A $ 1 \times 1 \times 1$ wire frame cube is to be made by gluing together pieces of wire. The wire can be bent to form corners of the cube. If exactly 12 units of wire is used to make the frame, what is the fewest number of pieces of wire that can be used to make the frame?

Sounds a bit like a graph theory problem.  I keep getting 3... can anyone get 2?  I'll be surprised if it's 1.

i'm pretty sure the answer is four
i'm not sure how to explain it though

If you "flatten out," so to speak, the cube, it looks like this...

[asy]size(100);
pair[] points;
points[1]=dir(60);
points[2]=dir(120);
points[3]=dir(180);
points[4]=dir(240);
points[5]=dir(300);
points[6]=dir(0);
draw(points[1]--points[2]);
draw(points[2]--points[3]);
draw(points[3]--points[4]);
draw(points[4]--points[5]);
draw(points[5]--points[6]);
draw(points[6]--points[1]);
draw(points[1]--points[3]);
draw(points[2]--points[4]);
draw(points[3]--points[5]);
draw(points[4]--points[6]);
draw(points[5]--points[1]);
draw(points[6]--points[2]);[/asy]

It's much easier to think in 2D than 3D.

Turns out the answer is 2.

i know for a fact that it is not 1
i'm pretty sure it's either 3 or 4
not sure which

Ignore my post, please, the diagram is completely wrong...

[img]http://img381.imageshack.us/img381/553/cubeay5.png[/img]

Hopefully this picture is interpretable  :). The answer is $ \boxed4$.

Hrm... after tinkering again, I don't know how I would have gotten three... I do have a different looking solution of 4 though:
[img]http://i59.photobucket.com/albums/g311/Seleucus/def.jpg[/img]

It occurs to me that the only way to make a corner is to join three or join a bend to an end.  But if you have three joining, you can replace a pair of ends with a bend, and reduce the number of breaks in the wire.

So, if you have to cut each corner of the cube into a bend and an end, you spawn at least 8 free ends.  These 8 free ends form the ends of at least 4 pieces of wire.


geometry
3D geometry
tetrahedron
analytic geometry
trigonometry

We consider a cube with sides of length 1. Prove that a tetrahedron with vertices in the set of the vertices of the cube has the volume $\dfrac 16$ if and only if 3 of the vertices of the tetrahedron are vertices on the same face of the cube.

[i]Dinu Serbanescu[/i]

Label one face of the cube $ABCD$ and the opposite face $A'B'C'D'$, such that $AA', BB', CC', DD'$ are the cube edges.

[b](1)[/b] Assume that 4 vertices of the tetrahedron are on the same cube face. Then they are coplannar and the tetrahedron volume is zero.

[b](2)[/b] Assume that 3 vertices of the tetrahedron are on the same cube face and the 4th one is not. WLOG, let the first 3 vertices be $A, B, C$. Regardles of the selection of the remaining vertex, the tetrahedron altitude to the plane $ABC$ is $h = 1$ and the tetradedron volume is

$V = \frac h 3\ |\triangle ABC| = \frac 1 3 \cdot \frac 1 2 = \frac 1 6$

[b](3)[/b] Assume that no 3 vertices of the tetrahedron are on the same cube face. Then each cube face must contain exactly 2 vertices. WLOG, consider the face $ABCD$.

[b](3a)[/b] Assume that 2 vertices in the face $ABCD$ are adjacent (they make an edge). WLOG, let the first 2 vertices be $A, B$. The only remaining 2 vertices, none of which makes 3 vertices in one cube face with $A, B$ are the cube vertices $C', D'$. Since the vertices $A, B, C', D'$ are coplannar, volume of the tetrahedron $ABC'D'$ is zero.

[b](3b)[/b] Assume that 2 vertices in the face $ABCD$ are not adjacent (they make a face diagonal). WLOG, let the first 2 vertices be $A, C$. The only remaining 2 vertices, none of which makes 3 vertices in one cube face with $A, C$ are either the cube vertices $A', C'$ or $B', D'$. Since the vertices $A, C, A', C'$ are coplannar, volume of the tetrahedron $ACC'A'$ is zero. This leaves only the tetrahedron $ACB'D'$. Putting the coordinate origin at the vertex $A$ and the positive x-, y-, z-axes along the edges $AB, AD, AA'$, the points $A, C, B', D'$ have the coordinates:

$A = (0, 0, 0),\ \ C = (1, 1, 0),\ \ B' = (1, 0, 1),\ \ D' = (0, 1, 1)$

The tetrahedron volume is then

$V = \frac h 3\ |\triangle AB'C| = \frac h 6\ |AB'| \cdot |AC| \sin \alpha = \frac h 6\ \vec{AB'} \times \vec{AC} = \frac 1 6\ \vec{AD'} \cdot (\vec{AB'} \times \vec{AC})$

$\vec{AB'} = (1, 0, 1),\ \ \vec{AC} = (1, 1, 0),\ \ \vec{AB'} \times \vec{AC} = (-1, 1, 1)$

$\vec{AD'} = (0, 1, 1),\ \ \vec{AD'} \cdot (\vec{AB'} \times \vec{AC}) = 2$

$V = \frac 1 3 \neq \frac 1 6$


The population of birds on Island $A$ is twice that of Island $B$. Each year, $\frac{1}{3}$ of island $A$'s population migrates from $A$ to $B$, and $\frac{1}{2}$ migrates from island $B$ to island $A$. This process is allowed to continue on indefinitely, and no birds are born or die. Eventually, what does the ratio of $\frac{\text{ Population of A}}{\text{ Population of B}}$ approach?

[hide="Follow-up question"]
Prove that the initial population ratio has no effect on the final ratio.[/hide]

[quote="i_like_pie"]Does the migration occur simultaneously, or is it A -> B and [i]then[/i] B -> A?[/quote]
Simultaneously.

Aren't you looking for a point where from 1/3 from Island A is the same as the 1/2 from Island B?

If that's the case, you simply say 1/3(3-x) = 1/2(x)  where the quantity on the left represents the output of birds from Island A and the right side represents the output of Island B. The 3 comes from the fact that Island A has twice as many birds as Island B, so 1 + 2 = 3 units of birds.  Solving gives you 9/5 = x = the units if birds on Island B.

The proportion is now (3-9/5)/(9/5) = 2/3

Also, since the proportion is determined by only by the initial number of total birds and the rate at which they leave, it doesn't matter the initia; conditions.[/img]


source: KCATM 2005 7-8

A man found that he could buy a cd player listed for 160 dollars with a cash discount of 4% at a local discount store. He found the same cd player at a department store that was originally listed at 200 dollars but had been marked down 15%. Additionally, the department store was running a 10% off sale for the holiday weekend. Which store had the better price and how much could he save over the other store?

u dont multiply by .85 and .9, just .9 kheg


geometry
circumcircle
ratio
angle bisector
geometry solved

Let $ABC$ be a triangle, and let $X$, $Y$, $Z$ be the feet of its angle bisectors (i. e. the points where the angle bisectors of the angles $CAB$, $ABC$, $BCA$ intersect the lines $BC$, $CA$, $AB$). The circumcircle of triangle $XYZ$ then cuts off a segment from each of the lines $BC$, $CA$, and $AB$. Prove that two of these segments' lengths add up to the third segment's length.

[i]Alternative formulation.[/i] Given a triangle $\triangle ABC$. Let the angle bisectors of the angles $CAB$, $ABC$, $BCA$ intersect the lines $BC$, $CA$, $AB$ at the points $X$, $Y$, $Z$, respectively. The circle through the points $X$, $Y$, $Z$ intersects the lines $BC$, $CA$, $AB$ at the points $X^{\prime}$, $Y^{\prime}$, $Z^{\prime}$ (apart from $X$, $Y$, $Z$), respectively. Prove that:
$XX^{\prime}=YY^{\prime}+ZZ^{\prime}$ or $YY^{\prime}=ZZ^{\prime}+XX^{\prime}$ or $ZZ^{\prime}=XX^{\prime}+YY^{\prime}$.

How about writing the powers of points A,B,C with respect to the circle? If you name the segments x, y, z you will get 3 equations with 3 unknown values(x, y, z). Assume a specific order for a, b, c- the sides of triangle ABC, and you MUST get what we want:x=y+z or another one like that. It's ugly, but I can't see any other ways. I can't post something like that right now, because it's WAY past my bed time :D .

Try it, however and let me know if it worked. Actually, it has to work, as long as you are patient enough to finish the calculations(which are horrible, by the way :D).

Yes , and to overcome the main incovenient in this problem , which is the many many cases , you use directed lengths . Computations turn out to be quite easy .

This problem was from Hungarian Olympiad 2000

well well sorry for my incompetence ,i am still not getting the solution.(i feel uncomfortable with directed lengths :D ).so cud ne1 be kind enuf so as to post the solution
best regards

I rewrite the problem with naming the points:

[i]Consider a triangle ABC; we direct the lines BC, CA, AB so that the segments BC, CA, AB will be positive and the segments CB, AC, BA will be negative. Now let the angle bisectors of the angles CAB, ABC, BCA meet the sides BC, CA, AB in the points X, Y, Z. The circle through these points X, Y, Z meets the lines BC, CA, AB again at the points X', Y', Z'. Then, prove that

  XX' + YY' + ZZ' = 0,

where the segments are directed.[/i]

In fact, I will prove a generalization of this, which is due to Victor Thebault:

[b]Theorem.[/b] Consider a triangle ABC; we direct the lines BC, CA, AB so that the segments BC, CA, AB will be positive and the segments CB, AC, BA will be negative. Also let x, y, z be three arbitrary reals. Now let X, Y, Z be three points on the lines BC, CA, AB such that BX : XC = y : z, CY : YA = z : x, AZ : ZB = x : y (with directed segments). The circle through these points X, Y, Z meets the lines BC, CA, AB again at the points X', Y', Z'. Then, prove that

  $ax\cdot XX^{\prime }+by\cdot YY^{\prime }+cz\cdot ZZ^{\prime }=0$,

where a = BC, b = CA, c = AB are the sidelengths of the triangle.

[i]Proof.[/i] Since BX : XC = y : z and BC = a, we can easily compute $BX=\frac{ay}{y+z}$, and similarly from AZ : ZB = x : y it follows that $ZB=\frac{cy}{x+y}$. But the points Z, Z', X and X' are concyclic; hence, the Intersecting Chords Theorem yields

  $BX\cdot BX^{\prime }=BZ\cdot BZ^{\prime }$,

so

  $BX\cdot \left( BX+XX^{\prime }\right) =BX\cdot BX^{\prime }=BZ\cdot BZ^{\prime }=ZB\cdot Z^{\prime }B=ZB\cdot \left( ZB-ZZ^{\prime }\right)$,

or

  $\frac{ay}{y+z}\cdot \left( \frac{ay}{y+z}+XX^{\prime }\right) =\frac{cy}{x+y}\cdot \left( \frac{cy}{x+y}-ZZ^{\prime }\right)$,

what simplifies to

  $\frac{a}{y+z}\cdot \left( \frac{ay}{y+z}+XX^{\prime }\right) =\frac{c}{x+y}\cdot \left( \frac{cy}{x+y}-ZZ^{\prime }\right)$,

or

  $\frac{a}{y+z}\cdot XX^{\prime }+\frac{c}{x+y}\cdot ZZ^{\prime }=\frac{c^{2}y}{\left( x+y\right)^{2}}-\frac{a^{2}y}{\left( y+z\right)^{2}}=\left( \frac{c^{2}}{\left( x+y\right)^{2}}-\frac{a^{2}}{\left( y+z\right)^{2}}\right) y$.

Dividing by y, we get

  $\frac{a}{y\left( y+z\right) }\cdot XX^{\prime }+\frac{c}{y\left( x+y\right) }\cdot ZZ^{\prime }=\frac{c^{2}}{\left( x+y\right)^{2}}-\frac{a^{2}}{\left( y+z\right)^{2}}$.

The right-hand side of this makes clear that if we sum up this equation and the two ones obtained from it by cyclic permutation, then we get 0. Hence we have

  $\left( \frac{a}{y\left( y+z\right) }\cdot XX^{\prime }+\frac{c}{y\left( x+y\right) }\cdot ZZ^{\prime }\right)+\left( \frac{b}{z\left( z+x\right) }\cdot YY^{\prime }+\frac{a}{z\left( y+z\right) }\cdot XX^{\prime }\right)+\left( \frac{c}{x\left( x+y\right) }\cdot ZZ^{\prime }+\frac{b}{x\left( z+x\right) }\cdot YY^{\prime }\right) =0$,

what can be rearranged into

  $a\left( \frac{1}{y\left( y+z\right) }+\frac{1}{z\left( y+z\right) }\right) \cdot XX^{\prime }+b\left( \frac{1}{z\left( z+x\right) }+\frac{1}{x\left( z+x\right) }\right) \cdot YY^{\prime }+c\left( \frac{1}{x\left( x+y\right) }+\frac{1}{y\left( x+y\right) }\right) \cdot ZZ^{\prime }=0$.

With $\frac{1}{y\left( y+z\right) }+\frac{1}{z\left( y+z\right) }=\frac{y+z}{yz\left( y+z\right) }=\frac{1}{yz}$ and similar, this simplifies to

  $a\cdot \frac{1}{yz}\cdot XX^{\prime }+b\cdot \frac{1}{zx}\cdot YY^{\prime }+c\cdot \frac{1}{xy}\cdot ZZ^{\prime }=0$.

Finally, by multiplication with xyz, we get $ax\cdot XX^{\prime }+by\cdot YY^{\prime }+cz\cdot ZZ^{\prime }=0$, and the Theorem is proven.

Now, we can solve your problem by setting x = 1/a, y = 1/b, z = 1/c (in fact, since an angle bisector of a triangle divides the opposite side in the ratio of the two adjacent sides, we have BX : XC = c : b = (1/b) : (1/c) = y : z and similarly CY : YA = z : x and AZ : ZB = x : y).

  Darij

Power of the triangle vertices A, B, C to the circle (XYZ) is

$Y'A \cdot YA = AZ' \cdot AZ,\ \ Z'B \cdot ZB = BX' \cdot BX,\ \ X'C \cdot XC = CY' \cdot CY$

Denote a = BC, b = CA, a = BC the triangle sides and x = XX', y = YY', z = ZZ' line segments directed the same way as the triangle sides. Then we have to show x + y + z = 0. Since the positions of the angle bisector feet X, Y, Z are known, the above 3 equations can be written as

$\frac{bc}{a+b}\left(\frac{bc}{a+b}-z\right) = \frac{bc}{c+a}\left(\frac{bc}{c+a}+y\right)$

$\frac{ab}{c+a}\left(\frac{ab}{c+a}-y\right) = \frac{ab}{b+c}\left(\frac{ab}{b+c}+x\right)$

$\frac{ca}{b+c}\left(\frac{ca}{b+c}-x\right) = \frac{ca}{a+b}\left(\frac{ca}{a+b}+z\right)$

Calculating x, y, z by normal elimination (for example, z from the 1st and 3rd equations in terms of x, y, respectively, yields another equation for x, y in addition to the 2nd one, etc.), we get

$2x = \frac{b(b+c)}{c+a}-\frac{c(b+c)}{a+b}-\frac{a(b-c)}{b+c}$

and by cyclic exchange,

$2y = \frac{c(c+a)}{a+b}-\frac{a(c+a)}{b+c}-\frac{b(c-a)}{c+a}$

$2z = \frac{a(a+b)}{b+c}-\frac{b(a+b)}{c+a}-\frac{c(a-b)}{a+b}$

It is clear that 2x + 2y + 2z = 0, for example, adding terms with the common denominator c + a,

$\frac{b(b+c)}{c+a}-\frac{b(c-a)}{c+a}-\frac{b(a+b)}{c+a}= \frac{b(b+c-c+a-a-b)}{c+a}= 0$


Let P be the midpoint of a chord AB of a given circle O. Circle O1
is tangent to the line AB at P and tangent to O. A line l through A,
different from the line AB, touches O1 and intersects O again at C. Let
Q be the midpoint of the segment BC and O2 be the circle tangent to BC
at Q and tangent to the segment AC. Prove that the circle O2 is tangent
to O.


Bobby is working on a science experiment, he accidently spills a chemical on his hand. He went from 10 fingers to 8 fingers.
Now, he has to count with base 8.
At school, after he waited a month, his teacher gave him an antidote.
But, he got used to counting with base 8. For his math, he accidentally did the problem in base 8, he needs to convert it to base 10. Help him convert it, he needs to convert $1254_8$ to base 10.

[hide]Expand the problem to look like:
$1\cdot8^3+2\cdot8^2+5\cdot8^1+4\cdot8^0$
Then, simplify to find the answer:
$512+128+40+4=684_{10}$[/hide]


我试了很多台电脑，总是打不开[url]http://www.kalva.demon.co.uk[/url],可怪的是让google翻译成中文后竟被我奇迹般地打开了， :D 可是google乱翻译的，看也看不懂，还是一场空。 :(
问谁能帮我？ :)

How do you let google translate for you?

I think we cannot access some of the sites because we live in China. I have been trying darij grinberg's homepage for several months, but haven't suceeded yet...

不一定，在中国有些地方可以打开这个网站，你能打开吗？
BTW：darij 的个人网站你能告诉我吗？

Hello,

here is a little experiment: http://noname.schlussda.com/darij . If things go right, this should work as a mirror of my website and it should be accessible from China (I've checked using a Chinese proxy). The mirror contains all the data from my website except of the solutions to the BWM (Bundeswettbewerb Mathematik) problems, which are in German anyway. I am not sure whether I will keep the mirror up to date all the time, so maybe some of the materials I will add to my website in future won't be mirrored (although this is not too probable).

If you want to access a link to my website, simply replace [color=green]de.geocities.com/darij_grinberg[/color] by [color=green]noname.schlussda.com/darij[/color] (so, for instance, http://de.geocities.com/darij_grinberg/Lamoen/Support.html will become http://noname.schlussda.com/darij/Lamoen/Support.html ).

Hope it will work well.

  Darij

Thanks darij, it works now!
It is really a great site!  :)  :roll:

To everyone interested in accessing my website from China, try out the following links:
http://www.stud.uni-muenchen.de/~darij.grinberg/
http://www.cip.ifi.lmu.de/~grinberg/
Hopefully, at least one of them will work. I don't know whether I will keep both of them up to date in future - most likely, I will make one of them my actual website and keep the other one as well as http://de.geocities.com/darij_grinberg/ as mirrors (which won't always be up to date).

  Darij


For a,b,c > 0 and abc=1 prove that

(1+a)(1+b)(1+c)<=5+a/b+b/c+c/a

Does anybody know a solution or where I can find one ?

Thanks, gelo

Let $ a\equal{}\frac{x}{y}, \ b\equal{}\frac{y}{z}$ and $ c\equal{}\frac{z}{x}, \ x,y,z>0.$

The inequality is equivalent with :

\[ 3\plus{}\displaystyle\sum_{cyc}\frac{yz}{x^2}\geq \displaystyle\sum_{cyc}\frac{y\plus{}z}{x}\Longleftrightarrow \sum_{cyc}(x^3y^3\minus{}x^3y^2z\minus{}x^3z^2y\plus{}x^2y^2z^2)\geq0\] , which is Schur for $ xy\ ,\ yz\ ,\ zx$


This format:
Age: 16
Grade: 11
Height: 5'8''
Weight: 140 lbs.

Age: 16
Grade: 12
Height: 5'10
Weight: 135

Age: 16
Grade: 11
Height: 5'7"
Weight: 110

Age: 13

Grade: 8th

Height: 5' 3"

Weight: 96 pounds.

Age: 15
Grade: 11
Height: 6'3"
Weight: 170 lbs.

Age: 17
Grade: 12
Height: 6'1''
Weight: 74,842 grams

I think 123 and furious are just playing around. Just kidding

I am sophmore gonna be in college
        age-19
        height -6.0
        weight- 80 or 82 kg, the machine goes on 2 before 0 when I stand on it, so I dont know for sure.

[quote="joml88"][quote="hello"]I think 123 and furious are just playing around. Just kidding[/quote]

What? :?  :?:  I don't get what you were trying to say at all.[/quote]

He thinks we're too tall to like math............  :lol:

I mean quite tall and thin, lol. But I believe you. LOL.

Age: 14
Grade: 9
Height: 5'5"
Weight: 110 lbs.

Bah you're all too skinny.. i'm skinny too, but...

Age: 13.7
Height: 5"2
Grade 8
Weight: 102 lbs.

So, I'm tall and rail thin.  :D
Age: 13
Grade: 9
Height: 5'7-8"
Weight: 118 lbs

Interesting....

  My profile is practically identical to Drunken Math's, except I'm an inch or two taller and a few pounds lighter.  :D

[quote="pkothari13"]hahaha yall are like stalking him or something... i'd be cautious if i were you solafidefarms...  :ninja:[/quote]

:P

:ninja: <-- that's NOT a ninja

They can easily find my name; it's my friends that I have fun stalking... ;)

I don't feel like doing profiles for my pets, as I do not weigh them ;)

[quote="solafidefarms"]OK, if you insist... Though I must guess as to height and weight:

Age: 4; Grade: K-4; Height: ~2'10"; Weight:~37lbs
Age: 2; Grade: K-2; Height: ~2'4"; Weight:~22lbs

I have 2 brothers too, but I don't think they want ya'll to know about them. :lol:[/quote]

HAHA, for some reason I found that to be hilariously funny.

they sound cute, I want a lil sis.

Age:12
Grade: 7
Height:5 foot exactly
Weight: 89 lbs.

Age: 12
Grade: 8
Height: 5"3'
Weight: 100

Age:14yrs

Height:5"9"ish

weight:*drum roll*.....46kg :D

I'm really light hearted :rotfl:

Age: 13
Height: 5' 1''
Grade: 8th
Weight: 40,369.688g or 89 lbs

[quote="solafidefarms"][quote="flipper2011"]Age: 16
Grade: 11
Height: 5'7 or 7.5
Weight: 110/115[/quote]

Are you sure that you're really that light? I've never known someone over the age of 1 who weighed less then a pound... :P[/quote]

You know what he means  :roll:  :P

Age: almost 14
Grade: 8th
Height: 5'7"
Weight: 115

Age:14
Grade:9 (freshman)
Height:5' 11''
Weight:180 lbs

Age: 14 (turning 15 on Jan 14)
Grade: 10
Height: about 177 cm (or 5' 9" to 5' 10" if you like)
Weight: a little short of 60 kg (130 pounds)
More information needed?
Shoe size: 11.5 (US)

my b'day's just a day b4!

age: 12
height: 5ft 1 in
weight: 100 lbs


How many integers $x$ satisfy $|5x-3|\leq 7$?

[hide]$|5x-3|\leq 7$
$-7\leq 5x-3\leq 7$
$-4\leq 5x\leq 10$
$-\frac{4}{5}\leq x\leq 2$
Therefore, there are $3$ values, $0$, $1$, and $2$, which satisfies the inequality.[/hide]


induction
superior algebra
superior algebra unsolved

Let Z_n^* be the group { m \in N | m < n, (m,n)=1 } with multiplication modulo n.

Determine Aut(Z_n^*).

Thanks for help.

By chinese remainder theorem, it is isomorphic to the group product of it's prime(power) factors. And they are isomorphic to cyclic groups each.

Okay, I see the isomorphism to the product of the prime power groups.

But how do we prove that Z_n^* is cyclic if n is a prime power?

Is it correct anyways?
What about Z_8^* for example?
I can't see that this is cyclic...
Z_8^* = { 1, 3, 5, 7 }
3^2 = 1
5^2 = 1
7^2 = 1

That's an exception.

It works for powers of odd primes. For powers of 2, you get a product of an order 2 group, and a larger cyclic group.

For odd primes $p$, the key is that the $p$-th powers mod $p^{n+1}$ are a complete set of residues mod $p^n$; this can be proved by induction. This almost works for $p=2$, but not for $n=2$.

I'm afraid I still don't fully understand...
Can you further elaborate on this?

How does the proof by induction work?

Is there a direct way to find the generators of $Z^*_{p^n}$?

All right- looking back, I said something wrong.

A better way to do the whole thing: $(a+bp)^{p^{n-1}}\equiv a^p+a^{p-1}bp^n\mod p^{n+1}$, for $p>2$ and $n\ge2$. This shows that there are at least $p$ $p^{n-1}$-th powers, which forces the group to be cyclic.


geometry
geometric transformation
homothety
geometry solved

let F_A, F_B,F_C be the points of tangency of the excircle with the nine-point circle. Prove A F_A, B F_B and C F_C pass through the internal center of similitude of the incircle and the nine-point circle.

This is a basic theorem - the 3 pairwise external homothety centers of 3 circles are collinear. Any 2 pairwise internal homothety centers of 3 circles and the external homothety center of the remaining circle pair are collinear. The vertex A is the external homothety center of the incircle $(I)$ and the excircle $(E_a)$ (the intersection of their common external tangents), the tangency point $F_a$ is the internal homothety center of the excircle $(E_a)$ and the 9-point circle $(N)$, so that the line $AF_a$ is one of the 4 homothety axes of the circles $(I),\ (N),\ (E_a)$ passing through the internal homothety center of the incircle $(I)$ and the 9-point circle $(N)$.

[color=red][[b]Moderator edit:[/b] This is also a particular case of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5364]the four circles theorem[/url] ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=74250]also see here[/url]), where $c_1$, $c_2$, $c_3$ are the three excircles of triangle ABC, and X is the nine-point circle.][/color]


Is there any way to simplify $ \sqrt{(x\minus{}y)(y\minus{}z)}\minus{}\sqrt{(w\minus{}y)(x\minus{}z)}$ to one radical?

If you don't care how dirty ( :D ) inside the radical is, note that

$ \sqrt{x}\minus{}\sqrt{y}\equal{}\sqrt{x\plus{}y\minus{}2\sqrt{xy}}$ and simplify.

hello, you are wrong it is $ \sqrt{x}\minus{}\sqrt{y}\ne \sqrt{(\sqrt{x}\minus{}\sqrt{y})^2}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, you are wrong it is $ \sqrt {x} \minus{} \sqrt {y}\ne \sqrt {(\sqrt {x} \minus{} \sqrt {y})^2}$.
Sonnhard.[/quote]
If $ x>y$ then it is true....

hello, it is also true, if $ x \geq y$.
Sonnhard.


function
algebra
domain
calculus
derivative
real analysis
real analysis solved

Let $f(x)$ be the real- valued function  which is differentiable on the opened-interval $(0,1).$ If $f'(x)$ is bounded, then prove that for the Cauchy sequence $\{a_n\}_{n=1}^{\infty}$ in the opened-interval $(0,1),\ f(a_n)$ will become Cauchy sequence as well.How about in case that $f$ is simply differentiable?

Suppose $|f'| \leq M$.
Given $\epsilon > 0$.  Since $\{a_n\}$ is Cauchy, then there exists $N$ such that for all $n,m>N$, $|a_n-a_m| < \frac\epsilon M$. Then $|f(a_n)-f(a_m)| = |f'(\xi)|\cdot|a_n-a_m| < M \cdot \frac\epsilon M = \epsilon$. So $f\{(a_n)\}$ is Cauchy too.

If $f'$ is unbounded, take $f(x)=\frac 1x$ for example. Let $a_n=\frac 1n$, then obviously it is Cauchy, however, $f(a_n)=n$ is not.

The general, and useful, theorem is this: if $f$ is [b]uniformly continuous[/b] on a domain $S$ and if $(a_n)$ is a Cauchy sequence, then $(f(a_n))$ is a Cauchy sequence. This works even in the general context of maps between metric spaces. This theorem then implies that $f$ can be uniquely extended to a continuous function on the closure of $S.$

Note that liyi's example of $\frac1x$ on $(0,1)$ cannot be extended to a continuous function on the closure of its domain.

The hypothesis of a bounded derivative implies, but is not equivalent to, the function being Lipschitz. In turn, being Lipschitz implies, but is not equivalent to, the function being uniformly continuous.

A capsule summary: continuous functions take convergent sequences to convergent sequences; uniformly continuous functions take Cauchy sequences to Cauchy sequences. (By "convergent" we mean that the limit exists and is in the domain of the function.)


abstract algebra
graph theory
group theory
superior algebra
superior algebra unsolved

Some friends of mine and I have been working, although not too hard, on the following problem:

Let $ G$ be a finite group and let $ A \subset G$ have the property that $ e \in A$ but if $ s \neq e \in A$ then $ s^{ \minus{} 1} \not \in A$.  Let $ A \cdot A \equal{} \{ a \cdot b : a, b \in A \}$.  Prove or disprove that

$ |A \cdot A| \ge 2|A| \minus{} 1$.

This is a very special case of an open graph theory question, but it's enough of a special case that I think it's known and doable.

You seem to have switched between multiplicative and additive notation.  Anyway, I assume this is an abelian group?

I'm not sure.  JSteinhardt originally stated this problem to me in an abelian group, but I can't find a reason why it wouldn't be true in a non-abelian group (although I have neither a proof nor a counterexample, of course).

Hmm.  JSteinhardt said he checked the statement for cyclic groups of order up to $ 30$, so I've stated it incorrectly.

Oh, sorry; I believe we actually have $ e \in A$.


Let a equation :$ X^2$+(2$ a$-9)$ X$+a-8=0; where $ a$$ /geq$2
 Determine $ a$ to large root of equation let max value.

[quote="hoangclub"][color=darkred]Let $ f(x)\equiv x^2\plus{}(2a\minus{}9)x\plus{}a\minus{}8\equal{}0$ be a equation , where $ a\ge 2$ .
Ascertain $ a$ so that the larger root of equation is maximum.[/color] [/quote]
[color=darkblue]Observe that for any $ a\ge 2$ the realizant $ \Delta\equal{}4a^2\minus{}40a\plus{}113>0\ .$

Prove easily that $ f(x)\equal{}0$ and $ x\in\mathcal R$ $ \implies x\ne \minus{}\frac 12\ .$ Therefore,

$ 2\le a\equal{}\frac {\minus{}x^2\plus{}9x\plus{}8}{2x\plus{}1}\implies$ $ (2x\plus{}1)(x^2\minus{}5x\minus{}6)\le 0$ $ \implies$

$ x\in (\minus{}\infty , \minus{}1]\cup \left(\minus{}\frac 12,6\right]\ .$ In conclusion, $ x_{\mathrm {max}}\equal{}6$ (for $ a\equal{}2$).[/color]


linear algebra
matrix
algebra unsolved
algebra

Let A be an m x n matrix, show that if A has a right inverse C and a left inverse B, then m=n, and C=B.  THus, only square matrices can have both left and right inverses, in which case they have a two-sided inverse.

Thanks!!!

It's the same situation as here ;-)
http://www.mathlinks.ro/Forum/viewtopic.php?t=87297

By that exercise, we have $m \leq n$ and $n \leq m$, so $m = n$. Besides, $B=B1 = BAC=1C = C$.

Just one thing to suedenation:
Please post Matrix and Field/Ring/whatever stuff in the Fora for Linear Algebra and Superior Algebra from now on, ok¿


probability
function
blogs
probability and stats

I apologize if this is the wrong place to post this - I couldn't find another statistics board!:

I first heard the concept of a power law distribution in a summer course I am taking, and it's becoming one of those things that I think I understand one minute and don't the next, so I was hoping if someone can help explain it to me in layman's terms (hopefully after that, I'll be able to understand the math).

The professor said something peculiar about how the probability of "observing an instance less frequent than the most frequently observed so far [i]declines at a fixed rate[/i]."

What makes sense to me is the scaling invariance and how scaling x by c scales f(x) by c^k. But when talking about a distribution, what is f(x)? And how do I find out that the probability does decrease at a constant rate as the professor said? I think I am having troubling discerning the differences between a power law function and a power law distribution.

I've read up on Wikipedia and descriptions of Zipf's and Pareto, but I think that has confused me more. Google shows mostly academic articles, so I am very lost and will appreciate any help!

Thanks in advance!

you can have a look at Terry Tao blog:
http://terrytao.wordpress.com/2009/07/03/benfords-law-zipfs-law-and-the-pareto-distribution/


I am starting a contest with Flash art.

The contest works like this:

1) In about a week I'll post a subject. All competitors will have a week to create a graphic, animation, game, etc. using Flash that relates to the subject in some way.
2) When you have created your graphic, animation, game, etc. PM it to me and when time runs out I will post all submitted artwork.
3) For a couple of days I will set up a poll to see whose artwork is the best. The artist will be proclaimed champion of the round.
4) The champion will then choose a subject for the next round. The champion [i]may not [/i]compete in that round but can compete in the round after that. [b]Regardless of who was the champion in the last round all entries will be PMed to me![/b]
5) And so on...

[size=200]The Flash/Art Contest Rules[/size]

0. Rule 1 will be enforced starting Round 4. All other rules will be enforced starting Round 3.

1. Participants must work from scratch, and may not copy any existing artwork/photos/anything.

2. Participants may not invite someone to AoPS for the sole reason of them voting for the aforementioned participant.

3. After winning first place in one round, you may not compete in the next. You can, however, vote in the next round.

4. I, NoSoupForYou, can, if I want, submit artwork of my own, just like any participant. However, I am a horrible artist, so I don't think you'll have to worry about that.

5. I, NoSoupForYou, can only vote in two cases: i) If there is a tie for first place at the end of the week. This is done so that there will never be a tie for first place. ii) If I have submitted artwork of my own, in which case I can vote for myself.

6. Ties for first place are not allowed, as mentioned in Rule 5, but ties for second and third are allowed.

NOTE: Since very few people here have Flash, I'm letting you use any graphics creation utility, such as Photoshop, Fireworks, or even Paint if you're desperate.

OK, who's interested?

Btw, mods, can you sticky this topic?

Interested. You guys will be blown away by the sheer 1337ness of bloody stick figures :P .

In response to a question that SirErnest PMed to me, you [i]can [/i] use PowerPoint, as long as it's artistic.

A+MATH just notified me that he's interested.

I'm anxious. when is the first first subject going to be posted?

[quote="NoSoupForYou"][b]I'll have to think of something good. Plus I'm going to be away this weekend for MATHCOUNTS, so it has to be after this weekend... Probably sometime next week.[/b][/quote]

I'll be doing it if f I find myself in a creative mode when the topic's posted..
Photoshop 7.0 all the way!!

a very good painttool software by adobe. goto adobe.com for more info.

Bah. PowerPoint is for noobs, and nobody uses it anyway :P . At least put it in a .gif, .jpg, or a .swf -_O .

[color=white][friendly reminder][/color]
Yay.  It's been 5 days. :)
[color=white][/friendly reminder][/color]

[quote="ccy"]*Poke*

 :P[/quote]

I'm sorry, I forgot!  :(

I'll try to do it today...

The winner for Round Eight is ... ccy!

ccy, you have one week to submit the topic for Round Nine.

Ok. The new topic is Holidays.

Good luck. :)

OK, round 9!

Topic: holidays

Entries due 12/21/05.

Have fun!

Lol.  That sure was a great topic. :D

Anyone still want to submit anything?

I hereby proclaim the AoPS Flash/Art Contest... dead.  :(

<cue funeral march>

OK, I know this topic is dead, but maybe this creepy snowman will help revive it!

EDIT: Actually, I think I should have made it b&w, the blue eyes scare me...

Someone has to start it, and there's no one here to start it, so you'll have to start it, because I'm way too busy.

How would you start it??? Submit a piece of work??? And can you do hand drawn stuff?? Like on paper then scanned?


Please help me on this question! :huh:
I am kind of stuck on this question:

ABCD is a square, and P is a point inside the square such that triangle ADP is equilateral. How big is angle PBC

I don't think this question will be hard for you  :) [/quote]

[hide] AP is congruent to AB, so triangle APB is isoceles. Angle BAP is 30, so angle $ ABP\equal{}APB\equal{}\frac {180\minus{}30}{2}\equal{}75$. Angle $ PBC\equal{}90\minus{}75\equal{}15$[/hide]


function
inequalities
inequalities proposed

let function y=1/(1+e^x)
we can use Jensen to prove it

____________
I will be out in next ten days

it's right for every xi >= n-1
but I don't think it can prove this problem
am I wrong?

can you give your $y''$

____________
I will be out in next ten days

sorry.I can't calculate it right now because I am in internet cafe.
but I also think y" >= 0 for x >= n-1
because use this inequality we can prove the opposite one if a <= 1/(n-1)

It's convex;

$y''=\frac {e^x(e^x-1)}{(1+e^x)^3}\geq 0$ for all $x\geq 0$

Yeah, I missed something.  If any of the $x_i$'s are less than one, the function loses its convexity.

I think I had post this problem:

If $k\ge n-1$ then:
$\sum{\frac{1}{k+x_i}}\le \frac{n}{1+k}$
For all prsitive $x_1,x_2,...,x_n$ with $x_1x_1...x_n=1$

And it equal with your problem.


Question: Let $ a$ and $ r$ be real numbers with $ r\neq1$. Prove by induction, that
$ a\plus{}ar\plus{}a{r}^2\plus{}a{r}^3\plus{}...\plus{}a{r}^{n\minus{}1}\equal{}\frac{\ a({r}^n\minus{}1)}{\ r\minus{}1}$ for $ n\equal{}1,2,3...$

My Answer: Let $ P(n): a\plus{}ar\plus{}a{r}^2\plus{}a{r}^3\plus{}...\plus{}a{r}^{n\minus{}1}\equal{}\frac{\ a({r}^n\minus{}1)}{\ r\minus{}1}$ for $ n\equal{}1,2,3...$

$ P(1):$ $ LHS\equal{}a$; $ RHS\equal{}\frac{\ a(r\minus{}1)}{\ r\minus{}1}\equal{}a$; LHS=RHS therefore $ P(1)$ true.

Assume $ P(n)$ is true.
$ P(n\plus{}1): a\plus{}ar\plus{}a{r}^2\plus{}a{r}^3\plus{}...\plus{}a{r}^{n}\equal{}\frac{\ a({r}^{n\plus{}1}\minus{}1)}{\ r\minus{}1}$
$ \equal{}>(a\plus{}ar\plus{}a{r}^2\plus{}a{r}^3\plus{}...\plus{}a{r}^{n\minus{}1}) \plus{}a{r}^n\equal{}\frac{\ a({r}^{n\plus{}1}\minus{}1)}{\ r\minus{}1}$
$ \equal{}>\frac{\ a({r}^n\minus{}1)}{\ r\minus{}1}\plus{}a{r}^n\equal{}\frac{\ a({r}^{n\plus{}1}\minus{}1)}{\ r\minus{}1}$
$ \equal{}>a({r}^n\minus{}1)\plus{}a({r}^{n\plus{}1}\minus{}{r}^n)\equal{}a({r}^{n\plus{}1}\minus{}1)$
$ \equal{}>a{r}^n\minus{}a\plus{}a{r}^{n\plus{}1}\minus{}a{r}^n\equal{}a{r}^{n\plus{}1}\minus{}a$
$ \equal{}>a{r}^{n\plus{}1}\minus{}a\equal{}a{r}^{n\plus{}1}\minus{}a$, which is true. Q.E.D

Is this right?

You've successfully demonstrated that assuming $ P(n)$, $ P(n \plus{} 1)$ implies something that is true.  This is not an overwhelmingly impressive result, since [i]every[/i] statement implies every true statement.  However, you're lucky: every single "=>" you wrote (btw, the code for that is \Rightarrow or \Longrightarrow to get $ \Longrightarrow$ -- using a lowercase first letter gets $ \longrightarrow$) actually denotes a reversible step and so could be replaced with "$ \Longleftrightarrow$" in which case the modified proof would show that assuming $ P(n)$, $ P(n \plus{} 1)$ is equivalent to a true statement.  And this last thing is what you need for the induction to work.

General principle: interesting arguments flow from trivial statements to interesting statements.  If your argument starts with an interesting statement and ends with a trivial statement, it may not be saying what you hope it was saying.

Since you're assuming
$ P(n)\equal{}\frac{a(r^{n}\minus{}1)}{r\minus{}1}$, then
$ P(n\plus{}1)\equal{}\frac{a(r^{n}\minus{}1)}{r\minus{}1}\plus{}ar^{n}\equal{}\frac{a(r^{n}\minus{}1)}{r\minus{}1}\plus{}\frac{ar^{n}(r\minus{}1)}{r\minus{}1}\equal{}$
$ \equal{}\frac{a(r^{n}\minus{}1\plus{}r^{n\plus{}1}\minus{}r^{n})}{r\minus{}1}\equal{}\frac{a(r^{n\plus{}1}\minus{}1)}{r\minus{}1}$,
which is what you want.


induction
combinatorics solved
combinatorics

:?  forgive me if this is the wrong place for a (statistics like) counting problem

suppose we have a sequence of length n+2 (n>=1 )
consisting of + and -
for example with n=5 :   ++-+-
now make a sequence of length n by looking at the inner n elements, and writing 1 if it differs from every neigbour, and 0 otherwise

here : 011

more examples :  +++ --->0
+-+-     ---> 11
++--     ---> 00

if the plus and minus signs occur independently with prob 1/2
what is the expected amount of 1's?

it should turn out to be n/4

what i discovered is that there is nasty dependence , for instance :
101 is impossible because the middle sign has to equal one of the outer signs , but the outer signs would both have to differ from all their neighbours?

any ideas? :?

I think induction on $n$ works to prove the expected amount $a_n=n/4$ :
$n=1$ is trivial.
There is a chance of $1/4$ that the last number is $1$ ( +-+ and -+- ) and so you have for the expected amount $a_n=a_{n-1}+1/4$.
Your observation that some sequences do not occur doesn't influence the expection, becaus the expected amount is additive also with dependent events.


Pascal\u0027s Triangle
algebra
binomial theorem

Treething, I like this problem because it can be done several ways, the most straight forward of which was your solution, which was of course correct.  Can you do this problem in such a manner that you only use one exponent right from the start, and without actually adding any of those numbers?

Both
[quote="jli"][quote="frost13"]If  $8^8(1) + 8^8(8) + 8^8(28) +8^8(56) +8^8(70) +8^8(56) +8^8(28) +8^8(8) +8^8(1) = 2^x$, find x.[/quote]

[hide]Use pascal's tiangle and you get $8^8(1+1)^8=8^82^8=16^8=2^32$

x=[b]32[/b][/hide][/quote]
and
[quote="Treething"][hide="solution"]
That equals (256)(8^8).
(2^8)(2^24)
2^32
[b]x = 32[/b]
[/hide][/quote]

Work. If you are doing a sprint round, you probably would have used Treething's method because you would have seen it quickly, but if you had higher numbers, such as the 13th power, Jli's method would have saved a lot of work... It just depends on the problem... and frost13 is right about the beauty of this problem...

The best part about this is the way I did this hasn't even been done yet.

Here is how I would do this if you're interested :

[hide] If you look at
$8^8(1) + 8^8(8) + 8^8(28) +8^8(56) +8^8(70) +8^8(56) +8^8(28) +8^8(8) +8^8(1)$

as being

$1(8^8)(8^0)+8(8^7)(8^1)+28(8^6)(8^2)+56(8^5)(8^3)+70(8^4)(8^4)+56(8^3)(8^5)+28(8^2)(8^6)+8(8^1)(8^7)+1(8^0)(8^8)$

Use Pascal's triangle and the binomial theorem to get  $(8 + 8)^8 = 16^8 = (2^4)^8 = 2^{32}$ so $x=32$

It definitely looks like more work, but most of that is the "how I looked at it" part.  The work didn't start until $(8+8)^8$

[/hide]


function
topology
real analysis
real analysis unsolved

Prove that if every real valued continuous function on the metric space $X$
[b]i[/b])takes its maximum value, then $X$ is compact.
[b]ii[/b])is uniformly continuous, then $X$ is complete.

What is wrong with the following? I "prove" that i) holds even if just every continuous function is bounded, but I doubt this is right.

Let $(\tilde{X},\rho)$ be the completion of $X$. Suppose $(x_{n})$ is a Cauchy sequence in $X$ which doesn't converge. Let $x_{0}$ be the limit in $\tilde{X}$. Now define $f: X \to \mathbb{R}$ by $f(x) = \frac{1}{\rho(\tilde{x},x_{0})}$. $f$ is a continuous unbounded function. Hence $X$ is complete.

Suppose $X$ is not totally bounded. Then find a sequence $(x_{n})$ of points with distance at least $\epsilon > 0$ between all the points. The balls $B(x_{n},\epsilon/4)$ all have positive distance to each other. For a fixed $n$ there is a continuous function $f_{n}: X \to \mathbb{R}$ taking the value $n$ on $x_{n}$ and the value 0 outside $B(x_{n},\epsilon/4)$. Now $\sum f_{n}$ is a continuous unbounded function. Hence $X$ is totally bounded.

Kalle's argument is correct.

Suppose $X$ is a space on which every (real-valued) continuous function is bounded. Then every continuous function on $X$ assumes its maximum.

Proof: Given a continuous function $f,$ let $M=\sup_{x\in X}f(x).$ If there is some $x$ for which $f(x)=M,$ we are done. Else consider $g(x)=\frac1{M-f(x)}.$ We have that $g$ is continuous but unbounded, which cannot be.

In his original post, The Captain meant for the last word in (ii) to be "compact," didn't he?

[quote="Kent Merryfield"]
In his original post, The Captain meant for the last word in (ii) to be "compact," didn't he?[/quote]
No, because then he made the same typo in the title too. Consider an infinite discrete metric space. Every function is uniformly continuous, but the space is not compact.


si $a$ y $b$ son dos de las raices de $x^{4}+x^{3}-1=0$.
probar que $ab$ es una raiz de $x^{6}+x^{4}+x^{3}-x^{2}-1=0$

Sean $a,b,c,d$ las raices del polinomio $f(x) = x^4 + x^3 -1$. Por Vieta, tenemos que

1. $a+b+c+d = -1 = abcd,$
2. $abc + abd + acd + bcd = 0 = ab + ac + ad + bc + bd + cd.$

Sea $p = a+b$ y $q = ab$. Entonces, las condiciones en (2) se pueden reescribir como:

$p = \frac{-q^2}{q^2 + 1}$ y $qp + qp^2 - q^2 +1 = 0$.

Finalmente, basta sustituir el valor de $p$ en la segunda ecuacion para obtener que $q^6 + q^4 + q^3 - q^2 - 1 = 0$, lo cual concluye la demostracion.


modular arithmetic
vector
number theory
relatively prime

Consider 97X+24Y+14Z,   26X+93Y+40Z,   35X+31Y+78Z.

Suppose that for some integer values of X, Y, Z  two of these numbers are divisible by 2000.

Prove that the third number is also divisible by 2000.

Note that
\begin{eqnarray*}
&& 1551(97x + 24y + 14z) + 1833(26x + 93y + 40z) + 397(35x + 31y + 78z) \\
&=& 212000x + 220000y + 126000z \equiv 0 \pmod{2000}.
\end{eqnarray*}

Since the coefficients 1551, 1833, and 397 are all relatively prime to 2000, the result follows.

[quote="nsato"]Note that
\begin{eqnarray*} & & 1551(97x + 24y + 14z) + 1833(26x + 93y + 40z) + 397(35x + 31y + 78z) \\
& = & 212000x + 220000y + 126000z \equiv 0 \pmod{2000}. \end{eqnarray*}
Since the coefficients 1551, 1833, and 397 are all relatively prime to 2000, the result follows.[/quote]

How did you find this? :|

Did you solve a system perhaps?

Certainly.  As far as linear combination goes, systems of linear equations $ \bmod 2000$ can be solved just as easily as systems of linear equations over the reals.

Edit:  [hide="Alternately"] All you are trying to prove is that the vectors $ \left< 97, 24, 14 \right>, \left< 26, 93, 40 \right>, \left< 35, 31, 78 \right>$ are linearly dependent over $ \mathbb{Z}_{2000}$.  You can do this by showing that the determinant

$ \left| \begin{array}{ccc} 97 & 24 & 14 \\ 26 & 93 & 40 \\ 35 & 31 & 78 \end{array} \right|$

is divisible by $ 2000$, which frees you of the need to actually find an explicit linear combination as nsato did, although that is the most concrete way to solve this problem. [/hide]


vector
algebra
system of equations
linear algebra
linear algebra unsolved

I have 2 subspaces of R^4 - M=span{(1,1,1,-1),(1,-1,1,1),(3,3,1,-1)} and N=span{(1,1,-1,1),(1,2,-1,2),(2,1,-1,2)}.

How to find a basis for the intersection of these subspaces?
And how to extend this basis to bases of M,N,and M+N?

I would appreciate every suggestion for approach to the problem!

You need to find systems of equations (call them A,B) describing both linear spaces (i.e. find a system of equations such that the space of its solutions is M or N), and then take the system consisting of equations from both systems (call it A*B). Then you simply find the basis of the space described by A*B.  Hope you know how to do it.

In order to extend the basis obtained form the previous step to the basis of M, just take any vector from M that is linearly independent of your basis (it's better to do this "by feel" instead of computing it), and continue until you get the basis of M. If M=spanC and N=spanD, then M+N is just span(C+D), so you can use the same method.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Assume that we have a subgroup $H$ of the group $G$ such that $(G:H)=4$. Show that there is some proper subgroup $H^{\prime}$ of $G$ such that $(G:H^{\prime})<4$.

Peter

of course, $H^{\prime}$ is supposed to be proper... :)

Nice problem :).

Let $H_G$ be the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=19724]normal interior[/url] of $H$ in $G$. We have $[G:H_G]=[G:H][H:H_G]=4[H:H_G]$. Since we have a monomorphism from $G/H_G$ to $S_4$ (see the link above), we can regard $G/H_G$ as a subgroup of $S_4$. If it's not a subgroup of $A_4$, then it has a subgroup of index $2$, and this corresponds to a normal subgroup $K$ of $G$, also of index $2$, which contains $H_G$, and we're done. This means that we can assume $G/H_G$ to be a subgroup of $A_4$.

Now, $|A_4|=12$, and $|G/H_G|$ is a multiple of $4$, so it's either $4$ or $12$ (the only multiples of $4$ which divide $12$). If it's $12$, then $G/H_G$ is the entire $A_4$, and it has a normal subgroup of index $3$ (formed by the products of two disjoint transpositions and the identity, so it has $4$ elements), which, again, corresponds to a normal subgroup of $G$ of index $3<4$, and we are, again, done.

If $|G/H_G|=4$, then $H$ is normal in $G$. If $x\in G\setminus H$ is such that $x^2\in H$, then the subgroup of $G$ generated by $x$ and $H$ satisfies the requirements.


[b]Prove the following algebraic inequality:   [/b]
$\prod\limits_{j = 1}^n {\left( {x_j  + \frac{1}{{x_j }}} \right)}  \ge \left( {n + \frac{1}{n}} \right)^n
$
[b]where [/b]$x_1 ,x_2 , \ldots ,x_n $
[b]are positive real mumbers with: [/b]
$\sum\limits_{j = 1}^n {x_j  = 1} $

If I made no mistake then you have only to apply Jensen and that is all.^You take the ln of both sides. Then the function ln(x+1/x) is convex. f''(x)=(-x^4+4x^2+1)/(x^3+x)^2. The inequality follows directly from the Jensen's inequality for this function.

Yes, although after applying jensen, you apply AM-HM to show
$\displaystyle\frac{\sum_{i=1}^{n}{\frac{1}{x_i}}}{n} \ge n$
(and he did show that f"(x)>0 for $x \in (0,1]$)
edit: ps. why isnt my sigma pretty? i used /displaystyle

GeoMar, I don't understand your comment with AM-HM. After you apply Jensen you have nf((x_1+x_2+..x_n)/n)=nf(1/n) which is the constant you need.

jensen says the natural log of the lhs is greater than or equal to

$\displaystyle n\ln{\frac{\sum_{i=1}^{n}{x_i}+\sum_{i=1}^{n}{\frac{1}{x_i}}}{n}}=n\ln{(\frac{1}{n}+\frac{\sum_{i=1}^{n}{\frac{1}{x_i}}}{n})}$ and that second part is what I used AM-HM for, to show it is greater than n... perhaps u have a way to not need to do that? I am not experienced with inequalities...

[quote="arthur"]you must show first that $\displaystyle f''(x)\geq0$.[/quote]

let me announce "the return of arthur"...

Prove the following algebraic inequality:
$\prod\limits_{j = 1}^n {\left( {x_j  + \frac{1}{{x_j }}} \right)}  \ge \left( {n + \frac{1}{n}} \right)^n $
where $x_1 ,x_2 , \ldots ,x_n $
 are positive real mumbers with:
$\sum\limits_{j = 1}^n {x_j  = 1} $

[u]Solution:[/u]
Consider:
$f:\left( {0,1} \right] \to \mathbb{R},f\left( x \right) = \ln \left( {x + \frac{1}{x}} \right)$

Then we have:
$f'\left( x \right) = \frac{{1 - \frac{1}{{x^2 }}}}{{x + \frac{1}{x}}}$ and    $f''\left( x \right) =  - \frac{{x^4  - 4x^2  - 1}}{{\left( {x^3  + x} \right)^2 }}$

It follows that:
$x \in \left( {0,1} \right] \Rightarrow 0 < x^2  \le 1 \Rightarrow  - 2 < x^2  - 2 \le  - 1 \Rightarrow $

$ \Rightarrow 1 \le \left( {x^2  - 2} \right)^2  < 4 \Rightarrow  - 4 \le \left( {x^2  - 2} \right)^2  - 5 <  - 1 \Rightarrow $

$ \Rightarrow x^4  - 4x^2  - 1 \le 0 \Rightarrow f''\left( x \right) \ge 0$

Conclusion:
$\forall x \in \left( {0,1} \right] \Rightarrow f''\left( x \right) > 0$

After applying Jensen we have:
$\frac{{f\left( {x_1 } \right) + f\left( {x_2 } \right) +  \cdots  + f\left( {x_n } \right)}}{n} \ge f\left( {\frac{{x_1  + x_2  +  \cdots  + x_n }}{n}} \right),\forall x_1 ,x_2 , \ldots ,x_n  \in \left( {0,1} \right] \Rightarrow $

$ \Rightarrow \frac{{\ln \left( {x_1  + \frac{1}{{x_1 }}} \right) +  \cdots  + \ln \left( {x_n  + \frac{1}{{x_n }}} \right)}}{n} \ge \ln \left( {n + \frac{1}{n}} \right),\forall x_1 ,x_2 ,,x_n  \in \left( {0,1} \right] \Rightarrow $

$ \Rightarrow \prod\limits_{j = 1}^n {\left( {x_j  + \frac{1}{{x_j }}} \right)}  \ge \left( {n + \frac{1}{n}} \right)^n ,\forall x_j  \in \left( {0,1} \right]$

Thanks for all your help !!!


[quote="sharkman"]Find the equation of a sphere with radius r and center P.

r = 3; P = (-1, 1, 2)[/quote]
[hide](x+1)^2+(y-1)^2+(z-2)^2=9[/hide]

[hide="Solution"]Using the formula for the equation of a sphere, we have:

$ (x\plus{}1)^2\plus{}(y\minus{}1)^2\plus{}(z\minus{}2)^2\equal{}9$[/hide]

The equation for a sphere is very similar to that of a circle.  It is...

[hide]$ (x\minus{}a)^2\plus{}(y\minus{}b)^2\plus{}(z\minus{}c)^2\equal{}r^2$
Where the center of the sphere is $ (a,b,c)$ and the radius is $ r$.[/hide]


MATHCOUNTS probability geometry rectangle ratio inequalities

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}


Consider a planet which is made of matter with the same average density as the earth.Assume that the atmospheric pressure on the planet's surface is the same as on the earth's surface.Assume for simplicity that the planet's atmospheric temperature is independent of height,and that it is equal to the temperature at the surface of the earth.And assume that the composition of the atmosphere on the planet is the same as on the earth.What should  the radius of the planet be,so that a light beam can travel in a circle around the planet,just above its surface?

Note:You will need to use the fact that the index of refraction depends on the density of air according to $n(\rho)=1+k\rho$,where k is a given constant.Give your answer in terms of $R_{E}$,$g_{E}$,$P_{E}$,$\rho_{E}$ and k.

$P_{E}$ is the atmospheric pressure in the Earth.


I need help with this problem. Could someone help me?

From the set (1/2, 1/3, ... , 1/100) are composed all possible subsets containing even number of elements and for each subset is calculated the product of its members. Find the sum of all products. :?:

I'm gonna make a list and brute force it :D

[quote="surge"]Seeing that there are $\displaystyle\sum_{n=1}^{50} \displaystyle{100\choose 2n}= 633825300114114700748351602687$ subsets, your plan may be a bit difficult.  :D[/quote]

I think that there must be some smart trick but couldn't make a guess. Go on!  :)

[quote="t0rajir0u"]This seems a little hard for Getting Started :huh:[/quote]

Could I move it in the proper place? How to do it? :(

Ask a mod to move it into the MathCounts.. .errr... Intermediate forum.

Oh right, I thought it was easier than it really is.

I moved it from Classroom to here, but I guess a mod from this forum should move it to Intermediate?

Well, I have [hide="an expression I cannot evaluate..."]
Let $P(x) = \prod_{k=2}^{100} \left( x - \frac{1}{k} \right)$, a polynomial of degree $99$.  The sum of all products of even subsets is given by the sum of all even symmetric sums of the roots, or the (alternating) sum of the odd coefficients.  This is given by

$| \Im( P(i) ) | = | \Im( \prod_{k=2}^{100} \left( i - \frac{1}{k} \right) ) |$,

Which I cannot find a neat evaluation for.   :huh:
[/hide]

[quote="Iron Fist"]Maybe he means product of all sums :P .[/quote]

No, sum of all products must be found :D

anyone? I'm interested in the solution, now.


calculus
integration
inequalities
real analysis
real analysis unsolved

We are given $f: \mathbb{R}\to [0,\infty)$, f continuous and $\exists \int_0^\infty f^2(x)dx\in \mathbb{R}$. Show that there exist and is finite $\int_0^\infty \frac{(\int_0^xf(t)dt)^2}{x^2}dx$.

Actually it is an implication of Hardy's inequality, directly,
if $1<p< \infty$ and $f \in L^p((0,\infty))$, relative to Lebesgue measure, and $F(x)=\frac{1}{x}\int_0^x f(t)dt$, then
$||F||_p \leq \frac{p}{p-1}||f||_p$.
In our case, put $p=2$ and we get $\int_0^\infty \frac{(\int_0^xf(t)dt)^2}{x^2}dx \leq 4 \int_0^\infty f^2(x)dx$.

I think it is not fair to propose this one as an olimipiad problem. My solution uses Cauchy and Cebishev inequalities:
Do you have a proof for Hardy's ineq, using more powerfull tools?

[quote="xirti"]I think it is not fair to propose this one as an olimipiad problem.
[/quote]
It was used in olympiad? which kind of olympiad
[quote="xirti"]
Do you have a proof for Hardy's ineq, using more powerfull tools?[/quote]
It was discussed on Mathlinks, i remember Liyi gave a proof: i'll try to find, if i find i'll post the link.

A Romanian Olympiad for 12-th degree. It is not very hard but, if someone knew Hardy's ineq (which is probably well-known for undergraduate student) could solve it imediatly.

Here is that proof for $p=2$ which i wrote about http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Hardy&t=35918#

This post is unreadable. Can you edit it?


geometry
trigonometry
geometric transformation
reflection
angle bisector
cyclic quadrilateral

[quote="the future"]
2. In square ABCD points P and Q are on BC and CD, respectively.  If angle(PAQ)=45 degrees, prove that PQ= PB + DQ.[/quote]

[hide="BEHOLD!"]
In the diagram below, the two squares are congruent, as are the triangles inscribed in them. We find that $ AQ \equal{} AQ$, $ AE \equal{} AP$, and $ \angle EAQ \equal{} x \plus{} y \equal{} 45^o \equal{} \angle PAQ$, so $ \triangle EAQ\cong \triangle PAQ$, and $ QE \equal{} DE \plus{} DQ \equal{} BP \plus{} DQ$, as desired.

$ \mathbb{QED}$

Aside: An interesting generalization can be found if you allow $ P$ and $ Q$ to be anywhere inside of the triangle while maintaining the original angle restriction; $ PQ$ can be found quite nicely in terms of $ BP$ and $ DQ$. A proof like the one above would then require a tessellation of four squares, arranged as one large square.
[/hide]

Whoa thanks, i never thought to do any reflections or anything like that.

for this problem

5. In isosceles triangle ABC, angle(A)=100 degrees. Let E be on AC such that BE is the angle bisector of angle(B). Show that AE+BE=BC.

i tried reflections and ended up that once  you reflect BE onto BC (construction 20-80-80 triangle BEX) and call that point X where E hits BC.  So now all we need to prove is that XC=AE.  That is very easy with trig because from the 100-40-40 isosecles triangle XEC, we havev XC=XE, and we basicaly need to prove that XE=AE.  With trig, since sin80=sin100 (laws of sines) its pretty trivial, but does anyone  have a more geometric solution to finish it off?

[quote="the future"]...but does anyone  have a more geometric solution to finish it off?[/quote]

[hide="I do."]
I will follow your lead in constructing point $ X$. Then we notice two things:

1) By the Angle Bisector Theorem, $ \frac{BC}{CE}\equal{}\frac{BA}{AE} \implies BA\cdot CE\equal{}BC\cdot AE$.

2) Looking at angles, $ \triangle XEC\sim \triangle ABC \implies \frac{XC}{CE}\equal{}\frac{BA}{BC} \implies BA\cdot CE\equal{}BC\cdot XC$.

So $ BC\cdot AE\equal{}BC\cdot XC \implies AE\equal{}XC$.

Thus $ XC\plus{}BX\equal{}BC \implies AE\plus{}BC\equal{}BC$, as desired.
[/hide]

[hide="1"]
Let $ AH$ meet $ BC$ at $ X$ and $ BH$ meet $ AC$ at $ Y$.  Now, $ \angle AYB\equal{}90\equal{}\angle AXB$ and $ \angle HYC\plus{}\angle HXC\equal{}90\plus{}90\equal{}180$, so $ AYXB$ and $ HYCX$ are cyclic.  Now, $ BH\equal{}AC$, $ \angle AXC\equal{}90\equal{}\angle HXB$, and $ \angle HBX\equal{}\angle HAY\equal{}\angle XAC$.  Thus, $ \triangle AXC\cong \triangle BXH\implies HX\equal{}XC$.  Since $ \angle HXC\equal{}90$, we have that $ \angle XHC\equal{}\angle HCX\equal{}45$.  Hence, $ \angle XAB\equal{}\angle BYX\equal{}\angle HYX\equal{}\angle HCX\equal{}45$.  Thus, $ \angle ABC\equal{}\angle ABX\equal{}180\minus{}\angle AXB\minus{}\angle BAX\equal{}45\equal{}\angle HCB$, as desired. [/hide]

[quote="the future"]In isosceles triangle ABC, angle(A)=100 degrees. Let E be on AC such that BE is the angle bisector of angle(B). Show that AE+BE=BC.[/quote]

We did this problem in WOOT.

[hide="hint"] Construct some isosceles triangles, and hunt for a cyclic quad.[/hide]

thanks guys for the responses

^hehe oh yea that would be a quicker finish

[hide="cyclic quads at the end"]I guess you would do what I did orignially by taking BE and constructing a segment BX such that BX=BE.  Now we get angle(EXB)=80 degrees, so that angle(EXC)=100 degrees, which means that angle(CEX)=40 degrees.

So we have CX=EX.  Now since angle(A)+angle(EXB)=180 degrees, we have a cyclic quadrilateral ABXE.  Since arc(AE)=arc(EX) -since BE is the bisector- EX=AE (chords to the same arc degree).  So that means AE=CX, which means that BX+AE=BC, and since BX=BE, BE+AE=BC.[/hide]
Of course, the way jvenezuela716 finished it was just as good.

Hmm for number 4, you can rewrite it as AC*AD=AB*AD+AB*AC, and since BD+AC, you can rewrite it again as BD*AD=AB*AD+AB*BD and you just get triangle ABD, but it seems  hard from here...I dont think this is the right way to go if you dont plan on using trig (more specifically that sin(2pi/7)sin(4pi/7)=sin(pi/7)sin(4pi/7)+sin(pi/7)+sin(2pi/7))...

umm nvm
[hide="4"]
So basically just consider that equation when AB*AC*AD is multiplied through.  So AC*AD=AB*AD+AB*AC.  Well,  this looks like Ptolemys but the corresponding segments dont make a cyclic quadrilateral.  So we have to look for one with the same length but just different segment names.  So we find cyclic quadrilateral ABCE (or anything in that shape).  We know that from Ptolemys, AB*CE+BC*AE=AC*BE.  Since CE=AC, BC=AB, and AE=AD, we can replace the RHS with AB*AC+AB*AD and since BE=AD, the LHS becomes AC*AD, and that equality satisfies our original one.[/hide]
i guess that only leaves 3, the ones that "impossible"

[quote="the future"]So im pretty bad at geometric proofs and I just found these old geometry questions and I need to set a foundation for doing harder proofs so could anyone try and do these?  And to prevent like a flood of posts i posted all these in one place.  You can solve it but hints would be especially nice (and please no trig)
Thanks.

1. In triangle ABC, the altitudes AD and BE intersect at H so that BH=AC.  Show that angle(BCH)=angle(ABC).

2. In square ABCD points P and Q are on BC and CD, respectively.  If angle(PAQ)=45 degrees, prove that PQ= PB + DQ.

3. In triangle ABC, AB=BC, angle(B)=20 degrees.  Let M be a point on AB such that angle(MCA) = 60 degrees, and N be a point on BC, such taht angle(NAC)= 50 degrees.  Find angle(NMC). (this ones pretty famous but I still get lost at where to construct the equilateral triangle)

4. Let ABCDEFG be a regular heptagon.  Show that (1/AB)=(1/AC)+(1/AD).

5. In isosceles triangle ABC, angle(A)=100 degrees.  Let E be on AC such that BE is the angle bisector of angle(B).  Show that AE+BE=BC.

For general geometry problems what do you guys look for first (similarity, cyclicity...).  Also, should try and use the question to try and find your answer (for example if it asks for to prove some angleA=angleB, do you guys try and use that to prove backwards as in proving the two triangles that contain those angles are similar)?[/quote]
[hide="2"]
Let $ AQ$ and $ BD$ meet at $ X$, $ AP$ and $ BD$ meet at $ X$, and $ Z$ be the point on $ PQ$ such that $ QZ \equal{} DQ$.  Now, let $ \angle DAQ \equal{} \alpha$, so $ \angle PAB \equal{} \angle DAB \minus{} \angle DAQ \minus{} \angle PAQ \equal{} 90 \minus{} 45 \minus{} \alpha \equal{} 45 \minus{} \alpha$.  Also, $ \angle XAP \equal{} 45 \equal{} \angle XBP$, so $ ABPX$ is cyclic.  Hence $ \angle ABX \equal{} 45 \equal{} \angle APX$.  Similarly, $ AYQD$ is cyclic, so $ \angle XQY \equal{} \angle ADY \equal{} 45$.  Furthermore, $ \angle AQD \equal{} 180 \minus{} \angle DAQ \minus{} \angle ADQ \equal{} 90 \minus{} \alpha$.  Also, $ \angle AXP \equal{} 180 \minus{} \angle ABP \equal{} 90$, so $ \angle QXP \equal{} 90$.  Similarly, $ \angle QYP \equal{} 90$, so $ XYPQ$ is cyclic.  Thus, $ 45 \minus{} \alpha \equal{} \angle BAP \equal{} \angle BXP \equal{} \angle YQP$.  This gives us that $ \angle PQC \equal{} 180 \minus{} \angle AQD \minus{} \angle AQY \minus{} \angle YQP \equal{} 2\alpha$, which results in $ \angle ZDQ \equal{} \angle DZQ \equal{} \frac {\angle PQC}{2} \equal{} \alpha \equal{} \angle DAQ$, which means that $ ZADQ$ is cyclic.  Hence, $ \angle DAZ \equal{} \angle PQC \equal{} 2\alpha$ and $ \angle QAZ \equal{} \angle ZDQ \equal{} \alpha$, so $ \angle QPC \equal{} 90 \minus{} \angle PQC \equal{} 90 \minus{} \angle ZAD \equal{} \angle ZAB$, which means that $ PBAZ$ is cyclic.  Finally, $ \angle ZBP \equal{} \angle ZAP \equal{} \angle QAP \minus{} \angle ZAQ \equal{} 45 \minus{} \alpha \equal{} \angle PAB \equal{} \angle BZP$, so $ BP \equal{} PZ$, and we conclude that $ PQ \equal{} PZ \plus{} QZ \equal{} QD \plus{} PB$, as desired. [/hide]

Problem 3 is quite frequently posted and well known; here is one of the links to it: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179738


MATHCOUNTS percent probability articles Support ARML

Well, I thought the show was pretty good, except they took my worst quote  :( . What did you all think?

Sadly, I don't have cable and therefore didn't watch myself make a fool of myself.  Could someone describe what happened?

I thought that it was good.  I never did mathcounts so that gave me a good idea of what it is like.  Those kids are really smart! and quick!

Ya.. except:

Solve the following spoiler problem in 3 seconds (including reading the problem):

[hide]What percent of 22.5 is 90?[/hide]

My answer: [hide]25[/hide]

Jeffrey's answer: [hide]400 -__-[/hide]

Which one were you probability?  I can't remember much of it(not with the names associated at least).

Well at least you got that one question right by buzzing in before you knew the answer!  At least you made it to semifinals!  At least you didn't miss 2nd place by a point written round (and I depended on written round)!  :-) ok I'll stop.

So what did happen?  They must have played those embarrassing interviews.  And then they must have done some other odd stuff...  Well you said it was better than last time (but that's not saying very much :P).

I was the 5th place guy. -_-

Yeah something like that.  Before each match up they showed a little profile of the two people in it and played a short tape of the person saying something about themself.  Most of the rest of show was the competition and they explained a few of the problems.  That's about it.  If you were there I don't think that you missed very much.

[quote="mathdool"]HEHEHE... I buzzed in multiple times before I knew the answer.

They took your quote about Lord of the Rings being philosiphical. lol.[/quote]

What???  I don't even have any vague impression of myself ever talking about LOTR.

This is what you said, (spoiler) [hide]My name is [blah] and i come from [blah]. I thought Lord of the Rings had like they had really good effects. His dark materials trilogy was philosophical...[/hide]

They screwed up on problem two in the round that danny faced jeffery.  It said :
how many ordered pairs (m,n) satisfy 2004^10*2004^20*2004^30*2004^40 = (2004^m)^n can.  In inches, what is the radius of the can?

I was disappointed by it, personally. They did a accurate job of showing what happened in the countdown round; I'll give them that, but they didn't really show what MathCounts was all about. Also, I could tell that they didn't spend much money on that program. I shouldn't complain though; it was definitely better than nothing.

My MathCounts coach videotaped it and showed it to us. One of my teammates was one of the interviewed people outside of the hotel o_o

Personally, I have always been surprised ESPN makes any effort whatsoever to cover MathCounts. I can't understand what logic goes into that, considering they are a business.

[quote="Treething"]I just realized that I'm better at Countdown round than almost all of the top 12 o_o . Now if I could actually make it there :P ...[/quote]

Not necessarily. There's the pressure of being on national television, and also the intensity of the competition. It's much easier to do it watching instead of actually being up there. That's a lot like being a monday-morning-quarterback. (sorta)

More than half of the countdown competitors at Nationals are dead before they even sit down to compete. You can tell just by watching them that they're completely out of it. Sometimes somebody gets on a roll, and then everybody is just scared to death.

It really is quite easy to beat them when you're watching them from the audience. Go up there and you'll see why.

Think about it this way. When you're sitting up there, you know that how fast you think in the next five seconds or so will determine the fate of two years of hard work. It's a nasty feeling.

haha, yes exactly! My friend Jeffrey Chen got 2nd in countdown and he said that he was so nervous he almost couldn't think. Add the sprint and target rounds before that and you see why it seems like you can beat most of them.

Good point. The nervousness of doing something stupid just mounts. It really helps if you're very difficult to embarrass.

I have personal experience with countdown round jitters and nervousness, except in states instead of nationals. When I had realized I made a stupid mistake on the tiebreaker question, I swore into the microphone. Not cool.

Haha. I faced a kid who had climbed from 8th to 3rd. This was before the new format. Intimidation anyone?

o_O

Did you win :P ?

[quote=yif man12]
Not necessarily. There's the pressure of being on national television, and also the intensity of the competition. It's much easier to do it watching instead of actually being up there. That's a lot like being a monday-morning-quarterback. (sorta)
[/quote]

Yep, I guess you're right...

Somebody I know, Shaunak Kishore, came in fourth in the written round but only eighth in countdown...

Yeah I won. And I would be the first to tell anyone that it's more than 50% luck after you get to that stage.

the thing about countdown rounds... especially the official countdown rounds (9th place vs. tenth place winner faces 8th winner face 7th) (the way it is at state and national, sometimes chapter).
i've never been in a countdown round... but I know from practicing them that they are extremely nerve racking...
aout ESPN: I guess they wanted to show sports obsessed kids that math is cool(and competitive)

[quote="darktreb"][quote]States or Chapter? I highly doubt that was in Nationals, but if it was, I applaud you.[/quote]

Nationals.[/quote]

Yep - it was pretty impressive.  MCrawford and I were on the stage (right behind you).

r u guyz all that goood? :( i think i know jeffery chen... he's really smart/quick


search
IMO Shortlist
inequalities unsolved
inequalities

$ x,y,z$ non-negative real number satisfyng $ x\plus{}y\plus{}z\equal{}1$ . find min and max:
$ P\equal{}xyz\plus{}xzt\plus{}xyt\plus{}yzt\minus{}2xyzt$

[quote="zaya_yc"]$ x,y,z$ non-negative real number satisfyng $ x \plus{} y \plus{} z \equal{} 1$ . find min and max:
$ P \equal{} xyz \plus{} xzt \plus{} xyt \plus{} yzt \minus{} 2xyzt$[/quote]

sorry . $ x,y,z,t$ non-negative real number  $ x\plus{}y\plus{}z\plus{}t\equal{}1$

[quote="zaya_yc"][quote="zaya_yc"]$ x,y,z$ non-negative real number satisfyng $ x \plus{} y \plus{} z \equal{} 1$ . find min and max:
$ P \equal{} xyz \plus{} xzt \plus{} xyt \plus{} yzt \minus{} 2xyzt$[/quote]

sorry . $ x,y,z,t$ non-negative real number  $ x \plus{} y \plus{} z \plus{} t \equal{} 1$[/quote]
1.min:
$ P \equal{} xyz \plus{} xzt \plus{} xyt \plus{} yzt \minus{} 2xyzt \equal{} xyz(1 \minus{} y) \plus{} yzt(1 \minus{} x) \plus{} ztx \plus{} txy \ge 0$
Equality holds for  $ t \equal{} z \equal{} 0,x \plus{} y \equal{} 1$ or any cyclic permutation.

[quote="FOURRIER"]The max is $ \frac {7}{2^7}$ for $ x \equal{} y \equal{} z \equal{} t \equal{} \frac {1}{4}$

Demo:

According to IMO 93, we have:

$ \sum xyz \le \frac {1}{27} \plus{} \frac {176xyzt}{27}$

Thus:

$ P \le \frac {1}{27} \plus{} \frac {176xyzt}{27} \minus{} 2xyzt$

and then apply am gm on $ xyzt$[/quote]
It's wrong.
Try $ x\equal{}y\equal{}z\equal{}\frac{1}{3},t\equal{}0$ we have $ P\equal{}\frac{1}{27}>\frac{7}{2^7}$

[quote="NguyenDungTN"]
It's wrong.
Try $ x \equal{} y \equal{} z \equal{} \frac {1}{3},t \equal{} 0$ we have $ P \equal{} \frac {1}{27} > \frac {7}{2^7}$[/quote]

No, It's not:

$ P \equal{} \frac {1}{27} < \frac {7}{2^7}$  since $ 2^7 \equal{} 128$ and $ 7*27\equal{} 189$

 :wink:

[quote="FOURRIER"]The max is $ \frac {7}{2^7}$ for $ x \equal{} y \equal{} z \equal{} t \equal{} \frac {1}{4}$

Demo:

According to IMO 93, we have:

$ \sum xyz \le \frac {1}{27} \plus{} \frac {176xyzt}{27}$

Thus:

$ P \le \frac {1}{27} \plus{} \frac {176xyzt}{27} \minus{} 2xyzt$

and then apply am gm on $ xyzt$[/quote]

No that imo result only holds for $ x\plus{}y\plus{}z\equal{}1$...but you use it four times...so $ x\plus{}y\plus{}z\equal{}y\plus{}z\plus{}t\equal{}...\equal{}1$...?!

[quote="Altheman"][quote="FOURRIER"]The max is $ \frac {7}{2^7}$ for $ x \equal{} y \equal{} z \equal{} t \equal{} \frac {1}{4}$

Demo:

According to IMO 93, we have:

$ \sum xyz \le \frac {1}{27} \plus{} \frac {176xyzt}{27}$

Thus:

$ P \le \frac {1}{27} \plus{} \frac {176xyzt}{27} \minus{} 2xyzt$

and then apply am gm on $ xyzt$[/quote]

No that imo result only holds for $ x \plus{} y \plus{} z \equal{} 1$...but you use it four times...so $ x \plus{} y \plus{} z \equal{} y \plus{} z \plus{} t \equal{} ... \equal{} 1$...?![/quote]
Problem from IMO shortlist 1997, Vietnam:
Let $ a,b,c,d \ge 0,a\plus{}b\plus{}c\plus{}d\equal{}1$. Prove that:
$ abc\plus{}bcd\plus{}cda\plus{}dab\le \frac{1}{27}\plus{}\frac{176}{27}abcd$ :|

[quote="Altheman"]
No that imo result only holds for $ x \plus{} y \plus{} z \equal{} 1$...but you use it four times...so $ x \plus{} y \plus{} z \equal{} y \plus{} z \plus{} t \equal{} ... \equal{} 1$...?![/quote]

No, it's with 4 variables.

[quote="NguyenDungTN"]
Problem from IMO shortlist 1997, Vietnam:
Let $ a,b,c,d \ge 0,a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove that:
$ abc \plus{} bcd \plus{} cda \plus{} dab\le \frac {1}{27} \plus{} \frac {176}{27}abcd$ :|[/quote]
http://www.mathlinks.ro/viewtopic.php?search_id=457550760&t=80894


What is the probability of being assigned a random phone number with all of the last four digits the same? Express your answer as a common fraction.

Only the last four digits are relevant. There are $ 10$ arrangements where the last four digits are the same, and $ 10^4$ possible last four digits, for an answer of $ \frac{10}{10^4}\equal{}\frac1{10^3}\equal{}\boxed{\frac1{1000}}$.

Or the 2nd to last has 1/10 of being same as last, also 3rd to last has 1/10, same w/ 4th to last, so (1/10)^3=1/1000.


For the SAT II Chemistry Exam, are we only given a simple Periodic Table?  How would you know the electron configuration of an element?  Are we given a table with acids and bases, or do we have to memorize them?  Also, how would we know the difference between hydrogen and deuterium?

Thanks in advance!

"How would you know the electron configuration of an element?"

Don't you know how to fill up the orbitals, by the correct order?

"how would we know the difference between hydrogen and deuterium?"

What do you mean?

[quote="Carcul"]"How would you know the electron configuration of an element?"

Don't you know how to fill up the orbitals, by the correct order?  [i]$ \rightarrow$ No - On the Periodic Table that I used at school, we were given all of the electron configurations.[/i]

"how would we know the difference between hydrogen and deuterium?"

What do you mean?  [i]$ \rightarrow$ Given the Periodic Table, how would we know exactly what deuterium is?[/i]
[/quote]

Thanks!

For most elements, you can determine their electron configurations just by looking at their positions on the periodic table. (The table is divided into s, p, d, and f blocks.) If you don't know how, you should learn because the periodic table on the SAT does not have that.

I also do not understand your second question. Deuterium is an isotope of hydrogen. It is not an element by itself.

One would distinguish deuterium from hydrogen by the number of protons. In symbols,

$ \mathrm{D \equiv ^2_1H}$

Deuterium is also denoted by 'D'.

The same goes for distinguishing tritium (also denoted by 'T')

$ \mathrm{T \equiv ^3_1H}$

For electron configurations, one should remember the three rules:
1. Aufbau principle or the building up principle
2. Hund's rule
3. Pauli exclusion principle
If you can draw the electron configurations, you should be able to write the configurations.
Just remember that there are a few anomalies...(Cr and Cu, for example).

I'd say that the absolute best way to practice for the SAT II Chemistry test would be to practice using the U.S. Chemistry Olympiad Local Section Multiple Choice tests. They are as hard as the hardest questions on the SAT II and harder. Plus, it could be a good way to study for the Chemistry Olympiad...  :maybe:  :maybe:

I somewhat agree with Jdlkem's advice. However, doing questions that you will never even SEE on the SAT is kinda pointless. I guess you should try some of the questions, but don't waste your time trying to solve a titration, or a buffer, or something that will not possibly show up on the test. I highly recommend reading over the Collegeboard's Chemistry SAT Test, as you get used to how they test you.


trigonometry
number theory
advanced fields
advanced fields unsolved

Show that $\cos \alpha \pi$ is either $0, \pm 1, \pm 1/2$ or irrational.

Presumably you meant to include the hypothesis "for $\alpha\in\mathbb{Q}.$"

Please don't [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63269]double post.[/url]

The problem could legitimately be considered either number theory or "advanced fields" but it belongs in only one place.


ratio
geometry
trigonometry
angle bisector
perpendicular bisector

\[ \displaystyle \text {Theorem: } 3=4\\ \text {Proof: }\\ \displaystyle a+b=c\\ (4-3)a+(4-3)b=(4-3)c\\ 4a-3a+4b-3b=4c-3c\\ \text {Add -4c to both sides}\\ 4a-3a+4b-3b-4c=-3c\\ \text {Add 3a to both sides}\\ 4a+4b-3b-4c=3a-3c\\ \text {Add 3b to both sides}\\ 4a+4b-4c=3a+3b-3c\\ \text {Factor out (a+b-c)}\\ 4 \times (a+b-c) = 3 \times (a+b-c)\\ 4=3\\ 3=4 \\ \]
 :rotfl:

\[ \displaystyle \text {Theorem: } 4 = 5\\ \text {Proof: }\\ -20 = -20\\ 16 - 36 = 25 - 45\\ 4^2 - (9 \times 4) = 5^2 - (9 \times 5)\\ 4^2 - (9 \times 4) + \frac{81}{4} = 5^2 - (9 \times 5) + \frac{81}{4}\\\\ (4 - \frac{9}{2})^2 = (5 - \frac{9}{2})^2\\\\ 4 - \frac{9}{2} = 5 - \frac{9}{2}\\\\ 4 = 5\\ \]
 :rotfl:

\[ \displaystyle \text {Does a Dollar equall Ten Cents?}\\ \text{Theorem: } 1\$ = 10 \ \text {cents} \\ \text {Proof: }\\ \text {We know that: }\\ \$1 = 100 \ \text {cents}\\ \text {If,}\\ \$1=1y\\ 1 \ \text {cent}=x\\ \text {Thus,}\\ 1y=100x\\ \text {Divide both sides by} \ 100\\ \frac{y}{100} = \frac{100x}{100}\\\\ \frac{y}{100} = x\\\\ \text {Take square root both side,}\\ \sqrt{ \frac{y}{100}} = \sqrt{x}\\\\ \frac{y}{10} = x \ (\text {Since} \ x = 1 \text {cent})\\\\ \text {Multiply both side by} 10,\\ y = 10x\\ \text {Substitute back in and you get: }\\ \$ 1 = 10 \ \text {cents}\\ \]
 :rotfl:

[hide="I got it"]
$a+b-c=0$!  :D

It is a classical trick!  :rotfl: [/hide]

Well , I also have a funny proof which states that the length any segment from a line is equal to the length of the entire line  :ninja:

Proof :For a triangle $ABC$ . $P$ is on $AB$ such that $\angle ACP=\angle B$ .Then $AP=AB$ .

So here is the working ..

Since $\triangle ACP$ is similar to $\triangle ABC$ , we have

$\frac{\triangle ABC}{\triangle ACP}=\frac{BC^2}{CP^2}$

Since both triangle have same heigth , then the ratio of their area is also the ratio of their base length

$\frac{\triangle ABC}{\triangle ACP}=\frac{AB}{AP}$

$\implies \frac{BC^2}{CP^2}=\frac{AB}{AP}$

or $\frac{BC^2}{AB}=\frac{CP^2}{AP}$ ....... (i)

But from cosine rules we have

$BC^2=AC^2+AB^2-2AC\cdot AB\cos A$
$CP^2=AC^2+AP^2-2AP\cdot AC\cos A$

Substitute them into (i) we get

$\frac{AC^2+AB^2-2AC\cdot AB\cos A}{AB}=\frac{AC^2+AP^2-2AP\cdot AC\cos A}{AP}$

simplify to get $\frac{AC^2}{AB}-AP=\frac{AC^2}{AP}-AB$

Which is also equivalent to $\frac{AC^2-AB\cdot AP}{AB}=\frac{AC^2-AB\cdot AP}{AP}$

Hence $AB=AP$  :D

Can you spot where goes wrong ?  :)

[quote="shyong"]Which is also equivalent to $\frac{AC^2-AB\cdot AP}{AB}=\frac{AC^2-AB\cdot AP}{AP}$ .[/quote]

similar triangles both sides are 0

Somewhat well-known

All triangles are isosceles

If the angle bisector at A and the perpendicular bisector of BC are
parallel, then ABC is isoceles.  On the other hand, if they are not
parallel, they intersect at a point, which we call P, and we can draw
the perpendiculars from P to AB at E, and to AC at F.  Now, the two
triangles labelled "alpha" in this figure have equal angles and share
a common side, so they are congruent.  Therefore, PE = PF.  Also,
since D is the midpoint of BC, it's clear that the triangles labelled
"gamma" are congruent, and so PB = PC.  From this it
follows that the triangles labelled "beta" are congruent by HL Theorem, so we have BE+EA = CF+FA, meaning the the triangle ABC
is isoceles.

[quote="shobber"]This is a famour logic problem. The key is that $P$ can't be inside the triangle.[/quote]

Not only that, E and F lies on the different side of line BC.

Proof that I'm bored right now: this post and the fact that I'm online.
*dry laff *cough cough* :rotfl:  :cursing:

Another good one: $90^\circ=100^\circ$ :D

x^2 = x+x+x+x+x+x.....('x' times)
Thus  d/dx(x^2) = d/dx(x)+d/dx(x)+d/dx(x)+d/dx(x)....('x' times)
Thus  2x = 1+1+1+1+1....('x' times)
Thus 2x=x
Thus 2=1
:rotfl:  :rotfl:  :rotfl:  :rotfl:

[hide]This is a damn simple problem but you can use it to fool the fools[/hide]


When I am as old as my father is now, my son will be seven years older than I am now. At present, the sum of the ages of my father my son and myself is 100. How old am I??

Your age - x years.
Your son's age - x-a years.
Your father's age - x+b years.

After b years, you'll be as old as your father.

We have an equation system:

(x-a)+b=x+7
(x-a)+x+(x+b)=100

From first equation we have:
b=a+7

(x-a)+x+(x+a+7)=100

3x+7=100

x=31

Adding on to the previous post...
You could make a table:
                       PAST            PRESENT          FUTURE
Your age
Father's age
Son's age

Then you set up the equations that way.

:ninja: [b]Thanks for giving the answer to that problem! You are intelligent![/b]


hello,

Let $ a \ge 0$ such that : $ a \minus{} a^3 \plus{} a^5 \ge 3$ , Prove that : $ a^6 \ge 5$

[hide]For $ a\equal{}0$ the initial condition isn't satisfied, so we take $ a>0$

$ (a^5\minus{}a^3\plus{}a)\left(a\plus{}{1\over a}\right)\equal{}a^6\minus{}a^4\plus{}a^2\plus{}a^4\minus{}a^2\plus{}1\equal{}a^6\plus{}1$

But $ (a^5\minus{}a^3\plus{}a)\left(a\plus{}{1\over a}\right)\geqslant 3\cdot 2\iff a^6\geqslant 5$[/hide]


function
logarithms
calculus
calculus computations

(1)
[b]       If f(x) = {(1+x)^1/x - e - ex/2} / x^2  .Then find Lim x-->0 f(x).
means We have to find the LImit of f(x) when x--->0
Here e = 2.71(exponent).[/b]

Did you mean to write: $ f(x) \equal{} \frac{(1\plus{}x)^{\frac{1}{x}} \minus{} e \minus{} \frac{e^x}{2}}{x^2}$ or $ f(x) \equal{} \frac{(1\plus{}x)^{\frac{1}{x}} \minus{} e \minus{} e^{\frac{x}{2}}}{x^2}$?

[quote="akech"]Did you mean to write: $ f(x) \equal{} \frac {(1 \plus{} x)^{\frac {1}{x}} \minus{} e \minus{} \frac {e^x}{2}}{x^2}$ or $ f(x) \equal{} \frac {(1 \plus{} x)^{\frac {1}{x}} \minus{} e \minus{} e^{\frac {x}{2}}}{x^2}$?[/quote]

no no It is e*x
[b]e multiplied by x.[/b]

Can you rewrite the function with a lot of parentheses please?

He meant almost what he wrote: $ f(x) \equal{} \frac {(1 \plus{} x)^{1/x} \minus{} e \plus{} \frac {ex}2}{x^2}.$ There was an incorrect sign in there.

$ \ln\left((1 \plus{} x)^{1/x}\right) \equal{} \frac1x\ln(1 \plus{} x) \equal{} \frac {x \minus{} \frac {x^2}2 \plus{} \frac {x^3}3 \plus{} \cdots}x \equal{} 1 \minus{} \frac x2 \plus{} \frac {x^2}3 \plus{} O(x^3)$ as $ x\to0.$

$ (1 \plus{} x)^{1/x} \equal{} e^{1 \minus{} \frac x2 \plus{} \frac {x^2}3 \plus{} O(x^3)} \equal{} e\cdot e^{ \minus{} \frac x2 \plus{} \frac {x^2}3 \plus{} O(x^3)}$

$ \equal{} e\left(1 \plus{} \left( \minus{} \frac x2 \plus{} \frac {x^2}3 \plus{} O(x^3)\right) \plus{} \frac12\left( \minus{} \frac x2 \plus{} \frac {x^2}3 \plus{} O(x^3)\right)^2 \plus{} O(x^3)\right)$

$ \equal{} e\left(1 \minus{} \frac x2 \plus{} \frac {x^2}3 \plus{} \frac {x^2}8 \plus{} O(x^3)\right)$

$ \equal{} e \minus{} \frac {ex}2 \plus{} \frac {11ex^2}{24} \plus{} O(x^3)$

Put that into the definition of $ f(x)$ and the limit, which is $ \frac {11e}{24},$ will fall right out.

[b][b] Thanks  Sir
ans is 11e/24.

is there is any other  method.
or how can i solve this question without using L-Hospital or expansion method. [/b][/b]


[color=blue]Given two intersecting circles [b]C1,C2[/b].(Let [b]C1[/b] be in the left)[b]C1[/b] intersects [b]C2[/b] in points [b]M,N[/b](let [b]M[/b] be in the top).[b]H[/b] is moving on [b]MN[/b].We draw a line such that passes [b]H[/b] and intersects [b]C1[/b] in points [b]A,B[/b](Let [b]A[/b] be in the left) and intersects [b]C2[/b] in points[b] C,D[/b](Let [b]C[/b] be in the left).
Then we draw a tangent line from [b]C[/b] that intersects [b]C1[/b] in points [b]E,F[/b](let [b]E[/b] be in the top).After it we draw another tangent line from [b]B[/b] that intersects [b]C2[/b] in points [b]H,I[/b](let [b]H[/b] be in the top).
[/color][color=indigo]Is [b]HECB[/b] a inscribed quadrilateral?Prove your notation. :spider:[/color]

I think it must be answer of Intersecting Circles
yeah?
if no what is it?

No,it is another cool problem.But it is more...

[quote="jensen"][color=blue]Given two intersecting circles [b]C1,C2[/b].(Let [b]C1[/b] be in the left)[b]C1[/b] intersects [b]C2[/b] in points [b]M,N[/b](let [b]M[/b] be in the top).[b]H[/b] is moving on [b]MN[/b].We draw a line such that passes [b]H[/b] and intersects [b]C1[/b] in points [b]A,B[/b](Let [b]A[/b] be in the left) and intersects [b]C2[/b] in points[b] C,D[/b](Let [b]C[/b] be in the left).
Then we draw a tangent line from [b]C[/b] that intersects [b]C1[/b] in points [b]E,F[/b](let [b]E[/b] be in the top).After it we draw another tangent line from [b]B[/b] that intersects [b]C2[/b] in points [b]H,I[/b](let [b]H[/b] be in the top).
[/color][color=indigo]Is [b]HECB[/b] a inscribed quadrilateral?[/color][/quote]

On my sketch, neither HECB nor HEAD is an inscribed quadrilateral, not even if the circles $C_1$ and $C_2$ have the same radius. But maybe I misunderstood the problem statement...

  darij

they dont have same radius ($ r_1$ is not equal to $ r_2$ in general)
i have solved it when CB=DA and here is a link to my [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47978]solution[/url]
can you prove your notation?

Khaknejad(I mean jensen :) ) ,darij said that,neither HECB nor HEAD is an inscribed quadrilateral,no matter what the cirlces are .

And about your solution:
http://www.mathlinks.ro/Forum/viewtopic.phpt=47978 : when $AD=BC$
First note that We always have $AD=(AB+BC+CD)> BC$. so we always have $AD \neq BC$.

[quote="jensen"]CDGH and CFED are inscribed quadrilaterals [/quote]
What is [b]G[/b] :?:
And we are not looking for these quadrilateral ,are we?

PS.Notation mean making signs for something(In persian=alamat gozari :) ).
It should be defined,not proved.

please forgive me may be i had a mistake in type i mean $AC=BD$ not $AD=BC$.thank you Arman.


I posted this problem a long time ago and i didn`t saw solution. It is still unsolved. That`s why i post it again.

Consider all arithmetical sequences of real numbers $(x_{i})_{i=1}^{\infty}$ and $(y_{i})_{i=1}^{\infty}$
with the common first term, such that for some $k > 1$,
$x_{k-1}y_{k-1}= 42$, $x_{k}y_{k}= 30$, and $x_{k+1}y_{k+1}= 16$.
Find all such pairs of sequences with the maximum possible $k$.

Good Luck!


[b]Problem 3.[/b] Two points $M$ and $N$ are chosen inside a non-equilateral triangle $ABC$ such that $\angle BAM=\angle CAN$, $\angle ABM=\angle CBN$ and \[AM\cdot AN\cdot BC=BM\cdot BN\cdot CA=CM\cdot CN\cdot AB=k\] for some real $k$. Prove that:

[b]a)[/b] We have $3k=AB\cdot BC\cdot CA$.

[b]b)[/b] The midpoint of $MN$ is the medicenter of $\triangle ABC$.

[i]Remark.[/i] The [b]medicenter[/b] of a triangle is the intersection point of the three medians:
If $A_{1}$ is midpoint of $BC$, $B_{1}$ of $AC$ and $C_{1}$ of $AB$, then $AA_{1}\cap BB_{1}\cap CC_{1}= G$, and $G$ is called medicenter of triangle $ABC$.

[i] Nikolai Nikolov[/i]

a) this is IMOSL 1998:
http://www.mathlinks.ro/Forum/post-124397.html#p124397

[size=59][color=olive]this is my 1000's post[/color][/size]

Nice problem. I will slightly rewrite it:

[color=blue][b]Problem.[/b] Two points M and N are chosen inside a triangle ABC such that < NAC = < MAB and < NBC = < MBA. Assume that $AM\cdot AN\cdot BC=BM\cdot BN\cdot CA=CM\cdot CN\cdot AB=k$ for some real k. Prove that:

[b]a)[/b] We have $3k=BC\cdot CA\cdot AB$.
[b]b)[/b] The midpoint of the segment MN is the centroid of triangle ABC.[/color]

[i]Solution.[/i] We begin, as in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124530#124530]http://www.mathlinks.ro/Forum/viewtopic.php?t=18479 post #3[/url], by proving that the point N is the isogonal conjugate of the point M wrt triangle ABC. Then we continue along the line of the first proof of Theorem 1 in that post, showing that

$\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$

and

$\frac{AM \cdot AN}{AB \cdot AC}+\frac{BM \cdot BN}{BC \cdot BA}+\frac{CM \cdot CN}{CA \cdot CB}= 1$.

Now, $\frac{k}{BC\cdot CA\cdot AB}=\frac{AM\cdot AN\cdot BC}{BC\cdot CA\cdot AB}=\frac{AM\cdot AN}{CA\cdot AB}=\frac{AM\cdot AN}{AB\cdot AC}$, and similarly $\frac{k}{BC\cdot CA\cdot AB}=\frac{BM\cdot BN}{BC\cdot BA}$ and $\frac{k}{BC\cdot CA\cdot AB}=\frac{CM\cdot CN}{CA\cdot CB}$. Hence,

$\frac{AM \cdot AN}{AB \cdot AC}+\frac{BM \cdot BN}{BC \cdot BA}+\frac{CM \cdot CN}{CA \cdot CB}= 1$

becomes

$\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}=1$,

so that $\frac{3k}{BC\cdot CA\cdot AB}=1$, and thus $3k=BC\cdot CA\cdot AB$. Thus, we solved part [b]a)[/b] of our problem. (This was basically Amir.S's solution.)

Now, let S be the midpoint of the segment MN. Then, part [b]b)[/b] requires us to show that this point S is the centroid of triangle ABC.

Using $\frac{k}{BC\cdot CA\cdot AB}=\frac{BM\cdot BN}{BC\cdot BA}$ and $\frac{k}{BC\cdot CA\cdot AB}=\frac{CM\cdot CN}{CA\cdot CB}$, we transform

$\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$

into

$\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$.

Similarly,

$\frac{k}{BC\cdot CA\cdot AB}+\frac{k}{BC\cdot CA\cdot AB}=\frac{\left[CMA\right]+\left[CNA\right]}{\left[ABC\right]}$.

Thus, $\left[BMC\right]+\left[BNC\right]=\left[CMA\right]+\left[CNA\right]$.

Now we need two lemmata (basic olympiad knowledge; if you don't know them, prove them as exercises):

[color=blue][b]Lemma 1.[/b] If P, Q, U and V are four points in the plane such that the points U and V lie in the same halfplane wrt the line PQ, and if M is the midpoint of the segment UV, then $\left[PMQ\right]=\frac12\left(\left[PUQ\right]+\left[PVQ\right]\right)$.[/color]

[color=blue][b]Lemma 2.[/b] If T is a point inside a triangle ABC such that [BTC] = [CTA] = [ATB], then this point T is the centroid of triangle ABC.[/color]

Now, applying Lemma 1 to the points B, C, M and N and the midpoint S of the segment MN, we get $\left[BSC\right]=\frac12\left(\left[BMC\right]+\left[BNC\right]\right)$. Similarly, $\left[CSA\right]=\frac12\left(\left[CMA\right]+\left[CNA\right]\right)$. Thus, $\left[BMC\right]+\left[BNC\right]=\left[CMA\right]+\left[CNA\right]$ yields [BSC] = [CSA]. Similarly, [CSA] = [ASB]. Hence, [BSC] = [CSA] = [ASB]. According to Lemma 2, this yields that the point S is the centroid of triangle ABC. Part [b]b)[/b] of the problem is solved.

  Darij

Similar to above, will just post anyways.

Let $\triangle M_aM_bM_c, \triangle N_aN_bN_c$ be the pedal triangles of $M, N,$ respectively. Observe that $M_bM_c \perp AN$, and so hence $[ AM_bNM_c] = \frac12 AN \cdot M_bM_c.$ However, observe by the Extended Law of Sines in $(AM)$ that $M_bM_c = AM \cdot \sin \angle A$. Therefore, we conclude that:
$$[AM_bNM_c] = \frac12 AN \cdot AM \cdot \sin \angle A = \frac{AM \cdot AN}{AB \cdot AC} \cdot \Delta,$$
where $\Delta$ denotes the area of $\triangle ABC.$ This implies that the area of $AM_bNM_c$ is one-third of the area of $\triangle ABC$, and we can obtain analogous results for $AN_bMN_c, BM_aNM_c, BN_aMN_c, CM_aNM_b, CN_aMN_b.$ Adding up $\frac{[AM_bNM_c]}{\Delta}, \frac{[BM_aNM_c]}{\Delta}, \frac{[CM_aNM_b]}{\Delta}$ gives us that:
$$\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = 1,$$
which is actually just [b] 1998 G4 [/b] as pointed out above. Anyhow, this trivializes part [b] (a) [/b].

Let's now turn to part [b] (b) [/b]. Let $X$ be the midpoint of $MN.$ Observe that
$$2[\triangle BXC] = [\triangle BMC] + [ \triangle BNC] = [\triangle BMN_a] + [\triangle CMN_a] + [\triangle BNM_a] + [\triangle CNM_a].$$
Now, notice that $[\triangle BNM_a] = \frac12 BN \cdot BM_a \cdot \sin \angle NBM_a = \frac BM \cdot BN_c \cdot \sin \angle MBN_c = [\triangle BMN_c],$ where we used that $BM_a = BM \cos \angle MBM_a, BN_c = BN \cos \angle NBN_c.$ Therefore, we have that
$$2 [\triangle BXC] = [\triangle BMN_a] + [\triangle CMN_a] + [\triangle BMN_c] + [\triangle CMN_b],$$
and so combining these triangles in pairs gives us that
$$2[\triangle BXC] = [BN_aMN_c] + [CN_aMN_b],$$
which we know is $\frac23 \Delta.$ Therefore, we have that $[\triangle BXC] = \frac13 \Delta,$ with analogous results for $[\triangle CXA], [\triangle AXB].$ This is easily enough to imply that $X$ is the centroid of $\triangle ABC$, as desired.

$\square$


Let $ a,b,c$ be positive real numbers such that $ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \equal{}1$ then find the maximum value of the follow expression
\[ P\equal{} \frac {a^{0.5}} {a^3\plus{}b^2}\plus{}\frac {b^{0.5}} {b^3\plus{}c^2}\plus{}\frac {c^{0.5}} {c^3\plus{}a^2}\]

by CS and AM-GM, $ \frac {a}{a^{3}} \plus{} \frac {1}{b^{2}}\ge\frac{4a^{0.5}}{a^{3}\plus{}b^{2}}$

applying this to the other two yields maximum $ \frac{1}{2}$

[quote="earldbest"]by CS and AM-GM, $ \frac {a}{a^{3}} \plus{} \frac {1}{b^{2}}\ge\frac {4a^{0.5}}{a^{3} \plus{} b^{2}}$

applying this to the other two yields maximum $ \frac {1}{2}$[/quote]

When does equality occur?

[quote="nguoivn"]It's obviously trues by Am-Gm  :wink:[/quote]

Can you share your proof, nguoivn?

[quote="RainL"]When does equality occur? [/quote]

Oh, sorry... Apparently, the method I used yields no equality (should there be?).  :maybe:

[quote="nguoivn"]It's obviously trues by Am-Gm  :wink:[/quote]

How about your proof by AM-GM, nguoivn? I want to see your solution. :)

what's your solution, I would love to see how equality occurs  :lol:


In a state with a $ 5\%$ sales tax, Joe buys a shirt for $ \$24.00$. Two weeks later, the same shirt goes on sale for $ 20\%$ less than its original price. How many dollars, including sales tax and rounded to the nearest hundredth, would Joe have saved by waiting to purchase the shirt on sale?

$ 20\% \cdot 24.00\cdot 105%$
is the answer.
simplified you get
5.04

Edit: thanks jxl I fail at latex

[quote="AceOfDiamonds"]$ 20\%\cdot24.00\cdot105\%$
is the answer.
simplified you get
$ 5.04$[/quote]
Fixed.


linear algebra
matrix
inequalities
superior algebra
superior algebra solved

Having $n \in N ,n>1$,and $A \in M_n(R)$,such that :$det(A+X)=detA+detX, \forall X \in M_n(R)$.
Prove that $A=O_n$.

From $\det(2A)=2^n\det A=2\det A$ we get $\det A=0$, so $\det(A+X)=\det X,\ \forall X\in\mathcal M_n(\mathbb R)\ (*)$.

Assume $A$ has a non-null column. We can choose $X$ so that its corresponding column is null, and the other columns add up to the corresponding columns of $A$ to give columns linearly independent with each other and with the one column we considered above. In other words, we can make $A+X$ invertible, although $X$ is non-invertible. This contradicts $(*)$, of course.

we can also use the fact that matrix A with rank $r$ can be represented as $A=P^{-1}I_{r}Q^{-1}$, where $I_r$-$n\times n$ matrix ${I_{r}}\oplus{O_{n-r}}$, if we take in $det(A+x)=detX$ $X=P^{-1}I_{n}Q^{-1}$ we obtain $det(I_r+I_n)=det(I_n)$, thus $2^r=1$ and it follows that $r=0$, which means that our matrix has zero rank.

More beautiful and general.

Having $n \in N,n>1$ ,and $A \in M_n(R)$,such that $det(A+X) \le detA+detX, \forall X \in M_n(R) $.
Prove that $A=O_n$.

Actually, that argument I had in post number $3$ works here as well. Assuming that $A$ has a non-null column $c_i$, take $X$ so that its $i$'th column is $-c_i$. Since $-c_i$ is non-null, the rest of the elements of $X$ can be chosen to make the determinant anything we want, meaning that $\det(A+X)=0$, and we can make $\det X>-\det A$. We don't even need the initial step showing that $\det A=0$.

$X=\frac{1}{2}A$ in the inequality $(3^n-2^n-1)det(A)\leq 0$ this imply  $det(A)\leq 0$

If n is even
$X=-A$ in the inequality we have $0\leq det(A)+det(-A)$ so
$det(A)\geq 0$

If n is odd
$X=-1/2A$ in the inequality we have $(2^{n-2}-1)det(A) \geq 0$
$det(A)\geq 0$

$det(A)=0$

the inequality reduce to (1) $det(A+X)\leq det(X)$

$A=P^{-1}J_rQ^{-1}$ where $r=rank(A)$
choose $X=P^{-1}Q^{-1}$ in (1), we have

$det(J_r+I_n)\leq 1$

$2^r \leq  1 $  wich is true only for $r=0$

$A=0$


[quote="Silverfalcon"]Hi,

Thank you lot of people for answering my poor written spanish writings. :D And I'm trying to learn new thing today. I hope this is right. Thanks for a lot of your help.

This is basically conversation thing I have from my spanish class about eating on the restaurant and stuffs.

C = El camarero
T = Tu

C: Buenas tardes. ¿Que le puedo traer?
T: Buenas tardes, pues, yo quisiera la carne de res.
C: Excelente. La sopa del dia es sopa de legumbres.
T: Muy bien, y una sopa de legumbres.
C: ¿Prefiere las zanahorias o el maiz?
T: Prefiero las zanharorias.
C: Quiere una ensalada?
T: Si, por favor.
C: ¿Y para tomar? ¿Que le puedo traer?
T: Me gustaria la limonada.
C: ¿Algo mas?
T: No, esta bien.

Also, one question.

If you say 1,256, would that be equal to mil doscientos cincuenta y seis?

4789 = cuatro mil setecientos ochenta y nueve
7785 = siete mil setecientos ochenta y cinco

Is this right? Because I have to hear these numbers and right it down on my paper tomorrow...

Listening to spanish is hard to me... Dunno why.. :( Gracias. MUCHAS GRCIAS!!! :lol:  :lol:[/quote]

Yes, they are right!

Best regards,