I don't know how to represent a decimal number to binary number. For example, in class, my teacher did the number 24
So he asked us the largest power of 2 in 24, it's 2^4. then he did: 1*2^4 + 1*2^3 + 0*2^2 + 0*2^1 +0*2^0 I understand about the base 2 and power, but I dont understand when and why sometimes it's 1 times blah blah, and sometimes it's 0 times blah blah. How do you figure out? Thank you.

Okay, say we want to translate [is that the proper word?] $ 81_{10}$ into base $ 2$.  Look at $ 81$.  It can be expressed as $ 8 \cdot 10^1 \plus{} 1 \cdot 10^0$.  This means that the number is in base $ 10$.  You have to do the same thing for base $ 2$.  However, the largest number allowed in base $ 2$ is the number $ 1$, so $ 81 \equal{} 1 \cdot 2^6 \plus{} 0 \cdot 2^5 \plus{} 1 \cdot 2^4 \plus{} 0 \cdot 2^3 \plus{} 0 \cdot 2^2 \plus{} 0 \cdot 2^1 \plus{} 1 \cdot 2^0$.  Then, you simply take the ones and zeros and "string" them, so we have $ 81_{10} \equal{} 1010001_2$.  Do the same thing for $ 24$.  Hope this helps!

I'm assuming you already know this but just in case:
In base two for example, the digit that normally represents the ones digit in base ten can either represent the value, 0 or 1 in base two. The next digit, the one that would normally represent tens in base ten represents values of twos. so there can either be 0 twos or 1 two. The next digit, which represents 100's or $ 10^2$'s represents values of $ 2^2$ so there can be either one $ 2^2$ or 0 $ 2^2$.
So to find 24 in base two, just find the series of powers of two that add up to 24.
Why he writes the "0 times blah blah" I think is just to show the other digits.

Your teacher presented you with a systematic procedure. Let's represent another number from decimal to binary, say $ 43_{10}$.

1) Find the largest power of 2 less than $ 43_{10}$. It's $ 2^5$. This is our first digit 1. Write $ 1\cdot2^5$.

2) Now, take the difference $ 43\minus{}2^5\equal{}43\minus{}32\equal{}11$.

3) Find the largest power of 2 less than $ 11$. That's $ 1\cdot2^3$. Now you wonder, what happened to $ 2^4$? That's because $ 2^4>11$, so it gets the digit 0 and write $ 0\cdot2^4$.

4) Take the difference $ 11\minus{}2^3\equal{}3$.

5) Repeat step 3: we get $ 1\cdot2^1$, giving $ 0\cdot2^2$.

6) Repeat step 4: $ 3\minus{}2\equal{}1$. We can stop now since we got $ 1\cdot2^0$.

7) Collect all the powers of 2 found and add them: we have the representation $ 1\cdot2^5\plus{}0\cdot2^4\plus{}1\cdot2^3\plus{}0\cdot2^2\plus{}1\cdot2^1\plus{}1\cdot2^0\equal{}101011_2$

This seems very exhaustive if you do it this way. There is an easier (if not shorter) way of doing it by performing repeated divisions by 2. Here is a way of doing it by finding the [u]remainder[/u] of each division (note that each remainder is a digit of the number in base 2 in [i]reverse order[/i]):

1) Divide 43 by 2. Quotient is 21 and remainder is 1 <---That's the [b]last[/b] digit!

2) Divide 21 by 2. Quotient is 10 and remainder is 1.

3) Divide 10 by 2. Quotient is 5 and remainder is 0.

4) Divide 5 by 2. Quotient is 2 and remainder is 1.

5) Divide 2 by 2. Quotient is 1 and remainder is 0.

6) Stop because quotient is already less than 2.

Thus, the number (starting with the last quotient and then each of the remainders) we have $ 101011_2$.

This method, if you write the number in a compact form, is equivalent to $ ((((1\cdot2\plus{}0)\cdot2\plus{}1)\cdot2\plus{}0)\cdot2\plus{}1)\cdot2\plus{}1$.

Hope this helps.

PS: When you have mastered converting decimals to binaries, and viceversa, try converting from decimals to base 3 numbers!

PSS: What is the base 3 representation of $ 43_{10}$?

[quote="cgyao15"]a helpful tip: Read Intro to NT.[/quote]

So... you are helping AoPS advertise xD

Anyway... in binary, you take the highest power of 2.

In the case of the number 24, it's 2^4 = 16.

Since it's 2^4, you have a certain amount of 2^4s, 2^3s, 2^2s, 2^1s, and 2^0 (1)s. As a result, you have five blanks: _ _ _ _ _

Now, you have one 2^4 (16) in 24. (If you have two, then it makes 2*16 = 32 which is greater than 24.) As a result, you put a 1 in the first blank: [u]1[/u] _ _ _ _ You subtract 16 from 24, which gives you 8. This is because you already used it.

Next, you do the same thing for 2^3, 8. You obviously have one 8 in 8, so the second blank is also a 1. This gives [u]1[/u] [u]1[/u] _ _ _. Now you subtract 8 from 8, giving you 0. When you have 0, you can just fill the rest of the blanks in with 0s. The answer is [u]1[/u] [u]1[/u] [u]0[/u] [u]0[/u] [u]0[/u], or 11000[size=59]2[/size].

I hope this helps! :)

P.S. I'm supposing that you don't have to turn it in anymore or anything, so that's why my example is with the actual number. You should try it with some other numbers too. For any other conversions [b]FROM[/b] base ten, it's similar; just use that number instead of two.


If $ a^{3}\minus{}3a^{2}b\equal{}\sqrt{402}$ and $ b^{3}\minus{}3ab^{2}\equal{}40$ then find tha value of $ a^{2}\plus{}b^{2}$

[quote="Elio (n)"]If $ a^{3} \minus{} 3a^{2}b \equal{} \sqrt {402}$ and $ b^{3} \minus{} 3ab^{2} \equal{} 40$ then find tha value of $ a^{2} \plus{} b^{2}$[/quote]

We have: $ \left\{ \begin{array}{l}
 a^3  \minus{} 3a^2 b \equal{} \sqrt {402}  \\
 b^3  \minus{} 3ab^2  \equal{} 40 \\
 \end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
 (a \minus{} b)^3  \equal{} \sqrt {402}  \minus{} 40 \\
 (a \plus{} b)^3  \minus{} 6ab(a \plus{} b) \equal{} \sqrt {402}  \plus{} 40 \\
 \end{array} \right.$

Difference: $ \left\{ \begin{array}{l}
 a \plus{} b \equal{} x \\
 a \minus{} b \equal{} y \\
 \end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
 a \equal{} \frac{{x \plus{} y}}{2} \\
 b \equal{} \frac{{x \minus{} y}}{2} \\
 \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
 y^3  \equal{} \sqrt {402}  \minus{} 40 \\
 x^3  \minus{} 6.\frac{{x^2  \minus{} y^2 }}{4}.x \equal{} \sqrt {402}  \plus{} 40 \\
 \end{array} \right.$

We can find (x;y) so find (a;b) so find $ a^2\plus{}b^2$  :wink:

This is the same as [url]http://www.mathlinks.ro/viewtopic.php?t=225240[/url] with different numbers. The complex number solution is very nice!

but l_Euler's solution on this thread is very complex and messy and brute force (much shorter than TZF's solution(s))

Two things:

[b](1) @dgreenb:[/b] Unfortunately, if you take a closer look, this one is not similar to the other one.

[b](2) @stevenmeow:[/b] That's because [i]this [/i]problem doesn't have a nice solution. For example, here's a typical $ b$ value in the solution set ((attached)). In fact, $ a^2\plus{}b^2$ is negative in at least one of the cases (I didn't bother to check all of the ordered pairs that solve this).

A different approach to find the solution may be the following one.

Let
$ \begin{array}{rcl} a^3 \minus{} 3a^2b & \equal{} & x \\
b^3 \minus{} 3ab^2 & \equal{} & y, \end{array}$
where $ x: \equal{} \sqrt {402}$ and $ y: \equal{} 40$. Then $ (a \minus{} b)^3 \equal{} x \minus{} y$ which implies that $ a^2 \plus{} b^2 \equal{} \sqrt [3]{(x \minus{} y)^2} \plus{} 2ab$ or equivalently $ ab \equal{} \frac {1}{2}((a^2 \plus{} b^2) \minus{} \sqrt [3]{(x \minus{} y)^2})$. For simplicity write $ z: \equal{} \sqrt [3]{(x \minus{} y)^2}$.

Next, multiplying the original equations we obtain
$ xy \equal{} (a^3 \minus{} 3a^2b)\cdot (b^3 \minus{} 3ab^2) \equal{} (ab)^2\cdot (10ab \minus{} 3(a^2 \plus{} b^2)).$
Using the fact that $ ab \equal{} \frac {1}{2}((a^2 \plus{} b^2) \minus{} z)$ this gives
$ ((a^2 \plus{} b^2) \minus{} z)^2\cdot (2(a^2 \plus{} b^2) \minus{} 5z) \equal{} 4xy$.
This is in fact a cubic equation with the unkown $ a^2 \plus{} b^2$. Solving this equation, we can obtain the values for the term $ a^2 \plus{} b^2$ without calculating $ a$ and/or $ b$.

[quote="M@ri@n"]A different approach to find the solution may be the following one.

Let
$ \begin{array}{rcl} a^3 \minus{} 3a^2b & \equal{} & x \\
b^3 \minus{} 3ab^2 & \equal{} & y, \end{array}$
where $ x: \equal{} \sqrt {402}$ and $ y: \equal{} 40$. Then $ (a \minus{} b)^3 \equal{} \sqrt [3]{(x \minus{} y)}$ [/quote]

Surely (a-b)^3=a^3-3a^2b+3ab^2-b^3=x-y  meaning $ (a\minus{}b)\equal{}\sqrt [3]{(x\minus{}y)}$

Clearly! Thanks!  :)

It is already corrected.


inequalities
function
inequalities unsolved

Theorem of Hardy,Littlewood,Polya

Let x_i (i=1,..,n) and y_i (i=1,..,n) be reals.
A necessary and sufficient condition for the inequality

                \sum(i=1,..,n) F(x_i)  \leq   \sum (i=1,..,n)F(y_i)

to hold for every real function F continuous and convex in some interval containing all the numbers x_i and y_i is that

  1)  x_1 \leq x_2 \leq ..... \leq x_n

  2)  y_1 \leq y_2 \leq ..... \leq y_n

  3)  x_k+...+x_n  \leq  y_k+..+y_n   for k=2,3,..,n

  4)  x_1+..+x_n = y_1+..+y_n

Using this theorem you can prove a lot of very hard trigonometric inequalities about triangles.

 Has anybody a proof of this very powerful theorem?

I think it's called also the "majorization inequality".

Probably google will help ....

>> F continuous and convex : isn't it a pleonasm  :D

That's the basis of my "Extended Jensen Inequality".

Assume f(x) is convex in [a,b] and x1,x2,...,xn are real numbers in [a,b]  satisfy Sigma x=n( \lambda a+(1- \lambda )b), then we have Sigma f(x) \leq n( \lambda f(a)+(1- \lambda) f(b)).

That is the so-called "Reversed Jensen Inequality".


Prove that in any triangle $ ABC$ there is the relations :

$ 1\blacktriangleright\ \ \sqrt {r_a}\plus{}\sqrt {r_b}\plus{}\sqrt {r_c}\ge3\sqrt {3r}$ .

$ 2\blacktriangleright\ \ \sum\sqrt [3]{a(b\plus{}c\minus{}a)^2}\ \ge\ a\plus{}b\plus{}c\ .$

$ \blacktriangleright$  [hide="solution to problem 1"]$ r \equal{} \frac {\Delta } {s}{\text{ ; }}r_a \equal{} \frac {\Delta } {{s \minus{} a}}{\text{ ; }}\frac {{r_a }} {r} \equal{} \frac {s} {{s \minus{} a}} \hfill \\
f(x) \equal{} \sqrt {\frac {s} {{s \minus{} x}}} {\text{ ; }}f''(x) > 0 \hfill \\
{\mathbf{By }}{\text{ }}{\mathbf{Jensen,}}\hfill \\
f(a) \plus{} f(b) \plus{} f(c) \geqslant 3f\left( {\frac {{a \plus{} b \plus{} c}} {3}} \right) \equal{} 3\sqrt 3 \hfill \\
\Leftrightarrow \sum\limits_{cyc} {\sqrt {\frac {s} {{s \minus{} a}}} } \geqslant 3\sqrt 3 \hfill \\
\Leftrightarrow \;\;\sqrt {r_a } \plus{} \sqrt {r_b } \plus{} \sqrt {r_c } \geqslant 3\sqrt {3r} \hfill$[/hide]

$ \blacktriangleright$[hide="solution to problem 2"]$ f(x) = \left\{ {4x(s - x)^2 } \right\}^{1/3} {\text{ ; }}f''(x) > 0 \hfill \\
\mathbf{By}\text{ }\mathbf{Jensen}, \hfill \\
f(a) + f(b) + f(c) \geqslant 3f\left( {\frac {{2s}} {3}} \right) = a + b + c \hfill \\
\Leftrightarrow \sum\limits_{cyc} {\sqrt [3]{{a(b + c - a)^2 }}} \geqslant a + b + c \hfill$
[/hide]

[quote="apratimdefermat"]$ \blacktriangleright$  [hide="solution to problem 1"]$ r = \frac {\Delta } {s}{\text{ ; }}r_a = \frac {\Delta } {{s - a}}{\text{ ; }}\frac {{r_a }} {r} = \frac {s} {{s - a}} \hfill \\
f(x) = \sqrt {\frac {s} {{s - x}}} {\text{ ; }}f''(x) > 0 \hfill \\
{\mathbf{By }}{\text{ }}{\mathbf{Jensen,}}\hfill \\
f(a) + f(b) + f(c) \geqslant 3f\left( {\frac {{a + b + c}} {3}} \right) = 3\sqrt 3 \hfill \\
\Leftrightarrow \sum\limits_{cyc} {\sqrt {\frac {s} {{s - a}}} } \geqslant 3\sqrt 3 \hfill \\
\Leftrightarrow \;\;\sqrt {r_a } + \sqrt {r_b } + \sqrt {r_c } \geqslant 3\sqrt {3r} \hfill$[/hide][/quote]
[hide="solution to problem 1; Without Jensen"]
$ r = \frac {\Delta } {s}{\text{ ; }}r_a = \frac {\Delta } {{s - a}}{\text{ ; }}\frac {{r_a }} {r} = \frac {s} {{s - a}}$
We are left to prove that \[ \sum \sqrt{\frac{s}{s-a}} \geq 3\sqrt{3}\]
By AM-GM, \[ \[\sum \sqrt{\frac{s}{s-a}} \geq 3 \sqrt[6] {\frac{s^3}{\prod (s-a)}}=3\sqrt[6]{\frac{s^4}{\Delta^2}}=3\sqrt[3]{\frac{s^2}{sr}}=3\sqrt[3]{\frac{s}{r}}\]
We know that $ r \leq \frac{s}{3\sqrt{3}}$. So,$ \sum \sqrt{\frac{s}{s-a}} \geq 3\sqrt{3}$, as desired.[/hide]


Here is a nice motto or essay topic, this was given this year at the local olympics in the Arges county! (English olympics)

"Every man is the architect of his own future"

Depends on your beliefs, if you believe in fate, then I would say the previous statement doesn't apply to you.


calculus
derivative
calculus computations

A girl of five feet of height is walking towards a light post, which is of $20$ feet of height, at speed of  $4$ feet/second. Her little brother is following her with a constant distance of $4$ feet between them. See the figure below.

Determine at which speed the end of the shadow is moving, that is $\frac{dy}{dt}$
[b]NOTE[/b]: When the girl is far from the light post, she controls the end of the shadow, while she is close to the post, her brother controls the end of the shadow.

If you are fed up, then [hide="Click here to see solution"]

First, let's see where (i.e. how much close to the light post) her little brother starts controlling the shadow speed. See the Figure below:

        [img]http://www.angelfire.com/blog2/nad_af/1.GIF[/img]

It is when both shadow tips are one over the other.
From the figure: $\frac{3}{y} = \frac{5}{y+4} = \frac{20}{x+y+4} \Longleftrightarrow \boxed{y=6} \wedge \boxed{x=30}$

$\therefore$ Her brother starts controlling shadow tip speed when $x < 30$ feet.

a) case 1, $x \geq 30$
[img]http://www.angelfire.com/blog2/nad_af/2.GIF[/img]
we want to find $\frac{d(x+y)}{dt}$

$\frac{d(x+y)}{dt} = \frac{d(x+y)}{dx} \cdot \frac{dx}{dt}$     (chain rule)

From the figure above: $\frac{20}{x+y}= \frac{5}{y} \Longleftrightarrow y=\frac{x}{3}$
$\therefore x+y = \frac{x}{3} + x = \frac{4}{3}x$

$\frac{d(\frac{4}{3}x)}{dt} = \frac{4}{3} \cdot 4 = \boxed{\frac{16}{3}feet/s}$

b) case 2, $x < 30$
[img]http://www.angelfire.com/blog2/nad_af/3.GIF[/img]
we want to find $\frac{d(x+y)}{dt}$

From the figure above: $\frac{3}{y}= \frac{20}{x+y} \Longleftrightarrow y=\frac{3}{17}x$

$\therefore x+y = \frac{3}{17}x + x = \frac{20}{17}x$

$\frac{d(\frac{20}{17}x)}{dt} = \frac{20}{17} \cdot 4 = \boxed{\frac{80}{17}feet/s}$
[/hide]


This shouldn't be too difficult.

Quadrilateral PQRS is inscribed in a circle with point A outside the circle. A-P-S and A-Q-R. Prove that (AP)(RS)=(AR)(PQ)

[hide="Hint 1"]
Angles of the Quadrilateral
[/hide]
[hide="Hint 2"]
Similarity
[/hide]

[quote="easyas3.14159..."]What does A-P-S and A-Q-R mean?[/quote]

P is between A and S and Q is between A and R.

[quote="easyas3.14159..."]What does A-P-S and A-Q-R mean?[/quote]

Collinearity and order of points - obviously, $A$ is the intersection of the lines $PS$ and $QR$ and the points are named in such way that $P$ is between $A$ and $S$ and $Q$ is between $A$ and $R$.


Construct triangle with vertices A, B and C if positions of points A, I (its incenter) and G (its gravity center) are given. (Standard notations).

http://www.mathlinks.ro/Forum/viewtopic.php?t=91729

see also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106337]this[/url]

Thank you!
  It is very interesting solution. It uses terms I don't know, but maybe "If you want to go to the Moon you need a rocket, nothing less is not useful".
  When I was 9-th grade (I took only 2 hours math at school weekly) I tried to solve this problem I though about algebraic approach. I found equations by virtuose using of it the triangle were constructible. But determining the count of solutions weren't an easy task.
  What do you think about this problem? Is it difficult? Do you know who is the autor of the problem?


On a map whose scale is $ 400$ miles to an inch and a half, a certain estate is represented by a rhombus having a $ 60^{\circ}$ angle. The diagonal opposite $ 60^{\circ}$ is $ \frac {3}{16}$ in. The area of the estate in square miles is:

$ \textbf{(A)}\ \frac {2500}{\sqrt {3}} \qquad\textbf{(B)}\ \frac {1250}{\sqrt {3}} \qquad\textbf{(C)}\ 1250 \qquad\textbf{(D)}\ \frac {5625\sqrt {3}}{2} \qquad\textbf{(E)}\ 1250\sqrt {3}$

[hide]By the properties of a rhombus, we know that the diagonals bisect the angles of the vertices from which they are drawn. We also know that adjacent angles are supplementary, so the angle adjacent to the $ 60^{\circ}$ angle is $ 180^{\circ}\minus{}60^{\circ}\equal{}120^{\circ}$. Thus the diagonal that is $ \frac {3}{16}$ inches on the map divides the $ 120^{\circ}$ angle into two $ 60^{\circ}$, dividing the rhombus into two congruent and equilateral triangles.

The actual length of the diagonal, which we will call x, is
$ \frac {\frac {3}{2}}{400}\equal{}\frac {\frac {3}{16}}{x}$. This simplifies out to $ x\equal{}50$.

By the formula for the area of an equilateral triangle, the area of each of the two triangles making up the rhombus is $ \frac {50^2\sqrt{3}}{4}$, which is $ 625\sqrt{3}$. Multiply this by $ 2$ for each traingle of the rhombus and you get $ 1250\sqrt{3}$, or $ E$.[/hide]


induction
geometry
3D geometry
pyramid
rectangle

[quote="elcric"]Prove the Fibonacci Sequence:

$ (F_{1})^{2} \plus{} (F_{2})^{2} \plus{} (F_{3})^{2} \plus{} \cdots \plus{} (F_{n})^{2} \equal{} F_{n}F_{n \plus{} 1}$, where $ F_{1}, F_{2}, \ldots, F_{n}$ are the elements of the Fibonacci sequence.

Thanks in advance![/quote]
[hide]We can prove this with induction, add $ F_{n\plus{}1}^{2}$ to both sides and the RHS becomes $ F_{n\plus{}1}(F_{n}\plus{}F_{n\plus{}1})\equal{}F_{n\plus{}1}F_{n\plus{}2}$, completing the induction.[/hide]

[quote="diophantient"][quote="elcric"]Prove the Fibonacci Sequence:

$ (F_{1})^{2} \plus{} (F_{2})^{2} \plus{} (F_{3})^{2} \plus{} \cdots \plus{} (F_{n})^{2} \equal{} F_{n}F_{n \plus{} 1}$, where $ F_{1}, F_{2}, \ldots, F_{n}$ are the elements of the Fibonacci sequence.

Thanks in advance![/quote]
[hide]We can prove this with induction, add $ F_{n \plus{} 1}^{2}$ to both sides and the RHS becomes $ F_{n \plus{} 1}(F_{n} \plus{} F_{n \plus{} 1}) \equal{} F_{n \plus{} 1}F_{n \plus{} 2}$, completing the induction.[/hide][/quote]

What does $ F_{n \plus{} 1}^{2}$ mean?  Thanks!

[color=green]Stop the pyramid quoting madness![/color]

$ (F_{n \plus{} 1})^{2}$

[hide="Combinatorial"]We will count the number of ways to tile a $ 1 \times n$ and $ 1 \times (n\plus{}1)$ rectangle with squares and dominoes. How many ways are there to do this?

Answer 1: $ F_n F_{n\plus{}1}$

Answer 2: Casework. Suppose the last tile of the $ n\plus{}1$ rectangle is a square. Then there are $ (F_n)^2$ ways to tile the remaining spaces. Suppose the last tile of the $ n\plus{}1$ rectangle is a domino and the last tile of the $ n$ rectangle is a square. Then there are $ (F_{n\minus{}1})^2$ ways to tile the remaining spaces. Suppose the last two tiles of the $ n\plus{}1$ rectangle are a square then domino, and the last tile of the $ n$ rectangle is a domino... and so on. This covers all possible cases and gets the answer $ (F_1)^2 \plus{} (F_2)^2 \plus{} (F_3)^2 \plus{} \ldots \plus{} (F_n)^2$.

Both answers are the same so $ (F_1)^2 \plus{} (F_2)^2 \plus{} (F_3)^2 \plus{} \ldots \plus{} (F_n)^2 \equal{} F_n F_{n\plus{}1}$ Done.[/hide]

[color=green]Stop the pyramid quoting madness![/color]

I still don't get it.  Could you explain a bit more, please?

@ MellowMelon - I'm trying to use induction.

Thanks!

[color=green]Stop the pyramid quoting madness![/color]
[hide="written in a more inductive manner"]
Ok, it's easy to prove the base case, n=1.
Now consider n=k.
We're given that $ (F_{1})^{2} \plus{} ... \plus{} (F_{k})^{2} \equal{} F_{k}F_{k \plus{} 1}$ and want to show that $ (F_{1})^{2} \plus{} ... \plus{} (F_{k \plus{} 1})^{2} \equal{} F_{k \plus{} 1}F_{k \plus{} 2}$
Add $ (F_{k \plus{} 1})^{2}$ to both sides of our equation to get
$ (F_{1})^{2} \plus{} ... \plus{} (F_{k \plus{} 1})^{2} \equal{} F_{k}F_{k \plus{} 1} \plus{} (F_{k \plus{} 1})^{2} \equal{} F_{k \plus{} 1}(F_{k} \plus{} F_{k \plus{} 1}) \equal{} F_{k \plus{} 1}F_{k \plus{} 2}$[/hide]

[color=green]Stop the pyramid quoting madness![/color]

How is $ F_{k \plus{} 1}(F_{k} \plus{} F_{k \plus{} 1}) \equal{} F_{k \plus{} 1}F_{k \plus{} 2}$?

Thanks.

[quote="elcric"]

How is $ F_{k \plus{} 1}(F_{k} \plus{} F_{k \plus{} 1}) \equal{} F_{k \plus{} 1}F_{k \plus{} 2}$?

Thanks.[/quote]

Fibonacci recursion, applied to $ (F_{k} \plus{} F_{k \plus{} 1})$


$f(xy)=f(x)+f(y)$ for all $x>0$ and $y>0$ and $f(x)$ is continuous.
Solve for $f(x)$

I know you're suppose to use logarithm on this one but how?

Put $ x= e^{u} , y = e^{v} $
Thus we get, $ f( e^{u+v} ) = f(e^u) + f (e^v) $
Letting, $ g(u) = f(e^u) $
we get,$ g(x+y) = g(x) + g(y) $ which can be solved easily!!!

I think $f(x)=\log{x}$ was just an example of a function that satisifes this identity. There are probably some others too, such as $f(x)=0$

Actually the general solution is $ f(x0 = c \log{x} $ where c is arbitrary and thus includes all possible solutions


Given trapezoid $ ABCD$ with $ AB\parallel{}CD, CD>AB$. Let the diagonals intersect at $ O$. Points $ P,Q$ are on sides $ AD,BC$ such that $ \angle APB\equal{}\angle CPD, \angle AQB\equal{}\angle CQD$. Prove that $ OP\equal{}OQ$.

Let the bisector of angle BPC meet BC at T , the bisector of angle AQD meet AD at S
Then PT ⊥ AD  SQ ⊥ BC
We have  CT/TB=PC/PB=CD/AB   and  DS/SA=DQ/QA=CD/AB

=>   CT/TB= DS/SA    =>  ST ∥ AB ∥ CD

DS/SA= CD/AB=OC/OA   => SO ∥ CD ∥ AB   => S,O,T are Collinear

OP=OQ=1/2ST


MATHCOUNTS
pigeonhole principle
Alcumus
FTW

While googling away to find free mathcounts tests, I found the 2003 State Sprint Round, which was ridiculously easy.  When I tried it, I was able to get 28/30, even though I typically only get in the 18-21 range on State Sprints(on a good day).  I couldn't find anything with google, so I wanted to know, were there tons of ties and super high scores for that year that were higher than usual or am I just hallucinating?   Take one of the last problems for instance:  Which number, when removed from the set
{1,2,3,4,5,6,7,8,9,10,11} increases the chance that when two numbers are selected, the sum is 12?  You just have to pair off the numbers to get 6.  You'd never see this now.

Just wondering.  Also, does Mathcounts test the pigeonhole principle on its test?  And if so, could you give me an example question?

Er, this problem may seem easy for you, maybe not for others, such as me.
When i first tried on Alcumus, i got the wrong answer. Then after 1/2 hour i figured it out.
Hey, no laughing, i was a my big drop. (on FTW)

Yes, but as a whole, the test was easier than usual.  At least in my opinion.

No one has told me the results of the 2003 State Sprint Round for a certain state(say, Indiana) to see the top score.  Ditto for the pigeonhole principle question.

Say there are _ items (in this case large,medium,and small)
The question says how many must you draw to insure you have n of itemZ
The answer is simply:
$ itemA \plus{} itemB \plus{}...\plus{} itemY \plus{} n$
So the answer is
$ 1337 \plus{} 1337 \plus{} n$
And in this case n is 2 so
$ \boxed{2676}$
Probably a very bad explanation srry...

[quote="Mewto55555"]Pigeonhole Example:
You have different sized balls. You have 1337 of each large,medium,and small. How many must you draw from a bag to ensure you have two large balls?[/quote]

Was this ever on a real MC Test?  And if so, what year and what kind?  Also, does anybody have a link to results to the 2003 competition, or did anybody who participated in the 2003 State Competition have any thoughts?

Yes, you're hallucinating.

I was 6 in '03, but I've taken most of the state tests for practice. It was just an easy year, period. MATHCOUNTS tests generally alternate in terms of difficulty, so you'll probably find '05 harder but, for example, '06 is easy. In fact, most if not all pre-2005 tests are a lot easier than what you'll find today. I generally get 43+ on the pre-2005s (higher when we hit the 90s :roll: but some of those were really, really easy) and 40-43 on post-2005s.

On the other hand, 5 out of 6 people rated the 2003 State Sprint Round "extremely easy" and the other one rated it "easy".  Maybe I liked it better because it had less of those annoying problems that take forever and more problems you can solve with easy tricks.


ABCD is a square.The diagonals AC and BD intersect at the point O.The bisector of  [geogebra]7281cb91771398811e04ac08104239fd06da2a4c[/geogebra] $ \angle$  BAC cuts BO at P and BC at Q.Prove that OP=$ \frac {1}{2}$ CQ.

This is what I did.I proved that the bisector of an angle in a triangle divides the opposite side in the ratio of other two sides.

 [geogebra]f01f11d026e419fd68c6685bb532dd4fc22ea075[/geogebra]

Let ABC be a $ \triangle$,where AD bisects $ \angle BAC$.

Reqired to prove : $ \frac {BD}{BC}$= $ \frac {AB}{AC}$.

Construction: CE||AD is drawn to meet BA produced at E.

Proof:

In $ \triangle$s ABD and BCE,

$ \angle BAD$=$ \angle BEC$,
$ \angle BDA$=$ \angle BCE$.
So, $ \triangle ABD$ $ \sim$  $ \triangle BCE$.

So, $ \frac {BD}{BC}$=$ \frac {BA}{BE}$

or $ \frac {BC}{BD}$=$ \frac {BE}{BA}$

or $ \frac {BC}{BD}$ - 1=$ \frac {BE}{BA}$ - 1

or $ \frac {DC}{BD}$=$ \frac {AE}{BA}$

or $ \frac {BD}{DC}$=$ \frac {BA}{AE}$

Again, $ \angle BAD$=$ \angle BEC$, $ \angle BAD$=$ \angle DAC$=$ \angle ACE$

i.e. $ \angle AEC$=$ \angle ACE$

i.e. AC=CE

So, $ \frac {BD}{DC}$=$ \frac {BA}{AC}$  [thus proved]

Coming to the main sum, AL=LO is cut off from AB.

In $ \triangle ALP$ and  $ \triangle APO$,

AP is common,$ \angle LAP$=$ \angle PAO$,AL=AO.

So, $ \triangle ALP$ $ \cong$ $ \triangle AOP$.

LP=PO

$ \frac {LP}{BQ}$=$ \frac{1}{\sqrt 2}$

BQ=$ \frac {QC}{\sqrt 2}$

This gives 2OP=QC

So, OP=QC/2

Can you please check proof and provide an alternative method..
Thank you.

since $ OP$ is not parallel to $ QC$
constructing $ OM \parallel AQ$ with $ M \in AQ$
$ M$ is mp. of $ AQ$
the proof is complete if we prove $ OM\equal{}OP$

$ \angle OMP\equal{} \angle DAM \equal{} \frac{45}{2} \plus{} 45$

$ \angle OPM \equal{} 180\minus{} \angle PBQ\minus{} \angle AQB$
           $ \equal{} 180\minus{}45\minus{}90\plus{}\frac{45}{2}$
           $ \equal{} 45\plus{}\frac{45}{2}$
since $ \angle OMP\equal{}\angle OPM$
$ OM\equal{}OP$
as, $ OM\equal{}\frac{CQ}{2}$
     $ OP\equal{}\frac{CQ}{2}$

That's an excellant solution.But, probably OM||AQ is not valid or required.

Thanks a lot again.

By [url=http://en.wikipedia.org/wiki/Menelaus%27_theorem]Menelaus[/url],
$ \frac {AO}{AC} \cdot \frac {CQ}{QB} \cdot \frac {BP}{OP} \equal{} 1$.
We know $ \frac {AO}{AC} \equal{} \frac {1}{2}$, so it remains to show $ BP \equal{} QB$, or $ \angle BPQ \equal{} \angle BQP$. But
$ \angle BQP \equal{} 90 \minus{} \angle BAQ \equal{} 90 \minus{} \frac {45}{2} \equal{} \frac {135}{2}$ and
$ \angle BPQ \equal{} \angle BAP \plus{} \angle ABP \equal{} \frac {45}{2} \plus{} 45 \equal{} \frac {135}{2}$

[i]That's an excellant solution.But, probably OM||AQ is not valid or required.
Thanks a lot again
[/quote][/i]

IF we choose M as mp. then OM \parallel AQ. as O is mp. of AC.

I do not understand.If I am not wrong,M cannot  not lie on AQ if OM||AQ.Similarly, M cannot be the mid-point of AQ and still OM||AQ.
Please help! :maybe:

He means $ OM \parallel BC$. His proof is very similar to mine.

M is the midpoint of AQ.
now in triangle AQC we have midpoints of two sides AQ and AC i.e. M and O.
OM parallel to QC or BC or AD

This problem can be brute-forced easily as follows:
scale the square to side length 1; $ QC \equal{} x,PO \equal{} y$
apply angle bisector thm we get

$ \sqrt {2}/x \equal{} 1/(1 \minus{} x)$

$ \sqrt {2}/(2y) \equal{} 1/(\sqrt {2}/2 \minus{} y)$

solving yields $ y \equal{} 1/(2 \plus{} \sqrt {2}),x \equal{} 2/(2 \plus{} \sqrt {2})$

Dear Mathlinkers,

If we consider the outer square BUVC of ABCD and the point R intersection of AQ and CU, we have a better outlook of the problem.
We have :	2.OP = CR.
Then DAP being D-isoceles by an angle chasing, CQR is C-isoceles by parallelism
and we are done…

Sincerely
Jean-Louis


For a long time I have been dabbling around with this one. It looks like a very natural generalization of Popoviciu, but I have no clue how to tackle the general case. (Darij, do you?) So I post it here :)

Let $ f$ be a convex function on an interval $ I$ and take $ n$ variables $ x_1,\ldots,x_n\in I$.
For $ k \equal{} 1,\ldots,n$ define the mean $ s_k$ by
\[ s_k \equal{} {n\choose k} ^{ \minus{} 1}\sum_{i_1 < \cdots < i_k}f\left(\frac {x_{i_1} \plus{} \cdots \plus{} x_{i_k}}{k}\right).
\]
Popoviciu states $ s_1 \minus{} 2s_2 \plus{} s_3\ge0$ for $ n \equal{} 3$ (and the same follows easily for all $ n > 3$, too).

I suggest the following generalization:

[b]Conjecture. [/b] Let $ 1\le m\le n$. Then $ \sum_{i \equal{} 0}^m( \minus{} 1)^i\binom{m}{i}s_{i \plus{} 1}\ge0$. ;)

Easy to see that it is sufficient to prove the case $ m \equal{} n$. :lol:

It is well known that the function $ i\mapsto s_i$ is not convex. (See
http://www.mathlinks.ro/Forum/viewtopic.php?t=98832 and
http://www.mathlinks.ro/Forum/viewtopic.php?p=293622.)
But this conjecture would mean a sort of convexity in a weaker sense ‘on the left end’ of the $ s_i$.  :)

[hide="Motivating remark"] By convexity, $ s_1\ge s_2\ge s_3$. Moreover, Popoviciu $ \Leftrightarrow s_1 \minus{} s_2\ge s_2 \minus{} s_3$.
Next step: As mentioned in the 2nd link above, $ s_2 \minus{} 2s_3 \plus{} s_4\ge0$ is wrong in general.  So it is reasonable to conjecture
$ s_1 \minus{} 2s_2 \plus{} s_3\ge s_2 \minus{} 2s_3 \plus{} s_4$, which is equivalent to the $ m \equal{} 4$ case
$ s_1 \minus{} 3s_2 \plus{} 3s_3 \minus{} s_4\ge0$.[/hide]

[hide=" An example for n=4"] Putting $ f(x) \equal{} e^{12x}, a_i \equal{} e^{x_i}$ yields

$ [12,0,0,0] \minus{} 3\cdot[6,6,0,0] \plus{} 3\cdot [4,4,4,0] \minus{} [ 3, 3, 3, 3]\ge0$ with variables $ a_1,\dots,a_4$, where [...] is the usual symmetric sum notation; equivalently

$ \sum a^{12} \plus{} 3\sum a^4b^4c^4\ge 2 \sum_{sym} a^6b^6 \plus{} 4a^3b^3c^3d^3$.
Note that equality occurs also for $ (1,1, 1, 0)$ and $ (1,1, 0,0)$ and permutations. Similarly for bigger $ n$. (Is there a nice proof for that one, like e.g. a not-too-ugly SOS?) [/hide]

I am sorry to disappoint you... but am I right that for $ n\equal{}4$ and $ f\left(x\right)\equal{}\left|x\right|$, this would yield

$ \left|a\right|\plus{}\left|b\right|\plus{}\left|c\right|\plus{}\left|d\right|$
$ \minus{}\left|a\plus{}b\right|\minus{}\left|a\plus{}c\right|\minus{}\left|a\plus{}d\right|\minus{}\left|b\plus{}c\right|\minus{}\left|b\plus{}d\right|\minus{}\left|c\plus{}d\right|$
$ \plus{}\left|b\plus{}c\plus{}d\right|\plus{}\left|a\plus{}c\plus{}d\right|\plus{}\left|a\plus{}b\plus{}d\right|\plus{}\left|a\plus{}b\plus{}c\right|$
$ \minus{}\left|a\plus{}b\plus{}c\plus{}d\right|\geq 0$ ?

Because this is wrong for $ a\equal{}\minus{}10$, $ b\equal{}c\equal{}1$, $ d\equal{}9$.

The behaviour of Popoviciu-type inequalities seems to be more erratic than we think...

  darij


Find all continuous function f:[0,1] --> [0,1] satisfying

f(x) =< x^2 . f( (x^3+2)/3  ) for all x in the segment [0,1]

Let $ x_{n \plus{} 1} \equal{} \frac {x_n^3 \plus{} 2}{3} \equal{} x_n \plus{} \frac {(x_n \minus{} 1)^2(x_n \plus{} 2)}{3}$, and $ y_n \equal{} 1 \minus{} x_n, y_{n \plus{} 1} \equal{} y_n \minus{} \frac {y_n^2(3 \minus{} y_n)}{3}$ and  let $ g(y) \equal{} f(1 \minus{} x)$.
Then $ g(y_{n \plus{} 1})(1 \minus{} y_n)^2 \equal{} g(y_n)$.
It give
\[ f(1) \equal{} g(0) \equal{} \lim_{n\to \infty} g(y_1)\prod_{i \equal{} 1}^n\frac {1}{(1 \minus{} y_i)^2}.
\]
Because $ y_n \equal{} \frac 1n \plus{} O(\frac {ln n}{n^2})$ we get
\[ \prod_i\frac {1}{(1 \minus{} y_i)^2} \to \infty.
\]
Therefore $ g(y_1) \equal{} 0$ or $ f(x)\equiv 0$. I use only  f bounded  near 1.

Why yn= 1/n + O(lnn/n^2) ??? I don't think u could show it easily, i didn't use sequence to solve my own problem

As f is continuous, there exists F such that  F'(x)=f(x) for all x in [0,1]

Consider G(x) = F((x^3+2)/3) - F(x), we have G'(x) >=0 for all x

Hence  0=G(1)>= G(x) >= F(2/3)- F(0) >=0  ( as F is a increasing function )

So we must have F(x)= F((x^3+2)/3) for all x. That yeilds F=const

--> f=0

Many times considered sequences $ y_{n\plus{}1}\equal{}f(y_n)$, were $ f(y)\equal{}y\minus{}a_2y^2\plus{}a_3y^3\plus{}...$
In these case consider $ z_n\equal{}\frac{1}{y_n}\minus{}n$. It give \[ z_{n\plus{}1}\equal{}z_n\plus{}\frac{2z_n\minus{}3}{3(n\plus{}z_n)^2\minus{}3(n\plus{}z_n)\plus{}1}\equal{}z_n\plus{}O(\frac{1}{n^2}.\]


Rwriting as $(p-q) \cdot (p-q-1) \cdot (p-q+1) = 2 \cdot q$

But $2 \cdot q$ has only $2$ prime divisor and $p$ and $q$ are both odd.

Only solution : $5,3$

Yes, that was truely an easy, but interesting problem.
Another solution is:
Let $p-q=n$ then $n^3=p+q$ , $2q=n^3-n=(n-1)n(n+1)$ , but
one of the numbers $(n-1), n, (n+1)$ is devisible by 3. Then $q=3$
$p=5$ is the only solution.

Here is another prolem which i want to post in this topic:
$p,q$-primes such that $q|p^2+1, p|q^2-1$. Prove that $p+q+1$ is not prime.

$p$ divides either $q+1$ or $q-1$.  If $p|q+1$, then $p + q + 1 = p + kp$ for some integer $k$, which is composite.

If $p|q-1$, then $p \equiv -1 \bmod q$ and $p^2 \equiv 1 \bmod q$.  Since $q|p^2+1$, we can write $p^2 \equiv -1 \bmod q$, which implies $q = 2, p = 1$, which contradicts the assumption that $p$ was prime, so $p + q + 1$ cannot be prime in any case.


Solve for all possible values of $x$:

$|x-2| + |x-5| = 4$

Look at the intervals.
First less than 2.
plug in 2, you get 3
plug in 1 you get 5,
so 1.5 works

2<x<5
all are 3

x>5
plug in 5, you get 3
plug in 6 you get 5
so 5.5 works

x=1.5,5.5


How many non-similar triangles can be drawn, such that their vertices are vertices of a regular 21-gon?

[hide]Draw the circumcircle of this $ 21$-gon, so that a unique triangle (that is not similar to any other triangle*) is determined for every solution of $ a\plus{}b\plus{}c\equal{}21$ for $ a\geq b\geq c\geq1$.

This is very similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245358]this problem[/url], except the equilateral triangle is possible.

I won't explain here, using a similar technique as in the aforementioned link, we see that there is $ 1$ equilateral triangle, $ 10\minus{}1\equal{}9$ isosceles triangles that are not equilateral (hence the $ \minus{}1$), and $ \frac{190\minus{}(1\cdot1\plus{}9\cdot3)}6\equal{}27$ scalene triangles, for a total of $ 1\plus{}9\plus{}27\equal{}\boxed{37}$ non-similar triangles.

*Two triangles are similar if and only if their respective angles are equal. In particular, the arcs that they subtend must be equal.[/hide]

I don't understand when the solution for the problem says

"We can count the number of isosceles triangles; the duplicated side length can be between 1 and 10 inclusive, except that if it is 7 then the triangle is actually equilateral. So there is 1 equilateral triangle, and 9 other isosceles triangles. "

How did they know that the duplicated side length can be between 1 and 10 inclusive, except that if it is 7 then the triangle is actually equilateral?


We can consider the problem about solving in positive integers the equation \[\left({\begin{array}{*{20}c}x+1 \\ y \\ \end{array}}\right)=\left({\begin{array}{*{20}c}x \\ y+1 \\ \end{array}}\right).\] Now I dont have enought time but I will post later that all the solutions of this equation are $x=F_{2k}F_{2k+1}-1$ and $y=F_{2k}F_{2k-1}-1.$


quadratics
analytic geometry
calculus
geometry
geometric transformation
reflection
Putnam

Let ABC be a right triangle(angleB=90) with AB=15. Let CED be a right triangle with angle E=90 and length of DE=10. Given B,C,E are collinear and BC+CE=BE=30. Find BC when length of AC+CD is minimized.

[hide]you know that if E is on line BC, then BC=15. If C was on line BE, then BE=15. If we made BC 15, then you would have a triangle with legs 15 and 15. That would make the hypotenuse 15 :rt2: . To minimize CD, CE would be very small, making CD a little more than 10. But if BE was 15 and C was between B and E, then you can set up an equation for AC and CD. AC :^2: =225+x :^2:  and CD :^2: =100+(15-x) :^2:

Expand out to CD :^2: =100+225-30x+x :^2:

Since you want to find the minimum value of AC+CD, you can combine these equations.

AD+CD= :sqrt: (225+x :^2: +325-30x+x :^2: )

that becomes  :sqrt: (2x :^2: -30x+550)

Since that's a quadratic equation, you can find the vertex since the x coordinate is -b/2a. This time, that's 7.5. Plug 7.5 in for x and get 437.5. Square root to get 20.9165. Now figure out 15 :rt2: to get 21.21. The first is larger, so that means BC=7.5[/hide]

C is between B,E for sure.
Several mistakes
BE=30, not 15

AC+CD does not equal to  :sqrt: (AC^2+CD^2)

is it [hide]18[/hide]? just a guess without doing anything

You don't need calculus or anything like that.

All you need is [hide]creativity[/hide]!!!!!!!!!

The key idea of the problem takes some of the simplest concept of geometry. That's the art of problem solving.

Hints:

1. [hide]What minimises the sum of lengths of two segments? Think geometric placement.[/hide]

2. [hide]You need to construct two lines.[/hide]

3. (practically giving it away) [hide]Use reflection.[/hide]

[hide]You reflect one of the "posts" across the "ground" and connect the "post_1" to "reflected post_2", then similar triangles.[/hide]

I forgot the letters in the picture... sorry.

[quote="gauss202"][quote="polymorphic"]true, but calculus is a nobrainer...[/quote]

I think this is what everyone means by people who don't belong on the GETTING STARTED board.[/quote]

Yes, that's right. There is a reason why mathematical problem-solving competitions all the way up to the International Mathematical Olympiad specify that problems can be solved by precalculus methods: the precalculus methods can be much more elegant and insightful.

As one of the three moderators here, I would request that participants on this forum NOT post calculus-based solutions to problems posted here (unless as an ADDITIONAL solution, and of course in spoiler). Putnam problems (which properly belong on the Advanced forum here) frequently involve tricky, challenging uses of calculus, so there is still a place for calculus problems on this site.

Use your brain to exercise the tools of noncalculus mathematics to their full power.

[quote="tokenadult"]

There is a reason why mathematical problem-solving competitions all the way up to the International Mathematical Olympiad specify that problems can be solved by precalculus methods: the precalculus methods can be much more elegant and insightful.[/quote]

No I disagree with that. Sometimes calculus methods can be more elegant or insightful - it all depends on the problem. But for this problem calculus is way overkill, and it obscures the lesson to be learned from the problem.

Also, you just has to know who their audience is when you're answering a question. No one should expect people in the Getting Started forum to be familiar with calculus.

[quote="tokenadult"]the precalculus methods can be much more elegant and insightful.[/quote]
[quote="gauss202"]Sometimes calculus methods can be more elegant or insightful[/quote]

I don't think you two are really disagreeing -- it is true that the vast majority of problems that appear on contests lack an elegant calculus solution, while there are many problems in other areas that have beautiful calculus solutions but only ugly non-calculus solutions.

Either way, calculus has no business in this forum.


[hide="Answer without Solution"]\[\displaystyle \frac {\frac {\frac {\frac {1}{2}}{3}}{4}}{\frac {5}{\frac {4}{\frac {3}{\frac {2}{1}}}}}=\frac {1}{45}\][/hide]

[hide="solution"]
*points to RC-7th's post*
1 + 45 = [b]46[/b]
:P
[/hide]

[quote="Silverfalcon"]\[\frac {\frac {\frac {\frac {1}{2}}{3}}{4}}{\frac {5}{\frac {4}{\frac {3}{\frac {2}{1}}}}}\]

This can be equal to $\frac {m}{n}$ where m and n are in lowest terms. Find m+n.[/quote]

[hide="my answer"]$\[\frac {\frac {\frac {\frac {1}{2}}{3}}{4}}{\frac {5}{\frac {4}{\frac {3}{\frac {2}{1}}}}}\]=\frac{1}{45}$

$1+45=46$[/hide]

[color=indigo]How do you get the answer? I got 25.875.[/color]   :blush:

[quote="jli"][hide="my answer"]\[\displaystyle \displaystyle \frac {\frac {\frac {\frac {1}{2}}{3}}{4}}{\frac {5}{\frac {4}{\frac {3}{\frac {2}{1}}}}}=\frac {1}{45}\]

$1+45=46$[/hide][/quote]

I don't think you meant to have the extra stuff.

@LadyKn1ght-Silverfalcon implied that m and n are integers, even if he didn't say so. Try changing your fraction so that m and n are integers.

[hide][color=indigo]I got it now. I had did the top half wrong and got 24/(15/8) instead of (1/24)/(15/8) which works out to be 1/45. The answer is 46.  :lol: [/color][/hide]

The way you do something like this is $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$ You just figure out where the middle is.

[quote="LuCy4EvA"]for the bottom half, do u work from your way up or down, like do u do 5/4 first or 2/1?[/quote]

The smallest fraction-bar is like an implied set of parentheses around that fraction.  Then the next-smallest fraction-bar is like another implied set of parentheses around that fraction.  As with ordinary operations, you work your way from the innermost parentheses to the outermost.

If you look at the TeX source for the fraction in the original problem, you will see how the \frac operator was nested.

To answer your question directly, the first operation to be performed in the denominator is $\frac{2}{1}.$


MATHCOUNTS
geometry
3D geometry
geometric transformation
reflection
function
binomial coefficients

The results for the Mock Mathcounts competition (Sprint and Target) are posted [url=http://mersenne.zapto.org/Submission]here[/url]!

Details regarding the countdown round will be posted shortly.

Could someone post a solution for #5 on target? I keep getting 30^2=900.

And also, in the correct answers page, your answer to #5 sprint (question about number of palindromic perfect cubes) is wrong (it should be 3 (or 4 if you included 1))

Dang, I'm facing Lazarus.
I got 900 for that too.
EDIT:Aww, I did reflect first, but messed up in calculations, so I just put 900.

[quote="andersonw"]And also, in the correct answers page, your answer to #5 sprint (question about number of palindromic perfect cubes) is wrong (it should be 3 (or 4 if you included 1))[/quote]

Oops that was a typo :blush:

[quote="unimpossible"][quote="andersonw"]Could someone post a solution for #5 on target? I keep getting 30^2=900.

reflect the point across both the x and y axises and draw the line.

And also, in the correct answers page, your answer to #5 sprint (question about number of palindromic perfect cubes) is wrong (it should be 3 (or 4 if you included 1))[/quote][/quote]

huh didn't mean to quote that.

you reflect the point across the x and y axises and draw the line

[hide="Solution for Target 5"]Note that if you reflect $ (5,12)$ across the origin, it will become $ (\minus{}5,\minus{}12)$. The distance from $ (8, 15)$ to $ (\minus{}5,\minus{}12)$ is the same as the distance that the problem asks for, because either way, the ant would pass through both axes. The shortest distance from $ (8, 15)$ to $ (\minus{}5, \minus{}12)$ would be $ \sqrt{(\minus{}13)^2\plus{}(\minus{}27)^2}$  by the distance formula. The answer is therefore $ \boxed{898}$.[/hide]

Actually, your bracket is seeded wrong.  It should be:
1-8 4-5 2-7 3-6 (I think...)

Anyway, cognos didn't make it so I have no reason to watch the countdown :( .

Can I see the results even though I didn't take the test?
What's the authentication code?

You don't need the authentication code.  Just click on the authentication link, and it will log you in if you are already logged into AoPS.
And you don't need to have taken the test.

MustIt speaks truth. THe authentication code part was made for those who are unable to use the auto-auth function, because of browser, crossdomain issues etc.

nooo! I'm 1 problem short to the CD round. Ah well, I still did much better than I expected. There was one question on the target round that I guessed on and got it right. And nice test by the way.

EDIT: I forgot to mention if we could get the answers that we put in. I think it would be helpful for us to check our mistakes and how we did it wrong. Thanks!

I don't think I have the answers you put in :(

Holy crap, there is no way I got 8 on the sprint and 0 on Target.

o.o"

I know I got 898.

Plus I checked my answers for sprint and I got the entire first page right (so that is greater than 8).

@catalyst: Yeah, I went thorugh the MC Teacher's guide handbook thing to go through the tiebreaker process.

@Lawrence Wu: I'll double check

[quote="CircleSquared"][quote="Lazarus"]hmm for #1 on the target round i wrote something along the lines of "too lazy"

i believe that should be correct ;] .

grats on everyone else who made the top eight, lol

and i'm planning on writing a mock aime. or something.[/quote]

"too lazy" i believe the correct answer was 1, as far as I know the two equations don't share a single root.

Congratulations on top mock scorer and perfect sprint, by the way.[/quote]
yeah i could only find one common zero when k was negative 1988+8891, but the root was positive so i put it as my answer.
lol when i finished #1 Target I was like "WHAT?! that's such an easily guessable answer." so i thought i must have missed it

[quote="returnofkiladi"][quote="#H34N1"][quote="returnofkiladi"]Please stop asking silly questions.
I need help on Sprint Problem# 11.
Plz help me.[/quote]

Its a basic counting identity that $ {n\choose 0} \plus{} {n\choose 1} \plus{} \dots {n\choose n} \equal{} 2^n$. Since $ {n\choose 0}$ was not included, the answer is $ 2^n \minus{} {n\choose 0} \equal{} \boxed{2^n \minus{} 1}$.

@moogra: The excel file is [url=http://mersenne.zapto.org/Submission/Scores.xls]here[/url].[/quote]

But I don't understand wat the
(n)
(0)
thing means...
How does that equal to 1.
I AM CONFUSED![/quote]
there is one way to choose 0 items...

[quote="SeventhSecret"][quote="Ihatepie"]how do I find out my results on the test?[/quote]
Right [url=http://mersenne.zapto.org/Submission/Scores.xls]here[/url][/quote]

Yeah I had trouble finding the stuff too.

[quote="CatalystOfNostalgia"]Also, Lazarus is in 10th grade, why is he in this at all?[/quote]

(5:20:33 PM) #H34N1: hey
(5:20:42 PM) Atheist Temple: sup
(5:20:44 PM) #H34N1: yo
(5:20:47 PM) #H34N1: i made a mock mathcounts
(5:20:48 PM) #H34N1: sprint round
(5:20:51 PM) #H34N1: do it

[quote="mathcrazed"]I'm still using my excuse that being on aim while taking a mathcounts test is not a good idea  :wink:   :ninja: . xD. yep. it's so awesome. Now i have to face bobby. this sux.[/quote]

Actually, I believe lazarus said he will miss it, so you'll face some other dude and I'll face *gulp* stevenmeow *shiver*

Lol would you rather face me... in a real countdown =)... andrenaline rushes are pretty amazing... hehehehe  :ninja:

I have a couple of questions.
For number 8 and 9, exactly how would you do them?

To #H34N1:

When I checked over my answers I saw that I got 8 wrong, although yours says that I got 10 wrong. When I checked it again, the two that I believe you miscorrected were #10 and #19.

On 10, I wrote 1.2, when the answer was 6/5, and on 19 I wrote 3.5, when the answer was 7/2. These are obviously the same answers.

About the target, could you guys go over the solutions for 1,2,5,6 and 7?

Thanks in advance.

I just did the Sprint and Target Rounds today.  Could someone please post the answer keys on the forum or PM them to me?  Thanks.

Can someone please show me how to do 14 and 18 please?

[quote="mathcrazed"]Target Round #1: $ 19988x^2 \plus{} kx \plus{} 8891$... Right? I think that's right but it doesn't matter...

and we have $ 8891x^2 \plus{} kx \plus{} 1988$... If they have the same roots then that means that for those two roots, $ 8891x^2 \plus{} kx \plus{} 1988 \equal{} 1988x^2 \plus{} kx \plus{} 8891\longrightarrow x \equal{} \pm1$

$ \boxed{1}$

Number 2 = too lazy to do -_-... But connect all the centers to get it... equilateral triangles and such

#3: Power of a point destroys this.

#4: Let $ a \equal{} x\sqrt {8}$.

Then we have that $ a \plus{} \frac {1}{a} \equal{} \sqrt {8}$
$ a^2 \minus{} \sqrt {8}a \plus{} 1 \equal{} 0$
$ a \equal{} \frac {\sqrt {8}\pm2}{2} \equal{} \sqrt {2}\pm1$

Divide by $ \sqrt {8}$, which I conveniently forgot to do -_-.
$ \frac {2 \plus{} \sqrt {2}}{4} \equal{} \frac {1}{2} \plus{} 4\sqrt {2}$.
Since one of them has the digit 6 in the 1994th place and the some digit later on must be greater than 0, then the other digit in the 1994th place must be $ 10 \minus{} 6 \minus{} 1 \equal{} 3$, because there will be carrying. xD

#5: It was on the forums before... use two reflections because you must touch two axes.

#6: Pattern  :wink:

#7: Trig. So I won't post.

#8: Let the number correct be $ c$ and the number wrong be $ w$.
Then $ c \plus{} w \equal{} 10$ and $ 5c \minus{} 2w \equal{} 29$
$ 2c \plus{} 2w \equal{} 20$

$ 7c \equal{} 49,c \equal{} 7$

SPRINT:

#8: Draw a tree diagram. Helps alot.

#9: AM-GM. Not mathcounts level... but it states $ \frac {a_1 \plus{} a_2 \plus{} \hdots \plus{} a_n}{n}\ge \sqrt [n]{a_1a_2\hdots a_n}$

arithmetic mean is always greater than or equal to the geometric mean, equality if all $ a_i$ are equal.

so... $ \frac {\frac {a}{2b} \plus{} \frac {b}{4c} \plus{} \frac {c}{8a}}{3}\ge \sqrt [3]{\frac {1}{64}}$
so... blahblahblah $ \ge \frac {3}{4}$
Minimum obviously occurs at $ \frac {3}{4}$[/quote]
ono on target #4 i forgot about carrying so i put 4


Could someone explain what is meant by homothecity? I really need to study it

[url]http://mathworld.wolfram.com/Homothecy.html[/url]

Look at that page, and have a look at the links.

I've bought myself Geometry Revisited. Is it enough really to make a beginning?

Most certainly! It's excellent.

You may want to look at Kedlaya's online geometry notes as well :)


Lat $ a,b,c>0$ such that $ abc\equal{}1$
  Prove that $ \sum\sqrt{a^2\plus{}3bc}\geq2\sqrt{3(a\plus{}b\plus{}c)}$

[quote="arqady"][quote="2424"]Lat $ a,b,c > 0$ such that $ abc \equal{} 1$
  Prove that $ \sum\sqrt {a^2 \plus{} 3bc}\geq2\sqrt {3(a \plus{} b \plus{} c)}$[/quote]
Because $ \sum_{cyc}\sqrt {a^2 \plus{} 3bc}\geq2\sqrt {3(a \plus{} b \plus{} c)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2 \plus{} 11ab \minus{} 12a) \plus{} 2\sum_{cyc}\left(\sqrt {(a^2 \plus{} 3bc)(b^2 \plus{} 3ac)} \minus{} 4ab\right)\geq0.$

It seems that the following inequality is also true.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that:
\[ \sum_{cyc}\sqrt {2a^2 \plus{} 9bc}\geq\sqrt {33(a \plus{} b \plus{} c)}\]
[/quote]
It's true by Am-Gm! :)


I'm sort of confused.

When sulfuric acid dissociates, doesn't it goes into H+ and HSO4-?

But in textbooks, the net ionic equations for reactions dealing with sulfuric acid just dissociates suluric acid into H+ and SO4-2. Why is that? And how do I know when to dissociate it into sulfate or bisulfate?

Which brings me to another question.

H+ + SO4-2 => H2SO4.

And my teacher says H+ + HPO4-2 => H2PO4- and not phosphuric acid. Why is that?

Thanks in adv. for the help!

When you have, say, a mol of sulfuric acid, not all of it will dissociate completely.  Some of it will remain $H_{2}SO_{4}$, some of it will dissociate once into $HSO_{4}^{-}$, and some of it will dissociate completley into $SO_{4}^{2-}$.

Experimentally, a mol of sulfuric acid dissolves into approximately 2 mols of ions, so (at least the way I was taught) we write sulfuric acid as dissociating once.  (It's a strong acid only for its first hydrogen ion!)

Your textbook is probably investigating a special case.  (In a reaction with another compound, there may be other factors contributing to $H_{2}SO_{4}$ dissociating completely.)

Your teacher is also correct about phosphoric acid; it's a strong acid, but only for its first hydrogen, so it tends (without outside forces interfering) to stay as $H_{2}PO_{4}^{-}$ in solution.

The real answer to your question (Chinaboy) is, it depends.  Your teacher is right that [i]sometimes [/i]the major reaction of $HPO_{4}^{-2}$ with acid will be $H^{+}+HPO_{4}^{-2}\rightarrow H_{2}PO_{4}^{-}$.  However, sometimes the major reaction will be $2H^{+}+HPO_{4}^{-2}\rightarrow H_{3}PO_{4}$.  What it really depends on is how much $H^{+}$ is present (that is, the concentration of hydrogen ions).

Naturally, as more $H^{+}$ is present, the $HPO_{4}^{-2}$ will tend to soak up more and more protons--if you've already covered it, that's basically Le Chatlier's Principle at work.  This should make sense: the more $H^{+}$'s that are available, the easier they become to pick up.  To recap: in acidic solutions (or, low pH/high concentrations of hydrogen ion) things (i.e. bases) tend to pick up more hydrogens than they have in basic solutions.  You can do all sorts of calculations, of course, to find exactly in what pH range each reaction occurs.

However, the case changes slightly with sulfuric acid.  Sulfuric acid is a STRONG acid, which by definition means that it completely loses its first proton in water, no matter what.  So you will never write $HSO_{4}^{-}+H^{+}\rightarrow H_{2}SO_{4}$ regardless of how low the pH is.

As a final note, t0rajir0u makes a valid point in his first paragraph.  In actuality/theory, all of these reactions are happening.  For example, there is some $H_{2}SO_{4}$ in all states of deprotonation at any given time.  However, in general there is only one or two states that are present in non-negligible amounts, so when doing calculations we usually ignore all but two species (e.g. $HPO_{4}^{2-}$ and $H_{2}PO_{4}^{-}$ while ignoring $PO_{4}^{3-}$ and $H_{3}PO_{4}$).

Oh and t0rajir0u, keep in mind that phosphoric acid is NOT a strong acid.  I think it has a pKa around 2.

Hope that helps.

[quote="gregx007"]phosphoric acid is NOT a strong acid.  I think it has a pKa around 2.[/quote]

pKa1 = 2.148
pKa2 = 7.199
pKa3 = 12.35


Find $\sin(20^{\circ})\sin(40^{\circ})\sin(60^{\circ})\sin(80^{\circ})$.  Show your solution without using a calculator.

[quote="Altheman"]does that have a nice value? it does for cosine [hint for that problem: creative use of double angle formula/telescoping][/quote]

[hide="For sine"]$\frac{3}{16}$[/hide]

[hide="For cosine"]$\frac{1}{16}$[/hide]

Could you post a solution?  I know the value already because I can punch it into my calculator.

[hide="I will try to do it then"]Using double angle formulas like heck,

$\sin(20^{\circ})\cdot \sin(40^{\circ}) \cdot \sin(60^{\circ}) \cdot \sin(80^{\circ})$

$\frac{\sqrt{3}}{2}\cdot \sin(20^{\circ})\cdot \sin(40^{\circ}) \cdot \sin(80^{\circ})$

$\frac{\sqrt{3}}{2}\cdot \sin(20^{\circ}) \cdot (2\cdot \sin(20^{\circ})\cos(20^{\circ}))\cdot (2\sin(40^{\circ})\cos(40^{\circ}))$

$\frac{\sqrt{3}}{2}\cdot \sin(20^{\circ}) \cdot (2\cdot \sin(20^{\circ})\cos(20^{\circ}))\cdot (2\sin(20^{\circ})\cos(20^{\circ}))(1-2\sin^{2}(20^{\circ})))$

$\frac{\sqrt{3}}{2}\cdot(4\sin^{3}(20^{\circ}))\cdot \cos^{2}(20^{\circ})\cdot (1-2\sin^{2}(20^{\circ}))$

$\frac{\sqrt{3}}{2}\cdot(4\sin^{3}(20^{\circ}))\cdot (1-\sin^{2}(20^{\circ}))\cdot (1-2\sin^{2}(20^{\circ}))$

I'm not sure if this is going to work, but use the fact that $3\sin(20^{\circ})-4\sin^{3}(20^{\circ}) = \frac{\sqrt{3}}{2}$ to finish it off.

The algebra seems too tedious for me; there has to certainly be a better way.[/hide]

It's good to have the problem solved.  Complex numbers is something I did not seriously think of using.

By the way, the second line needs a 2 in that parenthesis...

Still, any other methods are welcome.


How would one go about calculating the sum of tan(x)+tan2(2x)+tan4(4x)+...+1024tan(1024x)?

Is that general term $2^{k}\tan (2^{k}x)$ or $\tan^{2^{k}}(2^{k}x)$?

The term (2^k)*tan((2^k)*x), i.e. 2tan(2x), 4tan(4x), etc., to 1024tan(1024x) (the first one you posted).


calculus
integration
trigonometry
quadratics
algebra
calculus computations

Let $f(x)=t\sin x+(1-t)\cos x\ (0\leqq t\leqq 1)$.

Find the maximum and minimum value of the following $P(t)$.

\[P(t)=\left\{\int_0^{\frac{\pi}{2}} e^x f(x) dx \right\}\left\{\int_0^{\frac{\pi}{2}} e^{-x} f(x)dx \right\}\]

[quote="Jimmy"]I got max = $\frac{1}{4}e^{\pi/2}$ and min = $\frac{\sinh\left(\frac{\pi}{2}\right)}{2}$[/quote]
solution please  any idea

[color=darkred]Let $f(x)=t\sin x+(1-t)\cos x$ , where $t\in I=[0,1]$ . Find $P(I)$ , where $P(t)=\left[\int_0^{\frac{\pi}{2}} e^x f(x)\ \mathrm{dx} \right]\cdot\left[\int_0^{\frac{\pi}{2}} e^{-x} f(x)\ \mathrm{dx}\right]$ [/color]

[color=darkblue][b][u]Proof[/u].[/b] $\left\{\begin{array}{cc}
A=\int_0^{\frac{\pi}{2}} e^x\sin x\ \mathrm{dx} & B=\int_0^{\frac{\pi}{2}} e^x\cos x\ \mathrm{dx}\\\\
C=\int_0^{\frac{\pi}{2}} e^{-x}\sin x\ \mathrm{dx} & B=\int_0^{\frac{\pi}{2}} e^{-x}\cos x\ \mathrm{dx}\end{array}\right|\implies$ $\left\{\begin{array}{cc}
A+B=e^{\frac {\pi}{2}} & A-B=1\\\\
C-D=e^{-\frac {\pi}{2}} & C+D=1\end{array}\right|\implies$

$P(t)=\left[tA+(1-t)B\right]\cdot\left[tC+(1-t)D\right]$ and $P'(t)=2(A-B)(C-D)t+(AD+BC-2BD)\implies$

$P'(t)=(1-2t)e^{-\frac {\pi}{2}}$ . Hence $P\left(\frac 12\right)=\frac 14\cdot (A+B)(C+D)=\frac 14\cdot e^{\frac {\pi}{2}}$ and $P(0)=P(1)=\frac 14\cdot\left(e^{\frac {\pi}{2}}-e^{-\frac {\pi}{2}}\right)$ . In conclusion,

\[\blacktriangleleft\ \boxed{P(I)=\left[\frac 12\cdot\sinh\left(\frac {\pi}{2}\right)\ ,\ \frac 14\cdot e^{\frac {\pi}{2}}\right]}\ \blacktriangleright\] . [/color]


if i enroll for a class, and then the the whole class ends. like for example. if i enroll in intermediate counting, and then the class ends some months later. can i still read the transcripts?

please answer this quickly, the class deadline is today. and im leaving in like 30 minutes to a party. i could push that back to an hour i guess.

oh yeah, how bout the question packets. will i have access to those after the class also?

ack, cmon people. i need to go get ready, and when im back, i spect an answer

According to the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]Enrollment page[/url], it says "Class transcripts will be available [b]during [/b]the course for student review.", so no for the first one, but "Students will [b]receive [/b]six problem sets..." so yes for the second. The challenge sets however "extend the lessons taught [b]in the classes [/b]by offering more advanced problems.", so I guess it goes under "transcripts", so no.

I may be wrong, but some is better than none...well, I guess if I'm wrong it's "negative" :)

even if they're not kept up you could still print them out and keep them, or save them on your computer.

That's true. I mean, you could even steal it from the server, but who'd want to do that? :)

Oh, and btw, you could just sign up for now and drop out if the material you desire is unavailable.

...so you're going to party all night? *imagines Neal breakdancing with lotsa glowsticks*... [url=http://smiles03uk.tripod.com//sitebuildercontent/sitebuilderpictures/glowsticks.jpg]:lol:[/url]  [url=http://www.almostcool.org/pg/00pages/053000.html]:lol:[/url]  [url=http://www.jivemagazine.com/2000/12-06-2000-TimoMaas-Riviera/b1/big/36-glowsticks.JPG]:lol:[/url]  :cry:  [url=http://www.ravergeek.com/release/lightshow_release.jpg]:lol:[/url]  [url=http://www.glowsticking.com]:lol:[/url]

lol, yeah tare

yeah, i forgot about the drop option. lol

thanks guys

Syntax:

Yes you will ba able to view transcripts and all problem and challenge sets for several weeks (perhaps months even) after the course has ended...

woot... I'm in that class also... cya there syntax.

The transcripts are kept up for a certain period of time.  We could always leave them up longer if it is necessary, but I personally suggest to all the students in my classes to print out the transcripts and problem sets and keep them in a binder.  That way you end up creating a sort of personal problem solving textbook that you can always refer back to.

Also, the odds are very high that any given student will not be able to solve 100% of the problems in a course.  We recognize what kind of student we are teaching and feel like we would be failing if we didn't provide material difficult enough to make you think harder and longer than you have been asked to think in regular school work.  This makes it a very good idea to keep the material for future reference.

Mr. Crawford, arent the transcripts too long for us to print?...i mean printing 12 sets of bout 25 pages is a lot of printing to do isnt it?...and thats only transcripts


It looks good, but what would really help is if you get a recommendation from one of your professors.

[quote="mathemonster"]From a college admission standpoint how impressive is it to complete abstract algebra at a reputable university during your junior year? also if next fall i am in an almost grad NT class will that increase my chances? ive already taken intro to NT and honors linear so this semester i am opting for abstract instead of doing analysis in single variable and abstract first semester next year.
is this wise?[/quote]

Don't choose your classes based on what will look good for college admission.  Take the subject that interests you most.  (I'd vote for abstract algebra, which was easily one of my favorite classes at Princeton.)

I'll agree with Richard, and add this: find out who is teaching the class, and what his/her reputation is for providing stimulus and enlightenment. Don't hesitate to take the "wrong" class from the "right" instructor.

And if down the road, some advisor at your eventual undergraduate college says, "But you haven't had [i]our[/i] abstract algebra" (or analysis, or whatever) and asks you to re-take the course, then no harm has been done. You'll just wind up with a deeper understanding.

Ok thank you im most interested in abstract so thats what i will take. also, i already have a prof who can write me a rec (and is for RSI). thanks


inequalities
number theory proposed
number theory

[quote=broniran][hide="hint"]$ (n^2 \minus{} 2n \plus{} 9)^2 > n^4 \minus{} 4n^3 \plus{} 22n^2 \minus{} 36n \plus{} 18 > (n^2 \minus{} 2n \plus{} 8)^2$ and now it's up to check for which $ n$ those inequalities are true and check for those which they're not and you'll find out that there doesn't exist integer $ n$ with the needed property. :wink:[/hide][/quote]

But $n=1$ works!

[quote=jmerry]Well, yes. People make mistakes. Care to complete the solution?[/quote]

Sure.

Let $n^4-4n^3+22n^2-36n+18=k^2$. So $(n^2 - 2n + 9)^2-k^2=63$.  Factor this $(n^2-2n+9-k)(n^2-2n+9+k)=63$. Now use factors of $63$ and do casework to complete.

Not sure what broniran was trying to do though, maybe someone else can explain his/her solution?

[quote=gethd][quote=Fang-jh]Find all positive integers $ n$ such that $ n^4\minus{}4n^3\plus{}22n^2\minus{}36n\plus{}18$ is a perfect square.[/quote]
You can solve this in integers. If $n\in(-\infty,-4]\cup[6,+\infty)$, then $(n^2 - 2n + 9)^2 > n^4 - 4n^3 + 22n^2 - 36n + 18 > (n^2 - 2n + 8)^2$ (the first inequality is trivial; the second becomes quadratic after a simplification), so no solutions. It's only left to check $n\in\{-3,\ldots,5\}$, which only gives $n=-1,1,3$.[/quote]

Indeed and we exclude $-1$ since they wanted positive integers :)

edit: oops completely skimmed over that part :P, thanks.

[quote=gethd][b]You can solve this in integers.[/b] [/quote]
...is what he emphasized.


geometry
rectangle
conics
ellipse
parallelogram
geometric transformation
reflection

If a rectangle is inscribed in an ellipse ,prove that its sides must be parallel to the axes of the ellipse.

First of all, it's clear, by regarding the ellipse as the projection of a circle, that any parallelogram inscribed in the ellipse has the center of the ellipse as its own center. This must then be true for our rectangle as well.

Let $ABCD$ be the rectangle (so $AC,BD$ are diameters of the ellipse). Let $EF$ be the reflection of $AC$ in one of the axes of the ellipse. Then $AECF$ is also a rectnagle, so all the points $A,B,C,D,E,F$ lie on a circle. Since the intersection between a circle and an ellipse can have at most $4$ points, it means that $E,F$ belong to $\{A,B,C,D\}$, so they must actually coincide with $B,D$.


integration
trigonometry
calculus
calculus computations

[quote="mdk"]Let a,b be positive constants. Find the volume (in the first octant) which lies above the region in the xy-plane bounded by x = 0, x = pi/2, y = 0, y*sqrt(b^2(cosx)^2+a^2 (sinx)^2)) = 1, and below the plane z = y.[/quote]

$ V=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{b^{2}\cos^{2}x+a^{2}\sin^{2}x }}}\int_{0}^{y}dxdydz=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}\frac{1}{b^{2}\cos^{2}x+a^{2}\sin^{2}x}dx$
$ \tan\frac{x}{2}=t$ we got $ V=\int_{0}^{1}\frac{1+t^{2}}{b^{2}(1-t^{2})^{2}+4a^{2}t^{2}}dt$...


Let $ n \geq 3$ and $ A_1,A_2,...A_n$ be points on a circle.Find the greastest number of acute triangle having vertices in these points.

Let $ f(n)$ be the maximal number of acute triangles for $ n$ cocyclic points. We will prove that
$ f(2k) \equal{} \frac {k(k\minus{}1)(k\plus{}1)}{3}$ and $ f(2k\plus{}1) \equal{} \frac {(2k\plus{}1)k(k\plus{}1)} {6}$ for all positive integer $ k$.

Let's consider $ n$ cocyclic points, say $ A_1, \cdots, A_n$ in the clockwise sense. In the following, subscripts are considered modulo $ n$.

The idea is to evaluate the number of non-acute angles in the configuration, since any non-acute angle determines exactly one non-acute triangle and, conversely, each non-acute triangle determines one non-acute triangle.

For each $ t \in \{1, \cdots, [\frac {n\minus{}2} {2}] \}$, and each $ i \in \{1, \cdots, n \}$ the chord $ [A_iA_{i\plus{}t\plus{}1}]$ divides the circle into two arcs, one of them containing exactly $ t$ of the given points apart $ A_i$ and $ A_{i\plus{}t\plus{}1}$. Note that the other arc contains at least $ t$ points (apart $ A_i$ and $ A_{i\plus{}t\plus{}1}$ again). One of this arc is no more than half the circle, so that for each of the $ A_j$ in this arc, the angle $ A_iA_jA_{i\plus{}t\plus{}1}$ is at least $ \frac {\pi}{2}$. It follows that each of these chords determines at least $ t$ non-acute angles.

- If $ n\equal{}2k$ is even, then for each $ t$ such that $ 1 \leq t \leq k\minus{}2$, there are $ n$ such chords. And for $ t \equal{} k\minus{}1$ there are $ k$ such chords.
It follows that the number of non-acute angles is at least $ n(1\plus{} \cdots \plus{} (k\minus{}2))\plus{}k(k\minus{}1) \equal{} \frac {n(n\minus{}2)^2}{8}$.
This leads to at least $ \frac {n(n\minus{}2)^2}{8}$ non-acute triangles.
Since the total number of triangles is $ \frac {n(n\minus{}1)(n\minus{}2)}{6}$, we deduce that the number of acute triangles is at most $ \frac {n(n\minus{}2)(n\plus{}2)}{24}$.
Thus $ f(2k) \geq \frac {k(k\minus{}1)(k\plus{}1)}{3}$.

- If $ n\equal{}2k\plus{}1$ is odd, then for $ 1 \leq t \leq k\minus{}1$ there are $ n$ such chords.
As above, this leads to at least $ n(1 \plus{} \cdots \plus{} (k\minus{}1)) \equal{} \frac {n(n\minus{}1)(n\minus{}3)} {8}$ non-acute angles.
As above, we easily deduce that there are at most $ \frac {n(n\minus{}1)(n\plus{}1)} {24}$ acte triangles.
Thus $ f(2k\plus{}1) \leq \frac {(2k\plus{}1)k(k\plus{}1)} {6}$.

Now, it is a straightforward verification that if the the chosen points are the vertices of a regular $ (2k\plus{}1)$-gon then we have $ \frac {(2k\plus{}1)k(k\plus{}1)} {6}$ acute triangles.
For $ n\equal{}2k$, let's start from the vertices of a regular $ (2k\minus{}1)$-gon. And add any vertex between $ A_k$ and $ A_{k\plus{}1}$ which is not diametrally opposite to $ A_1$. It is easy to verify that this new vertex gives $ \frac {k(k\minus{}1)}{2}$ acute triangles, which we have to add to the $ f(2k\minus{}1)$ already found. This gives $ \frac {k(k\minus{}1)(k\plus{}1)}{3}$ acute triangles as desired.

Pierre.


Asymptote
geometry
analytic geometry
vector
circumcircle
incenter
complex numbers

How can I draw this by Asymptote? (Note that $ \angle A \equal{} 45^\circ$).
No matter what color you use.

To draw the picture is not a problem:
[asy]size(220);
pair A=D("A",(0,0),W), B=D("B",(1,0),plain.E),
C=D("C",IP(A--dir(45),B--B+dir(150)),N), E=D("E",foot(C,A,B)),
D=D("D",(B+C)/2,NE);
D(A--D--B--A--C--D);
D(C--E--D,red+dashed);
MA("30^\circ",C,B,A,0.08);
markscalefactor=0.005;
D(rightanglemark(C,E,A),green);
add(pathticks(C--D));
add(pathticks(D--B));[/asy]
(I used a few CSE5 and Olympiad commands here to shorten the code but I'm pretty sure you can do all that with geometry_dev in an even more elegant way)

The problem is that you should try to learn basic Asymptote yourself: we cannot program [b]every[/b] picture for you in full though we'll be happy to give you an advice on how to achieve this or that advanced effect. So, next time try to draw as much of the picture as you can yourself and post the result together with the question. We'll show how to correct things that look wrong and how to add ones that are missing. :) But don't expect us to do full drawings any more :P.

Of course you are right! I can't learn if I told you to draw all my picture. But here, I want to learn how can I do that. In fact, I don't need this picture.  :P

I did not learn until now CSE5, where can I learn it?

Finally, your code does not work with me  :D ...

The code won't work without Olympiad and CSE5 being imported. You want to learn how to draw something like that? As I said, the best way to go is to start small and first learn how to do triangles with given angles, how to drop altitudes, etc., etc. All that information can be found in the manual but let me give you a quick (but still rather long) guide to basic Asymptote commands based on the picture you presented. Since your post doesn't give me the slightest idea of what you already can do and what you still cannot (this is actually the main point of my complaint, sorry if I didn't make it clear enough), I'll start from the very beginning avoiding use of any packages.

The first thing to do is to choose the size of your picture. The normal range that gives you easy to view pictures is somewhere from 150 to 350. Let's choose 200. The command is

size(200);

Pairs. Those are the points on the plane and you can do all operations with them that you can do with complex numbers plus you can connect them to form paths. So, the first thing you want to do is to set the base points $ A$ and $ B$ on the same horizontal line (say, $ x$-axis). You do it by just specifying the coordinates:
[code]
pair A=(0,0), B=(10,0);
[/code]
Now you want to dot and label them.
[code]
dot(A); dot(B);
[/code]
(this makes small black circles at the locations of $ A$ and $ B$)
[code]
label("$A$",A,W); label("$B$",B,plain.E);
[/code]
(this puts the letters $ A$ and $ B$ west of $ A$ and east of $ B$. I used plain.E instead of just E because we'll have a pair named $ E$ in our drawing later.)

Now come the lines. In order to create a unit vector in the given direction, you can use dir(a) command where a is the angle (in degrees) your vector makes with the $ x$-axis. We need to go in the direction $ 45^\circ$ from $ A$ and $ 150^\circ$ from $ B$. The point $ C$ is just the intersection of the corresponding lines. To find that intersection, you can use the intersectionpoint() command; to make the lines (actually, segments), you can use -- operator (A--B gives you the segment with the endpoints $ A$ and $ B$). You need to make sure that your segments are long enough to actually intersect, so let's make them both 10 units long (as long as the base, which is apparently the largest side of the triangle:
[code]
pair C=intersectionpoint(A--A+10*dir(45),B--B+10*dir(150));
[/code]
Now we dot and label $ C$:
[code]
dot(C); label("$C$",C,N);
[/code]
(N stands for north, of course). We can now draw the triangle $ ABC$ by using the draw command:
[code]
draw(A--B--C--cycle);
[/code]
The point $ D$ is not a problem:
[code]
pair D=(B+C)/2; dot(D); label("$D$",D,NE);
[/code]
Draw the line $ AD$:
[code]
draw(A--D);
[/code]
Now we need to find $ E$. There are many ways available. I'll just notice that it has the same $ x$ coordinate as $ C$ but zero $ y$-coordinate. So, we can merely put
[code]
pair E=(C.x,0);
[/code]
Dot and label as usual:
[code]
dot(E); label("E",E,S);
[/code]
Now you want the broken line $ CED$ but you want it dashed and red rather than solid and black. Here you can use pens:
[code]
pen p=red+dashed;
draw(C--E--D,p);
[/code]
(of course, you could also just write draw(C--E--D,red+dashed);)

It is time to put those tick marks at $ CD$ and $ DB$. Those are just short segments. Let's make them of size $ 0.5$ ($ \frac{1}{20}$ of the base length. Again, there are plenty of ways to do it but, since you know their middlepoints and direction, you can just write
[code]
draw((C+D)/2-0.25*dir(60)--(C+D)/2+0.25*dir(60));
draw((B+D)/2-0.25*dir(60)--(B+D)/2+0.25*dir(60));
[/code]
Now let's mark the $ 30^\circ$ angle. We need the arc command that draws a circular arc. Again, since you know the angles explicitly, the easiest way is to write
[code]
draw(arc(B,1,150,180));
label("$30^\circ$",B+dir(165),dir(165),fontsize(8));
[/code]
Here $ 1$ is the radius of the arc and 150 to 180 is its angular span. B+dir(165) is just the middle point of the arc, dir(165) tells you in which direction to place the label, and fontsize(8) is there to make the label a bit smaller than usual to fit neatly between the lines.
Finally, let's put the right angle mark. It is nothing but a small broken line and, since we have all the coordinates available, we can just write
[code]
draw((E.x,0.5)--(E.x-0.5,0.5)--(E.x-0.5,0));
[/code]
I hope by now the meaning is obvious.

[asy]size(200);
pair A=(0,0), B=(10,0);
dot(A); dot(B);
label("$A$",A,W); label("$B$",B,plain.E);
pair C=intersectionpoint(A--A+10*dir(45),B--B+10*dir(150));
dot(C); label("$C$",C,N);
draw(A--B--C--cycle);
pair D=(B+C)/2; dot(D); label("$D$",D,NE);
draw(A--D);
pair E=(C.x,0);
dot(E); label("E",E,S);
pen p=red+dashed;
draw(C--E--D,p);
draw((C+D)/2-0.25*dir(60)--(C+D)/2+0.25*dir(60));
draw((B+D)/2-0.25*dir(60)--(B+D)/2+0.25*dir(60));
draw(arc(B,1,150,180));
label("$30^\circ$",B+dir(165),dir(165),fontsize(8));
draw((E.x,0.5)--(E.x-0.5,0.5)--(E.x-0.5,0));[/asy]

Of course, you may be quite dissatisfied with the necessity to take care or radii, positioning of the labels, coordinate computations and such, especially if you need to do that more than one time on the picture. The good news is that you can define various subroutines in Asymptote to do it and, maybe, an even better news is that other people have already done it and created various "packages" exactly for that purpose. Geometry_dev seems one of the best so far for drawing geometric diagrams but I'll let Ph.Ivaldi teach you how to use it himself :P. My CSE5 was created mainly in response to the 2000 symbol restriction for the codes on AoPS (which is gone now), though I still like it and use it myself. I don't know how many other people share my passion though. You can find the description in one of the stickies.

Meanwhile, as I said, you are welcome to ask questions here but, please, make it clear what you can do and are happy with, what you can do but are unhappy with, and what you cannot do at all when posting. This will spare both our and your time :).

Thanks fedja, that is amazing explaining :D . Now I understood these thinks very well. But is there any code to mark the angles better than drawing arcs ... ? The same things for marking the segments.

Is this code right for the picture? :?:  (I tried for a long time). :D
Also, I have another question: is there another style for dots ? like mmm... empty circle or some thing like that?
Final question  :D : If I want to name the final shape (all thing in a diagram), how can I do that?
[code]
size(200);
import olympiad;
import geometry_dev;
pair A,B,C,I,M,O,Y,L;
// drawing the triangle.
B=(0,0);C=(4,0);A=(1,3);
draw(A--B--C--cycle,linewidth(1bp));
dot(A);dot(B);dot(C);
label('$A$',A,N);label('$B$',B,W);label('$C$',C,E);
// circumcircle
O=circumcenter(A,B,C); dot(O); label('$O$',O,E);
draw(circumcircle(A,B,C));
// incenter
I=incenter(A,B,C); dot(I); label('$I$',I,W);
// make L
point[]temp = intersectionpoints(line(A,I), circumcircle(A,B,C));
L = temp[1]; dot(L); draw(A--L); label('$L$',L,S); draw(L--B);
// make M
temp = intersectionpoints(line(L,O), circumcircle(A,B,C));
M=temp[0]; dot(M); draw(B--M,dashed); draw(L--M,dashed); label('$M$',M,N);
// make Y
Y = intersectionpoint(perpendicular(I, line (A,C)), line(A,C));
dot(Y); label('$Y$',Y,NE); draw(I--Y); draw(I--O); draw(I--B);
// markangle
markangleradiusfactor = 1;
markangle (2, line(A,B), line(A,L));
markangleradiusfactor *= 3;
markangle (2, line(A,L), line(A,C));
markangle (2, line(B,L), line(B,C));
markangle(1, line(B,I), line(B,A));
markangleradiusfactor *= 3;
markangle(1, line(B,C), line(B,I));
//perpfactor *=5.5;
perpendicularmark(line(Y,I), line(Y,C));
[/code]


Let $ c$ be a fixed positive integer and $ \{x_n\}_{k\equal{}1}^{\infty}$ be a sequence  such that $ x_1\equal{}1$ and $ x_n\equal{}x_{n\minus{}1}\plus{}[\frac{2x_{n\minus{}1}\minus{}2}{n}]$ for $ n\geq 2$.Determine the explicit formula of $ x_n$ in terms of $ n$ and $ c$
I have read the official solution, but they are very natural and  difficult to understand.
I really want to know reasons, or a new resolution to this problem
can anyone help me?

[quote="math10"]Let $ c$ be a fixed positive integer and $ \{x_n\}_{k \equal{} 1}^{\infty}$ be a sequence  such that $ x_1 \equal{} 1$ and $ x_n \equal{} x_{n \minus{} 1} \plus{} [\frac {2x_{n \minus{} 1} \minus{} 2}{n}]$ for $ n\geq 2$.Determine the explicit formula of $ x_n$ in terms of $ n$ and $ c$
I have read the official solution, but they are very natural and  difficult to understand.
I really want to know reasons, or a new resolution to this problem
can anyone help me?[/quote]

There is surely a mistake in the problem statement since the solution seems obvious : $ x_n\equal{}1$ $ \forall n$

[quote="pco"][quote="math10"]Let $ c$ be a fixed positive integer and $ \{x_n\}_{k \equal{} 1}^{\infty}$ be a sequence  such that $ x_1 \equal{} 1$ and $ x_n \equal{} x_{n \minus{} 1} \plus{} [\frac {2x_{n \minus{} 1} \minus{} 2}{n}]$ for $ n\geq 2$.Determine the explicit formula of $ x_n$ in terms of $ n$ and $ c$
I have read the official solution, but they are very natural and  difficult to understand.
I really want to know reasons, or a new resolution to this problem
can anyone help me?[/quote]

There is surely a mistake in the problem statement since the solution seems obvious : $ x_n \equal{} 1$ $ \forall n$[/quote]
sorry,I had write wrong:$ x_1\equal{}c$

[quote="math10"]Let $ c$ be a fixed positive integer and $ \{x_n\}_{k \equal{} 1}^{\infty}$ be a sequence  such that $ x_1 \equal{} c$ and $ x_n \equal{} x_{n \minus{} 1} \plus{} [\frac {2x_{n \minus{} 1} \minus{} 2}{n}]$ for $ n\geq 2$.Determine the explicit formula of $ x_n$ in terms of $ n$ and $ c$
I have read the official solution, but they are very natural and  difficult to understand.
I really want to know reasons, or a new resolution to this problem
can anyone help me?[/quote]

Solution:$ x_n \equal{} \frac16\big((c \minus{} 1) n^2 \plus{} (3 c \minus{} 3) n \plus{} 2 c \plus{} 4\big)$

How I reached the solution:
Probably $ x_n$ is a polynomial in $ n$. So I tried to find a solution of the following Polynomial equation: $ P(1) \equal{} c$,   $ nP(n) \equal{} (n \plus{} 2)P(n \minus{} 1) \minus{} 2$  $ \forall b\in\mathbb{N}$. Firstly I tried to find a solution with $ deg P \equal{} 1$. $ P(n) \equal{} an \plus{} b$. The relation becomes:
$ an^2 \plus{} bn \equal{} (n \plus{} 2)(an \plus{} b \minus{} a) \minus{} 2 \equal{} an^2 \plus{} bn \plus{} an \plus{} 2b \minus{} 2a \minus{} 2$ or $ an \plus{} 2b \minus{} 2a \minus{} 2 \equal{} 0$.
So $ a \equal{} 0$, which is true only for $ c \equal{} 1$. So $ deg P\neq 1$. Let's try $ deg P \equal{} 2$.
$ P(n) \equal{} an^2 \plus{} bn \plus{} d$. The initial relation becomes:
$ an^3 \plus{} bn^2 \plus{} dn \equal{} (n \plus{} 2)(an^2 \minus{} 2an \plus{} a \plus{} bn \minus{} b \plus{} d) \minus{} 2 \equal{} an^3 \minus{} 2an^2 \plus{} an \plus{} bn^2 \minus{} bn \plus{} dn \plus{} 2an^2 \minus{} 4an \plus{} 2a \plus{} 2bn \minus{} 2b \plus{} 2d \minus{} 2$
or $ bn \plus{} 2a \plus{} 2d \equal{} 3an \plus{} 2b \plus{} 2$, because it's true for all positive integers $ n$, we must have $ b \equal{} 3a$ and $ a \plus{} d \equal{} b \plus{} 1$ so $ d \equal{} 2a \plus{} 1$ so $ P(n) \equal{} an^2 \plus{} 3an \plus{} 2a \plus{} 1$
It remains to find $ a$ according to $ c$. $ c \equal{} P(1) \equal{} 6a \plus{} 1$ so $ a \equal{} \frac {c \minus{} 1}6$. From here you get the upper solution.

It's easy to prove that this polynomial satisfy the problem condition by induction.


http://cache0.techcrunch.com/wp-content/uploads/2009/09/Googel-job-code.jpg

Does anyone know the details about this code breaking thing?
I use google search, and can't find any authentic source mentioning that google is hiring people who break this code. There are some minor sites that mention this code breaking stuff, but so far no major sites do a report on it.

Is it just a prank by MIT students or is it really true that google is hiring people who can break this code?

PS: the code is broken and it's very easy

the code is a telephone number:951-119-8957, call it and leave a message. maybe you can get the job. good luck!


Let ABC be a triangle and D$\in$(AB), E$\in$(AC). Link DE and let F$\in$(DE). Let AD/AB=x, AE/AC=y, DF/DE=z. Prove that:
(1)  the area of BDF = (1-x)yz{area of ABC}
(2)  the area of CEF = x(1-y)(1-z){area of ABC}
(3)  $\sqrt[3]{\{\text{area of }BDF\}}+\sqrt[3]{\{\text{area of }CEF\}}\leq \sqrt[3]{\{\text{area of }ABC\}}$.

[hide="No. 1"]The area of $ABC$ is half $AB$ times the altitude from $C$, which we shall call $h_{C}$.  Let us now drop altitudes from $E$ and $F$, and call their points of intersection with $AB$ $E'$ and $F'$, respectively.  Clearly, $\displaystyle y=\frac{AE}{AC}=\frac{EE'}{h_{C}}$, and $\displaystyle z=\frac{DF}{DE}=\frac{FF'}{EE'}$.  Thus $\displaystyle yz=\frac{FF'}{h_{C}}$, so multiplying that by the area of $ABC$ will give us half $AB$ times its altitude from $F$.  Now we note that $\displaystyle 1-x=\frac{AB-AD}{AB}=\frac{BD}{AB}$, and our proof is complete.

Q.E.D.[/hide]

We can prove No. 2 by rotating the diagram and using the same method.


algebra
polynomial
number theory
combinatorics proposed
combinatorics

Let $ p$ be a prime number, and $ n$ be an integer such that $ n\geq p$. Assume that $ a_1$, $ a_2$, ..., $ a_n$ are arbitrary integer numbers. For every $ k$, we denote by $ f_k$ the number of $ k$-subsets $ \left\{s_1,s_2,...,s_k\right\}$ of the set $ \left\{1,2,...,n\right\}$ such that the sum $ a_{s_1} \plus{} a_{s_2} \plus{} ... \plus{} a_{s_k}$ is divisible by $ p$. Prove that
$ p\mid \sum_{k \equal{} 0}^n\left( \minus{} 1\right)^kf_k$.
(Here, we understand $ f_0$ to be $ \equal{} 1$.)

Nice one!

Consider

$ A \equal{} \prod_{i \equal{} 1}^n (1 \minus{} \zeta_p^{a_i})\in \mathbb{Q}(\zeta_p)$

where $ \zeta_p$ is a primitive $ p$-th root of unity.

It is easily seen that

$ p\sum_{k \equal{} 0}^n ( \minus{} 1)^k f_k \equal{} \text{tr}_{\mathbb{Q}(\zeta_p)/\mathbb{Q}} A$.

Here $ \text{tr}_{\mathbb{Q}(\zeta_p)/\mathbb{Q}}$ denotes the trace from $ \mathbb{Q}(\zeta_p)$ to $ \mathbb{Q}$. Now, obviously $ (1 \minus{} \zeta_p^{a_i})$ is divisible by $ (1 \minus{} \zeta_p)$, thus by the assumption $ n\geq p$, $ (1 \minus{} \zeta_p)^p$ divides $ A$. However, $ (1 \minus{} \zeta_p)^{p \minus{} 1} \equal{} pe$ where $ e$ is a unit(I guess $ e \equal{} \zeta_p^{(p\minus{}1)/2}$). Thus $ 1 \minus{} \zeta_p$ divides $ \frac Ap$; it follows that $ 1 \minus{} \zeta_p$ divides $ \frac 1p \text{tr}_{\mathbb{Q}(\zeta_p)/\mathbb{Q}} A$. But since $ \frac 1p \text{tr}_{\mathbb{Q}(\zeta_p)/\mathbb{Q}} A$ is rational, we must have that it is divisible by $ p$. The conclusion follows.

I have just uploaded a more elementary (but longer) proof on my webpage:
[url=http://www.cip.ifi.lmu.de/~grinberg/StPeters2003.pdf]St. Petersburg 2003: An alternating sum of zero-sum subset numbers[/url].
(Only read until (3), and then from the end of p. 6 on if you know Lemma 2.)
So we have a proof by algebraic number theory and a proof by algebraic combinatorics now. What about one by homology sequences?

  darij

There is given a short algebraic and elementary proof in the book S-P 2003.Also there is combinatorics proof too,but it is a bit longer.

I don't have the book here, and I am not sure whether I have it in Karlsruhe (I definitely have 2002). Can you kindly look up whether the book tells something about where the problem comes from?

  darij

The problems author is Andrey Badzyan,and this problems was used as the last and hardest problem for the 10th grade.

[b]Generalization.[/b] Let $ p$ be a prime number, $ \alpha$ an nonnegative integer, $ c$ an integer and $ n$ be an integer such that $ n\geq p^{\alpha}$. Assume that $ a_1$, $ a_2$, ..., $ a_n$ are arbitrary integer numbers. For every $ k$, we denote by $ f_k$ the number of $ k$-subsets $ \left\{s_1,s_2,...,s_k\right\}$ of the set $ \left\{1,2,...,n\right\}$ such that the sum $ a_{s_1} \plus{} a_{s_2} \plus{} ... \plus{} a_{s_k}\minus{}c$ is divisible by $ p^{\alpha}$. Prove that
$ p\mid \sum_{k \equal{} 0}^n\left( \minus{} 1\right)^kf_k$.
(Here, we understand $ f_0$ to be $ \equal{} 1$.)

[quote="darij grinberg"][b]Generalization.[/b] Let $ p$ be a prime number, $ \alpha$ an nonnegative integer, $ c$ an integer and $ n$ be an integer such that $ n\geq p^{\alpha}$. Assume that $ a_1$, $ a_2$, ..., $ a_n$ are arbitrary integer numbers. For every $ k$, we denote by $ f_k$ the number of $ k$-subsets $ \left\{s_1,s_2,...,s_k\right\}$ of the set $ \left\{1,2,...,n\right\}$ such that the sum $ a_{s_1} + a_{s_2} + ... + a_{s_k} - c$ is divisible by $ p^{\alpha}$. Prove that
$ p\mid \sum_{k = 0}^n\left( - 1\right)^kf_k$.
(Here, we understand $ f_0$ to be $ = 1$.)[/quote]
The following proof is based on a solution given at the original book.
Proof: Denote $ S = \sum_{k = 0}^n\left( - 1\right)^kf_k$.
WLOG assume that all $a_i$ are $\geq 0$ and that $0\leq c<p^{\alpha}$.
Consider polynomial $ f(x) = (1 - x^{a_1})(1 - x^{a_2})\dots(1 - x^{a_n})$
We need to prove that the sum of coefficients by the $ x^c,x^{p^{\alpha} + c},x^{2p^{\alpha} + c}\dots$ (this sum is $S$) is divisible by $ p$.
Indeed since $ x^{kp^{\alpha} + c}\equiv x^c \mod x^{p^{\alpha}}-1$ and if $ x^{d}\equiv x^c \mod x^{p^{\alpha}}-1$ then $ d = kp^{\alpha} + c$,so $ S$ is equal to the coefficient by $ x^c$ in polynomial $ r(x)$,where
$ f(x) = q(x)(x^{p^{\alpha}} - 1) + r(x)$.
We will prove that all coefficients in $ r(x)$ are divisible by $ p$.Indeed,since $ n\geq p^{\alpha}$,then
$ f(x)$ is divisible by $ (x - 1)^{p^{\alpha}}$.
Then $ f(x) = (x - 1)^{p^{\alpha}}h(x) = (x^{p^{\alpha}} - 1)h(x) + ph(x)r(x)$.
The last equality follows from the well-known fact:
$ p\mid \binom{p^{\alpha}}{i}$,where $ 1\leq i\leq p^{\alpha} - 1$.
Now it is enough to prove that the coefficients in $ ph(x)r(x)$ modulo $ x^{p^{\alpha}} - 1$ are divisible by $ p$,which is trivially true.

[color=green]Very nice Erken!  :)

And this method (COMPLEX COMBINATORICS) is very useful and strong.[/color]

We need to prove  $(1-y^{a_1}),\hdots,(1-y^{a_n})=0$ in $\frac {\mathbb Z[y]}{(p,y^{p-1})}. $
$(1-y)^p=0\in\frac {\mathbb Z[y]}{(p,y^{p-1})}\,\text{if}\, p=2\,\text {also}. $
Note that, since $p\le n, $ $(1-y)^p\mid\text {left hand side}\text {in}\,\mathbb Z [y]. $


A square with side length $ n \ge 3$ is split into a unit grid, and every square within the grid is filled with either an apple or a banana. At each step, you may choose any square, and it and its immediate neighbours (including diagonal) are switched. Apple becomes banana and banana becomes apple. For which, if any values of $ n$ is it possible to get from any starting position to all other possible permutations of apples and bananas?

I have a marvelous solution to this problem, which this post is too small to contain, so I think I'll use [hide="Hide tags!"]The answer is all numbers 0 and 1 mod 3.

[u]Proof:[/u]If $ n\equiv 2\mod 3$, then consider the 1st, 2nd, 4th, 5th, 7th, 8th... squares in the top row.  Whenever any square is clicked, an even number of these special squares is flipped.  Thus the parity of the number of these squares that are bananas always stays the same, so an odd cannot get to an even and vice versa.  Hence it is impossible for $ n\equiv 2\mod 3$.

For $ n\equiv 0, 1\mod 3$, we see that if we can prove that from every position we can always get to all apples, then we can get to every position to every other position.  To show that we can always clear the board of all bananas, we need a lemma.

[u]Lemma:[/u]It is always possible to clear every row except the very bottom row of bananas.

Proof: Consider any row.  First we clear this row except the last square of bananas without disturbing any above row.  Consider the first banana in this row, and choose the square to its lower right.  This will change the banana to an apple and doesn't disturb any row above it or any square to the left of this new apple. We can do this over and over in this row, until either the row is completely filled with apples or the last square is a banana.  In the first case, we are done.

In the second case, however, the strategy is this: Choose the very first square in the row beneath the one with the banana.  Then continue the "Choose the lower-right square to the first banana in the row" strategy until the row is completely gone.  We know that the last banana will disappear because since $ n\equal{}3k\plus{}0, 1$, squares 1 and 2 will first be bananas, then 3 alone, then 4 and 5, then only 6, ... then 3k-2 and 3k-1, then 3k alone.  If $ n\equal{}3k$, then at the $ 3k\minus{}2$ and $ 3k\minus{}1$ stage, the next step will change $ 3k\minus{}2$, $ 3k\minus{}1$ and $ 3k$ all over to apples and the row will be only apples.  If $ n\equal{}3k\plus{}1$, then the 3k alone stage will have $ 3k$ and $ 3k\plus{}1$ as bananas and the next step will turn just these over, to apples.  Either way the row will turn to all apples.

We continue doing this until just the bottom row has bananas in it.  Now, if $ n$ is:

0 mod 3, use the 2nd to last row.
1 mod 3, use the last row.

By using the row, we mean: if there is a banana in the $ m$th square on the bottom row, choose the $ m\plus{}1$th square in the used row, for $ m$ from 1 to $ n\minus{}1$.  Then again, if the last square contains a banana at the end, choose the first square in the used row and do the whole thing over again.  Now the bottom row is all apples and we apply a "choose the upper right corner of the first banana in the lowest row" strategy, kind of like the first technique but upside down.  If you had to choose the first square in the last or 2nd to last row, then you will have to for every other time.  But from here, the strategy will make sure all the squares are apples.  Hence we can get to all apples from any position, and can get back again from all apples, so from any position we can get to any other position.[/hide]

It's a pretty long solution.  A picture is worth a thousand words, and a video is worth a thousand pictures, so here's a solution of a 12 by 12.  This is a [url=http://s69.photobucket.com/player.swf?file=http://vid69.photobucket.com/albums/i69/tenniskidperson3/the%20album/clip0002.flv&fs=1&os=1&ap=1]video[/url] of a TI-83+ program I wrote.

That's a nice solution. I had actually (finally) managed to solve the problem myself by the time you posted this, but I'm always interested in other ways of approaching a problem.

First I proved the impossibility of the 2 mod 3 case using a colouring/invariance proof (very similar to yours). This left me to prove that the other cases are possible.

The way I did it (no details, just a skeleton proof) was first to show that if you can switch any given single column on its own without affecting any other squares, then it is possible to swap any individual square (this is the tricky part). Since we can reach any permutation of apples and bananas by a series of single square swaps, being able to swap any given single column is equivalent to being able to reach all permutations.

Then I showed that on all grid sizes not congruent to 2 mod 3 it is possible to swap any given single column, and therefore possible to reach all permutations. I found the easiest way to do this was to construct an algorithm which achieved this effect for any grid size. This was much easier than the first part.

I believe that your solution is much quicker than mine, although a key characteristic of both solutions is that they both tell you that not only it is possible to reach all permutations, but also exactly how to reach them (you may not see this in the outline I have posted here, but my full solution certainly does).


complex numbers
geometry proposed
geometry

Let $z\in\mathbb{C}$ be such that points $z^{3},$ $2z^{3}+z^{2},$ $3z^{3}+3z^{2}+z$ and $4z^{3}+6z^{2}+4z+1$ are the vertices of an inscribed quadrangle at complex plane. Find $Re\,z$.
(V. Brayman)

Four complex numbers a,b,c,d  are on a circle iff the birapport
$\frac{c-a}{c-b}: \frac{d-a}{d-b}$ is real number

Here $a=3z^{3}+3z^{2}+z$, $b=2z^{3}+z^{2}$, $c=4z^{3}+6z^{2}+4z+1$ ,
$d=z^{3}$

The birapport here is $\frac{z^{2}+z}{(2z+1)^{2}}$, it is real iff it equal
to its conjugate, this leads to $Re(z)=-1/2$

--------------------------------------------

Is there any other geometric proof without using the birapport ?
thank's

The points $z^{4}+z^{3}=z^{3}(z+1),$ $z^{4}+2z^{3}+z^{2}=z^{2}(z+1)^{2},$ $z^{4}+3z^{3}+3z^{2}+z=z(z+1)^{3}$ and $z^{4}+4z^{3}+6z^{2}+4z+1=(z+1)^{4}$ are the vertices of an inscribed quadrangle at complex plane. Then the circles $\omega$ and $\omega'$ determined by the first three points and by the last three points coincide. The centres $z_{0},z_{0}'$ of these circles satisfy $z_{0}'=\frac{z+1}{z}\cdot z_{0}=z_{0}$ which imply $z_{0}=0.$ Now $|z^{3}(z+1)|=|z^{2}(z+1)^{2}|,$ $|z|=|z+1|$ and $Re\,z=-\frac{1}{2}.$ :)

thank you rogue for this elegant solution.


conics
hyperbola
symmetry
analytic geometry
graphing lines
slope
Asymptote

Prove why A^2 + B^2 = C^2 for hyperbolas, where A = semi-major axis and B is the semi-minor axis, and C = focal radius length.

A and C are usually used directly in the derivation of the formula, so they need no definition, we merely need to understand B, upon simplication, we get

(x^2)/(a^2)-(y^2)/(c^2-a^2)=1

in some sense, the B substitution is just to increase symmetry; the semi-minor axis does not really mean a whole lot...in contrast with the semi minor axis in say, an ellipse...it is more so a substitution to increase symmetry [note that b/a is the slope of the asymptote, but this result comes from algebra; we could just as well say that root(c^2-a^2)/a is the slope also]


Let $ \cos \alpha \equal{}\frac{\sqrt{6}}{3},\ \sin \alpha \equal{}\frac{\sqrt{3}}{3}\ (0\leq \alpha <2\pi)$, find the values of $ \sin 2\alpha ,\ \sin 4\alpha$. Then find natural number $ m$ such that $ \frac{\pi}{m\plus{}1}\leq \alpha <\frac{\pi}{m}$.

hello kunny, we get $ \alpha\equal{}\arctan\left(\frac{\sqrt{3}}{3\plus{}\sqrt{6}}\right)$ and
$ \sin(2\alpha)\equal{}\frac{2\sqrt{2}}{3}$
$ \sin(4\alpha)\equal{}\frac{4\sqrt{2}}{9}$
and
$ \frac{\pi}{6}\le \alpha<\frac{\pi}{5}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello kunny, we get $ \alpha \equal{} \arctan\left(\frac {\sqrt {3}}{3 \plus{} \sqrt {6}}\right)$ and
$ \sin(2\alpha) \equal{} \frac {2\sqrt {2}}{3}$
$ \sin(4\alpha) \equal{} \frac {4\sqrt {2}}{9}$
and
$ \frac {\pi}{6}\le \alpha < \frac {\pi}{5}$.
Sonnhard.[/quote]

I wonder why we get the range of $ \alpha$ , but not fix value(s) for $ \alpha$
when given the value of $ \tan\alpha \equal{}\frac {\sin\alpha}{\cos\alpha}\equal{}\frac{1}{\sqrt{2}}$

hello, but we have $ 0\le \alpha<2\pi$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, but we have $ 0\le \alpha < 2\pi$.
Sonnhard.[/quote]

Sorry .

How can we get $ \frac {\pi}{6}\le \alpha < \frac {\pi}{5}$?


abstract algebra
integration
function
calculus
calculus computations

Show that $ y_d(t)$ given in the following formula:
$ y_d\equal{}y_1(t)\int_{t_0}^{t}\frac{\minus{}y_2(s)f(s)}{W[y_1,y_2](s)}ds\plus{}y_2(t)\int_{t_0}^{t}\frac{\minus{}y_1(s)f(s)}{W[y_1,y_2](s)}ds$
may be written in the form:
$ y_d(t)\equal{}\int_{t_0}^{t}K(t,s)f(s)ds$
where K(t,s) is the kernel function (also called Green's kernel):
$ K(t,s)\equal{}\frac{y_1(s)y_2(t)\minus{}y_1(t)y_2(s)}{y_1(s)y'_2(s)\minus{}y_2(s)y'_1(s)}\equal{}\frac{y_1(s)y_2(t)\minus{}y_1(t)y_2(s)}{W[y_1,y_2](s)}$

Find K for the operators $ P(D)\equal{}D^2\plus{}1$, and $ P(D)\equal{}D^2\minus{}\frac{2}{t^2}$.

$ y_{1}(t)$ is a function of $ t$, so you can treat it as a constant and move it into the integral.  Then we get
$ y_d\equal{}\int_{t_0}^{t}\frac{\minus{}y_{1}(t)y_2(s)f(s)}{W[y_1,y_2](s)}ds\plus{}\int_{t_0}^{t}\frac{\minus{}y_{2}(t)y_1(s)f(s)}{W[y_1,y_2](s)}\ ds$
The rest is easy because the integrand is just $ K(s,t)f(s)$.  However, either you have an extra negative sign or you're missing one because you need that negative sign to get equality.


abstract algebra
superior algebra
superior algebra unsolved

Let $ p$ be a prime number and $ K$ a field, $ a \in K$ and $ L \equal{} K(\sqrt[p]{a})$. Is it true that the kernel of $ K^*/p \to L^*/p$ is generated by $ a$?


Hi Guys,
There is a problem in the Rice University Annual Mathematics Test which I do not understand. Please explain it to me.
What is the largest six-digit number whose digits are all different such that the sum of the squares of the digits in the even-numbered positions equals the sum of the squares of the digits in the odd-numbered positions?
Ans. 986702

Thanks,

[hide="Bad Solution"]
We start by noticing that the largest first four digits possible are $ 9867$ (since $ 9876$ would result in a odd-numbered sum of squares of at least $ 130$, which can't be attained by the even digits).

If we let the number be $ \overline{9867ab}$, then we must have $ 117\plus{}a^2\equal{}113\plus{}b^2\implies4\equal{}(b\minus{}a)(b\plus{}a)\implies(a,b)\equal{}(0,2)\implies\boxed{986702}$.
[/hide]

I thank you for providing an answer. Nevertheless, the answer did not make sense to me. I would deeply appreciate it if anyone else would care to provide an accurate and quick solution to the problem.

To reduce your casework, you could also note that the modulo $ 8$ residues of the perfect squares of the digits are

$ \{0^2,1^2,2^2,3^2,4^2,5^2,6^2,7^2,8^2,9^2\}\equiv\{0,1,4,1,0,1,4,1,0,1\}\pmod8$.

If we let the six-digit number be $ \overline{abcdef}$, then we must have

$ \{a^2,b^2,c^2,d^2,e^2,f^2\}\in\{0,1,4,1,0,1,4,1,0,1\}\pmod8$ (and $ a\neq b\neq c\neq d\neq e\neq f$!), so in order for

$ a^2\plus{}c^2\plus{}e^2\equal{}b^2\plus{}d^2\plus{}f^2$ (keep in mind that we can only use each member of the set $ \{0,1,4,1,0,1,4,1,0,1\}$ once), we quickly see that the following are the only possibilities:

$ \{a,c,e\}\equal{}\{0,0,1\}$ (order does not matter in these sets) and $ \{b,d,f\}\equal{}\{1,4,4\}$ or vice versa,
$ \{a,c,e\}\equal{}\{0,1,1\}$ and $ \{b,d,f\}\equal{}\{0,1,1\}$,
$ \{a,c,e\}\equal{}\{0,1,4\}$ and $ \{b,d,f\}\equal{}\{0,1,4\}$, and
$ \{a,c,e\}\equal{}\{1,1,4\}$ and $ \{b,d,f\}\equal{}\{1,1,4\}$.


search
inequalities proposed
inequalities

$ (ac\plus{}bd)(\frac{c^3}a\plus{}\frac{d^3}{b}) \ge (c^2\plus{}d^2)^2$
but $ \sqrt{(c^2\plus{}d^2)(a^2\plus{}b^2)} \ge ac\plus{}bd$
we get  $ (c^2\plus{}d^2)^2 \ge ac\plus{}bd$
So the problem is solved.

[quote="FOURRIER"][quote="Erken"]Already posted.Use search function. :wink:[/quote]
I didnt find it

Give me the link please.[/quote]
Here http://www.mathlinks.ro/viewtopic.php?t=166509
 :)