A square piece of paper, 4 inches on a side, is folded in half
vertically.  Both layers are then cut in half parallel to the fold.
Three new rectangles are formed, a large one and two small ones.
What is the ratio of the perimeter of one of the small rectangles to
the perimeter of the large rectangle?  Express your answer as a common fraction.

we know that both of them has 4 as one side length. The other is more difficult to find. One rectangle has width of 1 and the other has width of 2 since a fold connected them. If you don't see what I mean do it yourself. One rectangle has perimeter 4+4+1+1=10 and the other has 4+4+2+2=12. The ratio is 10/12=5/6

[hide=Hint]Folding the piece of paper in half, we get two rectangles with side lengths 2-4-2-4. We are then cutting off of the shape with a line parallel to the fold, a parallel cut means that we cut a line with the same slope as the fold, so when we cut the shape in half parallel to the fold, we cut it such that the long side of our rectangle is preserved but the short side is cut in half. Then, we have three rectangles, one with side 2-4-2-4 and the other two with sides 1-4-1-4. Then find the perimeter of both different rectangles and then the ratio.


Let $ r$ be an odd number, and $ f(r)$ denote the sum of the positive divisors of $ r$. Let $ q$ be a positive integer.

If $ (2^{q\plus{}1}\minus{}1)f(r)\equal{}2^{q\plus{}1}r\plus{}1$, prove that $ r\equal{}3(\mod 4)$

[hide="contradiction? what is wrong here"]
Let $ r=p_1^{f_1}p_2^{f_2}\cdots p_n^{f_n}$. Taking the equation $ \pmod 2$, we have:
\begin{eqnarray*}
f(r)&\equiv& 1\pmod2\\
&\equiv&(1+p_1^1+p_1^2+\cdots+p_1^{f_1})\cdots(1+p_n^1+p_n^2+\cdots+p_n^{f_n})\pmod2\\
&\equiv&\underbrace{(1+1+1+\cdots)}_{f_1+1\text{ times}}\cdots\underbrace{(1+1+1+\cdots)}_{f_n+1\text{ times}}\pmod2\\
&\equiv&(f_1+1)(f_2+1)\cdots(f_n+1)\pmod2\\
\end{eqnarray*}
If any one term of this product is even, then the whole thing will be, so for any $ k\le n\in\mathbb{Z}^+$, $ f_k$ will be even. Now, $ r$ must be a perfect square and therefore $ r\equiv1\pmod4$, which is a contradiction.
[/hide]

Ah, actually you are totally right. I'm a bit mistaken in formulating the problem. In fact, no $ r$ satisfies that equation. You have proved that $ r \equal{} 1\pmod 4$. Now, if you can prove that $ r \equal{} 3\pmod 4$, you'll get that no $ r$ satisfies that equation. So try it!  :)


What do you call an algebra formula that gives roots and has bite to it?


Solve:

How many ways it has to write number 20 like exact sum of three prime numbers?

Clearly 2 must be one of the primes.

So we must have a+b=18, a,b E P

This gives (5,13) , (7,11)

So two ways.

I'm not sure this problem is pre-olympiad level.


The length of each edge of a cube is increased by $ 20\%$. What is the percent increase in the surface area of the cube?

A well known fact is that area is proportional to the square of a length that defines it. If the side length of the cube is increased by $ 20$ percent, it is the same as the side length multiplied by $ 1.2$.

$ 1.2^2 \equal{} 1.44 \equal{} 144$ percent. Now we subtract $ 100$ percent to find the increase as a percent, so our answer is $ \boxed{44}$ percent.

Another way is to work with easy numbers. Say the edge of the cube has a length of 10. That would mean the surface area is $ 10 \times 10 \times 6$ or $ 600$.

10 increased by 20% is 12. The surface area of such a cube is $ 12 \times 12 \times 6 \equal{} 864$. $ 864 \div 600 \equal{} 1.44 \minus{} 1 \equal{} \boxed{44\%}$.


A scalene triangle with a 35 degree angle and side lengths 33 and 17 has triangles congruent to it pasted on its sides so that touching sides are congruent. To the new triangles, more triangles are pasted on in the same fasion. After one billion triangles are pasted, the smallest circle that contains all of them is drawn. Compute the smallest possible area of this circle.

Hide solutions.

Feel free to post any partial answers (got somewhere, but not finished).

You have also posted the problem here: http://www.mathlinks.ro/Forum/viewtopic.php?t=89314

Please delete the other one and avoid double-posting next time.


analytic geometry
geometry
geometric transformation
reflection

What is the sum of the coordinates of the reflection across the $ x$-axis of the point with coordinates $ (\minus{}3,4)$?

it will become $ (\minus{}3,\minus{}4)$, which has a sum of $ \boxed{\minus{}7}$.


The no. 139854276 contains all the digits from 1 to 9, and is a perfect square of 11826. What other nine digit numbers, with all their digits, are square ?
 :coolspeak:

[quote="Anandiscool"]Can you give me solution of any one of this answers coz I want to know the method, I have tried it but couldn't find the answer.[/quote]

Brute force : I computed all the squares of numbers from $ 10000$ to $ [\sqrt{1000000000}]$ and just kept those which fit the requirements.

... with some help from standard spreadsheet.  :wink: ...

And because of this last remark, this is surely not the method your contest / teacher was waiting for. If you get their solution one day, dont hesitate to post it, thanks.


Find the maximum anh minximum of:

$ S \equal{} \sin ^n x \plus{}\cos ^n x$ with $ n$ is natural number.

Help me!!!

hello, let ${ f(x)=\sin^n(x)\cos^n(x)=\frac{1}{2^n}\left(\sin(2x)\right)^n}$, then is
$ |f(x)|\le \frac{1}{2^n}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, let ${ f(x) = \sin^n(x)\cos^n(x) = \frac {1}{2^n}\left(\sin(2x)\right)^n}$, then is
$ |f(x)|\le \frac {1}{2^n}$.
Sonnhard.[/quote]
I'm sorry, the right problem is $ f(x) = \sin^n x + \cos^n x$

[hide="hint"]$ f'(x)\equal{}n \sin x \cos x (\sin^{n\minus{}2} x \minus{} \cos^{n\minus{}2} x)
\equal{} \frac{n}{2} \sin 2x (\sin^{n\minus{}2} x \minus{} \cos^{n\minus{}2} x)$[/hide]

The maximum is easily found by Jensen's as $ \frac {2}{(\sqrt 2)^n} \equal{} 2^{1 \minus{} \frac {n}{2}}$

edit: thats obviously wrong  :oops:

For the record, $ \sin ^n x\plus{} \cos ^n x$ does not have a consistent concavity over the interval we are considering, so Jensen's cannot be used.

Please help me to solve this problem!!! I cannot find solution!! :(

hello, by computing the first derivative you have $ f^'(x)=(\sin(x))^{n-1}n\cos(x)-n(\cos(x))^{n-1}\sin(x)$ you must solve the equation
$ (\sin(x))^{n-2}=(\cos(x))^{n-2}$.
Now you have to consider the cases $ n=1$ and $ n=2$.
For $ n>2$ and $ n$ odd you have to solve
$ \sin(x)=\cos(x) \Leftrightarrow \tan(x)=1 \Leftrightarrow x=\frac{\pi}{4}+k_1\pi$.
For $ n>2$ and $ n$ even we have
$ \tan(x)=\pm1 \Leftrightarrow x=\pm\frac{\pi}{4}+k_2\pi$
with $ k_1,k_2 \in \mathbb{Z}$.
Sonnhard.

Well as far as the maximum is concerned, we can say:

for n = 1, it is $ \sqrt 2$
for n = 2, it is a constant 1

for n>2 $ \sin^n x \plus{} \cos^n x \le \sin^2 x \plus{} \cos^2 x \equal{} 1$

Maybe we could hence condense for n>2,

$ |\sin^n x \plus{} \cos^n x| \le |\sin^n x| \plus{} |\cos^n x| \le \sin^2 x \plus{} \cos^2 x \equal{} 1$

Hence $ \minus{} 1 \le \sin^n x \plus{} \cos^n x \le 1$

For n = 1 $ \minus{}\sqrt 2 \le \sin x \plus{} \cos x \le \sqrt 2$

For n = 2 it is just 1


calculus integration geometry rectangle function algebra domain

[url=http://www.google.com]have fun![/url]

I suppose I could summarize though, an integral is basically a Riemann sum, which in turn is the sum of infinitesimal rectangles of equivalent width that make up the area under a function's curve to the y axis from a lower x limit to a higher x limit. The fundamental theorem of calculus also manages to show that integrals are anti-derivatives.

They don't! You have to do them your self! :) The principal rule of integration could be that a function $ F$ is called an anti-derivate (or an integral) of $ f$ if the following is true:

$ F'(x)\equal{}f(x),\,\forall x \in \mathbb{D}$

where $ \mathbb{D}$ is the domain of $ f$.

There are so called defined integrals (that have appropriate boundaries) and they can be solved as:

If the for some $ F$ and $ f$ the above is true then:

$ \int^b_a f(x)\,dx \equal{}\lim_{t\minus{}>b}F(t)\minus{}\lim_{u\minus{}>a}F(u)$

The limits are used for situations when there is no actual value in those points, but there is a limit! There are many techniques that may help you to solve integrals and can be found all over the web. You should probably (for some extra knowledge) see the topics of differential equations, Laplace transform, Fourier transform/integral/series and Taylor series. All of these are in close connection with integrals and derivations. Hope I've managed to help.


Call a k-digit positive integer a [i]hyperprime[/i] if all its segments consisting of $ 1, 2, ..., k$ consecutive digits are prime. Find all hyperprimes.

Since each digit of a hyperprime must be prime, a hyperprime has only the digits $ 2,3,5,7$. Each of these as a one digit number works. For a two digit number of higher, $ 2$ or $ 5$ can only exist as the leftmost digit of the number. The part of the number after the first digit is a hyperprime as well, so first we will examine just the hyperprimes consisting of only $ 3$'s and $ 7$'s. We cannot have two consecutive $ 3$'s or two consecutive $ 7$'s, since neither $ 33$ nor $ 77$ is prime, so this number must alternate $ 3$'s and $ 7$'s. Also, we cannot have more than one $ 7$ because then the segment $ 737 \equal{} 11\cdot 67$ is not prime. Thus the possible hyperprimes using only $ 3$'s and $ 7$'s are:

$ 3$
$ 7$
$ 37$
$ 73$
$ 373$.

We can check easily that these are indeed hyperprimes. Now we consider adding a $ 2$ or $ 5$ before either of these. $ 23$ and $ 53$ are both hyperprimes, but adding a $ 2$ or a $ 5$ to any of the other hyperprimes listed above would make the sum of its digits divisible by $ 3$ (so the new number itself will be divisible by $ 3$, and hence not prime.) Thus all the hyperprimes are:

$ 2,3,5,7,23,37,53,73,373$.


Determine all functions f:R->R such that the set:
\[\{\frac{f(x)}{x}|x\in \mathbb R , x\ne 0\}\]
is finite and
\[f(x-1-f(x))=f(x)-x-1\]for all $x \in \mathbb R$

Let $f$ be such a function. Let $E= \{x-f(x), x \in \mathbb {R} \}.$

First, we prove that $E$ is finite :
Suppose, for a contradiction, that $E$ is infinite. Then, there exists an infinite number of real  $k \neq 1$ such that for each, there exists $x_k$ for which $x_k - f(x_k)= k.$ For each of these $k$, we then have :
$\frac {f(k-1) } {k-1} = \frac {f(x_k-1-f(x_k))} {k-1} = \frac {f(x_k) - 1 - x_k} {k-1} = -1 - \frac 2 {k-1}.$
Thus, $\{ \frac {f(x)} x , x \in \mathbb {R} \}$ is infinite. A contradiction.

Therefore, we may consider $a\in \mathbb {R}$ such that $|a-f(a)|$ is maximal. Let $b = a-1-f(a).$ We have $b-f(b) = 2(a-f(a)).$ The maximality of $a-f(a)$ leads to $a-f(a)=0$. Then, using maximality again, we deduce that for all $x$, we have $f(x)=x.$

Conversely, $f(x)=x$ is clearly a solution of the problem.

Pierre.


vector
algebra
system of equations
Pythagorean Theorem
geometry

is there a misprint there?  in the picture there is one $M_2$ and two $M_1$'s but in your question you wrote $M_3$ > $M_2$.

Should the larger one be M3?
Just assuming... there's not a reason for larger objects to have greater mass...

Sorry, my mistake.

[b][size=150]The object connecting ${M}_2$ is ${M}_3$[/size]

EDIT: How did i double post?? Mod, please delete this one.

Sorry, my mistake.

[b][size=150]The object connecting $\text{M}_2$ is $\text{M}_3$[/size][/b]

[quote="shobber"]Subject:[b]Newton's Second Law.[/b]

   As shown in the figure, all surfaces are frictionless. $\text{M}_3 > \text{M}_2$. Find the acceleration of $\text{M}_1$[/quote]

[hide="See solution here"]

Taking the correction of the diagram into account, there are 3 equations and 3 unknowns to be solved simultaneously.

$T-m_3g=m_3a'\\
\\
T=m_2a'\\
\\
m_2a'+M_1a=0\text{     Newton's third law}\\
\\
$

Solving the system of equations for $a$:

$a=\frac{\frac{m_3}{M_1}g}{\frac{m_3}{m_2}-1}$
[/hide]

[quote="shobber"]well, your answer is not correct.  :(

Beware! My physics challenges are real challenges, they are not that easy to be 'killed'.[/quote]

Cursed sign errors. You're correct, I was mistaken. I shall leave it to my readers to discover where.

[quote="probability1.01"][quote="shobber"]Subject:[b]Newton's Second Law.[/b]

   As shown in the figure, all surfaces are frictionless. $\text{M}_3 > \text{M}_2$. Find the acceleration of $\text{M}_1$[/quote]

OK, here's my try

Force on the masses on the string thing is $M_1g$, which accelerates it at $\frac{M_1g}{M_1+M_2}$.  Now that means the tension in the string is $M_1 \Big( g - \frac{M_1g}{M_1+M_2} \Big)$.  Since there are perpendicular components of this on the pulley, 3rd law gives horizontal force on both systems of also $M_1 \Big( g - \frac{M_1g}{M_1+M_2} \Big)$.  And acceleration is then that quantity divided by $M_3$.  Then Pythagorean theorem gives...

$\sqrt{\frac{M_1^2g^2}{(M_1+M_2)^2}+ \Big( \frac{M_1^2g+M_1M_2g}{M_1+M_2} - \frac{M_1^2g}{M_1M_3+M_2M_3} \Big)^2}$

$=\sqrt{\frac{M_1^2g^2}{(M_1+M_2)^2}+\Big(\frac{M_1M_2g}{M_1M_3+M_2M_3}\Big)^2}$

$=\sqrt{\frac{M_1^2g^2M_3^2 + M_1^2M_2^2g^2}{(M_1M_3+M_2M_3)^2}}$

$=M_1g \sqrt{\frac{M_3^2+M_2^2}{(M_1M_3+M_2M_3)^2}}$

Well I'm not great at physics and this looks wrong :(.[/quote]

Me thinks, ye are trying too hard.  (Think 3 linear equations, and 3 unknowns. It's not as difficult as shobber believes it is.)

[quote="probability1.01"]Where does your first equation come from Dr. No?[/quote]

Ah, you spotted it. My first equation is Newton's 2nd law applied to block $m_3$.

There is tension in the chord connecting $m_3$ to $m_2$; I called it $T$.

$T$ acts in an upward direction on $m_3$. The weight of $m_3$, equal to $m_3g$ acts in a downward direction. Leading to a net force on $m_3$ equal to $m_3a'$ in a [b]downward direction[/b].  That's where I muffed it. I should have had a minus sign on the expression on the right-hand-side.

$T-m_3g=-m_3a'$

[quote="shobber"]Dr.No, i don't quite understand your third equation. Can you explain that??[/quote]

Sure, no problem. My third equation is Newton's 3rd law of motion applied to the system as a whole.

Newton's 3rd law: [i]For every action there is an equal and opposite reaction.[/i]

There are no external horizontal forces applied to the three blocks, therefore, momentum of the system is conserved, and the sum of internal horizontal forces must equal to zero. In other words, for every action there is an equal and opposite reaction.

Block $m_2$ moves toward the right with acceleration $a'$ under the tension on the chord. The sum of forces must equal to zero, and therefore, $m_1$ is subject to a force $m_1a$ toward the left such that $m_2a'+m_1a=0$.

[quote="shobber"]Dr.No:
The momentum of the system is not conserved since there is gravity acting on the system.[/quote]

shobber,

One assumes that $M_1$ itself rests upon a frictionless surface. Your problem definition implied this but did not explicitly state it. If this is so, then there is a [b]normal force[/b] directed upward from the surface upon which $M_1$ rests equal to the weight of $M_1+m_2$, and momentum is conserved.

[quote="kunny"]Maybe 垂直抗力?[/quote]
Don't know what it is...

And Kunny, may I ask, how many languages do you know?? It's impressive...

[quote="shobber"]The momentum of system is conserved if $\sum {F} = 0$.
In this case, gravity does work on $M_3$, so i don't think it's conserved.

By the way, what is [b]normal force[/b]??[/quote]

I was thinking of this on my way to and from church this morning. You are correct. The momentum of $m_3$ is [u]not conserved[/u]. The net force on $m_3$ is $m_3a'$ downward.

However, since acceleration has direction this brings up the point that both force and momentum are vectors; they have direction. So while not strictly conserved altogether, momentum is conserved in the horizontal direction. The net force on the system in the horizontal direction is zero independent of the fact that the net force on the system in the vertical direction is not zero.

Even so, I believe there is still a small error in my Newton's third law equation. As $m_2$ is accelerated toward the right, both $M_1$ and $m_3$ are accelerated toward the left, not just $M_1$ alone. Thus the correct equation is:

$m_2a'+(M_1+m_3)a = 0$   (Solve for $a$)

The normal force is the force with which a surface pushes on a body that is in contact with it.  The word "normal" in this context means perpendicular. (The normal direction is the direction of the gradient of a surface.)

As an example, think of standing on a flat floor. Your weight pushes you down, the floor pushes you up, and the net result is zero force. You neither sink into the floor nor fly into the air.  The normal force is the force of the floor pushing you up.

[quote]
The normal force is the force with which a surface pushes on a body that is in contact with it. The word "normal" in this context means perpendicular. (The normal direction is the direction of the gradient of a surface.)

As an example, think of standing on a flat floor. Your weight pushes you down, the floor pushes you up, and the net result is zero force. You neither sink into the floor nor fly into the air. The normal force is the force of the floor pushing you up.[/quote]
I see... Thanks!


Daca $a,b,c\in\mathbb{R}_{+},a+b+c=1$,atunci sa se arate ca

$\frac{a+b}{c+1}+\frac{b+c}{a+1}+\frac{c+a}{b+1}\geq\frac32$

Andrei Sorin Cosma

Sa notam $a+b=z,b+c=x,c+a=y$ si avem $\sum\frac{x}{y+z}\geq\frac{3}{2}$,care este foarte cunoscuta :)

nu e cam triviala?
[hide="solutie"]Sa notam cu $\ RHS$ membrul drept al inegalitatii.Atunci folosind Cauchy obtinem usor ca $\ RHS\geq\frac{2}{ab+bc+ca+1}$.Pentru a demonstra ca $\ RHS\geq\frac{3}{2}$ e necesar sa aratam ca $\ ab+bc+ca\leq\frac{1}{3}$, sau $\ ab+bc+ca\leq\ (a+b+c)^{2}$, care se reduce la $\ a^{2}+b^{2}+c^{2}\geq\ ab+bc+ca$.[/hide]

Eu am rezolvat-o ca Sorin.Si constientizez ca este simpla,tocmai de-asta i-am dat titlul asta.Iata o alta solutie[a autorului]:
Relatia se scrie sub forma $\frac1{c+1}+\frac1{b+1}+\frac1{a+1}\geq \frac94(*)$Si apoi folosim
$(a+1+b+1+c+1)(\frac1{a+1}+\frac1{b+1}+\frac1{c+1})\geq 9.$
Misto,nu?Singura chestie mai dificila ar putea fi demonstrarea echivalentei dintre egalitatea din enunt si $(*)$.


Let F be the field of all real numbers and let V be the set of all sequences (a1,a2,....a_n,...), a_i in F, where equality, addition, and scalar multiplication are defined component-wise.

(a) Prove that V is a vector space over F
(b) Let W={(a1, a2,....,a_n,...) in V | lim a_n = 0 as n-->inf}. Prove that W is a subspace of V.
(c) Let U={(a1,...,a_n,...) in V | summation of (a_i)^2 is finite, i evaluated from 1 to inf}. Prove that U is a subspace of V and is contained in W.

I know that in order for W to be a subspace of V, W must form a vector space over F under the operations of V. I've already proved (a). Do I need to know the limit of a_n to prove (b) or is that just for (c)? It seems like proving (b) is pretty similar to (a), right? Any tips on proving (c)?

[b](b)[/b] Yes, you have to use the fact, that $ \lim_{n \to \infty}a_{n}\equal{}0$.

In fact you have to prove that if $ A,A^{\prime} \in W$, then $ \lambda A\plus{}\mu A^{\prime} \in W$, where $ \lambda,\mu \in F$. The latter is true if $ \lim_{n \to \infty} \lambda a_{n}\plus{}\mu a^{\prime}_{n}\equal{}0.$ (and of course all components are from $ F$).

[b](c)[/b]
Similarly like in [b](b)[/b] you show that if $ A,A^{\prime} \in U$, then $ \lambda A\plus{}\mu A^{\prime} \in U$ $ \left(\right.$the limit of $ \lambda\sum_{i\equal{}1}^{\infty}a_{i}^{2}\plus{}\mu\sum_{i\equal{}1}^{\infty}(a^{\prime}_{i})^{2}$ must be finite$ \left.\right)$.

If $ \sum_{i\equal{}1}^{\infty}a_{i}^{2}$ converges, then $ \lim_{n \to \infty}a^{2}_{n}\equal{}0 \Rightarrow \lim_{n \to \infty}a_{n}\equal{}0$. Hence $ U \leq W$.


If you go to the blog page:
[url]http://www.artofproblemsolving.com/Forum/apsjour.php[/url],
the links "Go to page: 1, 2, 3, ..." point to the wrong page.

Confirmed. Fix is .... ugly, but will be done.


Let $a,b,c$ be given positive numbers and let $x,y,z$ be non-negative variables.
Find the minimum value $f(a,b,c)$ of the expression

$F(x,y,z)=\frac{ax}{y+z}+\frac{by}{z+x}+\frac{cz}{x+y}$.

[quote="Minh Can"]If a^(1/2), b^(1/2) and c^(1/2) are sides of a triangle. I think we can get some result.[/quote] If a^(1/2), b^(1/2) and c^(1/2) are not sides of a triangle, then the minimum value is $2\min\{\sqrt{ab},\sqrt{bc},\sqrt{ca}\}$ ;)

From your proposed answer I think my strategy might coincides with yours.. Let me play with it a bit. :P


So I wanted to find my old solution to
http://www.mathlinks.ro/Forum/viewtopic.php?t=155896
Strangely enough, searching for
*rstu* *smu*
yielded nothing, though the problem was posted with these two strings inside at
http://www.mathlinks.ro/Forum/viewtopic.php?t=16158
Even more strangely, editing the old post changed nothing (as opposed to my prediction - I expected that the reason was that the post was too old to be indexed).
Anyone knows where is the problem? Are LaTeX tags not indexed?

  Darij

[quote="buzzer11"]latex is most likely not indexed.[/quote]That's not the way it's supposed to be. $ test11235$

test11234

Well you guys are right there is a bug, however it's going to be hard to correct for the past entries ... (edit and resubmit should do the trick).
It doesn't affect all latex tags, but only the ones that start with a letter directly (like $ abc$).


calculus
vector
quadratics
trigonometry
analytic geometry
graphing lines
slope

[quote="shobber"]Let $P: (x_0, y_0)$ be a point out of the circle $x^2+y^2=R^2$. Find the equation of tangents through $P$ to the circle.[/quote]

Let the point that the line is tangent to the circle (on only the top) at be $\left(x_{1},y_{1}\right)$.  Then we have the following equation for the tangent line:

$y-y_{0}=\frac{-x_{1}x}{\sqrt{r^{2}-{x_{1}}^{2}}}\left(x-x_{0}\right)$

I'm only doing one line here because of this:

$\frac{y_{0}-\sqrt{r^{2}-{x_{1}}^{2}}}{x_{0}-x_{1}}=\frac{-x_{1}}{\sqrt{r^{2}-{x_{1}}^{2}}}$

That seems too messy to solve for x1 so I'm just going to wait and see how someone else does it...

The radial vector from the origin of the circle to $P$ is $(x_0,y_0)$. A point $Q = (x,y)$ lies on the tangent iff $PQ\cdot OP = (x - x_0, y - y_0)\cdot(x_0,y_0) = 0$, this is equivalent to $x_0x + y_0y = R^2$

edit: It's possible that this question is asking about something else. I don't think the formulation at the end is ver yclear.

[hide="Solution"]The tangent is perpedicular to the corresponding radius, and if the lines $y=k_1x+b_1$ and $y=k_2x+b_2$ are perpendicular, then $k_1k_2=-1$. The equation of the radius is $l: y={y_0\over x_0}x$. Knowing that, the equation of the tangent is $t: y-y_0=-{x_0\over y_0}(x-x_0)$ for $y_0\neq 0.$ In the point $(R,0)$ the tangent is $t: x=R$, and in the point $(-R,0)$ the tangent is $t: x=-R.$

All these cases can be unified in $t: y_0(y-y_0)+x_0(x-x_0)=0\cong t: xx_0+yy_0=R^2.$[/hide]

EDIT: This is wrong, since $P$ lies OUT of the circle...

So we get that that $P$ lies on the tangent line to the circle in $(x,y)$ iff $xx_0 + yy_0 = R^2$ and $x^2 + y^2 = R^2$. I see no way to describe such $(x,y)$ in a nice way.

[hide="General idea"]Take the general equation of the line through $P$, which is $y-y_0=k(x-x_0)$, and make the condition that it intersects the circle $x^2+y^2=R^2$ in exactly one point (i.e. plug $y$ in and equate the discriminant of the resulting quadratic with zero).[/hide]

[b]Theory[/b] : Let a circle $C$ with the general equation $x^2+y^2+mx+ny+p=0$ and the outer point $P(x_0,y_0)$.
Then the both tangents from the point $P$ to the circle $w$ have the equation :
$\left[xx_0+yy_0+\frac m2 (x+x_0)+\frac n2(y+y_0)+p\right]^2=$$(x^2+y^2+mx+ny+p)(x^2_0+y^2_0+mx_0+ny_0+p)\ .$
[b]A particular case[/b] : If the point $P$ belongs to the circle $C$, then the equation of the tangent
in the point $P\in C$ to the circle $C$ is : $xx_0+yy_0+\frac m2(x+x_0)+\frac n2(y+y_0)+p=0\ .$

[hide="hint"]think about the construction for the tangent through a point to a circle, just do that with algebraic equations[/hide]

the answer, and you learn this in 11th grade trigonometry i believe
[hide]
ok so (x,y) is a system that must make all x^2 + y^2 equal R^2.   Then, we know that the circle is centered at the origin so we have the circles center (0,0) and the point (x,y) anywhere on the circle.  we then take the slope of the line from the origin to this point  and this is y/x.  we then take the negative reciprocal to get the tangent line and that is -x/y.  We can then use the point slope formula as follows.

Y-Yo =-x/y(X-Xo)

hence, the equation satisfying all lines tangent to the circle that go through the point (Xo, Yo)  is

Y= -[(x)(X-xo)]/(y) +yo[/hide]


Find all bijective strictly increasing functions from $\mathbb{R}$ to $\mathbb{R}$ such that :
\[ f(x) + f^{ - 1} (x) = 2x \]

[quote="Lovasz"]Excuse me, $f^{-1}(x)$ means $\frac{1}{f(x)}$ or $f(\frac{1}{x})$ :blush:[/quote]
No,I think $f^{-1}(x)$ suche a function that$f(f^{-1}(x))=x$ :) .
This problem was be proposed in Armenian TST 2004(and I solved it :P  :D )


Let $p$ ia prime nubmer $p>2$.Prove that
$\llcorner (2+\sqrt{5})^{p}\lrcorner-2^{p+1}$ is divisible by $p$


Let $ k$ denote a positive constsant, and let $ F_1$ and $ F_2$ denote the points with rectangular coordinates $ (-k,0)$ and $ (k,0)$, respectively. A curve known as the lemniscate of Bernoulli is defined as the set of points $ P(x,y)$ such that $ (F_1P) \times (F_2P)=k^2$.

a. Show that the rectangular equation of the curve is $ (x^2+y^2)^2=2k^2(x^2-y^2)$.

b. Show that the polar equation is $ r^2=2k^2\cos 2 \theta$

This curve is Lemniscate, it may be wrong spell.


Actually, I would like to know this information as well, because at my school they are only allowing us to take AMC 10 A or AMC 12 A (based on your grade...but if you wanted to take 12A even if you are a freshman or a sophmore, you are allowed to). How would you sign up for 10 or 12B?

Did you contact any schools that you know that have AMC's? I think you should get your parents to do it. I'm not sure how but that's what I'll do.

I'm pretty sure you'll qualify so it really wouldn't influence you much. Anyway, two weeks aren't a lot of time either.

there might be a problem for ppl in my district are all taking A

must try harder...

it would if we want to take a 10 and a 12

hey...what is the AMC B day... unfortunately our school (like max's) only does the A rounds.
I think it would be good to take like the AMC10A and the 12B. what schools are there that would allow me to do them.


function
calculus
integration
advanced fields
advanced fields theorems

I've just read (for the 4th or 5th time I might add) a chapter on characteristics of P.D. Operators.  I understood the basics like writing down the principal part of the operator.  Then stick in some greek letter and set the resulting equation equal to zero.
The rest of the book seems well written.  Either the author does not really know what they (characteristics) are or the author just does not explain them in a fashion that I can understand.

$ P_m(x,D)\equal{}\displaystyle\sum_{\|\alpha\|\equal{}m} {a^{\alpha}(x) D^{\alpha}}\longrightarrow$ The Principal Part

$ P_m(x,\xi) \equal{}0 \longrightarrow$  The characteristic equation.

Could someone explain the underlying associated concepts. Regret embarrassing myself, but if you could use the most basic of mathmatical terms until I get the concept.

In short, what are characteristics and how do I use them?  I hate to move forward in the book without a fundemental grasp of the material.  Thanks in advance for your valuable time spent...

Partial differential equations are not a subject that can be properly understood with only "the most basic of mathematical terms". This is why they are normally studied at the graduate school level.

The underlying concept here is the [url=http://en.wikipedia.org/wiki/Fourier_transform]Fourier transform[/url].
[quote]Fourier transforms, and the closely related Laplace transforms are widely used in solving differential equations. The Fourier transform is compatible with differentiation in the following sense: if $ f(t)$ is a differentiable function with Fourier transform $ F(\xi)$, then the Fourier transform of its derivative is given by $ i\xi F(\xi)$. This can be used to transform differential equations into algebraic equations. Note that this technique only applies to problems whose domain is the whole set of real numbers. By extending the Fourier transform to functions of several variables, partial differential equations with domain $ \mathbb R^n$ can also be translated into algebraic equations.[/quote]

[quote="mlok"]This is why they are normally studied at the graduate school level. [/quote]

Well...  OK...  Bring it on then... let's hear it from a gradute school level then.  I've been through several graduate level courses. I don't see how the Fourier Transforms (mentioned in your post) are even related at this point though? The text I have doesn't get into a Fourier discussion as part of characteristic equations or related solution surfaces...?

I thought the following link was a reasonable treatment?

[url]http://rutherglen.ics.mq.edu.au/math336s207/notes/Bon%20Clarke/%20PDE%20Notes%20Ch%202.pdf[/url]

Thanks, again, in advance for your valuable time spent... [/url]

I'll try to give some insight...

Consider an "easy" PDE :

$ u_t \plus{} cu_x \equal{} 0,\newline u(0,x) \equal{} g(x), x\in\mathbb{R}$

where $ g$ is a "well-suited" function.

Rewrite the equation in a different way, you have $ (1,c) \nabla u \equal{} 0$, and you get that $ u$ is and constant function along lines, called characteristic's, with direction $ 1\choose c$, that is among $ x \minus{} ct \equal{} \mathrm{const}$. Then you get that your solution is $ u(t,x) \equal{} g(x \minus{} ct)$, so in a way observing how $ u$ behaves on hiperplane's ( in $ \mathbb{R}^2$, that's simply a line ) you can come up with a solution.

I need to point out that this "explanation" is very primitive (I after all, am having PDE this year, and I don't know it that well), and one shouldn't think that is the whole theory behind it, if you even consider quasilinear PDE you can easily see that it's much more complicated.

Can I ask what book are you using, Renardy, Rogers - Intro to PDE?

[quote="Mr.Doe"]
Rewrite the equation in a different way, you have $ (1,c) \nabla u \equal{} 0$, and you get that $ u$ is and constant function along lines, called characteristic's, with direction $ 1\choose c$
[/quote]

You rewrote the equation as a dot product with $ \stackrel{\rightarrow}{V} \equal{} \left\langle 1,c \right\rangle$

$ \vec V \cdot \nabla u \equal{} u_t \plus{} c u_x \equal{} 0$

From which you concluded V has a constant direction $ (1, c)$

I see that $ \vec V$is $ \bot$ to the grad u because the dot product is zero.
Which means that V is tangent U(x,t).  Since V is tangent to U(x,t) everywhere,  U(x,t) is a first integral of V.
First integrals have the form U(x,t) = C.

If I solve $ dt/1 \equal{} dx/c$ I get --> $ ct \equal{} x \plus{} C_1$ --> $ x \minus{} ct \equal{} C_1$

Use the function of the initial condition u(x,0) = g(x) to satisfy the initial condition.

$ g(x \minus{} ct) \equal{} C_1 \equal{} U(x,t)$

or simply $ U(x,t) \equal{} g(x \minus{} ct)$ is this implying that g(x) is constant as well?

Most excellent Mr. Doe.  I think I see where this is going. The foregoing can be extended to quasi-linear equations as well?

[quote="Mr.Doe"]
I need to point out that this "explanation" is very primitive (I after all, am having PDE this year, and I don't know it that well), and one shouldn't think that is the whole theory behind it, if you even consider quasilinear PDE you can easily see that it's much more complicated.

Can I ask what book are you using, Renardy, Rogers - Intro to PDE?[/quote]

I asked for as basic as I could get it.  I appreciate your time and response.
I'm using a book "Introduction to Partial Differential Equations with Applications", Zachmanoglou, Thoe


An easy question states: according to one well-known study, the total cost (in dollars) of opeating ahospital (of a particular type) can be aproximated by

$ C\equal{}4,700,000\plus{}0.00013x^2$ where $ X$ is the number of patient-days.

a) Derive a formula for the relationship between cost per patient-day and number of patient days. (I don't understand this. Isn't this already given to us by C?)

b) On the basis of the results of this study, how big must a hospital be (in terms of patient-days) to minimize the cost per patient-day? (this is the value of x that minimizes C correct?)

I think that $ C$ is the total amount of money spent to operate the hospital, so $ \frac{C}{x}$ must be the cost [i]per patient-day[/i].

Is this right?

[hide="How to answer b)"]We can use

$ \frac{C}{x}\equal{}\frac{4,700,000}{x}\plus{}0.00013x.$

Finding the minimum cost per patient-day is then a matter of using AM-GM.[/hide]


Let [tex](X,\,d)[/tex] be a compact metric space, where [tex]d[/tex] is the distance function. Let [tex]f: X\rightarrow X[/tex] be a mapping from [tex]X[/tex] to itself such that [tex] d(f(x),f(y)) < d(x,y)[/tex] for any two distinct points [tex]x,y[/tex] in [tex]X[/tex]. Prove that there exists an [tex]x\in X[/tex] such that [tex]f(x)=x[/tex].

Not convinced because it's seems too easy (and the usual trick is to consider xn=f(x(n-1)) :

[hide]g : X -> IR, g(x) = d(f(x),x). g is continuous on a compact so reaches its minimum (at x0 in E).

If  x0 != f(x0) than g(f(x0)) < g(x0) a contradiction. QED.

Your opinion ?[/hide]

Anyway it's false if X is not compact : f IR+ -> IR+, f(x) = sqrt (1+x) verifies the condition but has not fixed point.

belenos wrote:[hide]g : X -> IR, g(x) = d(f(x),x). g is continuous [/hide]

Prove it!

g continuous :

. d : ExE-> IR is always continuous (from triangle ineg)

. h : E -> ExE, g(x) = (f(x),x) is continous (obvious since f is continuous)

g = d o h is then continous.


real  x<y<z  number let x+y+z=6 and  xy+yz+zx=6  . Prove that  0<x<1<y<3<z<4

The assertion is false.  Counterexample:  $ x \equal{} 0, y \equal{} 3 \minus{} \sqrt {3}, z \equal{} 3\plus{} \sqrt {3} > 4$.

real  0<x<y<z  number let x+y+z=6 and  xy+yz+zx=6  . Prove that  0<x<1<y<3<z<4

OK, $ x\rightarrow 0$, $ y\rightarrow 3 \minus{} \sqrt {3}$, $ z\rightarrow 3 \plus{} \sqrt {3}$. I think it will still hold if $ x$, $ y$, $ z$ are close enough to these values.
For example, $ y \equal{} 1.26$, which gives $ x \plus{} z \equal{} 4.74$ and $ xz \equal{} 0.0276$. When we solve that like a quadratic equation, we have $ x\sim 0.0058$ and $ z\sim 4.7342$ (we can manually obtain the exact values). So, the changed assertion is also false.


A brand new type of ball pen is sold in a shop . There are five different colors of them all with same price . If the money you have is enough to buy $10$ of them (assume you use up all the money to buy $10$ of them), what is the probability that among the $10$ ball pen you bought , it contains all $5$ different colors ?

Probably wrong, but...

[hide]
The number of ways you can choose them is going to be $10 \cdot \binom 5 1 = 50$.  To have at least one of each color, we will assume just that.  Then we find how many ways we can pick 5 other ones, since they don't matter.  $5 \cdot \binom 5 1 = 25$.  So the probability is $\frac 1 2$... doesn't seem right... maybe I needed to use permutations instead.
[/hide]

um... here's my guess

$\frac{5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 5^5}{5^{10}}$?

In general, I think, if we have $n$ types of objects, the number of ways to choose $r$ objects is[hide]$\binom{n+r-1}{r}$[/hide]

Using this idea, my answer is:
[hide]$\frac{\binom{9}{5}}{\binom{14}{10}}=\boxed{\frac{18}{143}}$[/hide]

Edit- Oh, I didn't notice that Gyan actually gave us some hints that are related to my idea :roll:

The no. of ways to choose the 5 different coloured balls can be found by obtaining the coefficient of $x^(10)$ in \[ (x+x^2+x^3+x^4+x^5+x^6)^5  \]
$OR$
$(x+x^2+x^3+x^4+x^5+x^6+x^7+...)^5$   [its all the same]
The coefficient of $x^(10)$ in the above expression is $9C5 = 126$
Total no. of ways in which the ball pens can be selected = $50C10$
Thus the required probability is $126/50C10$

Hope it is right.  :roll: .
I saw the above solution. Why have you used the denominator as 14C10.

[quote="varun"]Total no. of ways in which the ball pens can be selected = $50C10$[/quote]

He didn't consider that solutions are different if they differ only in the order of the elements.

we must choose $x_1 \leq \ldots \leq x_{10} \in \{ 0,1,2,3,4 \}$. Set $y_i=x_i+i$.

Now we must choose $y_1 \leq \ldots \leq y_{10} \in \{1,\ldots,14 \}$. clearly, the map $x_i \to x_i + i$ is "bijective" (in the sense that a sequence that is a soln to the first one gives a soln to the second one, and conversely). hence we have $C_{14}^{10}$ possibilities.

Strangely enough, by running a program with a pseudorandom generator I get a probability of $50 \%$ which is quite different from the one he obtained.

Can anyone find the wanted ratio considering all the $10^5$ possibilities (meaning solutions are different if they differ in order)? It looks very hard, though. I wanted to say that some solutions can be obtained in many ways (like $0,1,\ldots,4,0,\ldots,4$), whilst others in fewer ways(like $1,1,1,\ldots,1$), i.e. they probabilities are not equally distributed

frt, do you understand what I mean (that is, in case it is not nonsense)

varun, please use brackets around exponents that are more than one character long.  For example, x^{k+2} gives $x^{k+2}$ whereas just throwing dollar signs around x^k+2 gives $x^k+2.$

[quote="perfect_radio"]Strangely enough, by running a program with a pseudorandom generator I get a probability of $50 \%$ which is quite different from the one he obtained.

Can anyone find the wanted ratio considering all the $10^5$ possibilities (meaning solutions are different if they differ in order)? It looks very hard, though. I wanted to say that some solutions can be obtained in many ways (like $0,1,\ldots,4,0,\ldots,4$), whilst others in fewer ways(like $1,1,1,\ldots,1$), i.e. they probabilities are not equally distributed

frt, do you understand what I mean (that is, in case it is not nonsense)[/quote]
$50 \% =\frac{1}{2}$ is what deej21 got.
I basically understand what you mean. Maybe I should reconsider my solution...

actually it's not exactly $50 \%$, it's more like $52,... \%$

but i must admit i have no idea how to count the ways to do it in this way  :(

hrm... I guess I got it with orders being involved...
I wonder if you can do it this way?
[hide="Another feeble attempt"]
000000000000000
you have to choose five spots to turn into "borders"
0|00|000|00|0|0
so $\binom{15}{5}$ would be the total number of ways???

and then take away five zeros and do the same thing with 10 zeros.
between each "border" you insert each of the five that you originally took away to get $\binom{10}{5}$
so $\frac{\binom{10}{5}}{\binom{15}{5}}$?
[/hide]

[quote="frt"]That's basically what I did, except that the number of borders was 4, not 5. That's how I got $\frac{\binom{9}{4}}{\binom{14}{4}}$. (I wrote $\frac{\binom{9}{5}}{\binom{14}{10}}$ in my original solution, but that's the same thing.)[/quote]

I guess you should be correct (I dont have the answer though) . What about prefect_radio's question which we consider the order ?  :ninja:

I think now I can answer the question in the proper way:

Total no. of ways in which we can buy 10 pens can be obtained by the coefficient of $x^{10}$ in $(x^0 + x^1 + \cdot\cdot\cdot +x^{10})^5$
which is $\binom{14}{10}$.

Now, no. of ways in which the pens can be chosen in the reqd. way= coefficient of $x^{10}$ in $(x^0 + x^1 + \cdot\cdot\cdot +x^6)^5$ which is 9C4.

Thus the probability = $\frac{\binom{9}{4}}{\binom{14}{10}}$

Hope, I made myself clear. :)

[quote="varun"]Total no. of ways in which we can buy 10 pens can be obtained by the coefficient of $x^{10}$ in $(x^0 + x^1 + \cdot\cdot\cdot +x^{10})^5$
which is $\binom{14}{10}$.[/quote]

How do you know that the number of ways is that coefficient? Is it some kind of theorem?

[quote="Gyan"]Hint Peek ONLY if you are really stuck! (Do not peek just to check your answer - or to find a formula - there will be none given!
[hide]Big Hint,   (None of  the answers given above are correct)  - So keep thinking -  Last chance to quit! Peek only if you are really stuck - Not  just to  check your answer.
[hide]Q1 Can you find the number of ways if you have to choose 10 pens?
Q2 Can you find the number of ways if you only have to choose 5 pens?  (Other 5, one of each color are already chosen for you.
Answer you are looking for  is  the ratio of these two...

So question is  how to  count ([b]correcty l[/b] :)  ) the answer to Q1.

Think for a while - [hide] How about how many ways  you  can drop 10 objects in 5 boxes?

Or how many 14 letter words can you make using 4 I's and 10 x's . (x's are objects, I's are "wall's" of boxes)
 [/hide] [/hide][/hide][/quote]
i think you like reveal hidden content,don't you Gyan?

[quote="shyong"]I wonder , is the total number of such permutation is just $\binom{14}{10}\times 10!$ ? And the total number of permutation of all color contains is $\binom{9}{4}\times 10!$ resulting the probability of permutation is also the same ?

 :?[/quote]

Did you bother reading this topic ... ?

(sorry for the dumb question)

[quote="perfect_radio"][quote="shyong"]I wonder , is the total number of such permutation is just $\binom{14}{10}\times 10!$ ? And the total number of permutation of all color contains is $\binom{9}{4}\times 10!$ resulting the probability of permutation is also the same ?

 :?[/quote]

Did you bother reading this topic ... ?[/quote]

well of course , since im the one who open this topic   :P   . What I do above is just to consider the question you come out with which is when the " order " is in consideration . :lol:

[b]perfect_radio[/b] I'll explain you that:

When we write $1+x + x^2 + x^3 +\cdot\cdot\cdot$ we just need the power of x. We have nothing to do with 'x'.
When we multiply, \[ (1+x + x^2 + x^3 +\cdot\cdot\cdot)*(1+x + x^2 + x^3 +\cdot\cdot\cdot)*(1+x + x^2 + x^3 +\cdot\cdot\cdot)*\cdot\cdot\cdot \]
Now the powers of x get added when we multiply them.
The power of x inside brackets represent the number of say objects we can choose of that type.

I think you might be confused, but put up your doubt and I'll explain better.
See for example,
[b]To find the no. of ways in which we can represent 20 as a sum of 3 natural numbers.[/b]
Now let the three numbers are $a,b,c$, then $1\leq a\leq 18$, $1\leq b\leq 18$, and $1\leq c\leq 18$.
So, th total number of ways is the coefficient of $x^{20}$ in the following product:
\[ (x+x^2+x^3+\cdot\cdot\cdot+x^{18})^3 \]

[b]See another important example[/b],
[b]To find the no. of ways in which we can represent 20 as a sum of 3 non-negative integers.[/b]
Here $0\leq a\leq 20$, $0\leq b\leq 20$, and $0\leq c\leq 20$.
We need the coefficient of $x^{20}$. [b]But[/b], now in the following prodeuct:
\[ (x^0+x+x^2+x^3+\cdot\cdot\cdot+x^{18})^3 \]
i.e.\[ (1+x+x^2+x^3+\cdot\cdot\cdot+x^{18})^3 \]

Hope I would have helped you.
For any thankx and all please pm me don't put up a new reply saying THANKS :rotfl:

[quote="jensen"]i think you like reveal hidden content,don't you Gyan?[hide]Q1 Can you find the number of ways if you have to choose 10 pens?
Q2 Can you find the number of ways if you only have to choose 5 pens?  (Other 5, one of each color are already chosen for you.
Answer you are looking for  is  the ratio of these two...

So question is  how to  count ([b]correcty l[/b] :)  ) the answer to Q1.

Think for a while - [hide] How about how many ways  you  can drop 10 objects in 5 boxes?

[hide]Or how many 14 letter words can you make using 4 I's and 10 x's . (x's are objects, I's are "wall's" of boxes) [/hide]
 [/hide] [/hide][/quote]
Yes, I think :) ... Please go ahead and peek..

Anyway, knowing  the "formula" for this type of problem should be a standard tool in your toolbox.

One way to think about the problem, is to  write it  down ....Use "p" for pens and separate  them by , say  a  ","    by the color.
So
[code]$ppp,pp,,p,pppp$ [/code]
means, 3 pens of first color, 2 of second color, 0 for the third color (none), 1 for fourth color and 4 of the 5th color.

You have 14 objects, can be arranged in  $14!$ ways, adjusting for $10$ "identical" $p$ 's and $4$ identical  [b] ","  [/b] we get the answer
$\frac{14!}{10! \ 4!}$       (or $14 \choose 10$)

In general the formula (when more than one object of each 'color' is  allowed) is something like $n+c-1\choose n$

Hope this helps.

[quote="shyong"]well of course , since im the one who open this topic   :P   . What I do above is just to consider the question you come out with which is when the " order " is in consideration . :lol:[/quote]

oops, sorry for the last remark  :blush:  i misunderstood your previous post. (i overlooked the $10!$) anyway, i think that the total number is in fact $5 \cdot 5 \cdot \ldots \cdot 5 = 5^{10}$.

Actually, here's what I obtained using a simple c++ program: (considering different orders)

Total no. of cases: $9765625=5^{10}$.

No. of favorable cases: $5103000=10^3 \cdot 3^6 \cdot 7$ (I still have no clue how to do it by hand)

Probability: $\displaystyle \frac{2^3 \cdot 3^6 \cdot 7}{5^7} = 0.5225472$.

varun's method is called Generating Functions.

[quote="perfect_radio"]
Actually, here's what I obtained using a simple c++ program: (considering different orders)

Total no. of cases: $9765625=5^10$.

No. of favorable cases: $5103000=10^3 \cdot 3^6 \cdot 7$ (I still have no clue how to do it by hand)

Probability: $\displaystyle \frac{2^3 \cdot 3^6 \cdot 7}{5^7} = 0.5225472$.[/quote]

Thanks for listing out the probablity . Actually , I come out with the idea is becasue last time I solved a simlar problem ( I think  :roll: )  which is you have $n$ book shelves and $m$ books . You want to put all those books into the book shelves where "order" is needed (assuming allow empty book shelf) and what is the total number of arrangement . I got something like $\binom{m+n-1}{n}\times m!$ so I thought this thing might have related to this question as well  :)


mathleague.org is initiating a mail-in Power Question contest this school year, information about which can be found at http://mathleague.org/power.htm.  The format of this contest is similar to the Power Question offered by ARML but is designed to be completed by a team of six (for large schools) or three (for small schools).   For this year only, mathleague.org is offering one free Power Question entry to each high school in California.  Schools wishing to participate should go to the website and follow the relevant instructions, and any questions should be directed to Tim Sanders at tim@mathleague.org.  Please forward this announcement to any students and teachers in California who you think may be interested in this program.  Deadline for entry is March 15..

The period at the end of your sentence got included in the link and so the link doesn't work right.  It should be: http://mathleague.org/power.htm

Are the answers to the first set of questions available?

Answer keys will be emailed to schools this week..


geometry
circumcircle
trigonometry
trig identities
Law of Sines

[quote="Valentin Vornicu"]Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$.

[i]Romania[/i][/quote]

$\frac{b+c}{\sqrt{bc}}\geq 2$
while $a\leq 2R$ (chord-diameter)

so $b=c$ and $A=90^0$

$a\geq 2R$ (chord-diameter)

so $b=c$ and $A=90^0$

[quote="José"]socrates wrote $a\leq2R$.
M4RIO wrote $a\geq2R$.

Who's wrong???[/quote]

Try this by yourself. Do some sketch....
I think i am not the liar..

 :D

[hide]Use Law of Sines: $a=2R\sin A$ and similar for other variables. So $\sin B+\sin C=2\sin A\sqrt{\sin B \sin C}\Rightarrow \sqrt{\sin B \sin C}=\frac{\sin B+\sin C}{2\sin A}$.

But $\sqrt{\sin B \sin C}\leq \frac{\sin B+\sin C}{2}$ (AM-GM) and $\frac{\sin B+\sin C}{2}\leq \frac{\sin B+\sin C}{2\sin A}$ since $0<\sin A\leq 1$.

Both equality conditions must be met, so $\sin A=1$, making $\angle A=90^\circ$, and $\sin B=\sin C$; both are acute, so both are equal. So the equation works if and only if $\Delta ABC$ is an isosceles right triangle.[/hide]

[quote="José"]socrates wrote $a\leq2R$.
M4RIO wrote $a\geq2R$.

Who's wrong???[/quote]

obviously $a \le 2R$ is correct because if you have a circle, the largest chord is the diameter

This thing follows directly from AM-GM:
$R(b+c)>=2R\sqrt{bc}>=a\sqrt{bc}$.
This means that $a=2R$ which means that $a$ is the diameter. Furthermore, the triangle is right. Also, equality holds in AM-GM when all the terms are equal so $b=c$. Therefore the triangle is isosceles right triangle which is of the for $(a,a,a\sqrt{2})$.
Nice problem.

using sine rule yields ,
$sinB+sinC=2sinA\sqrt{sinBsinC}\le 2\sqrt{sinBsinC}$
on the other hand , $sinB+sinC\ge 2\sqrt{sinBsinC}$
so , $sinB+sinC=2\sqrt{sinBsinC} \implies B=C $ and $A=\pi/2$

We know $R^2=a^2bc/(b+c)^2$
We have $R^2=a^2/4sin^2A$ thus $sin^2A=(b+c)^2/4bc$≥$1$
also we have $1$≥$sin^2A$
thus $A=90$ and $b=c$

hello, with
$R=\frac{abc}{4A}$
we get
$\frac{(b+c)bc}{4\sqrt{bc}}=A$
$\frac{b+c}{2\sqrt{bc}}=\sin(\alpha)\le 1$ thus
we obtain
$b+c\le 2\sqrt{bc}$ and we get
$b=c$ and $\alpha=90^{\circ}$
Sonnhard.

[quote=Valentin Vornicu]Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$.

[i]Romania[/i][/quote]

$a+b \geq 2 \sqrt{bc}$ we get $a\sqrt{bc} \geq 2R\sqrt{bc}$ => $a \geq 2R$ but $a \leq 2R$ and we get $a = 2R$ and $b = c$.


[quote="nebula42"]Let $p$ be any prime. Prove that \[{2p \choose p}\equiv 2 \pmod{p^{2}}.\] [/quote]

[hide] \[{2p \choose p}\equiv 2 \pmod{p^{2}}.\]  \[\sum_{k=0}^{p}\binom{p}{k}^{2}\equiv 2\pmod{p^{2}}\] All the terms in the sum are divisible by $p^{2}$ except the first and last. Since the first and last terms are both 1 then the LHS is 2[/hide]

[quote="srulikbd"]yes but why it's implies that \[{2p \choose p}\equiv 2 \pmod{p^{2}}.\] ?
thansk.[/quote]
$\sum_{0\leqslant k\leqslant n}\binom{n}{k}^{2}= \binom{2n}{n}$ for any $n\in\mathbb{N}^{+}$.  :D

A harder problem: Prove that $\binom{2p}{p}\equiv 2 \pmod{p^{3}}$ (without using Wolstenholme of course).


Ummmmmmmmmmmm thanks?  very.... informative? hahaha.

[quote="ZennyK"]What the . . . ? Thanks. Now I'm scarred for life.
That was like a really creepy take-off on ZOOM or something.[/quote]

Yea, I'm pretty sure it is.  But the old ZOOM though, like from the 70s.

That scared me!
it made me look like [hide="this"] :yoda: [/hide]

i will have to sleep with a bedlight FORVER!!!! :rotfl:  :roll:

woW..............
wowl.............

what was up with the kkk in orange and the random scene where they started like slaying each other?

Looked like a cult [size=150]:ninja: [/size]

Do they actually show this to children???? :stretcher:

[quote="ZennyK"]What the . . . ? Thanks. Now I'm scarred for life.
That was like a really creepy [b]take-off on ZOOM[/b] or something. (emphasis added)[/quote]
But given what's on children's TV I wouldn't be that surprised...

It's one way to teach kids pi.  Not my way, but i guess if forcing little kids to join the pi cult works then go for it.  :)

um...i don't get it? isn't that how all people learned about pi, or is it just me? :?  :|  :huh:

I liked it when the robot kept saying digits! 3.14159623........

Wow, those weird guys didn't blink. The robot was awesome though  :P .

Haha, check out this remix of the pi song (click the headphones next to "pi remix")

[url]http://www.newgrounds.com/audio/view.php?id=1399566&sub=54403[/url]

i wish i had a robot that shot lasers while singing the digits of pi.

[quote="nutz_for2.718281828"]I have never had trouble sleeping last week![/quote]
What?  :huh:

I watched the Pi movie for a few times and I had absolutely NO trouble sleeping last night!


Let $S = \lbrace A = (a_1, \ldots, a_s) \mid a_i = 0$ or $1, i = 1, \ldots, 8 \rbrace$.  For any 2 elements of $S$, $A = \lbrace a_1, \ldots, a_8\rbrace$ and $B = \lbrace b_1, \ldots, b_8\rbrace$.  Let $d(A,B) = \sum_{i=1}{8} |a_i - b_i|$. Call $d(A,B)$ the distance between $A$ and $B$.  At most how many elements can $S$ have such that the distance between any 2 sets is at least 5?

Since this has stayed empty for a while... time to pull out the bashing hammer.

WLOG $ (0,0,\dots,0) \in S$. Let $ \sigma(a_1,\dots,a_8) \equal{} a_1 \plus{} \dots \plus{} a_8$. Then for every other element $ T \in S$, $ \sigma(T) \ge 5$.

Claim 1: There is at most one element T of $ S$ with $ \sigma(T) \ge 6$.
Proof: Let $ Q \equal{} (1,1,1,\dots,1)$, and suppose $ T,T'$ satisfy $ \sigma(T), \sigma(T') \ge 6$. Then $ d(T,T') \le d(T,Q) \plus{} d(Q,T') \le 4$, a contradiction.

Claim 2: Suppose $ R,R' \in S$ satisfy $ \sigma(R) \equal{} \sigma(R') \equal{} 5$. Then the sets
$ \alpha(R) \equal{} \{k | R_k \equal{} 0\}$ and $ \alpha(R')$ are disjoint.

Proof: Suppose otherwise, with WLOG $ R_8 \equal{} R'_8 \equal{} 0$. Then if $ X \equal{} (1,1,1,1,1,1,1,0),$ we have $ d(R,R') \le d(R,X) \plus{} d(X,R') \equal{} 4$, a contradiction.

Corollary: There are at most two elements of $ S$ with $ \sigma \equal{} 5$.

Therefore, $ |S| \le 4$.

Lastly, we can choose $ S \equal{} \{(0,0,0,0,0,0,0,0), (1,1,1,1,1,1,0,0), (0,0,0,1,1,1,1,1), (1,1,1,0,0,0,1,1) \}$.

There is nice solution with double counting.Let elements of $S$ be $A_1,\dots,A_x$ and $A_i=(a_{i1}\dots,a_{i8})$. We count in two ways $T$,number of unordered pairs $(a_{ij},a_{tj})$ where $a_{ij}$ and $a_{tj}$ are different ; $T\ge5\binom{x}{2}$ because there are at least $5$ different pairs of numbers in any  two $A_i$ and $A_j$.On the other hand $8[x^2/4]\ge T $ because if we observe all $a_{ij}$ for fixed $j$ we see that there are at most $[x^2/4]$  different pairs ( and multiply by $8$ for all $j$-s).Now we have $8[x^2/4]\ge T \ge5\binom{x}{2}$ from which we get $x\le4$. Example is : $ S = \{(1,1,1,1,1,1,1,1), (1,1,1,0,0,0,0,0), (0,0,0,1,1,0,0,0), (0,0,0,0,0,1,1,1) \}$

Nice.  Also, I was hoping there is a geometric solution.

If you replace the $0$'s in the ordered sets with $-1$'s then the problem is equivalent to finding the number of ordered sets (a.k.a vectors) where the dot product between any two of them is less than $0$.  Which is equivalent to the vector angle between any two of them is greater than 90 degrees.  I don't know if there is a good way to finish this off geometrically though.


Two standard six-faced dice are rolled. Jean wins if the product of the two numbers rolled is ood or a multiple of three, otherwise Allen wins. What is the probability that Jean wins?

[hide]There are a total $36$ combinations of the numbers on the two dice.

The product will be odd if both numbers are odd, so there are $3\cdot3=9$ odd products.

The product will be a multiple of three if one of the numbers is either 3 or 6. Some multiples of 3 are also multiples of 6. When all is taken into account, there are $9$ products that are multiples of three.

We must subtract the products that are both multiples of three and odd because we have counted them twice. There are $3$ such products.

In all, there are $9+9-3=15$ products that are either multiples of three or odd.

The probability that Jean wins is $\boxed{\frac{15}{36}}.$[/hide]

[quote="davidli"]According to the answer key, your answer is incorrect.[/quote]
Use your common sense.
Look at that fraction again... :roll:

[hide]
odd=1/4 chance
mult of 3, excluding odds=1/6 chance

so, add those and the chance of winning for Jean is....
$\frac{5}{12}$

-jorian[/hide]

I'm not that dumb. $\frac{5}{12}$ is not what the answer key says.

What does it say? Does it have a solution?

[hide="I got a different answer... Am I right?"]
List them out:

Multiple of $3$:
At least one of the numbers is a multiple of $3$:
$3,1$
$3,2$
$3,3$
$3,4$
$3,5$
$3,6$
$6,1$
$6,2$
$6,3$
$6,4$
$6,5$
$6,6$

Odd:
It will be odd iff both numbers are odd:
$1,1$
$1,3$
$1,5$
$3,1$
$3,3$
$3,5$
$5,1$
$5,3$
$5,5$

Canceling all the repeats:
$3,1$
$3,2$
$3,3$
$3,4$
$3,5$
$3,6$
$6,1$
$6,2$
$6,3$
$6,4$
$6,5$
$6,6$
$1,1$
$1,5$
$5,5$

I counted and got $12$ cases. There are a total of $36$ cases.

Therefore, the answer is $\frac{12}{36}= \frac{1}{3}$.

Is $\frac{1}{3}$ what it says in your answer book?[/hide]

[hide="aha"]
@i like pie - 36 is the number of total combinations including repititions of products, so for the odd part you're right, but on the multiples of three, you undercounted.

Here's how I did it:

For the first die, there are two cases for a multiple of three - a three or a six.

If it's a three, the other die needs an even number to ensure that the resulting number is even (and thus not counted twice), for a probability of $\frac{1}{6}\cdot \frac{3}{6}= \frac{1}{12}$.

If it's a six, the other die can have anything for a probability of $\frac{1}{6}\cdot \frac{6}{6}= \frac{1}{6}$.

Adding all three probabilities (including the probability of it being odd), we get $\frac{1}{4}+\frac{1}{12}+\frac{1}{6}= \boxed{\frac{1}{2}}$. [/hide]

@humblemastermind: I listed all the cases to get $12$. How can there be $18$ cases, as you have said?

I think i_like_pie overcounted. You say he undercounted.

davidli, could you tell us which of the proposed answers is correct?

We have $3$ (actually $2$) answers so far:

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{5}{12}$ (which you said was incorrect)

Or, are none of the answers correct? :P

[quote="vishalarul"]Or, are none of the answers correct? :P[/quote]
Yes... None of our answers are correct. Davidli just posted his solution and the answer.

[quote="vishalarul"]@humblemastermind: I listed all the cases to get $12$. How can there be $18$ cases, as you have said?
[/quote]
Yeah, you didn't list all of them (where's $1,3$ on your list, for example?  :P )

[quote="bpms"]I personally get
[hide]$\frac{5}{12}$[/hide][/quote]
Me too. Is your reasoning the same?

[quote="davidli"][quote="bpms"][quote="davidli"]Everyone who's posted so far has been wrong. I just figured it out.

[hide="solution"]The product will be odd if both numbers on the dice is odd, so 9 of them.

The product is a multiple of 3 if one of the numbers on the dice is 3 or 6.

If the first die is 3, you have only 3 possibilities for the second die because you can't have 1,3,5 for the second die since that will be an odd product.

If the first die is 6, there are 6 possibilties for the second die.

If the second die is 3, you have only 2 possibilities for the first, it can only be 2 and 4 since 6 has already been the first.

If the second die is 6, you have 4 possibilities for the first die because 3 and 6 are already accounted for.

So the total number of odd/multiple of 3 is 24 and the probability is $\frac{24}{36}$ or $\frac{2}{3}$, which was what the answer key had.

I'm sure there's a faster way because this is not really efficient.[/hide][/quote]
You grossly overcounted since you counted many twice by at some points considering the first die and at others considering the second with many overlaps...
Where is this problem from?
I personally get
[hide]$\frac{5}{12}$[/hide][/quote]

I did not grossly overcount. It is you who grossly undercounted. This problem is from the 2002 Chapter team round, you can look up the answer yourself if you want. You should think about the problem more carefully before you accuse someone else of being wrong.[/quote]

Yes, davidli your answer is correct. This problem can be solved in a very fast way(using a grid). i_like_pie how did you only get 9 multiples of 3? Just because multiples of six are also multiples of three they shouldn't be excluded. Its asking for the probability of rolling a multiple of three so you should include multiples of 6s as well. :wink:

[quote="davidli"][quote="bpms"][quote="davidli"]Everyone who's posted so far has been wrong. I just figured it out.

[hide="solution"]The product will be odd if both numbers on the dice is odd, so 9 of them.

The product is a multiple of 3 if one of the numbers on the dice is 3 or 6.

If the first die is 3, you have only 3 possibilities for the second die because you can't have 1,3,5 for the second die since that will be an odd product.

If the first die is 6, there are 6 possibilties for the second die.

If the second die is 3, you have only 2 possibilities for the first, it can only be 2 and 4 since 6 has already been the first.

If the second die is 6, you have 4 possibilities for the first die because 3 and 6 are already accounted for.

So the total number of odd/multiple of 3 is 24 and the probability is $\frac{24}{36}$ or $\frac{2}{3}$, which was what the answer key had.

I'm sure there's a faster way because this is not really efficient.[/hide][/quote]
You grossly overcounted since you counted many twice by at some points considering the first die and at others considering the second with many overlaps...
Where is this problem from?
I personally get
[hide]$\frac{5}{12}$[/hide][/quote]

I did not grossly overcount. It is you who grossly undercounted. This problem is from the 2002 Chapter team round, you can look up the answer yourself if you want. You should think about the problem more carefully before you accuse someone else of being wrong.[/quote]
eep, sorry if I offended you I didn't mean to be offending. I thought you were making a mistake, but it looks like I'm wrong.

No, you didn't offend me. It's just that I didn't overcount. The order of the dice matters.

sonny, you have posted earlier that this problem can be solved very quickly using a grid. Could you show us how to do that?  :D

Make a grid with 1-6 going across and 1-6 going down. then check the appropriate boxes. annnnd, i think this topic is kinda old...


Prove that: $ \forall n\geq 3$ equation $ 2^n\equal{}7x^2\plus{}y^2$ have root $ x,y\in N$

[quote="nktp"]Prove that: $ \forall n\geq 3$ equation $ 2^n \equal{} 7x^2 \plus{} y^2$ have root $ x,y\in N$[/quote]
Already posted [url=http://www.mathlinks.ro/viewtopic.php?t=161309]here[/url]


My math teacher gave me some money to buy books at the camp, where will I be able to buy good books?

If this year is like the past two years (and I bet it is), you'll get two books for free.

Yea I have a friend that told me about that, but those I want to keep. I forgot to mention that my teacher wants the books for my school.

What types of books they gave you? Olympiad level or AoPS level?

there was stuff for every level from intermediate/aops to putnam

Not that it matters much, but here are some books that I used that could also be useful for a high school student who is sorta getting into problem solving: 1. AoPS books 2. ACoPS 3. math olympiad challenges/treasures.

i read some  REALLY harsh criticism on the book ACoPS

http://web.archive.org/web/20040703132631/www.kalva.demon.co.uk/books/zeitz1.html

What ACoPS stand for anyway?
And for "iYOA" can you send me what is written in that site as an e-mail since this site is blocked here (I don't know why) Thanks :)

ACoPS stands for Art and Craft of Problem Solving by Paul Zeitz.

And, the site says:

[quote]This is actually a disappointment to many, I suppose. This book is very light on Geometry, which is an important part of Olympiad competitions. However, it gives the novice some insights into combinatorics. Its coverage of Number Theory is atrocious. And I have expected the book to define injective, surjective, bijective. The book is far too thin to give complete exposure to beginners. [/quote]

I personally think the book is pretty good.

Well I never read this book but as I listen it's a very good book

Still nothing on "where" I can buy the books...

One of my friends last year also sought to buy books. I believe he was unsuccessful in doing so. You'll be best off talking to Dr. A once you get there.

Well ok I'll try that, see you all in texas, I leave today.

@altheman, I know you are competing in the pre-olympiad tournament, could you tell me during the camp the questions of each week?

not that you'll be able to see this reply but if you find me, I'd be glad to :)

I'm in that tournament too. Can you tell me the questions too? Will there be internet/computer access there so that I can send in my solutions (if I solved anything at all, which might not happen) each Friday?

What the hell? Zeitz is a great book. It actually has practice problems, which I vaguely remember as being somewhat simple so it's a great introductory book for concepts.

he wrote some comments about other books too and it seems as if he's quite picky
(the other comments were not as rough however)


function
calculus
derivative
calculus computations

f(x,y) = (y-3x^2)(y-x^2)

a) show that f has a relative minimum at the origin on every straight line through (0,0)

b) show that the origin is not a relative minimum of f

Consider first the lines given by $x=ay$, $a\in\mathbb R$. Then $f(x,y)=y^2-4a^2 y^3+3a^4 y^4$. This function, as a function of $y$, has zero first derivative at $0$ and positive second derivative ($2$) at $0$. Thus, it has a local minimum at $0$. The only remaining line is $y=0$, in which case $f(x,y)=3x^4$. This function, clearly, has a global minimum at $0$.

To see that $(0,0)$ is not a local minimum, consider $f(x,2x^2)$ for $x\to 0$.


$ (b,c)>0$ such that $ bc\leq 1$
Prove that
$ (b\plus{}c)(\frac{1}{b^2\plus{}1}\plus{}\frac{1}{c^2\plus{}1}) \plus{}\frac{1}{b\plus{}c} \geq \frac{5}{2}$

I see that you aren't interested  :maybe: !!
anyway here is my solution:
we put: $ x\equal{}b\plus{}c$
if : $ x \leq 2$:
we have: $ \frac{1}{b^2\plus{}1}\plus{}\frac{1}{c^2\plus{}1}\minus{}\frac{2\minus{}b}{2}\minus{}\frac{2\minus{}c}{2}\equal{}\frac{1}{2}(\frac{b(b\minus{}1)^2}{b^2\plus{}1}\plus{}\frac{c(c\minus{}1)^2}{c^2\plus{}1}) \geq 0$
so it remain to prove that:
$ x(\frac{4\minus{}x}{2})\plus{}\frac{1}{x}\minus{}\frac{5}{2}\geq 0$
wich is equivalent to:
$ \frac{(2\minus{}x)(x\minus{}1)^2}{2x} \geq 0$ wich is true
if : $ x\geq 2$:
we have $ \frac{1}{b^2\plus{}1}\plus{}\frac{1}{c^2\plus{}1} \geq \frac{1}{\frac{1}{c^2}\plus{}1} \plus{}\frac{1}{c^2\plus{}1}\equal{}1$  (because $ bc\leq 1$)
it remain to prove that:
$ x\plus{}\frac{1}{x} \geq \frac{5}{2}$
wich is equivalent to :
$ \frac{(2x\minus{}1)(x\minus{}2)}{2x}\geq 0$ wich is true
P.S:
this ineq can solve the following :
$ (a,b,c)\geq 0$ such that: $ ab\plus{}ac\plus{}bc\equal{}1$ then:
$ \sum{\frac{1}{a\plus{}b}} \geq \frac{5}{2}$ , ( notice that : $ a\equal{}\frac{1\minus{}bc}{b\plus{}c}$) :)

[quote="anas"]
P.S:
this ineq can solve the following :
$ (a,b,c)\geq 0$ such that: $ ab \plus{} ac \plus{} bc \equal{} 1$ then:
$ \sum{\frac {1}{a \plus{} b}} \geq \frac {5}{2}$ , ( notice that : $ a \equal{} \frac {1 \minus{} bc}{b \plus{} c}$) :)[/quote]

Great :)
I solved the following link just now.
http://www.mathlinks.ro/viewtopic.php?t=205057


rotation
geometry
geometric transformation
dilation
trigonometry
AMC
AIME

Isosceles $ \triangle ABC$ has a right angle at $ C$.  Point $ P$ is inside $ \triangle ABC$, such that $ PA \equal{} 11, PB \equal{} 7,$ and $ PC \equal{} 6$.  Legs $ \overline{AC}$ and $ \overline{BC}$ have length $ s \equal{} \sqrt {a \plus{} b\sqrt {2}}$, where $ a$ and $ b$ are positive integers.  What is $ a \plus{} b$?

[asy]pointpen = black;
pathpen = linewidth(0.7);
pen f = fontsize(10);
size(5cm);
pair B = (0,sqrt(85+42*sqrt(2)));
pair A = (B.y,0);
pair C = (0,0);
pair P = IP(arc(B,7,180,360),arc(C,6,0,90));
D(A--B--C--cycle);
D(P--A);
D(P--B);
D(P--C);
MP("A",D(A),plain.E,f);
MP("B",D(B),plain.N,f);
MP("C",D(C),plain.SW,f);
MP("P",D(P),plain.NE,f);[/asy]

$ \textbf{(A) } 85 \qquad \textbf{(B) } 91 \qquad \textbf{(C) } 108 \qquad \textbf{(D) } 121 \qquad \textbf{(E) } 127$

An illustration showing alan's transformation.

[hide]Instead of Law of Cosines, you can use the Pythagorean Theorem to find the length AB and then the length AC.

\begin{eqnarray*} (AB)^2 &=& 7^2 + (7+6\sqrt{2})^2 \\ (AB)^2 &=& 49 + (49+84\sqrt{2}+72) \\ (AB)^2 &=& 170 + 84\sqrt{2} \\ (AC)^2 &=& 85 + 42\sqrt{2} \end{eqnarray*}[/hide]

A brute-force solution:

[hide]Let $\alpha = \angle BCP$.  Then, by the Law of Cosines, we have

$7^2 = 6^2 + s^2 - (2)(6)(s) \cos \alpha$

$11^2 = 6^2 + s^2 - (2)(6)(s) \cos (\pi/2 - \alpha)$

Solving the first equation gives

$\cos \alpha = \frac{s^2 - 13}{12s}$.

Noting that $\cos (\pi/2 - \alpha) = \sin \alpha = \sqrt{1 - (\cos \alpha)^2}$, we have from the second equation

$85 = s^2 - 12s \sqrt{1 - \left(\frac{s^2 - 13}{12s}\right)^2} = s^2 - \sqrt{144s^2 - (s^2 - 13)^2}$

$\Rightarrow (85 - s^2)^2 = 144s^2 - (s^2 - 13)^2 \Rightarrow s = \pm \sqrt{85 \pm 42 \sqrt{2}}$

You know that the first $\pm$ must be positive, as $s$ must be positive.  The second $\pm$ must also be positive, as otherwise $85 - s^2 > 0$, which cannot occur.  So, we have $a + b= 85 + 42 = \fbox{127}$[/hide]

Yeah, I always think of Law of Cosines whenever I see these problems :lol:

[hide="hint"]rotate the triangle 90 degrees about C

[hide="another hint"]let the image of P be P', what is $\angle P'CP$?[/hide]

[/hide]

It can also be done using coordinates.
[hide]
P(x,y), C(0,0), A(s,0), B(0,s) where s is the leg length.

then

(1) $x^2+y^2=6^2$

(2) $(x-s)^2 + y^2 = 11^2$

(3) $x^2+ (y-s)^2 = 7^2$

After eliminating $x,y$, we get $(85-s^2)^2 + (13-s^2)^2 = 6^2 \cdot (2s)^2$

this is the same equation as appeared in the MathJam.

[/hide]

A question for the ones who got it, how do you figure out that you need rotations??

i had seen the problem that was a point inside an equilateral triangle, and the same idea applied to isosceles right triangles, also i made an almost identical problem in my mock aime, so it was pretty easy

Sorry to revive an old thread...how would you finish the problem?  I can't seem to find an elegant way (always end up with law of cosines)

EDIT: Ah..[hide]
So, we know that $ P'CP$ is a $ 45 \minus{} 45 \minus{} 90$ triangle.  If we extend $ BP$ to hit $ CA$, and we name this intersection point $ D$, then $ \angle CPD \equal{} 45$ because $ \angle CPP' \equal{} 45$ and $ \angle P'PB \equal{} 90$ (converse of the pythagorean theorem).

So, that means that the rotation of the triangles makes $ P'P$ an extension of $ B'P'$ because $ CP'P$ is also $ 45$.

So, use Pythagorean theorem on $ B'BP$ to get the hypotenuse, and divide by $ \sqrt {2}$ to get your answer.

[/hide]

Is that right?

sry 2 bump up the old thread again but another solution i found was

[hide]we can take rotate triangle $PCA$ so $AC$ lands on $BC$ and since we know $PC$ is $6$

the rotated P to C is also $6$ which gives us a $45-45-90$. now we have the length of rotated P to real P which is $6\sqrt{2}$ and the other lengths which are $11$ and $7$ which is a right triangle

thus we know angle $BPC$ is $135$ and it is just law of cosines from there
[/hide]

A different (non-trig, non-bash) solution based off of my solution to a [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_Alabama_ARML_TST_Problems/Problem_15]similar problem[/url]:

[hide="solution"]Note that the area of the triangle uniquely determines the lengths of legs $\overline{AC}$ and $\overline{BC}$, so we proceed to find the area of $\triangle ABC$.

Reflect $P$ across $BC$, $CA$, and $AB$ to points $D$, $E$, and $F$ respectively. Note that $[DBC]=[PBC]$, $[EAC]=[PAC]$, and $[FAB]=[PAB]$. Therefore $[AFBDCE]=2[ABC]$, so it suffices to find the area of $[AFBDCE]$.
Simple angle-chasing yields that $D$, $C$, and $E$ are collinear, so $[AFBDCE]$ reduces to $[AFBDE]$. We now have that

\[[AFBDE]=[AEF]+[BDF]+[DEF]\]

With some more angle-chasing we can show that triangles $\triangle AEF$ and $\triangle BDF$ are isosceles right triangles, with $AE=AF=AP=11$ and $BD=BF=BP=7$. Therefore $EF=11\sqrt{2}$ and $DF=7\sqrt{2}$. Note that $EF^2-DF^2=144=DE^2$, implying that $DEF$ is a right triangle. We can now compute the area of $[AFBDE]$:

\[[AFBDE]=\dfrac{AE^2}{2}+\dfrac{BD^2}{2}+\frac{DE\cdot DF}{2}=\dfrac{121}{2}+\dfrac{49}{2}+42\sqrt{2}=85+42\sqrt{2}\]

It follows that

\[[ABC]=\frac{85}{2}+21\sqrt{2}\Rightarrow s=\sqrt{85+42\sqrt{2}}\Rightarrow a+b=\boxed{127}.\][/hide]

Note that [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2630571&sid=08320096c073d9e53161cf6078104380#p2630571]2012 AIME I Problem 13[/url] can be solved in the same fashion.

[quote=alan][hide]rotate the triangle $45^\circ$ counterclockwise about vertex $B$ and dilate it so that $BC' = BA$
$BP' = 7\sqrt{2}$ and $PBP' = 45^\circ$ so $PP' = 7$
$AP' = 6\sqrt{2}$ and $AP'P$ is a right triangle because $(6\sqrt{2})^2 + 7^2 = 11^2$
apply law of cosines on $\triangle APB$
\begin{eqnarray*} (s\sqrt{2})^2 &=& 7^2+11^2-2(7)(11)\cos{\angle APB}\\ 2s^2 &=& 170-2(7)(11)\cos{90^\circ + \angle APP'}\\ s^2 &=& 85-(7)(11)(-\sin{\angle APP'})\\ &=& 85+(7)(11)\frac{6\sqrt{2}}{11}\\ &=& 85+42\sqrt{2}\\ s &=& \sqrt{85+42\sqrt{2}} \end{eqnarray*}
$85+42 = \boxed{127}$ (E)[/hide][/quote]

Magnificent solution. I've never seen anything similar to this before. Could someone explain the motivation behind this rotation? It feels kind of unwarranted to me...

P.S Brilliant solution, 1=2 as well!

Rotate $\triangle ABC$ $90^\circ$ counterclockwise about $C$, and let $B^\prime$ and $P^\prime$ be the images of $B$ and $P$, respectively.

[asy]
pair A,B,C,D,P,Q;
A=(10,0);
B=(0,10);
C=(0,0);
D=(-10,0);
P=(2.5,4);
Q=(-4,2.5);
draw(A--B--D--cycle,linewidth(0.7));
draw(B--C,linewidth(0.7));
draw(B--Q--C--P--cycle,linewidth(0.7));
draw(P--Q,linewidth(0.7));
label("$A$",A,S);
label("$B$",B,N);
label("$C$",C,S);
label("$B$'",D,S);
label("$P$'",Q,W);
label("$P$",P,E);
[/asy]

Then $CP^\prime = CP = 6$, and $\angle PCP^\prime =
90^\circ$, so $\triangle PCP^\prime$ is an isosceles right triangle. Thus $PP^\prime = 6\sqrt{2}$, and $BP^\prime = AP = 11$. Because $\left(6\sqrt{2}\right)^2 + 7^2 = 11^2$, the converse of the Pythagorean Theorem implies that $\angle BPP^\prime = 90^\circ$. Hence $\angle BPC = 135^\circ$. Applying the Law of Cosines in $\triangle BPC$ gives \[BC^2 = 6^2+7^2-2\cdot 6\cdot 7\cos 135^\circ
= 85+42\sqrt{2},\]and $a+b=\boxed{127}$.


abstract algebra
algebra
polynomial
function
domain
calculus
integration

If a,b are 2 algebric integers show that $Z[a,b]$ is a finitely generated module over $Z[a]$.Will $Z[a,b]$ be free then what will be its rank?
I think it will be the minimal polynomial satisfying b.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=72151 for the first part.

The freeness and the rank comes from the following generalistion:
Let $A$ be a principle ideal domain and $L$ a seperable extension of the field of fractions $K$ of $A$. Then the integral closure $B$ of $A$ in $L$ (and all nontrivial finetely generated $B$-submodules of $L$) are free $A$-modules of rank $n: =[L: K]$.
(note that $rank_{A}(B)=[L: K]$ is trivially correct since any basis of $B$ over $A$ is a basis of $L$ over $K$)

Proof of the last statement:
Take any basis $x_{i}$ of $L$ over $K$. WLOG $x_{i}\in B$.
By theorem 1 of http://www.mathlinks.ro/Forum/viewtopic.php?t=66031 we have $dB \subset A[x_{1},x_{2},...x_{n}] = Ax_{1}+Ax_{2}+...+Ax_{n}$, where $d$ is the extensions discriminant.
Let $M\neq \{0\}$ be any finetely generated $B$-submodule (special case $B=M$) and $y_{1},y_{2},...,y_{m}$ generating elements.
Now there is an $a \in A$ such that $aM \subset B$ and thus $daM \subset dB \subset Ax_{1}+Ax_{2}+...+Ax_{n}\cong A^{n}$.
Now the freeness over $A$ comes directly from the main theorem of modules over principle ideal domains (special case: every submodule of $A^{n}$ is itself free again).
And now $rank(M)=rank_{A}(daM) \leq rank_{A}(dB) \leq n=rank_{A}(A^{n}) \leq rank_{A}(M)$ gives the rest of the claim.


show that if x,y,z are positive integers, then (xy+1)(yz+1)(zx+1) is a perfect square if only if xy+1,yz+1,zx+1 are all perfect squares

if complex, give outline

is here place where question?

The question is placed correctly, but it was posted before a lot of times, see http://www.mathlinks.ro/Forum/viewtopic.php?t=4542 .


Find all the $ a,b\ge \minus{}\frac{5}{2}$ such that :

$ a\plus{}b\le 4$

$ \sqrt{2a\plus{}5}\plus{} \sqrt{2b\plus{}5} \ge 6$

Let $ x\equal{}2a\plus{}5, y\equal{}2b\plus{}5$, then the question becomes:

$ x,y \ge 0$, $ x\plus{}y \le 18$ and $ \sqrt{x}\plus{}\sqrt{y} \ge 6$.

It's easy to check that $ x,y \neq 0$, then we have

$ 9 \ge \frac{x\plus{}y}{2} \ge (\frac{\sqrt{x}\plus{}\sqrt{y}}{2})^2 \ge 9$

Equality holds $ \Leftrightarrow x\equal{}y$, which gives $ x\equal{}y\equal{}9$ and thus $ a\equal{}b\equal{}2$ as solution.


Let $ABC$ is triangles :$AB=c,BC=a,CA=b$ with $I$ is intercenter of $ABC$.Let$IA=d_a;IB=d_b;IC=d_c$.
Prove that:
$\sum \sqrt{a(bc-d_a)^2}\le \sqrt{6abc}$.

Prove that:
$2x^2+y^2+5z^2+6xy+7xz+2yz>0$.
With $x,y,z$ is real number $s.t$ $x+y+z<0,4xz>y^2$.


geometry
Hi
angle bisector
incenter
IMO 2004
Angle Chasing
geometry solved

I needed 10 minutes to sort the trash out of the configuration. After that everything was trivial.

If T is the point of intersection of the circumcircles of triangles BMR and CNR, then we must show that T lies on BC, i. e. that we have < BTR + < CTR = $180^{\circ}$. Since BMRT is a cyclic quadrilateral, we have < BTR = $180^{\circ}$ - < BMR, so < BTR = < AMR. Similarly < CTR = < ANR. So we have to show that < AMR + < ANR = $180^{\circ}$, or, in other words, we must prove that the quadrilateral AMRN is cyclic.

Well, the points M and N lie on the circle with diameter BC, and the center of this circle is obviously the midpoint O of BC. Thus, MO = NO, so that the triangle MON is isosceles, and hence the line OR, being the angle bisector of the angle MON, coincides with the perpendicular bisector of the segment MN. Hence, the point R lies on the perpendicular bisector of the segment MN. In other words, the point R is the point of intersection of the perpendicular bisector of the segment MN with the angle bisector of the angle MAN. But a well-known fact states that the perpendicular bisector of a side of a triangle meets the angle bisector of the opposite angle at a point on the circumcircle of the triangle. Hence, the point R lies on the circumcircle of triangle MAN, and therefore, the quadrilateral AMRN is cyclic.

  Darij

One can also prove the cyclicity with the following argument (for those unfamiliar with that well-known fact):

We know that $MR = NR$ and$ \angle MAR = \angle NAR$.

By sine law, ${{MR}\over{\sin MAR}}={{AR}\over{\sin AMR}}$ and ${{NR}\over{\sin NAR}}={{AR}\over{\sin ANR}}~.$

We thus obtain $\sin AMR = \sin ANR~.$ However $\angle AMR \neq \angle ANR$, therefore $\angle AMR + \angle ANR = 180^o$ and the quadrilateral $AMRN$ is cyclic.

For those unfamiliar with said well-known fact, it's trivial to derive:

In $\Delta AMN$, with circumcircle $\omega$, let the internal bisector of $A$ meet $\omega$ at $R'$.  Since $R'$ is the midpoint of arc$MN$ (not containing $A$), the perpendicular bisector of segment $MN$ also intersects $\omega$ at $R'$.  However, because the intersector of the internal bisector of $A$ meets the perpendicular bisector of $MN$ at a unique point, $R' = R$, implying that $ARMN$ is cyclic.

Hence, you didn't need to know this off the top of your head because it's simple to derive.

let's say it all can be done after some easy calculation and by applying inversion respect to a circumference having center in A and such that M'=B and N'=C...
not difficult, just some more elegant (apart fro ugly trigonometry)...
i'm wondering how many can have done it like me :)

[quote="Valentin Vornicu"]1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.[/quote]

Just a reaction from a victim of this problem. One half of all participants who had solutions to this problem got 6 of 7 points, the reasons being i. a.

- a lack of the proof that the common point lies on the [i]segment[/i] BC, not just on the [i]line[/i] BC;
- missing proofs for implicitely used statements about which point lies in which halfplane;
- a definition of the second common point of the two circumcircles without considering the (only theoretically) possible case when they touch.

Of course, if the coordinators have nothing better to do, they are allowed to bitch around about this, but I believe it is under the niveau of a reasonable IMO... just my two pennies. I have got 6 points on the problem, I was really under stress and randalated, but I still have 27 points and get silver...

  darij

[quote="darij grinberg"]a victim of this problem[/quote]

LOOOL  :cool:
Are you mad ????

6 is good ! and 27 is good too !! I wish I had 27  :lol: (All my team too  :cool: )

[quote="ltsin"]further more we can get that $R$ is the incenter of $\triangle MON$ and common point of the two circumcircles  $P$ lying on BC is on the bisector of $\angle BAC$[/quote]

these results are more interesting than the problem but still easy ... (and proving the first part gives another proof of the problem that automatically proves the intersection lies on the *segment* BC). Let the perpendiculars from C and B to AB and AC be C' and B' (i'm too lazy to think of M and N :) )
first let A' be the point where angle bisector of <BAC intersects BC. <C'RA'=180-<C'RA=180-<C'B'A (because MNAR is concyclic) = 180-<ABC (because B'C'BC is concyclic) so <C'RA' + <C'BA' = 180 proving the concylicity of RC'A'B.
To prove R is the incentre of B'C'O it suffices to show <B'C'O=<C'B'O=2*<RC'B'=<B'AC', i.e. <C'OB'=180 - 2*<BAC (since OB'=OC'). Let A'' be the foot of the altitude from A to BC --> since B'C'OA'' is concylic (nine point circle), <C'OB'=<C'A''B'=2*<C'A''A (because H is the incentre of A'B'C')
=2*<C'CA (since C'A''AC is concylic)=2*(90 - <BAC) = 180 - 2*<BAC as required.

[quote="darij grinberg"][quote="Igor"]LOOOL  :cool:
Are you mad ????

6 is good ![/quote]

Well, but don't forget I did never think to get a point rejected in a simple [b]geometry[/b] problem!

  dg[/quote]

Well, you had to...
How many people got 0 on this problem, eh ?

Antony

[quote="darij grinberg"][quote="Valentin Vornicu"]a small notice, now that everyone is cooled off ;) this problem was proposed by two Romanians, can you guess who? :D:D[/quote]

You and Branzei?

  darij[/quote] close enough. it's actually Dinu Serbanescu (Team Leader of the JBMO Team) and me :P:P

Consider the inversion $f$ with center $A$ and power $AM \times AB$. Then $f(M)=B$, $f(N)=C$.

Lets show that $f(R)=X$, with $X$ on the side $BC$. Let $\Gamma$ the circumcircle of $ABC$, $\Gamma_1$ the circumcircle of $AMN$ and $T=AR\cap \Gamma$.

As in the previous solutions, $MR=NR$ and $AMRN$ is cyclic. Also: $\triangle ANM \sim \triangle ABC$ and $R$, $T$ are  homologous (because $R$ is on $\Gamma_1$ and on the bisector of $\angle MAN$).  Then
\[\frac{AR}{AT}= \frac{AM}{AC}.\quad(1)\]
From similar triangles, it immediately follows that \[\frac{AX}{AC}=\frac{AB}{AT} .\qquad (2)\] ($\triangle AXB \sim \triangle ACT$).  From (1) and (2),
\[ AX\times AR=AB\times AM\] and then $f(R)=X$.

Is hellas in greece? cyclin then angle chasign

[hide=solution] Let $X$ be the second intersection of $(BMR)$ and $(CNR).$

[b] Claim 1: [/b] $A$ lies on line $RX.$
[b] Proof: [/b] By Power of a Point on $(BCNM),$ we have $AN \cdot AC = AM \cdot AB.$ Thus, $A$ has equal powers with respect to $(BMR)$ and $(CNR)$ which implies that $A$ is on the radical axis of these two circles which is line $RX.$ $\blacksquare$

[b] Claim 2: [/b] $AMRN$ is cyclic.
[b] Proof: [/b] By the Miquel's Point Theorem, $(BMX), (CNX),$ and $(AMN)$ are concurrent, which implies that $R$ is on circle $(AMN),$ so $AMNR$ is cyclic. $\blacksquare$

Now, angle chasing, we get
\begin{align*} \angle{RXC} + \angle{RXB} &= (180^\circ - \angle{RNC}) + (180^\circ - \angle{RMB}) \\
&= 360^\circ - \angle{RNC} - \angle{RMB} \\
&= 360^\circ - (180^\circ - \angle{ANR}) - (180^\circ - \angle{AMR}) \\
&= 360^\circ - 360^\circ + \angle{ANR} + \angle{AMR} \\
&=180^\circ,
\end{align*}
so $B,X,$ and $C$ are collinear. Since $X$ is on line $AR$ and $R$ is in the interior of $\triangle ABC,$ $X$ must be in the interior of $\angle ABC,$ so $X$ is on segment $BC,$ as desired $\square$

[/hide]

Let $E$ be intersection point of circles $BMRE$ and $ERNC$
Let $D$ be intersection of $OR$ and $MN$
Because $O$ is the center of the circle, $OD$ is the perpendicular bisector of $MN$
Construct circumcircle of $AMN$
The perpendicular bisector of $MN$ intersects the circle at the middle of arc $MN$. Let that point be $Z$
Angles $\angle MAZ$ and $\angle ZAN$ correspond to an arc of length half of $MN$, so $\angle MAZ = \angle ZAN$ and $AZ$ is the angle bisector of $\angle MAN$, so the angle bisector of $\angle MAN$ and the perpendicular bisector of $MN$ intersect at $Z$. They also intersect at $R$, so $Z=R$ and because $AMZN$ cyclic, $AMRN$ cyclic
Then use converse of Miquel to get that $E$ is on $BC$

Bruh this seriously took me half an hour because of the trash in my config that made it look like stuff I was given that I wasn't.
Anyways, since MON is isosceles it's angle bisector is equidistant from M and N and hence R on this line means MR=NR with <MAR=<NAR implies sinAMR/AR=sinMAR/MR=sinNAR/NR=sinANR/AR and hence either AMR=ANR or ANR=180-AMR. Indeed, if AMR=ANR then congruent triangles and basic trig yields ACcosCAM=AM=AN=ABcosBAN, contradiction. So note that AMRN is now cyclic, and now if BMR intersects BC at D, Miquel says that CNRD is cyclic, meaning that the intersections of the circles in question indeed satisfy the problem statement. $\blacksquare$ (Again, minimal latex since typing up this cr-- takes a long time)

[hide = solution]
Redefine $R$ as the midpoint of arc $MN$ not containing $A$. Since $OM = ON, RM = RN$ and $\angle RAB = \angle RAC$ this definition is equivalent to the original one. But the problem is trivial by Miquel now.
[/hide]

[asy]
            size(250,250);
            label("$B$", (0,0), W); label("$A$", (5,12), N); label("$C$", (16,0), E);

            label("$M$", (2.366864,5.608473), W); filldraw(circle((2.366864,5.608473),0.05),black); label("$N$", (8.69434,7.96981), NE); filldraw(circle((8.69434,7.96981),0.05),black); label("$R$", (6.261667663,4.8045573), SW); filldraw(circle((6.261667663,4.8045573),0.05),black); label("$O$", (8,0), S); filldraw(circle((8,0),0.05),black);

            draw((0,0)--(16,0)--(5,12)--(0,0));
            draw((2.366864,5.608473)--(6.261667663,4.8045573)--(8.69434,7.96981));
            draw((5,12)--(6.261667663,4.8045573)--(8,0));
            draw((8,0)--(8.69434,7.96981)--(2.366864,5.608473)--(8,0));
[/asy]

[color=#00f][b]1. $MR=NR$.[/b][/color]
Notice that since $M$ and $N$ are on the circle with diameter $BC$ and center $O$, we have that $OMN$ is isosceles with $OM=ON$. Since $OR$ is also the angle bisector of $\angle MON$, this implies that $R$ is on the angle bisector of $MN$, meaning that $MR=NR$.

[color=#00f][b]2. $AMRN$ is cyclic.[/b][/color]
Note that since by [color=#00f][b](2)[/b][/color], $MR=RN$ and $R$ is on the angle bisector of $\angle MAN$, we know that $R$ must be the arc midpoint of arc $MN$ not containing $A$ of circle $(AMN)$. Therefore $AMRN$ is cyclic.

[color=#00f][b]3. Finishing.[/b][/color]
Finally, if we let $R$ be the Miquel point, and $M$ and $N$ be the chosen points on $AB$ and $AC$, by the inverse of Miquel's theorem, we have that $(BMR)$ and $(CNR)$ must intersect at some point $X$ other than $R$ on $BC$, finishing the problem.

Let the angle bisector from $A$ hit $BC$ at $D$. We claim that the given circumcircles pass through that point.

We claim that the problem is equivalent to $AMRN$ cyclic

proof: We let $\angle NMO = x$ and $\angle BMO = y$. Since $OB = OM$ we have $\angle MBO = y$. Since $BCNM$ is cyclic we have $\angle NCO = \angle 180-x-y$. We can now see that $\angle BAC = x$ and $\angle DAC = \angle DAB = \frac{x}{2}$ This means that $\angle RDB = 180 - y - \frac{x}{2}$ and $\angle RDC = y + \frac{x}{2}$. We need therfore that $\angle RNC = 180 - y - \frac{x}{2}$ and $\angle RMB = y + \frac{x}{2}$. We need therefore that $\angle RMN = \angle RNM = \frac{x}{2}$ obviously this is equivalent to $AMRN$ cyclic.  $\square$

Since $OR$ is the angle bisector of $MON$ we can see that $MR = RN$. We look at the circumradii of $ANR$ and $AMR$. We can see from LOS that we have those circumradii equal since $\frac{RM}{2 \sin \frac{x}{2}} = \frac{RN}{2 \sin \frac{x}{2}}$. Using LOS again we need that $\frac{AR}{\sin \angle AMR} = \frac{AR}{\sin \angle ANR}$ This means we could either have $\angle AMR = \angle ANR$ or $\angle AMR + \angle ANR = 180$ (note the second implies cylic). The first condition is impossible because we have $AB \neq AC$. $\blacksquare$

[hide = remarks]The motivation for letting $D$ be the point that both circles pass through is realizing that $AR$ is the radical axis so if the circles both pass through a point on $BC$ then it must be $D$[/hide]

Observe that $R$ is the midpoint of minor arc $\widehat{MN}$ in $(AMN)$, as $AO$ is the $A$-symmedian of triangle $AMN$. Therefore as $AMNR$ is cyclic, by Miquel's theorem, as these circles meet at an internal point $R$, we must have that $(BMR)$ and $(CNR)$ meet at some point on the segment $\overline{BC}$, as desired. $\blacksquare$

[hide = Remark] I question if we really need the requirements that $AR$ even bisect $\angle ABC$ and similarly with $OR$ bisecting $\angle MON$. Indeed we have the beneath proof, which utilizes none of the statements in the problem, yet provides a general and coherent answer. Would this solution then be erroneous? Indeed we also must have $X$ must lie on [i]segment[/i] $\overline{BC}$ as opposed to just $BC$, as you would then trivially have a degenerate diagram.

Denote by $\measuredangle$ as the directed angle modulo $\pi$.

Let $X$ be the second point of intersection of $(BMR)$ and $(CNR)$ not equal to $R$. Observe;
\[\measuredangle BXR = \measuredangle AMR \text{\phantom{c} and \phantom{c}} \measuredangle CXR = \measuredangle ANR,\] so that $\measuredangle BXR = \measuredangle CXR$, so that $B, X, C$ are collinear. $\square$

I thought over it, and we do indeed require that $R$ has the problem condition, as we find out that $A, M, R, N$ are concylic, and [i]then[/i] we may proceed with the above angle chase. [/hide]

[hide][b]Claim:[/b] $AMRN$ is a cyclic quadrilateral
[b]Proof:[/b] We have $OM=ON$ and $\angle MOR=\angle NOR$
We obtain from these two facts that $OR$ is the perpendicular bisector of $MN$ and thus $OM=ON$.
Now by the Incenter-Excenter Lemma, we know that OR lies on $(AMN)$
                                                                                Now by Miquel's theorem, $(CNR)$ and $(BNR)$ must meet at some point on the line $BC$. Now as $R$ ,(The Miquel Point) lies inside $\triangle ABC$ we can say that the two aforementioned circles must meet somewhere on segment $BC$.
Our proof is thus complete. [/hide]

ONK.

Note that $ANM \sim ABC$ as we have $\angle AMN = 180 - \angle NMB = \angle NCB$.

Also it is easily deduced that $\angle MON = 180 - 2A$. Now the problem is solved if $AMRN$ is cyclic. To see this note that we have $\angle ROM = 90 - A$, so $\angle NMO = A$. Thus $MO$ is tangent to $(AMN)$. Similarly $NO$ is tangent to $(AMN)$. However then the angle bisectors of $MAN$ and $NOM$ concide on $(AMN)$ at the midpoint of $\widehat{MN}$, and thus we must have $R \in (AMN)$.

Now let $D = AR \cap BC$. Then we will show $D \in (RMN)$ and similarly $D \in (RNC)$. To see this note we have $$\angle MBD = \angle ANM = \angle ARM = 180 - \angle MRD$$ hence proven.

Solved with the nooks [b]peace09[/b].

Note that since $ON=OM$,  $R$ lies on the perpendicular bisector of $MN$. Since it also lies on the bisector of $\angle NAM$ it is the midpoint of $\overarc{MN}$ on $(AMN)$. Now since $AMRN$ is cyclic, by Miquels theorem $(MRB)$, and $(RNC)$ intersect on $BC$. $\square$

This problem is pretty easy with Miquel, trig, or a variety of other methods. But where's the modern art in that?

We use barycentric coordinates with $\triangle ABC$ as the reference. Let $AB = c$, $BC = a$, and $CA = b$. Let $A = (1,0,0),B = (0,1,0), C = (0,0,1)$. Let $S_A = \frac{b^2 + c^2 - a^2}{2} = bc\cos(A)$ (by Law of Cosines) and cyclically define $S_B,S_C$. Note that $\angle BMC = \angle BNC = 90^\circ$. Thus $BM = a\cos(B)$ and $AM = b\cos(A)$, so we get $M = (S_B:S_A:0)$ and $N = (S_C:0:S_A)$. Also, $O$ is the midpoint of $BC$ and is thus at $(0:1:1)$.

Our first order of business is to find $R$. Notice that, since $M,N$ lie on a circle with center $O$, the midpoint of $MN$ is collinear with $R,O$. Let the midpoint of $MN$ be $X$. Since the coordinates of $M$ we found above sum to $c^2$ and the coordinates of $N$ sum to $b^2$, the midpoint is
\[b^2(S_B:S_A:0) + c^2(S_C:0:S_A) = (b^2S_B + c^2S_C:b^2S_A:c^2S_A).\]
The coordinates here sum to $2b^2c^2$. Now, since $R$ lies on the angle bisector from $A$, the ratio of its $y$ and $z$ coordinates is $\frac{b}{c}$ (by the angle bisector theorem, $AR$ splits $BC$ into segments of length ratio $c$ to $b$). Then, since $R,X,O$ lie on a line so if the $x$-coordinate of $R$ is $x$, we get
\[\begin{vmatrix}
0 & 1 & 1 \\
b^2S_B + c^2S_C & b^2S_A & c^2S_A \\
x & b & c
\end{vmatrix} = 0.\] Evaluating by minors on the top row gives
\[(b^2S_B + c^2S_C)(b - c) + xS_A(c^2 - b^2) = 0\] so
\[x = \frac{b^2S_B + c^2S_C}{S_A(b + c)}.\]

Now, we will find the equation of circle $(BMR)$. Let it have equation
\[-a^2yz - b^2xz - c^2xy + (x + y + z)(ux + vy + wz) = 0\] (the generic equation for a circle). Plug in $B = (0,1,0)$ to get $v = 0$. Plug in $M = (S_B:S_A:0)$ to get $-c^2S_BS_A + c^2(uS_B) = 0$ so $u = S_A$. Now, plug in $R = \left(\frac{b^2S_B + c^2S_C}{S_A(b + c)}:b:c\right)$. We get
\begin{align*}
0 &= -a^2bc - bc(b + c)\frac{b^2S_B + c^2S_C}{S_A(b + c)} + (x + y + z)(S_Ax + wz) \\
bc\left(a^2 + \frac{b^2S_B + c^2S_C}{S_A}\right) &= \left(\frac{b^2S_B + c^2S_C + (b + c)^2S_A}{S_A(b + c)}\right)\left(\frac{b^2S_B + c^2S_C}{b + c} + wc\right) \\
bc\cdot \frac{a^2S_A + b^2S_B + c^2S_C}{S_A} &= \left(\frac{2b^2c^2 + 2bcS_A}{S_A(b + c)}\right)\left(\frac{b^2S_B + c^2S_C}{b + c} + wc\right) \\
\frac{(b + c)(a^2S_A + b^2S_B + c^2S_C)}{2(bc + S_A)} &= \frac{b^2S_B + c^2S_C}{b + c} + wc. \\
\end{align*}
Now, note that \[a^2S_A + b^2S_B + c^2S_C = \frac{-a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2}{2}.\]
Those who have expanded Heron's formula will recognize this. Let $s = \frac{a + b + c}{2}$. Then, this is just $8s(s - a)(s - b)(s - c)$, which is easy to prove by expansion. However, $bc + S_A = \frac{b^2 + 2bc + c^2 - a^2}{2} = 2s(s - a)$. Thus, we have:
\begin{align*}
	\frac{(b + c)(8s(s - a)(s - b)(s - c))}{4s(s - a)} &= \frac{b^2S_B + c^2S_C}{b + c} + wc \\
    2(b + c)(s - b)(s - c) &= \frac{b^2S_B + c^2S_C}{b + c} + wc \\
    (b + c)^2(a + b - c)(a - b + c) &= b^2(a^2 + c^2 - b^2) + c^2(a^2 + b^2 - c^2) + 2wc(b + c) \\
    (b + c)^2(a^2 - b^2 - c^2 + 2bc) &= -b^4 - c^4 + a^2b^2 + a^2c^2 + 2b^2c^2 + 2wc(b + c) \\
    (b + c)^2(a^2 - b^2 - c^2 + 2bc) &= a^2b^2 + a^2c^2 - (b^2 - c^2)^2 + 2wc(b + c) \\
    (b + c)^2(a^2 - b^2 - c^2 + 2bc) &= a^2b^2 + a^2c^2 - (b + c)^2(b - c)^2 + 2wc(b + c) \\
    (b + c)^2(a^2) &= a^2b^2 + a^2c^2 + 2wc(b + c) \\
    w &= \frac{a^2b}{b + c}.
\end{align*}

Now, we will find the intersection of this circle with $BC$, which is where $x = 0$. By scaling, assume $z = 1$ and plug $(0:y:1)$ into the circle equation and get
\[-a^2y + (y + 1)w = 0.\] Thus, $y = \frac{w}{a^2 - w} = \frac{b}{c}$. This means that the second intersection of $(BMR)$ with $BC$ is at $(0:b:c)$, which happens to be the intersection of $AR$ with $BC$ (wow, if only we could have found that synthetically). This is symmetric in $B,C$, so it must be the same for $(CNR)$ as well by symmetry. Thus, we're done.

By radical axis/radical center, we see that $A$ is the radical center of $(BCNM), (BMR)$ and $(CNR).$ Furthermore, we see that AR is the radical axis of the circles $(BMR)$ and $(CNR).$
Note that $AR$ bisects $\angle BAC.$ Let $AR$ intersect $BC$ at point $D.$ We wish to show that points $B, M, R, D$ and $C, N, R, D$ are concyclic.
By sum of angles on triangle $ABD,$ we get that $\angle BDA=180- \angle B -\frac{\angle A}{2}.$ Since $BOM$ is an isoceles triangle, $\angle BMO=\angle B.$
Therefore the problem reduces to showing that $\angle OMR=\frac{\angle A}{2}.$ Firstly, we can angle chase to get $\angle OMN=\angle ONM=\angle  A.$ Then by fact 5, $R$ lies on $(AMN)$ so by cyclic quads, $\angle NMR=\frac{\angle A}{2}.$ This means $\angle OMN=\frac{\angle A}{2}.$

Therefore, $\angle BMR+\angle BDR=180$ and $BMRD$ is a cyclic quadrilateral. By similar steps on points $C, N, R, D$ we get that $CNRD$ is a cyclic quadrilateral and we're done.

The diagram is harder to draw than the actual problem.

Observe that $M$ and $N$ are the feet of the altitudes from $C$ and $B$ into $AB$ and $AC$ respectively. It is well-known that $O$ is the intersection of the tangents of $(AMN)$ from $M$ and $N$, hence $R$ lies on $(AMN)$. Now define $D$ as the intersection of the angle bisector of $\angle BAC$ and $BC$, $H$ the orthocenter of $\triangle ABC$, and $K$ the foot of the altitude from $A$ into $BC$. We have

$$\angle AMR = 90^\circ - \angle HMR = 90^\circ - \angle KAR = 90^\circ - \angle KAD = \angle RDB.$$

But this implies quadrilateral $BMRD$ is cyclic, a similar argument shows $CNRD$ is cyclic as well. Hence, $(BMR)$ and $(CNR)$ have a common point, $D$ lying on $BC$, as desired.


A circle is inscribed in a triangle with sides length of 4 cm, 13cm and 15 cm. Find the area of the circle .

(A) 1.5π cm2 (B) 2.25π cm2 (C) 3π cm2 (D) 3.75π cm2 (E) 4π cm2

We have a=4, b=13 and c=15.

$ r \equal{} \frac {2P_{\Delta}}{a \plus{} b \plus{} c}$ where $ P_{\Delta} \equal{} \sqrt {p(p \minus{} a)(p \minus{} b)(p \minus{} c)} \Rightarrow P_{\Delta} \equal{} \sqrt {16(16 \minus{} 4)(16 \minus{} 13)(16 \minus{} 15)} \equal{} 24$,

therefore $ r \equal{} \frac {2 \cdot 24}{4 \plus{} 13 \plus{} 15} \equal{} \frac {3}{2}$, so  $ P \equal{} \pi r^2 \Rightarrow P \equal{} 2,25 \pi [cm^2]$


Um, I noticed that a number of the form $ 10^n\minus{}1$ ($ n$ an integer) has the same digit sum as its double ($ 999\cdot 2\equal{}1998$, $ 9999\cdot 2 \equal{} 19998$, etc.). So my guess would be $ 245\cdot 10^{10}\minus{}1$.

[quote="weyming001"]Hello....

I'm a newbie here..... I need help.....

Find the least positive integer N such that the sum of its digits is 100 and the sum of the digits of 2N is 110.

Thanks. :blush:[/quote]This is already solved in Olympiad Section:

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