trigonometry calculus integration HCSSiM

(Sum or Prove that)
\[ 1)\frac {{\left( \begin{array}{l} n \\
1 \\
\end{array} \right)}}{{\left( \begin{array}{l} n \\
0 \\
\end{array} \right)}} \plus{} 2\frac {{\left( \begin{array}{l} n \\
2 \\
\end{array} \right)}}{{\left( \begin{array}{l} n \\
1 \\
\end{array} \right)}} \plus{} ........ \plus{} n\frac {{\left( \begin{array}{l} n \\
n \\
\end{array} \right)}}{{\left( \begin{array}{l} n \\
n \minus{} 1 \\
\end{array} \right)}}
\]

\[ 2)\frac {1}{{1!\left( {n \minus{} 1} \right)!}} \plus{} \frac {1}{{2!\left( {n \minus{} 2} \right)!}} \plus{} ............. \plus{} \frac {1}{{\left( {n \minus{} 1} \right)!1!}}
\]

\[ 3)\left( \begin{array}{l} n \\
0 \\
\end{array} \right)^2 \minus{} \left( \begin{array}{l} n \\
1 \\
\end{array} \right)^2 \plus{} \left( \begin{array}{l} n \\
2 \\
\end{array} \right)^2 \plus{} .............. \plus{} ( \minus{} 1)^n \left( \begin{array}{l} n \\
n \\
\end{array} \right)^2 \equal{} \left\{\begin{array}{l} 0\_for\_odd\_n \\
\frac {{n!}}{{\left[ {\frac {n}{2}} \right]^2 }},for\_even\_n \\
\end{array} \right.
\]

\[ 4)\left( \begin{array}{l} n \\
1 \\
\end{array} \right)^2 \plus{} 2\left( \begin{array}{l} n \\
2 \\
\end{array} \right)^2 \plus{} 3\left( \begin{array}{l} n \\
3 \\
\end{array} \right)^2 \plus{} .............. \plus{} n\left( \begin{array}{l} n \\
n \\
\end{array} \right)^2 \equal{} \frac {{\left( {2n \minus{} 1} \right)!}}{{\left( {\left( {n \minus{} 1} \right)!} \right)^2 }}
\]

\[ 5)\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} 2\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} 3\left( \begin{array}{l} n \\
4 \\
\end{array} \right) \plus{} .............. \plus{} \left( {n \minus{} 1} \right)\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 6)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} \frac {1}{2}\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} \frac {1}{3}\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} ......... \plus{} ( \minus{} 1)^{n \minus{} 1} \frac {1}{n}\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 7)\left( \begin{array}{l} n \\
0 \\
\end{array} \right) \minus{} 2\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \plus{} 3\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} .......... \plus{} ( \minus{} 1)^n (n \plus{} 1)\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 8)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \plus{} 2\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} 3\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} ......... \plus{} n\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 9)\left( \begin{array}{l} n \\
0 \\
\end{array} \right)\left( \begin{array}{l} n \\
r \\
\end{array} \right) \plus{} \left( \begin{array}{l} n \\
1 \\
\end{array} \right)\left( \begin{array}{l} n \\
r \plus{} 1 \\
\end{array} \right) \plus{} ........... \plus{} \left( \begin{array}{l} n \\
n \minus{} r \\
\end{array} \right)\left( \begin{array}{l} n \\
n \\
\end{array} \right)\_for\_r < n
\]

\[ 10)\left( \begin{array}{l} n \\
0 \\
\end{array} \right) \minus{} \frac {1}{2}\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \plus{} \frac {1}{3}\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} ......... \plus{} ( \minus{} 1)^{n \minus{} 1} \frac {1}{{n \plus{} 1}}\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ \begin{array}{l} 11)if\_(1 \plus{} x \plus{} x^2 )^n \equal{} C_0 \plus{} C_1 x \plus{} C_2 x^2 \plus{} ........ \plus{} C_{2n} x^{2n} \\
prove\_tht \\
i)C_0 \equal{} C_0 \_for0 \le k \le 2n \\
ii)C_0 ^2 \minus{} C_1 ^2 \plus{} C_2 ^2 \minus{} ...... \plus{} C_{2n} ^2 \equal{} C_n \\
iii)C_0 \plus{} C_2 \plus{} C_4 \plus{} ............. \plus{} C_0 \equal{} 1 \plus{} C_1 \plus{} C_3 \plus{} C_5 \plus{} .......... \plus{} C_{2n \minus{} 1} \\
\end{array}
\]

\[ 12)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} 2^2 \left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} 3^2 \left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} ......... \plus{} ( \minus{} 1)^{n \plus{} 1} n^2 \left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 13)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \plus{} 2^2 \left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} 3^2 \left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} ......... \plus{} n^2 \left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ \begin{array}{l} 14)i)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} \left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} \left( \begin{array}{l} n \\
5 \\
\end{array} \right) \minus{} \left( \begin{array}{l} n \\
7 \\
\end{array} \right) \plus{} .......... \equal{} 2^{\frac {n}{2}} \sin \frac {{n\pi }}{4} \\
ii)\left( \begin{array}{l} n \\
0 \\
\end{array} \right) \minus{} \left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} \left( \begin{array}{l} n \\
4 \\
\end{array} \right)\_\left( \begin{array}{l} n \\
6 \\
\end{array} \right) \plus{} .......... \equal{} 2^{\frac {n}{2}} \cos \frac {{n\pi }}{4} \\
\end{array}
\]

\[ \begin{array}{l} 15)n\_is\_\_multiple\_of\_6,show \\
i)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} 3\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} 3^2 \left( \begin{array}{l} n \\
5 \\
\end{array} \right) \minus{} ........... \equal{} 0 \\
ii)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} \frac {1}{3}\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} \frac {1}{{3^2 }}\left( \begin{array}{l} n \\
5 \\
\end{array} \right) \minus{} ........... \equal{} 0 \\
\end{array}
\]

\[ \begin{array}{l} 16)if\_(1 \plus{} 2x \plus{} 2x^2 )^n \equal{} C_0 \plus{} C_0 x \plus{} C_0 x^2 \plus{} ......... \plus{} C_{2n} x^{2n} , \\
prove \\
C_2 \equal{} 4n^2 \minus{} 2^{2 \minus{} n} n\left[ {1 \plus{} 3\left( \begin{array}{l} n \minus{} 1 \\
2 \\
\end{array} \right) \plus{} .... \plus{} (2n \minus{} 1)\left( \begin{array}{l} n \minus{} 1 \\
n \minus{} 1 \\
\end{array} \right)} \right] \\
\end{array}
\]

\[ 17)\sum\limits_{k \equal{} 0}^n {\left( { \minus{} 1} \right)^k \left( \begin{array}{l} m \\
n \minus{} k \\
\end{array} \right)} \left( \begin{array}{l} m \\
k \\
\end{array} \right) \equal{} \left\{\begin{array}{l} ( \minus{} 1)^{\frac {n}{2}} \left( \begin{array}{l} m \\
n/2 \\
\end{array} \right),if\_n\_is\_even \\
0,if\_n\_is\_odd \\
\end{array} \right.
\]

[quote="apratimdefermat"]
\[ 2)\frac {1}{{1!\left( {n \minus{} 1} \right)!}} \plus{} \frac {1}{{2!\left( {n \minus{} 2} \right)!}} \plus{} ............. \plus{} \frac {1}{{\left( {n \minus{} 1} \right)!1!}}
\]
[/quote]
It's equal to
$ \frac{\binom{n}{1} \plus{} ... \binom{n}{n\minus{}1}}{n!}$
But
$ (1\plus{}1)^n \equal{} \binom{n}{0} \plus{} ... \binom{n}{n}$.
So
$ \frac{\binom{n}{1} \plus{} ... \binom{n}{n\minus{}1}}{n!} \equal{} \frac{2^n\minus{}2}{n!}$

[quote="apratimdefermat"]
\[ 3)\left( \begin{array}{l} n \\
0 \\
\end{array} \right)^2 \minus{} \left( \begin{array}{l} n \\
1 \\
\end{array} \right)^2 \plus{} \left( \begin{array}{l} n \\
2 \\
\end{array} \right)^2 \plus{} .............. \plus{} ( \minus{} 1)^n \left( \begin{array}{l} n \\
n \\
\end{array} \right)^2 \equal{} \left\{\begin{array}{l} 0\_for\_odd\_n \\
\frac {{n!}}{{\left[ {\frac {n}{2}} \right]^2 }},for\_even\_n \\
\end{array} \right.
\]

\[ 5)\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} 2\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} 3\left( \begin{array}{l} n \\
4 \\
\end{array} \right) \plus{} .............. \plus{} \left( {n \minus{} 1} \right)\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 6)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} \frac {1}{2}\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} \frac {1}{3}\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} ......... \plus{} ( \minus{} 1)^{n \minus{} 1} \frac {1}{n}\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 7)\left( \begin{array}{l} n \\
0 \\
\end{array} \right) \minus{} 2\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \plus{} 3\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} .......... \plus{} ( \minus{} 1)^n (n \plus{} 1)\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

\[ 8)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \plus{} 2\left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} 3\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} ......... \plus{} n\left( \begin{array}{l} n \\
n \\
\end{array} \right)
\]

[/quote]

$ (1\plus{}x)^n\equal{}^nC_0\plus{}^nC_1x\plus{}^nC_2x^2\plus{}^nC_3x^3\plus{}....\minus{}\minus{}\minus{}(*)$
$ (1\minus{}x)^n\equal{}^nC_0\minus{}^nC_1x\plus{}^nC_2x^2\minus{}^nC_3x^3\plus{}....\minus{}\minus{}\minus{}\minus{}(\sim )$
$ (x\minus{}1)^n\equal{}^nC_0x^n\minus{}^nC_1x^{n\minus{}1}\plus{}^nC_2x^{n\minus{}2}\minus{}^nC_3x^{n\minus{}3}\plus{}....\minus{}\minus{}\minus{}(@)$

3) It is equal to the coefficient of $ x^n$ in the expansion of $ (x^2\minus{}1)^n$ obtained by multiplying $ (*)$ and $ (@)$.

5) Divide $ (*)$ by $ x$ and differnetiate both sides and put x=1. Take care of the $ ^nC_0$ term.

6) Divide $ (\sim )$ by $ x$ and integrate both sides and put x=1 and take care of the constant of integration and $ ^nC_0$ term.

7) Multiply $ (\sim )$ by $ x$ and differentiate.

8) Differentiate $ (*)$

[quote="apratimdefermat"]
\[ \begin{array}{l} 15)n\_is\_\_multiple\_of\_6,show \\
i)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} 3\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} 3^2 \left( \begin{array}{l} n \\
5 \\
\end{array} \right) \minus{} ........... \equal{} 0 \\
ii)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} \frac {1}{3}\left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} \frac {1}{{3^2 }}\left( \begin{array}{l} n \\
5 \\
\end{array} \right) \minus{} ........... \equal{} 0 \\
\end{array}
\]

[/quote]

(i) It is equal to $ \frac {(1\plus{}\sqrt {3} i)^n\minus{}(1\minus{}\sqrt {3} i)^n}{2\sqrt {3} i}$, where $ i\equal{}\sqrt {\minus{}1}$

(ii) It is equal to $ \frac {(1\plus{}\frac {1}{\sqrt {3}} i)^n\minus{}(1\minus{}\frac {1}{\sqrt {3}} i)^n}{2\frac {1}{\sqrt {3}} i}$

[quote="apratimdefermat"]

\[ \begin{array}{l} 14)i)\left( \begin{array}{l} n \\
1 \\
\end{array} \right) \minus{} \left( \begin{array}{l} n \\
3 \\
\end{array} \right) \plus{} \left( \begin{array}{l} n \\
5 \\
\end{array} \right) \minus{} \left( \begin{array}{l} n \\
7 \\
\end{array} \right) \plus{} .......... \equal{} 2^{\frac {n}{2}} \cos \frac {{n\pi }}{4} \\
ii)\left( \begin{array}{l} n \\
0 \\
\end{array} \right) \minus{} \left( \begin{array}{l} n \\
2 \\
\end{array} \right) \plus{} \left( \begin{array}{l} n \\
4 \\
\end{array} \right)\_\left( \begin{array}{l} n \\
6 \\
\end{array} \right) \plus{} .......... \equal{} 2^{\frac {n}{2}} \cos \frac {{n\pi }}{4} \\
\end{array}
\]

[/quote]

Wrong.

(i) is equal to $ 2^{\frac n2}\sin \frac {n\pi}{4}$

4)\left( \begin{array}{l} n \\
1 \\
\end{array} \right)^2 + 2\left( \begin{array}{l} n \\
2 \\
\end{array} \right)^2 + 3\left( \begin{array}{l} n \\
3 \\
\end{array} \right)^2 + .............. + n\left( \begin{array}{l} n \\
n \\
\end{array} \right)^2 = \frac {{\left( {2n - 1} \right)!}}{{\left( {\left( {n - 1} \right)!} \right)^2 }}


Five students of a high school decided to have a competition amongst themselves about who would be able to consume the most hot dogs during an afternoon lunch. Each of the five students reside in separate rooms.

First Names: Graham, Iris, Gordon , Trina, Gertrude.
Last names: Sturgeon, Johannson, Goldberg, Bradbury, Woodward.
# Hot Dogs Eaten: 12, 16, 20, 22, 24.
Room Numbers: 201, 202, 203, 204, 205.

CLUES:

(i) Iris, whose last name is not Goldberg ate three-fourths as many hot dogs as Trina.
(ii) Gertrude is in Room Number 203.
(iii) Bradbury ate four more hot dogs than Graham, and two more than Woodward.
(iv) Goldberg’s room is between Graham’s and Gertrude's rooms.
(v) Gordon ate more hot dogs than Graham, who ate more than Goldberg.
(vi) The student from Room Number 204 ate 8 fewer hot dogs than the student in Room Number 201.
(vii) Sturgeon resides in Room Number 202.

From the clues given above, determine how many hot dogs each student ate, the first names and the  last names of each student and their room numbers.

[hide="answer"]
Graham Johannson, 20, rm. 205
Iris Sturgeon, 12, rm. 202
Gordon Bradbury, 24, rm. 201
Trina Goldberg, 16, rm. 204
Gertrude Woodward, 22, rm. 203
[/hide]


A rectangular field is 300 feet wide and 400 feet long. Random sampling indicates that there are, on the average, three ants per square inch through out the field. [12 inches = 1 foot.] Of the following, the number that most closely approximates the number of ants in the field is

$\textbf{(A)}\ \text{500 thousand} \qquad
\textbf{(B)}\ \text{5 million} \qquad
\textbf{(C)}\ \text{50 million} \qquad
\textbf{(D)}\ \text{500 million} \qquad
\textbf{(E)}\ \text{5 billion}$

[hide]The area of the field is 120,000 square feet. Since 1 square feet = 144 square inches, the area can also be written as 17,280,000 square inches. Multiplying this by three gives us something like 50,000,000, which is 50 million. [/hide]

[hide]Well, you find the number of square feet in the field. 300*400=120000. 12^3 is 1728 so you get 17280000. You have to multiply by 3 so you can get the number of ants. 17*3=51 so you get 51280000 which is close to 50 million[/hide]

[hide]
The field is 120000 square feet, there is 144 inches in every square foot.
120000*144=17280000. multiplying by three gives you 51840000.
So, the answer is C[/hide]


Convex quadrilateral $ABCD$ is cyclic in circle $(O)$, $P$ is the intersection of the diagonals $AC$ and $BD$. Circle $(O_{1})$ passes through $P$ and $B$, circle $(O_{2})$ passes through $P$ and $A$, Circles $(O_{1})$ and $(O_{2})$ intersect at $P$ and $Q$. $(O_{1})$, $(O_{2})$ intersect $(O)$ at another points $E$, $F$ (besides $B$, $A$), respectively. Prove that $PQ$, $CE$, $DF$ are concurrent or parallel.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=34102 .

  darij


Pretty sure the steel ball will, considering that the pillow just kind of floats around in the water.

"1.5 pound closed pillowcase 8" by 20" Is it just an empty pillow case or is it stuffed with feathers or sumthin??

It depends.  I was thinking originally that the steel ball never reaches land, that it first lands in the ocean and sinks, while the pillowcase floats to san francisco, but there's air resistance there too...

But the steel ball won't float, considering it's not as dense as water here...(hollow?)

I'm going to say the pillowcase, because they never dropped either out of the airplane because it's too cold?  :maybe: there's no definite answer

[quote="mathnerd314"][quote="Ihatepie"]Two people Bob and Joe are flying from Japan to San Francisco in a private jet. After 2 hours of the flight, they are 30,000 feet above sea level. Bob drops a 1 pound steel ball with a radius of 4 inches out the side 5 seconds after Joe drops a 1.5 pound closed pillowcase 8" by 20" out the side of the airplane. Not counting wind, only air resistance, will the steel ball or the pillowcase hit the ground first?[/quote]

[hide]1. All the air in the airplane would get sucked out and they would be DEAD.
2. Ground? What ground? We're over the Pacific Ocean.[/hide][/quote]
Yeah I tried to make it seem really hard but yet that was the simple answer

Oops...I assumed that by ground you meant the floor of the ocean!! gosh... :oops:

In space you can't drop anything?

or maybe

there is no ground in space?

[quote="moogra"]heres a similar problem:
It is a well-known fact that, if you drop a feather and a bowling ball at the same time, the bowling ball will hit the ground first. But this is only because it has less air resistance than the feather. So, if you are in a rocket that is in outer space, and you drop a bowling ball and a feather at the same time, which would hit the ground first?[/quote]

There's no sound. :?

[quote="kstan013"]If you're in a rocket and you drop them, they will never hit they ground, particularly because THERE IS NO GROUND!!!

Second of all, no one ever flies in a rocket. Rockets are the things that give the boost and are released after about 2 minutes of launch.

Third, if you mean the floor of the SHUTTLE, it will never it the ground because it will just float in space.

Try a better problem next time.  :rotfl:  :rotfl:  :rotfl:[/quote]

1. thats the answer
2. k
3. that's the point


Prove that
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq\frac{3}{2}(a\plus{}b\plus{}c\minus{}1)$,
where $ a, b, c$ are positive real numbers such that $ abc\equal{}1$. When does equality hold?

[quote="Sudoku"]Prove that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {3}{2}(a \plus{} b \plus{} c \minus{} 1)$,
where $ a, b, c$ are positive real numbers such that $ abc \equal{} 1$. When does equality hold?[/quote]

By denote $ a \equal{} \frac{{x^2 }}{{yz}},b \equal{} \frac{{y^2 }}{{zx}},c \equal{} \frac{{z^2 }}{{xy}}$ then the inequality becomes
\[ \frac{{x^3 }}{{y^3 }} \plus{} \frac{{y^3 }}{{z^3 }} \plus{} \frac{{z^3 }}{{x^3 }} \ge 3\left( {\frac{{x^2 }}{{yz}} \plus{} \frac{{y^2 }}{{zx}} \plus{} \frac{{z^2 }}{{xy}} \minus{} 1} \right)
\]
Denote again $ a \equal{} \frac{x}{y},b \equal{} \frac{y}{z},c \equal{} \frac{z}{x}$ then it becomes
\[ 2\left( {a^3  \plus{} b^3  \plus{} c^3 } \right) \plus{} 3abc \ge 3\left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right)
\]
By Schur and AM-GM inquality we have
\[ 2\left( {a^3  \plus{} b^3  \plus{} c^3 } \right) \plus{} 3abc \ge \left( {a^3  \plus{} b^3  \plus{} c^3 } \right) \plus{} \sum {ab\left( {a \plus{} b} \right)}  \ge 3\sum {a^2 b}
\]
:)


Naravno da znamo:

A. Ademović, A. Beširević, J. Radović, V. Uljarević, M. Ferizbegović, R. Darda.

Valjda nisam nešto fulao, a zadatke, rješenja i rang-liste očekujte na

http://www.pmf.unsa.ba/matematika/Udruzenje%20matematicara/index.htm

svaka cast svima posebno ratku i mini jer su 1. i 2. razred =)

:wow:  :coolspeak: (Prešao sam samo u smileye...)

Još da si moderator  :lol:

 :harhar:  :juggle:  :gleam:  :police:

ili u prijevodu: nja-nja-nja, ja se mogu igrati jer sam cool moderator  :lol:

[quote="hsiljak"]

Valjda nisam nešto fulao, a zadatke, rješenja i rang-liste očekujte na

http://www.pmf.unsa.ba/matematika/Udruzenje%20matematicara/index.htm[/quote]

   I checked the link and found nothing. Is there any site with BIH Olympiad problems?
   (Sorry for writing in English.)

I hoped the problems will be published on the site, and it turned out that I was hoping too much.

B&H Mathematical Olympiads 1996-2006 are available in

http://www.imomath.com/index.php?options=oth|Bih|Ne|Ne&ttn=Bosnia%20and%20Hercegovina&p=0 (in English)

The original texts of some MOs are available from

http://srb.imomath.com/zadaci/bihmo.doc

And the B&H MO 2008 is available here:

http://www.mathlinks.ro/resources.php?c=23&cid=147

Therefore, only IMO 2007 and 2009 are not online. Maybe some of the people who have them 'on paper' could upload the problems (I would if I had the texts).


floor function
number theory
least common multiple
relatively prime

If you want the number of integers divisible by $ n$ in general, use $ \left\lfloor\frac{10000}{n}\right\rfloor$, where $ \left\lfloor\cdot\right\rfloor$ denotes the greatest integer value function.

You take the number of integers that are divisible by $ 13$, and add them to the numbers that are divisible by $ 51$, and subtract the number of integers that divisible by both $ 51$ and $ 13$ from the previous sum.
$ \dfrac{10,000}{13}\equal{}769.23$ but we just want the $ 769$
$ \dfrac{10,000}{51}\equal{}169.08$ but we just want the $ 169$
We factor $ 13$ and $ 51$ to get $ 13\equal{}13$ and $ 51\equal{}3 \cdot 17$ So the least common multiple is $ 3 \cdot 13 \cdot 17\equal{}663$ There is only one number less than $ 10,000$, which is divisible by $ 663$ so we do $ 769\plus{}169\minus{}1\equal{} \boxed{938}$

[quote="AIME15USAMO"]There is only one number less than 10,000, which is divisible by 663[/quote]
Not quite. :wink:

[quote="AIME15USAMO"]
$ \dfrac{10,000}{13} \equal{} 769.23$ but we just want the $ 769$[/quote]
This is right

[quote="AIME15USAMO"]$ \dfrac{10,000}{51} \equal{} 169.08$ but we just want the $ 169$[/quote]
This is wrong... $ \dfrac{10,000}{51}\approx 196.08$, so we just want $ 196$.

[quote="AIME15USAMO"]
There is only one number less than $ 10,000$, which is divisible by $ 663$[/quote]
As JRav pointed out, this is also wrong  :( .  There are a total of $ \Big\lfloor\frac {10000}{663}\Big\rfloor \equal{} 15$ multiples of $ 663$ less than $ 10000$.

By [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]PIE[/url], the number of integers divisible by $ 13$ or $ 51$ from $ 1$ to $ 10000$ is $ 769 \plus{} 196 \minus{} 15 \equal{} \boxed{950}$.

[quote="yshridhar"]Find the number of integers divisible by 13 or 51 from 1 to 10,000
Regards
Sridhar[/quote]

[hide]$ \dfrac{10000}{13} \equal{} 769.23$ but we only consider the 769.
$ \dfrac{10000}{51} \equal{} 196.07$ but we only consider the 196.

Because 13 and 51 are relatively prime, the LCM of the two numbers is simply $ 51\cdot 13 \equal{} 663.$

There are $ \dfrac{10000}{663}$=15 multiples of 663 less than 10000.

After this, we simply perform the arithmetic 769+196-15=$ \fbox{950}.$
[/hide]

Thank you all for your solutions
Regards
Sridhar


If a,b>0 and a^n=a+1, b^2n=b+3a, where n is a natural number, n \geq 2, compare a with b.

I remember that this was also posed before ... we have that a>b..

well the proff is based on the fact that you presume that a<=b and in the end you'll get that a^(2n)/b^(2n)<=1.

this problem is simmilar to the following one which was given at the math district  olimpics in romania -- 2002:
Let a,b be two real numbers which satisfy the following relations: 3^a+13^b=17^a and 5^a+7^b=11^b. Prove that a<b.

you solve this by considering that a>=b and then you get that 13^a>=13^b which gives that 13^a>=17^a-3^a which gives you that 3^a+13^a>=17^a (1) and 5^a>=5^b which gives you that: 11^b-7^b>=5^b which gives you that 5^b+7^b<=11^b (2). But from (1) we get  that (3/17)^a+(13/17)^a>=1, but we know that the function h(x)=(3/17)^x+(13/17)^x is strictly decreasing and h(1)=16/17<1 and we
get that h(a)>h(1) from which we get that a<1 (3). from (2) we get that (5/11)^b+(7/11)^b<=1 and we know that the function p(x)=(5/11)^x+(7/11)^x is stricty decreasing and p(1)=12/11>1 so we get that p(b)<p(1) and from (3) and (4) we get that a<1<b, contradiction.

the same you do with your problem!!

cheers! :D :D

Doesn't this belong to Cristinel Mortici?

the one that ciprian posted doesn't belong to him , but the one I posted yes.

cheers!

just a quick notice: problems do not belong to anyone ... they may have been created  or proposed at a contest by someone.


geometry
geometric transformation
analytic geometry

$ |x\plus{}2|\plus{}|y\minus{}3|\equal{}1$ is an equation for a square. How many units are in the lengths of its diagonals?
[hide="Answer"]2 units[/hide]

[hide]
Since translation doesn't affect area, you could just solve $ |x|\plus{}|y|\equal{}1$, which I believe is a square centered at the origin with its vertices on the axes, with diagonal length 2.
[/hide]

if you solved abs(x)+abs(y)=1 the verticies of the square would be at +/-1,+/-1,so the squares side length would be two, but the diagonal would be 2sqrt2,which is what the question asks for

i hope i make nats  :lol:

2 things one, the answer is 2  :D

how would it work out with |x| + |y| = 1

if it is |1| + |-1|?
that equals 2, not one.

You would get the ordered pairs (1,0), (0,1), (-1,0) and (0,-1) for the square.   And, as infinity4ever stated that translation doesn't affect area (x+2 and y-3 just move the square around the coordinate plane), the sides would still have to be 1.  Therefore the diagonal is 2.


topology
real analysis
real analysis unsolved

Consider $R$ with the base for some topology be the collection of sets of the kind [a,b). Show that it is not second countable under this given topology.

Note:It is separable but not second countable.So it cannot be a metric space.

Let B be a base for the topology.

For each real number x, the interval [x, x + 1 ) is an open neighborhood of x, so there must be a base element [a, b ) in B such that x $\in$ [a,b) $\subseteq$ [x, x+1). we see that a = x.

We have proved that for each real number x, there is b > x such that [x,b) is in B. Therefore B is uncountable. QED


limit
logarithms
calculus
calculus computations

two questions

1) Find $\displaystyle\lim_{x\to\infty}\left(\frac{ln(x)}{ln(x+1)}\right)^{x}$

2) Let $f: R\rightarrow{R}$ be continuous, wiht $f(x)*f(f(x))=1$ for all real $x$. If $f(1000)=999$, find $f(500)$.

x:=1000  implies that $1000\cdot f(999)=1\Rightarrow \frac 1 {1000}\in f(\mathbb{R})$.
since $999=f(1000)\in f(\mathbb{R})$ and f is continuous we have that $f(\mathbb{R})$ is interval hence, $500\in f(\mathbb{R})$. For any $x\in f(R)$ we have that $xf(x)=1$ hence $f(500)=\frac{1}{500}$.

could you please explain how u solved the limit question.

I used the following facts:
1) If $f(x) \to 0$ then $(1+f(x))^{\frac 1 {f(x)}}\to e$
2) If $f(x)\to 0$ and $f(x)g(x)\to \ell$ then $\lim (1+f(x))^{g(x)}=\lim ((1+f(x))^{\frac 1 {f(x)}})^{g(x)f(x)}=e^{\lim f(x)g(x)}=e^\ell$
all limits are when $x\to \infty$.

could u use L'Hopital's Rule on the first part...i know it would be ugly, but it is doable righ?


Can anyone show me any others for this year because here is Tennessee's. (see attachment)

ours said Mickey Unlimited and had the shape of mickey's head on it...  disneyland donates cali's pins every year, y'see...

*crosses fingers and prays for a decent connecticut pin*

KS has a sunflower on it.  Each contestant gets about 75 pins to play the exchange game. :)

Georgia's gotten their pins, but the team members haven't seen them yet. (Though we might today; we're getting our team shirts today.)

At least we have ours this year...

OK good so you DID get the message that we are having practice today instead of tomorrow...

How do you know we're getting our shirts today??

Apparently as far as I know, our pins are similar to last year, with the pin in the shape of the state, but our state coordinator said that they were different from last year...iono...

We have to wait till thursday to get ours.

We got to see ours at practice today...they are still in the shape of our state (georgia) but instead of just having georgia written on it, it has the state seal on it...not the animal...yall know wut i mean

arizona had three different kinds,  a cactus, shape of arizona, and like a desert scene thing.  Anyways, I like the virgin islands one, the guam one, the illinois, and the georgia one.  Ya those were some of the cool ones I think especiall vi

[quote="314159265358979323846"]New York's this year are the same ugly I love NY pins we've had every year...

anyone know what Puerto Rico has this year?

I'm not going, but I'd love it if someone from puerto rico could mail me one of their awesome pins or give extras to my sister !!!  :)[/quote]

What I got was a yellow frog of some sort but I think there are different ones

[quote="yif man12"]i'm pretty sure puerto rico only had the koqui.
when my friend was reading the paper that it was on, i thought he said cocaine...[/quote]

lol

hey people, how much are those pins worth anyway?
can people sell them?
haha[/b]


Given a,b,c>0 :
          ab+bc+ca=1
Prove that:

[quote="duc 95"]this a file[/quote]

see here

http://www.irgoc.org/viewtopic.php?f=27&t=48&sid=fa1a0b11faf2d14f0e46107d8aa5f3ce&start=50[/url]

or

http://www.mathlinks.ro/Forum/viewtopic.php?t=313221[/url]

[quote="duc 95"]Can you solve it only by knowledge of secondary student[/quote]

I can not.    :blush:

[quote="duc 95"]I think this way is quite difficult and not natural.[/quote]

but I think this way is  useful.

[quote="duc 95"]Can you tell me what your idea is when you let a,b,c be... ?[/quote]

sorry  my  English is  poor,I can not answer you questions,you can come to here ask Professor Lu Yang .

[url]http://www.irgoc.org/viewforum.php?f=10&sid=90e19c99a8fa780feb87357f6f4d6a2e[/url]


I am new to TeXnicCenter, so this might be obvious:  How do you put equations on the same line as the text immediately following it?  I am trying to create a mock ARML, but the equations won't stay on the same line as the text, they keep going down to the next line.  If you don't understand my problem, see the PDF attachment.

I think that you put your expressions between four $\$$. You must put it between two dollars.

Example: Let $\$ z \$$ be a real number  =>  Let $z$ be a real number.
               Let $\$\$ z\$\$$ be a real number =>  Let \[ z \] be a real number.

I think that is your problem  :huh:


Im new to all this, but I do need help on the very first problem in the Mock AMC 12. Can someone answer detailedly? Here is the tag: http://www.artofproblemsolving.com/Forum/weblog.php?w=1597

Could you post a direct link to the specific post you are talking about? I can't really tell which one it is.

I'm flattered to see that that link directs to my blog.

Are you asking about Intricate Inquiry #5? P.S. There are no mock AMC 12 problems there. I shall post practice problems.


[quote="Allnames"]Find the greatest and the least value of expression :
\[ P \equal{} x^2 \minus{} xy \plus{} 3y^2
\]
where $ x,y$ are real number satisfying
\[ x^2 \plus{} xy \plus{} y^2\le 3
\]
[/quote]
I have got $ \max P\equal{}11\minus{}4\sqrt7$ and $ \min P\equal{}0.$ :maybe:

I think, your "maximum" is not maximal. The minimum can be found very simply, because
$ x^2 \minus{} xy \plus{} 3y^2 \equal{} \left (x \minus{} \frac {y}{2}\right )^2 \plus{} \frac {11}{4}y^2\ge 0$ and the equality holds iff $ x \minus{} \frac {y}{2} \equal{} 0$ and $ y \equal{} 0$ under the given copnstraint $ C(x,y): \; x^2 \plus{} xy \plus{} y^2\le 3$. Hence, we take $ x \equal{} y \equal{} 0$ which implies that $ \min_{C(x,y)} P \equal{} 0$.

The question of the maximal possible value of the term P under the constraint C(x,y) can be done with the following easy consideration:

$ \begin{array}{rcl} x^2 \minus{} xy \plus{} 3y^2 & \equal{} & (x^2 \plus{} xy \plus{} y^2) \plus{} 2(y^2 \minus{} xy)\le 3 \plus{} 2(y^2 \minus{} xy)\le 3 \plus{} 2(3 \minus{} x^2 \minus{} 2xy) \equal{} \\
& \equal{} & 9 \minus{} 2(x^2 \plus{} 2xy \plus{} y^2) \plus{} 2y^2 \equal{} 9 \minus{} 2(x \plus{} y)^2 \plus{} 2y^2. \end{array}$
The equality in the previous estimation occurs iff the equality occurs in the constraint C(x,y). Now, taking $ x \equal{} \minus{} y$ and $ x^2 \plus{} xy \plus{} y^2 \equal{} 3$ we obtain that
$ x^2 \minus{} xy \plus{} 3y^2\le 9 \plus{} 2(\sqrt {3})^2 \equal{} 15$ (by setting $ x \equal{} \minus{} y \equal{} \sqrt {3}$).

May be, I make a mistake. On the other hand, it works.

My mistake, M@ri@n. Thank you!
By the way, $ x^2 \minus{} xy \plus{} 3y^2 \equal{} 15$ for $ x \equal{} \minus{} y \equal{} \sqrt3.$  :P


Let $f: \Re \longrightarrow \Re$, such that:

$f(xy)(f(x) - f(y)) = (x-y)f(x)f(y),$ for all $x,y \in \Re$

and also $f$ is a increasing function.

Find all function $f$.

Carlos Bravo
Lima-PERU

$f$ is increasing, so isn't constant.
let $y = 1$ in the equation.
we get $f(x) = x.f(1)$.

the function $f(x) = ax$, with $a>0$ is good.  :)


group theory
abstract algebra
vector
superior algebra
superior algebra solved

Is this really easy? Maybe I'm missing something.

Let G be a group s.t. x^2=e for all x \in G. How can I prove that there is at least a surjective morphism f:G->Z_2 ? What I want to prove is that there is a subgroup H of G s.g. |G/H|=2. Then I want to use G/ker(f) isomorphic to im(f)=Z_2. I bet it's easy, but I don't know how to do it.. :(

Come on, dudes! Is it really hard, or am I missing something obvious?

It's not true if G={e}  :)

Is G supposed to be finite ? If G is finite is not too difficult.

I'll solve the problem for G<>{e}.
----------------------------------------------------------------------------------
First, a simple observation: G is commutative.
Proof of observation: xy*xy=e, so xxy*xyy=x*y, so y*x=x*y, for all x,y \in G.

----------------------------------------------------------------------------------
If G is finite, then it should be easy to see that G is isomorphic to Z2n for some n in N*.
0*Z2n-1 is a subgroup of Z2n and it checks the problem.

----------------------------------------------------------------------------------
----------------------------------------------------------------------------------
If (G,*) is infinite. Chose a in G\{e}.
Let S be the set of subsets of G\{a} [i](I edited this now from G to G\{a}, so be careful because it makes a big difference)[/i] that are stable with respect to *.

S isn't the empty set, because S includes {e,y} which is a subgroup of G for some y in the infinite set G\{a,e}, so it is stable by *.

Consider S ordered with respect to the inclusion.

----------------------------------------------------------------------------------
I'll use [b]Zorn's Lemma[/b]:
That is: [b]An inductively ordered set A, has a maximal element.[/b]

Lets translate:
An [b]ordered set[/b] (A,<) is a set such that < satisfies:

1)a<b and b<a then a=b;
2)a<b and b<c then a<c;
3)a<a; for all a,b,c \in A.
[i]Examples of order realtions:[/i]  \leq for reals, "|" for integers, inclusion for sets.
[i]Counterexample:[/i] < for reals.

Let S be a subset of A. Then an element m in A is called a [b]majorant[/b] for S if s<m for all s \in S.

An element M of A is called [b]maximal[/b], (if MM=a for a \inA);

[b]inductively ordered[/b]- every chain in A has a majorant in A;

[b]chain[/b]- totally ordered subset;

A subset S of A is called a [b]totally ordered subset[/b] if for all a,b in S, a<b or b<a.

[i]Note: It can be proven (don't try this at home :D) that Zorn's Lemma is equivalent to the axiom of choice and to Zermelo's Theorem.[/i]
----------------------------------------------------------------------------------

Let T a subset of S that is totally ordered. Let P =  \cup R \in T R.
I'll prove that P is a majorant for R. To do that, all I have to do is to check that P is in S (i.e. it is stable by * - because obviously R<P for all R in T).
Let a, b be in P. Then there exist R, R' in T such that a \in R and b\in R'. T is a chain so R<R' or R'<R ("<" means included in). WLOG, suppose R<R'. Then a,b \in R' which is stably by *, so a*b \in R'<P, so a*b \in P, so P is stable by *, so P is in S. So P is a majorant for T.
T was arbitrarily chosen, so every chain in S has a majornat in S, so S is inductive. [u]So, by Zorn, there exists a maximal element M of S[/u].

[u]Lets prove that M is subgroup of G[/u]. Let x in M. Then M is stable by *, so e=x*x \in M. So M has unity. G is commutative, so M is commutative.
x 2 =e, so every element of M is invertible. x*y \in M for all x,y \in M because M is stable. So M is a subgroup of G. [u]Actually every stable subset of G is a subgroup of G.[/u]

Here comes the maximality of M:
I'll prove G=M \cup aM, where aM={a*m| m \in M}.
Suppose G<>M \cup aM. Then there exists y in G\(M \cup aM).
Let M'=M \cup yM. [u]Lets prove M' is a subgroup of G. To do that, all we have to prove is that M' is stable.[/u]
Denote by AB={a*b| a \in A and b \in B}. We want to prove M'M'=M'.
M'M'=(M \cup yM)(M \cup yM)=(MM \cup MyM \cup yMM \cup yMyM)=(M \cup yM), because of the commutativity of G and y 2=e.
So M' is stable. But obviously, M' includes M, so M<M' and M<>M'. This contradicts the maximality of M.
So G=M \cup aM. This proves |G/M|=2.
Let f:G->{e,a} defined by f(x)=e, for x in M and f(x)=a, for x in aM. It is easy to see that f is a surjective morphism and {e,a} is obviously isomorphic to Z2.
-----------------------------------------------------------------------------------
Try to find a simpler proof. I couldn't.

The conclusion after this splendid solution: grobber, try to read the classical solution of that problem with |AUT(G)|=1. It's a "little" bit easier.

What's the classical solution? I found this problem in a book, but they don't have a complete solution. They just define a Z_2-vector space on G and they say "etc.", so I actually have no idea what they actually meant.

P.S. I was, of course, asking about the case when G isn't finite.

Thanks, and bye! :D

I thought I'd rekindle this candle, because it still looks like an interesting subject to me :).

The set $G$ can be organized as a vector space over $\mathbb Z_2$. I wrote this in my last post on the topic, but didn't know how to use it back then. This vector space must have a basis, $x_i,\ i\in I$ for some index set $I$. These $x_i$ are, in fact, an independent set of generators of $G$ as a group. Now take the subgroup generated by $(x_i)_{i\in I}\setminus\{x_a\}$ for some fixed $a\in I$. This subgroup has index $2$ in $G$.

It looks correct from here.. What about other angles? :)

[quote]Note: It can be proven (don't try this at home ) that Zorn's Lemma is equivalent to the axiom of choice and to Zermelo's Theorem. [/quote]
amfulger can you please tell me where can I find a solution of this note? or... maybe you can post a sol here
thanks

I think grobber can help you (us) with that. He mentioned me smth about a book that deals with this subject (among others).

I am looking at a proof of Zorn's theorem (assuming the Axiom of Choice) as we speak. The proof itself is not long (a little bit longer than half a page), but the refrences needed for it span over quite a few pages, so it would be hard for me to post it right now (especially since I didn't actually read this, so I might make some mistakes in the exposition; I'm unfamiliar with the notations and so on). Sorry for that..

The book (it's Romanian) is called "Fundamentele Matematicii", and the author is Izu Valsman.


geometry
geometric transformation
rotation
circumcircle
rectangle

[color=darkred]Let $ \triangle\ ABC$ with the orthocenter $ H$. Let  $ H_1$ and $ H_2$ be the feets of perpendiculars from $ H$ to the external and internal angle bisectors of $ \widehat {BAC}$. Let $ M$ be the midpoint of side $ [BC]$. Prove that $ M\in H_1H_2$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=169457 [/color]

[hide]
So $ AH_1HH_2$ is clearly a rectangle.  Then the angle of rotation which takes $ AH$ to $ AH_2$ is the one which takes $ AH_2$ to $ H_1H_2$.  But this is also the rotation which takes $ AH$ to $ AO$.  Therefore, $ AO\parallel{}H_1H_2$.  Let $ H_2$ be the midpoint of $ AH$.  Clearly $ H_1H_2$ goes through $ H_3$.  It now remains to show that $ MH_3\parallel{}AO$.  Let $ AO$ intersect the circumcircle of $ ABC$ again at $ A'$.  Then $ H, M, A'$ are collinear.  Therefore, $ MH_3$ is a midsegment of $ \triangle{AA'H}$, which gives $ AO\parallel{}MH_3$.  Thus, $ M \in H_1H_2$, as desired.[/hide]

[color=darkblue][b]Proof.[/b] This problem is nicely, but very easily! It is an aplication of the remarkable properties of a triangle w.r.t. the its circumcircle. See the triangle $ AHA'$, where $ [AA']$ is a diameter in the circumcircle $ C(O,R)$ of the triangle $ ABC$. Denote the midpoint $ S$ of the segment $ [AH]$. You know that $ M\in HA'$ (the its midpoint) and $ SM\parallel AO$ (the segment $ [SM]$ is the diameter of the Euler's circle w.r.t. the triangle $ ABC$). Denote the intersections $ \{\begin{array}{c}
D\in AI\cap BC\\\
L\in HH_2\cap AO\end{array}$. Since the ray $ [AD$ is the bisector of the angle $ \widehat {HAO}$, then $ H_2H=H_2L$. Thus, $ H_2\in SM$. But $ H_2\in H_1S$ (the point $ S$ is the center of the rectangle $ AH_1HH_2$). In conclusion, $ M\in H_1H_2$. [/color]

If $BB', CC'$ are the other altitudes, $MB'=MC'=MB=MC$, as radii of the circle $\odot (BCB'C')$, hence $SC'MB'$ is a kite and $MS\bot B'C'$, but $H_1, H_2$ are midpoints of the arcs $\stackrel{\frown}{B'C'}$ so $M, S, H_1, H_2$ are in line.

Best regards,
sunken rock


How can you include 3D graphs in $\LaTeX$?  Like for example the simple one $x=\frac{2}{y}=z^{3}$, or something like that?  Give at least one supporting package or url unless your post really tells me about what the program is about .

BIG IDEA!

Have you tried GNUPlot?  It can graph really cool things, and all you have to do is learn the code.  This is the Windows version:

[url=http://archives.math.utk.edu/software/multi-platform/gnuplot/msdos/gpt35win.zip]GNUPlot for Windows[/url]

You can also download it for Macs:

Getting gnuplot
Download [url=http://prdownloads.sourceforge.net/gnuplot/Gnuplot-4.0.0.dmg?use_mirror=umn]gnuplot, version 3.8i.0[/url] (this is the latest MacOSX compatible version at the time of the writing of this file).
Installing gnuplot
Open the Terminal, and become superuser by typing in su and typing-in your root password when asked to do so. Then cd to the folder where you downloaded gnuplot (for example /Users/goofy/downloads):
cd /Users/goofy/downloads

Unpack the archive:
tar -zxvf gnuplot-3.8i.0.tar.gz

Go into the newly created directory:
cd gnuplot-3.8i.0

Configure gnuplot (don't break the line):
./configure --prefix=/sw --without-gd --mandir='$\$${prefix}/share/man' --with-readline=gnu --enable-mouse --enable-pm3d --with-png --with-pdf --enable-filledboxes --enable-relative-boxwidth --enable-thin-splines
If you don't have libpng installed, remove the command --with-png. If you don't have teTeX installed or any Terminal-based TeX/LaTeX installation, remove the command --with-pdf.

When it's done, compile:
make

When it's done, install gnuplot:
make install

When it's done, resign from superuser:
exit

Now you should have a working installation of gnuplot in /sw/bin/gnuplot.


induction
combinatorial geometry
combinatorics proposed
combinatorics

Let $T$ a set with 2007 points on the plane, without any 3 collinear points.
Let $P$ any point which belongs to $T$.
Prove that the number of triangles that contains the point $P$ inside and
its vertices are from $T$, is even.

As $P$ is not contained inside any triangle with vertice $P$ we can  reformulate the problem:

Given an even number $n$ (say 2006) of points in the plain without any 3 collinear points, let $E$ be the union of all corresponding straight lines (determined by all pairs of points), and let $F : \mathbb{R}^{2}\setminus E \to \mathbb{N}$, where $F(x)$ is the number of triangles such that $x$ is contained in them. We have to show that $F$ takes even values only.

First some easy observations:
If $x$ is outside of the convex hull of the given points, then $F(x) = 0$.
$F$ is constant on any connected component of $\mathbb{R}^{2}\setminus E$.
The connected components are convex polygons.
We can move between connected components which are adjacent (in the sense that the intersection $e$ of their boundaries contains more than one point) by passing through an interior point of the line segment $e$. We call this a "move".
We can move from any connected component to any connected component by a sequence of moves.

Thus it suffices to show that for any move $F$ changes by an even number. But this is easy:

For convenience let us suppose that the move of point $x$ is from the left of line segment $e$ to the right of $e$.  Let $g$ be the straight line which contains $e$, and $P$ and $P'$ the two points which are on that line. For any triangle which is not formed by $P$, $P'$ and a third point the position of $x$ w.r.t. this triangle does not change at all. So it suffices to consider all triangles  formed by $P$, $P'$ and a third point.To the left of $g$ there are $l$ of the remaining points, to the right of $g$ there are $r$ points, and we have $l+r = n-2$, so either $l,r$ are both even numbers or $l,r$ are both odd.
Before the move $x$ is contained in every one of the left and none of the right triangles, and after the move $x$ is contained in every one of the right and none of the left triangles. So $F(x)$ changes by $r-l$ which is an even number.

we porve by induction on k the general case where 2007 is replaced with $2k+1$
$K=0$,trivial
suppose the assertion holds when $k=k$,let us consider $k+1$
by induction hypothesis,it suffices to prove the difference of the numbers of wanted triangles,when two other points are promoted,is even.
let $N_1,N_2$ be two newly promoted points,$N_1',N_2'$ be the reverse expansion of $PN_1,PN_2$.
we picked up a point $P$ arbitraily,and draw a line from any other point to P.we partition other points into $4$ classes A,B,C,D:
A:those in angle $N_1'PN_2'$
B:those in angle $N_1PN_2'$
C:those in angle $N_1'PN_2$
D:those in angle $N_1PN_2$
it's trivial to prove the following results:
(1)$N_1,N_2,A_i$ form |A| such triangles:
(2)$N_1,b_i,C_j$ and $N_2,D_i,A_j$ form or don't form such triangles simutaneously,so the number is even;
(3)only one of $N_1,D_i,A_j$ and $N_2,D_i,A_j$ is wanted;
(4)$N_2,A_i,B_j$ and $N_1,A_i,C_j$ are both wanted;
(5)no other triangle is formed.
so the difference (modulo 2) is $|A|+|A|(|B|+|C|+|D|)\equiv |A|-|A|^2\equiv 0(mod2)$
induction finished


[quote="RC-7th"]"If sat, categories will appear with a forum layer."

It appears in options>>board layout. I think "sat" should be "set".[/quote]It can be interpreted as the past tense of the verb sitting :) so I'm not really sure if it is or not a spelling error yet :)

I agree with RC-7th; we'll fix it (thanks for pointing out).

Also, you may find other weird grammar items like this - some of the modifications we used were created by people who are not native English speakers.  Please let us know when you see them!


number theory
least common multiple
MATHCOUNTS
modular arithmetic

I'm having lots of trouble with this one:

ind the smallest natural number that is divisible by 13 which when divided by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 each time leaves a remainder of 1.  If you get it can you show me how to get it.

it should be 479001601 that is obtained:
(12!) +1

thanks!  by the way what does your signature mean?

Not quite. While $12!+1$ is indeed divisble by 13, it is not the [i]smallest[/i] answer.

First, we will need to find the smallest possible number that when divided by 1-12 leave a remainder of 1. This is somehow related to the LCM (figure it out :))

Afterwards, we should note that 13 is a prime number. If a number is divisible by a composite number, say for instance 6, then it must be divisible by 2 and 3 which are factors of 6 (think about this a bit). However, what happens when we require the number is divisible by a [i]prime[/i] number?

[hide]Isn't it 277199? The LCM of all of the numbers is 27720. If you subtract 1, it doesn't come to a whole #. The smallest that it would come to a whole # is 27720 * 10 -1. This is equivalent to 21323 * 13. [/hide]

i don't agree with you
if you dived
277199 by 13 it's what u said but if you dived 277198 by 12 it comes a decimal number.....sorry  :?

i'm really sorry! :blush:
won't happen again.
and i think EUCLA is right

It's ok. I didn't mean to be offensive or anything. Hmm... I'll have to look at it again.  :)

Yeah, now I know I'm wrong. I divided it by 3 and got a remainder of 2.  :blush:

[quote="Naga"]LCM of 1 thru 12 =27720

Find the smallest multiple of 27720 that if you add 1 to, you get a multiple of 13.

The answer is 83161.[/quote]

Yeah, that's what I got. Here's how I did it:

[hide="Solution"]$k-1\equiv 0 \pmod {2}$
$k-1\equiv 0 \pmod {3}$
$k-1\equiv 0 \pmod {4}$
.
.
.

So without taking the fact that k is a multiple of 13 into account, $k-1=LCM(2,3,4,5,6,...,12)=27720.$

Now, $27720 \equiv 4 \pmod {13}$, so now we know that we must mulitiply 27720 by a number x, such that $4x \equiv 12 \pmod {13}$. The smallest value of x is 3.

$3(27720)+1=83161$.

Thus, the answer is 83161. [/hide]

[size=59]I edited my LaTeX.[/size]

[quote="elementsofmathkid"]I'm right???  And btw- [b][size=150]I AM A GIRL.... [/size][/b]

I just don't put that on the thing because then some people stereotype girls as not being as capable, and being dumb to some extent. That's just what I experience. Maybe that's just because though any contest or something I do usually has the rest of the team as guys... MATHCOUNTS, Science Bowl....[/quote]

There's a gender icon option in the Profile section, just to let you know :) .

[hide="Solution"]
2
6
12
60
60
420
840
2520
2520
27720

plus one.

27721, I think.
[/hide]
[/hide]


geometry
geometric transformation
reflection
trigonometry
dilation
geometry proposed

Given a quadrilateral ABCD,
Prove that the quadrilateral formed by the intersections other than the vertices A, B, C & D
of the circles with diameter AB, BC, CD & DA
is similar to the quadrilaterar ABCD.

What is the relation of similarity?

Anyone with a solution or should I post mine?

Take $ M$, $ N$ to be the midpoints of $ AB$, $ AD$.  Then the intersection of the circles with diameters $ AB$ and $ AD$ is the reflection of $ A$ over $ MN$.  That's also the projection of $ A$ onto $ BD$.  So we want to show that the quadrilateral formed by projecting $ A, C$ onto $ BD$ and $ B, D$ onto $ AC$ is similar to $ ABCD$.  Let $ AC \cap BD\equal{}P$.  Let the projections of $ A, B, C, D$ be $ A', B', C', D'$.  Let $ \theta$ be the angle of intersection of $ AC$ and $ BD$.  So then $ PA'\equal{}PA\cos \theta$, and similar statements hold for $ PB', PC', PD'$.  Notice that if $ A$ projects onto segment $ PB$, then $ B$ projects onto segment $ PA$, because one projects onto the other iff $ \angle{APB}$ is acute.  Therefore, it follows that $ A'B'C'D'$ is a reflection and then dilation of $ ABCD$ about a line through $ P$ and then through $ P$.  So the two are similar, as desired.


Να βρείτε τις συναρτήσεις $f: N \longrightarrow N$ με τις ιδιότητες:
(1)Όλοι οι φυσικοί αριθμοί ειναι τιμές της $f$
(2)Για κάθε ζεύγος φυσικών $m,n$ να ισχύει η ισοδυναμία $m/n \Longleftrightarrow f(m)/f(n)$

Ναι αλλά είναι η μόνη;

Πολύ ωραία άσκηση  :lol:
Από τη (2) παίρνουμε ότι $f(1)|f(m)$ για κάθε m φυσικό και αφού ο $f(m)$ παίρνει όλες τις τιμές στο $N$ το $f(1)=1$
Έστω τώρα ένας τυχαίος πρώτος αριθμός $p$ .
Τότε έστω $f(p)=p^{a}\cdot p_{1}^{a_{1}}....p_{k}^{a_{k}}$ όπου $p_{1},...,p_{k}$
πρώτοι αριθμοί
Σύμφωνα με την (2) υπάρχει $u$ ώστε $f(u)=p$
Τότε $f(u)|f(p)$  άρα $u|p$ άρα $u=p$
Συνεπώς $f(p)=p$
Τώρα λαμβάνονας υπόψη ότι καθε αριθμός μπορεί να γραφεί στη μορφή $m=p_{1}^{a_{1}}....p_{k}^{a_{k}}$
παίρνουμε $f(m)=m$ για κάθε φυσικό $m$

Σας άρεσε ?
Νίκο ποια είναι η λύση σου ?

Νομίζω ότι υπάρχουν περισσότερες από μια συναρτήσεις λύσεις.

Μια συνάρτηση επαληθεύει τις δοθείσες προυποθέσεις αν και μόνο αν:

Είναι πλήρως πολλαπλασιαστική, 1-1 συνάρτηση με $f(\mathbb{P}) = \mathbb{P}$.

Silouan για τη λύση σου έχω την εντύπωση ότι το $p | f(p)$ είναι λάθος.

Ίσως έχεις δίκαιο . :maybe:
Πάντως με τον τρόπο μου αποδεικνύεται ότι $f(P)=P$ και τα άλλα που έγραψες....

cmad,silouane ligo thelei....ante pame!!

σιγουρα το πεδιο ορισμου της συναρτησης του συνολο των φυσικων ?
η συναρτηση πρεπει να εχει μια μεταβλητη ?

[quote="kostas"]ποτε θα δεις αυτο το ποστ και να δωσεις τις λυσεις?  :spam:[/quote]

sorry kosta eixa bleksei tora me ton eukleidi kai to eixa ksexasei teleios...loipon molis gyriso apo to sxoleio tha ghrapso ti lysi!

Ουσιαστικά αν κοιτάξεις τα παραπάνω ποστ βρίσκεις τη λύση .
Λοιπόν όπως στο προηγούμενό μου ποστ βρίσκουμε ότι $f(1)=1$
Τώρα αν $f(k)=f(m)$ τότε $f(m)|f(k)$ kai $f(k)|f(m)$ άρα $m|k$  kai $k|m$ άρα $k=m$ άρα η $f$ είναι 1-1 αλλά η $f$ είναι και επί .Άρα υπάρχει αντίστροφη συνάρτηση .
Τώρα πλέον είναι σχετικά εύκολο να δείξουμε ότι $f(p^{k})=(f(p))^{k}$ αν $p$ είναι ένας πρώτος αριθμός .Επίσης δείχνουμε ότι
$f(P)=P$   όπου $p$ τοσύνολο των πρώτων αριθμών . Τέλος δείχνουμε ότι $f(a)f(b)=f(ab)$ .
Τελικά οι συναρτήσεις που ικανοποιούν το πρόβλημα έχουν τη μορφή : $f(1)=1$ και αν $n=p_{1}^{a_{1}}....p_{k}^{a_{k}}$
t;ote $f(n)=q(p_{1})^{a_{1}}....q(p_{k})^{a_{k}}$
όπου $q$ είναι μια αμφιμονοσύμαντη αντιστοιχία του συνόλου $P$

 Nick νομίζω σε έβγαλα από το μισό κόπο  :)
Και λέω μισό γιατί τον άλλο μισό θα τον έχεις αν ζητηθούν διευκρινήσεις..... :P

ποιες συναρτησεις εχεις βρει εσυ Σιλουανε?

[quote="silouan"]
Τελικά οι συναρτήσεις που ικανοποιούν το πρόβλημα έχουν τη μορφή : $f(1)=1$ και αν $n=p_{1}^{a_{1}}....p_{k}^{a_{k}}$
t;ote $f(n)=q(p_{1})^{a_{1}}....q(p_{k})^{a_{k}}$
όπου $q$ είναι μια αμφιμονοσύμαντη αντιστοιχία του συνόλου $P$ [/quote]

Το έγραψα παραπάνω . Οι συναρτήσεις όπως μπορείς να δεις είναι άπειρες ......Αυτή είναι η ομορφιά και η μαγκιά ταυτόχρονα της συγκεκριμμένης άσκησης.... :wink:

autos eisai.....nomiza oti den to eixes grapsei..sorry!

Μα και εγώ αυτό δεν είχα πει;


The number \[ \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006} \] can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers.  Find $a\cdot b\cdot c$.

[quote="Danbert"]This looks an awful lot like a USAMTS problem from this year or last...[/quote]
Could you happen to mean year 16, round 4, problem 2...? :whistling: (yes, I looked it up - no I don't have all past USAMTS problems memorized).  Basically that problem was really easy for anybody who did that round of the USAMTS, which disappoints me slightly, since there's a lot of overlap between the AMC and the USAMTS contests.

[hide="Solution"]
First we can write \[ \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006} = a\sqrt{2}+b\sqrt{3}+c\sqrt{5}.  \] Squaring both sides gives \[ 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006 = 2a+3b+5c + 2ab\sqrt{6}+2ac\sqrt{10}+2bc\sqrt{15}.  \] Since $a,b,c$ are rational, we can say that \[ ab = 52,\quad ac=234,\quad bc=72.  \] Multiplying all of these equalities together and taking the square root, we have \[ abc = \sqrt{(2^2\cdot 13)\cdot (2\cdot 3^2\cdot 13)\cdot (2^3\cdot 3^2)}=2^3\cdot 3^2\cdot 13 = \boxed {936}.  \][/hide]

[quote="lotrfan"]I haven't done much work with USAMTS, but i knew how to do this problem right away anyway, because i am familiar with the technique for expanding simpler double radicals like $\sqrt{3+2\sqrt{2}}$. I think in general one of the best things you can do for yourself when you prepare is to know where different formulas or tricks come from, so that when they show up in more convoluted forms on AIME's or AMC's, you can keep your head about it.[/quote]
I agree that it's a good general skill to have - when I did the USAMTS problem, I saw almost immediately how to do it without having any experience with radical simplification.  Still, it is interesting how similar it was to a USAMTS problem (and it just goes to show how many concepts tend to be "borrowed" from contest to contest, a good reason to do lots of them!).

[quote="chess64"]\[ abc = \sqrt{(2^2\cdot 13)\cdot (2\cdot 3^2\cdot 13)\cdot (2^3\cdot 3^2)}=2^3\cdot 3^2\cdot 13 = \boxed {936}.  \][/quote]

Little correction.

Thank you. I was wondering why the paranthesis was appearing so small... :D

[hide=Solution]
$2ab=104$ so $ab=52$
$2ac=468$ so $ac=234$
$2bc=144$ so $bc=72$

Multiply the $3$ equations together to get $(abc)^2=72\cdot52\cdot234$ so $abc=\boxed{936}$.
[/hide]

[hide = Solution]
So we can write the above expression equated to $a\sqrt{2}+b\sqrt{3}+c\sqrt{5}$. So we have....
 $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$  $=$ $(a\sqrt{2}+b\sqrt{3}+c\sqrt{5})^{2}$

So let's expand the right side first!

$(a\sqrt{2}+b\sqrt{3}+c\sqrt{5})^{2}$ $=$ $2a^2 + ab\sqrt{6} + ac\sqrt{10} + ab\sqrt{6} + 3b^2 + cb\sqrt{15}  + ac\sqrt{10} + bc\sqrt{15} + 5c^2$
$= (2a^2 + 3b^2 + 5c^2) + 2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15}.$
So we can make $4$ equations equation the corresponding terms from the right side and the left side.

$$2ab\sqrt{6} = 104\sqrt{6} \Rightarrow ab = 52$$
$$2ac\sqrt{10} = 468\sqrt{10} \Rightarrow ac = 234$$
$$2bc\sqrt{15} = 144\sqrt{15} \Rightarrow bc = 72.$$
$$(2a^2 + 3b^2 + 5c^2) =  2006. $$

Since we have $3$ variables we can just choose the first $3$ equations to work with. So instead of solving the variables, we can look at factors of $ab$ and $bc$. Since they have $b$ in common, $b$ has to be one of the common factors of $52$ and $72$ which is either $2$ or $4$. So we try out $b = 2$ and we find $c = 36$ and $a = 26$ and $(36)(26) \neq 234.$
So we try $b=4$ and find $c = 18$, $a = 13$. Those check out so our answer is $(13)(18)(4) = \boxed{936}$
[/hide]

Solved with [b]sarun[/b], [b]js104[/b], [b]crazyeyemoody[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b]

We have $ab=52$, $bc=72$, and $ca=234$ so $abc=\sqrt{52\cdot72\cdot234}=\boxed{936}$.


Want to show with proof whether the following series converges:

1 - 1/2 - 1/3 + 1/4 + 1/5 + 1/6 - 1/7 - 1/8 - 1/9 - 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 - ... = sum_n=1 to oo ( C(n)/n ), where C(n) = (-1)^(L-1), with L the unique positive integer such that L(L-1)/2 < n <= L(L+1)/2 .

I believe this series diverges because of the terms in the sequence, how do you show that this is the case?

It converges.

Consolidate each block of terms into one term. These new terms have alternating sign- how big are they?


Let $ ABCD$ be a trapezoid with $ \overline{AB}\equal{}14,\overline{BC}\equal{}7,\overline{CD}\equal{}4$ and $ \overline{DA}\equal{}5$. If $ E$ and $ F$ are the midpoints of the bases $ AB$ and $ CD$, respectively, find $ \overline{EF}$.

[hide="Solution"]By Euler's Theorem

$ 2\cdot CD\cdot AB \plus{} AD^2 \plus{} BC^2 \equal{} BD^2 \plus{} AC^2\Longrightarrow 186 \equal{} BD^2 \plus{} AC^2(*)$

Let $ P$ be the intersection of $ AC$ and $ BD$. Hence, the triangles $ PDC$ and $ PAB$ are similar. Thus

$ \dfrac{AC}{PC} \equal{} \dfrac{BD}{PD} \equal{} \dfrac{EF}{PF} \equal{} \dfrac{9}{2}\Longrightarrow PC \equal{} \dfrac{2}{9}\cdot AC,PD \equal{} \dfrac{2}{9}\cdot BD$ and $ PF \equal{} \dfrac{2}{9}\cdot EF$

Since $ PE$ is median of the triangle $ PDC$, we have

$ PF \equal{} \dfrac{1}{2}\cdot\sqrt {2\cdot(PC^2 \plus{} PD^2) \minus{} CD^2}\Longrightarrow PF \equal{} \dfrac{1}{2}\cdot\sqrt {\dfrac{8}{81}\cdot(BD^2 \plus{} AC^2) \minus{} 16}$.

Using $ (*)$, we get

$ PF \equal{} \dfrac{4}{3\sqrt {3}}$.

Therefore,

$ EF \equal{} \dfrac{9}{2}\cdot PF \equal{} \dfrac{9}{2}\cdot\dfrac{4}{3\sqrt {3}} \equal{} 2\sqrt3$.[/hide]

From F draw FM||AD and FN||CB. In triangle FMN FE is median so:
$ 4FE^2\equal{}2(FM^2\plus{}FN^2)\minus{}MN^2\equal{}2(25\plus{}49)\minus{}100\equal{}48$


A rectangular sheet of paper with dimensions 9 inches by 13 inches
is cut into $ n$ identical notecards, each of which is 3 inches by
2.6 inches. What is the maximum value of $ n$?

The total area of the rectangle is 13*9=117.
Each notecard is 2.6*3=7.8.
117/7.8= 15 notecards.

Take $ \lfloor \frac{9 \times 13}{3 \times 2.6} \rfloor$.

The answer is $ \boxed{15}$.

OMG...


LaTeX
Asymptote
MATHCOUNTS
FTW
geometry
geometric transformation
reflection

In this Challament, there will be 3 rounds. Time limits for each round will be posted later (measured in increments of days). Some of them will be original while others will be from other sources.

There are no multiple choices and [b]solutions are not required[/b]. There is no "guessing penalty". Each correct answer is 1 point and each incorrect/blank answer is 0. However, the final points are not just what you earned, they will be calculated based on the performance of others; so if you screwed up and someone else did too, [b] your scores will be pretty close to his/her [/b].

There are an unlimited amount of participants. Deadline for registration is [b] February 1, 2008 [/b]. CHEATING IS PROHIBITED. I trust y'all not to cheat.

To register, PM me and include your highest ForTheWin rating.

I need this so I can seperate you into 3 categories (this is only for the first round). Remember, [b] it is not unfair because I am calulating scores through a special process [/b]. Also, [b] you'll get an additional 1/2 for every solution you provide (you must use LaTeX and/or Asymptote) (for tiebreakers not your normal score) [/b].

I will PM you all the problems. DO NOT POST ANYTHING RELATED TO PROBLEMS/SOLUTIONS UNTIL I SAY IT IS OKAY TO DO SO. PM me the answers. I will PM your score.

I have special cutoff methods. [b] There is no fixed number of eliminations [/b].

 :)

akhil0422 is currently suspended. When an admin activates his account again, he will be able to freely chat. By the way, the answer to your question is MOEMS-MATHCOUNTS-AMC level.

Apparently, Challament = Challenge + Tournament.

:)

Heh.  When are problems coming?  Keep them easy.  Very easy.

Problems are coming FEBRUARY 2. I'm not suspended anymore! My Asymptote is screwed up though...

The two people that signed up so far are:

AIME15
007math

Hello everyone,

I think I'm going to have to cancel this because whenever I compile my "Problems File" from Texnic, it says no errors, warnings, whatsoever. However, when I clicked "view", Acrobat Reader says " there was an error in opening this file. Access is denied" I searched this on the Internet and many people have this problem too. I don't think there is a solution.

Either someone else has to make the problems or this Challament will be canceled.

Or, you could type them and post them in thread...

HELLO EVERYONE,

Thanks to AIME15, this Challament will not be canceled! I will post all the problems in the thread.

I need more people to join! Currently I only have 2!!!!

[hide="The current participants are"] [b] nikeballa96, 007math, and AIME15 [/b][/hide]

[b]THE DEADLINE FOR JOINING IS FEBBRUARY 1, 2008: THE PROBLEMS WILL BE POSTED ON THE NEXT DAY [/b]

I'll join. May I suggest making the deadline a bit sooner? We don't want people forgetting about this.

I'll join. This seems like a pretty cool thing.

[quote="akhil0422"]To register, PM me and include your highest ForTheWin rating. [/quote]

Yeah, uhh, read the first post guys :wink:

[hide="PARTICIPANTS"]

participants (highest ftw rating in parentheses)

nikeballa96 (1701)
Poincare
Twin Prime Conjecture
athunder
007math
AIME15 (2029, got this off leaderboard)
crazypianist1116

akhil, you can keep updating this list, with other peoples' highest ftw rating. it'll keep things organized.

[/hide]

btw, i agree with starting sooner. right now, this seems hot, but it'll die if you prolong it until february 1 st. maybe january 20th (i know its soon, but its reasonble)

UPDATE: THE DEADLINE HAS BEEN MOVED TO JAN. 20 (I can't edit my above posts).

As to athunder's question, you'll get the answer as you read the first post.

PARTICIPANTS:
nikeballa96 (1701)
Poincare
Twin Prime Conjecture
athunder (1365)
007math (1305)
AIME15 (2029)
crazypianist1116 (1200)
dragon96 (1579)

Poincare and Twin Prime Conjecture PLEASE PM ME YOUR HIGHEST FTW SCORES. IF YOU DON'T, I WILL GIVE YOU TWO OPTIONS:

a) disqualify you
b) disqualify you

I'm sorry about that athunder :P

I accidently missed it.

I don't really understand what you are saying. I'm just using FTW ratings so I can sort of even out the gap between people like AIME15 and people like me. I mean, it might be too easy for some people, yet too hard for others.

This is only for the first round though. Then I will average out the scores. All the other rounds will be the same test. Is that what you're talking about :huh: ?

Well, certain really beast people have never gone onto ftw (or almost never go on).

Instead of using ftw, you should use mathcounts/amc/aime/usamo scores...

i say amc 8 scores.
because mines a 25  :lol:

My FTW rating is just horrible and dreadful...I can still beat most people in the 1200s, but my rating doesn't reflect that.  You should use MathCounts scores, or AMC 8, etc.  FTW is just about fast reading, speed, and accuracy, and typing.  YOu could be beast at FTW, but just can't type as fast so you have a low rating.  And then there are those memers...

um.. ill join, my ftw score was 1305 ( back when i sort of cared!) ^_^

Mieh.

Certain people *coughmecough* who are usamo qualifiers have ftw ratings.

Idea: Use this:

1000*Usamo score + 100* AIME score + 10* AMC 10 score (+20 if its AMC12) + AMC 8 score.

Where usamo score is 0 if you haven't made, etc.

/me disagrees with pythag. amc 8 scores ftw.

No, let's keep it at FTW :)  The mod of the game should decide.

Or have an preliminary round dividing us into groups.  Most fair way, I think.  Why do you need us in groups?  You can always use random.org.

Thank you, I will use random.org.

The reason I will not use AMC scores is that some people may not be able to tell the truth. As for USAMO, I highly doubt most of the scores will be higher that 0.

BOGTRO is a new competitor!!!!

So y'all will be  :o .

just kidding :wink:

but bogtro is a new competitor

People could lie bout FTW rating also...

This Challament is officially CANCELLED. I am sick right now and am loaded with homework. I missed school becuase of my severe strep throat and fever.

For sure I will not be able to resume this until 2+ months. This is the busiest time of the year of the year for me; I have many things to study for. This Challament might resume, but people will probably disregard it because of what happened now; eh, your choice.

I am sorry.

[quote="iloveyouakhil"]you're so good at math akhil... OMGGG <33[/quote]

???
This is probably his multi...Lol.
Can a moderator delete this thread?


geometry probability trapezoid rectangle

A point $ P$ is selected at random from the interior of the pentagon with vertices $ A \equal{} (0,2)$, $B \equal{} (4,0)$, $C \equal{} (2 \pi \plus{} 1, 0)$, $D \equal{} (2 \pi \plus{} 1,4)$, and $ E \equal{} (0,4)$. What is the probability that $ \angle APB$ is obtuse?
[asy]
size(150);
pair A, B, C, D, E;
A = (0,1.5);
B = (3,0);
C = (2 *pi + 1, 0);
D = (2 * pi + 1,4);
E = (0,4);
draw(A--B--C--D--E--cycle);
label("$A$", A, dir(180));
label("$B$", B, dir(270));
label("$C$", C, dir(0));
label("$D$", D, dir(0));
label("$E$", E, dir(180));
[/asy]
$ \displaystyle \textbf{(A)} \ \frac {1}{5} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)} \ \frac {5}{16} \qquad \textbf{(D)} \ \frac {3}{8} \qquad \textbf{(E)} \ \frac {1}{2}$

I'm too lazy to post the full solution now, so I'll post three hints:

[hide="Hint 1"]Think circle.[/hide]

[hide="Hint 2"]If P lies on the circle with with AB as its diameter, APB is 90 degrees.[/hide]

[hide="Hint 3"]P must lie inside a circle with AB as a diameter.[/hide]

Elemennop wrote:[hide]

Through the shoelace theorem, we can find the area of the pentagon to be . An obtuse triangle is formed if the side opposite the obtuse angle is the diameter of the circle, and the point of the obtuse angle is inside the circle. So we only have to find the area of the semi-circle with AB as the diameter. We find the radius to be , so the area is . The probability is then .C.[/hide]

What exactly is the shoelace theorem, the way I solved for the area of the pentagon was just making it consisting of a right trapezoid and rectangle and got the area that way

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62543&start=20

See Elemenop and zannatrag's posts.

Diagram needs to be included with the problem on the contest. Here it is, as close as to the contest pdfs as possible (not necessarily to scale):
[asy]
size(200);
pair A, B, C, D, E;
A = (0,1.5);
B = (3,0);
C = (2 *pi + 1, 0);
D = (2 * pi + 1,4);
E = (0,4);
draw(A--B--C--D--E--cycle);
label("$A$", A, dir(180));
label("$B$", B, dir(270));
label("$C$", C, dir(0));
label("$D$", D, dir(0));
label("$E$", E, dir(180));
[/asy]


let L be a line and A,B are two points wich arent on L.

find point  P on L inwich we have:PA+PB=min

[quote="ashegh"]let $L$ be a line and $A,B$ are two points which aren't on $L$.

find point  $P$ on $L$ for which we have:$PA+PB\: \min$[/quote]

let $B'$ be the symmetric point of $B$ wrt $L$.
Then $RB=RB'$ for each point $R$ on $l$
so $RA+RB=RA+RB'\geq AB'$

Now we can say $P=l \cap AB'$

 :)  :)

[quote="ashegh"]hmmm... good solution ;)

i want to post a solution wich harmonic is used in it...[/quote]
i think in your own solution you used harmonic so i think if you send us your solution we will enjoy it very much(becasue i think no one can solve it like you. :D )

thanks alot dear jensen(khak nejad jigar).i will post it as soon as possible ;)

socrates solution is complete,but i want to add something tohis solution:

if we draw a prependicular from P to line L,this line will be the bisector of the angle APB(its clear...)

another solution for the ones who love harmonic:

ler C be the intersection of AB and line L.draw two arbitrary line,wich are paralell with each other.and go through

Aand B.and name them L'and L" respectivly.

let K be an arbitrary point on L'.let KC intersect L" in point S.choose S' on L" ,inwich BS=BS'

andlet KS' intersect AB in point D.

draw circle (C),with the diameter CD,and let it intersect L,inpoint P.

P is the wanted point. ;)

nice solution ashegh,really nice.(baba khafan!) :D


Currently the highest record is IMO 2008/6, 482 contestants scored 0 point on this problem.

Will there be a difficult problem, not solved by 500 contestants?

Who knows :wink:
Maybe yes, if there's a really hard problem, or bad generation of students competing.
Answer: Ask someone who can predict future :wink:

it's inevitable  :D

[hide]IMO 2009/6 (I should know, I'm one of them)[/hide]

Yes! The first problem!

540 zeroes, 25 >0 points.

Mean score is 0.168, mode and median 0.

Only 4.425% of competitors scored more than 0.

Some other problems:
06/6 5.42% scored more than 0, mean 0.187
07/6 9.04% scored more than 0, mean 0.152 (!!!)
08/6 9.91% scored more than 0, mean 0.260
09/6 4.42% (!!!) scored more than 0, mean 0.168

IMO 2009 Q6 as a mini-polymath project

[url]http://terrytao.wordpress.com/2009/07/20/imo-2009-q6-as-a-mini-polymath-project/[/url]


floor function
logarithms
Olimpiada de matematicas

Hola a todos.
Seria bueno empezar con el entrenamiento para la IMO 2009, muchos pensara que falta aun para dicha competencia pero la practica es necesaria para un progreso de Latinoamerica este año. Es por eso el de empezar temprano.

Problema 1.-
Sea $ A$ un conjunto infinito de numeros enteros positivos. Demostrar que existe un numero real positivo $ x$ tal que la parte entera de $ {x}^n$ este en $ A$ para infinitos valores de n.

Espero que lo resuelvan, a mi me gusto mucho este problema.

Chaufas.

Lema: Sea A conjunto infinito de enteros, entonces para todo $ b$ entero existe infinitos $ a \in A$ talque existe $ p$ entero que satisface $ (b\minus{}1)^p \le a \le (b\plus{}1)^p$.
demostracion:
sustituye $ b$ en vez de $ b\minus{}1$...
 $ (b)^p \le a \le (b\plus{}2)^p$, toma $ p\equal{} \lfloor \log_b a \rfloor$, entonces evidentemente  $ (b)^p \le a$.
la segunda parte de la desigualdad es equivalente a $ \log_b a \le \lfloor \log_b a \rfloor \log_b(b\plus{}2)$
esto es trivial si $ b$ es tan pequeño tal que $ \log_b(b\plus{}2) \ge 2$ y si $ \log_b(b\plus{}2) < 2$ entonces es lo mismo q tener
 $ \{\log_b a \}\le \lfloor \log_b a \rfloor \{\log_b(b\plus{}2)\}$ entonces podemos tomar un $ a$ suficientemente grande, tal que $ \lfloor \log_b a \rfloor \{\log_b(b\plus{}2)\}$  supera a 1 y la desigualdad seria evidente. como en $ A$ tiene infinitos elementos, el no esta acotado por lo tanto cualquier $ a_i \ge a$ satisface el desigualdad.

regresando al problema, tomamos el menor elemento de $ A$, $ a_0$ como $ b$. entonces existe subconjunto infinito de  $ A$ $ \{a_{i_0} .... a_{i_n}\}$, donde para cada una de ellos existe un $ p_j$ entero tal que $ (b\minus{}1)^{p_j} \le a_{i_j} \le (b\plus{}1)^{p_j}$. Entonces para cada uno de esos se tiene que el interseccion el intervalo de numeros reales $ [a_o,a_o\plus{}1)$ y todos los intervalos $ [\sqrt[p_j]{a_{i_j}},\sqrt[p_j]{a_{i_j}}\plus{}1)$ no es vacia, por lo tanto puedo tomar un $ x$ elemento de ese interseccion, y ese $ x$ claramente satisface el condicion del problema.


function
limit
real analysis
real analysis unsolved

Function $f: [a,b]\to\mathbb{R}$, $0<a<b$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Prove that there exists $c\in(a,b)$ such that  \[ f'(c)=\frac1{a-c}+\frac1{b-c}+\frac1{a+b}.  \]

Consider a function $g(x)$ such that $g'(x) = \frac{1}{a-x} + \frac{1}{b-x} + \frac{1}{a+b}$.  Integrating, $g(x) = \frac{1}{(a-x)^2} + \frac{1}{(b-x)^2} + \frac{x}{a+b} + C$.

Now consider the function $h(x) = f(x) - g(x)$.  We wish to show $\exists c : h'(c) = 0$.  Now, $h(x)$ is continuous and differentiable on $(a, b)$, but $\lim_{x \to a} h(x) = \lim_{x \to b} h(x) = - \infty$.

Intuitively, there must be some $h'(c) = 0$... a limit with the Mean Value Theorem?

i don't know, my intuition tells me $f$ (of $c$ or of $a$ or something) must appear somewhere in the relation, not just $f'$... but i could be wrong, i have no idea...for example, we can get one possible relation by doing the following.

let $g(x) = (a+b)(a-x)(b-x)f(x)+(a-x)(b-x)x$. since $g(a) = g(b) = 0$, by the MVT there's $c\in (a,b)$ s.t.

$0 = (a+b)(c-a)f(c)+(a+b)(c-b)f(c)+(a-c)(b-c)(a+b)f'(c)+(a-c)(b-c)+c(c-a)+c(c-b)$,

from which we get

$f'(c) = \frac{f(c)}{a-c}+\frac{f(c)}{b-c}+\frac{1}{a+b}+\frac{c(a-c)+c(b-c)}{(a-c)(b-c)(a+b)}$

which, yes, isn't very promising if the relation they ask us to prove is actually true.

We consider the function $g(x) =(x-a)(x-b) \, e^{f(x)-\frac{x}{a+b}}$
which is continuous on $[a,b]$ and differentiable in $(a,b)$.

It is $g(a)=g(b)=0$ therefore by Rolle's theorem there exists a $c \in (a,b)$ such that $g'(c)=0$.

But $g'(x)=[(x-b) + (x-a) + (f'(x)-\frac{1}{a+b})(x-a)(x-b)] \, e^{f(x)-\frac{x}{a+b}}$

so $g'(c)=0 \Longrightarrow (c-b) + (c-a) + (f'(c)-\frac{1}{a+b})(c-a)(c-b)=0$

or $f'(c) =\frac1{a-c}+\frac1{b-c}+\frac1{a+b}$ Q.E.D.


MATHCOUNTS
search
function
number theory

I'm not accusing anybody in particular of any of the following!

[b][u]Being on Topic[/u][/b]
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[*] Be sure to stay on topic in each specific thread; I've personally seen threads deviating greatly off topic. For example, a topic on Simple Arithmetic should not end in a display of Mathematica usage.
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[b][u]Thread Title[/u][/b]
[list][*]Make your title fairly descriptive and concise.  A good example is "MATHCOUNTS Number Theory"; a bad example is "HELP ME!!!!!!!!!!!!!!!!!!!!!!!
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[*][url=http://www.artofproblemsolving.com/Forum/search.php]Search[/url] the problem before you post it.[/list]

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[*]If you disagree with a reply, cite examples or give some sort of analysis supporting why you disagree.[/list]

[b][u]Basic Etiquette[/u][/b]
[list][*]Don't make a post to thank somebody who has helped you; even though it is respectful, it clutters up the forum.  Use the brand new thank function!
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If you feel like the problem is too easy or hard, please use the Too Easy/Hard buttons (yes, self explanatory). levans spent quite a bit of time developing that, and it would be great if the community adopted it!

Here is the list of [url=http://www.artofproblemsolving.com/Forum/memberlist.php?mode=group&g=79836]your moderators[/url], always ready to help.

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[size=50]Rules edited for modernization. - Dojo[/size]

[quote="AwesomeToad"]Would it be okay to post the [i]source[/i] in the subject and the math subject in the subject description.

Example:

2009 Workout 4 Problem 5 (Subject)
Counting/Probability (Subject description)[/quote]

At least in the olympiad forum, i think its encouraged if you do that to put as a subtitle stuff about the problem like "Process terminates" "Count number of black squares" etc