Walter gets up at 6:30 am, catches the school bus at
7:30 am, has 6 classes that last 50 minutes each, has 30
minutes for lunch, and has 2 hours additional time at school. He
takes the bus that arrives at 4:00 pm. How many minutes
has he spent on the bus?


let a,b,c be positive real number,prove that:
$ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \plus{} 1 \ge 4\sqrt {\frac{{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}}}$

[quote="Toan_VN_LC"]let a,b,c be positive real number,prove that:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 1 \ge 4\sqrt {\frac {{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217810


There are 10 1s and 10 0s, how many ways to distribute them so that they sequence does not have 3 consecutive 0s?

[hide="this is really a combinatorics problem..."]

anyway, there are $ 20!/(10!10!) \equal{} 184756$ arrangements possible, but some of these will contain 3 consecutive zeros.

So lets line up the 1s, and consider sticking either one 0 or 2 zeros in each gaps or on the ends:
1 1 1 1 1 1 1 1 1 1

So what we have is 11 spaces to start with, which we will stick pairs of zeros in, and then 11-(# of pairs) spaces left to stick in 10 - 2(# of pairs) 0's.
For no pairs of consecutive zeros: 11 take 10 = 11
For 1 pair of consecutive zeros: 11 take 1 * 10 take 8 = 11*45=495
for 2 pairs of consecutive zeros: 11 take 2 * 9 take 6 = 55**84=4620
for 3 pairs of consecutive zeros: 11 take 3 * 8 take 4 = 165*70=11550
for 4 pairs of consecutive zeros: 11 take 4 * 7 take 2 = 330*21=6930
for 5 pairs of consecutive zeros: 11 take 5 = 462

or equivalently $ \sum_{n \equal{} 0}^5 (_{11}C_n)( _{11 \minus{} n}C_{10 \minus{} 2n}) \equal{} 24068$

for a total of 24068 possibilities  :D
[/hide]

Sorry, I don't understand why they have to be in pairs...

If we have no 3 consecutive 0's, then we either have single zeros or pairs of zeros, as all other combinations are not allowed.


Find the minimum value of $ a\plus{} \frac{1}{(a\minus{}b)(b)}$ if $ a>b>0$

[quote="Zellex"]Find the minimum value of $ a \plus{} \frac {1}{(a \minus{} b)(b)}$ if $ a > b > 0$[/quote]

[hide="possible start"]
$ (a \minus{} b)(b)$ reaches its maximum when $ b \equal{} \frac {a}{2}$
this is easy to see when you compare $ (a \minus{} b)(b)$ ? $ \frac {a^2}{4}$
once you move the $ (a \minus{} b)(b)$ to the RHS, the equation becomes $ 0$ ? $ \frac {a^2 \plus{} 4ab \plus{} 4b^2}{4}$
obviously $ 0 \le \frac {(a \plus{} 2b)^2}{2}$
so, we put b as $ \frac {a}{2}$ and solve
$ a \plus{} \frac {4}{a^2}$

not sure where to go from here

EDIT: maybe try setting it to the am-gm minimum?

$ a \plus{} \frac {4}{a^2} \equal{} 2\sqrt {\frac {4}{a}}$
$ \frac {a}{2} \plus{} \frac {2}{a^2} \equal{} \sqrt {\frac {4}{a}}$
$ \frac {a^2}{4} \plus{} \frac {4}{a^4} \plus{} \frac {2}{a} \equal{} \frac {4}{a}$
$ \frac {a^6}{4a^4} \plus{} \frac {16}{4a^4} \minus{} \frac {8a^3}{4a^4} \equal{} 0$
$ a^6 \minus{} 8a^3 \plus{} 16 \equal{} 0$
$ a^3 \equal{} 4$

hmm
so if that's right the minimum is $ 2\cdot (4)^\frac {1}{3}$

[/hide]

[quote="MellowMelon"]Write $ a \equal{} c \plus{} b$ for a positive real $ c$. Then we want the minimum of $ c \plus{} b \plus{} \frac {1}{bc}$. By AM-GM, $ c \plus{} b \plus{} \frac {1}{bc} \geq 3 \sqrt [3]{c \cdot b \cdot \frac {1}{bc}} \geq 3$, and $ a \equal{} 2, b \equal{} 1$ achieves this lower bound.

(to above poster: it's easy to show $ 3 < 2 \cdot \sqrt [3]{4}$, so that's not the minimum)[/quote]

aha, you're right (and i'm wrong as always)
can someone explain why my way produces the wrong answer? (maybe i multiplied wrong?)

note: fixed latex.

So you have $ a \plus{} \frac {4}{a^2}$, and you want to use AM-GM.

The problem with your application of AM-GM is that you have not set a concrete lower bound. You may have that at one point AM = GM for $ a$ and $ \frac {4}{a^2}$, but that does not mean that anything is minimized because your lower "bound" itself varies with $ a$.

You can actually get a concrete lower bound with AM-GM if you rewrite this as:

$ a \plus{} \frac {4}{a^2} \equal{} \frac {a}{2} \plus{} \frac {a}{2} \plus{} \left( \frac {2}{a} \right) ^2$

Note that this is equal to $ 3 \cdot \frac {\frac {a}{2} \plus{} \frac {a}{2} \plus{} \left( \frac {4}{a^2} \right)^2 }{3} \geq 3 \cdot \sqrt [3]{\frac {a}{2} \cdot \frac {a}{2} \cdot \left( \frac {4}{a^2} \right)^2 } \equal{} 3$, which is the minimum.

Now you only have to check that you can have $ \frac {a}{2} \equal{} \left( \frac {2}{a} \right)^2$, which works just fine when $ a \equal{} 2$.

[b]Edit:[/b] [i]Haha, looks like kunny was after the same idea [/i]  :P

In your solution, you can proceed such like: $ a\plus{}\frac{4}{a^2}\equal{}\frac{a}{2}\plus{}\frac{a}{2}\plus{}\frac{4}{a^2}\geq 3\sqrt[3]{\frac{a}{2}\cdot \frac{a}{2}\cdot \frac{4}{a^2}}\equal{}3$. The equality holds when $ \frac{a}{2}\equal{}\frac{4}{a^2}\Longleftrightarrow a\equal{}2$.


Let $(S)$ denote the area of the figure $S$

Of course $(PEA)=(PEB)$, because $E$ is the midpoint of $AB$
Similarly for the other pairs of triangles

For any point $P$ in the interior of $ABCD$ we have:

$(PHAE) = (PHA)+(PEA)$
$(PEBF) = (PFB)+(PEB)$

$(PHAE) = (PEBF) \Rightarrow  (PHA)=(PFB)$ (the 'opposite' triangles have equal areas)

So it is $(PHA) = (PHD) = (PFB) = (PFC) \Rightarrow $

$(PAD) = (PBC)$

and similarly $(PAB) = (PCD) \ \ (1)$

Let $h_{AB}, h_{CD}$ be the distances from $P$ to $AB,CD$ respetively.

From $(1)$ we get $\frac{h_{AB}}{h_{CD}}= \frac{CD}{AB}= \lambda$

$a)$ If $AB \parallel CD$, then the locus of $P$ such that $(1)$ holds is a line $d$ parallel to $AB,CD,$ at the proper ratio $\lambda$

$b)$ If $AB \not \parallel CD$

Let $O \in AB \cap CD$. We will see that the locus of $P$ such that $(1)$ holds is a ray $Ox$, in the interior of $\angle AOD$

We draw two lines $m,n$ parallel to $AB,CD$ respectively, such that $\frac{d(m,AB)}{d(n,CD)}= \lambda.$
We choose these lines, such that they have common points with the angle $\angle AOD$

Let $T\in m \cap n$. Then $T$ is in the interior of $\angle AOD$

It's easy to prove (by similarity) that for every point $P$ on the ray $OT$ has the property

$\frac{d(P,AB)}{d(P,CD)}=\lambda$, hence $(1)$ holds

Similalry, we draw one more ray $OT'$ between the lines $BC,AD.$

The point $P$ can be found on the intersection of $OT$ and $OT'$, and it is clearly in the interior of $ABCD$

[u]Note[/u]
If $AB \parallel CD$, then, of course,  instead of $OT$ we use the line $d \parallel AB$, which defined above.
Similarly if $BC \parallel DA$

[quote="Johann Peter Dirichlet"]$ABCD$ is a quadrilateral,
$E,F,G,H$ are midpoints of $AB,BC,CD,DA$.
Find the point P such that
$area (PHAE) = area (PEBF) = area (PFCG) = area (PGDH)$.[/quote][color=darkblue][b]Lemma.[/b] Denote $\delta_{d}(X)$- the distance of the point $X$ to the line $d$. Given are two lines $d_{1}$, $d_{2}$ with the common point $A$ and a pozitive real number $k$.
Then the geometrical locus of the point $L$ for which $\frac{\delta_{d_{1}}(L)}{\delta_{d_{2}}(L)}=k$ is the reunion of two lines with its intersection in the point $A$.

[b]Proof of the proposed probem.[/b] Suppose w.l.o.g. that $X\in AB\cap CD$ and $Y\in BC\cap AD\ .$ Denote the area $S=[ABCD]\ .$
$[APE]=BPE]=[CPG]=$ $DPG]=x\ ,\ [APY]=[DPH]=[CPF]=[BPF]=y\ ;\ x+y=\frac{1}{4}\cdot S\ .$
$[PAD]=[PBC]=2y$ $\Longrightarrow$ $\frac{\delta_{BC}(P)}{\delta_{AD}(P)}=\frac{AD}{BC}\ \mathrm{(constant)}\ ;\ [PAB]=[PCD]=2x$ $\Longrightarrow$ $\frac{\delta_{AB}(P)}{\delta_{CD}(P)}=\frac{CD}{AB}\ \mathrm{(constant)}\ .$
[b]Construction of the point[/b] $P$. Draw the points : $U\in (YC)\ ,\ YU=BC$ ; $V\in (YD)\ ,\ YV=AD$ ; $R\in (XA)\ ,\ XR=AB$ ; $S\in (XD)\ ,\ XS=CD\ .$
Construct the middlepoints $E$, $F$ of the segments $[UV]$, $[RS]$ respectively. Then the point $P$ is the intersection of the lines $YE$ and $XF\ .$[/color]

What needs to be proved is "Area phae = area pebf = area pfcg = area pgdh =Area of ABCD/4"
      HE//DB//GF and EF//AC//HG by mid-point theorem
Let m, n be the mid-point of diagonals AC and BD respectively.
Clearly, Area of PHAE = Area of ABCD/4 when P=M as MHAE is similar to CDAB.
      So Area of PHE= Area of MHE . As they have common base HE, their heights to HE are the same.We can therefore construct a line L1 through M and parallel to HE (DB) and P must be on the line. Similarly P is on L1 to make Area of PFCG = Area of ABCD/4.
      Construct another line L2 through N and parallel to AC. Then ,P is also on L2 to make Area of PEBF=Area of PGDH=Area of ABCD/4.
      So P is the intersection point of L1 and L2

just construct the point : join $ {EF,FG,HG,HE}$;   draw $ {l1}\parallel{CD}$;$ {l2}\parallel{AD}$ meet at point $ {x}$;draw $ {q}\parallel{GH}$; THEN THE LAST STEP: draw a line $ {p}$ // $ {HE}$ such that the distance from $ {p}$ to $ {HE}$ is equal to the altitude of  $ {triangle FGC}$ ;Let $ {p}\cap{q}\equal{}{P}$       $ {P}$ is  the point we want .Easy to prove the area of triangles with the same color are same with S(HEP)=S(GFC) ;we get : $ {Area PHAE \equal{} area PEBF  \equal{} area PFCH \equal{} area PGDH\equal{}1/4S(ABCD)}$.  :)

$ \mbox{Let M and N midpoints of } [AC]\mbox{ and } [BD] \mbox{, respectively. If } \\
PM \parallel BD \mbox{ and }PN \parallel AC \mbox{, then: } S_{PHAE}\equal{}S_{PEBF}\equal{}S_{PFCH}\equal{}S_{PEDH}.$


Solve the following simultaneous equations on the set of integers:
                                                       ab + cd = -1
                                                       ac + bd = -1
                                                       ad + bc = -1

If you call them (1), (2), (3) in order as you write then

(1)-(2)=(b-c)(a-d)=0
(1)-(3)=(a-c)(b-d)=0
(2)-(3)=(c-d)(a-b)=0

It means that at least three of a,b,c,d are equal. By symmetry let a=b=c.
Then it turns to:
$a^2+ad=-1$
$a(a+d)=-1$
So solutions are obvious: a=b=c=1, d=-2 and a=b=c=-1, d=2. Also all cyclic with this are solutions. It's all solutions.

[quote="ondrob"]If you call them (1), (2), (3) in order as you write then

(1)-(2)=(b-c)(a-d)=0
(1)-(3)=(a-c)(b-d)=0
(2)-(3)=(c-d)(a-b)=0

It means that at least three of a,b,c,d are equal. By symmetry let a=b=c.
Then it turns to:
$a^2+ad=-1$
$a(a+d)=-1$
So solutions are obvious: a=b=c=1, d=-1 and a=b=c=-1, d=2. Also all cyclic with this are solutions. It's all solutions.[/quote]

I think you meant [b]a=b=c=1, d =-2[/b] and a=b=c=-1, d=2.

[/b]

Yes, Nukular, you're right. I find it very hard to write solutions without some small mistakes in typing :( .


Find the area of the square inscribed in the quarter circle of radius $ r$.

[hide]
Note that the square's diagonal is the circle's radius, so the answer is $ \boxed{r^2}{2}$
[/hide]

I think that the square's diagonal is strictly smaller than the radius of the circle...Just look at a correct picture...

hmm i think I have an answer: $ \frac {4r^{2}}{7\sqrt{5}}$ which is almost $ \frac{2r^{2}}{5}$

Well if it's like I think it is, then the two verticies of the square would bisect the radii of the quarter circle.
Meaning that the area of the square would be $ (\frac {r\sqrt {2}}{2})^2 \equal{} \frac {2r^{2}}{4} \equal{} \frac {r^2}{2}$

why would the two vertices of the square would bisect the radii of the quarter circle.

In my solution I constructed the whole circle with the squars is each quarter of circle, too. Thus we get $ 5$ complet squares, $ 4$ half squares( which gives an amount of $ 7$ squares) and 8 small "parts" of circle...(it's easy to see on a picture)

Couldn't the square be constructed in many different ways anyways?

I can see what you're saying, although I'm not completely sure if that's a square or just something very close to one.

the diagonal is r units long divide that by square root 2 and you will have a side lenght square the side length and you will obtain your answer :alien:  :starwars:  :spam: is actually a type of food

I believe the area of the square is:

(2R^2)/5

where R is the radius of the semi-circle.

tell me if I'm correct  :D

If the vertices of the square lies on The radii or the quarter-circumference, then area of the square = half the area of the square of side equal to the radius.

[quote="sunnykim"]One vertex on each radius and the other 2 vertices on the arc....Sorry for not mentioning it from the beginning, now I see how you imagined the picture...[/quote]

I don't think you have the right diagram sohumdatta.

[b]2[/b] vertices on the arc and one on each radius.  :wink:


inequalities
absolute value
triangle inequality

$\mid \mid x \mid - \mid y \mid \mid \   \leq \   \mid x - y \mid$

(Those are absolute value signs...I wasn't sure how to make them on LaTeX....)

How would I go about proving this inequality?

Did you try to split it into cases and prove each case:

Like $x$ and $y$ are both negative...

Okay, so the inequality $\vert z\vert-\vert y\vert\leq \vert z-y\vert $ is a direct consequence of the triangle inequality.

Similarly we have $\vert z\vert-\vert y\vert\geq -\vert z-y\vert $ is a direct consequence of the triangle inequality

But $-a\leq b\leq a$ means that $\vert b\vert \leq a$, and so we're done.

That way works too... haha *pretending like I actually saw that*


linear algebra
matrix
vector
IMC
college contests

Let $v_{0}$ be the zero ector and let $v_{1},...,v_{n+1}\in\mathbb{R}^{n}$ such that the Euclidian norm $|v_{i}-v_{j}|$ is rational for all $0\le i,j\le n+1$. Prove that $v_{1},...,v_{n+1}$ are linearly dependent over the rationals.

It follows that all scalar products $(v_{i},v_{j})$ are rational. Since the rational $(n+1)\times(n+1)$ matrix with entries $(v_{i},v_{j})$ is singular, let $(\alpha_{1},\ldots,\alpha_{n+1})$ be a non-zero rational vector in its nullspace. We have $\sum_{i=1}^{n+1}\alpha_{i}(v_{i},v_{j})=0,\ j=\overline{1,n+1}\ (*)$, and this is equivalent to $\sum\alpha_{i}v_{i}=0$ (because $(*)$ says that in the Euclidean space spanned by the $v_{i}$, the vector $\sum\alpha_{i}v_{i}$ is orthogonal to a set of generators, hence it must be zero).


In right triangle $ ABC$, $ M$ and $ N$ are midpoints of legs $ \overline{AB}$ and $ \overline{BC}$, respectively. Leg $ \overline{AB}$ is 6 units long, and leg $ \overline{BC}$ is 8 units long. How many square units are in the area of $ \triangle APC$?
[asy]draw((0,0)--(8,0)--(0,6)--cycle);
draw((4,0)--(0,6));
draw((0,3)--(8,0));
label("$A$",(0,6),NW); label("$B$",(0,0),SW); label("$C$",(8,0),SE); label("$M$",(0,3),W); label("$N$",(4,0),S); label("$P$",(8/3,2),SW);[/asy]

Hello
Carpet theorem
[hide="hint"]
ar of quad.$ MBNP$=ar. of $ \triangle APC$
[/hide]
Thank u.

It is well-known that the three medians of a triangle (illustrated by $ \overline{AN},\overline{BO},\overline{CM}$ above) divide the triangle into $ 6$ [url=http://en.wikipedia.org/wiki/Median_(geometry)#Equal-area_division]regions of equal area[/url].

Thus, the desired area of $ \triangle APC$ contains $ 2$ of the small areas, or $ \frac26\equal{}\frac13$ of $ \triangle ABC$'s area, and our answer is $ \frac13\left(\frac12\cdot6\cdot8\right)\equal{}\boxed8$.

$ \frac{AP}{PN} \times \frac{NC}{CB} \times \frac{BM}{MA} \equal{} \frac{AP}{PN} \times \frac{4}{8} \times \frac{3}{3} \equal{} 1$.  (by Menelaus' theorem)
Therefore $ AP: PN \equal{} 2: 1$.

So, $ \triangle APC \equal{} \frac{NC}{BC} \times \frac{AP}{AN} \times \triangle ABC \equal{} \frac{2}{4} \times \frac{2}{2\plus{}1} \times \left( \frac{1}{2} \times 8 \times 6 \right) \equal{} \boxed{8}$.

Why this when there is a much more elementary and faster, yet rigorous solution?


algebra
polynomial
induction
trigonometry
superior algebra
superior algebra unsolved

Prove that trigonometric polyomial
$\sum_{k=m}^{n}a_k(sin(\alpha))^k+b_k(sin(\alpha))^k$ has at least 2m and at most 2n zeros when $0\le\alpha<2\pi$

Do you mean $ \sum_{k=m}^{n}a_k(sin(\alpha))^k+b_k(cos(\alpha))^k $  :?

oh,... i'm sorry - it was an idiot typo, i meant
$\sum_{k=m}^{n}a_ksin(k\alpha)+b_kcos(k\alpha)$
I editted now in righ way-i hope it's ok(I'm sorry)

Upper bound can be easily proved by induction, can't it?

[quote="eugene"]oh,... i'm sorry - it was an idiot typo, i meant
$\sum_{k=m}^{n}a_kcos(k\alpha)+b_kcos(k\alpha)$[/quote]
The original problem had two sines, and this problem has two cosines?


geometry
geometric transformation
reflection
circumcircle
parallelogram
BritishMathematicalOlympiad

Let $ABC$ be a triangle with $AC>AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX=CA$ and $CY=BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that
$\angle BPC + \angle BAC = 180^{\circ}$.

[quote="Saul"]Let $ABC$ be a triangle with $AC>AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX=CA$ and $CY=BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that

$\angle BPC + \angle BAC = 180^{\circ}$[/quote]

Let's reflect this construction in the middle of $CB$.
Then $B \to C$; $C \to B$; $X \to X_{1}$; $Y \to Y_{1}$; $P \to P_{1}$.
Let $\omega$ be the circumcircle of the triangle $ABC$; $Q$ is the point of intersection of $AX_{1}$ and $\omega$ $\Rightarrow$ as $\angle BCX_{1} =\angle CBA$ $\Rightarrow$ $\angle ACX_{1} = 180^\circ - \angle CAB$; also we have $AC=CX_{1}$ $\Rightarrow$ $AX_{1}$ is a bissector of angle $\angle CAB$ $\Rightarrow$ $Q$ is a middle of arc $BC$ of $\omega$; $T=BA_{1} \bigcap AX_{1}$ $\Rightarrow$ as $A_{1}B \parallel CA$ $\Rightarrow$ $\angle ATB=\angle A_{1}TX_{1} = \angle A_{1}X_{1}T=\angle CAX_{1}=\angle TAB$ $\Rightarrow$ $BT=AB$ $\Rightarrow$ $T=Y_{1}$ and line $AT$ = line $X_{1}Y_{1}$ $\Rightarrow$ as $Q$ lies on the perpendicular bisector of side $BC$ $\Rightarrow$ $Q=P_{1}$ $\Rightarrow$ $\angle CPB=\angle CP_{1}B=\angle CQB=180^\circ - \angle CAB$ $\Rightarrow$ $\angle CAB +\angle CPB = 180^\circ$
what we need

Let's see this:
Suppose $O$ is the circumcentre of $\triangle XYA$.
You'll be delighted to see that $O$ lies on the perpendicular bisector of $BC$,
then,it is not difficult to prove quadrilateral $OXPB$ and $OPCY$ are cyclic.
Thus,the rest of the problem is trivial. :D
(I think these important steps are enough,so I haven't post the whole solution)

My friend [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=3230]jackhui[/url] (He's a silver medalist at IMO 2005!) has found a nice solution and I am helping him to post it here. (He seldom logs on Mathlinks.)

Let $\angle A=2\alpha$ and so $\angle AXY=\angle AYX=\alpha$. Now we construct a point $P'$ such that $P'$ is on $XY$ and $\angle P'BC=\alpha$. We continue to let $Q$ to be the intersection of the angle bisector of $\angle BAC$ and the perpendicular bisector of $BC$. Then it is easy to see that $Q$ is on the circumcircle of $\triangle ABC$ and so $\angle QBC=\angle QCB=\alpha$ and $\angle ACQ=180^\circ -\angle ABQ$. Now we construct a point $R$ outside $\triangle ABC$ such that $\triangle XBR\cong\triangle ACQ$. Since $A,B,Q,C$ are concyclic, $R,B,Q$ are collinear. Since $\angle RXP'=2\alpha =\angle P'BQ$, we know that $R,X,P',B$ are concyclic and so $RB=P'B\Rightarrow BQ=QC=RB=P'B$. Finally let $P$ be the intersection of $QM$ and $BP'$. $\triangle BMP\cong\triangle BMQ\Rightarrow BQ=BP$. Therefore, $P'$ and $P$ coincide and they are actually the point $P$ in the question.

I apologize for any steps missing. Please ask for clarification if needed. :)

[quote="leepakhin"]He's a silver medalist at IMO 2005!
[/quote]

Maybe i saw him there! :)

@GHAVR: Nice solution, but I have taken the freedom to edit your post, since writing plain text in LaTeX is pretty much a waste of space (I guess you used some program which automatically generates LaTeX code - in this case, please use this program for the formulas, and write the text normally).

Here is my solution (essentially the same as GHAVR's one, but in a simpler form):

Let the parallels to the lines CA and AB through the points B and C intersect at D. Then, the quadrilateral ABDC is a parallelogram, so that BD = CA, CD = BA and < BDC = < BAC. Since BD = CA and BX = CA, we get BX = BD, so that the triangle XBD is isosceles, and thus < BDX = < BXD. On the other hand, CD || AB yields < CDX = < BXD. Thus, < BDX = < CDX, so that the point X lies on the angle bisector of the angle BDC. Similarly, the point Y lies on the angle bisector of the angle BDC. Thus, the line XY is the angle bisector of the angle BDC. Hence, if this line XY meets the circumcircle of triangle BDC at a point Q (apart from D), then < BDQ = < CDQ; since in a circle, equal chordal angles correspond to equal chords, this yields BQ = CQ, and thus the point Q lies on the perpendicular bisector of the segment BC. So the point Q is the point of intersection of the line XY with the perpendicular bisector of the segment BC (thanks to AC > AB, such a point of intersection is unique). But this is exactly the way that we defined the point P. Consequently, the point Q must coincide with the point P, and since we know that the point Q lies on the circumcircle of triangle BDC, we can infer that the point P lies on the circumcircle of triangle BDC. Thus, < BPC + < BDC = 180°. Since < BDC = < BAC, this becomes < BPC + < BAC = 180°. That's all.

  Darij

Here is my solution(?):

  Line XY is symmetric to the angle bisector wrt radical line of two ex-circles.

  Radical line passes thru the midpoint of BC, angle bisector goes thru X - mid-arc BC,

  thus XY will intersect mid-perpendicular of BC in P symmetric to X,

  this makes BXCA inscribed.

   Thank you.

   M.T.

[b]Lemma (well-known).[/b] Let $ABC$ be a triangle. Denote: the middlepoint $M$ of the side $[BC]$; two mobile points $X\in BA$, $Y\in CA$ so that $BX=CY$ and the line $BC$ don't separate the points $X$, $Y$. Then the geometrical locus of the middlepoint of the segment $[XY]$ is the parallel line to the bisector of the angle $\widehat {BAC}$ and which passes through the point $M$.

Using the notations from the proposed problem, denote: $D\in BC$ so that $\widehat {BAD}\equiv \widehat {CAD}$; the middlepoints $U$, $V$, $M$ of the segments $[AX]$, $[AY]$, $[BC]$ respectively. Then $M\in UV$ and $AD\parallel UV\parallel XY$ a.s.o.

[quote="darij grinberg"]@GHAVR: Nice solution, but I have taken the freedom to edit your post, since writing plain text in LaTeX is pretty much a waste of space (I guess you used some program which automatically generates LaTeX code - in this case, please use this program for the formulas, and write the text normally).
[/quote]

OK  :)
(But I didn't use any program :what?:  I'm just a beginner in LaTeX (and i have some problems with my English :blush: ).
I'm realy sorry if it made you feel uncomfortable .
I'll try to follow your advice.

I just found a delightful solution to this problem. (Well, that's my opinion of course, but it's neat :))

Take point $Q$ on the angle bisector of $\angle A$ such that $AQ = PY$. It is easy to see that $AQ \parallel PY$, so $AYPQ$ is a parallellogram. Since $CY = AB$ and $\angle CYP = \angle QAB = \angle A/2$, we have $\Delta PCY \cong \Delta QBA$. Hence $BQ = CP = BP$, so $\Delta BPQ$ is isosceles and $\angle BPQ = \angle BQP$. Therefore,
$\angle BPC = 360^{\circ} - \angle CPY - \angle YPQ - \angle QPB = 360^{\circ} - \angle AQB - \angle YAQ - \angle PQB$
which is equal to $\angle AQP - \angle YAQ = (180^{\circ} - \angle A/2) - \angle A/2 = 180^{\circ} - \angle A.\ \blacksquare$

Let $(O_{B}),\ (O_{C})$ be 2 circles congruent to the triangle circumcircle (O) with circumradius R, $(O_{B})$ tangent to BC at B and $(O_{C})$ tangent to BC at C, their centers on the same side of the line BC as the point A. Let $(O_{B})$ meet AB again at X and let $(O_{C})$ meet AC again at Y. Since $(O_{B})$ is tangent to BC at B, the angles $\angle AOB = 2\angle ABC = \angle XO_{B}B$ are equal and the isosceles triangles $\triangle AOC \cong XO_{B}B$ with $OA = OC = O_{B}X = O_{B}B = R$ are congruent by SAS, so that BX = CA. In exactly the same way, CY = BA. Since $BC \ge 2R$ (with equality only for $\angle A = \frac{\pi}{2}$), the congruent circles $(O_{B}),\ (O_{C})$ intersect at P, Q (or at least touch at P) on the perpendicular bisector of BC. Let M be the midpoint of BC and let N be the midpoint of the arc BC of (O) opposite to A. The segments MP, MQ are geometric solutions of a quadratic equation $x^{2}-2Rx+\frac{a}{2}= 0$ (Descartes' method), hence

$x = R \pm \sqrt{R^{2}-\frac{a^{2}}{4}}= R \pm \sqrt{R^{2}-R^{2}\sin^{2}A}= R(1 \pm \cos A)$

Choose the signs, so that $MP = R(1-\cos A),\ MQ = R(1+\cos A).$ It follows that P is a reflection of N in BC and $Q \in (O).$ The perpendicular diagonals BC, NP of the quadrilateral BNCP cut each other in half at M, hence it is a rhombus, $BP = CP = BN = CN$ and $\angle BPC = \angle BNC.$ Since $(O_{B}) \cong (O)$ are congruent, the angles $\angle BXP = \angle BAN = \frac{\angle A}{2}$ are equal. Similarly, the angles $\angle CYP = \angle CAN = \frac{\angle A}{2}$ are also equal. Thus the lines XP, YP are both parallel to the internal bisector AN of the angle $\angle A.$ Since they pass through the same point P, they are identical, and XY cuts the perpendicular bisector of BC at P. Since the quadrilateral ABNC is cyclic,

$\angle CAB+\angle BPC = \angle CAB+\angle BNC = \pi$


:? I was wondering if there are any individual math competitions, i.e. where you can participate even if your school does not have a math team?

I am planning to take the AMC-12 next year, but I wanted to know if there are any others that I can take part in.

USAMTS -> http://www.nsa.gov/programs/mepp/usamts.html

Though it is more of a take-home thing than most other competitions like the AMC/AIME/USAMO/ARML/Mandelbrot/blah

Fierytycoon

Here's two more, although these are more international:

Tournament of Towns (IMToT): [url]http://www.amt.canberra.edu.au/imtot.html[/url] (the official site, although I find [url]http://www.math.toronto.edu/oz/turgor/index.php[/url] more helpful).

Math Olympiad Correspondence Program (MOCP): [url]http://www.cms.math.ca/Competitions/MOCP/[/url]

Oh, and one more.  Did I say two?  I lied.  The International Mathematics Talent Search, somewhat like the USAMTS, is at [url]http://www.cms.math.ca/Competitions/IMTS/[/url]

I'm not sure if everyone here has heard of the last two, so they may interest other people, also.

Would I be mistaken in saying the IMTS problems are the same as USAMTS problems? (Yes, exactly the same problems.)

No you wouldn't be mistaken. It says so right on the IMTS site itself.

Heehee...except the IMTS doesn't have any new problems on that site.  The only advantage is that it has problems from longer ago than the USAMTS site, which I believe only has back to Year 8.


Find 5 square numbers using each non zero digit once.

I gave up on this one.

[hide="Hint"]
Look for  a digit that doesn't appear often in squares.
[/hide]

[hide="Solution"]
1 9 25 36 784
[/hide]

Yes, your solution is correct, and I thank you for your time. Congrats on your first post! Welcome!

xtrememath8, 7 isn't a perfect square  :roll:

[quote="236factorial"]

xtrememath8, 7 isn't a perfect square  :roll:[/quote]

Wow, that was a stupid mistake.  :blush:

[quote="xtrememath8"][quote="chunkypotpan"][hide="Another way"]9, 25, 361, 784
It's not very different from Ritty's.[/hide][/quote]

Yeah, but he needs 5 squares[/quote]
Oh. Oops. I guess Ritty's solution is the only possible one.

You're missing the digit 7.

Basically, it's five squares using nine digits, and the square with a 7 in it must be a 3-digit number, as $7$ is not a perfect square, no numbers from 70-79 are perfect squares, and a perfect square can never end in a $7$. It isn't too hard to get from there; the combination of four squares and six digits must either be three one-digits and a three-digit, or two one-digits and two two-digits, and since the former is more restrictive (only three possible one-digit squares), you should start with the latter first.

:oops:  :blush:  :blush:  sorry
I forgot.


algebra
polynomial
number theory unsolved
number theory

Show that the polynomial

$ (x^2 \minus{} 13)(x^2 \minus{} 17)(x^2\minus{} 221)$ has no solutions in integers whereas it has solutions modulo p for all primes $ p$

It obviously has no integer solutions.
However, for mod p, consider Legendre symbol.
If $ \left(\frac{13}{p}\right)$ or $ \left(\frac{17}{p}\right)$ = 1, we are done. Otherwise, both symbols equal -1, and
$ \left(\frac{221}{p}\right) \equal{} \left(\frac{13}{p}\right) \left(\frac{17}{p}\right) \equal{} 1$, so we are done as well.


algebra
polynomial
geometry
3D geometry
quadratics

The polynomial $ f(x) \equal{} x^{4} \plus{} ax^{3} \plus{} bx^{2} \plus{} cx \plus{} d$ has real coefficients, and $ f(2i) \equal{} f(2 \plus{} i) \equal{} 0.$ What is $ a \plus{} b \plus{} c \plus{} d?$

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 16$

[hide]
If a polynomial with real coefficients has a complex roots, that root's conjugate is a root as well, so the roots are $2i$, $-2i$, $2+i$, and $2-i$

So the polynomial is
$f(x)=(x-2i)(x+2i)(x-2-i)(x-2+i)$
The sum of the coefficients is the polynomial evaluated at 1
$(1-2i)(1+2i)(1-2-i)(1-2+i)$
$(1-2i)(1+2i)(-1-i)(-1+i)$
$(1-4i^{2})(1-i^{2})$
$(5)(2)$
$10$

Since we aren't counting the coefficient of $x^{4}$, the answer is $10-1=\boxed{9}$
[/hide]

how did such a trivial problem get to #18...wtf...it is like: do you know conjugate root theorem? have you taken algebra 2?

[quote="mdk"]you'd be surprised how many people missed this question.[/quote]

... hey. It's not my fault I can't expand.

[quote="Allan Z"]lol.. I TI-89ed it, used EXPAND(f(x)) lol[/quote]

That's what I did...after the test.  If I had answered that problem and the problem before it using my calculator, I would have qualified and I wouldn't have needed to take the B during my break...  It was a really easy question.  Or, TI-89's can do too much for you. :P

It's actually not that hard to expand if you use Vieta's.

[hide="My Solution"]
The roots of the polynomial are $ 2i$, $ \minus{}2i$, $ 2 \plus{} i$, and $ 2 \minus{} i$. Note that these pairs of roots must be the roots of two quadratics with a leading coefficient of $ 1$ because $ f(x)$ has a leading coefficient of $ 1$. Consider the roots $ 2i$ and $ \minus{}2i$. The sum of these roots is $ 0$ and the product is $ 4$. Therefore, the quadratic $ x^2 \plus{} 4$ has these roots. Consider the roots $ 2 \plus{} i$ and $ 2 \minus{} i$. The sum of the roots is $ 4$ and the product of the roots is $ 2^2 \minus{} (i^2) \equal{}  4 \plus{} 1 \equal{} 5$. Therefore, the quadratic $ x^4 \minus{} 4x \plus{} 5$ has these roots. Thus, we have

$ f(x) \equal{} (x^2 \plus{} 4)(x^4 \minus{} 4x \plus{} 5)\implies f(1) \equal{} (5)(2) \equal{} 10$.

We are looking for $ f(1) \minus{} 1 \equal{} \boxed{9}$.
[/hide]

I solved it like tjhance but why are all the answer choices perfect squares? Makes me wonder if there is a slicker way to solve this...or it could be just there to throw people off. Anyone know if it's obvious it'll be a perfect square?

Here's my solution.

$\longrightarrow$ Since $f(x) \in \mathbb{R} [X]$ , we know that all complex roots exist in conjugate pairs.
Hence the roots of $ f(x) $ are $(2i), (2+i), (2-i)$ and $(-2i)$.
Now, let's expand $f(2i)=0$
$\implies f(2i) = 16 - 8ai - 4b +2ci + d = 0 \iff (16 +d) + 2ci = 4b + 8ai$
Hence we conclude that, $ 4b = 16 +d $ and $ c = 4a$.
 Now, by vieta
$ -a = (2i) + (-2i) + (2+i) + (2-i) \implies a = -4$
$ d = (2i)(-2i)(2+i)(2-i) = 20$.

Thus, $ a= -4 , c= -16 , d = 20 , b =9$ and so $\sum_{cyc} a = 9$.

[quote=mathgeniuse^ln(x)][quote="paladin8"][quote="Allan Z"]lol.. I TI-89ed it, used EXPAND(f(x)) lol[/quote]

Haha, darn. Didn't think of that. I used it on the cube truncating problem though :wink:.[/quote]

You are so lucky that tii-89 can do radicals.[/quote]

A ti-89 can do radicals? I never knew that.

[quote=Electron_Madnesss]INCONSEQUENTIAL[/quote]

Why? Why did you post another same solution and necro'ing an old post? Just solve and check your answer and move on! Why!!!!!


And if you are then say your name (unless your not comfortable doing so), if you've done MC before, what grade and school you go to.

I'll go first:

I'm Nick Sparkman
Yes i've done MC before
8th grade
Challenger

Um, if your going to post here, please at least answer the question.

ohh I will answer

let see

i am not competing in Alabama mathcounts, cause i don't live in mathcounts, but I did wish I competed in Alabama. Massachusetts is so hard to get into Nats.

Lets see

My name is Someone someone. Don't feel like releasing my name in the Alabama forum.
I have done MC for 2 years already
8th grade
Clarke

I'm not telling my name. However, I'll tell you i'm 14 and im a girl and im asian.
I've done MC for 2 years already and never made nats.
I'm from IL, in 8th grade.

*Actually, if I don't make Nationals, you'll never know my name. Ever.

WOAH, gauss is a girl....
anyways
im gene yu
im competing in alalalalalalabama mathcounts
never even made school before XD
trying for state team
12
in 8th grade


counting
distinguishability
combinatorics unsolved
combinatorics

Fifteen freshmen are sitting in a circle around a table, but the course assistant (who
remains standing) has made only six copies of today’s handout. No freshman should
get more than one handout, and any freshman who does not get one should be able to
read a neighbor’s. If the freshmen are distinguishable but the handouts are not, how
many ways are there to distribute the six handouts subject to the above conditions?

I think the result is 4320, but the solution says 125.  :?:


sphere Gauss function integration calculus

Find the net force that the southern hemisphere of a uniformly charged, insulating sphere exerts on the northern hemisphere. Express your answer in terms of the radius R and the total charge Q.

Well, this was kind of an interesting problem...
Look say that force that they exert on each other is $ F$ by magnitude. If you shrink down the radius of the southern hemisphere a little bit
the force between hemispheres won't be affected essentially. Now you complete the northern hemisphere to a full sphere, with the same surface charge density. Since the electric field in such a sphere is $ 0$ then it means that the force between the "shrunk-down" hemisphere and the completion of the northern hemisphere is $ F$. If we shrink  the northern hemisphere to the same size as we did for the southern one then we will have that a sphere of charge $ Q$ pushes the hemisphere of charge $ \frac{Q}{2}$ just on top it by force $ 2F$. It is easy to calculate that $ F=\frac{Q^{2}}{32\pi \epsilon_{0}R^{2}}$

OK. So let's denote northern hemisphere $ N$ and southern hemisphere $ S$. I take $ S$ and make its radius $ r$, where $ R-r<<R$, so the force between $ S$ and $ N$ is still $ F$. Then you add up a new hemisphere $ C$ which completes $ N$ to a full sphere of radius $ R$. $ C$, has the same surface charge density as $ N$ and $ S$. The new sphere contains $ S$ in its interior so the force on $ S$ is $ 0$. Thus, we can easily see that $ C$ acts on $ S$ by force of $ F$. By the way, let's mention that force between $ N$ and $ C$ is $ F$ too. Now if we shrink $ N$ to radius $ r$ as we did for $ S$, we get a sphere of radius $ r$ and $ C$ very close to its surface. From the argument we can see that the force on $ C$ is just $ 2F$. From this point on everything is simple. :)

[quote="crude"]OK. So let's we northern hemisphere $ N$ and southern hemisphere $ S$. I take $ S$ and make its radius $ r$, where $ R-r<<R$, so the force between $ S$ and $ N$ is still $ F$. Then you add up a new hemisphere $ C$ which completes $ N$ to a full sphere of radius $ R$. $ C$, has the same surface charge density as $ N$ and $ S$. The new sphere contains $ S$ in its interior so the force on $ S$ is $ 0$. Thus, we can easily see that $ C$ acts on $ S$ by force of $ F$. [/quote]
can u explain a bit more about these

Do you mean the last part, where we deduce the actual force, because I think I gave pretty detailed explanation of the solution...

i mean how reducing radius still keeps the force same...

Of course it changes the force... but since the change in radius is negligibly small, then the variation of force is also negligibly small...

Yes, $ R-r$ is the change of radius and $ R$ is the radius at which we vary it.

yes that's right.so how does that ever matter u need to calculate $ F$ ....how do arrive at that value

I would suggest you to draw carefully the diagram that is made up from my solution and use a couple of times Newton's third law.

[quote="crude"] If we shrink  the northern hemisphere to the same size as we did for the southern one then we will have that a sphere of charge $ Q$ pushes the hemisphere of charge $ \frac{Q}{2}$ just on top it by force $ 2F$. It is easy to calculate that $ F=\frac{Q^{2}}{32\pi \epsilon_{0}R^{2}}$[/quote]
can u explain that charge distribution and how to get
$ F=\frac{Q^{2}}{32\pi \epsilon_{0}R^{2}}$

So we have the a sphere and hemisphere on top of it. The surface charge density is uniform on both of those objects and is equal to $ \sigma =\frac{Q}{4\pi R^{2}}$, which is simply obtained by dividing the total charge by the surface area of the object. The electric pressure on hemisphere is simply $ E\cdot \sigma$, where $ E$ is the electric field at the surface of hemisphere and is equal to $ \frac{Q}{4\pi \epsilon_{0}R^{2}}$ so the electric pressure is $ p = E\cdot \sigma= \frac{Q^{2}}{16 \pi^{2}\epsilon_{0}R^{4}}$. If pressure $ p$ acts on hemisphere then it exert a force equal to $ p \pi R^{2}$, because if you close a hemisphere from the bottom and there is a uniform pressure acting from the inside then the total force on it is $ 0$. Thus, $ 2F = p \pi R^{2}$
$ F=\frac{Q^{2}}{32 \pi \epsilon_{0}R^{2}}$.

how is the electric field equal  to $ \frac{Q}{4\pi \epsilon_{0}r^{2}}$

It is simply the field of the sphere close to its surface, you can easily get it by Gauss' law...

but u cannot use that here because the question says
[quote="dominicleejun"]Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. Express your answer in terms of the radius R and the total charge Q.[/quote]
but the field formula u have used is the field due to all the charges but u cannot include the charges on one side of the hemisphere while calculating the force on it

[quote="crude"]By the way, let's mention that force between N and C is F too.[/quote]
Why? Could you explain that?

Because it is almost the same configuration as for $ N$ and $ S$, with slightly changed radii... therefore as in that configuration the force between $ N$ and $ C$ is $ F$.

of course it is an insulating solid sphere.

sorry, I thought it was not solid, but anyways I've solved for this one too, it is essentially the same...
So, let's say the force between hemispheres of radii $ R$ is $ F(R)$(of course it is also a function of $ Q$, but to avoid problems we consider it as a function of charge density $ \rho$ which is same for our problem). Let northern hemisphere be $ N$ and southern hemisphere $ S$. Define $ T_{N}$ as the shell of width $ dR$ ($ dR<<R$) , $ C_{N}$ is the solid core of northern hemisphere of radius $ R-dR\approx R$. The same way define the stuff for $ S$. The force between $ N$ and $ S$ can be decomposed into forces, between $ T_{N}$ and $ T_{S}$, $ T_{N}$ and $ C_{S}$, $ C_{N}$ and $ T_{S}$, and finally $ C_{N}$ and $ C_{S}$. The charges on both $ T_{S}$ and $ T_{N}$ are of first order magnitude, therefore the force between them is of second order therefore we will not take that force into consideration. The force between $ C_{N}$ and $ C_{S}$ is $ F(R-dR)$. The forces between $ C_{S}$ and $ T_{N}$, and, $ T_{S}$ and $ C_{N}$, is the same evidently and let it be equal to $ f$. Actually this infinitesimal force can be determined exactly by the same process I mentioned for the problem where spheres are not solid. Let me skip the first part of construction, because it's messy:
$ 2f=p\pi R^{2}= E \sigma \pi R^{2}= \frac{Q}{4\pi \epsilon_{0}R^{2}}\cdot \rho dR\cdot \pi R^{2}= \frac{\pi \rho^{2}R^{3}dR}{3 \epsilon_{0}}$. Also, on the other hand $ 2f = dF$ , which means that
$ F=\int dF= \int \frac{\pi \rho^{2}R^{3}dR}{3 \epsilon_{0}}= \frac{\pi \rho^{2}R^{4}}{12 \epsilon_{0}}= \frac{3Q^{2}}{64 \pi \epsilon_{0}R^{2}}$, which is the answer you desired.
I liked the first one because it involved more physics ... cool problem though

Let's consider generic shells with radius $ r_{1}$  and $ r_{2}$ and infinitesimal thickness $ dr_{1}$ and $ dr_{2}$, each in one of the hemispheres.

The force between the shells is

\[ d^{2}F ={{\rho^{2}\pi{r_{1}}^{2}}\over \epsilon_{0}}dr_{1}dr_{2}.\]

Witha double inegral, from 0 til R , we get

 \[ F ={{\rho^{2}\pi R^{4}}\over{3\epsilon_{0}}}={{3 Q^{2}}\over{16\pi\epsilon_{0}R^{2}}}\]

immanuel's explaination is right...but is there any non integration method

You're right crude.

So, dominicleejun, where from comes your answer?
[quote="dominicleejun"]
The answer is actually (3*k_e*Q^2)/(16R^2).
Gosh... the answer is so unlatex like... sorry.[/quote]

$ k_{e}$ is simply $ \frac{1}{4 \pi \epsilon_{0}}$

Of course!
I didn't notice the k.
Thanks.


If $ a,\ b,\ c \geq 1$, show that $ 4(abc+1) \geq (a+1)(b+1)(c+1)$. When do we have equality?


Find the sum of the roots of the polynomial x^2001 +
( - x)^2001.  Anyone have a good way to solve this problem?

Quote:Find the sum of the roots of the polynomial x^2001 +
( - x)^2001.

Solution:

[hide]Expand using the binomial theorem. Then we have 2001*1/2*x^2000 - 2001*2000/2*1/4*x^1999+... The sum of the roots is just (2001*2000/2*1/4)/(2001*1/2)=2000/4=500/[/hide]

How am I using it wrong? I just started writing the terms from the other end.  Can you explain?

Edit: I use the binomial theorem as

(x+y)^n = Sum (k=0 to n) nCk*x^(n-k)*y^k.

I think that you can make the change nCk to nCn-k to get some equivalent versions. But ultimately its all the same.

why do you have to multiply an extra 1/2 to the quotient? When I apply the formula, I get 1000.

I did this: (-1*2001*2000*1/2)/(-1*2001*1/4) =1000

Can someone explain what I am doing wrong?

[quote]sorry, i misread[/quote]

You were right, i put down 2001 instead of 2000 as the power of x for the first term. I noticed this right after my second post. The reason we got the same answer was that it was just a typo. Edited it right after.

[quote]I did this: (-1*2001*2000*1/2)/(-1*2001*1/4) =1000 [/quote]

on the numerator, you need to multiply by (1/2)^2, coming from the 1/2 term. on the bottom, i dont understand why you get 1/4. It should just be 1/2. By the way, as you posted it, the sum equals 4000, not 1000.

Oops, this is  what I really did: (-1*2001*2000*1/4)/(-1*2001*1/2) =1000
What I mean to ask now is why is there supposed to be an extra 1/2 in the numerator as in the solutions of you guys?

haha, wow, i feel so foolish! Thank you all for your solutions! I really appreciate them.


Jenna is at a fair with four friends. They all want to ride the roller coaster, but only three people can fit in a car. How many different groups of three can the five of them make?

Hello
[hide="solution"]
$ \binom{5}{3}\equal{}10$
[/hide]

Note: the notation above in combinations.

Let the people be

A,B,C,D,E

Clearly, there are $ 5$ choices for the first, $ 4$ for the second, $ 3$ for the third.

However, ABC is the same as BAC is it not?

ABC=ACB=BAC=BCA=CAB=CBA.

So we divide by $ 3! \equal{} 6$ to get $ 10$.

Note this is equal to $ \frac {5!}{3!2!}$, which is $ \binom{5}{2}$ by definition.

Hope this helped  :)

Actually, it's 5 choose 3

@bogtro answer is correct but it's 5 choose 3 (just correcting)

How do you do N choose K notation with Latex?

You click on the LaTeX he typed....

[code]
$\binom{5}{3}$
[/code]
gets $\binom{5}{3}$

${5\choose3}={5\choose2}$ by commutability so there really is no difference between writing the two.


[quote="Tony2006"]este fue un problema del selectivo de yucatan del 2004:

los reales $a$ y $b$ son tales que $a^3 - 3a^2 + 5a = 1$ y $b^3 - 3b^2 + 5b = 5$. Hallar $a + b$.

que pongan sus ideas o comentarios y por supuesto soluciones.[/quote]

[hide="también hay que"]factorizar y ver que las $2$ ecuaciones son iguales [/hide]

como que factorizar y ver que las dos ecuaciones son las mismas?
explicame por favor!!

[quote="José"]Las dos ecuaciones son de la forma $k^3-3k^2+5k$.[/quote]

Yo creo que si te decides a ayudar, entonces ayuda de verdad ,con comentarios como ese no dices nada.

$Tipe$

[quote="José"]Las dos ecuaciones son de la forma $k^3-3k^2+5k$.[/quote]

No entiendo como ayuda eso :D

Carlos br., Pascual 2005; pueden ayudar con este problema?

Un buen cambio de variable resuelve este problema, veamos cúal debe ser el cambio:

$a^3-3a^2+5a=(a^3-3a^2+3a-1)+(2a+1)=(a-1)^3+2(a-1)+3$
Sea $x=a-1$ , la condición $a^3-3a^2+5a=1$ es equivalente a $x^3+2x=-2$...(1).
Análogamente, si definimos $y=b-1$, la condición $b^3-3b^2+5b=5$ es equivalente a $y^3+2y=2$...(2)
Luego sumando (1) y (2): $x^3+y^3+2x+2y=0 \Leftrightarrow (x+y)(x^2+y^2-xy+2)=0$  , como el factor de la derecha es positivo siempre, entonces $x+y=0$, de donde $a+b=2$,

$Tipe$


Let $x,y \in Z$ such that $x^2+y^2+3\geq 6xy$.
Prove that $x^2+y^2\geq 6xy$


How many units are in the sum of the lengths of the three altitudes in a triangle with sides 7, 24, and 25? Express your answer as a decimal to the nearest hundredth.

[hide]We notice that $25^2 = 24^2+7^2$, so the triangle is right.

Hence the two altitudes are the perpendicular sides $7$ and $24$

If $x$ is the altitude corresponding to the hypotenuse , then from the formula of the area we get:

$2 \times Area = 25 \cdot x = 7 \cdot 24 \Rightarrow$

$x=6,72$

Hence the sum is $6,72+24+7 = 37,72$[/hide]

[hide="solution"]
Recognize right away that the triangle is right because $24^2+7^2=25^2$. That means that two of the altitudes are the two legs. The legs are the shorter sides in a right triangle, so they are $7 \text{ and } 24$. Also know that using the formula for the area of a triangle, $A=\frac 12 bh$, and using $24$ and $7$ for the bases, the area is $84$. So, applying the formula again, if you make the base the longest side, $25$, then $84=\frac 12 25h$. Simplify for $84=12.5h$ and divide both sides by $12.5$, and get $h=6.72$. So, add up the three bases, $6.72+24+7=\boxed{37.72}$[/hide]


analytic geometry
probability
parameterization
probability and stats

The six coordinates $ (X_i,Y_i)$,$ 1\le i \le 3$, of three points $ A,B,C$ in the plane are independent $ N(0,1)$. Show that the probability that $ C$ lies inside the circle with diameter $ AB$ is $ \frac{1}{4}$.

We have four events:

$ E_1:$ angle $ A$ is obtuse.
$ E_2:$ angle $ B$ is obtuse.
$ E_3:$ angle $ C$ is obtuse.
$ E_4:$ all three angles are acute.

These four events are mutually exclusive and almost exhaustive. (The only thing left out are the right triangles, which have probability zero.) Clearly, $ P(E_1)\equal{}P(E_2)\equal{}P(E_3),$ and we were asked for $ P(E_3).$ The claim is equivalent to claiming that $ P(E_4)\equal{}\frac14$ or even $ P(E_4)\equal{}P(E_1).$

- haven't gotten this to lead anywhere yet. One issue: If $ X$ and $ Y$ are independent standard normals and $ U\equal{}XY,$ what kind of distribution does $ XY$ have?

Hint:
[hide]the required probability equals:
$ P\left([X_3 \minus{} \frac {1}{2}(X_1 \plus{} X_2)]^2 \plus{} ([Y_3 \minus{} \frac {1}{2}(Y_1 \plus{} Y_2)]^2 \le \frac {1}{4}(X_1 \minus{} X_2)^2 \plus{} \frac {1}{4}(Y_1 \minus{} Y_2)^2 \right)\equal{}$
$ P(U_{1}^2\plus{}U_{2}^2 \le V_{1}^2\plus{}V_{2}^2)$,
where $ U_1,U_2,V_1,V_2$ are independent Normal random variables.[/hide]

The required probability equals
$ P\left([X_3 \minus{} \frac {1}{2}(X_1 \plus{} X_2)]^2 \plus{} ([Y_3 \minus{} \frac {1}{2}(Y_1 \plus{} Y_2)]^2 \le \frac {1}{4}(X_1 \minus{} X_2)^2 \plus{} \frac {1}{4}(Y_1 \minus{} Y_2)^2 \right) \equal{}$
$ P(U_{1}^2 \plus{} U_{2}^2 \le V_{1}^2 \plus{} V_{2}^2)$,
where $ U_1,U_2$ are $ N(0,3/2)$, $ V_1,V_2$ are $ N(0,1/2)$ and $ U_1,U_2,V_1,V_2$ are independent.
The answer is therefore
$ p\equal{}P\left(\frac{3}{2}(N_1^2\plus{}N_2^2) \le \frac{1}{2}(N_3^2\plus{}N_4^2)\right)$ where $ N_i$ are independent $ N(0,1)$
$ \equal{}P(K_1 \le \frac{1}{3}K_2)$ where $ K_i$ are independent chi-squared with parameter 2
$ \equal{}P\left(\frac{K_1}{K_1\plus{}K_2} \le \frac{1}{4}\right)\equal{}P\left(B \le \frac{1}{4}\right)\equal{}\frac{1}{4}$ where $ B$ is a beta with parameters 1,1.


find all polynomial   $P(x)$   such that :
              $P(x)P(x+1)=P(x+P(x))$ ; $\forall x \in R$

Let $degP(x)=m. \ If \ m>1 \ we \ have \ 2m=m^2, \ therefore \ m=2.$
If $m=1 we \ have \ left \ deg \ge2 \ right \deg \le1.$
Case 0. $m=0 \ \Longrightarrow P(x)=0 \ or \ P(x)=1.$
Case 1. $m=2 \ P(x)=ax^2+bx+c \Longrightarrow P(x)=x^2+bx+c, %Error. "have" is a bad command.
\ a=1 \ b \ and \ c \ any.$


Anyone watching hockey at the winter olympics? I think what is going on is crazy, bad teams are beating favorites, etc. How can the Swiss beat Canada and the Czecks, how can USA tie Latvia?

Ahhh i can't believe Canada lost....so sad...they'll come back, just watch!   :nhl:

Canada had over 17 power-play minutes.  Evidently, they were sloppy.

Are they playing now? Against what country? What's the result? :P

puuhuki, I cant believe Finland last today after such a great tournament! They were supposed to dominate Sweden. Oh well, congratulations to Finland for having a great tourney, I wish they would win the final game even though I wanted Russia to win the thing.

[quote="hello"]puuhuki, I cant believe Finland last today after such a great tournament![/quote]
Well, that is ice-hockey. Sweden is always a hard team to win and many times Sweden have won Finland. But anything can happen in sports. Before the tournament I thought Finland will be about fifth and before the final I thought Finland will be number one. So it is good that betting is not a hobby of mine.  ;)

*sigh*  Finland played some great games.  Canada on the other hand...oh so sad.  Oh well, we'll see what happens in Vancouver 2010!


Let x,y,z>0,prove that

$ \frac{(z\plus{}x)(x\plus{}y)}{z\plus{}2x\plus{}y}\plus{}\frac{(x\plus{}y)(y\plus{}z)}{x\plus{}2y\plus{}z}\plus{}\frac{(y\plus{}z)(z\plus{}x)}{y\plus{}2z\plus{}x} \geq  \frac{1}{2}(x\plus{}y\plus{}z)\plus{}\frac{yz}{y\plus{}z}\plus{}\frac{zx}{z\plus{}x}\plus{}\frac{xy}{x\plus{}y}.$

[quote="xzlbq"]Let x,y,z>0,prove that

$ \frac {(z \plus{} x)(x \plus{} y)}{z \plus{} 2x \plus{} y} \plus{} \frac {(x \plus{} y)(y \plus{} z)}{x \plus{} 2y \plus{} z} \plus{} \frac {(y \plus{} z)(z \plus{} x)}{y \plus{} 2z \plus{} x} \geq \frac {1}{2}(x \plus{} y \plus{} z) \plus{} \frac {yz}{y \plus{} z} \plus{} \frac {zx}{z \plus{} x} \plus{} \frac {xy}{x \plus{} y}.$[/quote]

S.O.S can solve it easy.

$ \sum{\frac {(z \plus{} x)(x \plus{} y)}{z \plus{} 2x \plus{} y}} \ge \frac {1}{2}(x \plus{} y \plus{} z) \plus{} \sum{\frac {xy}{x \plus{} y}}$

$ \iff \sum{\left(\frac {x \plus{} y}{4} \minus{} \frac {xy}{x \plus{} y}\right)} \ge \sum{\left(\frac {y \plus{} z}{2} \minus{} \frac {(z \plus{} x)(x \plus{} y)}{z \plus{} 2x \plus{} y}\right)}$

$ \iff \sum{\frac {(x \minus{} y)^2}{4(x \plus{} y)}} \ge \sum{\frac {y^2 \plus{} z^2 \minus{} 2x^2}{2(z \plus{} 2x \plus{} y)}}$

$ \iff \sum{\frac {(x \minus{} y)^2}{4(x \plus{} y)}} \ge \sum{\left( \frac {y^2 \minus{} x^2}{2(z \plus{} 2x \plus{} y)} \plus{} \frac {x^2 \minus{} y^2}{2(x \plus{} 2y \plus{} z)} \right)}$

$ \iff \sum{\frac {(x \minus{} y)^2}{4(x \plus{} y)}} \ge \sum{\frac {(x \minus{} y)^2(x \plus{} y)}{2(z \plus{} 2x \plus{} y)(x \plus{} 2y \plus{} z)}}$

$ \iff \sum{\left(\frac {1}{4(x \plus{} y)} \minus{} \frac {x \plus{} y}{2(z \plus{} 2x \plus{} y)(x \plus{} 2y \plus{} z)}\right)(x \minus{} y)^2} \ge 0$

$ \iff \sum{\frac {z^2 \plus{} 3z(x \plus{} y) \plus{} xy}{4(x \plus{} y)(z \plus{} 2x \plus{} y)(x \plus{} 2y \plus{} z)}(x \minus{} y)^2} \ge 0,$

clear.

But
$ {\frac {x \plus{} y \plus{} z}{t}} \plus{} 2\, \left( 3 \minus{} 2\,{t}^{ \minus{} 1} \right) \left( {\frac {yz }{y \plus{} z}} \plus{} {\frac {zx}{z \plus{} x}} \plus{} {\frac {xy}{x \plus{} y}} \right) \left( t \plus{} 2 \right) ^{ \minus{} 1}\leq {\frac {\left( z \plus{} x \right) \left( x \plus{} y \right) }{t x \plus{} y \plus{} z}} \plus{} {\frac {\left( x \plus{} y \right) \left( y \plus{} z \right) }{ty \plus{} z \plus{} x}} \plus{} {\frac {\left( y \plus{} z \right) \left( z \plus{} x \right) }{tz \plus{} x \plus{} y}}.$

$ 0.8071131713\equal{}1/12\,\sqrt [3]{199 \plus{} 24\,\sqrt {249}} \minus{} {\frac {47}{12}}\,{\frac {1}{ \sqrt [3]{199 \plus{} 24\,\sqrt {249}}}} \plus{} {\frac {7}{12}}\leq t.$

and,please:

$ 2\, \left( x\plus{}y\plus{}z \right) ^{2} \left( {\frac {zy}{y\plus{}z}}\plus{}{\frac {zx}{z\plus{}x
}}\plus{}{\frac {xy}{x\plus{}y}} \right) \leq  \left( {\frac { \left( z\plus{}x \right)
 \left( x\plus{}y \right) }{z\plus{}2\,x\plus{}y}}\plus{}{\frac { \left( x\plus{}y \right)  \left( y
\plus{}z \right) }{x\plus{}2\,y\plus{}z}}\plus{}{\frac { \left( y\plus{}z \right)  \left( z\plus{}x
 \right) }{y\plus{}2\,z\plus{}x}} \right) ^{3}
.$
BQ


Let $ f(x) \equal{} \frac {2x}{x^2 \minus{} 5x \plus{} 6}$

Find $ f^{(n)}(x)$ ?

This is the wrong forum for calculus problems. (Announcements: "PLEASE do not post calculus problems in this forum!")

Assuming $ f^{(n)}(x) \equal{} \frac {d^n}{dx^n}f(x)$
[hide="Big Hint on making it simpler"]$ f(x) \equal{} \frac {2x}{x^2 \minus{} 5x \plus{} 6} \equal{} \frac {2x}{(x \minus{} 3)(x \minus{} 2)} \equal{} \frac {2x}{x \minus{} 3} \minus{} \frac {2x}{x \minus{} 2} \equal{} \frac {6}{x \minus{} 3} \minus{} \frac {4}{x \minus{} 2}$[/hide]


[hide="hint"]
The two legs are $x$ and $48+x$. Then you use pythagorean theorum to solve
[/hide]

[hide][quote="linzk3"]If the height of the right triangle is 48 units larger than the base and the hypotenuse is 60 find the area[/quote]

a^2+(a+48)^2=60
solve, then multiply a and a+48 and divide by 2[/hide]


[quote="m1ght"]let a,b,c be the 3 sides of a triangle

prove that [color=red]a/(b+c)+b/(a+c)+c/(a+b)<2[/color][/quote]
We have: $ \frac{a}{b\plus{}c} <\frac{2a}{a\plus{}b\plus{}c} \Leftrightarrow a<b\plus{}c$, true!
Similarly, we are done!

Find the smallest real number $ K$ such that $ \sum \frac{a}{b\plus{}c}<K\,\forall\Delta ABC$. :P

[url=http://www.animath.fr/cours/inegalites.pdf]Take a look page 3.[/url]

Hint to another solution:

Use Ravi Substitution

plus: $ \sum \frac{x\plus{}y}{y\plus{}2z\plus{}x} \leq \sum \frac{x\plus{}y}{y\plus{}z\plus{}x}$

[quote="N.T.TUAN"]Find the smallest real number $ K$ such that $ \sum \frac {a}{b \plus{} c} < K\,\forall\Delta ABC$. :P[/quote]
I remember that darij has made a better bound than $ K\equal{}2$,but i can't find the link now. :(

[quote="stephen38"][quote="N.T.TUAN"]Find the smallest real number $ K$ such that $ \sum \frac {a}{b \plus{} c} < K\,\forall\Delta ABC$. :P[/quote]
I remember that darij has made a better bound than $ K \equal{} 2$,but i can't find the link now. :([/quote]

There cannot be any better bound - at least not a constant.

But a stronger inequality was posted by Monique and proved by me at http://www.mathlinks.ro/Forum/viewtopic.php?t=42896 .

  darij


geometry
circumcircle
inequalities unsolved
inequalities

Let triangle $ ABC$. Prove that
15R^2+2R\sum\sqrt{R^2+b^2+c^2-a^2}\leq 3(a^2+b^2+c^2)

hello, do you mean
$ 15R^2\plus{}2R\sum\sqrt{R^2\plus{}b^2\plus{}c^2\minus{}a^2}\leq 3(a^2\plus{}b^2\plus{}c^2)$
Is $ R$ the circumradius of the triangle?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean
$ 15R^2 \plus{} 2R\sum\sqrt {R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2}\leq 3(a^2 \plus{} b^2 \plus{} c^2)$
Is $ R$ the circumradius of the triangle?
Sonnhard.[/quote]
It's wrong. The counterexample see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216868


in atriangle with integral sides and integral area (a heronian triangle)prove the a median and a side can not be of the sam length

Clearly, if median equals side length then it is the connecting side. Now just use median formulas in conjunction with the side length to show they can't be equal

You just threw a whole lot of hand-waving at that question.

[quote="me@home"]Clearly, if median equals side length then it is the connecting side.[/quote]

I'm not exactly sure what this means, but I think it's wrong:  If $AB = 3$, $BC = 2$ and $CA = \sqrt{11}$, the median to side $BC$ has length 3.


abstract algebra
group theory
LaTeX
superior algebra
superior algebra unsolved

C_H(A) is the set of elements a belonging of A /ah=ha
Let A be a maximal abelian normal subgroups of H
then C_H(A)=A (WHY? HOW TO DEMONSTRATE?)
thank you!
SASTRA

it's very hard to understand you ... please work on your english, and try to use LaTeX!

I guess you mean the following:

if $H$ is a group, $A$ a subgroup of $H$, define
$C_{H}(A) = \{h \in H : \forall a \in A : ah = ha\}$
if $A$ is a maximal abelian normal subgroup of $H$, then $C_{H}(A)=A$.

but this is very very easy. notice that $A \subseteq C_{H}(A)$ since $A$ is abelian, and use that $A$ has a maximality property!


Find the value of $ 2\left(\frac{7!}{3!4!}\right)$


geometry
geometric transformation
reflection
function
complex analysis
complex analysis theorems

Let $ f$ be analytic on the open unit disk $ D$ and continuous on its closure $ \bar{D}$. If $ |f(z)|\equal{}1$ on the boundary of the disk $ \partial D$, prove that $ f$ is a rational function. I know that I have to consider the case when $ f\neq 0$ in $ D$. I think its a simple enough question (i.e. use Liouville's th. and the result follows immediately), but intuitively idk how such a nice result should follow from these simiple conditions!

Thanks in advance,

Los

The no-zeros case is pretty easy; $ |f|$ can't have a local minimum except where it's zero, so $ |f|\ge 1$ on the closed disk; together with the maximum principle, $ |f|$ is identically $ 1$ and $ f$ must be constant.

On the general problem, I'd use reflection. Extend $ f$ to the plane by $ f\left(\frac1z\right)\equal{}\left(\overline{f(\overline{z})}\right)^{\minus{}1}$; that extension is analytic outside the disk and matches $ f$ on the circle. Zeros inside reflect to poles outside; since there are only finitely many zeros in the disk, there are only finitely many poles and no other singularities outside. Even $ \infty$ can't be worse than a pole, so $ f$ is meromorphic on the extended plane, and must be rational.

Thanks!

The second case is like the theorem that says that all meromorph functions in $ \bar{\mathbb{C}}$ must be rational, except it´s for a disk!

cf. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134677

Los


[quote="JimYall"]hi i need help on these questions below... please be kind to help me out :'(

x = not times sign..

Question 1:

3y-7x=x
3x-1=y

Question 2:

y+2x=10+4y
4(x+y)=42-y[/quote]

[hide=" number 1"]$3y-8x=0$
$3x-y=1$
$-3y+9x=3$
$x=3$
$y=8$[/hide]

sorry i dont have time to do the other one I hope that helped though!

[quote="JimYall"]hi i need help on these questions below... please be kind to help me out :'(

x = not times sign..

Question 1:

3y-7x=x
3x-1=y

Question 2:

y+2x=10+4y
4(x+y)=42-y[/quote]
[hide="Hint for Question 1"]
$3y-7x=x$
$3x-1=y$

$3y-8x=0$
$3x-y-1=0$

$-8x+3y=0$
$3x-y=1$

You should be able to solve it from there! [/hide]

[hide="Hint for Question 2"]$y+2x=10+4y$
$4(x+y)=42-y$

$-3y+2x=10$
$4x+4y=42-y$

$2x-3y=10$
$4x+5y=42$

You should be able to solve it from there! [/hide]


[quote="catcurio"]Source: 2005 AMC-10B #11

The first term of a sequence is $2005$. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the $2005^{\text{th}}$ term of the sequence?

$\text{(A)} \ 29 \qquad \text{(B)} \ 55 \qquad \text{(C)} \ 85 \qquad \text{(D)} \ 133 \qquad \text{(E)} \ 250$[/quote]
[hide]
(E) $250$
Terms : 2005, 133, 55, 250, 133 ... etc.
2004'th term is 55, 2005th term is 250 [/hide]

[quote="catcurio"]Source: 2005 AMC-10B #11

The first term of a sequence is $2005$. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the $2005^{\text{th}}$ term of the sequence?

$\text{(A)} \ 29 \qquad \text{(B)} \ 55 \qquad \text{(C)} \ 85 \qquad \text{(D)} \ 133 \qquad \text{(E)} \ 250$[/quote]

I got 133 (D).

[quote="coffeym"][hide]Terms: $\mbox{2005, 133, 55, 250, 133...}$

If we ignore $\mbox{2005}$ then the 2005th term is really the 2004th.  This divides fully into 3, making the 2005th term equal to the 4th term:

$\Rightarrow\mbox{250 (E)}$[/hide][/quote]

It clearly says 2005 is the first term :)

He means the 2004th term of the 133, 55, 250,133, 55, 250,... sequence since we don't consider 2005 fue to teh fact that it is not a part of teh repeating sequence.

[hide]The sequence goes :

2005, 133, 55, 250, 133, 55, 250, 133, ...

Every three terms, after the first, repeat.

Since $2005\equiv1\pmod3$, it should be the same thing as the fourth term, since it is also 1 (mod 3).

So the 2005th term is 250. [/hide]


The geometr progress is $ \\S_{20}\equal{}230$ and $ \\S_{45}\equal{}1080$ then $ \\S_{80}\equal{}?$

this looks messy.
suppose the first term is a and the common ratio is r
then the terms are $ a, ar, ar^2, ar^3 \ldots$
we have $ S_{20}\equal{}ar^{19}\equal{}230$ and $ S_{45}\equal{}ar^{44}\equal{}1080$
dividing the first equation into the second gives
$ \frac{ar^{44}}{ar^{19}}\equal{}\frac{1080}{230} \implies r^{25}\equal{}\frac{1080}{230} \implies r\equal{}\sqrt[25]{\frac{108}{23}}$
we want $ S_{80}\equal{}ar^{79}\equal{}\left(ar^{44}\right) \cdot r^{35}\equal{}S_{45} \cdot r^{35}\equal{}1080 \cdot \left(\sqrt[25]{\frac{108}{23}}\right)^{35}\equal{}1080 \cdot \frac{108}{23}^{\frac75}$

this may be the final answer depending on what form you want it in


function
complex numbers
complex analysis
complex analysis unsolved

Let $ f: D(0,1)\rightarrow G$ be a biholomorph function and let $ A_r\equal{}\{z\in \mathbb{C}, |z|<r\}$ if $ G$ is convex , prove that for all $ 0<r<1$ ,$ f(A_r)$ is convex .

Ps , this is my homework

This is a corollary of the subordination principle.

[quote]subordination principle[/quote]

I couldn't find it. Could you at least give me a reference?  :(

It's really just the Schwarz lemma:
Let $ f,g$ be analytic on $ \mathbb{D}$ with $ f(0) \equal{} g(0)$, $ f$ univalent and $ g(\mathbb{D}) \subset f(\mathbb{D})$ Then for every $ r$, $ g(A_r) \subset f(A_r)$.

In this case, for $ z,w \in A_r$ let $ g(\zeta) \equal{} tf(\frac {z}{w} \zeta) \plus{} (1 \minus{} t)f(\zeta)$.

The more interesting proposition here is this:
Let $ f$ be analytic in $ \mathbb{D}$ with $ f(0) \equal{} 0$. Then $ f$ is injective with convex range iff $ f'(0) \neq 0$ and $ 1 \plus{} \frac {zf''(z)}{f'(z)}$ has positive real part in the disc. It follows by looking at the tangents of the curves $ f(\{ z \, : \, |z|\equal{}r \})$, the key observation being that since we have now proven the curve to be convex we know its tangent has a monotone argument. Differentiating the argument gives the condition.

can you please post a solution in more details.
thanks.


They are related to the [url=http://mathworld.wolfram.com/BernoulliNumber.html]Bernoulli numbers[/url].

You can also find the formula for $ k$ if you know the formulas for $ 1, 2, ..., k-1$:

$ (n+1)^{k+1}-n^{k+1}=\sum_{i=0}^{k}\binom{k+1}{i}n^{i}$
\[ \implies (n+1)^{k+1}=\sum_{j=0}^{n}\sum_{i=0}^{k}\binom{k+1}{i}j^{i}\]

but it appears there is a lot of computation involved for large $ k$.  But in general, this tells you it has to be a polynomial with degree $ k+1$ and leading coefficient $ \frac{1}{k+1}$.


[hide="Solution."]
The palindrome must be formed by an eight-digit number and its reverse.  Call the number $ abcdefgh$ and its reverse $ hgfedcba$.  The entire sixteen-digit number would then be equivalent to $ (10^7\plus{}1)a$ $ \plus{}$ $ (10^6\plus{}10)b$ $ \plus{}$ $ (10^5\plus{}100)c$ $ \plus{}$ $ (10^4\plus{}1000)d$ $ \plus{}$ $ (1000\plus{}10^4)e$ $ \plus{}$ $ (100\plus{}10^5)f$ $ \plus{}$ $ (10\plus{}10^6)g$ $ \plus{}$ $ (1\plus{}10^7)h$.  It can be verified that $ 10^7\plus{}1$, $ 10^6\plus{}10$, $ 10^5\plus{}100$, and $ 10^4\plus{}1000$ are all divisible by $ 11$; consequently, the entire palindrome must be divisible by $ 11$.  As a result, all the numbers from $ 10000000$ to $ 99999999$ form viable palindromes, yielding $ 90000000$ total palindromes.  Dividing this by $ 11$ yields a remainder of $ \boxed{2}$.[/hide]

[hide]Um...all 16 digit palindromes are divisible by 11, and there are 9*10^7(?, not sure...) 16-digit palindromes, and since 89999998 is divisible by 11, n=2[/hide]

[quote="Illos"]Let $ n$ represent the number of 16 digit palindromes divisible by 11. What is the remainder when $ n$ is divided by 11?[/quote]
[hide="Solution"]Let the number be $ \text{abcdefghhgfedcba}$, where each letter stands for a digit.

Since $ a \plus{} c \plus{} e \plus{} g \plus{} h \plus{} f \plus{} d \plus{} b$ always equals $ b \plus{} d \plus{} f \plus{} h \plus{} g \plus{} e \plus{} c \plus{} a$, all 16-digit palindromes are divisible by $ 11$.

The first digit can be $ 1$ through $ 9$, the next seven can each be $ 0$ through $ 9$, and the remaining $ 8$ will be determined by the first $ 8$.

There are $ 9\cdot10^7 \equal{} 90000000$ 16-digit palindromes, so the answer is $ 90000000\equiv\boxed{2}\pmod{11}$.[/hide]

[hide="a slightly faster way to determine the final solution mod 11"]

$ 9\cdot10^{7} \pmod{11}$
$ \equal{}\minus{}2 \cdot (\minus{}1)^7 \pmod{11}$
$ \equal{}\minus{}2 \cdot \minus{}1 \pmod{11}$
$ \equal{}2 \pmod{11}$
[/hide]


limit
trigonometry
LaTeX
function
calculus
calculus computations

hello
find: $ \lim_{x \rightarrow \pi}\frac{1}{x\minus{}\pi}(\sqrt{\frac{4cos^2x}{2\plus{}cosx}}\minus{}2)$

If $ x$ is close enough to $ \pi$,

$ \sqrt{\frac{4 \cos^2(x)}{2\plus{}\cos(x)}}\minus{}2$

can be simplified to

$ \minus{}2\left(\frac{\cos(x)}{\sin(\frac{x}{2})}\plus{}1\right)$

Now we have:

$ \minus{}2\lim_{x \rightarrow \pi}\frac{1}{x\minus{}\pi}\left(\frac{\cos(x)}{\sin(\frac{x}{2})}\plus{}1\right)$

Use L'Hopital's rule and get:

$ \minus{}2\lim_{x \rightarrow \pi}\frac{1}{(x\minus{}\pi)'}\left(\frac{\cos(x)}{\sin(\frac{x}{2})}\plus{}1\right)'\equal{}$

$ \equal{}\lim_{x \rightarrow \pi}\sin\left(\frac{x}{2}\right)\left(\cot\left(\frac{x}{2}\right)\plus{}\sin(x)\right)\equal{}0$

Helo
Thank you but can you do it without Hospital's Rule?

[quote="sami.d"]hello
find: $ \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \minus{} 2)$[/quote]

$ \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \minus{} 2)$
$ \equal{} \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}.\frac {\frac {4cos^2x}{2 \plus{} cosx} \minus{} 4}{\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2}$
$ \equal{} \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}.\frac {\frac {4cos^2x \minus{} 4cosx \minus{} 8}{2 \plus{} cosx}}{\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2}$
$ \equal{} \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}.\frac {4(cosx \plus{} 1)(cosx \minus{} 2)}{(2 \plus{} cosx)\left(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2\right)}$
$ \equal{} \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}.\frac {4(1 \minus{} cos(x \minus{} \pi))(cosx \minus{} 2)}{(2 \plus{} cosx)\left(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2\right)}$
$ \equal{} \lim_{x \rightarrow \pi}\frac {sin\frac {x \minus{} \pi}{2}}{\frac {x \minus{} \pi}{2}}.\frac {4sin\frac {x \minus{} \pi}{2}(cosx \minus{} 2)}{(2 \plus{} cosx)\left(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2\right)}$
$ \equal{}0$

Last two rows in ppt's solution should be:

$ \equal{} \lim_{x \rightarrow \pi}\frac {1}{\frac {x \minus{} \pi}{2}} \frac {4sin^2\frac {x \minus{} \pi}{2}(cosx \minus{} 2)}{(2 \plus{} cosx)\left(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2\right)}$
$ \equal{} \lim_{x \rightarrow \pi}\frac {sin\frac {x \minus{} \pi}{2}}{\frac {x \minus{} \pi}{2}} \frac {4sin\frac {x \minus{} \pi}{2}(cosx \minus{} 2)}{(2 \plus{} cosx)\left(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2\right)}$
$ \equal{} \lim_{x \rightarrow \pi} \frac {4sin\frac {x \minus{} \pi}{2}(cosx \minus{} 2)}{(2 \plus{} cosx)\left(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \plus{} 2\right)} \equal{} 0$

because $ \lim_{x \rightarrow 0}\frac {sinx}{x} \equal{} 1$.

Sorry, I had a mistake. I've just corrected it. andrejko, you are right.

actual most of such limits which are of the form
$ \lim_{x \to a} \frac{f(x)\minus{}f(a)}{x\minus{}a}\equal{}f'(a)$ can be directly done :)

[quote="pardesi"]actual most of such limits which are of the form
$ \lim_{x \to a} \frac {f(x) \minus{} f(a)}{x \minus{} a} \equal{} f'(a)$ can be directly done :)[/quote]

oho, look at that  :)

[quote="sami.d"][color=darkred]Find $ \lim_{x \rightarrow \pi}\frac {1}{(\pi - x)^2}\cdot(\sqrt {\frac {4\cos^2x}{2 + \cos x}} - 2)$ . [/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ L\equiv\lim_{x \rightarrow \pi}\frac {1}{(\pi - x)^2}\cdot(\sqrt {\frac {4\cos^2x}{2 + \cos x}} - 2)\stackrel{\begin{array}{c} t = \pi - x \\
\ (x = \pi - t)\end{array}}{\implies}$ $ L = \lim_{t\to 0}\frac {2}{t^2}\cdot(\sqrt {\frac {\cos^2t}{2 - \cos t}} - 1)$ $ \implies$

${ L = \lim_{t\to 0}\frac {2}{t^2}\cdot\frac {(\frac {\cos^2t}{2 - \cos t})^{\frac 12} - 1}{\frac {\cos^2t}{2 - \cos t} - 1}\cdot(\frac {\cos^2t}{2 - \cos t} - 1})$ $ \implies$ $ L = \lim_{t\to 0}\frac {1}{t^2}\cdot\frac {\cos^2 + \cos t - 2}{2 - \cos t}$ $ \implies$

$ L = \lim_{t\to 0}\frac {(\cos t - 1)(\cos t + 2)}{t^2}$ $ \implies$ $ L = - 3\cdot\lim_{t\to 0}\frac {1 - \cos t}{t^2}$ $ \implies$ $ L = - \frac 32$ .[/color]

Thank you, Virgil Nicula, but not$ \lim_{x \rightarrow \pi}\frac {1}{(\pi \minus{} x)^2}\cdot\left(\sqrt {\frac {4\cos^2x}{2 \plus{} \cos x}} \minus{} 2\right)$. It is $ \lim_{x \rightarrow \pi}\frac {1}{\pi \minus{} x}\cdot\left(\sqrt {\frac {4\cos^2x}{2 \plus{} \cos x}} \minus{} 2\right)$

[quote="ppt"]Thank you, Virgil Nicula, but not $ \lim_{x \rightarrow \pi}\frac {1}{(\pi \minus{} x)^2}\cdot\left(\sqrt {\frac {4\cos^2x}{2 \plus{} \cos x}} \minus{} 2\right)$. It is $ \lim_{x \rightarrow \pi}\frac {1}{\pi \minus{} x}\cdot\left(\sqrt {\frac {4\cos^2x}{2 \plus{} \cos x}} \minus{} 2\right)$[/quote]

$ \lim_{x \rightarrow \pi}\frac {1}{\pi \minus{} x}\cdot\left(\sqrt {\frac {4\cos^2x}{2 \plus{} \cos x}} \minus{} 2\right)\equal{}$ $ \lim_{x \rightarrow \pi}\ (\pi \minus{}x)\cdot\frac {1}{(\pi \minus{} x)^2}\cdot\left(\sqrt {\frac {4\cos^2x}{2 \plus{} \cos x}} \minus{} 2\right)\equal{}0\cdot\left(\minus{}\frac 32\right)\equal{}0$ .

[quote="hsiljak"][quote="pardesi"]actual most of such limits which are of the form
$ \lim_{x \to a} \frac {f(x) \minus{} f(a)}{x \minus{} a} \equal{} f'(a)$ can be directly done :)[/quote]

Good observation! I'll have to remember this trick, might be useful... :wink:[/quote]
Actually speaking of tricks have a look at third post [url=http://www.mathlinks.ro/viewtopic.php?t=186662]here[/url] ...a very very useful method. :)

I think this is simpler than everyone is making it.

Rule of limits: The limit of a product is the product of the limits hence,

$ \lim_{x \rightarrow \pi}\frac{1}{x\minus{}\pi}(\sqrt{\frac{4cos^2x}{2\plus{}cosx}}\minus{}2) \equal{}
\lim_{x \rightarrow \pi}\frac{1}{x\minus{}\pi}*\lim_{x \rightarrow \pi}(\sqrt{\frac{4cos^2x}{2\plus{}cosx}}\minus{}2)\equal{}
\lim_{x \rightarrow \pi}\frac{1}{x\minus{}\pi}*(\sqrt{\frac{4cos^2 \pi}{2\plus{}cos \pi}}\minus{}2)\equal{}
\lim_{x \rightarrow \pi}\frac{1}{x\minus{}\pi}*0$

so

$ \lim_{x \rightarrow \pi}\frac{1}{x\minus{}\pi}(\sqrt{\frac{4cos^2x}{2\plus{}cosx}}\minus{}2) \equal{} 0$

I hope I did this Latex right

Your solution is wrong. What's the answer of $ \lim_{x\rightarrow \pi} \frac{1}{x\minus{}\pi}?$

[quote="xxpor"]I think this is simpler than everyone is making it.

Rule of limits: The limit of a product is the product of the limits hence,

$ \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \minus{} 2) \equal{} \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}*\lim_{x \rightarrow \pi}(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \minus{} 2) \equal{} \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}*(\sqrt {\frac {4cos^2 \pi}{2 \plus{} cos \pi}} \minus{} 2) \equal{} \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}*0$

so

$ \lim_{x \rightarrow \pi}\frac {1}{x \minus{} \pi}(\sqrt {\frac {4cos^2x}{2 \plus{} cosx}} \minus{} 2) \equal{} 0$

I hope I did this Latex right[/quote]
no that's not right
$ \lim_{x \to a}f(x)g(x) \equal{} [\lim_{x \to a} f(x)][\lim_{x \to a} g(x)]$ if and only if
$ \lim_{x \to a}f(x)$ and $ \lim_{x \to a} g(x)$ exist and  are real numbers.
As in this case they arent's
Consider the famous limit $ \lim_{x \to 0} \frac {\sin x}{x} \equal{} 1$ using ur method this is $ [\lim_{x \to 0} \sin x][\lim_{x \to 0} \frac {1}{x}] \equal{} 0$
which is wrong :)

[quote="xxpor"]Rule of limits: The limit of a product is the product of the limits[/quote]

NO!  The limit of a product is the product of the limits, [i]provided both limits exist and are finite[/i]!!  $ \lim_{x \to \pi} \frac {1}{x - \pi}$ does not exist, so you most certainly can not apply this "rule."

Your latex is pretty good, but it would be even better if you used \cos for the trig function (likewise \sin, etc.) and used \cdot for multiplication.

Edit: pardesi got here first :)

A more general example than pardesi's: take [i]any[/i] function $ f(x)$ such that $ \lim_{x \to a} f(x) = L$ for any finite $ L$.  Then you want to claim that
\begin{align*} L & = \lim_{x \to a} f(x) \\
& = \lim_{x \to a} \frac {x - a}{x - a}\cdot f(x) \\
& = \left(\lim_{x \to a}\frac {1}{x - a}\right)\left(\lim_{x \to a}(x - a)f(x)\right) \\
& = \left(\lim_{x \to a}\frac {1}{x - a}\right) \cdot 0 \\
& = 0,\end{align*}
and this is clearly absurd.

[size=85]Edit: typo fixed.[/size]


Find all $ f$ $ :$ $ Z$ $ \longrightarrow$ $ Z$ that for every $ m,n$ $ \in$ $ Z$ :
$ f(m\plus{}f(n))$ $ \equal{}$ $ f(m) \plus{} n$


geometry
MATHCOUNTS
algebra
system of equations

[quote="Sadi"]Hope you will solve it easily :wink:[/quote]

[hide="me tried... :| "]umm...well, we know that
AB=AC, so that means that angles, ABC and angle ACB are congruent, because it is an isosceles triangle...
so, we know that the angle above the 50 degree angle is 30. we also know that the two legs have been divided into 4 congruent parts..
we also have two pairs of verticle angles...

ok, no idea. i will work on it...[/hide]

I'm not sure. I think it's...

[hide]$90?$[/hide]

Hmmm...this is as far as I got:
[hide="partial solution"]
I solved the triangle using only that the measures of angles in triangles add up to 180 and that supplementary angles add up to 180. For the top triangle (with one angle of 120), I called the two missing angles $\alpha$ and $\beta$, and the missing angles in the center triangle are $x$ and $y$. We want to find x. Using the angle measures for all of the angles except those four, we get the system of equations:
$x+y=110$
$x+\alpha=140$
$y+\beta = 130$
$\alpha+\beta=160$
However, we find that this is nothing more than an identity - you always get $0=0$. So I am either missing something about the side lengths or it works for any value of x.
[/hide]

the two equal sides are bisected? how do you know? :huh:

[hide]$60$? Its funny how everyone has a different answer.[/hide]

well, if the line bisects the two equal sides, then the answer would be 60 using similarity... :maybe:

Yeah if the line bisects the two equal sides than the answer has to be
[hide]$60$[/hide]

Unfortunately, the line doesn't bisect AB and AC. You can never assume that something is true just by looking at a picture. Many times it will just be a coincidence. However, if you find something that looks interesting/useful, try to draw a completely different picture, and see if it is still true.

And by the way, the answer is [i]not[/i] 60. If it were, that would mean that BC is parallel to the small line that just [i]looks[/i] parallel to it, but really isn't.

[hide="Hint"]Try to find the measure of every angle you can, and keep track of those angle measures on your diagram.[/hide]

[hide]I think i got it. 45? I'll post a solution if im right. [/hide]

[quote="gauss202"]This problem is way too hard for the Mathcounts Forum. It's a famous and old geometry problem of Olympiad level.[/quote]

Are you sure because I solved it with without any high level geometry but then again, I might be incorrect.

I'm pretty sure. But can you post an outline of your solution?

By the way, here's one of the threads where it was discussed before http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39387

[quote="sonny"]How do you post a picture on the forum?[/quote]
Upload it via [url]http://www.imageshack.us[/url] or some other site.

[quote="gauss202"]I'm pretty sure. But can you post an outline of your solution?[/quote]

I'm so sorry, my solution has a flaw in it. BTW thanks i_like_pie.

[hide="NO OLYMPIAD STUFF NECESSARY!!!"]
[img]5812[/img]

[/hide]

I'm sorry, I just do not see why that one angle in the middle triangle is 40 degrees...

@ sonny - just click the image

Thanks. By the way anirudh, 70+110+40=220 and the measure of a straight angle is always 180 so i think the solution is incorrect.

[img]5812[/img]

I fixed it.  :blush:

It was just a computational error. See if you can find any errors in this version. :D

Can you prove that it's the only solution?

$x=30^{o}$

I found a solution, which, unfortunately was the same as the 4th of the 8 solutions in http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf

[color=darkblue]In the my opinion, this problem is for you, Mathcounts ! It is a typical "slicing" problem.[/color][hide="Here is the my proof."] [color=darkred][b]Enunciation.[/b] Let $ABC$ be an $A$- isosceles triangle for which $A=20^{\circ}\ .$ Construct the points $D\in (AB)$ ,  $E\in (AC)$ so that $m(\widehat{ACD})=20^{\circ}$ and $m(\widehat{ABE})=30^{\circ}\ .$ Ascertain $m(\widehat{CDE})\ .$[/color]

$\boxed{\mathrm{Sketch\ : \ }\|\begin{array}{c}\ \mathrm{Define\ }F\in (AB)\\\\ m(\widehat{BCF})=20^{\circ}\ \end{array}\Longrightarrow CB=\begin{array}{c}\nearrow CE\\\ \searrow CF\end{array}\ ,\ FC=\begin{array}{c}\nearrow FE\\\ \searrow FD\end{array}\Longrightarrow m(\widehat{CDE})=30^{\circ}}\ .$

$\|\begin{array}{cccccc}1.\blacktriangleright & m(\widehat{CBE})=m(\widehat{CEB})=50^{\circ}& \Longrightarrow & \triangle BCE\mathrm{\ is\ C-isosceles}& \Longrightarrow & \boxed{CB=CE}\\\\ 2.\blacktriangleright & m(\widehat{CBF})=m(\widehat{CFB})=80^{\circ}& \Longrightarrow & \triangle BCF\mathrm{\ is\ C-isosceles}& \Longrightarrow & \boxed{CB=CF}\\\\ 3.\blacktriangleright & |\begin{array}{c}(1)\wedge (2)\Longrightarrow CF=CE\\\\ m(\widehat{ECF})=60^{\circ}\end{array}& \Longrightarrow & \triangle CEF\mathrm{\ is\ equilateral}& \Longrightarrow & \boxed{FE=FC}\\\\ 4.\blacktriangleright & m(\widehat{FCD})=m(\widehat{FDC})=40^{\circ}& \Longrightarrow & \triangle CDF\mathrm{\ is\ F-isosceles}& \Longrightarrow & \boxed{FC=FD}\\\\ 5.\blacktriangleright & |\begin{array}{c}(3)\wedge (4) \Longrightarrow FD=FE\\\\ m(\widehat{EFD})=40^{\circ}\end{array}& \Longrightarrow & |\begin{array}{c}\triangle DFE\mathrm{\ is\ F-isosceles}\\\\ m(\widehat{FDE})=70^{\circ}\\\\ m(\widehat{FDC})=40^{\circ}\end{array}& \Longrightarrow & \boxed{m(\widehat{CDE})=30^{\circ}}\end{array}$[/hide]

That's exactly the 4th solution of http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf

[hide="By the way"]Searching in the web for "[color=brown]geoprob.pdf[/color]" I found other forums that discussed about this problem.

I also found the 4th solution [url=http://agutie.homestead.com/files/LangleyProblem.swf]here[/url], in a really nice [color=blue]flash[/color] presentation

Check  [url=http://www.google.gr/search?hl=el&q=Langley+Problem]Langley's Problem[/url]

[/hide]
[color=red]EDIT: Oh, yes this flash is beutiful! :)[/color]

Thank you, Pontios, very nice "flash" proof !

virgil, how long did it take you to put that in latex.

it is very nice, ty,
-jorian

i still dont get anirudh's solution on why the inside angle must be 40


another one: [url]http://sheep-reaction.freeonlinegames.com/[/url]

my best score for this one is .1503 seconds


[quote="ductrung"]Prove that for any positive real numbers $ a,b$ and $ c$,
\[ \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\le\frac{1}{\sqrt[3]{abc}}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \]
[/quote]

Can you give a hint for solving this problem... It seems to be imposible that this inequality holds. Thanks

[quote="No Name"]
Can you give a hint for solving this problem... It seems to be imposible that this inequality holds. Thanks[/quote]

I believe it is still to early to give a hint :wink: . I will try to find some time and try it tomorrow, and I would not like to have a hint until that :) .

[quote="Gib Z"]I'm thinking AM GM, can anyone confirm?[/quote]

I tried, but I faild...
I think there should be used something stronger than AM-GM

use : a+b+c=1
 ab+bc+ca=q  and abc=r.
 links :

Since I hope to see an alternative solution, it is too early to give a hint (as Tiks said :-).
I don't mind doing it later. But don't give up too soon: it spans several months, from the time I conjectured it to the time I found a satisfactory solution. And by satisfactory I mean a solution without any advanced technique but totally classical (pen-and-paper) - only Cauchy-Schwarz is invoked. Good luck!

[quote=ductrung]Prove that for any positive real numbers $a,b$ and $c$,
\[\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\le\frac{1}{\sqrt[3]{abc}}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \][/quote]

This problem is easy. It is equivalent to
\[
f\left(a,b,c\right)=\frac{1}{abc}\left(\frac{1}{a^3+b^3}+\frac{1}{a^3+c^3}+\frac{1}{b^3+c^3}\right) -\frac{1}{a^6+b^3 c^3}-\frac{1}{a^3 c^3+b^6}-\frac{1}{a^3 b^3+c^6} \geq 0.
\]

Inequality is symmetric so we may assume that $a \geq b \geq c$. Then there exist non-negative reals $x$ and $y$ such that $b=c+x$ and $a=c+x+y$. After doing some algebra it is easy to verify that
\[
f\left(a,b,c\right) = f\left(c+x+y,c+x,c\right) \geq 0.
\]
Expression that we get when taking everything to the common fraction is a bit longer so I rather not post it here.

[quote=arqady]Dear [b]math_science[/b]! It does not help here even with computer. ;)[/quote]

I take it back that it was obvious. I have not seen some negative terms but I believe we can somehow deal with them. Anyway, solutions based on computers might be used just as a demonstration of the accuracy. Such proofs are not so interesting.

[quote=math_science]
but I believe we can somehow deal with them. [/quote]
I could not. :maybe:


I realize that this question may be premature, but after the AIME, how long will I have to wait for answers, and where will they be located? Will some one post them the next day on this forum, like with the AMC, or will I have to wait until they are on the AMC website a couple days later. Last year I didn't know about this site, so I had to wait in suspense for days until the answers were on the AMC websites. Also, when will we be allowed to discuss the problems and solutions. Thanks in advance for any responses!

I know that there's a math jam one to two days after the AIME that goes over all of the questions and answers to the test

And, of course, the Math Jams are almost always correct.
Correct me if I'm wrong, but this forum will remain closed for a while the AIME is in session, yes? When the forum gets unlocked, discussion of the problems are allowed.


geometry
geometric transformation
rotation

Suppose that each pair of eight tennis players (rotations) either played exactly one game last week or did not play at all.  Each individual player participated in all but 12 games.  How many games were played among the eight players.

[hide]Let the number of games played be $k$. Each player plays $k-12$ games, so in total, there are $8k-96$ total games. However, we double count because two people play in each game, so there are only $4k-48$ games in total. Hence, we have $k=4k-48\Rightarrow k=16$.[/hide]


inequalities
IMO Shortlist
inequalities proposed

Let $a,b,c$ be positive real numbers such that: $ab+bc+ca=1$. Prove the inequality:
\[ 3 \cdot \sqrt[3]{\frac{1}{abc}+6(a+b+c)} \le \frac{\sqrt[3]{3}}{abc}  \]

I remember an IMO shortlist problem like this:
show that if $ab+bc+ca=1$ then
$\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\le \frac{1}{abc}$  :D

the solution uses power mean and the above result ;)

notice that $ab+bc+ca\geq \sqrt{3abc(a+b+c)}$so $a+b+c\le \frac{1}{3abc}$ then call $abc=x\le \frac{1}{3\sqrt{3}}$
from here its an easy one-variable inequality...

Yeah, that's almost the same with the one in IMO2004 shortlist


A basketball player averaged 17 points in his first eight games and 24 points in his next
six games. What is his overall average for the 14 games?

The sum of his scores is $ 17\cdot 8\plus{}24\cdot 6\equal{}280$ (this follows from the definition of average), so the mean is $ \frac{280}{14}\equal{}\boxed{20}$.

Alternatively, the ratio of the number of elements in the first set to the second is $ \frac{8}{6}\equal{}\frac{4}{3}$, so the distance from the average of the second set is $ \frac{4}{3}$ is much as the distance from the average of the first set.

The total distance is just $ 24\minus{}17\equal{}7$, so $ x\plus{}\frac{4}{3}x\equal{}7 \Rightarrow x\equal{}3$, so $ 17\plus{}3\equal{}20$.

The second part was rather hard to explain?


algebra
polynomial
trigonometry
algebra unsolved

Trying to check something about a recurrence I found this (I'm not completely sure about its validity):
For any $ n \in \mathbb{Z}$, $ n \ge 2$ consider the polynomial
$ P_n : \equal{} \prod_{k\equal{}1}^{n\minus{}1} \left( t \minus{} \cos \frac{k\pi}{n} \right) \in \mathbb{R}[t]$,
and also $ P_1 \equiv 1$. Prove that the following identity holds for any positive integer $ n$:
$ P_{n\plus{}2} \equal{} t \cdot P_{n\plus{}1} \minus{} \frac{P_n}{4}$.

Yes your recurrence is true! The polynomial you defined $ P_n$ is in fact equal to $ 2^{n\minus{}1}U_{n\minus{}1}$ where $ U$ are the Chebyshev's polynomials of second kind.
They are special cases of the Jacobi polynomials, and are thus an othogonal sequence.
see for example [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]here[/url]


$ \text{Given the sequence}$ $ \{a_n\}$ $ \text{which is defined by}$ \[ a_1\equal{}1,\quad a_{n\plus{}1}\equal{}2a_n\plus{}2^n\ (n\equal{}1,2,3),\] $ \text{find the}$ $ n$$ \text{th term}$ $ a_n$ $ \text{and sum}$ $ \sum_{k\equal{}1}^na_k.$

hello, prove by induction that $ a_n\equal{}n\cdot2^{n\minus{}1}$, for the sum i have
$ \sum_{i\equal{}1}^na_i\equal{}2^n(n\plus{}1)\minus{}2^{n\plus{}1}\plus{}1$.
Sonnhard.


group theory
abstract algebra
factorial
calculus
integration
algebra
function

3)We have a group with 2002 elements, which has only one subgroup with 26 elements and only one subgroup with 77 elements, and they both are commutative. Show that the group is commutative.
4) Question How do you call an eucledian ring, a factorial ring... I saw notations life UFD... what do they mean?
Thanks

3) Since the two groups are the unique subgroups of a given order, they are both characteristic subgroups, hence in particular normal subgroups. Since they have trivial intersection and their product is the group, the group is the direct product of the two subgroups. The direct product of two abelian groups is abelian, so we are done.

EDIT: For 3), we use the following:

If $ M$ and $ N$ are normal subgroups with trivial intersection, and $ m \in M$ and $ n \in N$, then $ mn \equal{} nm$. Proof: $ mnm^{\minus{}1}n^{\minus{}1} \equal{} (mnm^{\minus{}1}n^{\minus{}1} \in N$, similarly $ m(nm^{\minus{}1}n^{\minus{}1}) \in M$. Thus $ mnm^{\minus{}1}n^{\minus{}1} \equal{} e$, so $ mn \equal{} nm$.

4) A Euclidean ring is a commutative ring with a norm satisfying certain nice properties that make it similar to the ring of integers. If you want the full list of properties of the norm, you can almost certainly find them on wikipedia or mathworld.

A factorial ring is a commutative integral domain where each element factors uniquely as a product of irreducibles, down to units (the integral domain part ensures that $ 0$ has "unique factorization", as well). Thus it is sometimes called a UFD, or unique factorization domain. Also, it should be noted that an integral domain is sometimes called an entire ring instead. It is a ring where $ ab \equal{} 0$ implies that $ a \equal{} 0$ or $ b \equal{} 0$.

$ 0$ is no product of irreducibles; and integral domains are assumed to be $ \neq 0$.


Hello,
On my website http://www.matharena.net/ I tried to host a mock AMC about a month ago, but no one participated.  I was just curious to see if anyone might be interested now.
Basically it would just be a one weekend type thing.  You have 75 minutes to solve the problems, but you choose a time over that weekend when you want to start.  The scores would then be emailed out a few days after.  I just thought it might be kind of fun if anyone is interested..

[quote="SnowStorm"]Hello,
On my website http://www.matharena.net/, I tried to host a mock AMC about a month ago, but no one participated.  I was just curious to see if anyone might be interested now.
Basically it would just be a one weekend type thing.  You have 75 minutes to solve the problems, but you choose a time over that weekend when you want to start.  The scores would then be emailed out a few days after.  I just thought it might be kind of fun if anyone is interested..[/quote]

I would be interested if you get rid of that comma in your url  :D

I would be interested. Noone participated in the last one? :huh: I would have, but I was in NY at the time... :(

definitely interested. already registered. :D so when is this starting?

I'm other admin of that site by the way.

Although this site and contest he mentioned is done all by himself, I did contributed some questions (that weren't in the contest I think).

Hopefully, I can get to add some problems. Maybe, we'll see. :)

I'm interested!  Just make sure to do it on the weekend, because otherwise 6 PM EST (which is when most of the Mock AMCs have been held) will be during school for those of us who live on the west coast.

[quote="sirorange"]I'd be interested if we didn't have to give out all that information.[/quote]

It's not that much information... but if you are really scared of someone stalking you because you live in the US and your title is Mister, you could BS something.

the link doesn't work...but i would like to participate, and if i had to give info, then i would just give fake stuff, because as a general rule, i never give out personal info

EDIT: i got the link to work...you ought to remove that extra comma

ok,
It still needs a little more work, but I'll keep you guys updated.  And if you don't want to give out that info, just make it up...

ok, it looks like a lot of people are interested.
I've made another post about it including the date.  Good luck.


Do I have enough information here to solve for the x, y, and z variables? Is there even an answer? If so, what are some possible ones? All of these conditions must be met for the correct answer, if there is one. Please help!

x, y, and z, are natural numbers
x is greater than or equal to 1
y is greater than or equal to 1
z is greater than (x +y)
(5x-y-z) is greater than or equal to 0
(2y-x-z) is greater than or equal to 1
(z-x-y) is greater than or equal to 1

I think not....

Suppose there is a solution.

From your last three inequalities, add the first and last to give

$4x - 2y \ge 1 \Rightarrow 4x - 2y \ge 2$ (since $4x -2y$ is even)
$\Rightarrow 2x - y \ge 1$

Adding the second of the inequalities gives :-

$x + y - z \ge 1 \Rightarrow z - x - y \le -1$

which contradicts the last one.

Therefore, there is no solution.


induction
geometry
3D geometry
linear algebra
matrix
algebra
factorization

An (classic) result about the Fibonacci sequence (not difficult to prove by induction) :

1) F(m+n) = F(m)F(n+1) + F(m-1)F(n)

Easy to deduce from this (or by induction) :

2) F(n)^2 + F(n+1)^2 = F(2n+1)

3) F(n+1)^2 - F(n)^2 = F(2n)

From 1) :

4) F(3n) = F(2n)F(n+1) + F(2n-1)F(n)

Injecting 2) and 3) in 4) gives the given formula, after some algebraic gymnastics.

I see... but, is there a way to do it without using those other identities???

PS I love Algebraic gymnastics... God I have no life...

[color=cyan]Well, I'm fairly sure you can do it using the fact that
F(n) =  (:phi:^n - (-:phi:)^(-n)) / :rt5:
where  :phi: = (:rt5: + 1) / 2
but I'm also sure that you really wouldn't want to unless you have, say, mathematica to simplify everything for you and then you can just check at the end that it works.

You can also mess around with the left side of the equation first and then use that substitution at the end to get it more nicely.  For example, if you factor (F(n+1):^3: - F(n-1):^3:) as a difference of cubes, then factor an
F(n) out of everything, then write (F(n+1):^2: + F(n-1):^2:) as
(F(n):^2: + 2F(n+1)F(n-1)) and then do the substitution, it should work out.  It's still very messy, though, and I suspect that you can keep manipulating it, without using much more than the definition of the numbers given, and it should work out.  Maybe.

Incidentally, this should work no matter what the values of F(1) and F(2) are, shouldn't it?[/color]

[quote="JBL"]Incidentally, this should work no matter what the values of F(1) and F(2) are, shouldn't it?[/quote]

No. That's not true. Take F_1 = a and F_2 = b. Then F_3 = a + b, F_4 = a + 2b, F_5 = 2a + 3b, F_6 = 3a + 5b. And the formula does not hold for n = 2. Recurrence relations can be tricky, they can depend greatly on what the initial conditions are.

[quote="Idiot_without_a_village"]I see... but, is there a way to do it without using those other identities???...[/quote]

We have a matrix identity [tex]\displaystyle \begin{pmatrix} F_{3n+1} & F_{3n} \\ F_{3n} & F_{3n-1}\end{pmatrix}=\displaystyle \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1}\end{pmatrix}^3[/tex], since [tex]\displaystyle \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1}\end{pmatrix} = \begin{pmatrix} 1 &1\\1&0\end{pmatrix}^n[/tex].

Therefore [tex]F_{3n}=(F_{n+1}^2+F_n^2)F_n + (F_{n+1}F_n+F_nF_{n-1})F_{n-1}[/tex] after multiplying the matrices on the right hand side.

Hence [tex]F_{3n} + F_{n-1}^3 = [/tex] [tex]F_n^3 + (F_{n+1}^2F_n + (F_{n+1}F_nF_{n-1}+(F_nF_{n-1}^2 +  F_{n-1}^3)))= F_{n}^3 + F_{n+1}^3[/tex].

(Well, we still need some kind of identity, a matrix identity here :-)).

An interesting question:  I've solved several Fibonacci problems by converting them to counting problems on a "zig-zag ladder," for lack of a better description.  For example, one such problem is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13654]here[/url].  Do you think that there is a reasonable way to use this method to construct a model for questions such as this one, or the other question fuzzylogic posted recently?


do you know how can we draw a line such that it is perpendicular to a fix line?(only with ruler and compasses)

[hide="Solution"]
what you do is,
take a point on the line and using the compass intersect the line twice when you get those points, putting the tip of the compass of both those points make arcs on both sides of the line so that the arcs intersect.
Join the intersectio. That is your perpendicular line.
[/hide]

nice solution,realy nice i enjoyed it so much. :D


calculus
integration
function
induction
real analysis
real analysis unsolved

7. Prove that if the real-valued function $f$ on the interval $[a,b]$ is bounded and is continuous except at a finite number of points, then $\int^{b}_{a}f(x)\,dx$ exists.

Here's my start:

Suppose that $f(x)$ is not continuous at $h$ number of points ${x_{1}}',{x_{2}}', \ldots,{x_{n}}'$. Let $a ={x_{0}}' <{x_{1}}' < \ldots <{x_{n}}' = b$. I don't really know what to do from here. It's pretty obvious that a finite number of discontinuities should not affect the existence of the integral, but how can I show this using precise mathematical notation?

ANy help is greatly appreciated.

Prove it for one discontinuity and use induction .  :)

Can you explain in more detail. Ok let $\eta$ be a fixed point of $[a,b]$, let $c \in R$, and set $f(x) = 0$ if $x\ne\eta$ and $f(x) = c$ if $x=\eta$.

Then we can prove for any Riemann Sum $S$ corresponding to a  partition of $[a,b]$ of width less than $\delta$ we have $|S|< 2|c|\delta$. So clearly, $f$ is integrable on $[a,b]$ with one point discontinuity.

But how can I use induction to extend this argument?

Using the upper/lower sums definition of the Riemann integral, we need to find a partition such that the upper and lower sums differ by at most $\epsilon$ (for any positive $\epsilon$).

Suppose that $|f(x)|<M$ for all $x$. Enclose the discontinuities in intervals of total length less than $\frac{\epsilon}{4M}$. The difference between the upper and lower sums on these intervals is less than $(M-(-M))\cdot \frac{\epsilon}{4M}=\frac{\epsilon}2$.
The rest of $[a,b]$ is a finite collection of (closed) intervals, and $f$ is continuous on this set. That makes $f$ integrable on this set, and we can partition it so that the difference between the upper and lower sums is less than $\frac{\epsilon}2$. Combining the two halves into one partition of $[a,b]$, we have the desired result.


suppose that [b]p[/b] is prime and odd number. prove that :
[b]1 p-1+2 p-1+...+(p-1) p-1
-p-(p-1)!    [/b]divisible by [b]p 2 [/b]

@ehsan2004!
Read http://www.mathlinks.ro/viewtopic.php?p=138915#138915
It will help you to name topics in right way next time.