This is a game to exercise your vocabulary and knowledge of the English language.  The way you play the game:  There will be a previously posted sentence (ONE SENTENCE) that you will lengthen.  Points will be awarded by dividing the number of words in the revision by the number of words in the orginal.(Round decimals up)(minimum 1 pt)
USE THOSE ADJECTIVES AND ADVERBS!!!!
EDIT: Original sentences may not be longer than ten (10) words and must be bolded.

Example: I am hanging ornaments on a tree.
I am attaching decorations to the branches of this Christmas tree.  Pts: 1

New one : He opened the door.

Have Fun!!!!

He crushed his toe when he opened the door.

The point is to restate the sentence without adding any irrelevant information.

He opened the door with the agility of a brown fox.

That is a good sentence, the only thing you forgot was to write the next one.

I'll do it:

It's Sunday tomorrow.

GSB: 6/4/2

The proper name of the day that we [i]homo sapiens[/i] call tomorrow is generally considered to be the word "Sunday" in  modern-day common English.

7, 8, or 9, depending on how you want to count.

BTW, bold anyone :P

[b]I think, therefore I am.[/b]

"I think, therefore I am" is a statement, containing five words of the english language.

He who hesitates is lost.

[hide="Scores"]james4l: 8 bluecarneal: 3[/hide]

The male specimen of homo sapiens who fails to react in time in turn is in a position unknown to its self.

With great power comes great responsibility.  :D

[color=#F8F8F8]@thread title: that's what she said...[/color]

If you really want, press Ctrl-A to find a message...

Whenever someone is bestowed upon them a quantity equivalent to the product of current and potential, that person is entitled to reciprocate back to the general societal populace by holding all responsibility over aforementioned population - we conclude this sentence with a green circle with two carets symbolizing eyes and a "D" partitioned into three sections, supposedly representing a mouth, to show that we are happy.

Six times nine equals to forty-two.

The numerical arithmetical, pointless, aimless, meaningless, unnecessary piece of information that the number six multiplied by a number greater than one, nine is however not equal and balanced to the numerical value of forty-two and is actually equivalent to the numerical value of fifty-four which equals two multiplied by three multiplied by three multiplied by three and it is also equal to sixty minus the value of six as well as fifty added to the numerical value of four.

That was 89 words whereas the original was 6...

Next sentence: [b]The fox was agile.[/b]

The fox was agle in avoiding the big brown bear.

I'll just post a new sentence.

[b]Bob likes pie.[/b]

The person with the boring name "Bob" likes a substance of pastry and other fruits, also known as the word pie.

21 words

[b]I like to stroll down the beach.[/b]

A curiously idiosyncratic action, which, when performed by me, causes pleasure to be induced upon myself, as I have learned from past experience, under normal, unexceptional circumstances, is for me to move, usually, but not always, by walking, usually on my feet, at a leisurely, unhurried pace like that of a turtle except possibly slightly faster, without any worries, or at least a seriously insignificant amount, along a horizontal (or at least slanted less than approximately thirty degrees, which also happens to be a reasonable bound for the temperature, at least taken in Celsius, which was invented in roughly 1745 and which Stigler's law also applies to) strip of land in close proximity to a large body of water in its liquid state (as opposed to its solid, gaseous, and plasma states, of course) and which is also very commonly adorned with extraordinary amounts of microscopic rocks, to the extent that, as opposed to most things commonly described as "microscopic", even when gathered in great amounts, they are visible to the naked eye, in the direction that decreases my distance from the center of the earth, or, if such a direction is unavailable, south, or the closer of the two directions offered by the strip, or, as a final resort, the direction in which more of the microscopic collection of rocks can be seen, and thus also allow more continuous movement without having to occupy simultaneously the same position as something else that is solid, rigid, and excessively large.

The stuff in the parentheses don't count, since they add a lot of extra information that nobody asked for and do not actually have anything to do with beaches. Not even close. But they look nice there.

Anyways, without the parenthesis stuff there appear to be precisely 200 words, which is strange since I was just thinking of fine-tuning it to make it precisely 200 words. But okay, I think the above is just blatant cheating.

29 points. I don't think anybody is going to bother to count. The scores, I mean.

[b]I have no life.[/b]

The species "human" who is genetically different from everyone else who is the current living soul inhabiting the physical body that is curently typing this message on this laptop (which currently has 5 plugs in the outlets, consisting of a non-wireless Inland mouse, a Micro Center Light blue metallic in a sort of cyan shade which has a memory of 2 bits multipled by the constant eight times one thousand three times in a row, a dirty beige cable which stretches for about two feet before hooking up to the printer which always has paper james for some reason and is always giving a warning about low black ink and where I printed out my last four notebook dividers for history, a bright yellow wire with a plastic end, which gives me access to the world wide web including Wikipedia, Yahoo, Google, artofproblemsolving, yougotrickrolled, and millions of other websites, and a thin charger that has a thickness of 96 mm to the nearest integer number of millimeters with error of plus minus 10 mm), which is a device modified to make the more antiquated desktop computer more recent, cheaper, more portable, having the ability to store more data, and more powerful, and has a polarized screen that I was looking through with my head tilted at 45 degrees to the right with my polarized shades clipped onto my glasses, creating an almost dark screen, does not have a soul contained within the set named "Soul" in the brain, which brings up a paradox because a soul which has no soul is like saying a person has no life, which contradicts itself like an oxymoron (a phrase that contradicts itself like "black light" and "jumbo shrimp" and "pretty ugly" etcetera) and therefore, this sentence itself may not be completely true because of such contradiction, whereas the Pledge of Allegiance, stating that "I pledge allegiance to the flag of the United States of America and to the Replublic for which it stands one nation under God indivisible with liberty and justice for all" (which consists a variety of grammatical and spelling errors), may be completely untrue depending on the person stating so, and this type of varying statement is called an opinion, but many people may argue with this and start complaining how this is a run on sentence, but if the first sentence of the Declaration of Independence (When in the Course of human events, it becomes necessary for one people to dissolve the political bands which have once connected them with one another, and to assume among the powers of Earth, the separate and equal station to which Laws of Nature and Nature's God entitle them, a decent respect to the opinions of mankind requires that they should declare the causes which impel them to the separation.) would also be a run on sentence, so try not to say so, and the entire point of this sentence is to state that "I have no life".

[b]That took me 30 minutes to type.[/b]

The irrepressibly long and completely unintelligible, even taking into consideration the geekiness, which is an adjective here not intended to cause the offendedness, which MIGHT be a coined-up-word-on-the-spot, and of course, being that every statement has two sides, even though certain, or perhaps many, people would dispute that, exclaiming that it is an overgeneralization and that every rule has exceptions, which brings to mind the curiously paradoxical question "Does 'Every rule has exceptions' have exceptions?", MIGHT not, depending heavily, however, on your choice of dictionary, of which there are literally, or not, the paradox contained of which I am, in fact, already discussing, and will thus refrain from reiterating for fear of infinite descent, which is also a very useful tool for mathematical proof of certain things, for instance the fact that a cube cannot be constructed from cubes of all distinct side lengths, without, of course, any inside holes or outside holes or any of various incompletenesses, which is another pathetic word that MIGHT be coined and MIGHT not, which we cannot discuss right now because of, again, fear of, you guessed it, infinite descent, thousands of varieties, of any of the people who are viewing this thread, even though the mere fact that you are READING this sentence and have gotten about halfway through is a good indication of your geekiness, of the normal AoPSer, since we like math so much, and math is the stuff that makes people near-sighted and far-sighted and realize with horror that their number one wish is to have published a paper jointly with Paul Erdos, whose dieresis I am going to omit, mass of garbled text I am referring to has captured my undivided attention for the duration of 16546737186000 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom, during which I made the conscious decision to continuously, as opposed to intermittently, being that there are, as explained above again, two sides to every statement, depress square-shaped plastic casings that convert my depressions, in the physical sense, to electrical pulses and then strange English characters inside a machine operated primarily by electricity and various magnetic forces, and then through a million, though probably much less, lines and routers to the Art of Problem Solving server or server cluster, whereupon it is displayed to anybody who would bother to check in a reasonably prominent position.

[b]If only creative writing assignments were like this![/b]


conics
parabola
algebra
system of equations
algebra unsolved

Let system of equations: $ \begin {cases} x^2\minus{}y\plus{}1\equal{}0 \\ x^2\minus{}y^2\plus{}(a\plus{}1)x\plus{}(a\minus{}1)y\plus{}a\equal{}0 \end {cases}$.

Find all $ a \in R$ for this system exist roots $ (x;y)$ with $ x,y \in R$.

The second equation writes $ (x \plus{} (a\plus{}1)/2)^2 \equal{} (y \minus{} (a\minus{}1)/2)^2$, i.e. $ y \equal{} \minus{}x \minus{} 1$ or $ y \equal{} x \plus{} a$ (geometric interpretation is intersection of the parabola $ y\equal{}x^2\plus{}1$ with the two lines above).
Now, $ x^2\plus{}1 \equal{} \minus{}x\minus{}1$ has no real solutions, while $ x^2\plus{}1 \equal{} x\plus{}a$ has real solutions iff the discriminant $ \Delta \equal{} 1 \minus{} 4(1\minus{}a) \equal{} 4a\minus{}3 \geq 0$, so $ a \geq 3/4$.


How many integer pairs $(a,b)$ for $-5\le a,b\le 5$ are there such that $ax^2+10x+b=0$ has two distinct real roots?

[hide]We need the discriminant to be positive, so 100-4ab is positive.  This disallows only (-5, -5) and (5,5).  There are 11*11=121 total ordered pairs, so 119 possibilities.[/hide]

But isn't $a=0$ not allowed? I got 109, becuase I forgot that $(-5,-5)$ wouldn't work either, but I guess it should actually be $(10)(11)-2=108$. You're probably right though.

No, you are right that it should be 108.  I forgot about a being nonzero.

[hide]$100-4ab>0$, so $ab<25$. Since $a$ and $b$ can be from -5 to 5, we have $11^2=121$ possibilities. However, $(a,b)\ne(5,5), (-5,-5)$, so there are 119 possibilities.
Also from the definition of the equation of a parabola, $a\ne0$, so there are 11 pairs such that a=0, so there are actually 119-11=[b]108[/b] pairs. [/hide]


complex numbers
complex analysis
complex analysis unsolved

[hide="1."]
Find the convergence radius of the series
\[ \sum_{n\equal{}1}^\infty \left( 1\plus{}\frac{1}{n} \right)^{n^2} \frac{x^n}{n!}.
\][/hide]

[hide="2."]
Let $ (a_n)_{n\equal{}0}^\infty$ be a sequence of complex numbers and denote
\[ b_n \equal{} a_0 \plus{} a_1 \plus{} \cdots \plus{} a_n.
\]

Prove that the series
\[ \sum_{n\equal{}0}^\infty \frac{a_n}{n!} \ z^n, \qquad \sum_{n\equal{}0}^\infty \frac{b_n}{n!} \ z^n,
\]
have the same radius of convergence.[/hide]

1. Root test.

$ a_n\equal{}\frac{\left(1\plus{}1/n\right)^{n^2}}{n!}$, so $ |a_n|^{1/n}\equal{}\frac{(1\plus{}1/n)^n}{(n!)^{1/n}}$. The numerator converges to $ e$ and the denominator diverges to infinity (use Stirling's approx or some other approx of $ n!$).

Thus $ R\equal{}\frac{1}{\limsup |a_n|^{1/n}}\equal{}\infty$.


calculus
derivative
ratio
analytic geometry
Pythagorean Theorem
geometry
calculus computations

A street light is mounted at the top of a 15 ft tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

I'm having troubles with this...here's what I figured out, assuming $ x$ is the distance between the pole and the man, and $ y$ the distance between the man and the tip of his shadow:

$ \frac{x\plus{}y}{y}\equal{}\frac{15}{6}$.
It is given that $ x\equal{}40$, $ \frac{dx}{dt}\equal{}5$ We want to find $ \frac{dy}{dt}$. However, when I do it, I get an answer of$ \frac{10}{3}$:

$ \frac{x\plus{}y}{y}\equal{}\frac{15}{6}\Leftrightarrow$
$ 6x\plus{}6y\equal{}15y\Leftrightarrow$
$ y\equal{}\frac{2}{3}x$

Taking the derivative in respect to $ t$ to both sides gives:

$ \frac{dy}{dt}\equal{}\frac{2}{3}\cdot\frac{dx}{dt}$

Clearly this is wrong, as I have no way of factoring in the fact that $ x\equal{}40$. There's something crucial missing, but I can't tell what. I think I need to get $ x$ to some power, but I don't know how else to relate $ x$ and $ y$ besides that equation gathered from similar triangles. Please help.

$ \frac {dy}{dt}$ is constant. It is the same whether $ x \equal{} 40$ or $ x \equal{} 4.$

Your statement that $ \frac {dy}{dt} \equal{} \frac {10}{3}$ is correct, but you have to add $ \frac {dx}{dt}$ to that to get the speed of the tip of the shadow, which is $ \frac {25}{3}.$

You don't "have" to get any particular form of an equation other than that to which a correct rendering of the problem leads. Go ahead and trust what you've done.

So...the edge of the shadow itself (the man) is moving at a constant rate of 5 ft/sec. Which means that the shadow too is moving a constant rate. However, it gets longer as the man moves farther away from the light, so...the rate that the shadow gets longer is $ \frac {10}{3}$ ft/sec? So you have to add those together...to find the rate of the tip of the shadow. So is the speed at any $ x$ always $ \frac{25}{3}$?

And thanks for your help by the way :D I have the answer and the method of arriving at the answer, now I'm trying to find what the answer is telling me.

EDIT: Another problem...here goes:

At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 pm?

Well...I think I've done a decent job on this one, but there is one aspect that confuses me, which I will explain when we reach it.

First, I drew a diagram. I assigned the distance traveled by boat A to $ x$, the distance travelled by boat B to $ y$, and the distance between them $ z$.

Here's what we know:
$ x = 140$
$ y = 100$
$ \frac {dx}{dt} = - 35$
$ \frac {dy}{dt} = 25$

Now, I don't know if this is true or not, but I think the two triangles formed are in fact similar, due to AA similarity (since both are right triangles, and one pair of homologous angles are opposite). So, I decided I needed one more variable, $ w$, for the other leg of boat A, and at the same time, the undefined leg of boat B. (Boat A's triangle consists of $ x$ and $ w$ as legs, while boat B's triangle consists of $ y$ and $ (100 - w)$ as legs).

Due to the similarity, I set up the following ratio:
$ \frac {x}{w} = \frac {y}{100 - w}\Leftrightarrow$
$ 100x - xw = yw\Leftrightarrow$
$ w = \frac {100x}{x + y}$

So, $ w = \frac {100x}{x + y}$ and $ 100 - w = \frac {100y}{x + y}$

Now, to find $ z$...Since $ z$ is comprised of the sum of the hypotenuses of the two triangles, I set up the equation that way (I left in $ w$ to make it easier to read for now):

$ z = (x^2 + w^2)^{\frac {1}{2}} + (y^2 + (100 - w)^2)^{\frac {1}{2}}$

I took the derivative of this ugly equation in respect to $ t$ and the derivative is so complicated and terrible that I think this can't be right:

$ \frac {dz}{dt} = (\frac {1}{2}(x^2 + (\frac {100x}{x + y})^2)^{ - \frac {1}{2}}\cdot[2x\cdot\frac {dx}{dt} + 2(\frac {100x}{x + y})\cdot[\frac {100\cdot\frac {dx}{dt}\cdot(x + y) - (\frac {dx}{dt} + \frac {dy}{dt})(100x)}{(x + y)^2}]]$
$ + \frac {1}{2}(y^2 + (\frac {100y}{x + y})^2)^{ - \frac {1}{2}}\cdot[2y\cdot\frac {dy}{dt} + 2(\frac {100y}{x + y})\cdot[\frac {100\cdot\frac {dy}{dt}\cdot(x + y) - (\frac {dx}{dt} + \frac {dy}{dt})(100y)}{(x + y)^2}]])$

Plugging my values into that gives $ - \frac {120}{13}$, while the answer is actually $ \frac {720}{13}$ The numbers are sort of similar, which is encouraging, but I feel there has to be a better way of doing this, one that is shorter, neater, and gives the right answer. Please help me find my error!

Are you sure of the answer? A simpler method gives a different answer. Consider the frame of reference of $ A$. The speed of $ B$ is $ 50 km/h$ If $ y$ is $ y$ coordinate of $ B$ then $ z \equal{} \sqrt {d^2 \plus{} y^2}$; $ d \equal{} 100km$. Hence; $ \frac {dz}{dt} \equal{} \frac {y}{z}\frac {dy}{dt} \equal{} \frac {100}{\sqrt {5}} km/h$

A rough estimate: We see that initially $ \frac{dz}{dt}$ is zero and as time progresses it increases to the limiting value of $ \frac{dz}{dt}\equal{}50km/h$

Um, the back of the book clearly says $ \frac{720}{13}$...so...I'd assume that's right. I don't really understand your work ...

The simpler argument works- if you notice that the relative speed is actually $ 60$ km/h.

haha, I get it. From the frame of reference where boat A isn't moving and boat B is moving south at 60 km/hr, the pythagorean theorem says that
$ z\equal{}(100^2\plus{}y^2)^{\frac{1}{2}}$ The derivative of this is...
$ \frac{dz}{dt}\equal{}\frac{1}{2}(100^2\plus{}y^2)^{\minus{}\frac{1}{2}}\cdot2y\cdot\frac{dy}{dx}$
At time 4:00, $ y\equal{}240$ and $ \frac{dy}{dt}\equal{}60$

Plugging in values gives:
$ \frac{dz}{dt}\equal{}\frac{720}{13}$!

Thanks a lot, I'll never overlook the advantage of frame of reference again :D Thanks for all your help...I really appreciate it.


inequalities
algebra
difference of squares
special factorizations
inequalities unsolved

Let a,b,c be non-negative numbers that a+b+c=1. Prove that $ \sqrt {a^2 \plus{} bc} \plus{} \sqrt {b^2 \plus{} ac} \plus{} \sqrt {c^2 \plus{} ab} < \equal{} \frac{3}{2}$

[quote="ghjk"][hide]My solution:
We have: $ a^2 \plus{} bc \leq a^2 \plus{} 2bc$(since $ a,b,c \geq 0$)
Similarly, we infer that $ LHS \leq \sqrt {a^2 \plus{} 2bc} \plus{} \sqrt {b^2 \plus{} 2ac} \plus{} \sqrt {c^2 \plus{} 2ab}$. $ (1)$
Using Cauchy-Schwartz's inequality we get:
$ RHS (1) \leq (a \plus{} b \plus{} c).\sqrt {3}$
Therefore, the $ LHS \leq \sqrt {3} < 2$(since $ a \plus{} b \plus{} c \equal{} 1$).
The equality doesn't occur though.
Am I missing something since if I don't, this problem is ridiculously trivial for MOP! :roll: [/hide][/quote]

You didn't miss anything... I missed something :blush:

check out the edited problem again please.

[quote="greentreeroad"]Let a,b,c be non-negative numbers that a+b+c=1. Prove that $ \sqrt {a^2 \plus{} bc} \plus{} \sqrt {b^2 \plus{} ac} \plus{} \sqrt {c^2 \plus{} ab} < \equal{} \frac {3}{2}$[/quote]
this inequality by hungkhtn :)
We must prove that:
$ 3(a\plus{}b\plus{}c) \ge 2(\sqrt{a^2\plus{}bc}\plus{}\sqrt{b^2\plus{}ca}\plus{}\sqrt{c^2\plus{}ab})$
Assume $ a \ge b \ge c \ge 0$
We have : $ \sqrt{b^2\plus{}ca}\plus{}\sqrt{c^2\plus{}ab} \le \sqrt{2(b^2\plus{}c^2)\plus{}2a(b\plus{}c)}$
So,we only prove that: $ 2\sqrt{2(2(b^2\plus{}c^2)\plus{}2a(b\plus{}c)}\plus{}2\sqrt{a^2\plus{}bc} \le 3(a\plus{}b\plus{}c)$
Put $ 2d\equal{}b\plus{}c$,this inequality is equivalent to:
$ (a\minus{}2d)^2\plus{}20bc \ge 12(a\plus{}2d)(\sqrt{a^2\plus{}bc}\minus{}a)$
On the other hand: $ \sqrt{a^2\plus{}bc}\minus{}a\equal{}\frac{bc}{a\plus{}\sqrt{a^2\plus{}bc}} \le \frac{bc}{2a}$
Finally,we  must prove that: $ (a\minus{}2d)^2\plus{}20bc \ge \frac{6(a\plus{}2d)bc}{a}$
$ <\equal{}> (a\minus{}2d)^2\plus{}2bc\plus{}\frac{12(a\minus{}d)bc}{a} \ge 0$
Which is obvious true
Equality holds when $ a\equal{}b,c\equal{}0$ or any cyclic permutations :)

[quote="ghjk"][hide]Okay, that makes sense to me now. Here's a hint for you in case you didn't get it:
Using Schur's inequality 3-degree and Cauchy-Schwartz's inequality after squaring both sides and homogenizing[/hide][/quote]

Could you show the details? Thanks.

I only spent about 10 hours on this problem ><

I kept trying to use difference of squares and then slightly loosening conditions, and it kept failing (despite my best efforts to weaken the ineq as little as possible)


P is a point inside a triangle $ABC$ and:
$PA\leq 2$ and $PB\leq 2$ and $PC\leq 2$

R is circumradius, proove that $R\leq 2$

I made the same mistake some days ago ..  :oops:

The red circle is centered at $P$ and includes all the points $A,B,C$

You can see that $R$ is greater than all distances $PA,PB,PC$

jerry144：How did you draw the picture?Did you use any software?Tell me,please. :roll:


superior algebra
superior algebra unsolved

If an+1 &bn+1 are both squares then prove true or false:8(a+b) divides n,if false,then give conditions when it will be true,its true for at leasy 2 cases

Try $ a \equal{} 3,b \equal{} 16,n \equal{} 5$.

I don't see any reason why this has to be true.

Btw what is this doing in the college section?

if 2n+1 & 3n+1 are perfect squares, 40|n
and if 3n+1 & 4n+1, 56|n
so i generally cinjectured this and asked for what a,b its true,i could not make the problem clear
the problem is: if an+1 & bn+1 are perfect squares,then 8(a+b)|n,find (a,b)


calculus
integration
limit
real analysis
calculus computations

if $a_n = \frac{1}{n} \sum_{k=1}^{n} \frac{1}{1 + (k/n)}$

Find $lim_{n \to  +\infty}a_n$ by interpreting $a_n$ as the Riemann sum of a definite integral.

How can i express it as the Riemann sum of a definite integral when k and n are together in the same term? i can't separate them....

$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}f\left(x_{k}\right)\Delta x_{k}=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\Delta x_{k}=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{n+k}$

I see what you mean about the $n$.  Something tells me that $\lim_{n\rightarrow\infty}a_{n}=0$.

I don't know why you want to bring the $\frac{1}{n}$ in.

$\lim_{n\to \infty} \frac{1}{n} \sum_{k=0}^n g(k/n) \rightarrow \int_0^1 g(x)dx$
That's the idea of Riemann summation.

The $k/n$ is the $x$ you're integrating over. That's where we get the bounds of our integration -- it varies between 0 and 1. We could have $k$ vary elsewhere or have other things in the summand that change the bounds of integration, if we wanted.

So here it's just $\int_0^1 \frac{1}{1+x} dx = \ln(2)$, which is what we'd expect since this is an oft-discussed limit.

[quote="violetcraze"]my idea of the reimann sum is something like amcavoy's.
what i don't understand here is how did you get the upper limit of $1$?     :?[/quote]

Because $x = \frac{k}{n}$ ranges from $\frac{1}{n} \to 0$ to $\frac{n}{n} = 1$.

$\Delta x_k = \frac{1}{n}$. You can bring it inside the sum if you want. You don't need to mess with the fraction, though.


I know there have been a lot of polls about scores and all but I want to know the highest score in your state.

Go OH...I am pretty sure I got a 38 or 37 because 4th got 37 and 2nd got 39

I have a higher sprint round score, so I would win the tiebreaker if I got 37, but I can't get 39 because I would win that tiebreaker as well

[quote="pgpatel"]38 in FL :(. We are [b]so[/b] screwed come nats.[/quote]

Ooh that's reassuring. In MO the first place guy got 37.

IN's top score was 40-some by Nathan Benjamin. It wasn't a 46 though.

I figured Nathan Benjamin didn't just because of the cartesian triangle prob.

[quote="kyyuanmathcount"]I would guess Lousiana or Indiana.[/quote]

I'm pretty sure no one here made a perfect either (Neal in particular), but I'll know the first placer's (Neal) score in a week or so to confirm this.

There better have been some perfect scorers....... last year was much more impossibly hard and there were still 2. :?:

[quote]There better have been some perfect scorers....... last year was much more impossibly hard and there were still 2. [/quote]
Wow- that's pretty amazing, last year's test was so much harder than this year's.

[quote="mathgeniuse^ln(x)"]Oh come on the test wasn't that hard.
I got a 40.  28 on Sprint and a 12 on Target.

Okay maybe the Target was difficult but not the sprint.

I think New York has the perfect scorer.[/quote]
I agree. Last year I got a 20/3...

I'm just saying that in comparison there was an obvious difference between the difficulty of the two tests, and for two people to get a perfect score on it is quite impressive.
BTW, I saw that New York's top score was a 43. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82236[/url]

[quote="mathgeniuse^ln(x)"]Oh come on the test wasn't that hard.
I got a 40.  28 on Sprint and a 12 on Target.

Okay maybe the Target was difficult but not the sprint.

I think New York has the perfect scorer.[/quote]

Yeah, last year's target was impossibly hard.

I got 4/8 right on the target and 25/30 on the sprint.
This year I did better on target but ... that's all I'm going to say. :ninja:

we received our team's individual scores and our team score right after the competition ended (same day).

[quote="binonunquineist"][quote="mathgeniuse^ln(x)"][quote="binonunquineist"]Oh. Um, I'm sorry. I was the first person to accidently vote for 46. Maybe other people decided to follow suit, I don't know.[/quote]

Smooth, thanks for fessing up though.

What was your score binonunquineist?[/quote]

42. Can anyone tell me what that would have gotten me in Indiana or California?[/quote]

what state r u from??

I been to mathcounts competitions from 6th grade to 8th grade.  I have came from failing chapter to making nats.  I have never ever received my score in any way, shape, or form...

really? I've always gotton my scores the same day. Sometimes, we get to see other people's scores too.

[quote="pgpatel"][quote="Phelpedo"]State scores don't always have a high correlation with national scores. Last year, Neal Wu only got a 40 at the state competition; he beat out Pardha (46 on state) even on the written round.[/quote] No, but if you get a 46 at states you know you will do well at nats. Similarly if you get a 46 on a school you'll do good at chapter and if you get a 46 at chapter you know you'll do well at state.[/quote]

If you get a 46 at School then you know you will do well at states, the school questions are much harder than the chapter ones.

I know. It's really weird. I got a 43 on school and a 45 at chapter.

Strange, I got a 46 on school and 45 on chapter.

I see no trend in my scores

43 on School (27/16)
44 on Chapter (28/16)
43 on State (29/14)

Okay, maybe I got a trend in sprint and target, but not overall.

Our school doesn't really DO school.

I got about 15 minutes to do the sprint, and got a 25, beating all the others who got 30 minutes.

We didn't do the target round...

Chapter 46 (30/16)

State  39 :ewpu: (27/12)

School isn't harder than chapter....

I nearly fell asleep on the school and still managed a 45. I got a 44 at chapter and actually didn't know how to do #30.

I don't think school was harder than chapter, but I made more mistakes. Actually, at chapter I didn't make any, I just couldn't do #30. Unfortunately, I can't say the same for state... :ninja:

Well, school countdown is DEFINITELY easier.

While in chapter, I actually have to take 5 seconds or so, I just buzz, then think in school and usually get them right.

I am usually good at countdown, but I choked at States Countdown.  Hopefully at Nationals, I don't screw up badly as I did in states.

In Chapters I took about 3 seconds per problem to solve during the countdown.

See ya at nats.

better than choking at both...(like me, missing two questions on both targets because of stupid mistakes is no small matter)

our team didn't do school.  Our team was selected by doing a bunch of national problems and as a result, the members that were able to solve the hardest problems weren't the best at chapter.  At least we got revenge at state though


Let $ b$ be an integer. Show that $ 8b\plus{}1$ is square if and only if $ b$ is of the form $ b \equal{} r(r\plus{}1)/2$ for some integer $ r$.

ლევანი: I don't think you proved the "only if" statement.

[hide="A Start"]
$ 8k+1=r^2\implies r=\sqrt{8k+1}$, so $ \sqrt{8k+1}$ must be an integer.

Case 1: The If statement
  $ \sqrt{8\left[\dfrac{r(r+1)}{2}\right]}=\sqrt{4(r^2+r)+1}=\sqrt{4r^2+4r+1}=\sqrt{(2r+1)^2}=2r+1$.

Case 2: The Only If statement
  Since $ r=\sqrt{8k+1}$, we let $ k=rx$. Expanding and re-arranging, we have $ r^2-8ir-1=0$. By the quadratic formula, we have:

\begin{eqnarray*}
r &=& \dfrac{8x\pm\sqrt{64x^2-4(1)(-1)}}{2}\\
&=& \dfrac{8x\pm \sqrt{64x^2+4}}{2}\\
&=& \dfrac{8x\pm 2\sqrt{16x^2+1}}{2}\\
&=& 4x\pm \sqrt{16x^2+1}
\end{eqnarray*}

Now I don't really know how to proceed. help?[/hide]

So you can take what I got in case 1 and plug it back in as equal to $ 8k\plus{}1$?

Yes, because if $ 8b\plus{}1$ is a square then it is a square of an odd number.

Can't we say

8b+1=x^2 so x is odd

Then, b=((x+1)/2*(x-1)/2)/2, which is of the desired form.

[quote="Cassi"]Let $ b$ be an integer. Show that $ 8b \plus{} 1$ is square if and only if $ b$ is of the form $ b \equal{} r(r \plus{} 1)/2$ for some integer $ r$.[/quote]
simply :

$ 8b\plus{}1\equal{}x^2$ ==> $ 8b\equal{}(x\minus{}1)(x\plus{}1)$ that mean $ (x\minus{}1\equal{}2k)$ or $ (x\plus{}1\equal{}2k)$

$ x\minus{}1\equal{}2k$ so $ x\plus{}1\equal{}2k\plus{}2$ ==> $ 8b\equal{}4k(k\plus{}1)$ ==>$ b\equal{}k(k\plus{}1)/2$

same thing for $ x\plus{}1\equal{}2k$  :wink:

$ 8b\plus{}1\equal{}x^2$, so $ x\equal{}2r\plus{}1$ for some $ r$. Substituting yields
$ b\equal{}\frac{(x\plus{}1)(x\minus{}1)}{8}\equal{}\frac{(2r\plus{}2)(2r)}{8}\equal{}\frac{r(r\plus{}1)}{2}$


How many primes amongst the positive integers, written as usual in base ten are such that their digits are alternating 1's and 0's, beginning and ending in 1?

[hide]let the number be $1+10^{2}+10^{4}+...+10^{2n}=\frac{10^{2n+2}-1}{99}=\frac{(10^{n+1}+1)(10^{n+1}-1)}{99}.$
In order for the number to be prime, one of the factors must be equal to 1 after the division by 99.
Since $10^{n+1}-1$ is a string of 9s, it is divisible by 11, and $10^{n+1}+1$ is not divisible by 9.
If the left factor is divisible by 11, the number is not prime, since $10...01$ cannot be equal to 1 upon division by 11 and $9999...$cannot be equal to 1 upon 9.
So the right factor is divisible by 11, and $99=10^{n+1}-1\implies n=1,$ if the number is to be prime.
so $101$ is the only solution
[/hide]


function
induction
algebra unsolved
algebra

Find all functions $f: \mathbb Q \to \mathbb Q$ such that for all rational $x$ and $y$,
\[ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)\plus{}2f(y).\]

[quote="amparvardi"]find all functions $ f: Q \rightarrow Q$ such that :

$ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x) \plus{} 2f(y)$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x) \plus{} 2f(y)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,x)$ $ \implies$ $ f(2x)\equal{}4f(x)$
$ P(2x,x)$ $ \implies$ $ f(3x)\equal{}9f(x)$
And an immediate induction using $ P(nx,x)$ shows that $ f(nx)\equal{}n^2f(x)$ $ \forall n\in\mathbb N$
$ P(0,x)$ $ \implies$ $ f(\minus{}x)\equal{}f(x)$ and so $ f(nx)\equal{}n^2f(x)$ $ \forall n\in\mathbb Z$

And so $ f(nx)\equal{}n^2f(x)$ $ \forall n\in\mathbb Q$

Hence the result : $ \boxed{f(x)\equal{}f(1)x^2}$ which, indeed, is a solution.


algebra polynomial Vieta algorithm geometry calculus

I've been looking for some other methods of finding roots to polynomials, thus far I've been using the Rational Zeros Theorem. Any other methods would be greatly appreciated.

Exact solutions are not generally possible for polynomials of degree $ 5$ or higher.  Exact solutions are known for polynomials of degree $ 4$ or lower, but for degree $ 3, 4$ those exact solutions are computationally intensive.  What you actually want to do, then, is [url=http://en.wikipedia.org/wiki/Polynomial_factorization]factor[/url].

Do you want approximate solutions?

check here for the useful vieta formulas.
http://en.wikipedia.org/wiki/Vieta_formulas  :wink:
also there is bolzano,single bound theorem and decent sing theorem that help you find roots of a polynomial but for sure not the exact roots as the RRT does.

what is decent sing theorem? can you explain please.

Perhaps he is referring to Descartes' Rule of Signs?

Useful, simple techniques for approximating roots include repeated application of the Intermediate Value Theorem, and Newton's Method.

These can algorithms can be iterated many times to produce a good approximate value, or can give you ballpark areas to assist you as you try to guess possibilities with the RRT.

If it's not to much trouble, can someone explain to me Newton's method or Laguerres method? I know there on wikipedia but I'm having trouble grasping it (all of my knowledge of calculus is self-taught). Anyway, thanks for your input.

Cubic roots can be found using [url=http://www.literka.addr.com/mathcountry/algebra/cubic.htm]Cardano's Formula[/url]


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What have you heard? What do you think? This could be particularly interesting, as I know there are many Iranians on this forum.

There's definitely some funny business going on... the fact that Moussavi got roughly the same share of the votes as everywhere else in his strongholds is rather striking.

I have been dazed by lots of either interesting or frustrarting news and events. I haven't been in such situation. There are a lot of gossips about this election. The situation here is really weird for us. Cell phone service have been cut in Tehran for some days and there was no immediate comment from Iran's Telecommunications Ministry. Facebook was blocked to prevent people from demonstrating ( it's really funny ,though). Lots of people clashed with police severely and so many people have been arrested, unfortunately, my cousins are included :( .
I've heard that Mir Hossein Mousavi (who had been expected to be the president) and Mehdi Karubi (the other Ahmadinejad's opponent) rejected the result as rigged and also they said, they won't surrender to this manipulation. As far as I know the majority of Irainian are opposing to Ahmadinejad's policies ,especially his economic & foreign policies. I do believe that his contemptuous attitude relating to the Iranian's opinions and other countries can cause serious problems. So all of my families and my friends voted to the other Ahmadinejad's opponents on Friday to prevent him from not to be re-elected.
It is quite clear for us that we are not living in a democratic country, we hoped that we were approaching to a democratic one but with this harsh situation, I don't know to what we can hope for?   :|

I am hearing a lot about rigged elections in the press these days, but I don't see any attempt at giving arguments or sources for this claim except that "everyone in Iran believes that". I am not saying there is no evidence, but I just have rarely seen such an amount of statements and such a lack of substantiation for it given in the European press or even on Al-Jazeera English. Even the War on Iraq had better "evidence"...

While I do see good evidence for the government to behave even more repressively than usually in the last few weeks, one crime doesn't necessarily prove another. So I'd like to ask you or other Iranians here on MathLinks (fortunately, we have quite a number of them here) whether you have more detailed information about this election than the Western media.

It is not like Ahmadinejad is supplying good evidence for the election to be clean either - what wouldn't be quite a bad idea after such actions as censoring social networking shortly before the elections... But I am interested in harder proofs. It sucks to have no halfway reliable data about any of the sides.

This is NOT personal. Here in Germany I and many people like me are in a similar situation, voting for a party which half of our circle of friends agrees with and left wondering why it gets 0.9%. But circles of friends are rarely representative for the population, and vote fraud requires better proofs... I just want to have more data. I am not questioning anyone's good intentions, particularly when opposing an autocratic government.

  darij

Evidence for the results being false:
- They were released too fast. 40 million paper ballots take longer than that to count. Also, Iran's rules include a three-day wait before the results can be certified, which was skipped.
- The results don't make sense in detail. Karubi certainly had significant support, and he was listed with a negligible total. Ethnic voting is common, and even Azeri cities were reported for Ahmedinejad. The official story is inconsistent; one version reported in Farsi has Ahmedinajad winning Tehran, while one reported in English has Mousavi winning Tehran but Ahmedinejad winning overall.
- Another set of numbers was leaked today, and it looks completely different: Ahmedinejad third at about 15%, with a runoff between Mousavi and Karubi indicated.

I prefer sourced evidence over just claimed evidence, but these are already some good points - thanks Jmerry!

[quote="jmerry"]Evidence for the results being false:
- They were released too fast. 40 million paper ballots take longer than that to count. Also, Iran's rules include a three-day wait before the results can be certified, which was skipped.[/quote]

Good point, but I'd like to have a source on that wait.

[quote="jmerry"]- The results don't make sense in detail. Karubi certainly had significant support, and he was listed with a negligible total. Ethnic voting is common, and even Azeri cities were reported for Ahmedinejad. The official story is inconsistent; one version reported in Farsi has Ahmedinajad winning Tehran, while one reported in English has Mousavi winning Tehran but Ahmedinejad winning overall.[/quote]

The latter sounds interesting, but again I'd better have the source. The former is a very good point. I don't understand how a politician can get from 17.24% to 0.9% without a major scandal, and ethnic voting by the Lur should have had more of an impact.

[quote="jmerry"]- Another set of numbers was leaked today, and it looks completely different: Ahmedinejad third at about 15%, with a runoff between Mousavi and Karubi indicated.[/quote]

Where? Link please.

  darij

You can find a debate at http://thelede.blogs.nytimes.com/2009/06/13/landslide-or-fraud-the-debate-online-over-irans-election-results/?ref=global-home about this very question. This page includes loads of sources.

I noticed that one NYTimes article mentioned that they had talked to an anonymous election official who told them that the numbers were entirely made up and that this rigging had been planned for weeks. (Edit: See http://www.nytimes.com/2009/06/14/world/middleeast/14memo.html, closer to the end.) I also heard that an ayatollah has declared the election was a fraud.

[quote="Ihatepie"]I've also heard that the 6 figures released by Iran's main news organization had the results at a near linear progression. (An r^2 value of .998)[/quote][url=http://www.fivethirtyeight.com/2009/06/statistical-evidence-does-not-prove.html]Misleading[/url]. If you release election results in a few batches, they'll probably look like that no matter what. You'd have to look at regional breakdowns to get a good picture, and I haven't seen any numbers for those.

There's not a lot of verified information out there; the crackdown has made it very hard for media to do anything, so even the primary sources tend toward hearsay.
I've been following through various blogs. One source with most of the pieces is [url=http://andrewsullivan.theatlantic.com/]here[/url].

Thanks for the link. [url=http://niacblog.wordpress.com/2009/06/13/election-liveblogging-saturday/]This[/url] in particular.

  darij

Some more reading material:
http://www.juancole.com/2009/06/stealing-iranian-election.html
http://tpmcafe.talkingpointsmemo.com/2009/06/13/irans_political_coup/?ref=fpblg
http://www.alarabiya.net/articles/2009/06/13/75853.html

The second page of the link Darij gave includes reference to alleged "real" election results -- they are, understandably, somewhat skeptical of whether these should be taken seriously.

Indeed, and it's completely different from the set of "real" numbers I saw earlier. When you're wondering whether a candidate got 1% or 6% or 30%, some of these numbers are coming out of thin air.

Check out the pictures on [url=http://en.wikipedia.org/wiki/2009_Iranian_election_protests]Wikipedia[/url] of people burning things in the streets of Iran in protest of the election results. I am very interested to see what will happen next.

I am from Azerbaijan, and I can confirm that the cities with predominantly Azeri population were reported to vote for Ahmadinejad, while the majority of the population in these cities openly opposes him. I have also heard that in addition to blocking facebook etc. internet has been slowed down, so that it will take 1 hour or so to load virtually any website through proxy.

I admit jmerry's statements. This may also help:

http://www.time.com/time/specials/packages/article/0,28804,1904645_1904644,00.html?cnn=yes

http://lotfan.org/

Omid, ehsan, thanks a lot for the links, and one more request: Could you please tell us how the news are covered in Iranian mass media (TV, newspapers, [i]not[/i] oppositional websites)?

To add to the link list,
http://www.rsf.org/An-election-without-free-flow-of.html
http://www.rsf.org/Massive-censorship-accompanies.html
The RSF being not quite a pro-American institution, I think these can halfway be trusted. Still not much real evidence for rigging, but tons of evidence for censorship and dictatorship. I am wondering whether the question about vote rigging has any relevance now - it seems that the point of no return is passé, and the only possible outcomes are either revolution or a police state under a militarist tyrant.

  darij

Right Iranian media calls the protesters  small groups of people tending to make the country unsafe and supported by foreign countries. They say that everybody should accecpt the results. Iran is under heavy censorship right now and Facebook and Youtube are blocked in Iran. Also Mobile phones are down.

http://www.physorg.com/news164174928.html

Yesterday Ahmadinejad celebrated his re-election in Valiasr Square which was heavily shown in Iranian media, Supreme leader has confirmed the election results and everybody has been called to support the elected president. A protest has started in Tehran at 4 PM (+3:30 GMT) and is announced illegal yesterday by Iranian media, and it is said that protesters will be stopped by the police.

Also last night police has attacked to Tehran and Isfahan university students and many students have bean beaten (and maybe killed)

These are some photos from yesterday:
http://www.bbc.co.uk/persian/iran/2009/06/090615_mm_ir88_after_election_riot.shtml

Another link about Domestic media:

http://news.bbc.co.uk/2/hi/middle_east/8100310.stm

Thanks very much to our Iranian members for the updates from the scene -- stay safe, and good wishes.

More analysis:

[url]http://www.fivethirtyeight.com/2009/06/iran-does-have-some-fishy-numbers.html[/url]

More pictures:

http://www.boston.com/bigpicture/2009/06/irans_disputed_election.html

I thought Ahmadi's comparison of the elections and ensuing protests to a soccer game was particularly ridiculous, arguing that the fans of the losing team are simply disappointed. And then he goes on to claim that the protests are proof that Iran is a free country with free speech and democracy. Genius!

http://twitpic.com/photos/madyar has some frightening photos (not safe for work or mind). This one in particular: http://twitpic.com/7f6tr (I don't know if the image links are persistent with only four identifying characters, but it shows a thin door to a dorm room full of bullet holes).

Omid sent me a good summary of the events: http://www.guardian.co.uk/commentisfree/2009/jun/14/iranian-election

Andrew Sullivan has been all over this story like white on rice.  In particular, he's culled together a cacophony of first-hand sources via youtube, twitter, etc.  As he himself emphasizes, one should consume such unfiltered information with one's brain turned on.

http://andrewsullivan.theatlantic.com/the_daily_dish/

I also heartily recommend the amazing pictures collected at Boston.com, already mentioned by Omid:

http://www.boston.com/bigpicture/2009/06/irans_disputed_election.html

@darij: The information that's out there (that I've seen, much of which has been linked to above) seems to imply that the election was stolen.  I'm not sure what your standard of proof is, but I would imagine that it's hard to PROVE that there is election fraud when the perpetrators control the electoral apparatus.  That's why it's natural for the burden of proof (not a high one) to fall on those claiming a fair election.  (Usually the burden is achieved by following certain procedures and providing a fair degree of transparency.)

To our Iranian members, I'm curious what you think the American response ought to be.  I (sort of) watched Obama's comments, and they seem reasonable to me at this early stage, but what do you say?  In any case, good luck and stay safe.

[quote="Kalle"]I thought Ahmadi's comparison of the elections and ensuing protests to a soccer game was particularly ridiculous, arguing that the fans of the losing team are simply disappointed. And then he goes on to claim that the protests are proof that Iran is a free country with free speech and democracy. Genius!
[/quote]

I have to say that your're right and we have got used to hearing such ridiculous speeches from Ahmadinejad during the last four years.

[quote="yenlee"]To our Iranian members, I'm curious what you think the American response ought to be.  I (sort of) watched Obama's comments, and they seem reasonable to me at this early stage, but what do you say?  In any case, good luck and stay safe.[/quote]

At first, I wish, I would have such a reasonable president like Obama. I heard his comments on BBC and I do think that U.S. and the other countries shouldn't approve this election until the government respect to the people's opinions and votes and they can officially prove that what the truth is, and U.S. would have to impose more sanctions on the government if these oppositions didn't work and we couldn't achieve our desire.

Anyway, unfortunately, 7 people were killed yesterday at night near the rally and I've heard that today more unauthorized rallies will be holding.

An interesting bit of news for today: Ahmedinejad is out of the country, in Russia. It's a meeting of an organization of central Asian states. He was originally scheduled to be there Monday, but delayed a day.

Proof enough?

[hide="click for image"][img]http://img196.imageshack.us/img196/1163/377pxfakeresultsiran.jpg[/img]
(Oh wait...maybe the vote was just in a quantum superposition! :P)[/hide]

In [url=http://www.cnbc.com/id/31391532/site/14081545]this video[/url], starting at around 12:15, Obama explains the reasoning behind his position.

Also see [url=http://www.msnbc.msn.com/id/26315908/#31396275]this video[/url], which, although biased in favor of Obama's approach, compiles various clips of statements both by Obama and by those who oppose his approach to the protests in Iran.

[quote="calc rulz"]Proof enough?[/quote]  Or, equally (if not more) plausibly, someone photoshopped those images and they have no determinative value whatsoever.  (Unrelatedly, you could have just put those comments in the same post!)

http://arxiv.org/pdf/0906.2789v1.pdf
Any qualified comments on this? I am somewhat skeptical about the argument that
[quote]The total numbers of votes in the 366 voting areas in the MOI’s data vary from about $ 10^4$ to $ 10^6$, i.e. two orders of magnitude. This suggests that Benford’s Law (Newcomb 1881; Benford 1938) may be applicable [...][/quote]
while Benford's law assumes logarithmic uniformity - even an approximate one. Could anyone here who has an idea on statistics give better arguments, or, preferrably, check this criterion against some other elections in history that are generally agreed to be non-rigged?

I am seeing two counterstatements:

http://www.heise.de/newsticker/foren/S-Carter-Center-Benford-s-Law-ist-nicht-auf-Wahlen-anwendbar/forum-160799/msg-16903490/read/
The crux of the argument seems to be that "In particular, Benford's Law is characteristic for scale-invariant data, while election machines are allocated to maintain a relatively constant number of voters per machine." Apparently, this is not completely on-topic, because they are talking about voting machines with a relatively constant expected number of votes counted per machine, while here we are comparing voting areas with "from about $ 10^4$ to $ 10^6$" inhabitants. But still the question remains whether this is enough. For instance, if this really means a lot of voting areas with $ 10^4$ inhabitants and a few main cities with $ 10^6$, then it clearly isn't.

http://www.heise.de/newsticker/foren/S-Ein-echter-Statistiker-hat-keine-solche-Auffaelligkeiten-gefunden/forum-160799/msg-16903346/read/
Sounds interesting, but let's keep in mind that the second digit Benford's law is way weaker than the first-digit version, i. e., the probabilities differ much less. How much? I'm leaving it to the statisticians.

EDIT: See also http://www.fivethirtyeight.com/2009/06/karroubis-unlucky-7s.html

To be honest, I don't really care anymore whether the elections were "clean". They were most likely already unfair given the censorship of oppositional web resources shortly before the elections, and the regime's reaction to the protest totally blew the pseudo-democratic cover. The point of this post was just to mention an argument that we ML'ers can probably evaluate and verify.

  darij

To JBL : I clearly remember the moment when Rezaies votes fell .

I'm curious about our Iranian members' opinions on [url=http://www.fivethirtyeight.com/2009/06/did-polling-predict-ahmadinejad-victory.html]this[/url] article from fivethirtyeight.com.

It discusses the results of a poll conducted about a month before the election.  Though only about a thousand people were polled, it seems those who were willing to speak to pollsters quite strongly supported Ahmadinejad (assuming the TFT took a statistically representative sample of the population).  If one is to take these results as being representative but nevertheless believes that Ahmadinejad would not have won legitimately, then one must conclude -- as the article does -- that an overwhelming majority of those who refused to answer or claimed they were undecided eventually voted for Mousavi.

One might interpret this to mean -- again, as the article does -- that people in Iran have a healthy fear of revealing their dislike for Ahmadinejad, even to a pollster over the phone.  This would strike me as surprising, especially in light of responses to, for example, poll questions like Q14e, Q14f, Q24a, and Q24b.  The last two, in particular, show a willingness to dissent far more serious than merely not voting for Ahmadinejad.

So here are my questions to our Iranian members: Do you think that there are many people in Iran who feel uneasy about voicing disapproval of the President?  If so, is this fear well founded?  If not, how do you interpret the results of this poll?  In particular, why might such a large percentage of those polled have been undecided so close to a highly anticipated election?

I apologize for not replying earlier. Here are the answers from my view point.

[quote="JoeBlow"]Do you think that there are many people in Iran who feel uneasy about voicing disapproval of the President?[/quote]

It's a sort of vexed question,though, and the answer depends on what kind of people you're referring to? By and large, I think many people feel uneasy from speaking out against Ahmadinejad's government, though he has got used to refusing this fact that his policy is not cracking down the oppositions during the last four years.

[quote="JoeBlow"]If so, is this fear well founded?[/quote]

yes, it has been evidently proven. Now I can mention to some cases like, closing a lot of news papers only because of critizing his policy, sending so many students, professors, journalists and researchers to prison because of exressing their opinions, etc. Even if he didn't perform like above cases, he would ignore lots of constructive criticisms from professionals and those who wanted to decrease the amount of destructions that his policies had brought for us.

[quote="JoeBlow"]In particular, why might such a large percentage of those polled have been undecided so close to a highly anticipated election?[/quote]

I don't know the reason, it's really odd for me, I can only justify myself with this that maybe because of the way of presidential election devised in Iran that people only two weeks before the election may find the chance to get familiar with the candidates.

Without diminishing the terror methods and the real repression, I point out there is a certain degree of misinformation.
1)There is a video where you can see real-time how one student is shot in front of the camera. It looks like a militia compound is attacked by students who throw rocks,  fire and bang in the gate.  Isn't it normal to be shot back then? Why young people risk their lives by attacking an armed compound with rocks?
CNN  showed only the part when the students are shot without showing the entire video.
2)Once it had a newscast called "Rift between the clerics"  showing a photo with 3 people wearing turbans along the protesters and other 2 photos with probably the same clerics, these times only one cleric per photo. That's the substance of the broadcast: 3 vague photos. The rest of the broadcast was just filling. Is this the big news, the rift?
3) One day, while a guy talked about the new developments in Iran CNN showed in background a video from 1 year ago  . The video is about a brutal beating. Another proof of misinformation because the video may be associated with the recent elections' protests. It is easy to manipulate the watchers.
4) Today, 9 July, on the post "Protesters in Iran met with tear gas", there's a line "Police blocked other roads leading to Tehran University, while some protesters set trash cans on fire to counter the effects of the tear gas. See the protests and violence through Iranians' own photos". I clicked on that link and there are 39 photos from Iran protest, none of them being shot today, as it would make sense.

I wonder why these guys from CNN  are paid when they select only what they like?  5-10 minutes of Youtube searches would have clarified the timing or the substance of the videos.
I can provide the Youtube links of the videos I talked about if requested. They are graphical and I didn't see the point of putting them here.

Addenda: About the video 1), one witness commented that Basij were shooting from the rooftops, so I assume the youngsters were seeking revenge by attacking their compound with rocks. A heroic, incredible act. I also playlisted some videos with rooftops shootings, where you can see clearly the snipers. So I tend to believe the witness.

If you watched the entire cnn report, they are mostly saying that due to the new Iranian policies concerning journalists, they have to rely on viewers. They then said that these are completely [b]unverified.[/b] So no, I don't think they are purposely misguiding the public.

At least you can still come here and other not so famous forum/chat websites


Prove that :$\forall x,y,z \in R$ and $n \in N:$
$(x+y)^n=\large\sum\limits_{k=0}^n \large{n\choose k}x(x-kz)^{k-1}(y+kz)^{n-k}$

[quote="Rust"]You are wrong.
$(x+y)^n=\large\sum\limits_{k=0}^n \large{n\choose k}(x-kz)^k(y+kz)^{n-k}$, because $x+y=(x-kz)+(y+kz)$.[/quote]

Trivially, It is true ! But it doesn'n mean my problem is wrong ! Do you think so ??

Well OK, but nevertheless the problem is wrong. Just consider the case
$n =1$ and you easily see that the equation does not hold
(as a equation of polynomials or rational functions).


linear algebra
matrix
linear algebra unsolved

Definition a matrix $X$ is  called a generalized inverse for $A\in M(n,C)$  if
$XAX=X, \; AXA=A$ and $AX,\; XA$ are Hermitian

Prove that any matrix $A \in M(n,C)$ there exist a unique generalized inverse $X$

Write $A$ as $A=BC$, where $B,C$ are $n\times r$ and $r\times n$ matrices respectively, both of rank $r$, where $r$ is the rank of $A$. Notice that $XAX=X$ gives $r(X)\le r$, while $AXA=A$ gives $r(X)\ge r$, so the rank of $X$ is also $r$.

$A^*=A^*X^*A^*$, together with $X^*A^*=AX$ and $A=BC$ gives $C^*B^*=C^*B^*BCX$. Multiply this to the left with $C$ and then $(B^*B)^{-1}(CC^*)^{-1}$ to get $CX=(B^*B)^{-1}B^*\ (\#)$. If in the procedure above we use the fact that $XA$ is hermitian (instead of $AX$), we get $XB=C^*(CC^*)^{-1}\ (\#\#)$.

From $(\#\#)$ and the fact that $X$ has rank $r$, just like $C^*$, we find that the space generated by the columns of $X$ is the same as the space generated by the columns of $C^*$, so we can write $X=C^*T$ for some complex matrix $T$. Replace $X$ in $(\#)$ and multiply to the left with $(CC^*)^{-1}$ to get $T=(CC^*)^{-1}(B^*B)^{-1}B^*$, so $X=C^*(CC^*)^{-1}(B^*B)^{-1}B^*$.


algebra
polynomial
calculus
derivative
superior algebra
superior algebra unsolved

Let $ L/K$ be a finite extension, and suppose that there is an irreducible polynomial $ f(x)\in K[x]$ that has a multiple root in $ L$.
Given that $ K$ has characteristic $ p$ with $ p>0$. Show that the degree $ [L: K]$ is divisible by $ p$.

What can we say about the derivative of $ f$?


Hi everyone.
I was wondering what CRUX is cuz i've seen it appear in many posts here.Is it a mths magazine? If so, in what language is it published and where can i find it?

CRUX is the short name of "Crux Mathematicorum". It is a Canadian magazine purely devoted to elementary mathematics and problem solving. The languages are English and French. The website of the magazine is http://journals.cms.math.ca/CRUX ; it contains all volumes of Crux from 1996 until now, but you need subscribing to gain access to them. (The 1991 and 1996-1999 volumes are free for public access.)

  Darij


calculus
integration
conics
parabola
calculus computations

What is the volume of the solid obtained by rotating the region between $x=y^{2}-1$ and $y=x$about the $y$-axis ?
A) $\pi$

B) $\frac{2\pi}{3}$

C) $\frac{7\pi}{15}$

D) $\frac{\pi}{2}$

E) $\frac{2\pi}{5}$

[u]My Solution Approach:[/u]
The parabola $x=y^{2}-1$ and the line $y=x$ intersect on the points; $(\frac{1-\sqrt5}{2}, \frac{1-\sqrt5}{2})$ and $(\frac{1+\sqrt5}{2}, \frac{1+\sqrt5}{2})$ by solving the equation $x^{2}-x-1=0$ obtained from these two functions. To make it easier let us take $\frac{1-\sqrt5}{2}\simeq{-0.618}$ and $\frac{1+\sqrt5}{2}\simeq{1.618}$ and after graphing by hand I see 3 different regions from left to right to compute the whole volume.

${V_{1}=\int_{-1}^{-0.618}(2pi.x)(2\sqrt{x+1})dx}$

$V_{2}=\int_{-0.618}^{0}(2pi.x)(\sqrt{x+1}+x)dx$

$V_{3}=\int_{0}^{1.618}(2pi.x)(\sqrt{x+1}-x)dx$

$V_{Total}=V_{1}+V_{2}+V_{3}$

I can solve the above integrals [i](and actually did but found a result different from those five choices)[/i] however I am not sure about my way of organizing these integrals, bounds, and regions, so would you please help me?

Thanks and regards

This is a badly stated problem. Both $V_{1}$ and $V_{2},$ which result from rotating something to the left of the axis, have a non-trivial overlap with the $V_{3},$ which comes from the right of the axis. In such an instance, I have a hard time seeing an unabiguous meaning for "volume of revolution." That being the case, I don't know what the "correct" answer is or whether there even is such a thing.

One particular comment: if you have a region to the left of the $y$ axis that you're rotating, the volume should actually be $\int_{-a}^{-b}(-2\pi x)(f_{2}(x)-f_{1}(x))\,dx.$ You don't seem to have included that minus sign and without that, you'll get a negative volume.

But seriously - if you're getting this problem from some other source, are you sure you copied it down correctly? Was there some extra qualifying language like "the region in the first quadrant" that you failed to copy?

Thanks for the answer and explanations.

The original question was in Turkish language and I tried to translate it into English using standard technical terms mentioned in Stewart's Calculus (Fifth Edition) book.

I'll try to resolve considering your particular comment.

No, there was no additonal statement such as "the region in the first quadrant" in the original wording of the problem.


inequalities
integration
inequalities unsolved

Let A = $ 1 \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {3}} \plus{} ... \plus{} \frac {1}{\sqrt {80}}$   find the integer n which satisfies the condition  n - 1 < A < n

What is the solution in the general case ?

Let $ S$ denote the sum in question.

In general, $ \int_1^k \frac{1}{\sqrt{x}} dx < 1\plus{}\frac{1}{\sqrt{2}}\plus{}\cdots\plus{}\frac{1}{\sqrt{k\minus{}1}} < 1\plus{}\int_1^k\frac{1}{\sqrt{x}}dx\minus{}\frac{1}{\sqrt{k}}$.

For $ k\equal{}81$, we get $ 16 < S <1\plus{}16\minus{}\frac{1}{9}$, i.e. $ 16<S<16\frac{8}{9}<17$, so $ n\equal{}17$.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321499[/url]


First I'd like to say hello to everyone as this is my first time here. Now to the question.

I'm coming up with a fund where I have a targeted value that I wish to reach at the end of a certain time period.  I want the amount to be paid each period to increase by a set percentage each time and contiue for however long the payment period is.

This may sound confusing in writing but let me just put an example to somewhat clarify what I'm trying to do.

[quote]"Bob" has a targeted value of 6,000 thousand at the end of a 30 month payment period.  He is starting off with 0. If the amount paid each month is increased by 175%, how much must he pay each period. The First payment would be 60.[/quote]

I was looking at formulas where this would apply and the only formula I could find similar is the compound interest formula. I thought of something myself but looked at it and it would take hours on end just to solve half of it. Here is what I came up with

P = target value
C = current amount
x1 = payment 1 (would be given, 60)
x2 = payment 2 where payment 2 = 1.75x1
x3 = payment3 where payment 3 = 1.75x2
etc.

P = x1 + x2 + x3 + ..... x30

P is the final value and each xn is the product of 1.75xn-1, where n is the current payment.

Some of this may be wrong as I was just taking a shot at it, so please anyone who could show me how it can be done would be great.

This is actually a good bit simpler than compound interest.  It's the sum of a geometric sequence.  We may write
$x_1 = a$ (the initial payment -- in your example, 60)
$x_n = r x_{n-1}$ (each payment is r times as large as the previous payment -- in your case, r = 1.75)
So $x_n = a\cdot r^{n - 1}$.
Then if we want to make k payments, the total sum is
$x_1 + x_2 + \ldots + x_k = a + ar + ar^2 + \ldots ar^{k-1}$.  The sum of such a geometric sequence is given by $a\frac{r^k -1}{r-1}$, which is what you are calling the target number, yes?

I'm not sure I understand that your mathematics in fact represents what you are saying about the fund, but the calculation you are trying to make can be simplified with the formula I just gave.

Thank you very much. I think I understand what's going on but I'm gonna give it a try in a bit and actually solve it and see if it seems right.  After I do it All I'll post the numbers I used etc. and hopefully it's correct, otherwise we can go from there on what I'm doing wrong.

Thank you very much.


$ \left(\frac{3!}{1!1!1!}\right)^3 \equal{} 6^3 \equal{} 216$
What do the three one factorials in the above equation mean?

It'll be better to post the whole problem
The problem might be:
How many ways can 3 sets of 3 different numbers be arranged, if the 3 sets of numbers have to be by themselves?

$ 1!\equal{}1$

$ 1! \equal{} 1$.
$ 1! 1! 1! \equal{} 1 \cdot 1 \cdot 1 \equal{} 1$.

The solution writer is likely referring to the more general formula which says that the number of way to arrange $ n$ things of which $ a$ are of one type, $ b$ are of another, ... is

$ \frac{n!}{a! b! ... }$


geometry
conics
hyperbola
Asymptote
geometric transformation
reflection
geometry proposed

See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]

Triangle $ABC$ and straight line $l$ are given at the plane. Construct using a compass and a ruler the straightline which is parallel to $l$ and bisects the area of triangle $ABC.$

Lines dividing the triangle area in half are tangent to 3 hyperbolic segments, each having 2 triangle side lines as asyptotes, and each touching 2 medians at their midpoints, the 2 medians not passing through the intersection of asymptotes. First, draw 3 lines parallel to the given line through the triangle vertices. One of these lines goes through the triangle interior. Eliminate the hyperbola having the intersection of asymptotes at this vertex. The focal points of the remaining hyperbolas are particulary easy to construct, when the triangle $\triangle ABC$ is projected by a parallel projection into an isosceles right triangle $\triangle A'B'C'$ with the angle $\angle A' = 90^\circ.$ One focal point of the normal hyperbola with the asymptotes A'B', A'C' is the midpoint F of B'C', the other one a reflection F' of F in A'. One hyperbola vertex is on the ray A'F at $A'V = A'F \frac{\sqrt 2}{2}$, the other is a reflection V' of V in A'. Circle (A') centered at A' and with radius A'V = A'V' is the hyperbola pedal circle. A normal to the line $l'$ (obtained from $l$ by the used parallel projection) through the focus F meets the pedal circle (A') at points P, Q. Parallels to $l'$ through P, Q are hyperbola tangents. One is tangent to the other hyperbola branch and it does not cut the triangle at all. If the other one touches the hyperbola branch passing through the triangle $\triangle A'B'C'$ on the hyperbolic segment MN, where M, N are midpoints of B'- and C'- medians (i.e., the constructed parallel to $l'$ intersects the segments A'B', A'C'), it divides in half the triangle area.

My solution is somewhat easier.  :)
It isn't difficult to prove that for at least one of the vertices (for example, $ B$) the line passing through this vertice and parallel to $ l$ contains points within the triangle and divides it into two parts. Let $ T$ be the point of intersection of this line and $ AC$, we assume that $ AT > TC$ (if $ AT \equal{} TC$ everything is clear), also let $ l_1$ be the line we need to construct.
Denote by $ X$ and $ Y$ the points of intersection $ l_1$ with $ AB$ and $ AC$ respectively. Apparently, $ 2*AX*AY \equal{} AB*AC$ and $ \frac {AX}{AY} \equal{} \frac {AB}{AT}$. Multiplying these two equations, we obtain $ 2AX^2 \equal{} \frac {AB^2 *AC}{AT}$, whence we can easily construct the length of $ AX$ and the line $ l_1$ therefore.

[quote="BlackMax"]
We obtain $ 2AX^2 \equal{} \frac {AB^2 *AC}{AT}$, whence we can easily construct the length of $ AX$ and the line $ l_1$ therefore.[/quote]

I suppose we first make a way to construct the lengt of $AX$ at some way, but how would you place it parallel with $l_1$, can you write out your solution complete, so we can see if it is simpler?

[quote="SCP"]I suppose we first make a way to construct the lengt of $AX$ at some way, but how would you place it parallel with $l_1$, can you write out your solution complete, so we can see if it is simpler?[/quote]

What exactly are you interested in, the way I construct $X$ or how to draw a parallel line? The latter is trivial when the line and one point is given. As to expressing $AX$, we know it equals $AX=\frac{1}{\sqrt{2}}AB\sqrt {\frac{AC}{AT}}$. Since we know how to construct a proportional segment if three others are given, and how to construct the geometrical mean, we rewrite the needed expression as $\frac{\sqrt{AB*AC}\sqrt{AB*AT}}{AT}$ and after a chain of transformations obtain a segment equivalent to $AX$.

[quote="SCP"][quote="BlackMax"]
$AX=\frac{\sqrt{AB*AC}\sqrt{AB*AT}}{AT}$ [/quote]

Yes, all you wrote, I agree, but you always tell, you can construct the length easily:
so you can construct length $\sqrt{AB*AC}$ and similar at an easy way (I don't know, but seems it will have an easy way, but how?) and how do you finish a length $xy/z$ if $x,y,z$ are already constructed?[/quote]

For example, if we have two segments with lengths $a$ and $b$, we can take another segment $AC=a+b$, the point $B$ on it such that $AB=a$, $BC=b$. Then by using $AC$ as the diameter to draw a circle containing $A$ and $C$ and intersecting it with a straight line through $B$ perpendicular to $AC$, we'll get two segments $\sqrt{ab}$ units long.

The second one uses the intercept theorem and drawing a parallel line through a given point.

Would I be at the contest writing the solution, I'd mention how they are obtained, just to stay on the safe side, but here I suppose that any user with basic experience in solving construction problems outside a school textbook must be familiar with common auxilliary lemmas.


modular arithmetic
superior algebra
superior algebra unsolved

[quote="heartwork"]Prove the following:

It is possible to partition the set $ Z_{3M}$ of all $ 3M$ residues modulo as {$ 0,1,...,3M \minus{} 1$} in $ M$ triplets of the form $ (a,b,a \plus{} b)$ for all positive integers not equal with $ 2$ - when $ M$ is either $ 1$ or at least $ 3$.

Here the sum (a+b) is also modulo $ 3M$.[/quote]

Let $ \{a_{1}, b_{1}, a_{1} \plus{} b_{1}\}, ... \{a_{M}, b_{M}, a_{M} \plus{} b_{M}\}$ such a partition. Taking the sum of all elements in these sets, we find :

$ a_{1} \plus{} b_{1} \plus{} a_{1} \plus{} b_{1} \plus{} ... \plus{}  a_{M} \plus{} b_{M} \plus{} a_{M} \plus{} b_{M} \equal{} O \plus{} 1 \plus{} 2 \plus{} ... \plus{} (3 M \minus{} 1) (mod 3 M)$.

The first member can be written $ 2(a_{1} \plus{}  ... \plus{} a_{M} \plus{} b_{1} \plus{}  ... \plus{} b_{M})$ and is thus even. The second member is $ (3M\minus{}1)3M/2$.
If $ M \equal{} 2 (mod 4)$, this value is odd, thus an even number is congruent modulo an odd number modulo the even number $ 3M$, which is absurd.

Thus, if I'm right, there is no solution if $ M \equal{} 2 (mod 4)$.
I didn't try to prove that there is a solution in the other cases, I'm not familiar with such problems.

Friendly,
P.

[quote="heartwork"]Prove the following:

It is possible to partition the set $ Z_{3M}$ of all $ 3M$ residues modulo as {$ 0,1,...,3M \minus{} 1$} in $ M$ triplets of the form $ (a,b,a \plus{} b)$ for all positive integers not equal with $ 2$ - when $ M$ is either $ 1$ or at least $ 3$.

Here the sum (a+b) is also modulo $ 3M$.[/quote]

As I said, I think this is wrong if $ M \equiv 2 \pmod{4}$.
I think it's true if $ M$ is odd. Then, take as partition the sets $ \{3a \plus{} 1, 3a \minus{} 1, 6a\}$, where $ a$ runs over the natural interval $ [0, M \minus{} 1]$.
By congruence modulo $ 3$, two sets $ \{3a \plus{} 1, 3a \minus{} 1, 6a\}$ and $ \{3b \plus{} 1, 3b \minus{} 1, 6b\}$ cannot have a common element unless $ 3a \plus{} 1 \equiv 3b \plus{} 1$ or $ 3a \minus{} 1 \equiv 3b \minus{} 1$  or $ 6a \equiv 6b \pmod{3M}$. Each of the first two of these three conditions imply the third, thus two sets $ \{3a \plus{} 1, 3a \minus{} 1, 6a\}$ and $ \{3b \plus{} 1, 3b \minus{} 1, 6b\}$ cannot have a common element unless $ 6a \equiv 6b \pmod{3M}$.
Since we assume $ M$ odd, this condition is equivalent to $ 3a \equiv 3b \pmod{3M}$, in other words $ a \equiv b \pmod{M}$.
If $ a$ and $ b$ are both in our natural interval $ [0, M \minus{} 1]$, this cannot happen unless $ a \equal{} b$, so our thesis is true if $ M$ is odd.

There remains the case where $ M$ is divisible by $ 4$...

Friendly,
P.

P.S. If $ M$ is divisible by $ 8$ and there is a solution for $ M/4$ instead of $ M$, then there is a solution for$ M$.
Proof : since there is a solution for $ M/4$ instead of $ M$ (thus a good partition for the remainders modulo $ 3M/4$), multiply all numbers in this solution by $ 4$ and you obtain a partition of the desired nature for the remainders modulo $ 3M$ which are divisible by $ 4$. There remains to find a partition of the desired nature for the remainders modulo $ 3M$ which are odd or $ \equiv 2 \pmod{4}$.
You can take the sets $ \{a, a \plus{} 3M/2, 2a \plus{} 3M/2\}$, where $ a$ runs over the odd numbers in the natural interval $ [1, 3M/2 \minus{} 1]$.

Hello.

A new fact : if $ M$ is as in the statement (i. e. there is a convenient partition of the set of residual classes modulo $ 3M$), if $ c$ is an odd natural number, then $ cM$ is as in the statelment, i. e. there exists a convenient partition of the set of residual classes modulo $ 3cM$.

Proof.

By hypothesis, we can find $ M$ sets of three integers :
$ \{a_{1}, b_{1}, a_{1} \plus{} b_{1}\}, ..., \{a_{M}, b_{M}, a_{M} \plus{} b_{M}\}$
such that :
1° for each $ i$ $ (1 \leq i \leq M)$, the three numbers $ a_{i}, b_{i}, a_{i} \plus{} b_{i}$ are distinct modulo $ 3M$;
2° if $ i \not \equal{} j$, no element of $ \{a_{i}, b_{i}, a_{i} \plus{} b_{i}\}$ is congruent modulo $ 3M$ with an element of $ \{a_{j}, b_{j}, a_{j} \plus{} b_{j}\}$.

For each $ i$ $ (1 \leq i \leq M)$ and each integer $ k$ such that $ 0 \leq k \leq c \minus{} 1$, let $ A_{i,k}$ denote the set
$ A_{i,k} \equal{} \{a_{i} \plus{} 3kM, b_{i} \plus{} 3kM, a_{i} \plus{} b_{i} \plus{} 6kM\}$.
Two elements of a same set $ A_{i,k}$ are always distinct modulo $ 3M$ and thus, a fortiori, distinct modulo $ 3cM$.
Assume that $ A_{i,k}$ and $ A_{i',k'}$ have a common element modulo $ 3cM$.
Then $ \{a_{i}, b_{i}, a_{i} \plus{} b_{i}\}$ and $ \{a_{i'}, b_{i'}, a_{i'} \plus{} b_{i'}\}$
have a common element modulo $ 3M$, thus $ i \equal{} i'$, thus $ A_{i,k}$ and $ A_{i,k'}$ have a common element modulo $ 3cM$. That is :
$ \{a_{i} \plus{} 3kM, b_{i} \plus{} 3kM, a_{i} \plus{} b_{i} \plus{} 6kM\}$
and $ \{a_{i} \plus{} 3k'M, b_{i} \plus{} 3k'M, a_{i} \plus{} b_{i} \plus{} 6k'M\}$
have an element in common modulo $ 3cM$.
Since $ a_{i}, b_{i}, a_{i} \plus{} b_{i}$ are distinct modulo $ 3M$, at least one of the three following conditions must be satisfied :
1° $ a_{i} \plus{} 3kM \equiv a_{i} \plus{} 3k'M \pmod{3cM}$,
2° $ b_{i} \plus{} 3kM \equiv b_{i} \plus{} 3k'M \pmod{3cM}$,
3° $ a_{i} \plus{} b_{i} \plus{} 6kM \equiv a_{i} \plus{} b_{i} \plus{} 6k'M \pmod{3cM}$.
Clearly, this amounts to saying that $ 6kM \equiv 6k'M \pmod{3cM}$, i. e. $ 2k \equiv 2k' \pmod{c}$. Since $ c$ is odd, this is equivalent to $ k \equiv k' \pmod{c}$, thus $ k \equal{} k'$ (since $ 0 \leq k, k' \leq c \minus{} 1$).
Thus the $ cM$ sets $ A_{i,k}$ furnish a convenient partition of the residual classes modulo $ 3cM$, which proves our thesis.

We have now two facts :
1° if the statement is true for $ M$ and $ c$ is an odd natural number, the statement is true for $ cM$;
2° if $ M$ is even and if the statement is true for $ M$, the statement is true for $ 4M$.
(For 2°, see a P.S. in a previous post.)

Now, the statement is true for $ M \equal{} 1$ (partition $ \{1, 2, 0\}$ of the residual classes modulo $ 3$) and for $ M \equal{} 4$ (partition $ \{4, 8, 0\},\ \{1, 2, 3\},\ \{5, 6, 11\},\ \{10, 9, 7\}$ of the residual classes modulo $ 12$).

Thus 1° and 2° give by recurrence : if the exponent of $ 2$ in the prime decomposition of $ M$ is even, then the statement is true for $ M$.
Remember that the statement is false if $ M \equiv 2 \pmod{4}$.

There remains the case where the exponent of $ 2$ in the prime decomposition of $ M$ is odd and $ > 1$.
I didn't feel up to checking the case $ M \equal{} 8$ (residual classes modulo $ 24$)...

Friendly,
P.

Hello.

There are solutions for $ M\equal{}8$, i.e. partitions as desired of the residual classes modulo $ 24$, for example :
$ \{7,17,0\},\ \{9,16,1\},\ \{11,15,2\},\ \{13,14,3\},\ \{8,20,4\},\ \{10,19,5\},\ \{12,18,6\},\ \{22,23,21\}$
Thus, if the preceding reasonings are right, there is a solution if and only if $ M \not\equiv 2 \pmod{4}$.

What do you think, Heartwork ? Do you know a more direct proof for the case where $ M$ is divisible by $ 4$ ?

Friendly,
P.

[quote="panurge"]Hello.

What do you think, Heartwork ? Do you know a more direct proof for the case where $ M$ is divisible by $ 4$ ?

Friendly,
P.[/quote]

It's still unclear for me the answer - not talking about a rigorous proof.
Please can you resume what you have proved?


hallo,
Let $n \in N$, $n \ge 3$, $X=\{1, 2, 3, ..., n^{2}\}$ and $A$ a subset of $X$. Prove that there are in $A$ or $X-A$ at least $n$ numbers belonging to an arithmetic progression.  :)

I think there is a missing hypothesis otherwise you could prove that the van der waerden number $w(n,2)$ is not greater than $n^{2}$, but it is known that $w(4,2) = 35 > 16$... :wink:

Pierre.


I'm really confused  :(  Any good ebook I could refer for IIT-JEE studies?  :ninja:

search solomons at http://www.4shared.com :)

[quote="kabi"]hello
If u r in xi go for Boyd with its solution(solutions are very nice) but if u r in xii go for Solomons only.
But be careful and first consult to ur teachers or seniors that what should u read from these book as both books contain much beyond what JEE demands .There is so much to read in it so be sure u stick to course of JEE only.
Thank u.[/quote]
I see what you mean. Its a pretty confusing book. I've got a second hand old edition. The prob is it doesn't start explaining concepts and mechanisms one by one. It rather focusses on a special family of compounds followed by another and so on  :|

Hello
Read those books I u want to read chemistry in deep.I know their contents is very much different from JEE pattern but the concepts given in these books r very much clear.
Almost every IITian would suggest u these two books.
But it is on u.
Thank u.

[quote="rituraj007"]:rotfl: i understand...i think u need sanyal's Organic mechanics..that book  along with ne other one....... like modern abc??(my fav)[/quote]
Can you elaborate a bit more on the exact names and writers of the books please? I tried looking for their ebook but was unable to find one. :(

u can buy it--just Rs75 or so
SN sanyal's-"Reactions,reagents and rearrangements"--Bharti bhavan publications :)

Is arihant organic chemistry good?
My tutor told me so but I want to reconsider before buying :wink:

the questions are good but answers all wrong

Arihant is another good 1--but as gaurav said-answers r  bloody wrong

[quote="Arnav...!!!"]i liked boyd...........

just read it lyk a novel.........trying to concentrate a bit more :blush:

if u have read the chapter before (a very gud knowledge is not reqd.)..........then u will find boyd very intresting.......

also...............is the solomons book the same..........

written by solomons and frhyle?????????[/quote]

Yes Arnav boyd is really very interesting

I am pleased to find chem is not as boring as i tht it wud be(Atleast until now) :rotfl:

But the ebook version i have of boyd is missing a few pages

So cud anyone plz post a link for an e book of boyd

Thanks,
Befuddlers

[quote="befuddlers"]

Yes Arnav boyd is really very interesting

I am pleased to find chem is not as boring as i tht it wud be(Atleast until now) :rotfl:

But the ebook version i have of boyd is missing a few pages

So cud anyone plz post a link for an e book of boyd

Thanks,
Befuddlers[/quote]

Why are u guys ignoring my request??????? :mad:

we dont read those books page by page :rotfl:  :P  :rotfl:

:rotfl:

http://www.4shared.com/file/32409442/231895d7/Morrison__Boyd--Organic_chemistry.html?err=no-sess

try it from here.............


number theory
least common multiple
floor function

Prove that there are no natural n satisfying:
[b]1/2 + 1/3 + ... + 1/n = k[/b] ; where k is an integer number

Ooh! I can actually solve this one! :)
[hide]Rewrite with  a common denominator equal to LCM of $1,2,3.....2^k...... n$ where $2^k$ is the highest power of two in the sequence. $k$ is a positive integer (Note: $n$ can also be $2^k$).
All the terms in the numerator will be even now but the term in the original sequence with a denominator of $2^k$, which will be odd, since the product of an odd integer and $2^k$ (which is the LCM) divided by $2^k$ is odd. So the numerator will be the sum of several integers and one odd integer, while the denominator will be even. So this sum can not be an integer for $n>1$.[/hide]

Well, I know my idea is right, but anyone want to help me make it more clear?

Here is a cool (sort of) solution using Bertrand's Postulate.  We have:
$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4}=\frac{3*4*5*...*k+2*4*5*...*k+...+2*3*4*...*(k-1)}{k!}$
Bertrand's Postulate states that for every $n\geq 1$, there is some prime $p$ with $n<p\leq 2n$.  Consider a prime $p$ with $\left\lfloor \frac{k}{2} \right\rfloor<p\leq k$. This prime divides all but one of the addends in the numerator of the RHS of the above.  Thus the whole numerator is not divisible by $p$. Clearly $p|k!$.  Thus the expression cannot be an integer.


linear algebra
matrix
vector
Putnam
linear algebra unsolved

Can anyone give me a hint for this question?

If A is an m X n matrix, B is an n X m matrix and n<m, then AB is not invertible.

Thank you^^

rank is indetermined in the question.  Or, do I need to know the rank of A?

Or rather: what do you know about the possible values of the rank of $ A$?

One way to prove that AB is not invertible is by showing that its columns, considered as vectors in R^m, are not linearly independent.

Let a1, a2, ..., an be the columns of A, and b1, b2, ..., bm be the columns of B. What is the k-th column of AB equal to? Well, it is just a linear combination of columns of A, with the coefficients being equal to respective entries of bk. In other words, it is (bk)1 a1 + (bk)2 a2 + ... + (bk)n an.

So, we have m columns in AB, each of which is a linear combination of n columns of A. Thus, the columns of AB cannot span the space of dimension more than n, and since n<m, it follows that the columns of AB are not linearly independent.

That is, in a form somewhat rawer and closer to the source, the argument that t0rajir0u has been talking about.

In effect, you said that $ \text{rank}(AB)\le n<m$ because $ \text{rank}(AB)\le\text{rank}(A)\le n,$ with $ n$ being the number of columns of $ A.$ That's been t0rajir0u's argument all along, and what you said is basically the same thing.

A place where this result pops up in a surprising way: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2102]Putnam 2003 B1[/url].


maybe i'm not getting the question, but arent those all integers? i.e., not decimals?  I must be missing something please explain

To be honest, i think that would be handier for the right-most digit! :)

[quote="tangak"]Find first
(1) decimal digit of 2^987654321
(2) two decimal digits of 2^987654321
(3) three decimal digits of 2^987654321
(4) n decimal digits of 2^98765321 (represent as function of n)[/quote]

anyone know how to do this?
i tried, but seems hard

[quote="joml88"][quote="amirhtlusa"][quote="tangak"]the questions are refering to the leftmost digits.[/quote]
[hide="hint"]
$mod 10$
[/hide][/quote]

mod 10 gives you the [i]right[/i]most digit.[/quote]
oh crap.
yeah you guys are right...that was a stupid mistake.
for the left most digit:
[hide]
let me think for a moment... ;) [/hide]

Ok since there is a general forula for the first n numbers, theres got to be something to do with the number 987654321. anyone anything particular about this number?

YES...i think i know the left most digit:
 :D
[hide="hint"]
ok, look at the following sequence, see if you can see the patter, the sequence is the leftmost digit of the $2^n$ powers:
$1,2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,...$
and so $1,2,4,8,1,3,6,1,2,5$ repeats, (that would be proveable!)
so find the $987654321$th term of the sequence
[/hide]
[hide="answer"]
$1$[/hide]
now for the first two terms
[hide]let me think for a second ;) [/hide]

I tried that but it dosnt actually become periodic.  :(
Going larger, it no work...

By [url=http://mathworld.wolfram.com/BenfordsLaw.html]Benford's Law[/url] , the probability the first digit is 1 is about 30%; the probability that it is 2 is 17.6% etc.  The probability that it is any one digit $d$ is $\log_{10}\left(1+{1\over d}\right)$

What I'm getting at here is that amirhtlusa might have been right!

My suggestion: instead of using mods, figure out how to use logarithms.  Seamus is correct that the pattern is not a simple repetition, and I do not believe that Amir's answer is correct.


vector
function
real analysis
real analysis unsolved

Let [i]f[/i]: R -> R^3 be a differentiable, vector-valued function
and [i]g[/i]: R -> R be an onto, and strictly increasing vector-valued function.
Let [i]h[/i]=[i] f[/i]([i]g[/i](x)): R -> R^3.
Show that the paths traced by [i]f [/i]and[i] h[/i] are equal and that
[i]h[/i]'(t) = [i]f'[/i]([i]g[/i]'(t))*[i]g[/i]'(t) where [i]h[/i] is called the reparameterization of[i] f [/i]by [i]g[/i].

Hmmm... Isn't it exactly the same problem (even with the same strange phrase about "increasing vector-valued function") as [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=40369[/url]?

Anyway, the whole thing follows immediately from the definitions and the chain rule (under the condition that $g$ is differentiable too, otherwise the equality $h'(x)=f'(g(x))g'(x)$ may be violated in the sense that the left side may exist without  existence of the right side). So I do not see what exactly is the difficulty with this problem :?.


calculus
integration
floor function
algebra unsolved
algebra

Let a sequence of integers $\{a_n\}$, $n \in \mathbb{N}$ be given, defined by

\[a_0 = 1, a_n= a_{n-1} + a_{[n/3]}\]

for all $n \in \mathbb{N}^{*}$.

Show that for all primes $p \leq 13$, there are infinitely many integer numbers $k$ such that $a_k$ is divided by $p$.
(Here $[x]$ denotes the integral part of real number $x$).

Generalization (by own) :

Let a sequence of integers $\{a_n\}$, $n \in \mathbb{N}$ be given, defined by

\[a_0 = 1, a_n= a_{n-1} + a_{[n/m]}\]

for all $n \in \mathbb{N}^{*}$.

Show that for all primes $p \leq m^2 +m +1$,if there exist $k_0$ such that $p|a_{k_0}$ then there are infinitely many integer numbers $k$ such that $a_k$ is divided by $p$.
(Here $[x]$ denotes the integral part of real number $x$)

[list]I guess for all $p$ there exist a $a_{k_0}$ . I try to show that the sequence is periodic , but I couldn`t . If someone show it , from $a_{-1} =0$ we get the result...[/list]

Dear mahanmath,
  It would be great if you could post your solution to your generalization.  (If the whole solution is too long, then hints would do fine.)

Thanks in advance!

[quote="mahanmath"]
if there exist $k_0$ such that $p|a_{k_0}$ then there are infinitely many integer numbers $k$ such that $a_k$ is divided by $p$. [/quote]

Do you know that this condition is necessary? That is, you can't just say there exists infinitely many $k$ such that $p|a_k$ for all $p\le m^2+m+1$? Either way, here is the proof

[hide="solution"]
Suppose $p|a_{k_0}$, then consider $S_1=\{a_{mk_0-1},\, a_{mk_0},\, a_{mk_0+1},\dots ,\,  a_{mk_0+m-1}\}$. We have $|S|=m+1$ and due to the recursive equation every element is equivalent $\mod p$. That is for $0\le i \le m-1$,

\[a_{mk_0+i}-a_{mk_0+i-1} \equiv a_{\left\lfloor \frac{mk_0+i}{m} \right\rfloor} \equiv a_{k_0} \equiv 0 \mod p\]

hence $a_{mk_0-1} \equiv a_{mk_0} \equiv a_{mk_0+1} \equiv ... \equiv a_{mk_0+m-1} \mod p$.

Now consider $S_2 = \{a_{m^2k_0+i} \; | i \in [-m, m^2-1] \}$. So $|S_2| = m^2+m$ and the difference between consecutive elements is an element of $S_1$ (from the reursion). Since all elements of $S_1$ are equivalent $\mod p$, it follows that $S_2$ increases linearly $\mod p$. But since $p \le m^2+m+1$, there are at most $m^2+m$ distinct, non-zero elements $\mod p$ and $S_2$ will cycle through all of them atleast once. Therefore if $a_{m^2k_0-m-1} \not \equiv 0 \mod p$ then there exists an element in $S_2$ that is divisible by $p$. Otherwise $a_{m^2k_0-m-1} \equiv 0 \mod p$ and we are done.

Since any element in $S_2$ is greater than $a_{k_0}$, we have found the next element that is divisible by $p$, and we can continue the process to infinity. So there are infinitly many values $k$ such that $p|a_k \qquad \blacksquare$
[/hide]

Now the question is, does there always exist a $k_0$ such that $p|a_{k_0}$ for all $p \le m^2+m+1$

Thank you !

BTW Your question is my question , and I want to see a proof for it !


induction logarithms geometry integration rectangle inequalities calculus

Hey guys! I'm going to think it's really difficult! :D
  Artiona

Well, you may try to prove a better and quite well-known estimate :
$\sqrt { \frac 4 5} \cdot e \cdot \sqrt n \cdot (\frac n e )^n < n!$

Pierre.

You need to use the fact that $(1+\frac{1}{n})^n < e$ for all positive integers n.  From that, we see that $(1+\frac{1}{n})^{3n+2} < e^3(1 + \frac{1}{n})^2 < 3^3$ for all n, so $(n+1)^{3n+2} < 3^3n^{3n+2}$.
Clearly, for n = 1, $(3n)! > n^{3n}$.
Now $(3(n + 1))! = (3n)!\cdot(3n+3)(3n+2)(3n+1) > 3^3(3n)!\cdot(n+1)n^2$.  Thus, by induction, if $(3n)! > n^{3n}$, we have $(3(n + 1))! > 3^3(3n)!\cdot(n+1)n^2 > 3^3(n+1)\cdot n^{3n+2} > (n+1)^{3n+3}$ and so it is proved by induction that  $(3m)! > m^{3m}$ for all positive integer m.  Your quesiton is the case m = 100.

In fact, $300! > 3 \times 10^{14} \cdot 100^{300}$

..well I think this would be more difficult for me than the given problem.If you've got a solution just post it!
Artiona
PS:This was my reply to Pierre. :) As it seems you were faster than me JBL :P

[quote="JBL"]You need to use the fact that $(1+\frac{1}{n})^n < e$ for all positive integers n.

I'm sorry, but i haven't got sufficient knowledge to understand each step of your reasonings which actually involve the Napier's number. ;)(I'll try hard to understand) :)
Have you got another idea?

I don't think that you are going to be able to find a solution that does not use the fact that (1 + 1/n)^n is "substantially" less than 3, although I could be wrong about that.

Let $C$ be the curve $y = \ln (x)$.
Let $S_n$ be the area under $C$ from $x=1$ to $x=n$. Thus $S_n = \int_1^n \ln(x) dx = n \ln (n) - n + 1.$

Using trapezes, you easy see that :
$S_n > \sum_{k=1}^{n-1} \frac { \ln (k) + \ln (k+1)} 2 = \ln (n! ) - \frac 1 2 \ln (n)$.

Now, let $K_n = S_n - \ln (n! ) - \frac 1 2 \ln (n) > 0 .$
Thus $K_n = \ln (\frac {n^n \sqrt n } {n! e^n} ) + 1.$ (1)

For $k=2, \cdots , n-1$, let's consider $n-2$ trapezes with parallel sides $x = k - \frac 1 2$ and $x = k + \frac 1 2$ and tangent to $C$ at $x = k$. Add a trapeze with parallel sides at $x=1$ and $x= \frac 3 2$ and tangent at $x = \frac 5 4$. And add a rectangle with height $\ln(n)$ and sides at $x=n-1$ and $x=n$.
The total area is greater than $S_n$ and leads to the inequality :
$S_n < \ln(2) + \cdots \ln(n-1) + \frac 1 2 \ln (n) + \frac 1 2 \ln (\frac 5 4).$
Thus $K_n < \frac 1 2 \ln ( \frac 5 4 ).$ (2)

From (1) and (2), you get the claimed inequality above....

Pierre.

...cool.It was exactly what I saw at the back of my book, but unfortunately I haven't learned yet integrals ;)  and other such things related.
Artiona


Ok... So... I registered for the October test date. I chose my top two test centers. And boom, without ANY notification whatsoever, I was placed in a test center far away without ANY WARNING at all (that wasn't either of my top two test centers)! It's as if I wasn't able to have any choice over where I tested or not.

So... I missed it. Now I have to reregister and pay a fee of 19. All for nothing.

(And, I bought Steward's Calculus for 15. Zumdahl Chemistry for 7. And these were cheaper than a single reregistration fee).

DOes it often seem that the Collegeboard wants money? I thought it was a nonprofit organization...?

Sadly, what alternatives are there? ACTs, but we still have to go through the Collegeboard for APs and SAT IIs. Meh...

What grade are you in?

When did you register for the test? First come first serve (I think).

Yes, it is first come, first serve. But couldn't Collegeboard at least warn us that we weren't going to get our top choices? It's as preposterous as IB refusing to let sophomores take their exams.

ANd I am a sophomore.

if you registered online, it should immediately tell you your testing location after registering. It is also on your ticket which they recommend you print out weeks in advance and then they remind you often closer to the testing date.

There is no reason for any confusion about testing location, how did you register?

[quote="Simfish"]
DOes it often seem that the Collegeboard wants money? I thought it was a nonprofit organization...?

[/quote]

I don't think it's a nonprofit organization.  Maybe that's why they have a ".com." address.  They give themselves the official-sounding name.

Money?  Of course they want money.  That's why they charge you so much to get AP grades over the phone.  And why do these tests cost so damn much?  Do you really think it costs $30 to print/grade the SAT tests?  Of course not.  Same with the APs.  Besides, they switched to the new SAT not to emphasize skills used in a classroom (or something like that) but because some schools threatened to stop accepting SAT.  And they jacked up the price to 40-something dollars (I think).  Only a for-profit organization would screw over students for market share.

http://www.yaledailynews.com/article.asp?AID=18766

I selected two testing sites (I registered online), and the Collegeboard instead selects another testing site for me. So that explains why I was enraged.

"The College Board is a national nonprofit membership association whose mission is
to prepare, inspire, and connect students to college and opportunity"

THe Collegeboard is nonprofit, according to collegeboard.com. And yet it still demains so much in registration fees.

Did you try calling the College Board to explain the situation?  Maybe they would give you a refund.

"non-profit" doesn't mean "non-money."  They have a massive operation and they have to pay people to run it, to print materials, to design tests, to publicize, to do education stuff, and on and on.  That all costs quite a bit and you pay for it in registration.

yeah, that's true.

i'm calling them tomorrow. Though I'll probably have to learn a lesson myself sooner or later.


A boat heads in the direction N 40 degrees W at a speed of 24 knots. The current runs at 6 knots in the direction N 25 degrees E. Find the resultant speed and direction of the boat.

I would put this in the physics forum...

[hide="Method"]Use either the tail to tip method for finding the resulting vector or use the tabular method for adding the x coordinates and y coordinates[/hide]

I think I got it.
[hide="solution?"]DAO=AOE=40 alt. angles // lines
BAO=115 angles on str. line=180
$BO^{2}=6^{2}+24^{2}-2*6*24*cos115$
$BO=27.087$
$sinAOB=sin115/27.087*6$
$AOB=11.58$
Therefore speed is 27.087knots, direction is N28.42W[/hide]


function calculus derivative trigonometry algebra polynomial limit

Is the following assertion true:
   If $ a_1,a_2,...a_n>0$ and f is a real function with all conditions of existence of derivatives assured and such that $ f^{(n)}(x)+a_{n}f^{(n-1)}(x)+...+a_2f'(x)+a_1f(x)$ goes to 0 when x goes to infinity, then $ f(x)$ goes also to 0?

It's definitely true for $n=1$, and it might be true for $n=2$, I have no idea what happens if $n=3$, but it definitely fails to be true for $n=4$, for example.

Take the functional equation $f^{(4)}(x)+\frac 12 f^{(3)}(x)+\frac 12 f''(x)+\frac 12 f'(x)+f(x)=0$. The function $f(x)=e^{\omega x}+e^{\bar \omega x}$ satisfies this equation, where $\omega=\frac 12+\frac{\sqrt 3}2\cdot i$, but it's clear that $f$ does not tend to $0$ as $x\to 0$, because $f(x)=2e^{\frac x2}\cos (\frac{\sqrt 3}2\cdot x)$.

I suppose a similar counterexample can be found for $n=3$. All we need to do is find a third degree polynomial with positive coefficients which has at least a root with positive real part.

Thank you very much, Moubinool, beautiful solution.

[quote="grobber"]I suppose a similar counterexample can be found for $ n=3 $. All we need to do is find a third degree polynomial with positive coefficients which has at least a root with positive real part.[/quote]
$\lambda^3+\lambda^2+4\lambda+30=(\lambda+3)(\lambda^2-2\lambda+10)$ has roots $\lambda=-3,\lambda=1\pm 3i$.

For quadratic polynomials, positive coefficients do imply roots with negative real parts.


$MD$ is a chord of length $\sqrt{2}$ in a circle of radius $1$ and L is chosen on the circle so that the area of triangle MLD is maximized. Find angle $MLD$.
[hide="My Solution"]

Let $O$ be the centre of the circle. Drawing $OM$ and $OD$ yields that $\angle MOD = \pi/2$ (!). Because of this, we have that $\angle MLD$ [b]can only be $\pi/4$ or $3\pi/4.$[/b] Clearly, to maximize the area of $MLD,$ point $L$ must be on the major arc $MD.$ We thus have that $\angle MLD$ is $\pi/4.$

[/hide]

This problem almost seems too easy... am I missing something?  :maybe:

[quote="sapphyre571"]All the answers and solutions should be up in the archive

http://web.mit.edu/hmmt/www/datafiles/solutions/[/quote]

http://web.mit.edu/hmmt/www/datafiles/solutions/1998.shtml

"Solutions for 1998 are MIA. They will be posted when available."

this is easy

yes that is right

fix the chord, take a parallel line to that chord, the intersection of the parallel and the circle is the last point, clearly, this is the height of the triangle, we want this to be as far away as possible, so clearly, this is when the line is tangent to the major arc MD, it follows that the angle is 45 [central angle to inscribed angle]