geometry ratio perimeter calculus integration circumcircle function

The first round in Turkish math. olymp.
There are 9 problems about geometry, 9 about combinatorics, 9 about algebra, and 9 about number theory.
Each high school participates with at most 6 students. Time is limited about 2.5 hrs.

The first problem in the PDF format (above) is incorrect. Correct one is below.

[b]6. Turkish Math Olympiad First Round - May 1998[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478062#p1478062]Problem 1[/url]

If $ \left|BC\right| \equal{} a$, $ \left|AC\right| \equal{} b$, $ \left|AB\right| \equal{} c$, $ 3\angle A \plus{} \angle B \equal{} 180{}^\circ$ and $ 3a \equal{} 2c$, then find $ b$ in terms of $ a$.

A) $ \frac {3a}{2}$  B) $ \frac {5a}{4}$  C) $ a\sqrt {2}$ D) $ a\sqrt {3}$ E) $ \frac {2a\sqrt {3} }{3}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478064#p1478064]Problem 2[/url]

Let $ A$, $ B$ be the number of digits of $ 2^{1998}$ and $ 5^{1998}$ in decimal system. $ A \plus{} B \equal{} ?$

A) 1998  B) 1999 C) 2000 D) 3996 E) 3998

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478065#p1478065]Problem 3[/url]

How many ways are there to divide a set with 6 elements into 3 disjoint subsets?

A) 90  B) 105  C) 120  D) 180  E) 243

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478067#p1478067]Problem 4[/url]

$ x,y,z\in \Re$, find the minimal value of $ f\left(x,y,z\right) \equal{} 2x^{2} \plus{} 5y^{2} \plus{} 10z^{2} \minus{} 2xy \minus{} 4yz \minus{} 6zx \plus{} 3$.

A) 0  B) 3  C) -3  D) 1  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478068#p1478068]Problem 5[/url]

$ ABCD$ is a cyclic quadrilateral. If $ \angle B \equal{} \angle D$, $ AC\bigcap BD \equal{} \left\{E\right\}$, $ \angle BCD \equal{} 150{}^\circ$, $ \left|BE\right| \equal{} x$, $ \left|AC\right| \equal{} z$, then find $ \left|ED\right|$ in terms of $ x$ and $ z$.

A) $ \frac {z \minus{} x}{\sqrt {3} }$ B) $ \frac {z \minus{} 2x}{3}$ C) $ \frac {z \plus{} x}{\sqrt {3} }$ D) $ \frac {z \minus{} 2x}{2}$ E) $ \frac {2z \minus{} 3x}{2}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478071#p1478071]Problem 6[/url]

Find the number of primes$ p$, such that $ x^{3} \minus{} 5x^{2} \minus{} 22x \plus{} 56\equiv 0\, \, \left(mod\, p\right)$ has no three distinct integer roots in $ \left[0,\left. p\right)\right.$.

A) 1  B) 2  C) 3  D) 4  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478074#p1478074]Problem 7[/url]

Find the minimal value of integer $ n$ that guarantees:
Among $ n$ sets, there exits at least three sets such that any of them does not include any other; or there exits at least three sets such that any two of them includes the other.

A) 4  B) 5  C) 6  D) 7  E) 8

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478077#p1478077]Problem 8[/url]

$ a_{1} \equal{} 1$, $ a_{n \plus{} 1} \equal{} \frac {a_{n} }{\sqrt {1 \plus{} 4a_{n}^{2} } }$ for $ n\ge 1$. What is the least $ k$ such that $ a_{k} < 10^{ \minus{} 2}$ ?

A) 2501  B) 251  C) 2499 D) 249  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478204#p1478204]Problem 9[/url]

$ C_{1}$ and $ C_{2}$ be two externally tangent circles with diameter $ \left[AB\right]$ and $ \left[BC\right]$, with center $ D$ and $ E$, respectively. Let $ F$ be the intersection point of tangent line from A to $ C_{2}$ and tangent line from $ C$ to $ C_{1}$ (both tangents line on the same side of $ AC$). If $ \left|DB\right| \equal{} \left|BE\right| \equal{} \sqrt {2}$, find the area of triangle $ AFC$.

A) $ \frac {7\sqrt {3} }{2}$  B) $ \frac {9\sqrt {2} }{2}$ C) $ 4\sqrt {2}$ D) $ 2\sqrt {3}$ E) $ 2\sqrt {2}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478205#p1478205]Problem 10[/url]

Let $ p$ and $ q$ be two consecutive terms of the sequence of odd primes. The number of positive divisor of $ p \plus{} q$, at least

A) 2  B) 3  C) 4  D) 5  E) 6

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478206#p1478206]Problem 11[/url]

If two faces of a dice have a common edge, the two faces are called adjacent faces. In how many ways can we construct a dice with six faces such that any two consecutive numbers lie on two adjacent faces?

A) 10  B) 14  C) 18  D) 56  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478208#p1478208]Problem 12[/url]

In a right triangle, ratio of the hypotenuse over perimeter of the triangle determines an interval on real numbers. Find the midpoint of this interval?

A) $ \frac {2\sqrt {2} \plus{} 1}{4}$ B) $ \frac {\sqrt {2} \plus{} 1}{2}$ C) $ \frac {2\sqrt {2} \minus{} 1}{4}$ D) $ \sqrt {2} \minus{} 1$ E) $ \frac {\sqrt {2} \minus{} 1}{2}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478210#p1478210]Problem 13[/url]

Let $ H$ be the intersection of altitudes in triangle $ ABC$. If $ \angle B \equal{} \angle C \equal{} \alpha$ and $ O$ is the center of circle passing through $ A$, $ H$ and $ C$, then find $ \angle HOC$ in terms of  $ \alpha$.

A) $ 90{}^\circ \minus{} \alpha$ B) $ 90{}^\circ \plus{} \frac {\alpha }{2}$ C) $ 180{}^\circ \minus{} \alpha$ D) $ 180{}^\circ \minus{} \frac {\alpha }{2}$ E) $ 180{}^\circ \minus{} 2\alpha$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478212#p1478212]Problem 14[/url]

Find the number of distinct integral solutions of  $ x^{4} \plus{} 2x^{3} \plus{} 3x^{2} \minus{} x \plus{} 1\equiv 0\, \, \left(mod\, 30\right)$ where $ 0\le x < 30$.

A) 0  B) 1  C) 2  D) 3  E) 4

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478213#p1478213]Problem 15[/url]

Twelve couples are seated around a circular table such that all of men are seated side by side, and every women are seated to opposite of  her husband. In every step, a woman and a man next to her are swapping. What is the least possible number of swapping until all couples are seated side by side?

A) 36  B) 55  C) 60  D) 66  E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478216#p1478216]Problem 16[/url]

If $ x^{2} \plus{} y^{2} \plus{} z \equal{} 15$, $ x \plus{} y \plus{} z^{2} \equal{} 27$ and $ xy \plus{} yz \plus{} zx \equal{} 7$, then

A) $ 3\le \left|x \plus{} y \plus{} z\right|\le 4$

B) $ 5\le \left|x \plus{} y \plus{} z\right|\le 6$

C) $ 7\le \left|x \plus{} y \plus{} z\right|\le 8$

D) $ 9\le \left|x \plus{} y \plus{} z\right|\le 10$

E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478217#p1478217]Problem 17[/url]

In triangle $ ABC$, internal bisector of angle $ A$ intersects with $ BC$ at $ D$. Let$ E$ be a point on $ \left[CB\right.$ such that $ \left|DE\right| \equal{} \left|DB\right| \plus{} \left|BE\right|$. The circle through $ A$, $ D$, $ E$ intersects $ AB$ at $ F$, again. If  $ \left|BE\right| \equal{} \left|AC\right| \equal{} 7$, $ \left|AD\right| \equal{} 2\sqrt {7}$ and $ \left|AB\right| \equal{} 5$, then $ \left|BF\right|$ is

A) $ \frac {7\sqrt {5} }{5}$ B) $ \sqrt {7}$  C) $ 2\sqrt {2}$ D) $ 3$  E) $ \sqrt {10}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478220#p1478220]Problem 18[/url]

Let $ p_{1} < p_{2} < \ldots < p_{24}$ be the prime numbers on the interval $ \left[3,100\right]$. Find the smallest value of $ a\ge 0$ such that $ \sum _{i \equal{} 1}^{24}p_{i}^{99!} \equiv a\, \, \left(mod\, 100\right)$.

A) 24  B) 25  C) 48  D) 50  E) 99

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478222#p1478222]Problem 19[/url]

There are 22 black and 3 blue balls in a bag. Ahmet chooses an integer $ n$ in between 1 and 25. Betül draws $ n$ balls from the bag one by one such that no ball is put back to the bag after it is drawn. If exactly 2 of the $ n$ balls are blue and the second blue ball is drawn at $ n^{th}$ order, Ahmet wins, otherwise Betül wins. To increase  the possibility to win,  Ahmet must choose

A) 2  B) 11  C) 12  D) 13  E) 23

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478225#p1478225]Problem 20[/url]

How many real solutions does the equation $ x^{3} 3^{1/x^{3} } \plus{} \frac {1}{x^{3} } 3^{x^{3} } \equal{} 6$ have?

A) 0  B) 2  C) 3  D) Infinitely many E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478226#p1478226]Problem 21[/url]

In an acute triangle $ ABC$, let $ D$ be a point on $ \left[AC\right]$  and $ E$ be a point on $ \left[AB\right]$ such that $ \angle ADB \equal{} \angle AEC \equal{} 90{}^\circ$. If perimeter of triangle $ AED$ is 9, circumradius of  $ AED$ is $ \frac {9}{5}$ and perimeter of triangle $ ABC$ is 15, then $ \left|BC\right|$ is

A) 5  B) $ \frac {24}{5}$  C) 6  D) 8  E) $ \frac {48}{5}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478227#p1478227]Problem 22[/url]

$ \left(x_{1} x_{2} \ldots x_{1998} \right)$ shows a number with 1998 digits in decimal system. How many numbers $ \left(x_{1} x_{2} \ldots x_{1998} \right)$ are there such that $ \left(x_{1} x_{2} \ldots x_{1998} \right) \equal{} 7\cdot 10^{1996} \left(x_{1} \plus{} x_{2} \plus{} \ldots \plus{} x_{1998} \right)$ ?

A) 0  B) 1  C) 2  D) 3  E) 4

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478228#p1478228]Problem 23[/url]

Ahmet and Betül play a game on $ n\times n$ $ \left(n\ge 7\right)$ board. Ahmet places his only piece on one of the $ n^{2}$ squares. Then Betül places her two pieces on two of the squares at the border of the board. If two squares have a common edge, we call them adjacent squares. When it is Ahmet's turn, Ahmet moves his piece either to one of the empty adjacent squares or to the out of the board if it is on one of the squares at the border of the board. When it is Betül's turn, she moves all her two pieces to the adjacent squares. If  Ahmet's piece is already on one of the two squares that Betül has just moved to, Betül attacks to his piece and wins the game. If Ahmet manages to go out of the board, he wins the game. If Ahmet begins to move, he guarantees to win the game putting his piece on one of  the $ \dots$  squares at the beginning of the game.

A) 0  B) $ n^{2}$  C) $ \left(n \minus{} 2\right)^{2}$ D) $ 4\left(n \minus{} 1\right)$ E) $ 2n \minus{} 1$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478229#p1478229]Problem 24[/url]

Let $ n\left(A\right)$ be the number of distinct real solutions of the equation $ x^{6} \minus{} 2x^{4} \plus{} x^{2} \equal{} A$. When $ A$ takes every value on real numbers, the set of values of $ n\left(A\right)$ is

A) $ \left\{0,1,2,3,4,5,6\right\}$

B) $ \left\{0,2,4,6\right\}$

C) $ \left\{0,3,4,6\right\}$

D) $ \left\{0,2,3,4,6\right\}$

E) $ \left\{0,2,3,4\right\}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478233#p1478233]Problem 25[/url]

In triangle $ ABC$ with $ \left|BC\right| > \left|BA\right|$, $ D$ is a point inside the triangle such that $ \angle ABD \equal{} \angle DBC$, $ \angle BDC \equal{} 150{}^\circ$ and $ \angle DAC \equal{} 60{}^\circ$. What is the measure of $ \angle BAD$?

A) 45  B) 50  C) 60  D) 75  E) 80

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478237#p1478237]Problem 26[/url]

How many ordered integer pairs $ \left(x,y\right)$ are there satisfying following equation: $ y \equal{} \sqrt {x \plus{} 1998 \plus{} \sqrt {x \plus{} 1998 \plus{} \sqrt {x \plus{} 1997 \plus{} \sqrt {x \plus{} 1997 \plus{} \ldots \plus{} \sqrt {x \plus{} 1 \plus{} \sqrt {x \plus{} 1 \plus{} \sqrt {x \plus{} \sqrt {x} } } } } } } }$

A) 0  B) 1  C) 1998 D) 3996 E) Infinitely many

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478241#p1478241]Problem 27[/url]

For which of the following$ n$, $ n\times n$ chessboard cannot be covered using at most one unit square piece and many L-shaped pieces (an L-shaped piece is a 2x2 piece with one square removed)?

A) 96  B) 97  C) 98  D) 99  E) 100

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478242#p1478242]Problem 28[/url]

How many distinct real roots does the equation $ \sqrt {x \plus{} 4\sqrt {x \minus{} 4} } \minus{} \sqrt {x \plus{} 2\sqrt {x \minus{} 1} } \equal{} 1$ have?

A) 0  B) 1  C) 2  D) 3  E) 4

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478245#p1478245]Problem 29[/url]

Let $ ABCD$ be convex quadrilateral with $ \angle C \equal{} \angle D \equal{} 90{}^\circ$. The circle $ K$ passing through $ A$ and $ B$ is tangent to $ CD$ at $ C$. Let $ E$ be the intersection of  $ K$ and $ \left[AD\right]$. If $ \left|BC\right| \equal{} 20$, $ \left|AD\right| \equal{} 16$, then $ \left|CE\right|$ is

A) 9  B) $ 6\sqrt {2}$ C) $ 4\sqrt {5}$ D) $ 7\sqrt {2}$ E) 10

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478246#p1478246]Problem 30[/url]

Let $ m \equal{} \left(abab\right)$ and $ n \equal{} \left(cdcd\right)$ be four-digit numbers in decimal system. If $ m \plus{} n$ is a perfect square, what is the largest value of $ a\cdot b\cdot c\cdot d$?

A) 392  B) 420  C) 588  D) 600  E) 750

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478248#p1478248]Problem 31[/url]

A two-player game is played on a chessboard with $ m$ columns and $ n$ rows. Each player has only one piece. At the beginning of the game, the piece of the first player is on the upper left corner, and the piece of the second player is on the lower right corner. If two squares have a common edge, we call them adjacent squares. The player having the turn moves his piece to one of the adjacent squares. The player wins if the opponent's piece is on that square, or if he manages to move his piece to the opponent's initial row or column. If the first move is made by the first player, for which of the below pairs of $ \left(m,n\right)$ there is a strategy that guarantees the second player win.

A) (1998, 1997)

B) (1998, 1998)

C) (997, 1998)

D) (998, 1998)

E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478249#p1478249]Problem 32[/url]

For every $ x,y\in \Re ^{ \plus{} }$, the function $ f: \Re ^{ \plus{} } \to \Re$ satisfies the condition $ f\left(x\right) \plus{} f\left(y\right) \equal{} f\left(x\right)f\left(y\right) \plus{} 1 \minus{} \frac {1}{xy}$. If $ f\left(2\right) < 1$, then $ f\left(3\right)$ will be

A) $ 2/3$

B) $ 4/3$

C) 1

D) More information needed

E) There is no $ f$ satisfying the condition above.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478250#p1478250]Problem 33[/url]

Chord $ \left[AD\right]$ is perpendicular to the diameter $ \left[BC\right]$ of a circle. Let $ E$ and $ F$ be the midpoints of the arcs $ AC$ and $ CD$, respectively. If $ AD\bigcap BE \equal{} \left\{G\right\}$, $ AF\bigcap BC \equal{} \left\{H\right\}$ and $ m(AC) \equal{} \alpha$, find the measure of angle $ BHG$ in terms of $ \alpha$.

A) $ 90{}^\circ \minus{} \frac {\alpha }{2}$ B) $ 60{}^\circ \minus{} \frac {\alpha }{3}$ C) $ \alpha \minus{} 30{}^\circ$ D) $ 15{}^\circ \plus{} \frac {\alpha }{2}$ E) $ \frac {180{}^\circ \minus{} 2\alpha }{3}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478253#p1478253]Problem 34[/url]

Let $ a,b,c,d$ be rational numbers with $ a > 0$. If for every integer $ n\ge 0$, the number $ an^{3} \plus{} bn^{2} \plus{} cn \plus{} d$ is also integer, then the minimal value of $ a$ will be

A) 1  B) $ \frac {1}{2}$  C) $ \frac {1}{6}$  D) Cannot be found   E) None

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478255#p1478255]Problem 35[/url]

What is the maximum number of subsets, having property that none of them is a subset of another, can a set with 10 elements have?

A) 126  B) 210  C) 252  D) 420  E) 1024

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1478256#p1478256]Problem 36[/url]

$ ABCD$ is a $ 4\times 4$ square. $ E$ is the midpoint of $ \left[AB\right]$. $ M$ is an arbitrary point on $ \left[AC\right]$.  How many different points $ M$ are there such that $ \left|EM\right| \plus{} \left|MB\right|$ is an integer?

A) 2  B) 3  C) 4  D) 5  E) 6

[hide="ANSWER KEY"]$ \begin{tabular}{|p{0.5in}|p{1.2in}|} \hline \multicolumn{2}{|p{1in}|}{\textbf{Answer Key}} \\
\hline NO. & Correct Answer \\
\hline \textbf{1} & \textbf{B} \\
\hline \textbf{}\textbf{2} & \textbf{B} \\
\hline \textbf{}\textbf{3} & \textbf{A} \\
\hline \textbf{}\textbf{4} & \textbf{B} \\
\hline \textbf{}\textbf{5} & \textbf{D} \\
\hline \textbf{}\textbf{6} & \textbf{D} \\
\hline \textbf{}\textbf{7} & \textbf{B} \\
\hline \textbf{}\textbf{8} & \textbf{A} \\
\hline \textbf{}\textbf{9} & \textbf{E} \\
\hline \textbf{}\textbf{10} & \textbf{C} \\
\hline \textbf{}\textbf{11} & \textbf{A} \\
\hline \textbf{}\textbf{12} & \textbf{C} \\
\hline \textbf{}\textbf{13} & \textbf{E} \\
\hline \textbf{}\textbf{14} & \textbf{B} \\
\hline \textbf{}\textbf{15} & \textbf{D} \\
\hline \textbf{}\textbf{16} & \textbf{C} \\
\hline \textbf{}\textbf{17} & \textbf{A} \\
\hline \textbf{}\textbf{18} & \textbf{C} \\
\hline \textbf{}\textbf{19} & \textbf{D} \\
\hline \textbf{}\textbf{20} & \textbf{E} \\
\hline \textbf{}\textbf{21} & \textbf{B} \\
\hline \textbf{}\textbf{22} & \textbf{E} \\
\hline \textbf{}\textbf{23} & \textbf{D} \\
\hline \textbf{}\textbf{24} & \textbf{D} \\
\hline \textbf{}\textbf{25} & \textbf{C} \\
\hline \textbf{}\textbf{26} & \textbf{A} \\
\hline \textbf{}\textbf{27} & \textbf{D} \\
\hline \textbf{}\textbf{28} & \textbf{A} \\
\hline \textbf{}\textbf{29} & \textbf{C} \\
\hline \textbf{}\textbf{30} & \textbf{D} \\
\hline \textbf{}\textbf{31} & \textbf{D} \\
\hline \textbf{}\textbf{32} & \textbf{A} \\
\hline \textbf{}\textbf{33} & \textbf{A} \\
\hline \textbf{}\textbf{34} & \textbf{C} \\
\hline \textbf{}\textbf{35} & \textbf{C} \\
\hline \textbf{}\textbf{36} & \textbf{E} \\
\hline \end{tabular}$[/hide]

[quote="xeroxia"]

[url=http://www.mathlinks.ro/viewtopic.php?t=273047]Problem 33[/url]

Chord $ \left[AD\right]$ is perpendicular to the diameter $ \left[BC\right]$ of a circle. Let $ E$ and $ F$ be the midpoints of the arcs $ AC$ and $ CD$, respectively. If $ AD\bigcap BE \equal{} \left\{G\right\}$, $ AF\bigcap BC \equal{} \left\{H\right\}$ and $ m(AC) \equal{} \alpha$, find the measure of angle $ BHC$ in terms of $ \alpha$.

A) $ 90{}^\circ \minus{} \frac {\alpha }{2}$ B) $ 60{}^\circ \minus{} \frac {\alpha }{3}$ C) $ \alpha \minus{} 30{}^\circ$ D) $ 15{}^\circ \plus{} \frac {\alpha }{2}$ E) $ \frac {180{}^\circ \minus{} 2\alpha }{3}$

[/quote]

It asks for $ BHG$ not $ BHC$.

Problem 2

Since $ 2^{1998}$ and $ 5^{1998}$ be multiplied, it became a 1999 digit number, which is $ 10^{1998}$.

So the answer is 1999.

answer is correct, solution is not.
If you click the "Problem" text, you will be redirected to the thread which the problem is discussed.

Problems added to the resources page: http://www.artofproblemsolving.com/Forum/resources.php?c=174&cid=241&year=1998.

Thank you! :)


[b][u]BMO TEAM[/u][/b]
[b]1)STELIOS DESPOTAKIS
2)NIKOS PAPAGIANNAKIS
3)NIKOS RAPANOS
4)KONSTANTINOS PSYCHAS
5)NIKOS BOUDAS
6)FRAGISKOS KOUFOGIANNIS[/b]

Συγχαρητήρια στους επιτυχόντες  :D  .Ποιοι είναι αναπληρωματικοί ??

συγχαρητηρια σε ολους και ιδιαιτερα στους βασικους της ομάδος!!

 :)

Οι αναπληρωματικοί είναι

         7) ΣΤΑΥΡΟΠΟΥΛΟΥ ΕΛΕΝΗ
         8) ΚΑΡΑΜΙΧΑΛΗΣ ΡΑΛΛΗΣ
         9) MΑΝΤΟΥΛΙΔΗΣ ΧΡΙΣΤΟΣ
     10) ΡΩΜΑΝΙΔΗΣ ΝΙΚΟΛΑΟΣ
     11) ΤΖΑΜΟΣ ΧΡΗΣΤΟΣ
     12) ΔΑΡΒΙΡΑ ΑΘΑΝΑΣΙΑ

ΕΧΟΥΜΕ ΜΗΠΩΣ ΚΑΙ ΤΟΥΣ ΜΙΚΡΟΥΣ?

Νίκο (Ρωμανίδη) αν και είσαι αναπληρωματικός σου στέλνω τα συγχαρητήριά μου όπως και στον φίλο  μου Χρίστο .....  :D
Μην ξεχνάτε ότι έχετε ακόμη μια ευκαιρία...... :D

Σε ευχαριστώ Σιλουανέ .
 Πληρώνω ένα καταστροφικό λάθος που έκανα στον Αρχιμήδη .
 Ελπίζω να αξιοποιήσω την τελευταία ευκαιρία !

[quote="nikos"]
 Πληρώνω ένα καταστροφικό λάθος που έκανα στον Αρχιμήδη .
 [/quote]

Όλοι έχουμε κάνει ένα τέτοιο μέσα σε Διαγωνισμό (το έχω πάθει και εγώ είναι το χειρότερο :o ). Εσύ τι έκανες αριθμητικό λάθος ή δεν κατέγραψες κάτι καλά?

Βλέπω ότι η εθνική ομάδα πληροφορικής μετακόμισε στην εθνική ομάδα μαθηματικών!!!

Σιλουανέ , στο 1ο θέμα αντί του

  "Πόσοι διαφορετικοί 5ψήφιοι ακέραιοι θετικοί αριθμοί υπάρχουν που καθένα απο τα ψηφία τους , εκτός του τελευταίου ,
   είναι μεγαλύτερο η ίσο του επομενου ψηφίου τους "

   εγώ έλυσα το

  "Πόσοι διαφορετικοί 5ψήφιοι ακέραιοι θετικοί αριθμοί υπάρχουν που καθένα απο τα ψηφία τους , εκτός του τελευταίου ,
   είναι μικρότερο η ίσο του επομενου ψηφίου τους " .

ΠΩΠΩ ΦΙΛΕ ΕΙΣΑΙ ΠΟΛΥ ΑΤΥΧΟΣ ΕΙΝΑΙ ΠΟΛΥ ΒΑΡΥ ΝΑ ΣΟΥ ΠΩ ΣΥΛΛΗΠΗΤΗΡΙΑ ΓΙ ΑΥΤΟ ΘΑ ΣΟΥ ΠΩ ΟΤΙ ΔΕΝ ΠΕΙΡΑΖΕΙ (ΑΝ ΚΑΙ ΞΕΡΕΙΣ ΟΠΩΣ ΚΑΙ ΞΕΡΩ ΟΤΙ ΠΕΙΡΑΖΕΙ ΚΑΙ ΠΟΛΥ ΓΙΑΤΙ ΚΙ ΕΓΩ ΕΙΧΑ ΚΑΝΕΙ ΠΕΡΣΙ ΚΑΤΙ ΠΑΡΟΜΟΙΟ) :mad:  :mad: ΚΙ ΟΤΙ ΑΥΤΟ ΔΕΝ ΠΡΠΕΙ ΝΑ ΣΕ ΑΠΟΓΟΗΤΕΥΣΕΙ ΟΚ?
ΧΑΙΡΕΤΩ

Χωρίς να γνωρίζω κανένα,συγχαρητήρια σε όλη τη βαλκανική ομάδα της Ελλάδας..!..Σας περιμένουμε στον Αγρό 27 Απριλίου-2 Μαϊου..!


function
real analysis
real analysis unsolved

Does there a function $ f: \mathbb{R} \to \mathbb{R}$ such that $ f((a,b)) \equal{} \mathbb{R}$ for every open interval $ (a,b)$.

I know it may discuss in the forum,but I can't find it,I am sorry :blush: .Can you help me?

I am sorry,again.
When I post it,I find the previous topic.
Everyone who interested in can visit
[url]http://www.mathlinks.ro/viewtopic.php?&t=31493[/url]


Prove that for every p (p is prime) there exist integers $ x, y$ such that $ p|x^2 \plus{} y^2 \plus{} 1$

[quote="t0rajir0u"]That should be $ x^2 \plus{} y^2 \plus{} 1$.  See its motivation [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].[/quote]
Yes, my mistake, thanks.

Hint: Of the numbers $ 0, 1, \ldots, p\minus{}1$, how many are squares modulo $ p$? How many are a square+1, modulo $ p$? Can these sets be disjoint?


yes you can.but you should read some analysis book  for the topic [b]definition by induction[/b].i think this way your doubt will be cleared :)


Suppose $\displaystyle y_i=\frac {1998}{x_i+1998}$
We have $y_1+y_2+...+y_n=1$ so $1-y_i =\sum_{j <> i} y_i$
Now use AM-GM and multiple all such inequalities You obtain the problem inequality.

Can you explain that solution in more detail?

If $y_{1},y_{2},\cdots,y_{n}>0$ and $\sum \frac{1}{y_{i}+1998}=\frac{1}{1998}$ then prove that: $\sqrt[n]{y_{1}y_{2} \cdot \ldots \cdot y_{n}} \geq 1998(n-1)$.

Solution:

we denote with:

$x_{i}=\frac{y_{i}}{1998}$ then we have: $\sum \frac{1}{1+x_{i}}=1$

Now we denote with $z_{i}=\frac{1}{1+x_{i}}$ then we have: $x_{i}=\frac{1-z_{i}}{z_{i}}$ with $z_{i}>0$ and $\sum z_{i}=1$.

Now $\displaystyle \prod x_{i}=\prod \frac{1-z_{i}}{z_{i}}=\frac{z_{2}+\cdots+z_{n}}{z_{1}} \cdot \ldots \cdot \frac{z_{1}+\cdots+z_{n-1}}{z_{n}} \geq \frac{(n-1)^{n}\sqrt[n-1]{(z_{1} \cdot \ldots \cdot z_{n})^{n-1}}}{z_{1} \cdot \ldots \cdot z_{n}}=(n-1)^{n}$

(the inequality comes from the relation AM-GM)

And now taking in terms of $y_{i}$ you get the desidered inequality.

I hope this is explained in detail.

I really don't think that this suitable for the begginers! Just my opinion.

Thanks Lagrangia! Your solution is clear and beautiful. The problem doesn't seem to be too hard (at least after I see the solution  :D ).

Well, thanks a lot. Allmost any inequality seems easy after seing a solution. Indeed, this problem is considered easy now because several variants on the same idea were given in different contests. But that's all.

Come to think of it, you are right. But I still appreciate someone taking the extra time to elaborate on the solution.


Doesn't replacement not matter unless you repeat the process?

[hide]There are $4$ different suits, so the probability of choosing a spade is $\frac14.$[/hide]

if there are jokers...then the answer's...[hide]$\frac{13}{54}$[/hide] :P

...what if there were info cards :rotfl:

-jorian

[quote="jhredsox"]if there are jokers...

...what if there were info cards[/quote]
A [b]standard[/b] deck of cards means 52 cards, 13 of each suit.

[quote]i_like_pie Doesn't replacement not matter unless you repeat the process?  [/quote]

My bad, I forgot to mention the probability of picking a spade three times without replacement. :blush:

[quote="ckck"]What is the probability of choosing a spade without replacement in standard deck of cards?[/quote]

$\frac{13}{52}*\frac{12}{51}*\frac{11}{50}=\frac{11}{850}$


logarithms
number theory unsolved
number theory

Let $S(n)$ denote the sum of digits of $n \in N$. Find all $n$ such that:

$S(n) = S(2n) = S(3n) =...= S(n^{2})$.


analytic geometry
geometry
graphing lines
slope

I came across this problem in my quest for math :rotfl: ...

[i]If a, b>0 and the triangle in the first quadrant bounded by the coordinate axes and the graph of ax+by=6 has area 6, then find ab.[/i]

My solution was completely algebraic, and I got 3 as my answer. I just wanted to see if other people get the same answer, because I think I may have made an error.

Since we know that it is bound by the first quadrant a cannot be 0. I am assuming that u wrote it in standard form. So the y intercept is 6 so the x intercept should be 2. 6*2/2 equals 6. So the slope has to be -1/3 in point slope form. y=-1/3x+6 so in standard this is x + 3y = 6 so ab is 1(3)=3[ANS.] so you were correct.
TEEHEE :rotfl:

When $ y\equal{}0$, $ ax\plus{}by\equal{}6$ becomes $ x\equal{}\frac{6}{a}$, so the $ x$-intercept is $ (\frac{6}{a}, 0)$.

When $ x\equal{}0$, $ ax\plus{}by\equal{}6$ becomes $ y\equal{}\frac{6}{b}$, so the $ y$-intercept is $ (0, \frac{6}{b})$.

Therefore, the area of the triangle (= half the base times the height)

$ \equal{}\frac{1}{2}\frac{6}{a}\frac{6}{b} \equal{} 6 \Rightarrow ab \equal{} 6$.

I think you mean $ ab\equal{}3$ andrewtom. But otherwise nice solution


factorial
summer program
PROMYS
Pascal\u0027s Triangle

1.  10a + b must be congruent to 0 mod a and b.  Thus
10a + b = 0 mod a
10a + b = 0 mod b

b = 0 mod a
10 a = 0 mod b.

This is true for all a = b, so 11, 22, ... 99.  That's nine.  Now we look for the numbers in which b and 10a share a factor.  This works when b is two times a or 5 times a.  For two times a, we have 12, 24, 36, and 48.  For 5 times a, we have 15.  That makes a total of 14 numbers.

[quote="keta"]I have a book that offers the following problems, but no answers, so can anyone solve for an answer and explain it?

1. Find the number of 2-digit positive integers that are divisible by both of their digits.

My Answer: [color=white]14[/color]

2. In how many ways can COMPUTER be spelled by moving either down or diagonally to the right in the figure below:

C
OO
MMM
PPPP
UUUUU
TTTTTT
EEEEEEE
RRRRRRRR

Sorry for the bad picture, but it's simply a triangle with 1 C on the top, 2 O's, 3 M's going down until 8'R's. The leftmost digits in each row are aligned, and the columns should be aligned as well, with the full word COMPUTER going down directly from the first C and diagonally right directly from the first C.

My Answer: [color=white]53488800[/color] Not sure if it's right or not![/quote]

2 is simply 8*7*....2*1 = 8!

1.

We have that a | 10a + b and b | 10a + b

This is convenient, because it is obvious that a | 10a and b | b

So we need to solve a | b and b | 10a, where it should be apparent that 1 <= a <= 9 and 1 <= b <= 9 and a <= b.

See if you can solve it from here.

blahblahblah,

How is number 2 8!? 8 factorial would mean that moving from C to the o's would yield 2 choices, the O's to the M's 3 choices, etc... However, don't we need to consider the cases such that the first O is allowed to move to any of the three M's below, but the second O cannot move to the first M, it can only move to the 2nd or third M. (the questions requires only moving down or diagonally to the right). [b] Similarily, these cases occur for each row down, so wouldn't there be more than 8! choices? [/b]

Fewer than 8!, in fact.  How many different options do you have for each move?

Apologies, I misread the problem.

Hint: Try to use recursion.

Hint even though this is about to be solved: [hide]How many choices do you have for every letter?[/hide]

I was too lazy to count the letters, but I'm assuming there's 1 C, 2 O's, 3 M's, etc.

JBL, that is correct, it is actually fewer than 8! choices.
Here's what I did, since some choices are limited, then from the first C to the two O's are two ways; from the first O to any M is 3 ways, and the second O to the M's are two ways, a total of 5 ways. Likewise, the First M to any P would be 4 ways, the second M to any P would be 3 ways, then 2 (a total of 4+3+2=9, and so on:

2*5*9*(5+4+3+2)*(6+5+4+3+2)*(7+6+5+4+3+2)+(8+7+6+5+4+3+2) ways. However, I am confused about whether all of these should be multiplied together or not, since not each of the same letters can be attributed to the same number of available "next letters".

[quote="JBL"]Oh, you've interpreted the question differently than I have!  blahblahblah and I have both read "diagonally to the right" to mean "exactly one down and exactly one to the right," or otherwise "moving down at a 45 degree angle with the vertical."

Your way is much harder.  Your expression for it is definitely not correct, however -- those things should not all be multiplied together.[/quote]

If you can move diagonally in that manner, it becomes REALLY ugly.

yeah, im pretty sure the final answer is [b]112 [/b]if i didnt make a mistake

The original interpretation (given in the problem) should yield [hide]27 = 128[/hide].

It makes sense that there should be more possibilities than [hide]128[/hide] if we take Miko's interpretation, so 112 isn't right.

For #2...

[hide]There was a similar problem to this on the PROMYS application. Basically, its just the 8th line of Pascal's triangle, all added together. See, try the number of ways to spell "C". There's one. To spell "CO", you have two ways, CO1 and CO2. To spell "COM", there are four ways, CO1M1, CO1M2, CO2M2, and CO2M3. So to get to each letter, it looks like this:

      C

  O1 O2

M1 M2 M3

  1

 1 1

1 2 1

See? So to spell COMPUTER, you have 7C0+7C1+...+7C6+7C7= 1+7+21+35+35+21+35+7+1=2(64)=128.... Which is the same as 2^7, so perhaps there are 2 ways to solve this.[/hide]

Yes, your 112 is not correct, either.  I'll keep thinking about doing it your way -- report back in a couple of hours.

The [hide]27[/hide] answer is easy to understand, you see...

[hide]You start with C. No mystery there. For each step, you have two options: either you walk straight down, or you walk down then right. So you need to make a decision with two choices. You make this decision SEVEN times. Therefore, the total number of ways you can spell out COMPUTER is 2 * 2 * 2 * 2 * 2 * 2 * 2 = 27 = 128[/hide].

Okay, having examined the way you interpreted the problem, I have come to the conclusion that there is no nice way to do it.  A simple "block-walking" argument will work nicely -- count how many ways there are to get to each of the O's, then to each of the M's from the previous row, and so on.  Very straight forward, computationally just adding lots of stuff.  I get as an answer [hide]1430[/hide], which we see is still well less than 8!, so we know our answer is theoretically correct.  Furthermore, a check of the first few values indicates that they are in fact the Catalan numbers.  You can read more about the Catalan numbers at MathWorld.  They often appear in "branching" problems, so it's not too surprising that they pop up with this problem.

Mathworld gives lots of ways that one can generate the Catalan numbers, as well as an explicit formula -- worth reading about.

[quote="rcv"]I agreee with JBL.  This is definately [tex]C_8=1430[/tex], the 8th Catalan Number.  [tex]C_n={2n\choose n}-{2n\choose n-1}[/tex]

If you follow the block-walking technique, the values you fill into each cell form a "Catalan Triangle".  See [url]http://mathworld.wolfram.com/CatalansTriangle.html[/url] for an illustration.

Our problem is a slightly disguised form of "Catalan Mountain Ranges".

Given the posted problem, you are allowed two moves.  "Down" and "Right".  Each time you make a move "Down", we mark the spot we stepped *from* as one of the letters of the word "computer".  To identify the last letter of the word, we add a row of 9 spaces beneath the row of R's.  When you get to the bottom row of spaces, you step to the far right.  With this technique, you will always step down exactly 8 times and you will step right exactly 8 times.  This first step is guaranteed to be "Down".  The final step is guaranteed to be "Right".  There are [tex]{14\choose 7}=3432[/tex] ways to arrange the intervening steps, which gives an upper bound on the number of valid paths.  However, there is the added restriction that you can never step right beyond the last letter in a row.

An example of Catalan Mountains can be found here: [url]http://www.iki.fi/~kartturi/matikka/tab9766.htm[/url].  Our steps "Down" correspond to an uptick on a Catalan Mountain.  Our steps "Right" correspond to a downtick on a Catalan Mountain.[/quote]

Yes.  One of the books I used to self-study discrete math this year introduced the Catalan numbers by asking how many ways there were to travel from one corner of an 8x8 chessboard to another without going over the diagonal joining the corners.

It can be seen relatively simply that this is an identical problem!

Aye, thanks guys! I had no idea at first about the Catalan numbers, but I will research them on Mathworld.com. My originial answer was WAY TOO high.

my earlier solution was...
*******ok, if you assume that the problem means any of the letters in the row below as long as they are not to the left of the prior letter your combinations should be... 1x(3+2)+(4+3+2)+(5+4+3+2)+(6+5+.....+2)+...+(8+...+2) and the number you get should be 112 i think
(if i can use a calc properly -_-) you dont multiply them together because youve already looked at every poss. combo for each number, youve already considered all your poss. choices, which is what you are in essence doing when you multiply... somebody check me plz on this********

i think that 2^7 is also correct if thats the way you interpert the prob, and i understand the other solutions, but can someone explain why my 112 is wrong the way i interperted it

[quote="Miko"]1x(3+2)+(4+3+2)+(5+4+3+2)+(6+5+.....+2)+...+(8+...+2) and the number you get should be 112 i think[/quote]

To be honest, I don't understand what you were trying to do, so its rather difficult to offer help.  Also, I get a value of 110 when I compute your expression.

Let me suggest you "try a simpler case".  Post you solution to the 3-row and the 4-row triangles.  Maybe you will see your own mistake or maybe someone here can help you from that.


geometry
trigonometry
3D geometry
sphere
pyramid
linear algebra
matrix

If there are all these formulae, there must be a way to derive them and prove them, right?
I'm certain that some of them will be advanced for this forum-unless we can use limits here. (Can we?)
In my book, they simply say "The proof of this formula is complex and will not be shown here." or "The proof of this formula requires the concept of limits and will not be given here."
So, if I wanted to see how certain formulas were derived/proved, where would I go? Is there a webbie or can I ask for certain ones on some forum?

Thanks in advance! :D

There are a lot of geometry formulas.  Which ones are you talking about?

In any case, the forums here are probably a good resource, although if the formula you're looking at has a name you probably want to look it up first.

[quote="7h3.D3m0n.117"]Some high school geometry textbooks limit some proofs because the easiest way to prove (and to save ink) would be to use material that will presented to you in future years. For example, Heron's can be proven easily with trigonometry, but it might take more than a page algebracally.[/quote]

Actually, I think there was a cool way to prove it algebraically (something using the Pythagorean Theorem if I remember correctly) that took a minimal amount of steps.

Also, searching "Derivation of [insert formula name]" might fetch you some results on Google, and if the theorem is a derivation of another theorem or formula (like the Extended Law of Sines) then it might be best to ask on the forums because we have t0rajir0u and Google doesn't.

Heh. Good point. Specifics, right.
Hmm. Well, I wanted to start with basic polyhedra and "circular solids" stuff.
A list. Right. I should have a list. (Sorry, rather frazzled at the moment.)
1) The lateral area of a cone is equal to one-half the product of the slant height and the circumference of the base.
2) The "total area" for a sphere is $ 4(pi)$$ r^2$ (Please excuse the crude LaTeX.)
3) The volume of a pyramid is equal to one third of the product of the height and the area of the base.
4) An explanaton of Cavalieris principle + The volume of a sphere is equal to four thirds of the product of $ pi$ and the cube of the radius.
5) Heron's Formula
6) (Still) Want to know how to derive the Shoelace Method from a matrix--although I think I've pretty much got that one down.
7) Brahmagupta's Formula.

@7h3.D3m0n.117
Hmm. I figured that. I still want to know, though...:P
And, well, I suppose since I'm "more advanced" than the typical student doing HS Geo., I think I can handle it. I hope...:|
And while we're at it, I'm going to add Heron's and another one to the list. :lol:

I believe two through four, such as Cavalieri's principle, require calculus to an extent.

For the first one, consider "unfurling" the lateral surface of the cone to form a sector of a circle. See if you can take if from there.  :)

The others, you indeed can find via google/wikipedia; proofs of Heron's formula have been posted quite often on these forums, so try the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=223098153&t=67486]search function[/url].

Heron's formula:

Let $ a,b,c$ be the sides of a triangle, and let $ A$ be the area of the
triangle.  Heron's formula states that $ A^2 = s(s-a)(s-b)(s-c)$,
where $ s = \frac{(a+b+c)}{2}$.

Here is a straightforward derivation.  Consider the general
triangle with edge lengths $ a,b,c$ shown below.

We have

$ a = u+v$
$ b^2 = h^2+u^2$
$ c^2 = h^2+v^2$

Subtracting the second from the third gives $ u^2-v^2 = b^2-c^2$.  Dividing both sides
by $ a = u+v$, we have $ u-v = \frac{b^2-c^2}{a}$.  Adding $ u+v = a$ to both
sides and solving for $ u$ gives

$ u=\frac{a^2+b^2-c^2}{2a}$

Taking $ h =\sqrt{b^2-u^2}$ (from one of the three above) we have:

$ A=\frac{ah}{2}=\frac{1}{2}\sqrt{(ab)^2-(\frac{a^2+b^2-c^2}{2})^2}$

which is equivalent to Heron's formula.  Factoring out 1/4, this gives
three different ways of expressing $ (2ab)^2 - (a^2+b^2-c^2)^2$ as a
difference of two squares.  Equivalently, it gives three different
factorizations of $ 16A^2$, each of the form

           $ 16A^2  =   [(a+b)^2 - c^2] [c^2 - (a-b)^2]$

Factoring each of these terms gives the explicitly symmetrical form

           $ 16A^2  =   (a+b+c)(a+b-c)(c-a+b)(c+a-b)$

so if we define $ s=(a+b+c)/2$ we can write the first equation as:

                     $ A = \sqrt{s(s-a)(s-b)(s-c)}$

P.S. $ \pi$ can be written in Latex by typing "\pi" under the dollar signs.
      This is an edited version of a proof on some site that I found a long time ago (6th grade?)

[quote="The_Scintillator"]2) The "total area" for a sphere is $ 4(pi)$$ r^2$ (Please excuse the crude LaTeX.)
3) The volume of a pyramid is equal to one third of the product of the height and the area of the base.
4) An explanaton of Cavalieris principle + The volume of a sphere is equal to four thirds of the product of $ pi$ and the cube of the radius.[/quote]

These are all essentially calculus results.  Cavalieri's principle is an intuitive statement that cannot be made rigorous without calculus, based as it is on a rather crude notion of adding up an infinite number of areas to form a volume.  You will learn that it is [b]always[/b] necessary to justify every intuitive notion you have with infinite sums:  for example, even a statement as intuitive as "changing the order of a summation doesn't change the summation" is not true for infinite sums.

[quote="The_Scintillator"]6) (Still) Want to know how to derive the Shoelace Method from a matrix--although I think I've pretty much got that one down.[/quote]

See [url=http://mathworld.wolfram.com/CrossProduct.html]Cross Product[/url].  This is an essentially algebraic result.

[quote="The_Scintillator"]7) Brahmagupta's Formula.[/quote]

http://en.wikipedia.org/wiki/Brahmagupta's_formula

You can see what I mean by we have t0rajir0u and Google doesn't...

There is a non-rigorous (with lots of assumptions) method of finding the area of a sphere without calculus.

[url=http://mathforum.org/library/drmath/view/55135.html]EZ PZ[/url]

And a more explained version of Cross Product:

[url=http://www.mathreference.com/la-det,cross.html]Not so easy...[/url]

You don't need calculus to find the volume of a pyramid...

http://www.math.ubc.ca/~cass/courses/m308/projects/seaton/Seatonhtml.html

[quote="happyme"]
There is a non-rigorous (with lots of assumptions) method of finding the area of a sphere without calculus.

[url=http://mathforum.org/library/drmath/view/55135.html]EZ PZ[/url]
[/quote]

I think he meant surface area, which indeed needs calculus (at least says my 7th grade math teacher when I asked her a while back)

@moogra
...

...Why do people insist on calling me a male? I resent that people would think that I posess teh "y" chromosome. Psh.

If you haven't guessed by now, I'm a girl. :lol:

And about the area of a sphere, I'd like to know both. But yeah, area was what I specifically asked for first. If both were possible, however, I'd definitely like both. :D

@t0rajir0u
[quote="t0rajir0u"]Arbitrary base. (In particular, circles.)[/quote]
[quote="t0rajir0u"]You will learn that it is always necessary to justify every intuitive notion you have with infinite sums: for example, even a statement as intuitive as "changing the order of a summation doesn't change the summation" is not true for infinite sums.[/quote]

I feel kind of silly asking this--but is there any way you can define the first quote and tell me what exactly you mean by the second quote? I think I get it, but I'm not entirely sure.

@at everyone
Thanks to everyone for all the links and advice thus far! :D
It's great that I can get all this help.

If we can assume that the surface area of a sphere with a given radius is constant, then a relatively simple derivation can be achieved by a pseudo-calculus result - dividing the sphere up into a number of pyramids that protrude from the center as an estimation and then substituting the variables... the result isn't entirely rigorous, though.


Let G a graph on 2m vertices such that the degree sum of any two nonadjiacent vertices is at least 2m-1. Prove that each perfect matching of G is contained in a Hamiltonian path of G.

Does anybody have anything on this? My attempts have been until now unsuccessful.


calculus
integration
real analysis
real analysis unsolved

given $ f\in \L^{1}(0,1)$
let $ g(x)\equal{}\int_{0}^{x}\frac{f(t)}{t}dt$ Prove that
 $ g\in L^{1}(0,1)$ and $ \int_{0}^{1}g(x)dx\equal{}\int_{0}^{1}f(x)dx$

thanks in advance

Clearly false as stated.

Just let $ f(t)\equal{}1.$ Then $ g(x)\equal{}\infty$ for all $ x\in (0,1).$

What did you mean to say?

Oop, I made a typo
let $ f\in L^{1}(0,1)$ for $ 1>x > 0$ let :

$ g(x) \equal{} \int_{x}^{1}\frac {f(t)}{t}dt$

prove that $ g\in L^{1}(0,1)$ and $ \int ^{1}_{0}g(x)dx \equal{} \int ^{1}_{0}f(x)dx$

thanks

In which case this bit of advice comes into play: if you ever write down a double integral (iterated integral), then you must intend to switch the order of integration.

Just write it out and see what happens.

I did but it does not come out :blush:
could you please explain ?

thanks

Note:

$ \int_0^1\int_x^1h(x,t)\,dt\,dx \equal{} \iint_{\{0\le x\le t\le 1\}}h\equal{}\int_0^1\int_0^th\,dx\,dt.$

I think, the following is exactly what Prof Kent Merryfield is saying.

$ \int^{1}_{0}g(x)dx \equal{}\int^{1}_{0}\left(\int_{x}^{1}\frac{f(t)}{t}dt\right) dx\equal{}\int^{1}_{0}\left(\int_{0}^{1}h(x,t)dt\right) dx$  where $ h(x,t)\equal{}\frac{f(t)}{t}\chi_{[x,1]} (t)\equal{}\frac{f(t)}{t}\chi_{[0,t]}(x)$     (a diagram would help to visualize)
$ \equal{}\int^{1}_{0}\left(\int_{0}^{1}h(x,t)dx\right) dt$ by Fibuni.
$ \equal{}\int^{1}_{0}\left(\int_{0}^{1}\frac{f(t)}{t}\chi_{[0,t]}(x)dx\right) dt$
$ \equal{}\int^{1}_{0}\frac{f(t)}{t}\left(\int_{0}^{1}\chi_{[0,t]}(x)dx\right) dt$
$ \equal{}\int^{1}_{0}\frac{f(t)}{t}.tdt$
$ \equal{}\int^{1}_{0}f(t)dt$


geometry
combinatorics unsolved
combinatorics

There is a picture graph of $1^2+2^2+\ldots{}+n^2=\frac{n(n+1)(2n+1)}{6}$ on this site.

Can anyone please explain it? I have no idea how the graph explains this.

Do you mean the graph concerning the area of a square?

I think he means the prove without words on aops...
- Every square shown at the bottom occurs three times in the figure.
- the side length from left to right is $1+2+3+...+n$, so equal $\frac{n(n+1)}{2}$
- the height of the figure is $2n+1$

conclusion: the sum of the first $n$ squares equals $\frac{1}{3}(1+2+3+...+n)(2n+1)=\frac{n(n+1)(2n+1)}{6}$


[quote="April"]Find the locus of excenters of right triangles with given hypotenuse.[/quote]
The locus are two arcs $ 45^o$ which are set up on the hypotenuse.

can  u  explain   how are the arcs setup ?

Sorry for reviving this old thread.
[quote=mathiee]can u explain how are the arcs setup ?[/quote]
The arcs are subtended on the either sides of the hypotenuse such that the given hypotenuse subtends $45^{\circ}$. And also the locus is points bounded by the extreme points on the hypotenuse.


Let $ K$ be fuerbach point of triangle $ ABC$ and $ A'$ be midpoint of $ BC$ and $ E,F$ be tangency point of incircle with $ AC$ and $ AB$ let $ L\equiv EF\cap A'K$ prove that circumcircle of $ \triangle KLB$ is tangent to $ BC$

Dear Amir and Mathlinkers,
1. let A* be the meetpoint of EF and BC,
D the point of contact of the incircle with BC, and D' the symmetric of D wrt the A-bissector of ABC.
2. DD' // EF
3. By a converse of the Reim's theorem, K, L, D and A* are concyclic.
3. (B, C, D, A*) is harmonic.
4. By the power theorem and the Newton's relation, A'B² = AL.AK
and we are done.
Sincerely
Jean-Louis


function
Functional Analysis
advanced fields
advanced fields unsolved

If $ B(X,Y) \equal{} \{f: X\longrightarrow Y \mid f$ linear, continuous and bounded$ \}$ is a Banach (normed and complete) space, then $ Y$ is a Banach space ($ X$ and $ Y$ are normed linear spaces).

But if $ X\not \equal{}0$, is it true? :maybe:

Let $ (y_n)_n$ be a Cauchy sequence in $ Y$. Knowing that $ X\not \equal{} 0$, for $ x_0\in X\setminus \{0\}$, we take $ f: X\longrightarrow \mathbb{R}$ s.t. $ f(x_0) \equal{} \| x_0\|$ and $ \|f\| \equal{} 1$ (this function exists, using a corollary of the Hahn-Banach theorem - probably for this we must add $ X$ complete in the statement, i'm not sure).

Then, we consider the sequence $ (u_n)_n \in \mathcal B(X,Y)$, $ u_n(x) \equal{} f(x)y_n$ and it is easy to prove that $ (u_n)_n$ is Cauchy. By the hypothesis, it is convergent. So $ u_n\longrightarrow u\in \mathcal B(X,Y)$ and, finally,  $ y_n\longrightarrow \frac {u(x)}{\|x_0\|}\in Y$.
This means $ Y$ is a Banach space.

    q.e.d.


Anyone interested in joining an AoPS group for the NCAA Men's Tourney?

I'll set up a group on ESPN.com if we get enough interest (probably 8+)

See the other thread in G/FF about the yahoo group i made...

Okay since it seems like the Yahoo tourney has already been established, let's ditch this one unless a decent sized amount of people are opposed to Yahoo.

Agreed; the Yahoo! one has been around for a (long) while.

(and I believe their fantasy sports are really interactive and good)


$ a,b,c \ge 0$ prove that:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 4 \ge ab \plus{} ac \plus{} bc \plus{} 2(a \plus{} b \plus{} c)$

We have

                 $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 4\ge ab \plus{} ac \plus{} bc \plus{} 2(a \plus{} b \plus{} c)(a,b,c\ge0)$ (*)
let

                $ a\equal{}\frac{ka}{a\plus{}b\plus{}c}, b\equal{}\frac{kb}{a\plus{}b\plus{}c},   c\equal{}\frac{kc}{a\plus{}b\plus{}c}$,

then
                $ (*) <\equal{}>[(k\minus{}1)^2\plus{}3]\sum a(a\minus{}b)(a\minus{}c)\plus{}(k\minus{}4)^2\sum(b\plus{}c)(a\minus{}b)(a\minus{}c)$

                             $ \plus{}2(k\plus{}6)(k\minus{}3)^2abc\ge0$.

Why can you know that we must set $ a\equal{}\frac{ka}{a\plus{}b\plus{}c}$ ...? Because there is no condition related to a, b, c. :?:

Here is another one  :)  :

$ a,b,c \ge 0$,prove that:

$ a^2\plus{}b^2\plus{}c^2\plus{}abc\plus{}2 \ge ab\plus{}ac\plus{}bc\plus{}a\plus{}b\plus{}c$

Let $ a \plus{} b \plus{} c \equal{} k$,then

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 2 \minus{} (ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c)$

$ \equal{}(k^2\minus{}k\plus{}2)\sum{a(a\minus{}b)(a\minus{}c)}\plus{} (8 \plus{} k^2 \minus{} 4k)\sum(b \plus{} c)(a \minus{} b)(a \minus{} c)$

$ \plus{} (54 \minus{} 27k \plus{} k^3)abc\ge0$.

[quote="rachid"]Here is another one  :)  :

$ a,b,c \ge 0$,prove that:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \plus{} 2 \ge ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c$[/quote]
I think, your inequality is true for all $ \{a,b,c\}\subset[\minus{}1,\plus{}\infty).$ :wink:


A 4m by 4m room may be covered by eight 1m by 2m mats in various ways. Three different ways are shown in the attachment.

How many ways are there altogether?

Source: Australian Math Competition, S30, 2001

[hide]I got 30.
If you take the mats 2x4 and 2x4, there are 5 ways in each block. Then you take the middle block and there is only one solution for the top 1x4 block and the bottom 1x4 block. $5\times5+5=30$.[/hide]

Unfortunately, there are more than 30 configurations.  :)

Is the best way to proceed here to enumerate all the possible configurations, or is there a better way? :?

Ok, now i counted 46. If that's right (which it probably isn't) or near right,  I'll post my solution.

There is a better way to systematicallly list all the solutions. However I wanted to see if you guys could come up with any easier way.  :)


calculus
integration
derivative
function
geometry
rectangle
calculus computations

Hi everyone,

Could you please tell me if the antiderivative of e^2x is e^2x/2 and if 2e^2x is e^2x/2? I got both of these answers by doing the opposite of taking the derivative. For example, the derivative of e^2x is e^2x(2), so wouldn't the antiderivative be e^2x/2?

Thank you very much

You've got two different functions and you're giving them the same antiderivative, which is problematic.  The first one is correct; that is,

$ \int e^{2x} \, dx \equal{} \frac{1}{2} e^{2x} \plus{} C$

hey are you online please help me!

Estimate the area under the graph of f(x) = e^-x^2
from x = -2 to x = 2 using four approximating rectangles and midpoints.

When you have a new question to ask, ask it in a new thread.  There was no need to revive this old one.


geometry
probability
logarithms
AMC 10
AMC

hw long did it take u guys to finish aops v1? do u think knowing v1 really well is enough to get 130+ on the AMC 10?

I didnt finish it (skipped the parts where the topics arent covered on the AMC), but I would say if you could understand a majority of the concepts of the book, you could easily get a 130+ on the AMC 10.  Infact, just by strictly reading that book, someone could probalby get a 150 (no concepts on the AMC 10 are above vol. 1) if they master all of the concepts.

Look at all of the geometry sections
the counting/probability sections
and the equation solving sections

The geometry section is incredibly important for the AMC, and the probability sections are also useful (if you're not going to buy the Probability books).

Pretty much everything could be on the AMC 10, with the possible exception of logarithms, which are on the AMC 12.

Yeah...I got a 138 on the AMC 10 before I'd even heard of AoPS, so yeah, easily 130+.

Actually, between the time of AMC 10A and B, I studied AoPS 1 (none of the geometry), just a few of the first 8 chapters. It helped not just on the AMC, but also on local math contests, so keep that in mind.


For a=0 all n are solutions so suppose $a\neq 0$. First notice that $X^{2}$ always divides$f(X)=(X^{2}+a^{2})^{n}-X^{2n}-a^{2n}$ because $0$ is a root
of $(X^{2}+a^{2})^{n}-X^{2n}-a^{2n}$ and its derivative. Now if n is solution then  all x such that $x^{2}+a^{2}=ax (1)$ are roots of $f$ and $f\^{\prime}$. From here we get that all such x satisfy $x^{n-1}=a^{n-1}$. From here we get $x=a\epsilon$ where $\epsilon$ is (n-1) root of unity. From
(1) we get $\epsilon^{2}-\epsilon+1=0(2)$ for now we have used only that x is a root of $f^\prime$ now since x is root of f we find$(\epsilon^{n})^{2}-(\epsilon^{n})+1=0$ so $\epsilon^{n}$ is a root of
(2). But this is true since $\epsilon^{n}=\epsilon$ so the only condition is for $\epsilon$ is to be (n-1) root of unity and from here we get that $n\equiv 1(mod 6)$ and these are the only solutions.


Are you sure that's correct?  I used my Poly root finder on my calculator and the too real roots added together was just over 2... :maybe:

I think he/she meant $ x_1\plus{}x_2>2$ :wink:


hey guys, just wondering if any of you could help me with the following problems:

1) Find X inside the hexagon:
7 ____8
2/  54   \ 9
1\____/4

3    ____6
12/   X     \ 2
5   \____/8

2) 2,7,11,20,38,69,127, X?
what should X be?

3) Someone asks 2 siblings (a brother and a sister), who is older
Sister: I am older
Brother: I am younger

If you know that atleast one of them was lying, who is older?

4) Two integers are chosen x,y such that 2 <= x <= y <= 20.
The sum x + y is given to a guy named A, and the product is given to a guy named B. They have the following conversation:
A: I see no way you can determine my sum.
B: Now I know your sum.
A: Now i know your product.
What are x and y?

For 1, shouldn't X just be 54? They are the same hexagons...

[hide="2"]The next term is the sum of the previous 3, so X=127+69+38=234.[/hide]

[hide="3"]If only one of them were lying, then either they would both be older than each other or both younger than each other, which would be false. Then they are both lying, so the brother is older.[/hide]


limit
trigonometry
calculus
calculus computations

Find the limit:
$ \lim_{x \to1} \frac{\sin (ax^2+bx+c)}{x^2-1} $
where $a,b,c$ are real numbers and $a+b+c=\pi$

I have said that I don't like to use L'Hospital's rule very much, but it does have [i]some[/i] uses! This is a $\frac00$ case, since $\sin\pi=0.$ By L'Hospital:

$\lim_{x\to1}\frac{\sin(ax^2+bx+c)}{x^2-1}=$

$= \lim_{x\to1}\frac{\cos(ax^2+bx+c)(2ax+b)} {2x}=\frac{(\cos\pi)(2a+b)}2=-a-\frac b2.$

[quote="cats..."]why dont you like l'hospitals rule?[/quote]
Because I regard power series as lying very near the heart of calculus. Because a deep understanding of rates of growth is one key to understanding calculus. Because I prefer insight to mechanical tests that hide the ideas they spring from.

L'Hospital's rule is a way of organizing and presenting knowledge related to the mean value theorem and Taylor's theorem. There are situations, such as this one, where it is the most efficiet approach, but there are many other situations that cry out for a direct application of power series, or of prinicples related to rates of growth.

Where can i find an article about power series ?

[tex]\mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {ax^2  + bx + c} \right)}}{{x^2  - 1}}\mathop  = \limits_{\mathop {a + b + c = \pi }\limits^ \uparrow  } \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {ax^2  + bx + \pi  - a - b} \right)}}{{x^2  - 1}} = [/tex]

[tex]\mathop  = \limits_{\mathop {\sin \left( {\pi  - \alpha } \right) = \sin \alpha ,\forall \alpha  \in \mathbb{R}}\limits^ \uparrow  } \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {a - ax^2  + b - bx} \right)}}{{x^2  - 1}}\mathop  = \limits_{\mathop {\left( {\frac{0}{0}} \right)}\limits^ \uparrow  } [/tex]

[tex] = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\sin \left( {a - ax^2  + b - bx} \right)}}{{a - ax^2  + b - bx}} \cdot \frac{{a - ax^2  + b - bx}}{{x^2  - 1}}} \right] = [/tex]

[tex] = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\sin \left( {a - ax^2  + b - bx} \right)}}{{a - ax^2  + b - bx}} \cdot \frac{{ - a\left( {x + 1} \right) - b}}{{x + 1}}} \right]\mathop  = \limits_{\mathop {\mathop {\lim }\limits_{\omega \left( x \right) \to 0} \frac{{\sin \omega \left( x \right)}}{{\omega \left( x \right)}} = 1}\limits^ \uparrow  } \frac{{ - 2a - b}}{2}[/tex]

I would add - and metru has demonstrated the point for me - that L'Hosptal's rule is never the only way to solve a problem. There are always other methods - sometimes Taylor series methods, sometimes (as in metru's solution) algebraic methods.


using only the digits 1 through 5 , how many even three – digit positive integers less than 500 can be written if digits can be used more than once ?

[hide="Solution"]
First digit has 4 choices, 1 to 4.
Second digit has 5
Third digit must be even, so it's 2 or 4.

$ 4\cdot 5\cdot 2\equal{}\boxed{40}$
[/hide]


How many distinct diagonals of a convex hexagon can be drawn?

Each of the 6 vertices can have a diagonal with all the other vertices except for three: itself and its two neighbors. This leaves 6-3=3 diagonals per vertice, or 6*3=18 all together. However, we have counted each diagonal twice, so our final answer is 18/2= $ \boxed{9}$.

I general, the number of diagonals for an n sided convex polygon is n(n-3)/2.


Let $ I$ be the center of the incircle of non-isosceles triangle $ ABC,A_{1} \equal{} AI\cap BC$ and $ B_{1} \equal{} BI\cap AC.$ Let $ l_{a}$ be the line through $ A_{1}$ which is parallel to $ AC$ and $ l_{b}$ be the line through $ B_{1}$ parallel to $ BC.$ Let $ l_{a}\cap CI \equal{} A_{2}$ and $ l_{b}\cap CI \equal{} B_{2}.$ Also $ N \equal{} AA_{2}\cap BB_{2}$ and $ M$ is the midpoint of $ AB.$ If $ CN\parallel IM$ find $ \frac {CN}{IM}$.

hint? it seems that I can't solve it without a hint :blush:

post a solution...:(....I can't solve it :blush:

I find it http://www.mathlinks.ro/viewtopic.php?p=1097205#1097205


The sum of the reciprocals of the roots of the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ is:

$ \textbf{(A)}\ \frac {1}{a} \plus{} \frac {1}{b} \qquad \textbf{(B)}\ \minus{} \frac {c}{b} \qquad \textbf{(C)}\ \frac {b}{c} \qquad \textbf{(D)}\ \minus{} \frac {a}{b} \qquad \textbf{(E)}\ \minus{} \frac {b}{c}$

The polynomial with reciprocal roots is $ cx^2\plus{}bx\plus{}a\equal{}0$, and the sum of these two roots, by Vieta's, is $ \boxed{\textbf{(E)}\ \minus{}\frac bc}$.


modular arithmetic
inequalities unsolved
inequalities

Find all positive integer solutions to the following equation:

$7^{a}-3^{b} = 4$

$(a,b)$ must be the same parity (checking the equation $mod$ $4$),so
if they are both even then $7^{\frac{a}{2}}+3^{\frac{b}{2}}=4$ and $7^{\frac{a}{2}}-3^{\frac{b}{2}}=1$
if they are odd then the equality is equivalent to $3(7^{\frac{a-1}{2}}-3^{\frac{b-1}{2}})(7^{\frac{a-1}{2}}+3^{\frac{b-1}{2}})=4(1-7^{\frac{a-1}{2}})(1+7^{\frac{a-1}{2}})$..and I'm too lazy to do the rest...

Taking the equation mod 8 we get [tex](-1)^a\equiv 3^b+4 \pmod{8}[/tex] so it must be [tex]a\equiv b\equiv 1 \pmod {2}[/tex]. Now suppose [tex]b>1[/tex] and so [tex]b \geq 3[/tex] Now we take the equation mod 27: [tex]7^a\equiv 4 \pmod {27}[/tex] that gives us [tex]a=9k-1[/tex]. But [tex]a[/tex] is odd so [tex]a=18k-1[/tex]. But we have [tex]3^b \equiv 3 \pmod{7}[/tex] and so [tex]b=6k+1[/tex]. Now we take the equation mod 19. We have:
[tex]7^{-1}-3*(3^6)^k \equiv 4 \pmod {19}[/tex] and so [tex]11-3*7^k \equiv 4 \pmod {19}[/tex] that have no solution. So we have [tex]b=1[/tex] and so [tex]a=b=1[/tex] is the only solution.

P.S.: Residues of 7 mod 27 --> 1,7,-5,-8,-2,13,10,-11,4,1 (period=9)
        Residues of 3 mod 7  --> 1,3,2,-1,-3,-2,1 (period=6)
        Residues of 7 mod 19 --> 1,7,11,1 (period=3)

Excellent! That was a really nice jumbling of mods. How do you do it? (I mean I don't yet get the logic about when to use which one.)

Yet another solution. If $b>1$, working mod 9 you get that $a\equiv 2$ mod 3. Working mod 13 you have that $3^b$ can only be 3, 1 or 9, so that either $7^a\equiv 7$ mod 13 (which implies $a\equiv 1$ mod 12, contradiction), or $7^a\equiv 5$ mod 13 (which implies $a\equiv 3$ mod 12, contradiction), or $7^a\equiv 13$ mod 13 (impossible).

One has the idea of working mod 9 because we need something which cannot be used to exclude the trivial solution with $b=1$, while working mod 13 is a good idea since 3 has a short period mod 13.

I suggest to move this topic to Number Theory.


linear algebra
matrix
vector
algorithm
LaTeX
algebra
system of equations

Suppose $ n$ points are given: $ (x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$, and we want to solve for $ a_i; i \equal{} 0,1,2,\ldots,k.$, so $ n$ and $ k$ are fixed numbers.
We have system of equations: $ XA \equal{} Y$, where $ A \equal{} [a_0,a_1,\ldots,a_k]^T$, and $ X \equal{} [x_{i,j}]$ is $ (k \plus{} 1)\times(k \plus{} 1)$ matrix with $ x_{i.j} \equal{} \displaystyle\sum_{m \equal{} 1}^n x_m^{i \plus{} j \minus{} 2}$, and for the vector $ Y$ we have $ Y \equal{} [ \displaystyle\sum_{m \equal{} 1}^n y_mx_m^{0}, \displaystyle\sum_{m \equal{} 1}^n y_mx_m^{1}, \ldots, \displaystyle\sum_{m \equal{} 1}^n y_mx_m^{k}]^T$, so in the $ i$-th place we have $ \displaystyle\sum_{m \equal{} 1}^n y_mx_m^{i\minus{}1}$.
What conditions are necessary for unique solution?

I'm, pretty sure we have a unique solution iff among $ n$ $ x_i$'s we have at least $ k \plus{} 1$ different ones, and in that case we take inverse of  $ A$ and solution follows, but i don't know how to get a solution when $ A$ is not invertible. Any algorithm maybe? Any help would be appreciated.

Edit: Sorry for my errors, i believe everything is ok now...

Here's a useful $ \text{\LaTeX}$ code for you: \times . I edited it into your post in the appropriate place, so you can see it.

As for the rest of it - I'm having some trouble reading it. In particular, I'm not sure which indices are supposed to start at zero, which at one, what's $ n,$ what's $ k,$ and what the actual sizes of matrices and vectors are. Could you please go back and proofread?


email
Putnam
geometry
3D geometry
\/closed

Google is now a WOOT sponsor.  In addition to the funds provided by D.E. Shaw group, this sponsorship will help establish WOOT as a valuable and productive program for the world's top students of mathematics.

Sponsorship information will be updated soon.

[quote="xxreddevilzxx"]cool!!!! how did you guys manage to have google as a supporter?[/quote]

For one thing, several of their top researchers and developers were medalists at the IMO.  I think some of those people in particular believe that WOOT is the kind of program that will help students develop deeper and highly useful problem solving skills.

Congratulations. I'm going to spread this news on an email list for parents or two.

Well, they had that whole e (the base of the natural log) contest a while ago.  I'm not too surprised.

Indeed, most go into academia, but there are more and more options outside academia now!

[quote="bubala"][quote="rrusczyk"][quote="chesspro"]Wow! This is awesome! I had no idea that people who worked for Google placed in the IMO.[/quote]

Indeed - Google and D.E. Shaw group both depend on master problem solvers for their success.  IMO participants who choose to turn their skills to real-world problems are obviously a great pool to find them![/quote]

Yeah, I was looking through the Putnam book.  At the back, the authors summed up the careers of every Putnam fellow.  Many turned to D.E. Shaw and Google.  However, most went into academia.[/quote]

bubala, what Putnam book are you refering to?

Does D.E. Shaw refer to Shaw U.S. the cable company? Probably not, I think that's a local cable company.

Wow, amazing. Google on of the most well known internet-related corporations sponsoring your program! You must be very pleased!  :lol:

I just wish one day I will reach the level in which I may participate in WOOT!

Also, I just want to thank you personally for devoting so much of your own time to help encourage students to understand and love math. (To all administrators)

Thanks much for the congrats - we are very happy with both D. E. Shaw group and Google!  D. E. Shaw group is a financial firm.  You can learn more about them [url=http://www.deshaw.com]here[/url].  They hire many IMO medalists to do quantitative analysis of financial markets.  You'll have a chance during the following year to attend a Math Jam to learn more.

D.E. Shaw gave everybody a free Rubik's Cube at MOP :)

And they were involved in ARML. I have a DE Shaw calendar in my wallet.

[quote="MCrawford"]Sponsorship information will be updated soon.[/quote]

The sponsors' logos are larger and more visible (due to the color and contrast) than the AoPS logo, when viewed on my browser.

[quote="fleeting_guest"]
The sponsors' logos are larger and more visible (due to the color and contrast) than the AoPS logo, when viewed on my browser.[/quote]

For now, that represents the excitement over WOOT.  Yeah, that's my story and I'm sticking to it.

[quote="MCrawford"][quote]The sponsors' logos are larger and more visible [...] than the AoPS logo, when viewed on my browser.[/quote]

For now, that represents the excitement over WOOT.[/quote]

The sponsors' logos are, currently, larger and more visible than the WOOT logo, on the page(s) where both appear.

[quote="Magnara"]Man, AOPS, you guys used to be cool, back before you sold out.

 :P  :P  :P[/quote]

Selling out only occurs when your mission and direction are compromised.  In that regard, I think we couldn't ask for better sponsorship relations.   :lol:

Yeah, they're being sponsored by Google, not McDonalds and Enron.


pigeonhole principle
number theory proposed
number theory

Given strictly increasing sequence $ a_1<a_2<\dots$ of positive integers such that each its term $ a_k$ is divisible either by 1005 or 1006, but neither term is divisible by $ 97$. Find the least possible value of maximal difference of consecutive terms $ a_{i\plus{}1}\minus{}a_i$.

I think it's 2010, but I could very well be wrong.

The maximal difference of consecutive terms is obviously minimized when the sequence contains every integer divisible by 1005 or 1006 but not 97. Define the length of the gap between two consecutive terms of the sequence as the number of integers between those two terms that are divisible by either 1005 or 1006, but also divisible by 97. If the length of the gap is 3 or more, by Pigeonhole we have two numbers that are excluded that are both divisible by, say, 1005 and also 97, which is impossible since then the two terms of the sequence that bound these numbers can't be consecutive (as we can find another multiple of 1005 between the two multiples of 97*1005). If the length of the gap is 0, clearly we can't do any better than 1005.

If the length of the gap is 1, we have three numbers, each of which are divisible by either 1005 or 1006 and the middle one is divisible by 97. The difference between the two "outer" numbers is clearly at most 1005*2=2010, and we see that this is achievable by making the middle one 97*1005*1006.

If the length of the gap is 2, we see that the four numbers (the middle two divisible by 97) must be of the form 1005x, 1006y, 1005(x+1), 1006(y+1) or 1006y, 1005x, 1006(y+1), 1005(x+1) (we can do some bounding stuff to see this, I'll omit it because it's annoying and pretty easy). But we see that the difference between the outer two is 2011-(difference between inner two), which is only bigger than 2010 if the inner two are equal, but then we get back to the previous case (and the first number has to be a multiple of 1005, not 1006). So we're done.

Yes, the answer is correct.

The solution is a bit shorter if you consider numbers of the form $ 1005\cdot1006\cdot97k$. There exists some $ k$ such that this expression is $ <a_1$, and it's easy to see the maximal difference is minimized around this number by picking $ 1005(1006\cdot97k\minus{}1)$ and $ 1005(1006\cdot97k\plus{}1)$.
Once we see that 2010 is the minimal difference, the restrictions on the sequence loosen... now it suffices to consider the sequence of all multiples of 1005 except those divisible by 97.


alright. plain and simple. I post n, then next person posts n-1, then n-2 and so on.

[b]1000[/b]


analytic geometry
MATHCOUNTS
number theory
prime numbers
prime factorization

this is the almost complete(I wont do the ones containing pictures) 2006 workout 7 for MC. I shall post answers after 5 replys with people saying "yes, answers" :D

1. Juan drove the 60 miles from Waco to Salado at 30 mph and drove back home to Waco over the same route at 60 mph. What was the average speed for the entire trip?

2. The harmonic mean is the reciprocal of the mean of the reciprocals of the numbers in a set. What is the harmonic mean of the set {1, 2, 2, 4, 4}?

[color=blue]Moderator kyyuanmathcount's note: #4 has been posted, see thread about PIN numbers.[/color]

6. The graph of the equation y = x^2 − 4x passes through the origin. For what other point on the graph of the equation are the values of the x -coordinate and y -coordinate equal?

7. A children‛s magazine states that 99% of people can‛t lick their own elbows, and 90% of people who read that statistic try to do it. Of 1000 randomly selected people who read that statistic, how many people can we assume will try to lick their own elbow and be able to do it? Express your answer to the nearest whole number.

8. The integer 2006 is the product of three prime numbers p , q and r . If p + q = r^2*(p + r ), what is the value of q ?

[hide="#1"]Takes 2 hours on the way there, 1 on the way back. So 3 hours for 120 miles. $\frac{120} {3}=\boxed {40}$ miles per hour.[/hide]

[hide="#2"]Take the reciprocals, making it $1,\frac{1} {2},\frac{1} {2},\frac{1} {4},\frac{1} {4}$

Average them, giving you $\frac{5} {10}$. Take that reciprocal to get $\frac{10} {5}$ or better known as $\boxed {2}$[/hide]

[hide="#6"]If you're trying to find the value where the x coordinate equals the y coordinate, you can simply state $x=y$. So substitute that in to get $y=y^2-4y$

Do some work to get $y(y-5)=0$ We already know they're equal to (0,0), so it's the other solution, making the answer $\boxed {(5,5)}$[/hide]

[hide="#7"]Of the 1000 people, we can assume only 10 can actually lick their elbow. 90% of those 10 will do it, so $\boxed {9}$[/hide]

[hide="#8"]It helps to know $2006=2*17*59$ The thing says $p+q=r^2*(p+r)$. Personally the way I would do it is substitute the numbers in, so after some trial and error, $\boxed {q=59}$[/hide]

[hide="#1"]
first trip was two hours. second was one hour. $\text{Average speed}=\frac{\text{total distance}}{\text{total time}}$, so then average speed is $\frac{120}{3}=40 \text{ mph}$[/hide]

where do you get all the handbooks? it seems like everyone has them...

EDIT: didn't see the number 1 before... did it because no one else did. whoops

[hide="number 1"]
You do the harmonic mean.
$\frac{1}{60}+\frac{1}{30}=\frac{1}{20}$
$\frac{1}{20}/2=\frac{1}{40}$
$\frac{1}{40}^{-1}=40$
Answer is 40
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{40}}}}}}}}}}}}}}}}$
[/hide]
[hide="number 2"]
$\frac 11 + \frac 12 + \frac 12 + \frac 14 + \frac 14=\frac 52$
$\frac 52 /5=\frac 12$
$\frac 12 ^{-1}=2$
Answer is 2
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{2}}}}}}}}}}}}}}}}$
[/hide]
[hide="number 4"]
9886 see other topic on this question
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{9886}}}}}}}}}}}}}}}}$[/hide]
[hide="number 6"]
For this we need $x=y$
Using Substitution, $x=x^2-4x$
$x^2-5x=0$
$x(x-5)=0$
$x=0,5$
They already tell us the $x=0$ is a solution.
Thus the answer is $x=5$
and the point is $(5,5)$
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{(5,5)}}}}}}}}}}}}}}}}$[/hide]
[hide="number 7"]
The total number of people is 1000
Thus the total amount of people who attempt to lick their elbow is 900
Only $1\%$ can do it, so
$900\cdot.01=9$ of the 900 people can actually do it.
Thus the answer is $9$
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{9}}}}}}}}}}}}}}}}$
[/hide]
[hide="number 8"]
The three prime factors of 2006 are 2,17,59.
By guess and check, we can determine that $q=59$
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{59}}}}}}}}}}}}}}}}$[/hide]

[quote="moogra"]
[hide]
1. 40 miles
2. 2
4. oh my god I saw that somewhere don't know how to do
6. 5,5
7. 9 (not sure because it is a whole number)
8. 59
[/hide][/quote]

Start showing your work and don't cheat!!!! :mad:  :mad:

You got to close the hide tag again so it will work.  Try previewing before submitting.

[quote="goldendomer"]
Start showing your work and don't cheat!!!![/quote]
if you steal peoples answers on a math forum, thats pathetic :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl: :rotfl:  :rotfl:

There really is no point cheating.  These forums are to help the young and furnish skills of the pros, so you don't really get anything except a rank(unimportant) and a post.  MATHCOUNTS is where they really test your skill, because it's impossible to cheat.

[quote="biffanddoc"]
if you steal peoples answers on a math forum, thats pathetic :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl: :rotfl:  :rotfl:[/quote]
You guys sound like my math teacher.
[hide]1. Juan drove a total of 120 mi. for the roundtrip. He took two hours to get to Salado and one hour to get back to Waco. 120/3=40 mph average.
2.Working backwards: The mean of the reciprocals of 1,2,2,4,4 is 1/1+1/2+1/2+1/4+1/4 all over 5, or 1/2. The reciprocal of 1/2 is 2.
4.With no cases, there is a total of 10000 possible PINs. Since case 2 counts the numbers in case 1 again, we can just subtract the number of cases for 2 and the number of cases for 3 to get our answer. The total number of cases for 2 is 90, since the repeated pair can be from 10-99. The number of cases for 3 is 14. 10000-90-14=9896.
6.If x=y, then substitute in y for x. Now, x=x^2-4x. Add 4x to both sides, then divide by x, to get x=5. The other point besides the origin is (5,5).
7.Of the 1000 people, 90%, or 900 people, will try to lick their elbow. 1% is expected to be able to lick their elbow, and 1% of 900=9.
8. The prime factorization of 2006 is 2*17*59, so we already have 3 numbers. We know r must be 2, or else r^2 would be too big. With a little trial and error, we see that q=59.[/hide]
If i start adding any click text to my hide thingy, it won't work...grr

[quote="easyas3.14159..."]4.With no cases, there is a total of 10000 possible PINs. Since case 2 counts the numbers in case 1 again, we can just subtract the number of cases for 2 and the number of cases for 3 to get our answer. The total number of cases for 2 is 90, since the repeated pair can be from [b]10-99[/b]. The number of cases for 3 is 14. 10000-90-14=9896.[/quote]
What about 00-09?

[quote="k8reindeer"]chinesechess, mathcounts handbooks can be found at
[url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=132]www.mathcounts.org/webarticles/anmviewer.asp?a=132[/url]
although that is only the current year's
I get previous years from my coach[/quote]

Thanks  :)

My math teacher got sent the handbook...but my coach didn't mention anything about it.
I saw somewhere that you get the handbook at the MATHCOUNTS competition or something like that...


modular arithmetic
number theory unsolved
number theory

Let $S\subseteq \{1,2,\cdots,n\}$ having following property: $S=\{x : (x,n)=1 \wedge (x+1,n)=1\}$ Prove, that : $\prod_{x\in{S}}\equiv{1}\pmod{n}$

[hide]If $x\in S$ then there're $x*$ and $y$ in $\{1,2,\ldots,n\}$ such that
\[xx*\equiv1\pmod n,\ \ (x+1)y\equiv1\pmod n\]
Thus $(x*+1)xy\equiv1\pmod n$ and $x*\in S$.
If $x*=x$ then $(x-1)(x+1)\equiv1\pmod n$. Since $x\in S$, we have $x=1$.
Thus $\prod_{x\in S}x\equiv1\pmod n$[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $\displaystyle G$ be a finite group of $\displaystyle n$ elements $\displaystyle ( n \geq 2 )$ and $\displaystyle p$ be the smallest prime factor of $\displaystyle n$. If $\displaystyle G$ has only a subgroup $\displaystyle H$ with $\displaystyle p$ elements, then prove that $\displaystyle H$ is in the center of $\displaystyle G$.

[i]Note.[/i] The center of $\displaystyle G$ is the set $\displaystyle Z(G) = \left\{ a \in G \left| ax=xa, \, \forall x \in G \right. \right\}$.

Let $g\in G$, and consider the automorphism $\phi_g: G\to G$ of $G$, defined by $\phi_g(x)=gxg^{-1},\ \forall x\in G$. The image of $H$ through $\phi_g$ is a subgroup of $G$ of order $p$, and since $H$ is the only such subgroup, $\phi_g(H)=H$. This means that the restriction of $\phi_g$ to $H$, which we'll also denote by $\phi_g$, is an automorphism of $H$.

The automorphism group of $H\cong\mathbb Z_p$ is a group with $p-1$ elements (an automorphism is uniquely determined by choosing among the $p-1$ possible images of the residue class of $1$), so the order of $\phi_g$ divides $p-1$. It's clear, however, that the order of $\phi_g$ divides the order of $g$ and thus $n$, so we find $\phi_g$ to have order dividing $(n,p-1)=1$, i.e. $\phi_g(h)=h,\ \forall h\in H$, and since $g$ was arbitrary, this says that $ghg^{-1}=h,\ \forall g\in G,\ \forall h\in H\Rightarrow H\le Z(G)$, as desired.

weird maybe not so correct solution...
at least i've tried my best
ok

it is known that if a group has only a subgroup
of a certain order then that subgroup
is for sure normal in the group.

also it is known that if a group has prime order than it is cyclic.
so it is clear that $xH=Hx,\forall x\in G$
                  and $<a> =H$
and if $|G|=p_1^{\alpha_1} \cdot p_2^{\alpha_2}\cdot ... \cdot p_k^{\alpha_k}$
with $p_1<p_2<...<p_k$
from the conditions of the problem we have $|H|=p_1$

now consider these following subgroups
$<ax>$ and $<xa>$
if we want to find out their orders they will be orders of $ax$ and $xa$
and $ord(ax)=(p_1,ord(x))$
and $ord(xa)=(p_1,ord(x))$
then we might have the feeling that they are isomorphic($<ax>$ and $<xa>$i mean).
actually they are.
consider $\phi : <ax> \to <xa>$ be an isomorphism
we can define $\phi$ like this $\phi((ax)^i)=(xa)^i$ then we will have
$\phi((ax)^i)=x(ax)^{i-1}a=x(ax)^{i} \cdot (ax)^{-1} a$
but we know that $(ax)^{-1}=x^{-1}a^{-1}$ so
$\phi((ax)^i)=x(ax)^ix^{-1}$
so $\phi$ can also be written $\phi(q)=xqx^{-1}$
but we also find that $\phi(x^{-1}qx)=q$ so we notice that
$\phi(\phi(x^{-1}qx))=x^{-1}qx$ otherwise written as $(\phi o \phi)(t)=t$ so
we find that $\phi^{-1}=\phi$
does this proove that $<ax>$ and $<xa>$ coincide?
(if this were to be true than it would mean that
$H\subseteq Z(G)$ wich wouldnt solve but a part of the
problem)

my feeling is that the answer is yes but im really not sure.

[quote="grobber"]
 the order of $\phi_g$ divides the order of $g$ and thus $n$,
[/quote]
may I know why?

[quote="spx2"]grobber why is it so obvious that $|\phi_g (H)|=p$ ?
its unclear to me.
also you say that "$Aut(H)\cong Z_p$ wich has $p-1$ elements" i dont understand
this(i knew $|Z_p|=p$) ,neither do i understand what is in the paranthesis.
how does $(n,p-1)\Rightarrow \phi_g(h)=h$ ?[/quote]
1. because H is normal (any conjugate is also a group of p elements)
2. for a cyclic group, the group of automorphisms is completely determined by T(a), where a is the generator of the cyclic group. Since T(a) can be a, ,..., $a^{p-1}$, the result follows.
3. $(n,p-1) = 1$ implies order of $\phi_g$ is 1 hence $\phi_g$ is the identity

For the homomorphism $\varphi: G\to Inn(G)$ we have $\varphi(g)=\phi_g$ thus $o(\phi_g)|o(g)|n$

ok i understand rafal :)
is what i wrote in my proof good or bad? is it even a proof ?

Consider the action of $G$ on $H$ by conjugation ($xzx^{-1}$). Let $z$ be an element of $H$. Since $H$ is normal $Orb(z)\subset H$. It is easy to see that $e$ is not in $Orb(z)$. But we know that $|G|=|Stab(z)||Orb(z)|$. But since $|Orb(z)|\le p-1$ and it divides $n$. We have that $|Orb(z)|=1$. Hence $z\in Z(G)$.

When $H$ is normal in $G$ we can say $G/C_{G}(H)$ embeds in $Aut(H)$ and here $|Aut(H)| \mid p-1$ but also $(|G|,p-1)=1 \implies G=C_{G}(H)$ and hence $H \subseteq Z(G)$


So far, the only solutions I've seen to Problem #5 on the '04 USAMO involved Holder's Inequality. Did anyone do it differently? (I thought maybe Power Mean, since the a+b+c term wasn't squared (no chance Cauchy's gonna work) and was on the less than side (not AM-GM there...)) I really don't think they would have put it on the test if that's the only way to do it...

Well, I can tell you this.  I have no idea what Holder's is, but I was able to prove it under certain scenarios.

I can prove it if

a, b, c >= 2

if

b, c >= 2 , a < 2

or if

c >= 2 , a,b < 2

I couldn't figure out the last case, so I BSed it.  But trying to break it down into different cases may work for you.

You bring up cauchy. Thing is, Holder is a much generalized form of cauchy, so its just good to know.

I don't know if there are any solutions w/o it that don't involve multiplying the whole thing out and delving into the depths of horrible, horrible algebra.

Oooh..... I didn't know how to do it, but I guessed that you had to use Cauchy and I wrote that. Do you think I'll get any credit? I doubt I will, but if I do, I'll laugh really hard.

Why couldn't you multiply a,b,and c by a constant k so that c=1, a>=b>=c?

I realize you can't do that. I just don't know why.

The inequality is not homogeneous in a,b,c. (i.e. the whole inequality is not in the same power in terms of variables, so you cant multiply by an arbitrary constant since it does change the whole inequality)

hmm... not too horrifying.

But remember, on an actual competition, you want to write forwards, not backwards.


Do you have, or have you had, braces?

(I've had braces for over half a year now- since Jan. Can't wait till they come off!- Still another year or so to wait.)

I have braces right now and I hate them
I've had them for a little more than a year now and I will get the wretched things off in a couple of months.

i have braces now ever since june 2004 (they hurt sometimes) :(

Yay for retainers!

(I think I've broken about 7 of them. :dry:)

I've had braces for more than year and half.  I get them off around January I think.

I've had them about a year. A little less than a year til they get out. I'm convinced that they're(the braces, that is) evil.

I had braces for 2 years then i got them off for half a year and now i have them on again for 3 years :(

No, my teeth are part of what makes me unique. Why should I want to make myself look like everyone else? (Though quite a few people have strongly encouraged me to get them) The same goes for my accent. If it goes away on its own, fine. But why should I invest work to make myself look or sound the same as everyone else??????

i've had my top braces for about 4 months now. gonna get the bottom ones sometime soon.  :(  and i will agree with the statement that they are pure evil.

I recently got my top-only braces off (I only had 1 out of 32 teeth that needed straightening!), but now I need full braces to fix my overbite! AAAARGH! (It's really bad, though, so I'm not going to avoid getting them, like, my bottom teeth irritate the roof of my mouth)

I had them, on the top, but I got them off at the end of 6th grade, so I was lucky.

Tylenol, the best thing ever if you wear braces.

Braces=time*money
time=money
Therefore Braces=money^2
Money=sqrt(evil)
Therefore Braces=evil
Q.E.D.

I thought it would fit more with braces than with girls.

;)

I have mine on until next March, then comes a retainer.

math92, I have a very similar case! I'm about to get them on in a week for another couple years.
solafidefarms, braces aren't meant to get rid of the unique qualities of people, they're meant for fixing issues with your bite, tooth orientation, and generally making life AFTER braces easier for your mouth. I hate em, but my speaking and bite will both be better afterwards, and that's what I'm trying to look forward to.

[quote="jhredsox"]lol, at first i actually WANTED them :)
-jorian[/quote]

lol why?

idk, i didn't know they were so annoying :P
-jorian

[quote="jhredsox"]are the things u have to put in ur teeth at night after u have braces annoying?

-jorian[/quote]

uh....no? i don't have to...it depends on your condition..for me i have a class 2 overbite so i don't have to where anything at night...but the doctor did mention somethign at night for the reverse overbite (i forgot the term)

they are only annoying if you have a VERY busy life. man my ipod just broke down so i am re installing all my software. this sucks. i have better things to do.

i need braces for a total of 2 1/2 years. only 1 year to go!  :roll:
i have a crossbite. my lower jaw is smaller than my upper jaw.
as for retainers, i have all you beat i think.  :lol:  i got one ran over by a car a few years ago.

wow 2 1/2 years i was 1 1/2 for overbite

hmmm.... braces... I never got them. Strange....

i have a regular overbite.

I only need it on for 17 months, which is 17/12 year.

Which isn't that bad. I have to wear headgear when i sleep though.

It hurts really bad a few days after you get them tightened.

It's been a month, so 16/12 or 4/3 years to go! yay

[quote="mustafa"][quote="jackdillon"]I swallow backwards so I need braces soon... :mad: not cool.[/quote]

How do you swallow backwards?[/quote]
My tounge pushes my teeth forward when I swallow :o It is extremely annoying!

a little less than a year for funcia to go!

an afterlife... would be how she might know  :)


inequalities geometry circumcircle trigonometry

Prove that in any triangle, 2Rr < ab/2, where a and b are the lengths of two of the sides, R is the length of the circumradius and r is the length of the inradius.

Edit: Inequality sign switched according to comments of Osiris, pointing out that the problem as originally stated was in direct opposition to the triangle inequality.  Oops.  Statement of question is now correct.

Hint: [hide]Try using area formulas to introduce some of the variables used here.[/hide]

I think the inequality sign should be reverted, since if this were true then we could get

[hide]2Rr > ab / 2

2Rr sin C > ab sin C / 2 (C is the angle formed by sides a and b)

ab sin C / 2 = K, area of triangle; c / sin C = 2R by sine law, so

cr > K = sr where s = (a + b + c) / 2, then

c > s

2c > a + b + c

c > a + b

Obviously wrong.[/hide]

hello, another approach, using the formulas $ A\equal{}\frac{abc}{4R}$, $ A\equal{}sr$ with $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$ we get $ 2\frac{abc}{4A}\frac{A}{s}<\frac{ab}{2}$ this is equivalent with $ c<a\plus{}b$ which is obviously true.
Sonnhard.


Calculate $ \sqrt{\sqrt[3]{0.000064}}$. Express your answer as a decimal to the nearest tenth.


[hide]The area of the whole circle is $9\pi$
The area of the white section is $\frac{1}{3}9\pi$
$9\pi-3\pi=6\pi$[/hide]

[quote="Quevvy"]If the diameter of the circle shown is 6 cm and the degree-measure of the central angle shown is 120, find the number of square cm in the area of the shaded region in terms of $\pi$.[/quote]
[hide]
the area of the sector is the angle in degrees times the area of the circle divided by 360
the sector is 2/3 of the circle. the area of the circle is $9\pi$. the area of the shaded area is $6\pi$
[/hide]

[hide]use the sector formula A/360*pi*r^2 which gets you 9*240/360pi which ends up as 6pi. [/hide]

how do you guys have it so that instead of puting pi u can just make the pi symbol and that other stuff

umm...it's called LaTex...look it up on the side of your screen..

Go here for the latex symbols : [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php[/url]

and for commands

for pi you just type in this [code]$\pi$[/code] and that gives you this: $ \pi$

[quote="spell it out 11"][quote="kimmystar94"][hide="wowsers"]6pi[/hide]

this was easy..[/quote]

actually that was hard.  It took me 30 mins.  I tried using caculus but I forgot how to do it..[/quote]

sorry...
all u needed 2 do was find the area and multiply by (360-120)/360 which simplifies to the answer...

[quote="youdonotknowme"][quote="spell it out 11"][quote="kimmystar94"][hide="wowsers"]6pi[/hide]

this was easy..[/hide][/quote]

actually that was hard.  It took me 30 mins.  I tried using caculus but I forgot how to do it..[/quote]

are you being sarcastic?[/quote]

No, Im a retard.

Actually, Im just kinda getting tired with 5 people posting the exact same solution (if you notice this topic was dead until somebody revived it just to post the exact same thing that was already posted)-before today it was lastposted feb. 6...

[quote="youdonotknowme"]well just don't read it[/quote]
but it gets "bumped" to the top..I cant avoid it without scrolling down.

[quote="youdonotknowme"]no one is forcing you to click the top few posts[/quote]

Yes, but I dont want to scroll down..too lazy, i usaually only read the top ones.

you won't even scroll down, well you are helpless


Suppose that $c$ is the greatest number among the nonnegative numbers $a,b,c$ satisfying the condition $a^2+b^2+c^2=8$, prove that $a+b+c\leqslant 4+ab$. Determine when there is equality.

WLOG suppose $b \ge a$.

If $b \ge 1$, $a+b+c \le ab+\sqrt{2(b^2+c^2)} \le ab+4$.

If $a \le b<1$, then $(a+b)+c \le (ab+1) + 3 = ab+4$.

Equality iff $a=0, b=c=2$.


I am having problems on exercises 8-7 and 8-9 in volume 1.  the problem i have in 8-7 is if  the number being square rooted is odd then a and b must either be odd or even.  in the even case its easy but in the odd i have no clue.  In 8-9 i don't really understand the question.  I looked at the answer and it said to get back to original number you need to multiply by a constant.  however if p/q is lowest terms and p/q=ap^2/aq^2 then that contradicts the condition that p/q is in lowest terms.  I need help :?

Not [i]everyone[/i] on the forum owns the AoPS books -- can you post the questions?

8-7:

Prove that the following are irrational by forming a contradiction when attempting to write it in lowest terms:

i.$\sqrt{3}$
ii. $\sqrt{p}$, if p is prime
iii. $\sqrt{n}$, where n an non-square integer

8.9:

Prove that every fraction that has a rational square root is of the form $\frac{ap^2}{aq^2}$ where a is an integer.

i thought most people had the books. thx for posting the probs.  in the future i will post the problems

[hide="Hint"]
You know the proof for $\sqrt{2}$, right? Well, instead of thinking 'odd and even' in that proof, think 'divisible by 2'. That may help you to prove the other ones.
[hide="In other words"]
For the $\sqrt{3}$ problem, don't think about odd/even, think about 'divisible by 3'.
[/hide][/hide]

let me see if m is a integer number and a/b=m and m=3 then a/b is not in lowest terms. because if b=1 then it would be a=m which is not  a fraction in lowest terms.

Er, 3/1 is in lowest terms, but you left out the part about it being the square root of 3 -- you just tried to prove that 3 isn't rational.

hmm i thats why this is in getting started lol.  i'll try the divisiblity by 3 thingy and report back


search
Functional Analysis
topology
advanced fields
advanced fields unsolved

Let T be a compact operator on a infinite dimensional space.
show that T cannot be bijectiv.
i need a little help.
Its easy to prove if we know that T^(-1) is bounded too,but i cant prove it in general.
thanks for help
tessa

In general, compact operator cannot be invertible,hence, it can not be bijective. The reason is that if it has bounded inverse $ T^{\minus{}1}$, then the product of compact operator and continuous should be compact, therefore $ T\cdot T^{\minus{}1}\equal{}Id$ has to be compact.But this is not the case for infinite dimensional space.

thanks for your answer!
i know this.
but the problem in your argument is that T^(-1) has to be bounded.but T being bijectiv says nothing about T^(-1) being bounded(We have no banach space,so the open mapping principle doesnt work here)

im not sure if this is really true.
if you think it isnt true if the space isnt banach,you may also write it here!

Hello tesssa,

I don't know if this argument is good enough, but I will try to do
this with a contradiction argument.

Therefore suppose $ T$ compact and at the same time being bijective.
Then from the Open mapping theorem we know that $ (T^{ \minus{} 1})^{ \minus{} 1} \equal{} T$
maps open set to open sets. In other words this means that the inverse
of $ T^{ \minus{} 1}$ maps open sets to open sets, which by definition shows that
$ T^{ \minus{} 1}$ is continuous (and therefore bounded since it is linear).

However, if you use the argument by leshik now, a contradiction is reached
and we are done.

It can be noted that I have assumed that the space is Banach since
the Open mapping theorem has this hypothesis.

Regards,
Hormander

In non-complete case the statement is false.

Namely, consider an injective compact operator in $ \ell_2$ (say, the diagonal operator $ T(x_1, x_2, \ldots) \equal{} (\frac{x_1}{2}, \frac{x_2}{4}, \ldots)$). Denote $ E \equal{} T(\ell_2)$. Then if we consider the same operator $ T$  acting from $ \ell_2$ to $ E$, it will be compact and bijective.

The same operator works if you want it to be an operator $ X\to X$. Just consider the space of all finite $ l^2$-sequences.

yes i search operators from X to X.
thanks for your answers!
but why is this operator compact,gockel?


Paula the painter had just enough paint for 30 identically sized rooms.  Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms.  How many cans of paint did she use for the 25 rooms?

$ \textbf{(A)}\ 10 \qquad
\textbf{(B)}\ 12 \qquad
\textbf{(C)}\ 15 \qquad
\textbf{(D)}\ 18 \qquad
\textbf{(E)}\ 25$


Let the sequence $u_n$ defined by:
$u_1=u_2=1$
$u_n=\frac{u_{n-1}^2+2}{u_{n-2}}$
Prove that all the terms of this sequence are integers.

Hmm..we may try a relation of $x_n, x_{n-1}...$ , then It's done easily, but I wonder why they can find it...


algebra
polynomial
number theory unsolved
number theory

Let $ p$ be a prime and $ a$ a positive integer not divisible by $ p$.Prove that $ \ x^p\minus{}x\minus{}a$ is irreducible

Captain Canada: could you please be more explicit?

Stronger thesis: you can prove that $ x^p\minus{}x\minus{}a$ is irreducible in $ \mathbb{Z}_p[x]$

My proof is porbably not the easiest possible. However, it consists of 2 steps:

1) Lemma: $ x^{p^n}\minus{}x$ is the product of irreducible polinomial (modulo p) whose degree divides n

2) $ x^p\minus{}x\minus{}a|x^{p^p}\minus{}x$ (where the division happens between polinomials mod p)

[quote="piever"]Stronger thesis: you can prove that $ x^p \minus{} x \minus{} a$ is irreducible in $ \mathbb{Z}_p[x]$[/quote]

Generalization: http://www.mathlinks.ro/Forum/viewtopic.php?t=181085 .

  darij


Do there exist any positive integers x,y,z such that 2548^x+(-2005)^y=(-543)^z
try to solve!

Please help me to solve this nt problem if u can