trigonometry
geometry
vector
power of a point
radical axis

[quote="Mihai Miculita"] [color=darkred]Prove that in a trapezoid $ ABCD$, $ AB\parallel CD$, the intesection $ O\in AC\cap BD$
belongs to the radical axis of the circles with the diameters $ [AD]$ , $ [BC]$ .[/color][/quote]

[color=darkblue][b][u]Proof I.[/u][/b] Denote the power $ p_w(X)$ of the point $ X$ w.r.t. the circle $ w$ . Observe that the radical axis $ d$ of the circles $ w_1$ , $ w_2$ with the diameters $ [AD]$ , $ [BC]$ respectively is perpendicularly on the middle line of the given trapezoid.

Denote the projections $ U$ , $ V$ on the sideline $ AB$ of the points $ D$ , $ C$ respectively and the projections $ P$ , $ R$ of the point $ O$ on the sidelines $ AB$ , $ CD$ respectively. Observe that $ PU \equal{} RD$ , $ PV \equal{} RC$ and $ U\in w_1$ , $ V\in w_2$ .

Therefore, $ O\in d$ $ \Longleftrightarrow$ $ P\in d$ $ \Longleftrightarrow$ $ p_{w_1}(P) \equal{} p_{w_2}(P)$ $ \Longleftrightarrow$ $ PU\cdot PA \equal{} PV\cdot PB$ $ \Longleftrightarrow$ $ RD\cdot PA \equal{} RC\cdot PB$ $ \Longleftrightarrow$ $ \frac {PA}{RC} \equal{} \frac {PB}{RD}$ , what is truly because $ \frac {PA}{RC} \equal{} \frac {OP}{OR} \equal{} \frac {PB}{RD}$ . In conclusion, the point $ O$ belongs to the radical axis of the circles $ w_1$ , $ w_2$ .

[b][u]Proof II.[/u][/b] I"ll use the same notations from the upper proof. Denote the middlepoints $ M$ , $ N$ of the sides $ [AD]$ , $ [BC]$ respectively. Therefore, $ O\in d$ $ \Longleftrightarrow$ $ p_{w_1}(O) \equal{} p_{w_2}(O)$ $ \Longleftrightarrow$ $ OM^2 \minus{} \left(\frac {AD}{2}\right)^2 \equal{}$ $ ON^2 \minus{} \left(\frac {CD}{2}\right)^2$ $ \Longleftrightarrow$ $ OA^2 \plus{} OD^2 \minus{} AD^2 \equal{}$ $ OB^2 \plus{} OC^2 \minus{} BC^2$ $ \Longleftrightarrow$ $ 2\cdot OA\cdot OD\cdot \cos\widehat {AOD} \equal{}$ $ 2\cdot OB\cdot OC\cdot\cos\widehat {BOC}$ $ \Longleftrightarrow$ $ OA\cdot OD \equal{} OB\cdot OC$ $ \Longleftrightarrow$ $ [AOD] \equal{} [BOC]$ $ \Longleftrightarrow$ $ [ACD] \equal{} [BCD]$ $ \Longleftrightarrow$ $ AB\parallel CD$ , what is truly.

[b][u]Remark.[/u][/b] The second proof can be applied also to the following problem : [/color]

[quote][color=darkred] [b]Prove that in a convex quadrilateral $ ABCD$ the intersection of the diagonals belongs to the radical axis of the circles with the diameters $ [AD]$ , $ [BC]$ if and only if the given quadrilateral is a trapezoid ( $ AB\parallel CD$ ) ![/b][/color] [/quote]
[color=darkblue][b]Please, I wait and another proofs of this nice and easy problem.[/b][/color] [color=darkred][b]Thank you ![/b] [/color]

Let $ P$ is the projection of $ D$ onto $ AC$, $ Q$ is the projection of $ C$ onto $ BD$, $ R$ is the projection of $ B$ onto $ AC$, $ S$ is the projection of $ A$ onto $ BD$
We want to prove that $ AO\cdot PO\equal{}BO\cdot QO$. (Equal power with respect to both circles with diameter $ AD$, $ BC$)
$ CDPQ$ is cyclic, therefore $ DO\cdot OQ\equal{}CO\cdot OP$, which implies $ \frac{PO}{QO}\equal{}\frac{DO}{CO}$.
$ \frac{DO}{CO}\equal{}\frac{BO}{AO}$ because $ AB \parallel{} DC$. $ \frac{PO}{QO}\equal{}\frac{BO}{AO}\implies AO\cdot PO\equal{}BO\cdot QO$ and done?

[quote="Mihai Miculita"] [color=darkred]Prove that in a trapezoid $ ABCD$, $ AB\parallel CD$, the intesection $ O\in AC\cap BD$
belongs to the radical axis of the circles with the diameters $ [AD]$ , $ [BC]$ .[/color][/quote]

[color=darkblue][b][u]Proof I.[/u][/b] Denote the power $ p_w(X)$ of the point $ X$ w.r.t. the circle $ w$ . Observe that the radical axis $ d$ of the circles $ w_1$ , $ w_2$ with the diameters $ [AD]$ , $ [BC]$ respectively is perpendicularly on the middle line of the given trapezoid.

Denote the projections $ U$ , $ V$ on the sideline $ AB$ of the points $ D$ , $ C$ respectively and the projections $ P$ , $ R$ of the point $ O$ on the sidelines $ AB$ , $ CD$ respectively. Observe that $ PU = RD$ , $ PV = RC$ and $ U\in w_1$ , $ V\in w_2$ .

Therefore, $ O\in d$ $ \Longleftrightarrow$ $ P\in d$ $ \Longleftrightarrow$ $ p_{w_1}(P) = p_{w_2}(P)$ $ \Longleftrightarrow$ $ PU\cdot PA = PV\cdot PB$ $ \Longleftrightarrow$ $ RD\cdot PA = RC\cdot PB$ $ \Longleftrightarrow$ $ \frac {PA}{RC} = \frac {PB}{RD}$ , what is truly because $ \frac {PA}{RC} = \frac {OP}{OR} = \frac {PB}{RD}$ . In conclusion, the point $ O$ belongs to the radical axis of the circles $ w_1$ , $ w_2$ .

[b][u]Proof II.[/u][/b] I"ll use the same notations from the upper proof. Denote the middlepoints $ M$ , $ N$ of the sides $ [AD]$ , $ [BC]$ respectively. Therefore, $ O\in d$ $ \Longleftrightarrow$ $ p_{w_1}(O) = p_{w_2}(O)$ $ \Longleftrightarrow$ $ OM^2 - (\frac {AD}{2})^2 =$ $ ON^2 - (\frac {CD}{2})^2$ $ \Longleftrightarrow$ $ OA^2 + OD^2 - AD^2 =$ $ OB^2 + OC^2 - BC^2$ $ \Longleftrightarrow$ $ 2\cdot OA\cdot OD\cdot \cos\widehat {AOD} =$ $ 2\cdot OB\cdot OC\cdot\cos\widehat {BOC}$ $ \Longleftrightarrow$ $ OA\cdot OD = OB\cdot OC$ $ \Longleftrightarrow$ $ [AOD] = [BOC]$ $ \Longleftrightarrow$ $ [ACD] = [BCD]$ $ \Longleftrightarrow$ $ AB\parallel CD$ , what is truly.

[b][u]Proof III (similarly to Minsoens').[/u][/b] Denote $ \{\begin{array}{c}
X\in AC\ ,\ DX\perp AC\\\
Y\in BD\ ,\ CY\perp BD\\\
Z\in CY\cap DX\end{array}$ . Observe that the points $ X$ , $ Y$ belong to the circle $ w$ with the diameter $ [CD]$ and $ X\in w_1$ , $ Y\in w_2$ . Since $ \widehat {OXY}\equiv\widehat {CXY}\equiv\widehat {CDY}\equiv\widehat {CDB}\equiv\widehat {DBA}$ , i.e. $ \widehat {OXY}\equiv\widehat {DBA}$ what means that the quadrilateral $ ABYX$ is cyclically (in the circle $ w'$ ) . Thus, $ p_{w'}(O)=OX\cdot OA=OY\cdot OB$ $ \implies$ $ p_{w_1}(O)=p_{w_2}(O)$ $ \implies$ the point $ O\in d$ .  Observe that the point $ O$ is the orthocenter of $ \triangle CDZ$ . Hence $ OZ\perp CD\parallel AB$ , $ d\equiv OZ$ and the point $ Z$ is the radical center of the circles $ w$ , $ w_1$ , $ w_2$ .

[b][u]Remark.[/u][/b] These proofs can be applied also to the following problem : [/color]

[quote][color=darkred] [b]Prove that in a convex quadrilateral $ ABCD$ the intersection of the diagonals belongs to the radical axis of the circles with the diameters $ [AD]$ , $ [BC]$ if and only if the given quadrilateral is a trapezoid ( $ AB\parallel CD$ ) ![/b][/color] [/quote]

[quote="Virgil Nicula"][quote="Mihai Miculita"] [color=darkred]Prove that in a trapezoid $ ABCD$, $ AB\parallel CD$, the intesection $ O\in AC\cap BD$
belongs to the radical axis of the circles with the diameters $ [AD]$ , $ [BC]$ .[/color][/quote][/quote]
[hide="Another Solution"]
Let the circles with diameters $ AD$ and $ BD$ meet $ AC$ at $ P$ and $ Q$ respectively.  Therefore, we have that $ \angle APO \equal{} 90 \equal{} \angle CQO$ and clearly, $ \angle AOP \equal{} \angle COQ$, which gives us that $ \triangle AOP\sim \triangle COQ\implies \frac {PO}{QO} \equal{} \frac {AO}{CO}$.  Also, since $ AB\parallel CD$, we have that $ \triangle AOB\sim \triangle COD\implies \frac {PO}{QO} \equal{} \frac {AO}{CO} \equal{} \frac {BO}{DO}\implies PO\cdot DO \equal{} QO\cdot BO$. Therefore, $ O$ has the same power with respect to the circles with diameter $ AD$ and $ BC$, so $ O$ lies on radial axis of the circles with diameter $ AD$ and $ BC$. [/hide]

This was also discussed at: [url]http://www.mathlinks.ro/viewtopic.php?p=818557#818557[/url].

There is an easy complex number solution. (It is equivalent to use vectors in this problem as well).

Affix $ O\equal{}0$ as the origin, call $ A \equal{} 2a$, $ B \equal{} 2b$, $ C \equal{} 2ka$, $ D \equal{} 2kb$ where $ k$ is a negative real number.

Now say that $ M$, $ N$ are the midpoints of $ AD$, $ BC$. Then $ M \equal{} a \plus{} kb$ and $ N \equal{} b \plus{} ka$. Also $ AD \equal{} 2kb \minus{} 2a$, $ BC \equal{} 2ka \minus{} 2b$. We must prove that
\[ OM^2 \minus{} \left(\frac {AD}{2}\right)^2 \equal{} ON^2 \minus{} \left(\frac {BC}{2}\right)^2
\]
The left hand side is equal to
\[ (a \plus{} kb)(\overline{a} \plus{} k\overline{b}) \minus{} (kb \minus{} a)(k\overline{b} \minus{} \overline{a}) \equal{} 2k(a \overline{b} \plus{} b\overline{a})
\]
It is clear that expression is symmetric in $ a$, $ b$, so the left and right sides are equal. Hence we have that $ O$ has equal power with respect to both of the circles and we are done.


Sorry if this is in the wrong section, but it looked like the best choice...

I am 15 and in Calc BC at my high school, and I've always thought of myself as pretty advanced with math, but when I come to this page I don't know half the stuff people are talking about.  Are there any good learning-oriented books I could get to teach myself the math topics discussed on this site? (not just raw problem books, but to teach me the material, although having problems in it to help learn would be preferred)

Exactly the same situation here..
I don't know about books, but I've started taking the introductory classes here...
they're helping. :)

My parents bought the Art of Problem Solving books available on this site for me.  I am working on them.

[url]http://www.artofproblemsolving.com/Books/AoPS_B_V1.php[/url]

I also took the MATHCOUNTS and AMC-10 classes.  Although I took Geometry last year, my teacher did not focus too much on problem solvings.  So, I am taking the Intro to Geometry class right now.


[quote="Kyron"]Purely out of interest, what good does it do to specify $ c$ as a positive number? It is not used in the equations. Should it be there or is the fact that you included at a typo?  :huh:[/quote]
The sigma symbol indicates a cyclic sum, which means the variables are cycled (replacing $ a$ with $ b$, $ b$ with $ c$, $ c$ with $ a$) until all combinations have been used, and then they are summed.
So, I think the LHS of the inequality is
$ \frac {a^2 \plus{} b^2}{ab} \plus{} \frac {b^2 \plus{} c^2}{2bc} \plus{} \frac {c^2 \plus{} a^2}{2ca}$.

But then I'm also confused about the problem, as $ a\equal{}b\equal{}c\equal{}1$ doesn't work.

[quote="i_like_pie"][quote="archimedes1"][quote="Kyron"]Purely out of interest, what good does it do to specify $ c$ as a positive number? It is not used in the equations. Should it be there or is the fact that you included at a typo?  :huh:[/quote]
The sigma symbol indicates a cyclic sum, which means the variables are cycled (replacing $ a$ with $ b$, $ b$ with $ c$, $ c$ with $ a$) until all combinations have been used, and then they are summed.
So, I think the LHS of the inequality is
$ \frac {a^2 \plus{} b^2}{ab} \plus{} \frac {b^2 \plus{} c^2}{2bc} \plus{} \frac {c^2 \plus{} a^2}{2ca}$.[/quote]
Why $ 2bc$ and $ 2ac$? Shouldn't it be $ bc$ and $ ac$?[/quote]
Ok  :oops:  but still,, $ a\equal{}b\equal{}c\equal{}1$ doesn't work.

Sorry, I made a computation error; it is fixed now.

[quote="CatalystOfNostalgia"]$ a,b,c$ are positive numbers. Show that

$ \displaystyle\sum\dfrac{a^{2} \plus{} b^{2}}{ab}\ge2 \plus{} \dfrac{4\displaystyle\sum a^{2}}{\displaystyle\sum ab}$.[/quote]
Hence:
$ \sum\dfrac{a^{2} \plus{} b^{2}}{ab}\minus{}6\equal{}\sum \frac{(a\minus{}b)^2}{ab}$
$ \frac{4(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\minus{}4\equal{}\frac{2((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2)}{ab\plus{}bc\plus{}ca}$
So:
$ LHS\minus{}RHS\equal{}S_c{(a\minus{}b)^2}\plus{}S_{a}{(b\minus{}c)^2}\plus{}S_{c}{(a\minus{}b)^2}$
$ S_c\equal{}\frac{1}{ab}\minus{}\frac{2}{ab\plus{}bc\plus{}ca}$
$ S_b\equal{}\frac{1}{ca}\minus{}\frac{2}{ab\plus{}bc\plus{}ca}$
$ S_a\equal{}\frac{1}{bc}\minus{}\frac{2}{ab\plus{}bc\plus{}ca}$
Assume: $ a \ge b \ge c$, we have:
$ S_b \ge 0, S_a \ge 0$ and:
 $ S_c\plus{}S_b\equal{}\frac{1}{ab}\plus{}\frac{1}{ca}\minus{}\frac{4}{ab\plus{}bc\plus{}ca} \ge \frac{4}{ab\plus{}ca}\minus{}\frac{4}{ab\plus{}bc\plus{}ca} >0$
It follows that:
$ S_c{(a\minus{}b)^2}\plus{}S_{a}{(b\minus{}c)^2}\plus{}S_{c}{(a\minus{}b)^2} \ge (S_c\plus{}S_b)(a\minus{}b)^2\plus{}(S_a\plus{}S_b)(b\minus{}c)^2 \ge 0$ ( because $ (a\minus{}c)^2 \ge (a\minus{}b)^2\plus{}(b\minus{}c)^2$)

Very nice solution!

I had in mind the following:

[hide]To rewrite as $ \sum\dfrac{(a \plus{} b)^{2}}{ab}\ge\dfrac{4\sum a^{2}}{\sum ab}$, which follows from Cauchy.[/hide]


Ok..

I have two types of the problems that when I see it, I can't do it.

A) Like $x^2-5x+8$ is perfect square on certain $n$ values. Find the sum of all possible values.

B) Function f(x) = ax^4-bx^3+cx^2+dx-e, f(2) = 3, and find f(5) =

Or something like that.

How do you solve something like them?

A) n = -1 or n = -4

Check a few base cases, otherwise:

(x-3)^2 < x^2 - 5x + 8 < (x-2)^2 and vice versa if n is negative.

B) As Scotty would say, I canna work miracles, Captain!

Huh?

How did you get that?

I mean I have NO CLUE on how to do this...

Can you show the process?

$ x^2-5x+8 $
how is this a perfect square?

the idea is that for sufficiently large positive or negative n, n is essentially equal to its first and second coefficients, so that:

x^2-6x < x^2-5x < x^2 - 4x for positive x, and flip the signs for negative x.

Now complete the square on the two bookends of this inequality.

We get x^2-6x+9 < x^2-5x < x^2 - 4x + 4 for large enough x>0, or:

(x-3)^2 < something < (x-2)^2.

But there are no perfect squares between two consecutive perfect square, so we are done.


integration
real analysis
real analysis unsolved

Suppose $ f\in L^{1}([0,1])\setminus L^{2}([0,1])$. Prove that there exists a complete orthonormal basis $ \phi_{n}$ for $ L^{2}([0,1])$ such that for each $ n,\phi_{n}$ is continuous and moreover $ \int_{0}^{1}f(x)\phi_{n}(x)dx \equal{} 0$.


ABCDEFGH is a regular octagon inscribed in a circle.point P bisects arc HG in point Q show that $ <PQG\equal{}78.75$

[quote="aadil"]
...point P bisects arc HG in point Q...[/quote]

Can you please explain that? Where is point Q?

Upon inspection, it appears that $ P$ is the center of the octagon/circle. Nevertheless, this is easy to prove; $ \triangle PQG$ is an isosceles triangle since both $ Q$ and $ G$ are on the circumference of the circle and $ P$ is the center, and $ \angle QPG \equal{} \frac {180^\circ}8 \equal{} 22.5^\circ$, thus $ \angle PQG \equal{} \frac 12(180^\circ \minus{} 22.5^\circ) \equal{} \frac 12\cdot 157.5^\circ \equal{} 78.75^\circ$, as desired.


vector
linear algebra
linear algebra unsolved

Let $V$ be a vectorspace and $V^{*}$ the set of all linear transformations from $V$ to $\mathbf{R}$.

a) Show that $V^{*}$ is a vectorspace.

b)Assume $\mbox{dim}V$ is finite. Show that $\mbox{dim}V=\mbox{dim}V^{*}$
(consider bases in each space)

c)Show that when $\mbox{dim}V<\infty$ there exsist a natural isomorfism $\Phi$ between
$V$ and $V^{**}=(V^{*})^{*}$ defined by
\[(\Phi(\mathbf{v}))T=T(\mathbf{v})\]
for all $\mathbf{v}$ in $V$ and $T$ in $V^{*}$

[color=red]Please help me with these questions[/color]

a),b) and c) are routine!

a) You need check all axioms on vector space.
b) If $e_{1},e_{2},...,e_{n}$ is a basic of $V$ then $f_{1},f_{2},...,f_{n}$ is a basic of $V^{*}$, where $f_{i}(e_{j})=\delta_{i,j}$, $\delta_{i,j}$ is Kronecke's symbol.
c)Check direct!

Please I am not as familiar with this as you seem to be.
Could you please give som more explicit help?
Especially on c)    ( and b) ).
Thanks anyway :)


let $a,b,c$be positive  numbers such that$abc \leq 8$.
prove that :
$\frac{1}{a^2-a+1} + \frac{1}{b^2-b+1} + \frac{1}{c^2-c+1} \geq 1$

It is very easy.
First put $a=\frac 2x$ $b=\frac 2y$ $c=\frac 2z$
We have $abc \leq 8$ so $xyz \geq 1$
We have $a^2-a+1= \frac{x^2-2x+4}{x^2}$
So $(\sum \frac{1}{a^2-a+1}) = (\sum \frac{x^2}{x^2-2x+4}) \geq \frac{(\sum x)^2}{\sum x^2 -2 \sum x+12}$
So we have to prove that $\frac{(\sum x)^2}{\sum x^2 -2 \sum x+12} \geq 1$
expand them we will get that $\sum xy +\sum x \geq 6$ with is true due to AM-GM

[quote="Detective"]Of course it's very easy when the official solution is in front of you! ;)[/quote]
Sorry, I don't understand what you mean. In fact, I take me a lot of times to try and think and it is also not much difficult. And as I said before, my major is chemistry, I solve math just for fun. I don't have any math's book nor any documents involved in math. How can I see the" official sol" ?
 By the way, in this inequality, we can use estimate like:

  $\frac{1}{a^2-a+1} \geq \frac{3-a}{3}$ ... Sum it up but we can't not go further. But if we have $a+b+c \leq 6$ then we done.
 So if you replace $abc \leq 8$ by $a+b+c \leq 6$ then you have the same answer.

[quote="e.lopes"]Gibbenergy, this problem is from Mathlinks Contest, and the official solution is the same of your.  ;)

(by the way, with the same letters)  :D[/quote]

   Really! Ha ha... I didn't know it . Since I am only in ML for couple months and only check ML at weekend , I don't really see much in ML.
   About the letters, I always use $x,y,z$ and$a,b,c$ for inequal, if it is the same, there is no surprise.  :)

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=56194 and http://www.mathlinks.ro/Forum/viewtopic.php?t=35975 .

  darij


What is the sum of the largest and smallest prime factors of 990?

$ 990\equal{}2*3^2*5*11$

smallest prime factor=2
largest prime factor=11

2+11=13

answer : 13

[size=200]990
[/size]
$\text{Largest Prime Factor}\rightarrow11$
$\text{Smallest Prime Factor}\rightarrow2$

$11+2=\boxed{13}$


geometry
geometric transformation
reflection
circumcircle
conics
parabola
geometry proposed

See http://www.mathlinks.ro/Forum/viewtopic.php?t=60529

Let the triangle $ABC$ and the points $Q\in (BC),E\in AB,F\in AC$ such that $Q\in (EF)$. I note the reflection $A'$ of the point $A$ w.r.t. to the circumcenter $O$ and the point $S\in BA'$ such that $OS\perp BC$. Prove that $QE=QF\Longleftrightarrow SQ\perp EF$.

I don't think so.
Consider the lines L intersecting BC at Q, AB at E, AC at F such as Q is the midpoint of EF. If ABC is not isosceles, the perpendicular at Q to such a line L will not go to a fixed point.
More precisely, the only point P such as, if Pa,Pb,Pc are the projections of P upon the sidelines BC, CA, AB, Pa is the midpoint of PbPc is the second common point of the circumcircle with the A-Apollonius circle or, equivalently, the second common point of the circumcircle with the A-symedian or, equivalently the point of the circumcircle whose Simson-line is perpendicular to the A-median.
It follows that the lines L envelope the parabola with focus P and directrix the perpendicular at H (orthocenter) to the A-median and that the perpendicular at Q to the lines L envelope the parabola with focus P and directrix the A-median.

Let's reformulate your generalization.
Given a point A' and a real k (not 0), prove that (1)<=>(2)
(1) For any point Q on the line BC, the perpendicular at Q to A'Q intersects AB and AC at E and F such as EQ/QF = k(signed distances)
(2) k = tan B/tan C and A' is the antipode of A on the circumcircle

Thanks. I am sorry, you are right ! I will correct soon (may be differently of your).

Another possible generalization :
Find u and a point P such as QE = QF <=> <PQF = u (mod Pi)
(P will be the point above and 2 cot u = cot B - cot C)

The same deed I searched and I, but I didn't find (more correctly, I made a mistake).  I renounced to generalize because the both your generalizations are nicely.


geometry
symmetry
Euler
circumcircle
ratio
similar triangles
geometry proposed

Let ABC be a triangle with orthocenter H and let P and Q be the feet of B and C on AC and AB respectively. If R and S are the intersection of the bisectrix of angle < QHC with  AC and AB respectively, prove that the orthocenters of ABC, ARS and APQ are collinear.

PS
I found a synthetic prove of above result which explore other nice properties of the involved figures.

As far as I know the problem is original

[quote="sprmnt21"]Let ABC be a triangle with orthocenter H and let P and Q be the feet of B and C on AC and AB respectively. If R and S are the intersection of the bisectrix of angle < QHC[/quote]

Did you really mean < QHC ? As for me, < PHC would make sense...

[quote="sprmnt21"]with  AC and AB respectively, prove that the orthocenters of ABC, ARS and APQ are collinear.[/quote]

Thanks, looks really interesting.

  Darij

[quote="darij grinberg"]

Did you really mean < QHC ? As for me, < PHC would make sense...

[/quote]

Sorry for the big confusion. Here follows the correct (I hope) description.

P and R are on AC as Q and S are on AB and <PHR = <CHR = <QHS = <BHS.

Sorry again, for the misunderstanding.  :blush:

Actually, after drawing it, I think $R,S$ are the intersections of $AC,AB$ with the bisector of $\angle PHQ$. It can't be $CHQ$ (because those three points are collinear), and I don't think it's $PHC$ because it should have some sort of symmetry, right? Anyway, I checked it with dynamic software and it works for $\angle PHQ$.

I missed to clarify what follows

"let P and Q be the feet of the altitudes from B and C to AC and AB respectively"

I hope this is enough to make all consistent. If not I will try to send a picture.

All we need to do is show that $\frac {CP}{PR}=\frac {BQ}{QS}$, and this is true because the following are pairs of similar triangles: $(HQS,HPR),\ (HQB,HPC)$.

I agree. In order to prove the claim, all we need is THALES' THEOREM.

... and perhaps Euler could shorten our efforts.

Now prove tha also the circumcenters and the centroids of the three triangles ABC, APQ and ARS are collinear.

Let ABC be a triangle with orthocenter H and let P and Q be the feet of B and C on AC and AB respectively. If R and S are the intersection of the bisectrix of angle  <PHC with AC and AB respectively, prove that the orthocenters, circumcenters and centroids of ABC, ARS and APQ are collinear.

one can prove the following lemma

Given a quadrilateral ABCD, let X,  Y be points on AB, CD respectively so that AX/XB = DY/YC and let Z, W be points on AD, BC respectively so that AZ/ZD=BW/WC. Then the point P=XY.ZW divide ZW with the same ratio as X divide AB (or Y divide DC) and divide XY as Z divide AD.

From this lemma we have that if L,M,N are the mid points of BC, RS and PQ, they are collinear. Then from this and the fact that tha centroid of a triangle divide a median in a fixed ratio(2:1), follows that the centroids G, J, I of ABC, ARS and APQ are collinear.

From Euler theorem we have that the orthocenter, centroid and circumcenter of a triangle are collinear and the segment between them have a fixed ratio (1:2).

Combining this facts all togheter (and using the converse of the lemma) we can prove the claim.


counting
distinguishability
rotation
combinatorics unsolved
combinatorics

We have N (identical) balls to be placed into K boxes (empty boxes allowed).
The number of ways this can be done is well known: (N+K-1) choose (N-1)

However, what I'm trying to figure out (and have been stuck with) is the following.

What is the solution to the above problem, if I impose the added constraint of circular invariance. What I mean by that is: Let the boxes be numbered 1,2,...K
Then a choice of boxes b1 b2 b3 .. is not distinguishable from the choice (b1+i) (b2+i) (b3+i) .... , where the addition is modulo K   { b1,b2,b3,.. and, i, are all integers in [1,K] }
For ex: choosing the boxes  1 1 is identical to choosing 2 2, or, 3 3, ..., or, K K .
choosing boxes  1 2 is identical to choosing 2 3, or, 3 4, ...., or K 1.

Similarly, choosing boxes 2 6 5 is same as choosing 1 5 4, or, 3 7 6, and so on...

I'm in particular interested in K=8, for which I computed the solution manually for 2 balls.
The solution is 5, and one set of solutions is

Both balls in box 1, OR, 1 ball each in box 1 and 2, or, 1 and 3, or, 1 and 4, or, 1 and 5.

For 3 balls, I tried manually, and the answer is about 15 (I say about 15 since I might just have made a mistake of 1 or 2 here or there while counting all possibilities.)

Any help would be greatly appreciated.

:D  Hi, jaspreet_singh.
I could say that you are facing to the BIG problem.
However, I will show you some easy case that may enlighten your idea.  :wink:
The main thing is if you count normally, i.e. no circular condition, there are lots of over-counting cases.
The thing I'm going to show you here is the case that the amout of boxes be a prime number $ \equal{} p$.

What benefits us here is no matter how you put the balls into the boxes and you rotate it all $ p$ ways,
each of $ p$ ways will be different!

Hence, we can count the number normally and then divide it by $ p$, that's the answer.
In particular, for $ n$ balls and $ p$ boxes,
\[ \frac{\binom{p\plus{}n\minus{}1}{n}}{p}\].

For example,
If we have 2 balls to be put in 3 boxes, we have 2 different ways.
If we have 2 balls to be put in 5 boxes, we have 3 different ways.

[quote="jaspreet_singh"]Both balls in box 1, OR, 1 ball each in box 1 and 2, or, 1 and 3, or, 1 and 4, or, 1 and 5. [/quote]
{1,2} is as same as {1,5} and so {1,3},{1,4}  :D


calculus
integration
geometry
rotation
function
calculus computations

A bowl has the shape of the graph of $ y\equal{}x^4$ between the points $ (1,1)$ and $ (\minus{}1,1)$ rotated about the y-axis. When the bowl contains water to a depth of h units, it flows out rhough a hole in the bottom at a rate (volume/time) proportional $ \sqrt{h}$, with constant proportionality 6.
a) Shows that the water level falls at a constant rate.
all I got so far:
$ \displaystyle\int^{\minus{}1}_1 \pi x^4\,dx$ (I tried to slice it and then take the integral...)
[hide="ANSWER"]
$ \frac{dh}{dt} \equal{} \frac{\minus{}6}{\pi}$[/hide]
b) Find how long it takes to empty the bowl if it is originally full to the brim.
[hide="ANSWER"]
$ t\equal{}\frac{\pi}{4}$[/hide]

I'm a bit lost. Help please!

[quote="jeez123"]A bowl has the shape of the graph of $ y \equal{} x^4$ between the points $ (1,1)$ and $ ( \minus{} 1,1)$ rotated about the y-axis. When the bowl contains water to a depth of h units, it flows out rhough a hole in the bottom at a rate (volume/time) proportional $ \sqrt {h}$, with constant proportionality 6.
a) Shows that the water level falls at a constant rate.
all I got so far:
$ \displaystyle\int^{ \minus{} 1}_1 \pi x^4\,dx$ (I tried to slice it and then take the integral...)
[hide="ANSWER"]
$ \frac {dh}{dt} \equal{} \frac { \minus{} 6}{\pi}$[/hide]
b) Find how long it takes to empty the bowl if it is originally full to the brim.
[hide="ANSWER"]
$ t \equal{} \frac {\pi}{4}$[/hide]

I'm a bit lost. Help please![/quote]

a) Consider that the volume as a function of $ h$ is given by rotating the curve $ x \equal{} y^{\frac {1}{4}}$ from $ y \equal{} 0$ to $ y \equal{} h$ about the y-axis: using the disc method we have that

 \[ V(h) \equal{} \int_{0}^{h} \pi \big(y^{\frac {1}{4}}\big)^2dy \equal{} \pi\int_{0}^{h}y^{\frac {1}{2}}dy \equal{} \frac {2\pi}{3}h^{\frac {3}{2}}\]

Now differentiate w.r.t. $ t$ to get
\[ \frac {dV}{dt} \equal{} \frac {d}{dt}\left(\frac {2\pi}{3}h^{\frac {3}{2}}\right) \equal{} \pi\sqrt {h}\,\frac {dh}{dt}
\]
But the problem states that $ \frac {dV}{dt} \equal{} \minus{} 6\sqrt {h}$, so
\[ \pi\sqrt {h}\,\frac {dh}{dt} \equal{} \minus{} 6\sqrt {h}\Rightarrow \boxed{\frac {dh}{dt} \equal{} \minus{} \frac {6}{\pi}}
\]

b) The bowl will be drained when $ h(t)\equal{}0$, we know that $ \frac {dh}{dt} \equal{} \minus{} \frac {6}{\pi}$, and we can set $ h(0)\equal{}1$ since the bowl is initially filled to the brim. We have that

\[ h(t)\equal{}\int\frac {dh}{dt}\, dt \equal{} \minus{}\int \frac {6}{\pi}dt \equal{} \minus{}t\frac {6}{\pi}\plus{}C\]

Also, plugging-in $ h(0)\equal{}1$ gives the value $ C\equal{}1$ and hence $ h(t)\equal{} 1\minus{}t\frac {6}{\pi}$. Setting $ h(t)\equal{}0$ and solving for $ t$ gives the time for the bowl to drain as $ \boxed{t\equal{}\frac{\pi}{6}}$.
I realize this is different than the solution you posted, but I think it is correct.


Find the first $3$ terms, in ascending powers of $x$, in the expansion of

$\frac{1+x}{\sqrt{1-x}}$

Is there supposed to be an exponent on the term?

Yeah, you can't really have 3 terms when you give the actual expression that doesn't have three terms :D

that was pretty confusing (i mean what I typed)

[quote="coffeym"]No the question is right...the bottom line is a square root, convert that to a power of $\frac{1}{2}$.  I think you'll find the denominator can be expanded...and then multiplied by the numerator... :roll:[/quote]

But when you expand it like that there are three monomials under the radical but that radical is one term only.  So basically it is terms within terms.

[quote="coffeym"]No the question is right...the bottom line is a square root, convert that to a power of $\frac{1}{2}$.  I think you'll find the denominator can be expanded...and then multiplied by the numerator... :roll:[/quote]

Are you talking about using the binomial theorem?

In other words, $(1-x)^{\frac{1}{2}}={1/2\choose0}1^{0.5}(-x)^0+{1/2\choose1}1^{-0.5}(-x)^1+...$

Is that what you want us to use?

Yes 236factorial, thats exactly what the question was geared towards  :lol: Although, you might have some difficulty taking that method of expansion, as $n$ is not an integer. Try
$(1-x)^\frac{1}{2}=1-\frac{x}{2}-\frac{x^2}{8}+...$

[quote="t0rajir0u"][quote="coffeym"]in ascending powers of $x$[/quote]

In other words, he wants the Taylor series.[/quote]

Is that high school level material?

Note: I just realized the title of the topic - Binomial theorem!  :roll: Maybe if people read that first....


superior algebra
superior algebra theorems

Hi every body,
I want to know more about completion and Hensel's lemma in Commutative Algebra. So, Please recommend me some books to read. I have D.Eisenbud and R.B.Ash's Lecture Notes.
Thank you very much.

Unless you are looking for rather advanced stuff, you might like [i]Local Fields[/i] (Cassels).

Thks. I found somethings in Bourbaki'Commutative Algebra as well.


[hide] :) is in $\dfrac 15$ of them for a total of 4444 posts,  :( is in $\dfrac 16$ of them for a total of $3703 \dfrac 13$, and  :rotfl: is in $\dfrac {1}{11}$ of them for a total of $2020$. Therefore, the expected value is $2020$, as it is the least.[/hide]

biff....I said how many were likely to have all 3...I don't know what you did.

I don't really see what you mean as the expected value. I assumed that all of the ones with  :rotfl: had :( which all had :) .

What I mean by expected value of the number with all 3 is the probability that all three are in a post multiplied by the total number of posts.

[quote="mathnerd314"][quote="solafidefarms"]It has been found by analysis of 10 Games and Fun Factory posts that the most frequently used smilie is :) followed by :( and :rotfl: in that order. A post may have a maximum of 1 of each smile. :) occurs in 1/5th of the posts, :( occurs in 1/6th of the posts, and :rotfl: occurs in 1/11th of the posts. What is the expected value of the number of posts with all of those 3 emoticons if G&FF currently has 22220 posts? You may use :) and :( in a post; they are not mutually exclusive.

:) :( :rotfl:

Billy[/quote]

I'm confused, solafidefarms. First you said that you only looked at 10 posts. Then you said that 1/6 have :( and 1/11 have :rotfl: , neither of which go into 10. First you said that a post may have a maximum of 1 of each smile. What's a smile? Is a smile defined as just a :) or all three of :) :( :rotfl: ? If you can use both :) and :( in a post, can you also use both :) and :rotfl: or both :( and :rotfl: ? Thanks for the clarification.[/quote]

Hmmmm..........A smile is any smilie. Yes, you may use any combination: :), :(, :rotfl:, :) :(, :) :rotfl:, :( :rotfl:, or :) :( :rotfl:.

The 10-posts was an editorial comment; it does not mean I actually analyzed 10 posts. I'm sure though you could find 10 posts with a total of 10/11ths of a :rotfl: and 1 2/3rds of a :(.

Billy


induction
combinatorics unsolved
combinatorics

The jury of an Olympiad has $30$ members in the beginning. Each member of the jury thinks that some of his colleagues are competent, while all the others are not, and these opinions do not change. At the beginning of every session a voting takes place, and those members who are not competent in the opinion of more than one half of the voters are excluded from the jury for the rest of the olympiad. Prove that after at most $15$ sessions there will be no more exclusions. (Note that nobody votes about his own competence.)

Specially for Pascual :)

By induction: if we have $2n$ men, then thre are at most $n$ voitings.

Suppose the contrary. Let $a_1$, $a_2$, ..., $a_{n}$ be number of people excluded after 1st, 2nd, ..., 15th voiting respectively. Let $b_i$ be number of competences from non-excluded people after 1st, ..., (i-1)th voiting, which posses man excluded in $i$-th voiting.

If $a_1\geq 2$ then, obviously, we are done. So we will assume $a_1=1$.
Consider the man $A$, which was excluded in $i$-th voiting. Since he wasn't excluded in $i-1$th voiting we conclude \[b_i+a_{i-1}\geq \frac{2n-a_1-a_2-...-a_{i-2}}{2},\] but as he was excluded in $i$th voiting we can write \[b_i<\frac{2n-a_1-a_2-...-a_{i-1}}{2}.\]
It follows \[\frac{2n-a_1-a_2-...-a_{i-1}-1}{2}\geq b_i \geq \frac{2n-a_1-...-a_{i-2}-2a_{i-1}}{2}.\qquad \eqno (1)\]
It is clear there are $i$ s.t. $n\geq i-1\geq 2$ and $a_{i-1}=1$ (otherwise after $n$ votings there is at most 1 man in jury, and he cannot be excluded). Let $m>2$ be minimal such $i$. From $(1)$ we conclude $b_{m}=\frac{2n-a_1-a_2-...-a_{m-1}-1}{2}$, so $a_1+a_2+...+a_{m-2}$ is even. Due to choice of $m$ it means $a_1+...+a_{m-2}\geq 2(m-2)$. And, obviously, we are done.


can someone please draw the structure of $ \text{cis 1,3\minus{}dimethyl cyclohexane}$

Is your difficulty in drawing a structure for a substituted cyclohexane ring, or is it in knowing if both methyl groups will be equatorial, axial, or one equatorial and the other axial?

I believe that the methyl groups are placed on the first and third carbons of the six carbon single bonded ring.

EDIT: Here we go:

[quote="7h3.D3m0n.117"]I believe that the methyl groups are placed on the first and third carbons of the six carbon single bonded ring.[/quote]

I believe Pardesi already know that, since that's part of the information contained in the name. I think what Pardesi wants is this: both groups are equatorial. Look the following picture:

Ah I didn't see the "cis" part of the original post. Never mind me, I don't know what I'm talking about.

yse carcul that's what i wanted
but why is this so.
because in $ \text{cis 1,2 dimetyl cyclohexane}$ we have $ \text{ axial \minus{} equatorial}$ but here why did it change or more so generally
how do u know which is $ \text{cis}$ and which is $ \text{trans}$ in cyclic ocmpounds

[quote="Pardesi"]why did it change or more so generally how do u know which is cis and which is trans in cyclic ocmpounds[/quote]

Regarding cyclohexane rings, do the following:

Step 1 - draw the skeleton of a cyclohexane ring (without the equatorial-axial bonds) (see picture 1);

Step 2 - let's add the axial bonds (see picture 2): start in one of the "extreme" carbons (carbons a or d) and draw the axial bonds alternating up and down from carbon to the next carbon - for example, let's start in a: since this "extreme" carbon is pointing up, we draw its axial bond pointing also up, as shown; in next carbon, b, we draw it pointing down; and in carbons c, d, e, and f we draw the axials pointing up, down, up, down, respectively.

Step 3 - now let's add the equatorial bonds (see picture 3): starting again in carbon a, since the axial bond is already pointing up, then the equatorial bond must point down; now we draw in the next carbons alternating up and down as was done before - for carbons b, c, d, e, and f draw the equatorials respectively up, down, up, down, and up.

Step 4 - draw the groups in the cyclohexane ring in the respective carbons: if two groups are cis, then we must draw them in the bonds on their carbons that point both up or down; if they are trans, then draw one of them in a bond pointing down and the other in a bond pointing up (picture 4 exemplifies this with 1,2-dichlorocyclohexane).

Step 5 - in the remaining bonds put hydrogen atoms.

Remarks. 1) Take into account that eq-eq configurations are more stable than eq-ax configurations, and these are more stable then ax-ax configurations. 2) Note that the bonds are drawed in each carbon in the way shown because of the tetrahedral configuration of tetravalent carbon. 3) Click in the picture bellow to get a better one in a new window.

wow that was a good tutorial   :coool:  :omighty:
thanks a lot carcul

You are welcome. For cyclobutane and cyclopentane rings you can use a similar method (after drawing the most stable conformation of these rings).


calculus
integration
function
vector
real analysis
real analysis unsolved

Is $ C[0,1]$ under the uniform metric path connected (and therefore connected)?

I wrote something that shows it is polygonally connected and thus even locally connected. If I am right, I will have a nice application of Baire category theorem.

This is the one I'm stumped on: If $ A \equal{} \{ f \in C[0,1]|0 < f(x) < 1\}$is a subset of $ C[0,1]$ under the mean square deviation integral metric, what is the closure of A? I've thought about it for some weeks now but I don't see it having a 'nice' form. There are all sorts of functions in a neighborhood of a point of A.

Every topological vector space is path connected. Use the segment between the points as the path.

[quote="§outh§tar"]This is the one I'm stumped on: If $ A \equal{} \{ f \in C[0,1]|0 < f(x) < 1\}$is a subset of $ C[0,1]$ under the mean square deviation integral metric, what is the closure of A? [/quote]
By "mean square deviation integral metric" I'm guessing you mean the $ L^2$ norm. That is, $ d_2(f,g)\equal{}\|f\minus{}g\|_2$ where $ \|f\|_2\equal{}\left(\int_0^1|f(x)|^2\,dx\right)^{\frac12}.$

The set you describe has as its closure in $ C[0,1]$ in the uniform norm the set $ B\equal{}\{ f \in C[0,1]|0 \le f(x) \le 1\}$ It also has $ B$ as its closure [b]in[/b] $ \mathbf{C[0,1]}$ with respect to the $ 2$-norm.

On the other hand, if you were asking about its completion in the 2-norm, you're asking an entirely different question. The completion of $ C$ in the $ L^2$ norm is the space $ L^2[0,1],$ and you have to develop Lebesgue measure and integration to properly describe this set. (Its elements aren't even exactly functions - they're equivalence classes of functions, in which two functions are equivalent if they're equal almost everywhere.)

We can recognize the completion of $ A$ in the $ L^2$ norm as the closure of $ A$ in $ L^2,$ which is the set of functions $ f\in L^2$ such that $ f(x)\in[0,1]$ for almost every $ x.$

Hi Kent, I was referring to the closure in the 2 norm. I haven't studied Lebesgue theory yet.

I see that B is contained in the closure of A under the 2 norm, but how do we know there is no other continuous function in that closure?

Suppose there is a continuous function $ f$ in the 2-norm closure of $ A$ that is not in $ B.$ That means that there is some point $ x_0\in[0,1]$ for which either $ f(x_0)>1$ or $ f(x_0)<0.$ Suppose WLOG that $ f(x_0)>1.$ Then there is a $ \delta>0$ such that $ f(x)>1\plus{}\frac{f(x_0)\minus{}1}2$ for $ |x\minus{}x_0|<\delta.$ If $ g\in A$ then there is an interval of length at least $ \delta$ on which $ |f(x)\minus{}g(x)|\ge |f(x)\minus{}1|\ge\frac{f(x_0)\minus{}1}2.$

That gives us that $ \|f\minus{}g\|_2^2\ge\delta\left(\frac{f(x_0)\minus{}1}2\right)^2.$ Since this is a fixed positive number, we have that $ f$ is a positive distance away from $ A$ and is thus not in the closure of $ A.$


A screen covering the front of a fireplace has wires 1mm in diameter, spaced in a mesh that has 5mm square openings between the wires. A spart 2mm in diameter pops out of the fire and heads directly toward the screen. What's the probability that the spark misses the wires?

[hide="solution"]
Around each 5 mm opening, there is a "frame" of .5 mm wire.  This is repeated throughout the screen.  So it's really a 5x5 mm opening within a 6x6 space.  Since the spark is 2 mm in diameter, the center must be 1 mm away from the wire, forming a 4x4 square to represent the space that will allow passage.  So probability:

$\frac{4^2}{6^2} = \frac{16}{36} = \frac{4}{9}$[/hide]


Yes, it's out..

Being a Waterloo resident, I assume I'm the first one to get it!!
Interesting point: Ian's not running it. It's professor Koeller this time. Wonder what happened?

This is only the first round of invitations though... ie those based on PCF.  The ones based on Euclid and CMO and the other oddball ones aren't out yet.

I haven't got any invitations yet.  If you aced Fermat, theoretically you are in right? :?   Also, how do they notify you? by mail or email?  I have never been to the Waterloo camp before, how many people do they invite each year?

Edited: Never mind guys,
I am in

[quote="billzhao"]I got my invitation today.

This year, the seminar will take place one week right before my final exams (stupid school board delaying everything by a week), where as last year, it coincided with them. Well, good thing I don't have any science courses this semester.[/quote]

Didn't you IMO people get to go to another camp to prepare for internationals last year?

[quote="Elyot"]Didn't you IMO people get to go to another camp to prepare for internationals last year?[/quote]

That's after Waterloo Seminar.

What exactly is the date of the seminar?  (*crossing my fingers* - yes, I'm counting an egg that looks sort of infertile, but you never know, it might hatch.)

[quote="lightrhee"]If I got a gold in Fermat, I'm in, right? :?
How long will the invitation take to get to Edmonton?[/quote]

Couple days?

Course you're in, lol that's where they give the golds out!

[quote="lightrhee"]I am invited! I got the invitation at my school today. Now, I need to find a way to get there(most likely plane).[/quote]

I told ya, couple days.

Most likely plane eh?  No.....I thought you'd walk.   :D

[quote="blackegg2"]hey there everyone! i'm new to the board as i was searching for seminar related infomation after receiving my invitation today :P mm i guess i'm really really lucky that i got in this year! (because of fermat) good luck everyone!

btw this is irene, ^_^;; nice to meet all of you[/quote]
Nice to meet you,too Irene!
Same here, I got in by accident I think
just as I mentioned, I used to go to physics contests then one day wrote Fermat and get the invitation.
I'd like to meet those smart people there.
See you there :D


Chelsea has made 6 of 17 free throw attempts. How many
consecutive free throws must she make to raise her
percentage of free throws made to exactly 50$ \%$?

17-6=11.

she now failed 11 times.

if she wants her percentage to 50%, make : fail = 1:1.

fail = 11, so make should be 11, too.

thats total 22 times.

22-17=5.

answer : 5


function
floor function
inequalities
induction
limit
algebra
algebra unsolved

Prove that there is no function $ f: \mathbb{R^{ \plus{} }}\mapsto \mathbb{R^{ \plus{} }}$ such that for all $x,y\in\mathbb{R^{ \plus{} }}$,
\[f(x)^{2}\geq f(x \plus{} y)[f(x) \plus{} y].\]

[quote="cnyd"]Prove that there is no function, $ f: \mathbb{R^{ \plus{} }}\mapsto \mathbb{R^{ \plus{} }}$,

$ \forall x,y\in\mathbb{R^{ \plus{} }},f(x)^{2}\geq f(x \plus{} y)[f(x) \plus{} y]$[/quote]

Are brackets in $ [f(x) \plus{} y]$ just parenthesis "$ (f(x) \plus{} y)$" or the marks of floor function "$ \lfloor f(x) \plus{} y\rfloor$" ?

[quote="cnyd"]Prove that there is no function, $ f: \mathbb{R^{ \plus{} }}\mapsto \mathbb{R^{ \plus{} }}$,

$ \forall x,y\in\mathbb{R^{ \plus{} }},f(x)^{2}\geq f(x \plus{} y)[f(x) \plus{} y]$[/quote]

Let suppose that there exist a function $ f(x): \mathbb{R^{ \plus{} }}\mapsto \mathbb{R^{ \plus{} }}$ that satisfy the given condition.

Let $ x_0 > 0.$ $ f(x_0) > 0$ and $ \forall y\in \mathbb{R^ \plus{} }$ we should have
\[ y \le \frac {f(x_0)}{f(x_0 \plus{} y)} \left [f(x_0) \minus{} f(x_0 \plus{} y)\right] \;\;\;(*)\]

where $ f(x_0) \minus{} f(x_0 \plus{} y) > 0$ since that it must be $ f(x_0)f(x_0 \plus{} y) < f(x_0)^2.$

The inequality $ (*)$ is a contradiction. So there is no function $ f(x): \mathbb{R^{ \plus{} }}\mapsto \mathbb{R^{ \plus{} }}$ that verifies the given property.

[quote="prester"][quote="ocha"][quote="prester"]
where $ f(x_0) \minus{} f(x_0 \plus{} y) > 0$ since that it must be $ f(x_0)f(x_0 \plus{} y) < f(x_0)^2.$

This is a contradiction. [/quote]

sorry, I don't see the contradiction, can you please explain. thanks[/quote]

The contradiction is the inequality above $ y \le ...$[/quote]

I cannot delete my previous post.  However, [i]ocha[/i] is right. There is no contradiction since the right hand side of the inequality is a function of $ y$ and it is increasing. So the inequality is $ y \le g(y)$ $ \forall y \in \mathbb{R^\plus{}}$ and there is no contradiction... :(  :blush:

[hide="my solution"]

From the condition it is obvious that $ f$ is strictly decreasing, futhermore if we fix $ x$ and take the limit as $ y \to \infty$ then $ f(x) \to 0$

We have $ f(x)^{2}\geq f(x \plus{} y)(f(x) \plus{} y) \Longrightarrow f(x)\ge f(x \plus{} y) \plus{} \frac {1}{\frac {1}{f(x)} \plus{} \frac {1}{y}} \qquad(1)$

Also by AM-GM: $ f(x \plus{} y) \plus{} \frac {f(x) \plus{} y}{4} \ge \sqrt {f(x \plus{} y)(f(x) \plus{} y)} \equal{} f(x) \qquad(2)$

So $ f(x \plus{} y) \plus{} \frac {f(x) \plus{} y}{4} \ge f(x) \ge f(x \plus{} y) \plus{} \frac {1}{\frac {1}{f(x)} \plus{} \frac {1}{y}}$

If we set $ y \equal{} f(x)$ then we get $ f(x \plus{} f(x)) \equal{} \frac {1}{2}f(x) \qquad(*)$

If now set $ x\to x \plus{} f(x)$ in $ (*)$ we get $ f(x \plus{} \frac {3}{2}f(x)) \equal{} \frac {1}{4}f(x)$

By the same substitution we get by induction $ f(x \plus{} 2f(x) \minus{} \frac {1}{2^n}f(x)) \equal{} \frac {1}{2^n}f(x) \qquad (**)$

If we send $ n\to \infty$ in $ (**)$ then we get $ f(x \plus{} 2f(x)) \equal{} 0$

This is an obvious contradiction so we cannot have any functions that satisfy the condition  :)[/hide]

[i]Here is the solution I read in somewhere that I can't remember  :blush: :

  We have     $ \forall x,y\in\mathbb{R^{ \plus{} }},f(x)^{2}\geq f(x \plus{} y)[f(x) \plus{} y] (1)$

   From the hypothesis , we can deduce that $ f$ is a strictly decreasing function

   In $ (1) \rightarrow f(x \plus{} y) \leq \frac { f(x)^{2}}{f(x) \plus{} y}$

 Fix $ x$ , and let $ y$ come to infty , we can see easily that : $ \lim_{x \to \plus{} \infty } f(x) \ \equal{} \ 0$

    On the other hand , change $ y$ to $ f(x)$ , we have : $ f(x \plus{} f(x)) \le \frac {f(x)}{2}$

  For each positive real number $ x_0$ , construct the sequence $ (x_n)$ as follows :

   $ x_{n \plus{} 1} \equal{} x_n \plus{} f(x_n)$

  So , $ f(x_{n \plus{} 1}) \le \frac {f(x_n)}{2} \rightarrow f(x_{n}) \le \frac {f(x_0)}{2^n} \ \ \forall n \ \in \ \mathbb{N}$

   $ x_{n \plus{} 1} \ \equal{} \ x_0 \plus{} f(x_0) \plus{} f(x_1) \plus{} .... \plus{} f(x_n) \le x_0 \plus{} f(x_0) \left( 1 \plus{} \frac {1}{2} \plus{} ... \plus{} \frac {1}{2^n} \right) < x_0 \plus{} 2f(x_0)$

    Thus , the  sequence $ (x_n)$  is up  bounded by the constent $ m \ \equal{} \ x_0 \plus{} 2f(x_0)$

   So,    $ 0 < f(m) \le f(x_n) \le \frac {f(x_0)}{2^n} \ \ \forall n \ \in \ \mathbb{N}$

  But , $ \lim_{n \to \plus{} \infty} \frac {f(x_0)}{2^n} \equal{} 0$ , contradiction :wink:

       We conclude that there in no function $ f$ satisfies the condition [/i]

[quote="pco"][quote="ocha"] ... $ f(x \plus{} 2f(x) \minus{} \frac {1}{2^n}f(x)) \equal{} \frac {1}{2^n}f(x) \qquad (**)$

If we send $ n\to \infty$ in $ (**)$ then we get $ f(x \plus{} 2f(x)) \equal{} 0$ [/quote]

You need continuity to get this conclusion.
And we dont have ...  :([/quote]

[b]EDIT: [/b] [i] on second thought wouldn't monotonicity be enough the get the conclusion?[/i]

Thanks PCO. Can it still be fixed though? Say we take $ x_0$ as a constant, then we still get

$ f(x_0 \plus{} 2f(x_0) \minus{} \frac {1}{2^n}f(x_0)) \equal{} \frac {1}{2^n}f(x_0)$

So we can send $ n\to \infty$ and get $ f(x_0 \plus{} 2f(x_0)) \equal{} 0$

If we let $ x \equal{} x_0 \plus{} 2f(x_0)$ in our original inequality we get

$ 0 \equal{} f(x_0 \plus{} 2f(x_0))^2 \ge yf(x_0 \plus{} 2f(x_0) \plus{} y)$

Since $ y\not \equal{} 0$ we have $ f(x_0 \plus{} 2f(x_0) \plus{} y) \equal{} 0 \forall y \in \mathbb{R}$ which is absurd


calculus
integration
limit
calculus computations

Find:

$ \lim_{n\to\infty} \sum_{k = 1}^{n} \frac {k + n}{n^{2}}e^{\frac {k}{n}}$

offfff.... sorry - wrong division of forum - it should be posted in unsolved problems - sorry again - but please help me anyway;

$ \lim_{n\to\infty} \sum_{k = 1}^{n} \frac {k + n}{n^{2}}e^{\frac {k}{n}}$

This can be written as

$ \lim_{n\to\infty} \sum_{k = 1}^{n} \frac {k + n}{n}e^{\frac {k}{n}} \frac{1}{n}$

= $ \lim_{n\to\infty} \sum_{k = 1}^{n} (\frac {k}{n} + 1) e^{\frac {k}{n}} \frac{1}{n}$

= $ \int_0^1 e^x (1+x) \ dx$

hope you can proceed now...


Solve the inequality $ x\cdot [x]\leq\sqrt{2008\minus{}x^2}.$
(here $ [x]$ denotes the integer part of $ x$.)

x^2 <=2008/(1+[x]^2)

By brute force, the answer is
-7<x<7


function
induction
algebra unsolved
algebra

Prove that $ \forall n \in \mathbb{N}$   $ \exists k\in \mathbb{N}$ so that $ \left(1\plus{}\sqrt{2}\right)^n\equal{}\sqrt{k}\plus{}\sqrt{k\plus{}1}$.

Notice that the function $ y \equal{} \sqrt {x} \plus{} \sqrt {x \plus{} 1}$ is one to one for $ y > \equal{} 1$ and $ x > \equal{} 0$
If $ y \equal{} \left(1 \plus{} \sqrt {2}\right)^n$, $ y > \equal{} 1$. Hence there is always a $ k\in\mathbb{R}$ that satisfies the equation.
Now we proceed by contradiction: assume $ k\not\in\mathbb{N}$ for some $ k$.
Inverting each side, we find $ \left(1 \minus{} \sqrt {2}\right)^n \equal{} \sqrt {k} \minus{} \sqrt {k \plus{} 1}$. Hence $ k \equal{} \frac {\left(3 \plus{} 2\sqrt {2}\right)^n \plus{} \left(3 \minus{} 2\sqrt {2}\right)^n\plus{}2}{4}$. This is clearly an integer from the recurrence relation; hence we have proven contradiction, and the result follows.

It must is $ (\sqrt{2}\minus{}1)^n\equal{}\sqrt{k\plus{}1}\minus{}\sqrt{k}$, i think.

$ \left(1\plus{}\sqrt{2}\right)^n\equal{}a\plus{}b\sqrt{2}$ where $ a,b\in\mathbb{N}$
$ a\plus{}b\sqrt{2}\equal{}\sqrt{k}\plus{}\sqrt{k\plus{}1}\equal{}\sqrt{a^2}\plus{}\sqrt{2b^2}$
So if $ |a^2\minus{}2b^2|\equal{}1$ we are done. (And it is true :lol: )
This can be proved by induction:
If $ n\equal{}1$, then $ a\equal{}1$, $ b\equal{}1$ so $ |a^2\minus{}2b^2|\equal{}1$
Assume that for $ n$ it is true. Prove for $ n\plus{}1$:
$ (a\plus{}b\sqrt{2})(1\plus{}\sqrt{2})\equal{}(a\plus{}2b)\plus{}(a\plus{}b)\sqrt{2}$
$ (a\plus{}2b)^2\equal{}a^2\plus{}4b^2\plus{}4ab$
$ 2(a\plus{}b)^2\equal{}2a^2\plus{}2b^2\plus{}4ab$
The difference is: $ a^2\minus{}2b^2$ which is obviously $ \pm 1$ by the condition.
So it is true for all $ n$.


AMC
AIME
Harvard
college
MIT
trigonometry
complex numbers

Never mind. *blinks* Time to get back to work.

Tare wrote:mathfanatic wrote:Source: Harvard-MIT Math Tourney - Team 2001

Express (sin10 + sin20 + sin30 + sin40 + sin50 + sin60 + sin70 + sin80)/(cos5*cos10*cos20) without using trig functions.
[hide]10+20+30+40+50+60+70+80=360. sin 360=0, so 0/cos5*cos10*cos30=0[/hide]
I just learned Trig, so I'm not quite sure...

No.  You can't take things out of the sine, then add them, then bring them back in.  It's like this:

1^2 + 2^2 + 3^2 + 4^2 ≠ (1+2+3+4)^2

Since 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30, whereas (1+2+3+4)^2 = 10^2 = 100

Oh, yeah.

You have to use that really long way to do it...

[hide]sin10+sin20=2sin15*cos-5[/hide] and on and on and on...

*sigh* Is there an easier way to do this?

I would like to point out that this problem is quite difficult and involves some rather obscure identities. See if you can find an identity for sin(a)+sin(b) and cos(a)+cos(b). Alternatively, you can consider the sines as the imaginary part of e^i*theta, which leads to a lot of messy algebra. See if you can use either of these ideas on this problem.

Never mind, I'm not [b]that[/b] into Trigonometry yet...

[quote]Source: Harvard-MIT Math Tourney - Team 2001

Express (sin10 + sin20 + sin30 + sin40 + sin50 + sin60 + sin70 + sin80)/(cos5*cos10*cos20) without using trig functions.[/quote]

Well:

sin10 + sin20 = 2*sin15*cos(-5)
sin30 + sin40 = 2*sin35*cos(-5)
sin50 + sin60 = 2*sin55*cos(-5)
sin70 + sin80 = 2*sin75*cos(-5)

Since cos(-x) = cos x, we can rewrite all the cos(-5)s as cos(5)

The sum of these is 2*cos5*(sin15+sin35+sin55+sin75)
= 2*cos5*[(sin15+sin75)+(sin35+sin55)]
= 2*cos5*[2sin90*cos(-30)+2sin90*cos(-10)]
= 2*cos5*[2*1*:rt3:/2 + 2*1*cos10]
= 2*cos5*[:rt3: + 2cos10]

Putting this over cos5*cos10*cos20 gives:
2*cos5*[:rt3: + 2cos10]/(cos5*cos10*cos20)
= (2:rt3: + 4cos10)/(cos10*cos20)

I'm not sure how to simplify this any more, other than the cos10...any ideas?

Combine sin(10) and sin(80), etc. at the beginning. See where that gets you.

Oh, right, okay..I see...

Quote:Source: Harvard-MIT Math Tourney - Team 2001

Express (sin10 + sin20 + sin30 + sin40 + sin50 + sin60 + sin70 + sin80)/(cos5*cos10*cos20) without using trig functions.

Spoiler:

[hide]sin10 + sin80 = 2*sin45*cos(-35)

sin20 + sin70 = 2*sin45*cos(-25)

sin30 + sin60 = 2*sin45*cos(-15)

sin40 + sin50 = 2*sin45*cos(-5)

Adding gives:

2*sin45*(cos5+cos15+cos25+cos35)

= 2*:rt2:/2*(cos5+cos15+cos25+cos35)

= :rt2:(2cos20*cos(-15)+2cos20*cos(-5))

= :rt2:(2cos20(cos5+cos15))

= :rt2:(2cos20(2cos10*cos(-5))

= 4*:rt2:*cos20*cos10*cos5

Putting this over cos20*cos10*cos5 produces 4*:rt2: Answer.[/hide]

[quote="ComplexZeta"]I would like to point out that this problem is quite difficult and involves some rather obscure identities. [/quote]

Are those identities really obscure? I tend to use them alot...

Well, they aren't extremely obscure, but I have to derive them every time I use them. I don't think they're usually taught in trig classes.

[quote="ComplexZeta"] I don't think they're usually taught in trig classes.[/quote]

Definitely not in mine.  When I saw the problem, I had to look at my big [i]CRC Math Tables & Formulae[/i] book to find those identities...

But are they usually taught in problem solving classes?

Some trig classes do teach them...or at least they're in the textbooks...

If you know the identities for sin(A  :pm:  B) and cos(A :pm: B), then you know the identities some of you used in this solution.  For example:

sin 40 + sin 50 = sin(45 - 5) + sin(45 + 5) =
sin 45 cos 5 - sin 5 cos 45 + sin 45 cos 5 + cos 45 sin 5 = 2 sin 45 cos 5.

Notice we strategically pick our 45 and 5 in the relatively obvious way: one is right between 40 & 50 and the other is the common difference from this midpoint to the other two.  We think to do this because we know we'll generate a couple terms that will cancel (the sin 5 cos 45 terms in our expansion).

I used to have those formulas memorized, too.  And would always forget them, get signs wrong, etc.  Would spend time trying to guess what they were supposed to be.  Then I realized the above, and the formulas are clear results of the addition/subtraction formulas of sine and cosine.


Since my calculator sucks, I only use it to do calculations (And sometimes graphing), I know some people have a couple they use, (ie. one for graphing, one for algebra solver, etc.)

Edit: Can a mod add the option "I use everything!" ? It didn't show up =\

I have a ti-84 plus silver, and I use it for graphing/calculations. It doesnt have a good algebra solver :(

I use a TI-84+ Silver, and I sometimes use the graphing part (but not often) And the only algebra solver it has is quadratics, I entered in a value for X, and it doesn't work anymore.   :(

Calculators are too complicated for a simple mind like me  :P .

I mostly use calculations, but sometimes table things, I just discovered the table function this year XD.

I have an TI-84+ SE, and I use it for programs more than actual calculations. :D
Most of my programs aren't even to do math (i.e. stupid counting programs, games, fun ways to calculate $\pi$, stuff like that, and a few do math).

I only use a calculator for neccessary calculations during tests and contests.  If you use your calculator too much, then you will become dependent.  However, I do frequently use my calculator to create programs outside of competitions (like games).  I have a TI-84+ Silver Edition.

i barely use my ti-84 + silver, cuz its WAY too slow. i only use a calculator when it makes me go faster.
next year it wont matter anyways. we can't use calculators on any mao test. :)

If you use a graphing calculator extensively, how did you learn how to use it (well)?  Were you taught in a math course, or did you learn on your own, or what?  It seems like it would take some training to be able to use a calculator like a TI-84 effectively for Mathcounts.

It isn't that hard to figure out how to use a graphing calculator (well). I got one last year at state, and I spent countless days playing with it. After a while you can pretty much figure out everything, and now I can use mine quite well.

I consider it good practice to use it only for necessary calculations. I can do all the rest. :)


CONGRATULATIONS SUPER!!

i want to congratulate you but im banned for 10 minutes for flooding...  :rotfl:

btw, YEA, WHATS A LLAMEL??

Congrats, Super!!! Nice rating!!!

And what [b]is[/b] a llamel?

thanks everyone

What a llamel is is highly classified information, that I shall never reveal. I'm secretive like that. Sort of like a ninja..  :ninja:  :ninja:  :ninja:

i thought a llamel was just lame with 2 fancy l's around it or absolute values idk which one =]

There are actually two meanings, and someone got part of one.  :o

but in super's avatar...

isnt it a llama? (...or a camel)

oooooo maybe piggy is onto something, inferring from what super said,

or maybe super is just talking about the camel/llamel thing.

and btw CONGRATS, SUPER!!!!!!!!!!!!!!!!!!!!!! on 1560+.

We should all support superllamel and help him get to 1600+!

1563 so far.                                :oops:   Congrats!

yay 1571!!

o and was i kind right about the lame thing? or was it the camel/llama thing? :D

Well, in his user name, there is no | as an absolute value symbol and I as a capital i. So I think we got his leg at the camel/lamell thing. :D

Yeah, this rating system is less moderate than the other one.

well, at first i was $ 1458\minus{}1298\equal{}160$
but now i am $ 1337\minus{}1458\equal{}{\infty}$

recently, kl made me 1330. :(

From my current at the time of the rating change to my new highest rating, 267 points. My current rating at this moment is inaccurate, due to a very bad day yesterday.

[quote="Dojo"]bobbymathking123 1661[/quote]
... :| [size=0].[/size]

1=2, that's definitely spam...

anyway, i'm averaging around 1600, but i'm advancing on 1700 now!
so, i'd say i moved up about 300
yay! :winner_first:

I shifted around 200 points also by playing 20+ games per day......

If i play like johnny (i am pretty sure) everyday (around 60 games) i should soon make 1600s (with some m3m advantage; i can finally mem (even though it is a bad thing to do))

FTW helped me gain speed  (after MC was over :| ) i used to be 5 seconds fast, now without memming, i get to 2.3 seconds ( i will never catch up to mewto...... darn)

current rating: 1495

[quote="kl2836"]it used to be godly to get to 1500, much less 1600. after this new rating its pretty easy to get to 1600. when i used to 1550, i won two cds against 1430 person and had no rating change. now, with a 1600 rating, when i beat a 1250 i get like 5 rating.[/quote]
what do you mean by "5 rating"?

[quote="Mewto55555"]congrats superllamel on your 1553 rating! Now tell us, whats a llamel? :?:[/quote]

I [i]hate[/i] to revive threads, but I have some pertinent info.

http://www.urbandictionary.com/define.php?term=llamel

It's a cross between a camel and a llama. They are very cool, however they enjoy spitting alot. They like to express themselves through the art of dance. Llamels can be found roaming the mirc channels.

So, the other day, I was at a bbq, and this llamel was eating all teh ribs!

wow. first of all, way to revive.
second of all, thats HILARIOUS. :rotfl:
EDIT: pointless revive. just read through the rest of this thread...that was stated (inferred) like 493029492039 times.

Except someone else already mentioned this and superllamel said that it wasn't his definition.

Dang......  as I read this I think, ppl like Superllamel were the highest raters with a rating of 1553 last year...
This is amazing...

my highest right now (1608)


A Mersenne prime is a prime number that can be written in the form $ 2^p\minus{}1$, where $ p$ is a prime number. Find the smallest Mersenne prime that is greater than 100.

$ 2^{p}\minus{}1>100$
$ 2^{p}>101$
$ p>\left \lfloor \frac{\log101}{\log2} \right \rfloor\equal{}6$

Since $ 2^{7}\minus{}1\equal{}127$ is prime, $ \boxed{127}$ is the first Mersenne prime $ >100$.

All the logarithm stuff is awesome but one could just find that $ 2^7 \minus{} 1$ is greater than $ 100$ and $ 2^5 \minus{} 1$ the next prime power of $ 2$ down is less than $ 100$.

How do you know off the top of your head that 127 is a prime?

Check if any primes that are less than the $\sqrt{127}$ are divisible by $127$. Or just memorize your primes(if u want).

You can just assume that 127 is a prime, using the square root trick ( Take a number ( n ), and then find out that all numbers less than sqrt(n)  are not divisible to then tell you if it prime. This is because this trick eliminates all factors efficiently )

You may also just find the first number / power of 2 -1 bigger than 100 if you want to do quickly, and do it quickly if you know your powers of two. This is because it works out nicely.


those brackets don't mean floor do they?


Find all positive integers with two or more digits such that if we insert a $0$ between the units and tens digits we get a multiple of the original number.

[hide]CAll the original number 10x+y
The transformed number is 100x+y, where y is a digit.
Therefore, k(10x+y)=100x+y.
k cannot be greater than 10.
k=10, 100x+10y=100x+y or y=0.  Therefore, for any multiple of 10, this is true.
k=9, 90x+9y=100x+y or 8y=10x.  (4,5) is a solution or 45.
k=8, 80x+8y=100x+y or 7y=20x which has no solution besides (0,0) where y is a digit.
k=7, 70x+7y=100x+y or y=5x which has the solution (1,5) or 15.
k=6, 60x+6y=100x+y or y=8x which has the solution (1,8) or 18.
k=5, 50x+5y=100x+y or 4y=50x which no longer has any solutions for digit y.
Thus the solutions are all multiples of 10, 45, 15, and 18.[/hide]


How to prove (mathmatically) that choosing an even number of elements from n elements is the same as the number of options to choose an odd number of elements out of n (>0).

Thanx

The even case is $S_e = \sum C_n^{2k}$ whereas the odd case is $S_o = \sum C_n^{2k+1}.$
Thus $S_e - S_o = \sum (-1)^kC_n^k = (1-1)^n = 0$ from the binomial theorem.

Pierre.

you can see this combinatorially as follows:
if $n$ is odd then choosing an even set of $\{1,2,\ldots,n\}$ is the same as rejecting an odd set,so the number of odd sets and number of even sets is the same.
suppose $n$ is even.fix an element $x$ then any set either contains $x$ or it doesn't. if the chosen set contains $x$ and is even,then we are basically choosing an odd set from $\{1,2,\ldots,n\}\setminus\{x\}$,thus the number of odd sets containing $x$ is the same as the number of even sets containing $x$.
likewise if the chosen sets don't contain $x$ then again the number of such odd sets is the same as the number of such even sets. and so we are through.


inequalities
quadratics
number theory unsolved
number theory

Solve in integers $x$ and $y$ the equation $x^3-y^3=2xy+8$.

if you substitute $a=x-y$ and $b=xy$ then $b=\frac{a^3-8}{2-3a}$ and from $3a-2 | a^3-8$ and $3a\equiv 2 \mod 3a-2$ we get $2^4 \cdot 13 \equiv 0 \mod 3a-2$ and it all comes to cases banging which I left to the reader  :P

Multipying the equation by $27$ and using that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$, it follows that

\[ (3x)^3+(-3y)^3+(-2)^3-54xy=8\cdot 27-8  \]

or

\[ (3x-3y-2)(9x^2+9y^2+4+6x-6y+9xy)=208=2^4\cdot 13  \]

and we've arrived to the cases Megus said we should consider.

i have a different approach:

we have $x>y$ but if $x\geq y+2$ then $2xy+8\geq 6y^{2}+12y+8\rightarrow x>3y$. from our initial equation we have $y^{3}+8=x(x^{2}-2y)>3y.(9y^{2}-2y)=27y^{3}-18y$ which is really absurd cuz $y\geq 1$. so $x=y+1$ . ...

Case1:x=0 or y=0, solution (0,-2) and (2,0)
Case 2: x>0, y>0
Then x>y, we have
$ x^3 \minus{} y^3 \equal{} (x \minus{} y)(x^2 \plus{} xy \plus{} y^2) \equal{} 2xy \plus{} 8 \ge x^2 \plus{} xy \plus{} y^2 \ge 3xy$, so $ xy \le 8$. Since we cannot have $ x\ge 3$, $ y \ge 3$, we only need to test out y=1,2, no solution.
Case 3: x<0, y<0
Then let x=-p, y=-q, we get case 2
Case 4: x>0, y<0
Then let y=-q, we have $ x^3 \plus{} q^3 \equal{} 8 \minus{} 2xq$, so $ x \le 2$, $ q \le 2$, no solution
Case 5: x<0, y>0
Let x=-p, again we have $ p^3 \plus{} y^3 \equal{} 8 \minus{} 2py$, so $ p \le 2$, $ y\le 2$ , no solution.

[quote="Xixas"]Solve in integers $ x$ and $ y$ the equation $ x^3 \minus{} y^3 \equal{} 2xy \plus{} 8$.[/quote]

we have that $ (x\minus{}y)(x^2\plus{}xy\plus{}y^2)\equal{}2(xy\plus{}4)$ if $ x\minus{}y\geq{2}$, we have that $ 4\geq{x^2\plus{}y^2}$,
which, when testing, gives us $ (2,0),(0,\minus{}2)$. Notice that $ x^2\plus{}xy\plus{}y^2$ is always positive.So if
$ (x\minus{}y)$ is negative than we would have $ xy\leq{\minus{}4}$, absurd ,just break into cases: $ x\minus{}y$ positive and $ x\minus{}y$ negative and then just use inequalities:

1:$ x\minus{}y\geq{1}$
                                      $ (x\minus{}y)(x^2\plus{}xy\plus{}y^2)\equal{}2(xy\plus{}4)$

 LHS positive ,so  we need only consider $ x\minus{}y\equal{}1$   ( $ x\minus{}y\geq{2}$ has already been done.)
 to do this case , just note that we have that $ (x\minus{}y)^2\plus{}x^2\plus{}y^2\equal{}8$ which is absurd.

2:$ x\minus{}y\leq{\minus{}1}$
                          $ (x\minus{}y)(x^2\plus{}xy\plus{}y^2)\equal{}2(xy\plus{}4)$

LHS is negative, so $ xy\leq{\minus{}4}$
if $ x\minus{}y\leq{\minus{}2}$, we´ll have $ \minus{}2(x\plus{}y)^2\geq8$,absurd.
so we need only consider $ x\minus{}y\equal{}\minus{}1$, which leads to $ \minus{}(x\plus{}y)^2\equal{}xy\plus{}8$, and then $ 3y^2\minus{}5y\equal{}7$ , since $ x\minus{}y\equal{}\minus{}1$.Easy to check that it is absurd.

So the only possible solutions are $ (2,0)$ and $ (\minus{}2,0)$ :wink:

Denote $ x\equal{}k\plus{}y$. Then solve as quadratic toward $ y$ and we get $ D\equal{}\minus{}\left(3k\minus{}2\right)\left(k^{3}\plus{}2k^{2}\minus{}32\right)$ which is very esy to check when is positive.

[quote=scorpius119][quote="zweig"]If $x = y$, we have no solution in Z, then let us consider $x \neq y$.

Setting $d = x-y$, $p = xy$, we have

$d(d^{2}+3p) = 2(4+p)$

From this relation, we have these 2 systems to solve
\[(1)\left\{\begin{array}{rcl}d&=&2\\ d^{2}+3p&=&4+p \end{array}\right. \]

\[(2)\left\{\begin{array}{rcl}d^{2}+3p&=&2\\ d&=&4+p \end{array}\right. \]
(1) gives $d = 2$, $p = 0$, so $(x,y)=(2,0)$

(2) gives no solution in Z.

Finaly, only $(x,y) = (2,0)$ is solution

-------------------------

Is it correct ?  :oops:[/quote]

Unfortunately, we're not given statements like $d$, $d^{2}+3p$, or $4+p$ prime. The factors might break up another way.

[hide]
So letting $a=x-y$ and $b=xy$, we have $a(a^{2}+3b)=2b+8$, or $b(3a-2)=8-a^{3}$. Isolating $b$, we get
\[b=\frac{8-a^{3}}{3a-2}\]
(We know that $3a-2$ can't possibly be equal to 0.) Now note that $a^{2}+4b=(x+y)^{2}$ which is a perfect square, so
\[\frac{32-2a^{2}-a^{3}}{3a-2}\]
is a perfect square. In particular, it's nonnegative! If $a\geq 3$, then $3a-2>0$ and $32-2a^{2}-a^{3}\leq 32-5a^{2}<0$ which is bad. If $a\leq 0$, then $3a-2<0$ and $32-2a^{2}-a^{3}>0$, which is also bad! (To see this, the cases $a=0,-1$ are obvious, while $a\leq-2$ implies $-2a^{2}-a^{3}\geq 0$.)

That only leaves $a=1$ and $a=2$. If $a=1$, we get $b=7$ and $(x+y)^{2}=29$ which is not a perfect square. This means $a$ must equal 2. Now we get $b=0$ so the only solutions are $\boxed{(2,0),\ (0,-2)}$. Both work.
[/hide][/quote]

good solution

Let $x = \frac{u+v}{2}, y = \frac{u-v}{2}$ where $u,v$ are integers.
Then $\left(\frac{u+v}{2}\right)^3-\left(\frac{u-v}{2}\right)^3 = 2\left(\frac{u+v}{2}\right)\left(\frac{u-v}{2}\right)+8$ which is equivalent to
\begin{align*}
u^2(3v-2) &= -v^3+2v^2+32 \\
u^2 &= \frac{-v^3+2v^2+32}{3v-2}
\end{align*}
Clearly, $\frac{-v^3+2v^2+32}{3v-2} \geq 0$ only when $v = 1,2$.
Checking, $v = 2$ produces $u = \pm 2$.
Thus the solutions are $(2,0)$ and $(0,-2)$.

If $x-y=D$ then we have:

$x^3-(x-D)^3-2x(x-D)-8=0\implies(3D-2)x^2-(3D^2-2D)x+(D^3-8)=0\implies x=\frac{D}{2}\pm\frac{\sqrt{-3D^4-4D^3+4D^2+96D-64}}{2(3D-2)}$

\[-3D^4-4D^3+4D^2+96D-64\geq 0\implies (3D-2)(D^3+2D^2-32)\leq 0\implies 0.\bar{6}=\frac{2}{3}\leq D\leq\frac{-2+2 \sqrt[3]{53-6 \sqrt{78}}+2 \sqrt[3]{53+6\sqrt{78}}}{3}\approx 2.62919\]

Although we don't actually need to solve the cubic in closed form. Instead, we can easily show the cubic only has one real root and then show it is n the $(2,3)$ interval using IVT.

Therefore, since we need $D\in\mathbb{Z}$, we are forced into $D\in\{1,2\}$.

If $D=1$ then $x=\frac{1\pm\sqrt{29}}{2}$, which is no good.

If $D=2$, then $x\in\{0,2\}$ and so we get the pairs $(0,-2)$ and $(2,0)$.


Solve the inequality

  x(x+1)(x+2)(x+3) \geq 9/16

You are right. It is an inequation and not an inequality.

This is easy when you notice that x(x + 1)(x + 2)(x + 3) = (x2 + 3x + 1)2 - 1.

Then you have to consider the cases

(i) x2 + 3x + 1 \geq 5/4
(i) x2 + 3x + 1 \leq -5/4

which is straightforward.

Here is an idea (I was to lazy to do the computations)
  We denote x 2 +3x=t and we have:
  t 2 +2t-9/16 \geq 0 and from here it's easy :)


For any natural number $n, 19<n<27$, how many of the number $484_{n}$ is a perfect square in base 10?

I have a feeling that this is all going to go terribly wrong, but here's my best shot...

[hide]
$484_n$ can be written as $4n^2+8n+4$. Since we're looking for squares in base-10, let $4n^2+8n+4=k^2$.

If we factor the left side, we have $4n^2+8n+4=(2n+2)(2n+2)$, which is essentially a perfect square. So, since $19 < n < 27$, there are 7 choices of $n$.[/hide]

[hide]the number $484_n$ is always a perfect squre in base $10$, so it will work for all natural numbers in the inerval $(19,27)$, the answer is $7$. [/hide]

So my answer didn't go all horribly wrong!!!  :lol:

Btw, amirhtlusa, many users have said this before. Please hide your answers.

i remember that problem. except it was like $121_{n}$, which was even easier to figure out.

[quote="white_horse_king88"]There was a problem almost identical to this on the 2003 National Countdown on ESPN.

I remember the contestant getting it almost immediately.

I also remember Mr. Rusczyk explaining it to the newcaster who was like, "Um, sure. Ok - I'll pretend I understood that."[/quote]

lol. Yeah...for some reason I didn't trust myself when I was answering the Q. I dunno why.

amirhtlusa, beautiful post! ;)


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Let $a$, $b$, $c$ be positive integers such that $g.c.d.(a,b,c)=1$, and let $k$ be any nonzero integer. Prove that there exists integers $x$, $y$, such that $g.c.d.(x,y)=1$ and $g.c.d.(ax^2+bxy+cy^2, k)=1$.

Assume there is some prime $p$ s.t. $p|ax^2+bxy+cy^2,\ \forall x,y$. We immediately get a contradiction if $p>2$ (because we can fix $x\ne 0$ in $\mathbb Z_p$ and regard that as an identically null polynomial in $y$; it has more roots than its degree, so $a=b=c=0$ in $\mathbb Z_p$, meaning that $p|(a,b,c)=1$, false). We must thus check the case $p=2$, and this can be checked "manually" :).

I'm sorry, but I don't see how the original problem implies that there exists a $p$ that divides $f(x,y)=ax^2+bxy+cy^2$ for all $x$, and $y$. Since $k$ can be a product of many primes, and $gcd(f(x,y), k)$ can be a different prime each time.

For each $p_i$ in the composition of $k$ we can find $(x_{p_i},y_{p_i})$ s.t. $f(x_{p_i},y_{p_i})$ is not divisible by $p_i$.  Using Chinese residue theorem we will find $(x,y)$ s.t. $x\equiv x_{p_i}\pmod{p_i}$ and $y\equiv y_{p_i}\pmod{p_i}$. Then $f(x,y)$ is not divisible by $p_i$ for all $i$.


A cuboid has volume 42, surface area 80 and the sum of its edge lengths is 46. What are the dimensions of the cuboid and how is this linked to the roots of cubic equations?

Let $ x,y,z$ be the side lengths of the cuboid. Then we have

$ xyz\equal{}42$
$ 2(xy\plus{}yz\plus{}zx)\equal{}80$
$ 4(x\plus{}y\plus{}z)\equal{}46$

Hence by Vieta's formulas, x,y,z are the roots of $ r^3\minus{}23r^2/2\plus{}40r\minus{}42$.


Let x,y,z,p>0 and $ \frac{x}{y} \le \frac{y}{z} \le 1$ Prove that:
(p+1)([tex]\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x} ) \ge \frac{x}{z}+ \frac{y}{x}+ \frac{z}{y} + $ \frac{3p(a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}$

Latexed;

Let $ x,y,z,p > 0$ and $ \frac {x}{y} \le \frac {y}{z} \le 1$ Prove that:
$ (p \plus{} 1)\left(\frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \right) \ge \frac {x}{z} \plus{} \frac {y}{x} \plus{} \frac {z}{y} \plus{} \frac {3p(x^2 \plus{} y^2 \plus{} z^2)}{xy \plus{} yz \plus{} zx}$

[hide]
The restriction is the same as $ x\le y\le z$ so

The first part of the inequality is
$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge \frac {x}{z} \plus{} \frac {y}{x} \plus{} \frac {z}{y}$

Multiply through $ xyz$ and subtract to get
$ xy(y \minus{} x) \plus{} yz(z \minus{} y) \plus{} zx(x \minus{} z) \ge 0 \Longleftrightarrow (x \minus{} y)(y \minus{} z)(z \minus{} x) \ge 0$
This is true since $ x\le y\le z$

Now we must prove
$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge \frac {3(x^2 \plus{} y^2 \plus{} z^2)}{xy \plus{} xz \plus{} xz}$

This is stronger than cauchy, but if we multiply through $ xy \plus{} yz \plus{} zx$ we get

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} \sum x^2(\frac {y}{z} \plus{} \frac {z}{x}) \ge 3(x^2 \plus{} y^2 \plus{} z^2)$

From AM-GM $ \frac {y}{z} \plus{} \frac {z}{x} \ge 2$, so this is true

And this completes the proof  :lol:

[\hide]


Find the smallest value of $ a^{2}\minus{}6a\plus{}5$

Turning point is $ x\equal{}\frac{\minus{}b}{2a}\equal{}\frac{6}{2}\equal{}3$ so at $ x\equal{}3$ is the minimum, which is $ 3^{2}\minus{}6(3)\plus{}5\equal{}9\minus{}18\plus{}5\equal{}\minus{}4$.

Ok . That's it           :)

My solution:

$ a^{2} \plus{} 6a \plus{} 5\Leftrightarrow (a \plus{} 3)^{2} \minus{} 4$ for $ a \equal{} \minus{} 3$ the smallest value is $ \minus{} 4$


We have a circle and 1991 points around it numbered from 1 to 1991. Starting from 1, we delete every 2'nd and 3'rd point (for example, the first 4 points to be deleted are 2,3,5,6). We repeat this until there is only 1 point left. What's the label of that point?

should be 1894 if I counted correctly

(not gonna redo it if I'm wrong :D)

Same here. I was very nervous aboput this: I thought you guys would find something else and prove me wrong :D.

Ok, Peter(VDD)...since YOU first give the result, YOU have to write a complete solution  :D , in order to we may move this topic into the solved section....

Pierre.

Pfff....since we cannot trust Peter, I will write a complete proof before moving it...

Let the origin be between 1 and 1991. A round is supposed to start from the origin and end at the origin.
So at the first round we only keep the number of the form  1 + 3a, where a = 0,...,663.
But, the second round start with erasing 1 (that is a = 0), and keeping a = 1.
Thus, we keep the a of the form a = 1+3b, where b = 0,1,...,220.
Round n3 start with keeping b = 0, and then we keep only the numbers of the form b = 3c, where c = 0,...,73.
Round n4 start we erasing c = 0 and keeping c = 1. Thus we keep the numbers of the form c = 1 + 3d, where d = 0,...,24.
Round n5 start with erasing d = 0 and d = 1. Thus we keep the numbers of the form  d= 2 + 3e, where e = 0,...,7.
Round n6 start with erasing e = 0 and keeping e = 1. Direct checking shows that we finish with e = 7.
Thus, the last number is x = 193 + 243e = 1894.

Pierre.

heh sorry, i didn't get back to this one :D

(arne just linked me here)

sorry for the lack of response, I didn't notice the reply then :?


function
vector
induction
real analysis
real analysis unsolved

a)Find all functions $f\in  C^0(R^2,R)$ which satisfy the following property:for every $u \perp v,f(u+v)=f(u)+f(v)$
b)(unsolved)Find all functions $f$ from $Z^2$ to $R$ which satisfy the same property.

For b):
 First of all, we have all linear functionals, restricted to $ \mathbb{Z}^{2} $,
and also Euclidean norm squared : $ f(x,y) = x^{2} + y^{2} $. There is also
the following function: $ f(x,y) = (-1)^{x} - (-1)^{y} $. Let us check that this function
satisfies the condition: if we have $ u =(x_{1},y_{1}) $, $ v =(x_{2},y_{2}) $, such
that $ x_{1}x_{2} + y_{1}y_{2} = 0 $, we have to check that
$ (-1)^{x_{1}} - (-1)^{y_{1}} + (-1)^{x_{1}} - (-1)^{y_{1}} =
(-1)^{x_{1}+x_{2}} - (-1)^{y_{1}+y_{2}} $, which is equivalent to
$ (-1)^{x_{1}} + (-1)^{x_{2}} - (-1)^{x_{1}+x_{2}} =
(-1)^{y_{1}} + (-1)^{y_{2}} - (-1)^{y_{1}+y_{2}} $. Now, the quantity
$ (-1)^{m} + (-1)^{n} - (-1)^{m+n} $ is equal to $ 1 $ if one of $ m,n $ is even,
and to $ -3 $ if both of them are odd. If one of $ x_{1},x_{2} $ is even, then because of
$ x_{1}x_{2} + y_{1}y_{2} = 0 $, one of $ y_{1},y_{2} $ is even, so we have equality
and if both of $ x_{1},x_{2} $ are odd, then also $ y_{1},y_{2} $ are odd,
so again we have equality.
 So we have four functions $ f_{1}(x,y) = x $, $ f_{2}(x,y) = y $,
$ f_{3}(x,y) = x^{2} + y^{2} $ and $ f_{4}(x,y) = (-1)^{x} - (-1)^{y} $.
If for each such $ f_{i} $ we consider a vector $ v_{i} =
(f_{i}(1,0),f_{i}(0,1),f_{i}(-1,0),f_{i}(-1,0)) \in \mathbb{R}^{4} $, then
it is easy to see that $ v_{1},v_{2},v_{3},v_{4} $ are independent so they form a basis
of $ \mathbb{R}^{4} $.
 Now assume that we have some function $ f : \mathbb{Z}^{2} \rightarrow \mathbb{R} $, which
satisfies that for every $ u,v \in \mathbb{Z}^{2} $ and orthogonal, we have
$ f(u+v) = f(u) + f(v) $, and also $ f(1,0) = f(0,1) = f(-1,0) = f(0,-1) = 0 $.
Take $ u=(x,0) $, $ v=(0,y) $ and we get that
$ f(x,y) = f(x,0) + f(0,y) $. Denote $ g(x) = f(x,0) $, $ h(y) = f(0,y) $.
Then $ f(x,y) = g(x) + h(y) $. We have that $ g(0)= g(-1)=g(1)= h(0) =h(-1)=h(1)= 0 $.
Now, applying once more $ f(u+v) = f(u) + f(v) $ for $ u=(a,b) $ and $ v=(-b,a) $, we get
that $ g(a-b) + h(b+a) = g(a) + h(b) + g(-b) + h(a) $. Now, if for some $ k \geqslant 1$,
we know that $ g(a) = h(a) = 0 $ for $ | a | \leqslant k $, then applying
$ g(a-b) + h(b+a) = g(a) + h(b) + g(-b) + h(a) $ for $ a=\pm k , b= \pm 1 $
and for $ a = \pm 1 , b = \pm k $, we get that $ g(k+1) = g(-k-1) = h(k+1) = h(-k-1) = 0 $,
so by induction we get that $ g \equiv h \equiv 0 $, so $ f \equiv 0 $.
So if we have such a function such that $ f(1,0) = f(0,1) = f(-1,0) = f(0,-1) = 0 $,
then $ f(x,y) \equiv 0 $. Therefore any such function is equa to some combination of
$ f_{1},f_{2},f_{3},f_{4} $, because for such $ f $ we can always find
$ a_{1},a_{2},a_{3},a_{4} \in \mathbb{R} $, such that for $ g(v) =
f(v) - a_{1}f_{1}(v) - a_{2}f_{2}(v) - a_{3}f_{3}(v) - a_{4}f_{4}(v) $ we have
$ g(1,0) = g(0,1) = g(-1,0) = g(0,-1) = 0 $, so $ g \equiv 0 $.

I hope it is correct.


Name the bond present in 1 molecule of water & give reason 4 ur ans.

[quote="JRav"]It's a covalent bond, and specifically, a hydrogen bond.  If you have two non-metal elements bonded, it's most likely a covalent bond.  A hydrogen bond is a bond between hydrogen and F, O, and N.[/quote]

A hydrogen bond is a secondary force of attraction, not an actual "bond". It is a dipole-dipole attraction among H and {F,O,N}, both belonging to different molecules.

[quote="Carcul"]There are no hydrogen bonds in a single molecule of water.

"If you have two non-metal elements bonded, it's most likely a covalent bond."

Are there examples of bonds between non-metals that are not covalent?[/quote]

I'm new to this but would it be organic compounds?  I still think it's safe to go with the majority of the time that the bond between 2 non-metals or 2 metals is covalent.

[quote="Carcul"]You may want to revise the last words of your last post.[/quote]

Indeed.  No idea why I wrote that.

[quote="Mathematicswizard"]Generally ionic compounds are solid. Does the nature of bonding in water change when it is frozen to form ice? :?:[/quote]

The intramolecular bond is still covalent and the intermolecular force is also still hydrogen bonding (so, no). However the water molecules do arrange into hexagonal sheets.


calculus integration geometry vector inequalities quadratics trigonometry

Let $ a,\ b$ be real numbers satisfying $ \int_0^1 (ax\plus{}b)^2dx\equal{}1$.
Determine the values of $ a,\ b$ for which $ \int_0^1 3x(ax\plus{}b)\ dx$ is maximized.

[hide]
We wish to maximize $ f(a,b) \equal{} \int_{x \equal{} 0}^1 3x(ax \plus{} b) \, dx \equal{} a \plus{} \frac {3b}{2}$

subject to the constraint $ \int_{x \equal{} 0}^1 (ax \plus{} b)^2 \, dx \equal{} \frac {a^2}{3} \plus{} ab \plus{} b^2 \equal{} 1$.

But

$ 1 \equal{} \frac {a^2}{3} \plus{} ab \plus{} b^2 \equal{} \frac {1}{3}\left(a^2 \plus{} 3ab \plus{} 3b^2\right) \equal{} \frac {1}{3}\left(a \plus{} \frac {3b}{2}\right)^2 \plus{} \frac {b^2}{4}$,

so in particular

$ f(a,b) \equal{} \sqrt{3(1 \minus{} (b/2)^2)}$.

Since the square of a real number is never negative, it is clear that $ f$ is maximized when $ (b/2)^2$ is minimized, hence $ b \equal{} 0$, from which it follows that $ f(a,b)_{\rm max} \equal{} a$, $ a^2/3 \equal{} 1$ and $ a \equal{} \pm \sqrt {3}$.  We discard the negative root since this yields $ f < 0$.  Therefore, the second integral is maximized when $ a \equal{} \sqrt {3}$, $ b \equal{} 0$, with maximum area $ \sqrt {3}$.[/hide]

Consider $ L_2[0,1]$ and vectors $ f(t)\equal{}at\plus{}b,g(t)\equal{}3t$, and $ e_1(t) \equal{} 1,e_2(t)\equal{}t$.  Let $ E\equal{}<e_1,e_2>$
Then we need to find $ \max_{f\in E,\|f\|\equal{}1} \left(f,g\right)$.
Obviously $ \left(f,g\right)\le \|f\|\cdot \|g\| \equal{} \sqrt{\int_0^1 (3t)^2\ dt} \equal{} \sqrt{\frac93}\equal{}\sqrt{3}$.
when we take $ f\equal{}\frac{g}{\parallel{}g\parallel{}} \in E$ we get $ (f,g) \equal{} \frac{(g,g)}{\parallel{}g\parallel{}}\equal{}\parallel{}g\parallel{}.$
So $ \max_{f\in E,\|f\|\equal{}1} \left(f,g\right) \equal{} \sqrt{3}$ when $ f(t)\equal{}\sqrt{\frac13} 3t \equal{}\sqrt{3}t$

Yes, Cauchy-Schwarz's inequality can solve the problem easily, although  Japanese Highschool students could understand atomicwedgie's solution.

i try to look at the problem  from general position  :)

[quote="atomicwedgie"]
We wish to maximize $ f(a,b) \equal{} \int_{x \equal{} 0}^1 3x(ax \plus{} b) \, dx \equal{} a \plus{} \frac {3b}{2}$

subject to the constraint $ \int_{x \equal{} 0}^1 (ax \plus{} b)^2 \, dx \equal{} \frac {a^2}{3} \plus{} ab \plus{} b^2 \equal{} 1$. [/quote]

After this, we are to find the condition for which the curve $ a^2 \plus{} 3ab \plus{} 3b^2 \equal{} 3\ \cdots [*]$ has more than one intersection points with the line $ a \plus{} \frac 32 b \equal{} k$ in the $ a$-$ b$ plane.

Alternative Solution 1:

Elliminating $ b \equal{} \frac {2k \minus{} 2a}{3}$ from these equations gives $ a^2 \minus{} 2ka \plus{} 4k^2 \minus{} 9 \equal{} 0$, thus the above condition can be obtained by $ D\geq 0\Longleftrightarrow D/4 \equal{} k^2 \minus{} (4k^2 \minus{} 9)\geq 0$, yielding $ \minus{} \sqrt {3}\leq k\leq \sqrt {3}$. The equality holds when $ k \equal{} \sqrt {3}$, or the quadratic equation has a double root $ a \equal{} \sqrt {3}$, then from $ [*]$, we have $ b \equal{} 0$.

Alternative Solution 2:

The points on the curve $ a^2 \plus{} 3ab \plus{} 3b^2 \equal{} 3$ can be parameterized by $ a \equal{} \sqrt {3}\cos \theta \minus{} 3\sin \theta ,\ b \equal{} 2\sin \theta$.

$ \therefore a \plus{} \frac 32b \equal{} \sqrt {3}\cos \theta\leq \sqrt {3}$.

by cauchy shwarz's inequality,
            (integral(ax+b)^2dx)(integral(x^2)dx)>=(integral(x(ax+b))dx)^2
and by the condition, the first term is 1, second term is 1/3
so we got the maximum of (integral(x(ax+b))dx)^2 and then the answer.
In here the integral interval is [0,1].


inequalities
function
inequalities proposed

Let $ f(n)$ be defined for $ n \in \mathbb{N}$ by $ f(1)\equal{}2$ and $ f(n\plus{}1)\equal{}f(n)^2\minus{}f(n)\plus{}1$ for $ n \ge 1$. Prove that for all $ n >1:$

$ 1\minus{}\frac{1}{2^{2^{n\minus{}1}}}<\frac{1}{f(1)}\plus{}\frac{1}{f(2)}\plus{}...\plus{}\frac{1}{f(n)}<1\minus{}\frac{1}{2^{2^n}}$

[quote="moldovan"]Let $ f(n)$ be defined for $ n \in \mathbb{N}$ by $ f(1) \equal{} 2$ and $ f(n \plus{} 1) \equal{} f(n)^2 \minus{} f(n) \plus{} 1$ for $ n \ge 1$. Prove that for all $ n > 1:$

$ 1 \minus{} \frac {1}{2^{2^{n \minus{} 1}}} < \frac {1}{f(1)} \plus{} \frac {1}{f(2)} \plus{} ... \plus{} \frac {1}{f(n)} < 1 \minus{} \frac {1}{2^{2^n}}$[/quote]

Notice that $ \frac{1}{f(n\plus{}1)\minus{}1}\equal{}\frac{1}{f(n)^2\minus{}f(n)}\equal{}\frac{1}{f(n)\minus{}1}\minus{}\frac{1}{f(n)}$ or $ \frac{1}{f(n)}\equal{}\frac{1}{f(n)\minus{}1}\minus{}\frac{1}{f(n\plus{}1)\minus{}1}$.

So, we have the telescoping series that

$ \frac{1}{f(1)}\plus{}...\plus{}\frac{1}{f(n)}\equal{}\frac{1}{f(1)\minus{}1}\minus{}\frac{1}{f(n\plus{}1)\minus{}1}\equal{}1\minus{}\frac{1}{f(n\plus{}1)\minus{}1}$.

So, we must show that $ 2^{2^{n\minus{}1}}\plus{}1<f(n\plus{}1)<2^{2^n}\plus{}1$

We can prove this by strong induction. Easy to show this for n=2.

Assume true for n=k, so $ f(k\plus{}1) \le 2^{2^{k}}$, so $ f(k\plus{}2)\equal{}2^{2^{k\plus{}1}}\minus{}2^{k}\plus{}1<2^{2^{k\plus{}1}}\plus{}1$

And $ f(k\plus{}1)>2^{2^{k\minus{}1}}\plus{}1$ implies that $ f(k\plus{}2)>2^{2^{k}}\plus{}2^{2^{k\minus{}1}}\plus{}1>2^{2^{k}}\plus{}1$.

So, we are done.


Calculate the sum in terms of $n$ and $cosx$
$\dbinom{n}{0}+\dbinom{n}{4}+\dbinom{n}{8}+\dbinom{n}{12}+...$


[quote="filomen"]If $ a,b,c > 0$, then
$ \frac {4a^2 \plus{} 18bc}{b^2 \plus{} c^2} \plus{} \frac {4b^2 \plus{} 18ca}{c^2 \plus{} a^2} \plus{} \frac {4c^2 \plus{} 18ab}{a^2 \plus{} b^2}\geq19$[/quote]

not true...

try $ a\equal{}b\equal{}100,c\equal{}1$ .

[quote="filomen"][quote="filomen"]If $ a,b,c > 0$, then
$ \frac {4a^2 \plus{} 18bc}{b^2 \plus{} c^2} \plus{} \frac {4b^2 \plus{} 18ca}{c^2 \plus{} a^2} \plus{} \frac {4c^2 \plus{} 18ab}{a^2 \plus{} b^2}\geq19$[/quote]

Correct:

$ \frac {5a^{2} \plus{} 18bc}{b^{2} \plus{} c^{2}} \plus{} \frac {5b^{2} \plus{} 18ca}{c^{2} \plus{} a^{2}} \plus{} \frac {5c^{2} \plus{} 18ab}{a^{2} \plus{} b^{2}}\geq19$[/quote]

The following inequality is equivalent to

$ \minus{} 23b^2c^2 a^2 \plus{} 18 b c^3 a^2 \plus{} 18 b c a^4 \plus{} 18 b^3 c a^2 \plus{} 18 c a^3 b^2 \plus{} 18 c a b^4 \plus{} 18 c^3 a b^2 \plus{} 18 a b^3 c^2 \plus{} 5 a^6 \plus{} 5 b^6 \plus{} 5 c^6 \minus{} 14 b^4 c^2 \minus{} 14 b^2 a^4 \minus{} 14 b^4 a^2 \minus{} 14 c^4 a^2 \minus{} 14 c^4 b^2 \minus{} 14 c^2 a^4 \plus{} 18 b^3 c^3 \plus{} 18 c^3 a^3 \plus{} 18 a^3 b^3 \plus{} 18 a b c^4 \plus{} 18 a^3 b c^2\geq 0$

 is equivalent to

$ 5\sum{(a^4(a \minus{} b)(a \minus{} c))} \plus{} 5\sum{(a^3(b \plus{} c)(a \minus{} b)(a \minus{} c))} \minus{} 9(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \plus{} 5abc\sum{(a(a \minus{} b)(a \minus{} c))} \plus{} 36abc\sum{((b \plus{} c)(a \minus{} b)(a \minus{} c))} \plus{} 124b^2c^2a^2\geq 0$

we only to prove

$ 5\sum{(a^3(b \plus{} c)(a \minus{} b)(a \minus{} c))} \minus{} 9(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\geq0$

or

$ 3\sum{(a^3(b \plus{} c)(a \minus{} b)(a \minus{} c))} \minus{} 9(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} 3(\sum{(a^3(b \plus{} c)(a \minus{} b)(a \minus{} c))} \minus{} 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2)\geq0$

without loss of generality, we can assume that $ a\geq b \geq c$

substitution  $ a \equal{} x \plus{} y \plus{} c,b \equal{} y \plus{} c$ ($ x,y\geq 0$)

$ \sum{(a^3(b \plus{} c)(a \minus{} b)(a \minus{} c))} \minus{} 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\geq0$

Therefore, we just need to prove

$ (2 x y \plus{} 2 x^2 \plus{} 2 y^2) c^4 \plus{} (3 x y^2 \plus{} 13 x^2 y \plus{} 6 x^3 \plus{} 2 y^3) c^3 \plus{} (21 x^2 y^2 \plus{} 6 x^4 \plus{} 21 y x^3) c^2 \plus{} (21 x^3 y^2 \plus{} 14 y^3 x^2 \plus{} 11 y x^4 \plus{} 2 x^5) c \plus{} x^5 y \plus{} y^2 x^4\geq 0$


Let $ABC$ be an acute triangle with orthocenter $H$. Let $M$, $N$ and $R$ be the midpoints of $AB$, $BC$ an $AH$, respectively. If $A\hat{B}C=70^\large\circ$, compute $M\hat{N}R$.

[hide]$M,N,R$ all lie on the nine point circle of $\Delta ABC.$ Also, the nine point circle intersects side $AC$ at the midpoint $X$ and foot of altitude $F.$
since $X,R,M,N$ all lie on the same circle,
$\angle RNM=\angle RXM.$
$\Delta AXM\sim\Delta ACB$ and $R$ is the orthocenter of $\Delta AXM$ so
$\angle RXM=\angle HCB=20.$
Thus, $\angle MNR=20.$[/hide]

That is a very elegant solution!

A few months ago I saw a solution that don't use the nine point circle, but the circuncircle of $ABC$, but I can't remember it, and I also can't find it myself. Can anyone post a similar solution?

Let $ABC$ be an acute triangle with orthocenter $H$. Let $M$, $N$ and $R$ be the midpoints of $AB$, $BC$ an $AH$, respectively. If $\angle ABC=70^ \circ$, compute $\angle M N R$.


[quote="manocanh"]$x,y,z,t >0$ and $x^2 + y^2 + z^2 + t^2 = 2005$
find minimum value of expression
$F = \frac{x}{2005\sqrt{2005} + yzt} + \frac{y}{2005\sqrt{2005} + ztx} + \frac{z}{2005\sqrt{2005} + txy} + \frac{t}{2005\sqrt{2005} + xyz}$[/quote]

nobody   :(   :(

It's easy.
Use Cauchy-Schwarz's in Engel form

I get the minimum is $\frac 1{2005}$ for $x,y,z,t \ge 0$
Proof:
The equality can be hold when $x=\sqrt{2005},x=z=t=0$
Without loss of generality,$x \ge y \ge z \ge t$
thus we get:$F\ge \frac{x+y+z+t}{2005\sqrt{2005}+xyz}$.
So it suffies to prove:
\[ \frac{x+y+z+t}{2005\sqrt{2005}+xyz} \ge \frac 1{2005} \]
square both side,you will find it is not hard.
Hope I am right. :)


Let $ OAB$ and $ OA'B'$ two equilateral triangles with the same orientation,and S the center of weight for the triangle $ OAB$ and M,N are the midle of the sides A'B and AB'.Prove that $ \Delta SMB' \sim  \Delta SNA'$

[hide="Solution"]Let $ 0,a,a\epsilon,b,b\epsilon$ the afixez of the points  $ O,A,B,A',B'$
\[ N \equal{} \frac {b\epsilon}{2}
\]

\[ S \equal{} \frac {a \plus{} a\epsilon}{3}
\]

\[ A' \equal{} b
\]

\[ M \equal{} \frac {b \plus{} a\epsilon}{2}
\]

\[ \vec{SM} \equal{} \frac {3b \minus{} 2a \plus{} a\epsilon}{3}
\]

\[ \vec{SM} \cdot \epsilon \equal{} \frac { \minus{} a \plus{} (3b \minus{} a)\epsilon}{6}
\]

\[ \vec{SB} \equal{} \frac { \minus{} a \plus{} (3b \minus{} a)\epsilon}{3}
\]

\[ \vec{SM} \equal{} \frac {\vec{SB'}}{2}
\]
In the same way $ \vec{SA'} \equal{} 2\vec{SN}$[/hide]


Uh-oh!

It seems like the entire forum hierarchy (FF 3.0.8, Win XP Professional, Mathcounts topic)

Even here there seem to be extraneous "-->" tags.

Happened while working on code. Not a permanent bug. Closed report.


Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, and the second from {A,I,O}, and the third from {D,M,N,T}.
        When Flatville needed more liscense plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?

(A) 24
(B) 30
(C) 36
(D) 40
(E) 60

The answer is (D) 40, but I do not know how to solve this problem. Can someone help me?

The total amount of possible license plates before adding the letters is $5*3*4=60$. To make a maximum product you need the numbers to be the closest together. That would be making them 5 letters to choose from, 5, 4. You could get that by either adding 2 to the second set or 1 to each the second and third set. $5*5*4=100$. $100-60=\boxed{40}$


[quote="grobber"]Let $(A_i)_{i\in I}$ ($I$ is an index set) be a collection of real sets, all of measure zero, totally ordered with respect to inclusion, such that their union is measurable.

Is it true that $\bigcup_{i\in I}A_i$ has measure zero?[/quote]

I have an idea, but I'm not sure whether it can be called a proof...

If the answer to your question is yes, that is, $\bigcup_{i\in I}A_i$ must have measure zero, then:
Consider $P=\{A\subseteq{R}|m(A)=0\}$.
Every totally ordered subset $(A_i)_{i\in{I}}$ of P has an upper bound $\bigcup_{i\in{I}}A_i$, which by your assertion is also in $P$.
Then by Zorn's lemma, there is an element $B\in{P}$ such that no element in $P$ is strictly larger than $B$.
But that is impossible, since obviously $B\neq{R}$, we can find $x\in{R\setminus{B}}$, and $B\cup{\{x\}}$ still have measure zero and is strictly larger than $B$, which is a contradiction.
So the measure of $\bigcup_{i\in{I}}A_i$ may have positive measure, though I have not worked out an example yet...


probability
MATHCOUNTS
percent
FTW
AMC 10

um what do you mean does countdown count?

Some places have countdown affecting their ranking.

No, I really wish it did.

I would've gotten first at State instead of Second last year.

don't think so in MA....
but it would probably help me if there was...

Last year in States, when I beat Allen Yuan, I did like the countdown. Otherwise, I don't like it being official. It's too... random. Also, I suck at it.

Yeah perfect628, I remember that. I was glad about it too or else I would've gotten in 5th place instead of 4th... But then again I would've probably actually tried in the CD if it was official.

I suck and have never made states cd (and never really will, cause this is my last year), but it is quite exciting from a spectators point of view :rotfl:

I'm in CT, it doesn't count and I'm glad it doesn't. Even though I knew it didn't count, I did horrible in Chapter Countdown from all of the nerves. So imagine what it would be like with them.

if i have my uber adrenaline rush at countdown at states. I'll be beast!. but probably not.  I hope it doesn't count or I'm screwed xD

well luckily I'm in an easy state  :lol: so I should be able to easily pull ahead and make it into nats  :). Unlike some of those unlucky people on Texas and California.

Even though I hate countdown round, I'm still glad it's done officially here.

[quote="frost13"][quote="perfect628"]Well, I was 7th(don't ask...) last year, and got to 4th with official-style CD(i.e. ladder), in a state with unofficial CD. You can imagine what my opinion was when I was in MC.[/quote]

Hey perfect, how did you do in countdown in 7th grade?  I can't remember if you were 2nd or 4th at state(Carl 1st, George 3rd I think)[/quote]

I was 2nd, and George wasn't there, he took the test beforehand. I lost to that lexington kid who pwned everyone, so 3rd after CD..

[quote="nick42"]here in nc, it's just for the glory and possibly some random trophy.  :P[/quote]

Yeah, same for CT.

yes and im glad...

i usually play cd in ftw...

OMG fantasy you're in ca mathcounts? im dead.......
wut grade r u in nxt year?

[color=blue]* AIME15 swears loudly[/color]

oh well...

oh geezzz, you have to face the really tough competition with bshen and many others from sugarland...
they all pwn at cd :D

[quote="isabella2296"]Does it count in New Jersey?[/quote]
YES IT DOES!! :o

and fantasy lover lives in nj where it counts!

CD does not count in my state. I really dont care if it counts or doesnt because I am usually not good enough to get in the CD round so it really doesnt matter for me.

I actually would like CD to count cuz i pwn at it.

I wish cd counted in my state, then i would've been first a state 3 yrs in a row... then again, my chapter in 6th and 7th would've sucked...


Given 2 points $ A(3,\ 1),\ B(1,\ 4)$ and the circle $ (x\minus{}1)^2\plus{}(y\plus{}2)^2\equal{}4$. When the point $ P$ moves on the circle, find the maximum and minimum area of $ \triangle{PAB}$.

[hide]Here's the outline of my solution(sorry I don't have time to post full solution):
Since P(x,y) lies on the circle, we have:
(x-1)^2+(y+2)^2=4
Now, we see that $ AB\equal{}\sqrt{13}$, which means we just need to find the maximum and minimum of the perpendicular drawn from a point P.
On the other hand, the distance from a point P to the line AB is:
$ \frac{y\plus{}3/2.x\minus{}11/2/}{sqrt{13/4}}$ since the equation of the line AB is: $ y\equal{}\minus{}3x/2\plus{}11/2$
Therefore, now we just need to find the maximum and minimum of $ /y\plus{}3x/2\minus{}11/2/ with (x\minus{}1)^2\plus{}(y\plus{}2)^2\equal{}4$
Calculate y in terms of x and note that $ 3>\equal{}x\geq 1 and 0\geq y\geq \minus{}4$. It's done with derivative and calculus.
My result is the minimum value occurs at $ x\equal{}\minus{}1/2$. I haven't calculated y  yet.
This problem is nice, but the result seems to be so ugly(square root of something+something).If the proposer changes the number a little bit, then it's a great problem for university entrance exam. By the way, this way doesn't need to use calculator, which is a great thing. :)  [/hide]

[hide="Solution"]
Note that the line is outside of the circle. The minimum and maximum areas occur when the point on the circle is on the line through the center of the circle perpendicular to the line through $ A$ and $ B$, which is $ y \equal{} \minus{} \frac32x \plus{}\frac{11}2\implies3x \plus{} 2y \minus{} 11 \equal{} 0$. The distance from the center $ (1, \minus{} 2)$ to the line $ 3x \plus{} 2y \minus{} 11 \equal{} 0$ is $ \frac {|3(1) \plus{} 2( \minus{} 2) \minus{} 11|}{\sqrt {3^2 \plus{} 2^2}} \equal{} \frac {12}{\sqrt {13}}$, so the minimum and maximum heights are $ \frac {12}{\sqrt {13}}\pm2$, giving areas $ \frac12\cdot\sqrt {13}\cdot\left(\frac {12}{\sqrt {13}}\pm2\right) \equal{} \boxed{6\pm\sqrt{13}}$.
[/hide]

Edit: Thanks, kunny. :oops:

The $ y$ intersept of the line $ AB$ is incorrect.

[quote="math154"][hide="Solution"]
Note that the line is outside of the circle. The minimum and maximum areas occur when the point on the circle is on the line through the center of the circle perpendicular to the line through $ A$ and $ B$, which is $ y \equal{} \minus{} \frac32x \plus{} \frac {11}2\implies3x \plus{} 2y \minus{} 11 \equal{} 0$. The distance from the center $ (1, \minus{} 2)$ to the line $ 3x \plus{} 2y \minus{} 11 \equal{} 0$ is $ \frac {|3(1) \plus{} 2( \minus{} 2) \minus{} 11|}{\sqrt {3^2 \plus{} 2^2}} \equal{} \frac {12}{\sqrt {13}}$, so the minimum and maximum heights are $ \frac {12}{\sqrt {13}}\pm2$, giving areas $ \frac12\cdot\sqrt {13}\cdot\left(\frac {12}{\sqrt {13}}\pm2\right) \equal{} \boxed{6\pm\sqrt {13}}$.
[/hide]

Edit: Thanks, kunny. :oops:[/quote]

You are welcome. :lol:
That's correct and geometrical solution without using calculus.

My solution is using Trianglular function. :oops:

[url=http://en.wikipedia.org/wiki/Triangular_function]Some calculus thing that I don't understand?[/url]

[quote="brightzhu"]What is the Triangular function?[/quote]

That was Trigonometric function. :blush: