What does  $\sum \limits _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right)^{2}$ equal?

We know that $\left(\begin{array}{c} {n} \\ {k} \end{array}\right)=\left(\begin{array}{c} {n} \\ {n-k} \end{array}\right)$ , so we can rewrite the previous equation as:

$\sum \limits _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right)\left(\begin{array}{c} {n} \\ {n-k} \end{array}\right)$

*Insert explanation for how to end with next step here*

This tells us that:

$\sum \limits _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right)\left(\begin{array}{c} {n} \\ {n-k} \end{array}\right) =\sum \limits _{k=0}^{n}\left(\begin{array}{c} {n+n} \\ {k+n-k} \end{array}\right) =\sum \limits _{k=0}^{n}\left(\begin{array}{c} {2n} \\ {n} \end{array}\right)$

Okay where you see the insert here is where the basis for the counting argument comes up.

AoPS addresses this using Pascal's triangle...  I've also heard a method using a supercommittee.  Could someone help explain how you derive $\sum \limits _{k=0}^{n}\left(\begin{array}{c} {2n} \\ {n} \end{array}\right)$ without direct use of pascals triangle?

[hide]Consider an $n*2$ grid. The number of ways to fill in $n$ spaces on this grid is then $\binom{2n}{n}$. The identity works like this: Let $k$ be the number filled in in the first row. $k$ can be any integer from $0$ to $n$, inclusive. The number of ways to fill in the grid such that $k$ are filled in the first row is $\binom{n}{k}^2$ (first choose $k$ to fill in from the first row, then choose $k$ not to be filled in from the second row.) Summing from $k=0$ to $n$ gives
\[ \binom{2n}{n}=\sum_{k=0}^n \binom{n}{k}^2 \][/hide]

Well, this is a particular case of the Vandermonde's identity ;)
So, if you know this formula, it's fairly easy to derive the asked result!

Vandermonde's identity :
$\sum \limits _{k=0}^{r} \binom {m}{k} \binom {n}{r-k} = \binom {m+n}{r}$

So, here, it's just the case where : $m=n=r$ :)


Let us have 6 circles such that the first three go through the same point and the second three go through the same point. Can anybody give me some suggestions on ways how to prove that all of the 6 circles go through the same point...i've tried with chord/string 4-angles, and maybe with the power of a point, and homothety. With what can i try next?

I think the problem statement needs to be a little clearer.  Does the second set of three circles all go through the same point as the first set of three circles or what?  Or maybe you mean that if you have a set of 6 circles, if you choose any three then they will all go through one point.

Like this o_O ?

[img]http://img.photobucket.com/albums/v259/drspyder/aops_circles.gif[/img]

I meant just like Treething showed. Now i have to prove that the two point are actually the same, but with what "tool"?


LaTeX
number theory
relatively prime
number theory unsolved

Here $M=2 \cdot 3 \cdot p(1) \cdot p(2) \cdot p(3) \cdot p(4) \cdot ... \cdot p(n)$ is the product of consecutive prime numbers. $p(n)$ is $n$-th prime number. Then the minimal reduced residue system modulo $M$ is:
 $1<a(1)< a(2) <a(3)<...<a(n) <M$,  That is to say every $a(i)$ is relatively prime to $M$.

Obviously, $a(1)$ is $(n+1)$th prime number.
Could we prove $a(1)-1$ is bigger than difference of any other two consecutive elements of $a(i)$:
$a(i+1)-a(i)<a1-1$?

[i]moderator: $\LaTeX$-edit[/i]

There is a little mistake in my description.
The conclusion  should be a(i+1)-a(i)<=a(1)-1 or a(i+1)-a(i)<a(1)

For example, for M = 2*3*5 = 30, Its reduced residue system:
a: 1 7 11 13 17 19 23 29. Their consecutive difference:
d: 6 4 2 4 2 4 6
The maximal value is 6.

For M = 2*3*5*7 = 210,
a:
1 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 121 127 131 137 139 143 149 151 157 163 167 169 173 179 181 187 191 193 197 199 209
Their consecutive difference:
d:
10 2 4 2 4 6 2 6 4 2 4 6 6 2 6 4 2 6 4 6 8 4 2 4 2 4 8 6 4 6 2 4 6 2 6 6 4 2 4 6 2 6 4 2 4 2 10
The maximal value is 10.
If we can plot the d in a figure, we will find it's a symmetrical figure.
It may be a easy problem, but how to prove it?


Let t be a real number: 0<t<1. Find the largest x such that for every 3 positive real numbers a,b,c is satisfied inequality:

(a/(b+c))^t + (b/(a+c))^t + (c/(a+b))^t >= x.

@ Johan Gunardi
i didn't quite understand.could you please explain

No, i think he's wrong..
The sign is reversed if you use Power Mean, because $ 0<t<1$


Hi! Can someone please help me with this problem?

Let $ a$, $ b$, $ c$ be sides of a triangle with $ a\plus{}b\plus{}c\equal{}2$. Prove that $ ab\plus{}bc\plus{}ca\minus{}abc>1$.

Thanks!

[quote="itiselizabeth"]Hi! Can someone please help me with this problem?

Let $ a$, $ b$, $ c$ be sides of a triangle with $ a \plus{} b \plus{} c \equal{} 2$. Prove that $ ab \plus{} bc \plus{} ca \minus{} abc > 1$.

Thanks![/quote]
$ \Leftrightarrow (1\minus{}a)(1\minus{}b)(1\minus{}c)>0$, it is obvious.

Wow! Thank you! :D

I can’t believe I missed something as simple as that. :oops:

Let $ a,b,c$ be the sides of a triangle with $ a\plus{}b\plus{}c\equal{}2$; Prove that:
\[ \frac{28}{27}\geq ab\plus{}bc\plus{}ca\minus{}abc>1\]

[hide="Hint:"] Inequality  $ \implies(1\minus{}a)(1\minus{}b)(1\minus{}c)\leq \frac{1}{27}$
It's easy AM-GM on $ (1\minus{}a); (1\minus{}b); (1\minus{}c)$[/hide]


quadratics
algebra
number theory unsolved
number theory

$n,m \in N$ , for each n find the mumber of integer roots :  $x^2+2^nx+1= 3^m$

This is a quadratic equation in $x$. If the equation has integer roots, the discriminant will be a perfect square... Is that enough (as a hint)?


calculus
integration
trigonometry
geometry
calculus computations

Compute

\[ \begin{array}{l} \int {\sin ^{ - 1}( x)} dx \\ \int {y^3 \sqrt{y^2 - a^2 } dy} \\ \end{array}  \]

$I=\int {x^3 \sqrt{x^2 - a^2 } dx} = \frac{\sqrt{x^2-a^2}}{5}(x^4-\frac{a^2x^2+2a^4}{3})+C$
Method to solve:
You amplify with $\sqrt{x^2-a^2}$. You will obtain $\int {x^3 \sqrt{x^2 - a^2 } dx}= \int \frac{x^5}{\sqrt{x^2-a^2}}dx-a^2\int \frac{x^3}{\sqrt{x^2-a^2}} dx$ and you make the same thing with the resulting integrals.
$\int \frac{x^5}{\sqrt{x^2-a^2}}dx= x^4\sqrt(x^2-a^2)-4\int {x^3 \sqrt{x^2 - a^2 } dx} = x^4\sqrt(x^2-a^2)-4\;I+C$. Note that $(\sqrt{x^2-a^2})'=\frac{x}{\sqrt{x^2-a^2}}$
$\int \frac{x^3}{\sqrt{x^2-a^2}} dx=\frac{1}{3}(x^2\sqrt{x^2-a^2}+2a^2\sqrt{x^2-a^2})+C$
Combining those two you will obtain the result.

[quote="nemesis"]$\int {\sin ^{ - 1}( x)} dx = \frac{1}{2} ln(\frac{cos\;x-1}{cos\;x+1})+C$
Method to solve: changing the variable.
You chose u(x)=cosx so u'(x)=-sinx. You apply the fallowing trick:
$\int {\sin ^{ - 1}( x)} dx = -\int \frac{-sinx}{1-cos^2x}dx$
You will obtain:
$-\int \frac{u'(x)}{1-u^2(x)} du$
And now you change the variable by applying the formula :
$\int \frac{1}{t^2-a^2}dt =\frac{1}{2}ln(\frac{t-a}{t+a})dt$ you wil obtain that result[/quote]
thanks for reply nemesis
$\int {\sin ^{ - 1}( x)} dx = -\int \frac{-sinx}{1-cos^2x}dx$
Are you considering that $\sin ^{ - 1}( x)=\frac{1}{sin(x)}$?
because $sin ^{ - 1}( x)=arcsin(x)=$inverse sine.

For $\int\arcsin x\,dx,$ try setting $\theta=\arcsin x$ so that $x=\sin \theta$ and $dx =\cos \theta\,d\theta.$

Then we have $\int\theta\cos\theta\,d\theta.$ Continue from there by integration by parts.

Alternative (but related):

$\int_0^a\arcsin x\,dx$ is the area under the curve $y=\arcsin x$ for $x\in[0,a].$ That fits into a box that is $[0,a]\times[0,\arcsin a]$ The area in the box but not under the curve is the area to the left of the curve, and that can be written as

$\int_0^{\arcsin a}\sin y\,dy=1-\cos(\arcsin a)=1-\sqrt{1-a^2}.$

Thus, $\int_0^a\arcsin x\,dx=a\arcsin a-1+\sqrt{1-a^2}.$

(I was tacitly assuming that $a\in[0,1].$)


calculus
integration
limit
floor function

If 3.9~ stands for the infinite decimal fraction 3.9999....... , then which of the
following statements are true?
I. 3.9~ < 4
II 3.9~ = 4
III. The integral part of 3.9~ is 3
IV. The integral part of 3.9~ is 4

Ouch, tough call.  It can be easily proved that $ 3.\bar{9}\equal{}4$.  Therefore, I is not true and II is true.  The ones I'm not sure about are III and IV.  Since it is equal to 4, then I would think IV is true, although this might be subject to interpretation.  FYI, here's a proof:
$ x\equal{}3.\bar{9}$ so $ 10x\equal{}39.\bar{9}$.  Then by subtracting, we get $ 9x\equal{}36\Rightarrow x\equal{}4$.

I can only say that  $ 3.999.......\equal{}\frac{39\minus{}3}{9}\equal{}4$ so II is true
Reguard to the integral part of $ 3.999....$ I am not very sure but I think culd be $ 3$  :maybe:

I think that the main problem is that theinteger part is not continuous. So if we denote by $ a_n$ the number truncated $ 3.\overline{9}$ after the n-th digit behind the point we have
\[ 3 \equal{} \lim_{n\to\infty} \left\lfloor a_n \right\rfloor \neq \left\lfloor \lim_{n\to\infty} a_n \right\rfloor \equal{} 4
\]
and it is up to personal preference which one of the two values to take. I would go with the second one, since after all $ 3.\overline{9}$ is the limit, and it's somehow logical that it is inside the braces.

And, 'II and III only' is not an answer choice in the poll ;)


Mathnerd314 has a little brother, mathnerd315 (not really). Mathnerd315 is in third grade (again, this is all made up) and has to do multiplication worksheets for his math homework. Unfortunately, while trying to get his worksheet out of his folder, mathnerd315 accidentally ripped off part of one of the problems. :oops_sign: All that is visible is:

$2? \times 6? =$

where the ? represents a digit that was on the ripped part of the paper.

Mathnerd315 takes an estimate and assumes that both of the ? digits are fives. He successfully performs the multiplication with these two numbers. What is the probability that his guess is within 10% of the actual result? (Both ? digits are any integers between 0 and 9, not necessarily distinct.)

I paraphrased this one a lot. The original problem was about a half-blind mathematician who couldn't see the right side of numbers :roll:

Oh by the way, no calculators. I found this problem rather tedious without a calculator, and I was hoping there was an easier way. :ninja:

Anybody? :help:

[size=59]Sorry for double posting[/size]

[quote="me@home"]Some ripped worksheet you've got there... what happenned, a chainsaw or a lawnmower? Normally I'd expect a chunk out of the paper, but here there appears to be a jagged pattern where the right digits of numbers are coincidentally cut off... :o  :roll:[/quote]

That's because the problem is written [b]vertically[/b]:
[code]
  2?
x 6?
----
????
[/code]
:harhar:

Does anyone know of any other solutions without multiplying all of them out? Remember, I would have to do this in five minutes without a calculator.

[quote="mathnerd314"]
That's because the problem is written [b]vertically[/b]:
[code]
  2?
x 6?
----
????
[/code]
:harhar:
[/quote]
Guess what? It should be[code]
  2?
x 6?
---?
????
[/code]
:harhar: right back

Look, if you think this problem is implausible, the original problem was about a math professor who wore an eye patch on his right eye and couldn't see the right half of numbers. :roll: I paraphrased it both because I didn't want to violate copyright or anything :police: and because I thought this version would be marginally more reasonable :P

That post was borderline :spam:


pigeonhole principle
combinatorics unsolved
combinatorics

thats right.
otherwise it would be obviously wrong.

hi

well everything is not that obvious

r u sure integers is the only condition

\[ x_1+x_2 = y_1+y_2 = 2 < 4 \]

\[ x_1 = x_2 =1 \]
\[ y_1 = 4, y_2 = -2 \]

wat terms wud u omit

now this example forces me to beleive that they r natural numbers  :P
its always better to tell the conditions altogether

yeah theyre all positive as well.
btw i solved it so you might as well moved to proposed problems section.

[quote="epitomy01"]prove by induction.
say x1+x2 .. . + x_m = y1+y2 .. + y_n < mn
Prove we can cancel some but not all terms from the equality to obtain another equality.[/quote]

WLOG m >= n.

Induct on the sum.

Base case: Sum is m.

Proof of base case: We have $x_i$ = 1. Now suppose n = m. Then we have $y_j$ = 1 and we simply remove $x_1$ and $y_1$ to obtain equality.

Suppose n < m. Then for all j we have $1 \le y_j < m$. Simply pick one and then choose $y_j$ elements of x and you have equality.

Inductive step:

Suppose it is possible if the sum is k. Then prove it is possible if the sum is k+1, if k+1 < mn.

Note that we can reach the sum of k+1 by adding 1 to both sides of any of mn configurations.

Now we use proof by contradiction. Assume that the k+1 case is impossible. This implies that for all mn of these configurations, previously the changed term was cancelled on one side but not on the other side.

At this point it looks like some application of pigeonhole or something should be applied. Maybe someone else can work it out.


$ 2009$ boys and $ 2009$ girls attend several groups. It is known that any of them is attending no more than $ 100$ groups and that any boy in a couple with any girl attend at least one common group. Prove that there is a group containing at least $ 11$ boys and $ 11$ girls.

Let's translate our problem to graph languge.We have tripartite graph.Denote set vertices of these parts as $ B(boys)$,$ L(ladies)$ and $ G(groups)$.Define as a "WING"  path $ b \in B  \to g \in G \to l \in L$.It is known that number of wings at least $ 2009*2009$.Let's denote as $ x_i,y_i$ the number of edges from $ g_i \in G$ to vertices of $ B$ and $ L$ respectively.Assume in $ G$ there are $ k$ vertices.Notice that $ x_1\plus{}x_2\plus{}...\plus{}x_k$ no more than $ 2009*100$ also $ y_1\plus{}y_2\plus{}...y_k$ no more than $ 2009*100$(Notice that each student attend at most $ 100$ groups).
Notice that number of wings equal to $ x_1*y_1\plus{}x_2*y_2\plus{}...\plus{}x_k*y_k$ which is at least $ 2009*2009$
Now assume contradiction there is no $ g \in G$ such that both $ x,y \ge 11$.
As we know $ x_1\plus{}y_1\plus{}x_2\plus{}y_2\plus{}...\plus{}x_k\plus{}y_k \le 2009*200$ let's multiplicate $ 10$ to both sides
We get $ 10*x_1\plus{}10*y_1\plus{}...\plus{}10*x_k\plus{}10*y_k \le 2009*2000$
Notice that $ 10*x_i\plus{}10*y_i \ge x_i*y_i$ by our assumptation
so we have $ 2009*2000 \ge x_1*y_1\plus{}...\plus{}x_k*y_k \ge 2009*2009$
Contradiction.

So finally, $ 2001$ students were quite enough the statement to be true. :)

Well, that was apparently easy to conjecture that something has to be counted here, but the main problem was to guess the objective. I guess, the jury intended to direct the participants onto this hurdle, though the problem is lacking for smth, and it's apparently too tough to be given at the pre-last round of NO, especially for 9th graders.

Skillful solution, [b]birzhan[/b], congratulations.


probability
geometry
3D geometry
sphere
analytic geometry
graphing lines
slope

I'm posting this mostly for the Bellevue High School math club, although of course everyone is free to look over these if they are of help.  :)

[hide="1)  What is the sum of the reciprocals of the five smallest positive prime numbers?"] First of all, the five smallest positive prime numbers are $ 2, 3, 5, 7, 11$ ($ 1$ is not a prime!).  This means that the denominator of the resulting summation divides $ 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \equal{} 2310$, so immediately we can eliminate answers $ b, c$, and answer $ a$ is $ \frac {1}{2} \plus{} \frac {1}{3}$, which is also wrong.  We are left with $ d$ or $ NOTA$.  Calculation reveals that

$ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \plus{} \frac {1}{11} \equal{} \boxed{\frac {2927}{2310} }$ [/hide]
[hide="2)  What is 104 x 96?"] $ 104 \cdot 96 \equal{} (100 \plus{} 4)(100 \minus{} 4) \equal{} 100^2 \minus{} 4^2 \equal{} 10000 \minus{} 16 \equal{} \boxed{ 9984 }$ [/hide]
[hide="3)  If three distinct positive integers have a product of 770 and the greatest difference between any two numbers is 4, what is the sum of the three integers?"] Inspection: $ 7 \cdot 10 \cdot 11 \equal{} 770$ and $ 11 \minus{} 7 \equal{} 4$.  Their sum is $ \boxed{ 28 }$.

A more systematic approach:  if the smallest of the integers is $ k$, then the largest is $ k \plus{} 4$, so we are looking for two divisors of $ 770$ that are $ 4$ apart.  We can do this either by listing the divisors of $ 770$ (say, by prime factorizing it) or by testing numbers that are $ 4$ apart.  Either way, the above reveals itself quickly. [/hide]
[hide="4)  If x is a positive integer and $ x^{2/3} > 4$, which of the following must be true?"] The choices are $ x > 2, x < 2, x > 8, x < 8$.  For positive integers (and for that matter, positive reals), if $ a > b$ then $ a^2 > b^2$, and so forth.  Hence

$ x^2 > 64 \Leftrightarrow$
$ \boxed{ x > 8 }$ [/hide]
[hide="5)  Given that m and n are each integers between -20 and +20 and 2m + 3n = 15, what is the sum of all possible values of m?"]  Note that

$ n \equal{} \frac {15 \minus{} 2m}{3} \equal{} 5 \minus{} \frac {2m}{3}$

so $ n$ is an integer if and only if $ 3$ divides $ m$.  This occurs when $ m \equal{} \minus{} 18, \minus{} 15, \minus{} 12, ... \minus{} 3, 0, 3, ... 18$, and each of these produces valid values of $ n$.  Since the negative of each value of $ m$ also appears as a possible value of $ m$, their sum is $ \boxed{ 0 }$. [/hide]
[hide="6)  In terms of pi, what is the volume of a circular cylinder with a radius of 11cm and a height of 3cm?"] $ V_{cyl} \equal{} \pi r^2 h \equal{} \boxed{ 363 \pi }$ [/hide]
[hide="7)  What is 24^2 + 48^2?"]  If you memorized your squares, you should know that $ 24^2 \equal{} 576$.  But what about $ 48^2$?  Well,

$ 24^2 \plus{} 48^2 \equal{} 24^2 \plus{} (2 \cdot 24)^2 \equal{} 24^2 \plus{} 4 \cdot 24^2 \equal{} (1 \plus{} 4) \cdot 24^2 \equal{} 5 \cdot 576 \equal{} \boxed{ 2880 }$ [/hide]
[hide="8)  An unfair coin has a 3/4 probability of showing heads on any given toss.  This unfair coin is tossed along with a fair coin.  What is the probability that both coins show heads or both coins show tails?"] The probability that both coins show heads is $ \frac {3}{4} \cdot \frac {1}{2} \equal{} \frac {3}{8}$.  The probability that both coins show tails is $ \frac {1}{4} \cdot \frac {1}{2} \equal{} \frac {1}{8}$.  Their sum is $ \boxed{\frac {1}{2} }$. [/hide]
[hide="9)  Adam leaves his house, jogs on a flat, level road, up a hill, then back the same way.  When he comes back, he has found his job lasted exactly one hour.  Adam can jog at 8 mph on flat surfaces, 6 mph uphill, and 12 mph downhill.  How many miles did Adam jog?"] Say that the level road has length $ L$ and the hill has length $ H$.  Adding together the amount of time Adam spent on each portion of his jog,

$ \frac {L}{8} \plus{} \frac {H}{6} \plus{} \frac {H}{12} \plus{} \frac {L}{8} \equal{} 1 \Leftrightarrow$
$ \frac {L}{4} \plus{} \frac {H}{4} \equal{} 1 \Leftrightarrow$
$ L \plus{} H \equal{} 4$

Adam jogged the level surface and the hill twice, so the total length of his jog is $ 2H \plus{} 2L \equal{} \boxed{ 8}$. [/hide]
[hide="10)  What is the total surface area of a sphere with a radius of 12cm?"] $ SA_{sphere} \equal{} 4 \pi r^2 \equal{} \boxed{ 576 \pi }$ [/hide]
[hide="11)  How many ways can you arrange the letters in SACRAMENTO?"] "a" is the only repeated letter; otherwise, the letters can be permuted arbitrarily, so the number of arrangements is therefore $ \frac {10!}{2!} \equal{} \boxed{ 1814400 }$ (remember that $ 9! \equal{} 362880, 10! \equal{} 3628800$). [/hide]
[hide="12)  What is the equation of a line in slope-intercept form that passes through the origin and is perpendicular to the line 2x - 3y = 4?"]  Answers $ c, d$ do not pass through the origin (their $ y$-intercept is nonzero), so they can be immediately eliminated.  The slope of the original line is $ \frac {2}{3}$, so the slope of a line perpendicular to it is $ \minus{} \frac {3}{2}$; the answer is therefore $ \boxed{ y \equal{} \minus{} \frac {3}{2} x }$. [/hide]
[hide="13)  If a + b = 18 and b + c = 25 and a + c = 17 then find a + b x c."] Adding up all three equations produces $ 2a \plus{} 2b \plus{} 2c \equal{} 60 \Leftrightarrow a \plus{} b \plus{} c \equal{} 30$ (this is a well-known strategy).  Subtracting each equation in turn, we have $ c \equal{} 12, a \equal{} 5, b \equal{} 13$, hence $ a \plus{} b \times c \equal{} 5 \plus{} 13 \times 12 \equal{} \boxed{ 161 }$.

Motivation:  Given a triangle with side lengths $ x, y, z$, the points at which the incircle of the triangle intersects the side lengths divides the side lengths into lengths $ a, b, c$ such that $ x \equal{} a \plus{} b, y \equal{} b \plus{} c, z \equal{} c \plus{} a$. [/hide]
[hide="14)  A complex prime is any integer greater than 1 that has exactly three positive integral factors.  The factors are one, the number itself, and the square root of the number.  Which of the following is a complex prime number?"]  The choices are $ 225, 169, 144, 81$.  The problem hints at the form a "complex prime" should take; namely, it is the square of a prime.  The square of a prime $ p^2$ has only the divisors $ 1, p, p^2$, hence satisfies the problem condition.  $ \boxed{169} \equal{} 13^2$ is therefore a "complex prime." [/hide]
[hide="15)  On a math exam the lowest possible score is a 0 and the highest possible score is 100.  The average of the 4 test scores was an 86.  What is the lowest possible score for one of the four exams?"] If the average of the scores was $ 86$, the total of the scores is $ 4 \cdot 86 \equal{} 344$.  To minimize one of the scores, we should maximize the others.  Since the highest possible other scores are $ 100 \plus{} 100 \plus{} 100 \equal{} 300$, the lowest possible fourth score is $ \boxed{44}$.

Alternately:  each student got an average of $ 14$ points off, so in total they got $ 56$ points off.  To minimize one person's score, that person has to have gotten all of those points off; that score is therefore $ 100 \minus{} 56 \equal{} 44$. [/hide]
[hide="16)  What is the area of an isosceles right triangle that has a hypotenuse with length 8 sqrt2?"] An isosceles right triangle has side lengths in the ratio $ 1 : 1 : \sqrt {2}$.  The other two side lengths are $ 8, 8$; these are the legs of the right triangle, so the area is $ \frac {8 \cdot 8}{2} \equal{} \boxed{ 32 }$. [/hide]
[hide="17)  If x^2 + y = 0, which of the following must be true?"] The choices are $ x^2 \equal{} y, xy \equal{} y, x^3 \equal{} y^2, x^4 \equal{} y^2$.  One way to handle this is by a numerical example.  Say that $ x \equal{} 2, y \equal{} \minus{} 4$.  Then:

- $ x^2 \equal{} 4 \neq y \equal{} \minus{} 4$
- $ xy \equal{} \minus{} 8 \neq y \equal{} \minus{} 4$
- $ x^3 \equal{} 8 \neq y^2 \equal{} 16$
- $ x^4 \equal{} 16 \equal{} y^2 \equal{} 16$

To make sure the last alternative works:  $ x^2 \equal{} \minus{} y$, so squaring both sides we have $ \boxed{ x^4 \equal{} y^2 }$ as desired. [/hide]
[hide="18)  How many of the capital letters A, B, C, ... Z in the alphabet have horizontal line symmetry?"] On the assumption that the font is sans serif, the letters B, C, D, E, H, I, K, O, X satisfy this requirement.  (N and S have rotational symmetry, which is not the same!  Horizontal symmetry refers to a [b]reflection[/b] across a horizontal line.)  There are therefore $ \boxed{9}$ such letters. [/hide]
[hide="19)  What is the sum of the following arithmetic sequence?  5, 9, 13, ... 401 "] The sum of an arithmetic sequence with first term $ A$, last term $ L$, and $ n$ terms is given by

$ n \cdot \frac {A \plus{} L}{2}$

Think of this as replacing every term by the average of the terms.  This keeps the sum the same, and for an arithmetic sequence the average of every term is just the average of the first and last terms.  Now, the common difference is clearly $ 4$, and $ 401 \minus{} 5 \equal{} 396$, so there are $ \frac {396}{4} \plus{} 1 \equal{} 100$ (not $ 99$!) terms.

Don't forget the $ 1$!  My computational trick for never being off by $ 1$ is to use numerical examples.  For example, in the two-term arithmetic sequence $ 5, 9$, the common difference is $ 4$, so there are $ \frac {9 \minus{} 5}{4} \plus{} 1 \equal{} 2$ terms, which is true (not $ 1$ term!).

Having established that, the summation is $ 100 \cdot \frac {5 \plus{} 401}{2} \equal{} \boxed{ 20300 }$. [/hide]
[hide="20)  If x > y > 0, which of the following is less than y/x?"]  The choices are $ \frac {x}{y}, \frac {1}{\frac {x}{y} }, \frac {y}{2x}, \frac {2y}{x}$.

Now, if $ x > y > 0$ then dividing by $ x$ doesn't change the sign of the inequalities, so $ 1 > \frac {y}{x} > 0$.  On the other hand, $ \frac {x}{y} > 1$, so the first two options are clearly wrong.  The third answer, $ \boxed{\frac {y}{2x} }$, is clearly between $ \frac {y}{x}$ and $ 0$. [/hide]
[hide="21)  For any positive odd integer n, how many even positive integers are less than n?"] Another good time to bust out my computational trick for never being off by $ 1$.  There are $ 2$ even [i]positive[/i] integers ($ 0$ is neither positive nor negative!) less than $ 5$ and $ 3$ less than $ 7$, so $ \boxed{\frac {n \minus{} 1}{2} }$ seems reasonable.

A demonstration:  the even positive integers less than $ 2k \plus{} 1$ are $ 2 \cdot 1, 2 \cdot 2, ... 2 \cdot k$, and there are $ k$ of these. [/hide]
[hide="22)  The probability that Xavyr will get an A in math is .2, the probability that Dahir will get an A in math is .4, and the probability that Nyssa will get an A in math is .9.  Given that these are all independent events, what is the probability that exactly one of them will get an A in math?"] The probability that Xavyr gets an A but Dahir and Nyssa do not is $ .2 \cdot .6 \cdot .1 \equal{} .012$.
The probability that Dahir gets an A but Xavyr and Nyssa do not is $ .8 \cdot .4 \cdot .1 \equal{} .032$.
The probability that Nyssa gets an A but Xavyr and Dahir do not is $ .8 \cdot .6 \cdot .9 \equal{} .432$.

Summing the probabilities together, the probability that exactly one of the three students gets an A is $ \boxed{ .476}$. [/hide]
[hide="23)  What is the largest prime factor of 2007?"]  $ 2007 \equal{} 3^2 \cdot 223$ (this should be memorized); the largest prime factor is $ \boxed{ 223 }$.

If you haven't memorized this, the answer choices are $ 11, 13, 17, 19$.  None of these actually divide $ 2007$ (which doesn't take too long to check), so the answer is clearly $ NOTA$. [/hide]
[hide="24)  Who said, If I have seen further than others it was by standing upon the shoulders of Giants?"]  Math history is surprisingly interesting.  If you have time, check out a few books.  In any case, this quote is due to $ \boxed{\text{Isaac Newton}.}$.  [/hide]
[hide="25)  How many positive even three-digit numbers are there where the sum of the units digit and the tens digit is 10?"] If the number is even, its last digit is even.  Since $ 10$ is not a digit, the units digit can be $ 2, 4, 6, 8$, with corresponding tens digits $ 8, 6, 4, 2$.  The first digit can be any nonzero digit $ 1, ... 9$, for a total of $ \boxed{ 36 }$ possibilities. [/hide]
[hide="26)  If A = 50^10, B = 40^20, and C = 20^30, list A, B, C in order from biggest to smallest."] $ B \equal{} (40^2)^{10} \equal{} 1600^{10}, C \equal{} (20^3)^{10} \equal{} (8000)^{10}$.  Clearly $ \boxed{ C > B > A }$. [/hide]
[hide="27)  At a certain restaurant, there are 25 tables and each table has either exactly 2 or exactly 4 chairs.  If there is a total of 86 chairs accompanying the 25 tables, how many tables have exactly 4 chairs?"] The standard method is a system of equations.  Say there are $ a$ tables with 2 chairs and $ b$ tables with $ 4$ chairs.  Then

$ a \plus{} b \equal{} 25 \implies 2a \plus{} 2b \equal{} 50$
$ 2a \plus{} 4b \equal{} 86 \implies 2b \equal{} 36 \implies \boxed{ b \equal{} 18 }$.

An equivalent solution with words (which is actually, in my opinion, faster for mental math):  if all of the tables had $ 2$ chairs then there would only be $ 50$ chairs total.  The difference is $ 36$ chairs, which needs to be made up by adding $ 2$ chairs to each table at a time; this has to be done $ 18$ times, to produce $ 18$ tables with $ 4$ chairs each. [/hide]
[hide="28)  The lines y = m_1 x + 4 and y = m_2 x + 3 will intersect to the right of the y-axis if and only if:"] The choices are $ m_1 \equal{} m_2, m_1 > m_2, m_1 < m_2, m_1 \neq m_2$.  Graphically, $ m_1 x \plus{} 4$ has a $ y$-intercept that is above $ m_2 x \plus{} 3$, so the intersection will occur to the right of the $ y$-axis if the second line increases faster; in other words, if $ \boxed{ m_1 < m_2 }$.

Algebraically, the right of the $ y$-axis is where $ x$ is positive, so we want a solution to

$ m_1 x \plus{} 4 \equal{} m_2 x \plus{} 3 \Leftrightarrow$
$ x \equal{} \frac{1}{m_2 \minus{} m_1}$

which is positive if and only if $ m_2 > m_1$ as before. [/hide]
[hide="29)  If |x| represents the greatest integer less than or equal to x, what is the solution set to the equation 1 - 2|x| = -3?"] Clarification:  $ [x]$ is also known as the [b]floor function[/b], and is to my mind better written $ \lfloor x \rfloor$.  The greatest integer less than or equal to $ x$ is $ x$ if $ x$ is an integer, and the "integer part" of $ x$ otherwise.  For example, $ \lfloor 5 \rfloor \equal{} 5$ and $ \lfloor 4.9 \rfloor \equal{} 4$.

In any case, the above equation can be written $ \lfloor x \rfloor \equal{} 2$, which occurs when $ \boxed{ 2 \le x < 3 }$.  The graph of the floor function looks like this (look at the endpoints carefully!):

[img]http://www.mathwords.com/f/f_assets/floor_graph.gif[/img] [/hide]
[hide="30)  Two identical spheres of radius 6 intersect so that the distance between their centers is 10.  The points of intersection of the two spheres form a circle.  What is the area of this circle?"] [img]http://www.etsu.edu/physics/etsuobs/starprty/120598bg/spheres.gif[/img]

The above diagram is a useful starting point.  Consider the center $ C$ of the circle.  Since the distance between the centers of the spheres is $ 10$, $ C$ is a distance of $ 5$ away from both centers.  Consider, however, a point $ P$ on the circle.  since $ P$ is on both spheres, it is a distance of $ 6$ away from both centers.

We conclude that $ CP \equal{} \sqrt{6^2 \minus{} 5^2} \equal{} \sqrt{11}$ by the Pythagorean theorem.  But this is just the radius of the circle; its area is therefore $ \boxed{ 11 \pi }$. [/hide]

[hide="31)  What is the domain of the function f(x) = sqrt(x^2 - x - 12)?"] For our purposes, square roots of negative numbers are undefined; we therefore want to solve the inequality $ x^2 - x - 12 \ge 0$.  Well, if

$ (x - 4)(x + 3) \ge 0$

then either both terms are non-positive (which occurs when $ \boxed{x \le - 3}$) or both terms are non-negative (which occurs when $ \boxed{x \ge 4}$).  In other words, the graph of $ y = (x - 4)(x + 3)$ dips below the $ y$-axis when $ - 3 < x < 4$.  [/hide]
[hide="32)  Simplify the following:  2/(3+1/(2+3/(4+2/3)))"] We'll take this step by step.  $ 4 + \frac {2}{3} = \frac {14}{3}$.  Then $ \frac {3}{\frac {14}{3} } = \frac {9}{14}$.  Then $ 2 + \frac {9}{14} = \frac {37}{14}$.  Then $ \frac {1}{\frac {37}{14} } = \frac {14}{37}$.  Then $ 3 + \frac {14}{37} = \frac {125}{37}$.  Then $ \frac {2}{\frac {125}{37} } = \boxed{\frac {74}{125} }$. [/hide]
[hide="33)  Given that the digits 1-9 appear exactly once in the cryptarhythm a4b + 2c5 = d1e, what is the value of a x b - c + d x e?"] Cryptarhythms may look daunting to the beginning problem solver who expects all problems to have formulaic solutions.  We don't have to rely entirely on guess-and-check, though; we have certain clues.  For example, if $ b < 5$ then the second column tells us that $ c = 7$ (because $ 4 + 7 = 11$), and if $ b > 5$ then $ c = 6$.  This is only two cases to check.  On the assumption that $ b < 5$, we have $ c = 7, a + 3 = d, b + 5 = e$.  Since the digits $ 1, 2, 4, 5$ are already used, assigning $ b = 3$ gives $ e = 8$, and then the only digits left are $ a = 6, d = 9$.  Hence

$ a \times b - c + d \times e = 6 \times 3 - 7 + 9 \times 8 = \boxed{ 83 }$.[/hide]
[hide="34)  How many five-digit integers exist using the digits 1, 2, 3, 4, and 5 if each digit may be used exactly once in the number?"] Tom Norris intended the word "integer" to mean either positive or negative integers, although in this case I don't think that use of the word is valid; the problem gives no hint that the addition of a negative sign is permissible.

The intended solution:  clearly there are $ 5! = 120$ possible positive integers that can be made (there are $ 5$ choices for which digit to use in the first spot, $ 4$ for the second, and so forth).  However, since integers can be positive or negative, there are $ \boxed{240}$ total integers that can be made.  [/hide]
[hide="35)  A square is inscribed inside a circle with an area of 100pi cm^2.  What is the number of square centimeters in the area of the square?"] The area of the circle is $ 100 \pi = \pi r^2$, so $ r = 10$ and $ d = 20$.  The diameter of the circle is the diagonal of the square, so the square has side length $ 10 \sqrt {2}$ and area $ \boxed{ 200 }$. [/hide]
[hide="36)  How many distinct integers are solutions for x in the following equation?  7 < sqrt(x) ≤ 10"] We are free to square this inequality; we get $ 49 < x \le 100$, hence $ x \in \{ 50, 51, ... 100 \}$.  There are $ 100 - 50 + 1 = \boxed{51}$ possibilities.  Remember my trick for never being off by $ 1$!  For example, between $ 50$ and $ 55$ there are $ 6$ numbers, not $ 5$:  we have $ 50, 51, 52, 53, 54, 55$.  [/hide]
[hide="37)  Given that 3^n divides 15!, what is the greatest possible integral value of n?"] The following is standard.  Because there are $ \lfloor \frac {15}{3} \rfloor = 5$ multiples of $ 3$ between $ 1$ and $ 15$ inclusive, we know that we have at least $ 5$ powers of $ 3$.  However, there is also one multiple of $ 9$, which contributes two powers of $ 3$, so we must add it in again; in total, $ n = \boxed{6}$.

This technique is usually used to find the number of zeroes at the end of $ k!$ (which is the greatest possible value of $ n$ such that $ 5^n$ divides $ k!$).  First count the multiples of $ 5$, then the multiples of $ 25$, and so forth. [/hide]
[hide="38)  A space diagonal of a cube is sqrt(147) units long.  How many square units are in the total surface area of the cube?"] It is easy to verify from the Pythagorean theorem that the length of the space diagonal of a cube is $ s \sqrt {3}$, where $ s$ is the side length of the cube.  Since $ \sqrt {147} = \sqrt {3 \cdot 49} = 7 \sqrt {3}$, we conclude that $ s = 7$.

Then the surface area is given by $ SA_{cube} = 6s^2 = \boxed{ 294 }$ (remember, there are six square faces to a cube). [/hide]
[hide="39)  Express 1.412121212... as a common fraction."] [b]Lemma:[/b]  $ \frac {1}{r} + \frac {1}{r^2} + \frac {1}{r^3} + ... = \frac {1}{r - 1}$

Proof:  This is a slight restatement of the usual geometric series summation formula, but I find it useful for mental math.

[b]Corollary:[/b]  $ 0.ababab... = \frac {ab}{99}$

Proof:  $ 0.ababab... = \frac {ab}{100} + \frac {ab}{100^2} + \frac {ab}{100^3} + ... = \frac {ab}{100 - 1} = \frac {ab}{99}$ by the lemma.

Hence $ 1.4121212... = 1.4 + \frac {12}{990} = \frac {7}{5} + \frac {2}{165} = \frac {231 + 2}{165} = \boxed{\frac {233}{165} }$. [/hide]
[hide="40)  What is the mean of the following data:  2, 3, 4, 5, 6, 100?"] $ \frac {2 + 3 + 4 + 5 + 6 + 100}{6} = \frac {120}{6} = \boxed{20}$. [/hide]
[hide="41)  If x, 3x+3, and 5x+5 are three non-zero consecutive terms of a geometric sequence, the sum of these three terms is?"] The common ratio of the geometric sequence is $ \frac {3x + 3}{x} = \frac {5x + 5}{3x + 3} = \frac {5}{3}$, hence $ 5x = 9x + 9 \Leftrightarrow x = - \frac {9}{4}$.  The sum of the terms is $ 9x + 8 = \boxed{ - 12.25}$. [/hide]
[hide="42)  How many subsets does a set with n distinct elements have?"] The choices are $ n^2, {2n \choose n}, n!, n + 1$.  Before we discuss the actual answer to this question, let's talk about the process of elimination.

Consider a set with two elements $ A, B$ (that is, the case $ n = 2$).  It has the following subsets:

$ \emptyset, \{ A \}, \{ B \}, \{ A, B \}$

(remember that the empty set is considered a subset of every set!).  Well, $ 2^2 = 4$ so we can't eliminate the first choice, but $ {4 \choose 2} = 6, 2! = 2, 2 + 1 = 3$, so the other choices are obviously wrong.  Now consider a set with three elements $ A, B, C$ (the case $ n = 3$).  It has the following subsets:

$ \emptyset, \{ A \}, \{ B \}, \{ C \}, \{ A, B \}, \{ B, C \}, \{ A, C \}, \{ A, B, C \}$

However, $ 3^2 = 9$.  Our answer is clearly $ \boxed{NOTA}$, whatever that is, just by examining two simple cases!

Of course, in the computations for these cases you might notice something interesting.  It appears that the number of subsets of a set with $ n$ distinct elements is $ \boxed{ 2^n }$.  We will prove this as follows:

Imagine you are constructing a generic subset.  One way to do this is to go through every element of the original set in turn and decide whether to put it in or not.  Thus each element is either in a given subset or not in a given subset; this is $ 2$ possibilities for each element, so the total number of combinations of being in and not being in is $ 2^n$.

This can also be proven by induction.  If a set with $ k$ distinct elements has $ 2^k$ distinct subsets, then when we add a new element we introduce exactly as many subsets as we had; half which contain the new element and half which don't.

Furthermore, it is possible to see directly the form that the subsets will take:  for distinct elements $ A, B, C, ...$ the subsets are represented by the terms in the expansion of

$ (1 + A)(1 + B)(1 + C)...$

where $ 1$ represents the empty set.  For $ n = 2$, we get $ (1 + A)(1 + B) = 1 + A + B + AB$, precisely the subsets we listed above; similarly, for $ n = 3$ we get $ (1 + A)(1 + B)(1 + C) = 1 + A + B + C + AB + BC + CA + ABC$.  Clearly this expansion has $ 2^n$ terms in general.  :)

The lesson to take from counting problems is to organize what you've counted carefully.  In this case, our choice to organize from subsets with the least elements to subsets with the least elements led to something familiar. [/hide]
[hide="43)  If the expansion of (2b^2 - 3b^{-3})^n is expanded in decreasing powers of b, and the fifth term does not contain a factor of b, what is the value of n?"]  Recall that the binomial theorem states that

$ (x + y)^n = {n \choose 0} x^n + {n \choose 1} x^{n - 1} y + {n \choose 2} x^{n - 2} y^2 + ...$

Since $ x = 2b^2$ and $ y = - 3b^{ - 3}$, it is clear that the terms are already in order of decreasing powers of $ b$ as we have written them.  The fifth term in this expansion is $ {n \choose 4} (2b^2)^{n - 4} ( - 3b^{ - 3})^4$ (notice the $ 4$s instead of $ 5$s; remember, we start at $ 0$).  In order for this term to not have a factor of $ b$, we require that the exponent of $ b$ is $ 0$; in other words,

$ 2(n - 4) - 12 = 0$

so $ n = \boxed{ 10 }$. [/hide]
[hide="44)  The graph of the curve represented by x = sec theta, y = cos theta is:"] The choices are a line, a hyperbola, an ellipse, and a circle.  Now, since $ \sec \theta = \frac {1}{\cos \theta}$ we can write this curve as $ y = \frac {1}{x}$, which is in fact a hyperbola (just rotated $ 45^{\circ}$).  [b]However:[/b]  we are constrained by the fact that $ - 1 \le y \le 1$.  The graph is in fact a cross section of a hyperbola, which is not a full hyperbola (just as a semicircle is not a circle).  The answer is $ \boxed{ NOTA }$.

[hide="Discussion of rotation"] A counterclockwise $ 45^{\circ}$ rotation takes a point $ (x, y)$ to the point $ ( \frac {x - y}{\sqrt {2} }, \frac {x + y}{\sqrt {2} }$ (which can be proven either by an explicit geometric construction or through polar coordinates).  Substituting into the equation $ xy = 1$, we have the "rotated" equation

$ x^2 - y^2 = 2$

which is the more familiar equation of a hyperbola. [/hide] [/hide]
[hide="45)  If the graph of a frequency distribution is skewed to the left, which of the following must be true?"] The choices are $ \text{mean} < \text{median} < \text{mode}$, $ \text{mode} < \text{median} < \text{mean}$, $ \text{median} < \text{mode} < \text{mean}$, $ \text{mode} < \text{mean} < \text{median}$.

A distribution is "skewed to the left" if its left tail is longer than its right tail; in other words, it has a hump on the right.  The mode of the distribution is its peak.  The median of the distribution is the point at which a vertical line divides the distribution in half, which is clearly to the left of the mode, hence we can elminate choices $ b, d$.

In general, a distribution that is skewed to the left has more values that are very small compared to values that are large; this tends to skew the mean farther to the left than the median, so the correct answer is $ \boxed{ \text{mean} < \text{median} < \text{mode} }$.

(Personally, I dislike this question since it is actually possible to come up with a distribution that appears skewed to the left but such that the mean is greater than the median.  Go figure.)  [/hide]
[hide="46)  How many non-congruent right triangles are there such that the perimeter in cm and the area in cm^2 are numerically equal?"] Pick any right triangle.  Given that the ratio of its perimeter to its area is $ r$, scale the triangle by $ r$.  This scales the perimeter by $ r$ and the area by $ r^2$, so the perimeter and area are numerically equal.  We can do this for a right triangle with arbitrary angles, so the answer is $ \boxed{\text{infinitely many.} }$.

Alternately:  recall that $ A = rs$, where $ r$ is the inradius and $ s$ is the semiperimeter (half the perimeter).  If $ A = 2s$ then $ r = 2$, and there are infinitely many right triangles with inradius $ 2$.  (Construction:  consider a circle $ C$ of radius $ 2$ and a point $ M$ such that the two tangent lines of $ C$ passing through $ M$ are perpendicular.  Pick a point $ P$ on either tangent line and construct the tangent line of $ C$ passing through $ P$; it passes through the other tangent line at $ Q$, and then $ \triangle MPQ$ is a triangle with inradius $ 2$.) [/hide]
[hide="47)  How many ordered triples (a, b, c) of positive integers satisfy the simultaneous equations ab + bc = 44, ac + bc = 23?"]  This is one of those problems that will clearly require casework.  There are several ways to go about this problem.  The fastest is to factor the second equation; it becomes $ c(a + b) = 23$.  Now, $ c | 23$, a prime, so either $ c = 1$ or $ c = 23$.

[b]Case:[/b]  $ c = 1$.  We have $ ab + b = 44, a + b = 23$.  Substituting $ a = 23 - b$, we have

$ 23b - b^2 + b = 44 \Leftrightarrow$
$ b^2 - 24b + 44 = 0 \Leftrightarrow$
$ (b - 22)(b - 2) = 0$

When $ b = 2$ we have $ (a, b, c) = (21, 2, 1)$ and when $ b = 22$ we have $ (a, b, c) = (1, 22, 1)$.

[b]Case:[/b]  $ c = 23$.  We have $ ab + b = 44, a + b = 1$.  Substituting $ a = 1 - b$, we have

$ b - b^2 + b = 44 \Leftrightarrow$
$ b^2 - 2b + 44 = 0$

which has no integer roots (the determinant is negative; alternately, $ (b - 1)^2 + 43$ has no [b]real[/b] roots either).

There are therefore $ \boxed{2}$ solutions. [/hide]
[hide="48)  Six distinct integers are picked at random from the integers 1, 2, 3, ... 10.  What is the probability that, among those selected, the second smallest is 3?"] If the second smallest of the six integers is $ 3$, then the smallest is either $ 1, 2$, and the larger three are between $ 4$ and $ 10$ inclusive.  There are $ 2$ ways to pick the smallest of the six integers and $ {7 \choose 3} = \frac {7 \cdot 6 \cdot 5}{3!} = 35$ ways to pick the larger three (remember, distinct integers, order doesn't matter, and there are $ 10 - 4 + 1 = 7$ integers in that range).  The total number of possibilities is $ 2 \cdot 35 = 70$.

The total number of way to pick six distinct integers from ten is $ {10 \choose 6} = {10 \choose 4} = \frac {10 \cdot 9 \cdot 8 \cdot 7}{4!} = \frac {10 \cdot 9 \cdot 7}{3} = 10 \cdot 3 \cdot 7 = 210$.  I have done this calculation explicitly to demonstrate to you the easy way of evaluating binomial coefficients:  first we must remember that picking six elements out of ten is the same thing as picking four elements out of ten (since we are just picking the elements we leave out; alternately, both combinations are equal to $ \frac {10!}{4!6!}$), and second we cancel out factors as an alternative to division.

Our probability is therefore $ \frac {70}{210} = \boxed{\frac {1}{3} }$. [/hide]
[hide="49)  For how many integers x does a triangle with side lengths 10, 24, and x have all its angles acute?"] Say we have $ \triangle ABC$ so that $ AC = 24, CB = 10, AB = x$.  We should recall the Pythagorean triple $ 10, 24, 26$; this tells us that when $ x = 26$ the triangle is right (hence not acute).  When $ x$ is slightly less than $ 26$ angle $ C$ will decrease, so $ \triangle ABC$ will be acute, so we must have $ x < 26$.

However, if we keep increasing $ x$ then $ \angle B$ increases.  Specifically, when $ 21 < x = \sqrt {24^2 - 10^2} = \sqrt {476} < 22$ angle $ B$ becomes right (by the Pythagorean theorem; memorize your squares!).  Hence we must have $ x \ge 22$.

Combining our two restrictions, $ x = 22, 23, 24, 25$; this is $ \boxed{4}$ possibilities. [/hide]
[hide="50)  What is the smallest positive prime factor of 2007?"]  This is your reward for being a smart test-taker and making sure you [b]read every question![/b]  :)

$ 2007 = 3^2 \cdot 223$, so the smallest prime factor is $ \boxed{3}$.  One of the answer choices is $ 1$, but remember that [b]$ 1$ is neither prime nor composite.[/b]  In ring theory terms, it is a [i]unit[/i], which means its multiplicative inverse is also an integer.  The only units in the integers are $ 1, - 1$. [/hide]


Here is the question: Consider a triangle and take an inscribed circle. Join each vertex with the point of tangency at the opposite side. Prove that three lines you constructed pass through one point?

In what proportion the point of intersection divides three segments?

to daemon4eva

Be ABC the triangle, D, E and F the tangency points of incircle with the sides BC, CA and AB respectively. As usually, note a, b and c the sides BC, CA, AB and 2p=a+b+c.
Since AE=AB=p-a, BF=BD=p-b and CD=CE=p-c, by reciprocal of the theorem of Ceva [ (AF/BF).(BD/CD).(CE/AE)=1 ] we get AD, BE and CF concurrent.

Let's note P their common point. Using the [b]Van Aubel[/b] formula [ AP/PD = AF/BF + AE/CE ].
After calculations, we shall get AP/PD = a(p-a)/[(p-b)(p-c)] and simmilarily for the other segments.

I would like to suggest that the homeworks to be placed on a special page.

Good luck!
sunken rock

A blind man sees better the details


ive heard u guys talk about all these tests. My school has a math team but we don't do that. How?

first, what grade are u in, what math class are u in, and what competition(s) does your team do?

-jorian hoover


[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {1}{a \plus{} \sqrt []{2(a^2 \plus{} bc)}} \plus{} \frac {1}{b \plus{} \sqrt []{2(b^2 \plus{} ca)}} \plus{} \frac {1}{c \plus{} \sqrt []{2(c^2 \plus{} ab)}} \geq\ \frac {3}{a \plus{} b \plus{} c}$[/quote]

Cauchy-Schwarz kills it.  :)  The following inequalities are also true:

1/ $ \frac {1}{a \plus{} \sqrt {3a^2 \plus{} bc}} \plus{} \frac {1}{b \plus{} \sqrt {3b^2 \plus{} ca}} \plus{} \frac {1}{c \plus{} \sqrt {3c^2 \plus{} ab}} \geq\ \frac {3}{a \plus{} b \plus{} c}$

2/ $ \frac {1}{a \plus{} \sqrt {a^2 \plus{} 3bc}} \plus{} \frac {1}{b \plus{} \sqrt {b^2 \plus{} 3ca}} \plus{} \frac {1}{c \plus{} \sqrt {c^2 \plus{} 3ab}} \geq\ \frac {3}{a \plus{} b \plus{} c}$

I think the general inequality holds for all $ k\ge0$

 $ \frac {1}{a \plus{} \sqrt {ka^2 \plus{} bc}} \plus{} \frac {1}{b \plus{} \sqrt {kb^2 \plus{} ca}} \plus{} \frac {1}{c \plus{} \sqrt {kc^2 \plus{} ab}} \ge\ \frac {9}{(1\plus{}\sqrt{k\plus{}1})(a \plus{} b \plus{} c)}$

Besides, we have

 $ \frac {1}{a \plus{} \sqrt {2(a^2 \plus{} bc)}} \plus{} \frac {1}{b \plus{} \sqrt {2(b^2 \plus{} ca)}} \plus{} \frac {1}{c \plus{} \sqrt {2(c^2 \plus{} ab)}} \le \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {1}{a \plus{} \sqrt []{2(a^2 \plus{} bc)}} \plus{} \frac {1}{b \plus{} \sqrt []{2(b^2 \plus{} ca)}} \plus{} \frac {1}{c \plus{} \sqrt []{2(c^2 \plus{} ab)}} \geq\ \frac {3}{a \plus{} b \plus{} c}$[/quote]
[quote="dduclam"]
Cauchy-Schwarz kills it.  :)  The following inequalities are also true:
1/ $ \frac {1}{a \plus{} \sqrt {3a^2 \plus{} bc}} \plus{} \frac {1}{b \plus{} \sqrt {3b^2 \plus{} ca}} \plus{} \frac {1}{c \plus{} \sqrt {3c^2 \plus{} ab}} \geq\ \frac {3}{a \plus{} b \plus{} c}$[/quote]
Can U post your proof ? I haven't solved it yet. Thank you   :oops:
And I think the stronger is trues:
$ \frac{1}{\sqrt[]{5a^2\plus{}bc}} \plus{} \frac{1}{\sqrt[]{5b^2\plus{}ca}} \plus{} \frac{1}{\sqrt[]{5c^2\plus{}ab}} \geq\ \frac{9}{\sqrt[]{6}(a\plus{}b\plus{}c)}$


Equate what this problem is asking for to another problem (simpler) that asks for the same thing - this is in very very layman terms...

In a bijection, construct a one to one ratio between two sets...

That problem doesn't make sense as is.  If you are asking to find a set $A$ for which $f$ is a bijection between $[-1,\infty)$ and $A$, then we can take $A=[-1,\infty)$ again, since $f$ is everywhere increasing and fixes $x=-1$.(I used calculus there).

[quote="Fermat -Euler"]Find the bijection of $f(x)=x-1+\sqrt[3]{x^3+1}$ for all $x\geq -1$[/quote]
find $f^{-1}$

I see a solution, is the answer an elegant one? Ive got a brute force solution...

[quote="maokid7"]i still do not understand what the original question is asking.  :([/quote]
Let $f(x)=x-1+\sqrt[3]{x^3+1}$ for all $x\geq -1$
find $f^{-1}$
I think it's clair now ;)

If $x\neq 0$

Solving for x by the quadratic formulas boils down really well to:

$x=\frac{3(f(x)+1) +- (f(x)+1)}{4}$

Dont kill me, but if we get the turning points of the original function, we find one at -1. Since this function is defined for $x\geq-1$, we are ok, the graph is injective. Pluging in a value for x, we get that we should choose the + sign, So $f^{-1}(x)=x+1$, or if x=0, $f^{-1}(x)=0$

I may have screwed up here, so please check! I a bit shaky on the injective part, if someone would like to verify...


Let the gcd of $a,b$ be $1$.
Find all possible values of gcd$(a^2+b^2,a+b)$

Is it 1,2. :rotfl:
Actually the same problem was asked a few days ago I remember in which we had to prove that it is 1 or 2.
I actually don't remeber it I'll try.

I tried some numbers and I got 1 and 2 also, but I can't tell if there's more.  :ninja:

Hint: Is it true that gcd$(a,b) = 1 \leftrightarrow$ gcd$(a^2,b^2)=1$?


trigonometry calculus integration AMC AIME USA(J)MO USAMO

Because I totally suck at trig, I need to focus more on it, but I'm not sure exactly which of the many formulas I absolutely need to know. Do you have any suggestions?

I have memorized the double angle formulas for sin and cos, as well as sin(x+y), sin(x-y), cos(x+y), cos(x-y) formulas, but don't know what else I need to memorize.

Should I memorize tan, cot, sec, csc for $ \pi / 4$ , $ \pi / 6$, $ \pi / 3$ , so on, rather than derive them each time?
Are the half-angle formulas really worth memorizing? What about sum to product formulas? Tan addition and subtraction formulas? etc, etc, etc.

Which ones do you find most useful?

P.S.
I don't understand how to get this:

$ \sqrt{\frac{1\minus{}\cos78}{1\plus{}\cos78}}\equal{}\tan\frac{78}{2}\equal{}\tan36$

The expression you gave is a consequence of the sine and cosine half-angle formulas.  Specifically, divide the sine half-angle formula by the cosine half-angle formula.

As to your main question, it depends what your goals are.  If you just want to do well in class (presumably pre-calculus), it would be good to memorize the standard values of $ \sin, \cos$ as well as perhaps a few values of $ \tan$ (it's not too hard to remember that $ \tan \frac {\pi}{6} \equal{} \frac {1}{\sqrt {3} }, \tan \frac {\pi}{3} \equal{} \sqrt {3}$).  You're not likely to need much else except for possibly the half-angle formulas (which I like to derive from the double-angle formulas anyway).

If you want to prepare yourself for the future (that is, a calculus course), you'll definitely need the half-angle formulas (they are used to calculate certain integrals), and the product-to-sum and sum-to-product formulas might also come in handy.

If you want to prepare for an Olympiad, the sum-to-product and product-to-sum formulas are indispensable.  I wouldn't bother actually memorizing them since they're easy to derive from the angle summation identities, but keep them in mind.

Edit:  In terms of what you [b]absolutely[/b] need to know, the standard values, the Pythagorean identity, and the angle addition formulas are it (along with the fact that $ \sin (\minus{}x) \equal{} \minus{} \sin x, \cos (\minus{}x) \equal{} \cos x$).  Everything else you can derive from those.

thanks for your reply! I hope to learn enough to do the majority of the questions on the AMCs, AIME, and possibly USAMO somewhere in the future.

In that case, the most important identity you want to get a good grasp of is actually

$ e^{ix} \equal{} \cos x \plus{} i \sin x$

Why?

- $ e^{i(x \plus{} y)} \equal{} e^{ix} e^{iy}$.  This allows you to deduce both angle summation formulas simultaneously.
- $ e^{inx} \equal{} (e^{ix})^n \Leftrightarrow \cos nx \plus{} i \sin nx \equal{} (\cos x \plus{} i \sin x)^n$.  This is known as de Moivre's formula, and it allows you to deduce the $ n$-tuple angle formulas using the Binomial Theorem.
- $ 1 \plus{} (e^{ix}) \plus{} (e^{ix})^2 \plus{} ... \plus{} (e^{ix})^n \equal{} \frac {e^{(ix)(n \plus{} 1)} \minus{} 1}{e^{ix} \minus{} 1}$.  This allows you to calculate $ \cos 0 \plus{} \cos x \plus{} \cos 2x \plus{} ... \plus{} \cos nx$ and $ \sin 0 \plus{} \sin x \plus{} \sin 2x \plus{} ... \sin nx$.
- And more!

thanks, that should come in handy, but some questions:
[quote="t0rajir0u"]
- $ e^{inx} \equal{} (e^{ix})^n \Leftrightarrow \cos nx \plus{} i \sin x \equal{} (\cos x \plus{} i \sin x)^n$.  This is known as de Moivre's formula, and it allows you to deduce the $ n$-tuple angle formulas using the Binomial Theorem.
[/quote]
What do you mean by the n-tuple angle formulas?, and $ e^{inx} \equal{} (e^{ix})^n \Leftrightarrow \cos nx \plus{} i \sin x$ was meant to be $ e^{inx} \equal{} (e^{ix})^n \Leftrightarrow \cos nx \plus{} i \sin nx$, right?

[quote="t0rajir0u"]
$ 1 \plus{} (e^{ix}) \plus{} (e^{ix})^2 \plus{} ... \plus{} (e^{ix})^n \equal{} \frac {e^{(ix)(n \plus{} 1)} \minus{} 1}{e^{ix} \minus{} 1}.$ This allows you to calculate $ \cos 0 \plus{} \cos x \plus{} \cos 2x \plus{} ... \plus{} \cos nx$ and $ \sin 0 \plus{} \sin x \plus{} \sin 2x \plus{} ... \sin nx.$
[/quote]
umm... could you explain this further?

So for the AMCs and AIME, I should probably memorize the half-angle formulas (so I would be able to recognize it when a problem resembles their form, such as in the example problem I gave), the basic values of sin, cos, tan, sec, csc, cot, learn how to derive the sum-product formulas quickly and get used to using them.

Are there any modifications/ additions that I should probably add?
What about the power-reduction formulas?

Thanks so much for your help!

[quote="AnonAmoz"]What do you mean by the n-tuple angle formulas?, and $ e^{inx} \equal{} (e^{ix})^n \Leftrightarrow \cos nx \plus{} i \sin x$ was meant to be $ e^{inx} \equal{} (e^{ix})^n \Leftrightarrow \cos nx \plus{} i \sin nx$, right? [/quote]

Yes.  The $ n$-tuple angle formulas are a generalization of the double angle formulas:  you can quickly calculate the triple angle formulas, the quadruple angle formulas, and so forth, without having to repeatedly expand using the angle-summation identities.

[quote="AnonAmoz"][quote="t0rajir0u"]
$ 1 \plus{} (e^{ix}) \plus{} (e^{ix})^2 \plus{} ... \plus{} (e^{ix})^n \equal{} \frac {e^{(ix)(n \plus{} 1)} \minus{} 1}{e^{ix} \minus{} 1}.$ This allows you to calculate $ \cos 0 \plus{} \cos x \plus{} \cos 2x \plus{} ... \plus{} \cos nx$ and $ \sin 0 \plus{} \sin x \plus{} \sin 2x \plus{} ... \sin nx.$
[/quote]
umm... could you explain this further? [/quote]

This is a little more obscure.  Certain AIME/Olympiad type problems will ask you to evaluate a sum like

$ \sin x \plus{} \sin 2x \plus{} ... \plus{} \sin 44x$.

These sums might be quite difficult to evaluate until you realize that the above summation is just the imaginary part of

$ 1 \plus{} e^{ix} \plus{} e^{2ix} \plus{} e^{3ix} \plus{} ... \plus{} e^{44ix}$

which is just a geometric series with sum

$ \frac {e^{45ix} \minus{} 1}{e^{ix} \minus{} 1}$

The imaginary part of this value is not too hard (but it is a little tedious) to calculate, and then we have found our summation.  This is not a technique you'll use often, but it will crop up every once in awhile.  The more general idea is that

$ \sin x \equal{} \frac {e^{ix} \minus{} e^{ \minus{} ix}}{2i}$
$ \cos x \equal{} \frac {e^{ix} \plus{} e^{ \minus{} ix}}{2}$

In other words, in some fundamental way, sine waves are essentially exponential in nature.  Later on, when you study calculus, this is a trick that trivializes integrals involving sines and cosines.  (You might find this form more useful than Euler's formula; the product-to-sum rules are a direct consequence of using this form.)

[quote="AnonAmoz"]So for the AMCs and AIME, I should probably memorize the half-angle formulas (so I would be able to recognize it when a problem resembles their form[/quote]

This is of more use than you think.  Some AIME/Olympiad problems disguise the half-angle formula or similar in a recursion, and being able to recognize that fact is extremely useful.

[quote="AnonAmoz"]What about the power-reduction formulas?[/quote]

I'm not familiar with the term, but I looked these up, and these are just the square of the half-angle formulas.  It's actually this form that will be most useful in calculus.  My point, though, is that if you know the half-angle formulas you know these.

[quote="t0rajir0u"][/quote][quote="AnonAmoz"]
[quote="AnonAmoz"][quote="t0rajir0u"]
$ 1 + (e^{ix}) + (e^{ix})^2 + ... + (e^{ix})^n = \frac {e^{(ix)(n + 1)} - 1}{e^{ix} - 1}.$ This allows you to calculate $ \cos 0 + \cos x + \cos 2x + ... + \cos nx$ and $ \sin 0 + \sin x + \sin 2x + ... \sin nx.$
[/quote]
umm... could you explain this further? [/quote]

This is a little more obscure.  Certain AIME/Olympiad type problems will ask you to evaluate a sum like

$ \sin x + \sin 2x + ... + \sin 44x$.

These sums might be quite difficult to evaluate until you realize that the above summation is just the imaginary part of

$ 1 + e^{ix} + e^{2ix} + e^{3ix} + ... + e^{44ix}$

which is just a geometric series with sum

$ \frac {e^{45ix} - 1}{e^{ix} - 1}$

The imaginary part of this value is not too hard (but it is a little tedious) to calculate, and then we have found our summation.  This is not a technique you'll use often, but it will crop up every once in awhile.  The more general idea is that

$ \sin x = \frac {e^{ix} - e^{ - ix}}{2i}$
$ \cos x = \frac {e^{ix} + e^{ - ix}}{2}$
[/quote][quote="t0rajir0u"][/quote]
I didn't understand how to calculate the imaginary part of that sum. (I'm not so great at other areas of math, either.  :blush: )

First, an exercise:  find the imaginary part of $ \frac{2 \plus{} 3i}{3 \plus{} 4i}$.

To simplify, our goal is to "rationalize" the denominator (get rid of the imaginary part), which we do by multiplying by its conjugate, $ 3 \minus{} 4i$.  In other words,

$ \frac{(2 \plus{} 3i)(3 \minus{} 4i)}{(3 \plus{} 4i)(3 \minus{} 4i)} \equal{} \frac{18 \plus{} i}{25} \equal{} \frac{18}{25} \plus{} \frac{1}{25} i$

Hence the imaginary part is $ \frac{1}{25}$.

Similarly, write

$ \frac{e^{45ix} \minus{} 1}{e^{ix} \minus{} 1} \equal{} \frac{ (\cos 45x \minus{} 1) \plus{} i (\sin 45x)}{ (\cos x \minus{} 1) \plus{} i (\sin x) }$

and multiply by the conjugate of the denominator.  (Like I said, tedious.)  :)

Thanks so much for helping me!

I understand how to rationalize it when in a fraction, normally, I just don't understand how to do it when $ i$ is in the exponents.   :huh:

In other words I don't know how to tell the imaginary part of
$ \frac {e^{45ix} \minus{} 1}{e^{ix} \minus{} 1}$
from the real part of it, since $ i$ is in the exponents?  :(

[quote="t0rajir0u"][quote="t0rajir0u"]$ \frac {e^{45ix} \minus{} 1}{e^{ix} \minus{} 1} \equal{} \frac { (\cos 45x \minus{} 1) \plus{} i (\sin 45x)}{ (\cos x \minus{} 1) \plus{} i (\sin x) }$[/quote]

This is a consequence of Euler's formula $ e^{ix} \equal{} \cos x \plus{} i \sin x$, which I gave above.[/quote]

ok, so basically to find $ \sin{x} \plus{} \sin{2x}.... \sin{44x}$, you have to find the imaginary part of $ \frac {e^{45ix} \minus{} 1}{e^{ix} \minus{} 1}$, which can be determined through another use of Euler's formula which gives $ \frac { (\cos 45x \minus{} 1) \plus{} i (\sin 45x)}{ (\cos x \minus{} 1) \plus{} i (\sin x) }$ which can then be multiplied by the denominator's conjugate to separate the imaginary part from the real part?
Thanks for your patience with me!

[quote="Altheman"]$ \frac {1}{e^{ix} \minus{} 1} \equal{} \frac {1}{e^{\frac {ix}{2}}*(2i\sin \frac {x}{2})}$[/quote]
I'm not sure where that is from. Could you elaborate on it?


Hi,

Can somebody post the solution for the chess problem in the following URL

[url]http://oraclearticles.googlepages.com/recyclebin[/url]

Black to move. Mate in two steps.

I was told that there is more than one solution to this... ??.

[hide="Solution 1"]

1...Ne2+
2. Nxe2  Qxh2 checkmate
[/hide]

[hide="Solution 2"]
1...Ne2+
2. Kh1   Qxh2 checkmate
[/hide]

I think the 'other' solution is [hide="this"]
1...Nh3+ 2Kh1 Qxh2 mate or 1...Nh3+ 2gxh3 Qxh2 mate[/hide]

In my opinion, the two solutions by 7h3.D3m0n.117 should count as one solution because they are just two different ways for white to respond

Well somehow I find this way too easy........

[hide="One way"] 1... Ne2 2. Ne2 * 2... Qh2#

* 2. Kh1 2... Qh2#[/hide]

[hide="Second way"] 1... Nh3 2. gh ** 2... Qh2#

** 2. Kh1 2... Qh2#[/hide]

Way too easy... got it in 3 seconds... eh maybe that's cause I'm

1815

Here's the image of the above puzzle for those unwilling to download.

[hide="A Quick Finish"][list]1. Nc6+, Ka8
2. Rxa6+, bxa6
3. Rb8++[/list][/hide]

I see a mate in 1...

-eidt-
that's a different problem

[hide]errr...to the original one, knight to E2 queen to H2?[/hide]

Incredibly easy - Ne2+.  The rest follows.

[quote="program4"]lol i have one[/quote]
1.Nc6+ Ka8
2.Rxa6+ bxa6
3.Rb8 checkmate

My puzzle:
1.e4 e5
2.Nf3 d6
3.Nc3 Nc6
4.Bc4 Bg4
5.h3 Bh5
6.d3 h6
7.b3 g5

What is whites best move?

[quote="ffao"]Legall's mate
8.Nxe5 for a free pawn.[/quote]
Yes, that is correct, because it creates a discovered attack against the bishop.

It's great to see all of these chess problems being posted, but please try to keep it to one problem per thread to avoid possible confusion and what not. Also, try to use the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193263]marathon thread[/url] if possible (if there's no problem currently unsolved, or if you can solve the previous one) to keep everything in one place. :)


calculus integration function trigonometry inequalities algebra polynomial

Show that the bessel function:
$ J_n(z) \equal{} \frac {1}{\pi}\int_0^\pi \cos(z \sin(\theta) \minus{} n \theta)\,d\theta$
is, for 'n' even:
$ J_{n}(z) \equal{} \frac {2 ( \minus{} 1)^n}{\pi}\int_0^1\frac {\cos(z y)\,T_n(y)}{\sqrt {1 \minus{} y^2}}\,dy$
where $ T_n(y)$ is the chebyshev polynomial given by
$ T_n(y) \equal{} \cos(n\,\arccos(x))$

I've gotten as far as
$ J_{n}(z) \equal{} \frac {2 ( \minus{} 1)^n}{\pi}\left(\int_0^1\frac {cos(z y)T_n(y)}{\sqrt {1 \minus{} y^2}}\,dy \plus{} \int_0^1\frac {\sin(z\,y) \sin(n\,\arccos(y))}{\sqrt {1 \minus{} y^2}}\,dy\right)$
using the substitution
$ \theta \equal{} arccos(y)$

It appears that all I need is to prove that, for even 'n'
$ \int_0^1\frac {\sin(z y) \sin(n\,\arccos(y))}{\sqrt {1 \minus{} y^2}}\,dy \equal{} 0$
but how?
Help, please. :blush:

[quote="Simo_the_Wolf"]Note that $ \sin(z \sin( \theta) )$ is even wrt $ \pi/2$ in $ [0, \pi ]$ while $ \sin(n\theta)$ is odd thus changing variable you get the thesis.[/quote]
Thanks!
But, if I substitute $ y \equal{} \sin(\theta)$ then I get $ \sin(n\,\arccos(\sin(\theta)))$ and not what you said $ \sin(n\,\theta)$
Or am I looking at it the wrong way?  :maybe:

yeah i'm sorry...

ok so you'll get $ \sin(n \arccos( \sin \theta ) ) \equal{} \sin( n( \frac{\pi}2 \minus{} \theta) )$

now if $ n\equal{}2k$ and $ k$ is odd you'll get $ \sin(k\pi \minus{} n\theta)\equal{} \sin(n\theta)$
if $ k$ is even you'll get $ \sin(k\pi\minus{}n\theta)\equal{}\minus{} \sin(n\theta)$
Ang again you have the same integral


Demonstrati ca \[ a^rb^s\plus{}b^rc^s\plus{}c^ra^s>a^sb^r\plus{}b^sc^r\plus{}c^sa^r\] unde \[ a>b>c>0, r>s>0\]


inequalities
function
calculus
inequalities unsolved

hello, to 2) we have $ \frac{x^2\plus{}2y^2\plus{}2x\minus{}6y\plus{}5}{xy}\geq2$ this is equivalent with $ (x\plus{}1\minus{}y)^2\plus{}(y\minus{}2)^2\geq0$, the equal sign holds if $ x\equal{}1$ and $ y\equal{}2$.
Sonnhard.

[quote="FantasyLover"]Find the maximum value of $ \frac {1}{1 \plus{} |x|} \plus{} \frac {1}{1 \plus{} |y|} \plus{} \frac {1}{1 \plus{} |z|}$ for real numbers $ x,y,z$ where $ 3x \plus{} 2y \plus{} z \equal{} 1$.

Find the minimum value of $ \frac {x^2 \plus{} 2y^2 \plus{} 2x \minus{} 6y \plus{} 5}{xy}$ where $ x,y$ are positive integers.

Find the maximum value of $ 26x^3 \plus{} 27xyz$ for real numbers $ x, y, z$ where $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$

Find the minimum value of $ \frac {(x \plus{} y^2)}{\sqrt {x^2 \minus{} 4} \plus{} \sqrt {y^2 \minus{} 9}}$ for real numbers $ x, y$ where $ x > 2, y > 3$

I have been trying to solve these inequlities, but failed. :(

For your information, this is not from my homework; this is from former Korean Math Olympiad for Middle Schoolers.

Thanks in advance!  :)[/quote]

$ 26x^3 \plus{} 27xyz$≤$ 26x^3 \plus{} x^3 \plus{} 27y^3 \plus{} 27z^3$=81 :lol:

hello, for the function $ f(x,y)\equal{}\frac{x\plus{}y^2}{\sqrt{x^2\minus{}4}\plus{}\sqrt{y^2\minus{}9}}$ i have found no local mimimum, could it be this is a typo and the function is $ f(x,y)\equal{}\frac{(x\plus{}y)^2}{\sqrt{x^2\minus{}4}\plus{}\sqrt{y^2\minus{}9}}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, for the function $ f(x,y) \equal{} \frac {x \plus{} y^2}{\sqrt {x^2 \minus{} 4} \plus{} \sqrt {y^2 \minus{} 9}}$ i have found no local mimimum, could it be this is a typo and the function is $ f(x,y) \equal{} \frac {(x \plus{} y)^2}{\sqrt {x^2 \minus{} 4} \plus{} \sqrt {y^2 \minus{} 9}}$?
Sonnhard.[/quote]

Oops, I have made a typo. :oops:

The function is indeed $ \frac{(x\plus{}y)^2}{\sqrt{x^2\minus{}4}\plus{}\sqrt{y^2\minus{}9}}$.

Thanks for pointing out. :)

hello, with the help of calculus i have proved that $ \frac{(x\plus{}y)^2}{\sqrt{x^2\minus{}4}\plus{}\sqrt{y^2\minus{}9}}\geq10$, the equal sign holds if $ x\equal{}2\sqrt{2}$, $ y\equal{}3\sqrt{2}$.
Sonnhard.

I have simplified #4:

Find the minimum value of the function $ \frac {(\sqrt {a^2 \plus{} 4} \plus{} \sqrt {b^2 \plus{} 9})^2}{a \plus{} b}$ where $ a, b > 0$.

(Here, I defined two new variables as $ a \equal{} \sqrt {x^2 \minus{} 4}$ and $ b \equal{} \sqrt {y^2 \minus{} 9}$)

Solution without calculus, anyone?

This is an olympiad for Middle Schoolers, you know :wink:

Thanks everyone :)

Two problems left:

Find the minimum value of $ \frac{(\sqrt{a^2\plus{}4}\plus{}\sqrt{b^2\plus{}9})^2}{a\plus{}b}$ where $ a,b>0$.

Find the maximum value of $ \frac{1}{1\plus{}|x|}\plus{}\frac{1}{1\plus{}|y|}\plus{}\frac{1}{1\plus{}|z|}$ for real numbers $ x,y,z$ where $ 3x\plus{}2y\plus{}z\equal{}1$.

Appreciate any help! :)


What is the most powerful/useful programming language, in general, and particular to a biology-major? The answers I have heard range from Java, C++, Visual Basic and even assembly language!

I'd really want to know what you guys think as I'm deciding which language(s) to learn this summer.

Thanks!

Well, what exactly do you mean by "powerful"?

Powerful by its capabilities...yeah, it's a pretty loose term, so I guess whatever definition you guys have is fine.

Also, which is the most useful?

Every language has its strengths and weaknesses. There are three general classes, though:
1) low-level (close to the system): Assembly language
2) mid-level (intermediate): usually C, Pascal are considered pretty medium-level languages
3) high-level (abstracted away from the system, to a degree): Python, Java (arguable), C++ (also arguable), Ruby, Perl, etc.

Low level languages have very high control over the system, but are really quite difficult to code for (_nobody_ codes serious applications in Assembly!) and are also usually not portable to other operating systems/processors.

Mid-level languages are still (arguably) difficult to code in and may require some training to code in (e.g. to use pointers in C, one must learn how a pointer works, etc.).

High-level languages offer the least control over the system, but have easier learning curves and simpler code design. For example, I can usually make a working program in Python about 4-6x faster than in C, not due to familiarity but because Python is far easier to code in.

The above illustrates the "system control power" vs. the "ease of coding power": you really can't have both.
In addition, IMHO, the medium and high-level languages are more useful because they carry over to many systems and are easier to code in.

As far as capabilities go, all (serious) languages come with enough functions and libraries that they can do anything that you might need. Perl, for example, has the extensive CPAN library of code, Java has it's huge standard library and C++ has the STL (in a similar fashion, Python has Cheese Shop).

For biology, I imagine that you would be more interested in a simple language, rather than having more control, so I would suggest the "trio" Perl/Ruby/Python. All are interpreted, scriptable languages which come with a rich feature set. I personally use Python, but good things can be said for all three.

(btw: with higher system control (lower levels of abstraction) come the benefit of having faster code; Assembly is vastly faster than C if coded correctly; Perl is much slower than C).

If you compile with optimization, C++ and C are not that much slower than assembly.

I would say it depends on your needs. If you need something to run fast because you are doing protein folding, C/C++ are basically musts, as the only thing faster is assembly, which as discussed earlier is infeasible for complex programs.

If you just need to write random scripts, Python is pretty easy.

Can you further describe what you plan to do as a biologist that requires computers?

Then again, it may be that there is no best language, but you should use a combination of everything. Like, write the most time consuming pieces of code in assembly (assuming they aren't too complex), and the next most time consuming code (more complex parts) in C/C++. Then write a python/perl/ruby script to put it all together, and it will run decently fast.

For example (nothing related to biology): I had a friend who was writing an anagrammer, and he needed to load a dictionary file. He did this in python, but when he wrote the section to load the dictionary file in C++, the program ran much faster, but writing the overall program was just about as easy, since loading a file is rather simple.

@JSteinhardt: Of course that's true; C and C++ have had decades to improve upon optimization techniques. However, there are still things (such as a highly demanding image processing loop, or vectorized code (SSE[2/3]), etc.) which assembly does best.

Also, I agree with Hamster1800 in that there really is no best language. It's whatever happens to suit your needs. Even then, there might be no best language.

Finally, the more advanced features of high-level languages, things like generator functions, simple syntax for things like hashed maps and arrays (it takes much more code to use the hash_map class in C++ than it does for Python), and easy iteration ("for variable in iterable:" as opposed to "for(vector<string>::iterator i=v.begin();i!=v.end();i++)") can make life so much easier for a programmer that a potential speed benefit in another language could be lost by debugging and longer/more complex code.

Thanks for the responses.

Well, this year I worked on a project that involved image-based tissue classification, and quite luckily, necessitated Java, the only language I know. In the future I'm looking into completing similar projects, like tissue scans.

Really, I took a Java course by chance, so I'd like to reduce the possibility of having to learn an entire language in order to work on a single project.

I could pose my question a third way, then: what's the most popular language among research scientists?

I wouldn't ask that either; there are many options.

Python is used in scientific work often (see SciPy), and from what I hear Java is on a (very) slow decline in scientific work, with many preferring to go to C++ (but I'm not a research scientist, so this is basically from what I hear at the Computer Science Dept. at University, and this is not a representative sample!)

Well, the research people usually need to have their programs work fast. The best performance you can get is actually with Fortran - I'm not kidding, most computational physicists use it, or at least the ones I know. This is because you can get a compiler optimized for your particular processor / chipset, it can automatically parallelize your code and so on. But I wouldn't really recommend learning it unless you think you're going to need your programs to run super fast. Also, I'm not sure whether you can get the compiler for free.

I don't have any experience with Python/Perl/Ruby but I heard a lot of good things about them so maybe you could try that.

As for C and C++, they are somewhat similar to Java but they also introduce a lot of new concepts (mainly pointers) which may take quite a while to get hang of. I wouldn't recommend these two languages that much, programming people like to use them but I think they are step back from for example Java. For a biologist that is. Consider them only if your Java code runs too slow or if you plan to write only small and simple programs that don't require too much maintenance.

The .NET laguages - C#, VB.NET etc. - are conceptually similar to Java, I personally like them but they're mostly good for creating nice user interfaces, working with databases and other useless stuff so again probably not the best choice, especially since you already know Java.

Conclusion - I didn't help you too much, did I? :) But from what I read, Python and Perl seem to be quite popular among biologists, there are even some very thick books called like "Perl Programming for Biologists", "Beginning Perl for Bioinformatics" and so on, so I would probably try these. And other people ^^^ recommend them too.

I think you need to state in your criteria whether you'll want graphical applications. For example, programming math problems usually doesn't require GUIs (graphical user interfaces). But end-users are generally more comfortable with GUIs. Also, if you're going to want a graphs, charts, etc., then you're going to want graphical applications. Every language supports GUIs but some languages document this aspect of programming better, as well as making GUIs easier to program.

My first recommendation would be Java. Just to warn you, Java's my best language, so I'm speaking with bias here. But objectively speaking, it's object-oriented, it's platform-independent, it's a good stepping stone, it's the AP Comp. Sci. language, it's very well-rounded, and it has excellent documentation on its libraries and facilities. My second recommendation would be Python, which is very similar to Java. I haven't done much in Python, so I can't speak too much for it.   But what I do know is that Java and Python are among the most sought-for languages in the software industry. Java and Python are also both good at facilitating GUI development.

I also think it would be good to spend a little time getting to know the basics of C, especially the difference between C pointers and Java references.

I really don't think Perl is such a good choice -.- It's not a good stepping stone for other languages because it's so different from other languages. I must say however, Perl is very fun =p

I also think there's no point in learning assembly if you want to make applications useful for biology. It's not like you'll be writing device drivers or reverse-engineering anything for your research on biology =p

Well, I've decided to learn C++, which should only take a day since it's so similar to Java. I might move onto Perl later since it seems widely used in bioinformatics.

What about Visual Basic? I've been doing some research, and that seems quite popular due to its "liberality".

Now I have another question: what are the easiest languages to learn (with Java background)?

Visual Basic is popular because it's easy to use but I wouldn't recommend it, simplicity is a nice thing but only for people who don't care how the program works as long as it works. I just don't like it. Unless you're talking about Visual Basic .NET, that's quite an improvement over the previous versions, but it's also pretty much equivalent to C# (same functions, only different syntax) and C# might be easier to learn since it's much more similar to Java than VB. Both good for creating windowy applications easily. C# is actually more similar to Java than C is, Java and C# are almost the same except for the fact that all the library functions are different. (Yeah, sure, there are many other differences but I was able to start coding in Java after about 5 minute tutorial and prior knowledge of C and C#.)

EDIT: However, if you don't really want to create some advanced applications, Visual Basic might work for you too since it's quite easy to learn.

If you have a lot of time to invest in learning programming, I would pick up an assembly language. It's always useful to have assembly at your disposal when you need to process a lot of data, as is sometimes the case in bioengineering and what not. Mid-level languages offer some control over data handling, but you'd be working with limited data types etc. C, C++, and C# wouldn't be too difficult to learn, but they're not all that much different from compiled Java. Bytecode is rather slow because it is interpreted, but remember that Java can be compiled natively as well.

Perhaps you should learn assembly and then C++ or Delphi (er... Pascal) if you are up to it (it shouldn't be too hard). Then you can integrate assembly blocks where it's important, while using a somewhat higher-level language for things like initializing the program and rendering the interface.

Of course, if you're not prepared to invest quite a good chunk of time learning assembly, then C++ or any similar language would be a good option.

Just to note, the Biological Engineering program at MIT requires an intro computer programming class which is taught in python.  Actually, to tell the truth, Python is being incorporated much more into the CS curriculum as well...  This actually might be more because Python is a really good language to get started with programming and yet powerful as well.

Of course, C++ and Java are widely used and very useful to know.

Once you learn your first programming language, you'll be able to pick up others rather quickly for the most part.  For example, I just learned python this past semester.  Then for the summer I got a job which requires me to program in C++ all day.  It only took me a few days to get a grasp of the language (still haven't become fully comfortable with pointer and such, but I've got a basic grasp on those things).

@_Daniel_: With all respect, I think your recommendation is quite shortsighted.
Assembly is nearly useless, to be frank. Writing long programs in Assembly is nearly impossible, due to the sheer scale (would you like to write 100000+ instructions in assembly, and keep the thing maintainable? I didn't think so...) and a good deal of the most used assembly bits (e.g. the heavily optimized string routines in C, or the GFX code in most games) lives in the libraries, where dedicated assembly hackers do the work for you so you don't have to ;)
C, C++ and C# are quite different from compiled Java. C in particular is almost guaranteed to be faster than compiled Java, and also a lot smaller in terms of the standard library. C++ has (most) of the features of Java, but without the crazy overhead (yes, I'm well aware that natively compiled Java has less overhead, but it's still there), and will probably run faster too; C# is compiled to MSIL (Microsoft Intermediate Language) which is basically bytecode, but optimized to run reasonable fast.
I personally would not learn assembly, and that's the recommendation I would make to anyone listening. It is completely non-portable, and if something happens (e.g. Apple's switch to Intel, your organization suddenly needing to support another chipset, etc. etc.) your code would need to be completely rewritten, as opposed to merely ported (porting is most certainly easier, and in some cases, e.g. Python/Java, it consists of finding an interpreter or compiler and off you go).
@cooljoe: I wouldn't recommend Perl, despite CPAN. It is more suited as a CGI language or a quick scripting language, and less so as a serious biology language (this is IMHO, so take it any way you like ;) )
Visual Basic is a definite no from me, as it is non-portable and will most certainly not be an easy language to (reliably) maintain (again, IMHO and personal experience).
The easiest language to learn with a Java background is arguably C#, and (very close second) C++. You should know that, again, C# is not very portable (though Mono works, it is far from perfect), while C++ is much more well supported and portable.

The best are Maya and Java. Ruby is good too.

[quote]Assembly is nearly useless, to be frank. Writing long programs in Assembly is nearly impossible, due to the sheer scale (would you like to write 100000+ instructions in assembly, and keep the thing maintainable? I didn't think so...)[/quote]
Well, I certainly wouldn't recommend that assembly be used for every project, or every part of most projects. A major plus for assembly is that it can be integrated into projects which use another language like Pascal, C, etc. for the more lengthy and less important tasks like initialization and drawing (the example from my last post).

[quote]C, C++ and C# are quite different from compiled Java. C in particular is almost guaranteed to be faster than compiled Java, and also a lot smaller in terms of the standard library. C++ has (most) of the features of Java, but without the crazy overhead (yes, I'm well aware that natively compiled Java has less overhead, but it's still there), and will probably run faster too; C# is compiled to MSIL (Microsoft Intermediate Language) which is basically bytecode, but optimized to run reasonable fast.[/quote]
I'm aware of the speed deficit from array index checking, auto garbage collection, platform flexibility, etc. But any speed comparison is rather ill-founded without considering what operations are being done and what compiler (especially the Java compiler, as they are less matured) is being used. Most of the typical speed issues can be eliminated with dynamic optimization techniques.

Basically, I agree that C remains the local king of general efficiency, but the differences can be made pretty negligible with effective optimization.

[quote]It is completely non-portable, and if something happens (e.g. Apple's switch to Intel, your organization suddenly needing to support another chipset, etc. etc.) your code would need to be completely rewritten, as opposed to merely ported (porting is most certainly easier, and in some cases, e.g. Python/Java, it consists of finding an interpreter or compiler and off you go).[/quote]
There are plenty of solid assembly conversion programs available on the web. Besides, easy porting by simply using a different compiler is only a practical expectation if you have a rather simple program which doesn't use any platform-specific software - might work for some experimental, research-oriented programs related to biology, but calling APIs or any other platform-specific resources (e.g., DirectX) will lead to serious compatibility issues.

Converting assembly loses all the benefits of using hand-tuned assembly in the first place, so I see no reason at all why that would be a good idea.

Well, I suppose you could continue working on the pre-conversion assembly code if it needed further modification. It would be a bit of a hassle, and you certainly have something of a point with the compatibility drawback, but I don't think it's a major issue. Assemblers are becoming less popular over time, but are still relatively well-used in fields like bioengineering. It's also a great way to learn about rudimentary low-level processing - not as important for a bio major as a CS or EE major, but still important for properly assessing operation efficiency and it's great general knowledge outside of biology.

Lambda Calculus!!!

In lambda calculus, the only datatype is the lambda function. A lambda function is a function that takes a lambda function and returns.... a lambda function.

Some random examples:
La: a
The identity function.

La: Lb: Lf: f a b
The CONS function from Scheme. It creates the tuple (a, b).

Lf: Lx: x
The Church Numeral for 0.

Ln: Lf: Lx: f ((n f) x)
The successor function. Takes the Church Numeral representing a number m and returns the Church Numeral for (m + 1).

Lf: (Lx: f (x x)) (Lx: f (x x))
The Y-combinator. It has the fixed-point property Yf = f(Yf). When evaluated lazily, it allows for recursive lambda functions.

Learning is half the func. Multiplication takes about 100 steps. Yet lambda calculus is as powerful a language (computationally speaking) as any ever written.

http://en.wikipedia.org/wiki/Lambda_calculus


group theory
abstract algebra
limit
Galois Theory
superior algebra
superior algebra unsolved

Let $ k$ be a field whose all finite extensions are cyclic. and has an extension of degree n for all n.
what can we say about $ Gal (k^a/k)$

it's clear that the algebraic closure $ k^a$ is separable here, so that we can talk about the profinite group $ G \equal{} Gal(k^a / k)$. applying the hypothesis to the compositum of two fields and using that cyclic groups have only one subgroup for a given order, we see that $ k^a$ contains exactly one cyclic extension of $ k$ of degree $ n$. this yields an isomorphism $ G \to \lim_{\leftarrow} \mathbb{Z}/n \equal{} \widehat{\mathbb{Z}}$.

then another standard question:
if $ k$ is as above show that there is an automorphism of $ k^a$ that fixes our field.

this comes down to the fact that $ 1$ is a topological generator of $ \widehat{\mathbb{Z}}$. so if $ k_n$ is the cyclic extension of degree $ n$ over $ k$ and $ \sigma_n$ generates its galois group, we may arrange that $ n | m \Rightarrow \sigma_m$ extends $ \sigma_n$, so that the $ \sigma_n$ induce a global isomorphism $ \sigma$ with fixed field $ k$.


geometry
3D geometry
geometric transformation
reflection
Putnam
geometry solved

Let $Q$ be a point inside a cube. Prove that there are infinitely many lines $l$ so that $AQ=BQ$ where $A$ and $B$ are the two points of intersection of $l$ and the surface of the cube.

The lines are precisely those determined by the point $Q$ and a point of the intersection between the cube and the reflection of the cube in $Q$, and it's clear that this intersection has infinitely many points.

The problem is just a slight modification of the [url=http://www.kalva.demon.co.uk/putnam/putn77.html]Putnam 1977 B4 Problem[/url], which is fairly well known. Quite an easy problem for a National Math Olympiad (also the TST). :10:


[hide]$  f(x)\equal{}\frac{x^{3}\minus{}3x^{2}\minus{}5}{3x\minus{}4}$
All I had to do was find 'All the intercepts', but along the way I decided I wanted to get all of the roots. Soon after I realized that this is no open and shut case.

Find all roots to $ x^{3}\minus{}3x^{2}\minus{}5\equal{}0$?
[hide="I tried"]to find a radical that met the real zero, and came up with $ 2\sqrt{3}$, but that isn't exact.
Is there some method to solve this type of problem?
(not a cubic factoring formula)[/hide][/hide]

Unfortunately, it looks like the best way to solve this one is by Cardano's formulas. Inspection suggests that it has only one real root, and it is probably not a pretty one (i.e. there's probably a ton of radicals). Cardano's can be found on most math reference sites (PlanetMath, Mathworld...), so you can search for it easily if you don't know about it yet.

Yeah, the discriminant is 1215.  :read:
1 real zero, 1 set of complex conjugates, as I had surmised from the graph.

Wow. I am only going to do this once. Ti-basic here we come!


We just have to rewrite the numbers as powers of $ 2$.

$ 2^{2^{x}} \equal{} 16^{16^{16}}\implies 2^{2^{x}} \equal{} 2^{2^2\cdot 2^{64}} \equal{} 2^{2^{66}} \implies x \equal{} \boxed{66}$.

Sorry sunehra but how you did this $ 2^{2^{x}}\equal{} 2^{2^{2}\cdot 2^{64}}$?

Thanks

[hide="Maybe this is simpler to understand."]
$ 16\equal{}2^4$.

$ 2^{2^x}\equal{}2^{4\cdot {16^{16}}}$.

Taking base $ 2$ logarithms of both sides:

$ 2^x\equal{}4\cdot 16^{16}\equal{}2^2\cdot 2^{4\cdot 16}\equal{}2^{4\cdot 16\plus{}2}$.

Taking base $ 2$ logarithms again,

$ x\equal{}4\cdot 16\plus{}2\equal{}\boxed{66}$.
[/hide]


probability
geometry
Putnam
college contests

[i]Two real numbers $ x$ and $ y$ are chosen at random in the interval $ \left(0,1\right)$ with respect to the uniform distribution.  What is the probability that the closest integer to $ x/y$ is even?  Express the answer in the form $ r\plus{}s\pi$, where $ r$ and $ s$ are rational numbers.[/i]

I don't know if this (partial) proof is entirely correct, but I'm having trouble proceeding from the last step:

Suppose $ y>2x$ so that the closest even number to $ x/y$ is $ 0$.  The probability of $ x<1/2$ is $ 1/2$, and the probability that $ y>2x$ is again $ 1/2$.  Therefore, the probability that $ y>2x$ is $ \left(1/2\right)^{2}\equal{}1/4$.

Now let $ 2n$ be the even number where $ n\in\mathbb{N}$.  Then we need

\[ 2n\minus{}\frac{1}{2}<\frac{x}{y}<2n\plus{}\frac{1}{2}\implies\frac{2x}{4n\plus{}1}<y<\frac{2x}{4n\minus{}1}.
\]
How do we find the probability that $ y$ is on such an interval?

That's the hard part. It's an infinite sum.

You should draw the picture- shade the areas you want, and don't shade the rest.

[hide="Answer"]$ \frac{5}{4}\minus{}\frac{\pi}{4}$[/hide]


prove that sum of any twin primes (except the first two, 3,5) is always divisible by 12.

Too trivial for this section.
$(6k-1)+(6k+1)=12k$

we know that any number greater than 3 which is prime can be written as $2k+1$.....
now it is given that they are twin primes. so another is $2k+3$
now if $2k+1$ is prime then k must not belong to the following sequence
$1,4,7,10.....$
similarly from $2k+3$
we can say that it is prime only when k not belong to following sequence
$3,6,9,12.....$
Now we can say that k belong to following sequence(because k is an integer)
$k = 2,5,8,.....$
general term is $3n-1$. so $k=3n-1$ nust hold.....
which show twin primes is of type $(6n-1,6n+1)$
hence the addition of the above is 12n which shows that addition of sum of twin prime is divisible by 12.


If you've read Knuth's book, he gives the mathematical definiton of a computational method as a quadruple $(Q,I,\Omega, f ) $ s.t. $ I \ and \Omega \subset Q $ . Also he says that $ \forall q \in \Omega, f(q) = q $ . Then he says that Q is equivalent to the "State of computation", I - Input ,$ \Omega $ - Output, $f$- computational method. Finally, if we have $x \in I$, we take the iterations $ f(x_{k}) = x_{k+1} $. till $x_{k} \in \Omega $ , after which, according the above definiton, the iteration stops. An Algorithm is a finite computational method etc....

I have some questions regarding this.
1) Why is every element of $\Omega$ fixed w.r.t.ti the method $f$. Is it to make sure that once we reach an element in $\Omega$ by a no. of iterations from I, the iterations stop and we get the result???
2) What is a "state of computation"?? How are the elements of the above given quadruple directly related to things we understand(e.g. $I$ is equiv. to Input as given above)?? Is this definiton of computational method universal??
3)Why do we establish a series of iterations? Can't have have a simple functions as we know ???

As I see it, the "state of computation" is just some state in which is for example the computer u're using to do the computation. So if you wrote down the exact content of your RAM and Hard Drive and processor registers and everithing in your computer in some moment during the computation, that could be considered a "state of computation". In more abstract terms, the state of computation could be the content of all the variables in your program plus the actual position in the program plus the state of the I/O devices if you use them in your program. $x_{k+1}=f(x_k)$ means proceeding one step in the computation - it can be one command in your program, or one function call, or one instruction for the processor, or it could be the whole computation so that $\forall i\in I,f(i)\in\Omega$. It's there just to generalize the definition a little bit. The statement $\forall q\in\Omega,f(q)=q$ only means that after you get the right answer, there are no more things to do. The next step is just waiting there in the same state.


floor function
vector
combinatorics proposed
combinatorics

The National Foundation of Happiness (NFoH) wants to estimate the happiness of people of country. NFoH selected $n$ random persons, and on every morning asked from each of them whether she is happy or not. On any two distinct days, exactly half of the persons gave the same answer. Show that after $k$ days, there were at most $n-\frac{n}{k}$ persons whose “yes” answers equals their “no” answers.

Let $m$ be the sum, over all pairs of days, of the number of people who give the a different answer on the two days. Then

$m = \frac{nk (k-1)}{4}$.

For any person who gives an equal number of yeses and nos, the number of pairs of days on which he gives a different answer is $\frac{k^{2}}{4}$.  So the maximum number of such people is:

$\lfloor \frac{\frac{nk (k-1)}{4}}{\frac{k^{2}}{4}}\rfloor$ = $\lfloor n-\frac{n}{k}\rfloor$.

Let $ v_i$ be the vector that $ v_i\equal{}(a_{i1},\ldots,a_{in})$ where $ a_{ij}\equal{}1$ if the j-th person's answer on the i-th day is yes and $ a_{ij}\equal{}\minus{}1$, otherwise. The giving condition implies that $ v_i \cdot v_j\equal{}0$ and clearly $ v_i^2\equal{}n$.
Let the i-th person's answer (1 or -1) adds up to $ x_i$. If there is more than $ n\minus{} \frac{n}{k}$ people whose yes's equal to no's then
$ k^2 \cdot \frac{n}{k} < (x_1^2\plus{}\ldots\plus{}x_n^2)\equal{}|v_1\plus{}\ldots\plus{}v_k|^2\equal{}v_1^2\plus{}\ldots\plus{}v_k^2\equal{}kn$. Contradiction.


When you log in. Do you all have links: Favorites_Profile_Preferences_This one for messages_Log out.I know this sounds foolish but please answer me, why is the same whether I click Profile or Preferences?

Actually they shouldn't be the same: preferences should lead here: http://www.mathlinks.ro/Forum/usercp.php?mode=prefs ... it's a small typo :P

When I click Preferences it leads here: [url]http://www.mathlinks.ro/Forum/usercp.php[/url]. Why is that? Can you fix it?


$ ABC$ is a acute triangle.the incircle of $ ABC$ triangle is tangent to $ BC$ at $ K$. $ M$ is the midpoint of $ AD$ altitude. $ KM$ Line intersect incircle at $ N$ again.prove that the circumcircle of $ BCN$ is tangent to incircle of $ ABC$ at $ N$.

We can calculate that IaI/IaA=r/AM=2a/(a+b+c)
Ia,D,M,N are collinear
∠AFD=∠ABIa=90+∠B/2  => BIa∥FD
∠AFN=∠FDN=∠BIaN  => F,N,Ia,B are concyclically
Analogously , E,N,Ia,C are concyclically
∠BNIa=∠BFIa=∠CEIa=∠CNIa
So ND is the bisector of ∠BNC  =>  I’D’⊥BC  => ∠I’D’N=∠IDN=∠DNI=∠D’NI=∠D’NI’
⊙I and ⊙I’are tangent at N


I have been trying this problem
let S=(1,2,3,...,100) .
let A be a permutation of the elements of S. multiply the i'th element of S with the i'th element of A..
for i=1,2,...100
reduce all the products so obtained mod 101..
Prove that one cannot recover a permutation of S in this manner.

I hope i haven't messed up the problem statement.. i dont have the book right now.. just remember the problem

regards

Consider the product of all the elements.


The remainder when x^3+3x^2-2x-2 is divided by x+(square root 3)

could you explain this problem please.

It's still synthetic / long division, just with a square root instead of an integer.  The details of the calculation don't change.

Well, what prevents you from simply brute-forcing (dividing) $x-\sqrt{3}$ into $x^{3}+3x^{2}-2x-2$? You would get radical coefficients, but...

Well, $x^{2}-3 = (x+\sqrt{3})(x-\sqrt{3})$ is clearly divisible by $x+\sqrt{3}$, so let's find the remainder when we divide by $(x^{2}-3)$ first.

$x^{3}+3x^{2}-2x-2-(x+\sqrt{3})^{3}$ $=x(x^{2}-3)+3x^{2}-2x-2$ $=x(x^{2}-3)+3(x^{2}-3)+7+x$

So we're left with $7+x$ so far. Now we can look away from the quadratic and simplify it further:

$7+x=(x+\sqrt{3})+(7-\sqrt{3})$, so the remainder is $\boxed{7-\sqrt{3}}$

Or you can just use the Remainder Theorem. $P \left(-\sqrt{3}\right) = \boxed{7-\sqrt{3}}$

[hide]Well, it's just $f(-\sqrt{3}))$ which is $7-\sqrt{3}$.[/hide]