Well, this problem was one i just recently remembered, and thought you guys might want to try.

This is an easier version, but it's still slightly challenging:

In a game, if you roll two dice and the sum of the top faces is 7 or 11, you win, but only if it is on the first try.  If you roll anything else, the only way to win is to roll two fours (don't have to be consecutive), but if you roll a sum of 7 after the first one, you lose the game.  Given that the first roll's sum is 4, what is the probability that you win by getting another 4?

[hide]well, there is a 6/36 chance of getting a 7, 3/36 chance of getting a 4..

so (3/36)+(27/36)-cant get a 4 or 7-(3/36)+(27/36)^2(3/36)... infinite geometric series..so we sum that up and get 1/3?[/hide]

Madskillz1:
[hide]Correct, but could you explain the reasoning better, I didn't quite get it[/hide]

ALright..first of all the question is too comlicated.  the question can be rephrased as what is the probability that you get a 4 before you get a 7 (you dotn need to worry about the 4 on the first toss because it already stated that the first toss is a 4)

so [hide]
you can get a 4 by 1,3 2,2 3,1
you can get a 7 by 1,6 2,5 3,4 4,3 5,2 6,1

so there are 36 total differnet possibilities
on one roll of two dice there is a 3/36 chance you get a 4, 6/36 chance you get a 7

now you want to find the probability that you get a 4 before a 7..so chance you get the 4 on the first try+chances on second try+chances on third try..and so on

so (3/36)-1st try
(27/36)(3/36)-2nd try..the first is 27/36 because you cannot get a 7(you will lose) and you cannot get a 4 (then you will already have won)..so just keep on doing that until infinity and use the infinite geometric series[/hide]

Ok i see, but now i know why i didn't get it ... because they weren't simplified fractions -_-, and i was like whoa... why aren't they simplified, and i thought you had a reason, but i guess it really doesn't matter.  :lol:


1. Find all solutions of $ (x^2\plus{}7x\plus{}6)^2 \plus{} 7(x^2\plus{}7x\plus{}6)\plus{}6 \equal{} x$
2. Let ABC be a triangle where AB = 25 and AC = 29.  $ C_1$ is a circle that has AB as a diameter and $ C_2$ is a circle that has BC as a diameter.  D is a point on $ C_1$ such that BD = 15 and CD = 21.  $ C_1$ and $ C_2$ always intersect at B, however, they also intersect at another point.  Let this point be called E.  Find all possible values of DE.

#1 reminds me of HMMt 2006 Algebra question 8: $ z^4\plus{}4z^2\plus{}6\equal{}z$

[hide="Sketch of solution"]
Observe that a solution to $ x^2\plus{}7x\plus{}6\equal{}x$ is a solution to the given equation.

Then, we can subtract $ x$ from the given equation and factor out $ (x^2\plus{}7x\plus{}6\minus{}x)\equal{}(x^2\plus{}6x\plus{}6)$ and solve the other quadratic.[/hide]


conics
parabola
analytic geometry
quadratics
Vieta

Tangents are drawn from the point $T(2,-1)$ to the parabola $x^{2}=4y$. P and Q are the points of contact of the tangents.

a) Find the equation of the chord PQ
$xx_{0}= 2a(y+y_{0})$
$y=x+1$

b) Show that the x-coordinate of P and Q are the roots of the quadratic $x^{2}-4x-4$

From part a) x = -1 $\Rightarrow$ subbing into quad. = 0 (approx.)

c)Find the sum of the roots of the equation in part b)

Using quad. formula roots are -1 and 5
therefore sum = 4

[b]d) Hence, find the midpoint M of the chord PQ, and show that TM is // to the axis of the parabola[/b]

How is part (d) related to part (c)

Well, the sum of the roots divided by 2 will be the $x$ coordinate of the midpoint of $PQ$, so you can easily find the $x$ coordinate using [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formula]Vieta's Formula[/url].


quadratics
inequalities proposed
inequalities
109

[quote="Ji Chen"]If $ a,b,c,d,e$ are real numbers, then

$ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2) \plus{} (\sqrt {5} \minus{} 1)(ac \plus{} bd \plus{} ce \plus{} da \plus{} eb)$

$ \geq(\sqrt {5} \plus{} 1)(ab \plus{} bc \plus{} cd \plus{} de \plus{} ea).$[/quote]
$ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2) \plus{} (\sqrt {5} \minus{} 1)(ac \plus{} bd \plus{} ce \plus{} da \plus{} eb)\minus{}(\sqrt {5} \plus{} 1)(ab \plus{} bc \plus{} cd \plus{} de \plus{} ea)$
$ \equal{}\frac{\sum_{cyc}((\sqrt{5}\plus{}1)(b\plus{}e)\minus{}(\sqrt{5}\minus{}1)(c\plus{}d)\minus{}4a)^2}{20} \geq 0$

Let $ u,t$ be reals. Then

$ (a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}e^{2})\plus{}t(ac\plus{}bd\plus{}ce\plus{}da\plus{}eb) \plus{}u(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)\geq 0$

holds for all reals $ a,b,c,d,e$ if and only if

$ 1\plus{}t\plus{}u\geq 0$, $ 4\minus{}t\minus{}u\minus{}\sqrt{5}|t\minus{}u|\geq 0$, $ 4\minus{}t\minus{}u\plus{}\sqrt{5}|t\minus{}u|\geq 0$.

[quote="Parkdoosung"]Let $ u,t$ be reals. Then

$ (a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2} \plus{} e^{2}) \plus{} t(ac \plus{} bd \plus{} ce \plus{} da \plus{} eb) \plus{} u(ab \plus{} bc \plus{} cd \plus{} de \plus{} ea)\geq 0$

holds for all reals $ a,b,c,d,e$ if and only if

$ 1 \plus{} t \plus{} u\geq 0$, $ 4 \minus{} t \minus{} u \minus{} \sqrt {5}|t \minus{} u|\geq 0$, $ 4 \minus{} t \minus{} u \plus{} \sqrt {5}|t \minus{} u|\geq 0$.[/quote]
just need
$ 1 \plus{} t \plus{} u\geq 0$, $ 4 \minus{} t \minus{} u \minus{} \sqrt {5}|t \minus{} u|\geq 0$
because
$ 4 \minus{} t \minus{} u \minus{} \sqrt {5}|t \minus{} u|\geq 0$
$ \rightarrow4 \minus{} t \minus{} u \plus{} \sqrt {5}|t \minus{} u|\geq 0$

For 6 variables, we have

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \plus{} f^2 \plus{} 2(p \plus{} q \minus{} 1)(ad \plus{} be \plus{} cf)$

$ \geq p(ab \plus{} bc \plus{} cd \plus{} de \plus{} ef \plus{} fa) \plus{} q(ac \plus{} bd \plus{} ce \plus{} df \plus{} ea \plus{} fb)$

holds for all real numbers $ a,b,c,d,e,f$ if and only if $ q\leq 1, 3p \plus{} q\leq 4,p \plus{} q\geq 0.$

A nice special case is $ q\equal{}0,p\equal{}\frac{4}{3}:$

$ 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \plus{} f^2) \plus{} 2(ad \plus{} be \plus{} cf) \minus{} 4(ab \plus{} bc \plus{} cd \plus{} de \plus{} ef \plus{} fa)$

$ \equal{} \frac {1}{6}\sum_{cyc}(3a \plus{} d \minus{} 2b \minus{} 2f)^2\geq 0.$


superior algebra
superior algebra solved


geometry
circumcircle
geometric transformation
reflection
parallelogram
geometry unsolved

Let $ ABC$ be a triangle and $ H$ its orthocenter. The circle with $ AH$ as a diameter intersects the circumcircle of $ BHC$ at $ P$. Prove that $ AP$ passes through the midpoint of $ BC$.

[asy]defaultpen(fontsize(8));
size(250);
pair A=(6,6), B=(8,0), C=(0,0), H, Q, A1=(2,-6);
H = extension(A,foot(A,B,C),B,foot(B,A,C));
Q = foot(H,A,A1);
draw(A--B--C--A);draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B));draw(A--A1--H);draw(C--A1--B);draw(H--Q);
draw(rightanglemark(A,Q,H,8));
label("$A$",A,(0,1));label("$B$",B,(1,0));label("$C$",C,(-1,1));
label("$A'$",A1,(-1,-1));label("$H$",H,(1,1));label("$Q$",Q,(-1,1));[/asy]Let $ A'$ be a point on the opposite side of $ BC$ as $ A$ such that $ \triangle A'CB \cong \triangle ABC$, and let $ Q$ be the projection of $ H$ onto $ AA'$.
Since the reflection of $ H$ over $ BC$ is on the circumcircle of $ \triangle ABC$, we have that $ A'$ is on the circumcircle of $ \triangle BCH$.
$ \angle A'CH \equal{} \angle A'CB \plus{} \angle BCH \equal{} \angle ABC \plus{} \angle HAB \equal{} 90$. Therefore $ A'CQHB$ is cyclic. Also $ Q\equal{}P$.
Since $ ABA'C$ is a parallelogram, we have that $ AQ\equal{}AP$ passes through the midpoint of $ BC$.


algebra
polynomial
geometry
3D geometry
number theory

Find all $ a,b,c,d$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}0$ and $ a^3\plus{}b^3\plus{}c^3\plus{}d^3\equal{}24$.

I got until \[ bc(c\plus{}b)\plus{}cd(c\plus{}d)\plus{}bd(b\plus{}d)\plus{}2cbd\equal{}8\] and from there to \[ b(d\plus{}\frac{b}{2})^{2}\plus{}c(b\plus{}\frac{c}{2})^{2}\plus{}d(c\plus{}\frac{d}{2})^{2}\plus{}2bcd\minus{}\frac{b^{3}}{4}\minus{}\frac{c^{3}}{4}\minus{}\frac{d^{3}}{4}\equal{}8\]but now I'm totally stuck.

hello, from the first equation we get
$ b\equal{}\minus{}a\minus{}c\minus{}d$
inserting this in the second equation expanding and simplifying we have
$ \minus{}3a^2c\minus{}3a^2d\minus{}3ac^2\minus{}6acd\minus{}3ad^2\minus{}3c^2d\minus{}3cd^2\minus{}24\equal{}0$
solving this for $ a$ we get
$ a_1\equal{}\frac{1}{2}\frac{\minus{}(d\plus{}c)^2\plus{}\sqrt{(d^2\minus{}c^2)^2\minus{}32c\minus{}32d}}{c\plus{}d}$
$ a_2\equal{}\minus{}\frac{1}{2}\frac{(d\plus{}c)^2\plus{}\sqrt{(d^2\minus{}c^2)^2\minus{}32c\minus{}32d}}{c\plus{}d}$
and so on.
Sonnhard.

Let $ a,b,c,d$ be the roots of $ t^4\plus{}mt^3\plus{}pt^2\plus{}qt\plus{}r \equal{} 0$.

Let $ S_n \equal{} a^n\plus{}b^n\plus{}c^n\plus{}d^n$. We are given $ S_1 \equal{} 0$ and $ S_3\equal{}24$.

Using [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_sums]Newton's Sums[/url]:
$ S_1 \plus{} m \equal{} 0 \Rightarrow m \equal{} 0$

$ S_3 \plus{} m\cdot S_2 \plus{} p \cdot S_1 \plus{} 3q \equal{} 0 \Rightarrow q \equal{} \minus{}8$

Thus, $ a,b,c,d$ must be the roots of a polynomial in the form: $ t^4 \plus{} pt^2 \minus{} 8t \plus{} r$.

Beyond this, there are no other restrictions.

Er...that doesn't really help. Does anyone have a full solution with some definite answers?

Or should I ask this in like...collegiate topics or something?

Were there any restrictions on $ a,b,c,d$ such as $ a,b,c,d$ must be integers, reals, etc.?

Cause the roots of any polynomial in the form $ t^{4}\plus{}pt^{2}\minus{}8t\plus{}r$ with $ p,r \in \mathbb{C}$ satisfy $ a\plus{}b\plus{}c\plus{}d \equal{} 0$ and $ a^3\plus{}b^3\plus{}c^3\plus{}d^3\equal{}24$.

Given that it is a number theory problem, perhaps $ a,b,c,d$ were intended to be integers?    -- heh its been confirmed, so

[hide]\[ (a\plus{}b)^3 \equal{} \minus{}(c\plus{}d)^3\ \Longrightarrow\ \sum a^3 \equal{} \minus{}3ab(a\plus{}b) \minus{} 3cd(c\plus{}d)\ \Longrightarrow\ (cd\minus{}ab)(a\plus{}b) \equal{} 8\] Suppose wlog $ s \equal{} a \plus{} b > 0$. Then $ (cd\minus{}ab)s \equal{} (c(\minus{}s\minus{}c) \minus{} a(s\minus{}a))s \equal{} (a\plus{}c)(a\minus{}c\minus{}s)s \equal{} 8$.

If $ s\equal{}1$ then $ a\plus{}c, a\minus{}c\minus{}1$ have opposite parities, so $ \{a\plus{}c, a\minus{}c\minus{}s\} \equal{} \pm \{1,8\}$, which gives the solutions $ \{a,b,c,d\} \equal{} \{5,3,\minus{}4,\minus{}4\}$. If $ s\equal{}2$ then $ a\plus{}c,a\minus{}c\minus{}2$ have the same parities so $ \{a\plus{}c,a\minus{}c\minus{}2\} \equal{} \pm\{2,2\}$, which gives the solutions $ \{a,b,c,d\} \equal{} \{3,\minus{}1,\minus{}1,\minus{}1\}$. If $ s\equal{}4$ again the parities of $ a\plus{}c,a\minus{}c\minus{}4$ are the same, so their product cannot be $ 2$. If $ s \equal{} 8$, we get the same solutions from before. Hence $ \{a,b,c,d\} \equal{} \boxed{\{\minus{}4,\minus{}4,3,5\},\{\minus{}1,\minus{}1,\minus{}1,3\}}.$[/hide]

Could you explain it in a much more simple way? I don't know what parities are and I don't know how you got $ (a\plus{}b)^{3}\equal{}\minus{}(c\plus{}d)^{3}\ \Longrightarrow\ \sum a^{3}\equal{}\minus{}3ab(a\plus{}b)\minus{}3cd(c\plus{}d)\ \Longrightarrow\ (cd\minus{}ab)(a\plus{}b) \equal{} 8$. For a dumb person like me, it makes no sense at all.

Sorry I had it typed up in summary, saw that you fixed the problem statement and immediately posted. With details:

When experimenting with fitting the $ a^3 + b^3 + c^3 + d^3$ expression, rather than starting with it and substituting $ a = - (b + c + d)$, it turns out to be easier to start with separately cubing $ a + b,c + d$ as follows:
\begin{align*} & a + b + c + d = 0 \,\Longrightarrow\, a + b = - (c + d) \\
\Longrightarrow\ \ & (a + b)^3 = a^3 + b^3 + 3ab(a + b) = - c^3 - d^3 - 3cd(c + d) = - (c + d)^3 \end{align*} Moving the cubes to one side, and substituting $ a^3 + b^3 + c^3 + d^3 = 24$ gives
\[ 24 = - 3ab(a + b) - 3cd(c + d)\ \Longrightarrow\ - 8 = ab(a + b) + cd(c + d)\] Well $ c + d = - (a + b)$, so we can factor the last statement as $ - 8 = ab(a + b) - cd(a + b) = (ab - cd)(a + b)\ (*)$, and we've written $ - 8$ as a product of a number of integer terms; it looks like casework will suffice now. Unfortunately, there's still way too many variables and too much freedom in that statement, so we need to find a more systematic method of characterizing the quadruples $ (a,b,c,d)$. A simple approach is to define a new variable $ s$ to be $ a + b$. Note also $ - s = c + d$, which means we can rewrite $ (*)$ with three free variables (as $ b = s-a, d = -s-c$):
\begin{align*}8 & = (cd - ab)(a + b) = [c( - s - c) - a(s - a)]s \\
& = [ - sc - c^2 - as + a^2]s \\
& = [(a + c)(a - c) - s(a + c)]s \\
& = (a + c)(a - c - s)s.\end{align*}
We're almost done! In particular, because $ a,b,c,d,s$ are all integers, then $ s$ has to divide into $ 8$, eg $ s \in \pm \{1,2,4,8\}$. We could check all of them, but a shortcut is to notice that if $ s = a + b < 0$, then $ c + d > 0$ and we could have 'swapped' the variables $ (c,d) \mapsto (a,b)$ because all of them were symmetric in the original question. That is, we can make the assumption that $ s > 0$ and cut half of the cases. (why can't $ s = 0$?)

$ \bullet$ If $ s = 1$, then the question becomes $ (a + c)(a - c - 1) = 8$. Well $ 8$ factors as $ (\pm 1) \cdot (\pm 8), (\pm 2) \cdot (\pm 4)$. Now notice that if $ a + c$ is even, then $ a - c$ is even, and if $ a + c$ is odd, then $ a - c$ is odd (just try cases; if $ a$ is even and $ c$ is even, etc). Then $ a + c$ and $ a - c - 1$ have opposite parities, eg if one is odd then the other has to be even and vice versa. Then obviously we can discard $ 2 \cdot 4$ as a possible factorization, so we have to solve the four 2-by-2 systems
\begin{align*} a + c & = [1, - 1,8, - 8] \\
a - c - 1 & = [8, - 8,1, - 1] \end{align*} respectively. Solving them by any manner (take the sum and difference of the equations, etc), and then substituting back into $ s = a + b = - c - d = 1$ gives us the desired quadruples. It turns out that all of the solutions given by these systems is some permutation of the solution set $ \{a,b,c,d\} = \{ - 4, - 4,3,5\}$.

$ \bullet$ If $ s = 2$, we try the same argument. We get $ (a + c)(a - c - 2) = 4$, but this time $ a + c, a - c - 2$ have the same parities (eg if $ a + c$ is even, then $ a - c - 2$ is even, or both are odd). The possible factorizations of $ 4$ are $ (\pm 1)(\pm 4), (\pm 2)(\pm 2)$, but we discard $ 1 \cdot 4$, so $ \{a + c,a - c - 2\} = \pm\{2,2\}$. This gives us another 4 systems, all of which yield some permutation of the solution set $ \{a,b,c,d\} = \{3, - 1, - 1, - 1\}$.

Now try $ s = 4,8$ using very similar arguments. We'll find no new solutions, hence the answer follows. Of course you can verify that indeed $ \{ - 4, - 4,3,5\}, \{3, - 1, - 1, - 1\}$ work.

Cubing $ a \equal{} \minus{} (b \plus{} c \plus{} d)\implies \minus{} a \equal{} b \plus{} c \plus{} d$ actually isn't too bad if we remember the identity

$ (b \plus{} c \plus{} d)^3 \equal{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 3(b \plus{} c)(c \plus{} d)(d \plus{} b)$

$ \implies(b \plus{} c \plus{} d)^3 \minus{} b^3 \minus{} c^3 \minus{} d^3 \equal{} 3(b \plus{} c)(c \plus{} d)(d \plus{} b)$, so

$ \minus{} a^3 \equal{} (b \plus{} c \plus{} d)^3\implies \minus{} 24 \equal{} \minus{} (a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \equal{} 3(b \plus{} c)(c \plus{} d)(d \plus{} b)$

$ \implies(b \plus{} c)(c \plus{} d)(d \plus{} b) \equal{} \minus{} 8$.

We can finish from here easily with casework. In fact, if we substitute $ s\equal{}a\plus{}b$ in $ (a\plus{}c)(a\minus{}c\minus{}s)s\equal{}8$, then we get

$ (a\plus{}c)(\minus{}b\minus{}c)(a\plus{}b)\equal{}8\implies(a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}\minus{}8$, which is equivalent to this statement.


[hide]1. Rewrite in the following way
\[4\cos^{2}\frac{x+y}2-4\cos\frac{x-y}2\cos\frac{x+y}2+1=0 \]
The rest is trivial
2.$x=\frac{\pi}2-u,\ y=\frac{\pi}2-v$, then $\cos u+\cos v-\cos(u+v)=\frac32$[/hide]

I think it has no real  solutions  :maybe:
Because  adding the first two gives
$cos\ x+sin\ x+cos\ y+sin\ y=3$ which is impossible since
$cos\ x+sin\ x+cos\ y+sin\ y\leq 2\sqrt{2}<3$
and we are done

I think those are 2 independent equations(not system)

[quote="April"]Solve the equation:

$1.\quad\cos x+\cos y-\cos{(x+y)}=\frac{3}{2}$

$2.\quad\sin x+\sin y+\cos{(x+y)}=\frac{3}{2}$.[/quote]
I think it easy
Put $a=\frac{x+y}{2}$; $b=\frac{x-y}{2}$$\to$ convemient....


Hi,

I'm looking for good statistics books for an Intro Stats class.

Any suggestions?


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

$\textbf{(A)}\ 392  \qquad\textbf{(B)}\ 420  \qquad\textbf{(C)}\ 588  \qquad\textbf{(D)}\ 600  \qquad\textbf{(E)}\ 750$

[hide]$ 101\overline{ab} \plus{} 101\overline{cd} \equal{} 101(\overline{ab} \plus{} \overline{cd}) \equal{} t^2\implies\overline{ab} \plus{} \overline{cd} \equal{} 101k^2$

$ 20\leqslant\overline{ab} \plus{} \overline{cd}\leqslant 198\implies k \equal{} 1\implies 10(a \plus{} c) \plus{} b \plus{} d \equal{} 101$

We don't want either of the four digits to be zero, since we're trying to maximize their product. Hence $ 2\leqslant b \plus{} d\leqslant 18\implies 83\leqslant 10(a \plus{} c)\leqslant 99\implies a \plus{} c \equal{} 9\implies b \plus{} d \equal{} 11$

Now we want to make $ a,c$ and $ b,d$ as close to one another as possible, since then we'll maximize the product keeping the sum constant. So, $ \{a,c\} \equal{} \{4,5\}$ and $ \{b,d\} \equal{} \{5,6\}$, hence $ P_{max} \equal{} 600$, and the answer is $ D$.[/hide]


search
calculus
real analysis
geometry
geometric transformation
invariant
college

This post seems somewhat misplaced here ... I'm going to send it off to "Other Problem Solving Topics."

[quote="DovesnUnicorns"]Hi all!

Can someone explain Fourier Transformations and apply that to determination of crystal structures in materials based on x-ray diffraction patterns?

Thanks very much![/quote]
WHOA! That is beyond my abilities to do, but that certainly is no small request! That's a subject for an entire textbook, not a post on a message board. I hope you find the answers, though.

What confuted said.

I teach a semester-long course (at the upper division undergraduate level) in Fourier Analysis. Maybe at the very end of the semester, I might wave my hands briefly in the direction of X-ray crystallography. I'm not going to try to parcel out that course in small chunks here.

I use my own notes but there are books. For a textbook I like, I would suggest the one written by David Kammler. You should be able to search for it given the author's name. You do need a solid understanding of calculus, differential equations, and probably linear algebra to understand a book at this level.

Here are a few other key words that might form the basis of some searches:

Complex exponential.
Roots of unity.
Convolution.
Theorem: convolution on one side is pointwise multiplication on the other side.
Translation (or shift) invariant linear operators.
Poisson summation formula.

I ran a search on David Kammler and found his book, "A First Course in Fourier Analysis."  However, many libraries in the San Francisco Bay Area do not carry this book.  Perhaps you know where I can get this book?

Thanks!

Look for it in the UC Berkeley library. If you can't get check-out privaliges, you can try photocopying it a few chapters at a time as you read it. Or, if you want to buy it, you can try Amazon.com or the UC Berkeley bookstore.


geometry
circumcircle
angle bisector
geometry unsolved

In a convex quadrilateral $ABCD$ we have $AB\cdot CD=BC\cdot DA$ and $2\angle A+\angle C=180^\circ$. Point $P$ lies on the circumcircle of triangle $ABD$ and is the midpoint of the arc $BD$ not containing $A$. It is known that the point $P$ lies inside the quadrilateral $ABCD$. Prove that $\angle BCA=\angle DCP$

[i]Proposed by S. Berlov[/i]

Let $F \in BD \cap AP$. Then $AF$ is the angle bisector of $\angle BAD \Rightarrow \frac{FD}{FB}= \frac{AD}{AB}\Rightarrow$

$\frac{FD}{FB}= \frac{CD}{CB}\Rightarrow CF$ is the angle bisector of $\angle DCB$

Let $w=(O,R)$ be the circumcircle of $\triangle ABD$

$2A+C=180 \Rightarrow A<90$

So $\angle BOD = 2A$

as a result, $C$ is on the circumcircle $k$ of $\triangle ODB$ (on the arc $BD$ which doesn't contain $O$)

[hide="Lemma"]
[quote]$A$ is a point on the exterior of $w=(O,R)$ and $M$ is the point of intersection of $w$ and the ray $OA$.
Let $B,C \in w$ and $D \in OA \cap BC$. [u]Then:[/u]

[color=blue]$O,A,B,C$ lie on the same circle $\Leftrightarrow D$ is inverse of $A$ w.r.t. $w\Leftrightarrow M$ is the incenter of $\triangle ABC$[/color]
[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=108097[/hide]

From the lemma we get that the points $F,C$ are inverse points w.r.t. $w$

Again according to the lemma, since $F \in (AP) \Rightarrow$ the angle bisector of $\angle ACP$ is the ray $CO$

Let $\phi = \angle OCA = \angle OCP$ and $\theta = \angle OCD = \angle OCB$

$\angle BCA = \angle BCO+\angle OCA = \theta+\phi$

$\angle DCP = \angle DCO+\angle OCP = \theta+\phi$

I think that my solution is simpler:
Let $X$ be a point lying in exterior of $ABCD$ s.t. $CX=CD$, $\angle BCX=\angle A$. We have:
$\triangle DCX \sim \triangle DPB$ and $\triangle BCX \sim \triangle BAD$, so
$\triangle DCP \sim \triangle DXB \sim \triangle ACB$.Thus
$\angle BCA=\angle DCP$. :D  :play_ball:  :wow:


trigonometry
inequalities
inequalities unsolved

Let $ \triangle ABC$ and $ x$ be a nonegative number.Prove that:
         $ a^{x}cosA\plus{}b^{x}cosB\plus{}c^{x}cosC \leq \frac1{2}(a^{x}\plus{}b^{x}\plus{}c^{x})$

[quote="math10"]Let $ \triangle ABC$ and $ x$ be a nonegative number.Prove that:
         $ a^{x}cosA \plus{} b^{x}cosB \plus{} c^{x}cosC \leq \frac1{2}(a^{x} \plus{} b^{x} \plus{} c^{x})$[/quote]
If $ x\geq1$ we obtain:
$ \sum_{cyc}a^x\cos A\leq\frac1{2}(a^{x} \plus{} b^{x} \plus{} c^{x})\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^{x \plus{} 2} \minus{} a^xb^2 \minus{} a^xc^2 \plus{} a^xbc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^{x \plus{} 2} \minus{} a^{x \plus{} 1}b \minus{} a^{x \plus{} 1}c \plus{} a^xbc) \plus{} \sum_{cyc}(a^{x \plus{} 1}b \plus{} a^{x \plus{} 1}c \minus{} a^xb^2 \minus{} a^xc^2)\geq0,$
which is true by Schur and Miurhead.

Yes, the problem is easy for the appearance. :lol:

Without Loss of Generality, $ a\geq b\geq c\Longleftrightarrow A\geq B\geq C\Longleftrightarrow \cos A\leq \cos B\leq \cos C$, thus

$ a^{x}\cos A \plus{} b^{x}\cos B \plus{} c^{x}\cos C\leq a^{x}\cos B \plus{} b^{x}\cos C \plus{} c^{x}\cos A$

$ a^{x}\cos A \plus{} b^{x}\cos B \plus{} c^{x}\cos C\leq a^{x}\cos C \plus{} b^{x}\cos A \plus{} c^{x}\cos B$

$ a^{x}\cos A \plus{} b^{x}\cos B \plus{} c^{x}\cos C\leq a^{x}\cos A \plus{} b^{x}\cos B \plus{} c^{x}\cos C$

Adding the inequalities gives $ 3(a^{x}\cos A \plus{} b^{x}\cos B \plus{} c^{x}\cos C)\leq (a^{x} \plus{} b^{x} \plus{} c^x)(\cos A \plus{} \cos B \plus{} \cos C)$

$ 3(a^{x}\cos A \plus{} b^{x}\cos B \plus{} c^{x}\cos C)\leq \frac{3}{2}(a^{x} \plus{} b^{x} \plus{} c^x)$ which leads to the result. Q.E.D.

my solution similar to  solution by arqady.
 solution of kunny are also very interesting.


Circle $ O$ is inscribed in a trapzoid $ ABCD$, $ \angle{A}$ and $ \angle{B}$ are all acute angles. A line through $ O$ intersects $ AD$ at $ E$ and $ BC$ at $ F$, and satisfies the following conditions:

(1) $ \angle{DEF}$ and $ \angle{CFE}$ are acute angles.

(2) $ AE\plus{}BF\equal{}DE\plus{}CF$.

Let $ AB\equal{}a$, $ BC\equal{}b$, $ CD\equal{}c$, then use $ a,b,c$ to express $ AE$.

The answer is $\frac{a+c-b}{2}$ and $\frac{2a^2-3ab+2ac-bc}{2(a-c)}$.


Assume that $\sqrt{x}+\sqrt{y}\leqq {k\sqrt{x+2y}}$ for all positive real numbers   $x,y$.
Find the minimum value of $k$.

The required minimum value is $k=\frac{\sqrt6}{2}$.

In order to prove this, it is enough to show that the inequality $\sqrt{x}+\sqrt{y}\leq \frac{\sqrt6}{2}\sqrt{x+2y}$ holds for every two positive real numbers x and y, and that equality is attainable (and, more precisely, it must be attainable for $\sqrt{x+2y}\neq 0$, so that the constant $\frac{\sqrt6}{2}$ cannot be improved).

[i]Proof:[/i] We have to prove the inequality $\sqrt{x}+\sqrt{y}\leq \frac{\sqrt6}{2} \sqrt{x+2y}$. Substituting $\sqrt{x}=a$ and $\sqrt{y}=b$, we can rewrite this as $a+b\leq \frac{\sqrt6}{2} \sqrt{a^2+2b^2}$. After squaring, this takes the form $\left(a+b\right)^2 \leq \frac32 \left(a^2+2b^2\right)$. But this follows from $
\frac32 \left(a^2+2b^2\right) - \left(a+b\right)^2 = \frac12 \left(2b-a\right)^2 \geq 0$. This proves our inequality. Equality holds for 2b - a = 0, i. e. for 2b = a, i. e. for 4y = x; thus, equality is attainable with $\sqrt{x+2y}\neq 0$; and hence, the proof is complete.

  Darij

Thank you,darij.

 In this case, I think the inequality in question isn't symmetric about $x$ and $y$,so your solution is not complete.What do you think on my opinion?

kunny

Why should it necessarily be symmetric?

  Darij

To darij:
How could you get the value of ${k=\frac{\sqrt{6}}{2}}$?

kunny

To Myth:

I couldn't your solution immediately. Now I can understand it.Indeed your solution is elegant, so I have the total of 7 solutions including yours,thank you.

kunny

I want to generalize this problem.

Let $x_k (k=1,2,\cdots, n) $ be positive real numbers and $n$ be positive integer.
Find the minimum value of $c_n$ in terms of $n$ which satisfies the following inequality.

\[\sqrt{x_1}+\sqrt{x_2}+\cdots +\sqrt{x_n}\leqq{c_n}\sqrt{x_1+2x_2+\cdots +nx_n}\]

Thanks again, Myth. :)  :)

Here is my solution.

In using [b]Buniakovsky's inequality[/b],

${[(\frac{1}{\sqrt{1}})^2+(\frac{1}{\sqrt{2}})^2+\cdots+(\frac{1}{\sqrt{n}}})^2][(\sqrt{x_1})^2+(\sqrt{2x_2})^2+\cdots+(\sqrt{nx_n})^2]$
$\geqq[\frac{1}{\sqrt{1}}\sqrt{x_1}+\frac{1}{\sqrt{2}}\sqrt{2x_2}+\cdots+\frac{1}{\sqrt{n}}\sqrt{nx_n}]^2$,which follows [b]Myth's[/b] solution.

kunny

[quote="Myth"]And now short solution ;)
\[\sqrt{x}+\sqrt{y}=3\cdot\frac{\sqrt{x}/2+\sqrt{x}/2+y}{3}\leq 3\cdot \sqrt\frac{x/4+x/4+y}{3},\]
so \[\sqrt{x}+\sqrt{y}\leq \sqrt{\frac{3}{2}}\sqrt{x+2y}\]
and equality holds iff $\sqrt{x}/2=y$.
:D :D[/quote]

Your solution is substantially not shorter than mine. It is more general, of course. I knew weighted AM-GM would work, but I was too lazy to work out the factors...

As for Kunny's question how I guessed the required value of k, I just checked some values on a computer algebra system. Clearly I know the systematic method and I tried it, but made mistakes in calculations several times and finally gave up and did it by trial'n'error...

  darij


I I'm trying to start doing olympiad problems and normally I try them for 30 mins-maybe 45 mins and then look at the solutions. Is this time about right or should I spend more?

In the USAMO and IMO you have an average of 90 minutes to do the problems so you should give yourself more time.

you will probably learn more from solving a problem yourself than looking at the solution, even if it takes you a week.

i dont know whether I am elligible to answer this post ....but I want to pour my views.......
If you dont get a problem after 30-45 mins please dont look at solutions ......
I used to be like this .......but my mentor and my fellow friends said me that was wrong.....I stopped ......
I said that I am very tempted to look at the solution .....they said -That means you just want to look at solutions ......you dont want to be thrilled by mathematics .....you just want to answer a question and you are not expecting challenges and thrill ....you just want to be like this -If given a question you want to answer.....so you do math only when some question is given ...

After this I changed .
It is not a problem even if you sit on a problem for weeks and weeks .......see I presume that you do math only because you like it ........so whenever you try a problem you will relish each and every moment ......in the process of investigation you may also discover beautiful things ........
So when you try a problem be happy that you have not yet solved it and be very happy if you have solved it ......Dont have a motive that math is just problem solving .......do it for pleasure ......so you will do it for a longer and longer  time  ......
But my mentor said me that he will be happy if he has not a solved a problem and will be very sad if he has solved it  :w00tb: ....because there wont be anymore challenge and thrill left ............
But this is too much of a philosophy for me  :stretcher: .....I maintain this when you try a problem be happy that you have not yet solved it and be very happy if you have solved it :D

Well, I suppose srinath is right.

If you were doing intensive olympiad training though, I would say spend no less than 2 hours per problem.

If it is, in a sense, just to get better/become more interested, then spend as much time as you need. Because in all the genuine hours that you spend on it, you are sure to understand more concretely any new ideas, problem solving techniques, and maybe discover something beautiful.

I guess even if you don't have that kind of time, 1.5 hours should still be the minimum.

I disagree with you guys :( ...
I think of course that you must spent plenty of time in EVERY problem you will try.BUT i strongly believe that you can learn A LOT by seeing the solution of others especially in these great site where brilliant minds giving their solutions. :wink:

Dimitris

i agree seeing solutions is not at all wrong but it isnt correct to do so without giving the problem a single try that is more like trying mug things!


Join Happy Face Tourney [size=200]NOW![/size] [size=150]NOW![/size] [size=100]NOW! [/size][size=75]NOW![/size] [size=59]NOW![/size] [size=42]NOW![/size]     [size=25]NOW![/size]

Please sign up to the coolest tournament ever, hosted by your beloved (^_^)

(this is a serious tournament btw)

This will be a team tournament, pick your teammates, and if a team is way too strong or too weak, i will have to switch players to make it more even.

4 person to a team.

Heres how it goes:

Everyone from team $ X$ plays everyone from team $ Y$

Person $ X_i$ from Team $ X$ goes against Person $ Y_i$ from Team $ Y$.

The team with the most wins wins!

The winning team move on to the next round.

Please play responsibly.

[size=50](^_^) will not be liable for any injury of any nature that will most likely result from the tournament. The tournament is not available in all states. Please contact your local (^_^) for more information. [/size]

Many many ppl for 1 group, fasee waisee?  :D :lol:

JUsut of curiosity, do you think a team:

1.RomanianGenius
2. Mewto55555
3. Superllamel
4. FantasyLover

is too strong?

lol

Don't ask questions like that. The answers are bad.  :D  :D

lol the tourney's name is really happy face  :rotfl:

Team Y:
Skaldjfhrulez
Skaldjfhrulez imaginary friend #1
Random Guy
Bob

[quote="RomanianGenius"]Team SuperRomanianFantasy55555

1. RomanianGenius
2. Mewto55555
3. FantasyLover
4. Superllamel

Guys, if you want to change the name tell me, but that's what mew and I agreed on.[/quote]

lol awesome name

think its mostly gonna be SuperRomanianFantasy55555 vs plastic

[quote="budi713"]Team plastic:
budi713
stevenmeow
ra5249
(^_^)

did i get ra's number rite?[/quote]

so u signed everyone up without them knowing it huh?  :lol:

you will not need a team to sign up for this. most people will not already have their teammembers established, and it is not recommended that you do. it'll be better if you sign up as individuals and i will hopefully group you into somewhat equal team.

I'll go and sign up...hopefully I remember about this...

well, they didn't post, so
i'm gonna do the same thing as i did with my summer tourny
if all games can be completed by July 20th, then the tourny can continue
but i doubt it

well, as i said
deadline is july 20th
only 4 more days

Also FTW is being slow for me...can't compete now.

nope
noone's playing their games, so this tourney's basically dead

aw =[
this tourney is now officially dead

srry, this tourney's basically dead
so, no games will go on

has it already started?
iwanna join but i dont have a team

o i didnt read all the stuff
so its too late?
o well
 :ninja:  :ninja:  :ninja:


Hi everyone,
Im currently the math club president, and I wanna assign practice problems for my members during this summer through e-mail, as well as sending the solutions. So do you have any suggestions about how to write the fraction sign, the root sign, draw diagrams and stuff like that through e-mail?. Can I use LATEX to do that through e-mail?
Thanks in advance. :)

If you want to typeset mathematics, send that as an attachment. Plain text is the only universally readable email format.

Since plain text is pretty useless once you know a fair amount of maths, i guess you either need to send your $\LaTeX$ source by mail and ask your members to build it, or an easier thing would be to build it yourself and attach the DVI files.... (they're quite small compared to PDF)

As far as the pictures are concerned, you just need to send them as attachments,mostly in bitmap or JPG formats.....

The members of a math club won't have the software to build from source- you might be able to expect that from a group of math grad students, but not this club. I know everyone can read PDFs- I'm not sure about DVIs.

If I were doing it, I'd attach PDFs- that's what everything defaults to on a Mac.

It is possible to send serious mathematics in plain text pseudo-markup, but that's best used for draft arguments rather than finished documents.

Also, $\LaTeX$ is not the only environment you could typeset problems- they can be done reasonably well if you know the tricks in a word processor.

Thanks for all of your replies.
Can you put LATEX in your Word document, since Im worried that putting them in PDF would make the members' mail box heavier,which makes them delete those mails after doing it ( but I wan them to keep those for review) ? I think DVI is a little bit unpopular (is it right ?), and they wouldnt mind spending their time studying LATEX or getting to know how to use DVI. Convincing them to solve 25 problems a week was already really hard for me.

[quote]
Also, \LaTeX is not the only environment you could typeset problems- they can be done reasonably well if you know the tricks in a word processor.[/quote]
Do you mean that I draw the lines for fraction and use subscript and superscript for exponents and stuff? I dont know any more tricks in M.Word.

I also have another question: do you know which program should I use to draw diagrams ( the one that can make the midpoints, midlines for you) ?

If it's just equations, you can actually use equation editor while creating a word document.....

And yes, DVI is not really all that popular............if your full club's using windows, though, I  dont think viewing DVI is any more difficult than downloading a viewer (like YAP ) and then opening the files....

Thanks for reply.
With your help, I'm done with writing equations and stuff via e-mail already. So do you know any program/software that can help us draw diagrams (with the function of determining/locating midpoints, midlines, or angle bisectors exactly? Thanks in advance.


Consider a triangle $ABC,$ $\angle A, \angle B, \angle C < 90$.
Determine the position of point $T$ $(Toricelli's$ $point)$ in the interiour of the triangle so that $TA+TB+TC$ is minimum.

Classic one...Rotate triangle $ATB$ around point $B$ with $60^{0}$.
You'll eventualy obtain the broken line $A^{/}-T^{/}-T-C$, with the properties that $BT^{/}=TT^{/}$; $A^{/}T^{/}=AT$. So the minimum is obtained in the case in which the broken line is straight. In that case the angles $ATB$, $ATC$, and $BTC$ all have $120^{0}$.

[quote="Stormraven"]In that case the angles $ATB$, $ATC$, and $BTC$ all have $150^{0}$.[/quote]
That must be just a slight typo. I got $360^\circ/3=120^\circ$. :roll:

O, of course...What an embarassing mistake  :blush:
It was really obvious... :blush:

I think this point may be called "Fermat's point"


Harvard
college
geometry
geometric transformation
reflection

Here is a list of books about the college admission process in the United States, including comments from me or from other readers who have emailed me about these books.

Chuck Hughes, [url=http://www.amazon.com/exec/obidos/tg/detail/-/007141259X/learninfreed]What It Really Takes to Get Into Ivy League and Other Highly Selective Colleges[/url] (McGraw-Hill, 2003) (ISBN 0-07-141259-X). This book is written by a former Harvard admissions officer, who was an undergraduate student at Harvard admitted to play on Harvard's hockey team. Hughes includes detailed lists of the top high school academic competitions in his book, and generally gives a realistic perspective on Ivy League admission for a wide variety of applicants. Hughes's book is particularly useful for getting applicants to consider what they would do if their dream major or intended sport in college somehow doesn't work out. A parent comments that this book really lives up to its name.

Elizabeth Wissner-Gross, [url=http://www.amazon.com/What-Colleges-Other-Parents-Know/dp/1594630313/learninfreed]What Colleges Don't Tell You: (And Other Parents Don't Want You to Know)[/url] (New York: Hudson Street Press, 2006) (ISBN 1-59463-031-3). Detailed guide by a parent for parents about the admission process at highly selective colleges and how to make the high school years count to have the best chance of admission.

Richard Montauk and Krista Klein, [url=http://www.amazon.com/How-Get-Into-Top-Colleges/dp/0735204098/learninfreed]How to Get Into the Top Colleges[/url] (Prentice Hall, rev. ed. 2006) (ISBN 0-73520409-8). This book was recommended to me by other readers. The book is based of interviews with college admissions officers and quotes many of their responses. The first person to recommend this book to me particularly liked that aspect of each college being able to speak for itself, with differences among colleges showing in the differing responses. Two readers recommended this book to me because of its section on homeschooled applicants.

Stephen Kramer and Michael London, eds., [url=http://www.amazon.com/exec/obidos/tg/detail/-/0743280679/learninfreed]The New Rules of College Admissions: Ten Former Admissions Officers Reveal What it Takes to Get Into College Today[/url] (Fireside, 2006) (ISBN 0-74328067-9). A new book including ten chapters on different aspects of the college selection, admission, and financial aid process by ten former officers of highly regarded colleges.

Andrew Allen, [url=http://www.amazon.com/exec/obidos/tg/detail/-/059519897X/learninfreed]College Admissions Trade Secrets: A Top Private College Counselor Reveals the Secrets, Lies, and Tricks of the College Admissions Process[/url] (New York: Writers Club Press, Second Edition 2001)(ISBN 0-595-19897-X). This book claims, as the hard-hitting subtitle shows, to have "inside information," and indeed the author doesn't write to please the egos of anyone else in the college admission process. Some of Allen's specific recommendations are idiosyncratic, including his opinions about which colleges are most worth applying to, but all are worth thinking about. I rank this book as highly as I do because the author stresses the important point of becoming a well-educated person, wherever you go to school.

Michele Hernandez, [url=http://www.amazon.com/exec/obidos/tg/detail/-/034545409X/learninfreed]Acing the College Application: How to Maximize Your Chances for Admission to the College of Your Choice[/url] (New York: Ballantine Books, 2002) (ISBN 0-345-45409-X [pbk.]). Hernandez's book is a very useful book, especially for students applying to the most selective universities. This book is highly valued by parents for its specific nuts-and-bolts advice. Hernandez is somewhat controversial among college admissions officers as one of the first authors of the "kiss and tell" genre of college admission guides (she used to be an admissions officer at Dartmouth), but except for a small number of idiosyncratic specific recommendations about filling out applications and debatable statements about the generality of admissions officers, most of her book tells it like it is and encourages applicants to prepare well in their high school years.

Joyce Slaton Mitchell, [url=http://www.amazon.com/Winning-Heart-College-Admissions-Dean/dp/1580086705/learninfreed]Winning the Heart of the College Admissions Dean: An Expert's Advice for Getting into College[/url] (Berkeley: Ten Speed Press, rev. ed. 2005) (ISBN 1-58008-670-5). This book was recommended to me by other readers. One admissions officer commented that the title is all wrong: you DON'T want to try to win the heart of the director of admissions. But the advice inside is mostly good, and covers most aspects of the college selection and application process.

Cafi Cohen, [url=http://www.amazon.com/exec/obidos/tg/detail/-/0761527540/learninfreed]Homeschoolers' College Admissions Handbook: Preparing Your 12- to 18-Year-Old for a Smooth Transition[/url] (Prima, 2000) (ISBN 0-76152754-0). This is the best book to get for homeschoolers preparing to go to college. Cafi Cohen's book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0913677116/learninfreed]And What about College?[/url] is also good, and about equally up to date.

Jay Matthews, [url=http://www.amazon.com/exec/obidos/tg/detail/-/0761536957/learninfreed]Harvard Schmarvard: Getting Beyond the Ivy League to the College That is Best for You[/url] (Three Rivers Press, 2003) (ISBN 0-76153695-7). Jay Matthews is a journalist who writes an education column for the Washington Post. He transferred from Occidental to Harvard to finish his undergraduate degree, so he is a rare Harvard graduate who can make a head-to-head comparison between Harvard and another college. Matthews's book contains interesting information about the highly competitive admission situation facing young people who go to the best high schools--their problem is looking better than their classmates. This book is well-liked by parents advising their children on the pressure-cooker aspects of college admission.

Edward B. Fiske with Bruce G. Hammond, [url=http://www.amazon.com/Guide-Getting-Right-College-Second/dp/140220230X/learninfreed]The Fiske Guide to Getting into the Right College[/url] (New York: Times Books, 2nd revised edition, 2004) (ISBN 1-40220230-X). This book has a lot of information about selecting colleges but less about making a strong application than some of the other books. Fiske, a former journalist for the New York Times, is well-liked by admissions officers, in large part (I think) because he doesn't recommend that everyone apply to the same small list of top schools.

Jacques Steinberg, [url=http://www.amazon.com/exec/obidos/tg/detail/-/0142003085/learninfreed]The Gatekeepers: Inside the Admissions Process of a Premier College[/url] (New York: Viking, 2002) (ISBN 0-670-03135-6). This book scares parents of bright children who are not "underrepresented" in terms of college admission diversity policies--the conclusion they reach after reading the book is that it is more important to have an admissions "hook" than to be an academically able applicant. The "premier college" profiled in the book is Wesleyan, a strong liberal arts college for preprofessional aspirants but with a weak math program, and thus not a college of interest to the young people I know best locally, who would prefer Carleton among colleges in that category. One admissions officer comments that the book is well-written and fair, but that it may overstate the influence of private school counselors. I think the books that focus more specifically on the top national universities are more useful for applying to that echelon of school than this book. A Wesleyan alumna commented to me that this book is a depressing read about her alma mater.

Richard Moll, [url=http://www.amazon.com/exec/obidos/tg/detail/-/0140513035/learninfreed]Playing the Selective College Admissions Game[/url] (New York: Penguin, 2nd rev. ed. 1994). Richard Moll writes more from the point of view of someone who still is an admissions officer and sometimes is less skeptical about the process than some of the authors no longer in that occupation. He is well respected by other admissions officers, including some who have never read his book. One admissions officer told me that this book is out of date insofar as it doesn't reflect the impact of early admission processes in universities today.

I appreciate comments from other participants here about what materials you find most helpful for preparing a college application and getting ready (academically) for college.

Is there any good ebook/unpaid resource to know about admissions to MIT/UC Bekley etc. for an International student? Please PM if possible.
Thanks!

A Discussion Forum
http://www.collegeconfidential.com/
http://talk.collegeconfidential.com/

I have another good book called [url=http://www.amazon.com/High-School-Success-College-Application/dp/1449584381]High School Success and College Application[/url] by Anthony Freeman. It's a brief, but concise book on tips to getting into a top college.

A helpful, recent book (2012) that is full of college admission information is [url=http://www.amazon.com/The-College-Solution-Everyone-Looking/dp/0132944677]The College Solution[/url] by Lynn O'Shaughnessy.

The main sections of the book are: 1) Shrinking the Cost of College, 2) Increasing Your Admission Chances, 3) Knowing Your Academic Choices, 4) Evaluating the Academics, 5) Admission Nuts and Bolts, and 6) Borrowing for College.

[url=http://www.amazon.com/Paying-College-Without-Edition-Admissions/dp/0804124361]Paying for College Without Going Broke[/url] is a solid introduction to the financial issues surrounding attending college.  This book discusses short and long-term financial strategies, plus it walks the reader through filling out the common forms (e.g. FAFSA and Profile). Additionally, the authors discuss how to interpret the schools' aid offers and other common areas of concern.

[url=http://www.amazon.com/The-Habits-Highly-Effective-Teens/dp/076241474X]The 7 Habits of Highly Effective Teens[/url] helps you develop good habits as a teenager and become a better person. Indirectly, the book can also help you prepare for college.


Prove that the sum
$\sqrt{1001^{2}+1}+\sqrt{1002^{2}+1}+\cdot\cdot\cdot+\sqrt{2000^{2}+1}$
is irrational.

it is pretty obvious that none of the numbers under the radical, are perfect squares. so isn't that pretty easy to prove then??? ;)

Well, no.  You can't just say sums of irrationals are irrationals.  It's actually untrue.  You'd have to prove it for this case.

[quote="amirhtlusa"][quote="paladin8"]Yeah, you can't do that. $\pi$ and $1-\pi$ are both irrational but their sum is $1$.[/quote]
$1-\pi $ is a negative number, but you have a all positive sequence above[/quote]

then $4-\pi$ and $\pi$ should add up to a rational number. ;)

Well, in general, irrational+irrational=rational isn't necessarily true.

Consider, for example $\sqrt{2}, 10-\sqrt{2}$.
Both numbers are irrational, but they sum to $10$.

Paladin's statement isn't affected by one of his numbers being negative. He could have easily made it $10-\pi$, although I chose $\sqrt{2}$ over $\pi$ for irrational over transcendental. (Not that this matters either.)

EDIT: You beat me to it, xxreddevilzxx.  :P


induction
number theory unsolved
number theory

find positive integer n so that has a positive integer k such that
                       2k 2 +2004k+3 = 3^n *h (h is a positive number) :)

There is a solution for n=1, we show n<2. supose there is a solution for n>=2, then it is also a solution for n=2 so we just have to show n=2 has no solution.
first we try to transform the equation modulo 9, after multiplying by 8 it becomes: (16k 2 +2004*8+2004 2 )-(2004 2 -3*8), but the sum in the first parentesis is (4k+2004) 2 , so replace 4k+2004 by t, then if there is a solution for t, there is a solution for k,  and viceversa. then our ecuation becomes:

t^2=2004 2 -24, but lt implies the rhs to be a square mod 9, but it is congruent to 3 and 3 is not a square mod 9, so there is no solution.

We conclude n=1, and taking k=3 gives a satisfactory h

Pascual

yeah
in this case: (4k+2005) 2 =2005 2 -24, now we have something of the form t 2 =a (mod 3^n), since it has a solution for n=1, we prove it has a solution for every n by induction, supose y is a solution of    n=k, that is: y 2 =a(mod 3^k), now put y'=y+x3^k, and:
y' 2 =y 2 +x*2*3^k (mod 3^(k+1)), now we want
y 2 +x*2*3^k =a (mod 3^(k+1)), so this is equivalent to: 2x=(a-y 2)/3^k (mod 3), but because 2 is relative prime to 3, and the rhs is an integer by oinduction, for a suitable x, y' is a solution for k+1 so the induction is complete!.......Also this is trivial by hensel lemma


function
limit
algebra
polynomial
real analysis
real analysis unsolved

Let $ f(x)$ be a real function such that
\[ \lim\limits_{n\to\infty} f(x)\equal{}\infty,    \text{and }   \lim\limits_{n\to\infty} f'(x)\equal{}0\]
Prove that $ f(x)$ can't be represented as quotient of two polynomials, (i.e. $ f(x)$ isn't
rational function).
Give some examples of such functions.

Remember that the limit of a quotient of polynomials behaves in the same way as
 the quotient of their highest order terms. Assume your function is such a quotient
and arrive at a contradiction.

 For examples of such functions, consider x^a for suitable a...


In the plane let a finite family of circles be given satisfying the condition: every two circles, either are outside each other, either touch each other from outside and each circle touch at most 6 other circles. Suppose that every circle which does not touch 6 other circles be assigned a real number. Show that there exist at most one assignment to each remaining circle a real number equal to arithmetic mean of 6 numbers assigned to 6 circles which touch it.

We can phrase it in terms of graphs. Consider a planar graph $G$ whose vertices represent the circles, and in which two vertices are adjacent iff the corresponding circles touch each other. We are interested in the functions $f$ defined on the set $V$ of vertices of $G$, having the property that $f(x)=\frac{f(x_1)+\ldots+f(x_6)}6$ whenever $x_1,\ldots,x_6\in V$ are the neighbours of $x\in V$.

Suppose now that for a given function $f_1$ defined on the set $V_1\subseteq V$ of vertices of degree $<6$ we have two different extensions $f,\tilde f$ to the entire $V$. This means that $\tilde f-f$ is a non-zero extension of the function which is identically $0$ on $V_1$. It thus suffices to assume that $f_1\equiv 0$ and prove what we want in this particular case.

Let $f$ be an extension of $0_{V_1}$ to $V$, and let $H$ be a connected component of the graph $\widetilde G$ we're left with after eliminating all the vertices in $V_1$ and the edges adjacent to them. Suppose the minimum of $f$ on $H$ together with its neighbours in $V_1$ is $m<0$ and it's reached in some $x\in H$. Then $f=m$ in all neighbours of $x$, none of which can belong to $V_1$ because $f|_{V_1}\equiv 0>m$. Then $f=m$ on all the neighbours of the neighbours of $x$, and so on. Clearly, this process cannot stop, so we have a contradiction with the hypothesis that the graph is finite. We do something similar to prove that the maximum of $f$ on $H$ cannot be $>0$, and from these results we get $f|_H\equiv 0$. The same holds for all components of $\widetilde G$, and this simply means that $f\equiv 0$, i.e. exactly what we wanted.

P.S.

In order to get the contradiction above, we used the fact that eventually, we must hit a vertex of degree $<6$, because a finite planar graph cannot have all vertices with degree $6$. This follows from Euler's formula $v-e+f=2$, where $v,e,f$ are the numbers of vertices, edges, and faces respectively of the planar graph. If all vertices have degree $6$, we get $e=3v,\ f\le\frac{2e}3=2v\Rightarrow 2=f-2v\le 0$, contradiction.


For a positive integer $a$, if for a prime $p$, there exists an odd number $b$ such that $p|2^{2^{a}b}-1$, then we say $p$ is in the set $S_{a}$.

Show that for any $a$, there are infinitely many primes not in $S_{a}$.

i think you need use
lema:$2^{2^{n}}-1$ has infinite prime factor!(it is very easy :lol: )

Note that all the Fermat number are relaively prime,namely$(2^{2^{m}}+1,2^{2^{n}}+1)=1$
then for $a$ Let $p$ is arbitary prime factor of $2^{2^{a+k}}+1,\forall k\geq 1$ then $p \not\in S_{a}$

[quote="Summerburn"][quote="Hawk Tiger"]then for $a$ Let $p$ is arbitary prime factor of $2^{2^{a+k}}+1,\forall k\geq 1$ then $p \not\in S_{a}$[/quote]

Eh?

If $p$ divides $2^{2^{a+k}}+1$, then it doesn't divide $2^{2^{a}}+1$, but it might still divide $2^{2^{a}b}-1$[/quote]
Notice that b is odd.

there are infinitely primes $p\not\in S_{a}$
__________________________________
we can prove that there are infinitely primes $p\in S_{a}$
[quote]$\forall (p_{1},p_{2})\ prim$ $GCD(2^{2^{a}p_{1}}-1,2^{2^{a}p_{2}}-1)=2^{GCD(2^{a}p_{1},2^{a}p_{2})}-1=2^{2^{a}}-1$[/quote]

we take $X_{b}=\frac{2^{2^{a}b}-1}{2^{2^{a}}-1}$
$A=\{X_{p}\ /\ p\ prim\}$
$\forall (c,b)\in A^{2},\ c\neq b\ ;\ gcd(c,b)=1$
then there are infinitely primes $p$ such that there exists $c\in A$ ($c=X_{q}$, $q$ prim )such that $p|c$ $\Rightarrow$ $p|2^{q2^{a}}-1$

then there are infinitely primes $p\in S_{a}$

[quote="aviateurpilot"]sorry  :blush: ()

$A=\{p\ such\ that\ there\ existe\ k\ such\ that\ p|2^{2^{a+k}}-1\}$
$B=\{p\ such\ that\ p|2^{2^{a}}-1\}$
$C=\{p\not\in S_{a}\}$

we have $card(A)=+\infty$ and $card(B)=c$
then $card(A-B)=+\infty$
and we have $A-B$ in $C$ , then $card(C)=+\infty$[/quote]
is another solution


Prove that there are infinitely many positive integers $ n$ for which there exists a partition of $ \{1,2,...,3n\}$ into subsets $ \{a_1,a_2,...,a_n\},\{b_1,b_2,...,b_n\},\{c_1,c_2,...,c_n\}$ such that $ a_i\plus{}b_i\equal{}c_i$ for all $ i$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=287757
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224827
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218089
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192117
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184260

Keywords: partition, partitioning, partitioned, 3n, triples


Έστω $ a \plus{} b \plus{} c \equal{} 3 ,a, b, c \ge 0$ νδο:

$ \frac{a \plus{} 3}{3a \plus{} bc}\plus{}\frac{b \plus{} 3}{3b \plus{} ca}\plus{}\frac{c \plus{} 3}{3c \plus{} ab} \ge 3$
Ποιος διαβάζει τώρα τεχνολογία επικοινωνιών :rotfl:  :?:

[quote="Dimitris X"]Έστω $ a \plus{} b \plus{} c \equal{} 3 ,a, b, c \ge 0$ νδο:

$ \frac {a \plus{} 3}{3a \plus{} bc} \plus{} \frac {b \plus{} 3}{3b \plus{} ca} \plus{} \frac {c \plus{} 3}{3c \plus{} ab} \ge 3$
Ποιος διαβάζει τώρα τεχνολογία επικοινωνιών :rotfl:  :?:[/quote]

H anisotita einai tis morfis $ \sum\frac{x\plus{}y}{xy}\geq 3$ me $ \sum x\equal{}6$ opou profanos isxuei


inequalities
calculus
search
Cauchy Inequality

Find the minimum value of $ \sqrt{x^2 \plus{} y^2}$ if $ 5x \plus{} 12y \equal{} 60$.

Applying Cauchy inequality, we get:
$ 5x\plus{}12y\le\sqrt{(5^2\plus{}12^2)(x^2\plus{}y^2)}\implies \sqrt{x^2\plus{}y^2}\ge\frac{60}{13}$

What's values of $ x,\ y$ for which $ \sqrt{x^2\plus{}y^2}$ is maximal? :wink:

Isn't the minimum $ (x,y)\equal{}(0,0)$?
Considering it's a square root.

[hide="Proof (multivariable calculus)"]
If $ z\equal{}\sqrt{x^2\plus{}y^2}$, we have:

$ \frac{\partial z}{\partial x}\equal{}\frac{1}{2\sqrt{x^2\plus{}y^2}}\cdot 2x\equal{}0\implies x\equal{}0$.

$ \frac{\partial z}{\partial y}\equal{}\frac{1}{2\sqrt{x^2\plus{}y^2}}\cdot 2y\equal{}0\implies y\equal{}0$.

$ z\equal{}\sqrt{0\plus{}0}\equal{}0$.

So the minimum is at $ \boxed{(0,0,0)}$.
[/hide]

[quote="kunny"][quote="Kyoshiro"][quote="kunny"]What's values of $ x,\ y$ for which $ \sqrt {x^2 \plus{} y^2}$ is maximal? :wink:[/quote]
There is no maximum value of $ \sqrt {x^2 \plus{} y^2}$ :huh:[/quote]

 :blush: not maximal but minimal[/quote]
The equality occurs when $ x \equal{} 5y/12$. Plugging this into the equation: 5x+12y=60, we easily solve for y: $ y \equal{} 720/169$. So $ x \equal{} 300/169$

[quote="grn_trtle"]Isn't the minimum $ (x,y) \equal{} (0,0)$?
Considering it's a square root.

[hide="Proof (multivariable calculus)"]
If $ z \equal{} \sqrt {x^2 \plus{} y^2}$, we have:

$ \frac {\partial z}{\partial x} \equal{} \frac {1}{2\sqrt {x^2 \plus{} y^2}}\cdot 2x \equal{} 0\implies x \equal{} 0$.

$ \frac {\partial z}{\partial y} \equal{} \frac {1}{2\sqrt {x^2 \plus{} y^2}}\cdot 2y \equal{} 0\implies y \equal{} 0$.

$ z \equal{} \sqrt {0 \plus{} 0} \equal{} 0$.

So the minimum is at $ \boxed{(0,0,0)}$.
[/hide][/quote]
Your equality occurs when x=y=0, which is unreasonable since 5x+12y=169.

Sorry, I was looking at kunny's post, and didn't notice the $ 5x\plus{}12y\equal{}60$ requirement.

Check out link below:

[url]http://www.mathlinks.ro/viewtopic.php?search_id=1103376538&t=202613[/url]

Here is my solution:

$ (5x \plus{} 12y)^2 \plus{} (12x \minus{} 5y)^2 \equal{} 169(x^2 \plus{} y^2)$

$ \therefore x^2 \plus{} y^2 \equal{} \frac {1}{169}\{60^2 \plus{} (12x \minus{} 5y)^2\}$, yielding the desired minimum value is $ \sqrt{\frac {3600}{169}}\equal{}\frac{60}{13}$

The equality holds when $ 12x \minus{} 5y \equal{} 0$ and $ 5x \plus{} 12y \equal{} 60\Longleftrightarrow x \equal{} \frac {300}{169},\ y \equal{} \frac {720}{169}$

$ \sqrt {x^2 \plus{} y^2}$ is the distance of de point $ (x,y)$ to origin $ (0,0)$. The minimum value of $ \sqrt {x^2 \plus{} y^2}$ with $ 5x \plus{} 12y \equal{} 60$ is de distance of the straight line $ 5x \plus{} 12y \equal{} 60$ to origin, that is:
$ d \equal{} \frac {|5\cdot 0 \plus{} 12\cdot 0 \plus{} 60|}{\sqrt {5^2 \plus{} 12^2}} \equal{} \frac {60}{13}$.

The point of minimum is the solution of $ 5x \plus{} 12y \equal{} 60$ with $ 12x \minus{} 5y \equal{} 0$.


geometry
3D geometry
algebra unsolved
algebra

do there exist two real numbers x and y such that: their sum is 1 ,sum of their squares is 2 and sum of their cubes is 3. i mean
x+y=1 ; x2 + y2=2 ; x3+ y3 =3


Numere naturale care au un numar impar de divizori primi diferiti se coloreaza cu [color=yellow]galben[/color], iar cele care au un numar par de divizori cu [color=blue]albastru[/color]. Exista o progresie aritmetica infinita care contine numere naturale de aceeasi culoare?

E suficient sa ne uitam la o progresie $ ar\plus{}b$ cu $ (a,b) \equal{} 1$ si $ r\in\mathbf{N^{*}}$.Vom demonstra ca sunt numere de culori diferite in aceasta progresie.Din teorema lui Dirichlet avem ca sunt o infinitate de numere prime,deci o infinitate de numere galbene in progresie.De asemenea tot din teorema lui Dirichlet avem ca in progresia $ ar\plus{}1$ sunt o infinitate de numere prime.Fie $ ar_{1}\plus{}1 \equal{} p$ si $ ar_{2}\plus{}b \equal{} q$ cu $ p$ si $ q$ prime atunci $ pq \equal{} a(r_{1}r_{2}\plus{}r_{1}b\plus{}r_{2})\plus{}b$ deci $ pq$ face parte din prima progresie si este albastru. :lol:


you konw that we can use tex on this forum
for example $y=x^2$
but how can we use tex on msn, and make nice mathematical notations on msn
thx

I don't think it is. The client you are using would need to take some special steps to have LaTeX enabled. It's not something you can randomly use on any service. It'd be cool if it was though. :)

I have an idea. You know that you can put in smileys? Copy and paste the mathemathical symbols into the smileys list and use them to write calculations.

it would be great if latex could be implemented on msn.. physics and mathematical related chat would be made much more simpler.
hmm..
but it would take too much of one's time defining each note as a smiley isn't it? haha..

Well, you could have a symbol for pi, angle and so on without a fatal time loss.

Can't you just type the LaTeX notation and let the other person interpret it as a compiler normally would?

That assumes the other person is familiar with $\LaTeX$ :P

You could use the $\text{\TeX}$er and copy, paste...

Yes... Meaning also that the reciever has to copy and paste... kinda complicated.
The only really easy way would be to make the LATEX a set of letters (like verdana, times new roman and so on) and set the text on latex when you want to use it. Then shut it off when you want to write normally.

http://www.student.cs.uwaterloo.ca/~kzorin/texim.php

You might need to change ur chat client tho.


Solve the system equation:
$ 1)\left\{ \begin{array}{l}x^3  - 8x = y^3  + 2y \\  x^2  - 3 = 3(y^2  + 1) \\ \end{array} \right.$

$ 2)\left\{ \begin{array}{l} x^4  - x^3 y + x^2 y^2  = 1 \\ x^3 y - x^2  + xy =  - 1 \\  \end{array} \right.\#$

hello, solving the first equation for $ y$ we get
$ y_1\equal{}\frac{1}{3}\sqrt{3x^2\minus{}18}$
$ y_2\equal{}\minus{}\frac{1}{3}\sqrt{3x^2\minus{}18}$
setting this in the second equation we get
$ x^3\minus{}8x\equal{}\frac{1}{27}(3x^2\minus{}18)^{3/2}\plus{}\frac{2}{3}\sqrt{3x^2\minus{}18}$
squaring both sides and bringing the right-hand side to the left  and factoring  we get
$ x^2(3\plus{}x)(\minus{}3\plus{}x)(13x^2\minus{}96)\equal{}0$
By the same procedure we get with $ y_2$ the product
$ x^2(3\plus{}x)(\minus{}3\plus{}x)(13x^2\minus{}96)\equal{}0$
The solutions are
$ (3,1)$, $ (\minus{}3,\minus{}1)$, $ (0,\sqrt{2}i)$, $ (0,\minus{}\sqrt{2}i)$
Sonnhard.

hello, solving the second equation for $ y$ we get $ y\equal{}\frac{x^2\minus{}1}{x(x^2\plus{}1)}$ and setting this in the first equation and factoring we get $ x^2(x\minus{}1)(x\plus{}1)(x^4\plus{}2x^2\plus{}3)\equal{}0$.From here you will get all solutions.
Sonnhard.


$ a^{2}\plus{}b^{2}\leq 1$
Consider a sequence as follow
$ x_{n\plus{}1}\equal{} f(x_{n})$ where  $ f(x) \equal{} x\plus{}a\text{cos}x\plus{}b\text{sin}x$
Prove that   $ x_{n}$ converge.

$ f(x)\equal{}x\plus{}csin(x\plus{}\alpha),\ |c|<1$, therefore $ f'(x)>0$ and sequenct had limit $ \minus{}[\frac{\minus{}x_{1}\minus{}\alpha}{\pi}]\minus{}\alpha.$


function
trigonometry
linear algebra
matrix
geometry
circumcircle
geometric transformation

$ \fbox{1}$ [url=http://www.mathlinks.ro/viewtopic.php?t=202401]determinant including the trigonometric function[/url]

Let $ \cos A_i \neq \cos A_j \ ( \ \{ i, j \in \mathbb{Z} | 1 \le i < j \le n \} \ ), \ \sin A_k \neq 0 \ ( \ \{ k \in \mathbb{Z} | 1 \le k \le n\} \ )$.

$ \begin{vmatrix} 1 & 0 & 1 & 0 & \cdots & 1 & 0 \\
\cos A_1 & \sin A_1 & \cos A_2 & \sin A_2 & \cdots & \cos A_n & \sin A_n \\
\cos (2A_1) & \sin (2A_1) & \cos (2A_2) & \sin (2A_2) & \cdots & \cos (2A_n ) & \sin (2A_n) \\
\cos (3A_1) & \sin (3A_1) & \cos (3A_2) & \sin (3A_2) & \cdots & \cos (3A_n) & \sin (3A_n) \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
\cos ((2n \minus{} 1)A_1) & \sin ((2n \minus{} 1)A_1) & \cos ((2n \minus{} 1)A_2) & \sin ((2n \minus{} 1)A_2) & \cdots & \cos ((2n \minus{} 1)A_n) & \sin ((2n \minus{} 1)A_n) \end{vmatrix}$

$ \equal{} \left( \prod_{1 \le i < j \le n} (2(\cos A_i \minus{} \cos A_j))^2 \right) \left( \prod_{k \equal{} 1}^n \sin A_k \right)$

$ \fbox{2}$ [url=http://www.mathlinks.ro/viewtopic.php?t=220525]find an area of rectangular[/url]

On each side of rectangular $ ABCD$, We take four point $ P$, $ Q$, $ R$, $ S$.
($ P$ on $ AB$, $ Q$ on $ BC$, $ R$ on $ CD$, $ S$ on $ DA$)
Let, $ O \equal{} PR \cap QS$.
When an area of $ \triangle PQO$ and $ \triangle RSO$ is equal($ 30 cm^2$), find an area of rectangular $ ABCD$.
($ \overline{AS} \equal{} 8cm$, $ \overline{PB} \equal{} 10cm$, $ \overline{QC} \equal{} 12cm$, $ \overline{RD} \equal{} 8cm$)

$ \fbox{3}$ [url=http://www.mathlinks.ro/viewtopic.php?t=222174]center of quadrant[/url]

Find center of quadrant without using calculus.

[asy]pair O=(0,0), A=(3,0), B=(0,3), G=(4/pi, 4/pi);
draw(O--A, black);
draw(O--B, black);
draw(Arc(O, 3, 0, 90), black);
dot(G);
label("(0, 0)", O, SW);
label("(a, 0)", A, SE);
label("(0, a)", B, NW);
label("(0, a)", B, NW);
label("G", G, SW);[/asy]

[hide="Answer"]$ G \equal{} \left( \frac {4a}{3\pi}, \frac {4a}{3\pi} \right)$[/hide]

$ \fbox{4}$ [url=http://www.mathlinks.ro/viewtopic.php?t=223818]Harmonic series[/url]

$ a$, $ n$ are positive, $ \sum_{k \equal{} 1}^a \frac {1}{k} > n$
Find smallest a.

$ \fbox{5}$ [url=http://www.mathlinks.ro/viewtopic.php?t=223821]Comic Market 75 Circle Cut Problem[/url]

$ \left\{\begin{array}{l} \angle BAC \equal{} 90^\circ \\
\angle ABC \equal{} 45^\circ \\
\text{midpoint of }\overline{AC}\text{ is }D \\
\text{midpoint of }\overline{BC}\text{ is }E \\
BA \equal{} BF \\
G \equal{} \text{arc }AF \cap \overline{ED} \end{array}\right.$

Find $ \angle GCB$.

[asy]size(300);
pair A=(0, 10), B=(-10, 0), C=(10, 0), D=(5, 5), E=(0, 0), F=(-10+10*sqrt(2), 0);
draw(A--B--C--cycle);
draw(D--E);
draw(Arc(B, 10*sqrt(2), 0, 45));
pair G=intersectionpoint(D--E, Arc(B, 10sqrt(2), 0, 45));
draw(C--G);
label("A", A, N);
label("B", B, SW);
label("C", C, SE);
label("D", D, NE);
label("E", E, SW);
label("F", F, SE);
label("G", G, NW);
fill(C--Arc(C, 2, 150, 180)--cycle, gray);
label("?", (8,1), SW);[/asy]

[hide="Answer"]$ \angle GCB \equal{} 30^\circ$[/hide]

I think that there is a lot of different solution.
Have Fun! :jump:

[quote="Fermat's Little Turtle"]What do you mean by the center of the quadrant? Center of gravity assuming its of uniform density?[/quote]
Yes, that's right!! :)

$ \fbox{6}$ [url=http://www.mathlinks.ro/viewtopic.php?t=214550]Isosceles triangle[/url]

The center of the green circle is $ A$.
The center of blue circle is $ B$, $ C$. And there is it on green circumference.
Let $ D, E \equal{} \text{blue circle}\cap\text{blue circle}$. And there is $ D$ on green circumference.
Let $ F \equal{} \text{blue circle (center is } B\text{)}\cap\text{green circle}$.
Let $ G \equal{} \text{blue circle (center is } C\text{)}\cap\text{green circle}$.
Let $ H \equal{} \text{blue circle (center is } B\text{)}\cap\text{straight line }FG$.
Let $ I \equal{} \text{blue circle (center is } C\text{)}\cap\text{straight line }FG$.
$ \fbox{Prove EH \equal{} EI.}$

[quote="THECHAMP"]Find the intersection.[/quote]
Which problem is it?
Please do the comment in the place of the link which there is beside a problem number.

[quote="sunken rock"]E is the symmetrical of D about BC, therefore B, E and G are collinear, same for C, E anf F, hence E is the circumcenter of the triangle DHI.

Best regards,
sunken rock[/quote]

Yes, E is the symmetrical of D about BC.
But I think that it is not easy to prove B, E and G are collinear.

And, please do the comment in the place of the link which there is beside a problem number.

$ \fbox{7}$ [url=http://www.mathlinks.ro/viewtopic.php?t=240894]Kubota's theorem[/url]

I introduce an interesting theorem.

Do you know Kubota's theorem?
This theorem seems to be comparatively famous for the projective geometry.

If there is a ruler of length d, we can draw an arbitrary straight line to link two points without using the compasses.
(i.e. With only a ruler of limited length, we can draw a figure of an arbitrary straight line to link two points. )

The details please look at the following site.
http://www13.plala.or.jp/epsilons/Kubota%27s%20thm/kubota.html
(This site is Japanese. Please read after translating it in translation sites.)

$ \fbox{8}$ [url=http://www.mathlinks.ro/viewtopic.php?t=242431]Bisector of triangle area[/url]

There is a point not on the circumference(top and side) of triangle.
Construct a bisector of triangle area through this point.

There is a lot of construction methods.
Please find it a lot  :)


Well here's a poll on how long USA will keep its power.

I won't tell you my thoughts on when the expected "fall of USA" will come, as that might swing you in one direction or the other, but USA is definitely showing cracks. Rather than shown as an overshadowing peacemaker, now it is being portrayed in many countires as a busybody warmaker who can't even solve its own problems (i. e. Katrina disaster etc) yet tries to solve others' for them, under unwelcome circumstances. With incapable people like Bush succeeding to the presidency, it further weakened the country with pointless money spending. Now, power hungry politicians promise us false hopes with their big slogans that tell lies. Also, to protect their popularity, they spend money more on themselves rather than the country they should be worrying about. And if this was the end of the list of problems I would keep my mouth shut. But that is not the case. According to the statistics, the percentage of unemployed people have grown, and the monetary value of the US dollar has fallen to the ground to at least 30~50 cents in other countries. Wars and such further undermine our economy, and biased as my opinion could be, our troops have still not completely returned from Iraq. Well I won't say more, as this is chiefly a poll, and allow YOU to contemplate the fate of USA.

Those poll choices are horrible.  Specifically how can you have "within fifty years" and jump to "never"???  No one is naive enough to put 'never' but there definitely are people who would think more than fifty years.

in my thoughts, in my lifetime is a 100 years. Anyways, I think it won't go much more than 100 years.
Besides, I couldn't change the poll...maybe you can add the option of "
more than 100 years", joml88? ;)
and no offense, but "Those poll choices are horrible" isn't really "constructive critiscim", and you are a mod; you ought to set an example.

Your description (and what I think) sounds like a textbook in the 22nd or 23rd century. "The US had been a super power for 100 years, but at the beginning of the 21st century corruption began to infiltrate the government. Leaders gave false promise to the people, and soon the military took over....."

I put 50 years but I'd think 100 years is more like it :)

LOL I decided to be dull and boring this case to make the reader convinced about the utter fall of the USA. :rotfl:

[quote="joml88"]I don't have moderator powers in the Round Table forum.

I'm saying that your poll options are way too biased.  You are basically saying that the US WILL, without a shadow of a doubt, fall from power within 50 years.

Also, are you telling me that the average life expectancy is around 115 now?

Look, my point is that if you're conducting a poll and if you actually care about what people really think you wouldn't bias your poll.  Your post obviously shows that you think the U.S. is going to fall relatively soon.

You said "I won't tell you my thoughts on when the expected "fall of USA" will come,"  but then you basically went on and told us your views anyways (it definitely wasn't a fair and balanced perspective of the current state of U.S.).

My criticism was constructive.  It had the intent of teaching you so that you will try to make better, balanced polls in the future. :) (the smiley lightens up the mood of the post so you know it's nothing personal ;) )[/quote]
First of all, I don't want it to fall.
Second of all, that's not the whole story of the US; I forgot to mention the various powerhouses (Alaska etc) and military USA has at its disposal. But, we shouldn't ignore the fact that USA's economy isn't the once stable economy, and also that it's response systems to disasters etc is not the once impregnable wall it seemed portray.
Third, I let the reader choose when it would fall (admittedly, I forgot the option of more than a 100 years, but hopefully a mod will change it).
Fourth, I am a little offended by what you called "an unfair and unbalanced perspective of the current state of US".  :mad: It just listed all the problems that USA has today and then it let the reader decide that those reasons are reasonable or not.
It had good points, and they were facts. How many times did we hear that the troops were coming home etc? Those politicians aren't all corrupt, but some major ones are, causing further problems. If you have a question about one or more facts on there, you can post them.
Fifth, you might consider revising the standards of your "constructive critiscm" because all it did was call my poll and my posts horrible.
Sixth, I'm as patriotic as you or anyone else on this forum; don't make me out being a redneck. I don't think USA will fall within 50 years, but soon enough that I might live to see it.
Seventh, you seem a little miffed at me bringing this subject of US falling up, but it helps to discuss it.
Now now, not trying to offend joml88 or anyone here, and go on polling and posting people! :lol:

I take it you are under the age of eighteen.

Have you ever lived in any country outside the United States?

[quote="sugarwergoingdown"]
Sixth, I'm as patriotic as you or anyone else on this forum; don't make me out being a redneck.[/quote]

Rednecks are likely more patriotic than your 'average' american, and states which might be considered 'redneck' are typically states where the Republicans clean up.

[quote="tokenadult"] I take it you are under the age of eighteen. [/quote]

That upper bound can doubtless be tightened considerably.

edited to correct quote, my apologies.

I'm not miffed because of the topic itself.  I'm miffed at how you said "I won't tell you my thoughts" and then you went on and told us the 'facts'.  But they weren't the facts, they were your views (at least your view of how things are going in America).

[quote="sugarwergoingdown"][quote="blahblahblah"][quote="sugarwergoingdown"]
Sixth, I'm as patriotic as you or anyone else on this forum; don't make me out being a redneck.[/quote]

Rednecks are likely more patriotic than your 'average' american, and states which might be considered 'redneck' are typically states where the Republicans clean up.

[quote="sugarwergoingdown"] I take it you are under the age of eighteen. [/quote]

That upper bound can doubtless be tightened considerably.[/quote]
I never sed that, how come u erased the other person's name and put mine in it?
i can do that too.

[quote="blahblahblah"] I take it you are under the age of eighteen. [/quote][/quote]

  I suddenly feel inspired by this thought-envoking attempt to divert attention.

Yes, I was the one who guessed the OP is under eighteen.

[quote="sugarwergoingdown"][/quote][quote="sugarwergoingdown"][quote="sugarwergoingdown"][quote="sugarwergoingdown"]
Sixth, I'm as patriotic as you or anyone else on this forum; don't make me out being a redneck.[/quote]

Rednecks are likely more patriotic than your 'average' american, and states which might be considered 'redneck' are typically states where the Republicans clean up.

[quote="sugarwergoingdown"] I take it you are under the age of eighteen. [/quote]

That upper bound can doubtless be tightened considerably.[/quote]
I never sed that, how come u erased the other person's name and put mine in it?
i can do that too.

[quote="sugarwergoingdown"] I take it you are under the age of eighteen. [/quote][/quote][quote="sugarwergoingdown"]

  I suddenly feel inspired by this thought-envoking attempt to divert attention.[/quote]

oh man, that was fun

um, yes, but blahblahblah obviously just made an error (in my effort to keep this post from being TOTAL spam)

Imho, it all depends on what it means to "fall from power." Power isn't an easy term to define. Are we talking about military power? Influence in international diplomacy or politics? Technological innovation?

Generally, I find fire-and-brimstone speeches about the doom of the United States to be generally blown out of proportion. True, there are certain situations the US is involved in that I don't like (or even despise.) But I could create a similar apocalyptic analysis for the US in the '30s, '50s, '70s (Vietnam, anyone?), and probably '80s. A good idea is to also compare the US with other countries in the world before we throw out the "fallen" label anywhere.

[quote="sugarwergoingdown"]well guys share some of your own point of views with us all!!! :D[/quote]

  I take it most viewers are hesitant from engaging in the matter becuse of the absurdity and ambiguity of the topic itself... :roll: ...... but this of course, is only my opinion.

hmm not much discussion has been done about the original question
to the OP, next time, appease everyone by creating more poll options, if you like

back on topic : yeah any empire falls eventually
but will there be a new superpower, who will it be
also what is a super power, a feared country, or a wealthy country, but no offense, but even my country would then be a super power  ;)

looking at population, only india and china have more people
i doubt india will take over
china, with recent economical changes might become the superpower, but they are still technology behind usa, they are only a 'mediocre' nuclear power, while russia and usa are big, their space program is far behind the other two, but they are improving (my american uncle constantly tells me not the underestimate the chinese, and he was right when they actually launched yang liwei into orbit and back)

now what about russia?  the commonwealth of independent states is only a loose confederation, with ukraine, georgia, kyrgizistan (sorry if english name is wrong) turning away, so their population is now only half of that of usa
technologically they are still a huge player however...

and then there is europe, where i live
western european countries have never been so aligned, certainly with since 2002 our common euro money
europe seems to be turning away from usa, disagreeing on many levels, like the war in iraq and also on religion
usa doesn't really like the euro, nor does it like our recently started galileo program
certainly economically, europe might be the next super power

[quote="fredbel6"]certainly economically, europe might be the next super power[/quote]I doubt it. Soon Romania will enter the European Union and you will feel a huge economical drawback :) Like, imagine 10 Alaska would suddenly join the 51 states :D

romania and bulgaria are the next to join in 2007 , but that is not completely certain yet, right?
valentin are you actually saying that you don't trust your own country in contributing to the eu?

[quote="fredbel6"]romania and bulgaria are the next to join in 2007 , but that is not completely certain yet, right?[/quote]It's certain, and it has been for years now.

It's not that I don't trust, but the [b]real[/b] reason for Romania to be invited is certainly not the Economy (which will draw EU back a lot), but rather the extremely important geo-political-stategic influence of Romania in the Natural Gas distribution, and in the Black Sea / Middle East areas in general.

In plus Romania has been the first country to sign the agreement with US regarding extradition of US Military accused of crimes on Romanian soil (actually happened already - a good thing too because here the guy would have probably gotten away with the crime, whereas in US it seems he'll be staying a few decades behind the bars), and made all sorts of [b]very[/b] generous concessions to US (both political, economical (Bechtel - billion $\$$ contracts) and military - Iraq troops, and US military bases).

Furthermore why would you think EU still voted in a huge majority for Romania to get in last year, when the [b]very first[/b] thing that the newly ellected Romanian president said was that "Bucharest-London-Washington" is the main directon of external politics for Romania from now on?

It's a simple as this: EU wants US's foothold in Central Europe and it's ready to pay whatever it costs to get it.

[quote="probability1.01"]First, the U.S. will gradually fall out of power in the next 50 years or so. Then, power will shift to Asia, where China will eventually lead with its huge population. After a short-lived period (~25 years) of Chinese superpower, China's problems will catch up and the world will fall into turmoil for a while, compounded by all these environmental issues and such. After that, who knows...[/quote]Keep in mind that the number of people in China's work force is going to decrease in a few decades, because the majority of the population now is like 25-55 years old.

Heard this over my dad's conversation a while ago. :P

true
it had to make a maple file on a project at uni using markov chains about that
wasn't  that fun, don't even remember what exactly I found because we couldn't model it well anyway

Still the chinese have a one child law, that is especially enforced in the cities, less in rural areas

dunno exactly how much 'illegal' children they make but if every two people make only one your population dies out

the whole thing is so weird, mao himself at first promoted many children

It's not really that weird. For most people you can only have one child. If you're an only child as a result of the one child law, you can have two children. Basically it's making the population a slowly decreasing monovariant . They do this because China has a population problem, but obviously they're not going to keep that up until the population dies out. Certainly stifling your freedom, but that's a completely different issue that goes with pretty much everything controversial about China.

Just wondering, what happens if you get twins?  :P

Or more?.....  :P

i think you have to choose between them, or pay some kind of tax... but i'm not sure.  what i've also heard about chinesse population is that the new generations are mostly men, so that is also going to accelerate their population decrease.

[quote="bandrak"]i think you have to choose between them, or pay some kind of tax... but i'm not sure.[/quote]
In fact the mother is forced to eat the redundant ones.
Someone would believe anything because they think they were in a country with freedom of press.

Here's a good article:

[url]http://en.wikipedia.org/wiki/One-child_policy[/url]

[quote="Wikipedia"]
Many couples who want more than one child are able to side-step this policy by using fertility drugs to result in a multiple birth. Fines are not charged when a couple has more than one child at the same time.
[/quote]

It depends. If the current trend continues, I expect fairly soon. Bush certainly doesn't seem to care about how we stand in the eyes of others.

Well what do all of you think about INDIA rising to POWER??!! :D  :D

tokenadult , who is the man on your pic?

hmm didn't mao initially promote making a huge family?

My avatar is mathematician Israel M. Gelfand, one of my favorite authors of mathematics textbooks.

People from USA have a problem: The love too much they´re country, but in the wrong way (I´m talking in general, not in especific cases). As a result, they accept everything Bush says, because if not, you´re not AMERICAN. Let say something true: Bush controls the World. 100% sure. 100% true. Americans think they´re the best, but that´s not true. Do you think that destroying Irak, and killing Irak´s people is correct, just for oil and money??? mmmmmmmmmm, I don´t think so.

I have nothing against Americans, but mostly of them think that they´re the best, because of being Americans ---> FALSE!!!.

Actually, less than half of Americans "accept everything Bush says"--I know very few people who do, and I live in a city in which Republican Party candidates usually win most elections. If your statement is that most Americans still think that the United States has something to offer the world as an example (however imperfect, as all human examples are) of freedom and democracy, then I think that is indeed something most Americans agree about. Some aspects of United States foreign policy are very long-lasting and will likely still look much the same to observers in other countries even if people with different points of view on politics (from Bush) gain control of the presidency and of Congress.

[quote="tokenadult"]Actually, less than half of Americans "accept everything Bush says".[/quote]
I agree.
The point is, how many american people would/could say NO to him? and how many would/could try to do something to stop him?
The US is the only superpower today, it can smash anything if necessary, yet I haven't seen any of its leaders, i.e.,  the presidents, who had ever shown a little bit cautiousness on using the power. They always said, I want to fight, let US fight. And what did american do? They paid the tax as usual, the newspapers began to prove the righteousness of what the president is doing, for example, those enemies eat human beings alive, they're not demoncratic, their officers are so corrupt, they have death penalty, so on and so on. so the US should, cut the food supply(like what happened in Iraq), millions of people died of famine, this is just the insignificant cost to reach the glorious goal. and then, kill the rest with the weapons of mini destruction.
My conclusion is, there are only four possibilities,
1)Though always claim that they want to liberate the world from evil, the american people never take the pain of the others into consideration.
2)They will do anything the presidents ask them to do, no matter what they are.
3)They don't want to follow, but they have no choice.
4)They don't want to follow, and they could stop the president, but it's not worth doing that, it's either costly or dangerous, or there are plenty of more important or more beneficial stuff waitting for them.
Any other ones?

Kill several at the cost of your life, you will be called terrorist, kill millions, you're called the liberator.

[quote="Buffalo"]The point is, how many american people would/could say NO to him? and how many would/could try to do something to stop him?
The US is the only superpower today, it can smash anything if necessary, yet I haven't seen any of its leaders, i.e.,  the presidents, who had ever shown a little bit cautiousness on using the power. They always said, I want to fight, let US fight.[/quote]
I don't think anyone can take you seriously after such an ignorant statement.

Explain how George H.W. Bush and Bill Clinton were not militarily cautious.

maybe in my lifetime. The US is going onlya little downhill, not that much

[quote="kheg_k"]i said never, even though i believe it will fall in the mid-late 3000's.
just like the byzantium and roman empires[/quote]

Except not really, because both of those empires fell from what we might consider "superpower" status in under 500 years. Actually, Byzantium was never all that strong anyway.


I have already solved it, but the solution is rather long to be written here.
So, here is the problem.
In the triangle $ ABC$ $ AB \equal{} 13 cm$, $ BC \equal{} 14 cm$, $ AC \equal{} 15 cm$. Point $ B$ lies on the line which is perpendicular to the side $ AB$. This line intersects the bisector $ CD$ in point $ E$. Find the length of $ BE$.

[quote="skalgar"] This line intersects the extended bisector $ CD$ of the angle $ C$ in point $ E$. Find the segment $ BE$.[/quote]

-why extended? I think it's the original bisector

-The segment $ BE$ is right between the points $ B$ and $ E$... Let's find the length of $ BE$.

OK. I edited my post. BTW, this problem is a translation.

Draw $ CF \perp AB, F \in AB$, find BF and AF from $ BF^{2} \minus{} AF^{2} \equal{} BC^{2} \minus{} AC^{2}$, then find CF from $ CF^{2} \equal{} BC^{2} \minus{} BF^{2}$, next find BD from $ \frac {BC}{AC} \equal{} \frac {BF}{AF}$ and DF from $ DF \equal{} BF \minus{} BD$ and finally BE from $ \frac {BE}{CF} \equal{} \frac {BD} {DF}$.
Good luck

Best regards,
sunken rock


analytic geometry
calculus
integration
combinatorics unsolved
combinatorics

$ n$ points in the plane are such that for any 3 points we can find a cartesian coordinate system in which the points have integral coordinetes. Show that there is a cartesian coordinate system in which all $ n$ points have integral coordinates.

This has been posted (but not answered) twice recently.  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199142]Here[/url] was the first instance.

But I need the solution.
Is there any one??

[quote="Erken"]The main idea is to show that if in some cartesian system $ C$,three non-collinear points $ A,B,C$ have integer coefficients, then any other point $ D$ has rational coefficients in $ C$.The last idea can be proved easily if we consider at first cartesian system for $ A,B,C$ and then another coordinate system for $ B,C,D$.Hence if the change the unit length in $ C$ and obtain $ C'$ then all distnces in $ C'$ between the points will be integer.[/quote]

Could you please show the "main idea" in more detail?

"Main idea":
If in cartesian system $ \zeta_1$three non-collinear points $ A,B,C$ have integer coordinates and in another cartesians coordinate systems $ \zeta_2$ and $ \zeta_3$ points $ C,B,D$ and $ A,B,D$ have integer coordinates,respectively,then $ D$ has rational coordinates in $ \zeta_1$.
Proof of the "main idea" :lol::
If $ d_1$ is the length of $ BC$ in $ \zeta_1$ and $ d_2$ is the length of $ BC$ in $ \zeta_2$ then
 $ (\frac {d_1}{d_2})^2\in\mathbb{Q}$                              $ (*)$
,because $ d_1^2,d_2^2\in\mathbb{Z}$.
It is known that $ \overrightarrow{BC}\cdot\overrightarrow{BD}\in\mathbb{Z}$ in $ \zeta_2$.So from $ (*)$ we conclude that $ \boxed{\overrightarrow{BC}\cdot\overrightarrow{BD}\in\mathbb{Q}}$ in $ \zeta_1$.
In a similar way we show that $ \boxed{\overrightarrow{BA}\cdot\overrightarrow{BD}\in\mathbb{Q}}$ in $ \zeta_1$.

Thus if $ BC \equal{} (x_c,y_c),BA \equal{} (x_a,y_a),BD \equal{} (x_d,y_d)$ in $ \zeta_1$ then
$ x_c\cdot x_d \plus{} y_c\cdot y_d\in\mathbb{Q}$ and $ x_a\cdot x_d \plus{} y_a\cdot y_d\in\mathbb{Q}$,hence $ x_d,y_d\in\mathbb{Q}$.
Hence $ D$ has rational coordinates in $ \zeta_1$ and the "main idea" is proved.

[quote="nguyentrang"][quote="Erken"]

Thus if $ BC \equal{} (x_c,y_c),BA \equal{} (x_a,y_a),BD \equal{} (x_d,y_d)$ in $ \zeta_1$ then
$ x_c\cdot x_d \plus{} y_c\cdot y_d\in\mathbb{Q}$ and $ x_a\cdot x_d \plus{} y_a\cdot y_d\in\mathbb{Q}$,hence $ x_d,y_d\in\mathbb{Q}$.
[/quote]

Hmm... what about something like $ (x_c,y_c) \equal{} (1,2), (x_a,y_a) \equal{} (2,4)$ and $ (x_d,y_d \equal{} 2\sqrt {2}, \minus{} \sqrt {2})$?

Sorry if I misunderstand something.[/quote]
I guess that you was inattentive,at the beginning i suppose that $ A,B,C$ are not collinear.

Yes, I was inattentive. Sorry Erken  :oops:

[quote="Erken"]Do you completely understand the proof?[/quote]

I think so  :)

It is very nice.

[quote="Erken"]The main idea is to show that if in some cartesian system $ C$,three non-collinear points $ A,B,C$ have integer coordinates, then any other point $ D$ has rational coordinates in $ C$.The last idea can be proved easily if we consider at first cartesian system for $ A,B,C$ and then another coordinate system for $ B,C,D$.Hence if the change the unit length in $ C$ and obtain $ C'$ then all distnces in $ C'$ between the points will be integer.[/quote]

That doesn't make very much sense :D

You can look in his other post for more detail  :D

[quote=nguyentrang] [b]Russia 2002 Grade 11 #2: [/b] Several points are given in the plane. Suppose that for any three of them, there exists an orthogonal coordinate system (determined by the two axes and the unit length) in which these three points have integer coordinates. Prove that there exists an orthogonal coordinate system in which all the given points have integer coordinates.[/quote]

[quote=Erken]The main idea is to show that if in some cartesian system $ C$,three non-collinear points $ A,B,C$ have integer coordinates, then any other point $ D$ has rational coordinates in $ C$.The last idea can be proved easily if we consider at first cartesian system for $ A,B,C$ and then another coordinate system for $ B,C,D$.Hence if the change the unit length in $ C$ and obtain $ C'$ then all distnces in $ C'$ between the points will be integer.[/quote]

This problem says that the coordinate system must have unit length. But this solution said that one must change the unit length in $C$ to obtain integer coordinates. Can one do this?

Sorry, but what does unit length mean? Does it mean fixed no matter which coordinate system or just each square has equal length? So does the proof by Erken work for the Russia problem? Thank you very much for all your help!


Suppose that $ a$ and $ b$ are digits, not both nine and not both zero, and the repeating decimal $ 0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

[hide]${ 0.\overline{ab}} = \frac {ab}{99}$, so the denominator in lowest terms must be a positive divisor of $ 99$, one of $ 1, 3, 9, 11, 33, 99$. However, $ 1$ isn't possible, as ${ 0.\overline{ab}} < 1$, for the digits are not both nine. So, there are five different denominators possible, and the desired answer is C.[/hide]

In order to consider the words"in lowest terms", we should eliminate 9's factors, 1, 3,9. Because of that, there are only 3 solutions, 11,33,99.

[hide]
Denote the decimal $ x$

Then $ x \equal{} \dfrac{ab}{99}$

Thus, the factors of 99 that will reduce the fraction are 3,6,9,11,33,66

$ 3/99$ and $ 6/99$ yield the same denominator, as does $ 33/99$ and $ 66/99$, so there are $ 4$ factors that yield a unique denominator.

There is also the case where the denominator does not reduce, so there are $ 5$ total.
[/hide]

@svnfmy - consider $ 0.\overline{33} \equal{} 0.33333333.....$
Carrying out, we get $ \frac {33}{99} \Longrightarrow \frac {1}{3}$, which is perfectly legitimate.  :wink:

@feuxfollets - the factors of $ 99$ are $ 3, 9, 11, 33, 99$ (let's not include one, because ab cannot equal $ 99$)

$ 6, 66$ are NOT factors of $ 99$  :wink:

@towersfreak2006 - good job!

[b]Solution:[/b] $0.\overline{ab} = \frac{10a+b}{99}$. It seems that the only possible denominators are the factors of $99$ (when the fraction is reduced). However, since both $a$ and $b$ can't be $9$, $1$ is not a possible denominator. So our answer is the number of factors of $99$ minus $1$ (of them) which gives us our answer of C.


A driver approaches a toll booth and randomly selects two coins from his pocket. If the pocket contains 2 quarters, 2 dimes, and 2 nickels, what is the probability that the value of the two coins he selects will be at least enough to pay the 30-cent toll? Express your answer as a common fraction.

Posted here already:[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=driver+toll+booth&t=27630]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=driver+toll+booth&t=27630[/url]

Could anyone post their solutions here, though? They never reached a conclusion on the other thread.

[hide]$\frac{1}{15}$ chance of two quarters, two dimes, or two nickels.
$\frac{4}{15}$ chance of quarter dime, quarter nickel, and dime nickel.

Quarter dime, quarter nickel, and two quarters work giving you a total probability of $\frac{9}{15}$ or $\frac{3}{5}$[/hide]

Edit: Added hide tags, simplified fraction  :blush:

[quote="Ignite168"]$\frac{1}{15}$ chance of two quarters, two dimes, or two nickels.
$\frac{4}{15}$ chance of quarter dime, quarter nickel, and dime nickel.

Quarter dime, quarter nickel, and two quarters work giving you a total probability of $\frac{9}{15}$[/quote]

You might want to reduce.

All the driver needs is to get at least one quarter so...

2/3*3/5=2/5

1-2/5=3/5


let $ B_{\epsilon}(x)$ denote the open ball of radius $ \epsilon$ centered at $ x$. for every $ A\subseteq X$, $ A \equal{} \bigcup _{x\in A} B_{\frac {1}{2}}(x)$, and since every point of $ A$ is contained in exactly one ball, this cover has a finite subcover only if $ A$ is finite. conversely, if $ A$ is finite, it is obvious that every cover has a finite subcover (this is true for any space).

[quote="hebby"]Let X be a set donated by the discrete metrics
d(x,y) = 0 if x=y,
            1 if x̸=y.

Show that a subset Y of X is compact iff this set is finite[/quote]
The topology induced on the set by the above defined metric is called the discrete topology because as you can see the individual points of the set are open sets.This property(discrete topology) induced by the metric is the one which forces compact subsets of the space to be finite because as [b]pluristiq[/b] pointed out an open cover of the set is the union of all the points belonging to the set and no subset of this cover can cover it.


trigonometry
inequalities
function
inequalities unsolved

For every $ a,b,c\in [0,\frac{1}{\sqrt{3}}\}$, prove that

$ \frac{a\plus{}b}{1\minus{}ab}\plus{}\frac{b\plus{}c}{1\minus{}bc}\plus{}\frac{c\plus{}a}{1\minus{}ca}\leq \frac{2(a\plus{}b\plus{}c\minus{}abc)}{1\minus{}ab\minus{}bc\minus{}ca}$

Let $ a\equal{}\tan x$ etc. and WLOG let $ a\leq b\leq c$. Then $ 0\leq x\leq y\leq z<\frac{\pi}{6}$ and the inequality becomes $ \tan (y\plus{}z)\plus{}\tan (x\plus{}z)\plus{}\tan (x\plus{}y)\leq 2\tan (x\plus{}y\plus{}z)$, which follows from Majorization on the function $ f(t)\equal{}\tan t$ which is concave up on $ [0,\frac{\pi}{2})$.


Find all fourth degree polynomial $ p(x)$ such that the following four conditions are satisfied:
(i) $ p(x)\equal{}p(\minus{}x)$ for all $ x$,
(ii) $ p(x)\ge0$ for all $ x$,
(iii) $ p(0)\equal{}1$
(iv) $ p(x)$ has exactly two local minimum points $ x_1$ and $ x_2$ such that $ |x_1\minus{}x_2|\equal{}2$.

$ p(x)\equal{}ax^4\plus{}bx^2\plus{}1$, $ p'(x)\equal{}x(4ax^2\plus{}2b)\equal{}0\to b\equal{}\minus{}2a,a>0$, because $ x_1,x_2\equal{}\pm 1$.
Therefore $ p(x)\equal{}a(x^2\minus{}1)^2\plus{}1\minus{}a\to 0<a\le 1$.

How did you prove that: x1,x2=+1 or -1??