geometry
geometric transformation
reflection
floor function

A beam of light strikes $\overline{BC}$ at point $C$ with angle of incidence $\alpha=19.94^\circ$ and reflects with an equal angle of reflection as shown.  The light beam continues its path, reflecting off line segments $\overline{AB}$ and $\overline{BC}$ according to the rule: angle of incidence equals angle of reflection.  Given that $\beta=\alpha/10=1.994^\circ$ and $AB=AC,$ determine the number of times the light beam will bounce off the two line segments.  Include the first reflection at $C$ in your count.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=452[/img]

Hint:

[hide] Instead of the light 'bouncing' off the side, reflect one side over another and let the light continue in a straight line.[/hide]

[hide="Solution"]

It's not hard to find how many times it reflects going toward the vertex because it turns around after the angle becomes greater than $90^\circ.$  You can use the equal angles and everything to figure that out.  Now, difficulty arises when coming out.  How do you know when it is 'out'?  At this point with this method you can only make an educated guess.  So we should try to find a better method.

We have all these equal angles and reflections etc.  Is there another way we can view this and still have equal angles and the relflections?  Aha! We could reflect $AB$ across $BC$.  The equal angles are preserved because they are now vertical angles.  This already looks better than the other method for we are know dealing with a straight line.  Imagine we did this every time there was a reflection.  Now we can think to draw a circle through the endpoints of these segments with center at $B$ (remember they are equal segments so this is perfect!).  Now how do we find out how many reflections it makes?  This is where the picture comes in.  The beam enters in at $C$ and leaves at $Z$ (I use $Z$ just to denote the point where it leaves, the fact that $Z$ is the 26th letter of the alphabet means nothing).  Connect the points to make $\triangle ABZ$.  This is an isosceles triangle (two radii) so thus $\angle BAZ=\angle BZA=19.94^\circ$ and $\angle ABZ =140.12^\circ.$  To find the number of reflections we just find $\left\lfloor\frac{140.12}{1.994}\right\rfloor+1=\fbox{71}.$  (the +1 is because we have to count the point where it enters the circle).[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=648[/img][/hide]

sorry to be picky, but from a physics point of view, the angle of incidence is the angle that the beam makes with the line normal to the surface

eeeeerm.....one more question, how is [b]AB=AC[/b] ??? :fool:

EDIT: Never mind
 :blush:

this is a pretty nifty problem with reflections...

there's a similar problem in the art and craft of problem solving
where it's billiard balls bouncing into a corner.

Here's an additional challenge for you: can you give a general solution to this problem, given that the angle between the walls is B, and the initial angle of incidence is A; and the walls now extend infinitely?


group theory
abstract algebra
advanced fields
advanced fields unsolved

Let $G$ be a group. We say that $f: G\to G$ is a [i]square root[/i] of $G$ if $f(x)^{2}=x,\ \forall x\in G$.

Prove that a compact connected Lie group $G$ has no continuous square roots.

We reduce the problem to the circle.
$G$ contains an $S^{1}$.  Being a Lie group, near 1, square roots exist, and being connected, the local square root map is unique, and therefore sends this $S^{1}$ to itself. The square root map is, near the identity, a homomorphism of $S^{1}$, so by multiplication and continuity we see it must extend to larger and larger neighborhoods, eventually (by connectivity and continuity of $f(x)$) covering all $S^{1}$, which is impossible.

I may be misinterpreting what you wrote, but it looks like you're assuming $f(1)=1$. This is not given. $f(1)$ can be an element of order two, far away from $1$.

P.S.

I'm reposting this message because I wanted to ask the additional question: what happens if instead of a Lie group we consider a compact connected group (with more than one element, of course :); the same observation goes for the initial problem)? Does the statement remain true?


limit
trigonometry
calculus
calculus computations

Prove that $ \lim_{n\to \plus{} \infty}{n\left[\sum_{k \equal{} 1}^{n}{\left(1 \plus{} \frac {k}{n}\right)\sin{\frac {k\pi}{n^2}}} \minus{} \frac {5\pi}{6}\right] }\equal{} \pi.$

[quote="Ji Chen"]Prove that $ \lim_{n\to + \infty}{n[\sum_{k = 1}^{n}{(1 + \frac {k}{n})\sin{\frac {k\pi}{n^2}}} - \frac {5\pi}{6}] } = \pi.$[/quote]

This proof can be divided into two parts.note that when $ n$ is large enough,$ \sin \frac{k\pi}{n^2}<\frac{k\pi}{n^2}$.donote $ L= {n[\sum_{k = 1}^{n}{(1+\frac{k}{n})\sin{\frac{k\pi}{n^{2}}}}-\frac{5\pi}{6}] }$.It follows that

$ L<{n[\sum_{k = 1}^{n}{(1+\frac{k}{n})\frac{k\pi}{n^{2}}}}-\frac{5\pi}{6}] }={n[\sum_{k = 1}^{n}{\frac{k\pi}{n^2}+{\frac{\pi}{n^3}\sum_{k = 1}^{n}k^2-\frac{5\pi}{6}]=\frac{(n+1)\pi}{2}+\frac{(n+1)(2n+1)\pi}{6n}-\frac{5n\pi}{6}=\frac{(6n+1)\pi}{6n}\rightarrow\pi$(when $ n\rightarrow\infty$)

[quote="Ji Chen"]Prove that $ \lim_{n\to + \infty}{n\left[\sum_{k = 1}^{n}{\left(1 + \frac {k}{n}\right)\sin{\frac {k\pi}{n^2}}} - \frac {5\pi}{6}\right] } = \pi.$[/quote]

second part of proof

note that  $ \sin \frac{k\pi}{n^2}>\frac{k\pi}{n^2}-\frac{k^3\pi^{3}}{6n^6}$
denote  $ L={n\left[\sum_{k = 1}^{n}{\left(1+\frac{k}{n}\right)\sin{\frac{k\pi}{n^{2}}}}-\frac{5\pi}{6}\right] }$
$ L>{n\left[\sum_{k = 1}^{n}{\left(1+\frac{k}{n}\right)(\frac{k\pi}{n^2}-\frac{k^3\pi^{3}}{6n^6})-\frac{5\pi}{6}\right] }}$

$ ={\frac{\pi(360n^5-35\pi^2n^2-27\pi^2n^4-60\pi^2n^3+60n^4+2\pi^2)}{360n^5}}\rightarrow \pi(n\rightarrow \infty)$


May I ask what the definition of a geometric progression is and if the set of {0, 0, 0, 0, ...} is considered a geometric progression?

Yes, that is. The definition is that there must exist a real $ r$ such that $ a_k\equal{}a_1r^{k\minus{}1}$ for all $ k$.


Let ABCDEF be a hexagon such that $AB.CD.EF=BC.DE.FA$  and $\angle A+\angle C+\angle E=\angle B+\angle D+\angle F=360$. Then prove that
$AB.CE.DF=DE.AC.BF$

See the links at http://www.mathlinks.ro/Forum/viewtopic.php?t=19479 .

  darij


The number $n$ is randomly selected from the set $\{1,2,\ldots,10\},$ with each number being equally likely. What is the probability that $2n-4>n$? Express your answer as a common fraction.

[hide]Solving, we know $n>4$, so the probability is $\frac{3}{5}$.[/hide]

[hide]
solve the equation and you get n>4
that means you want to find the probablity that you will pick a number greater than 4
the probablity is 6/10 or 3/5[/hide]

[hide]
1. Solve the equation.
2n-4>n
n-4>0
n>4
2. Find the probability that the number you choose is greater than 4.
{[color=red]1, 2, 3, 4[/color], [color=green]5, 6, 7, 8, 9, 10[/color]}
3. 6 out of 10 are greater than 4. Simplify that to get $\frac{3}{5}$[/hide]

[quote="xdl615x"]Please Please solve it once! If an explanation is not clear then try to explain it to the person! I just think there's too much unnecessary posting going on.[/quote]
What do you mean? It is solved! "Click and reveal hidden content" <-----You dont understand what this means?

He means that too many people are providing answers.
Read this
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43075[/url]


function
real analysis
real analysis unsolved

Is the sequence of functions $ f_1$, $ f_2$, $ f_3$ on $ [0,1]$ uniformly convergent if:

$ f_n(x)\equal{}\frac{x}{1\plus{}nx^2}$? If $ f_n(x)\equal{}\frac{nx}{1\plus{}nx^2}$? If $ f_n(x)\equal{}\frac{nx}{1\plus{}n^2x^2}$

Any help would be appreciated. Thanks

1. Yes.
Notice that $ 1 \plus{} nx^2 \ge 2\sqrt{n} x$.
2. No.
It converges to the function $ \frac {1}{x}$, the difference $ |f_n \minus{} f| \equal{} \frac {1}{x(1 \plus{} nx^2)}$. Uniform convergence would mean that if $ n$ is large enough, this difference is small no matter what $ x$ is. But for fixed $ n$ as $ x$ goes to 0 the thing is unbounded.
3. No.
It converges to the zero function. And for any fixed n, choose $ x \equal{} \frac {1}{n}$ which makes the thing $ \frac {1}{2}$ and cannot get close to 0.


Yes!!Got it  :showoff:

[hide="Solution"]
Take log both the sides to get
$ \ln(x \plus{} 1) \plus{} \ln(x \plus{} 2) \plus{} \ln(x \plus{} 3).... \plus{} \ln(x \plus{} n) \equal{} \ln(A_{0} \plus{} A_{1}x \plus{} A_{2}x^{2} \plus{} ... \plus{} A_{n}x^{n})$

Differenciate both the sides to get
$ \left(\frac {1}{x \plus{} 1} \plus{} \frac {1}{x \plus{} 2} \plus{} .... \plus{} \frac {1}{x \plus{} n}\right) \equal{} \frac {A_{1} \plus{} 2A_{2}x \plus{} ... \plus{} nA_{n}x^{n \minus{} 1}}{A_{0} \plus{} A_{1}x \plus{} A_{2}x^{2} \plus{} ... \plus{} A_{n}x^{n}}$.

Now put $ x \equal{} 1$ we get $ \left(\frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n \plus{} 1}\right) \equal{} \frac {A_{1} \plus{} 2A_{2} \plus{} .... \plus{} nA_{n}}{A_{0} \plus{} A_{1} \plus{} .... \plus{} A_{n}}$

Also put $ x \equal{} 1$ in the given statement we get $ A_{0} \plus{} A_{1} \plus{} A_{2} \plus{} ... \plus{} A_{n} \equal{} (n \plus{} 1)!$
Hence Proved
[/hide]

wow ! it was good i could;ve never thought about it !

[hide="Induction"]This is true when $ n\equal{}1$. Assume that it is true for $ n$, so

$ \sum_{q\equal{}0}^{n\minus{}1}(q\plus{}1)A_{q\plus{}1}\equal{}(n\plus{}1)!\left(\frac12\plus{}\cdots\plus{}\frac1{n\plus{}1}\right)$.

Then let $ \prod_{i\equal{}1}^{n\plus{}1}x\plus{}i\equal{}\sum_{j\equal{}0}^{n\plus{}1}B_jx^j$ so we have $ B_{k\plus{}1}\equal{}A_k\plus{}(n\plus{}1)A_{k\plus{}1}$ where $ A_{n\plus{}1}\equal{}0$, and

$ \sum_{k\equal{}0}^n(k\plus{}1)B_{k\plus{}1}\equal{}\left(\sum_{l\equal{}0}^n(l\plus{}1)A_l\right)\plus{}(n\plus{}1)\left(\sum_{m\equal{}0}^n(m\plus{}1)A_{m\plus{}1}\right)$

$ \equal{}\left(\sum_{p\equal{}0}^nA_p\right)\plus{}(n\plus{}2)\left(\sum_{q\equal{}0}^{n\minus{}1}(q\plus{}1)A_{q\plus{}1}\right)$

$ \equal{}(n\plus{}1)!\plus{}(n\plus{}2)!\left(\frac12\plus{}\cdots\plus{}\frac1{n\plus{}1}\right)$

$ \equal{}(n\plus{}2)!\left(\frac12\plus{}\cdots\plus{}\frac1{n\plus{}2}\right)$.[/hide]


If $ a,b,c>0$ and $ \frac {1} {a}\plus{}\frac {1} {b}\plus{}\frac {1} {c}\equal{}3$ then:

$ 1\le\frac {a^2} {a^2\plus{}a\plus{}1}\plus{}\frac {b^2} {b^2\plus{}b\plus{}1}\plus{}\frac {c^2} {c^2\plus{}c\plus{}1}\le \frac {a\plus{}b\plus{}c} {3}$

[hide="Left side"]For the left side we have $ \sum \frac {1}{(1/a)^2 \plus{} (1/a) \plus{} 1} \ge 1$, true by Jensen on $ f(x) \equal{} \frac {1}{x^2 \plus{} x \plus{} 1}$[/hide]

For the right side, we have $ (a\minus{}1)^2\ge 0\implies a^2\plus{}a\plus{}1\ge 3a\implies \sum\frac{a^2}{a^2\plus{}a\plus{}1}\le\sum\frac a3$.


[quote="risinglegend"]wad is
#1. (1/4)^-6 * (1/4)^3 * (1/4)^5
#2. (9/25)^-3/2 - (81/16)^-3/4
nid hlp on this two q's ty. i dun understand???[/list][/quote]

[hide]This is pretty simple once you learn some rules about exponents.

1: When something is raised to a negative power, it is the same as saying "the recipricol of it raised to the positive power" like this:

$x^{-2}= \frac{1}{x^{2}}$
and
$x^{-8}= \frac{1}{x^{8}}$

2: when something is raised to a fractional power, then the denominator is the "root" and the numerator is the "exponent" like this:

$x^{\frac{3}{2}}= \sqrt{x^{3}}$
and
$x^{\frac{7}{2}}= \sqrt{x^{7}}$

and one can combine those concepts together...

$x^{\frac{-5}{2}}= \frac{1}{\sqrt{x^{5}}}= \frac{\sqrt{x}}{x^{3}}$

hope that helped...  :) [/hide]

[quote="risinglegend"]wad is
#1. (1/4)^-6 * (1/4)^3 * (1/4)^5
#2. (9/25)^-3/2 - (81/16)^-3/4
nid hlp on this two q's ty. i dun understand???[/list][/quote]
Ok a general rule if two numbers have different exponents and the same bases, you can add both exponents.

[hide="#1"]So $\frac{1}{4}^{-6}\cdot\frac{1}{4}^{3}\cdot\frac{1}{4}^{5}$. So adding all bases you finally have $\frac{1}{4}^{2}=\frac{1}{16}$.[/hide]

[hide="#2"]In this one if you have a negative exponent you flip the fraction and then solve. So flipping 9/25 to 25/9 and taking the 3/2 root. This can be simplified to $\frac{25}{9}^\frac{1}{2}\cdot2$. So taking the square root because thats what 1/2 means you get 5/3. Then you take the third root of that to get 125/27. Doing the same method for the first one you get 8/27.[/hide]

[quote="myspacesucks"]Actually, you are all wrong on number 2, the answer is 9 which I will get to later.  If you have a negative exponent, then the number is like so: x^-1=1/x^1.  Therefore, (9/25) ^-3/2 is
sqrt 225 ^3 and (81/16) ^-3/4 is quadruple root 1296^3.  The exponential 3's cancel off because you are subtracting so you get sqrt 225 - quadruple root 1296.  Therefore it is 15-6 (you can check) or 9.[/quote]
I don't know what you did, but I'm pretty sure that Quevvy and me are correct. Since $\frac{9}{25}^{-\frac{3}{2}}=\frac{25}{9}^{\frac{3}{2}}$. So this equals $\frac{25}{9}^{\frac{1}{2}\cdot3}=\frac{5}{3}^{3}=\frac{125}{27}$. And the second one doing the same method is $\frac{81}{16^{-\frac{3}{4}$=$\frac{16}{81}^{3}{4}$=$\frac{16}{81}^{\frac{1}{4}\cdot3}$=$\frac{2}{3}^{3}=\frac{8}{27}$. So subtracting $\frac{125}{27}-\frac{8}{27}=\frac{117}{27}=\frac{13}{3}$.
Edit: Just press quote to see what it says. Somehow this isn't working.

Actually, you are wrong on the first part only , the negative fraction doesn't make 9/25 become 25/9, it makes it 9/1/25, which is the same as saying 9*25 or 225. It then becomes sqrt of 225 ^3

[quote="ckck"][quote="myspacesucks"]Actually, you are all wrong on number 2, the answer is 9 which I will get to later.  If you have a negative exponent, then the number is like so: x^-1=1/x^1.  Therefore, (9/25) ^-3/2 is
sqrt 225 ^3 and (81/16) ^-3/4 is quadruple root 1296^3.  The exponential 3's cancel off because you are subtracting so you get sqrt 225 - quadruple root 1296.  Therefore it is 15-6 (you can check) or 9.[/quote]
I don't know what you did, but I'm pretty sure that Quevvy and me are correct. Since $\frac{9}{25}^{-\frac{3}{2}}=\frac{25}{9}^{\frac{3}{2}}$. So this equals $\frac{25}{9}^{\frac{1}{2}\cdot3}=\frac{5}{3}^{3}=\frac{125}{27}$. And the second one doing the same method is $\frac{81}{16}^{-\frac{3}{4}}$=$\frac{16}{81}^{\frac{3}{4}}$=$\frac{16}{81}^{\frac{1}{4}\cdot3}$=$\frac{2}{3}^{3}=\frac{8}{27}$. So subtracting $\frac{125}{27}-\frac{8}{27}=\frac{117}{27}=\frac{13}{3}$.
Edit: Just press quote to see what it says. Somehow this isn't working.[/quote] You forgot a bracket and a fraction, like I did when I was typing mine up. It took so long... and I'm pretty sure you and I are right...
Edit: I checked it on a calculator, and it is $\frac{13}{3}$


trigonometry
inequalities proposed
inequalities

let a,b,c be positive numbers such that: a^2+b^2+c^2+abc=4.Find the best k constant such that :
  4/(4-a^2) + 4/(4-b^2) + 4/(4-c^2) + kabc ≥ 4 + k     :lol:
Good luck.

[quote="nguoivn"]Thank you for your gift, my friend.
I just go home after a nice Valentine day. Now, I really very very happy   :lol:
With your problem, the best constant $ k \equal{} 1$.
It's obviously trues because:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ 2$
 :)[/quote]
VERY NICE SOLUTION. :)

[quote="karis"]Let $ a;b;c;d\in[1;2]$.Prove that:
       $ \frac {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}{(ac \plus{} bd)^2}\leq\frac {25}{12}$[/quote]
$ \frac {(a^2 \plus{} b^2)(c^2 \plus{} d^2)}{(ac \plus{} bd)^2}\leq\frac {25}{16}$ is true too. :wink:

[quote="kunny"]If we set $ a \equal{} 2\cos A,\ b \equal{} 2\cos B,\ c \equal{} 2\cos C\ \left(0 < A,\ B,\ C < \frac {\pi}{2}\right)$, can we solve the original problem?[/quote]
I didn't try this way. But the following substitution solves this problem easy enough.
$ a \equal{} \sqrt {\frac {4yz}{(x \plus{} y)(x \plus{} z)}}...$ :wink:

[quote="kunny"]If we set $ a \equal{} 2\cos A,\ b \equal{} 2\cos B,\ c \equal{} 2\cos C\ \left(0 < A,\ B,\ C < \frac {\pi}{2}\right)$, can we solve the original problem?[/quote]
 I don't think. I SET a=2sin(A/2),b=2sin(B/2),c=2sin(C/2) .

$ (a\minus{}2b)(b\minus{}2a)\geq0,$ which gives $ 2(a^2\plus{}b^2)\leq5ab.$
Similar $ 2(c^2\plus{}d^2)\leq5cd.$
Id est, $ 4(a^2\plus{}b^2)(c^2\plus{}d^2)\leq25abcd\leq\frac{25(ac\plus{}bd)^2}{4}.$

[quote="kunny"]How do you notice such a substitution?[/quote]
The substitution arquady made reminds me of the HALF ANGLE rule, but he might have noticed it in some other way...
NICE SUBSTITUTION, brilliant, arquady :lol:


geometry
3D geometry
tetrahedron
geometry unsolved

Let $ABCD$ be a tetrahedron and let $S$ be its center of gravity. A line through $S$ intersects the surface of $ABCD$ in the points $K$ and $L$. Prove that \[\frac{1}{3}\leq \frac{KS}{LS}\leq 3\]

Suppose $K$ lies on the surface $BCD$. Let $AS$ intersect $BCD$ at the centroid $G$ of $\triangle BCD$. Then we can easily show by Menelaus that $\frac{AS}{SG}=3$. Now note that the fraction $\frac{KS}{SL}$ is minimal when $L$ is as far away as $BCD$ as possible, which is point $A$. Hence $\frac{KS}{SL}\geq \frac13$. Similarly by swapping $K$ and $L$, we get $\frac{KS}{SL}\leq 3$.


limit
real analysis
real analysis unsolved

Let $ a_{ij} \equal{} \left\{ \begin{array}{lr} 0 & i < j \\
\minus{} 1 & i \equal{} j \\
2^{j \minus{} i} & i > j \\
\end{array} \right.$

 Find  $ \displaystyle\sum_{i \equal{} 1}^\infty\displaystyle\sum_{j \equal{} 1}^\infty a_{ij}$

Are you sure that's the sum you want? The values of the inner sum grow rapidly, and it diverges to $ \infty$.

If you reverse that power of $ 2$, this is a nice example of a sum which converges to different vlues depending on the order.

If you just do the order as written, it converges to 0.

I thought $ \displaystyle\sum_{j \equal{} 1}^\infty\displaystyle\sum_{i \equal{} 1}^\infty a_{ij}\equal{}0$. Can you prove $ \displaystyle\sum_{i \equal{} 1}^\infty\displaystyle\sum_{j \equal{} 1}^\infty a_{ij}\equal{}0$?

I don't fully understand double sum notations, but would this be
$ (a_{11} \plus{} a_{12} \plus{} a_{13} \plus{} ... \plus{} a_{1\infty}) \plus{} (a_{21} \plus{} a_{22} \plus{} a_{23} \plus{} ... \plus{} a_{2\infty}) \plus{} (a_{31} \plus{} a_{32} \plus{} a_{33} \plus{} .. \plus{} a_{3\infty}) \plus{} .....$??

Thanks :)

And if so, I think its $ \minus{}2$.

$ \displaystyle\sum_{j \equal{} 1}^n \displaystyle\sum_{i \equal{} 1}^m a_{ij}\equal{}\displaystyle\sum_{i \equal{} 1}^m \displaystyle\sum_{j \equal{} 1}^n a_{ij}$ for all n,m :) , it's easy to prove it by suming the table $ n \times m$ by two methods. (by collumns and by rows).

[quote="Sylwek"]$ \displaystyle\sum_{j \equal{} 1}^n \displaystyle\sum_{i \equal{} 1}^m a_{ij} \equal{} \displaystyle\sum_{i \equal{} 1}^m \displaystyle\sum_{j \equal{} 1}^n a_{ij}$ for all n,m :) , it's easy to prove it by suming the table $ n \times m$ by two methods. (by collumns and by rows).[/quote]

if $ a_{ij}\geq0 \ \forall i,j$, right?

Sylwek: That equation is correct- but it doesn't apply here. You need $ m$ and $ n$ to be finite to say that unconditionally. Alternatively, either absolute convergence (Fubini's theorem) or positivity (Tonelli's theorem) give that conclusion.
Euclid's Turtle: There are no terms with infinite index. A sum to $ \infty$ features an implicit limit; $ \sum_{j\equal{}1}^{\infty} a_j\equal{}\lim_{n\to\infty} \sum_{j\equal{}1}^n a_j$, just like an improper integral.

No, $ a_{ij}$ needn't be positive. Try to prove it by making table and suming it by rows and by collumns.

[b]jmerry[/b], yes, You're right.


[COLOR="Sienna"]1/ [/COLOR]with a acute triangle, find Min of expression:
$ T=cotA+cotB+cotC+tan \frac{A}{2}+tan \frac{B}{2}+tan \frac{B}{2}$

[COLOR="Sienna"]4/[/COLOR] prove: we have a equilateral t. if:

$ \{\begin{array}{l} tanA+tanB-tanC=\sqrt{3} \\2(sin2A+sin2B-sin2C)=\sqrt{3} \end{array}  $

[COLOR="Blue"]III-solve equation[/COLOR]

[COLOR="Sienna"]5/[/COLOR]$ 2cos^3x+cos2x+sinx=0$

[COLOR="Sienna"]6/[/COLOR]$ 4cos^23x+cos^2x=cos3xcos^4x$

[COLOR="Sienna"]7/[/COLOR] $ cosx-cos4x+4cos3x+4=0$

[COLOR="Sienna"]8/[/COLOR]$ cosx-cos2x+cos3x=1/2$

[COLOR="Sienna"]9/[/COLOR] $ sin4x - 4sin x- (cos4x-4cosx) =1$

This is definitely not HSB. Can someone move this to intermediate?

Also, you make colors by looking at the top by font color and scrolling down.

[color=blue]example[/color]


Let $ ABC$ be a triangle.$ D$ and $ E$ ies on sides $ BC$ and $ CA$,respectively,such that $ BD\equal{}AE$.Segments $ AD$ and $ BE$ meets at $ P$.The bisector of angle $ BCA$ meets segment $ AD$ and $ BE$ at $ Q$ and $ R$,respectively.Prove that $ \frac{PQ}{AD}\equal{}\frac{PR}{AC}$

AE=EF=BD  => ∠EAF=1/2∠CEF=1/2∠C=∠QCA
⊿PQR∽⊿ADF


Let $\displaystyle \mathcal P$ be a plane. Find all functions $\displaystyle f : \mathcal P \to \mathcal P$ which have the following property: if $\displaystyle A,B,C,D$ are the vertices of a convex quadrilateral, then $\displaystyle f(A),f(B),f(C),f(D)$ are the vertices of a concave quadrilateral.

[i]Dinu Serbanescu[/i]

It is impossible. If ABCDE is a convex pentagon, then some four of five points f(A), f(B), f(C), f(D),  f(E) are vertices of a convex (or degenerated) quadrilateral - a contradiction.


We consider two non intersecting circles $ C_1(O_1,R_1)$ and $ C_2(O_2,R_2)$ and let one of their external common tangent touches $ C_1,C_2$ at $ A,B$ respectively .Now consider one of their internal common tangents which touches $ C_1,C_2$ at $ K,L$ and intersects $ AB$ at the point $ S$ . Let now $ AK,BL$ are intersecting at the point $ Q$ . Prove that the points
$ O_1,Q,O_2$ are collinear

[hide][b][size=100][color=DarkBlue]LEMMA. – A trapezium $ ABCD$ with $ AB\parallel CD$ is given and let $ E$ be, an arbitrary point on $ BC.$ Prove that the point $ Z$ of intersection of the lines through $ B,\ C$ and parallel to $ DE,\ AE$ respectively, lies on $ AD.$[/color][/size][/b]

The proof of this [b][size=100]Lemma[/size][/b], is very easy by [b][size=100]Thales theorem[/size][/b].

$ \bullet$ Because of now, the line segments $ SO_{1},\ SO_{2}$ are the angle bisectors of the angles $ \angle ASK,\ \angle BSL$ respectively, we have that $ SO_{1}\perp SO_{2}$ $ ,(1)$

From $ (1)$ and because of $ SO_{1}\perp AK$ and $ SO_{2}\perp BL,$ we have that $ SO_{1}\parallel BL$ and $ SO_{2}\parallel AK.$

Hence, in the orthogonal trapezium $ O_{1}ABO_{2}$ with $ AO_{1}\parallel BO_{2},$ based on the above [b][size=100]Lemma[/size][/b], we conclude that the point $ Q\equiv AK\cap BL,$ lies on the line segment $ O_{1}O_{2}$ and the proof is completed.[/hide]
 Kostas Vittas.

Observe that $O_{1}A\perp AB\perp O_{2}B,QA\perp O_{1}S\perp O_{2}S\perp QB$, so $O_{1}A\parallel O_{2}B,QA\parallel O_{2}S,O_{1}S\parallel QB$, and by the converse Pappus theorem for points $\left\{O_{1},Q,O_{2}\right\}$ and $\left\{A,S,B\right\}$ we obtain the required assertion.


calculus
integration
logarithms
calculus computations

[quote="kcn2rivers"]Edit: nevermind, I figured it out : $ \int \ln u du \equal{} u\ln u \minus{} u \plus{} C$.[/quote]
That particular formula is derived by means of an integration by parts.

You could have done the integration by parts in the beginning, without using the substitution at all.

Your next exercise: $ \int x^2e^x\,dx$

[quote="kcn2rivers"]What am I doing wrong here with finding $ \int x e^x dx$ ?

My attempt:
Let $ u \equal{} e^x$. $ \ln u \equal{} x$. $ du \equal{} e^x dx$.  So $ \int x e^x dx \equal{} \int \ln u du \equal{} \frac {1}{u} \plus{} C \equal{} \frac {1}{e^x} \plus{} C$.
The answer is supposed to be $ e^x (x \minus{} 1) \plus{} C$.

Edit: nevermind, I figured it out : $ \int \ln u du \equal{} u\ln u \minus{} u \plus{} C$.[/quote]
[hide="It may be solved"]
By Parts,
$ \int x e^x dx \equal{} \int xd e^x \equal{} xe^x \minus{} \int e^xdx \equal{} xe^x \minus{} e^x \plus{} c$

$ \int x^2 e^x dx \equal{} \int x^2d e^x \equal{} x^2e^x \minus{} \int e^xdx^2 \equal{} x^2e^x \minus{}2\int xe^x dx$
$ \equal{}x^2e^x\minus{}2(xe^x \minus{} e^x)\equal{}x^2e^x\minus{}2xe^x \plus{}e^x\plus{}c$

Finally , we can try $ I_n\equal{}\int x^n e^x dx$
[/hide] :starwars:


Hey, a day has gone. Does anyone have answer for a,b, or [b]c[/b]????
 :o


videos
quadratics
inequalities
geometry
3D geometry
calculus
Gauss

Now you can divide by zero!

http://www.bbc.co.uk/berkshire/content/articles/2006/12/06/divide_zero_feature.shtml

And I bet you thought education couldn't have gotten any worse in the U.S...

Wait did you hear about the "we don't teach long division" thing? See [url=http://www.nytimes.com/2006/11/14/education/14math.html?ex=1165640400&en=9026890c2e53d4d8&ei=5070]here[/url]:
[quote] “When my oldest child, an A-plus stellar student, was in sixth grade, I realized he had no idea, no idea at all, how to do long division,” Ms. Backman said, “so I went to school and talked to the teacher, who said, ‘We don’t teach long division; it stifles their creativity.’ ” [/quote]
It's a sad state of affairs. People from other countries: do you guys have these kinds of problems too? (I mean, people who are "educated" are taught horribly, not people aren't educated at all -- of course the latter happens in a lot of countries.)

Umm wait a minute. Did that man start making some null symbol for $\frac{0}{0}$ or was he trying to tell us it equals 1? I wish that video was better quality.

   I find it strange that any children in public schools aren't taught long division. If they aren't I wouldn't suppose it is an issue with creativity but one of their attitude toward math-making people think it is too hard to teach.

   Of course, I am from the USA so I can't say anything about the math in other countries.

Yeah! this   report  is even better than past BBC or Other Media  gems:

Like:  Men live on sunlight alone (Story abut a guy who for last 70 years  apart for looking at sun for 5 minutres daily,  never had anything for food - and how NASA is going to learn form him etc ... NO, it was not  a April-fool-joke kind of thing.... story was reprinted in MANY "respected" news outlets.

Or: BBC's [url=http://news.bbc.co.uk/2/hi/health/2284783.stm]Blondes 'to die out in 200 years'[/url]

 [quote] Of course, I am from the USA so I can't say anything about the math in other countries.[/quote]
As Carl Sagan said, math is same everywhere in the universe, let alone other countries outside US :)[/quote]

[quote="Ignite168"]Why is this a sad state?[/quote]
I think schools in the USA are less and less inclined to teach kids about how to reason analytically. They only want to teach synthetic reasoning. But analytic reasoning is the only thing people can unconditionally agree on -- synthetic arguments are interpretations of observation, which generally reasonable people can disagree about even if they share all of the same evidence, whereas analytic arguments are either right or wrong. So anyway my point is that if kids are not taught to deal well with analytic things in school, they won't know how to deal with analytic things in the real world, and they will treat them like synthetic ones. And then what are you supposed to do when you are arguing with this sort of person about an analytic thing? Normally you would just share all of your evidence and then agree, but this person will not be convinced! How would you like it if you and some other person perform the same mathematical calculation and get different answers, and after checking your work against theirs you see you did it right, but they try to defend themselves by saying that their method is just a different way of looking at things? Consider the $\frac{0}{0}$ thing, for example...

Of course my philosophy of knowledge is not universally accepted, but these are my personal opinions and you can take them as you will.

I'm not able to see the video, so can someone explain to me how this works?

All that's hapenned is that the guy gave a name to 0/0.  But what can you actually do with it?

He makes these three definitions:
\begin{eqnarray*}\frac{1}{0}&=& \infty\\ \frac{-1}{0}&=&-\infty\\ \frac{0}{0}&=& \Phi = \rm{nullity}\end{eqnarray*}
He says that nullity is a new number which lies off of the real number line.

Then he goes on to "solve a 1200 year old problem that Pythagoras and Newton couldn't solve" (Pythagoras lived B.C. btw...):
\begin{eqnarray*}0^{0}&=& 0^{1-1}\\ &=& 0^{1}\cdot 0^{-1}\\ &=& \left(\frac{0}{1}\right)^{1}\cdot \left(\frac{0}{1}\right)^{-1}\\ &=& \frac{0}{1}\cdot \frac{1}{0}\\ &=& \frac{0\cdot 1}{1\cdot 0}\\ &=& \frac{0}{0}\\ &=& \Phi \end{eqnarray*}

(May be some one whould rewrite $0$ as $(2\cdot 0)$ in one of the factors and get the  newer, better, and improved value for $0^{0}=2\Phi$ :)

Or look at this

link:[url=http://www.digitaljournal.com/article/68947/British_Professor_How_to_Divide_By_Zero_Riddle_Solved_A_New_Number_Introduced_Nullity_]link[/url] and see the image above "The problem of nothing seems to be solved"
 [img]http://www.digitaljournal.com/images/photo/Cardano2.jpg[/img]

Looks like someone copied Cardano cubic  equation method or something like that ..:) :lol:  :lol:  :lol: and hoping that no one will notice.

One of my math teachers told me that they came up with the idea of complex numbers like this:

"We can't take the square root of -1.  But what if we could?  Well, let's call that $i$, and we'll see what happens..."

Maybe new things will come out of $\Phi$... you never know.  And that might be taught in schools 100 years from now.

That's what I love about math - new things are coming in all the time.  And we may never know everything about math.

I remember a story my current math teacher told me:
His professor said that if an entire blackboard was all the math in the world, he would know the contents of a dot.  And his students, well, let's just say he needed to sharpen his pencil for that.

[url=http://www.bookofparagon.com/Mathematics/PerspexMachineIX.pdf]Here[/url]'s  where he lays out his theory in detail.

[quote="lingomaniac88"]Maybe new things will come out of $\Phi$... you never know.  And that might be taught in schools 100 years from now.[/quote]
The main differences between $i$ and $\Phi$:
- adjoining $i$ conserves all the properties one wants (like being a field).
- you can calculate with $i$ in a consistent way, but not with $\Phi$.
- is there even a formalisation of what it does¿

If you really want infinite numbers, learn surreal analysis etc., but ignore that spammers that don't more math than a six year old child.
And the idea of $\Phi$ is not new, nor has it any sense.

Conslusion: some stupid medias as always portray someone that hasn't even a clue of anything.

Are you sure that calculations with this "phi" give inconsistent results?  Showing that, say, phi = 2*phi will not work, since you can also reason from the definition of phi that it works like zero.  I think producing a result like 0 = 1 would be the best way to show that this is all bunk, but I haven't been able to do that yet.

I agree with the reasoning about the lack of any algebraic properties being preserved, though.  The reals with phi appended don't even form a group, as phi doesn't have an additive inverse.  For perspective, appending $i$ to the reals not only keeps it a field, but it makes it an algebraically closed field, which is in a way even better.

Would this be correct with his rules?
$1 = \frac\Phi\Phi=\frac{0^{0}}{0^{0}}= 0^{0-0}= 0^{0}= \Phi$

[quote="Sly Si"]Are you sure that calculations with this "phi" give inconsistent results?[/quote]
Well, at first one needs even to get a definition of what calculating with $\Phi$ is; more exactly, a formal definition only using the axioms and already existing structures is what is needed (I bet that guy has no idea of this).

The next thing is: what's even the sense of introducing this stupid symbol¿ The reason with the CPU etc. is just stupid, math is not possible with it, and no other reasons were given.

[quote="yasiru89"]... the poor unrecognised bloke who introduced the NaN error...[/quote]
Let's give credit where credit is due: the current IEEE floating point standard was primarily developed by [url=http://en.wikipedia.org/wiki/William_Kahan]William Kahan[/url], a professor of mathematics and computer science. I took numerical analysis from him- he understands very well that the "numbers" a computer works with are not a field in any implementation. With careful use of $\pm\infty$ and NaN, we can sometimes handle errors and achieve a meaningful result.

That's the problem with "nullity": everything it's actually good for has already been implemented.

He is just defining $\frac00$ and use it to define $0^{0}$
So actually, he didn't solve the 1200-year-old problem.
Anyone can define $0^{0}$ and claim that he has found out $0^{0}$ :D

[quote="modeler"]He is just defining $\frac00$ and use it to define $0^{0}$[/quote]

He defined $\frac{0}{0}$ and then [i]found[/i] $0^{0}$ using that.

He hasn't really found $0^{0}$ (of course) but it's not true that he merely defined it to be some term.  He defined something else which then allowed him to find it :D

And by defining $0^{0}=\frac{0}{0}$, he hasn't really done anything new.

Hey look, I define $\frac{0}{0}=\text{cheesecake}$. Wow! I discovered a new theory! Let's teach it in schools!

[quote="joml88"][quote="mathnerd314"]I find the title of this thread interesting. "Divide by zero!".

Anyone can divide by zero!. In fact, zero! just equals one, and you can divide by one.[/quote]

I would have titled it "Divide by zero!." or "Divide by zero factorial." if that's what I meant.[/quote]

I know; it was just my excruciatingly bad attempt at an excruciatingly bad pun.

And what about his "universal perspex machine"? Any comments on that?

[quote="daermon"][quote="1=2"]But, then again,

$\phi=\frac{0}{0}$

$n\phi=\frac{n*0}{0}=\frac{0}{0}=\phi$

Where n is any real number. But then, we divide by nullity to get $n=1$. So multiplication is undefined in nullity?

Also,

$\phi^{2}=\phi$, so $\phi=1$. So I'm beginning to think that multiplication is undefined with nullity.[/quote]
i'm sure most of what you just did is against the rules in transreal arithmetic :wink:[/quote]

And that's exactly the problem. There are so many rules legislating what you can or can't do with nullity. It's an exception to most of the traditional rules of arithmetic, [b]just like NaN is[/b]. Which renders his concept of nullity utterly useless.

Sure, he can bring up examples about pacemakers dividing by zero. (To the best of my knowledge, they wouldn't anyway.) But he hasn't gotten rid of the undefined; he's just given it a cool-sounding name.

In fact, you can't divide by zero in any field. If you could, there would exist a number $a$ such that $0a=1$.

By the definition of 0, $1+0=1$, so $0=1-1$.
Then $0a=(1-1)a$.
Distributing, $0a=1a-1a=0$.
Thus we are left with the absurdity $0=1$.

Thus we conclude that the "transreals" are not a field, which is a problem...

the thing is, there's so many things you can't use nullity with (such as all this multiplication with it being undefined and stuff) that's there isn't anything to do with it.

somebody was comparing this to complex numbers somewhere... but the difference is, complex numbers, have a lot of great usage. nullity... just doesn't.

wow...ok nullility is just a symbol for undefined

[quote="tjhance"] the difference is, complex numbers, have a lot of great usage. nullity... just doesn't.[/quote]

Agreed.  Much of what he's trying to do would be served equally well if common math packages / programming languages just had a special flag that flagged the value of a variable as undefined if a division by zero occurs (which some packages / languages do), which would be like nullity only without the unnecessary extra word that doesn't describe a useful concept.

[quote="t0rajir0u"]...if common math packages / programming languages just had a special flag...[/quote]
They do (or they have special "numbers"). They have several, and some computations are allowed: $\frac{1}{+0}=\infty$, and $\frac{-1}{\infty}=-0$, for example. Division by zero shouldn't kill everything- it's possible for that division to appear as an intermediate stage in an otherwise reasonable calculation, which later does something like dividing by that $\infty$ to bring it back.

Incidentally, it does make sense to have multiple zeros- then we can handle $\infty$ and $-\infty$, with results like $\infty+\infty=\infty,\infty+(-\infty)=\text{NaN}$.

Why would you get money for giving useless information? If your pacemaker would divide by zero, it would show nullity, and THEN you're dead.

$\frac{0}{0}$ isn't undefined, it's $\phi$. That's the fundamental assumption of his theory.

But why doesn't anyone attack how he defines division? That errs me much. Find an inconsistency there.

Hmmn... and more... he's a freaking cheater. If you look at his axioms ([url]http://www.bookofparagon.com/Mathematics/PerspexMachineVIII.pdf[/url]), he doesn't include the axiom that
$0a=0$ (it's implied in his multiplcation tables at [url]http://www.bookofparagon.com/Mathematics/PerspexMachineIX.pdf[/url]). He also doesn't include $a^{-1}=\frac{1}{a}$, but that also seems to be implied. Also, you don't know if $a\frac{b}{c}=b\frac{a}{c}$. But, if all these stated properties are true, THEN you'd be able to find an inconsistency, quite easily.


LaTeX
vector
calculus
calculus computations

$ (\vec{i} + \vec{j}) \times (\vec{i} \times \vec{j})$

isn't this just:

${ \vec{i} \times \vec{i} + 2 \times \vec{i} \times \vec{j} + \vec{j} \times \vec{j}}$

$ \text{\LaTeX}$ hint: use \times, not x.

Note: the cross product is not commutative.

so if it's not commutative then how would I solve this problem?

It looks almost the same, but that's $ i\times j\plus{}j\times i$, not $ 2(i\times j)$. You should have the cross products of the standard basis vectors memorized.

[quote="jmerry"]It looks almost the same, but that's $ i\times j \plus{} j\times i$, not $ 2(i\times j)$. You should have the cross products of the standard basis vectors memorized.[/quote]

so is $ i\times j \plus{} j\times i$ equal to 0 as both are opposite of each other regarding their magnitude?
and $ i\times i$ is simply just 0 right?

I think I just misread the question, so what is it now? I've revised the question

[quote="EquinoX"]so is $ i\times j \plus{} j\times i$ equal to 0 as both are opposite of each other regarding their magnitude?[/quote]
Yes, the cross-product is anti-commutative: $ i\times j \equal{} \minus{} j\times i\to i\times j \plus{} j\times i \equal{} 0$. I don't see what you mean by "regarding their magnitude".
[quote]and $ i\times i$ is simply just 0 right?[/quote]
Yes. (This is easily seen from the anti-commutativity: $ i\times i \equal{} \minus{} i\times i\to i\times i \equal{} 0$)

Of course $ j\times j \equal{} 0$ too, so now you have the answer.

So the answer to $ (\vec{i} \plus{} \vec{j}) \times (\vec{i} \times \vec{j})$ is just 0?

No. That was the answer to your original question $ (i\plus{}j)\times (i\plus{}j)$. None of the posts in this thread address your new question. Have you even tried it yet?

[quote="jmerry"]No. That was the answer to your original question $ (i \plus{} j)\times (i \plus{} j)$. None of the posts in this thread address your new question. Have you even tried it yet?[/quote]

Here's what I had so far:

(i + j ) x  (i x j) = ((i+j) . j)(i)) - ((i+j).i)(j))

That's more complicated than it needs to be. Don't bother with identities- just calculate as it's written.


limit
algebra
polynomial
calculus
inequalities
calculus computations

I  can't find simple  solution  for  the problem:
$ L\equal{} \lim\limits_{x  \to 0}  \frac{ e^x\minus{}1\minus{}x}{x^2}$
C1:   Use  Taylor's :
 $ e^x \equal{}1\plus{}x\plus{}\frac{x^2}{2} \plus{}...$
Hence,$ L\equal{}\lim\limits_{x \to 0} \frac{x^2/2}{x^2} \equal{}\frac{1}{2}$
   C2: $ L\equal{} \lim\limits_{x \to 0} \frac{ e^x\minus{}1}{2x}\equal{} \lim\limits_{x \to 0} \frac{e^x}{2} \equal{}\frac{1}{2}$

Can u help me!

imo the 'problem' is disingenuous since any alternative problem will be repeating the same construction of a taylor polynomial

Perhaps it will work, when you write the whole term as

$ \frac{\frac{\exp{x}\minus{}1}{x}\minus{}1}{x}$

For teh first term you can use the mean value theorem, the rest should be easy.

EtotheX

[quote="EtotheX"]Perhaps it will work, when you write the whole term as

$ \frac {\frac {\exp{x} \minus{} 1}{x} \minus{} 1}{x}$

For teh first term you can use the mean value theorem, the rest should be easy.

EtotheX[/quote]
But really i think  that,  mean  value theorem  like as L'Hospital rule.

 Clearly,  we have:  $ \frac{e^x\minus{}1}{x} \equal{}e^c$,where $ c \in (0,x)$
i wanna  see solution with  basical  knowledge.

Thanks.

You can do this without any calculus by first proving using the definition of $ e$ the inequality
\[ 1 \plus{} x \plus{} \frac {x^2}{2}\leq e^x \leq 1 \plus{} x \plus{} \frac {x^2}{2} \plus{} cx^3
\]
for some constant $ c$.  An outline of this proof is in steps 1-3 of post #3 of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228058]this[/url] thread.  But why in the world would you want to?


function
analytic geometry
graphing lines
slope
inequalities
algebra unsolved
algebra

Prove that there is no function $f: \mathbb{R^{+}}\rightarrow\mathbb{R^{+}}$ such that for all $x,y > 0$,
$f (x+y) \geq \ f(x) + y f( f(x))$.

I think it's from a TST in 2001, not 2002, but I guess it's not that important :).

Clearly, $f$ is increasing, and in fact increases to infinity, which means that so does $f\circ f$. Take some $x_0$ such that $(f\circ f)(x_0)>1$. Then, since the slope between the points $(x_0,f(x_0)),(x_0+y,f(x_0+y))$ is $>1$ for all $y>1$, we can find some $x$ such that $f(x)>x+1$.

In the initial inequality replace $y$ with $f(x)-x$ for $x$ as above. We get $(f\circ f)(x)=f(x+f(x)-x)\ge f(x)+[f(x)-x](f\circ f)(x)$, which, since $f(x)-x>1$ and $f(x)>0$, is $>(f\circ f)(x)$, and we have our contradiction.

My problems:
Prove that:there is no function $f: R\to R$such that:
$f(0)>0$and$f(x+y)\geq f(x)+yf(f(x)) \forall x,y \in R$


Here's question 10 on Euclid 1998. Any solutions would be greatly appreciated:

10. The equations x^2+5x+6=0 and x^2+5x-6=0 each have integer solutions whereas only one of the equations in the pair x^2+4x+5=0 and x^2+4x-5=0 has an integer solution.

a) Show that is x^2+px+q=0 and x^2+px-q=0 both have solutions, then it is possible to find integers a and b such that p^2=a^2 + b^2. (i.e. (a,b,p) is a pythagorean triple).

b) Determine q in terms of a and b

I read the solutions by the Euclid contest, but they didn't make much sense to me.  :D

[quote="Inspired By Nature"]Here's question 10 on Euclid 1998. Any solutions would be greatly appreciated:

10. The equations x^2+5x+6=0 and x^2+5x-6=0 each have integer solutions whereas only one of the equations in the pair x^2+4x+5=0 and x^2+4x-5=0 has an integer solution.

a) Show that is x^2+px+q=0 and x^2+px-q=0 both have solutions, then it is possible to find integers a and b such that p^2=a^2 + b^2. (i.e. (a,b,p) is a pythagorean triple).

b) Determine q in terms of a and b

[/quote]
[hide="solution"]
I'll assume you meant for part a) that both equations had integer solutions.
For part a), we have
$p^2-4q = r^2$ for some integer $r$
$p^2+4q= s^2$ for some integer $s$

Adding the two equations gives
$2p^2 = r^2 + s^2$
Note that we can write $r^2+s^2$ as
$\frac{1}{2} ((s+r)^2 + (s-r)^2)$
Therefore, the previous equation becomes
$p^2 = (\frac{s+r}{2})^2 + (\frac{s-r}{2})^2$
where $a = \frac{s+r}{2}$ and $b = \frac{s-r}{2}$
For part b), subtract $p^2-4q = r^2$ from $p^2+4q= s^2$ to get
$8q = (s+r)(s-r)$
since $s+r=2a$ and $s-r = 2b$,
we subsitute back into the original equation to obtain
$q=\frac{ab}{2}$
[/hide]


induction
Pascal\u0027s Triangle
combinatorics unsolved
combinatorics

Let $S$ be a string of $99$ characters, $66$ of which are $A$ and $33$ are $B$. We call $S$ [i]good[/i] if, for each $n$ such that $1\le n \le 99$, the sub-string made from the first $n$ characters of $S$ has an odd number of distinct permutations. How many good strings are there? Which strings are good?

I am pretty sure the one and only "good" string is $64$ $A$'s followed by $32$ $B$'s, and then $AAB$.

zgorkster, is it just a guess, or you can prove it?

Let's generalize a little bit...

[i]Under which conditions, given $a$ and $b$ integers, there is exactly one good string with $a$ "A"s and $b$ "B"s? Is it possible to have $a, b$ such that there is more than one good string?[/i]

Here's my solution although it's rather long.  Your question is very hard I think cmmmrc. I think that a better solution (if there is one) might be required to answer your question.

[b]Solution:[/b]

We must prove the following Lemma:

[i]For all $i\in\{1,2,\dots,2^n-1\}$, we have that ${2^n\choose i}$ is even.[/i]

Proof:

By Lucas' Theorem, we have the result immediately:

${2^n\choose i}\equiv {1\choose 0}{0\choose i_k}\dots {0\choose 1}\dots {0\choose i_0}\equiv 0$ mod $2$.

where $i=i_k i_{k-1}\cdots i_0$ is the binary representation of $i$.

Knowing this, we can begin the problem.

Now, the number of distinct permutations of a substring is simply ${n\choose b}$, where $n$ is the length of the substring and $b$ is the number of $B$'s in these $n$ letters.

So if a $B$ is in the first $64$ letters, then the number of permutations of the substring with length $64$ is ${64\choose x}$, where $x\in\{1,2,\dots,33\}$.  But by our Lemma, the number must be even, and thus the first $64$ letters must all be $A$.

Now, there can only be $2$ more $A$'s in the string, so we have three cases.  If the $65$th and $66$th letters are both $A$'s, then the rest must be $B$'s.  However, if $n=68$, we have ${68\choose 2}$ permutations, which is even.  So the first $66$ letters cannot all be $A$'s.

Now, if exactly the first $65$ letters are $A$'s, then the $66$th is a $B$ and for $n=66$ we have ${66\choose 1}$ permutations, an even number.

Thus the $65$th letter must be $B$.

Now comes the tedious part.

Now for $x\le 33,$ let $x$ be the number of $B$'s and assume that ${64+x\choose x}$ is odd.  Then if we add a $B$ to the sequence, then we have ${65+x\choose x+1}$ permutations, which will remain odd since the greatest power of $2$ dividing $x+1$ divides $65+x$ as well (since $x\le 33$).  But if we add an $A$ to the sequence, we have ${65+x\choose x}=\frac{65+x}{65}{64+x\choose x}$ which will remain odd iff $x$ is even.  Therefore, we can only add an $A$ if $x$ is even.

Now, consider the case that added an $A$ when $x$ is even.

By the same method, we can show if ${65+x\choose x}$ is odd (as it should be if $x$ is even), then by adding a $B$ we have ${66+x\choose x+1}=\frac{66+x}{1+x}{65+x\choose x}$ is either not an integer or even.  But, if we add an $A$ after the other $A$, we have that ${66+x\choose x}=\frac{66+x}{66}{65+x\choose x}$ is an odd integer iff $x$ is divisible by $4$.

So the two $A$'s must be together, and $4|x$.  Hence the two $A$'s cannot be the last two of the sequence ($4$ does not divide $33$).

Now, we must add a $B$ after the final two $A$'s are gone, but if one $B$ is used, we are still ok, since ${67+x\choose x+1}=\frac{67+x}{1+x}{66+x\choose x}$ which is still odd for $x$ divisible by $4$.  However, if we add another $B$ we have that ${68+x\choose x+2}=\frac{68+x}{x+2}{67+x\choose x+1}$ which cannot be even if $4|x$ and ${67+x\choose x+1}$ is odd.

Therefore, the only string that works (and barely fits) is the string with $64$ $A$'s, $32$ $B$'s, and finally $AAB$.

And finally, its 1 in the morning and if I made any stupid mistakes please correct me.

Edit: I changed my proof for the Lemma, it's much simpler & thanks for seeing that mistake cmmmrc  :lol:

[quote="Cmmmrc"]Let's generalize a little bit...
[i]Under which conditions, given $a$ and $b$ integers, there is exactly one good string with $a$ "A"s and $b$ "B"s? Is it possible to have $a, b$ such that there is more than one good string?[/i][/quote]

That's actually quite easy to answer:
There is one good string, if {a+b choose a} is odd, and
there is no good string, if {a+b choose a} is even.

The proof can be done by induction:
If {a+b choose a} is even, there is no good string, since the
entire string allows an even number of permutations.

If x={a+b choose a} is odd, then exactly one of y={a+b-1 choose a}
and z={a+b-1 choose a-1} is odd  (since y+z=x).  Either way you
get a unique possibility for extending the string to length a+b.
End of proof.

Another way of looking at it: You move through the odd entries of
Pascal's triangle, downwards, step by step.

Yes. It is basically the same idea I had. I also read carefully zgorkster's and it is correct too. I have found just a typo: somewhere there is a
$\binom{66+x}x=\frac{66+x}x \cdot \cdots$ and it should be $\frac{66+x}{66}$, but ...peanuts... the proof holds.


Fie ABC un triunghi oarecare in care $ |AB|\neq |AC|$ si O centrul cercului circumscris triunghiului. Tangenta in A la cercul circumscris triunghiului ABC intersecteaza dreapta BC in punctul D. Perpendiculara ridicata in varfuril B pe dreapta BC intersecteaza mediatoarea laturii [AB] in punctul E, iar perpendiculara ridicata in varful C pe dreapta BC intersecteaza  mediatoara laturii [AC] in punctul F. Demonstrati ca punctele D, E si F sunt coliniare!

[b]1).[/b] Notam cu $ \{M\}=AB\cap OE$(mijlocul laturii [AB]) si cu $ \{N\}=AC\cap OF$ (mijlocul laturii [AC]).
$ \angle {DAC}\equiv \angle {ABD}\Rightarrow \triangle{DAC}\sim  \triangle{ABD}\Rightarrow \frac{|CD|}{|AD|}=\frac{|AC|}{|AB|}=\frac{|AD|}{|BD|}\Rightarrow \\ \Rightarrow \frac{|CD|}{|BD|}=(\frac{|AC|}{|AB|})^2.$   (1)
[b]2).[/b] Pe de alta parte:
$ (BC\perp CF)\&(AB\perp OF)\Rightarrow \angle C\equiv \angle NFC\Rightarrow \\ \Rightarrow\frac{|AB|}{2.R}=sin C=sin{\angle {NFC}}=\frac{|CN|}{|CF|}=\frac{|AC|}{2.|FC|}\Rightarrow \frac{|FC|}{R}=\frac{|AC|}{|AB|}.$  (2)
In mod analog se arata, ca: $ \frac{|BE|}{R}=\frac{|AB|}{|AC|}.$  (3)
[b]3).[/b] Impartind acum relatiile (2) si (3), membru cu membru, obtinem: $ \frac{|FC|}{|BE|}=(\frac{|AC|}{|AB|})^2.$   (4)
[b]4).[/b] Din relatiile (1) si (4) $ \Rightarrow \frac{|CD|}{|BD|}=\frac{|FC|}{|BE|}\& [m(\angle{FCD})=m(\angle{EBD})=90^0]\Rightarrow$
$ \Rightarrow \triangle{FCD}\sim \triangle{EBD\Rightarrow \angle {FDC}\equiv \angle{EDB}\Rightarrow}$ [b]punctele D, E si F sunt coliniare.[/b]


find the compounds A,B,C,D,E
isobutyraldehyde+t-butyl amine----> A + EtMgBr ---->C +D
C +  ------->E+ ------------>F
pardesi are u sure its t-butyl amine?


Take a look at the number theory article:

http://www.artofproblemsolving.com/Wiki/index.php/Number_theory

I started creating lists of topics, using tiered bullets.  Of course, there will eventually be more text at the beginning, but historical text can be left for a "history of number theory" page.  The goal is to make the content useful to students studying those topics.

Are the tiered bullets a good style of organization?  Are they going to be too limiting in terms of the verbage we'll be able to add?  Will it not matter because that verbage will be included in the articles linked from the number theory article?

[quote="MCrawford"]Are the tiered bullets a good style of organization? [/quote]

Yes.

[quote="MCrawford"]Are they going to be too limiting in terms of the verbage we'll be able to add?[/quote]

No.

[quote="MCrawford"]Will it not matter because that verbage will be included in the articles linked from the number theory article?[/quote]

Yes.


Frankinfueter and Albatross went to the store and got 36 lemon drops, and 24 gum drops for them to eat.  They both agreed to have now more than 4 pieces of candy a day, but they can have less.  Albatross is a picky eater, so he will only eat gum drops.  Frankinfueter will eat both gum drops and lemon drops, trying to keep the amount of each type as equal as possible.  Albatross and Frankinfueter both like the candy they're eating, and will eat as much candy as they can.

How many days will Albatross eat candy, and how many days will Frankinfueter have candy?

[quote="Treething"][hide="answer"]
Albatross will have 4 days of candy.
Frankinfueter will have 7 days of candy.
[/hide][/quote]

Incorrect

isnt there a contradiction in the problem

[hide]Albatross will have 4 days of candy.    Frankinfueter will have 11 days of candy.[/hide]

@ hunter34 - Why? When you think something should be done, please provide a reason. I see no point in this topic being locked.

[u]I know the reason.[/u]

Since Albatross and Frankinfueter will eat the maxium # of candies, the amount would be 4 for each.

Since Frakinfueter eats 2 lemon drops and 2 gum drops, (according to the question) because he would equivalize the amount of certain kinds that he eats.

And Albatross will eat 4 gum drops everyday.

So gum drops would be gone in 4 days,($ \ 24/4$).

And 28 lemon drops would be left since Frankinfueter constantly ate 2 lemon drops with 2 gum drops.($ \ 36 \minus{} (4*2))$

Frankinfueter will eat and finish the lemon drops in 7 days.($ \ 28/4$)

Answer: Albatross: 4 days, Frakinfueter: 11 days (4 days + 7days)

Correction:

[quote="i_like_pie"]@ hunter34 - Why? When you think something should be done, please provide a reason. [b]This topic should be locked. No reviving of old threads/[/b][/quote]


im going to take either the PSAT or the SAT this december cuz im going into 7th grade. (i dunno which test im gonna take. mebbe some1 can tell me. i put my grade level there) i dunno which books are good and everything. mebbe some1 can share some of their experiences? nd which type of problem is usually the experimental? well ty and i hope some ppl can help me out. :D

You can only take the PSAT in October.  I'm assuming you're taking this for a talent search, so it'd probably be ACT or SAT.  But yea, I'd just say try some of the problems, and just get used to the test.  The point of taking this in 7th grade is to just see how smart you are, not to get into college.

But I did the talent searches in 6th, 7th, and 8th grades.  In 6th and 8th grades, I took the ACT in a small school, and there were only at most 3 or 4 other middle schoolers taking it with me.  I don't tihnk the high schoolers noticed that I was there.  And then in 7th grade, I took the SAT, and there were a lot of people taking it, so they had a room for just middle schoolers.  SO that was pretty cool.  But to me, the test felt just like the practice ones I took at home.

And it's really hard to tell which one is experimental.  The problems are just like the regular ones.

cool...i need it for CTY.
for CTY nd GATE nd stuff like that, do u need the SAT or the PSAT?
ty 4 answering so far.

CTY used to only take the SAT as a basis for eligibility. Now their [url=http://www.jhu.edu/gifted/summer/oselig.html]current rules[/url] allow other tests for eligibility, but the SAT or ACT are still the standard tests to take through CTY's [url=http://www.jhu.edu/gifted/ts/grades78.html]talent search[/url].

I was wondering about that phrase in the earlier reply too. Should that have been "in a small school"?

Yea, sorry about that.  I don't really know how that hapenned  :blush:  :blush:

I took the SAT and ACT last year for Midwest Talent Search.(In 6th) I didn't think  it was going to be the 'real thing' because my dad thought he was signing me up for an easier version of the test.

[quote="eryaman"][quote="h_s_potter2002"]In 6th and 8th grades, I took the ACT in a small girl,[/quote]Whaaa...? :o[/quote]

:rotfl: Oh my goodness :roll:

Anyway, I took SAT in 7th grade for CTY. First in January, I started preparing a week before the test and got a 1250. Then I decided to take it again in June and got a 1340. Basically all I did was go through books by Kaplan and stuff.

Study hard. STUDY HARD! If you get a score you want to keep though, you'll have to call them up and tell them so, because they delete scores of 7th or lower graders about 6 months later.

Now is when you need to start studying. Get several SAT prep books and study them several hours a day. Get to where you know every word in those books(or at least all but maybe 30). Being an AoPSer, you will naturally be pretty good at math. Keep doing the Mathcounts and lower forums. The math in the books though is important. Learn trig. Get to where you take little time on those problems. Get good speed-reading skills. Always answer every question. When you're done studying, you should be able to get done with a section in about 2/3s the time allotted to it. Then go back and check. I have no advice on the Essay part; I took the SAT as a 7th grader before that part was added in. I did all the things that I recommended and got a 720V,730M on the real SAT.

im going to have to disagree with the "study hard" theory. as a seventh grader (for TIP) i went through ONE old version of the sat with my friends who were also taking it. that was it, and i did swell. also, "study hard" is not really helpful in all cases: i have a friend who is a musical genius (she can pick up any instrument and be able to play melodies on it in less than an hour, and harmonies with melodies in less than a day) and very intellegent scholastically, as well. however, she does not take tests well. she has studied for the SAT for three years, taking practice test after practice test. she knows the SAT backwards and fowards, and yet her scores now are pretty comporable to the ones she had before studying. studying has its place for real tests, but for the SAT, i would recommend reading in your spare time. as a member of this site, i believe you can figure most math things out. but READ!!!! reading boosts your word bank, your reading comprehension, your inituitive grammar knowledge, AND your speed.

Well, if you want my advise, I'd say not to study, except for maybe take a practice test.  There is a reason for taking these tests at 7th and 8th grades, and that's not to try to get a good score for college.  The point of the tests is to see how good you really are at reading and math, since a lot of 7th and 8th grade tests aren't very good at seeing the difference between some in the 97th percentile, and someone at the 99th percentile.  I don't really like how for the CTY classes and stuff, you have to have you SAT score above a certain amount, because then it just causes kids to try and study too much for the test, just so that they can take a class.

But yea, the only thing you'll get by studying really hard is not having to study as much when you get to 11th grade for college.  Of course, you might get a really good score in 11th grade without studying, so if it were me, I wouldn't spend too much studying.  But the awards and things don't really matter that much in the long run.

But I did the Northwestern talent search, where the high scorers just got medals.  I've heard that TIP and the Johns Hopkins ones give money or something.  I dunno about this though.

[quote="eryaman"][quote="h_s_potter2002"]In 6th and 8th grades, I took the ACT in a small girl,[/quote]Whaaa...? :o[/quote]

lmao

wow shravas....I wonder what you were really thinking of when you typed that message...


Find all $(x, y, z, n) \in {\mathbb{N}}^4$ such that $ x^3 +y^3 +z^3 =nx^2 y^2 z^2$.

here is my solution;
i won't consider trivial cases:
we assume that $ x > y \ge z$
note that we have
$ x^2 | y^3 \plus{} z^3$
also we have
$ \frac {x}{y^2z^2} \plus{} \frac {y}{z^2x^2} \plus{} \frac {z}{x^2y^2} \equal{} n$
by using inequality...
$ \frac {x}{y^2z^2} \plus{} \frac {2}{x} > \frac {x}{y^2z^2} \plus{} \frac {y}{x^2} \plus{} \frac {z}{x^2} \ge n \implies \frac {x}{y^2z^2} > n \minus{} \frac {2}{x} \ge n \minus{} 1 \implies x > (n \minus{} 1)y^2z^2$
so we get
$ x^2 > (n \minus{} 1)^2y^4z^4$.
now the rest is easy by knowing that $ x^2 | y^3 \plus{} z^3 \implies (n \minus{} 1)^2y^4z^4 < x^2 \le y^3 \plus{} z^3$.

Without loss of generality, let $ z$ be the greatest of the variables. Then $ nx^2y^2z^2 \equal{} x^3 \plus{} y^3 \plus{} z^3 \leq 3z^3$, so $ nx^2y^2 \leq 3z$. Since $ x^2y^2 \leq nx^2y^2$, we have $ x^2y^2 \leq 3z$. On the other hand, $ z^2 | x^3 \plus{} y^3$, so $ x^3 \plus{} y^3 \geq z^2 \geq \frac {x^4}{y^4}{9} \geq \frac {x^3}{y^3}{9}$. Thus, $ 9x^3 \plus{} 9y^3 \geq x^3y^3$, so $ 81 \geq (x^3 \minus{} 9)(y^3 \minus{} 9)$. It is therefore to sufficient to check all $ (x, y)$ in $ (1, 2, 3, 4) \times (1, 2, 3, 4)$; the only $ x$ and $ y$ that yield solutions are $ (1,1)$, $ (1,2)$, and $ (2,1)$.

Hence, our solution set with $ x \leq y \leq z$ is $ (1, 1, 1, 3), (1, 2, 3, 1)$.


assum $a,b,c,d\in\mathbb{N}$ where $ab=cd$  can $S$ be a prime number, if $S=a+b+c+d$?

[hide="click to reveal hidden content"]
In order for a + b + c + d to be odd one or three of the numbers would have to be even because if we have an even number of odds the sum would be even and therefore not prime.  However if one of the numbers is even one of the other numbers must be even in order for ab=cd to be true.  Since it is immpossible to have an and odd number of odds the sum S cannot be odd.  Since each of the numbers must be at least odd the answer is not 2 and is therefore not prime.
[/hide]

why can't 3 of them be even and one odd? i think it's false

7 x 12 = 14 x 6 is a non-working example of 3 even, 1 odd.

But that's not saying that none of those examples work...

We've established that a+b+c+d can be odd. But to prove a+b+c+d to be prime... we do not know enough about the number to do any kind of proof.

Lets say a is odd, because we know either 1 or 3 of the variables must be odd for the sum to be odd, and 3 cannot be odd because ab = cd.  So a is odd, and b, c, and d are even. b+c+d is even and therefore not prime. Can we PROVE that an even number + an odd number is prime and not just show it?

So if we are to prove the problem true by finding 1 working example (seems to be the easiest way right now):
Let P be any prime. Let a be odd.
Then find P-a= b+c+d such that ab=cd, or if it is easier for you, a/c = d/b -or- a/d = c/b.
I will admit this is essentially a trial and error approach. Pick a random a (odd) and b (even). Find all the even factors of P-a-b and then look for a pair such that ab = cd.

Given this, any such prime would probably be big because the list of even factors of P-a-b becomes bigger. Maybe really big. Probably not mersenne big, simply because we can't readily work with those numbers, not knowing what they are exactly or having the processing power to test them.

I'm sorry if I've simply stated the obvious. I thought I might tie up some loose ends and make it easier for those not quite at the level required (like myself) to understand whats going on.

so you're saying such prime can exist? ;)
[hide="hint"]
this very easy...you guys are making this much more complicated ;)
use what it is given and try to write $a+b+c+d$ in another form, then you'll be able to say whether this can be a prime or not.
[/hide]

[hide="big hint"]
we know: $d=\frac{ab}{c}, d\in\mathbb{N}\Rightarrow c|a$ or $c|b$
$S=a+b+c+d=a+b+c+\frac{ab}{c}\Rightarrow ...$[/hide]

:( see this is why I didn't solve it. I don't think like that yet.

d = ab/c.
a + b + c + ab/c = S
(ca + cb + c^2 + ab)/c = S

If a or b is odd, then S is even. (c and b are even, so ca, cb, c^2, and ab are even)
If c is odd, then ca is even, cb is even, ab is even but c^2 may be odd. Hmm....

AAAAGH!!!
S must be a natural number because a, b, c, and d are natural numbers, and the natural numbers enjoy closure of addition.
For S to be prime, ca + cb + c^2 + ab must be an odd multiple of a prime by a multiple of c.
                ca + cb + c^2 + ab = Sc
Sc is the odd multiple. S is the prime, c is the odd multiplier.
I've been playing with this equation for a while now, and I must be burnt out.
a(c + b) + c(b + c) = Sc
(a + c)(b + c) = Sc
[(a + c)(b + c)]/c = S
If c is odd, then S is odd. a + c is odd. b + c is odd. Odd divided by odd is odd.
(a + c)(b + c) has to be an odd multiple of a prime by c.

This cannot be. An odd multiple of a prime's factors are 1, the multiplier, the prime and the number. a + c must be the prime, and b + c must be the multiplier, or vice versa. But because the multiplier must be equal to c, and b is a positive integer, b + c does not equal c. So S is either not a natural number or not prime because it is divisible by c (if b is a multiple of c).

yes, but you can show the whole thing mathematically,
now what if, b is not a multiple of c??

Er... I did not mean for that comment to apply to both cases.

Case 1:  B and A are not multiples of C
(a + c)(b + c) is not divisible by c.

Case 2: B or A is a multiple of C:
[(a + c)(b + c)]/c also has the factors of B + C because B is xC. b + c = c( x + 1) where x is the multiplier of C to get B.
It is still divisible by at least x + 1 after dividing by c  and therefore not prime.

I don't see anything wrong with ChaoSpectre's proof up to (a+c)(b+c)/c=S, so I'll work with that.
As he said, (a+c)(b+c) has to be Sc, which is prime * number.  WLOG, assume that c is the odd number.  Then a+c and b+c are both odd numbers.  Obviously, S has to be a factor of one of those two because it's a prime.  That's not possible, because each of these is less than S (a+b+c+ab/c=S).  Therefore, we've reached a contradiction, and the original assumption that a prime S did exist is false.

would be interesting...show us what you did...
I haven't looked at this problem much...

ok, here goes:
$ S=a+b+c+d=a+b+c+\frac{ab}{c}\Rightarrow \frac{c^2+ac+bc+ab}{c}=\frac{(a+c)(b+c)}{c}=$
$(\frac{a}{c}+1)(b+c) (I)$
$=(\frac{b}{c}+1)(a+c) (II)$
if $a|c$ then the paranthesis in $I$ would both be integers, so $S$ won't be a prime, if $c|b$ then paranthesis in $II$  are both integers so $S$ can't be a prime...
thus: $S$ can never be a prime

[quote="amirhtlusa"][hide="big hint"]
we know: $d=\frac{ab}{c}, d\in\mathbb{N}\Rightarrow c|a$ or $c|b$ [/hide][/quote]I don't think you can say that. For example, $15\mid20\cdot18$ but $15\nmid20$ and $15\nmid18$. So in your proof above, neither of the left-hand parentheses is necessarily integral.

edit: removed the last part.. i was just kidding  :blush:

[quote="AntonioMainenti"][quote="amirhtlusa"][hide="big hint"]
we know: $d=\frac{ab}{c}, d\in\mathbb{N}\Rightarrow c|a$ or $c|b$ [/hide][/quote]I don't think you can say that. For example, $15\mid20\cdot18$ but $15\nmid20$ and $15\nmid18$. So in your proof above, neither of the left-hand parentheses is necessarily integral.

If $S$ were to be prime, then either $1$ or $3$ of $a,b,c,d$ must be odd. But then we could not have $ab=cd$.[/quote]
Ty AntonioMainenti, that conclusion ONLY works when $c$ is a prime number but nowhere in the problem assumed that $c$ was prime  :(

[quote="JBL"]The proof can still be made simple, however, since both $a + c$ and $b + c$ are larger than $c$, so no matter how the factors of $c$ divide into their product, we will still have on the LHS a product of two integers, each greater than 1.[/quote]

That seemed the easiest sort of approach...checking with factors.  A number cannot have a prime number larger than it as a factor.  So, S cannot go into a+c or b+c

You guys probably had something like this in mind already but here it is..

$c\mid{a}b$, so let $c=uv$ such that $a=mu$ and $b=nv$, all natural numbers. Then $S=a+b+c+\frac{ab}{c}=mu+nv+uv+mn=(m+v)(n+u)$.

[quote="ChaoSpectre"]Just to clarify, we are assuming that c must be odd in order for S to be odd and therefore possibly a prime, so a and b MUST be the product of two numbers because they're even...

Right?[/quote]Well, if $S$ is to be prime, one of the numbers $a,b,c,d$ must be odd, I agree. I never had that in mind though for $c$ in particular. My definitions for $a,b,c$ as the product of pairs natural numbers doesn't exclude the possibility that some of them are just $1$'s. Maybe I don't understand your question..

>_<
Right... WLOG... WLOG...

$a, b, c$ and $d$ are interchangeable. So one of them has to be odd in order for $S$ to be odd. Following the example you presented earlier, if you express $d$ as $\frac{ab}{c}$, in having $c$ as the divisor for the equation, $c$ must be  the odd number for $S$ to be odd, right?

I hope this is any clearer. And I must begin learning how to post in Latex.

[quote="ChaoSpectre"]>_<
Right... WLOG... WLOG...

Following the example you presented earlier, if you express $d$ as $\frac{ab}{c}$, in having $c$ as the divisor for the equation, $c$ must be  the odd number for $S$ to be odd, right?[/quote]I don't see why that must be the case. My solution (really just a slight correction of amirhtlusa's) doesn't depend on the numbers' parity.

yes, in my solution when i  said, when "$ d=\frac{ab}{c}, d\in\mathbb{N}\Rightarrow c|a$ or $c|b$" that is not true... what AntonioMainenti said is perfectly valid and correct...
it's the same idea...

it doesn't matter whether c goes into a or b.  And the terms are interchangeable anyways.  If you wanted it to be, c could be even in that equation.


Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

I believe you are posting old AMC problems. Why not just use the resources?

[hide]

Given two circles, the maximum intersections is 2. The number of ways to choose 2 circles out of the 4 is $ \binom{4}{2} \equal{} 6$. So, the number of intersections is $ 2\cdot 6 \equal{} 12$

[/hide]

Here is another way to look at it:

[hide]
Draw one circle. There are now 0 points of intersection.
Draw another circle.  There are now at most 2 points of intersection (it enters and exits the existing circle).
Draw another circle. There are now at most 2+4=6 points of intersection (it goes through both circles, entering and exiting).
Draw another circle. There are now at most 2+4+6=12 points of intersection (it goes through all three circles, entering and exiting).
There are a total of 12 points of intersection.
[/hide]

And if there are n circles, maximum number of intersection points is $ \binom n2 \cdot 2\equal{}n(n\minus{}1)$


You are right but there is nobody post solution for both of them mine and yours  :D


Euler
geometry
circumcircle
geometry proposed

Let $ABC$ be an acute triangle and $F$ its Fermat point, that is, the interior point of $ABC$ such that $\angle AFB = \angle BFC = \angle CFA = 120^\circ$. For each one of triangles $ABF$, $BCF$ and $CAF$, draw its Euler line, that is, the line connecting its circumcenter and its centroid.

Prove that these three line pass through one same point.

Nice problem!

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=6188[/url]


$ABCDE$ is a convex pentagon and:
\[BC=CD, \;\;\; DE=EA, \;\;\; \angle BCD=\angle DEA=90^{\circ}\]
Prove, that it is possible to build a triangle from segments $AC$, $CE$, $EB$. Find the value of its angles if $\angle ACE=\alpha$ and $\angle BEC=\beta$.

The first part is possible to prove using trigonometry: set $BC=CD=a\wedge DE=EA=b\wedge \angle BDA=\gamma$. It's possible to write equations for $AC$, $CE$, $EB$ using law of cosines and only mentioned symbols. Now you just have to prove that triangle triangle inequality for segments of these lengths.


(This is essentially a repeat of an earlier announcement, with some updated information.)

The world premiere of the movie [i]Hard Problems[/i] will be at the 2008 Joint Mathematics Meetings in San Diego on Tuesday, January 8 at 6:00 p.m.

Attached to this message is a poster for the movie.

Students in high school (or younger) who happen to be in the area can get in for 5 dollars (if you register in advance by December 14 -- that's this Friday) or 10 dollars if you register at the door.  This registration fee allows you to attend the entire 4-day Meetings January 6-9.  Although the meetings are largely intended for professional mathematicians, there are some events for students, including the always popular "Who Wants to be a Mathematician" game show.

As a (slight) added bonus, you can meet some the Art of Problem Solving staff: we'll be at the meetings and will have a booth as part of the Exhibits Hall.

See http://www.ams.org/amsmtgs/2109_intro.html for more details and the advance registration forms.

When is it going to be available to people who aren't on the West Coast?

Is there a guarantee that the movie will eventually be in theaters across the country, or is it just a one time event?

I don't know what the theatrical release plans are, but it will be available on DVD.  Details to be announced.

Beginning in late January or early February the DVD will be available for sale from the Mathematical Association of American (go to the Bookstore/publications area of the website) and also from the AMC.  I will announce here at AoPS when it is available.  That will be the easiest way to obtain a copy for viewing.  Final price details are being set, but it should be affordable, I think.

The director/producer George Csicery is entering this documentary in film festivals.  Your best chance for seeing it in a theater setting will be to watch for film festivals where it will appear.  If, after screenings in film festivals it becomes popular as did "Spellbound", it might appear in the kind of theater that specializes in foreign films, documentaries, independent films, etc., as did Spellbound.  In spite of it being what I consider an exciting adventure, don't expect to see it in the local 16-plex. :)

Steve Dunbar
MAA Director, American Mathematics Competitions


Vieta
number theory unsolved
number theory

Find all prime numbers $ p,q < 2005$ such that $ q | p^{2} \plus{} 8$  and $ p|q^{2} \plus{} 8.$

[b]Solution:[/b]
Clearly $ p\equal{}q\equal{}2$ satisfies to the conditions.Now assume that $ p,q\geq 3$.

As we know $ \frac{(p^2\plus{}8)(q^2\plus{}8)}{pq}$ is an integer. It follows that $ \frac{8(p^2\plus{}q^2\plus{}8)}{pq}$, recalling that $ p,q$ are

odd, it can be inferred that $ \frac{p^2\plus{}q^2\plus{}8}{pq}$ is an integer as well. So the problem reduces to finding all odd solutions $ (p_n,q_n)$, that are less than $ 2005$ and satisfy to condition, that:
\[ \frac{p^2\plus{}q^2\plus{}8}{pq}\equal{}k
\]
for some fixed $ k$.
Assume that $ (p_0,q_0)$ is the solution, so that the sum $ p\plus{}q$ reaches its minimum, and $ p_0\geq q_0$.
If $ p_0\equal{}q_0$, then, obviously $ p_0\equal{}q_0\equal{}1$. Now assume $ p_0>q_0$.
It follows that

$ p_0^2\minus{}p_0(q_0k)\plus{}q_0^2\plus{}8\equal{}0$

As we know, by Vieta's formula:

$ p'\equal{}\frac{q_0^2\plus{}8}{p_0}<\frac{p_0^2\minus{}2p_0\plus{}9}{p_0}<p_0$, as long as, $ p_0\geq 5$, so this case leads to contradiction.

So we might assume that $ p_0\leq 4$. If $ p_0\equal{}3$, then $ q_0\equal{}1$, but this case is invalid.

So if $ (p_0,q_0)$ is a solution than $ p_0\equal{}q_0\equal{}1$ and $ k\equal{}10$.

All solutions determined by this method are in the sequence $ q_{n\plus{}1}\equal{}p_n, p_{n\plus{}1}\equal{}10q_n\minus{}p_n$ are $ (1,1),(9,1),(89,9),(881,89)$.

Now let's prove that all solutions of the equation $ p^2\minus{}10pq\plus{}q^2\plus{}8\equal{}0$ are in the described sequence. Let $ (p',q')$ be the odd solution with minimal sum that is not in the sequence. By the explicitly described method, we could proceed to the another solution with less sum and that will either lead to the contradiction of minimality or to the contradiction of the assumptation that this solution is not in the sequence.

So all solutions are:
\[ (2,2),(881,89),(89,881)
\]

well, it is because my assumptation about the case $ p_0 \equal{} 3, q_0 \equal{} 1$ and consequently $ k \equal{} 6$ was wrong and this case is valid.

I have eventually realized that but it was too late to fix the solution.

is it clear now?


Let $ABC$ a triangle and let $D,E,F$ points such that :

-  $B$ and $E$ are NOT in the same side of $AC$
-  $D$ and $C$ are NOT in the same side of $AB$
-  $A$ and $F$ are NOT in the same side of $BC$
-  the triangles $ADB,CEA,CFB$ are similar

Prove that triangles $BDF$ and $FEC$ are also similar and that the segments $[AF]$ and $[DE]$ intersects in their midpoints.

Are you sure that this triangles are not isosceles, because i have already solved some problem like this but it was long time ago and I think that those triangles were isosceles, but never mind I'll try

Yes, I'm sure that there isn't isocele triangles

The problem is wrong even if the triangles are isosceles (even if they are the "Napoleon" triangles, i. e. isosceles triangles with bases BC, CA, AB and base angle 30°).

  darij

[quote="M@re"]I share darij's opinnion, especialy part with midpoints[/quote]

Obviously, because everyone who doesn't share Darij's opinion on a geometry problem must be a fool to believe that Darij is wrong! :D

I don't think so....
I think that darij is fool :) :)
he thinks he's so smart and that he knows everything about geometry :D :D

Darij, just kidding no hard filings...


Dear Mathlinkers,
let ABCD be a quadrilateral, (1), (2) the Euler's circle of ABD, ABC,
I, J, L the midpoints of AB, BC, DA,
L' the second meetpoint of the parallel to BC through L with (1)
J' the second meetpoint of the parallel to AD through J with (2).
Prove : I, J' and L' are collinear.
Sincerely
Jean-Louis

Let $ P$ be the second intersection point of Euler circles of $ \triangle ABD$ and $ \triangle ABC$. It's known that this is Poncelet point of $ ABCD$ hence it lies also on Euler circles of $ \triangle BCD$ and $ \triangle CDA$. Let $ Q, R$ be midpoints of $ AC$, $ BD$. We have $ \angle JPL\equal{}\angle JPM\plus{}\angle MPL\equal{}\angle JRM\plus{}\angle MQL\equal{}\angle MCJ\plus{}\angle LDM\equal{}\angle DCB\plus{}\angle ADC\equal{}\angle (AD,BC)\equal{}\angle (JJ',LL')$ where $ M$ is midpoint of $ CD$, hence $ \angle L'IJ'\equal{}\angle L'IP\plus{}\angle PIJ'\equal{}\angle L'LP\plus{}\angle PJJ'\equal{} 0(mod \pi)$ hence $ L',I,J'$ are collinear.

Let $ X \equiv BC \cap AD$ and let $ (3)$ be Euler circle of $ \triangle ABX,$ passing through the midpoints $ I, M, N$ of $ AB, AX, BX.$ Let $ E \in BCX$ be foot of the common A-altitude of $ \triangle ABC, \triangle ABX$ and $ F \in ADX$ foot of common B-altitude of $ \triangle ABD, \triangle ABX.$ Then $ E \in (1) \cap (3)$ and $ F \in (2) \cap (3).$ Let $ I'$ be the 2nd intersection of $ IJ'$ with $ (3).$ By definition, $ J'J \parallel AD \parallel IN.$ By Reim's theorem for $ (1), (3)$ intersecting at $ I, E$ $ \Longrightarrow$ $ J'J \parallel I'N$ $ \Longrightarrow$ $ I' \equiv I$ are identical and $ IJ'$ is tangent of $ (3)$ at $ I.$ In exactly the same way, $ IL'$ is tangent of $ (3)$ at $ I$ $ \Longrightarrow$ $ I, J', L'$ are collinear.

Dear Vladimir,
nice for using the Reim's theorem...
For me, it avoid to use angles.
Sincerely
Jean-Louis

Dear Mathlinkers,
an article concerning the ‘’Euler-Poncelet’’ point has been put on my website
http://perso.orange.fr/jl.ayme    vol. 8 Le point d’Euler-Poncelet d’un quadrilatère  p. 79.
You can use Google translator
Sincerely
Jean-Louis


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_{ \minus{} \pi}^{\pi} \frac {\sin nx}{(1 \plus{} 2009^x)\sin x}\ dx\ (n\equal{}0,\ 1,\ 2,\ \cdots)$.

[quote="Timestopper_STG"]$ I\equal{}\int_{\minus{}\pi}^{\pi}\frac{\sin nx}{(1\plus{}2009^{x})\sin x}dx$

$ I\equal{}\int_{\pi}^{\minus{}\pi}\frac{\sin n(\minus{}u)}{(1\plus{}2009^{\minus{}u})\sin (\minus{}u)}d(\minus{}u)\equal{}\int_{\minus{}\pi}^{\pi}\frac{2009^{u}\sin nu}{(1\plus{}2009^{u})\sin u}du$

$ I\equal{}\int_{\minus{}\pi}^{\pi}\frac{2009^{x}\sin nx}{(1\plus{}2009^{x})\sin x}dx$

$ I\equal{}\frac{1}{2}\int_{\minus{}\pi}^{\pi}\left[\frac{\sin nx}{(1\plus{}2009^{x})\sin x}\plus{}\frac{2009^{x}\sin nx}{(1\plus{}2009^{x})\sin x}dx\right]$

$ I\equal{}\frac{1}{2}\int_{\minus{}\pi}^{\pi}\frac{\sin nx}{\sin x}dx\equal{}\int_{0}^{\pi}\frac{\sin nx}{\sin x}dx$

$ \therefore I\equal{}0$ if $ 2|n$ else I got stuck. :P[/quote]

Let $ I_n\equal{}\int_{\minus{}\pi}^{\pi}\frac{\sin nx}{(1\plus{}2009^{x})\sin x}dx=\int_{0}^{\pi}\frac{\sin nx}{(1\plus{}2009^{x})\sin x}+\frac{\sin (-nx)}{(1\plus{}2009^{-x})\sin (-x)}dx$

$=\int_{0}^{\pi}\frac{\sin nx}{\sin x}\biggl(\frac{1}{1\plus{}2009^{x}}+\frac{2009^{x}}{1\plus{}2009^{x}}\biggr)dx=\int_{0}^{\pi}\frac{\sin nx}{\sin x}dx$

Now Consider

$I_n-I_{n-2}=\int_{0}^{\pi}\frac{\sin nx-\sin (n-2)x}{\sin x}dx$ for $n\ge 2$

$I_n-I_{n-2}=\int_{0}^{\pi}\frac{2\cos (n-1)x\sin x}{\sin x}dx=\int_{0}^{\pi}2\cos (n-1)x\ dx=0$

$I_n = I_{n-2}\ \forall\ n\ge 2 $

$\implies I_0=I_2=I_4=\cdots= 0$

and

$I_1=I_3=I_5=\cdots= \pi $


geometry
3D geometry
tetrahedron
linear algebra

Let we have a space $ V^{3}$ and the linear transformation: $ A\vec{x}\equal{}\vec{x}\plus{}\left\langle \vec{x},\vec{p}\right\rangle\vec{q},$ where $ \vec{p}\equal{}(1,2,1), \vec{q}\equal{}(1,0,\minus{}1)$. Show that the volume of any tetrahedron is the same to the volume of its image with $ A$.

$ <p,q>\equal{}0$ then $ A\equal{}id\plus{}N$ where $ N^2\equal{}0$; then $ det(A)\equal{}det(id)\equal{}1$ and the  volume is $ A$-invariant.

Thank you, loup blanc.

But does there exist any solution concerning scalar triple product?

I mean showing $ (\vec{AB}\times\vec{AC}))\cdot\vec{AD} = (A(\vec{AB})\times A(\vec{AC}))\cdot A(\vec{AD})$, where $ A, B, C$ are $ D$ the vertices of the tetrahedron. I have tried this, but there are too many terms on the right hand side.

We have
$ \vec{x}\equal{}\left[\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}\end{array}\right] \Rightarrow
A\vec{x}\equal{}\left[\begin{array}{c}
2x_{1}\minus{}x_{3}\\
x_{1}\plus{}x_{2}\minus{}x_{3}\\
x_{1}\end{array}\right] \Rightarrow
A\equal{}\left[\begin{array}{ccc}
2&0&\minus{}1\\
1&1&\minus{}1\\
1&0&0
\end{array}\right]\Rightarrow \det{A}\equal{}1$


The real numbers a, b and c are such that a + b + c = 15 and ab + bc + ca = 75.

(a) Calculate a² + b² + c².
(b) Calculate (a – b)² + (b – c)² + (c – a)².
(c) Find the values of a, b and c.

hello, for a) we get from $ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2ac\plus{}2bc\equal{}225$
$ a^2\plus{}b^2\plus{}c^2\equal{}225\minus{}2(ab\plus{}bc\plus{}ac)\equal{}225\minus{}2\cdot75\equal{}225\minus{}150\equal{}75$.
Sonnhard.

hello for b) $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 2(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} ac \minus{} bc) \equal{} 2(75 \minus{} 75) \equal{} 0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello for b) $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 2(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} ac \minus{} bc) \equal{} 2(225 \minus{} 75) \equal{} 300$.
Sonnhard.[/quote]

a^2 + b^2 + c^2 = 225 ?

is the correct a^2 + b^2 + c^2 = 75 ?

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 75$ is correct answer.

[quote="Dr Sonnhard Graubner"]hello for b) $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 2(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} ac \minus{} bc) \equal{} 2(225 \minus{} 75) \equal{} 300$.
Sonnhard.[/quote]

Please try again. :)

hello, sorry, it was a typo from me.
Sonnhard.

For b) we get $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 2a^2 \plus{} 2b^2 \plus{} 2c^2 \minus{} 2ab \minus{} 2bc \minus{} 2ca \equal{}$
                            $ 2(a^2 \plus{} b^2 \plus{} c^2 \minus{} (ab \plus{} bc \plus{} ca)) \equal{} 2(75 \minus{} 75) \equal{} 0$

EDIT: Looks like I was beaten to it... twice!

Yeah, but c i don't now.

my b:

b) (a – b)² + (b – c)² + (c – a)².

(a² -2ab + b²) + (b² - 2bc + c²) + (c² -2ca + a²)

2a² + 2b² + 2c² - 2ab - 2bc - 2ca
2(a² + b² + c² - ab - bc - ca)
2(75 - ab - bc - ca)
2(75 - 75)
2*0
0

For real numbers $ x,\ y,\ z$, $ \boxed{x^2 \plus{} y^2 \plus{} z^2\equal{}0\Longleftrightarrow x \equal{} 0\ and y \equal{} 0\ and z \equal{} 0}$.

hello, and now we have $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$ and the equal sign holds iff $ a \equal{} b \equal{} c$ and the result is $ a \equal{} b \equal{} c \equal{} 5$.
Sonnhard.

For part c)

$ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 0$ implies that $ (a \minus{} b)^2 \equal{} (b \minus{} c)^2 \equal{} (c \minus{} a)^2 \equal{} 0$ since squares cannot be negative. This implies that $ a \equal{} b , b \equal{} c , c \equal{} a$ , or rather, $ a \equal{} b \equal{} c$

Since $ a \plus{} b \plus{} c \equal{} 15$ , the only possible solution is $ (a,b,c) \equal{} (5,5,5)$

Checking with the original conditions verifies that it is a genuine solution.

EDIT: Too slow, beaten to it again.

hello kunny, please correct your post, i can't read it.
Sonnhard.

That's right.
When I saw the problem first, it occured to me that $ \boxed{3(a^2 \plus{} b^2 \plus{} c^2)\geq (a \plus{} b \plus{} c)^2\geq 3(ab \plus{} bc \plus{} ca)}$, the equality holds when $ a \equal{} b \equal{} c$.

Thanks to everybody here. Question solution:

a) a²+b²+c² = 75
b) 0
c) 5,5,5

Thanks.

hello kunny, where are your nice problems today?
Sonnhard.

You'll notice that this is a special case of [url=http://en.wikipedia.org/wiki/Maclaurin%27s_inequality]Maclaurin's inequality[/url].


I have a problem.. a huge problem, 'cause I don't really know if there's a good mathematical solutions to it. And I'm not sure if my "idea" is correct either...

But I'll try to explain:

I'm doing some personal research about moneymanagement in the stockmarket, where I use a lot of probability-caluclations.. but it has been a while since I took classes in probablity theory, so it's not the best today :-)

Let say we flip a coin. Head we win, tail we lose (the same amount). The probability for 10 tails in a row is 0,5^10... (this about my level of probability-thinking :-))

What I want to do, is to get some kind of information, like plotted in a graph, about the outcomes of series of tosses..

Let us say we have an equity of 100k. Every time it's a head, we win 1%... and everytime it's a tail, we lose 1%. Eg. we get 10 tails, 2 head, and the 5 tails again.. were now down:

0,99^10 * 1,01^2 * 0,99^5

Now can i measure the probability (at an approximately level is enough) the chance for my equity to be down this much?

And let's say the wins is twice the losses, how will that change?

Sorry that I don't know how to express my thinking more matemathically..

Any one that knows what I mean.. and can give some tips/solutions/recommendations etc..? I would really appreciate it!

Let's say your proportional loss/win per throw is equal to $p$ (in your example $p=0.01$). Then if we have $a$ heads and $b$ tails, the total win/loss will be $(1+p)^{a}(1-p)^{b}$, and if it's given that the total loss should be $L$, then we have to solve the equation

$(1+p)^{a}(1-p)^{b}=1-L$.

Generally, this is difficult to solve, but approximately, for $p\ll 1$, we can take $(1+p)^{a}\approx 1+ap$ and $(1-p)^{b}\approx 1-bp$, giving

$1+(a-b)p-abp^{2}=1-L$

Again, for $p\ll 1$ we can neglect the last term on LHS, which gives

$b-a={L\over p}$

Hence, if we have $a$ heads, we need to have $a+{L\over p}$ tails to obtain the approximate loss of $L$. So, we have to have the series of $2a+{L\over p}$ throws and among them exactly $a$ heads. The probability of this is $\frac{{{2a+{L\over p}}\choose a}}{2^{2a+{L\over p}}}$, and then it should be summed up as an infinite series for all the possible values of $a$: $P=\sum_{a=0}^{\infty}\frac{{{2a+{L\over p}}\choose a}}{2^{2a+{L\over p}}}$, but I'll leave that to someone else :)


Let $ x$ be a real number such that $ \sec x \minus{} \tan x \equal{} 2$. Then $ \sec x \plus{} \tan x \equal{}$

$ \textbf{(A)}\ 0.1 \qquad
\textbf{(B)}\ 0.2 \qquad
\textbf{(C)}\ 0.3 \qquad
\textbf{(D)}\ 0.4 \qquad
\textbf{(E)}\ 0.5$

We have $ \sec x \minus{} \tan x \equal{} 2$, so multiplying both sides by $ \cos x$ and rearranging, $ 1 \minus{} 2\cos x \equal{} \sin x$.

Squaring both sides,

$ 1 \minus{} 4\cos x \plus{} 4\cos^2x \equal{} \sin^2x \equal{} 1 \minus{} \cos^2x\implies5\cos^2x \minus{} 4\cos x \equal{} 0$

$ \implies\cos x \equal{} \frac45$ (since $ \tan x$ is defined, $ \cos x\ne0$) and $ \sin x \equal{} 1 \minus{} 2\cos x \equal{} \minus{} \frac35$.

Thus, $ \sec x \plus{} \tan x \equal{} \frac54 \minus{} \frac34 \equal{} \frac12\implies\boxed{\textbf{(E)}\ 0.5}$.

[hide="easier"]
$ \sec{x}\plus{}\tan{x}\equal{}y$

$ (\sec{x}\plus{}\tan{x})(\sec{x}\minus{}\tan{x}) \equal{} 2y \equal{} \sec^2{x} \minus{} \tan^2{x} \equal{} 1 \implies x \equal{} \frac{1}{2}$[/hide]

[hide]
$ \sec^2 x \minus{} \tan^2 x\equal{}1$

$ \dfrac{\sec^2 x \minus{} \tan^2 x}{\sec x \minus{} \tan x}\equal{}\dfrac{1}{2}\implies$
$ \sec x \plus{} \tan x \equal{}\boxed{0.5}\implies \boxed{E}$[/hide]


geometry
circumcircle
symmetry
perpendicular bisector
geometry unsolved

Let $P$ be a point in the interior of the equilateral triangle $\triangle ABC$ such that $\sphericalangle{APC}=120^\circ$. Let $M$ be the intersection of $CP$ with $AB$, and $N$ the intersection of $AP$ and $BC$. Find the locus of the circumcentre of the triangle $MBN$ as $P$ varies.

Let $O$ be the center of $ABC$. The locus is a segment situated on the perpendicular bisector of $BO$, but I'll just show that the circumcenter of $BMN$ lies on this perpendicular bisector. The bounds of the locus can easily be determined by the restriction of $P$ to the interior of $ABC$.

$BNPM$ is cyclic, because $\angle MBN=\frac\pi 3,\angle\MPN=\frac{2\pi}3$. From here we get $\angle BMC=\angle CNA$. We also have $\angle MBC=\angle NCA$, and $BC=CA$, so $MBC,NCA$ are congruent. In particular, $NC=MB$. This means that $\angle MON=\frac{2\pi}3$, so $O$ lies o the circumcircle of $BMN$, and thus the center of this circle lies on the perpendicular bisector of $BO$.

Let $ O$ be the center of $ [ABC]$. From $ \angle APC \equal{} 120^{\circ}$, we get $ P$ lies on the circumcircle of $ [AOC]$, and $ \angle MPN \equal{} 120^{\circ}$, so $ P$ also lies on the circumcircle of $ [BMN]$.
Lemma: O lies on the circumcircle of $ [BMN]$.
Proof:
If $ P \equal{} O$, it's trivial, so suppose $ P\ne O$ and, WLOG, $ P$ belongs to the small arc $ AO$.
So we have $ \angle POB \equal{} \angle AOB \minus{} \angle AOP \equal{} 120^{\circ} \minus{} \angle ACP \equal{} \angle AMC \equal{} \angle PNB$, so $ O$ lies on the circumcircle of $ [BMN]$, QED.
Let $ D$ and $ E$ be the circumcenters of $ [AOB]$ and $ [BOC]$. By symmetry, $ DE$ is the perpendicular bisector of $ [OB]$, therefore the circumcircle lies on line $ DE$.
Now we shall prove the desired locus is the line segment $ [DE]$, without points $ D$ and $ E$.
Let $ Q$ be a point on line $ DE$ and let $ P$ be one of the intersections of the circumference with center $ Q$ and radius $ \overline{QO} \equal{} \overline{QB}$ with the circumcircle of $ AOC$.
Note $ P$ must lie inside $ [ABC]$. If $ Q \equal{} D$ we have $ P \equal{} A$. If $ Q \equal{} E$, we have $ P \equal{} B$. If $ Q$ doesn't lie on line segment $ [DE]$, $ P$ lies outside $ [ABC]$.
Let $ M \equal{} CP\cap AB$ and $ N \equal{} AP\cap BC$. Now it suffices to prove $ M$ and $ N$ lie on the circumference with center $ Q$ and radius $ \overline{QO} \equal{} \overline{QB}$.
We have $ \angle MPN \equal{} \angle APC \equal{} 120^{\circ}$, so $ B,M,N,P$ lie on the same circumference.
If $ Q$ is the midpoint of $ [DE]$, then $ O \equal{} P$ and $ Q$ is also the midpoint of $ OB$, so we have what we want.
If $ Q$ is not the midpoint of $ [DE]$, then $ O\ne P$.
Using the Lemma, $ O$ lies on the circumference which contains $ B,M,N,P$, so that circumference is the circumcircle of $ [BPO]$, and it's center is $ Q$, so we have what we want.


I got 748 but I would like to make sure I'm right.
Thanks to anyone that can double check this, I'm not sure I understand quite how I got to this number.

Well, the answer to that question lies in knowing 2 * 5 = 10, which has a zero. Since 2's are aplenty, you just need to find all multiples of 5 which are in 3003!. Lets start with 5^1, there are [3003/5] = 600 of those. 5^2 = 25, there are [3003/25] = 120 of those. 5^3 = 125, there are [3003/125] = 24 of those. 5^4 = 625, there are [3003/625] = 4 of those. However there aren't any 5^5 = 3125 in 3003! after accounting for all the previous powers of 5. 600 + 120 + 24 + 4 = 748, and you were correct!

well either he knew that or he counted them manually :D

As a follow-up, how would you determine the first rightmost digit that isn't 0? or maybe two of them, just for fun...

What do we do if the question was asking for another number besides zero?

What do you mean?  Are you talking about something like:

What is the last digit of 7^80?

For something like this you find a pattern.  We only care about the last digit so we only look at the last digit.  We are essentially doing math mod 10 (for this example).

The last digit of powers of 7 are as follows
7, 9 ,3, 1,7,9,3,1
It is pretty obvious it will repeat.  So we divide 80 by 4 and see what the remainder is 80/4 comes out evenly (remainder 0).  So we match that up with one of the smaller ones that we know the last digit of.  it is congruent to 7^4 so the last digit is 1.

No, that's not what I wanted. I wanted to find the rightmost digit in 3003! that isn't zero. I tried finding it myself but then it got complicated.

I was talking to kool_dudy.
[edit] You can do your problem if you factor out all the tens and then find the last digit of the other number.  This would get really mess though.

[quote="white_horse_king88"][quote="mikewd"]I got 748 but I would like to make sure I'm right.
Thanks to anyone that can double check this, [b]I'm not sure I understand quite how I got to this number[/b].[/quote]

Well, sorry if I answered him. I think that implies that he was doubtful, so I guess I just reassured him.[/quote]

heh :) I don't mean that you shouldn't have answered ;)
I meant: how can he possibly get that number without knowing how? :?

This is what I mean.  My example.

[i]Find the number of fours in 3003![/i]

Is it possible to solve?

Of course it is POSSIBLE to solve!  You just have to right the number out and count the number of 4's in it  ;)   .  Seriously though, I don't know a quick solution to that and I doubt that there is.

Lol...but of course we can solve it that way but that would just take a few weeks to compute that number.

hmm.

that sounds tough

how many 4s in 3003! :?

haku-your solution for the 4's in 2003! is definitely not right. Perhaps you don't understand the question. The question is asking for the number of digits that are 4, not the number of factors of 4.

Or else your solution obviously fails for 16, with two factors of 4, but contains no digits that are 4.


geometry
circumcircle
geometry proposed
U sviti matematyky

Let $ H$ be the orthocenter of acute-angled triangle $ ABC.$ Circle $ \omega$ with diameter $ AH$ and circumcircle of triangle $ BHC$ intersect at point $ P\ne H.$ Prove that the straight line $ AP$ pass through the midpoint of $ BC.$

[quote="rogue"]Let $ H$ be the orthocenter of acute-angled triangle $ ABC$. M is the midpoint of side BC. P is on the median AM such that $ HP\perp AM$. Prove that P lies on the circumcircle $ (\omega)$ of triangle BHC.[/quote]
$ D\equal{}BH\cap AC, E\equal{}CH\cap AB, F\equal{}AH\cap BC$.
$ AP.AM\equal{}AH.AF\equal{}AE.AB\equal{}AD.AC$, hence BMPE and CMPD are two cyclic quadrilateral.
Then, $ \widehat{BPC}\equal{}\widehat{BPM}\plus{}\widehat{CPM}\equal{}\widehat{BEM}\plus{}\widehat{CDM}\equal{}\widehat{ABC}\plus{}\widehat{ACB}\equal{}180^o\minus{}\widehat{BAC}\equal{}\widehat{BHC}$
So B,H,P,C lie on a circle.


Is 444 a palindrome? Or 111? or any number with the same value for all its digits?

I thought it was, but the answers #465 in AOPS Vol. 1 only find the palindromes in xyx format.

Maybe my book is outdated?

By definition, a palindrome is a number that is the same when read from left to right as it is when read from right to left (and words, like racecar), so 444 and 111 are palindromes.

xxx is included in numbers in the form xyx, because y could equal x.

@ tz:

The two most memorable which I can remember:

Mr. Owl ate my metal worm.

A man, a plan, a canal, Panama.

And the one I made up:
Live evil :)

I can't find palindromes in AoPS 1. What chapter is it in?

If you look at my first post, it says its question number 495. It's under the Learning how to count chapter

[quote="tz"]Does anyone know any word or phrase palindromes?[/quote]
Rats live on no evil star.
Go hang a salami, I'm a lasagna hog!
God, a red nugget, a fat egg under a dog.

Searching on Google is probably the best way. :D

[quote="xdl615x"]Let me rephrase the question:

Can someone open up to the AOPS vol. 1 book and make sure that was a mistake on the book?[/quote]
Well, it doesn't say $h \ne t$, in $hth$. So I think it's right. Unless we have different editions and the answer given isn't 90.

[quote="xdl615x"]Let me rephrase the question:

Can someone open up to the AOPS vol. 1 book and make sure that was a mistake on the book?[/quote]

Go Hang A Salami I am A Lasagna Hog.

I heard this one form somewhere.

[quote="tz"]Does anyone know any word or phrase palindromes?[/quote]

We had a President called MENEM, and we have a state called NEUQUEN

[quote="b-flat"]Maybe I'm not thinking of the correct thing, but in AoPS volume 1 (I checked this), problems are expressed something like 25-6, where the first number is the chapter, and the second number is the problem in the chapter.[/quote]

That only applies for examples and exercises. There are end of the chapter problems.

I am greatly annoyed by the amount of unnecessary and off-topic posts on this thread. I'd greatly appreciate it if this was locked.

Thanks to those of you who actually came to answer my question rather than increase their post count.

[quote="xdl615x"]I am greatly annoyed[/quote]

"greatly annoyed"????? Come one man! don't be so orthodox, you said that as if someone have made something bad to you. If you think about what you have written, you'll see you were wrong.

[quote="José"]"greatly annoyed"????? Come one man! don't be so orthodox, you said that as if someone have made something bad to you. If you think about what you have written, you'll see you were wrong.[/quote]These comments should never be posted here. If you have a complain, send a PM.

Admins/Mods, please lock this topic.