ratio
geometry
geometric transformation
rotation
circumcircle
inequalities

Given a triangle ABC,
draw squares ABIH, BCED, CAGF,
then connect segments GD, IF, HE,
their intersections forms a triangle XYZ.

also connect GB, IC, EA,
their intersections forms triangle TUV.

Prove that, XYZ and TUV are similar and the ratio of their area is 2.

And then, prove that $\mathrm{Area}\left( \Delta XYZ\right) \leq \left( 4-2\sqrt{3}\right) \mathrm{Area}\left( \Delta ABC\right) $

(The second one was the original problem,The first one
is the noticed property  )

Here is not a solution, but some stuff to think about:

At first, I try to make the notations as symmetric as possible:

X = intersection of HE and GD;
Y = intersection of GD and IF;
Z = intersection of IF and HE;
X' = intersection of IC and EA;
Y' = intersection of EA and GB;
Z' = intersection of GB and IC.

I use X', Y', Z' instead of T, U, V.

Now, your first problem is to show that the triangles XYZ and X'Y'Z' are similar and the ratio of their areas is 2. Actually, one can identify the center of similitude: If A', B', C' are the centers of the squares BCED, CAGF, ABIH, then the lines AA', BB', CC' concur at one point W - the so-called [i]first Vecten point[/i] of triangle ABC. Now, I conjecture that the triangle X'Y'Z' is the image of triangle XYZ in the spiral similitude with center W, rotation angle 45 and stretch factor $\frac{\sqrt{2}}{2}$. In other words, the points X', Y', Z' are the centers of the squares erected on the segments WX, WY, WZ (in the correct direction).

I am still searching for a proof of this but it doesn't seem very hard. We know quite much about X, X', .... At first, the lines DA, IC, HE and BB' concur at one point and make angles of 45 with each other at that point. This can be easily proven synthetically (the point where they concur is the point of intersection of the circumcircles of the squares ABIH and BCED different from B !). Hence we know that the lines YZ, ZX, XY are perpendicular to AA', BB', CC', respectively, and thus, triangles XYZ and A'B'C' are homothetic.

Can anybody find a complete proof based on this?

Thanks!

  Darij

Hi Darij:
Denote P,Q,R the cut points of the lines IF-BG-CH,EH-AD-CI,DG-AE-BF,
further X'',Y'',Z'' the cut points of the lines AD-BF, BF-CH,AD-CH
respectively.
We have:
<Y'PY=<YPY''=<YRY''=<Y'RY= 45
<X'QX=<XQX''=<X''RX=<XRX'= 45
<ZPZ'=<ZPZ''=<ZQZ''=<Z'QZ= 45
The three group of points: PY'YY''R,QX''XX'R and PZ''ZZ'Q are cyclics.
Then: <QX''R=<QXR, <PY'R=<PYR, <PZ'Q=<PZQ
So:XYZ and X'Y'Z' are similar.
From  my sketch:
1) X''Y''Z'' and X'Y'Z' are congruents.
2)The triangle bounded by the lines XX',YY',ZZ' is similar to ABC
triangle.
But I haven't a proof. :(
Regards
Juan Carlos

This setting of the problem helps prove Darij's conjecture: the first Vecten point $V$ is also on the three circles, and it's easy to show that $\angle VYY'=\angle VYY=\angle VXX'=\angle VXX=\angle VZZ'=\angle VZZ=45^{\circ}$, so the triangles $VXX', VXX', VYY'$ etc. are right isosceles triangles, so $Y',\ Y$ are obtained from $Y$ by a spiral similarity of ratio $\displaystyle \frac{\sqrt 2}2$ and angle $\pm 45^{\circ}$.

Moreover, we can now prove your two assertions (for the second one I still have to show something, but I don't think it's that hard):

1) Clearly, $VY'Z'$ is obtained from $VYZ$ by a rotation of center $V$ and angle $90^{\circ}$, so $Y'Z'=YZ$ and we do the same for the other pairs of sides.

2) $X'X\perp VX$ and the other similar relations hold, so it would be enough to show that $BC\perp VX$. I'm still trying to prove this, but I don't think it's that tough.

Dear Grobber:
Thanks for your remarks.
So, we have VX''XX', VY''YY', VZ''ZZ' are squares.
The points (Q,R);(R,P);(P,Q) are on the circumcircircles (A'),(B'),(C')
of the squares BCED, CAGF, ABIH resp.,now:
wrt circle (A'): <VRQ=<CBQ
wrt circle of points VQX''XX'R: <VXQ=<VRQ
then:<VXQ=<CBQ.
As QX perpendicular to BQ(BV), then: VX is perpendicular to BC.
Also,we can to consider that:
The triangle XYZ is the image from X'Y'Z' by two transformations:
1.Rotation with center V and angle 45.
2.Homothety with center V and k=sqrt(2).
Therefore:  [XYZ]=2[X'Y'Z'].
Greetings
Juan Carlos

[Moderator edit: Proving the inequality $\mathrm{Area}\left( \Delta XYZ\right) \leq \left( 4-2\sqrt{3}\right) \mathrm{Area}\left( \Delta ABC\right)$ was the 3rd problem in the 5th round of the 6th MathLinks contest.]

Here is my solution
[color=green]Problem:3. Let ABC be a triangle, and let $ABB_2A_3, BCC_3B_1$ and $CAA_1C_2$ be squares constructed
outside the triangle. Denote with $S$ the area of the triangle $ABC$ and with $s$
the area of the triangle formed by the intersection of the lines $A_1B_1$, $B_2C_2$ and $C_3A_3$.
Prove that
$s\leq (4-2\sqrt{3})S$.[/color]
Solution:Let squares $BCC_3B_ 1,ACC_2A_1,ABB_2A_3$ inscribes in the circles  $C_1(O_1,a/(\sqrt{2})), C_2(O_2,b/(\sqrt{2})), C_3(O_3,c/(\sqrt{2}))$ respectively.
Let $A'\in{CA_3}\cap{BA_1},B'\in{CB_2}\cap{AB_1},C'\in{AC_3}\cap{BC_2}$.
We have that $AC=AA_1,AB=AA_3$ and $AB\perp{AA_3},AC\perp{AA_1}$=>
$CA_3=BA_1$ and $CA_3\perp{BA_1}$=>$\angle A_1A'C=90^0$=>
$A'\in{C_2}\cap{C_3}$=>
$\angle B_2A'B=45^0,\angle C_2A'C=45^0,\angle BA'C=90^0.$=>
$\angle B_2A'B+\angle C_2A'C+\angle BA'C=180^0.$=>$A'\in{B_2C_2}$
Also we can get to the similar results for points $B'$ and $C'$.
We know this theorem.
                                                        Theorem.1
Let $w_1,w_2,w_3$ three circles in the plan that
$(A',A'')\in{w_1}\cap{w_2}$
$(B',B'')\in{w_1}\cap{w_3}$
$(C',C'')\in{w_3}\cap{w_2}$
Then the line $A'A'',B'B''$and $C'C''$ intersecting in same point.

From theorem.1 =>$O\in{AA'}\cap{BB'}\cap{CC'}$
Let $B'''\in{A_1B_1}\cap{B_2C_2}$ and $B_0\in{AC}$,$B'''B_0\perp{AC}$.=>
 $\angle AB_0B'''=\angle AA'B'''=90^0$=> cuadrilateral$ABB_0B'''$ is cyclic.
$\angle CB_0B'''=\angle CC'B'''=90^0$=>cuadrilateral$CC'B_0B'''$is cyclic too.
But $AA'C'C$ inscribed in $C_2$=>from theorem.1 we can say that  $O\in{AA'}\cap{CC'}\cap{B_0B'''}$  =>$B'''O\perp{AC}$
To the similar result we can get for $C'''$ and $A'''$($A'''\in{A_1B_1}\cap{A_3C_3}$, $C'''\in{C_2B_2}\cap{A_3C_3}$)
We have that $OB'''\perp{AC}$ and $OA'''\perp{CB}$=>$\angle B'''OA'''=180^0-C$=>
$s=1/2(OA'''*OB'''*sinC+OA'''*OC'''sinB+OB'''*OC'''*sinA)$
Let $\angle OAC=a_1,\angle OAB=a_2$,
$\angle OBA=b_1,\angle OBC=b_2$,
$\angle OCB=c_1,\angle OCA=c_2$
Let’s calculi $OA''',OB'''$and $OC'''$
Let$X\in{OB'''}\cap{A_1C_2}$=>$\angle B'''C_2A_1=90^0-\angle CC_2A'=90^0-a_1$
Similar $\angle B'''A_1C_2=90^0-c_2$
=>$B'''X=A_1C_2/(ctg{\angle B'''A_1C_2}+ctg{\angle B'''C_2A_1})=b/(tg{a_1}+tg{c_2})$(b=AC)
Also $OB_0=b/(ctg{a_1}+ctg{c_2})$=>
$OX=OB_0+B_0X=OB_0+b=OB'''+B'''X$=>
$OB'''=b+OB_0-B'''X=b+b/(ctg{a_1}+ctg{c_2})-b/(tg{a_1}+tg{c_2})= b(1+tg{a_1}tg{c_2}/(tg{a_1}+tg{c_2})-1/(tg{a_1}+tg{c_2}))=b(1-ctg{(a_1+c_2)})\Rightarrow$
$OB'''=b(1-ctg{(a_1+c_2)})$similar
$OC'''=c(1-ctg{(a_2+b_1)})$
$OA'''=a(1-ctg{(b_2+c_1)})$
=>$s=1/2(OA'''*OB'''*sinC+OA'''*OC'''sinB+OB'''*OC'''*sinA)= S((1-ctg{(a_1+c_2)}) (1-ctg{(a_2+b_1)})+ (1-ctg{(a_2+b_1)}) (1-ctg{(b_2+c_1)})+ (1-ctg{(a_1+c_2)}) (1-ctg{(b_2+c_1)}))$
We can easy see that if $m+n+p=180^0$=>$ctgm*ctgn+ctgm*ctgp+ctgn*ctgp=1$so
if $x=ctg{(a_1+c_2)}$
$y=ctg{(a_2+b_1)}$
$z=ctg{(b_2+c_1)}$  =>$xy+xz+yz=1$ and $s/S=(1-x)(1-y)+(1-x)(1-z)+(1-y)(1-z)$=>
$s/S=4-2(x+y+z)$ but $(x-y)^2+(x-z)^2+(y-z)^2\geq 0$<=>$(x+y+z)^2\geq 3(xy+xz+yz)$=>
$x+y+z\geq \sqrt3$=>$s\leq S(4-2\sqrt3)$

                                                                                [b]Tigran Sloyan[/b]

This is my solution and I used the theorem of Weitenbock and I also have a proof of the theorem. :)

Guys who else solved this problem diuring the 5th round? :blush:

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 41
Sincerely
Jean-Louis


logarithms
advanced fields
advanced fields unsolved

how do you prove the number of primes less than x is greater than or equal to c*logx?  I

This follows from [url=http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate]Bertrand's postulate[/url] (the article gives a complete proof). I don't know if this estimate admits a simpler proof than the postulate itself.

$ \prod_{p\le N}\left(1\minus{}\frac1p\right)^{\minus{}1}>\sum_{k\equal{}1}^N\frac1k>\ln N$.

If there are $ \pi(N)$ primes less than or equal to $ N$, that product is at most $ \prod_{j\equal{}2}^{\pi(N)\plus{}1}\left(1\minus{}\frac1j\right)^{\minus{}1}\equal{} \prod_{j\equal{}2}^{\pi(N)\plus{}1}\left(\frac{j\minus{}1}j\right)^{\minus{}1}\equal{}\pi(N)\plus{}1$

Therefore $ \pi(N)\plus{}1>\ln N$, and we could refine this fairly easily. This is of course a gross underestimate; the Prime Number Theorem says the $ \pi(N)$ is closer to $ \frac{N}{\ln N}$.


topology
advanced fields
advanced fields unsolved

If we define $ X \equal{} \{\frac {1}{n} : n\in \mathbb{N} \} \cup \{0\}$

with Topology

$ \tau \equal{}\{U : 1\in U,X\minus{}U \text{ \ is finite }\}$ $ \cup$ $ \{ V : \text{ V is an open in the subspace X of }\ (\mathbb{R},\tau_u) \}$

i.e.  There are two types of open sets .

The question :

Is there a sequence has limit two points ? I think yes. what is it? its limits?

Thank you in advanced

[quote="LydianRain"]I think it's confusing to talk about sequences.  The whole space is countable, so it's clearly the same as asking what the limit points of the whole space are.

There are three kinds of points in this space.  For each point $ x$, answer the question: Is $ \{x\}$ open?

It's very difficult to read your post; please use the "Preview" button.[/quote]

there wasn't any problem when I posted it . Anyway, I wrote it in another way.

$ \{x\}$ in general is not open. namely, $ \{0\}$ is not open since it's cluster point of the space.

Also, the space is $ T_1$ (Clearly) , the open set which contains $ 0$ must has infinitely many points of Space $ X$ . Moreover, $ \{1\}$ is not open since the complement of it is not finite.

However, I think if we have the sequence $ x_n \equal{} \frac {1}{n}$  then there is two points of its limit.

What do you think ?

thank you in advanced.

[quote="Hidden Scofield"][quote="LydianRain"]However, I think if we have the sequence $ x_n \equal{} \frac {1}{n}$  then there is two points of its limit.[/quote][/quote]

What are they?

Again, I think that bringing in sequences (or the $ T_1$ axiom, for that matter) is unnecessary notation.  As far as I can tell, you've already proved that this sequence has exactly two limit points.

I mentioned $ T_1$ to prove that $ \{x\}$ is not open set in general.

Anyway, Do you think this sequence has the limits $ 0$ and $ 1$ at the same time ?

Thank you in advanced.


Tyrone announces, "I just found $5.00. I now have five times more money than if I
had lost $5.00." How many dollars did Tyrone have before finding the $5.00?
Express your answer to the nearest hundredth.

[hide="Solution"] We can set up an equation with $x$ being the amount of money Tyrone had before. $x+5=5(x-5) \longrightarrow x+5=5x-25 \longrightarrow 4x=30 \longrightarrow x=7.5$ So Tyrone had $\boxed{\$7.50}$ :D[/hide]

[hide]5-(-5)=10 represents 4 times the number of dollars as if he lost 5 dollars.
If he lost 5 dollars, he would have 2.50.
2.50+5=7.50 dollars before finding the 5 dollars.[/hide]

[quote="jhollenbeck"][hide]750[/hide][/quote]

A point missing here!!!! or maybe it's 750 cents!

[quote="mad_skillz_aops"]It's not 750 cents, because the problem says "How many dollars..."[/quote]

True. Then he's answer is wrong

[hide][quote="math92"]Tyrone announces, "I just found 5.00. I now have five times more money than if I
had lost 5.00." How many dollars did Tyrone have before finding the 5.00?
Express your answer to the nearest hundredth.[/quote]

x+5=5(x-5)
x+5=5x-25
4x=30
x=30/4
x=15/2
x=7.50[/hide][/hide]

[hide]
5(x-5)=x+5
5x-25=x+5
4x=30
x=7.50
[hide]

EDIT: I fixed it :blush:

[hide] you set up an equation based on the words;
$5+x=5(x-5)$. solving, we get
$5+x=5x-25$, which becomes
$30=4x$, which becomes
$\frac{30}{4}=x$, which simplifies to
$7\frac{2}{4}$, which simplifies to
$7\frac{1}{2}$, or
$\boxed{7.5}$.
[/hide]

Its easy, x-5 has to be equal to x/3, so the only answer can be 7.50.

i thnk someone should just lock this already...


function
Euler
number theory proposed
number theory

For which positive whole numbers is it true that:
$\sum_{d|n}\mu (d) \varphi (d) =0$
$\mu (d)$ is the möbius function, $\varphi (d)$ is the euler function

$\sum_{d|n}\mu (d) \varphi (d) =0$ if and only if n is even.

Put $F(n)=\sum_{d|n}\mu(d)\cdot \varphi(d)$. Then by $\mu(n)\cdot \varphi(n)$ is multiplicative we have $F(n)$ is too. Therefore $F(n)=\prod_{i=1}^{r}(1-\varphi(p_{i}))$, where $n>1$ and $n=p_{1}^{k_{1}}...p_{r}^{k_{r}}$ and $F(1)=1$.

Final, $F(n)=0$ iff $n>1$ and there exists $i$ such that $\varphi{(p_{i})}=1$ or $n$ is a even positive integers.


A ball with radius 1 is placed tangent to a wall (so that it touches the wall and the floor. The wall makes a right angle with the floor). Another ball, just large enough to fit under the first ball, is then placed under the first ball, so that it is tangent to the ball, the wall, and the floor. Another ball is placed tangent to the second ball, the wall, and the floor. This is continued indefinitely, using an infinite amount of balls.

What is the area (two dimensional) underneath the first ball, but outside of the smaller balls?

It took me a while to figure out what you meant. You said balls, so I immediately assumed this was a 3 dimensional problem.

However, I couldn't work out how that could be meaningful, as it is really a volume underneath the ball, and this volume is not closed. I thought maybe the floor underneath the ball, but this is only obstructed by infinitesimal points.

However, this makes sense 2 dimensionally, with circles, a graphic is attached. Repeating the circles forever, and shading in the areas, find the shaded area?

Gah! I can't attach a graphic on this forum. Off to photobucket...

[img]http://i169.photobucket.com/albums/u236/Mozza314/mathcirclesproblem/1.gif[/img]

yes, they are balls, but it is meant to be a 2-D problem, basically.

i_like_pie, your answer looks right.

[quote="i_like_pie"][hide="Solution"]After some simple calculations, it can be determined that the radius of the smaller circle (if we look at in in 2D) is $ 3-2\sqrt{2}$. The radius of the next one will then be $ \left(3-2\sqrt{2}\right)^{2}$, and so on.

The area of the first small circle will be $ \left(3-2\sqrt{2}\right)^{2}\pi$. The area of the second will be $ \left(3-2\sqrt{2}\right)^{4}\pi$, etc. The sum of all the areas will be $ \frac{\left(3\sqrt{2}-4\right)\pi}{8}$.

The area bordered by the wall, the floor, and the big ball is $ \frac{1-\pi}{4}$. Therefore, the answer is $ \frac{1-\pi}{4}-\frac{\left(3\sqrt{2}-4\right)\pi}{8}\approx\boxed{0.095}$.[/hide][/quote]

My solution is very different, and has a different answer to yours. There's some stuff that's the same though...:

[hide="Solution"]

Firstly, may I point out that the area bordered by the wall, the floor, and the big ball cannot be $ \frac{1-\pi}{4}$ because this is negative $ \approx{-0.535}$ .

Anyway, my final solution came out as:

$ \frac{4\sqrt{2}+8+(6-7\sqrt{2})\pi}{32}\approx{0.04394}$

And the working...:

let $ a$ be the larger circle, with equation $ (x-1)^{2}+(y-1)^{2}= 1$
let $ b$ be the first smaller circle with equation of the form $ (x-r)^{2}+(y-r)^{2}= r^{2}$

circle $ b$ is a tangent to circle $ a$, so the distance between their centres is r+1.

$ \sqrt{(1-r)^{2}+(1-r)^{2}}= r+1$

$ \sqrt{2}(1-r) = r+1$

$ \sqrt{2}-\sqrt{2}r = r+1$

$ (1+\sqrt{2})r = \sqrt{2}-1$

$ r = \frac{\sqrt{2}-1}{\sqrt{2}+1}$

$ =\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2}+1)(\sqrt{2}-1)}$

$ =3-2\sqrt{2}$

Because placing a circle underneath $ b$ is similar to placing $ b$ under $ a$, this similarity can be exploited by saying that the radius of each successive circle is $ 3-2\sqrt{2}$ times the previous circle:

$ r_{n+1}=(3-2\sqrt{2})r_{n}$
$ r_{n}=(3-2\sqrt{2})^{n}$

Where $ a$ has radius $ r_{0}$, $ b$ has $ r_{1}$, and so on.

Now each area must be expressed in terms of the radius of the larger circle:

[img]http://i169.photobucket.com/albums/u236/Mozza314/mathcirclesproblem/234.gif[/img]

$ \alpha_{2}= (1-\frac{\pi}{4})r_{n}^{2}-\pi[(3-2\sqrt{2})r_{n}]^{2}$

$ =[1-\frac{\pi}{4}-\pi(17-12\sqrt{2})]r_{n}^{2}$

$ =(1-\frac{1}{4}\pi-17\pi+12\sqrt{2}\pi)r_{n}^{2}$

$ =[1+(12\sqrt{2}-17\frac{1}{4})\pi]r_{n}^{2}$

Now take away the lower left area, calculated from the radius of the smaller circle:

[img]http://i169.photobucket.com/albums/u236/Mozza314/mathcirclesproblem/5.gif[/img]

$ \alpha_{3}= [1+(12\sqrt{2}-17\frac{1}{4})\pi]r_{n}^{2}-(3-2\sqrt{2})\alpha_{1}$

$ =[1+(12\sqrt{2}-17\frac{1}{4})\pi]r_{n}^{2}-(3-2\sqrt{2})(1-\frac{1}{4}\pi)r_{n}^{2}$

$ =[...]r_{n}^{2}-(3-\frac{3}{4}\pi-2\sqrt{2}+\frac{\sqrt{2}}{2}\pi)r_{n}^{2}$

$ =[...]r_{n}^{2}+[(\frac{3}{4}-\frac{\sqrt{2}}{2})\pi+2\sqrt{2}-3]$

$ =[...]r_{n}^{2}+[2\sqrt{2}-3+(\frac{3}{4}-\frac{\sqrt{2}}{2})\pi]$

$ =[2\sqrt{2}-2+(11\frac{1}{2}\sqrt{2}-16\frac{1}{2})\pi]r_{n}^{2}$

Since the 2 areas in $ \alpha_{3}$ are congruent (flipped over the line y=x), and we want 1 of those areas, we can simply divide $ \alpha_{3}$ by 2.

[img]http://i169.photobucket.com/albums/u236/Mozza314/mathcirclesproblem/6.gif[/img]

$ \alpha_{4}= [\sqrt{2}-1+(5\frac{3}{4}\sqrt{2}-8\frac{1}{4})\pi]r_{n}^{2}$

And since this is the final area, it will be expressed as:

$ A_{n}= [\sqrt{2}-1+(5\frac{3}{4}\sqrt{2}-8\frac{1}{4})\pi]r_{n}^{2}$
$ =[\sqrt{2}-1+(5\frac{3}{4}\sqrt{2}-8\frac{1}{4})\pi](3-2\sqrt{2})^{2n}$

And the total area underneath all the circles is the sum of all these areas from $ A_{0}$ onwards:

$ A=\sum_{n=0}^{\infty}A_{n}$

$ =\sum_{n=0}^{\infty}\{[\sqrt{2}-1+(5\frac{3}{4}\sqrt{2}-8\frac{1}{4})\pi](3-2\sqrt{2})^{2n}\}$

$ =[\sqrt{2}-1+(5\frac{3}{4}\sqrt{2}-8\frac{1}{4})\pi]\sum_{n=0}^{\infty}(17-12\sqrt{2})^{n}$

$ =[...]\sum_{n=0}^{\infty}( \frac{1}{\frac{1}{17-12\sqrt{2}}})^{n}$

$ =[...]\sum_{n=0}^{\infty}\frac{1}{( \frac{1}{17-12\sqrt{2}})^{n}}$

Use the identity:

$ \sum_{n=0}^{\infty}\frac{1}{x^{n}}= \frac{1}{x-1}$

Note: I'm not sure if this is a recognised identity. I assure you it's true though, I can prove it if needed...

$ A=[...]\sum_{n=0}^{\infty}\frac{1}{( \frac{1}{17-12\sqrt{2}})^{n}}$

$ =[...]( 1+\sum_{n=1}^{\infty}\frac{1}{( \frac{1}{17-12\sqrt{2}})^{n}})$

$ =[...]( 1+\frac{1}{\frac{1}{17-12\sqrt{2}}-1})$

$ =[...](1+\frac{1}{\frac{1+12\sqrt{2}-17}{17-12\sqrt{2}}})$

$ =[...](1+\frac{17-12\sqrt{2}}{12\sqrt{2}-16})$

$ =[...](1+\frac{(17-12\sqrt{2})(12\sqrt{2}+16)}{32})$

$ =[...](\frac{32+204\sqrt{2}+272-288-192\sqrt{2}}{32})$

$ =[...](\frac{8+51\sqrt{2}+68-72-48\sqrt{2}}{8})$

$ =[...](\frac{4+3\sqrt{2}}{8})$

$ =[\sqrt{2}-1+(5\frac{3}{4}\sqrt{2}-8\frac{1}{4})\pi](\frac{4+3\sqrt{2}}{8})$

$ =\frac{1}{8}[(\sqrt{2}-1)(4+3\sqrt{2})+(5\frac{3}{4}\sqrt{2}-8\frac{1}{4})(4+3\sqrt{2})\pi]$

$ =\frac{1}{32}[(4\sqrt{2}-4)(4+3\sqrt{2})+(23\sqrt{2}-33)(4+3\sqrt{2})\pi]$

$ =\frac{1}{32}[(16\sqrt{2}+24-16-12\sqrt{2})+(92\sqrt{2}+138-132-99\sqrt{2})\pi]$

$ =\frac{1}{32}[4\sqrt{2}+8+(6-7\sqrt{2})\pi]$

$ =\frac{4\sqrt{2}+8+(6-7\sqrt{2})\pi}{32}$

$ \approx0.043944675$

[/hide]

Now that took a long time... mainly 'cause I've never used BBCode before... oh well, good practice :D

PS: I had to combine 3 of my diagrams and the text that was between them into one picture, because I was only allowed 3 images and I wanted 5. Having a larger image then made it shrink when displayed here. I hope it's readable enough.

PPS: I found out it was actually photobucket that shrunk my picture. I went around and signed up to a few different photo hosting sites, but none of them would allow me to post the picture here. If you know of a good photo hosting site that won't shrink my picture, please let me know and I'll change it.

[quote="vishalarul"][quote="i_like_pie"][hide="Solution"]After some simple calculations, it can be determined that the radius of the smaller circle (if we look at in in 2D) is $ 3-2\sqrt{2}$. The radius of the next one will then be $ \left(3-2\sqrt{2}\right)^{2}$, and so on.

The area of the first small circle will be $ \left(3-2\sqrt{2}\right)^{2}\pi$. The area of the second will be $ \left(3-2\sqrt{2}\right)^{4}\pi$, etc. The sum of all the areas will be $ \frac{\left(3\sqrt{2}-4\right)\pi}{8}$.

The area bordered by the wall, the floor, and the big ball is $ \frac{1-\pi}{4}$. Therefore, the answer is $ \frac{1-\pi}{4}-\frac{\left(3\sqrt{2}-4\right)\pi}{8}\approx\boxed{0.095}$.[/hide][/quote]

[hide="Correction"]That $ \frac{1-\pi}{4}$

should be

$ 1-\frac{\pi}{4}$[/hide]
[quote="mozza314"][hide="In your solution, you said"]Use the identity:

$ \sum_{n=0}^{\infty}\frac{1}{x^{n}}= \frac{1}{x-1}$

Note: I'm not sure if this is a recognised identity. I assure you it's true though, I can prove it if needed... [/hide][/quote]

[hide="That is incorrect"]$ \sum_{n=0}^{\infty}\frac{1}{x^{n}}= \frac{1}{1-x}\quad \text{for }|x|<1$[/hide]

Why don't you guys use Asymptote? :roll:[/quote]

Ah yes it [i]should[/i] be $ 1-\frac{\pi}{4}$. Remember the square we are concerned with has radius $ 1$, and then the quarter circle that must be subtracted from that area is $ \frac{1}{4}\cdot \pi$.

[quote="vishalarul"][quote="i_like_pie"][hide="Solution"]After some simple calculations, it can be determined that the radius of the smaller circle (if we look at in in 2D) is $ 3-2\sqrt{2}$. The radius of the next one will then be $ \left(3-2\sqrt{2}\right)^{2}$, and so on.

The area of the first small circle will be $ \left(3-2\sqrt{2}\right)^{2}\pi$. The area of the second will be $ \left(3-2\sqrt{2}\right)^{4}\pi$, etc. The sum of all the areas will be $ \frac{\left(3\sqrt{2}-4\right)\pi}{8}$.

The area bordered by the wall, the floor, and the big ball is $ \frac{1-\pi}{4}$. Therefore, the answer is $ \frac{1-\pi}{4}-\frac{\left(3\sqrt{2}-4\right)\pi}{8}\approx\boxed{0.095}$.[/hide][/quote]

[hide="Correction"]That $ \frac{1-\pi}{4}$

should be

$ 1-\frac{\pi}{4}$[/hide]
[quote="mozza314"][hide="In your solution, you said"]Use the identity:

$ \sum_{n=0}^{\infty}\frac{1}{x^{n}}= \frac{1}{x-1}$

Note: I'm not sure if this is a recognised identity. I assure you it's true though, I can prove it if needed... [/hide][/quote]

[hide="That is incorrect"]$ \sum_{n=0}^{\infty}\frac{1}{x^{n}}= \frac{1}{1-x}\quad \text{for }|x|<1$[/hide]

Why don't you guys use Asymptote? :roll:[/quote]

[hide="Made a mistake, but look closely at my solution, I still used a true identity"]

$ \sum_{n=0}^{\infty}\frac{1}{x^{n}}= \frac{1}{x-1}$

The problem here is only the starting term, it's not n=0, but n=1:

$ \sum_{n=1}^{\infty}\frac{1}{x^{n}}= \frac{1}{x-1}$

If you look closer at my solution, you'll see I've changed my sum from n=0 to n=1 by taking out the n=0 term and adding it separately.

[quote="I"]

$ A=[...]\sum_{n=0}^\infty\frac{1}{(\frac{1}{17-12\sqrt{2}})^{n}}$

$ =[...]( 1+\sum_{n=1}^\infty\frac{1}{(\frac{1}{17-12\sqrt{2}})^{n}})$

[/quote]

[/hide]


prove that an integer which has three hundred $1$ and three hundred $2$ and three hundred $3$ can't be a perfect square !!!  :P  ;)

[quote="jastrzab"]but is there any special order of these numbers, or it doesn't matters :huh:[/quote]
i wanted to say there is no special order !! :)

:oops_sign:  :oops_sign:
Sorry :D  I dont know ,what was i thinking about
Back to the paper

[quote="lomos_lupin"][color=red]Dont read,Its flawed[/color]
Let $\mathcal E$ be such a integer.
Assume that problem is false:
Simply we have $3 \mid \mathcal E$ and Cause $\mathcal E$ is a perfect square then
$9 \mid \mathcal E$ ,But we know ,Sum of the Digits of $\mathcal E$ in decimal Is $600$ which is not divisable by$9$,Contradiction,So $\mathcal E$ is not a Perfect square.

PS.[quote]God is nearer to us than our vein.[/quote]
I basiclly disagree but here is not a place for arguement.You can express you opinions in this topic:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=306743#p306743[/url][/quote]
i won't read is it good ? ;)  but try to read the questions correctly !! ;)

[quote="Diarmuid"]I think maybe you have misread, lomos.

The number has 300 each of 1, 2 and 3, and so the sum of it's digits is 1800.[/quote]
that's compeletly correct !! thanks for your attention !! but what about the solution !?? :lol:

i don't think that nobody wants to solve it but why ? come on gentelmans !!  :oops:


Hallo, Leute!
Also, ich bin im Moment in Berlin, und wüsste gern ob es hier irgend eine Buchhandlung gibt, wo man mathematische Bücher finden kann (ich meine solche für Olympiaden mit Problemen und so)
Danke an allen!

Eigentlich kenne ich keinen speziellen Buchladen mit Mathematikliteratur in Berlin. Die Bücher, die es in den Buchläden gibt, kann man in der Regel auch online, z.B. bei amazon, bestellen. Suchst du nach etwas Speziellem?

ich war letzte Monate in Berlin und hab auch kein Mathe Buch gefunden.... :(

War gestern in Berlin und habe dort nur das Matheon gefunden :D ;)


Let $ a$, $ b$, $ c$ reals with $ a\plus{}b\plus{}c\equal{}\minus{}2$

What is the maximal value of the following?

$ |a|\plus{}|b|\plus{}|c|\minus{}|a\plus{}1|\minus{}|b\plus{}1|\minus{}|c\plus{}1|$

[hide]Note that decreasing any of the variables causes the expression to increase or stay the same. If one of the variables is less than $ \minus{}1$, say $ a$, then increasing $ a$ to $ \minus{}1$ does not change $ |a| \minus{} |a\plus{}1|$. Increase another variable by the amount $ a$ was decreased to keep the constraint, and the expression either increases or stays the same. So WLOG $ a,b,c \geq \minus{}1$.

If any of the variables, say $ a$, is greater than $ 0$, then $ |a| \minus{} |a\plus{}1| \equal{} \minus{}1$. Since $ |b| \minus{} |b\plus{}1| \leq 1, |c| \minus{} |c\plus{}1| \leq 1$, the maximum we can get for the expression is $ 1$. This can be achieved with $ a \equal{} b \equal{} c \equal{} \minus{}2/3$ anyway. So WLOG $ a,b,c \leq 0$.

Since $ \minus{}1 \leq a,b,c \leq 0$, this gives $ |a| \minus{} |a\plus{}1| \equal{} \minus{}1 \minus{} 2a$ and the same for $ b,c$. Then the expression becomes $ \minus{}3 \minus{} 2a \minus{} 2b \minus{} 2c \equal{} \minus{}3 \minus{} 2(\minus{}2) \equal{} 1$. So $ 1$ is the maximum, achieved by any $ a,b,c \in [\minus{}1,0]$.[/hide]


Starting with the third term in the sequence 1, 5, 6, 11, 17,...,
each term is the sum of the previous two terms in the sequence.
What is the seventh term in the sequence?

Simply continue the sequence.
1,5,6,11,17,28,[b]45.[/b]


ARML trigonometry AMC USA(J)MO USAMO geometry AIME

Who will be coming to the ARML practice?  I plan to come, and I know that Brendan is coming.

I am going, and hopefully I will not get a 1 on individual again, because hopefully I will not have a life-threatening sickness during the practice. :wink:

Dude also I hope I make A Team!

Okay, here is a list of ARMLers who have been invited to Penn State.

2006 A-Team Members:
1. Yakov Berchenko-Kogan (Yasha) **
2. John Berman
3. Vivek Bhattacharya
4. Drew Boyuka
5. Jeremy Hahn
6. Mikhail Lavrov (Misha) **
7. Joseph Lozier (Joe) **
8. Arnav Tripathy **
9. Daniel Vitek
10. Bangchen Wang
11. Ray Wang

2006 B-Team Members
1. Hyunji Bae (Jane)
2. Lizi Chen
3. Yutong Chen
4. Yevgeniy Grechka (Yev)
5. Damien Jiang*
6. Steven Ji
7. Kevin Lang
8. Bryce Taylor

New Invitees
1. Logan Bailey
2. Haoyu Chen
3. Jenny Chen
4. Brendan Fletcher
5. Andrew Hertzberg
6. Joseph Keefer
7. Seoungjun Lee*
8. Emil Mayev
9. Pom McCabe
10. John Pardon*
11. Dylan Peifer*
12. Haoyang Ren
13. William Schlieper
14. Nicholas Tobey (Nick)
15. Fei Fei Wang

* denotes invitees who are marked as unconfirmed in Mr. Noland's May 27 email.  ** denotes A-Team members who are seniors this year.  I suspect that Seoungjun Lee and John Pardon will not go to ARML, even though the latter is a former MOSPper.  I also suspect that Damien has since confirmed (OR WILL VERY SHORTLY - hint hint).  The jury is out on Dylan Peifer.  not_trig, what do you mean by this life-threatening illness? Stupidity?  Oh no!  Just kidding.  You will probably make A Team this year as one of the new competitors.

My predictions for who will fill the four slots on the A Team:
1. Steven Ji - He did very well at ARML last year; expect him to be bumped up.
2. Damien Jiang - we need a freshman.  Look what happened last year when we had two.  Plus he made red MOSP.  Good job.  Needs to cut down on the annoying factor, however.
3. Probably Kevin Lang.  He qualified for USAMO, although I don't know how well he did.  In addition, he did fairly well on B Team last year.  He has potential.
4. This is really hard to decide.  I would say John Pardon for sure if he comes, which is not likely.  If he does not come, then  the best I can do is narrow it down to  candidates: Hyunji Bae,  Yutong Chen, Yevgeniy Grechka, Bryce Taylor, or William Schlieper.  If somebody else does extremely well at practice, however, then they could very well make the spot.  Logan Bailey is an unknown; he did very well at the Comprehensive Math Contest, but is this a fluke or an undiscovered talent?  As quite a few people seem to be fond of saying, "The race is very close."  Hyunji needs to shine at practice to make it; so do Yutong and Yevgeniy.  Bryce may be put on B Team to get some experience in leadership; he will need it in a few years.  William Schlieper is good; he has finished one spot above me in Regionals Algebra I 2003, Regionals Geometry 2004, Regionals Algebra II 2005, and Regionals Comprehensive 2007.  He needs to make up for a poor showing at States Comprehensive 2007 and in the AMC contests with an excellent performance at practice to emerge from relative anonymity and make it to the A Team.  Jenny Chen did extremely well at State Comprehensive, but I think she will be on B Team for the same reason that I think may apply to Bryce.  Bryce and Jenny should be four-time A-Teamers once they enter high school.

dude zoo tycoon sucked.  take it from one who knows.  :)

uh? not stupidity

I had chest pain and shortness of breath, and had to go to urgent care immediately afterwards :wink:

EDIT: plus, you should denote those who are seniors that are on the $B$ Team and/or new members, as well (Yev, Lizi, Yutong, Steven, and the "new people.")

EDIT AGAIN: What did Kevin get on USAMO?

[color=red]b-flat[/color]

[quote="Teki-Teki"]dude zoo tycoon sucked.  take it from one who knows.  :)[/quote]

As if I don't know... I have TWO copies of it.

I guess if we're making predictions, I may as well say what I think, though to be honest I'm not nearly as knowledgable in the state arena as the national.

John Pardon is definitely not going. And I think he's only a MOSP invitee, as he didn't go last year. Damien never got a form, but did end up confirming somehow.

1. Steven Ji - as teki said, he did really well last year (a 6?).
2. Damien/Tianyi Jiang - err, he made mosp... that is self explanatory. But what is this about "look what happened last year when we had two (freshmen)"? Umm what? I definitely agree that he needs to cut down on aforementioned property though.
3. Bryce Taylor - "I think she will be on the B team for the same reason that may apply to Bryce" What is this reason? I was on the A team in eighth grade, and I know Arnav and Anders were as well, but there are probably others I simply don't know about. Whatever the case, he is certainly at that level. And I don't think he would be put on the B team just for leadership reasons, anyway. Leadership doesn't do anything for ARML.
4. Hyunji Bae - this position is really up for grabs, but I think she has the best chance. Yevgeniy may also be able to make it, or Kevin Lang or Jenny Chen. I know nothing about William Schlieper. Jenny will do extremely well in high school, but she may not have mastered the more conceptual contests yet, as she does not do as well at AIME and the like as subject tests, AMC, etc. However, her high placement at comprehensive is very in her favor on that venue. Of the other three, it's hard to say anything for or against them, but Jane just seems more active in math, and she qualified for the USAMO last year, which is probably better than qualifying this year (even though she didn't). In other news, the sickle and hammer shirts have got to go. (ooh! next year our state shirt could be a sickle and hammer).

Note to not_trig: For the record, I never directly refer to AoPS people with their real names. You may as well be any of the people mentioned above. I don't think teki did either in that post.

[quote="jb05"]I'm coming! Maybe I'll get the latest edition of zoo tycoon? That would be PRETTY EXCITING.

EDIT: Can someone please explain why answering the question deserves a post rating of two? I don't like this system because tons of posts are tagged as spam that are completely in context and just answering a simple question.

Not to complain... just... it's kind of annoying.[/quote]

I don't know.  I personally just ignore posts such as that (and the one I am posting right now) for ratings because they can't be good, but aren't bad.

As for zoo tycoon, the second one was better than the first (graphics wise, etc), but free terrain changes???  Also, it crashed a lot for me (I uninstalled it to make room).

jvp went to nonreal MOP with the rest of us in 04

also you guys are sort of weird for making predictions for like ... A team ... 15 people ... wtf

like that is even more silly than making predictions for IMO team

Yeah, contrary to appearances, I actually thought the same thing... mine was mostly just a carried-away response to teki's.

Maybe I'll just give up on the post rating thing.

[quote="MysticTerminator"]anyway I hear anders was quite the character[/quote]

Huh? Why does this pertain to anything? (Except possibly that the race is very close?)

well, he was from NC

and we were talking about NC ARML

so uh yeah

But a more pressing issue is... who is on each relay team and does anybody have any strategies for relays? Either we distribute the people who we know will do well among a few teams (how many?) and leave the others without much hope, or we gamble and distribute everybody evenly. The first strategy is definitely better for the B team, and probably for the A team as well.

Here's my opinion:
1. Misha, Vivek, Arnav
2. Ray, Yasha, Berman
3. Boyuka, Vitek*, Jeremy
4. Bryce, Joe, Damien
5. ?, Steven Ji, Bangchen Wang

Assuming that Bryce, Damien, and Steven Ji make it.

*Uggh, why must we differentiate between Vivek/Vitek... and the same school too!

yo ok I don't actually know how good people are but usually jeremy is on my team so I can be all like OH MAN IT'S JEREMY HAHN WHAT IS GOING ON JEREMY HAHN

Why only four positions open on the A team?

Seems to me that once you get into the NC ARML A team, you are always in the NC ARML A team. Is this true, good, and fair?

Anders was on the IMO team in 2003 (Gold medal) but did not get into the team in 2004. He won the State Comprehensive three years in a row (2001, 2002, and 2003) but was lost in 2004 (to Arnav) also.

Wherefore not, as a new site must needs reconsideration of the annual epic voyage of the North Carolina ARML teams to Penn State?

@CatalystOfNostalgia: No kidding, man.  That post was from so long ago, no wonder your name includes "Nostalgia".  Go away, you Andover Yankee.  Most Andover people are bad at math (compared to, say, Enloe or East Chapel Hill people, and especially when compared to NC as a whole), and most Yankees shouldn't mess in Southern affairs.  Look at the aftermath of the Hartford Convention.

[quote="Teki-Teki"] Most Andover people are bad at math (compared to, say, Enloe or East Chapel Hill people[/quote]

2007 USAMO Qualifiers
________________________________________________________
Steven Bartz	11	Phillips Academy	Andover
Tony Feng	10	Phillips Academy	Andover
Kwonyong Jin	10	Phillips Academy	Andover
Pichayut Jirapinyo	12	Phillips Academy	Andover
Meng Li	11	Phillips Academy	Andover
Benjamin Niedzielski	11	Phillips Academy	Andover
James Yang	11	Phillips Academy	Andover
Tianyuan Zheng	12	Phillips Academy	Andover

________________________________________________________
Jeremy Hahn	11	East Chapel Hill HS	Chapel Hill
Arnav Tripathy	12	East Chapel Hill HS	Chapel Hill
________________________________________________________
Vivek Bhattacharya	11	William G Enloe HS	Raleigh
Mikhail Lavrov	12	William G Enloe HS	Raleigh
Daniel Vitek	10	William G Enloe HS	Raleigh
________________________________________________________

William G Enloe HS : 2487 students

East Chapel Hill HS : 1735 students

Phillips Academy: 1087 students

Aha, but how about the number of MOPpers? :wink:

Specifically, Blue and Black MOPpers.

EDIT: also, CatalystOfNostalgia, maybe you shouldn't be messing around in other people's forums making fun of them, eh? Because Arnav is in MA and could come tackle you, eh? Eh?  :moose: Ah, good times... But seriously, don't rub it in that we lost one of our top 3 freshmen.

yeah, could we have some scores associated to that list of USAMO qualifiers? how about for the last few years, too? it might provide a more concrete measurement of vitek's assertion of how good people are at math as opposed to how numerous they are.

[quote="Teki-Teki"]

@CatalystOfNostalgia: No kidding, man. That post was from so long ago, no wonder your name includes "Nostalgia". Go away, you Andover Yankee. Most Andover people are bad at math (compared to, say, Enloe or East Chapel Hill people, and especially when compared to NC as a whole), and most Yankees shouldn't mess in Southern affairs. Look at the aftermath of the Hartford Convention.

[/quote]

So when did
[quote]
Yes I agree but UNFORTUNATELY one of them will not be a four-time A-Teamer on your team.
[/quote]
qualify as "making fun" of anybody? Or is it just your "thing" to justify being a total jerk with the accomplishments of other inhabitants of the same state?

Anyway, I wasn't aware that Enloe had large number of MOP qualifiers last year, blue, black, or red. And I, for one, think it's shameful that a little school can't compete with the entire state of North Carolina  :o . [/sarcasm]

PS. I'm pretty sure that the Kentucky and Virginia Resolutions preceded the Hartford Convention.

GUYS! Stop being mean!

@Teki: I have no problem with yankees messing in southern affairs. If they hadn't, NC would still have slaves and all be crazy rednecks. Also, yankees are awesome.

@Phelpedo: What's wrong with the Kentucky / Virginia Resolutions?

@MysticTerminator: I thought you are leaning toward Andover side since the president of Harvard was graduated from Phillips Academy. Also Phillips Academy was founded by a young Harvard graduate.

[quote="Phelpedo"]Um, they're like the same thing as the Hartford Convention?[/quote]

Hartford convention - New England threatens to secede because they don't like the war of 1812

Kentucky/Virginia resolutions - The south refuses to abide by Adams's authoritarian rule (i.e. alien/sedition acts)

Very different.

Okay, but history is boring. Especially US history. Let's talk about math.

[quote="MysticTerminator"]how numerous they are.[/quote]

[quote="jb05"][quote="Phelpedo"]Um, they're like the same thing as the Hartford Convention?[/quote]

Hartford convention - New England threatens to secede because they don't like the war of 1812

Kentucky/Virginia resolutions - The south refuses to abide by Adams's authoritarian rule (i.e. alien/sedition acts)

Very different.
[/quote]

Sorry, I can't resist.

Hartford convention - New england refuses to abide by Madison's authoritarian rule (i.e. endorsing the expansionist tendencies of land-hungry southern War Hawks at the expense of the New England economy).

Kentucky/Virginia resolutions - The south threatens to secede because they don't like Federalist political superiority in the government.

Hopefully, this demonstrates that there are multiple sides of the issue. You can have your own opinion, but don't presume that it should be considered the correct or sole interpretation.

1. I have no affiliation with respect to any of the New England preparatory schools (Exeter, Andover, Milton, et cetera). At this point, I am much more loyal to UNC than to Harvard.

2. I am admittedly far removed from any study of American history, but I believe the main problem with the Hartford convention was not so much that they disagreed with the expansionist policies of the War Hawks, but rather that they explicitly mentioned secession; I do not believe that the Kentucky and Virginia resolutions did so, rather just championing states' rights over the national authority.

3. I believe the numbers you posted are for matriculation rates at various colleges, not for scores on the USAMO. I still do not see the direct correlation between how good people are at math and the posted statistics. Indeed, while I would certainly agree that Andover has many positive qualities, I would be tempted to agree that at this point, many of the top North Carolinians might have an advantage over Andover students of similar prestige in terms of mathematical depth and technique, which is I believe roughly what Vitek was asserting.

4. I very much doubt that most people in the Chapel Hill school district consider themselves as from the south, and as such, the issue of Yankees interfering in Southern affairs loses some meaning.

[quote="Phelpedo"]Hartford convention - New england refuses to abide by Madison's authoritarian rule (i.e. endorsing the expansionist tendencies of land-hungry southern War Hawks at the expense of the New England economy).

Kentucky/Virginia resolutions - The south threatens to secede because they don't like Federalist political superiority in the government.

Hopefully, this demonstrates that there are multiple sides of the issue. You can have your own opinion, but don't presume that it should be considered the correct or sole interpretation.[/quote]

The Kentucky/Virginia resolutions were one (albeit rare) instance wherein the South did not threaten to secede. Also, the Hartford convention had very little to do with the president or misuse of power. It was just about ending the war and having the federalists take over the government. Not that I disapprove of this - greater representation of the north would serve the government quite well nowadays.

[quote="jb05"] Also, the Hartford convention had very little to do with the president or misuse of power. It was just about ending the war and having the federalists take over the government. [/quote]

I wonder if this is an unbiased opinion. The Hartford Convention attacked the war as unfair because it was passed by a South-dominated President/Congress and favored the South at the expense of New England - which, I might add, is in explicit violation of the Constitution (I'm not saying that I agree with their reasoning). Of course, I recognize that it was at least partly motivated by Federalist bitterness. Also, I've always thought that the Kentucky/Virginia Resolutions implied, even if they did not explicitly state it, a threat of secession. Well, that's that - you've added some more dimension to my perspective of these events. I think I'll stop my ranting here so I don't get accused of taking this thread too far off track.

@Arnav: I don't know much about the top North Carolinians so I can't really speak on the matter. I acknowledge that Phillips Andover isn't exactly a perennial powerhouse at the USAMO. I won't give disclose scores of specific people without their permission, but some were fairly good, in my opinion at least: there would be a 17 and a couple 14s next to some of those names.


solve (sin3x)(sin8x)=(sin4x)(sin5x), show steps please...

Ans: 10 degrees.

$\sin 3x\sin 8x=\sin 4x\sin 5x\Longleftrightarrow \cos 5x-\cos 11x=\cos x-\cos 9x\Longleftrightarrow$

$\cos x-\cos 5x=\cos 9x-\cos 11x\Longleftrightarrow \sin 3x\sin 2x=\sin 10x\sin x\Longleftrightarrow$

$\sin x=0\ \vee\ 2\sin 3x\cos x=\sin 10x\Longleftrightarrow \sin x=0\ \vee\ \sin 4x=$$\sin 10x-\sin 2x\Longleftrightarrow$

$\sin x=0\ \vee \ \sin 4x=2\sin 4x\cos 6x\Longleftrightarrow \sin x=0\ \vee \ \sin 4x=0\ \vee \ \cos 6x$ $=\frac 12\Longleftrightarrow$ \[ x\in \left \{ \ \frac{k \pi }{4},\ 2k\pi \pm \frac{\pi }{3}\ |\ k\in Z\ \right \}.  \]

I got it!

(sin3x)(sin8x) = (sin4x)(sin5x)

2(sin3x)(sin4x)(cos4x)- (sin4x)(sin5x)=0

sin4x=0   or      2(sin3x)(cos4x)- (sin5x)=0

                            sin(7x)-sin(x)-(sin5x)=0

                             2cos(6x)sin(x)-sin(x)=0

                            sinx=0  or      2cos(6x)=1


function
calculus
integration
LaTeX
geometry
geometric transformation
algebra

"n" is a positive integer and f:[0,1] -> R a continuous function st:
integral[(x^k)f(x)]dx = 1 for k = 0, 1, ..., n-1.
Prove the existence of such function and prove that
integral[(f(x))^2]dx >= n^2.
The limits of integration are 0 and 1 for all integrals.

$\LaTeX$ translation:
[quote="joao1"]Let $n$ be a positive integer and $f$ be a continuous function from $[0,1]$ to $\mathbb{R}$ such that $\int_{0}^{1}x^{k}f(x)\,dx=1$ for $k=0,1,\dots,n-1$.

Prove that such an $f$ exists, and also that $\int_{0}^{1}(f(x))^{2}\,dx\ge n^{2}$ for any such $f$.[/quote]

This is about the inner product structure on $L^{2}[0,1]$; the inner product is $\langle f,g\rangle=\int_{0}^{1}f(x)g(x)\,dx$. If we apply Gram-Schmidt to the linearly independent set $1,x,x^{2},\dots$, we get an orthonormal sequence of polynomials $p_{0}(x)=1,p_{1}(x),p_{2}(x),\dots$, with $p_{0}$ of degree $k$.

Using these $p_{i}$, we can construct a unique polynomial $q(x)=a_{0}+a_{1}p_{1}(x)+a_{2}p_{2}(x)+\cdots+a_{n-1}p_{n-1}(x)$ of degree at most $k-1$ satisfying the relation; $\int_{0}^{1}x^{k}q(x)\,dx=c_{0}^{k}a_{0}+c_{1}^{k}a_{1}+\cdots+c_{k}^{k}a_{k}$, where $x^{k}=\sum_{j=0}^{k}c_{j}^{k}p_{j}(x)$. Since the last coefficient $c^{k}_{k}$ of each row is nonzero, the system can be solved routinely.
This is existence; we can also show that $f-q$ is orthogonal to $q$ if $f$ is any other solution, and thus $\|f\|^{2}\ge \|q\|^{2}$. We need only show that $\|q\|^{2}=\int_{0}^{1}(q(x))^{2}\,dx\ge n^{2}$.
That's a nontrivial computation, and I'm not going to do it right now.

This will be a long post.

As fedja noted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=147066]here[/url], the reduction of the problem I posted above is equivalent to finding the sum of the elements of the inverse of the Hilbert matrix. Consequently, the following will focus on the Hilbert matrix and its inverse, along with the related polynomials on $[0,1]$.

We will start all sums at zero; the rows and columns of our matrices are indexed by $\{0,1,\cdots,n-1\}$.

Using the inner product notation above, we have that the $ik$ entry $\frac1{i+k+1}$ of the Hilbert matrix $H$ is $\langle x^{i},x^{k}\rangle.$ If $f(x)=a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}$ and $g(x)=b_{0}+b_{1}x+\cdots+b_{n-1}x^{n-1},$ this gives $\langle f,g\rangle=\sum_{i,k}a_{i}b_{k}h_{ik}=a^{T}Hb,$ where $a$ and $b$ are the column vectors of coefficients. Suppose $q$ is a polynomial of degree at most $n$ satisfying $\langle q,x^{k}\rangle=1$ for each $k,$ and let $a=(a_{0},a_{1},\dots,a_{n-1})^{T}$ be its coefficient vector in the standard basis. Our inner product relation becomes
$(0,\dots,0,1,0,\dots,0)\cdot H\cdot (a_{0},\dots,a_{k-1},a_{k},a_{k+1},\dots,a_{n-1})= 1$, or equivalently $I\cdot H\cdot\begin{pmatrix}a_{0}\\ a_{1}\\ \vdots\\ a_{n-1}\end{pmatrix}=\begin{pmatrix}1\\ 1\\ \vdots\\ 1\end{pmatrix}=j.$ (In all following computations, $j$ is this vector.)
Multiplying by $H^{-1}$, $a=H^{-1}j,$ and $\langle q,q\rangle=(H^{-1}j)^{T}H(H^{-1}j)=j^{T}H^{-1}j.$ For any matrix $M,$ $j^{T}Mj$ is the sum of the entries of $M$, and we have the equivalence noted by fedja.

Now, we need to compute some details of $H^{-1}.$ To do this, we work with the $p_{i}$, which are scaled Legendre polynomials. For each $k,$ $p_{k}(x)=\frac{\sqrt{2k+1}}{k!}\frac{d^{k}}{dx^{k}}(x-x^{2})^{k}.$ Orthogonality can be proven by integrating by parts repeatedly, differentiating the polynomial of lower degree and integrating the polynomial of higher degree; the boundary terms vanish at each step. The normalization constant also comes from an integration by parts: $\int_{0}^{1}\left(\frac{d^{k}}{dx^{k}}(x-x^{2})^{k}\right)^{2}\,dx=(2k)!\int_{0}^{1}x^{j}(1-x)^{j}\,dx,$ and this latter integral can be computed by integrating by parts another $j$ times.

Let $P$ be the upper triangular matrix with $ik$ entry the $x^{i}$ coefficient of $p_{k}.$ We have $P^{T}HP=\begin{pmatrix}\langle p_{0},p_{0}\rangle&\langle p_{0},p_{1}\rangle&\cdots&\langle p_{0},p_{n-1}\rangle\\ \langle p_{1},p_{0}\rangle&\langle p_{1},p_{1}\rangle&\cdots&\langle p_{1},p_{n-1}\rangle\\ \vdots&\vdots&\ddots&\vdots\\ \langle p_{n-1},p_{0}\rangle&\langle p_{n-1},p_{1}\rangle&\cdots&\langle p_{n-1},p_{n-1}\rangle\end{pmatrix}=I.$

Therefore, $H=(P^{T})^{-1}IP^{-1}=(PP^{T})^{-1},$ and $H^{-1}=PP^{T}.$ We get $j^{T}H^{-1}j=(j^{T}P)(P^{T}j)=\|P^{T}j\|^{2}.$ The $k$th entry of $P^{T}j$ is the sum of the entries in the $k$th column of $P,$ which is $p_{k}(1).$ By the symmetry of our definition, $p_{k}(1)=(-1)^{k}p_{k}(0)=(-1)^{k}\sqrt{2k+1}.$ Therefore, $j^{T}H^{-1}j=\sum_{k=0}^{n-1}2k+1=n^{2}$ and we have solved the problem.

That's not all, of course: we can get plenty of information on $H^{-1}$ from $P$. The upper left entry of $H^{-1}$ is the sum of the squares of the entries in the first row of $P;$ the $k$th entry in the first row is $p_{k}(0)=\sqrt{2k+1},$ and the upper left entry is also equal to $n^{2}.$
We can show that the entries of $H^{-1}$ are all integers: since $\frac1{k!}\frac{d^{k}}{dx^{k}}$ maps polynomials with integer coefficients to other polynomials with integer coefficients, $P=\begin{pmatrix}q_{0}&q_{1}&\cdots&q_{n-1}\end{pmatrix}\cdot \begin{pmatrix}1&0&\cdots&0\\ 0&\sqrt{3}&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\sqrt{2n-1}\end{pmatrix},$ where each $q_{k}$ is an integer vector (which we will also interpret as the integer polynomial $\frac1{\sqrt{2k+1}}p_{k}$). This makes $PP^{T}=\begin{pmatrix}q_{0}&q_{1}&\cdots&q_{n-1}\end{pmatrix}\cdot \begin{pmatrix}1&0&\cdots&0\\ 0&3&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&2n-1\end{pmatrix}\cdot\begin{pmatrix}q_{0}^{T}\\ q_{1}^{T}\\ \cdots\\ q_{n-1}^{T}\end{pmatrix},$ a product of integer matrices.
We can go a step further and write the exact entries of $H^{-1}$ as sums; $(x-x^{2})^{k}=\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}x^{i+k},$ and $q_{k}(x)=\frac1{k!}\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}\frac{(k+i)!}{i!}x^{i}= \sum_{i=0}^{k}(-1)^{i}\binom{k}{i}\cdot\binom{k+i}{i}x^{i}.$ Let $(q_{k})_{i}$ be the coefficient of $x^{i}$ in $q_{k}.$
The $ik$ entry of $H^{-1}$ is then equal to $\sum_{l=0}^{n-1}(q_\ell)_{i}\cdot (2\ell+1)\cdot (q_\ell)_{k}=(-1)^{i+k}\sum_{\ell=0}^{n-1}(2\ell+1)\binom{\ell}{i}\cdot\binom{\ell}{k}\cdot\binom{\ell+i}{i}\cdot\binom{\ell+k}{k}.$ For many entries, most terms in this sum are zero. In an extreme case, the lower right entry of $H^{-1}$ has just one nonzero term in the sum; that entry is $(2n-1)\binom{2n-2}{n-1}^{2}.$

Since $P$ is triangular, the determinant of $H^{-1}$ is relatively easy to calculate; it's $\prod_{i=0}^{n-1}(2i+1)(q_{i})_{i}^{2}=\prod_{i=0}^{n-1}(2i+1)\binom{2i}{i}^{2}.$
If we write $\binom{2i}{i}=2^{2i}\cdot\frac{1\cdot 3\cdot 5\cdots (2i-1)}{2\cdot 4\cdot 6\cdots (2i)},$ the determinant becomes $\det(H^{-1})$
$=2^{4+8+\cdots+(4n-4)}\cdot 1^{2n-1}\cdot 2^{-(2n-2)}\cdot 3^{2n-3}\cdot 4^{-(2n-4)}\cdots (2n-1)^{1}\cdot (2n)^{0}$
$=2^{2n^{2}-2n}\prod_{k=1}^{2n}k^{(-1)^{k+1}(2n-k)}.$


The lengths of the radii of two concentric circles differ by 1 centimeter. How many centimeters are in the positive difference of the circumferences of the two circles? Express your answer in terms of $ \pi$.

The diameters differ by $ 2$ centimeters, so the circumferences differ by $ 2\pi$


Find the maximum and minximum of

$ S \equal{} \cos ^2 A \plus{} \cos ^2 B \plus{} \cos ^2 C$

$ S \equal{} \cos ^2 A \plus{} \cos ^2 B \minus{} \cos ^2 C$

$ S \equal{} \cos 5A \plus{} \cos 5B \minus{} \cos 5C$

$ S \equal{} \cos nA \plus{} \cos nB \minus{} \cos nC$ $ (n \in R)$

 with $ A,B,C \in R$ such that $ A \plus{} B \plus{} C \equal{} \pi$.

I think that's very nice!!!

Can you use trionmial to find the solution of problem?  :)  :)

[hide="Q2:"]
$ \begin{array}{l}
 \begin{array}{*{20}c}
   {S \equal{} \cos ^2 A \plus{} \cos ^2 B \minus{} \cos ^2 C \equal{} \frac{{1 \plus{} c{\rm{os}}2A}}{2} \plus{} \frac{{1 \plus{} c{\rm{os}}2B}}{2} \minus{} \frac{{1 \plus{} c{\rm{os}}2C}}{2}}  \\
   { \equal{} \frac{1}{2} \plus{} \frac{1}{2}(c{\rm{os}}2A \plus{} c{\rm{os}}2B \minus{} c{\rm{os}}2C)}  \\
\end{array} \\
 S' \equal{} c{\rm{os}}2A \plus{} c{\rm{os}}2B \minus{} c{\rm{os}}2C \equal{} 2.c{\rm{os}}(A \plus{} B)c{\rm{os}}(A \minus{} B) \minus{} 2c{\rm{os}}^2 C \plus{} 1 \\
  \equal{}  \minus{} 2c{\rm{os}}^2 C \minus{} 2\cos C\cos s(A \minus{} B) \plus{} 1 \leftrightarrow 2c{\rm{os}}^2 C \plus{} 2\cos C\cos (A \minus{} B) \plus{} S' \minus{} 1 \equal{} 0 \\
 \Delta ' \ge 0 \leftrightarrow \cos ^2 (A \minus{} B) \minus{} 2(S' \minus{} 1) \ge 0 \leftrightarrow S' \le \frac{{\cos ^2 (A \minus{} B)}}{2} \plus{} 1 \le \frac{3}{2} \\\end{array}$

So $ S \le \frac{1}{2} \plus{} \frac{1}{2}.\frac{3}{2} \equal{} \frac{5}{4}$[/hide]

Please solve continuely!!!

[hide="Solution1"]
$ {S \equal{} \cos ^2 A \plus{} \cos ^2 B \plus{} \cos ^2 C \equal{} \frac{3}{2} \plus{} \frac{1}{2}(c{\rm{os}}2A \plus{} c{\rm{os}}2B \plus{} c{\rm{os}}2C)}$

$ \equal{} \frac{3}{2} \plus{} \frac{1}{2}\left[ {2c{\rm{os}}(B \plus{} C)c{\rm{os}}(B \minus{} C) \plus{} 2c{\rm{os}}^2 A \minus{} 1} \right]$

$ \equal{} \frac{3}{2} \plus{} \frac{1}{2}\left[ {2c{\rm{os}}^2 A \minus{} 2\cos A\cos (B \minus{} C) \minus{} 1} \right]$

$ \begin{array}{l}
  \leftrightarrow c{\rm{os}}^2 A \minus{} \cos A\cos (B \minus{} C) \plus{} 1 \minus{} S \equal{} 0 \\
  \Rightarrow \Delta ' \ge 0 \leftrightarrow \cos ^2 (B \minus{} C) \minus{} 4(1 \minus{} S) \ge 0 \\
  \leftrightarrow \cos ^2 (B \minus{} C) \ge 4(1 \minus{} S) \leftrightarrow 1 \ge 4(1 \minus{} S) \leftrightarrow S \ge \frac{3}{4} \\
 \end{array}$[/hide]

Can you give me an other solution!!!

Can you give me the solutions of Q3 and Q4? Thanks very much!!! :oops:

For $ S \equal{} \cos ^2 A \plus{} \cos ^2 B \plus{} \cos ^2 C, \cos^2 x \le 1 \implies x \equal{} n\pi; n \in \mathbb{Z}.$
By inspection $ A \equal{} B \equal{} 0; C \equal{} \pi$ (or other permutations) to give a maximum of $ 3$.

For
$ S \equal{} \cos ^2 A \plus{} \cos ^2 B \minus{} \cos ^2 C,$ $ \cos^2 A \plus{} \cos^2 B \ge 0 \implies A \equal{} B \equal{} (n\plus{}1)\pi/2; \cos^2 C \le 1 \implies C \equal{} n\pi$.
Take $ n\equal{}0$ to get a minimum of $ \minus{}1$.

[quote="mathwizarddude"]For $ S \equal{} \cos ^2 A \plus{} \cos ^2 B \plus{} \cos ^2 C, \cos^2 x \le 1 \implies x \equal{} n\pi; n \in \mathbb{Z}.$
By inspection $ A \equal{} B \equal{} 0; C \equal{} \pi$ (or other permutations) to give a maximum of $ 3$.

For
$ S \equal{} \cos ^2 A \plus{} \cos ^2 B \minus{} \cos ^2 C,$ $ \cos^2 A \plus{} \cos^2 B \ge 0 \implies A \equal{} B \equal{} (n \plus{} 1)\pi/2; \cos^2 C \le 1 \implies C \equal{} n\pi$.
Take $ n \equal{} 0$ to get a minimum of $ \minus{} 1$.[/quote]

I think that wrong method!  :?:

You quintuple-posted. That is something that is done by silly people, which makes you a very silly person. Just like anonynynyneeeemus who just fell off an aerobics ball. And will do so again very shortly. Never mind.

For the first one, max(S)=3, min(S)=3/4.
For the second one, max(S)=1, min(S)=-1.

More answers to come when I get a big brother.

[quote="mathwizarddude"]So what is the ideal method that you are looking for?  Calculus?[/quote]

I often use the trinomial to solve the following problems that. I solved Q1 anh Q2, do you agree?


algebra
polynomial
number theory unsolved
number theory

Natural number k>1 is given and integers a1, a2, ... , an satisfying:

a1 + 2^i *a2 + 3^i * a3 +...+ n^i * an = 0
for i=1,2, ... , k-1. Prove that k! divides:

a1 + 2^k * a2+ 3^k *a3 +....+ n^k *an

 :starwars:

We consider polinomial $ P(x) \equal{} (x\minus{}1)(x\minus{}2).....(x\minus{}k)$ .Easily see that $ P(x)$ is divisible by $ k!$ for all
$ x \in Z$ . Supposed that $ P(x) \equal{} x^k \plus{} T_{k\minus{}1}.x^{k\minus{}1} \plus{}...\plus{}T_1.x \plus{} k!$
Then we have $ a_i.g(i)\equal{}a_i.i^k \plus{} T_{k\minus{}1}.a_i.i^{k\minus{}1} \plus{}...\plus{} T_1.a_i.i \plus{}k!.a_i$ for $ i\equal{}1,2,...,n$
Hence  $ a_1g(1)\plus{}a_2g(2)\plus{}...\plus{}a_ng(n) \equal{} (a_1\plus{}a_2.2^k\plus{}...\plus{}a_n.n^k) \plus{} k!.(a_1\plus{}a_2\plus{}...a_n)$
And now,you can easily see that QED.


trigonometry
calculus
calculus computations

Why would you even want to use L'Hopitals rule? Differentiating these expressions seems to be a headache. Sometimes I think that when people say that they can't use L'Hopitals that they haven't learned taylor expansions yet, because there is no sane reason for wanting to use L'Hopitals rule in that case. Did you learn Taylor yet, chy1013m1?

anik_andrew: When you did $\frac{1}{\cos x}$, did you taylor expand $\frac{1}{1+u}$ and insert $u = -x^2/2 + o(x^3)$, or is there a different method? I once saw Kent Merryfield mention long division, but I don't precisely how he meant that.

Yes Kalle, i did exactly as you are writed :)

although i did learn tylor expension in high school, right now its not taught in the course
in university yet, so i cannot use it. Also, i said cannot use the L Hopital rule cuz it says on
the question sheet..

thanks anyway


calculus
integration
complex analysis
real analysis
real analysis unsolved

Calculate integral:

$ \int_{0 \leq x \leq y \leq \pi} ln|sin(x\minus{}y)|dx dy$

I don't know how to do it. The hint is to use in some way the Vandermonde determinant for n-th roots of unity i. e. the Vandermonde determinant of $ {1, \epsilon, {\epsilon}^{2}, \cdots, {\epsilon}^{n\minus{}1}}$, where $ \epsilon$ is $ e^{\frac{2\pi}{n}}$, so probably it's more of a complex analysis problem.

Any ideas?


linear algebra
matrix
function
algebra
polynomial
geometry
linear algebra unsolved

Does anybody have a proof that $(AB)^{*}=B^{*}A^{*}$ for [b]singular[/b] matrices?

$A^{*}$ is the transpose of the cofactor matrix of $A$ (so called [i]adjugate[/i] or [i]classical adjoint[/i] or whatever).

I have a proof, but you will not like it.  :ninja:

Both sides depend continuously on the entries of $A$ and $B$. They are equal for invertible matrices, which form a dense subset of all square matrices. Thus, they are equal for all square matrices.

Both sides are not merely continuous functions of the entries of $A$ and $B,$ they are polynomial functions of those entries. Hence we don't even need that the set on which they are identical is dense - a sufficiently large and appropriately arranged finite set of values should suffice.

What about matrices in other fields than $\mathbb{C}$? Surely there is a different way of establishing such an algebraic equality

[quote="Kent Merryfield"]a sufficiently large and appropriately arranged finite set of values should suffice.[/quote]
But I don't know enough algebraic geometry to decide which sets are sufficiently large and appropriately arranged.  :(

I agree with Kalle that there should be a proof for matrices over a general commutative ring.

If you try to compute, say, the elements in the first row and the first column on both sides, you'll see that what you need to prove is a partial case of Binet-Cauchy formula, whose various proofs can be found in many places (look it up on PlanetMath, for example).


1. $K$ takes $30$ minutes less time than $M$ to travel a distance of $30$ miles. K travels $\frac{1}{3}$ mile per hour faster than $M$. If $x$ is $K$'s speed in miles per hour, then find $K$'s time for the distance in terms of $x$.

2. Find the ordered pair $(x,y)$ that is the solution of the system $\frac{x+2y}{xy}=\frac{11}{12}$ and $\frac{2x-3y}{xy}=\frac{2}{3}$.

3. When 3 numbers are added two at a time, the sums are $29$, $46$, and $53$. What is the sum of all three numbers?

[quote]2. Find the ordered pair $(x,y)$ that is the solution to the system:

$.\qquad\frac{x+2y}{xy} = \frac{11}{12}\qquad\frac{2x-3y}{xy} = \frac{2}{3}$[/quote][hide]
$\frac{x+2y}{xy} = \frac{1}{12}\;\;\;\Rightarrow\;\;\;\frac{1}{y} + \frac{2}{x} = \frac{11}{12}$

$\frac{2x-3y}{xy} = \frac{2}{3}\;\;\;\Rightarrow\;\;\;\frac{2}{y} - \frac{3}{x} = \frac{2}{3}$

Let $a = \frac{1}{y},\;\;b = \frac{1}{x}$
and we have the system: $\begin{Bmatrix}a + 2b = \frac{11}{12}\\  \\ 2a - 3b = \frac{2}{3}\end{Bmatrix}$

Solving, we get: $\begin{Bmatrix}b = \frac{1}{6}\;\;\Rightarrow\;\;x = 6\\ \\  a = \frac{7}{12}\;\;\Rightarrow\;\;y = \frac{12}{7}\end{Bmatrix}$[/hide]

Solution to question 1
[hide]30/x[/hide]

[quote="Soroban"][quote]3. When 3 numbers are added two at a time, the sums are 29, 46 , and 53 .
What is the sum of all three numbers?[/quote][hide]
Let the numbers be $a,\;b,\;c.$

Then we have: $\begin{array}{ccc}a + b = 29\\ b + c = 46\\ a + c = 53\end{array}$

Add the three equations: $3a + 3b + 3c = 128$

Therefore: $a + b + c = \frac{128}{3}$[/hide][/quote]

Adding the 3 equations, you get $2a + 2b + 2c = 128$, which means the answer is $ \boxed{64} $

[quote="keta"]1. $K$ takes $30$ minutes less time than $M$ to travel a distance of $30$ miles. K travels $\frac{1}{3}$ mile per hour faster than $M$. If $x$ is $K$'s speed in miles per hour, then find $K$'s time for the distance in terms of $x$.

[/quote]

[hide="Solution"]K travels 30 miles. K's rate is $x$. Therefore K's time is distance/rate = $\boxed{\frac{30}{x}}$. All the information about M is entirely useless.[/hide]

Why are we all reviving a post from 9 years ago?

New users don't realize that this topic was from 9 years ago because they don't realize / look at the time on the right hand top corner or every post.

Come on.


Find all functions $f: \mathbb{R^{+}}\to \mathbb{R^{+}}$such that

$f(xf(y))f(y)=f(x+y)$ for all $x,y>0$

This is problem 5 from the second day of IMC 2000


is there a quick way to figure out if a number is prime without having to divide it by a lot of numbers? Like is 137573 prime???

Normally when I try to determine if a number is prime I divide it by the primes less than or equal to its square root.

The only way I can think of is square-rooting it and then dividing by all the primes up until that number.  There might be an easier way but I don't know it either.

Are you allowed to use a calculator?  :lol:

Well, you see, once you prove the Riemann hypothesis (which I notice you're currently working on) you will have an easier time :D

Amazingly, there is a deterministic polynomial time algorithm for testing primality!! http://www.cse.iitk.ac.in/news/primality.pdf . However, it's only really useful for theory or big numbers. You'll probably want to use Rabin-Miller for most circumstances.

[quote="worthawholebean"]That just confused me.

So the Rabin-Miller algorithm only tells you if it's composite, and only sometimes?[/quote]

If you're 99% sure it isn't composite, it probably is prime.

[quote="NoSoupForYou"][quote="chess64"]Ah. Now I know exactly how to perform the Rabin-Miller test. Thankyou  :lol: lol[/quote]

Isn't " :lol: lol " a bit redundant?[/quote]

No :lol: lol


ratio induction function algebra polynomial search

I saw this formula in a precalc book I was reading and it's for the nth term of the Fibonnaci sequence. Does anyone know how to derive this formula?  It seems strange that a recursive sequence has a formula.

$\frac 1{\sqrt{5}}\times(\frac {1+\sqrt{5}}{2})^n-\frac 1{\sqrt{5}}\times(\frac {1-\sqrt{5}}{2})^n$

I dont understand the need for that formula.. isnt fibbonaci = next number is the sum of the prev 2? 1,1,2,3,5,8,13,21,34 etc.

It has something to do with the golden ratio I think, because the ratio between Fibonnaci numbers get closer and closer to the golden ratio.

There are techniques that help us build formulas out of linear recursions.

I'm going to move this to the Intermediate forum as this is a problem at the high end of the Intermediate level.

well, $\frac{1+\sqrt{5}}{2}$ is the golden ratio, the number that the quotient of consecutive fibonacci numbers approaches as the numbers increase. Math world : [url=http://mathworld.wolfram.com/BinetForms.html]Binet Forms[/url] has some info on this formula :)

The formula is important, because we would like a way to figure out what $F_n$ is without first knowing what $F_1, F_2,...,F_n$ are. In the same way, we would like to know if $n$ is prime without first checking divisibility by $1,2,3,...$.

One way to prove the formula is induction - see if you can work out the details.

Of course, this gives little in the way of reasoning.

We could use generating functions to determine that formula, but forget that for now.

Another technique is using characteristic equations. We assume that the formula can be written in the form $F_n=\alpha a^n + \beta b^n$. Then $a,b$ are the roots of $x^2-x-1$, which comes from $F_n-F_{n-1}-F_{n-2}=0$. (Why? Try to plug in that relation into our assumed formula and see what happens.) We just need to calculate the values of $\alpha, \beta$ which is easy since we know a whole bunch of values which we can plug in.

I don't know any other way to motivate the formula other than by using generating functions.  Everything else mentioned above seems to assume some prior knowledge of algebraic facts regarding this kind of recursion.

Hmm... The 3 ways I know to derive this formula are:

1.  Characteristic polynomial approach.
2.  Generating Functions
3.  Unhumanly good pattern recognition :lol:

The characteristic polynomial approach seems to me tied to the generating functions approach in the sense of identifying the characteristic polynomial, or can you motivate the derivation in some other manner (aside from unhumanly good or lucky pattern recognition)?

[quote="fourierseries"]z transform.[/quote]
Can you explain what that means?

We discussed the formula in the "Other Problem Solving" forum. We constructed a polynomial to get. Also kunny provided another neat solution.
You can try to search it there.

its actually not hard to prove it.
here it is:
just use polynomials.
let $F(x)=a_0+a_1x+a_2x^2+....$
here $a_j$ are the fibonacci numbers. note that we don't care what $x$ is, we just want to use $F(x)$ to find the $a_j$
$xF(x)=a_0x+a_1x^2+a_2x^3+a_3x^4....$
$x^2F(x)=a_0x^2+a_1x^3+a_2x^4+a_3x^5....$
now if you group these you get:
$F(x)-xF(x)-x^2F(x)=a_0+(a_1-a_0)x+ (a_2-a_1-a_0)x^2+(a_3-a_2-a_1)x^3+(a_4-a_3-a_2)x^4+.....$
the importance of the fibonacci sequence is that $(a_2-a_1-a_0), (a_3-a_2-a_1), (a_3-a_2-a_1),....=0$ so the above's equation is so simple.
$F(x)-xF(x)-x^2F(x)=a_0+(a_1-a_0)x$
we also know that $a_0=a_1=1$ so $(1-x^2-x)F(x)=1$ or
$F(x)=\frac {1}{1-x-x^2}$. now if you factor out the denominator you get:

$F(x)=\frac {1}{[1- \frac {-2x}{1-\sqrt{5}}][1-\frac {-2x}{1+\sqrt{5}}]}$
now with use of some algebra you get:
$F(x)=\frac {5+\sqrt5}{10}*[\frac {1}{[1+\frac {2x}{1-\sqrt{5}}]}]+\frac {5-\sqrt5}{10}*[\frac {1}{[1+\frac {2x}{1+\sqrt{5}}]}]$
the above parts that are in a bracket can be rewritten by the summation formula:
$F(x)=\frac {5+\sqrt{5}}{10}.\sum_{n=1}^{\infty} [\frac{-2x}{1-\sqrt{5}}]^{j}+\frac {5-\sqrt{5}}{10}.\sum_{n=1}^{\infty}[\frac{-2x}{1+\sqrt{5}}]^{j}$
now if you group the similar powers of $x$  you get:
$F(x)=\sum_{n=1}^{\infty}[\frac {5+\sqrt{5}}{10}.[\frac{-2x}{1-\sqrt{5}}]^{j}+\frac {5-\sqrt{5}}{10}.[\frac{-2x}{1+\sqrt{5}}]^{j}]$
now remember we started the problem by saying
$F(x)=a_0+a_1x+a_2x^2+....$
so now we have two differen formulas for $F(x)$ these two MUST equivelant to each other, especially the coefficents of the same powers of $x$. so we conclude that:
$a_j= \frac {5+\sqrt{5}}{10}.[\frac{-2x}{1-\sqrt{5}}]^{j}+\frac {5-\sqrt{5}}{10}.[\frac{-2x}{1+\sqrt{5}}]^{j}$
now we are almost there, all we need to do is to replace some of the stuff up there by equivalent of them.
$\frac {5+\sqrt{5}}{10}=\frac {1}{\sqrt{5}}.\frac {1+\sqrt{5}}{2}$ also $\frac {5-\sqrt{5}}{10}=\frac {-1}{\sqrt{5}}.\frac {1-\sqrt{5}}{2}$ and $\frac {-2}{1-\sqrt{5}}=\frac {1+\sqrt{5}}{2}$, and $\frac {-2}{1+\sqrt{5}}=\frac {1-\sqrt{5}}{2}$
if we replace these values for the $a_j$ formula we get:
$a_j=\frac {[\frac{1+\sqrt{5}}{2}]^j- [\frac{1-\sqrt{5}}{2}]^j}{\sqrt{5}}$

for more info
take a look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=prove++fibonacci&t=34840]here[/url].

yea the proof for binets is in the chapter on series and sequences in teh AOPs book volume 2


calculus
LaTeX
real analysis
real analysis unsolved

Comparison Lemma
;Let the sequence $ \{a_n\}$ \text {converge to the number $ a$. Then the sequence}$ \{b_n\}$  converges to the number $ b$ if there is a nonnegative number $ C$ and an index $ N_1$  such that

$ \|b_n \minus{} b\|\leq C\|a_n \minus{} a \|$  for all indices $ n \geq N_1$.

 I dont understand this Lemma. I have a sequence $ \{a_n\}$ that converges, so what. Why does $ \{b_n\}$ converge. I can find a $ \{b_n\}$ that doesnt converge when $ \{a_n\}$ converges. What property does $ \{b_n\}$ have that is related to $ \{a_n\}$ that we can use the Comparison Lemma. I really would like to know if anyone can explain this. I know this is important because the Comparison Lemma is used to prove " a set S is dense in R if and only if every number x is the limit of a sequence in S.

This comes from an Advanced Calculus by Patrick M Fitzpatrick

[quote="ayatollah85"]What property does $ \{b_n\}$ have that is related to $ \{a_n\}$ that we can use the Comparison Lemma.[/quote]
[quote="ayatollah85"]There is a nonnegative number $ C$ and an index $ N_1$ such that $ |b_n \minus{} b|\leq C |a_n \minus{} a |$ for all indices $ n \geq N_1$.[/quote]

Unrelatedly, it would be nice if you would edit your post for readability by removing the unnecessary LaTeX commands and placing dollar signs around the equations.


i am sponsoring a chess tournament thing.  i need at least 8 people to join, but hopefully more are interested. we could either put our moves on this string, but i think it would be better if we played over pm's because it might be called spamming. i will play, so that's one person. if there are any questions, post them here and i will answer them, or at least try to.

hopefully we get some more people. so far we only have 3. i guess 4 would work, too, but i still like the bigger ones.

I'm in. Just pm me when it starts, or I'll forget. :D

[quote="Adunakhor"]Would putting this in the Fun and Games Forum get more people? That forum seems to get a lot more traffic than this one.[/quote]

I'll put a thread up there pointing people to this forum. that should work okay.

now that we have 4 people playing, i will wait for 2 more days to see if anyone else wants to join.

I would like to play but when will the game start? I might be not able to play cause I do not know if there is any internet cafe around the place I am going to enjoy the summer :)

[quote="4everwise"]Would you guys like to play on US Chess Live?

You could just record the moves after each game and post it.  :)[/quote]

that would be great. the games would go a lot faster.
unfortunately, i don't have us chess live, and i don't think i will anytime soon.  i used to have it.  :(
I will make a list of people who have chess live and their usernames when registration is over.  so, when you sign up, could you tell me if you have it? because that would be good.

the games will start in a maximum of 4 days.

I can play if it's not too late to sign up. But I do not have Chess Live :(

okay, we have 8 people now, but more are allowed to join, the more, the merrier (and more complicated, which is fun :) )

so far i have:
Captain Sugar (me) ^
me@home
Adunakhor ^
4everwise*
Ilfirin
GoBraves
rem ^
Author Message
PenguinIntegral
nutz_for2.718281828
ShoeFactory

so far, i only know that 4everwise has us chess live, so please tell me if you have it or not, because that would be convenient. i put a little star by his name to show he has it.

edit: i will put one of these things ^ by people who don't have chess live.

Yeah, maybe u should. let s see what captain sugar says.

it seems like this chess tournament has long started. i didn't know about it. i would really like to play in the next one. thx.

i guess the bracket now is:

Adunakhor vs. 4everwise
shoefactory vs. splashD

consolation bracket:
nutz_for2.718281828 vs. Ignite 168
bubka vs. rem

I have started using FICS and babaschess. Now I am ready. Someone PM rem.

I want to play. What happened to this anyway? It just suddenly dissappeared?  :diablo:  :gathering:

oh i see.

Well... on the plus side.. Welcome back!

Umm... Can this Begin again? I'm gonna start a new thread.

I play chess on yahoo every once in a whle.

[quote="anirudh"]Yes. Please. I'm AMAZING at chess (Rating somewhere around the 1850's) but haven't played for a LONG time. I really want to play.[/quote]

play online?

Yes.The message is too small. Please make the message longer before submitting.

yahoo chess is actually pretty cool.

and you can undo and stuff when practicing  :lol:

YAHOO CHESS=HORRIBLE!!!

US CHESS LIVE=AMAZING


Take 10 points inside the circle with diameter 5. Prove that for any these 10 points there exist two points whose distance is less than 2.

[quote="pleurestique"]i think the circle is too big. maybe you mean diameter 5.[/quote]

Yes, you are  right. I made a typo.

kunny

in that case... divide the circle into 9 pieces... the disc of radius 1 concentric to the given circle, and 8 congruent pieces surrounding it cut out by 8 equally spaced radii in the given circle. the distance between two points in one of the outer pieces is at most $\sqrt{1^2+(5/2)^2-(5/2)\sqrt{2}} = \sqrt{\frac{29-10\sqrt{2}}{4}} < 2$. so we can shrink the inner circle ever so slightly so that the distance between any two points in any of the 9 regions is strictly* $<2$.

Let the points be $ A_1, A_2, A_3, \cdots A_{10}$ in that order in clockwise direction.

Suppose all of $ A_1A_2, A_2A_3, \cdots A_{10}A_1$ are greater than or equal to $ 2,$

for $ i\in \{1, 2, \cdots 10\},$

$ 4 \le \left(A_iA_{i \plus{} 1}\right)^2 \equal{} \left(\frac {5}{2}\right)^2 \plus{} \left(\frac {5}{2}\right)^2 \minus{} 2\left(\frac {5}{2}\right)^2\cos(\theta)$ where $ \theta$ is the angle subtended by $ A_iA_{i \plus{} 1}$ and $ A_{11}$ is considered to be $ A_1$

So, rearranging gives $ \cos(\theta) \le \frac {17}{25}$

Now, consider $ 25\sqrt {5} > 50 > 43 \equal{} 17\times 4 \minus{} 25$

$ \Longrightarrow \frac {1 \plus{} \sqrt {5}}{4} > \frac {17}{25}$

So, $ \cos(\theta) \le \frac {17}{4} < \frac {1 \plus{} \sqrt {5}}{4} \equal{} (\cos 36^{\circ})$

So, $ \theta > 36^{\circ}$

So, sum of the angles subtended by $ A_1A_2,A_3A_4,\cdots A_{10}A_1 > 10\times 36^{\circ} \equal{} 360^{\circ}$

which contradicts angle around a point,

and so, our assumption is false $ \Longrightarrow$ there exists two points with distance less than $ 2$

[quote="pleurestique"]The distance between two points in one of the outer pieces is at most $\sqrt{1^2+(5/2)^2-(5/2)\sqrt{2}} = \sqrt{\frac{29-10\sqrt{2}}{4}} < 2$[/quote]

How do you find this distance?


Problem:
------------------------------------------------------------------------------------------------------
Let have a sequence of K positive finite integers, starting with 1 and ending with K (each element of this sequence represents a free slot) and an arbitrary multiset of L integers of size 1..K.

Using a greedy algorithm, try to fill slots of this sequence with multiset elements, so that the sequence will be the most similar to entered multiset. Similarity is defined as a minimal sum of
absolute values of subtraction of filled sequence slots and elements, which were put in sequence slots.

Examples:

K=3
Multiset = {1,1,1}
Filled_sequence= 1[1] 2[1] 3[1]
Minimal sum = abs(1-1) + abs(2-1) + abs(3-1)= 3

K=5
Multiset = {1,2,3,4,5}
Filled_sequence= 1[1] 2[2] 3[3] 4[4] 5[5]
Minimal sum = abs(1-1) + abs(2-2) + abs(3-3) + abs(4-4) + abs(5-5) = 0

K=7
Multiset = {2,2,4,4,6,6,6}
Filled_sequence= 1[2] 2[2] 3[4] 4[4] 5[6] 6[6] 7[6]
Minimal sum = abs(1-2) + abs(2-2) + abs(3-4) + abs(4-4) + abs(5-6) + abs(6-6) + abs(7-6) = 4

K=7
Multiset = {1,1,7,7}
Filled_sequence= 1[1] 2[1] 3[-] 4[-] 5[-] 6[7] 7[7]
Minimal sum = abs(1-1) + abs(2-1) + abs(6-7) + abs(7-7) = 2
------------------------------------------------------------------------------------------------------

I could think of some greedy approaches, but cannot really determine if they really work, or even if they really are greedy. Could you write me a way, or at least idea how to implement this?

Thanks, I am going to try it now and then I'll tell you, if it works.

[quote="m00li"][hide]arrange the elements of the multiset in an ascending order.

begin
1. check the first(leftmost) element of the multiset.
2. if the first element is <= the leftmost empty cell position
    then fill it
    else if it is >=  the rightmost empty cell
          then fill the rightmost empty cell
          else if there is an empty cell position = current element
                then fill that cell
                else find the closest empty cell with cell position = current element
                      if the choice of the closest cell is unique
                         then fill that cell
                         else fill the left choice
3. remove this element from the multiset.
4. if the multiset is empty
           then end
           else go back to 1[/hide]

well the indentation has gone for a toss and i dont know how to fix it[/quote]

I found out that this solution is wrong.
Example:

Your algorithm:

K=3
Multiset = {1,1,2,5,6,6,7,7}
Filled_sequence Sum
1 1
2 1 1
3 2 1
4 7 3
5 5 0
6 6 0
7 6 1
8 7 1

Sum is 7

However, better solution exists -
1 1
2 1 1
3 2 1
4 5 1
5 6 1
6 6 0
7 7 0
8 7 1

Sum is 5

So, again please, what is a right greedy solution?

[quote="proggged"]what is a right greedy solution?[/quote]  The "right solution" (the one that minimizes the total sum) may not be a greedy algorithm.


[b]1. [/b] Prove the following: (a) If $ x \neq 0$ and $ xy \equal{} xz$ then $ y \equal{} z$. (b) If $ x \neq 0$ and $ xy \equal{} x$ then $ y \equal{} 1$. (c) If $ x \neq 0$ and $ xy \equal{} 1$ then $ y \equal{} 1/x$. (d) If $ x \neq 0$ then $ 1/(1/x) \equal{} x$.

a) $ y \equal{} y\times1 \equal{} y(x\times\frac {1}{x}) \equal{} (yx)\frac {1}{x} \equal{} (xy)\frac {1}{x}$
$ \equal{} (xz)\frac {1}{x} \equal{} (zx)\frac {1}{x} \equal{} z(x\times\frac {1}{x}) \equal{} z\times1 \equal{} z$.

b) replace $ (xy)$ with $ x$ in the the above equation and you should get $ 1$

c) replace $ (xy)$ with $ 1$ in the the above equation and you should get $ \frac{1}{x}$


complex analysis
function
integration
calculus
derivative
complex analysis unsolved

Prove that if $ f(z)$ is holomorphic function in a neighborhood of a closed disk $ \overline{D}$, then  the function $ \varphi(z) \equal{} \displaystyle\int_{\partial D} \frac {f(\zeta)d\zeta}{\zeta \minus{} z}$ is also holomorphic on $ \mathbb{C}\setminus \overline{D}$.

Its derivative is $ \int_{\partial D}\frac{f(\zeta)}{(\zeta\minus{}z)^2}\,d\zeta$. That integral makes sense, because the integrand is continuous on the circle. It's uniformly bounded on compact sets, so we can interchange the integral and derivative.

Note that we don't actually need much of anything about $ f$; just that it's continuous on $ \partial D$. Even $ L^1$ would be good enough. If $ f$ is actually holomorphic in the interior of $ D$, then $ \varphi\equal{}2\pi i f$ in the interior and zero in the exterior.


[url=http://www.malevole.com/mv/misc/killerquiz/]Can you tell the difference?[/url]

notorious mass murderer or innocent singing sensation?

Wow, that was wierd! I got a 7 out of 10.

I got 7/10. Its the normal looking ones, oddly enough, that are the killers.

Then again, with those Mac people, you never can tell......

Just learn to pick out nerds from...well...non-nerds. :D

i got 6/10.  i'm pretty bad.  i'm going to do it again.

that' quiz is awesome, i also got a 7/10

Yay! I got 9/10.  I guess I'm just good at telling who's a nerd.

7/10  :D

I hope I don't grow up to be a serial killer o_O

i got 8/10, going strictly by glasses=programmer; no glasses=serial killer

[quote="mathclass"]8/10. Some of those people SCARE ME.[/quote]
You aren't the only one...

On a completely unrelated note, where can I find underground bunkers, preferably with submachine guns?

I got 8/10  wtf...how are you sopposed to know that the guy with the huge beard was actually a cartoon artist

I got 7/10, i just said those smart looking people were murders and et c.

[quote="G-UNIT"]Funny this sould be brought up.  Did you know that the majority of serial killers are mentally gifted and in some cases prodigious?  The uni bomber happened to be a child prodigy (uni bomber in reference to the individual who murdered by mailing bombs) and get this, he loved math!  If im not mistaken, the only person to graduate from harvard and end up as he has.   During the BTK killers interview, physiatrists found he deeply-thought out all murders, and by the segments aired on TV, he showed proficient levels of english.  So if tommorow or the day after that, you feel the uncanny sensation to pick up a knife in your hand and sincerely wet it......it can only mean one thing......you have the unusual desire for cooking![/quote]
I want my sanity back.

Just a 8/10. Of course they didn't have any windoze coders on there because they might qualify for both. Each year they send millions of innocent cilivians into their deaths with bad programs.

Is everyone here a linux fanboy?

Yes, some of mictosoft sucks, but nowadays it isn't that bad.

Execpt for the Wind32 API. That makes little kids cry.

I'm actually not that into linux. Just when replying to this my comp crashed.  :D

8/10... I went on just reverse-reverse-reverse psychology to predict what they would be... now i feel the desire to k-- cook.

And the reason that they are all linux guys is that it says "programming language INVENTOR" or something similar to that.

5/10, how do you know if they are a killer or not?
Some of the things they did scare me...

I got 10/10, just basing on the following:
- The 2 pics that seem to date from before the programming era, are serial killers
- for the others, the size of the brains probably tells you enough :)

W@@T first try 10/10.

I guess i really know my killers...


FTW
MATHCOUNTS
Gauss
geometry
AoPS Books

Is there a way for me to prepare for the AMC 10/12 rather than do old practice tests, or take the AOPS class.

I'm going to 8th grade at Kennedy jr. high.
Any suggestions?

If you'd like to take my place as the manager of this FTW tourney, feel free to. I'm done with Mathcounts so I probably won't go on FTW much, so I'm not doing it. You can simply take my spot.

Also, I guess you should make a new signup list, because some of those people on the old list are probably inactive.

lol @ gauss's wrong post.  Oh and gauss lol I remember that time in your signature.

Well that's basically what you do to prepare for AMC 10/12.  If you're bad in any specific areas you should probably go through them in the intro book too.  Maybe you could get volume 1 if you wanted practice problems.  Other than that there isn't really anything else to do.

I don't really know about anything you could do at kennedy since I never went there but maybe someone else from IL could say something about that...?

Uh originally when you posted this it used to be titled "IL Pwnage Meet" so I posted before you changed the topic title.

Older AMC tests and AoPS 1 are probably your best bet. Maybe the other AoPS books like the Intros would help, but I'm not familiar with those, and I don't know if you have the time to run through the topics (Algebra > Geo > Combo > NT for AMC's, generally).


geometry
trigonometry
projective geometry
trapezoid
perpendicular bisector
cyclic quadrilateral
angle bisector

Let M be the midpoint of the side BC of a triangle ABC, and let D and E be the orthogonal projections of this point M on the sides AB and CA. The points D and E clearly lie on the circle with diameter AM. Now, let the tangents to this circle at the points D and E meet each other at a point P. Prove that this point P lies on the perpendicular bisector of the segment BC.

  Darij

Let $F$ be the intersection of $PM$ with the circle $(ADE)$. $FM$ is the $F$-symmedian in $FED$ (which, by the way, is similar to $ABC$). This means that $\widehat{(FM,BC)}=\widehat{(AK,DE)}=\frac{\pi}2$, where $K$ is the Lemoine point of $ABC$, and the last equality holds because $M$ lies on the isogonal cevian of $AK$.

[quote="grobber"]$FM$ is the $F$-symmedian in $FED$ (which, by the way, is similar to $ABC$). This means that $\widehat{(FM,BC)}=\widehat{(AK,DE)}=\frac{\pi}2$,[/quote]

Sorry, but I really don't understand why $\widehat{(FM,BC)}=\widehat{(AK,DE)}$. Could you explain please?

I have two solutions of the problem: a simple trigonometric and a nice synthetic one.

  Darij

That's because we have some similar figures over there, and $\widehat{(FM,DE)}=\widehat{(AK,BC)}$, and we can form a cyclic quadrilateral with these four lines (it seems Ok right now, but it could be wrong :)).

Grobber, I apologize that the below sounds rather harsh, but I really don't understand your argument. I fear that what you have posted is not your solution, and your solution is not what you have posted.

You begin with writing "That's because we have some similar figures over there". Actually, could you please be more exact about which figures are similar? I know, you probably refer to your statement before that the triangles FED and ABC are similar (actually indirectly similar). But when you made that statement, you didn't prove it (you just wrote "in $FED$ (which, by the way, is similar to $ABC$)"). Actually, I don't think that this statement would be so immediate to prove. (In any case, don't you see that if you can prove this statement, then the problem becomes almost trivial?)

BTW, I think that the orientation of your angles is incorrect: Since the triangles FED and ABC are indirectly similar, we have < (FM; DE) = - < (AK; BC), and this, after some easy transformations, yields < (FM; BC) = - < (AK; DE). From here on, everything works as you wrote. But it remains to show that triangles FED and ABC are indirectly similar...

  Darij

Sorry for this.. That is my solution, and the only reason why I'm being so vague is that I have erased my drawing (a dynamic sketch), and I didn't want to draw another one all over again :). I'll see what I can do, Ok? :)

It is obvious that they are similar, it's just a bit boring to write it out in detail:

Since $S_{ABM}=S_{ACM}$ we have $\frac {MD}{ME}=\frac {AC}{AB}\ (*)$. On the other hand, from the similarity of $PDM,PFD$ and $PEM,PFE$ we get $\frac{FD}{MD}=\frac{PD}{PM},\ \frac{FE}{ME}=\frac{PE}{PM}$, and since $PD=PE$, we have $\frac{FD}{FE}=\frac{MD}{ME}=\frac{AC}{AB}\ (\#)$ (the last one follows from $(*)$). Now $\angle BAC=\angle EFD$, and this, together with $(\#)$ gives the inverse similarity.

Do we really need the triangle similarity?

  Simple angle chase will do the job.   Angles B and C are outside  the circle
  and are measured by semi-difference of their arcs.

   Pestich.

To Pestich:

[quote="pestich"]Do we really need the triangle similarity?

Simple angle chase will do the job.[/quote]

I don't think so. How do you want to involve in the angle chasing the fact that M is the midpoint of the segment BC ? On the other hand, if you don't use this fact, you can't prove anything...

Sorry if I misunderstood you.

To Grobber:

[quote="grobber"]It is obvious that they are similar, it's just a bit boring to write it out in detail:[/quote]

Thanks a lot. Well, it is not obvious, but anyway it is rather easy (even I would have found it if you explicitely had said that you prove this using FD / FE = AC / AB; I was trying to prove it in other ways all the time).

Now, now as you know that the triangles FED and ABC are indirectly similar, it is easy to complete your solution: We have < FED = - < ABC. Now, < FMD = < FED (since the points D, E, F and M lie on one circle); thus, < FMD = - < ABC. Now,

  < (MP; BC) = < (MP; MD) + < (MD; AB) + < (AB; BC)
             = < FMD + < (MD; AB) + < ABC
             = - < ABC + < (MD; AB) + < ABC
             = < (MD; AB) = 90

(since $MD \perp AB$). Thus, $MP \perp BC$. And this means that the point P lies on the perpendicular bisector of the segment BC.

Thank you for this proof. It looks as a kind of middle between the two proofs I have.

Here is the [i]first proof[/i] - alas, it is damn situation dependent:

Let U be some point on the perpendicular bisector of the segment BC. We want to prove that the point P lies on the perpendicular bisector of the segment BC; to this end, it will suffice to show that the point P lies on the line MU. In other words, it is enough to prove that the lines DP, EP and MU concur at one point.

Since the tangents from a point to a circle are equal in length, we have PD = PE, so that the triangle DPE is isoscles. Thus, < PED = < EDP. Also, we have < DMU = < ABC and < UME = < BCA since the edges of these angles are pairwisely perpendicular. We also have < MEP = < MAE and < PDM = < MAD as secant-tangent angles. Thus, with help of the Sine Law,

$\frac{\sin \measuredangle DMU}{\sin \measuredangle UME}\cdot \frac{\sin \measuredangle MEP}{\sin \measuredangle PED}\cdot \frac{\sin \measuredangle EDP}{\sin \measuredangle PDM}$
$=\frac{\sin \measuredangle ABC}{\sin \measuredangle BCA}\cdot \frac{\sin \measuredangle MAE}{\sin \measuredangle EDP}\cdot \frac{\sin \measuredangle EDP}{\sin \measuredangle MAD}$
$=\frac{\sin \measuredangle ABC}{\sin \measuredangle BCA}\cdot \frac{\sin \measuredangle MAE}{\sin \measuredangle MAD}=\frac{\sin \measuredangle ABM}{\sin \measuredangle MCA}\cdot \frac{\sin \measuredangle MAC}{\sin \measuredangle MAB}$
$=\frac{\sin \measuredangle ABM}{\sin \measuredangle MAB}\cdot \frac{\sin \measuredangle MAC}{\sin \measuredangle MCA}=\frac{AM}{BM}\cdot \frac{CM}{AM}=\frac{CM}{BM}=1$.

Hence, by the trigonometric version of the Ceva theorem, it follows that the lines DP, EP and MU concur. Thus, the problem is solved.

The [i]second proof[/i] is situation independent, but it uses some theoretical preknowledge:

We will work with directed angles modulo 180.

Since the point P is a vertex of the tangential triangle of triangle MDE, the line MP is a symmedian of triangle MDE. If we call L the midpoint of the segment DE, then the line MD is a median of triangle MDE. Since a median and the corresponding symmedian of a triangle are symmetric to each other with respect to the corresponding angle bisector, we have < PME = < DML.

Let K be the symmedian point of triangle ABC, and let X, Y, Z be the orthogonal projections of this symmedian point K on the sides BC, CA, AB. It is well-known that the point K is then the centroid of triangle XYZ. Thus, the line XK passes through the midpoint X' of the side YZ of this triangle.

Now, since < AYK = 90 and < AZK = 90, the points Y and Z lie on the circle with diameter AK. Hence, < KYZ = < KAZ. In other words, < KYZ = < KAB. Similarly, < MDE = < MAC. But the line AM is a median in triangle ABC, while the line AK is a symmedian; the median and the symmedian are symmetric to each other with respect to the respective angle bisector, and hence we have < KAB = - < MAC. This, together with < KYZ = < KAB and < MDE = < MAC, gives < KYZ = - < MDE. Similarly, < KZY = - < MED. Thus, the triangles KYZ and MDE are inversely similar. Now, the point X' is the midpoint of the side YZ of the first triangle, while the point L is the midpoint of the side DE of the second triangle. Thus, the points X' and L are corresponding points in the triangles KYZ and MDE. Since corresponding points in indirectly similar triangles form oppositely equal angles, we thus have < DML = - < YKX'. In other words, < DML = - < (KY; KX) = - < (CA; BC), what is because the lines KY and KX are perpendicular to the lines CA and BC, respectively.

Since < PME = < DML, we thus have < PME = - < (CA; BC). Hence,

  < (MP; BC) = < (MP; ME) + < (ME; CA) + < (CA; BC)
             = < PME + 90 + < (CA; BC)
             = - < (CA; BC) + 90 + < (CA; BC) = 90.

Thus, $MP \perp BC$. In other words, the point P lies on the perpendicular bisector of the segment BC. And we are done again.

So now we have three different solutions... something I really couldn't expect from this little problem!

  Darij

So what about the angle chasing, dear pestich? I tried to find somenthing like this but unfortunately.....

I'm still chasing those unruly angles.

  Pestich.

[quote="darij grinberg"]

So now we have three different solutions... something I really couldn't expect from this little problem!

  Darij[/quote]

here it is a fourth way. Only a sketch and a joke  ;) .

let's "complete" the traingle AKL of which M is the orthocenter, where K=AC^DM and L=AB^ME. The  circle through E, D and P is the nine-point-circle of AKL. P is of course the mid-point of KL and at the same time the center of c(EDKL).
Now we can use the well-known(?) fact of the butterfly theorem applying it to the chord through C,M,B and conclude that PM is the perpendicular bisector of CB.

Wow. That was cool, Rocco. I probably should not only collect applications of the Pascal, Brianchon, Desargues and Pappos theorems, but now also of the Butterfly theorem  :D

  Darij

[quote="darij grinberg"]

I probably should not only collect applications of the Pascal, Brianchon, Desargues and Pappos theorems, but now also of the Butterfly theorem  :D

  Darij[/quote]

very interesting theorem indeed.

Some time ago I posted to Alexander Bogomolny a very short projective proof of this theorem, I was very proud of that results, but Alexander "kindly" answered me that the proof I posted was already known.

Now I would like to provide a specific (hopefully non already known, at least by Alexander) proof for our particular case. Or to be more precise I will prove a little bit more general case of which our question is a corollary.

Let AB be the diameter of a circle c with center O and let C a point on the plane of c. Let E=c^BC and D=c^AC, if M and N are the intersections of the orthogonal line through C to the line OC with AE and BD respectively, then CM = CN.

Proof

If H is the orthocenter of CBN, then AOC and ABH are similar and therefore AC=CH. Since AMC and HNC, as AM//HN, are similar then CM = CN.

[quote="sprmnt21"]Let AB be the diameter of a circle c with center O and let C a point on the plane of c. Let E=c^BC and D=c^AC, if M and N are the intersections of the orthogonal line through C to the line OC with AE and BD respectively, then CM = CN.

Proof

If H is the orthocenter of CBN, then AOC and ABH are similar and therefore AC=CH. Since AMC and HNC, as AM//HN, are similar then CM = CN.[/quote]

Nice proof! Here is my proof (which I found long ago):

Since the points D and E lie on the circle with diameter AB, we have < ADB = 90 and < AEB = 90.

Let U and V be the orthogonal projections of the points A and B on the line MN. Then, the lines AU, BV and OC are parallel to each other, since they are all perpendicular to the line UV. Hence, the quadrilateral AUVB is a trapezoid, and since the line OC passes through the midpoint O of its leg AB and is parallel to its bases AU and BV, this line OC is a midparallel of the trapezoid. In other words, the point C is the midpoint of the leg UV, and it follows that CU = CV.

Now, since the triangles AUC and NDC are similar (as < ACU = < NCD and < AUC = < NDC = 90), we have CA : CU = CN : CD. Thus, $CA \cdot CD = CU \cdot CN$. Similarly, $CB \cdot CE = CV \cdot CM$. But $CA \cdot CD = CB \cdot CE$ by the intersecting chords theorem. Thus, $CU \cdot CN = CV \cdot CM$. Since CU = CV, we can conclude that CM = CN, and we are done.

  Darij

Could anybody explain me what about direct and inverse simillarity?  :(  I haven't found any paper about this  :(

thanks

José Carlos

[hide="Here is a short metrical proof."]Denote: $x=m(\widehat {BAM})$, $y=m(\widehat {CAM})$; the circle $w$ with the diameter $[AM]$ and the intersections $P_1$, $P_2$ of the perpendicular bisector of the side $[BC]$ with the tangents to the circle $w$ at the points $D$, $E$ respectively. We observe that $MD=m_a\sin x$ and $ME=m_a\sin y$, i.e. $\frac{MD}{ME}=\frac{\sin x}{\sin y}\ (1)$. From the Sinus' theorem applied in the triangles $MDP_1$, $MEP_2$ results:
$\frac{MP_1}{\sin x}=\frac{MD}{\sin (B-x)}$ and $\frac{MP_2}{\sin y}=\frac{ME}{\sin (C-y)}$. Therefore, the point $P$ belongs to the perpendicular bisector of the side $[BC]$ $\Longleftrightarrow$ $MP_1=MP_2$ $\Longleftrightarrow$ $\frac{MD\sin x}{\sin (B-x)}=\frac{ME\sin y}{\sin (C-y)}$ $\Longleftrightarrow$ $\frac{\sin (B-x)}{\sin (C-y)}=\frac{\sin ^2x}{\sin ^2y}$, what is truly, from the relation $(1)$ and the identity posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=76955 .[/hide]


The ratio of the number of games won to the number of games lost(no
ties) by the Middle School Middies is $ 11/4$. To the nearest whole
percent, what percent of its games did the team lose?

Games won: 11x
Games lost: 4x
Games total: 15x
So, the percentage games lost is 4x/15x*100=400/15=26.66...% => [b]27%[/b]

You could also just note that the total games played is $ 4 \plus{} 11 \equal{} 15$, so the answer is $ 4 \div 15 \equal{} 26.666666... \equal{} \boxed{27\%}$.

Not necessarily izzy. The total number of games played is a [b]multiple[/b] of 15. YOur strategy worked here, but won't always.