2 \theta)))| \equal{} 1$ UNIFORMLY even though
the absolute value of  $ e^{i R^2 e^{i 2 \theta}}$ is NOT equal to 1?

And if so, IN WHAT CASES THEN DO A NUMBER OF THE FORM $ e^{i \theta}$ actually have an absolute magnitude of 1? A theta value of $ R^2 cos(2 \theta)$ seems to work (correct me if I'm wrong), but a theta value of $ R^2 e^{i 2 \theta}$ doesn't seem to work. I'd like to know why (besides the "you need to simplify it" explanation - which does not help with my intuition here).
==

So I want to evaluate $ \int_0^{\infty} e^{i z^2}$. One of the central steps is to evaluate the integral around the contour line $ z \equal{} R e^{i \theta}$. So I substitute in $ z \equal{} R e^{i \theta}$ so that I get $ e^{i R^2 e^{i 2 \theta}} \equal{} e^{i R^2 cos(2 \theta) \minus{} R^2 sin(2 \theta)}.$. I have to show convergence to 0. This equals $ e^{i R^2 cos(2 \theta)} e^{ \minus{} R^2 sin(2 \theta)} \equal{} (cos(R^2 cos(2 \theta)) \plus{} i sin(R^2 cos(2 \theta))) e^{ \minus{} R^2 sin(2 \theta)}$. So my sources say that the absolute value of this becomes $ e^{ \minus{} R^2 sin(2 \theta)},$ implying that $ |(cos(R^2 cos(2 \theta)) \plus{} i sin(R^2 cos(2 \theta)))| \equal{} 1$ throughout the entire interval. Of course, the absolute value of $ e^{i \theta}$ is 1 as long as the number is of this form. But then, why can't the absolute value of $ e^{i R^2 e^{i 2 \theta}}$ also be 1 throughout the range? What properties make a the absolute value of a number of the form $ e^{i \theta}$ NOT converge to 1 for ANY value of $ \theta$?

====

Another question: Since I have to evaluate       $ \int_0^{\infty} e^{i z^2} dz$, where $ dz \equal{} i e^{i \theta} d \theta$ - I have to multiply this equation by that factor. I probably have to do this for the final solution. But since this is bounded by i, and my only objective is to show that this integral converges to 0, this factor of dz should be effectively negligible, right?

==

On your first question- of course not. That absolute value relation only holds when the argument of the sines and cosines is pure real. We do have that $ \sin^2 t\plus{}\cos^2 t\equal{}1$ everywhere, but that's different.

The integral you're trying to find is along the $ x$-axis, so there are no extra multiples- it's just $ \int_0^{\infty} e^{iz^2}\,dz$ there.

On the circular arc $ z\equal{}Re^{i\theta}$, we can estimate
$ \left|\int_{Re^{i\alpha}}^{Re^{i\beta}}e^{iz^2}\,dz\right|\le \int_{Re^{i\alpha}}^{Re^{i\beta}}\left|e^{iz^2}\right|\,|dz|\equal{} \int_{\alpha}^{\beta}\left|\exp(iR^2e^{2i\theta})\right|\cdot \left|iRe^{i\theta}\right|\,d\theta$
$ \left|\int_{Re^{i\alpha}}^{Re^{i\beta}}e^{iz^2}\,dz\right|\le \int_{\alpha}^{\beta}\exp\left(\minus{}R^2\cdot\text{Im}(e^{2i\theta})\right)\cdot R\,d\theta$

Now, that $ \sin 2\theta$ comes from the imaginary part. The $ \alpha$ and $ \beta$ you want are zero and $ \frac{\pi}{2}$- the first quadrant piece.
Proving this integral goes to zero is tricky, because the function we're integrating only goes to zero away from the endpoints. Integration by parts is the most elementary way of dealing with this.

[quote="Simfish"]MAIN QUESTION:

Does      $ |(cos(R^2 cos(2 \theta)) + i sin(R^2 cos(2 \theta)))| = 1$ UNIFORMLY even though
the absolute value of  $ e^{i R^2 e^{i 2 \theta}}$ is NOT equal to 1?

And if so, IN WHAT CASES THEN DO A NUMBER OF THE FORM $ e^{i \theta}$ actually have an absolute magnitude of 1? A theta value of $ R^2 cos(2 \theta)$ seems to work (correct me if I'm wrong), but a theta value of $ R^2 e^{i 2 \theta}$ doesn't seem to work. I'd like to know why (besides the "you need to simplify it" explanation - which does not help with my intuition here).
[/quote]

You forgot the $ e^{ - R^2Sin(2\theta)}$ part:

$ e^{iR^2e^{i2\theta}} = e^{iR^2[Cos(2\theta) + iSin(2\theta)]}$

$ = e^{iR^2Cos(2\theta) - R^2Sin(2\theta)}$

$ = e^{ - R^2Sin(2\theta)}\left[Cos(R^2Cos2\theta) + iSin(R^2Cos2\theta)\right]$

so the absolute value of $ e^{iR^2e^{i2\theta}}$ is $ |e^{ - R^2Sin(2\theta)}|$ and that's going to be all kinds of values depending on R and $ \theta$.

Now, $ |e^{iw}| = |Cos[w] + iSin[w]|$  which is 1 when $ w$ is real but when w is a complex number it's:

$ e^{i(x + iy)} = e^{ - y + ix}$ and so:

$ |e^{i(x + iy)}| = |e^{ - y}|$

So now you see when the complex component of w is zero, then the absolute value of $ e^{iw}$ is 1.  In your case,

$ w = R^2e^{i2\theta} = R^2[Cos(2\theta) + iSin(2\theta)]$

so the complex component is in general not zero so the absolute value of your expression won't in general be 1.

[quote="jmerry"]
Proving this integral goes to zero . . . [/quote]

First I'd make the substitution $ \phi\equal{}2\theta$ and obtain:

$ \frac{R}{2}\int_0^{\pi} Exp[\minus{}R^2 Sin(\phi)]d\phi$

Now we wish to find an upper bound for this integral.  Note the plot below:

[img]http://img87.imageshack.us/img87/3495/plotaik6.jpg[/img]

On Blue:  $ Sin(\phi)\geq \frac{2}{\pi} \phi$

On Green:   $ Sin(\phi)\geq \minus{}\frac{2}{\pi} \phi\plus{}2$

Then I can write:

$ \left|\frac{R}{2}\int_0^{\pi} Exp[\minus{}R^2 Sin(\phi) d\phi\right|
\leq \frac{R}{2}\left(\left|\int_0^{\pi/2} Exp[\minus{}R^2\frac{2\phi}{\pi}]d\phi\right|\plus{}
\left|\int_{\pi/2}^{\pi}Exp[\minus{}R^2(\minus{}\frac{2}{\pi}\phi\plus{}2)]d\phi\right|\right)$

evaluating the integrals I obtain:

$ \left|\int_0^{\pi/2}R Exp[\minus{}R^2 Sin(2\theta)]d\theta\right|\leq \frac{R}{2}\left[\frac{\pi}{2R^2}(1\minus{}e^{\minus{}R^2})\plus{}\frac{\pi}{2R^2}(1\minus{}e^{\minus{}R^2})\right]$

or:

$ \left|\int_0^{\pi/2}R Exp[\minus{}R^2 Sin(2\theta)]d\theta\right|\leq \frac{\pi}{R}\left(1\minus{}e^{\minus{}R^2}\right)$

which goes to zero as $ R\to\infty$


As I know, of the 5 fellows, there'll be one receiving the putnam scholarship to Harvard, but I can't find past scholars. Anyone knows? and how would they decide who'll receive the scholarship?

[quote="transseries"](At least long ago, when I was Putnam scholar)
I think admission to graduate study at Harvard is in practice never denied to that winner.  Even before he completes his bachelor's degree: when I was at Harvard, another Putnam scholar began there immediately without completing his bachelor's degree.[/quote]

What?!  You can do that?  Can you go to grad school in math without completing a bachelor's degree first?

Yes, it occasionally happens, but for obvious reasons it's quite rare.

[quote="sexynsmartjenny"]What?!  You can do that?  Can you go to grad school in math without completing a bachelor's degree first?[/quote]

You can if the graduate school admits you.

[quote="ohhotgirl"][quote="transseries"](At least long ago, when I was Putnam scholar)
The advisors/professors of the 5 Putnam Fellows write recommendations, and one is selected based on that.

I think admission to graduate study at Harvard is in practice never denied to that winner.  Even before he completes his bachelor's degree: when I was at Harvard, another Putnam scholar began there immediately without completing his bachelor's degree.

In fact the fellowship (one year tuition plus generous allowance for living and expenses at Harvard) need not be used for graduate study in mathematics.  It has sometimes been used for undergraduate study (by a student already at Harvard), and it has at least once been used for Harvard Medical School.

The volumes published by the MAA on the Putnam have listings of the winners at the back, and the holders of the Putnam Fellowship are noted in the first volume (1938--1964) but not in the later ones.[/quote]

Thank you for the info! I ask about this since I am quite confident to become a fellow this year or next year, and going to harvard is my dream. getting the scholarship helps my admission to harvard much easier.[/quote]

wow, what is your name?

If you have the ability to become a fellow, I don't think there is any difficulty for you to get into Harvard.

Putnam fellows never need to worry about grad admissions

[quote="JBL"][quote="EulerFish"]Putnam fellows never need to worry about grad admissions[/quote]  What basis do you have for making this statement?  (My guess: none whatsoever.)[/quote]

I don't mean that the title "putnam fellow" can guarantee admissions.

What I tried to say was that if someone has the ability to be a putnam fellow, he should have made a background strong enough to earn admissions from top grad schools without much difficulty.

[quote="jmerry"]No, it doesn't work that way. The Putnam is a very strong positive factor, but the top schools will care if you screw up otherwise- and people do. Like me.[/quote]
How did you "screw up"?

[quote="ohhotgirl"]

Thank you for the info! I ask about this since I am quite confident to become a fellow this year or next year, and going to harvard is my dream. getting the scholarship helps my admission to harvard much easier.[/quote]

Did you take putnam last Saturday?

It is amazing to heard someone saying "confident to become a fellow".


$m, m+1 , ... , m+n$ are consecutive natural numbers and sum of numbers on digits $m$ and $m+n$ are divisible 8, but  sum of numbers on digits other natural numbers are not divisible 8. Find maximum value of $n$.

$n\le 15$
Let $d(n)$ be the sum of the digits of n.
Note that if the number last digit of $k$ isn't $9$ we have that
$d(k)+1=d(k+1)(\mod 8)$.
If the last $t$ digits of $k$ are $9$ we have that:
$d(k)+1-t\equiv d(k+1) \left(\mod 8\right)$ because $9=1 (\mod 8)$
If the last digit of $m$ is $0$ or $1$ we have that
$d(m)= d(m+8) (\mod 8)$
If the last digit of $m$ is $a\not=0,1$ then $n \le 10-a+7 \le 10-2+7 = 15$ Because the first digit  $n+k$ that ends at $0$ is at least $1$ modulo $8$ and one number between $d(n+k), d(n+k+1)...,d(n+k+7)$ is $0$ modulo $8$.
Now we note that if $m=9999992$ the secuence $d(m),d(m+1),...,d(m+15)$ modulo 8 is:
$0,1,2,3,4,5,6,7,1,2,3,4,5,6,7,0$ and we are done.


[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=128763]Done[/url]  ($a=b=1$ satisfy  $a\not =0,b>0,a+b\not =1$)

Reviewing old problems, any solutions here???

[quote=N.T.TUAN]Find all continuous functions $f : \mathbb{R}\to\mathbb{R}$ such that for all $x \in\mathbb{ R}$,
\[f (f (x)) = f (x)+x.\][/quote]

See http://artofproblemsolving.com/community/c6h1344617p7316804 subcase 7.1.1


Let $ a, b, c$ be positive real numbers. Prove that: $ \frac{a^2}{2a^2\plus{}ab\plus{}c^2}\plus{}\frac{b^2}{2b^2\plus{}bc\plus{}a^2}\plus{}\frac{c^2}{2c^2\plus{}ca\plus{}b^2} \le \frac{3}{4}$
 :maybe:

[quote="nguoivn"]Let $ a, b, c$ be positive real numbers. Prove that: $ \frac {a^2}{2a^2 \plus{} ab \plus{} c^2} \plus{} \frac {b^2}{2b^2 \plus{} bc \plus{} a^2} \plus{} \frac {c^2}{2c^2 \plus{} ca \plus{} b^2} \le \frac {3}{4}$
 :maybe:[/quote]

assume $ a \equal{} max{(a,b,c)}$

The following inequality is equivalent to

$ (2b^2 \plus{} 7bc \plus{} 2ca)(a \minus{} b)^2(a \minus{} c)^2 \plus{} ((12acb^2 \plus{} 10ac^2b\minus{} 22b^2c^2)$$ \plus{} ac^3 \plus{} b^4 \plus{} 8b^3c \plus{} 7ab^3 \plus{} 4bc^3 \plus{} 3c^4)(a \minus{} b)(a \minus{} c)$

$ \plus{} ((3a \minus{} c)b^3 \plus{} (15ca \minus{} 10c^2)b^2 \plus{} (15ac^2 \minus{} c^3)b \plus{} 3ac^3)(a \minus{} b)^2\geq 0$

obvious true.


What is the units digit of the prime number $ 2^{44,497}\minus{}1$?

The units digit of powers of 2 repeats as follows:

2,4,8,6,2,4,...

44497=1 mod 4, so the answer is 2-1=1.


geometry
geometric transformation
rotation
ratio
symmetry
reflection
similar triangles

This is Tournament of the Towns 2004 problem (8 points)
But I think it is easier than that one with geometrical locus.

Angle AOB and COD are combined with a rotation such that
ray OA is combined with a ray OC, OB with OD.
In them are inscribed two circles which intersect in points E and F.
Prove that angle AOE  is equal to the angle DOF.

Just wondering, had all the countries finished Tournament of Towns yet? I'm not sure how this contest is conducted internationally.

This is a sketch. Some details are left out.

I'll assume WLOG (because I don't want to name other points) that $A,B$ are the points of tangency between $OA,OB$ and the circle inscribed in $\angle AOB$, and the same for $C,D$. In this case we replace the word "rotation" in the text with "spiral similarity". I'm sure you won't mind :).

Let $\mathcal C_1$ be the circle inscribed in $\angle AOB$ and $\mathcal C_2$ the other one, and let $r_1,r_2$ be their respective radii, and $p_1,p_2$ the powers of $O$ wrt these two circles (respectively). Also, let $P$ be the second pt of intersection between $OE$ and $\mathcal C_2$, while $Q$ is the second pt of intersection between $OF$ and $\mathcal C_1$. Assume for now that we have shown $\frac{OE}{OQ}=\frac{OP}{OF}=\frac{r_1}{r_2}\ (*)$. In this case we find the triangles $QCF,EBP$ to be similar, and we’re pretty much done.

In order to prove $(*)$, let $X$ be the intersection between the tangent at $E$ to $\mathcal C_1$ and the tangent at $F$ to $\mathcal C_2$. Let $Y$ be the second intersection between $XF$ and $\mathcal C_1$, and $Z$ the second intersection between $XE$ and $\mathcal C_2$. Try to show that $(1)\ \frac{XE}{XF}=\frac{r_1}{r_2}$ and $(2)\ XEF$ is similar to both $OEQ$ and $OPF$.

Another solution:

Again assume that $A, B, C, D$ are points of the relavant tangencies. Let $O_1, O_2$ be the centres of the two circles respectively ($O_1$ for the circle in $\angle AOB$).

Since $\angle AOB= \angle COD$, we have $\angle AOO_1 = DOO_2$. Similar triangles give $\frac{O_1O}{O_2O} = \frac{r_1}{r_2}$ (the $r$'s are the radii of the circles)

Let the internal angle bisector of $\angle O_1OO_2$ touch $O_1O_2$ at $X$. Then all four ratios

\[
\frac{O_1O}{O_2O}, \frac{O_1E}{O_2E}, \frac{O_1X}{O_2X}, \frac{O_1F}{O_2F}
\]
are equal to $r_1/r_2$. By the Apollonius circle, the points $O, E, X, F$ are concyclic (or collinear in the trivial case $r_1=r_2$.) By symmetry, $XE=XF$, so $\angle XOE = \angle XOF$. After knowing this, the rest is just trivial angle calculations about the point $O$.

Wow, billzhao you have a very pretty solution, nobody from our country
didn't solve it in a such way!  Very good idea with Apollonius circle.

 Actually all who solved it used the same idea that Grobber used,
For example I just explained solution using inversion and symmetry
- similar to Grobber spiral similarity.

This might come as a shock to you, but I for one was very much thinking about an Apollonius circle :).

I also solved it wth apollonius circle during the test! gteat billzhao!

Here's my solution, in the test. I found it quite easy for an 8 points problem.
 Let the circles C1 and C2 be tangent at A,B,C and D to the lines. Then consider the inversion with center O and ratio OA.OD . Clearly it takes the circle C1 to a circle equal to C2 but reflected across the angle bisector of AOD, and the same with C2 to C2' equal to C1 but reflected. Suposse E maps to E' . It is obvious now to reflect the inverted figure across the angle bisector of AOD. C2' maps to C2'',  C1' maps to C1'' and E' to E'' . Compare the obtained figure with the original one. It is obvious that C1'' = C2 , C2'' = C1, therefore E'' = F, and we know that the line OE'' (the same as OF) is the reflection of the line OE' (the same as OE), so OF is the reflection across the angle bisector of AOD of the line OE, and the conclussion follows.
hope u could understand what i did..


Hi,
I have a deadLine untill Thursday to find out the solutions to the following Questions.
Can Anyone Help me??? please??

1) A cylinder with Relative Density=0.5 and height=10m is holding vertical with the bottom base touching the surface of water.
If we leave the cylinder to move vertical towards water, Find the maximum depth in which the upper base of cylinder will reach.

2) Flow of Water in an experimental tube with diametre 100mm has pressure fall 10.000 Pascal with n(a coefficient)=0.01. Find the length of the tube when Turbulent appears with Re=1000000

3) Water is flowing in a Venturi Meter of 400mm X 100mm with discharge 72000lit/h and  pipe difference 250mm.
If the 'narrowing' is 200mm X 50 mm,
find the coefficient z of the meter.

4) An Open reservoir fill with water has at it's side walls and in 2meters depth from the surface, Orifice(hole) with diametre=25mm.
a) Find the Velocity of Water at the Exit of the Orifice
b) At the Orifice we Adjust a Pipe with diametr=25mm and length=9meters. Find the velocity at the exit of the Pipe.
c) How much % has the Velocity increase or Decrease??
Generally, take the water flow normal.

5) A cube with edge of 10 cm and Relative Density p=1.2 is equalising in 2 non-mixable fluids of relative density a=0.6 and b=1.6.
Find the parts of the Cube's Edge that are Immersed for each case.

PLEEEEEAAAASSSSEEEEEEE!!!!!

YO new member.

FOr the first question,i'm not really getting the wordings.Could u be more clear?

I think the fourth one has something to do with Toricelli's equation.Directly substitute the values i guess.

[quote="shadysaysurspammed"]YO new member.

FOr the first question,i'm not really getting the wordings.Could u be more clear?
quote]

Well, Excuse my English,
What The Problem means is that We have a Cylinder that we are holding above the surface of water (the down part of cylinder touches the water) and the question is :
If we leave it to fall towards the water , the upper part of cylinder in what depth will reach (I suppose before changing it's vertical direction to some angle).
So, We have it's 'Relative Density' =0.5 and the cylinder height=10meters.

For All I know, the 'Relative Density' =( Weight of Object / weight of moved water )

I don't know if this is the right Solution but I think something like :

Weight of Cylinder = Weight of Moved Water   ==>
9,81(gravity) X 0.5 (Relative Density) X 1000  = 9810 X(10m X h)     [where h=what we are looking for].

so solving this , should be something like : h= 0.05 m

But I guess this is wrong cause the Result (h=0.05m) seems too small as a number. If I could guess I'd say h=50meters but who knows!

That's why I want your help.!!

Please, I've LOST all my notes and books in a fire!!! [and I need answers till Thursday]

Thanks again.

[quote="shadysaysurspammed"] What starts with "f" and ends in "uck"?

answer-Firetruck!!(heheh,i know what u were thinkin:)) [/quote]Really? I thought it was "Friar Tuck," but then that must be because I just watched "Robin Hood (Prince of Thieves)" the other night on TV :)


geometry inradius inequalities circumcircle trigonometry function incenter

Notation "p" is generally accepted for semiperimeter. Am I right?

I'm more accustom to having $s$ as the semiperimeter and $p$ as the perimeter.

But that goes along with Darij's [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=138794#138794]comment[/url].

[quote="nickolas"]In any triangle ABC  proove  [tex] \frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\geq\frac{p}{2r} [/tex]

[i]Edited by Myth[/i][/quote]

Myth, you hardly made the problem easier to understand. I have always been used to denoting the semiperimeter by s (if my memory doesn't cheat me, this notation goes back to Euler). Could you better just write explicitely what is meant by $m_a$, ..., p, r, as I proposed in another thread? Thanks.

  Darij

I hear that $s$ is a semiprimetr for the first time. In all russian books I saw notation $p$ is used (see geometr's bible Prasolov's "Planimetry").

[quote="nickolas"]In any triangle ABC  proove  [tex] \frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\geq\frac{p}{2r} [/tex][/quote]

At first, the $\geq$ sign should actually be an $\leq$ sign here, and also, I rewrite the problem using the notations I prefer:

[color=blue]In any triangle ABC with semiperimeter $s=\frac{a+b+c}{2}$, inradius r and medians $m_a$, $m_b$, $m_c$, we have $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s}{2r}$.[/color]

[i]Proof.[/i]

[color=red][b]EDIT:[/b] The proof below is WRONG. Thanks to Mak for pointing this out. In fact, the "well-known inequalities" $m_a\leq\sqrt{s\left(s-a\right)}$, $m_b\leq\sqrt{s\left(s-b\right)}$, $m_c\leq\sqrt{s\left(s-c\right)}$ which I used are incorrect; actually, one has $m_a\geq\sqrt{s\left(s-a\right)}$, $m_b\geq\sqrt{s\left(s-b\right)}$, $m_c\geq\sqrt{s\left(s-c\right)}$.

Sorry to everybody for the wrong proof. I will try to find a correct one.[/color]

Using the well-known inequalities $m_a\leq\sqrt{s\left(s-a\right)}$, $m_b\leq\sqrt{s\left(s-b\right)}$, $m_c\leq\sqrt{s\left(s-c\right)}$, we find

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{\sqrt{s\left(s-a\right)}}{a}+\frac{\sqrt{s\left(s-b\right)}}{b}+\frac{\sqrt{s\left(s-c\right)}}{c}$.

Hence, in order to prove the inequality

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s}{2r}$,

it will be enough to establish the sharper inequality

$\frac{\sqrt{s\left(s-a\right)}}{a}+\frac{\sqrt{s\left(s-b\right)}}{b}+\frac{\sqrt{s\left(s-c\right)}}{c}\leq\frac{s}{2r}$.

Upon division by $\sqrt{s}$, this inequality becomes

$\frac{\sqrt{s-a}}{a}+\frac{\sqrt{s-b}}{b}+\frac{\sqrt{s-c}}{c}\leq\frac{s}{2r\sqrt{s}}$.

Now, let's transform the right hand side of this inequality: On the one hand,

(s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s.

On the other hand, by a well-known formula for the inradius of a triangle,

$r=\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}$,

so that

$r\sqrt{s}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}\sqrt{s}=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.

Hence, the fraction on the right hand side of our inequality can be transformed as follows:

$\frac{s}{2r\sqrt{s}}=\frac{\left(s-a\right)+\left(s-b\right)+\left(s-c\right)}{2\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}$
$=\frac{\left(s-a\right)}{2\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}+\frac{\left(s-b\right)}{2\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}+\frac{\left(s-c\right)}{2\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}$
$=\frac{\sqrt{s-a}}{2\sqrt{\left(s-b\right)\left(s-c\right)}}+\frac{\sqrt{s-b}}{2\sqrt{\left(s-c\right)\left(s-a\right)}}+\frac{\sqrt{s-c}}{2\sqrt{\left(s-a\right)\left(s-b\right)}}$.

Hence, the inequality that we must prove rewrites as

$\frac{\sqrt{s-a}}{a}+\frac{\sqrt{s-b}}{b}+\frac{\sqrt{s-c}}{c}$
$\leq\frac{\sqrt{s-a}}{2\sqrt{\left(s-b\right)\left(s-c\right)}}+\frac{\sqrt{s-b}}{2\sqrt{\left(s-c\right)\left(s-a\right)}}+\frac{\sqrt{s-c}}{2\sqrt{\left(s-a\right)\left(s-b\right)}}$.

But this follows from the following three inequalities:

$\frac{\sqrt{s-a}}{a}\leq\frac{\sqrt{s-a}}{2\sqrt{\left(s-b\right)\left(s-c\right)}}$;
$\frac{\sqrt{s-b}}{b}\leq\frac{\sqrt{s-b}}{2\sqrt{\left(s-c\right)\left(s-a\right)}}$;
$\frac{\sqrt{s-c}}{c}\leq\frac{\sqrt{s-c}}{2\sqrt{\left(s-a\right)\left(s-b\right)}}$.

So, to complete the proof, it remains to verify these three inequalities. We will only verify the first inequality,

$\frac{\sqrt{s-a}}{a}\leq\frac{\sqrt{s-a}}{2\sqrt{\left(s-b\right)\left(s-c\right)}}$;

the other two are just its cyclic permutations.

The proof of the inequality $\frac{\sqrt{s-a}}{a}\leq\frac{\sqrt{s-a}}{2\sqrt{\left(s-b\right)\left(s-c\right)}}$ is very easy: First, we divide it by $\sqrt{s-a}$ and obtain

$\frac{1}{a}\leq\frac{1}{2\sqrt{\left(s-b\right)\left(s-c\right)}}$;

this is obiviously equivalent to

$a\geq 2\sqrt{\left(s-b\right)\left(s-c\right)}$.

Now we square this:

$a^2\geq 4\left(s-b\right)\left(s-c\right)$.

But

$4\left(s-b\right)\left(s-c\right)=\left(2\left(s-b\right)\right)\cdot\left(2\left(s-c\right)\right)=\left(c+a-b\right)\cdot\left(a+b-c\right)$
$=\left(a-\left(b-c\right)\right)\cdot\left(a+\left(b-c\right)\right)=a^2-\left(b-c\right)^2$,

so that $a^2\geq 4\left(s-b\right)\left(s-c\right)$ rewrites as $a^2\geq a^2-\left(b-c\right)^2$, what is trivial since $\left(b-c\right)^2\geq 0$.

Proof complete.

Note that in the initial inequality $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s}{2r}$, equality holds if and only if the triangle ABC is equilateral. This is easy to see from the above proof.

  Darij

In  fact , this is  a prove  for  the ineq

[tex] \frac{w_a}{a}+\frac{w_b}{b}+\frac{w_c}{c}\leq \frac{s}{2r} [/tex]
where  [tex] w_a , w_b , w_c [/tex] are  angle  bisectors  and  [tex]  s  [/tex] is semiperimeter

[quote="darij grinberg"]Hey, but the proof is wrong!![/quote]
At least, it is a good edification for long non-conceptual proofs. :?

One man asked me some days ago to prove $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\geq\frac{3\sqrt 3}{2}$.

I proved it. And now I can't find the topic, which contain this inequality, though I remember we discussed it. I want to know what solution was proposed here.
Does anybody know?

[quote="Myth"]One man asked me some days ago to prove $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\geq\frac{3\sqrt 3}{2}$.

I proved it. And now I can't find the topic, which contain this inequality, though I remember we discussed it. I want to know what solution was proposed here.[/quote]

Hmm, I don't remember whether there was a separate topic for this inequality, but it is a special case of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21016]the general geometric inequality[/url] stating that $\frac{MA}{a}+\frac{MB}{b}+\frac{MC}{c}\geq\sqrt3$ for every point M in the plane of the triangle ABC (in your case, one has just to take M = centroid of triangle ABC).

By the way, are there any advances on the initial inequality $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s}{2r}$? I have showed it for all acute-angled triangles ABC, but I can't prove it for obtuse-angled ones...

  darij

[quote="darij grinberg"][color=blue]In any triangle ABC with semiperimeter $s=\frac{a+b+c}{2}$, inradius r and medians $m_a$, $m_b$, $m_c$, we have $\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s}{2r}$.[/color][/quote]

After giving a wrong proof of this inequality in post #9, I have now succeeded to find a correct one, although I find it quite ugly. Anyway, here it goes:

First let's consider the case when the triangle ABC is acute-angled or right-angled. Let D be the midpoint of its side BC. Then, the segment AD is the A-median of triangle ABC and thus has the length $AD=m_a$. On the other hand, if O is the circumcenter of triangle ABC, then $OD\perp BC$ (since the point O, being the circumcenter of triangle ABC, must lie on the perpendicular bisector of its side BC). Thus, in the right-angled triangle ODC, we have $OD=OC\cdot\cos\measuredangle DOC$. But since the point O is the circumcenter of triangle ABC, we have < BOC = 2 < BAC = 2A by the central angle theorem, and also we have OA = OB = OC = R, where R is the circumradius of triangle ABC. Thus, the triangle BOC is isosceles with OB = OC, and hence the line OD, being the altitude from its apex O (since $OD\perp BC$), must bisect its angle < BOC. Hence, $\measuredangle DOC=\frac{\measuredangle BOC}{2}=\frac{2A}{2}=A$. Thus, $OD=OC\cdot\cos\measuredangle DOC=R\cdot\cos A$.

Now, by the triangle inequality in the (possibly degenerate) triangle AOD, we have

$AD\leq OA+OD=R+R\cdot\cos A=R\cdot\left(1+\cos A\right)=R\cdot 2\cos^2\frac{A}{2}=2R\cos^2\frac{A}{2}$.

By the extended law of sines, $2R=\frac{a}{\sin A}$; thus,

$AD\leq 2R\cos^2\frac{A}{2}=\frac{a}{\sin A}\cdot \cos^2\frac{A}{2}=a\frac{\cos^2\frac{A}{2}}{\sin A}$
$=a\frac{\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{a}{2}\cdot\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}=\frac{a}{2}\cdot\cot\frac{A}{2}$.

Since $AD=m_a$, this becomes $m_a\leq\frac{a}{2}\cdot\cot\frac{A}{2}$. Dividing this by a, we get $\frac{m_a}{a}\leq\frac12\cdot\cot\frac{A}{2}$. By the half-angle laws, $\tan\frac{A}{2}=\frac{r}{s-a}$, so that

$\frac{m_a}{a}\leq\frac12\cdot\cot\frac{A}{2}=\frac12:\tan\frac{A}{2}=\frac12:\frac{r}{s-a}=\frac{s-a}{2r}$.

Similarly, $\frac{m_b}{b}\leq\frac{s-b}{2r}$ and $\frac{m_c}{c}\leq\frac{s-c}{2r}$. Thus,

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s-a}{2r}+\frac{s-b}{2r}+\frac{s-c}{2r}=\frac{3s-\left(a+b+c\right)}{2r}=\frac{3s-2s}{2r}=\frac{s}{2r}$.

Thus, the inequality in question is proved for the case of an acute-angled or right-angled triangle ABC.

The case when the triangle ABC is obtuse-angled is much more tricky. In fact, assume WLOG that A is the obtuse angle. Then, we have $OD=-R\cdot\cos A$ rather than $OD=R\cdot\cos A$, and the inequality $m_a\leq\frac{a}{2}\cdot\cot\frac{A}{2}$ doesn't hold. But rather rough estimates can be applied:

First, we have A > 90°; in other words, $\measuredangle CAB>90^{\circ}$. Thus, the point A lies inside the circle with diameter BC. The center of this circle is the midpoint D of the segment BC, and the radius of this circle is $\frac{BC}{2}=\frac{a}{2}$. Thus, $AD<\frac{a}{2}$. Since $AD=m_a$, this becomes $m_a<\frac{a}{2}$. Hence, $\frac{m_a}{a}<\frac12$.

For the next estimate, we will use vectors. If E is the midpoint of the segment CA, then $\overrightarrow{BE}=\frac{\overrightarrow{BC}+\overrightarrow{BA}}{2}$, so that

$BE=\left|\overrightarrow{BE}\right|=\left|\frac{\overrightarrow{BC}+\overrightarrow{BA}}{2}\right|=\frac12\cdot\left|\overrightarrow{BC}+\overrightarrow{BA}\right|\leq\frac12\cdot\left(\left|\overrightarrow{BC}\right|+\left|\overrightarrow{BA}\right|\right)$       (by the triangle inequality)
$=\frac12\cdot\left(BC+BA\right)=\frac12\cdot\left(a+c\right)$.

Since the point E is the midpoint of the side CA of triangle ABC, the segment BE is the B-median of this triangle; thus, its length is $BE=m_b$, and hence this inequality becomes $m_b\leq\frac12\cdot\left(a+c\right)$. Similarly, $m_c\leq\frac12\cdot\left(b+a\right)$.

Thus,

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}<\frac12+\frac{\frac12\cdot\left(a+c\right)}{b}+\frac{\frac12\cdot\left(b+a\right)}{c}$
$=\frac12\cdot\left(1+\frac{a+c}{b}+\frac{b+a}{c}\right)=\frac12\cdot\left(1+\left(\frac{a+b+c}{b}-1\right)+\left(\frac{a+b+c}{c}-1\right)\right)$
$=\frac12\cdot\left(\frac{a+b+c}{b}+\frac{a+b+c}{c}-1\right)=\frac12\cdot\left(\frac{2s}{b}+\frac{2s}{c}-1\right)=\frac{s}{b}+\frac{s}{c}-\frac12$.

Let F be the area of triangle ABC; then, by a formula for the area of a triangle, $F=\frac12bc\sin A$. Since the angle A is obtuse, sin A < 1, and thus this becomes $F<\frac12bc$. On the other hand, by another formula for the area of a triangle, F = rs. Thus, $rs<\frac12bc$. In other words, 2rs < bc. This yields $\frac{s}{b}<\frac{c}{2r}$ and $\frac{s}{c}<\frac{b}{2r}$. Thus,

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}<\frac{s}{b}+\frac{s}{c}-\frac12<\frac{c}{2r}+\frac{b}{2r}-\frac12=\frac{c+b}{2r}-\frac12$
$=\frac{\left(a+b+c\right)-a}{2r}-\frac12=\frac{2s-a}{2r}-\frac12=\frac{s+\left(s-a\right)}{2r}-\frac12=\frac{s}{2r}+\frac{s-a}{2r}-\frac12$.

But since A > 90°, we have $\frac{A}{2}>\frac{90^{\circ}}{2}=45^{\circ}$, and thus, since the tangens function is monotonically increasing on the interval [0°; 90°[, we conclude that $\tan\frac{A}{2}>\tan 45^{\circ}=1$. But by the half-angle laws, $\tan\frac{A}{2}=\frac{r}{s-a}$. Hence, $\frac{r}{s-a}>1$, so that s - a < r. Thus,

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}<\frac{s}{2r}+\frac{s-a}{2r}-\frac12<\frac{s}{2r}+\frac{r}{2r}-\frac12=\frac{s}{2r}+\frac12-\frac12=\frac{s}{2r}$.

So the required inequality is proven for the case of an obtuse-angled triangle ABC, too. This completes the proof of the inequality.

Finally, let's remark that the inequality we just proved,

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s}{2r}$,

can be rewritten in the form

$m_ah_a+m_bh_b+m_ch_c\leq s^2$,

where $h_a$, $h_b$, $h_c$ are the three altitudes of triangle ABC. In fact, this is an equivalent transformation, since the area F of the triangle ABC satisfies $F=\frac12ah_a=\frac12bh_b=\frac12ch_c=rs$, so that $a=2F:h_a$, $b=2F:h_b$, $c=2F:h_c$ and $r=F:s$, and thus

$\frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}\leq\frac{s}{2r}$
$\Longleftrightarrow\ \ \ \ \ \frac{m_a}{2F:h_a}+\frac{m_b}{2F:h_b}+\frac{m_c}{2F:h_c}\leq\frac{s}{2F:s}$
$\Longleftrightarrow\ \ \ \ \ \frac{m_ah_a}{2F}+\frac{m_bh_b}{2F}+\frac{m_ch_c}{2F}\leq\frac{s^2}{2F}$
$\Longleftrightarrow\ \ \ \ \ m_ah_a+m_bh_b+m_ch_c\leq s^2$.

In the form $m_ah_a+m_bh_b+m_ch_c\leq s^2$, the inequality was posted by Manlio at http://www.mathlinks.ro/Forum/viewtopic.php?t=5208 .

  Darij

I read this post this noon and have a solution which is considerably longer than Darij 's. However, as this solution is replete with properties, so I post it here.
    In order to prove this, let 's first say a little bit about about a bijection $f$ that maps from the set of all triangles $X$ to itself (the set $X$ doesn't have any two congruent triangles in it ). Let 's map a triangle with sides $a, b, c$ to the triangle with side $\frac{2}{\sqrt{3}}m_{a}, \frac{2}{\sqrt{3}}m_{b}, \frac{2}{\sqrt{3}}m_{c}$. Then of course $f = f^{-1}$. Notice that $f$ preserves the area of triangles.
    Now let me introduce some lemmas that are very useful :
    [b]Lemma 1[/b]  Let $ABC$ be a triangle. Then there exists a triangle $XYZ$ such that $sin^{2}{X}= \frac{1}{2}sinA(\sum tg{\frac{A}{2}}, ...$
    [b]Lemma 2 [/b] Let $XYZ$ be a triangle and $x, y, z$ be real numbers. Then $\sum yz sin^{2}{X}\leq \frac{1}{4}(x+y+z)^{2}$
    From lemma 1, it follows that
    [b]Lemma 3 [/b] Let $XYZ$ be a triangle and $x, y, z$ be real numbers. Then $\sum yz sin{X}\leq \frac{(x+y+z)^{2}}{2.(\sum \tg \frac{A}{2})}$
    Let 's stop there and progress our proof for a moment. Call $R'$ the circumradius of the image triangle $XYZ$ of $ABC$ under $f$. Then $\frac{2}{\sqrt{3}}m_{a}= 2R' sin{X}, ....$. By applying lemma 3 we obtain
    ${m_{a}.bc = \sqrt{3}R'(\sum bc.sinX) \leq \frac{\frac{\sqrt{3}}{2}}R'(a+b+c)^{2}}{\sum tg{\frac{X}{2}}}= \frac{2\sqrt{3}R'p^{2}}{\sum tg{\frac{X}{2}}}$.
    We need to show that
  $\frac{2\sqrt{3}R'p^{2}}{\sum tg{\frac{X}{2}}}\leq abc.\frac{p}{2r}$
    Notice that $abc = 4RS$, and thus it is equivalent to proving $R' \leq \frac{\sum{tg\frac{X}{2}}}{\sqrt{3}}.R$
    Carrying out the bijection $f$, this inequality is equivalent to
       $R \leq \frac{\sum{tg \frac{A}{2}}}{\sqrt{3}}.R'$  [color=blue](*) [/color]
    We now go on introducing the lemmas
    [b]Lemma 4 [/b] $4Rm_{a}\geq b^{2}+c^{2}$
    It is easy to prove if one compares the diameter and the chord formed by the $A$ median of the circumcircle. Therefore it follows that
    [b]Lemma 5[/b]  $2R(m_{a}+m_{b}+m_{c})\geq a^{2}+b^{2}+c^{2}$
    Lemma 6 is a direct corollary of this inequality if we carry out the bijection $f$
    [b]Lemma 6[/b] $\frac{m_{a}.m_{b}.m_{c}}{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}\geq r$, here $r$ is the inradius of $ABC$.
     [b]Lemma 7[/b] $a+2+b^{2}+c^{2}= 2(p^{2}-r^{2}-4Rr)$
     [b]Lemma 8[/b] $\sum tg{\frac{A}{2}}= \frac{4R+r}{p}$
     By replacing $R' = \frac{2m_{a}.m_{b}.m_{c}}{3\sqrt{3}S}\geq \frac{2(m_{a}^{2}+m_{b}^{2}+m_{c}^{2}). r}{3\sqrt{3}S}= \frac{(a^{2}+b^{2}+c^{2}).r}{2\sqrt{3}S}= \frac{(p^{2}-r^{2}-4Rr).r}{\sqrt{3}S}$ and using lemma 4 to the inequality [color=blue](*)[/color], we obtain we need to prove
      $p^{2}(R+r) \geq 16R^{2}r+r^{3}+8Rr^{2}$ [color=blue](**)[/color]
     And here is the final lemma (which I don't prove here, as a special case of Leibnitz 's theorem)
     [b]Lemma 9 [/b] Let $I, G$ be the incenter and the centroid of $ABC$. Then $IG^{2}= p^{2}+5r^{2}-16Rr \geq 0$
     [color=blue](**)[/color]  follows immediately as $R \geq 2r$.
     PS This may be the longest post I ever reply in ML.  :lol:


hmm I remember saying  no questions about the problems :D I belive that the questions are pretty well put :) read the questions carefully and take your time :)

suppose I think the question 3 is wrong (the one with BMN isosceles)?  could you please double check that question?   :D

[quote="Valentin Vornicu"]I've checked the problem and I don't think there is anything wrong with it! :) anyway it's good that you asked :D[/quote]

then i must be really stupid because I still cannot draw a proper diagram to that question for x /= 1    :(  :(  :(

... I will answer this question next Monday :D - anyway the problem's fine, I can assure you of that.


Before I start, let me just say I'm not necessarily bad at grammar, or overly obsessed with it (or the SAT, for that matter). There just always seems to be ONE question I get wrong in the writing section every time, and I get frustrated with 48C, 1W, 0O test after test. So here's one more (and hopefully the last):

A major cause of stress in school is [u]where seniors must manage not only academic requirements and sports schedules, but also[/u] standardized testing and college applications, during the first semester.

(A) where seniors must manage not only academic requirements and sports schedules, but also
(B) seniors need to manage not only academic requirements and sports schedules, but also
(C) where seniors must manage not only academic requirements and sports schedules, and also
(D) when seniors must manage both academic requirements and sports schedules, but also
(E) the management by seniors of not only academic requirements and sports schedules, but also

[hide="My reasoning..."]
Okay, so I know the correct format is "not only... but also", so I can eliminate (C), which says "and also".
Furthermore, "a major cause" cannot be a "where" or "when", so eliminate (A) and (D). I'm down to (B) and (E).
Now take out all the fillers and look at the skeleton sentence:
(B) A cause is seniors need to manage not only requirements but also testing.
(E) A cause is the management of not only requirements but also testing.
The management causes the stress, not the seniors, so the answer must be (E)![/hide]

[hide="But I was wrong!"]
The answer is (B). Can someone please explain?[/hide]

I think you're parsing (E) wrong.  Your interpretation is that "the management causes the stress, not the seniors."  But if we strip (E) down to a basic copular sentence, we get "A cause of stress is [the] management" which is certainly OK.

My best guess as to why (E) is wrong is that it's in the passive voice-- except actually it's not, it just looks suspiciously like it.

I find the lack of an overt complementizer in (B) pretty jarring, but maybe that's just me.

[quote="Osud"]I think you're parsing (E) wrong.  Your interpretation is that "the management causes the stress, not the seniors."  But if we strip (E) down to a basic copular sentence, we get "A cause of stress is [the] management" which is certainly OK.[/quote]

I'm not sure what you mean, I never said (E) was [i]not[/i] okay...

[quote="Osud"]My best guess as to why (E) is wrong is that it's in the passive voice-- except actually it's not, it just looks suspiciously like it.[/quote]

The book says that (E) is a "very awkward passive phrase". But is (B) any less awkward?

[quote]The book says that (E) is a "very awkward passive phrase".[/quote]

What your book says is completely wrong.  (E) is emphatically [b]not[/b] a passive sentence.

Here's how the passive works.  In converting a sentence from active to passive, a few things happen:
(1)  The active verb is converted to a form of "to be" plus a participle;
(2)  The direct object of the active verb is promoted to subject position;
(3)  The subject of the active verb is demoted to an optional "by" phrase.

So, if your active example is "seniors manage academic requirements and sports schedules," then the passive form of the sentence will be "academic requirements and sports schedules are managed by seniors."  We've converted the active verb "manage" to the passive "are managed," moved "academic requirements and sports schedules," the direct object of the "manage," to subject position, and moved the subject of the active verb "manage" to an optional "by" phrase, "by seniors."

"The management by seniors of academic requirements and sports schedules" is not a passive phrase.  This is a noun phrase headed by the noun "management."  The fact that this noun phrase has a "by" phrase containing what would have been the subject had "management" been an active verb does nothing to change the fact that nouns cannot be either active or passive and that the real verb of the sentence, "is," is active, not passive.

Agreed that that construction is very awkward, though.

[quote]I'm not sure what you mean, I never said (E) was not okay... [/quote]
Sorry, I wasn't clear about that.  You said in your earlier post that you read the sentence to mean that "The management causes the stress, not the seniors," I responded with an alternate meaning, and by "OK" I meant that my parse of the sentence was in line with what you thought the sense of the sentence should have been.  That make sense?

[quote="Osud"]
[quote]I'm not sure what you mean, I never said (E) was not okay... [/quote]
Sorry, I wasn't clear about that.  You said in your earlier post that you read the sentence to mean that "The management causes the stress, not the seniors," I responded with an alternate meaning, and by "OK" I meant that my parse of the sentence was in line with what you thought the sense of the sentence should have been.  That make sense?[/quote]

Thanks for your explanation, makes sense now!

As for the passive stuff, yeah, I see what you mean.. with management as the noun, the phrase is not really passive at all. But do you agree with them that (B) is the correct answer? And if it is, is awkwardness the only thing wrong with (E)? And [i]if it is[/i], isn't (B) awkward too???  :lol:

I would actually say that (B) is worse than (E), and here's why.  Imagine you have read only this much of the sentence:

[quote]A major cause of stress in school is[/quote]

What do you naturally expect to come next?  A noun phrase, of course.  So then you hit the next word:

[quote]seniors[/quote]

And, since it's a noun, you naturally interpret it as the noun phrase that's the complement of the verb.  But wait!  There's more!

[quote]need to manage not only academic requirements[/quote]

You start interpreting that as part of a noun phrase headed by "seniors," but that's ungrammatical if all of that stuff goes in the noun phrase.  So you have to go back and interpret the whole thing starting with "seniors" as a complementizer phrase introduced by a silent complementizer.  Usually in English silent complementizers are used when omitting an overt complementizer won't create a garden-path sentence (so-called because it leads you down a garden path instead of the 'real' interpretation).

Even if grammar books have nothing against garden-path sentences, it is very bad form to write a sentence which many people will parse wrong on the first reading.

If it were up to me, I would pick (D).  The "both...but also" isn't great, but at least it is easy to read and it's immediately obvious what the sentence means.

Actually, if it were me, I'd just say "Managing standardized testing and college applications, on top of academic requirements and sports schedules, is a major cause of stress in school for seniors."  Or, if you really wanted to put "a major cause of stress in school" at the front of the sentence, I would say, "It is a major cause of stress for seniors to manage standardized testing and college applications on top of academic requirements and sports schedules."  Just delete the prepositional phrase "in school," it's not really adding anything here.  What else would it be, a major source of stress in church?  In video gaming?  In Sub-Saharan Africa? :lol:

I agree. When I first read the question, I really thought none of the choices made good sentences.

But thanks a lot for your input, I feel better about the question now that we've thrown it through a grinder. :)

Cracking the SAT 2009 Edition. The Princeton Review.

For the SAT, I wouldn't trust a test-prep book besides the College Board's own Study Guide.  You can use the other books if you have run out of other material, but don't be surprised to see errors.  In math, I have caught numerous errors by the outside companies.

Disclosure: I work for a tutoring company.

If you add "that" to (B), it becomes an obvious right choice. "That" can be sometimes omitted but here the omission creates a terribly awkward sentence, much worse than (E), so I suspect it was just a typographical error. (E) changes the meaning of the sentence: the cause is not "management by seniors" but the "necessity for seniors to manage" if you want that type of construction.

Fedja wrote:
[quote]If you add "that" to (B), it becomes an obvious right choice. "That" can be sometimes omitted but here the omission creates a terribly awkward sentence, much worse than (E), so I suspect it was just a typographical error.[/quote]

That's what I was trying to get at with the business about silent vs. overt complementizers, but didn't think to give an example with an overt complementizer... thanks, fedja!   :blush:

buzzer11 wrote:
[quote]I agree. When I first read the question, I really thought none of the choices made good sentences.[/quote]

I think your instincts are spot-on, it's just that the SAT requires rote application of a set of prescriptive rules instead of a more holistic approach to style, grammar, and readability. This is what I meant earlier when I said not to take the stuff too seriously-- if you puzzle over something for a while and all of the answers seem a bit off to you, then stick with your instinct.  The SAT is a big deal at the time you take it, but you have a whole life of writing ahead of you.  You shouldn't try to permanently override your instincts about writing and language for the sake of one test.

[quote]But thanks a lot for your input, I feel better about the question now that we've thrown it through a grinder. [/quote]

Sure!  Glad I could help.  (I'd never thought of myself as a grinder before...  :gleam:)

A, C, and D are automatically eliminated, because you know it shouldn't begin with "where" or "when".

The management is not the cause of stress. The cause of stress is either the things that the seniors have to manage or the fact that they have to manage them.

Hence, the best answer is B.

Having said that, I don't really like Princeton Review. Their practice problems are usually significantly easier than the actual test, and are sometimes weird. Kaplan has the right difficulty, but is riddled with technical errors and typos. Barrons is a bit harder than the real thing. For subject tests in science, Barrons introduces more material than is required for the actual tests, and the questions are not as understanding-of-theory type as the real ones, but perhaps provide the best practice.

Also, I don't particularly agree with the above poster. When I took the SAT's, I went off gut feeling for every single writing question, and it worked just fine. In the past, when I tried consciously invoking particular grammar rules, I got bogged up and confused, and ended up not only getting questions wrong, but also not finishing on time.

Why can't management be the cause of the stress? When I read the question that was what I thought the question was implying. The seniors were stressed because they have to [i]manage[/i] a lot.

In terms of the "gut feeling" idea, I usually go with strict grammar rules and not my gut and I do fine most of the time. I guess it depends on the person.

I've reduced B and E by eliminating some adjectives and unnecessary prepositional phrases. I have placed in parentheses what is communicated as the cause of stress.

(B) A cause of stress is ([implicit*: the fact that] seniors need to manage various tasks).

(E) A cause of stress is (the management) by seniors of various tasks.

The management is not a cause of stress -- management is generally a good thing. Causes of stress might be the tasks themselves, the fact that they have to manage these tasks (as in part B), or inability to manage.

I totally agree that the assumption of (*) should not be necessary. However, in the SAT's, the first thing you must do is see if the sentence means what it wants you to mean, and then see if it is properly expressed -- it is not just about finding a sentence that is gramatically correct! Sometimes they will try to trick you by making a few gramatically correct responses, only one of which has the desired meaning.

Oh. When I read (B) I just read "the cause of stress is --> seniors" and so I eliminated that one. :(

And that [i][b]is[/b][/i] a bad mistake ;)

[quote="TZF"](B) A cause of stress is ([implicit*: the fact that] seniors need to manage various tasks). [/quote]

[quote="fedja"]And that is a bad mistake  :wink: [/quote]

Then how do we know to interpret (B) implying an object?