The integer N is positive. There are exactly 2005 ordered pairs (x; y) of positive integers satisfying 1/x+1/y=1/N. Prove that N is a perfect square.

Doing some algebra, we find

[tex]N^2 = (x-N)(y-N)[/tex].

Since [tex]x-N [/tex] can be any factor of [tex]N^2[/tex], we know that [tex]N^2 [/tex] has 2005 factors. Let [tex]\displaystyle N^2 = \Pi p_k^{2e_k}[/tex], wnere [tex]p_k[/tex] is the kth prime. Then the number of factors in [tex]N^2[/tex] is [tex]\displaystyle \Pi(1+2e_k)=2005=5*401[/tex]. Seeing the factorization of 2005, we deduce that one of the following cases must be true.

Case 1:
For some [tex]a[/tex], [tex]1+2e_a=5 \Leftrightarrow e_a =2[/tex].

Case 2:
For some [tex]a, b[/tex], [tex]1+2e_a=5 \Leftrightarrow e_a =2[/tex] and [tex]1+2e_b=401 \Leftrightarrow e_b =200[/tex].

Since [tex]\displaystyle N = \Pi p_k^{e_k}[/tex] and all [tex]e_k[/tex] must be even, N must be a perfect square.


Prove that in any triangle ABC
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\leq\frac{1}{a\plus{}b\minus{}c}\plus{}\frac{1}{b\plus{}c\minus{}a}\plus{}\frac{1}{c\plus{}a\minus{}b}$,
where a,b,c are the sides of the triangle.

Let $ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y,\ x,y,z>0$. The inequality becomes
$ \frac{1}{y\plus{}z}\plus{}\frac{1}{x\plus{}z}\plus{}\frac{1}{x\plus{}y}\leq\frac{1}{2}\left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)$
We have $ \frac{x\plus{}y}{2}\geq\frac{2xy}{x\plus{}y}\Rightarrow\frac{1}{x\plus{}y}\leq\frac{1}{4}\left(\frac{1}{x}\plus{}\frac{1}{y}\right)$
$ \frac{1}{y\plus{}z}\leq\frac{1}{4}\left(\frac{1}{y}\plus{}\frac{1}{z}\right)$
$ \frac{1}{x\plus{}z}\leq\frac{1}{4}\left(\frac{1}{x}\plus{}\frac{1}{z}\right)$
Adding the last three inequalities we obtain the original inequality.

We don't need let $ a\equal{}y\plus{}z,b\equal{}z\plus{}x,c\equal{}x\plus{}y$
We have:
$ \frac{1}{a\plus{}b\minus{}c}\plus{}\frac{1}{b\plus{}c\minus{}a}\geq\frac{4}{a\plus{}b\minus{}c\plus{}b\plus{}c\minus{}a}\equal{}\frac{2}{b}$
$ \frac{1}{b\plus{}c\minus{}a}\plus{}\frac{1}{c\plus{}a\minus{}b}\geq\frac{2}{c}$
$ \frac{1}{c\plus{}a\minus{}b}\plus{}\frac{1}{a\plus{}b\minus{}c}\geq\frac{2}{a}$
$ \Rightarrow$ DONE. :)

[quote="ADN"]We don't need let $ a \equal{} y\plus{}z,b \equal{} z\plus{}x,c \equal{} x\plus{}y$
We have:
$ \frac{1}{a\plus{}b\minus{}c}\plus{}\frac{1}{b\plus{}c\minus{}a}\geq\frac{4}{a\plus{}b\minus{}c\plus{}b\plus{}c\minus{}a}\equal{}\frac{2}{b}$
$ \frac{1}{b\plus{}c\minus{}a}\plus{}\frac{1}{c\plus{}a\minus{}b}\geq\frac{2}{c}$
$ \frac{1}{c\plus{}a\minus{}b}\plus{}\frac{1}{a\plus{}b\minus{}c}\geq\frac{2}{a}$
$ \Rightarrow$ DONE. :)[/quote]
Very nice and smart solution ADN  :)


Let \[ \frac{1 \plus{} a_1x \plus{} a_2x^2 \plus{} \ldots}{1 \plus{} b_1x \plus{} b_2x^2 \plus{} \ldots} \equal{}  1 \plus{} c_1x \plus{} c_2x^2 \plus{} \ldots,\] where $ 0 \leq a_i \leq 1, 0 \leq b_i \leq 1.$ Prove that $ |c_n|$ does not exceed the $ n\minus{}$th Fibonacci number.

If we let $ b_i \equal{} 0, \forall i$, then $ a_i \equal{} c_i$ and so $ 0 \leq c_i \leq 1$. I am confused...


Let $ P(x,y)$ be a polynomial in two variables $ x,y$ such that $ P(x,y)\equal{}P(y,x)$ for every $ x,y$ (for example, the polynomial $ x^2\minus{}2xy\plus{}y^2$ satisfies this condition). Given that $ (x\minus{}y)$ is a factor of $ P(x,y)$, show that $ (x\minus{}y)^2$ is a factor of $ P(x,y)$.

Let $ (x, y)$ be a factor of $ P(x, y)$. There exists a polynomial $ Q(x,y)$ such that $ P(x,y) \equal{} (x \minus{} y).Q(x,y)$.

$ \underline{Claim : } \ (x\minus{}y)$ is a factor of $ Q(x,y)$. To prove our claim we use the method of proof by contradiction. Let $ (x\minus{}y)$ is not a factor of $ Q(x,y)$. Thus there exists a $ x'$ such that $ Q(x',x') \neq 0 \ ... \ (1)$

On the other hand by assumption we have :
$ P(x,y) \equal{} P(y, x) \Rightarrow (x \minus{} y).Q(x,y) \equal{} (y \minus{} x).Q(y,x) \Rightarrow Q(x,y) \equal{} \minus{}Q(y,x) \Rightarrow Q(x,y) \plus{} Q(y,x) \equal{} 0.$

For $ y \equal{} x \equal{} x'$ the last equation turns to $ Q(x',x') \plus{} Q(x',x') \equal{} 0$. Therefore $ Q(x',x')  \equal{} 0$ which contradicts with $ (1)$. So $ (x \minus{} y)$ is a factor of $ Q(x,y)$ which means that $ (x\minus{}y)^2$ is a factor of $ P(x,y)$.

I think in your second paragraph instead of saying $ Q(x,y) \plus{} Q(y,x) \equal{} 0$ you must say $ Q(x,y) \plus{} Q(y,x) \equal{} 0$ OR $ x \equal{} y$, because you cancelled an $ x \minus{} y$ from both sides.

EDIT: Right, $ Q$ is polynomial. Sorry. :D

Hi MellowMelon,

You are right however it doesn't change anything in solution since $ Q(x,y)$ is a polynomial. Therefore $ Q(x,y) \plus{} Q(y,x) \equal{} 0$ for all $ x, y$ and this is why I didn't mention that $ x \equal{} y$.

Disclaimer: In this proof, we are assuming that all coefficients of $P(x,y)$ are integers, if not, just multiply by some $k$ (which won't change the $x$ and $y$ factors in the polynomial)

Claim (1): $P(x,x)=0$
Proof: $P(x,x)=(x-x)Q(x,x)=0$

Because of this, we can split $P(x)$ into different subpolynomials where each term in that polynomial can be written as $ax^ky^{n-k}$ for some n and k.

Now, if the subpolynomial can be written as $a_nx^n+a_{n-1}x^{n-1}y...a_0y^n$, we get that $a_n+a_{n-1}+...+a_0=0$

Because this polynomial is symmetrical, we get that $a_k=a_{n-k}$

Claim (2): Each of these "subpolynomials" have $(x-y)^2$ as a factor.

Proof: We can split the subpolynomial up into $a_n$ $x^n+y^n$s, $a_{n-1}$ $x^{n-1}y+xy^{n-1}$s, etc.

Subclaim (1): We can pair the subpolynomial up into factors of the form $x^ay^b-x^cy^d-x^dy^c+x^by^a$, where $a>c>d>b$, and $a+b=c+d=n$.
Proof: Apparent by the fact that $a_n+a_{n-1}+...+a_0=0$

Now we need to determine that $x^ay^b-x^cy^d-x^dy^c+x^by^a$ has 2 factors. We can factor this as $x^cy^b(x^{a-c}-y^{d-b})-x^by^c(x^{a-c}-y^{d-b})=(x^cy^b-x^by^c)(x^{a-c}-y^{d-b})=x^by^b(x^{c-b}-y^{c-b})(x^{a-c}-y^{d-b})$.

As $c-b=c-b$ and $a-c=d-b$, we get that these 2 factors of the pair have a factor of $x-y$ each, resulting in this mini-pair having $(x-y)^2$ as a factor, resulting in the subpolynomial having $(x-y)^2$ as a factor, resulting in $P(x,y)$ having $(x-y)^2$ as a factor, proving the claim.

Note: For even $n$, just set expressions like $x^py^p$ as its own symmetrical pair (which works)


vector
linear algebra
matrix
modular arithmetic
absolute value
combinatorics unsolved
combinatorics

There are n natural numbers written along a circle. We write the absolute value of their difference between every two consecutive numbers, and then erase the original numbers. We may repeat this procedure a finite number of times. Determine those values of n for which all numbers of along the circle will be eventually zero, regardless of the particular choice of the original numbers !

See also: IMO1986_3 / USAMO2003_6

I was looking through some of the old problems, and I found this.

Since the maximum of the numbers on the circle decreases at each step (or stays constant), it means that the numbers are in a bounded interval. This means that the problem is equivalent to finding those $n$ for which no matter how large $k$ is, we can make all numbers divisible by $2^k$. This allows us to work in $\mathbb Z_2$, and we don't need the absolute value anymore.

At each step, the new vector formed by the numbers around the circle (in counterclockwise direction) is $\left(\begin{array}{c}a_1'\\a_2'\\\vdots\\a_n'\end{array}\right)=(I_n+A_n(2))\left(\begin{array}{c}a_1\\a_2\\\vdots\\a_n\end{array}\right)$, where $A_n(k)$ is the $n\times n$ matrix with elements in $\mathbb Z_2$, having $0$ everywhere except for $(1,k),(2,k+1),\ldots,(n,k+n-1)$, where indices are taken $\pmod n$ (with this notation, $I_n=A_n(1)$).

It's clear that getting a  only $0$s no matter what the initial vector is is equivalent to getting $0$s in $\mathbb Z_2$ no matter what the initial vector is, because after all the numbers are divisible by $2^t$ we can disregard $2^t$ (i.e. we can divide all of them by $2^t$), and we can make these new numbers divisible by some other power of two, and we can continue like this indefinitely.

The problem is thus reduced to the following: which are those $n$ for which we can surely find a $u$ s.t. $(I_n+A_n(2))^u=O_n$ (all of this in $\mathbb Z_2$). Since $I_n+A_n(2)$ is an $n\times n$ matrix, it's nilpotent iff its $n$'th power is null, so, again, we can simplify the problem like this:

Which are those $n$ for which, in $\mathbb Z_2$, $(I_n+A_n(2))^n=O_n$?

I don't think I'll provide all the necessary details, but by using the fact that $A_n(2)^k=A_n(k+1)$ and $(X+Y)^{2^t}=X^{2^t}+Y^{2^t}$ for any matrices $X,Y$ with elements in $\mathbb Z_2$ which commute, we can prove that the $n$s we are looking for are precisely the powers of $2$.


$ \mbox{Fiecare dintre ecuatiile cu coeficienti pozitivi: }$
$ ax^2 \plus{} 2bx \plus{} c \equal{} 0,\ bx^2 \plus{} 2cx \plus{} a \equal{} 0,\ cx^2 \plus{} 2ax \plus{} b \equal{} 0 \mbox{ are radacini reale.} \\
\mbox{Ce valori pot avea coeficientii }a, b \mbox{ si } c?$

[quote="kunny"]Could anyone translate the problem into English?[/quote]

$ a,b,c \geq 0$ and each equation has real roots.What values can take $ a$, $ b$ and $ c$ ?

Hum, $ D_{1} /4 \equal{} b^2 \minus{} ac\geq 0$ and $ D_{2}/4 \equal{} c^2 \minus{} ab\geq 0$ and $ D_{3}/4 \equal{} a^2 \minus{} bc\geq 0$.

I think that something condition of the original context is insufficient.

$ (a^2 - bc\ge 0,b^2 - ac\ge 0,c^2 - ab\ge 0)\Leftrightarrow(a\ge\sqrt {bc}, b\ge\sqrt {ac},c\ge\sqrt {ab})\Leftrightarrow \\
\Leftrightarrow a = b = c.$
$ \mbox{Intr - adevar, daca de exemplu: } a\le b\le c, \mbox{ atunci: } \\
 \begin{array}{c}{a\le b\le \sqrt {bc}\le c} \\
{\sqrt {bc}\le a}\end{array}\} \Rightarrow a\le b\le \sqrt {bc}\le a\Rightarrow \\
\Rightarrow a = b = \sqrt {bc} = a\Rightarrow a^2 = b^2 = bc\Rightarrow a = b = c.$


Let $ ABC$ be a triangle with $ BC \equal{} 2007$, $ CA \equal{} 2008$, $ AB \equal{} 2009$. Let $ \omega$ be an excircle of $ ABC$ that touches the line segment $ BC$ at $ D$, and touches extensions of lines $ AC$ and $ AB$ at $ E$ and $ F$, respectively (so that $ C$ lies on segment $ AE$ and $ B$ lies on segment $ AF$). Let $ O$ be the center of $ \omega$. Let $ \ell$ be the line through $ O$ perpendicular to $ AD$. Let $ \ell$ meet line $ EF$ at $ G$. Compute the length $ DG$.


AMC
AIME
calculus
geometry
inequalities
modular arithmetic
symmetry

[hide="1st place"]
beta (13)
[/hide]

[hide="2nd place"]
SplashD, Altheman (11)
[/hide]

[hide="3rd place"]
Xantos C. Guin (10)
[/hide]

[hide="Honorable mention"]
calc rulz (9)
[/hide]

[hide="Participants"]
1337h4x, Altheman, arjunlandes, beta, bmps, calc rulz, Calculus, ccy, ch1n353ch3s54a1l, E^(pi*i)=-1, junggi, lotrgreengrapes7926, mustafa, SplashD, Ubemaya, Xantos C. Guin, Xaero
[/hide]

[hide="Answers"]
1. 116
2. 010
3. 251
4. 011
5. 880
6. 004
7. 387
8. 008
9. 209
10. 042
11. $\frac{3+\sqrt{129}}{20}$ (sorry)
12. 009
13. 016
14. 288
15. 012
[/hide]

[hide="Most Common Answers"]
1. 116
2. 010
3. 251
4. 011
5. 880
6. 004
7. 387
8. 000, 008
9. 209
10. 042
11. n/a (wrong form)
12. 000
13. 016
14. 288
15. 012

which would score like... 14!! lol
[/hide]

[hide="Number of submissions"]
17
[/hide]

[hide="All scores"]
2, 3, 3, 3, 4, 4, 4, 4, 5, 6, 6, 7, 9, 10, 11, 11, 13
[/hide]

[hide="Mean"]
5.83
[/hide]

[hide="Median"]
5
[/hide]

Congratulations! What did you guys think of the difficulty? And are there any other results you want to see?

[quote="junggi"]Dude are you KIDDING ME
Two stupid mistakes took away my 3rd place  :(
wahhh
anyway, congrats to the top scorers  :)
And nice test, chess64  :)[/quote]

i had a stupid mistake too...i bet others did as well...

@altheman: yea, I know.... :(  :(

[hide="10"]$((1+1)^{2007}+(1+w)^{2007}+(1+w^{2})^{2007})/3$
where $w$ is 3rd root of unity other than 1.[/hide]

I forgot what I did for 8...lemme try to redo it
edit:
[hide="8"]Let $a_{2}=a_{1}+b_{1},a_{3}=a_{1}+b_{1}+b_{2}+,...,a_{10}=a_{1}+b_{1}+\cdots+b_{9},$
$2007=a_{1}+b_{1}+\cdots+b_{10}.$
Since consecutive $a_{i}$ have different parity, all $b_{i}$ are odd. $a_{1}$ is also odd.
So substituting $b_{i}=2b_{i}'-1$ where $b_{i}'$ is a positive integer(and same for $a_{1}$), we get
$1009=a_{1}'+b_{1}'+\cdots+b_{10}'$
And any set of solutions to this equation gives valid sequence, and any valid sequence satisfies this equation after the substitutions.
The number solns to this is $\binom{1008}{10}$ so $m=1008=8\mod1000$
I think that's right?[/hide]

[quote="PenguinIntegral"]I'd like to know how to do 10, 14, and I don't understand 8 very well. The others I can do or are geometry, and we all know geometry is a tool of Satan. :wink:

EDIT: BTW I have to credit chess for kicking my butt back into mathematics after a poor fall 2006...[/quote]

excuse me, geometry totally pwns, and the geo on this test was not so difficult
--
again, how does one do #3???

i got pwnd on that one counting problem, even number of even partitions...also what type of partition is that...i am a newb at counting, so does partitions count
1,2 and 2,1 as distinct?

[quote="junggi"]Dude are you KIDDING ME
Two stupid mistakes took away my 3rd place  :(
wahhh
[/quote]

I had about 6 stupid mistakes :mad:

[quote="junggi"]@altheman: yea, I know.... :(  :(

[hide="10"]$((1+1)^{2007}+(1+w)^{2007}+(1+w^{2})^{2007})/3$
where $w$ is 3rd root of unity other than 1.[/hide]

\[/hide][/quote]
Yeah I remembered that much from "102 Combintorical Problems", but I didn't see a nice way to simplify it, maybe I just lack basic complex number skills?

Complex number seive is sooo fun.

EDIT: The solution for 8 was cool, I don't know why I missed that. How'd you come up with the idea of writing them with differences from the first number, or should I be trying that every time I have a sequence of numbers as a basic strategy?

I think I see how to do 14, gimme a sec...

EDIT:
Write the equation as k^4-144=577x, or $(k^{2}+12)(k^{2}-12)=577x$, 577 is prime, so now find all sols to that...

[quote="PenguinIntegral"][quote="junggi"]@altheman: yea, I know.... :(  :(

[hide="10"]$((1+1)^{2007}+(1+w)^{2007}+(1+w^{2})^{2007})/3$
where $w$ is 3rd root of unity other than 1.[/hide]

\[/hide][/quote]
Yeah I remembered that much from "102 Combintorical Problems", but I didn't see a nice way to simplify it, maybe I just lack basic complex number skills?

Complex number seive is sooo fun.[/quote]
[hide="hmm?"]Expand with binomial theorem and bunch of stuff cancels with $1+w+w^{2}$....
like
$1+\binom{2007}{1}+\binom{2007}{2}+\binom{2007}{3}+\cdots+$
$1+\binom{2007}{1}w+\binom{2007}{2}w^{2}+\binom{2007}{3}w^{3}+\cdots+$
$1+\binom{2007}{1}w^{2}+\binom{2007}{2}w^{4}+\binom{2007}{3}w^{6}+\cdots+$
Add vertically and boom[/hide]
[quote="PenguinIntegral"]
EDIT: The solution for 8 was cool, I don't know why I missed that. How'd you come up with the idea of writing them with differences from the first number, or should I be trying that every time I have a sequence of numbers as a basic strategy?[/quote]
I dunno I think I saw someone do that to that kind of inequalities before

[quote]Write the equation as k^4-144=577x, or (k^{2}+12)(k^{2}-12)=577x, 577 is prime, so now find all sols to that...[/quote]
how would we get the solns  :blush:

[quote="junggi"]Smart stuff[/quote]

Oh yeah 1+w+w^2, heh  :maybe:

god I suck at math......

For 14, isn't there something about squares in prime moduli that could help us?

[quote="beta"]Woohoo I got number 1 wrong yet again!!![/quote]
WTF Beta...

Help us on 14....

#14 probably takes the least calculation of all the problems.

[hide="hint"]$a^{2}\equiv (x-a)^{2}\pmod{x}$[/hide]

[quote="Altheman"]wait how did my score go down? i just sent you the radical answer...i don't see how that can be counted wrong...[/quote]

Your answer was floor(7+sqrt(43)) = 13.

oh **** i found BD...that was pretty stupid...but oh well

[quote="Altheman"]all you have to do is be like... lets suppose a is a solution, say it is mod m, m-a is a solution, and there are 4 solutions [that theorem extends to prime mods], so the answer is just 2m/4 or m/2...which is like omg pwnd

but the idea we were supposed to do was like multiply by 4/4, then you can do some tricky stuff...after solving the equation and using i...[/quote]

How do you know there are 4 solutions?

Just wiki'd all the problems. Might do some solutions later.

Well I suck at number theory so I didn't get any of what you just said...

Regarding question 10, what is the question asking for? I interpret the problem as 2007C0+2007C1+2007C3+2007C5+...+2007C2007 and I this led me to 2^2006-2006.

[quote="Elemennop"][quote="Ubemaya"][quote="Fraxia"]For $10$, how do you evaluate
\[\frac{2^{2007}+2(-1)^{669}}{3}\bmod{1000}? \]
It seems like this should be the easier part of the problem, but I don't know what to do  :blush:[/quote]

Well I got $\frac{2^{2007}-2}{3}\bmod{1000}$ Wait that's the same thing nevermind.

I took it mod 3000 and divided it by 3. To find it mod 3000 I said took it mod 375, used Totient, and multiplied it by 8. So $2^\phi(375) = 375x+1$, $2^{200}\times 8 = 3000x+8$, so $2^{203}$ is congruent to $8$ mod 3000, so $2^{2}00$ is congruent to 1 mod 3000.

So $2^{2007}$ is congruent to $2^{7}$ mod 3000, which is 128, then subtract 2 and divide by 3 to get $42$.

Is there a better way to do this problem?[/quote]

um you can divide by 3 w/o any repercussions cause (3,1000)=1 so just do 2^2007 - 2 = 2^7 -2 (by totient) = 126 mod 1000 so divide by3 and you get 42 mod 1000.[/quote]
Or you could notice that the answer to life, the universe, and everything else is 42 and just plug that in to the answer box.

Heh, for some reason I didn't see that $(1000, 3) = 1$, so I couldn't see how to do the division in the congruences  :oops: . Thanks for the help.

[quote="chess64"][quote="scorpius119"]for #15, i get $x=6\sqrt{43}$ which would make the answer 12...  :maybe:[/quote]

Bah! I accidentally did floor(43+sqrt(6)) = 045... omg, i'll fix that in the results soon.[/quote]

Oh...that explains why I couldn't find the mistake...I was a little confused about that one...

[quote="bpms"][quote="Ubemaya"]I don't get why having $(1000, 3)$ means you can do division without taking it in mod 3000...

where do you learn number theory from?[/quote]
aops vol. 2 is decent on number theory.[/quote]
Meh I say just get a decent college text and get that learning out of the way (obviously not so close to the AMC, though). Better just to learn the advanced stuff now, so you can recognize when certain cool theorems trivialize problems...

Though AoPS 2 covers that exact question.


Find all prime numbers [b]p[/b] and [b]q[/b], such
[b]pq is a divisor of 2^p + 2^q[/b]

That is wrong. A multiple of 4 can have odd divisors as well!

For example, $p = 2$, $q = 3$ is a solution.

Sorry for the wrong solution. Perhaps I was too sleepy (it's almost 2am now... . :sleeping:  )
I hope I don't make any stupid mistakes this time.
Since $2^p+2^q$ is even, one of p and q must be 2.
Assume wlog $p=2$, then $2q|4+2^q$
$q|2+2^{q-1}$
$q=2$ is obviously a solution. Now assume $q$ is an odd prime. By Fermat's Little Theorem,
$q|2^{q-1}-1$
$\Rightarrow q|(2+2^{q-1})-(2^{q-1}-1)\Rightarrow q|3\Rightarrow q=3$
Therefore, $(p,q)=(2,2),(2,3),(3,2)$

Still wrong I'm afraid - $30$ is even, but has two distinct odd prime factors ($3$ and $5$) - as do most other even numbers.

Supose $2<p<q$, so $pq|2^{q-p}+1$ (the other cases are trivial). I'll prove this is not possible. In general, if $p|2^{n}+1$, then $p|2^{2n}-1 \Rightarrow p|2^{(2n,p-1)}-1$. If $(2n,p-1)=(n,p-1)$, then $p|2^{(n,p-1)}-1|2^{n}-1 \Rightarrow p|2$, false. Hence $(2n,p-1) \neq (n,p-1) \Rightarrow \exp_{2}(p-1) > \exp_{2}n$. Back to the problem, it follows that $\exp_{2}(p-1) > \exp_{2}(q-p)$ and $\exp_{2}(q-1)>\exp_{2}(q-p)$. This is clearly imposible because for $a=p-1$ and $b=q-1$ we have $\exp_{2}(b-a)<\exp_{2}a,\exp_{2}b$.
I think the problem is from Bulgaria and I'm quite sure it has already been posted on the Forum.

This problem was given in a competition in Bulgaria in the late 90s and the answers are
$ (2;2) (2;3) (3;2) $ as far as I remember. :)


analytic geometry
calculus
linear algebra

Hello everyone,

For the following problem, I have uploaded a self-drawn diagram and shown my work below.

Thank you.

---

11. Find the equation of the line ($ \ell$) through the point $ (\minus{}5, \minus{}4, 2)$ that intersects the line at $ \vec{r} \equal{} [7, \minus{}13, 8] \plus{} t[1, 2, \minus{}2]$ at 90°.

---

[b]Diagram: [/b]
[img]http://i39.tinypic.com/10eqwpv.png[/img]

[b]Calculations:[/b]

$ \vec{OP} \times \vec{r} \equal{} [\minus{}5, \minus{}4, 2] \times [1, 2, \minus{}2] \equal{} [4, \minus{}8, \minus{}6] \equal{} \vec{a}$

$ \vec{a} \times \vec{r} \equal{} [2, \minus{}4, \minus{}3] \times [1, 2, \minus{}2] \equal{} [14, 1, 8]$

So according to my work, equation of $ \ell \equal{} [\minus{}5, \minus{}4, 2] \plus{} t[14, 1, 8]$.

However, the correct answer is [hide] $ \ell \equal{} [\minus{}5, \minus{}4, 2] \plus{} t[14, \minus{}5, 2]$[/hide].

hello, le $ F(x_F,y_F,z_F)$ be the intersection point of the line $ \ell$ through the point $ P( \minus{} 5, \minus{} 4,2)$ perpendicular to the line $ \vec(r)$ then we have the following condictions
$ \vec {PF} \equal{} \left( \begin{array}{c} x_F \plus{} 5 \\
y_F \plus{} 4 \\
z_F \minus{}2\end{array} \right) \circ \vec {a} \equal{} \left( \begin{array}{c} 1 \\
2 \\
\minus{} 2 \end{array} \right) \equal{} 0$
and
$ \vec {OF} \equal{} \left( \begin{array}{c} x_F \\
y_F \\
z_F \end{array} \right) \equal{} \left( \begin{array}{c} \minus{} 7 \\
\minus{} 13 \\
8 \end{array} \right) \plus{} t\cdot \left( \begin{array}{c} 1 \\
2 \\
\minus{} 2 \end{array} \right)$
frome these two conditions we get the following equation system
$ x_F \equal{} \minus{} 7 \plus{} t$
$ y_F \equal{} \minus{} 13 \plus{} 2t$
$ z_F \equal{} 8 \minus{} 2t$
from here we get the equation to calculate $ t$
$ ( \minus{} 7 \plus{} t \plus{} 5)\cdot1 \plus{} ( \minus{} 13 \plus{} 2t \plus{} 4)\cdot2 \plus{} (8 \minus{} 2t \minus{} 2)\cdot( \minus{} 2) \equal{} 0$ and with this $ t$ the the coordinates of $ F$.
Sonnhard.

Here's how I would do it.

Find the intersection of the two lines as follows. The distance between $ [7, \minus{} 13, 8] \plus{} t[1, 2, \minus{} 2]$ and $ [ \minus{} 5, \minus{} 4,2]$ is $ \sqrt {(12 \plus{} t)^2 \plus{} ( \minus{} 9 \plus{} 2t)^2 \plus{} (6 \minus{} 2t)^2}$. Because the two lines intersect perpendicularly, the intersection must occurs when that distance is minimized. Using calculus, you can find that the distance is minimized when $ t \equal{} 2$. So, the intersection is $ [9, \minus{} 9,4]$ and so the direction of the desired line is $ [9, \minus{} 9,4] \minus{} [ \minus{} 5, \minus{} 4,2] \equal{} [14, \minus{} 5,2]$.

Note that there is really no reason to take the cross product of $ \vec{OP}$ and $ \vec{r}$. You're diagram is a bit deceptive because the line $ [7, \minus{} 13, 8] \plus{} t[1, 2, \minus{} 2]$ does not go through the origin. Also, you are using $ \vec{r}$ in different ways, which might have messed you up. Make sure to use clear notation.


As I type $ x^{x^{x^\ldots }}\in\mathbb{R}$ shouldn't the R be bigger?

Erm, no it shouldn't (sorry to contradict!). If you compile [code]$x^{x^{x^\ldots }}\in\mathbb{R}$[/code] in a document you get errors and the log says:
[code]! Missing { inserted.
<to be read again>
                   \mathinner
l.15 $x^{x^{x^\ldots
                    }}\in\mathbb{R}$
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)

! Missing } inserted.
<inserted text>
                }
l.15 $x^{x^{x^\ldots}}\in\mathbb{R}$

I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.[/code]
But $ x^{x^{x^{\ldots} }}\in\mathbb{R}$ works fine using {\ldots} both in a document and in AoPS. So AoPS is behaving correctly :)

I thought that $ x^\ldots$ should work without braces ...

The problem is that it isn't like x^\alpha where \alpha is a single command. \ldots is a TeX macro which is defined as \mathinner{\ldotp\ldotp\ldotp} so x^\ldots is expanded to x^\mathinner{\ldotp\ldotp\ldotp}. This would be read as x^\mathinner followed by {\ldotp\ldotp\ldotp} which confuses TeX and hence the error involving \mathinner.

Ah, ingenious. The thing is that my latex parser should recognize the \ldots and put it into bracers, that's why I said it's a bug.

[quote="MeKnowsNothing"]As I type $ {x^{x^{x^\ldots }}}\in\mathbb{R}$ shouldn't the R be bigger?[/quote]

wait so does this solve the problem?

You have to put the \ldots in brackets. You put the whole thing in brackets which gets parsed (but still generates a TeX warning, only that this time it guesses correctly what we want)


advanced fields
advanced fields unsolved

Is the comb space $\{0\}\times I \cup \bigcup_{n=1}^\infty(\{\frac{1}{n}\}\times I) \cup I \times \{0\}$ ($I = [0,1]$) a retract of $I \times I$?

No. A retract of a locally connected space must be locally connected.


A parallelogram ABCD has perimeter 52, and there are drawn lines from D such that DE is perpendicular to AB, DF is perpendicular to BC. DE = 5, DF = 8, what is BE + BF?

[hide="Hint"]If $AB = x$ then $AD = 26 - x$. Similar triangles.[/hide]


Find all pairs $ (n,m)$ of positive integers $ \ge 3$ such that a regular $ n$-gon can be partitioned into some congruent regular $ m$-gons.

This problem is not Pre-Olympiad material, so I'll provide some hints instead of a solution.

[hide="Hint 1"] The interior angle of the $ m$-gon needs to divide the interior angle of the $ n$-gon. [/hide]
[hide="Hint 2"] Dispose of the trivial case $ n \equal{} m$.  What happens when $ m \equal{} 3$?  What is the interior angle of an $ m$-gon when $ m \ge 4$?  What is the maximum value of the interior angle of the $ n$-gon?  [/hide]


conics
linear algebra
matrix
geometry
angle bisector
geometry proposed

([url=http://img89.imageshack.us/img89/3531/img4xl9.jpg]figure[/url])Let a triangle ABC and call $ E$ and $ F$ two point on BC. The angle bisector of $ \angle BEA$ meet AB in X and AC in J, the angle bisector of $ \angle CFA$ meet AC in Y and AB in K, the angle bisector of $ \angle AEC$ meet AC in R and AB in U, the angle bisector of $ \angle AFB$ meet AB in S and AC in V; $ B_1$ is teh feet of the angle bisector from B and $ C_1$ from C.
Denote these point:

$ \displaystyle \begin{bmatrix} W_1 : ER \cap FY & \ \ W_2 : FS \cap EX \\ W_3 : BB_1 \cap FY & \ \ W_4 : CC_1 \cap EX \\ W_5 : BB_1 \cap ER & \ \ W_6 : CC_1 \cap FS \end{bmatrix}$ $ \displaystyle \begin{bmatrix} X_1 : XY \cap BC & \ \ X_2 : UV \cap BC \\ K_1 : XE \cap AF & \ \ K_2 : YF \cap AE \\ Y_1 : W_1W_2 \cap BC & \ \ Y_2 : K_1K_2 \cap BC \end{bmatrix}$ $ \displaystyle \begin{bmatrix} P : XE \cap YF & \ \ T : UV \cap XE \\ Q : AY_1 \cap UV & \ \ H : XY \cap TY_1  \end{bmatrix}$

Call $ \zeta$ a generic conic and $ \xi$ a generic line, prove that:

$ 1 \blacktriangleright \square$ $ \begin{Bmatrix} E \cup F \cup X \cup Y \cup R \cup S \end{Bmatrix} \in \zeta$

$ 2 \blacktriangleright \square$ $ \displaystyle \bigcup_{i\equal{}1}^{6} W_{i} \in \zeta$

$ 3 \blacktriangleright \square$ $ X_1 \equiv X_2$ e $ Y_1 \equiv Y_2$

$ 4 \blacktriangleright \square$ $ \begin{Bmatrix} A \cup W_1 \cup W_2 \end{Bmatrix} \in \xi$

$ 5 \blacktriangleright \square$ $ \begin{Bmatrix} K \cup Q \cup H \cup J \end{Bmatrix} \in \xi$

$ 6 \blacktriangleright \square$ $ \begin{Bmatrix} A \cup Y \cup X_1 \cup Y_1 \cup T \cup P \end{Bmatrix} \in \zeta$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that a group $ G$ has classes whose orders as follows :

$ C\equal{}\{ 1 , 2 , 15 , 1 , 23 , 14 , 1 \}$

1- Find the order of the center of the group .

2- Give the possible orders of all proper normal subgroups of $ G$ .

 :D

1) Hint: what are elements of the center conjugate to? :)

2) The Sylow theorems give you pretty immediately that there's a normal subgroup of order $ 19$, and that if there were one of order $ 3$, the group would be abelian.

3) Your group doesn't actually exist!  A nonabelian group of order $ 57\equal{}3\cdot 19$ must have $ 1 \plus{} \frac{19\minus{}1}{3} \plus{} (3\minus{}1)\equal{}9$ conjugacy classes, with orders $ 1$, $ 3$, and $ 19$, if I'm not mistaken...

4) If you meant that [i]some[/i] of the classes of $ G$ have those orders, then you can't fully answer 2 but you can form all possible sums of elements of $ C$ for some information.

[quote="LydianRain"]1)
3) Your group doesn't actually exist!  A nonabelian group of order $ 57 \equal{} 3\cdot 19$ must have $ 1 \plus{} \frac {19 \minus{} 1}{3} \plus{} (3 \minus{} 1) \equal{} 9$ conjugacy classes, with orders $ 1$, $ 3$, and $ 19$, if I'm not mistaken...
[/quote]

Sorry , I meant some of them ( not all).

Give mean all the possible sum of $ C$,  :wink:

 :D

[quote="Hidden Scofield"]Give mean all the possible sum of $ C$[/quote]

I could not parse this.

Anyway, clearly the group has order divisible by $ 2\cdot 3\cdot 5\cdot 7\cdot 23$... but I see no reason that the answers to 1) and 2) should be unambiguous given just this information and a few of the class sizes.

Somebody should feel free to prove me wrong, of course.  It would be a very interesting problem if it were true.


Hello!
I was wondering what would happen to the resting membrane potential if K+ ions could no longer leak across the membrane, but the Na+/K+ ATPase pump continued to function? Would it remain at -70mV regardless?
Thanks! :)

[quote="rituraj007"]atleast u kept the forum alive :rotfl:  :P[/quote]
That's it.   :D  This is the first post of the biology forum in 2009. It seems that maths genius doesn't have too much interests in biology.

LOL! i think we should keep this forum alive

i'm sure people can get a lot of help here =)
if they visited/posted more often... :blush:

Biomathematics, anyone? It's one of the newest and most active research in an interdisciplinary subject.


How many ways can a committee of four members be selected from a club consisting of 6 members?


AMC 10
AMC 10 A
AMC 12 A
AMC 10 B
AMC 12 B
AMC

Is it possible to take both the AMC 10 A and the AMC 12 A of one year? How about both the AMC 10 B and the AMC 12 B of one year?

[quote="E^(pi*i)=-1"]No.  Some of the problems are the same.[/quote]

Yes, I knew that, but I was just hoping that I could have more chances at a 150. lol

I am going to be very aggressive about locking topics that don't have a lot of content, so that this board doesn't get too out of control.


calculus
integration
Euler
limit
function
real analysis
real analysis solved

Calculate  \int from 0 to 1 from {1/x}dx. ({} is the fractional part)

This one is very well known, you 'll find it in any text book about Euler
constant  \gamma

A(n) =  \sum _(k = 0 to n-1) \int _[k/n,(k+1)/n](1/x-E(1/x))dx + b(n)
where 0 \leq b(n) =  \int _[0,1/n]f(x)dx  \leq 1/n since 0 \leq f(x) \leq 1
 \lim b(n) = 0
According that E(1/x) is n/(k+1) on [k/n,(k+1)/n]

A(n) = ln n -  \sum _(k =1 to n) 1/k + b(n) + 1

 \lim A(n) = 1 -  \gamma
------------------------------------------------------------------
But we must also prove that f(x) = 1/x - E(1/x) is integrable on [0,1]

Gn(x) = C_[1/n,1](x)f(x)  is integrable on [0,1]

where C_[1/n,1] is the caracteristic function on [1/n,1] wich means
C_[1/n,1](x) is 1 if x \in [1/n;1] and 0 elsewhere

0 \leq G_1 \leq G_2 \leq ... \leq Gn \leq G_(n+1) \leq f  \leq 1

Gn converge simply to f when n-->+oo
with monotone convergence theorem f is integrable on [0,1]


inequalities
LaTeX
inequalities unsolved

suppose that 0=X1<X2<....<X2n+1=1 are real numbers.and also Xi+1 -Xi<h for i=1,...,2n. prove that:
the sum of X2i(X2i+1 - X2i-1) for i=1,...,n is more than 1-h/2 and less than 1+h/2.

Yes, you misunderstood my notation.the corect form is the form that you have written at the end of you massage.

Let me...

[quote="P.H"]Suppose that $0=x_{1}<x_{2}<\cdots<x_{2n+1}=1$ are real numbers and also $x_{i+1}-x_{i}<h$ for $i=1,2,\ldots,2n$. Prove that: \[{1-h \over 2}\le \sum_{i=1}^{n}x_{i}(x_{i+1}-x_{i-1}) \le{1+h \over 2}.\] [/quote]

1+h/2 means half of 1+h and 1-h/2 means half of 1-h.

[quote="P.H"]suppose that 0=X1<X2<....<X2n+1=1 are real numbers.and also Xi+1 -Xi<h for i=1,...,2n. prove that:
the sum of [b]X2i(X2i+1 - X2i-1)[/b] for i=1,...,n is more than 1-h/2 and less than 1+h/2.[/quote]

I think it should be $\sum_{i=1}^{n}X_{2i}(X_{2i+1}-X_{2i-1})$ or it would be too easy  :P

Well I think it is better for P.H to learn how to write in latex rather than for other to understand what he wrote..


If $ a \equal{} b \equal{} 2c$ and $ abc \equal{} 864$ , then the value of $ a \plus{} b \plus{} c$ is equal:

a) 16
b) 23
c) 30
d) 32
e) 34

$ a\equal{}12, b\equal{}12, c\equal{} 6\implies a\plus{}b\plus{}c\equal{}30$

Here my solution :

$ a.b.c \equal{} 864$
$ 2c.2c.c \equal{} 864$
4c³ = 864
c³ = 216
$ c \equal{} 6$

$ b \equal{} 12$
$ a \equal{} 12$

$ a\plus{}b\plus{}c \equal{} 30$

Thanks ;)


modular arithmetic
number theory
prime numbers
number theory unsolved

[quote="nukeemaway"]What is the maximum size of a set whose property is that the sum of any three elements of the set is a prime number?[/quote]

Looking at values of primes modulus 3, we get :

If the set contains $ 3$ :
- it must not contains both primes equal to $ 1$ and $ 2\pmod 3$
- it must contain at most two primes equal to $ 1\pmod 3$
- it must contain at most two primes equal to $ 2\pmod 3$
So it has at most 3 elements $ \{0,1,1\}$ or $ \{0,2,2\}$

If the set does not contain $ 3$ :
- it must contain at most two primes equal to $ 1\pmod 3$
- it must contain at most two primes equal to $ 2\pmod 3$
So it has at most 4 elements $ \{1,1,2,2\}$

And such a set with 4 elements exists : $ \{5,7,11,9421\}$
Hence the answer : $ 4$

[quote="basil mohammed"]how did u construct that set  :|[/quote]

Brute force : spreadsheet with the list of prime numbers up to 10000  :blush:

[quote="kevinatcausa"]Nothing in the question said that the elements of the original set had to be prime, so $ \{5,7,11,25\}$ is a mite bit simpler to find (I guess this also means you need to worry about the possibility that one of the prime sums is actually equal to $ 3$ -- what if the set has a negative element?).[/quote]

 :blush:   I thought that all elements must be primes  :blush:

Thanks for this much simpler example

Looks like $ \{\minus{}15, 3, 15, 19, 79\}$ works as a set of size $ 5$ (I found this set by starting off with three multiples of $ 3$ adding to $ 3$, then brute-forcing for a $ 4^{th}$ element.  At this point the only possibility modulo $ 15$ for the fifth element was $ 4$, and $ 79$ was the next prime satisfying this condition).

[quote="kevinatcausa"]Looks like $ \{ \minus{} 15, 3, 15, 19, 79\}$ works as a set of size $ 5$ (I found this set by starting off with three multiples of $ 3$ adding to $ 3$, then brute-forcing for a $ 4^{th}$ element.  At this point the only possibility modulo $ 15$ for the fifth element was $ 4$, and $ 79$ was the next prime satisfying this condition).[/quote]

Very nice !!!!


geometry
circumcircle
ratio
trigonometry
radical axis
power of a point
geometry proposed

Given $ \triangle ABC$, median $ AM$. Let $ D,E$, respectively, be the projections of $ B,C$ onto $ AC, AB$. Let $ P \equiv DE\cap AM$. Suppose $ AM \equal{} \frac {BC\sqrt {3}}{2}$. Prove that: $ P$ is the midpoint of $ AM$. :lol:
[b]Remark[/b]: I suggest that we should use some pure geometry method to prove this :blush:

$ O, G, H$ are circumcenter, centroid, orthocenter of $ \triangle ABC,$ $ D \in BC, E \in CA, F \in AB$ are feet of altitudes $ AH, BH, CH$ (change of notation). $ AM^2 \equal{} \frac {_3}{^4}BC^2 \Longrightarrow AB^2 \plus{} CA^2 \equal{} 2BC^2.$ It follows that $ AO^2 \minus{} AG^2 \equal{} AO^2 \minus{} \frac {_1}{^3}BC^2 \equal{} OG^2$ $ \Longrightarrow$ $ OG \perp AG.$ $ (N)$ is the 9-point circle through $ D, E, F, M.$ $ (Q)$ is a circle with diameter $ AH,$ $ E, F, G \in (Q).$ $ (S)$ is a circle with diameter $ MH,$ $ D, G \in (S).$ Pairwise radical axes $ EF, GH, MD$ of $ (N), (Q), (S)$ are concurrent at their radical center $ R \in BC$ $ \Longrightarrow$ the cross ratio $ (A, G, P, M) \equal{} \minus{} 1$ is harmonic $ \Longrightarrow$ $ P$ is the midpoint of $ AM.$

[hide="Synthetic and elementary solution"]Let $ F$ a point on side $ BC$ such that $ AF$ is perpendicular to $ BC$. Also let $ AF$, $ BE$ and $ BD$ intersect at point $ H$. Note that $ \angle AEH\plus{}\angle ADH\equal{}90^{\circ}\plus{}90^{\circ}\equal{}180^{\circ}$, so $ A$, $ E$, $ H$ and $ D$ lie on a circle. Therefore \[ \angle AED\equal{}\angle AHD\equal{}90^{\circ}\minus{}\angle HAD\equal{}90^{\circ}\minus{}\angle FAC\equal{}\angle ACF\equal{}\angle ACB.\] We also we have \[ \angle ADE\equal{}\angle AHE\equal{}90^{\circ}\minus{}\angle HAE\equal{}90^{\circ}\minus{}\angle FAB\equal{}\angle ABF\equal{}\angle ABC.\] Thus $ \angle AED\equal{}\angle ACB\equal{}\angle AD'E'$ and $ \angle ADE\equal{}\angle ABC\equal{}\angle AE'D'$, and hence triangle $ EPE'$ is similar to triangle $ D'PD$. So we have $ EP: PE'\equal{}D'P: PD$, or \[ EP\cdot PD\equal{}D'P\cdot PE'\quad\ldots(1)\]

Let $ \frac{AP}{AM}\equal{}x$. Since triangle $ AE'P$ is similar to triangle $ ABM$ and triangle $ AD'P$ is similar to triangle $ ACM$, we have $ \frac{D'P}{MC}\equal{}x\equal{}\frac{PE'}{BM}$. But $ BM\equal{}MC\equal{}\frac{BC}2$, so $ D'P\equal{}PE'\equal{}\frac{BC}2\cdot x$, and \[ D'P\cdot PE'\equal{}\frac{BC^2}4\cdot x^2\quad\ldots(2)\]

Now consider the circle $ \omega$ whose diameter is $ BC$. Since $ \angle BEC\equal{}\angle BDC\equal{}90^{\circ}$, then $ \omega$ passes through $ E$ and $ D$. Let the line $ MP$ intersects $ \omega$ at $ R$ and $ S$, with $ RP<PS$. Note that $ MP\equal{}AM\minus{}AP\equal{}AM\minus{}x\cdot AM\equal{}AM(1\minus{}x)\equal{}\frac{\sqrt3}2BC(1\minus{}x)$. Also note that \[ RP\equal{}RM\minus{}MP\equal{}\frac{BC}2\minus{}\frac{\sqrt3}2BC(1\minus{}x),\] and \[ PS\equal{}PM\plus{}MS\equal{}\frac{\sqrt3}2BC(1\minus{}x)\plus{}\frac{BC}2.\] By the power of a point theorem, we have \[ EP\cdot PD\equal{}RP\cdot PS\equal{}\left(\frac{BC}2\minus{}\frac{\sqrt3}2BC(1\minus{}x)\right)\left(\frac{\sqrt3}2BC(1\minus{}x)\plus{}\frac{BC}2\right)\equal{}\frac{BC^2}4(1\minus{}\sqrt3(1\minus{}x)(1\plus{}\sqrt3(1\minus{}x)\equal{}\frac{BC^2}4\left(1\minus{}3(1\minus{}x)^2\right)\quad\ldots(3)\]

Substitute (2) and (3) to equation (1), we have \[ \frac{BC^2}4\left(1\minus{}3(1\minus{}x)^2\right)\equal{}\frac{BC^2}4\cdot x^2.\]So $ 1\minus{}x^2\equal{}3(1\minus{}x)^2$, thus $ 1\plus{}x\equal{}3(1\minus{}x)$, and finally $ x\equal{}\frac12$. Therefore $ \frac{AP}{AM}\equal{}\frac12$, and point $ P$ is the midpoint of $ AM$ as desired.[/hide]

[quote="Johan Gunardi"][hide="Synthetic and elementary solution"]Let $ F$ a point on side $ BC$ such that $ AF$ is perpendicular to $ BC$. Also let $ AF$, $ BE$ and $ BD$ intersect at point $ H$. Note that $ \angle AEH \plus{} \angle ADH \equal{} 90^{\circ} \plus{} 90^{\circ} \equal{} 180^{\circ}$, so $ A$, $ E$, $ H$ and $ D$ lie on a circle. Therefore
\[ \angle AED \equal{} \angle AHD \equal{} 90^{\circ} \minus{} \angle HAD \equal{} 90^{\circ} \minus{} \angle FAC \equal{} \angle ACF \equal{} \angle ACB.
\]
We also we have
\[ \angle ADE \equal{} \angle AHE \equal{} 90^{\circ} \minus{} \angle HAE \equal{} 90^{\circ} \minus{} \angle FAB \equal{} \angle ABF \equal{} \angle ABC.
\]
Thus $ \angle AED \equal{} \angle ACB \equal{} \angle AD'E'$ and $ \angle ADE \equal{} \angle ABC \equal{} \angle AE'D'$, and hence triangle $ EPE'$ is similar to triangle $ D'PD$. So we have $ EP: PE' \equal{} D'P: PD$, or
\[ EP\cdot PD \equal{} D'P\cdot PE'\quad\ldots(1)
\]
Let $ \frac {AP}{AM} \equal{} x$. Since triangle $ AE'P$ is similar to triangle $ ABM$ and triangle $ AD'P$ is similar to triangle $ ACM$, we have $ \frac {D'P}{MC} \equal{} x \equal{} \frac {PE'}{BM}$. But $ BM \equal{} MC \equal{} \frac {BC}2$, so $ D'P \equal{} PE' \equal{} \frac {BC}2\cdot x$, and
\[ D'P\cdot PE' \equal{} \frac {BC^2}4\cdot x^2\quad\ldots(2)
\]
Now consider the circle $ \omega$ whose diameter is $ BC$. Since $ \angle BEC \equal{} \angle BDC \equal{} 90^{\circ}$, then $ \omega$ passes through $ E$ and $ D$. Let the line $ MP$ intersects $ \omega$ at $ R$ and $ S$, with $ RP < PS$. Note that $ MP \equal{} AM \minus{} AP \equal{} AM \minus{} x\cdot AM \equal{} AM(1 \minus{} x) \equal{} \frac {\sqrt3}2BC(1 \minus{} x)$. Also note that
\[ RP \equal{} RM \minus{} MP \equal{} \frac {BC}2 \minus{} \frac {\sqrt3}2BC(1 \minus{} x),
\]
and
\[ PS \equal{} PM \plus{} MS \equal{} \frac {\sqrt3}2BC(1 \minus{} x) \plus{} \frac {BC}2.
\]
By the power of a point theorem, we have
\[ EP\cdot PD \equal{} RP\cdot PS \equal{} \left(\frac {BC}2 \minus{} \frac {\sqrt3}2BC(1 \minus{} x)\right)\left(\frac {\sqrt3}2BC(1 \minus{} x) \plus{} \frac {BC}2\right) \equal{} \frac {BC^2}4(1 \minus{} \sqrt3(1 \minus{} x)(1 \plus{} \sqrt3(1 \minus{} x) \equal{} \frac {BC^2}4\left(1 \minus{} 3(1 \minus{} x)^2\right)\quad\ldots(3)
\]
Substitute (2) and (3) to equation (1), we have
\[ \frac {BC^2}4\left(1 \minus{} 3(1 \minus{} x)^2\right) \equal{} \frac {BC^2}4\cdot x^2.
\]
So $ 1 \minus{} x^2 \equal{} 3(1 \minus{} x)^2$, thus $ 1 \plus{} x \equal{} 3(1 \minus{} x)$, and finally $ x \equal{} \frac12$. Therefore $ \frac {AP}{AM} \equal{} \frac12$, and point $ P$ is the midpoint of $ AM$ as desired.[/hide][/quote]
This first part, pretty nice, but the second part is so algebric... :lol: (I think)
[b]Here is my own solution:[/b]
Denote $ R$ the radius of $ (BEDC)$. It is easy to notice that: $ \mathcal {P}(A/(BEDC)) \equal{} AM^2 \minus{} R^2 \equal{} 3R^2 \minus{} R^2 \equal{} 2R^2$. Consider the inversion through pole $ A$, power $ 2R^2$, we get, $ I(A,2R^2): E\mapsto B$, $ D\mapsto C$. Therefore, $ I(I,2R^2): ED\mapsto (ABC)$. $ I(I,2R^2): P'\mapsto P$, hence, $ P'\in (ABC)$. But, $ \mathcal {P}(M/(ABC)) \equal{} \overline {MP'}.\overline {MA} \equal{} \overline {MB}.\overline {MC} \equal{} R^2$,$ (1)$. Now, $ I(I,2R^2): BC\mapsto (AED)$, $ M\mapsto M'$, hence, $ M'\in (AED)$. Therefore, $ \overline {AM}.\overline {AM'} \equal{} 2R^2$, $ (2)$.From $ (1),(2)$, we get, $ \frac {MP'}{M'A} \equal{} \frac {1}{2}$, hence, $ 2MP' \equal{} M'A$, $ (3)$. Now, let $ K\equiv ED\cap BC$. It is easy to see that $ KH$ is the polar of $ A$ wrt $ (BEDC)$, therefore, $ KH\bot AM$, which implies the fact that $ K,H,M'$ are collinear, which leads to the result that $ (B,H,M',C)$ is the set of concyclic points. Hence, $ \angle {M'BC} \equal{} \angle {M'HC} \equal{} \angle {EAM'}\equiv \angle {BAP'} \equal{} \angle {BCP'}\equiv \angle{MCP'}$, hence, $ M'B//CP'$, which implies that: $ MM' \equal{} MP'$, $ (4)$. From $ (3),(4)$, we get, $ M'A \plus{} M'M \equal{} 2MP' \plus{} M'M$, or $ AM \equal{} 3MP' \equal{} 3MM' \equal{} \frac {3}{2}M'P'$. Finally, we get, $ MP \equal{} 2R^2.\frac {M'P'}{AM'.AP'} \equal{} 2R^2.\frac {2}{3}.AM.\frac {1}{\frac {2}{3}AM.\frac {4}{3}AM} \equal{} 2R^2.\frac {2}{3}.R\sqrt {3}.\frac {9}{8.3R^2} \equal{} \frac {R\sqrt {3}}{2} \equal{} \frac {AM}{2}$, which leads to the result of the problem.
Our proof is completed :D

[quote="mathVNpro"] [color=darkred]Let $ ABC$ be a triangle . Denote : the midpoint $ M$ of the side $ [BC]$ ; the projections $ D$ , $ E$ of $ B$ , $ C$ onto the opposite sides

respectively ; the intersection $ P\in DE\cap AM$ . Prove that $ AM \equal{} \frac {a\sqrt 3}{2}\Longleftrightarrow b^2 \plus{} c^2 \equal{} 2a^2\ \Longleftrightarrow\ PA \equal{} PM$ .[/color] [/quote]

[color=darkblue][b][u]Proof 1 (metric)[/u].[/b] $ \frac {PA}{PM} \equal{} \frac {EA}{EM}\cdot\frac {\sin\widehat {PEA}}{\sin\widehat {PEM}} \equal{} \frac {b\cdot \cos A}{\frac a2}\cdot\frac {\sin C}{\sin A} \equal{} \frac {2bc\cdot\cos A}{a^2} \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{a^2}$ . Therefore,

$ PA \equal{} PM\ \Longleftrightarrow\ b^2 \plus{} c^2 \minus{} a^2 \equal{} a^2\ \Longleftrightarrow\ b^2 \plus{} c^2 \equal{} 2a^2$ . Is evidently that $ AM \equal{} \frac {a\sqrt 3}{2}\ \Longleftrightarrow\ b^2 \plus{} c^2 \equal{} 2a^2$ .[/color]

[color=darkred][b][u]Lemma 1[/u].[/b] Let $ ABC$ be a triangle. Consider the points $ X\in (BC)\ ,\ Y\in (CA)\ ,$

$ Z\in (AB)$ and $ T\in AX\cap YZ$ . Then $ \boxed {\ \frac {AB}{AZ}\cdot XC \plus{} \frac {AC}{AY}\cdot BX \equal{} \frac {AX}{AT}\cdot BC\ }$ .

[b][u]Particular case[/u].[/b] If $ XB \equal{} XC \equal{} \frac a2$ , then $ TA \equal{} AX$ $ \Longleftrightarrow$ $ \frac {AB}{AZ} \plus{} \frac {AC}{AY} \equal{} 4\ (*)$ .[/color]

[hide="Proof."]
[color=darkblue]The relation from the conclusion of this lemma is well-known. Try to prove alone ![/color][/hide]

[color=darkblue][b][u]Proof 2 (metric)[/u].[/b] $ AM \equal{} \frac {a\sqrt 3}{2}\Longleftrightarrow b^2 \plus{} c^2 \equal{} 2a^2$ and  $ PA \equal{} PM\stackrel {(*)}{\Longleftrightarrow}\frac {AB}{AE} \plus{}$ $ \frac {AC}{AD} \equal{} 4\Longleftrightarrow$

 $ \frac {c}{b\cdot \cos A} \plus{}$ $ \frac {b}{c\cdot \cos A} \equal{} 4\Longleftrightarrow$ $ b^2 \plus{} c^2 \equal{} 4bc\cdot\cos A\Longleftrightarrow$ $ b^2 \plus{} c^2 \equal{} 2\left(b^2 \plus{} c^2 \minus{} a^2\right)\Longleftrightarrow b^2 \plus{} c^2 \equal{} 2a^2$ .[/color]

[color=darkred][b][u]Lemma 2[/u].[/b] The division $ (A,B;C,D)$ is harmonically $ \Longleftrightarrow$ $ \frac {\overline {BA}}{\overline {BD}} \equal{} 2\cdot\frac {\overline {CA}}{\overline{CD}}$ .[/color]

[hide="Proof."]
[color=darkblue]The division $ (A,B;C,D)$ is harmonically $ \Longleftrightarrow$ $ \frac {\overline {CA}}{\overline{CB}} \equal{} \minus{} \frac {\overline {DA}}{\overline {DB}}$ $ \Longleftrightarrow$ $ \frac {a \minus{} c}{b \minus{} c} \plus{} \frac {a \minus{} d}{b \minus{} d} \equal{} 0$ $ \Longleftrightarrow$ $ (a \plus{} b)(c \plus{} d) \equal{} 2(ab \plus{} cd)$ .

$ \frac {\overline {BA}}{\overline {BD}} \equal{} 2\cdot\frac {\overline {CA}}{\overline{CD}}$ $ \Longleftrightarrow$ $ \frac {a \minus{} b}{d \minus{} b} \equal{} 2\cdot\frac {a \minus{} c}{d \minus{} c}$ $ \Longleftrightarrow$ $ (a \minus{} b)(d \minus{} c) \equal{} 2(d \minus{} b)(a \minus{} c)$  $ \Longleftrightarrow$ $ (a \plus{} b)(c \plus{} d) \equal{} 2(ab \plus{} cd)$ .[/color][/hide]

[color=darkblue][b][u]Proof 3 (synthetic)[/u].[/b] Let $ G$ , $ H$ be centroid and orthocenter of $ \triangle ABC$ . Define $ F\in AH\cap BC$ , $ R\in DE\cap AH$ , $ L\in DE\cap BC$ .

The division $ (A,H;R,F)$ is harmonically. From the second lemma obtain $ \frac {HA}{HF} \equal{} 2\cdot \frac {RA}{RF}$ (a characterization of any harmonical division !).

and  $ LH\perp AM$ are well-known (prove easily !).  Denote $ S\in LH\cap AM$ . Therefore, the division $ (P,M;A,S)$ is harmonically and

$ \frac {SA}{SM} \equal{} 2\cdot\frac {PA}{PM}$ . In conclusion, $ PA \equal{} PM\Longleftrightarrow S\equiv G$ $ \Longleftrightarrow GH\perp GA\Longleftrightarrow$ $ GO\perp GA$ $ \Longleftrightarrow$ $ OA^2 \equal{} GA^2 \plus{} GO^2$ $ \Longleftrightarrow$

$ R^2 \equal{} \frac {4m_a^2}{9} \plus{} R^2 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{9}$ $ \Longleftrightarrow$ $ 4m_a^2 \equal{} a^2 \plus{} b^2 \plus{} c^2\Longleftrightarrow$ $ 2\left(b^2 \plus{} c^2\right) \minus{} a^2 \equal{} a^2 \plus{} b^2 \plus{} c^2$ $ \Longleftrightarrow$ $ b^2 \plus{} c^2 \equal{} 2a^2$ . [/color]


let $a,b,c$ be positive reals all greater than $1$ with sum $9$
Show that
$\sqrt{ab+bc+ca}\le \sqrt{a}+\sqrt{b}+\sqrt{c}$

I found that it's not as easy as it looks!!!

I don't think it's a good idea, but I try.Because a>1,b>1,c>1 we must prove that ab+bc+ca<9=a+b+c.We take now a=tanA,b=tanB and c=tanC, where A,B,C are the angles of an acute trinagle such that m(<A)>45,m(<B)>45 and m(<C)>45.(because we have a,b,c>1).In each triangle exists the identity tanAtanB+tanBtanC+tanCtanA=1.
Then we must prove that tanA+tanB+tanC>1.We suppose that can exists a triangle wich has all its angles <60.Results the sum of the angles is <180-false.So, we have at least an agle whith m(<X)>60.,X=A,B,C.But tan60>1-qed

I really don't understand your solution :)

[quote="Albanian Eagle"]let $a,b,c$ be positive reals all greater than $1$ with sum $9$
Show that
$\sqrt{ab+bc+ca}\le \sqrt{a}+\sqrt{b}+\sqrt{c}$[/quote]

Applying [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=400655#p400655]http://www.mathlinks.ro/Forum/viewtopic.php?t=4900 post #13[/url] Theorem 1 to the nonnegative reals a - 1, b - 1, c - 1 (nonnegative because a, b, c are all > 1), we get

$\sqrt{1+\left(a-1\right)}+\sqrt{1+\left(b-1\right)}+\sqrt{1+\left(c-1\right)}\geq\sqrt{15+\left(b-1\right)\left(c-1\right)+\left(c-1\right)\left(a-1\right)+\left(a-1\right)\left(b-1\right)}$
$\Longleftrightarrow\ \ \ \ \ \sqrt{a}+\sqrt{b}+\sqrt{c}\geq\sqrt{18-2\left(\underbrace{a+b+c}_{=9}\right)+\left(bc+ca+ab\right)}$
$\Longleftrightarrow\ \ \ \ \ \sqrt{a}+\sqrt{b}+\sqrt{c}\geq\sqrt{bc+ca+ab}$,

qed.. To be honest, I don't see a simpler proof...

  Darij

Guys can you prove the following ineq.(of cours if it's true)
$63(x^2y^2+x^2z^2+y^2z^2)\leq 5(x^4+y^4+z^4)+30xyz(x+y+z)$
$+14(xy(x^2+y^2)+xz(x^2+z^2)+yz(y^2+z^2))$
for $x,y,z\geq 0$ :? ?
If yes,then I can show you another way,to solve this problem.

canyou prove it now with a,b,c greater than $\frac{1}{3}$ ;)


geometry
power of a point
radical axis
geometry unsolved

Since $ MN$ passes through the midpoint of the common tangent $ PQ$...

Best regards,
sunken rock

Very easy :)

Let be $ K$, intersection of $ MN$ and $ PQ$. Since radical axis $ KM$, so $ |KP| \equal{} |KQ|$. Hence $ Area(MNP) \equal{} Area(MNQ)$

[quote="AndrewTom"]Two intersecting circles $ C_{1}$ and $ C_{2}$ have a common tangent which touches $ C_{1}$ at $ P$ and $ C_{2}$ at $ Q$. The two circles intersect at $ M$ and $ N$, where $ N$ is nearer to $ PQ$ than $ M$ is. Prove that the triangles $ MNP$ and $ MNQ$ have equal areas.[/quote]

As it was said above only see that the radical line MN goes through the middle of PQ .

[b] @ AndrewTom this kind of problems below to the intermediate topic forum, plz don not post such easy problems here[/b]

Thanks Tiger100. AS you find these problems so easy, perhaps you wouldn't mind answering some of my other posts in this forum: @$ 30^{\circ}$ as a unit@ is bothering me. Thanks.


At the beginning of my document I set

\parskip=12pt

This allows me to naturally space my text and get new paragraphs.  This works great except if I put an enumerate section in which causes additional white space that I don't want.  See below:

[img]http://img.skitch.com/20090130-qspna11jkx3r1kpmsh2cimgegt.jpg[/img]

Which was created with the following code:

[code]%18
\probr{A renter with \$1,104 has a one-year lease.  The landlord is willing to accept two payment options:
\goarabic
\begin{enumerate}
\item \$1,104 now; or
\item \$100 paid at the beginning of each month for 12 months
\end{enumerate}
What monthly interest rate would be required for the two options to be equivalent?}
{0.012}{0.013}{0.014}{0.015}[/code]

If I add a \parskip=0pt immediately before the enumerate and a \parskip=12pt immediately after the question mark, then it solves the problem, but I was looking for a more elegant solution.

Thanks!

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=enumitem]enumitem package[/url] will do this for you using \begin{enumerate}[topsep=0pt, partopsep=0pt]
[code]\usepackage{enumitem}
...
A renter with \$1,104 has a one-year lease.  The landlord is willing to accept two payment options:
\begin{enumerate}[topsep=0pt, partopsep=0pt]
\item \$1,104 now; or
\item \$100 paid at the beginning of each month for 12 months
\end{enumerate}
What monthly interest rate would be required for the two options to be equivalent?
{0.012}{0.013}{0.014}{0.015}[/code]

You could even replace enumerate by enumerate* if you define
[code]\newenvironment{enumerate*}
  {\begin{enumerate}[topsep=0pt, partopsep=0pt]}
  {\end{enumerate}}[/code]then use
[code]A renter with \$1,104 has a one-year lease.  The landlord is willing to accept two payment options:

\begin{enumerate*}
\item \$1,104 now; or
\item \$100 paid at the beginning of each month for 12 months
\end{enumerate*}
What monthly interest rate would be required for the two options to be equivalent?
{0.012}{0.013}{0.014}{0.015}[/code]


[quote="lotrgreengrapes7926"]If there are an even number of floors, there isn't really a middle floor.  :maybe: :D
Wait, I'm confused. Is the lobby not a floor?[/quote]

Yeah, I'm thinking of British rules, where it's Lobby, then 1, then 2, etc.

[quote="bpms"][quote="Ignite168"]Dianti is in a elevator. (how ironic) He is on the 26th floor, goes up to the top floor, then goes down to the middle floor of the building (not including the basement), goes to the 36th floor, goes to the basement (1 floor below the lobby), and then goes to the lobby. If the building has 100 floors, how many total floors has Dianti traveled?[/quote]
Whats so ironic?[/quote]

Dian Ti means elevator in Chinese. Want to try solving the problem, it's not that hard.

[quote="bpms"][hide="is it?"]180[/hide][/quote]
Incorrect, and next time, please provide a solution so people can see where you went wrong.

[quote="lotrgreengrapes7926"][quote="Ignite168"]Dianti is in a elevator. (how ironic) He is on the 26th floor, goes up to the top floor, then goes down to the middle floor of the building (not including the basement), goes to the 36th floor, goes to the basement (1 floor below the lobby), and then goes to the lobby. If the building has 100 floors (lobby doesn't count, so 101 floors really), how many total floors has Dianti traveled?[/quote]
[hide] Going from the 26th to the 100th floor, Dianti travels $100-26=\boxed{74}$ floors. Then he goes to the middle of the building, to the 36th floor, and to the basement. Since the middle of the building and the 36th floor are along the way to the basement, we can just say he travels $100-(-1)=\boxed{101}$ floors, since we can call the basement the $-1$th floor. Then, he goes up $\boxed{1}$ floor to the lobby, for a total of $74+101+1=\boxed{\boxed{176}}$ floors. :D [/hide]
And you should fix your signature, Ignite. :D[/quote]
The lobby doesn't count, remember? Anyway, I'm making a bug report about my signature in a second.


In the figure, the outer equilateral triangle has area $ 16$, the inner equilateral triangle area $ 1$, and the three trapezoids are congruent.  What is the area of one of the trapezoids?

$ \textbf{(A)}\ 3\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 7$

[hide]

The area of the big triangle is $ 16$, subtracting the area of the small triangle will tell us the area of all three trapezoids, or $ 15$.  Since all the trapezoids are congruent, dividing by $ 3$ reveals the answer to be $ \boxed{\textbf{(C)}\ 5}$[/hide]


A piece of graph paper is colored with a crayon. The upper left square is colored pink, and then every other square is colored pink. Every square that is not pink is colored yellow. The squares are all then cut out into individual pieces and placed in a hat. What is a reasonable estimate of the probability that a pink square will be removed at random from the hat?

[hide]1/2.

The exact probability varies on the configuration of the board.  for example, an 8x8 chessboard has 32 of each color (exactly 1/2), but an 8x7 chessboard has more of one color than the other.[/hide]

[hide]1/2 since every other box is colored.[/hide]

[quote="colts18"]The upper left square is colored pink, and then every other square is colored pink. Every square that is not pink is colored yellow.[/quote]
Every other square is pink.

lol its awkwardly worded.  every OTHER as in like every alternating one, as opposed to each single other one.... im not sure why interpreted it as the first meaning since its ambiguous.

The problem never said that there was an even number of pink squares. If the first one was pink, and the last one was pink (odd number of squares) there would be one more pink than yellow, the probability not 1/2

If a board is $x$ by $y$, then the probability of getting a pink would be
$\frac{\left[\frac{xy+1}2\right]}{xy}$


geometry
trigonometry
geometric transformation
reflection
geometry proposed

Let a quadrilateral  $ABCD$ [color=red]inscribed in a circle[/color].. On rays  $AB , AD$  we take points  $P , Q$ respectively such that  $AP = CD$ and $AQ = BC$ . If $N$  is the middle of $BD$ , prove that $PQ = 2CN$

 [u]Babis[/u]

Please read the problem again. I forgot something. I really apologize!
 [u]Babis[/u]

Denote a = AB, b = BC, c = CD, d = DA the quadrilateral sides, e = AC , f = BD its diagonals and p = PQ, m = CN. By the cosine theorem for the triangle $\triangle BCD$ with the sides BC = b, CD = c and the angle $\angle C$,

$f^2 = b^2 + c^2 - 2bc \cos \widehat C$

By the cosine theorem for the triangle $\triangle APQ$ with the sides AP = c, AQ = b and the angle $\angle A = 180^\circ - \angle C$,

$p^2 = b^2 + c^2 + 2bc \cos \widehat C = b^2 + c^2 + 2bc \cdot \frac{b^2 + c^2 - f^2}{2bc} =$

$= 2b^2 + 2c^2 - f^2$

On the other hand, m = CN is the C-median of the triangle $\triangle BCD$, hence,

$m^2 = \frac 1 4 \left(2b^2 + 2c^2 - f^2\right) = \frac{p^2}{4}$

$CN = m = \frac p 2 = \frac{PQ}{2}$

Let $R$ be the reflection of the point $B$ w.r.t. the point $C$. $\triangle PAQ\equiv \triangle DCR$ $\Longrightarrow$ $PQ=DR$; $NB=ND$, $CB=CR$ $\Longrightarrow$ $CN\parallel DR$, $DR=2\cdot CN$. Therefore, $PQ=2\cdot CN$.

[u]Remark.[/u] In my opinion, this Stergiu's problem and the Siluan's problem from
http://www.mathlinks.ro/Forum/viewtopic.php?t=67124
are some problems for beginners and not for the the Olympiad Section !

To Virgil:

 Yes , these are not IMO level problems. But both are olympiad problems , even not for final round.
 Any way, you must forgive us. From your point of vew - and I agree with it - you are right. These problems could be posted in the preolympiad section.
 But sometimes I prefare this section , because everybody likes more to solve an easier problem than not to solve a difficult problem !

 [color=brown]Merry Christmas to all of you !!![/color]

   Babis

Thanks !. $\overline {\underline {\left| \ Happy\ New\ Year\ !\ \right| }}$ to all of you, [i]Stergiu[/i] !.


The thermodinamical state equation is $PV=kT^{3}$.The internal energy of this matter changes like this:$U=BT^{3}ln(\frac{V}{V_{0}})$.$k$ and $V_{0}$ are known constants.Find the constant $B$.

is this hypothetical?i mean do we know that PV=yRT and U=yCvT?and is U the variation of internal energy or is a state energy?

Thaakisfox,your answer is correct. :) Can you post the solution?

there must exist an entropy function.
So the small changes of the $S(U,V)$ entropy function:
$dS=\frac{1}{T}dU+\frac{p}{T}dV$
on constant temperature we divide by $dV$ on constant volume by $dT$ we get:
$\bigg (\frac{\partial S}{\partial V}\bigg )_{T}=\frac{1}{T}\bigg (\frac{\partial U}{\partial V}\bigg )_{T}+\frac{p}{T}$
and
$\bigg (\frac{\partial S}{\partial T}\bigg )_{V}= \frac{1}{T}\bigg (\frac{\partial U}{\partial T}\bigg )_{V}$
from the equality of the secon derivatives of the entropy:
$\frac{\partial}{\partial T}\bigg (\frac{1}{T}\frac{\partial U}{\partial V}\bigg )+\frac{\partial}{\partial T}\bigg (\frac{p}{T}\bigg ) =\frac{\partial}{\partial V}\bigg (\frac{1}{T}\frac{\partial U}{\partial T}\bigg )$
we know that:
$\frac{\partial^{2}U}{\partial T \partial V}=\frac{\partial^{2}U}{\partial V \partial T}$
and
$-\frac{1}{T^{2}}\bigg (\frac{\partial U}{\partial V}\bigg )_{T}+\frac{\partial}{\partial T}\bigg (\frac{p}{T}\bigg )=0$
so we get:
$\bigg (\frac{\partial U}{\partial V}\bigg )_{T}=T \bigg (\frac{\partial p}{\partial T}\bigg )_{V}-p$      $(1)$
it's given that $U=BT^{3}ln \frac{V}{V_{0}}$
so :
$\bigg (\frac{\partial U}{\partial V}\bigg )_{T}=\frac{BT^{3}}{V}$
we also know that:
$pV=kT^{3}$
so:
$T \bigg (\frac{\partial p}{\partial T}\bigg )_{V}=\frac{3kT^{3}}{V}$
and
$p=\frac{kT^{3}}{V}$
Writing these in $(1)$:
$\frac{BT^{3}}{V}=\frac{3kT^{3}}{V}-\frac{kT^{3}}{V}$
From here:
$B=2k$.
The trick was deriving the
$\bigg (\frac{\partial U}{\partial V}\bigg )_{T}=T \bigg (\frac{\partial p}{\partial T}\bigg )_{V}-p$
equality and from here it was childs play.
this equality can be used in most these kind of problems.


if $ x\geq 0; y\geq 0; z\geq 0$ and $ x\plus{}y\plus{}z\equal{}2$
prove that::
$ 3xyz\geq 2(xy\plus{}yz\plus{}zx\minus{}1)$

[quote="mhar_teens"]if $ x\geq 0; y\geq 0; z\geq 0$ and $ x \plus{} y \plus{} z \equal{} 2$
prove that::
$ 3xyz\geq 2(xy \plus{} yz \plus{} zx \minus{} 1)$[/quote]
Use the homogenization and Schur. :wink:

[quote="mhar_teens"]if $ x\geq 0; y\geq 0; z\geq 0$ and $ x \plus{} y \plus{} z \equal{} 2$
prove that::
$ 3xyz\geq 2(xy \plus{} yz \plus{} zx \minus{} 1)$[/quote]

It equalivents

\[ r \ge \frac{{2\left( {q \minus{} 1} \right)}}{3}
\]

By Schur we have

\[ r \ge \frac{{8q \minus{} 8}}{9}
\]

Note that if $ q \le 1$ then it's done.

It remains to prove that

\[ \frac{{8q \minus{} 8}}{9} \ge \frac{{2\left( {q \minus{} 1} \right)}}{3}
\]

which is clearly true.

[quote="mhar_teens"]if $ x\geq 0; y\geq 0; z\geq 0$ and $ x \plus{} y \plus{} z \equal{} 2$
prove that::
$ 3xyz\geq 2(xy \plus{} yz \plus{} zx \minus{} 1)$[/quote]

$ \leftrightarrow 2(xy \plus{} yz \plus{} zx) \minus{} 3xyz\le 2$

setting $ f(x,y,z) \equal{} 2(xy \plus{} yz \plus{} zx) \minus{} 3xyz$

$ f(x,z \plus{} y,0) \minus{} f(x,y,z) \equal{} yz(3x \minus{} 2)$

assume $ a \equal{} max(a,b,c)$ $ \rightarrow x\ge \frac {3}{2}$

$ \rightarrow f(x,z \plus{} y,0)\ge f(x,y,z)$

we needs prove that $ f(x,z \plus{} y,0)\le 2 \leftrightarrow 2x(z \plus{} y)\le 2 (*)$

that we is had $ 2x(z \plus{} y)\le 2(\frac {x \plus{} y \plus{} z}{2})^2 \equal{} 2$ $ (Am \minus{} Gm)$

$ \rightarrow (*) right$

 done~!

is there any more solution(s) using this::
$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)[x^2 \plus{} y^2 \plus{} z^2 \minus{} (xy \plus{} xz \plus{} yz)]$

because this is the hint of the problem..
thanks..

[quote="Nawahd Werdna"]
Now, it remains to prove that $ 3xyz \ge \dfrac{2}{3}$
[/quote]
Try $ x\equal{}0.$ :wink:

[quote="Nawahd Werdna"]
But, by AMGM, $ \dfrac{x \plus{} y \plus{} z}{3} \ge \sqrt [3]{xyz}$
Thus, $ xyz \ge \dfrac{8}{27}\ge \dfrac{2}{9}$, and we are done. QED.[/quote]
Also, the last line should be $ xyz \le \frac{8}{27}\equal{}\frac{2}{9}$, which won't work with want you want to prove.

solved..
$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)[x^2 \plus{} y^2 \plus{} z^2 \minus{} (xy \plus{} xz \plus{} yz)]$

we get $ 3xyz \equal{} x^3 \plus{} y^3 \plus{} z^3 \minus{} (x \plus{} y \plus{} z)[x^2 \plus{} y^2 \plus{} z^2 \minus{} (xy \plus{} xz \plus{} yz)]$
$ 3xyz \equal{} x^3 \plus{} y^3 \plus{} z^3 \minus{} 2[x^2 \plus{} y^2 \plus{} z^2 \minus{} (xy \plus{} xz \plus{} yz)]$

from the problem...
$ 3xyz \minus{} 2(xy \plus{} yz \plus{} zx \minus{} 1)\geq 0$
$ \equal{} x^3 \plus{} y^3 \plus{} z^3 \minus{} 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 2$
***$ 2 \equal{} x \plus{} y \plus{} z$
$ \equal{} x^3 \plus{} y^3 \plus{} z^3 \minus{} 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} (x \plus{} y \plus{} z)$
$ \equal{} x(x \minus{} 1)^2 \plus{} y(y \minus{} 1)^2 \plus{} z(z \minus{} 1)^2\geq 0$


Each triangle in this figure is an isosceles right
triangle. The length of $ \overline{BC}$ is 2 units. What
is the number of units in the perimeter of
quadrilateral $ ABCD$? Express your answer
in simplest radical form.

[asy]draw((0,0)--(10,0)--(0,10)--cycle);
draw((10,0)--(0,10)--(10,20)--cycle);
draw((2,0)--(2,2)--(0,2));
draw((2,12)--(4,10)--(2,8));
label("$A$",(0,0),S);
label("$B$",(10,0),S);
label("$C$",(10,20),N);
label("$D$",(0,10),W);[/asy]

$ DC\equal{}\frac{2}{\sqrt{2}}\equal{}\sqrt{2}$

$ AB\equal{}AD\equal{}\frac{\sqrt{2}}{\sqrt{2}}\equal{}1$

Thus, the perimeter of $ ABCD$ is $ 1\plus{}1\plus{}\sqrt{2}\plus{}2\equal{}\boxed{4\plus{}\sqrt{2}}$


What are those ? marks right above your avatar(if you have one)?

They signify what rank you are in and it corresponds with the title above it such as P vs. NP. It goes from 0 to 7 depending on how much you post.

They don't really matter much, just decoration.

i read somewhere on the phpBB site (its what powers this forum) that the ranks in every forum will be taken out. and it said it would take affect immediately. but how come it didnt affect this forum?

Do they have control over already-existing pHpBBs?

I think it's only referring to their own forum, not to other pHpBBs.

[quote="Syntax Error"]http://www.phpbb.com/news.php?id=15

that shows what im saying. but at the end i get confused, because doesnt it contradict what they just said?[/quote]
[quote="phpBB administration"]All aforementioned changes affect only the forums here at phpBB.com. It will in no way affect your own forum, or any forum using phpBB, as it is not a change of the Core-Code of phpBB. Simply a change in our forums administration structure.[/quote]

You know, it's not half-bad of an idea. It'll certainly reduce spamming and new people won't have to feel bad and pressured and such.


How can I prove $\sqrt{x}$ is continuous on $[0, \infty)$?
Using $\delta, \epsilon$ definitions?

Here's my attempt:

Let $x_{0}\in [0, \infty)$. Fix $\epsilon > 0$. $\exists \ \delta > 0$ such that if $|x-x_{0}| < \delta$, then $|\sqrt{x}-\sqrt{x_{0}}|< \epsilon$.

I rewrite $\sqrt{x}-\sqrt{x_{0}}= \frac{x-x_{0}}{\sqrt{x}+\sqrt{x_{0}}}$ through rationalizing.

I try to place some kind of an upper bound on the expression $|x-x_{0}| < \delta$. I notice that $\frac{x-x_{0}}{\sqrt{x}+\sqrt{x_{0}}}< \frac{x-x_{0}}{\sqrt{x_{0}}}$, for $x_{0}\ne 0$. That is my epiphany. But where should I go from here? I am stuck with trying to manipulate this expression into something to do with $|x-x_{0}| < \delta$.

Thanks.

Merci mlok

I see how you got that:

$\frac{x-x_{0}}{\sqrt{x}-\sqrt{x_{0}}}< \frac{\delta}{\sqrt{x}-\sqrt{x_{0}}}< \frac{\delta}{\sqrt{x_{0}}}< \epsilon$. So pick $\delta = \epsilon \sqrt{x_{0}}$

With, of course, a separate argument using $\delta=\epsilon^{2}$ needed to prove continuity at zero.

And yet a little more argument, starting with the same ingredients, needed to prove uniform continuity.


Let a,b,c>0 prove that  $ (1\plus{}abc)(\frac{1}{a(1\plus{}b)}\plus{}\frac{1}{b(1\plus{}c)}\plus{}\frac{1}{c(1\plus{}a)}\ge3$ :(

[quote="Pirkuliyev Rovsen"]Let a,b,c>0 prove that  $ (1 \plus{} abc)(\frac {1}{a(1 \plus{} b)} \plus{} \frac {1}{b(1 \plus{} c)} \plus{} \frac {1}{c(1 \plus{} a)}\ge3$ :([/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161059


$Let$ $a,b,c$ $be$ $nonnegative$ $numbers$ $and$ $t \in \mathbb{R^{+}}$. $Prove$ $that$
$\sum_{cyc}(2a^{t}-b^{t}-c^{t})(a-b)(a-c)\ge 0$


I think , it deserves to keep this problem in your archiev. New solutions are welcome(if not yet posted)!

 http://www.mathlinks.ro/viewtopic.php?t=176836

Let $ X$ be the isogonal point of $ Z$ wrt $ \triangle DBA$.Then $ \angle XBA \equal{} \angle EBC$ and $ \angle XDA \equal{} \angle EAC$ and $ \triangle DBA \sim \triangle ABC$. So $ X,E$ are corresponding points of centrall similar $ \triangle DBA \sim \triangle ABC$. Hence $ \angle ZAB \equal{} \angle XAD \equal{} \angle ECA$.
$ \angle BAC \equal{} 90$ and $ \angle ZAB \equal{} \angle ECA$ imply $ CE\perp AZ$.

The crux idea come from Leonhard Euler when he consider an isogonal point.
My approach:
let E* the isogonal of X wrt ABC
1 the circonscript circle of ABC,
P the second point of intersection of AZ and 1
Q the second point of intersection of BE and 1
R the second point of intersection of BE* and 1
S the foot of the perpendicular issue from E* on AC.
We have :
QR is parallel to AC and CX is perpendicular to DS.
From the Pascal theorem applicated to RQBCAPR, it follow that P, D and R are colinear.
The points D, E*, S, R and C are cocyclic.
By an angle chasing or by the Reim's theorem, it follow that DS is parallel to AP i.e. to AZ
.... and the desired result is obtain.
Jean-Louis


Hallo,

Waar kan ik de VWO (Vlaamse Wiskunde Olympiade) resulaten vinden ? (2003-2004-2005-2006)
Op de site kan je ze aanvragen maar dit kost geld.
Ik moet enkel de juiste antwoorden weten, er hoeft geen uitleg bij.
Heeft iemand deze soms toevallig en wil die deze even inscannen en doormailen ?

verplaats deze post als hij hier niet juist staat.

staan de oplossingen niet meer in de engelse versies van de pdf-documenten die de vragen bevatten?

de pdf's die ik daar vind zijn enkel nederlandstalig en daar staan geen oplossingen bij

Nee idd, ze hebben die weggedaan met de vernieuwing van de site blijkbaar. Dan heb ik ze niet meer.


analytic geometry
graphing lines
slope
Gauss

[hide="hint"]
ok find the slope between those two points, that is the value of k.  Then find the equation with slope k that passes through a point that you gave lets say $ A$.  The y intercept of that line will be b!
[/hide]

Hmmm.... can you give the the total solution.
Just think for the 2 points
$ A(1;3)$ so $ x_A\equal{}1$ and $ y_A\equal{}3$
$ B(3;\minus{}2)$ so $ x_B\equal{}3$ and $ y_A\equal{}\minus{}2$

Replace in the equation $ y\equal{}kx\plus{}b$ x and y first with $ x_A;y_A$ and $ x_B;y_B$ than  solve the system.  :)

umm...are you trying to challenge us or are we supposed to help you cause you dont understand how to do it.  And btw...systems are kinda unnecessary!

[hide="solution(easy way!)"]
um ok os what i do is find the slope of the line that passes throught these points, this is $ \frac{y_2 \minus{} y_1}{x_2 \minus{} x_1} \equal{} \frac{\minus{}5}{2} \equal{} k$ so now we have half of the equation because we know $ k$ so we plug the x and y coordinates into this equation and solve for $ b$ so we have $ \minus{}2 \equal{} \frac{\minus{}2}{5} \times 3 \plus{} b$ solving this gives us that $ b \equal{} \frac{\minus{}4}{5}$ so we have that $ \boxed{k \equal{} \frac{2}{5} \text{ and that } b \equal{} \frac{\minus{}4}{5}}$[/hide]

[quote="Elio (n)"]The equation of a line is $ y \equal{} kx \plus{} b$ , if in this line are include the 2 points $ A(1,3)$ and $ B(3,\minus{} 2)$ find $ k$ and $ b$.[/quote]

[hide="Method 2"]
Plug the points into y=kx+b
3=k+b
2=3k+b
Now solve
2k=-1
k=-1/2
b=3.5[/hide]

Moogra and zolojetto have different answers. Moogra's is $ \boxed {\text {k}\equal{}\minus{}\frac{1}{2} \text {and that b}\equal{}\frac{7}{2}}$, and zolojetto's is $ \boxed {\text {k}\equal{}\frac{2}{5} \text {and that b}\equal{}\minus{}\frac{4}{5}}$. Their equations are
$ y\equal{}\minus{}\frac{1}{2}x\plus{}\frac{7}{2}$
$ y\equal{}\frac{2}{5}x\minus{}\frac{4}{5}$.
Plugging in $ A$, only moogra's works, so his is the correct answer.

You can also use point-slope form for this.

First we find slope.

slope (or k)=$ \frac {3 \minus{} ( \minus{} 2)}{1 \minus{} 3}$
=$ \minus{} \frac {5}{2}$

Then we plug x and y into formula (plug in values of only one point into x' and y'):

$ y \minus{} y' \equal{} k(x \minus{} x')$

So if I choose (1,3) as our point:

$ y \minus{} 3 \equal{} \minus{} \frac {5}{2}(x \minus{} 1)$

$ y \minus{} 3 \equal{} \minus{} \frac {5}{2}x \plus{} \frac {5}{2}$

$ y \equal{} \minus{} \frac {5}{2}x \plus{} \frac {11}{2}$

@moogra You plugged in 2 instead of -2 so your 2nd set of points does not check out. Mine does.
@mrrubix So you were wrong too.

Aplly  $ \frac {x\minus{}x_A}{x_A\minus{}x_B}\equal{}\frac {y\minus{}y_A}{y_A\minus{}y_B}$
we have:
$ \frac {x\minus{}1}{\minus{}2}\equal{}\frac {y\minus{}3}{5}$ $ \Leftrightarrow$ $ y\equal{}\minus{}\frac {5}{2}x \plus{} \frac {11}{2}$
So that $ \boxed {\text {k}\equal{}\minus{}\frac{5}{2} \text  {and that b}\equal{}\frac{11}{2}}$.

fishythefish and thanhnam2902 are correct.

[hide="Solution"]The slope of the line ($ k$) is $ \frac {3 \minus{} \left( \minus{} 2\right)}{1 \minus{} 3} \equal{} \frac {5}{ \minus{} 2} \equal{} \minus{} \frac {5}{2}$.
Then, using the point-slope form of a line, we have:
\[ y \minus{} y_1 \equal{} k\left(x \minus{} x_1\right)
\]
Plugging in $ \left(x_1, y_1\right) \equal{} \left(1, 3\right)$, we have:
\[ y \minus{} 3 \equal{} \minus{} \frac {5}{2}\left(x \minus{} 1\right)
\]
Rearranging, we find that:
\[ y \equal{} \minus{} \frac {5}{2}x \plus{} \frac {11}{2}
\]
Then, $ \boxed{k \equal{} \minus{} \frac {5}{2} \text{ and } b \equal{} \frac {11}{2}}$.[/hide]

Ok you guys. When the first correct solution is posted, then you don't need to post it 3 more times.

[quote="Elio (n)"]The equation of a line is $ y \equal{} kx \plus{} b$ , if in this line are include the 2 points $ A(1;3)$ and $ B(3; \minus{} 2)$ find $ k$ and $ b$.[/quote]
[hide]k+b=3
3k+b=-2
-2k=5
k=-2.5
b=5.5[/hide]
Easy solution is dealt with a system of equations. :D


geometry
ratio
calculus
geometric sequence

A sequence of polygons is derived as follows. The first polygon is a regular hexagon of area 1. Thereafter each polygon is derived from its predecessor by joining two adjacent edge midpoints and cutting off the corner. Show that all the polygons have area greater than 1/3.

If the way I interpreted the problem is correct then I must admit that the area of the subsequent polygons are forming a Geometric progression with a ratio of 3/4 hence the area of the fifth polygon is less than 1/3....

Let the hexagon be ABCDEF. First note that any polygon formed in the described manner will be convex. Second, note that you can never entirely get rid of a side of the original hexagon, since we're connecting midpoints and so we only take away half of the remaining part of that edge. So, in the final polygon, there is a point originally on AB and a point originally on BC connected with a convex boundary. But any such boundary lies outside line AC. So, if we draw AC, BD, CF, etc. to form a smaller hexagon, any polygon formed from the cutting will be outside this smaller hexagon. But the smaller hexagon has area 1/3!

If I have interpreted correctly, polygons of the series aim to the circumference inscribed in the original hexagon. Doing some calculus, we know that this circumference has area 1/12*pi*√3 > 1/3.
Can this proof be correct?


probability
combinatorics proposed
combinatorics

[color=orange]dear friend  [/color]
[size=150][color=green]prove that [/color][/size]

rewrite the statement as $\sum_{k=1}^n (\frac{k}{n})(\frac{n\cdot(n-1)\cdots (n-k+1)}{n^k}) =1$.
 it is easy to interpret both sides as probabilities.consider an experiment of drawing(say different numbers from 1 to $n$) and making samples and we stop if there is a repeat in some draw .the probability that the experiment eventually stops, eventually of course is $1$(the RHS); on the other hand, the repeat could occur after $k$ earlier draws(i.e. on the $k+1^{th}$ draw), and the probability of this is precisely the expression on the LHS.


[quote="zaya_yc"]$ a,b > 0$     and     $ a^{9} \plus{} b^{9} \equal{} 2$             prove:

$ a^{3} \plus{} b^{3}\geq2ab$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176919


1 Show that $ \frac {1}{\sqrt {4n}} \le \left(\frac {1}{2}\right)\left(\frac {3}{4}\right) ...\left(\frac {2n \minus{} 1}{2n}\right) < \frac {1}{\sqrt {2n}}$

Let $ x_n\equal{} \left(\frac {1}{2}\right)\left(\frac {3}{4}\right) ...\left(\frac {2n \minus{} 1}{2n}\right)\equal{}\frac{(2n)!}{(2^n n!)^2}\equal{}\frac{\theta_{2n}}{\theta_n^2\sqrt{\pi n}},$
were $ 1\le \theta_n\equal{}\frac{n!e^n}{\sqrt{2\pi n}n^n}<1\plus{}\frac{1}{12n}.$ For $ n\equal{}1$ it is easy to chek.
When $ n\ge 2$ it is equvalent to $ (1\plus{}\frac{1}{12n})^4<\frac{4}{\pi}, (1\plus{}\frac{1}{6n})<\sqrt{\frac{\pi}{2}}.$


How do you put an image as the background of a blog? I see there is an option under the blog preferences but when I enter a URL the image does not appear as the background.  :oops: Are there only specific files that can be used? What about picture size?

I tried CSS, but it didn't work.  Is it a bug or am I doing something wrong?

I had this trouble too. It should go like:

url(The url)

Sweet! Thanks hunter34.  :)

Though my blog still needs a lot more modifications.  :lol:


geometry
rhombus
trigonometry
geometry unsolved

An isosceles triangle $ ABC$ , with $ AB \equal{} AC$ , has a point $ D$ on $ AC$ such that $ AD \equal{} BC$ . Find angle $ B\hat{D}C$ , if $ B\hat{A}C \equal{} 20^\circ$ .

This is basic geometry, no crazy theorems please :)

Thanks in advance for the help.

And sorry if this is in the wrong section, it seems pretty difficult, that's why I posted it here. I apologize if I am mistaken.

This is Langrey's problem. In Japan, it is called Flanklin's kite.

[url]http://agutie.homestead.com/FiLEs/LangleyProblem.html[/url]

Construct the equilateral triangle BCM, with M inside the triangle ABC, see that triangles ABD and BAM are congruent, (s.a.s.), but AM is bisector of the angle BAC, hence angle ABD = 10 degs and angle BDC = 30 degs.

Best regards,
sunken rock

Thanks for the help guys

I don't think that is langley's problem, different bounds and different angle, though it is on the same triangle.

sunken, how did you find this solution, what caused you to think of constructing an equilateral triangle there? What a beautiful solution, all you use is s.a.s. I can hardly believe how easy you've made it.
Gotta love geometry :)

[quote="Langrey or Franklin's kite"] [color=darkred]Given are $ \triangle ABC$ with $ AB \equal{} AC$ , $ A \equal{} 20^{\circ}$ and a point $ D\in (AC)$ so that $ AD \equal{} BC$ . Find $ m(\widehat {BDC})$ .[/color] [/quote]

[color=darkblue][b][u]Proof I[/u].[/b] Construct the equilateral triangle $ ADE$ so that the line $ AC$ separates $ E$ , $ B$ .Observe that $ ABE\equiv BAC$

and $ ADB\equiv EDB$ . Thus $ m(\widehat {BDE}) \equal{} 150^{\circ}$ and $ m(\widehat{BDC}) \equal{} m(\widehat {BDE}) \minus{} 120^{\circ}$ , i.e. $ m(\widehat{BDC}) \equal{} 30^{\circ}$ .[/color]

You guys really are good, thank you Virgil Nicula.

Never thought this problem could have so many solutions.

[color=darkblue][b][u]Proof II ("slicing")[/u].[/b] Construct the points $ \left\|\begin{array}{ccc} M\in (AC) & ; & m(\widehat {CBM}) \equal{} 20^{\circ} \\
 \\
N\in (AB) & ; & m(\widehat {BMN}) \equal{} 60^{\circ} \\
 \\
D'\in (AC) & ; & m(\widehat {AND'}) \equal{} 20^{\circ}\end{array}\right\|$ . Observe easily that

$ AD \equal{} BC \equal{} BM \equal{} BN \equal{} MN \equal{} ND' \equal{} AD'$ , i.e. $ AD' \equal{} AD$ what means $ D'\equiv D$ .

Thus, from $ NB \equal{} ND$ and $ m(\widehat {BND}) \equal{} 160^{\circ}$ obtain $ m(\widehat{BDN}) \equal{} 10^{\circ}$ , i.e. $ m(\widehat {BDC}) \equal{} 30^{\circ}$ .[/color]

[color=darkblue][b][u]Proof III ("slicing")[/u].[/b] Construct the rhombus $ ABEF$ so that $ E\in BC$ and the line $ AC$ separates $ B$ , $ E$ . Observe easily that

$ AC \equal{} AB \equal{} AF \equal{} FE \equal{} FD \equal{} DE \equal{} BE$ and $ CE \equal{} CD$ $ \implies$ $ ED \equal{} EB$ and $ m(\widehat {BED}) \equal{} 40^{\circ}$ $ \implies$ $ m(\widehat{BDC}) \equal{} 30^{\circ}$ .[/color]

[color=darkred]Maybe someone'll realize the draws for both "slicing" proofs. [b]Thanks.[/b][/color]

[color=darkblue][b][u]Proof IV (trig.)[/u].[/b] Denote $ m(\widehat {BDC}) \equal{} x$ . Thus, $ AD \equal{} BC$ $ \implies$ $ \frac {\sin \widehat {ADB}}{\sin \widehat {ABD}} \equal{} \frac {AB}{AD} \equal{} \frac {AB}{BC} \equal{} \frac {\sin\widehat {ACB}}{\sin\widehat {BAC}}$ $ \implies$

$ \frac {\sin x}{\sin (x \minus{} 20)} \equal{} \frac {\sin 80}{\sin 20}$ $ \implies$ $ \boxed {\ \begin{array}{c} \sin 20\sin x \equal{} \cos 10\sin (x \minus{} 20) \\
\ \cos (x \minus{} 20) \minus{} \cos (x \plus{} 20) \equal{} \sin (x \minus{} 10) \plus{} \sin (x \minus{} 30) \\
\ \cos (x \minus{} 20) \minus{} [\cos (x \plus{} 20) \plus{} \cos (100 \minus{} x)] \plus{} \sin (30 \minus{} x) \equal{} 0 \\
\ \cos (x \minus{} 20) \minus{} 2\cos 60\cos (40 \minus{} x) \plus{} \sin (30 \minus{} x) \equal{} 0 \\
\ \cos (x \minus{} 20) \minus{} \cos (40 \minus{} x) \plus{} \sin (30 \minus{} x) \equal{} 0 \\
\ 2\sin 10\sin (30 \minus{} x) \plus{} \sin (30 \minus{} x) \equal{} 0 \\
\ \sin (30 \minus{} x)[2\sin 10 \plus{} 1] \equal{} 0 \\
\ \boxed {x \equal{} 30}\end{array}\ }$[/color]

This is Japanese famous geometrist, Seimiya Toshio's paper.

Regrettably I can't get the paper.

[url]http://www.asahi-net.or.jp/~NJ7H-KTR/e_mokuji00-01.html[/url]

@Virgil:

Pls, find attached the needed drawing, I hope it doesn't bother you, nor other mathlinkers as being made together, this way it may easily generate new problems.

Best regards,
S. Fulger.


The number of contests besides math contests you have/will participate in this year.

oops

The message is too small. Please make the message longer before submitting.

USA Computing Olympiad
(regional, state, not national) National History Day
Science Fair (school/regional/state/ISEF)
USA Bio Oly
50 million essay contests ;)

Definitely 5, just off the top of my head.

There's also a bunch of other science contests for middle school too, like science bowl, science olympiad, science fair and exploravision.

I don't think you can do USABO in mid-school though. Doesn't matter

Various Debate and Scholar's bowl tournaments are basically it for me  (though they're all over now) so I guess i'll call that two...

DOE Science Bowl
USNCO exam
F=ma exam

meh

Academic Bowl
Game show called The Challenge

Does it have to be academic?

Oh, wow, that's a toughie. :roll:

Ummm, scibowl, scioly, some karate and tennis and other sports stuff.....
used to do lots of music and chess......

Ummm, how about [size=150]LOTS OF MATH CONTESTS[/size]

[size=59]but not too many other things[/size]

Can these contests have math as a part of it but not all of it?

Ok, so here's my revised list:

Academic Bowl
Game show called The Challenge:
The Challenge Qualifiers
The Challenge Connecticut round of 16
The Challenge Ct quarter finals
The Challenge Ct semi finals
(hopefully ct finals and then tri state quarters, semis, and finals)
Soccer tournament
FCIAC indoor track
FCIAC track (hopefully)
tons of track meets
soccer games

Wait, is the AIME in a math competition?  :wink:


I have a quiz tomorrow (hehe,,, again) and I want to see how I'm doing now...

1. [tex]8^{log_8 3}+^{log_8 2} [/tex]
2. [tex]9^{log_9 12}+^{log_9 4}[/tex]
3. [tex]log_2 10 - log_2 t = log_2 2[/tex]

If log_2 3 = 1.585 and log_2 7 = 2.807
4. [tex]log_2 27[/tex]
5. [tex]log_2 36[/tex]
6. [tex]log_2 48[/tex]
7. [tex]log_2 \frac {36}{49}[/tex]
8. [tex]log_2 3 + log_2 7 = log_2 x[/tex]
9. [tex]log_3 14 + log_3 y = log_3 42[/tex]
10. [tex]log_{10} m = \frac {1}{2} log_{10} 81[/tex]
11. [tex]log_{10} 7 + log_{10} (n-2) = log_{10} 6n [/tex]
12. [tex]log_{10} (m+3) - log_{10} m = log_{10} 4[/tex]
13. [tex]3 log_5 x - log_5 4 = log_5 16[/tex]

Find the log of a number and round to four decimal places and state the characteristic.

14. 53.7
15. 2.25
16. 4.322

Find the antilog of a number
17. 0.2586
18. 1.0024
19. -2.0112
20. 8.1342 - 10
21. 4.9243

Find the log of a number and state the characteristic and mantissa.
22. 13.54
23. 10^-7
24. 0.265
25. 6678

So do you want answers?  Btw I am unfamiliar with the characteristics of a log and the mantissa.  So I would appreciate if you told me what those are.

Also, for the first two are the second logs supposed to be raised or is that an error?

Some more TeX hints:

1. Log has a special symbol, \log. This removes the italics that is present when you directly type in log.

2. The first line, for example, should be retyped as

[tex]8^{\log_8 3 + \log_8 2}[/tex]

Um... Like look on some questions on here (it would be awesome if someone can look on all) and see if I know what I'm doing or not... I think I'm doing pretty good though...

I just scanned through it, and #1 and #2 are not correct. Recall that

[tex]\log_a b + \log_a c = \log_a (bc)[/tex].


Find all $ a$, that the equation
$ (a\minus{}1)^2 \cdot x^4 \plus{} (a^2\minus{}a)\cdot x^3 \plus{} 3\cdot x \minus{} 1 \equal{} 0$
has only one solution... and find this solution.

[u]Case1:[/u]$ a \equal{} 1$ then
$ x \equal{} \frac {1}{3}$
[u]Case 2[/u] $ a$ is different from 1.
We has $ f(0) < 0$
$ \lim_{x\to \plus{} \infty} \equal{} \plus{} \infty$
$ \lim_{x\to\ \minus{} \infty} \equal{} \plus{} \infty$
So it  solution on $ x_0\in( \minus{} \infty,0),x_1\in (0, \plus{} \infty)$
So $ a \equal{} 1,x \equal{} \frac {1}{3}$


Hi-

I haven't been eligible for this competition for a few decades now, but my nephew is in high school and I was curious enough to work out the first set of problems from his problem set.

I didn't find the official solution to the "groovy numbers" problem really satisfying.  While certainly correct, it relies on algebraic manipulation with no clear motivation.  I've attached an alternate solution in case anyone is interested, which proves a more general result and derives the groovy result as a consequence.  (As an added benefit, this solution has relatively little algebraic manipulation.)

Enjoy,
Dan
(IMO, 1978)

very nicely done.  I actually had the idea of trying to show $ o^2\minus{}e^2 \equal{} 1$ but then really botched it... tried to brute force it with odds.  Anyway, I think the idea you are getting at is very similar to the official solution, but it is just expressed in a very different manner.

Ah, now I see what my solution should have looked like (I had the same idea, only with $ a\equal{}n\plus{}1,b\equal{}n$, but [b]much[/b] less clearly written).


geometry
3D geometry
sphere
geometry solved

A sphere is touched by all the four sides of a (space) quadrilateral. Prove that all the four touching points are in the same plane.

I've seen this in like hundreds of places! :)

We just use Menelaus in space: Let $M,\ N,\ P,\ Q$ be points on the sides $AB,\ BC,\ CD,\ DA$ of a space quadrilateral respectively. Then $M,\ N,\ P,\ Q$ are coplanar iff $\frac{MA}{MB}\cdot \frac{NB}{NC}\cdot \frac{PC}{PD}\cdot \frac{QD}{QA}=1$. We apply this to our quadrialteral considering that $AM=AQ,\ BN=BM,\ CP=CN,\ DQ=DP$.


Let $ P$ be a convex $ n$ polygon each of which sides and diagnoals is colored with one of $ n$ distinct colors. For which $ n$ does: there exists a coloring method such that for any three of $ n$ colors, we can always find one triangle whose vertices is of $ P$' and whose sides is colored by the three colors respectively.

For all odd number $ n$ that $ n>1$

First of all since there are $ C^n_3$ ways to choose 3 among $ n$ colors, and $ C^n_3$ ways to choose 3 vertices to form a triangle, so if the question's condition is fulfilled, all the triangles should have different color combination among each other. (a 1-1 correspondence)

(From now on i shall include "sides" in "diagonals" for convenience.)

As each color combination is used in exactly one triangle, for each color, there should be $ C^{n\minus{}1}_2$ triangle which has one side in this color.

Now there are $ C^n_2$ diagonals in the polygon and for each diagonal, there are $ n\minus{}2$ triangles with this diagonal as side. If $ n$ is even, $ (C^n_2)/n$ is not an integer, so there exist at least one color where the number of diagonals with this color is less than $ (C^n_2)/n$, i.e. there are less than $ (C^n_2) \times (n\minus{}2)/n \equal{} C^{n\minus{}1}_2$ triangles with a side in this color, contradiction.

So $ n$ is odd.

Now gives a construction method for all $ n$ is odd.

As the orientation of the vertices doesn't really matters, we assume that the polygon is a regular $ n$ polygon.

First color the $ n$ sides of the polygon in the $ n$ distinct colors.

Then for each side, color those diagonals that are parallel to this side into the same color.

In this way, for each color, there are $ (C^n_2)/n$ diagonals colored in this color, notice that each of these diagonals are of different length...(i)

Besides, for any two triangles with all vertices in $ P$, I shall prove that they should have different color combination.

Suppose the contrary, i.e. they have exactly the same three colors as their sides.

For two of the 3 colors, the corresponding sides in the two triangles should have the same included angle since two sides are of the same color iff they are parallel.

However, notice that the third side of each triangle, i.e. the side opposite to that included angle, should have the same length. (Since a regular n-gon is circumscribed by a circle, in a circle, same angle correspond to same side)

But this is obviously impossible since they have the same color, contradict to (i).

This completes the proof.

First of all, $ n$ cannot be even. [hide="Why?"]For if any two distinct triangles get different color triples, there must be exactly $ {n - 1 \choose 2}$ triangles that get color $ i$, and since every segment is a part of exactly $ n - 2$ triangles, there must be $ \frac{n - 1}{2}$ segments of color $ i$. This is an integer only if $ n$ is odd.[/hide]

For odd $ n$, consider the following coloring: label the vertices with $ 1, \ldots, n$ and color segment $ ij$ by the $ (i + j \bmod n)^{\mathrm{th}}$ color.

[hide="Why this works?"]
If for two triangles $ i_1, j_1, k_1$ and $ i_2, j_2, k_2$ we had $ i_1 + j_1 \&\equiv_n i_2 + j_2$ and $ j_1 + k_1 \&\equiv_n j_2 + k_2$ and $ k_1 + i_1 \&\equiv_n k_2 + i_2$ then we would have had $ (i_1 + j_1) + (j_1 + k_1) - (i_1 + k_1) \equiv_n (i_2 + j_2) + (j_2 + k_2) - (i_2 + k_2)$ and hence $ n \,|\, 2(j_1 - j_2)$.

As $ n$ is odd, also $ n \,|\, j_1 - j_2$, and as $ |j_1 - j_2| < n$, it must be $ j_1 = j_2$. Similarly $ i_1 = i_2$ and $ k_1 = k_2$, so the triangles were not different after all and the coloring is as desired.
[/hide]

Easy problem!
at first step we will prove that the number of vertexes of $G$(which is $n$) can not be even. suppose  the number of vertexes of $G$ is even. the number of edges which colored by color $i$ is less than $n/2$(which is a  contradiction)  since if its equal $n/2$ , we have $n/2*(n-2)$ triangles such that one of the edges of this triangle is $i$.in the other hand the number of such triangles is at most $(n-2)/2*(n-1)$ which is a contradiction.

so we easy to see if $n$ is odd so we have such graph(about this segment we can help to [hide]Vizing's theorem__> http://en.wikipedia.org/wiki/Vizing's_theorem[/hide])

An equivalent formulation of this problem appeared in Hong Kong 2002. See [url=https://artofproblemsolving.com/community/c6h127860p725458]here[/url].

tfw only the construction part of the problem is slightly nontrivial :oops:

We claim only odd $n$ work.

Consider a $K_n$ with edges colored in $n$ colors such that for any $3$ colors, there exists a triangle whose edges are of the three colors. Observe that for any three colors, there exists a unique triangle whose edges are of the three colors. For every color $c$, there are exactly $\binom{n-1}{2}$ triangles containing an edge of color $c$, and each edge of color $c$ is contained in exactly $n-2$ triangles. Thus there are exactly $\frac{n-1}{2}$ edges of each color so $n$ is odd.

Now the construction. Let $n=2k+1$. Consider the $K_n$ with vertices $v_1,v_2\ldots,v_{2k+1}$. Edges $v_{i+j}v_{i-j-1}$ where $0\leq j\leq k-1$ are colored with color $i$ (indices taken cyclically). Note that edge $v_{i+j}v_{i-j-1}$ is the same as edge $v_{i+(2k-j)}v_{i-(2k-j)-1}$. For any three colors $a,b,c$, we need to find a unique solution to
\begin{align*}
a+x&\equiv b-y-1\pmod{2k+1}\\
b+y&\equiv c-z-1\pmod{2k+1}\\
c+z&\equiv a-x-1\pmod{2k+1}
\end{align*}
(other systems of congruences will yield the same triangle). The unique solution is
\begin{align*}
x&\equiv b-c+k\pmod{2k+1}\\
y&\equiv c-a+k\pmod{2k+1}\\
z&\equiv a-b+k\pmod{2k+1}
\end{align*}
which corresponds to a unique triangle.

We will see the problem as a graph :P
Let $C_1, C_2, ..., C_n$ the n colors
Let $k$ the number of edges color $C_1$
By the condition of the graph: there are at least ${n \choose 3}$ triangles with different colorationes, but we have exactly ${n \choose 3}$ triangles in the graph $\rightarrow$ Each coloration appears exactly one time and each triangle has a coloration of 3 different colors ....$(F_1)$
Let $A_1 =$ {$w_1, w_2$}, $A_2 =$ {$w_3, w_4$}, ...., $A_k =$ {$w_{2k-1}, w_{2k}$} the edges which are color $C_1$
By $(F_1)$ : $A_1, A_2, ..., A_k$ are disjoint two to two ....$(F_2)$
Let $T$ the set of triangles which have exactly one edge which is color $C_1$
By $(F_2)$: Every $A_i$ has $n-2$ edges which can make a triangle of  $T$, so $\lvert T \rvert=k(n-2)$ ....$(i)$
By $(F_1)$: Each triangle of $T$ has a different coloration that depends by the 2 edges which are not color $C_1$ ($n-1$ possible colors), so $\lvert T \rvert= {n-1 \choose 2}$ ....$(ii)$
By $(i)$ and $(ii)$: $k(n-2) = \frac{(n-1)(n-2)}{2} \rightarrow k = \frac{n-1}{2} \rightarrow n-1$ is even $\rightarrow$ [b]$n$ is odd[/b]
[b]For n odd the problem works with this example:[/b]
Let $v_1, v_2, ..., v_n$ the vertices of the graph
We will work with the subscript of the vertices and the colors in module $n$
Color the edge $v_p v_q$ of the color $C_{\frac{p+q}{2}}$ ($\frac{p+q}{2}$ is always possible in module $n$ because n is odd)
[b]Proof of that it works:[/b]
For each three colors $C_{x_1}, C_{x_2}$ and $C_{x_3}$:
The numbers $a \equiv x_1+x_3-x_2\pmod{n}, b \equiv x_1+x_3-x_2\pmod{n}$ and $c \equiv x_1+x_3-x_2\pmod{n}$ satisfy
$\frac{a+b}{2} \equiv x_1\pmod{n}$
$\frac{b+c}{2} \equiv x_2\pmod{n}$
$\frac{c+a}{2} \equiv x_3\pmod{n}$
Then the vertices $v_a, v_b$ and $v_c$ form a triangle with edges which are of $C_{x_1}, C_{x_2}$ and $C_{x_3}$ colors
Therefore, the answer is n odd

The number of distinguishable triangles with edges all different color is $\tbinom{n}{3}$ and so is the number of $C_3$ subgraphs in a $K_n$.
Also, every edge are included in a same amount of triangles, so each color must be used on the same amount of edges. This way we know that even $n$ won't work.
Now we show that it is possible for odd $n$. Label the polygon as $A_1A_2\cdots A_n$. Assign one of the color to each of the $A_i$s. Now if $A_i$ is colored as color X, the all edges in the form $A_{i-j}A_{i+j}$ where $1\leq j\leq \frac{n-1}{2}$ would be colored as color X (indices taken mod $n$).