for those who took the Mathcounts competition this year, can u please tell me ur
1. grade
2. ur scores for chapter, state (if availiable) , and national (if avaliable) tests.
3. ur ranking for chapter, state, and national competitions
4. state

thanx
(just doing this for fun)

mine is
1. 8th grade
2. 44 for chapter and 43 for state
3. 2nd in chapter and 7th in state
4. California

1. 8th grade
2. 30th at chapter and 7th at states
3. Chapter: 36 State: 37
4. Massachusetts

1. 8th grade
2. 9th Chapter
3. 2nd Chapter, forgot what in state..but not top ten :-x IT WAS SOOO EASY!!
4. Pencil vania hehehhe

scores                        rank
Chapter      43                                1st...2nd last year

State          40                                4th...38th last year

Nationals    34                                59th (really really sucky by my standards)

1. 7th
2. 44, 40, 30(extremely bad)
3. 2nd, 2nd, 118th(extremely bad)
4. Oklahoma

[quote="Nexmus"]Wow, this was one of the first posts in the Mathcounts forum. :omighty:[/quote]

KID, DONT REVIVE OLD TOPICS!
ITS GOING TO GET DELETED ANYWAYS>>>>>

[quote="Nexmus"]dude, I was just joking. LOL :rotfl:[/quote]

Dude, get some friends.  And Im not joking.

No flaming.

No reviving old topics.

Locked  :harhar:


function
number theory proposed
number theory

[quote=Tintarn][quote=rogue]For every positive integer $n$ let $\varphi(n)$ be the Euler’s function defined as follows: $\varphi(n) =n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\ldots(1-\frac{1}{p_m}),$ where $p_1, p_2, \ldots, p_m$ are all pairwise distinct prime divisors of $n.$ For example, $\varphi(1)=\varphi(2)=1,$ $\varphi(3)=\varphi(4)=2.$ Find all $n \le 1000$ such that $\varphi(n) = \varphi(n + 1).$[/quote]
See [url=https://artofproblemsolving.com/community/c6h160894]here [/url]for a solution.[/quote]

But.. there is not a solution? It's said that it's the only case when n is even, he didn't look at n-1 e.g.


AMC
AIME
Hi
geometry
inequalities
area of a triangle
Heron\u0027s formula

Given that $x$, $y$, and $z$ are real numbers that satisfy:

\[ x=\sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} \]
\[ y=\sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}} \]
\[ z=\sqrt{x^2-\frac{1}{36}}+\sqrt{y^2-\frac{1}{36}} \]

and that $x+y+z=\frac{m}{\sqrt{n}}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n$.

I don't have latex on this computer, so I won't bother trying to type it up. To start off, all you have to do is go through a standard squaring procedure to get rid of the radicals (move one radical to one side, square, then isolate the radical again and square). This should give you some fairly workable equations to solve.

[quote="mkkool64"]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.

[hide]

Consider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.

[/hide][/quote]
wow, that is so simple.  There's barely any algebra.

I found the same solution as mkkool64. Here's the complete thing:

[hide]Since the area is constant, you get that $y=\frac54 x$ and $z=\frac64 x$. So $s=\frac{x+y+z}2=\frac{15}8 x$, and $\displaystyle\left(\frac{x}8\right)^2=s(s-x)(s-y)(s-z)=\frac{15}8x\left(\frac78x\right)\left(\frac38x\right)\left(\frac58x\right)$. Cancelling, we get $15\cdot7\cdot3\cdot5x^2=64\implies x=\frac8{15\sqrt7}$. Now, since $x+y+z=\frac{15}4x=\frac{15}4\frac8{15\sqrt7}=\frac2{\sqrt7}$, so our answer is $2+7=\fbox{009}$.[/hide] If this is the solution that goes in the contests section, I could edit it to be more comprehensive.

Try looking at the transcript in the AIME ii math jam.

How does a solution get chosen to be a link next to the amc/aime questions?

yes, that is exactly what I want to know, do we have to join the math jam to have our solution up there?

I think it's a mod (joml88 maybe) that chooses them for each contest. Not that it really matters, just as long as it gets done.

Generally, either 4everwise or I put the questions/solutions in the database for the AIME.  It has nothing to do with the mathjam.  We just go through the particular thread and find the post with the best solution and then input that one into the database.  If another proof uses an alternate method, we'll input that one as well.  We highly prefer solutions to be as polished as possible and for the post to have nothing but the solution.

Since my spring break just started, I'll try and find some time to input the AIME II 2006 solutions into the database...if they're not in after a few days, someone remind me :P

[quote="matt276eagles"]I've noticed that 4everwise has a tendency of choosing his own solutions...  :P[/quote]

Of course :D  But in all fairness, his solutions usually are very well written, clear, concise, and correct :)

4everwise also has the tendency to post about 10 problems in seperate topics all at once in the GS forum and post solutions if noone posts the answer in 15 mins.

This is very educational, so I don't mind.

[quote="mkkool64"]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.

[hide]

Consider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.

[/hide][/quote]

*brings up a dead topic* [ but in all fairness, the given solution is not complete, so i should revive discussion]

i found the same solution, except there is one thing i cannot get, how do you know that x,y,z satisfy the triangle inequality, i.e. $x+y\ge z$, and symmetric?

In response to Altheman: Taking the first equation, we have $x=\sqrt{y^{2}-\frac1{16}}+\sqrt{z^{2}-\frac1{16}}<\sqrt{y^{2}}+\sqrt{z^{2}}=y+z$, which is the triangle inequality. Similarly, $y<x+z$ and $z<x+y$. Thus, a triangle can be constructed with sides of lengths $x,y,z$.

[quote="mkkool64"]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.

[hide]

Consider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.

[/hide][/quote]

at my internship i gave this prob to one of the engineer's wives who was also an engineer. She came up with that solution.

[quote=nat mc][hide]Each is in the general form
$a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}$ Isolate a radcal and square.

$b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2$ Cancel, isolate the radical, and square

$a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2$

$a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2$
Everything is cyclic but the last term, so

$-4x^2\cdot\frac1{16}=-4y^2\cdot\frac1{25}=-4z^2\cdot\frac1{36}$ so

$x: y: z=4: 5: 6$ so $x=\frac{4y}5$ and $z=\frac{6y}5$ Plug it into the middle equation

$\frac{256y^4+625y^4+1296y^4}{625}=\frac{800y^4}{625}+\frac{1800y^4}{625}+\frac{1152y^4}{625}-\frac{100y^2}{625}$

$1575y^4=100y^2$ $y$ is not $0$

$y^2=\frac4{63}$ $y=\frac2{3\sqrt7}$

$x+y+z=3y=\frac{2}{\sqrt7}$ The answer is $\boxed{009}$[/hide][/quote]

Sorry for the bump but what being cyclic mean?

It means the terms are symmetric and only the variables are switched out.
I think it can be represented with $\sum_{cyclic}$ or something similar

[quote=gary2003][quote=mkkool64]I was talking to my friend about his.  I think we found an amazing solution but I never actually worked out the numbers.

[hide]

Consider a triangle with sides x,y,z and altitudes of 1/4, 1/5, 1/6 to each side, respectively.  It is then clear that a triangle with these side lengths would satisfy the three given equations.  Then you can write x * 1/4 = y *1/5 = z * 1/6 = 2 *AREA = 2*Heron's Formula.  Solve for y and z in terms of x, then plug this into Heron's formula and solve for x (the algebra isn't too hard)  Then easily solve for y and z.

[/hide][/quote]
How come the equations would be satisfied[/quote]
For the first equation, if you split the side with length $x$ into two segments split by the altitude called $m$ and $n$ with $m$ being adjacent the side with length $y$ and $n$ being adjacent to the side with length $z$, it follows that $m^2+\frac{1}{16}=z^2$ and $n^2+\frac{1}{16}=y^2$ because of the Pythagorean theorem. Rearranging, you get $m=\sqrt{z^2-\frac{1}{16}}$ and $n=\sqrt{y^2-\frac{1}{16}}$.
 Because $m+n=x$, $x=\sqrt{z^2-\frac{1}{16}}+\sqrt{y^2-\frac{1}{16}}$.
You can use this method on the other sides, and you get the other two equations.

[quote=NewBeginning]Wow, what was the motivation behind the triangle solution?[/quote]

There are a lot of "Algebra" problems that contain these Pythagorean terms; generally, if you can find a geometric interpretation, the problem will solve itself after that. Here, the main goal is to recognize that the terms are square roots of differences of squares, which are legs of a right triangle.

Another problem with this approach is [url=http://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_15]1991 AIME Problem 15[/url]

Problem
Given that $x, y,$ and $z$ are real numbers that satisfy:

$$x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}$$$$y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}$$$$z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}$$and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$

Solution
Let $XY = z$, $XZ=y$, and $YZ = x$ in $\Delta XYZ$ and $h_x$ be the altitude from $X$ to $YZ$, $h_y$ be the altitude from $Y$ to $XZ$, and $h_z$ be the altitude from $Z$ to $XY$. Then, by Pythagoras' theorem, $x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}$. We find that $h_x=\frac{1}{4}$, $h_y=5$, and $h_z=6$. Because the area of a triangle is constant, $x\cdot h_x = y\cdot h_y = z \cdot h_z \Longrightarrow \frac{x}{4} = \frac{y}{5} = \frac{z}{6}=2[ABC]$. Alternatively, we can find the area of $\Delta XYZ$ by using Heron's formula: $s=\frac{x+y+z}{2}=\frac{8A+10A+12A}{2}=15A$ and $A=[ABC]=\sqrt{s(s-x)(s-y)(s-z)}=$ $\sqrt{(15A)(15A-x)(15A-y)(15A-z)}$=$\sqrt{15A(15A-8A)(15A-10A)(15A-12A)} =$ $\sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = $ $15A^2\sqrt{7}$. Hence, since $A=15A^2\sqrt{7}$, $A=\frac{1}{15\sqrt{7}}$ and $x+y+z=2s=15A=\frac{1}{\sqrt{7}}$. $m+n=1+7=\boxed{8}$

@above,  I don't think that $x+y+z$ is equal to $s$, maybe $2s$? ;)

Consider triangle $ABC$ with $AB=x,BC=y,CA=z$, and let $h_A,h_B,h_C$ be the altitudes from $A,B,C$, respectively. The problem statement implies $h_A=\tfrac14,h_B=\tfrac15,h_C=\tfrac16$. If $K$ denotes the area of $\triangle ABC$, then $xh_A=yh_B=zh_C=2K$, so $x:y:z=4:5:6$.

Let $x=4t,y=5t,z=6t$. The area of a triangle with sides of length $4,5,6$ is $\tfrac{15}4\sqrt7$ by Heron's formula, which implies $K=\tfrac{15}4\sqrt7t^2$. Then \[\frac15=h_B=\frac{2K}y=\frac{3\sqrt7}2t\;\Longrightarrow\; t=\frac2{15\sqrt7},\]
so the perimeter of $\triangle ABC$ is $15t=\tfrac2{\sqrt7}\;\Longrightarrow\;2+7=\boxed{009}$.

We can geometrically interpret this as a triangle with altitudes $\frac{1}{4}, \frac{1}{5}, \frac{1}{6}$. Let the sides be $8p, 10p, 12p$. Then the area is $\sqrt{15p \cdot 7p \cdot 5p \cdot 3p}$ from Heron, so $\sqrt{15p \cdot 7p \cdot 5p \cdot 3p} = p$. This means $p=\frac{1}{15\sqrt7}$, for an answer of $30 \cdot \frac{1}{15\sqrt7} = \frac{2}{\sqrt7} \implies \boxed{9}$.

ugh why does there always have to be an annoying quick solution after you're done bashing

These looks really like pythagorean theorems, which is clear that $x, y, z$ are the three sides of a triangle with height $\frac{1}{4}, \frac{1}{5}, \frac{1}{6}$. This gives $x:y:z=4:5:6$. This is basically a similar triangle from a triangle with sides $4, 5, 6$ which has an area of $\frac{15\sqrt{7}}{4}$. Now, look at one height of the triangle. One height is $\frac{3\sqrt{7}}{2}$. Therefore, the answer is $(4+5+6)\frac{2}{3\sqrt{7}\cdot 5}=\frac{2}{\sqrt{7}}\to \boxed{009}$

[hide=solution]Note that $x,y,z$ form the side lengths of a triangle with heights of length $\frac{1}{4},\frac{1}{5},\frac{1}{6}.$ This means that $x:y:z=4:5:6.$ We can easily find the lengths of the heights of a triangle with side lengths $4,5,6$ and scale appropriately to find $x,y,z.$ The final answer is $(x,y,z)=\left(\frac{8\sqrt{7}}{105},\frac{10\sqrt{7}}{105},\frac{12\sqrt{7}}{105}\right).$[/hide]

not too bad :P there's some symmetry so we really only need to worry about equation 1
[hide=writeup]It's evident that $x$, $y$, and $z$ are all positive. Let's focus on the first equation for now, and eliminate the square roots.
\[x-\sqrt{y^2-\frac{1}{16}}=\sqrt{z^2-\frac{1}{16}}\]
square both sides and cancel the $\frac{1}{16}$ term that results
\[x^2-2\sqrt{y^2-\frac{1}{16}}+y^2=z^2\]
now move the terms without radicals to one side and the term with a radical to the other, and square yet again, then move the terms all to one side:
\[x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2+\frac{1}{4}x^2=0.\]
Similarly 'simplify' the other two equations to get:
\begin{align*}x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2+\frac{4}{25}y^2&=0
\\ x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2+\frac{1}{9}z^2&=0\end{align*}
So we obtain that $\frac{1}{4}x^2=\frac{4}{25}y^2=\frac{1}{9}z^2$, that is, $\frac{1}{2}x=\frac{2}{5}y=\frac{1}{3}z$. Hence $y=\frac{5}{4}x$ and $z=\frac{3}{2}x$. Plugging this into the first equation (or actually any equation) results in
\[x=\sqrt{\frac{25}{16}x^2-\frac{1}{16}}+\sqrt{\frac{9}{4}x^2-\frac{1}{16}}.\]
Solving this, $x=\frac{8}{15\sqrt{7}}$. So $(x, y, z)=\left(\frac{8}{15\sqrt{7}}, \frac{2}{3\sqrt{7}}, \frac{4}{5\sqrt{7}}\right)$.
$x+y+z=\frac{2}{\sqrt{7}}$. Finally, $m+n=\boxed{009}$. $\blacksquare$[/hide]

[hide=solution]
Look at the equations, and how closely they resemble the Pythagorean Theorem. Note that $x$, $y$, and $z$ could represent the sides of a triangle with altitudes of $\frac{1}{4}$, $\frac{1}{5}$, and $\frac{1}{6}$ on the $x$, $y$, and $z$ sides. This satisfies the equation.
The formula for the area of a triangle gives $\frac{1}{4}x=\frac{1}{5}y=\frac{1}{6}z$, so $y=\frac{5}{4}x$ and $z=\frac{3}{2}x$.
Plugging this back into the equation for $x$ gives
\[x=\sqrt{\frac{25}{16}x^2-\frac{1}{16}}+\sqrt{\frac{36}{16}x^2-\frac{1}{16}},\]
so
\[x=\frac{1}{4}\sqrt{25x^2-1}+\frac{1}{4}\sqrt{36x^2-1}.\]
Multiplying both sides by $4$ and moving $\sqrt{25x^2-1}$ to the other side gives
\[4x-\sqrt{25x^2-1}=\sqrt{36x^2-1}.\]
Squaring both sides gives
\[(4x-\sqrt{25x^2-1})^2=16x^2-8x\sqrt{25x^2-1}+25x^2-1=36x^2-1,\]
so simplifying gives $5x^2=8x\sqrt{25x^2-1}$. Dividing both sides by $x$ gives $5x=8\sqrt{25x^2-1}$, so squaring both sides again gives
\[25x^2=64(25x^2-1),\]
so $x^2=\frac{64}{63\cdot 25}$.
Hence, $x=\frac{8}{3\cdot 5\sqrt{7}}=\frac{8}{15\sqrt{7}}$. This gives $y=\frac{2}{3\sqrt{7}}$ and $z=\frac{4}{5\sqrt{7}}$, so $x+y+z=\frac{2}{\sqrt{7}}$, which gives an answer of $\boxed{009}$.
[/hide]

Alg mani FTW
Interpret as triangle ABC side lengths x=BC,y=CA,z=AB and altitudes 1/4,1/5,1/6, respectively. Then $x/4=y/5=z/6$ and simple substitution solves it.

Consider an acute triangle with sides $x$, $y$, and $z$, with corresponding altitudes $\frac 14$, $\frac 15$, and $\frac 16$. This cleanly satisfies the equations in the problem.

From the height ratios, we have $x : y : z = 4 : 5 : 6$. Note the height to the side with length 4 in a 4-5-6 triangle has length $\frac{15\sqrt{7}}{8}$, so the scale ratio is $\frac{2\sqrt{7}}{105}$, and our solution is
\[(x,y,z)=\left(\frac{8\sqrt{7}}{105},\frac{2\sqrt{7}}{21},\frac{4\sqrt{7}}{35}\right) \implies x+y+z = \frac{2}{\sqrt{7}} \implies \boxed{009}.~\blacksquare\]

Notice that $x,y,z$ are the sides of a triangle with heights $\frac{1}{4}, \frac{1}{5}, \frac{1}{6}$, where the side lengths and height correspond. Then, we have

\[\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\]

so $y=\frac{5}{4}x$ and $z=\frac{3}{2}x$. This means $s=\frac{x+y+z}{2}=\frac{15}{8}x$. By Heron's formula we have

\[2\sqrt{s(s-x)(s-y)(s-z)}=\frac{x}{4}\]
\[\implies \frac{30x^2\sqrt{7}}{64}=\frac{x}{4}\]

Thus, $x=\frac{8}{15\sqrt{7}}$, so $y=\frac{5}{4} \cdot \frac{8}{15\sqrt{7}}$ and $z=\frac{3}{2} \cdot \frac{8}{15\sqrt{7}}$. Simplifying, we get

\[(x,y,z)=\left(\frac{8\sqrt{7}}{105}, \frac{2\sqrt{7}}{21}, \frac{4\sqrt{7}}{35} \right)\]


advanced fields
advanced fields unsolved

Let $f \in X^{*}-\{0\}$. Put $\alpha = inf\{\parallel x\parallel : f(x) = 1\}$. Prove: $\parallel f\parallel = 1/ \alpha$

Maybe telling us the definition of $X^{*}$ would help? :huh:

I guess that $X$ is a Banach space and $X^{*}$ is its dual ..

Set $L : = f^{-1}(0)$, then
\[\|f\| = \sup_{x \in X}\frac{\left\langle x,f \right\rangle}{\|x\|}= \sup_{x \in L}\frac{\left\langle x,f \right\rangle}{\|x\|}= \frac 1{\inf_{x \in L}\|x\|}.\]


Estimate the following to a nearest ten:
\[ \frac {0.5 \times 0.12 \times 2500} {31 \times 0.09}\]


superior algebra
superior algebra theorems

Someone is expert of the theory of group?
Please help-me
thanks
sastra

You can ask as many questions as you wish on the forum. As long as you don't do it like you did a lot of times before:
- do not post any question more than once
- do not write words in capitals, especially "HOW TO DEMONSTRATE

More than half of capitals, and see my comment there:
http://www.mathlinks.ro/Forum/viewtopic.php?p=733515#733515
DoubleThread of the first one:
http://www.mathlinks.ro/Forum/viewtopic.php?p=733516#733516
Just a impudence:
http://www.mathlinks.ro/Forum/viewtopic.php?p=734532#734532

All three locked now, and be sure that locking and deleting of any multithread will be done from now, for all threads.


Let $A',B',C'$ denote the apices of the equilateral triangles $\triangle A'BC,$ $\triangle B'CA,$ $\triangle C'AB$ erected outwardly. Their centers are $X,Y,Z,$ respectively, thus $Np \equiv AX \cap BY \cap CZ$ is the 1st Napoleon point of $\triangle ABC.$

The point $Z$ is constructible and $C$ is then on $ZNp,$ such that $\tfrac{BX}{BC}=\tfrac{\sqrt{3}}{3}$ and $\angle XBC=30^{\circ}.$ Hence, $C$ is found as the intersection of $ZNp$ with the line image of $ANp$ under the spiral similarity with center $B,$ coefficient $\tfrac{\sqrt{3}}{3}$ and  rotational angle $30^{\circ}$ counterclockwise.


logarithms
function
calculus
derivative
calculus computations

${e^{y}>\frac{16 (1-y)}{(2-y)^{4}}}$

${(0<y<1)\land \left(e^{y}>1\right)}$

${y>\log \left(\frac{16 (1-y)}{(2-y)^{4}}\right)}$
${y>\log (16)+\log \left(\frac{1-y}{(2-y)^{4}}\right)}$
${y-\log \left(\frac{1-y}{(2-y)^{4}}\right)>\log (16)}$

${\frac{\partial \left(y-\log \left(\frac{1-y}{(2-y)^{4}}\right)\right)}{\partial y}=0}$
${\frac{y}{y^{2}-3 y+2}=0}$
${y=0}$

${\left(y-\log \left(\frac{1-y}{(2-y)^{4}}\right)\text{/.}y\to 0\right)=\log (16)}$

Since $y \neq 0$ and this the minimum in this interval, and the function is crescent. so it's true

Correction: $\frac{d}{dy}\left(y-\ln\left(\frac{1-y}{(2-y)^{4}}\right)\right)= \frac{y^{2}}{(1-y)(2-y)}.$

In other words, $y^{2}$ in the numerator, not $y.$

The conclusion still holds.

Two bits of notation: there's no need to use the partial derivative symbol $\partial$; this is an ordinary derivative, not a partial derivative. And "crescent" in English would be "increasing."

That for $x\in \left(0,\frac12\right)$ is sharpen than $e^{x}>x+1$. It's nice isn't it?

i use partial because im typing on mathematica

i use D[f[x],x]

if i use Dt[f[x],x]   mathematica will differentiate all constants too

Dt[a]  + Dt[c]   etc...

and this sucks...

and sorry dor the y^2


geometry
geometric transformation
reflection
trapezoid
perpendicular bisector
angle bisector
geometry unsolved

let $ABC$ be triangle  where $\angle A=80^o$ and $\angle B=\angle C$ .
 let $M$ be a point in triangle $ABC$  satisfying $\angle MAC=20^o$and $\angle MCA=30^o$
 find value of$\angle MBC$

Are you sure about this problem? The value of $\angle{MBC}$ is NOT an integer...

i'm sory  .
 i have a mistake  in my post
     i  have just complete it

Reflect $M$ through $AC$ to $M'$, then $\angle{M'CA}=\angle{MCA}=30^o$. Since $\angle{ACB}=50^o$, so $\angle{BCM'}=80^o$.

Because $\angle{MAC}=\angle{M'AC}=20^o$, and $\angle{BAC}=80^o$, so $\angle{BAM'}=100^o$. Now we see that $\angle{BAM'}+\angle{BCM'}=180^o$, so $A,B,C,M'$ are cyclic points. Therefore $\angle{M'BA}=30^o$ and $\angle{M'BC}=20^o$.

Since: $\angle{AM'M}=\frac{180^o-\angle{M'AM}}{2}=\frac{180^o-40^o}{2}=70^o$, and $\angle{AM'B}=\angle{ACB}=50^o$, so $\angle{BM'M}=20^o=\angle{BCM}$, thus $BM'=BC$. From here we can get $\triangle{BM'M} \cong \triangle{BCM}$, hence $\angle{M'BM}=\angle{CBM}$. But because $\angle{M'BM}+\angle{CBM}=20^o$, so finally we have $\angle{MBC}=10^o$.

it,s very nice .  i can't  solve this problem  but you are very mart
 thanks veru much ,shobber!!!

Reflect $M$ in perpendicular bisector of $BC$ in $M'$, so triangle $\triangle AM'M$ is $40^\circ-70^\circ-70^\circ$; construct equilateral triangle $\triangle MM'P$ ($P$ inside $\triangle AMM'$, on perpendicular bisector of $BC$, hence $\angle PAM = \angle CAM$ and $APMC$ is cyclic ($\angle APM=150^\circ$, so $\angle APM + \angle ACM=180^\circ$), $AM$ angle bisector of $\angle PAC$, so $PM = MC$; consequently $CM = MM' = M'B$, $BCMM'$ is isosceles trapezoid (i.e. cyclic), and $BM$ is angle bisector of $\angle CBM'=20^\circ$, with $\angle CBM=10^\circ$.

Best regards,
sunken rock


William's father was older than his grandfather. How did that happen ?

[hide]William's father was really old. William's grandfather is not his father's father, he is his mother's father. He was younger than his son-in-law.[/hide]

...

The grandfather is on William's mother's side, so William's mother is really young compared to his father.    :blush:

The grandfather is on William's mother's side, so William's mother is really young compared to his father.

Or...

William's father was really old. William's grandfather is not his father's father, he is his mother's father. He was younger than his son-in-law.

[quote="Stank"]Or maybe he was born on a leap day.[/quote] :rotfl:  :rotfl:  :rotfl:

a couple has a childat around 20. that child marries someone around 40 when the child was around 20.
or it could be a stepgrandfather


calculus
integration
calculus computations

[b](1)  Integrate
  e^x*(x^3-x+2)/(1+x^2)^2 with respect to x
here e^x is multiplied by(x^3-x+2).[/b]

Actually,the answer is $ \frac{e^{x}(1\plus{}x)}{1\plus{}x^2}$

As there have been no further replies, I will offer my naive approach.

With the quotient rule in mind, I assume (admiting it's quite an assumption and perhaps unwarranted) that the answer must be of the form

$ \frac{e^{x}f(x)}{1\plus{}x^{2}}$.

Differentiating, $ f(x)$ must be linear: $ ax\plus{}b$.
.
Substituting and comparing coefficients, $ a\equal{}b\equal{}1$ and so $ I\equal{} \frac{e^{x}(1\plus{}x)}{1\plus{}x^{2}}$.

Hope this is of some use.

[quote="AndrewTom"]admiting it's quite an assumption and perhaps unwarranted[/quote]
Well, there are pretty much two ways it could go: either the assumption is right, or it's not and there's no elementary antiderivative at all. If we just wrote something similar at random, we would most likely be in the second case.

hmmmm.. a bit confuse of the equations.  :(

Thanks jmerry.

Not sure what you mean, emjhay, so I'll try to write out all the steps:

$ I = \int {\frac {e^{x}(x^{3} - x + 2)}{(1 + x^{2})^{2}} dx}$.

Because the denominator is $ (1 + x^{2})^{2}$, it seems that the integrand could be derived (no pun intended)  by applying the quotient rule to $ y = \frac {e^{x}f(x)}{(1 + x^{2})}$, giving

$ \frac {(1 + x^{2}(e^{x}f'(x) + e^{x}f(x)) - 2xe^{x}f(x)}{(1 + x^{2})^{2}}$

= $ \frac {e^{x}((1 - 2x + x^{2})f(x) + (1 + x^{2})f'(x))}{(1 + x^{2})^{2}}$.

By comparison with the given integrand, $ f(x)$ must be linear, that is, $ f(x) = ax + b$ and so $ f'(x) = a$.

Substituting, we get

$ ax + b - 2ax^{2} - 2bx + ax^{3} + bx^{2} + a + ax^{2} = x^{3} - x + 2$.

Therefore $ a = 1$ and $ b = a = 1$, so that $ f(x) = 1 + x$, and

$ I = \frac {e^{x}(1 + x)}{1 + x^{2}}$.

I hope that this is of some help to you, emjhay; I tried rearranging $ x^{3} - x + 2$ (without differentiating the answer given by centry57 and using this to get a rearrangement), various substitutions and integration by parts but got nowhere.


[color=green]Does anybody have a fast way (less than 2 minutes time) to do this problem?

Suppose a coin purse contains 30 coins which are either nickels, dimes, and/or quarters. How many different combinations of these coins are there whose value is $5 ?[/color]

if you can do mental math quickly and are organized and work fairly fast it is possible to do it within the 2 min time limit :?

This is took me about 3 minutes to work out . . . [hide]

Let $N$ = number of nickels,
$D$ = number of dimes,
$Q$ = number of quarters.

There are 30 coins:
$.\qquad[1]\;\;N + D + Q = 30$

Their total value is 500 cents:
$.\qquad\;5N + 10D + 25Q = 500\;\;\Rightarrow\;\;[2]\;\;N + 2D + 5Q = 100$

Subtract [1] from [2]: $D + 4Q = 70$

If $Q = 13,\; D = 18\;\;$ . . .  and there are more than 30 coins.

We have the following solutions:
Q = 14, D = 14, N = 2
Q = 15, D = 10, N = 5
Q = 16, D = 6, N = 8
Q = 17, D = 2, N = 11[/hide]

There is a fairly obvious minimum on the number of quarters. After that, the problem simplifies significantly, at least in terms of the size of the numbers. The problem itself, of course, is simply tedious and not complicated.

There is a fairly obvious minimum on all of the three coins. I think you mean maximum :P .

Soroban's solution is very nice.  Thanks.

The difficulty is to understand that Q<18, then to start from Q=17 to get the desired solutions.


induction
inequalities
inequalities unsolved

Given a sequence$ \{a_n\}$. such that $ a_1\equal{}1,  a_{n\plus{}1}\equal{}(1\plus{}\frac{1}{n^2\plus{}n})a_n\plus{}\frac{1}{2^n},  n\in N^*$
(1).prove that  $ a_n\ge 2$       $ (n\ge 2)$
(2).Prove that$ a_n<e^2$         $ (n\ge 1)$  if $ ln(1\plus{}x)<x$ for any $ x>0$, where $ e\equal{}2.71828...$

hello, to 1) we prove by induction that $ a_n\geq 2$ for all $ n\ge2$.
At first let $ n \equal{} 2$, then $ a _2 \equal{} (1 \plus{} 1/2)a_1 \plus{} 1/2 \equal{} 2\geq2$. Now we assume that $ a_n\geq2$, then we get
$ a_{n \plus{} 1} \equal{} \left(1 \plus{} \frac {1}{n^2 \plus{} n}\right)a_n \plus{} \frac {1}{2^n}\geq\left(1 \plus{} \frac {1}{n^2 \plus{} n}\right)2 \plus{} \frac {1}{2^n}\geq2 \Leftrightarrow \frac {2}{n^2 \plus{} n} \plus{} \frac {1}{2^n}\geq0$ and this is true.
Sonnhard.

[quote="ye109"][quote="sxgfly"]Given a sequence$ \{a_n\}$. such that $ a_1 \equal{} 1, a_{n \plus{} 1} \equal{} (1 \plus{} \frac {1}{n^2 \plus{} n})a_n \plus{} \frac {1}{2^n}, n\in N^*$
(1).prove that  $ a_n\ge 2$       $ (n\ge 2)$
(2).Prove that$ a_n < e^2$         $ (n\ge 1)$  if $ ln(1 \plus{} x) < x$ for any $ x > 0$, where $ e \equal{} 2.71828...$[/quote]
1)$ a_n\geq a_{n \minus{} 1} \geq \cdots a_2 \equal{} 2$
2)we prove $ a_n < e^{2 \minus{} \frac {2}{n}}$
for $ n \equal{} 1$ right
if for $ n \equal{} k$ is right
when $ n \equal{} k \plus{} 1$
$ a_{k \plus{} 1} \equal{} (1 \plus{} \frac {1}{k^2 \plus{} k})a_k \plus{} \frac {1}{2^k} < (1 \plus{} \frac {1}{k^2 \plus{} k})e^{2 \minus{} \frac {2}{k}} \plus{} \frac {1}{2^k}$
$ < e^{2 \minus{} \frac {2}{k}}(1 \plus{} \frac {1}{k^2 \plus{} k} \plus{} \frac {1}{2^ke^{2 \minus{} \frac {2}{k}}})$
$ < e^{2 \minus{} \frac {2}{k}}e^{\frac {2}{k(k \plus{} 1)}} \equal{} e^{2 \minus{} \frac {2}{k \plus{} 1}}$[/quote]

How to obtain the last inequality!  can you tell me?


let $ a,b$and $ c$ be positive reals numbers , prove that :
$ Min((b\minus{}c)^2,(c\minus{}a)^2,(a\minus{}b)^2))\le\frac{a^2\plus{}b^2\plus{}c^2}{5}$

Wolog $ a\le b\le c$, and let $ b \equal{} a \plus{} x, c \equal{} b \plus{} y$ then as $ a,b,c$ are positive reals, $ b \equal{} a \plus{} x > x$.
Then:
$ \min((b \minus{} c)^2,(c \minus{} a)^2,(a \minus{} b)^2))\le\frac {a^2 \plus{} b^2 \plus{} c^2}{5}$
$ \Leftrightarrow \min (x^2,y^2)\le \frac {(b \minus{} x)^2 \plus{} b^2 \plus{} (b \plus{} y)^2}{5}$ (as $ x^2,y^2\le(x \plus{} y)^2$)
$ \Leftrightarrow \min (x^2,y^2)\le \frac {b^2 \plus{} x^2 \plus{} y^2 \plus{} 2by \plus{} 2b^2 \minus{} 2bx}{5}$
$ \Leftrightarrow \min (x^2,y^2)\le \frac {b^2 \plus{} x^2 \plus{} y^2 \plus{} 2by \plus{} 2b(b \minus{} x)}{5}$  (*)
Now:
$ b^2 \plus{} x^2 \plus{} y^2 \plus{} 2by \plus{} 2b(b \minus{} x) > b^2 \plus{} x^2 \plus{} y^2 \plus{} 2by$ (as $ b > x$)
$ > 2x^2 \plus{} y^2 \plus{} 2xy > 5\min (x^2,y^2)$ and so (*) is true, and the original inequality follows.

[quote="tchebytchev"]let $ a,b$and $ c$ be positive reals numbers , prove that :
$ Min((b \minus{} c)^2,(c \minus{} a)^2,(a \minus{} b)^2))\le\frac {a^2 \plus{} b^2 \plus{} c^2}{5}$[/quote]
When you reduce the same quantity c=min(a,b,c)>=0 then we have:LHS is the same, but the RHS will decrease.Therefore, we  need to prove when c=0! Then it becomes so easy! :)


Ten teams participated in a football competition where each team played against every other team exactly one .When the comprtition was over ,it turned out that for every three team $A;B;C$,if $A$ defeated $B$  and $B$  defeated $C$ then $A$  defeated $C$.
prove that there were 4 teams $A: B: C$ such that $A$ defeated $B$ ;$B$  defeated $C$ ;$C$  defeated $D$ or such that each match between them was a draw

If a team won no games, we call it "rank 0."  If a team won games only against rank 0 teams, we call it rank 1.  If a team won games only against rank 0 and rank 1 teams, we call it rank 2.  And so on.  If there are no chains of length 3 (that is, involving 4 teams) then every team is of rank 0, 1 or 2.  But there are 10 teams, so at least one of these ranks must have 4 teams.  The 4 teams at this rank must have all draws between them, since no team beats other teams of its rank.


My school registered for the AMC A test in December, we got a notification that we would receive our AMC stuff by early January. However, that hasn't arrived yet (as of 1/21) so I was wondering if any schools have received their AMC A stuff?

i'm taking AMC 10b and i just got my packet on friday

We are caught up on A-date shipping, having now sent approximately 3000 packages since January 2.
If your contest manager provided an email  address, the contest manager will receive an automatically generated email with a tracking number.  Use the tracking number to find the current location of your package.  If you are within the contiguous 48 states, shipping should take 5 days or less.  If you are in Alaska or Hawaii, it will take about 5-7 days.  Outside the US, it is variable, about 5-7 days depending on distance and location with the country.

If you believe your order or your package has been lost, contact the AMC office by any of the means printed on all of our materials.

Steve Dunbar
MAA Director, American Mathematics Competitions


quadratics
number theory unsolved
number theory

I think you should better define what is the Pisano period... ;)

Pierre.

The m pisano Period ($\pi (m)$) are the period of Fibonacci sequance mod m
for example: $\pi(4) = 6$
since the residue of Fibonacci mod 4 are [1,2,3,1,0,1],1,2,......
from what I know, there is only one prime {3}
freom $\pi(2)=[1,0,1],1,0...........$ =3
because pisano Period are even if m>2

[quote="wonbert"]Is it trival Pisano Period are even if m>2?[/quote]

I don't know if you'll consider it trivial, but it certainly isn't hard.

Let $u=\frac{-1+\sqrt 5}2$ and $\bar u=\frac{-1-\sqrt 5}2$. It sufices to prove it for prime $m$'s. If $p=5$, check it by hand. If $p\ne 5$, we work in the field formed by adjoining $\sqrt 5$ to $\mathbb Z_p$ (this field either has $p^2$ elements if $5$ is not a quadratic residue modulo $p$, or $p$ elements if $5$ is a quadratic residue modulo $p$).

We want to show that a $t$ satisfying $(u^{n+t}-u^n)-(\bar u^{n+t}-\bar u^n)=0,\ \forall n\ (*)$ must be even (remember, $(*)$ must hold in the mentioned field). $(*)$ translates to $u^n(u^t-1)-\bar u^n(\bar u^t-1)=0,\ \forall n$. By considering two consecutive values of $n$, we get a system of two equations with the unknowns $u^t-1$ and $\bar u^t-1$, and we conclude that these must both be $0$. Since $\bar u=\frac {-1}u$, the conclusion that $t$ must be even follows quickly.


Let$0\leq p_i \leq 1,\forall i = 1.,2,...,n$.Prove that:$\exists x \in [0,1]$ such that:
$\sum_{i=1}^n \displaystyle\frac {1}{|x-p_i|} \leq 8n(1+\frac{1}{3} +.....+\frac{1}{2n-1})$

Hi pigfly!

I think, there are more sharp estimate.
Divide [0,1] into $2n$ equal segments. From Pigeon-hole principle we obtain that at least n of these segments don't contain $p_i$ for all $i=1..n$. Let $x_i$ be midpoints of these segments.
We have
\[\sum_{j=1}^n\sum_{i=1}^n\frac{1}{|x_j-p_i|}=\sum_{i=1}^n\sum_{j=1}^n\frac{1}{|x_j-p_i|}\].
It is easy to show that
\[\sum_{j=1}^n\frac{1}{|x_j-p_i|} < 2\cdot\left(\frac{1}{1/2n}+\frac{1}{3/2n}+\dots+\frac{1}{(2[n/2]+1)/2n}\right)=4n\left(\frac{1}{1}+\frac{1}{3}+\dots+\frac{1}{2[n/2]+1}\right).\]
Hence we conclude that there exists $x_j$ s.t.
\[\sum_{i=1}^n\frac{1}{|x_j-p_i|} < 2\cdot\left(\frac{1}{1/2n}+\frac{1}{3/2n}+\dots+\frac{1}{(2[n/2]+1)/2n}\right)=4n\left(\frac{1}{1}+\frac{1}{3}+\dots+\frac{1}{2[n/2]+1}\right).\]


function
logarithms
geometry
geometric transformation
rotation
homothety
advanced fields

Function f(z) is regular in the strip $|Rez|<\frac{\pi}{4}$ and satisfy |f(z)|<1 and f(0)=0. Prove that |f(z)|<|tg(z)|

the trick here is to transform the unit disc via a homography into the upper halfplane ,later by taking the squre to transform the halfplane into the whole C but a halfline and then by taking a logarithm to transform it into the strip  -pi<im z< pi then transform this strip by rotation and homotety into the strip from the problem. by combining these functions and applying f to what's derived you receive a function from the unit disc into the unit disc (since |f(z)|<1) take care the zero to be transformed into zero. now just apply shwarz lemma

2qw:Could you please help me with understanding, how i can get |tgz| using you solution(Shwarz lemma says that |f(z)|<|z| and as far as i understand function tgz doesn't include in our transformation functions)


function
linear algebra
matrix
pigeonhole principle
combinatorics proposed
combinatorics

A problem I came up with a short while ago. I don't know the answer yet, but then I haven't worked on it much yet. Here it is.

There are $ n$ players at a Go tournament (as you may know, there are no draws in modern Go tournaments). The tournament is in a round-robin format, meaning that each player in the tournament plays a game against each other player in the tournament, receiving 1 point for a win and 0 points for a loss, for a total of $ n \minus{} 1$ games per player. The tournament ends with Victor as the decisive winner (he had a higher score than anyone else, no SOS involved). What is the largest number of games that Victor could conceivably have lost, as a function of $ n$?

For instance, for $ n \equal{} 5$, it's possible that he lost up to one games, using the following scoresheet (Victor on first row):
\[ \begin{bmatrix} X & 1 & 1 & 0 & 1 \\
0 & X & 1 & 1 & 0 \\
0 & 0 & X & 1 & 1 \\
1 & 0 & 0 & X & 0 \\
0 & 1 & 0 & 1 & X \end{bmatrix}
\]

I'm not sure how hard this is, but I suspect not very.

[b]Edit:[/b] gave a more compelling example.

it looks easy :)

Victor won n-k-1 games and lost k games; then, he earned n-k-1 points.

at all other games there were (n-2)+(n-3)+...+1= (n-1)(n-2)/2 points for n-1 people... but all them plated against Victor, then there are  (n-1)(n-2)/2 +k points for n-1 people.

Victor won more than the half of the games (if not, at least one rival earnt more points than Victor or tied with him, because dividing (n-1)(n-2)/2 +k points to n-1 people gives one with at least [(n-2)/2]* +1 points, following the Pigeonhole principle, and it is more than [(n-1)/2]*), then k < (n-1)/2

NOTE: when I write [x] i mean "the smallest integer higher than or equal to x"

Then, the max number of points for n-1 people is (n-1)(n-2)/2 +[(n-3)/2]. This number is smaller than the number
(n-1)(n-2)/2 +[(n-2)/2]. And this is:

n(n-2)/2, for n even (case 1)
or
((n-1)^2)/2, for n odd. (case 2)

then, if we divide for n-1:
for case 2, we have n-2, but this number is bigger than the original maximum, and we have the demostration!

and, for case 1, we have, for AM-GM:

for case 1 we have that AM1=n-1, but we have not equality, then GM1 is smaller than it.

AM1>GM1

Then, (n-1)^2>n(n-2), and dividing for n-1, we have it again!

Then, the minimum value of k is

                                                                      k=[(n-3)/2]

 :lol:


Prove that all rhombuses (rhombi  :D ) are parallelograms.

[quote="236factorial"]Must they be?[/quote]

[hide="Yes."]
But what is your definition of rhombus? If you include in the definition the fact that the sides are parallel, then we're done. Definition please. :D
[/hide]

Rhombus-a quadrilateral with all four sides equal  :D

[hide="my proof"]
Draw a diagonal acroll the rhombus. This creates two triangles. Since all the sides of the rhombus are equal, and the diagonal is congruent to itself, the two triangles are congruent by SSS. Then, using CPCTC (Corresponding Parts of Congruent Triangles are Congruent) and the base angles theorem, combined with the transitive property of congruence, we know that the four angles that have the diagonal as a side are congruent. Then, by the converse of the alternate interior angles theorem, we have that the opposite sides are parallel.
[/hide]

Good proof  :)

I failed to notice because that the diagonal was congruent to itself, maybe next I'll work on cleaner paper  :roll:  :blush:


probability
ratio
floor function
factorial

1. How many terminal zeroes are there in the combination of 383 taken 144?

2. A, B, and C are playing a game with a coin by flipping it several times. If A flips heads, then the coin goes to B, if he flips tails, it goes to C. If B flips heads, then the coin goes to C, if he flips tails, it goes to A. If C flips heads, then the coin goes to A, if he flips tails, it goes to B.After each flip, the person now holding the flips it again, and so on. Suppose A makes the first flip of the coin, what is the probability that after the sixth flip, the coin goes to B? (Is there any other way other than using the tree method??)

3. In a large party, the ratio of boys to girls is 5:4 . Each boy shook hands with all the girls, and also half of the boys. Each girl shook hands with all the boys, and also half of the girls. If there were 5929 handshakes made, and no pair of person shook hands more than once, how many people were there?

4. Jonah is inside a whale at sea. At the dawn of each day, there is a 25% chance that he will write in his diary, and at noon, there is a 1/3 chance that the whale will swallow him. If he remains stuck indefinitely, what is the probability that Jonah will be swallowed without writing anything in his diary?

5. In how many ways can you arrange the letters of NUMERO such that no two vowels are together?

1. Do you mean $ \binom{383}{144}$? If yes: Let $ f(x)\equal{}$number of zeroes in which $ x$ ends.
$ f(383)\equal{}\lfloor \frac{383}{5}\rfloor \plus{}\lfloor \frac{383}{25}\rfloor \plus{}\lfloor \frac{383}{125}\rfloor\equal{}94$
Do the same for $ f(144)$ and $ f(383\minus{}144)$.
Answer: $ 3$

Do you know how to do it? I think that it is $ 3$, but correct me!

I dont know. These problems are the ones I was not able to answer in a competition. The quizmaster said that it is not as easy as what i think which i answered as 3.

[quote="Bugi"]Let $ f(x) \equal{}$number of zeroes in which $ x$ ends.
$ f(383) \equal{} \lfloor \frac {383}{5}\rfloor \plus{} \lfloor \frac {383}{25}\rfloor \plus{} \lfloor \frac {383}{125}\rfloor \equal{} 94$
Do the same for $ f(144)$ and $ f(383 \minus{} 144)$.
Answer: $ 3$[/quote]

First of all, what you wrote as the definition of $ f$ is not what you meant.  Second of all, $ 10/2 \equal{} 5$ is the ratio of a number that ends in 1 zero over a number that ends in 0 zeroes, yet the quotient has 0 (not 1 - 0 = 1) zeroes -- yet according to your solution, this is not possible!  So, perhaps time to reconsider some of your (implicit) assumptions!  :)

[quote="orangefronted"]4. Jonah is inside a whale at sea. At the dawn of each day, there is a 25% chance that he will write in his diary, and at noon, there is a 1/3 chance that the whale will swallow him. If he remains stuck indefinitely, what is the probability that Jonah will be swallowed without writing anything in his diary?[/quote]

4) $ \frac14$ chance that he writes. $ \frac34\cdot\frac13 \equal{} \frac14$ chance that he gets eaten without writing anything. There's a $ \frac{\frac14}{\frac14\plus{}\frac14}\equal{}\frac12$ chance that he gets swallowed without writing anything.

[quote="ManuelS"]1) 10 = 2(5)
floor(383/2) = 191
or 383! = k(2^191)
floor(383/5) = 76
so there are 76 terminal zeros in (383!) .

floor(144/2) = 72   ,  floor(144/5) = 28
floor(239/2) = 119 ,   floor(239/5) = 47

76 - (28 + 47) = 1
There is 1 terminal zero in 383C144 or 383C239[/quote]

You fail to consider that repeated powers in the factorials (such as $ 125 \equal{} 5^3$) should be counted multiple times.

as has been shown before by Bugi, $ 5^3 \parallel{} \binom{383}{144}$

Now we must find $ k$ such that $ 2^k \parallel{} \binom{383}{144}$, and our answer will be $ min(3,k)$.

$ \lfloor \frac{383}{2} \rfloor \equal{} 191$, $ \lfloor \frac{191}{2} \rfloor \equal{} 95$, $ \lfloor \frac{95}{2} \rfloor \equal{} 47$, $ \lfloor \frac{47}{2} \rfloor \equal{} 23$, $ \lfloor \frac{23}{2} \rfloor \equal{} 11$, $ \lfloor \frac{11}{2} \rfloor \equal{} 5$, $ \lfloor \frac{5}{2} \rfloor \equal{} 2$, $ \lfloor \frac{2}{2} \rfloor \equal{} 1$
so $ 383! \equal{} 2^{375}a$ where $ a$ is odd

$ \lfloor \frac{144}{2} \rfloor \equal{} 72$, $ \lfloor \frac{72}{2} \rfloor \equal{} 36$, $ \lfloor \frac{36}{2} \rfloor \equal{} 18$, $ \lfloor \frac{18}{2} \rfloor \equal{} 9$, $ \lfloor \frac{9}{2} \rfloor \equal{} 4$, $ \lfloor \frac{4}{2} \rfloor \equal{} 2$, $ \lfloor \frac{2}{2} \rfloor \equal{} 1$
so $ 144! \equal{} 2^{142}b$ where $ b$ is odd

$ \lfloor \frac{239}{2} \rfloor \equal{} 119$, $ \lfloor \frac{119}{2} \rfloor \equal{} 59$, $ \lfloor \frac{59}{2} \rfloor \equal{} 29$, $ \lfloor \frac{29}{2} \rfloor \equal{} 14$, $ \lfloor \frac{14}{2} \rfloor \equal{} 7$, $ \lfloor \frac{7}{2} \rfloor \equal{} 3$, $ \lfloor \frac{3}{2} \rfloor \equal{} 1$
so $ 239! \equal{} 2^{232}c$ where $ c$ is odd

Thus we have $ k\equal{}375\minus{}142\minus{}232\equal{}1$, so $ min(1,3) \equal{} 1$.

Wow, i didn't noticed :blush:
I guess i though it was simpler.

I've worked simpler problems such as how many trailing zeros a factorial has, in which the 2's don't matter.  If you start thinking only a trailing zero on bottom can get rid of a trailing zero on top, then it is quite easy to forget about the 2's.  In short: be careful with your assumptions.

@manuel: well, I guess you were just lucky

Now #5:

[hide="Hint"]Use a ball and divider argument[/hide]

[hide]
Let the balls be consonants and the dividers be balls.  So we have $ 3$ balls and thus $ 4$ places to put the dividers.  Since we have $ 3$ dividers, we have $ \binom{4}{3} \equal{} 4$ ways to choose where to place the dividers.  But the dividers (vowels) and the balls (consonants) are not indistinguishable.  We have $ 3! \equal{} 6$ possible permutations of the vowels, and $ 3! \equal{} 6$ possible permutations of the consonants, giving a total of $ 6*6*4 \equal{} 144$ possible arrangements.
[/hide]

The correct answers are as follows:
1. 1
2. 21/64
3. 126
4. 1/2
5. 144


trigonometry
geometry
circumcircle
geometry proposed

$\triangle ABC,\ A>90^{\circ}: \ P,N\in (BC),\ PA=PB,\ NA=NC;$

$[AM]\cap BC\ne\emptyset,\ MB=MC,\ m(\widehat {BCM})=A-90^{\circ}.$

[u]Ascertain the measures of the MNP's angles and prove that[/u] $\frac{MN}{\cos B}=\frac{NP}{\sin A}=\frac{PM}{\cos C}.$

M isnt fixed.then by moving it,the angle BCM wont be fixed.

plz get information about it ;)

$MB=MC,\ m(\widehat {CBM})=m(\widehat {BCM})=90^{\circ}-A\Longrightarrow$the point $M$ is fixed !.

.The line $BC$ separates the points $A,M$. The all points are fixed !.

$A>90^{\circ}\Longrightarrow P\in (BN)$. Indeed,

$P\in (BN)\Longleftrightarrow a>\frac{c}{2\cos B}+\frac{b}{2\cos C}\Longleftrightarrow$

$2a\cos B\cos C>c\cos C+b\cos B\Longleftrightarrow$

$2\sin A\cos B\cos C>\sin C\cos C+\sin B\cos B \Longrightarrow$

$4\sin A\cos B\cos C>\sin 2C+\sin 2B\Longleftrightarrow$

$4\sin A\cos B\cos C>2\sin (B+C)\cos (B-C)\Longleftrightarrow$

$2\cos B\cos C>\cos (B-C)\Longleftrightarrow$

$2\cos B\cos C>\cos B\cos C+\sin B\sin C\Longleftrightarrow$

$\cos (B+C)>0\Longleftrightarrow \cos A<0.$

The three interior angles of $\triangle MNP$ can be found by angle chasing. Then the desired equality can be obtained by applying sine law to this triangle.

The point $M$ is the circumcenter of the triangle $ABC$ !


we have $102$ coins $100$ regular and $2$ unregular
find that the $2$ coins are heavier or lighter than regulars by 3 weightings :?:

Hit me if I'm wrong, but:

1)Weigh a regular coin
2)Weigh first unregular coin
3)Weigh second unregular coin.

Hannes

I think we don't know which coins are the irregular ones...

[quote="M4RI0"]Is it posible?[/quote]
Since we just want to find out if they are heavier or not, and not to find out which ones, I think yes (I haven't thought about it till now).

Divide the coins into 3 groups of 34. Of these three groups, weigh every combination of two (first and second, first and third, second and third).

If the two unusual coins are in two different groups, these will balance. Then consider whether the other group, consisting of only regular coins, are heavier or lighter than these two.

If the two unusual coins are in the same group, the other two will balance. Then consider whether they are heavier or lighter than this one.

First, you only need 2 weighings to carry that out.  Second, that doesn't work because it presupposed I know the answer.  For instance: suppose I do what you say, and I find that two piles are heavier than the third.  Are the coins heavy or light?

let it be known.

divide the 102 coins into 2 piles, A and B, each of 51 coins (duh). compare these two piles.

1) if they're different, this means the irregular coins are either both in A or both in B. choose either one of the two piles--say A!!--and divide it into 3 piles, X, Y, Z, each of 17 coins. compare X to Y and X to Z. there are two possible results:

(a) X equals Y and X equals Z, in which case the irregular coins must have been in pile B; since A differed from B, then the result we got when we compared A to B tells us if the irregular coins are lighter or heavier;
(b) if X differs from either Y or Z, then the irregulars must both be in A and again, since A differed from B, we know. (and we also know (and we also know))

2) if A = B, this means there's one irregular coin in each of A and B. choose either pile, say A!! divide it into 3 piles, X, Y, Z, each of 17 coins. compare X and Y and X and Z. keeping in mind that there's exactly one irregular coin somewhere among X, Y, Z, we have 3 possible results:

(a) X differs from Y and X differs from Z. in that case, the irregular is clearly in X, and the comparison of X and Y tells us if it's light or heavy.
(b) X differs from Y and X equals Z. in that case, the irregular's in Y and again we know.
(c) X equals Y and X differs from Z. in that case, the irregular's in Z and again, ya know.

-letting it be, Holografio Insinuendo


function
algebra
domain
calculus
derivative
calculus computations

If a function haas a local max or a local min value at an interior point c of its domain, and if the first derivative, f', exists at c, then f'(c)=

c
f
-c
1
none of these

$ f'(c)\equal{}0$.  In order for a function $ f$ to have a critical point, its derivative must equal 0 at that point.

[quote="JRav"]In order for a function $ f$ to have a critical point, its derivative must equal 0 at that point.[/quote]

[u]Or[/u] the derivative be undefined at that point! (in this case, of course, he specified that the derivative exists - but in general it is something to be considered).

Also note that being a critical point of f does not gaurantee that f has a max or a min at that point. It is a necessary, but not sufficient, condition.

Obviously, but I'm going based on the info given.


Two brothers sold a herd of sheep which they owned. For each sheep, they received as many dollars are the number of sheeps there are originally in the herd. The money was taken and divided in the following manner. The older brother takes 10 dollars, then the younger takes 10 dollars, after the elder takes again, and the younger and so on. At the end of the division, the younger brother, whose turn it was found there were less than 10 dollars left, so he took what remained. To make the division fair, the older brother then gave the younger his penknife. How much is the penknife worth????
Have fun  :D Wont let me use the dollar sign...

[hide] If the brothers had $ x$ sheep in their flock, then they made $ x^2$ dollars. Now, whenever the older brother and younger brother make a division, there are dividing twenty dollar holdings. Thus, on the last division, the older brother made $ 10$ dollars and the younger brother made $ y$ dollars, where $ y < 10$. However, if considered $ \bmod20$, it suffices to find the units digit, given that the tens digit is odd. We see that $ 4^2$ and $ 6^2\equiv16\pmod{20}$, and these are the only digits (we need only cosider units digits in our test). In either case, the last digit is $ 6$, and the penknife is thus worth $ 10 \minus{} 6 \equal{} \boxed{4}$ dollars.[/hide]


calculus
integration
number theory
prime factorization

Can you say -2 is a factor of 4? Can there be negative factors of positive numbers?

Also when a problem asks you to find the zeros of an equation which of the following way to write it is correct? (or multiple ones):
a.) x=a,b,c,...
b.) x={a,b,c,...}
c.) a,b,c,...
d.) {a,b,c,...}
e.) (given equation)=0 when x=a,b,c,...

[quote="Tare"]Can you say -2 is a factor of 4? Can there be negative factors of positive numbers?[/quote]

Yes, you can.

[quote]It depends on your definition of factors. It is meaningful to define numbers with negative factors, but it is more useful to only work with the positive ones. Usually, people stick to a definition of positive factors.[/quote]

Ok, yeah, this is often true, but you can still say that -2 divides 4.

I would say that 4 has 6 factors: 3 positive factors and 3 negative factors.

In that case why not say it has an infinite number of factors.

In number theory you only consider positive integral factors, because the negatives are considered associates. Basically meaning they obviously correspond to the positive ones. This also makes the statement of the fundamental theorem of arithmatic - that each number has a unique prime factorization - meaningful.


geometry AMC AIME circumcircle trapezoid trigonometry symmetry

(This is the edited version of problem #10 on Mock AIME 2.) ABCDE is a cyclic pentagon with BC = CD = DE. The diagonals AC and BE intersect at M. N is the foot of the altitude from M to side AB.  We have MA = 25, [b]MD = 113[/b], and MN = 15.  The area of triangle ABE can be expressed as m/n where m and n are relatively prime positive integers. Determine the remainder obtained when m+n is divided by 1000.

(It is elsewhere on forums, but nobody has yet solved it.)

Edit- Changed MD = 39 to MD = 113.

Source of the problem: [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic+pentagon&t=14899]Mock AIME 2 Contest
[/url]

Let $(O)$ be the circumcircle of the pentagon $ABCDE$ with center $O$ and radius $R = OA = OB = ...$. Since $BC = CD = DE$, the quadrilateral $BCDE$ is an isosceles trapezoid and $BE \parallel CD$. Assume that this isosceles trapezoid is fixed and the point $A$ moves on the minor arc $BE$. The angle $\alpha = \measuredangle BAC \equiv \measuredangle BAM$ then spans the major arc $BC$ of the circumcircle $(O)$ and it is constant. The area of the triangle $\triangle ABM$ is

$|\triangle ABM| = \frac 1 2 AB \cdot MN = \frac 1 2 AB \cdot AM \sin \alpha$

$\sin\alpha = \frac{MN}{AM} = \frac {15}{25} = \frac 3 5$

$\cos\alpha = \sqrt{1 - \sin^2{\alpha}} = \frac 4 5$

Since the triangle $\triangle CDE$ is isosceles with the base $CE$ and the lines $BE \parallel CD$, we have

$\measuredangle DEC = \measuredangle DCE = \alpha$

$\measuredangle CBE = \measuredangle DEB = \measuredangle DEC + \measuredangle DCE = 2 \alpha$

$\cos{\measuredangle CBE}  = \cos{2 \alpha} = \cos^2{\alpha} - \sin^2{\alpha} = \frac{7}{25}$

$\sin{\measuredangle CBE}  = \sin{2 \alpha} = 2 \sin \alpha \cos \alpha = \frac{24}{25}$

$BE = BC \cos{2 \alpha} + CD + DE \cos{2 \alpha} = BC (1 + 2 \cos{2 \alpha}) =$

$= BC \left( 1 + \frac{14}{25} \right) = \frac{39}{25} BC$

Since the circle $(O)$ is also the circumcircle of the triangles $\triangle BCD, \triangle BCE$,

$BC = 2 R \sin \alpha = \frac 6 5 R$

$BD = 2 R \sin {(180^o - 2\alpha)} = 2 R \sin {2\alpha} = \frac{48}{25} R = \frac 8 5 BC$

Let $P, Q$ be the feet of normals from the vertices $C, D$ to the diagonal $BE$. Then

$CP = DQ = BC \sin{2 \alpha} = \frac{24}{25} BC$

$BP = EQ = BC \cos{2 \alpha} = \frac{7}{25} BC$

$BQ = EP = BD \cos{\alpha} = \frac{32}{25} BC$

Using Pythagorean theorem for the right angle triangles $\triangle CPB, \triangle CPM$,

$MC^2 = CP^2 + MP^2 = CP^2 + (BM - BP)^2 = BC^2 - \frac{14}{25} BC \cdot BM + BM^2$

Using Pythagorean theorem for the right angle triangles $\triangle DQB, \triangle DQM$,

$MD^2 = DQ^2 + MQ^2 = DQ^2 + (BQ - BM)^2 = \frac{64}{25} BC^2 - \frac{64}{25} BC \cdot BM + BM^2$

By the intersecting chords theorem, we have

$MA \cdot MC = MB \cdot ME = BM \cdot (BE - BM)$

$\frac{MA}{MD} = \frac{BM \cdot (BE - BM)}{MC \cdot MD} = \frac {25}{39}$

Denoting $x = \frac{BM}{BE}$ and substituting $BC = \frac{25}{39}BE$ into the equations for $MC^2, MD^2$,

$\left(\frac{MC}{BE}\right)^2 = \left(\frac{25}{39}\right)^2 - \frac{14}{39}\ x + x^2$

$\left(\frac{MD}{BE}\right)^2 = \left(\frac{40}{39}\right)^2 - \frac{64}{39}\ x + x^2$

and we get the following equation for $x$:

$x^2 \left(1 - x\right)^2 = \frac {25^2}{39^2} \left[\left(\frac{25}{39}\right)^2 - \frac{14}{39}\ x + x^2\right] \left[\left(\frac{40}{39}\right)^2 - \frac{64}{39}\ x + x^2\right]$

Changing the variable to $x = y + \frac 1 2$, in order to achieve some symmetry, and using $25^2 + 25 \cdot 39 = 40^2$,

$\left(y^2 - \frac 1 4\right)^2 = \frac {25^2}{39^2} \left[\left(\frac{25}{39}\right)^2 + \frac 1 4 - \frac{7}{39} + \frac{25}{39}\ y + y^2\right] \left[\left(\frac{25}{39}\right)^2 + \frac 1 4 + \frac{25}{39} - \frac{32}{39} - \frac{25}{39}\ y + y^2\right]$

$\left(y^2 - \frac 1 4\right)^2 = \frac {25^2}{39^2} \left[\left(\left(\frac{25}{39}\right)^2 + \frac 1 4 - \frac{7}{39} + y^2\right)^2 - \left(\frac{25}{39}\right)^2 y^2\right]$

Thus the odd powers of $y$ disappeared and we obtained a quadratic equation for $y^2$.

$16 \cdot 39^4 (39^2 - 25^2)\ y^4 - 39^2 [8(39^4 + 2929 \cdot 25^2) - 16 \cdot 25^4]\ y^2 + 39^6 - 25^2 \cdot 2929^2 = 0$

$2^4 \cdot 7 \cdot 39^4 y^4 - 2^8 \cdot 39^2 \cdot 821\ y^2 - 17 \cdot 19 \cdot 109 \cdot 409 = 0$

The roots of this quadratic equation are

$y^2 = \frac{2^5 \cdot 821 \pm \sqrt{2^{10} \cdot 821^2 + 7 \cdot 17 \cdot 19 \cdot 109 \cdot 409}}{2^2 \cdot 7\cdot 39^2} = \frac{2^5 \cdot 821 \pm \sqrt{3^4 \cdot 5^{10}}}{2^2 \cdot 7 \cdot 39^2} =$

$=\frac{2^5 \cdot 821 \pm 3^2 \cdot 5^5}{2^2 \cdot 7 \cdot 39^2} = +\frac{19 \cdot 409}{2^2 \cdot 39^2}$ [color=white].[/color] or [color=white].[/color] $-\frac{17 \cdot 109}{2^2 \cdot 7 \cdot 39^2}$

Since $y^2 > 0$, the negative root is not acceptable and $y = \pm \frac{\sqrt{19 \cdot 409}}{2 \cdot 39} \doteq \pm 1.13017$. Returning to the variable $x$,

$x = \frac{BM}{BE} = \frac 1 2\left(1 \pm \frac{\sqrt{19 \cdot 409}}{39}\right) \doteq +1.63017$ [color=white].[/color] or [color=white].[/color] $-0.63017$

Consequently, the point $M$ cannot lie within the segment $BE$ and the defined pentagon $ABCDE$ does not exist. Rest assured that the results of all equations, including the roots of the final quadratic equation, check with the Sketchpad. :D

You are right, Yetti. I apologize to anyone who agonized over this fallacious problem. I don't know if I'll ever  trust myself when writing another geometry problem.

Fortunately, this one is not over-constrained. Rather, it's a matter of bounds. $MA \le \frac{81}{125}R\cdot\sec{\frac{\alpha}{2}} = .6147R$, but $MD \ge \frac{6}{5}R\cos{2\alpha} = 1.1520R$. In fact, these limiting cases do not correspond, so $\frac{MA}{MD} < \frac{6.147R}{1.1520R} = .5336$, but $\frac{25}{39} = .6410$, impossible. It is easily seen, however, that $\frac{MA}{MD}$ can become arbitrarily small; simply translate $A$ towards $B$. The trapezoid $BCDE$ remains fixed w.r.t. its circumcircle, but $M$ also approaches $B$ as $MD$ approaches $BD$. A quick lower estimate: $MD < 2R$, so it is possible to obtain $\frac{MA}{MD} = \frac{.6147R}{2R} = .3074$ and any lower, positive ratios.

In my sloppy effort to find nice numbers, I chose $MD = 39$, which fails to be defined.  I invite you to try this problem with $MD = 113$. If the area of $ABE$ is expressed as $\frac{m}{n}$ with two relatively prime positive positive integers $m$ and $n$, then what remainder is obtained when $m+n$ is divided by 1000?

[quote="Mildorf"]Fortunately, this one is not over-constrained. Rather, it's a matter of bounds...[/quote]
Using the cosine theorem for the triangles $\triangle BCM, \triangle EDM$

$MC^2 = BC^2 - 2 BC \cdot BM \cos{2 \alpha} + BM^2$

$MD^2 = BC^2 - 2 BC \cdot (BE - BM) \cos{2 \alpha} + (BE - BM)^2$

Dividing these equations by $BE$, denoting  $x = \frac{BM}{BE}$ and substituting $\frac{BC}{BE} = \frac{1}{1 + 2 \cos{2 \alpha}}$, we get

$\left( \frac {MC}{BE} \right)^2 = \frac{1}{(1 + 2 \cos{2 \alpha})^2} - \frac{2 \cos{2 \alpha}}{1 + 2 \cos{2 \alpha}}\ x  + x^2$

$\left( \frac {MD}{BE} \right)^2 = \frac{1}{(1 + 2 \cos{2 \alpha})^2} - \frac{2 \cos{2 \alpha}}{1 + 2 \cos{2 \alpha}}\ (1 - x) + (1 - x)^2$

By the intersecting chords theorem, we have

$MA \cdot MC = MB \cdot ME = BM \cdot (BE - BM)$

$\frac{MA}{MD} = \frac{BM \cdot (BE - BM)}{MC \cdot MD}$

Squaring this equation, reducing the right fraction by $BE$ and substituting $x = \frac{BM}{BE}$ and the above expressions for $\frac{MC}{BE}, \frac{MD}{BE}$ in terms of $x$, we get the following quartic equation for $x$, where we denoted $p = \frac {MA}{MD}$:

$x^2 (1 - x)^2 = p^2 \left[ \frac{1}{(1 + 2 \cos{2 \alpha})^2} - \frac{2 \cos{2 \alpha}}{1 + 2 \cos{2 \alpha}}\ x + x^2\right] \times$

$\times \left[ \frac{1}{(1 + 2 \cos{2 \alpha})^2} - \frac{2 \cos{2 \alpha}}{1 + 2 \cos{2 \alpha}}\ (1 - x)  + (1 - x)^2\right]$

Because of the symmetry of the isosceles trapezoid $BCDE$, this quartic equation can always be symmetrized into a quadratic equation for $y^2$ by the substitution $y = x - \frac 1 2$:

$\left( \frac 1 4 - y^2 \right)^2 = p^2 \left[ \frac{1}{(1 + 2 \cos{2 \alpha})^2} -  \frac{2 \cos{2 \alpha}}{1 + 2 \cos{2 \alpha}} \left( y + \frac 1 2 \right ) + \left(y + \frac 1 2 \right)^2 \right] \times$

$\times \left[ \frac{1}{(1 + 2 \cos{2 \alpha})^2} + \frac{2 \cos{2 \alpha}}{1 + 2 \cos{2 \alpha}} \left( y - \frac 1 2 \right) + \left(y - \frac 1 2 \right)^2 \right] = $

$= p^2 \left[ \left( \frac{5 - 4 \cos^2{2 \alpha}}{4(1 + 2 \cos{2 \alpha})^2} + y^2 \right)^2 - \frac{y^2}{(1 + 2 \cos{2 \alpha})^2} \right]$

Using $\tan{\alpha} = \frac{MN}{MA} = \frac 3 4$ and

$\cos{2 \alpha} = cos^2{\alpha} - sin^2{\alpha} = \frac{1 - \tan^2{\alpha}}{1 + \tan^2{\alpha}} = \frac{MA^2 - MN^2}{MA^2 + MN^2} = \frac{4^2 - 3^2}{4^2 + 3^2} = \frac{7}{25}$

we can calculate the coefficients of the quadratic equation for $y^2$:

$q = \frac{5 - 4 \cos^2{2 \alpha}}{(1 + 2 \cos{2 \alpha})^2} = \frac{5 \cdot 25^2 - 4 \cdot 7^2}{(25 + 2 \cdot 7)^2} = \frac{2929}{4 \cdot 39^2} = \frac{29 \cdot 101}{2^2 \cdot 3^2 \cdot 13^2}$

$r = \frac{1}{1 + 2 \cos{2 \alpha}} = \frac{25}{39} = \frac{5^2}{3 \cdot 13}$

The quartic polynomial

$P(y) = \left( \frac 1 4 - y^2 \right)^2 - p^2 \left[ (q + y^2)^2 - r^2 y^2 \right] =$

$= (1 - p^2) y^4 - \left( 2p^2 q - p^2 r^2 + \frac 1 2 \right) y^2 + \frac{1}{16} - p^2 q^2$

is an even function of $y$, its graph having the shape of a bottle bottom. Since the derivative is equal to zero at $y = 0$, $P(-1/2) = P(1/2) < 0$ and $P(y) \rightarrow +\infty$ for $y \rightarrow \pm \infty$, it has a maximum at $y = 0$. Hence, it always has 2 real roots symmetrical WRT zero outside of the interval $(-1/2, +1/2)$. Depending on the ratio $p = \frac{MA}{MD}$, it also can have 2 real roots inside of this interval. The upper limit of this ratio can be obtained when $P(y)$ has a double root equal to zero, i.e., when

$P(0) = \frac{1}{16} - p^2 q^2 = 0$

$p = \frac{MA}{MD} = \frac{4}{q} = \frac{39^2}{2929} \doteq 0.51929$

If $MA = 25$, then the segment $MD$ has to have the length at least $MD = \frac {MA}{p} = \frac{25 \cdot 2929}{39^2} = \frac{5^2 \cdot 29 \cdot 101}{3^2 \cdot 13^2} \doteq 48.143$.

[quote="Mildorf"]In my sloppy effort to find nice numbers, I chose $MD = 39$, which fails to be defined.  I invite you to try this problem with $MD = 113$.[/quote]
The nice integral lengths of the segment $MD$ are those, for which the discriminant of $P(y)$, considered as a quadratic polynomial of $y^2$, is a full square. It is a full square for $MD = 39$, but this value is below the lower limit for $MD$. The next (and the last!) nice integral length is $MD = 113$. Then

$a = (1 - p^2) = \frac{2^4 \cdot 3 \cdot 11 \cdot 23}{113^2}$

$b = -\left( 2p^2 q - p^2 r^2 + \frac 1 2 \right) = -\frac{2^3 \cdot 7^2 \cdot 26111}{3^2 \cdot 13^2 \cdot 113^2}$

$c = \frac{1}{16} - p^2 q^2 = \frac{7^2 \cdot 11 \cdot 19 \cdot 41 \cdot 59 \cdot 61}{3^4 \cdot 13^4 \cdot 113^2}$

The discriminant is then

$D = b^2 - 4ac = \left( 2p^2 q - p^2 r^2 + \frac 1 2 \right)^2 - 4(1 - p^2) \left( \frac{1}{16} - p^2q^2 \right) = \left( \frac{2^8 \cdot 5^5 \cdot 7}{3^2 \cdot 13^2 \cdot 113^2} \right)^2$

and the roots of the quadratic equation for $y^2$ are

$y^2 = \frac{-b \pm \sqrt D}{2a} = \frac{2^3 \cdot 7^2 \cdot 26111 \pm 2^8 \cdot 5^5 \cdot 7}{2^5 \cdot 3^3 \cdot 11 \cdot 13^2 \cdot 23} = \frac{7^2 \cdot 26111 \pm 2^5 \cdot 5^5 \cdot 7}{2^2 \cdot 3^3 \cdot 11 \cdot 13^2 \cdot 23} =$

$ = \frac{7 \cdot 11 \cdot 19 \cdot 41}{2^2 \cdot 3^2 \cdot 13^2 \cdot 23}$ [color=white].[/color] or [color=white].[/color] $\frac{7 \cdot 59 \cdot 61}{2^2 \cdot 3^3 \cdot 11 \cdot 13^2}$

As a check,

$y_+^2 \cdot y_-^2 = \frac{7 \cdot 11 \cdot 19 \cdot 41}{2^2 \cdot 3^2 \cdot 13^2 \cdot 23} \times \frac{7 \cdot 59 \cdot 61}{2^2 \cdot 3^3 \cdot 11 \cdot 13^2} = \frac{7^2 \cdot 19 \cdot 41 \cdot 59 \cdot 61}{2^4 \cdot 3^5 \cdot 13^4 \cdot 23}$

$\frac c a = \frac{7^2 \cdot 11 \cdot 19 \cdot 41 \cdot 59 \cdot 61}{3^4 \cdot 13^4 \cdot 113^2} \times \frac{113^2}{2^4 \cdot 3 \cdot 11 \cdot 23} = \frac{7^2 \cdot 19 \cdot 41 \cdot 59 \cdot 61}{2^4 \cdot 3^5 \cdot 13^4 \cdot 23}$

We need only the smaller root (the one with the minus sign), because the larger one leads to $y$ outside of the interval $(-1/2, 1/2)$, i.e., to the positions of the point $M$ outside of the segment $BE$;

$y_-^2 = \frac{7 \cdot 59 \cdot 61}{2^2 \cdot 3^3 \cdot 11 \cdot 13^2} \doteq 0.12548 < \frac 1 4$

as required (checks with the Sketchpad). Returning to the variable $x$,

$x(1 - x) = \frac 1 4 - y_-^2 = \frac{3^3 \cdot 11 \cdot 13^2 - 7 \cdot 59 \cdot 61}{2^2 \cdot 3^3 \cdot 11 \cdot 13^2} = \frac{2 \cdot 5^5}{3^3 \cdot 11 \cdot 13^2}$

Thus we found the value of

$x(1 - x) = \frac{BM \cdot ME}{BE^2} = \frac{MC \cdot MA}{BE^2}$

[quote="Mildorf"]If the area of $ABE$ is expressed as $\frac{m}{n}$ with two relatively prime positive positive integers $m$ and $n$, then what remainder is obtained when $m+n$ is divided by 1000?[/quote]

The area of this triangle has to be expressed with the help of this ratio and some of the known lengths $MA, MN, MD$.

$|\triangle ABE| = |\triangle ABM| + |\triangle EBM| = \left(\frac{MB}{MC}\right)^2 |\triangle CEM| +  \left(\frac{MA}{MB}\right)^2 |\triangle CBM|$

because the quadrilateral $ABCE$ is cyclic and the triangle pairs $\triangle ABM \sim \triangle CEM$ and $\triangle EBM \sim \triangle CBM$ are similar. The triangles $\triangle CEM, \triangle CBM$ have the same altitudes from the common vertex $C$ to the line $BE$ and consequently, their areas are in the ratio of their bases $ME, BM$. Hence

$|\triangle ABE| = \left( \frac{BM \cdot ME}{MC^2} + \frac{MA^2}{BM^2} \right) |\triangle CBM| =  \left( \frac{MA}{MC} + \frac{MA^2}{BM^2} \right) |\triangle CBM|$

where we used $BM \cdot ME = MA \cdot MC$. Again, the triangles $\triangle CBM$ and $\triangle CBE$ have the same altitudes from the common vertex $B$ to the line $BE$ and consequently, their areas are in the ratio of their bases $BM, BE$. Hence

$|\triangle ABE| = \left( \frac{MA}{MC} + \frac{MA^2}{BM^2} \right) \frac{BM}{BE} |\triangle CBE| = \left( \frac{MA \cdot BM}{MC \cdot BE} + \frac{MA^2}{BM \cdot BE} \right) |\triangle CBE|$

The area of the triangle $\triangle CBE$ is equal to

$|\triangle CBE| = \frac 1 2 BE \cdot BC \sin{2 \alpha} = \frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} BE^2$

$|\triangle ABE| = \frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} \left( \frac{MA \cdot BM}{MC \cdot BE} + \frac{MA^2}{BM \cdot BE} \right) BE^2 = \frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} \left( \frac{BE \cdot BM}{MC \cdot MA} + \frac{BE}{BM} \right) MA^2$

Using $BM \cdot ME = MA \cdot MC$ again,

$|\triangle ABE| = \frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} \left( \frac{BE}{ME} + \frac{BE}{BM} \right)\ MA^2 = \frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} \frac{BE \cdot BM + BE \cdot ME}{BM \cdot ME}\ MA^2 =$

$= \frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} \frac{BE^2}{BM \cdot ME}\ MA^2 = \frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} \frac{BE^2}{MC \cdot MA}\ MA^2$

which is the desired equation for the area of the triangle $\triangle ABE$. Substituting the values:

$\frac{\sin{2 \alpha}}{2(1 + 2 \cos{2 \alpha})} = \frac{24}{2(25 + 14)} = \frac{4}{13}$

$|\triangle ABE| = \frac{4}{13} \cdot \frac{3^3 \cdot 11 \cdot 13^2}{2 \cdot 5^5} \cdot 25^2 = \frac{2 \cdot 3^3 \cdot 11 \cdot 13}{5} = \frac{7722}{5} = \frac m n$

$m + n \mod 1000 = 7727 \mod 1000 = 727$

All prime factorizations check all right and the result checks with the Sketchpad. :lol: It is a very nice problem but it seems too difficult for the (Mock) AIME contest.

Regards, yetti.

[quote="yetti"]$|\triangle ABE| = \frac{12}{39} \cdot \frac{3^3 \cdot 11 \cdot 13^2}{2 \cdot 5^5} \cdot 25^2 = \frac{2 \cdot 3^4 \cdot 11 \cdot 13}{5} = \frac{23166}{5}$

I still have to check all those prime factorizations, because the result does not check with the Sketchpad. It is a nice problem but to difficult for the (Mock) AIME contest.

yetti[/quote]

I'll save you the time: you added a factor of 3 right there.

My solution:
[hide]Pythagoras gives $AN = 20$.  We draw $BD$ and $AD$, and construct the altitute $MP$ to $AD$, with $P$ on $AD$, and altitude $MM'$ to $AE$, with $M'$ on $AE$. Because $BC = CD = DE$, angles $BAC$, $CAD$, and $DAE$ are congruent.  Because $P$ is on $AD$, triangles $MNA$ and $MPA$ are congruent by AAS, so $MP = 15$ and $PA = 20$, from which Pythagoras gives $PD = 112$, implying $AD = 132$.

Let $\alpha = m\angle{BAC}$, so $m\angle{MAE} = 2\alpha$, and $m\angle{NAE} = 3\alpha$.  Because we have $\sin{\alpha} = \frac{3}{5}$ and $\cos{\alpha} = \frac{4}{5}$, we compute $\sin(2\alpha) = \frac{24}{25}$, and $\sin(3\alpha) = \frac{117}{125}$.  We find that $MM' = 24$ using $\sin(2\alpha) = \frac{24}{25}$.  By a simple Law of Sines argument $DE$ : $EB$ : $BD = 25$ : $39$ : $40$.

Let $[ABE]$ = the area of $ABE$.
We have $[ABE] = 1/2(15 \cdot AB + 24 \cdot AE)$.

Ptolemy on $ABDE$ yields $AB \cdot DE + AE \cdot BD = AD \cdot BE$.  Using the abundance of facts that we have ascertained previously, this gives:

$AB \cdot 25x + AE \cdot 40x  =  132 \cdot 39x$
$25AB + 40AE  =  39 \cdot 132$
$15AB + 24AE  =  \frac{39 \cdot 132 \cdot 3}{5}$

Finally, $[ABE] = \frac{1}{2} \cdot (15AB + 24AE) = \frac{1}{2}\frac{39 \cdot 132 \cdot 3}{5} = \frac{7722}{5}$. Therefore, the answer is $722 + 5 = 727$.[/hide]

[quote="Mildorf"][quote="yetti"]$|\triangle ABE| = \frac{12}{39} \cdot \frac{3^3 \cdot 11 \cdot 13^2}{2 \cdot 5^5} \cdot 25^2 = \frac{2 \cdot 3^4 \cdot 11 \cdot 13}{5} = \frac{23166}{5}$

I still have to check all those prime factorizations, because the result does not check with the Sketchpad. It is a nice problem but too difficult for the (Mock) AIME contest.

yetti[/quote]

I'll save you the time: you added a factor of 3 right there.[/quote]
Hey, that was just a few seconds before I hit the <Submit> button on my final edit.

[quote="Mildorf"]My solution:
<snip>
[/quote]
I suspected that there would be a simpler solution.


Do not forget to post the problems of IMC 2008.

Thank you

But where are the problems 4-6 of the first day?

I typed problems 1-3 yesterday, but I didn't have the paper with me, so I only typed those problems of which I was sure I remembered accurately.


geometry
geometric transformation
reflection
inequalities
triangle inequality

Please help me in solving the following two problems which appear to interrelated.

1. C and D are two points on the same side of a straight line AB.Find a point X on AB such that the angles CXA & DXB are equal.

2. C and D are two points on the same side of a straight line AB and P is any point on AB. Show that PC + PD is least when angles CPA and DPB are equal.
Regards
Pinaki Mishra

[hide="answer to 1"]
Connect C to B and D to A. The intersection of these two lines is the point X. Then CXA and DXB are vertical angles, and therefore are equal.
[/hide]

[hide="2"]
Let $D'$ be the point obtained by reflecting $D$ over $AB$.
$PD=PD'$, so $PC+PD=PC+PD'$.
Clearly, $PC+PD'$ is minimized when $C$, $P$, $D'$ are collinear (which can be proven by triangle inequality).
If $P$ is on $CD'$, then $\angle CPA=\angle D'PB=\angle DPB$.
Therefore, $PC+PD$ is minimized when $\angle CPA=\angle DPB$.
[/hide]

[hide="1"]Reflect D over AB to D'. X lies on the intersection of CD' and AB, since then $\angle DXB=\angle D'XB$, since they are relflections, and $\angle D'XB=\angle CXA$ from vertical angles, implying $\angle DXB=\angle CXA$.[/hide]

[hide="2"]Reflect D over AB to D'. PC+PD=PC+PD', and PC+PD' is minimum when P lies on CD', since,if P does not lie on PC,  from triangle inequality, PC+PD'>PC. $\angle CPA=\angle D'PB$ from vertical angles, and $\angle D'PB=\angle DPB$ since they are reflections $\implies \angle CPA=\angle DPB$.[/hide]


Hi all,

I want to create a small text at the bottom of the first page of a document, like the attachment below. Could you please tell me how to do that? Someone suggested me to use \footnote{} but it will automatically create a number above the text, and I don't like that, I just wanted to create exactly the same as shown in the attachment.

Thanks for your time :-)

[code]\makeatletter
\newcommand\unmarkfootnote[1]{%
  \begingroup
    \let\@makefntext\noindent
    \footnotetext{#1}%
  \endgroup}
\makeatother[/code]Then use \unmarkfootnote{text} for your unnumbered footnote.

Incredible!!! This is exactly what I want to do!!! Thanks a lot, dear [b]stevem[/b], you are the  :first:


inequalities
trigonometry
inequalities unsolved

Given $a_i, b_i\geq 0$ and and $p,q$ reals with $\frac 1p+ \frac 1q=1$.
Prove that $\sum_{i=1}^k \sum_{j=1}^k \frac {a_i b_j}{i+j} \leq \frac {\pi} {\sin {\frac {\pi}p}} (\sum_{i=1}^k a_i^p)^{\frac 1p} (\sum_{i=1}^k b_i^q) ^{\frac 1q}$


Please help me with these two questions...I've tried everything, but am clueless. I have the chart, but can't figure out the answers. Any help?

1. Assume that the heights of 18 year old males are normally distributed with a mean of 69 inches and a standard deviation of 6 inches.

a) What % of 18 year old males are between 69 and 75 inches tall?
b) If 1000 18 year old males are selected at random how many will be less than 72 inches tall?

2. A placement exam is given to all entering students at a certain College. The scores are normally distributed with a mean of 80 and a standard deviation of 8.
a) What percent of the students scored between 74 and 86
b) What percent of the students scored above 92?

I assume that when you say 'chart' you are referring to a Normal Distribution Table.

For the first problem ,we have to convert the info given into standard-normal form, i.e., calculate the z-scores.
We are given that $\mu = 69 $ , $\sigma = 6$. The z-score for some value of a random variable $X$ representing height is given by $z= \frac{x- \mu}{ \sigma}$. for $x=69$ we get a z-score of $z= \frac{69-69}{6} =0$, and for our transformed random variable $Z$, $P(Z<0)=.5000$ (read this off of your table).
For $x=75$, we get a z-score of $z= \frac{75-69}{6} = \frac{6}{6} =1$, and $P(Z<1)=.8413$. Now we want to find;
$P(69<X<75)=P(0<Z<1)=P(Z<1)-P(Z<0)=.3413$. I.e., $34.13\%$.

Part(b) is an easy application of expectation it seems.

Problem 2 is the same idea as 1(a)-- transform $X$ to $Z$, and calculate the z-scores.


Someone in my club would like to know about the commitment needed for AYS. If you or someone you know has participated, I'd greatly appreciate being able to pass on the contact information. Use a PM to share emails or phone numbers.

Kiwi!

Billy

Do you mean ASYO (Atlanta Symphony Youth Orchestra)?

You mean Atlanta Youth Symphony Orchestra, right?  If so, the usual acronym is ASYO.  If you use this acronym, everybody who does these things will know exactly what you are talking about.  If you use a different acronym, most people probably won't.

I go to school with several participants this year.  It is too late to audition for this year, unless your friend has already made it, in which case (s)he should probably do it.  There's a website at [url]http://www.atlantasymphony.org/communityandeducation/youthorchestra/default.aspx[/url];
it has a [url=http://www.atlantasymphony.org/communityandeducation/youthorchestra/asyorehearsalsasyo.aspx]rehearsal schedule[/url].  If your friend is wondering about outside-of-rehearsal commitment, his/her private teacher should be able to give a good answer.  Anyway, making ASYO is a non-trivial feat---there are people I know who didn't get in and probably . . . should have?  (I kind of felt I needed to be able to do math team stuff on Saturdays; I'm doing the [url=http://www.music.emory.edu/eyso]Emory Youth Orchestra[/url].  It is also too late to audition for Emory this year, unfortunately, along with most other youth orchestras in the area.)

. . . Or did you by chance mean the [url=http://www.music.emory.edu/AYWS/]Atlanta Youth [i]Wind[/i] Symphony[/url] at Emory?  They rehearse every Monday night at Emory for two hours.  Most of what I said about where to find information for ASYO holds here.


I know I saw a proof on here a while back (probably 4 or 5 months ago now) that proved that 1+1=3. I was just wondering if someone knew where I could find it. I googled it, and also searched through the art of problem solving forum search, but got now results. Thanks

Josh

hmm....

my solution would be ,,,

1 drop of water + 1 drop of water = 1 drop of water

applying this rule

1 drop of water / 3 = 3 drops of water

 :D

Do a search for 1=2.  There have been numerous threads on this.  Like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=1%3D2&t=33143]this[/url]  one for example.

well, you can prove it by that negative thingy...

assume: x /= 3
multiply by 0: 0 /= 0

but 0 does = 0

so x=3, where x is anything (or everything, since everything is a thing)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group. $ H$ is called a characteristic subgroup of $ G$ if $ f(H)\equal{}H$ for any automorphism $ f$ on $ G$. Show that if $ H$ is a characteristics subgroup of $ N$ and $ N$ is normal subgroup of $ G$, then prove or disprove that $ H$ is a normal subgroup of $ G$. Thanks in advance.

For every $ a\in G$ consider automorphism $ f_a(x): \equal{}axa^{\minus{}1}$. Since $ N$ is normal so $ f|_{N}$ is an automorphism of $ N$. So $ H$ is normal subgroup of $ G$.


trigonometry geometry parallelogram AoPSwiki AMC AIME search

having trouble with these:

1. find arccotan(3)
2. find arcsc(1)

can someone explain these well?

also, can someone please post problems in which you need to use stewart's theorem in order to solve? I need to get better at applying stewart's theorem.

For the first, memorize your common angles. The second one is $ \csc^{\minus{}1}1\equal{}\frac{\pi}{2}.$ You can use a calculator to get the answer to the first.

Don't really know what you want me to explain...:maybe: I guess you find an angle $ \theta$ such that $ \csc\theta\equal{}1.$

First of all, do you know what Stewart's Theorem is?

[hide="Problems with Stewarts Theorem"]
[hide="1)"]
Show that if AD is an angle bisector of $ \triangle ABC$ then \[ AD^2\plus{}(BD)(CD)\equal{}(AB)(AC).\][/hide]

[hide="2)"]
Segment $ CX$ divides $ AB$ such that $ AX\equal{}2BX\equal{}4.$ If $ AC\equal{}7$ and $ BC\equal{}5,$ find $ CX.$[/hide][/hide]

maybe I should specify... I mean problems involving Stewart's Theorem that are NOT in AoPS volume 2. :P

It was the only resource I had at the moment :P

[hide="1)"]
In an isosceles triangle with two sides of length $ 17,$ a line with length $ 16$ is drawn from the vertex to the base. If one segment of the base, as cut by the line, exceeds the other by $ 8,$ find the lengths of the base.[/hide]

[hide="2)"]
Prove that the sum of the squares of the distances from the vertex of the right angle, in a right triangle, to the trisection points along the hypotenuse, is equal to $ \frac{5}{9}$ the square of the length of the hypotenuse.[/hide]

[hide="3)"]
Prove that a triangle with two equal angle bisectors must be isosceles.[/hide]

[hide="4)"]
Prove that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals.[/hide]

[hide="5)"]
Two shorter sides of a triangle measure $ 9$ and $ 18.$ If the internal angle bisector drawn to the longest side has length $ 8,$ find the length of the longest side.[/hide]

I don't believe these are in Volume 2, but I never bothered to check.

[quote="RoFlLoLcOpT"]It was the only resource I had at the moment :P

[hide="1)"]
In an isosceles triangle with two sides of length $ 17,$ a line with length $ 16$ is drawn from the vertex to the base. If one segment of the base, as cut by the line, exceeds the other by $ 8,$ find the lengths of the base.[/hide]

[hide="2)"]
Prove that the sum of the squares of the distances from the vertex of the right angle, in a right triangle, to the trisection points along the hypotenuse, is equal to $ \frac {5}{9}$ the square of the length of the hypotenuse.[/hide]

[hide="3)"]
Prove that a triangle with two equal angle bisectors must be isosceles.[/hide]

[hide="4)"]
Prove that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals.[/hide]

[hide="5)"]
Two shorter sides of a triangle measure $ 9$ and $ 18.$ If the internal angle bisector drawn to the longest side has length $ 8,$ find the length of the longest side.[/hide]

I don't believe these are in Volume 2, but I never bothered to check.[/quote]

some were, but these are great! Thanks! Where did you get these anyways?

Does anyone besides me have an unreasonable hatred for Stewart's Theorem and think that it does not deserve to be called a theorem?

Well it did have to be proven, did it not?


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
AMC 8
AMC 10

Vote above, talk below.

I got a 41 on chapter (27 sprint, 7 target) I am sad, I goofed up on four problems with silly mistakes :mad:  :(

Since no one's posting, I'll post. I got 29/16=45... I forgot what y=x^3 looked like :P

I feel like invading your forum.

I got a 38(22/16), I know really bad. Placed like I forgot. Then in states I did so much better. :lol:

LOL stargirl it seemed nice you're posting :lol:

I will post..... I got 36 and tied for 2nd :rotfl: but after tiebreakers got 5th :(

[quote="cognos599"]29/16[/quote]
shouldn't you be hanging out with yo little indiana homies

shouldn't you be hanging out with your little california homies? you do realize this is the illinois forum......

dang isnt this really late to keep this active huh
30/16

[cough cough] i got a better score on my chapter then on my school mc but i still made state because my team is amazing (river forest roosevelt) :blush:

don't laugh... :blush:

these are my sprint round scores i cant remember my other ones :blush:

School  - 13
Chapter- 17
State   - 13
 :blush:

:rotfl:  :rotfl:  :rotfl:

(lol, jk)

Here are my FULL scores:

Chapter: 41
State: 30

This is the first year that I have been doing any math competitions. Even though there was not a large break between school and chapter competitions I learned a huge amount.
After I had the thrill of going to MC state, I decided to join AoPS. It's pretty cool.

Do AMC 8, and AMC 10/12 A/B if possible.
Make sure you don't overdevote yourself to MC, or in high school you might fry.

[quote="dynamo729"]The year was 1991 (The only year AL won woohoo). I would venture a guess and say bobby or darryl would probably get a perfect if they were to take the same test as MCrawford.[/quote]
Do you know what the person in second got in 1991?

no.

btw did you notice one person on the poll voted 9 or below as chapter score? who the heck was that? :o

[quote="gauss1181"]Aryth did YOU do that as a joke? :o[/quote]
Yah remember for the state forum he voted 40-46.   :rotfl:

lol did he really score that high? :huh:

Lol for reference this was 9 years ago :)


let $ (x,y,z) \geq 0$ , such : $ xyz\equal{}1$
prove that $ (xy\plus{}yz\plus{}xz)^5 \geq ( (x\plus{}y)^2\plus{}(y\plus{}z)^2\plus{}(x\plus{}z)^2)^2 \frac{27}{16}$

[quote="anas"]let $ (x,y,z) \geq 0$ , such : $ xyz \equal{} 1$
prove that $ (xy \plus{} yz \plus{} xz)^5 \geq ( (x \plus{} y)^2 \plus{} (y \plus{} z)^2 \plus{} (x \plus{} z)^2)^2 \frac {27}{16}$[/quote]
Try $ a\equal{}4$ and $ b\equal{}c\equal{}\frac{1}{2}.$ :wink:

u are right  :blush:  , I made a mistake in my calculation..., but with the same conditions the following is true  :D :

$ 16(xy\plus{}yz\plus{}xz)^4(x^2\plus{}y^2\plus{}z^2) \geq  27 ((x\plus{}y)^2\plus{}(x\plus{}z)^2\plus{}(y\plus{}z)^2)^2$ , I hope that u give a solution with SOS  :)

[quote="anas"]u are right  :blush:  , I made a mistake in my calculation..., but with the same conditions the following is true  :D :

$ 16(xy \plus{} yz \plus{} xz)^4(x^2 \plus{} y^2 \plus{} z^2) \geq 27 ((x \plus{} y)^2 \plus{} (x \plus{} z)^2 \plus{} (y \plus{} z)^2)^2$ , I hope that u give a solution with SOS  :)[/quote]
It easy follows from $ (xy \plus{} xz \plus{} yz)^2\geq3(x \plus{} y \plus{} z)xyz.$ :wink:

yes u are right , but I hope to see a solution with SOS :)


geometry
angle bisector
geometry proposed

Problem 2.
Let $ABC$ the triangle, $D$ the point in $BC$ that $BD*AB=DC*AC$.
$M$ and $N$ is midpoints of segments $AD$ and $BC$ respectively .We know that
$\angle{MBC}+\angle{MCA}=\angle{MBA}+\angle{MCB}$.Prove that circumcentre  of triangle $BMC$ lie on $MN$.

[quote="pavel kozlov"][quote="Tiks"].Prove that circumcentre  of triangle $BMC$ lie on $MN$.[/quote]
This big statement is equivalent $\angle{BMC}=90^o$.[/quote]
I am not sure !:) ,also can be $BM=CM$ :P  :rotfl: .

[quote="Sathej"]Doesn't the statement BD. AB = DC . AC,
imply that AD is the angular bisector.

Sanjith.[/quote]
No!for angle bisector need this statement $BD*AC=DC*AB$ ;) .
Guys this easy problem still unsloved!:P  :D .

Tiks can you post the solution for this problem

Davron

It's impossible...................... :o
I created this problem one month ago .Is it possible to find this in Armenia :o
I have no comments...........What can I say ?Maybe two people in the world had the same idea.....

Ok. A hint is to draw the angle bisector $AK$ . Then  $BK=DC$ and we can see that
$AK//MN$  and  after some angle chasing or using isogonal point we find that  $\widehat{ BMC}=90$  and we are done

Thank you very much! But could you write your full soluiton explaining each point please ...
Because i am preaparing for the IMO this way...but i could understand everything my self but i need them exlained to learn methods of proof

Thanks

Davron

Ara, Tiko inch es erexeqi vra kayfavat linum. Meghqen :( . Chnayac vor kardaci ceric cer es postd, mtqov es artahaytutyun@ ancav: vax hors arev. Ara, es prchot@ ova :huh:  karoxa hesa tam oberkot fra. :mad: Vechni tuftum a u bltcnum vonc vor debil-degenrat lini. Gagiki asac super buta  :rotfl:


inequalities
continued fraction
algebra unsolved
algebra

Prove that  there is a positive integer  $n$  such that

$sin\ n<10^{-k}$  for any positive integer  $k$  and  $sin\ n>0$

:P [b]Observe that: sin(4) = -0.7568 < 0 < 10^(-k) for any k. Thus n = 4 did the job and is the smallest positive integer you get.[/b]

Sorry I missed a condition .It should be  $sin\ n>0$

Silouan:
[b]
The TI-83 graphing calculator did the finding and no computer involved and that is a proof for your proposal.

Truthfully it is not worth to use heavy machine to find a much longer proof. You just need to realize that if the

inequality holds for all k > 0 and by letting k tends to infinity it must hold as well. Thus the natural n you seek must

be such that sin(n) <= 0. One easy way out to see n = 4 is the first of such n's is to draw a unit circle and mark on

this circle 1, 2, 3, 4, 5,... radians and find the mark that gives you sin negative. This shouldn't be a difficult

task.

 :) [/b]

See my correction of the problem above .

According to Hurwitz's theorem, there are infinitely many pairs of integers $(m,n)$ such that $\epsilon: =|\pi-n/m|<1/(\sqrt{5}m^2)$. Because $\pi>0$, we may assume that $m,n>0$. Now $n=m\pi-m\epsilon$, so
\begin{eqnarray*} |n\text{ sin }n|&=&|n\text{ sin }(m\pi-m\epsilon)|\\ &=&n|\text{sin }m\epsilon|\leq nm|\epsilon|<\frac{n}{\sqrt{5}m}\\ &=&\frac{m\pi-m\epsilon}{\sqrt{5}m}\\ &=&\frac{\pi-\epsilon}{\sqrt{5}}<\frac{\pi}{\sqrt{5}}+\frac{1}{5m^2}. \end{eqnarray*}
Thus, for any $c>\pi/\sqrt{5}$ there are infinitely many positive integers $n$ such that $|n\text{ sin }n|<c$. Now by choosing $c=10$ you can find infinitely many integers such that $|\text{sin }n|<10^{-k}$ as $n>10^{k+1}$.

Actually this is not a valid proof. We don't know if $\text{sin }n>0$. At least I gave some ideas which might lead to a solution.

All that needs to been proven is that there are positive integers $m$ and $n$ with
$0 < n-m \pi < 10^{-k}.$
Any all but finitely many of the convergents $n/m$ to the continued
fraction of $\pi$ satisfy $0 < |n-m \pi| < 10^{-k}$. These convergents
alternate between being greater and less than $\pi$.


$K$ a field. If $-1$ is a sum of squares in $K$, let $f(K)$ be the minimal number $n$ such that there exists $(x_1,..,x_n) \in K^n$ satisfying $x_1^2+x_2^2+..+x_n^2=-1$.
Prove that $f(K)$ is a power of $2$.

Your problem is so beautiful.It's true for many fields such that:C , Z/p (p is a prime ) ,....But I didn't find the solution base on the solution for Z/p.If you have the solution can you post it ?

[quote="alekk"]$K$ a field. If $-1$ is a sum of squares in $K$, let $f(K)$ be the minimal number $n$ such that there exists $(x_{1},..,x_{n}) \in K^{n}$ satisfying $x_{1}^{2}+x_{2}^{2}+..+x_{n}^{2}=-1$.
Prove that $f(K)$ is a power of $2$.[/quote]

This is a consequence of the problem posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70349]here[/url]. I'll restate it: if $m$ is a power of $2$, then the set of non-zero elements of $K$ which are sums of $m$ squares is a group under multiplication.

Denote $f(K)$ by $n$, and let $k$ be the unique non-negative integer for which $2^{k}\le n<2^{k+1}$. Notice that $n+1$ is the smallest $t$ for which there exist $t$ non-zero squares in $K$ which add up to $0$. Let $x_{1}^{2}+\ldots+x_{2^{k+1}}^{2}=0$ be a non-trivial solution (i.e. not all $x_{i}$ are zero). Since $2^{k}<n+1$, we have $a=x_{1}^{2}+\ldots+x_{2^{k}}^{2}\ne 0$. But $x_{2^{k}+1}^{2}+\ldots+x_{2^{k+1}}^{2}=-a$, so both $a$ and $-a$ are sums of $2^{k}$ squares. According to the problem cited above, $-1$ must then be a sum of $2^{k}$ squares as well (because it's in the multiplicative subgroup of $K^{*}$ generated by $\pm a$), so $n\le 2^{k}$. Combined with $2^{k}\le n$, this gives the desired result: $n=2^{k}$.

P.S.

That other problem is stated for fields of characteristic different from $2$, but that's all we need anyway. In fact, fields of finite characteristic are not interesting here: in such a field, $-1$ is either a square or a sum of two squares, so there's nothing to prove.

[quote="alekk"]$K$ a field. If $-1$ is a sum of squares in $K$, let $f(K)$ be the minimal number $n$ such that there exists $(x_{1},..,x_{n}) \in K^{n}$ satisfying $x_{1}^{2}+x_{2}^{2}+..+x_{n}^{2}=-1$.
Prove that $f(K)$ is a power of $2$.[/quote]

This was considered very surprising when Pfister discovered it.  The proof went through several simplifications before reaching the textbooks.  So it can be quite difficult as an exercise if proposed without context or hints.


algebra polynomial AMC USA(J)MO USAMO ARML

Let $f$ be a polynomial of degree 98, such that $f(k)=\frac 1k$ for $k=1,2,3\ldots99.$  Determine $f(100).$

[quote="zabelman"]Ah, a classic... where does this problem come from?[/quote]

I'm pretty sure this problem has been posted on this board at LEAST 3 times, in one form or another :).

This one is from USAMTS(round 4), but I have seen a similar problem from USAMO and BMO.


The surface area and volume of a prisim r given find the surfase area and volume of a prisim with (a) double ,(b) triple,(c) half the dimensions of the given prisim?

1a)                    S.A= 23 ft sq
1b)                    Volume= 12 ft cubed
1c)

2a)                     S.A = 61.7 metrs sq

2b)                     volume+ 28.2 meters cubed
2c)


Especially when viewing marathon threads or threads with many hide tags, it is sometimes difficult to find the instances of the searched term. For instance, searching "balls and urns" returns a top result of the MATHCOUNTS Marathon thread. Good luck finding where balls and urns is explained in the 230+ pages. :P

I'm wondering if there could be a feature that allows the viewer to "jump" to the first instance of the searched term, opening hide tags if necessary and such. That would be pretty helpful, especially when searching in longer threads.

You can search by posts instead of threads, which almost always does the job.


Να δείξετε ότι:

$\frac{lnx}{x}<\frac{lny}{x}+\frac{1}{y}$ με $0<y<x$.

BOUZOUKMAN,sigura ine etsi i ekfonisi? :maybe:
grafete isodinama  $\frac{In\frac{x}{y}}{x}<\frac{1}{y}$
 diladi $In\frac{x}{y}<\frac{x}{y}$
  $\frac{x}{y}<e^{\frac{x}{y}}$ pu isxii apo tin gnosti $e^{a}\geq a+1>a$
ali8ia to $y<x$ pu xriazete?

 :D

Από υπόθεση: $\ln x>\ln y$. Έχουμε:

$\frac{\ln x}{x}<\frac{\ln y}{x}+\frac{1}{y}\implies\frac{\ln x}{x}<\frac{y\ln y+x}{xy}$

Αφού $x>0$, διαιρώ με $x$ και έχω:

$\ln x<\frac{y\ln y+x}{y}\implies y\ln x<y\ln y+x\implies y\ln\frac{x}{y}<x\implies \ln\frac{x}{y}<\frac{x}{y}$

που ισχύει.

Oupssssssss... με πρόλαβε ο Ράλλης !  :P


analytic geometry
number theory
least common multiple
geometry
geometric transformation
vector
homothety

prove that there dont exist a polygon like $A_1A_2A_3A_4...A_{2n+1}$ such that $|A_iA_{i+1}|=1$ ,$A_{2n+2}=A_1$ such that the coordinates of $A_i$be rational.

I don't think it's true...What about the square with vertices $(0,0),(1,0),(1,1),(0,1)$?

Pierre.

I am not sure but I *think* this might be some sort of descent?

Ok. So, assume that such a polygon does exist. With an appropriate change of scale (just multiply by the lcm of all the denominators of the coordinates of the vertices), we deduce that all the vertices may be assumed to be integer points, and its side is a positive integer.
Thus, there exists a $k$-gon with equal sides with integer length and integer points as vertices.
Among all of these polygons, let's choose one with the smallest side length, say $d>0$ (recall that $d$ is an integer).
Wlog (use translation), we may assume that $A_1 = O$, the origin (I keep the notation $A_1, A_2, \cdots$).
Let the vector $A_iA_{i+1}$ has coordinates $(a_i,b_i).$
Then $a_i^2 + b_i^2 = d^2$ for each $i$. (1)
Moreover $\sum_{i=1}^k a_i = \sum_{i=1}^k b_i = 0.$
It follows that $(\sum_{i=1}^k a_i )^2 + ( \sum_{i=1}^k b_i )^2 = 0$, which leads to $2\sum_{i \neq j} (a_ia_j + b_ib_j) = -kd^2.$
Since $k$ is odd, it forces $d$ to be even. Thus $d^2 = 0 \mod [4]$ and, using (1), we deduce that both $a_i$ and $b_i$ are even, for each $i$. Thus, we may use an homothety with ratio $\frac 1 2 $ to construct another $k$-gon with equal side and integer points as vertices, but with side length $\frac d 2$, which contradicts the minimality of $d$. Hence result.

Pierre.


algebra
polynomial
calculus
calculus open

x' = f_a(x)
no e.q. pt if a<0
1 e.q. pt if a=0
4 e.q. pts if a>0

Thanks in advance.....

What are we assumed to do? To find an example of $f_a$ smoothly depending on $a$ with this property?  :?

thanks fedja....

[quote="fedja"]Well, here is a polynomial: $f_a(x)=(6a^2+1)(a-x^2)+a x^4$.[/quote]


ratio
complex analysis
complex analysis unsolved

For what values of $ z$ is $ \sum_{k\equal{}1}^\infty(\frac{z}{1\plus{}z})^n$ convergent?

That's a geometric series. Where is the ratio less than $ 1$ in absolute value?


superior algebra
superior algebra unsolved

Hi
When do 2 elt's in SO(3) commute? Intuitively, I think two elements should commute iff the eigenvectors with eigenvalue 1 of both matrices are the same. I can prove that if this is the case then they commute, but I can't prove it the other way round.
Thanks

[hide="Hint"]The matrices are simultaneously diagonalizable.[/hide]

1) The matrices are simultaneously diagonalizable by an unitary matrix.
2) If $ r_1\equal{}Rot(D_1,\pi),r_2\equal{}Rot(D_2,\pi)$ and $ D_1,D_2$ are orthogonal, then $ r_1r_2\equal{}r_2r_1$.


inequalities
calculus
integration
trigonometry
function
calculus computations

Let $I_n=\int^{\pi/4}_0\tan^n\theta d\theta$ ,prove that

$\frac{1}{2(n+1)}<I_n<\frac{1}{2(n-1)}$ for natural number , $n>1$

Im stuck at this one , anyone help?

1. I(n+2) <  I(n)  < I(n-2)

2. I(n) = 1/(n-1) - I(n-2)

3. I(n+2) = 1/(n+1) - I(n)

Use 2. and 3. to find I(n-2) and I(n+2) in terms of I(n) and write these expressions into inequality 1 which you can then adjust to the required form.

Equations 2 and 3  come from the  standard reduction formula for the tangent function raised to a positive integral power
.
Inequality 1 is true because (TanX)^(n+2)  < (TanX)^n  < (TanX)^(n-2) in the given range of integration.


how many circles do two points determine?

let's assume that these two points lie on a line.
Therefore, all possible centers of the circles lie on the perpendicular bisector of the line, because the radius to the two points have to be equal (equidistant to the two points)

However, the point has to lie within the circle created when the diameter is the distance between the two points.

Therefore, there is an infinite amount of circles, but the center has to lie on the perpendicular bisector, and the center has to be within the semicircle made when those points diametrically determine a circle.

As the $ 15$th definition in the $ 1$st book of Euclid's [u][i]Element[/i][/u], he writes:
[i]"A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure equal one another."[/i]
Two points determine only $ 1$ line.  Therefore, to determine this "$ 1$ point" that all of these straight lines fall upon, you must need at least 3 points.  In other words, there are $ \boxed {infinite}$ circles that go through 2 points.


If I already sent in the resume for my application (my school makes us send in a resume, which they send along with the counselor recommendation) and I want to add something, how do I do that? Do I e-mail them or send them a letter? Also, what should I say? The main reason I'm doing this is that I found out that I was a semi-finalists in Siemens and I definately want to let my college know.

I would just lay out the fact that you were just notified that you are a Westinghouse-Siemens semifinalist, and send a postal mail letter with a photocopy of the notification to the school. Ask your school to be sure what kosher procedure is. I rather suspect that all of the schools to which you are applying receive a copy of the national list of semifinalists, but it can't hurt to be sure that the admission office knows of your selection.

Congratulations. Good luck on your applications.

You could also call up the admissions office of whatever colleges you've applied to and politely ask them how you should tell them.  Also, can't you just include it in the portion of your application that you send out personally?

Thanks! Also, do you know if we should send the entire research report to them along with our application? Do they even read it?

You could, but I doubt it's likely to make much difference.  I'd think they'll probably take Siemen's word that it's a good paper.  No harm in sending a copy along, I guess.

Having worked in college offices years ago, my tendency is to minimize the amount of extra papers included with an application. Send what they ask for. Noting that you are a Siemens-Westinghouse semifinalist is likely to be enough; if you knew a faculty member in the school you are applying to who would really be interested in reading the paper, and flagging your application for the admission office's attention, that would be one thing, but the admission office can simply take your work about being a semifinalist, again because there is a verifiable [url=http://www.siemens-foundation.org/docs/semi_finalist_list_2004-05.pdf]national list of semifinalists[/url].

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]books I recommend about college admission[/url] explain more about what large piles of paper admissions offices have to deal with.