A quadrilateral that has consecutive sides of length $ 70, 90, 130$ and $ 110$ is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length 130 divides that side into segments of length $ x$ and $ y$. Find $ |x\minus{}y|$.

[hide]Uh, label the quadrilateral $ ABCD$ such that $ A$ is between $ 70$ and $ 90$, $ B$ is between $ 90$ and $ 130$, and etc. Let $ O$ be the center of the incircle.

Looking at the quadrilateral, if we draw the radii of the incircle to $ AB,BC,CD,$ and $ DA,$ and labeling the points of intersection $ E,F,G,H$ respectively, we get $ AE \equal{} AH, DH \equal{} DG,$ and etc.  If we label $ BE \equal{} BF \equal{} x$, then $ CF \equal{} CG \equal{} 130 \minus{} x$, and $ DG \equal{} DH \equal{} x \minus{} 20$, and $ AH \equal{} AE \equal{} 90 \minus{} x$.

Now, since ABCD is cyclic, $ \angle DAB \equal{} \angle GOF$ because $ \angle DAB \equal{} 180 \minus{} \angle DCB$.  But, $ \angle GOF \equal{} 180 \minus{} \angle DCB$.

Now, if we connect $ AO$, $ \triangle AHO \equal{} \triangle AEO$ because the hypotenuse and the legs are congruent.  This implies that $ \angle HOA \equal{} \angle EOA$ Similarly, if we connect $ CO$, $ \triangle COF \equal{} \triangle CGO$.  Likewise, $ \angle GCO \equal{} \angle FCO$

So, we have two similar triangles by AA similarity, $ \triangle CFO$ and $ \triangle OEA$. Using ratios, we get that $ \frac {r}{90 \minus{} x} \equal{} \frac {130 \minus{} x}{r}$.

So, $ r^2 \equal{} 90(130) \minus{} 220x \plus{} x^2$.

Now, repeat for the the other two vertices to get $ r^2 \equal{} x^2 \minus{} 20x$.

Solving for x, we get $ x \equal{} \frac {117}{2}$.  So the answer is $ \boxed{13}$.[/hide]

Uh, if someone could tell me if my solution is correct/incorrect, it'd be nice (I know if no one objects to it, I assume it's correct, but it's be nice if someone just posted, "It seems fine to me"...so I know that at least SOMEONE agrees with me.)

Hmm, I followed it, makes sense...I didn't really go through the ratio solving stuff so I'll assume it's correct, 13 is one of the choices haha. Nice solution  :P

That is the nicest solution of course. We can alternately solve the problem using areas and a bit of trig.


if someone can help me with the oim problem 5 please write the solution or
or maybe some ideas
thank u


function
algebra
domain
ratio
inequalities
arithmetic sequence
geometric sequence

There are some questions I need to ask for algebra II homework I have today..

What does Composition of Function means? It's like [i]Suppose f and g are functions such that the range of g is a subject of the domain of f. Then the composite function, f of g, can be described by the equation f of g of x.

[/i]Another questions is: [i]Inverse Functions and relations: Two functions, f and g, are inverse functions if and only if both their compositions are the identity function. That is, f of g of x = x and g of f of x= x.[/i]

well, i'm taking algebra II right now and i know the answer to the second question. the inverse of a function is so that if you submit the inverse in for x in the original equation, you will get f(x)=x. The usual method is reversing the x and y in the equation. For example, if you needed to find the inverse of y=8x-16, you flip x and y
x=8y-16
8y=1+16
y=x/8+2
Check by plugging that into the x for the original equation.
y=8(x/8+2)-16
y=x+16-16
y=x

The first explanation is just a fancy way of saying that f of g is f(g(x)).

Example: If f(x) = x^2 + 2, and g(x) = x + 3, then f(g(x)) = (x+3)^2 + 2 = x^2 + 6x + 11.

yep.. i got it... wasn't complicate as i thought, though...
Um.. another question.. i heard arithmetic sequence is constant increase or decrease such as adding or subtracting while geometric sequence is constant multiplication or division... Then what's geometric mean and arithmetic mean?
And can anyone post the formula for geo and arthimetic sequence formula?
I have geo formula which is:
L = F * M^(N-1) where L is the last term, F is the first term, M is the constant ratio, N is the number of terms..
But I don't know arthimetic formula so... I'm just studying extra on these stuffs which my math class doesn't cover on...

Arithmetic mean is  :Sigma: /n, where n is the number of terms.  Geometric mean is  :prod: ^ (1/n).  And just for kicks, harmonic mean is n/(sum of the recipricols).  At least, I'm almost completely sure of that.
It's also worth noting that AM>GM>HM
So, for the set {1,2,3,4,5}
Arithmetic is: (1+2+3+4+5)/5=3
Geometric is: (1*2*3*4*5)^(1/5), which I don't feel like calculating
Harmonic is:5/(1/1+1/2+1/3+1/4+1/5), which I also don't feel like doing.


Okay, I thought this problem was extremely simple, but I guess I was being stupid. Can someone help me realize where I went wrong? Here's the problem:

What is the maximum rectangular area in square meters that can be enclosed by 240 meters of fencing if the fence were to be constructed next to a river so that no fence was needed on the river boundary?

[hide="What I did"] I made a square, since in previous problems of maximum rectangular area the square always gave the most. One of the sides was already provided by the river, so each side is 240/3 = 80. $ 80^2 \equal{} \boxed {6400}$   [/hide]

They say the answer is...
[hide] 7200 [/hide]

How did they arrive at this answer? Thanks in advance.

Where you went wrong was in assuming that the rectangle has fixed perimeter. In fact the perimeter varies based on the length of the fence on the side oposite the river. We are trying to find the maximum value of $ xy$ given that $ 2x\plus{}y\equal{}240$.

Using AM-GM:

$ \sqrt{2xy} \le \frac{2x\plus{}y}{2}$

$ \sqrt{2xy} \le 120$

$ 2xy \le 14400$

$ xy \le 7200$

So the maximum area is 7200. Equality occurs when $ 2x\equal{}y$, or put another way, the side oposite the river is 120 and the other two sides are 60.

You could just call the sides perpendicular to the river $ x$ and the other side $ 180\minus{}2x$. Multiplying the two, you can then use $ \frac{\minus{}b}{2a}$ to find the max.


If x1,x2,..,xn are real numbers and n natural >=3 then the following inequality holds:

x1^2+...xn^2-x1x2-x2x3-...-xnx1>=n/(2(n-1)) * max((x1-x2)^2, (x2-x3)^2,..,(xn-x1)^2));


ratio
complex analysis
complex analysis unsolved

I want to show that for $ 4$ distinct points $ z_1,z_2,z_3$ and $ z_4$ in $ \mathbb{C}_{\infty}$ we have that the cross ratio $ (z_1,z_2,z_3,z_4)$ is a real number if and only if all points lie on a circle and the proof goes as follows:

Let $ S: \mathbb{C}_{\infty} \to \mathbb{C}_{\infty}$ be defined by $ Sz \equal{} (z,z_2,z_3,z_4)$ then $ S^{ \minus{} 1}(\mathbb{R})$= all the $ z$ such that $ (z_1,z_2,z_3,z_4)$ is a real number

Then they say: "we will be finished if we can show that the image of $ \mathbb{R}_{\infty}$ under a Mobius transformation is a circle"

Not really following there

(conway page 49)

If the image of $ \mathbb{R}_\infty$ under $ S^{\minus{}1}$ is a circle $ \Gamma$, then it follows that the image of $ \Gamma$ under $ S$ is $ \mathbb{R}_\infty$.  Since Möbius transformations are bijective on $ \mathbb{C}_\infty$, you have the result you're looking for if $ \Gamma$ is in fact the circle determined by $ z_2, z_3, z_4$.  But it's trivial to verify that these points lie on $ \Gamma$, and since these points are distinct, it uniquely determines $ \Gamma$.

Conway seems to frequently leave out rather large steps like this to see if the reader is paying attention. :)


I had been solving this problem for quite a while and I gave up, so I post it here. Please help. Thanks.

PROBLEM

Arrange the following in descending order.

A. 3^60         B. 5^48          C. 6^36             D. 7^24

[hide="Hint"]Remember that $(a^b)^c = a^{bc}$[/hide]

Yes, exactly, everytime you see a problem like this always remember the rules for exponents, and in this case, you should find a greatest common factor between the exponents, and divide them by it, and raise it to that power. Hmoonster did it perfectly :first:

Is this problem from a competition of some sort of is it a problem your math teacher gave you?

I've gotten a few of these from the SDMC competitions before.


Find all functions $ f: R\rightarrow R$ satisfying
$ f(x\plus{}f(y))\equal{}x\plus{}f(f(y))$
for all real numbers x,y with the additional constraint that $ f(2004)\equal{}2005$

Denote $ f(0)\equal{}c$. Then $ y\equal{}0$ implies $ f(x\plus{}c)\equal{}x\plus{}f(c)$ for all $ x$. Hence $ x\equal{}\minus{}c$ yields $ c\equal{}\minus{}c\plus{}f(c)$ and thus $ f(c)\equal{}2c$. Hence $ f(x\plus{}c)\equal{}x\plus{}2c$. Now $ x\plus{}c\mapsto x$ implies $ f(x)\equal{}x\plus{}c$. Since $ f(2004)\equal{}2005$ we conclude that $ c\equal{}1$ and thus $ f(x)\equal{}x\plus{}1$.


I've only found two cases so far and I'm not sure they are correct.

Case 1: Each person chooses a different combination of categories.

This suggests that there are 10! possibilities for the team members to choose their three categories.  As every category choice is different, parity is preserved.

Case 2:  There are essentially two groups of five people, and in each group, there are three people in each category.  In other words, two people make the same choice for the category.

This implies that we still have 10! choices for categories, but there are some combinations.  Since each of the five categories is chosen twice, we need to divide 10! by 2! five times, or by 32.

Thus, the total probability is 10! + 10!/32, or  3,742,200.

Feel free to prove me wrong.

Well, there's a little button on the bottom right of the post where you can delete a post, but you need a mod to delete a whole topic, I think...


geometry
trigonometry
perpendicular bisector
angle bisector
trig identities
Law of Sines
AMC

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects $ \angle ABC$. If AD=9, DC=7, what is the area of ABD?
A. 14
B. 21
C. 28
D. $ 14\sqrt{5}$
E. $ 28\sqrt{5}$
Find the number of ordered pairs of real numbers (a,b) such that $ (a\plus{}bi)^{2002}\equal{}a\minus{}bi$
A. 1001
B. 1002
C. 2001
D. 2002
E. 2004

[hide="2nd one"]Write $ z \equal{} a\plus{}bi$ and multiply both sides by $ z$. We get $ z^{2003} \equal{} |z|^2$. Taking the magnitude of both sides shows $ |z| \in \{0,1\}$. If $ |z| \equal{} 0$ then $ z \equal{} 0$. If $ |z| \equal{} 1$ then $ z^{2003} \equal{} 1$ and $ z$ is a $ 2003$rd root of unity; there are $ 2003$ of these. So $ 2003 \plus{} 1 \equal{} \boxed{2004}$.[/hide]

That's what I got, using
[hide]Exponential form[/hide], but the AMC website says the answer is
[hide]B[/hide].  :huh:

[quote="E^(pi*i)=-1"]the AMC website says the answer is
[hide]B[/hide].  :huh:[/quote]
Thats what I've confused on too...

[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2002-12a/02answers12a.html]The official answers[/url] say
[hide]
E
[/hide]
actually. Perhaps you read the answer to question 25 by an accident?

Edit: (Post #200  :lol:). I just realized that the topic mistakenly says the 'last 2 problems', but in fact the two problems are #23 and #24 respectively (I have the packet right in front of me). That should clarify things.

The first one:
[hide]
Let the midpoint of $ BC$ be $ M$.  Ley $ BM \equal{} x$ and $ AB \equal{} y$.  Then by the Angle Bisector Theorem, $ \frac {2x}{7} \equal{} \frac {y}{9}$, or $ 18x \equal{} 7y$.

Now, clearly $ BDC$ is isosceles, because of its perpendicular bisector dropped from $ D$.  Thus, $ \angle CBD \equal{} \angle DCB$.  Let this angle be $ \theta$.  Then, by the Law of Sines on triangle $ ABC$, we have $ \frac {\sin \theta}{y} \equal{} \frac {\sin 2\theta}{9 \plus{} 7}$.  This becomes $ y \sin 2\theta \equal{} 16 \sin \theta$.  Expanding using the Double-Angle Identity for sine, we have $ 2y \sin \theta \cos \theta \equal{} 16 \sin \theta$, or $ y \cos \theta \equal{} 8$.  From right triangle $ DMC$, we see that $ \cos \theta \equal{} \frac {x}{7}$.  Thus, $ \frac {yx}{7} \equal{} 8$, giving us $ xy \equal{} 56$.

Now that we have two equations, we can solve for positive $ x$ and $ y$.  We get $ y \equal{} 12$ and $ x \equal{} \frac {14}{3}$.

Now we can find the area of triangle $ ABD$.  This is $ \frac {(AB)(DB) \sin \angle ABD}{2} \equal{} \frac {7y \sin \theta}{2}$.  We have that $ \sin \theta \equal{} \frac {\sqrt {5}}{3}$ from the Pythagorean identity, and we know $ y \equal{} 12$  so our final area is EDIT (oops) $ 14\sqrt {5}$.[/hide]

[quote="archimedes1"]The first one:
[hide]
Let the midpoint of $ BC$ be $ M$.  Ley $ BM = x$ and $ AB = y$.  Then by the Angle Bisector Theorem, $ \frac {2x}{7} = \frac {y}{9}$, or $ 18x = 7y$.

[hide="diff answer"]Now, clearly $ BDC$ is isosceles, because of its perpendicular bisector dropped from $ D$.  Thus, $ \angle CBD = \angle DCB$.  Let this angle be $ \theta$.  Then, by the Law of Sines on triangle $ ABC$, we have $ \frac {\sin \theta}{y} = \frac {\sin 2\theta}{9 + 7}$.  This becomes $ y \sin 2\theta = 16 \sin \theta$.  Expanding using the Double-Angle Identity for sine, we have $ 2y \sin \theta \cos \theta = 16 \sin \theta$, or $ y \cos \theta = 8$.  From right triangle $ DMC$, we see that $ \cos \theta = \frac {x}{7}$.  Thus, $ \frac {yx}{7} = 8$, giving us $ xy = 56$.

Now that we have two equations, we can solve for positive $ x$ and $ y$.  We get $ y = 12$ and $ x = \frac {14}{3}$.

Now we can find the area of triangle $ ABD$.  This is $ \frac {(AB)(DB) \sin \angle ABD}{2} = \frac {7y \sin \theta}{2}$.  We have that $ \sin \theta = \frac {\sqrt {5}}{3}$ from the Pythagorean identity, and we know $ y = 12$  so our final area is $ 28\sqrt {5}$.[/hide][/quote]

Since the perpendicular bisector meets at D we know that $ BD = CD$ so $ \triangle{BDC}$ is isosceles. Now that means that BD=7. By the angle bisector theorem

$ \frac {BC}{AB} = \frac {CD}{AD} = \frac {7}{9}$

Now let $ BC = 7x$ and let $ AB = 9x$. Now we also know that $ \angle{ACB} = \angle{DBC} = \angle{DBA} = \theta$ . Also $ \angle{ADB} = 2\theta$ since $ \angle{BDC} = 180 - 2\theta$. Notice that $ \angle{ABC} = 2\theta$ as well so $ \triangle{ABC}~\triangle{ABD}$ So

$ \frac {7}{9x} = \frac {7x}{16}$

Solving that gives that $ x = \frac {4}{3}$ so $ AB = 9x = 12$

Now $ \triangle{ABD}$ has side lengths 9 7 and 12 so by Heron's the answer is $ 14\sqrt {5}$ or D[/hide]

[hide=" 23"]$ BD \equal{} 7$ from isosceles congruent right triangles.

Let E be base of perp bisector. Let $ BE \equal{} EC \equal{} x$. Let $ \angle ABC \equal{} 2\theta$.

From Angle Bisectors: $ AB \equal{} 18x/7$.

From Stewart's (hey, it ain't bad, everything ends up as being squares): $ x \equal{} 14/3$.

Then $ ED \equal{} 7\sqrt {5}/3$. Thus $ \sin \theta \equal{} \sqrt {5}/3$.

From area formula for triangle $ ab\sin c/2$, $ [ABD] \equal{} 14\sqrt {5}/3$. D[/hide]

[hide="24"]$ z \equal{} |z|/z^{overbar}$

Write as $ re^{ix}$ form. We want number of solutions (r,x) for

$ r^{2002}e^{i2002x} \equal{} r^2e^{ \minus{} ix}$

or $ r^{2000} \equal{} e^{ \minus{} i2003x}$. So solutions x are x such that the RHS is 1 in [0,2\pi), and for any of these, r=1 is the only positive solution. These are $ 2\pi k/2003$ for k = 0,...,2002. 2003 solutions. or r=0, then a=b=0, so 2004 solutions. E?[/hide]

btw, #25 of this contest was much more easier than the two questions posted here

[quote="pianoforte"][quote="unimpossible"]for #23 if you dont know much trig you could also express everything in terms of AB and disect the triangle into 3 parts, add them up and set them equal to the area of the entire trianlge. and you can find x, which leads you to area of triangle ABD

(a LOT of messy algebra though)[/quote]

Uh, eww... hey, mine practically didn't use any trig at all[/quote]

Mine used no trig, just one theorem which im sure most people over here know about.


Professor Math wants to formulate a math potion. He mixes in 2.5 ounces of [i]x[/i], 1.75 ounces
of [i]y[/i], and 3 ounces of [i]z[/i]. However he notices that for every 1.5 ounces of [i]z[/i] he puts,
.5 of an ounce of [i]x[/i] evaporates (assume none of it condenses), and 25% of an ounce of [i]y[/i] becomes a part of [i]z[/i]. The flask he is using can hold 8 ounces.

Using the provided, what percent of the potion would be left if professor math drinks 5.75 ounces of the
potion?

*note some details are irrelevant.
EDIT: It's supposed to be 5.75 ounces, not 6.75

[quote="evilhamster"]There are 3 ounces of z, so the amount of x goes down by 1. Thus we are left with 1.5 ounces of x, 1,75 ounces of y, and 3 ounces of z. Add them up, and there are 6.25 ounces...?
Did you have a typo?[/quote]
whups it is supposed to be 5.75 ounces drunk, sorry!
That pretty much gave you the answer  :wink:

Finishing off the solution:
If there are $ 6.25$ ounces total and $ 6.25\minus{}5.75\Rightarrow 0.5$ ounces left, then our answer is
\[ \frac{0.5}{6.25}\equal{}\boxed{8\text{ percent}}
\]

hi bioharzardman and the answer is the same as the guy after the guy above me so it's the same thing like .5/6.25=8 percent

So, after 3 ounces of z 1 ounse of x evaporates. So there is 1.5 ounces of X left. It doesnt matter about 25% of Y going to Z because it just matters overall.. So... 1.5+1.75+3= 6.25 He then drinks 5.75 so he has .5 ounces left...  or 8%  :ninja:


geometry
geometric transformation
rotation
linear algebra
matrix
rotation matrix

I found my value of t which is square root of 117, but I have no idea why we use the rotation matrix.

I do not find any pattern for general formula when multiplying the rotation matrices by themselve several times

(This shouldn't be a new thread)

Don't think algebra; write the rotation using trigonometry.
The angle isn't a nice angle, so we don't get a nice algebraic formula.

Using the rotation matrix, we obtain

For M,

a11= t cos x
a12 = -sin x
a21 = sin x
a22 = cos x

For M^2,

a11= (t^2)(2cos^2 x - 1)
a12= -(t^2)sin 2x
a21= (t^2)sin 2x
a22= (t^2)(2cos^2 x - 1)

For M^3,

a11 = (t^3)(4cos^2 x -3)
a12= (t^3)(sin x)(4sin^2 x - 3)
a21= -(t^3)(sin x)(4sin^2 x - 3)
a22= (t^3)(4cos^2 x -3)

I don't see any pattern to generate formulas for M^n, please help


$a,$ $b$ and $c$ are lengths of the sides of a right-angled triangle ( $c$  - length of the hypotenuse ).
Prove that $c^{3}+(a+b)(c^{2}+ab)\geq(2+3\sqrt2)abc.$

Nice!  :lol:

Inequality is homogenous, so we can assume that $c=1$.
Now, substituting $x=a+b$ and $y=ab$, and replacing $1$ by $x^{2}-2y$, the inequality reduces to \[1+x(1+y) = x^{2}-2y+x(1+y) \geq (2+3\sqrt{2})y\] Now, we know that $x^{2}\geq 4y$, so we only have to prove the following inequality: \[2\sqrt{y}(1+y) \geq 3\sqrt{2}y\] And this reduces to: \[\sqrt{y}(2y-3\sqrt{2}\sqrt{y}+2) \geq 0\] The second term is a quadratic equation, with zeros at $\frac{1}{\sqrt{2}}$ and $\sqrt{2}$, but this is always nonnegative, because $y \leq \frac{1}{2}$


Given segments of lenght $a$ and $b$, construct a segment of length $\sqrt{ab}$.

[hide]Perhaps this is kind of long but here's my solution:

Notice how $\sqrt{ab}$ is the geometric mean between $a$ and $b$.  That reminds me of a right triangle with an altitude to the hypotenuse.  Also, any triangle inscribed inside a circle with one side as the diameter is right.

So, I would take the two segments, join their ends to make one long segment, draw a circle with the long segment as the diameter, and draw another segment from the point of intersection of the original segments directly up until it reaches the circle.

That last segment would have a length of $\sqrt{ab}$.[/hide]


geometry
circumcircle
ratio
analytic geometry
cyclic quadrilateral
geometry proposed

Let a triangle $ABC$ and let $AD,BE,CZ$ three cevians which intersect in $H$ in the interior of $ABC$ .Let $M$ the midpoint of $EZ$ and $P$ the point of intersection of the lines $EZ$ and $BC$ . The line $PH$ intersects the lines $AB$ and $AC$ at the points $K$ and $L$ respectively .Prove that the line $AM$ and the parallels from $K$ and $L$ to $CZ$ and $BE$ respectively ,intersect in one point .

I refuse to call the foot of the C-cevian Z and I will call it F. Using a parallel projection, project the $\triangle ABC$ with the interior point H into an acute $\triangle A'BC$ with the interior orthocenter H'. That is, project 2 of the concurrent cevians AD, BE, CF of $\triangle ABC,$ say AD, BE, into altitudes A'D, BE'. The 3rd concurrent cevian CF' follows the suit. Midpoint M of EF goes into the midpoint M' of E'F' and parallels to CF, BE through K, L go into parallels to CF', BE' through K', L'. Then drop the primes from the notation.

Let the parallelels to CF, BE through K, L meet at X. Let (O) be the circumcircle of the projected $\triangle ABC$ and let $R \in BC$ be the foot of the A-symmedian. Let AO meet BC at X and (O) again at Y. Let the A-altitude AH meet EF at Q and (O) again at Z, ZH = 2 DH. AO, AH are isogonals with respect to $\angle A,$ hence $YZ \parallel BC.$ The circumcircle of $\triangle AEF$ passes through H, because the quadrilateral AEHF cyclic on account of the right angles. The oposite similarity of $\triangle AEF \sim \triangle ABC$ yields $\frac{AH}{AQ}= \frac{AY}{AX}= \frac{AZ}{AD}.$ Then

$1+\frac{DQ}{AQ}= \frac{AD}{AQ}= \frac{AZ}{AH}= 1+\frac{ZH}{AH}= 1+2\ \frac{DH}{AH}$

so that the cross-ratio $\frac{DQ}{AQ}\cdot \frac{AH}{DH}= 2.$ This cross-ratio is preserved in the central projection of AD on AB, AC from P, therefore

$\frac{CE}{AE}\cdot \frac{AL}{CL}= \frac{BF}{AF}\cdot \frac{AK}{BK}= 2$

$\frac{b-AE}{AE}= \frac{CE}{AE}= 2\ \frac{CL}{AL}= 2\ \frac{b-AL}{AL}$

$AL = \frac{2b \cdot AE}{b+AE},\ \ AK = \frac{2c \cdot AF}{c+AF}$

$\frac{AK}{AL}= \frac{c}{b}\cdot \frac{AF}{AE}\cdot \frac{b+AE}{c+AF}= \frac{b+AE}{c+AF}$

Let normals to AB, AC from the foot R of the A-symmedian meet AB, AC at U, V. The right angle triangles $\triangle RBU \sim \triangle CBF,\ \triangle RCV \sim \triangle BCE$ are similar. R divides BC in the ratio $\frac{RB}{RC}= \frac{c^{2}}{b^{2}},$ so that $RB = \frac{ac^{2}}{b^{2}+c^{2}},\ RC = \frac{ab^{2}}{b^{2}+c^{2}}.$ It follows that

$\frac{BU}{BF}= \frac{RB}{BC}= \frac{c^{2}}{c^{2}+b^{2}},\ \ \frac{CV}{CE}= \frac{RC}{BC}= \frac{b^{2}}{c^{2}+b^{2}}$

$\frac{AU}{AV}= \frac{c-BU}{b-CV}= \frac{c(b^{2}+c^{2})-c^{2}\cdot BF}{c(b^{2}+c^{2})-b^{2}\cdot CE}=$

$= \frac{c(b^{2}+c^{2})-c^{2}(c-AF)}{c(b^{2}+c^{2})-b^{2}(b-AE)}= \frac{c}{b}\cdot \frac{b^{2}+c \cdot AF}{c^{2}+b \cdot AE}$

Using the similarity $\triangle AEF \sim \triangle ABC,$ we can switch $b \cdot AE = c \cdot AF$ in the numerator and denominator, which leads to

$\frac{AU}{AV}= \frac{c}{b}\cdot \frac{b^{2}+b \cdot AE}{c^{2}+c \cdot AF}= \frac{b+AE}{c+AF}= \frac{AK}{AL}$

As a result, the triangles $\triangle AUV \sim \triangle AKL$ are centrally similar with similarity center A and $UV \parallel KL.$ But then the triangles $\triangle RUV \sim \triangle XKL$ are also centrally similar, having parallel sides. As the lines $KU \equiv AB,\ LV \equiv AC$ meet at A, A is the similarity center, so that RX passes through A. In other words, X lies on the A-symmedian of $\triangle ABC,$ which cuts the side EF of the oppositely similar $\triangle AEF$ at its midpoint M.

If H is outside of $\triangle ABC,$ it is possible to transform it with a parallel projection into a triangle $\triangle A'B'C'$ with orthocenter H' only if H is not in one of the plane sectors with the triangle excircles. But the problem proposition holds regardless of the position of H and if H is in a plane sector with one of the excircles, the above proof fails.

WLOG, assume H is in the plane sector with the excircle $(I_{c})$ opposite to C. Consider the $\triangle EBC$ with the interior point F and project it to an acute $\triangle E'BC$ with the interior orthocenter F'. Then drop the primes from the notation. From the solution in the previous post, we know the cross-ratio $\frac{PQ}{EQ}\cdot \frac{EF}{PF}= 2,$ which is preserved in the central projection of EP to AD, so that $\frac{DQ}{PQ}\cdot \frac{AH}{DH}= \frac{PQ}{EQ}\cdot \frac{EF}{PF}= 2,$ but now we do not need this.

The angle $\angle EAF$ is now right, because $BF \equiv BA$ is the B-altitude of $\triangle EBC.$ Likewise, the angle $\angle KXL$ is now right, because BE is the side and $CF \equiv CH$ the C-altitude of $\triangle EBC,$ $CF \perp BE,$ and $KX \parallel CF,\ LX \parallel BE.$ From the cyclic quadrilateral AKXL with right angles at the vertices A, X, $\angle KAX = \angle KLX.$ As $LX \parallel BE$ and $H \in BE,$ the angles $\angle KLX = \angle PHB$ are equal. The triangles $\triangle PBH \sim \triangle EBC$ are oppositely similar, because P, H are the feet of the E-, B-altitudes of $\triangle EBC,$ so that $\angle PHB = \angle BCE.$ The hypotenuse $EF \equiv EP$ of the right triangle $\triangle AEF$ is the E-altitude of $\triangle EBC,$ $EF \perp BC.$ Therefore, the A-altitude of $\triangle AEF$ is parallel to BC, hence it forms an angle with AE equal to $\angle BCE.$ This means that the A-altitude of $\triangle AEF$ is isogonal of AX with respect to the angle $\angle EAF,$ so that AX passes through its circumcenter, identical with the midpoint M of its hypotenuse EF.

[size=117][color=darkred]Indeed, very, very nice the [b]Kostas Vittas[/b]' problem ! I like it and wish to see a pure synthetical proof.
As well, very nice [b]Yetti[/b]'s proof ! I"ll present a metrical proof and later a vectorial proof of this "jewel". Come on ![/color][/size]

[quote="silouan"][color=darkred]Let a triangle $ABC$ and let $AD,BE,CF$ three cevians which intersect in $H$ in the interior of $ABC$ .
Let $M$ the midpoint of $EF$ and $P$ the point of intersection of the lines $EF$ and $BC$ .
The line $PH$ intersects the lines $AB$ and $AC$ at the points $K$ and $L$ respectively .
Prove that the line $AM$ and the parallels from $K$ and $L$ to $CF$ and $BE$ respectively ,intersect in one point.[/color][/quote]

[color=darkred][b][u]Strategy.[/u][/b] First I"ll estabilish the position of the intersection $T\in AM\cap BC\ ,$ i.e. the ratio $\frac{TB}{TC}\ .$ Second I"ll specify the intersection $U\in AT\cap KV\ ,$ where $V\in BC$ and $KV\parallel CF\ ,$ i.e. the ratio $\frac{UA}{UT}\ .$ Analogously, I"ll specify the intersection $U'\in AT\cap LW\ ,$ where $W\in BC$ and $LW\parallel BE\ ,$ i.e. the ratio $\frac{U'A}{U'T}\ .$ Therefore, the lines $AM$ , $KV$ , $LW$ are concurrently if and only if $U\equiv U'\ .$ Denote $AF=x$ , $AE=y$ and $S\in AM\cap CF\ .$[/color]

$\blacktriangleright$ [color=darkblue]Apply the [u]Menelaus ' theorem[/u] to the transversal $\overline{AMS}$ in the triangle $CFE\ :$ $\frac{AE}{AC}\cdot\frac{SC}{SF}\cdot\frac{MF}{ME}=1$ $\implies$ $\boxed{\ \frac{SC}{SF}=\frac{b}{y}\ }\ .$ Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{AST}$ in the triangle $BFC\ :$ $\frac{AF}{AB}\cdot\frac{TB}{TC}\cdot\frac{SC}{SF}=1$ $\implies$ $\frac{TB}{TC}=\frac{AB}{AF}\cdot\frac{SF}{SC}$ $\implies$ $\frac{TB}{TC}=\frac{c}{x}\cdot\frac{y}{b}$ $\implies$ $\boxed{\ \frac{TB}{cy}=\frac{TC}{bx}=\frac{a}{bx+cy}\ }\ .$

$\blacktriangleright$ The points $A$ , $K$ [u]are conjugate w.r.t. the pair of points[/u] $\{B,F\}\ ,$ i.e. $\boxed{\ \frac{KB}{KF}=\frac{AB}{AF}\ }$ $\implies$ $\frac{KB}{c}=\frac{KF}{x}=\frac{c-x}{c+x}$ $\implies$ $KA=KF+AF$ $\implies$ $KA=\frac{x(c-x)}{c+x}+x$ $\implies$ $KA=\frac{2cx}{c+x}$ $\implies$ $\frac{KA}{KB}=\frac{c+x}{c(c-x)}\cdot \frac{2cx}{c+x}$ $\implies$ $\boxed{\ \frac{KA}{KB}=\frac{2x}{c-x}\ }\ .$ Because $KV\parallel CF$ obtain $\frac{VB}{VC}=\frac{KB}{KF}$ $\implies$ $\frac{VB}{VC}=\frac{AB}{AF}$ $\implies$ $\frac{VB}{c}=\frac{VC}{x}=\frac{a}{c+x}$ $\implies$ $VT=VB-TB=\frac{ac}{c+x}-\frac{acy}{bx+cy}$ $\implies$ $VT=\frac{acx(b-y)}{(c+x)(bx+cy)}$ $\implies$ $\frac{VB}{VT}=\frac{(c+x)(bx+cy)}{acx(b-y)}\cdot\frac{ac}{c+x}$ $\implies$ $\boxed{\ \frac{VB}{VT}=\frac{bx+cy}{x(b-y)}\ }\ .$ Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{KUV}$ in the triangle $ABT\ :$ $\frac{VT}{VB}\cdot \frac{KB}{KA}\cdot\frac{UA}{UT}=1$ $\implies$ $\frac{UA}{UT}=\frac{VB}{VT}\cdot\frac{KA}{KB}=\frac{bx+cy}{x(b-y)}\cdot\frac{2x}{c-x}$ $\implies$ $\boxed{\boxed{\ \frac{UA}{UT}=\frac{2(bx+cy)}{(c-x)(b-y)}\ }}\ \ (1)\ .$

$\blacktriangleright$ The points $A$ , $L$ [u]are conjugate w.r.t. the pair of points[/u] $\{C,E\}\ ,$ i.e. $\boxed{\ \frac{LC}{LE}=\frac{AC}{AE}\ }$ $\implies$ $\frac{LC}{b}=\frac{LE}{y}=\frac{b-y}{b+y}$ $\implies$ $LA=LE+AE$ $\implies$ $LA=\frac{y(b-y)}{b+y}+y$ $\implies$ $LA=\frac{2by}{b+y}$ $\implies$ $\frac{LA}{LC}=\frac{2by}{b+y}\cdot\frac{b+y}{b(b-y)}$ $\implies$ $\boxed{\ \frac{LA}{LC}=\frac{2y}{b-y}\ }\ .$ Because $LW\parallel BE$ obtain $\frac{WC}{WB}=\frac{LC}{LE}$ $\implies$ $\frac{WC}{WB}=\frac{AC}{AE}$ $\implies$ $\frac{WC}{b}=\frac{WB}{y}=\frac{a}{b+y}$ $\implies$ $WT=WC-TC=\frac{ab}{b+y}-\frac{abx}{bx+cy}$ $\implies$ $WT=\frac{aby(c-x)}{(b+y)(bx+cy)}$ $\implies$ $\frac{WC}{WT}=\frac{ab}{b+y}\cdot\frac{(b+y)(bx+cy)}{aby(c-x)}$ $\implies$ $\boxed{\ \frac{WC}{WT}=\frac{bx+cy}{y(c-x)}\ }\ .$ Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{LU'W}$ in the triangle $ACT\ :$ $\frac{WT}{WC}\cdot\frac{LC}{LA}\cdot\frac{U'A}{U'T}=1$ $\implies$ $\frac{U'A}{U'T}=\frac{WC}{WT}\cdot\frac{LA}{LC}=\frac{bx+cy}{y(c-x)}\cdot\frac{2y}{b-y}$ $\implies$ $\boxed{\boxed{\ \frac{U'A}{U'T}=\frac{2(bx+cy)}{(c-x)(b-y)}\ }}\ \ (2)\ .$

$\maltese$ From the relations $(1)$ and $(2)$ obtain $\frac{UA}{UT}=\frac{U'A}{U'T}$ $\implies$ $U\equiv U'\ ,$ i.e. the lines $AM$ , $KV$ , $LW$ are concurrently.

[b][u]Remark.[/u][/b] Soon I"ll present here and a vectorial method.[/color]

[color=darkblue][b]A vectorial proof.[/b] Let $(\alpha ,\beta ,\gamma )$ be the normal barycentrical coordinates of the point $H$ w.r.t. the triangle $ABC\ ,$ i.e. $\boxed{\ H(\alpha ,\beta ,\gamma)\Longleftrightarrow\alpha+\beta+\gamma =1\ \mathrm{and}\ H=\alpha A+\beta B+\gamma C\ }\ .$

$\{\begin{array}{ccc}D\in AH\cap BC & \Longleftrightarrow & (1-\alpha )D=\beta B+\gamma C\ .\\\\ E\in BH\cap CA & \Longleftrightarrow & (1-\beta )E=\gamma C+\alpha A\\\\ F\in CH\cap AB & \Longleftrightarrow & (1-\gamma )F=\alpha A+\beta B\ .\end{array}$

$P\in EF\cap BC\Longleftrightarrow (\beta-\gamma )P=\beta B-\gamma C\ .$

$2M=E+F\Longleftrightarrow 2(1-\beta )(1-\gamma )M=\alpha (1+\alpha )A+\beta (1-\beta )B+\gamma (1-\gamma )C\ .$

$\blacktriangleright\ K\in PH\cap AB$ $\Longleftrightarrow$ exist $\{u,x\}\subset R$ so that $K=uA+(1-u)B=xH+(1-x)P$ $\Longleftrightarrow$ $K=uA+(1-u)B=x(\alpha A+\beta B+\gamma C)+\frac{1-x}{\beta-\gamma }\cdot (\beta B-\gamma C)$ $\Longleftrightarrow$

$\{\begin{array}{c}u=\alpha x\\\\ 1-u=\beta x+\frac{\beta }{\beta-\gamma }\cdot (1-x)\\\\ 0=\gamma x-\frac{\gamma }{\beta-\gamma }\cdot (1-x)\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}x=\frac{1}{1+\beta-\gamma }\ ;\ (1-x=\frac{\beta-\gamma }{1+\beta-\gamma })\\\\ u=\frac{\alpha}{1+\beta-\gamma }\ ;\ (1-u=\frac{2\beta }{1+\beta-\gamma })\end{array}$ $\Longrightarrow$ $\boxed{\ (1+\beta-\gamma )K=\alpha A+2\beta B\ }\ .$

$\blacktriangleright\ L\in PH\cap AC$ $\Longleftrightarrow$ exist $\{v,y\}\subset R$ so that $L=vA+(1-v)C=yH+(1-y)P$ $\Longleftrightarrow$ $L=vA+(1-v)C=y(\alpha A+\beta B+\gamma C)+\frac{1-y}{\beta-\gamma }\cdot (\beta B-\gamma C)$ $\Longleftrightarrow$

$\{\begin{array}{c}v=\alpha y\\\\ 0=\beta y+\frac{\beta }{\beta-\gamma}\cdot (1-y)\\\\ 1-v=\gamma y-\frac{\gamma}{\beta-\gamma }\cdot (1-y)\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}y=\frac{1}{1+\gamma-\beta }\ ;\ (1-y=\frac{\gamma-\beta}{1+\gamma-\beta})\\\\ v=\frac{\alpha}{1+\gamma-\beta }\ ;\ (1-v=\frac{2\gamma}{1+\gamma-\beta})\end{array}$ $\Longrightarrow$ $\boxed{\ (1+\gamma-\beta )L=\alpha A+2\gamma C\ }\ .$

I used often the well-known equivalence $mA+nB+pC=0\Longleftrightarrow m=n=p=0$ which means " the barycentrical coordinates are uniqually for a point ". Denote the point $R$ for which $RK\parallel FC$ and $RL\parallel EB\ .$ Therefore, exist $\{\lambda ,\rho\}\subset R$ so that $R=K+\lambda (C-F)=L+\rho (B-E)$ $\Longleftrightarrow$ $\frac{\lambda}{\gamma (1-\gamma )}=\frac{\rho}{\beta (1-\beta )}=\frac{2}{(1+\beta-\gamma )(1+\gamma-\beta)}$ $\Longrightarrow$ $\boxed{\ (1+\beta-\gamma )(1+\gamma-\beta )R=\alpha^{2}A+2\beta (1-\beta )B+2\gamma (1-\gamma )C\ }\ .$

In conclusion, the points $A$ , $M$ , $R$ are concurrently $\Longleftrightarrow$ $|\begin{array}{ccc}1 & 0 & 0\\\\ x_{M}& y_{M}& z_{M}\\\\ x_{R}& y_{R}& z_{R}\end{array}|=0$ $\Longleftrightarrow$ $\frac{y_{M}}{y_{R}}=\frac{z_{M}}{z_{R}}\ ,$ what is truly.
[b]Remark.[/b] I noted $(\ x_{S}\ ,\ y_{S}\ ,\ z_{S}\ )$- the normal barycentrical coordinates of the point $S$ w.r.t. the triangle $ABC\ .$[/color]

A short proof for the case when the $\triangle ABC$ with a point H can be transformed by a parallel projection into a $\triangle A'BC$ with the orthocenter H' (the other case already has a short proof). Dropping the primes from the notation after performing the parallel projection.

Quadrilateral BCEF is cyclic on account of the right angles $\angle BEC,\ \angle CFB,$ with the circumcenter N at the midpoint of BC. The orthocenter H of $\triangle ABC$ is the diagonal intersection of BCEF, hence PH is a polar of A with respect to (N), so that $AN \perp PH.$ Let the parallels to CF, BE through K, L meet at X. Let R be the foot of the A-symmedian of the $\triangle ABC$ and U, V the feet of normals from R to AB, AC. Let RU, RV meet AC, AB at S, T. SU, TV are 2 altitudes of $\triangle AST,$ R its orthocenter, AR the remaining altitude, $AR \perp ST.$ Quadrilateral STUV is cyclic on account of the right angles $\angle SUT,\ \angle TVS,$ hence UV is antiparallel to ST with respect to the $\angle A.$ Since AR, AN are isogonals with respect to the angle $\angle A$ and $AR \perp ST,$ it follows that $AN \perp UV.$ As a result, $UV \parallel PH \equiv KL,$ which means that the triangles $\triangle RUV \sim \triangle XKL$ are centrally similar, having parallel sides. Since UK, VL meet at A, A is their similarity center, so that RX passes through A. In other words, X lies on the A-symmedian of the $\triangle ABC,$ which cuts the side EF of the oppositely similar $\triangle AEF$ at its midpoint M.


In how many ways can we pick three different numbers out of the group
1,2,3,...,100
such that the largest number is larger than the product of the two smaller ones? (The order in which we pick the numbers does not matter) Source: Intro to Counting and Probability by Dave Patrick.
Um when giving a solution please make it simple. Thanks

[quote="life=tennis+math"]Ok first of all order doesn't matter, does it? Just making sure otherwise my solution's completely incorrect.
[hide]
You really need to notice patterns. In the first one (starting number as $1$) its decreasing by $1$, starting with $98$, which you can find the sum as $4851$. The next one decreases by $2$ starting with $94$ sum of $2352$, next decreases by $3$ starting with $88$ sum of $1157$. Anyway, after about $5-10$ minutes, you should be done and have obtained the sums:
$4851, 2352, 1157, 840, 600, 310, 161, 64, 10$. Sum them all to obtain $10345$. There's probably a pattern somewhere with the sums, which could help solving it faster, but my way gets me around $8$ minutes. Anyway, answer $\boxed{10345}$[/hide][/quote]

i double check your list....
i get: $4851$, $2256$, $1335$, $840$, $525$, $310$, $161$, $64$, $11$..

and sum up to be.. $\boxed{10353}$. I repeated this trial for 5 times get the same answer =/

at least the last one: 10.. its wrong for sure..
if you get 9,10.. you can have 91 to 100. -> 10 ways
if you get 9,11.. you can have 100 -> 1 way

[quote="DanK"]Is there a way to do this with combinations?  That was my first thought, but I can't see a way to make it work...[/quote]

Ditto.  :)


I'm sure that this problem is wrong !
 Try this one : $\sqrt{\frac{ab+ac+bc}{3}} \leq \sqrt[3]{\frac{(a+b)(a+c)(b+c)}{8}}$

[quote="tranthanhnam"]let $a,b,c>0$ , prove that :
            $\sqrt [3]{\frac{(a+b)(b+c)(c+a)}{8}} \geq \frac{\sum\sqrt{ab}}{3}$[/quote]

MathLinks Contest, 3th Edition, round 1, problem 2. See http://www.mathlinks.ro/Forum/viewtopic.php?t=4526 .

  darij

[quote="tranthanhnam"]Use:  $\sum\sqrt{ab} \leq \sqrt{3(ab+bc+ca)}$
         by chur's eneq :
           $x(x-y)(x-z)+y(y-z)(y-x) + z(z-x)(z-y) \geq 0$
            and$(ab+bc+ca)^2 \geq 3abc(a+b+c)$
             this problem will  complete[/quote]

 Tranthanhnam , you should have noticed that : your problem is edited .  :o


The deck of a ship is 100 feet above sea level.

a) How far can a person see from the deck?

b) How far can a person see from the bridge, which is about 150 feet above sea level.

NOTE: Use the fact that the radius of the Earth is 3960 miles.

Please, provide me with the necessary steps needed to find the answer.

[hide]draw the big circle that is the earth, with the radius of 3960 miles. one point on the circle (the ship) sticks out 100 feet (you don't have to draw it to scale.  draw the line from the top of the ship tangent to the circle. this is your line of sight, looking at the horizon. Tangent lines mean right angles. Then it's the pythagorean theorem:
$(3960 mi.)^{2}+(d)^{2}=(3960 mi.+100ft.)^{2}$
do some unit conversions and solve for d.[/hide]


All triangles have the same value, and all circles have the same value. What is the sum of two circles?\\
$ \Delta \plus{} \bigcirc \plus{} \Delta \plus{} \bigcirc \plus{} \Delta \plus{} \bigcirc \plus{} \Delta \plus{} \Delta\equal{} 21$\\
$ \bigcirc \plus{} \Delta\plus{}\bigcirc\plus{}\Delta\plus{}\bigcirc \plus{} \Delta \plus{} \bigcirc \plus{} \bigcirc \equal{} 16$\\
$ \bigcirc \plus{} \bigcirc \equal{}$ ?

Adding the two equations, we get that

$ \bigcirc\plus{}\Delta\equal{}\frac{37}{8}$

$ \implies \Delta\equal{}\frac{37}{8}\minus{}\bigcirc$

Thus, $ 2\bigcirc\equal{}16\minus{}\frac{37}{8}\equal{}\boxed{\frac{91}{8}}$.


algebra
polynomial
quadratics
group theory
abstract algebra
Galois Theory
superior algebra

Determine, if there are rational numbers c,d so that the polynomails x^4 -4x-1 and x^4+ax^2+b have same splitting field over Q(rational numbers), If yes find a and b.

The splitting field of the second polynomial is always the composition of two quadratic extensions. I suspect this is not the case for the first one (in particular the splitting field of $ x^4 \minus{} 4x \minus{} 1$ probably doesn't have degree $ 4$, i.e. adding a single root doesn't make the polynomial split).

EDIT: To add on, since $ x^4\minus{}4x\minus{}1$ is irreducible, if its splitting field is the composition of two degree $ 2$ extensions, then it must split as

$ (x^2\minus{}(p\plus{}q\sqrt{r})x\minus{}(s\plus{}t\sqrt{u}))(x^2\minus{}(p\minus{}q\sqrt{r})x\minus{}(s\minus{}t\sqrt{u}))$

which one can probably check is impossible.

$ x^4\minus{}4x\minus{}1$ has exactly two real roots. Conjugation fixes those two roots and interchanges the two complex roots, and it must be one of the automorphisms in the Galois group.
A Galois group acts transitively on the roots, and can be embedded in the symmetric group on those roots. Any transitive subgroup of $ S_4$ that contains a transposition either has order $ 8$ or is the whole group. If you don't want to work that hard, you can still easily show that such a group can't have order $ 4$.


In past years, what has the qualifying score for AIME from USAMTS been? I saw that last year's was a 68 or higher after 3 rounds. Is it always after 3 rounds?

yes, always after 3 rounds and 68 should be around the cutoff. as of what ive heard from others on this forum.


[hide]A group of 49 people has a mean income of exactly 25,000. Given that one new person with an income of 62,000 is included in the group, by how many dollars does the group's mean income increase?

$\displaystyle\frac{49*25000+62000}{50}-25000= 740$[/hide]

[hide]not going to do arithemetic, but here's how to do the problem:
so 49 people have the income of 25,000 so the sum of their total income combined is 49*25000.... when the next person is added, the new total income is going to be: 49*25000 + 62000. so we jsut divide this by 50 to get our answer[/hide]

If we let the new person have a baseline average of 25,000, then there is 37,000 remaining to be split evenly among all 50 people.  This raises everyone's average by 37,000/50 = 740.

Shake, don't forget to subtract the original income to get the increase :P .

A group of 49 people has a mean income of exactly 25,000. Given that one new person with an income of 62,000 is included in the group, by how many dollars does the group's mean income increase?

[hide]62000-25000 = 37000 37000/(49+1) = 740[/hide]


topology
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advanced fields unsolved

Let $T$ be a homeomorphism of a compact metric space $X$ onto itself and let $x \in X$. If the orbit $\{T^{n}(x) : n \in Z\}$ is a closed subset of $X$, then $x$ is a periodic point, i.e. there is $n>0$ with $T^{n}(x)=x$.

As for the subject title, periodic orbits of a dynamical system are sometimes called "closed orbits", as if they are like "closed curve".

This makes the orbit a countable compact Hausdorff space, and the restriction of $T$ a self-map of this space. Hence, it suffices to prove that if $X$ is a countable compact Hausdorff space and $f: X\to X$ is continuous, $f$ has a periodic point. There are two proofs of this statement [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=936223177&t=108969]here[/url].


trigonometry
geometry
parallelogram
AoPS Books

I was reading about parallelograms in one of my AoPS books, and it said that one of the formulas for finding the area of one is (AB)(BC)sinB, but I don't understand how to use this formula using trigonometry with a quadrilateral. Please help.

You can split a quadrilateral into two triangles.

[geogebra]435c3e842dcabe21c92e64178dcb41a9b1268724[/geogebra]

Notice that the altitude can be found by BCsin(B), so our area is h(AB)=AB BC sinB

Yes, it is. Also, you can divide into 2 triangles (Isabella2296's way), and the area of each one of them is AB*BC*sinB/2

Alright, thanks.
That makes a lot of sense.
Thanks mewto, isabella, and Bugi.


A number $n$ is said to be [i]nice[/i] if it exists an integer $r>0$ such that the expression of $n$ in base $r$ has all
its digits equal. For example, 62 and 15 are $\emph{nice}$ because 62 is 222 in base 5, and 15 is 33 in base 4. Show that 1993 is not [i]nice[/i], but 1994 is.

n is nice if and only if $n=\frac{a(r^k-1)}{r-1},a<r$ Therefore n is nice if r>n (k=1,a=n). n is nice for r=n-1 (k=2,a=1).  :D
If condition is $0<r\le n-2$, then if n=ab,a<b-1 n is nice (r=b-1,k=2). For example 1994=2*997 is nice a=2, r=998. If n s prime it is nice only if n is general mersene prime: $n=\frac{r^p-1}{r-1}, p$ p is odd prime. (Mersene primes for r=2).  It is easy cheked, that 1993 is not generalised Mersene prime, therefore it is not nice.

A number is nice if it can be writen as $q\frac{r^m-1}{r-1}$ for some $q\in \{0..9\}$ and r integer,$q\le r$.

Since 1993 is prime we would have that $q=1$ or $q=1993$ and it is easy too see that neither of these cases works. For 1994 $q$ can be equal to $1,2,997,1994$ then check each of these cases.

Where it is written $ r>0$, it should be written $ 0<r<n\minus{}1$.


Gegeben fuenf Liniensegmente der Laengen a, b, c, d, e mit der Eigenschaft, dass sich aus drei beliebig ausgewaehlten solchen Segmenten stets ein Dreieck bilden laesst. Man zeige, dass mindestens ein solches Dreieck spitzwinklig ist. [i]Meinen Beweis, der eine uebermaessig komplizierte Fallunterscheidung und Trigonometrie verwendete, habe ich vergessen. [/i]

[quote="Pi-J"]oBdA $e \geq d \geq c \geq b \geq a$
Angenommen, alle Dreiecke sind stumpf- oder rechtwinklig. Dann ist $e^{2}\geq d^{2}+c^{2}\geq (c^{2}+b^{2})+(a^{2}+b^{2}) \geq (a^{2}+b^{2})+2b^{2}+a^{2}\geq 3b^{2}+2a^{2}> (a+b)^{2}$
Widerspruch.[/quote]

6>5>4
6^2>5^2+4^2 ???

Ein Dreieck mit Seiten $4,5,6$ ist spitzwinklig.

I apologyze, I just can speak some german. Now I see that It was a geometry problem.


function
algebra
domain
algebra proposed

Find all functions $ f : \mathbb R \longrightarrow \mathbb R$ such that for every $ x\in \mathbb R$,
\[ f\left(\frac{1+x}{1-x}\right) = af(x),\]
where $a$ is a real constant.

[quote="behdad.math.math"]you are right dear pco.I saw this problem in a book and now i am sure that this problem is wrong.

but if $ f : R \longrightarrow R$ ,then?[/quote]
Let $ E(x)$ be the statement $ f(\frac {1 \plus{} x}{1 \minus{} x}) \equal{} af(x)$
Obviously, this statement can not be verified for $ x \equal{} 1$. I'll consider that it is verified for any other $ x$.

Let $ g(x) \equal{} \frac {1 \plus{} x}{1 \minus{} x}$. We have : $ g(g(g(g(x)))) \equal{} x$

So :

1) If $ a \equal{} 0$, we have the trivial solution $ f( \minus{} 1) \equal{} c$ and $ f(x) \equal{} 0$ $ \forall x\neq \minus{} 1\in\mathbb{R}$

2) If $ a\neq 0$, we have $ f(g(g(g(g(x)))) \equal{} a^4f(x) \equal{} f(x)$ And so :

2.1) Either $ f(x) \equal{} 0$ $ \forall x\notin\{ \minus{} 1,0, \plus{} 1\}$ and we have the trivial solution :
$ f( \minus{} 1) \equal{} c$
$ f(0) \equal{} ac$
$ f( \plus{} 1) \equal{} a^2c$
$ f(x) \equal{} 0$ $ \forall x\notin\{ \minus{} 1,0, \plus{} 1\}$

2.2) Either $ f(u)\neq 0$ for some $ u\notin\{ \minus{} 1,0, \plus{} 1\}$ and so we have $ a \equal{} 1$ or $ a \equal{} \minus{} 1$
$ g(x)$ has the following other properties :

$ g(]0, \plus{} 1[) \equal{} ] \plus{} 1, \plus{} \infty[$
$ g(] \plus{} 1, \plus{} \infty[) \equal{} ] \minus{} \infty, \minus{} 1[$
$ g(] \minus{} \infty, \minus{} 1[) \equal{} ] \minus{} 1,0[$
$ g(] \minus{} 1,0[) \equal{} ]0, \plus{} 1[$

So we can easily build solutions $ f(x)$:

Let $ h(x)$ be any real function defined in $ (0,1)$

$ \forall x\in(0,1)$ $ f(x) \equal{} h(x)$
$ \forall x\in(1, \plus{} \infty)$ $ f(x) \equal{} ah(\frac {x \minus{} 1}{x \plus{} 1})$
$ \forall x\in( \minus{} \infty, \minus{} 1)$ $ f(x) \equal{} h( \minus{} \frac {1}{x})$
$ \forall x\in( \minus{} 1,0)$ $ f(x) \equal{} ah(\frac {1 \plus{} x}{1 \minus{} x})$

$ f( \minus{} 1) \equal{} c$
$ f(0) \equal{} ac$
$ f( \plus{} 1) \equal{} c$
And it's easy to verify that this mandatory form is a solution.

===================================
Synthesis :

Since cases 1 and 2.1 are the same, we have :

S1) For any value $ a$, we have the solution :
$ f( \minus{} 1) \equal{} c$
$ f(0) \equal{} ac$
$ f( \plus{} 1) \equal{} a^2c$
$ f(x) \equal{} 0$ $ \forall x\notin\{ \minus{} 1,0, \plus{} 1\}$

S2) For $ a \equal{} 1$ or $ a \equal{} \minus{} 1$, we have the solutions :
Let $ h(x)$ be any real function defined in $ (0,1)$

$ \forall x\in(0,1)$ $ f(x) \equal{} h(x)$
$ \forall x\in(1, \plus{} \infty)$ $ f(x) \equal{} ah(\frac {x \minus{} 1}{x \plus{} 1})$
$ \forall x\in( \minus{} \infty, \minus{} 1)$ $ f(x) \equal{} h( \minus{} \frac {1}{x})$
$ \forall x\in( \minus{} 1,0)$ $ f(x) \equal{} ah(\frac {1 \plus{} x}{1 \minus{} x})$

$ f( \minus{} 1) \equal{} c$
$ f(0) \equal{} ac$
$ f( \plus{} 1) \equal{} a^2c \equal{} c$


Can anyone post a proof to prove there are infinitely many primes of the form

i)  4m+1
ii) 10m+9
If anybody knows a proof for cases other than the ones above and for 4m+3 and 6m+5, please post them too

If $ p_1,p_2, \ldots p_n$ are the primes of the form $ 4k\plus{}1$, then consider $ 4p_1^2p_2^2 \cdots p_n^2 \plus{} 1$. If of the form $ 4k\plus{}3$, then consider $ 4p_1p_2 \cdots p_n \minus{} 1$. If of the form $ 6k\plus{}5$, then consider $ 6p_1p_2 \cdots p_n \minus{} 1$.

In general, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1718136044&t=201236]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1457003416&t=106257]here[/url].

I am sorry, I don't understand the proof for $ 4k\plus{}1$.

My apologies, I thought I linked to it but I didn't. Let $ P \equal{} 2p_1p_2 \cdots p_n$, and suppose $ P^2 \plus{} 1 \equiv 0 \pmod{4k \plus{} 3}$ for some prime of the form $ 4k \plus{} 3$. Then $ P^2 \equiv \minus{} 1 \pmod{4k \plus{} 3}$, but by Fermat's Little Theorem, $ 1 \equiv P^{4k \plus{} 2} \equiv (P^2)^{2k \plus{} 1} \equiv \minus{} 1 \pmod{4k \plus{} 3}$, contradiction. It follows that if $ P^2 \plus{} 1$ is divisible by a prime, it is of the form $ 4k \plus{} 1$.

I do not yet have an easy proof for the $ 10k \plus{} 9$ case, but there should be one.


Let $a,b,c\geq0$ and $a^{2}+b^{2}+c^{2}=4$. Prove that:
$a+b+c\geq ab+bc+ca$

This inequality is false!
 Take for example a=sqrt(2),b=1,c=1 then you get a+b+c=sqrt(2)+2
 and ab+bc+ca=2sqrt(2)+1 which is bigger

[quote="cefer"]May be it was like this one?

Let $a,b,c \geq0$ and $a^{2}+b^{2}+c^{2}=3$. Prove that:
  $a+b+c \geq ab+bc+ca$[/quote]
I do not agree with you as this your inequality is obvious.

But hein's inequality is false? :(
What do you think? How is the correct version? :maybe:

Obviously, if $a^{2}+b^{2}+c^{2}=4$, then homogenize and take $a=b=c=1$, it's false.
And when $a^{2}+b^{2}+c^{2}=3$, it's true by, Muirhead's or $a^{2}+b^{2}+c^{2}\ge ab+bc+ca$ and $(a+b+c)^{2}\ge 3(ab+bc+ca)$.


a,b,c>0 , a 2 +b 2 +c 2 =1 . Prove  that  a+b+c+1/(abc)  \geq 4 \sqrt 3

let abc=t^3
and f(a,b,c)=abc(a+b+c)-4 \sqrt 3*abc +1
we have to prove f(a,b,c) \geq 0.
note that f(a,b,c) \geq 3t^4-4 \sqrt 3*t^3+1
so we have to prove 3t^4-4 \sqrt 3*t^3+1  \geq 0 which is easy by derivatives.

In fact, this one can be easily proved by AM-GM inequality.
    \sum a +1/(abc) >= 4[ 1/27(a+b+c)/(abc)^3]^1/4 >=4 \sqrt 3 :D


How many interior diagonals does an icosahedron have?

An icosahedron is consisted of 20 triangular faces, with 5 of them meeting at a vertice. Since a triangle has 3 corners, and 5 meet at a vertice, we can do 20*3/5 to get 12 vertices. Now for each vertice, there are 6 other vertices it can't meet to make a diagonal. This leaves 6. So we do 12*6 to get the total number of diagonals. We overcounted, so divide by two to get 36.


calculus function integration trigonometry algebra polynomial

Does anyone know where I can find good non-routine calculus problems?  I want to see the true powers of calculus.  If someone can refer me to a book, website, or even just post questions, it'd be helpful.  THanx

Here's my all time favorite calc question. It's from the 2003 Rice Math Tournament.

Sammy the Owl is making a one-eyed Jack'o'Lantern. He bought a perfectly spherical pumpkin of radius 1 foot. He first carves out the inside so that the inner radius is 10 inches. He then drills a perfectly circular hole of radius 1 inch for the eye. He drills straight toward the center of the pumpkin at the "equator." What is the total volume in cubic inches of pumpkin that he has removed?

(reality check: [hide]the piece that he removes for the eye is NOT a cylinder[/hide])

[quote="Ragingg"]Does anyone know where I can find good non-routine calculus problems?  I want to see the true powers of calculus.  If someone can refer me to a book, website, or even just post questions, it'd be helpful.[/quote]

Try the [url=http://www.ms.uky.edu/~carl/ma123/kob98/kob98htm/intro1.html]Kentucky edition of Neal Koblitz's online calculus text[/url] for an online resource, and the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0914098896/learninfreed]Calculus[/url] third edition by Michael Spivak for a print resource.

Hope this helps!

Although this is not a modern contest-type of problem, one of the most famous problems to show off the power of calculus is the Brachistochrone problem. The problem was issued as a challenge to the mathematicians of Europe by John Bernoulli is the 1690s. The problem was to find among all smooth curves lying in a vertical plane and connecting a given higher point A to a given lower point B, the curve along which a particle will slide under the force of gravity in the shortest possible time.

For simplicity's sake, try to solve this problem for A = (0,0) and B = (:pi:, -2).

A hint to get started (and to avoid working out the physics of the problem): [hide]Assume the graph of the function y = f(x) is the curve of fastest descent from A to B. f must satisfy the differential equation y[1 + (dy/dx)2] = c, for some positive constanc c.[/hide]

[quote="gauss202"]Although this is not a modern contest-type of problem, one of the most famous problems to show off the power of calculus is the Brachistochrone problem.[/quote]

Ooh.  I know this one.

It's a fairly straightforward exercise in calculus of variations - using conservation of energy, we need to minimize the integral of:

sqrt (1+y':^2:)/ sqrt (y)

We apply Beltrami to obtain:

y(1+y':^2:) = K, as you said earlier.

Separation gives: dy/dx = sqrt([K-y]/y)
It follows that x = int(y/[k-y])dy

The trig substitution y= k sin :^2:t simplifies everything nicely, and a bit of routine calculus gives us the equations of a cycloid.

I don't know if you would consider it difficult or not but problems from the calculus-help.com website which includes the Calculus Bowl competition.

[quote="JS1527"]There are a lot of hard proving-type questions w/calculus.  For example:

We call x=a a double zero of the polynomial P(x) if it appears twice as a factor, that is, if P(x)=(x-a) :^2: Q(x), where Q(x) is a polynomial.
(a) If x=a is a double zero of P(x), show that P(a)=P'(a)=0.
(b) If P(x) is a polynomial and P(a)=P'(a)=0, show that x-a is a double zero of P(x).[/quote]

Err .... that's not particularly hard.

(a) It is immediately apparent that P(a) = 0.  But P' = (x-a)^2*Q'+2(x-a)*Q so it is obvious that P'(a) = 0.

(b) Let P(x) = an*Product[(x-ri)] by the fundamental theorem of calculus.

Wolog, let a = r1 E R so that P(a) = 0

Then we apply the Leibniz product rule to show that:

P(x) = an*Sum[(x-r2)....(x-rn)d/dx (x-r1))] with an != 0
When we let x = r1 as before, this is equal to 0 by hypothesis.

It is immediately obvious that all of the terms EXCEPT for an*(x-r2)....(x-rn) go to zero.
But since P(a) = 0, this term must equal 0 and hence one of the factors r2.....rn = r1.  It then follows that there is a double root at a = r1.


function
real analysis
real analysis unsolved

Prove that, if $ f: \mathbb{R}\to\mathbb{R},\: f(x)\in\mathbb{Q}\Leftrightarrow f(x\plus{}1)\not\in\mathbb{Q}$ than $ f$ must have at least one point of discontinuity.

the number of irrational values in, say,[1,2) is equal to the number of rational values of the function in [0,1) thus countable, contradiction.

I don't understand the above :(, but I think I finally remembered the nice solution to this one:
$ f(x \plus{} 1) \minus{} f(x)$ takes only irrational values :o, but in most cases $ f(x \plus{} 2) \minus{} f(x)$ has some rational values... :dry:

However, as you can see, "I'm" using much, much more than the mere countability of $ \mathbb Q$ :huh:

I see: if $ f$ is continuous, then by the Intermediate Value theorem $ f(x \plus{} 1) \minus{} f(x)$ is either constant or attains a rational value somewhere. And if $ f(x \plus{} 1) \minus{} f(x)\equiv \alpha\notin\mathbb Q$, then $ f(x \plus{} 2) \minus{} f(x)\equiv 2\alpha\notin \mathbb Q$, which is a contradiction when $ f(x)$ is rational.

So, is it possible to replace $ \mathbb Q$ with an arbitrary countable (dense?) subset of $ \mathbb R$?

[quote="Albanian Eagle"]why don't you understand my argument? :)
if the function takes values in interval I for [0,1). then the number of it's rational values in I is clearly countable. (i.e. x with f(x) rational)
but in [1,2) the irrational values will correspond to the rational values of the function in [0,1), but the irrational values in I_2 cannot be a countable set  :wink:[/quote]

If  $ f(x)\in [0,1)$ doesn't mean that $ f(x\plus{}1)\in [1,2)$ we are working with the values of the function not with the argument!


search
ratio
number theory unsolved
number theory

Anybody know where can I find the complete proof for a theory(I don't know the name)
that for any prime p>2: the numerator of 1/1+1/2+1/3+...+1/p-1 can be divided by p^2
and that of 1/1^2+1/2^2+...+1/(p-1)^2 can be devided by p?

Thanks

it's been posted a few times before. search for Wolstenholme's Theorem.

Your condition is definitely wrong! It is right when p>=5(p is a prime)
Hint: Using the permutation of ratio when (1,2,...p-1) is divisible by p!


trigonometry
Product
algebra
Calculate
IMO Shortlist
IMO Longlist
complex numbers

Without using tables, find the exact value of the product:
\[P = \prod^7_{k=1} \cos \left(\frac{k \pi}{15} \right).\]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Let $ P \equal{}\prod^{7}_{k \equal{} 1}\cos\left(\frac{k\pi}{15}\right)$ and  $ Q \equal{}\prod^{7}_{k \equal{} 1}\sin\left(\frac{k\pi}{15}\right)$ obviously $ Q\neq0$,

by $ \sin\frac{k\pi}{15}\cos\frac{k\pi}{15}\equal{}\frac{1}{2}\sin\frac{2k\pi}{15}$ we have: $ PQ \equal{}\frac{1}{2^{7}}\prod^{7}_{k \equal{} 1}\sin\left(\frac{2k\pi}{15}\right)$

but $ \sin\frac{8\pi}{15}\equal{}\sin\frac{15\pi\minus{}7\pi}{15}\equal{}\sin\left(\pi\minus{}\frac{7\pi}{15}\right) \equal{}\sin\frac{7\pi}{15}$

$ \sin\frac{10\pi}{15}\equal{}\sin\frac{5\pi}{15}\;\;,\sin\frac{12\pi}{15}\equal{}\sin\frac{3\pi}{15}\;\;,\sin\frac{14\pi}{15}\equal{}\sin\frac{\pi}{15}$

As a consequence:  $ \prod^{7}_{k \equal{} 1}\sin\left(\frac{2k\pi}{15}\right) \equal{}\prod^{7}_{m \equal{} 1}\sin\left(\frac{m\pi}{15}\right) \equal{} Q$

and so $ PQ \equal{}\frac{1}{2^{7}}Q$ and $ \boxed{P \equal{} 2^{\minus{}7}}$.

[b]Solution.[/b] Let $ z \equal{} e^{i\frac{2\pi}{15}}$ be the 15-th root of unity.  Clearly, $ \cos\frac{k\pi}{15}\equal{}\frac{z^{\frac{k}{2}}\plus{}\overline{z}^{\frac{k}{2}}}{2}\equal{}\frac{z^{k}\plus{}1}{2z^{\frac{k}{2}}}$. Hence,
\[ P\equal{}\prod_{k\equal{}1}^{7}\cos\frac{k\pi}{15}\equal{}\frac{1}{2^{7}z^{14}}\prod_{k\equal{}1}^{7}(z^{k}\plus{}1).\]
Using $ \sum_{k \equal{} 0}^{14}z^{k}\equal{} 0$, we obtain that $ P \equal{}\frac{1}{128}$. $ \Box$

Uhhh...it does not follow that immediately. $ \sum_{0}^{14}x^{k}\not|\prod_{1}^{7}x^{k}\plus{}1$

You need a stronger factorization of $ x^{15}\equal{} 1$. Since $ 15$ is not prime, not all roots are primitive.


geometry
geometric transformation
analytic geometry
algebra
polynomial
superior algebra
superior algebra open

Let $A_1,A_2,\ldots,A_n$ be $n$ points inside a circle. For each $i\neq j$ we denote by $B_{ij}$ the intersection of the ray emerging from $A_i$ and passing through $A_j$ with the given circle. Let $z_{i,j}\in \mathbb{C}$ be the complex affix of the point $B_{ij}$, for all $i\neq j$, and let $P_i(x)= (x-z_{i,1})(x-z_{i,2})\cdots (x-z_{i,i-1})(x-z_{i,i+1}) \cdots (x-z_{i,n}) \in \mathbb{C}[X]$.

Prove that $P_1$, $P_2$, $\ldots$, $P_n$ are linear independent over the field of complex numbers.

Sorry, but one question:
what means 'affix' and what $z_k$ stands for¿

[quote="grobber"]What are $z_1,\ldots,z_n$? :? You've defined $z_{ij},\ i\ne j,\ i,j\in\overline{1,n}$.[/quote]Sorry, I hurried a little bit with the translation :)
Affix is the corresponding complex number of that point in the plane (ie if the point has the coordinates $(a,b)$ then its affix is $z=a+bi$, where $i^2=-1$).

there is a similar problem i think here http://www.mathlinks.ro/Forum/viewtopic.php?t=73954

is there any link between them ?

[quote="anubis9999"]i'll bite:

is this necessarily true? consider the points An on a line..
you then get only rays intersecting the circle at one of two points (we'll say their affixes are z1 and z2), and then each Pn = (x - z1)^m (x - z2)^k for some int m, k and i believe its possible to show that for some i != j, Pi = Pj..

i guess i might be a little confused. this problem interests me, so any input is welcome! thanks[/quote]

if $A_1, A_2, ..., A_n$ is their order on the line and $(A_iA_j$ ($i<j$) intersects the circle in the point of affix $z_1$ and $(A_iA_j$ ($i>j$) intersects the circle in the point of affix $z_2$ then the polynomial
$P_1 = (z-z_1)^{n-1}(z-z_2)^0$
$P_2 = (z-z_1)^{n-2}(z-z_2)^1$
...
$P_i = (z-z_1)^{n-i}(z-z_2)^{i-1}$

so we won't get that $P_i=P_j$ for $i\neq j$ :)


algebra polynomial trigonometry inequalities

I'm not really sure if this is the correct thread so we'll see.

I would like to find a polynomial such that one of its solutions yields the value of $ \sin{1^{\circ}}$.  For example, I have already worked out, and it's easy enough to see, that
$ 4x^{3}\minus{}3x\plus{}.5\equal{}0$
yields $ \sin{10^{\circ}}$ as one of its solutions.

It seems that one should try to come up with expressions for $ \sin{5x}$ or something like that and then go from there.  Any help?

$ \sin 90 x \equal{} 1$... $ x \equal{} 1$ is a solution.

Use the $ \sin nx \equal{} ...$ expansion to get the polynomial.

Try $ x \minus{} \sin 1^{\circ} \equal{} 0$.  :P

[hide="Discussion"] What you really want, in all likelihood, is a polynomial with rational or integer coefficients.  Altheman's technique almost works; unfortunately, the expansion of $ \sin nt$ into powers of $ \sin t$ is possible if and only if $ n$ is odd, so instead try $ \left( \sin 45t \right)^2 \equal{} \frac{1}{2}$.

Expanding the above by de Moivre's is straightforward but annoying.  Alternately, $ 45 \equal{} 3^2 \cdot 5$, so consider the polynomials $ f_3(x), f_5(x)$ such that $ f_3(\sin t) \equal{} \sin 3t$ and $ f_5(\sin t) \equal{} \sin 5t$ (these polynomials are easy enough to calculate, so I will leave it as an exercise).  Then the polynomial we want is

$ \left( f_5(f_3(f_3(x))) \right)^2 \equal{} \frac{1}{2}$.

You can avoid the use of a fifth-degree polynomial by instead using the known value of $ \sin 18^{\circ}$.  This brings up an interesting point:  to solve algebraically for the value of $ \sin 1^{\circ}$, it turns out that it is impossible for us to avoid having to solve a general cubic.  In other words, a one-degree angle is not [url=http://planetmath.org/encyclopedia/ConstructibleAnglesWithIntegerValuesInDegrees.html]constructible[/url]. [/hide]

Can't we use Cardano for cubics?
[img]http://www.csh.rit.edu/~topher/math/sin1.gif[/img]

[quote="t0rajir0u"]Try $ x \minus{} \sin 1^{\circ} \equal{} 0$.[/quote]

[hide="well,"]$ \dfrac{4x^3 \minus{} 3x \plus{} .5}{x \minus{} \sin 1^{\circ} } \equal{} 4x^2 \plus{} 4x\sin{1} \plus{} 4\sin^2{1}$[/hide]

right? :)

I agree, solving for $ \sin{5x}$ is indeed easy.  Is it possible to go from $ \sin{45}$ to $ \sin{9}$ to $ \sin{3}$ to $ \sin{1}$ or some other similar pathway?

I was hoping for something similar to my example for $ \sin{10}$.  Is it possible to get something like that?  I don't care if it's 10th power or 15th power or whatever.  Any polynomial of the form
$ f(x)\equal{}a_{n}x^{n}\plus{}\cdots\plus{}a_{1}x\plus{}a_{0}$
would do.

[quote="JRav"]I agree, solving for $ \sin{5x}$ is indeed easy.  Is it possible to go from $ \sin{45}$ to $ \sin{9}$ to $ \sin{3}$ to $ \sin{1}$ or some other similar pathway?[/quote]
This is essentially what I did in my solution.  What's unclear?

In terms of constructibility, the usual path is to construct $ \sin 15^{\circ} \equal{} \sin (45^{\circ} \minus{} 30^{\circ})$ and to construct $ \sin 18^{\circ}$ to construct $ \sin 3^{\circ} \equal{} \sin (18^{\circ} \minus{} 15^{\circ}$.  It's a rather fascinating fact that statements about geometric constructions can be written in terms of statements about minimal polynomials, and there is rather a lot of interesting theory there.

[quote="JRav"]I was hoping for something similar to my example for $ \sin{10}$.  Is it possible to get something like that?  I don't care if it's 10th power or 15th power or whatever.  Any polynomial of the form
$ f(x) \equal{} a_{n}x^{n} \plus{} \cdots \plus{} a_{1}x \plus{} a_{0}$
would do.[/quote]
I already gave you a polynomial.  Are you asking for its coefficients?  The standard way to construct them would be to write $ \sin 1^{\circ} \equal{} \cos 89^{\circ}$ and then the polynomial we want is (or rather, divides) $ T_{90}(x) \equal{} \cos 90^{\circ} \equal{} 0$, where $ T_{90}(x)$ is the ninetieth [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomial of the first kind[/url].  Again, there are ways to decrease the degree of the polynomial we need significantly; we can get degree $ 36$ by using $ T_{18}(x) \equal{} \minus{} \cos (18^{\circ})$ and perhaps we can do even better.  We should be able to get degree $ 18$.


${ (\sqrt[n]{n!}-1})<(n/2-1)$
since $ n!<(n/2)^n$
and then
[hide="  :blush: Lazy"]
unmathematically speaking, as $ n$ gets bigger, powers LHS increases only by multiplication, while RHS increases by power, so after trying out 3,4, the rest is obvious.
[/hide]
I'm sure a rigorous proof wont be hard to find

[quote="RobRoobiks"]${ (\sqrt [n]{n!} - 1}) < (n/2 - 1)$
since $ n! < (n/2)^n$
and then
[hide="  :blush: Lazy"]
unmathematically speaking, as $ n$ gets bigger, powers LHS increases only by multiplication, while RHS increases by power, so after trying out 3,4, the rest is obvious.
[/hide]
I'm sure a rigorous proof wont be hard to find[/quote]

My bad ArrangeMyTan.
I meant
$ n! < (\frac {n+1}2)^n$
${ (\sqrt [n]{n!} - 1}) < (\frac {n+1}2 - 1)$
But the rest kinda goes the same


calculus
integration
calculus computations

Convert to taylor series, integrate termwise, etc.


I learned recently that we had a math coach who was an Austrailian IMOer in her past. But now she's moving to Russia, so we are back to not having any former IMOers [to my knowledge] in GA.

Am I mistaken; is there another, and did someone else know about her before?

Regarding IMOers in Georgia, there was Boris, who wasn't exactly on the IMO, but an alternate. Also, Harrison and Miles (maybe Miles) have good chances this year.

i dont get it

OOOOOO imo yea

i get it...but...that ones hard amirite?

yay cool

btw i didnt notice you placed well also, good job

Yea, we briefly met for about 13 seconds or so. Not that I timed it or anything.

We have one that I know of. One of my students father was a member of Vietnam's team about 20 years ago. He was one of 3 perfect scores! I'm hoping that he will be able to come and talk to the ARML team during a practice this (next) year.

[quote="Karth"]Seva go away.

@John: Nice job at today's tournament.  :)[/quote]

why

waht else am i supposed to do with ALL THIS FREE TIME THAT I HAVE>?

what college do you go to now? Just wondering  :)

[quote="Elemennop"]plumber's trade school[/quote]

mario-in-training? :wink:

dude there have been multiple IMOers from NC so ha


Thirty-five percent of the girls at Athletic High School tried out for the soccer team. Given that $ 55 \frac{5}{9}$ percent of the student body of 900 is male, how many girls tried out for the soccer team?

$ 55 \frac{5}{9}$ of the 900 students are male, so $ (\frac{500}{9})(9)\equal{}500$ students are male. That means there are 400 female students. Of them, 35% of the females tried out for the soccer team, which is equal to $ \frac{35}{100}(400)\equal{}35*4\equal{}\boxed{140}$ females.


calculus
integration
trigonometry
function
abstract algebra
logarithms
absolute value

Hello all. Does anyone know if this one has a nice solution?

$ \int_{\minus{}\infty}^{\infty} \frac{|\sin x|}{1\plus{}x^2} \, dx.$

Convergence is clear since it's monotonic and bounded above by
$ \int_{\minus{}\infty}^{\infty} \frac{1}{1\plus{}x^2} \, dx \equal{} \pi.$

Other than that though, I didn't really get anywhere. I came across it while playing with some positive functions, so I don't know if it has a nice solution or not.

Cheers,
Bentn

What do you mean by a solution? Isn't that absolute convergence enough for you or you want to have some numerical result?

[quote="Bentn"]Hello all. Does anyone know if this one has a nice solution?

$ \int_{ \minus{} \infty}^{\infty} \frac {|\sin x|}{1 \plus{} x^2} \, dx.$

Convergence is clear since it's monotonic and bounded above by
$ \int_{ \minus{} \infty}^{\infty} \frac {1}{1 \plus{} x^2} \, dx \equal{} \pi.$

Other than that though, I didn't really get anywhere. I came across it while playing with some positive functions, so I don't know if it has a nice solution or not.

Cheers,
Bentn[/quote]

I could easily be wrong, but I doubt this has a closed form solution. The reason being the absolute value. That makes things more difficult. Obviously ommitting the absolute value makes the integral equal zero.

as I know for this integral there is no closed form.   :maybe:

If you meant certainty, it was misplaced. Sometimes these things are accessible to the right tricks.

I get a value of $ 2\left(e \minus{} e^{ \minus{} 1}\right)\arctan(e^{ \minus{} 1})$.

Expand $ |\sin x|$ in its Fourier series: $ |\sin x| \equal{} \frac {2}{\pi} \plus{} \sum_{n \equal{} 1}^{\infty}\frac2{\pi}\left(\frac1{2n \plus{} 1} \minus{} \frac1{2n \minus{} 1}\right)\cos(2nx)$
Next, use that $ \int_{\mathbb{R}}\frac {\cos ax}{1 \plus{} x^2}\,dx \equal{} \pi e^{ \minus{} |a|}$ (doable by complex analysis or other methods):
$ \int_{\mathbb{R}}\frac {|\sin x|}{1 \plus{} x^2}\,dx \equal{} \int_{\mathbb{R}}\frac {2}{\pi(1 \plus{} x^2)} \plus{} \sum_{n \equal{} 1}^{\infty}\frac2{\pi}\left(\frac1{2n \plus{} 1} \minus{} \frac1{2n \minus{} 1}\right)\frac {\cos(2nx)}{1 \plus{} x^2}\,dx$
$ \equal{} 2 \plus{} 2\sum_{n \equal{} 1}^{\infty}\int_{\mathbb{R}}\frac1{\pi}\left(\frac1{2n \plus{} 1} \minus{} \frac1{2n \minus{} 1}\right)\frac {\cos(2nx)}{1 \plus{} x^2}\,dx$ (We have absolute convergence to justify the interchange)
$ \equal{} 2 \plus{} 2\sum_{n \equal{} 1}^{\infty}\left(\frac1{2n \plus{} 1} \minus{} \frac1{2n \minus{} 1}\right)e^{ \minus{} 2n}$
$ \equal{} 2 \plus{} \frac23e^{ \minus{} 2} \plus{} \frac25e^{ \minus{} 4} \plus{} \cdots \minus{} 2e^{ \minus{} 2} \minus{} \frac23e^{ \minus{} 2} \minus{} \frac25e^{ \minus{} 4} \minus{} \cdots$
$ \equal{} 2\frac1{e^{ \minus{} 1}}\arctan(e^{ \minus{} 1}) \minus{} 2e^{ \minus{} 1}\arctan(e^{ \minus{} 1})$

[Edit- fixed mistaken factor of 2 in Fourier series]

Oh  :oops: , I'm sorry for my stupid mistake, I just was thinking about $ \int_{0}^{\infty} \frac {\sin x}{ 1 \plus{} x^{2}}$
I have forgotten about module $ |\sin x|$

For $ \alpha > 0$ the function $ |\sin\alpha x|$ has the Fourier series $ \frac {2}{\pi} \plus{} \frac {2}{\pi}\sum_{n \equal{} 1}^\infty \left(\frac {1}{2n \plus{} 1} \minus{} \frac {1}{2n \minus{} 1}\right)\cos 2n\alpha x$ .

So $ \int_{ \minus{} \infty}^\infty \frac {|\sin \alpha x|}{1 \plus{} x^2}\, dx \equal{} 2 \plus{} 2\sum_{n \equal{} 1}^\infty \left(\frac {1}{2n \plus{} 1} \minus{} \frac {1}{2n \minus{} 1}\right)\, \underbrace{\frac {1}{\pi} \int_{ \minus{} \infty}^\infty \frac {\cos 2n\alpha x}{1 \plus{} x^2}\, dx}_{e^{ \minus{} 2n\alpha}}$

$ \equal{} 2 \plus{} 2\sum_{n \equal{} 1}^\infty \frac {(e^{ \minus{} \alpha})^{2n}}{2n \plus{} 1} \minus{} 2\sum_{n \equal{} 1}^\infty \frac {(e^{ \minus{} \alpha})^{2n}}{2n \minus{} 1} \equal{} 2\sum_{n \equal{} 0}^\infty \frac {(e^{ \minus{} \alpha})^{2n}}{2n \plus{} 1} \minus{} 2\sum_{n \equal{} 0}^\infty \frac {(e^{ \minus{} \alpha})^{2n \plus{} 2}}{2n \plus{} 1}$

$ \equal{} 2\, (e^\alpha \minus{} e^{ \minus{} \alpha})\sum_{n \equal{} 0}^\infty \frac {(e^{ \minus{} \alpha})^{2n \plus{} 1}}{2n \plus{} 1} \equal{} 4\sinh\alpha\, \,\text{artanh}\, e^{ \minus{} \alpha} \equal{} 2\sinh\alpha\,\log\coth\frac {\alpha}{2}$ .

Similarly $ \int_{ \minus{} \infty}^\infty \frac {|\cos \alpha x|}{1 \plus{} x^2}\, dx \equal{} 4\cosh\alpha\, \arctan e^{ \minus{} \alpha}$ .

That's mostly redundant, but you did make me spot my error. I lost a factor of 2 in converting from $ \int_0^{\pi}\sin x\cos(2nx)\,dx$ to a Fourier coefficient.

Actually, I was originally looking for a series representation when I thought of using the Fourier series. I just got lucky with a closed form.

Edit:
A new correction: that's not the arctangent series. $ \sum_{n \equal{} 0}^{\infty}\frac1{2n \plus{} 1}x^{2n \plus{} 1} \equal{} \int_0^x \frac1{1 \minus{} t^2}\,dt \equal{} \frac12\ln(1 \plus{} x) \minus{} \frac12\ln(1 \minus{} x)$ is the inverse of the hyperbolic tangent, and a correct closed form for the original series is $ (e \minus{} e^{ \minus{} 1})\ln\frac {1 \plus{} e^{ \minus{} 1}}{1 \minus{} e^{ \minus{} 1}}$
Thanks again to Doe John.

Amazingly for $ \alpha>0$ the integrals $ \int_{\minus{}\infty}^\infty \coth\frac{\pi x}{2}\,\,\,\frac{\sin\alpha x}{1\plus{}x^2} \, dx \; , \; \int_{\minus{}\infty}^\infty \frac{|\sin\alpha x|}{1\plus{}x^2}\, dx$ are equal.


$P$ is a point in the interior of the quadrilateral $A_{1}A_{2}A_{3}A_{4}$ that does not lie on either diagonal. The points $B_{i}$ lie in the interior of each line segment $A_{i}P$. Let $C_{ij}$ be the intersection of the lines $A_{i}B_{j}$ and $A_{j}B_{i}(1 \leq i<j \leq4)$. Prove that the line segments $C_{12}C_{34}$, $C_{13}C_{24}$, $C_{14}C_{23}$ are all concurrent.
Merry christmas and Happy new year :)  :)  :)

What a beautiful problem!

[color=blue][b]Problem.[/b] Let $A_1$, $A_2$, $A_3$, $A_4$, P be five points in the plane. Let $B_1$, $B_2$, $B_3$, $B_4$ be points on the lines $PA_1$, $PA_2$, $PA_3$, $PA_4$, respectively.

Denote $C_{12}=A_1B_2\cap A_2B_1$, $C_{13}=A_1B_3\cap A_3B_1$, $C_{14}=A_1B_4\cap A_4B_1$, $C_{23}=A_2B_3\cap A_3B_2$, $C_{24}=A_2B_4\cap A_4B_2$, $C_{34}=A_3B_4\cap A_4B_3$, where the abbreviation $g\cap h$ means the point of intersection of two lines g and h.

Prove that the lines $C_{12}C_{34}$, $C_{13}C_{24}$, $C_{14}C_{23}$ are concurrent.[/color]

[i]Solution.[/i] First, we are going to use directed lengths. Define $a_1=\frac{PB_1}{B_1A_1}$, $a_2=\frac{PB_2}{B_2A_2}$, $a_3=\frac{PB_3}{B_3A_3}$, $a_4=\frac{PB_4}{B_4A_4}$.

Then, we are going to use vectors. Let O be an arbitrary point in the plane. In future, we will identify each point X in the plane with the vector $\overrightarrow{OX}$; so, an assertion of the kind $X=\lambda Y+\mu Z$ (with X, Y, Z being points and $\lambda$ and $\mu$ being reals) will mean that $\overrightarrow{OX}=\lambda\cdot\overrightarrow{OY}+\mu\cdot\overrightarrow{OZ}$.

Since the point $B_1$ lies on the line $PA_1$ and divides the segment $PA_1$ in the ratio $a_1=\frac{PB_1}{B_1A_1}$, we can write this point vectorially as $B_1=\frac{a_1A_1+P}{a_1+1}$. Now, consider the point $\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}$. Since

$\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}=\frac{\left(a_1A_1+P\right)+a_2A_2}{a_1+a_2+1}=\frac{\left(a_1+1\right)\frac{a_1A_1+P}{a_1+1}+a_2A_2}{\left(a_1+1\right)+a_2}=\frac{\left(a_1+1\right)B_1+a_2A_2}{\left(a_1+1\right)+a_2}$.

Hence, this point $\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}$ lies on the line $A_2B_1$ (and divides the segment $A_2B_1$ in the ratio $\frac{a_1+1}{a_2}$). Similarly, we can show that this point $\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}$ lies on the line $A_1B_2$. Hence, the point $\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}$ is the point of intersection of the lines $A_1B_2$ and $A_2B_1$; thus, this point $\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}$ must be the point $C_{12}$.

So we have found that $C_{12}=\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}$. Similarly, we can show that $C_{34}=\frac{a_3A_3+a_4A_4+P}{a_3+a_4+1}$. Now, define a point $Q=\frac{a_1A_1+a_2A_2+a_3A_3+a_4A_4+2P}{a_1+a_2+a_3+a_4+2}$. Then,

$Q=\frac{a_1A_1+a_2A_2+a_3A_3+a_4A_4+2P}{a_1+a_2+a_3+a_4+2}=\frac{\left(a_1A_1+a_2A_2+P\right)+\left(a_3A_3+a_4A_4+P\right)}{\left(a_1+a_2+1\right)+\left(a_3+a_4+1\right)}$
$=\frac{\left(a_1+a_2+1\right)\frac{a_1A_1+a_2A_2+P}{a_1+a_2+1}+\left(a_3+a_4+1\right)\frac{a_3A_3+a_4A_4+P}{a_3+a_4+1}}{\left(a_1+a_2+1\right)+\left(a_3+a_4+1\right)}=\frac{\left(a_1+a_2+1\right)C_{12}+\left(a_3+a_4+1\right)C_{34}}{\left(a_1+a_2+1\right)+\left(a_3+a_4+1\right)}$.

Consequently, the point Q lies on the line $C_{12}C_{34}$ (and divides the segment $C_{12}C_{34}$ in the ratio $\frac{C_{12}Q}{QC_{34}}=\frac{a_3+a_4+1}{a_1+a_2+1}$). Similarly, the point Q lies on the lines $C_{13}C_{24}$ and $C_{14}C_{23}$.

Hence, the lines $C_{12}C_{34}$, $C_{13}C_{24}$, $C_{14}C_{23}$ are concurrent, namely they concur at the point Q. Problem solved.

  Darij

Indeed, a very nice problem !

[hide="For Darij."][i]Darij[/i], in my opinion, if we will choose the origin $P$ of the vectorial system, then:

$B_k=m_k\cdot A_k,\ k\in \overline {1,4}\ ;\ C_{ij}=n_{ij}\cdot A_i+n_{ji}\cdot A_j$,

where $n_{ij}=\frac{m_i(1-m_j)}{1-m_im_j},\ m_im_j\ne 1,\ 1\le i<j\le 4$ a.s.o.[/hide]

One has to 'space out' and view point P as a vertex of a pyramide.

   Then the planes containing the C-lines will intersect in one point.

   Thank you.

   M.T.

It is a good opportunity for a practice, in perspective triangles.

$($ In my drawing, $A_{1}A_{2}= 4.7,$ $A_{2}A_{3}= 13.9,$ $A_{1}A_{3}= 15.5,$ $A_{3}A_{4}= 15.8,$ $A_{1}A_{4}= 7.4,$ $PA_{1}=6.1,$ $PA_{2}= 4.5,$ $PB_{1}= 3.5,$ $PB_{2}= 1.8,$ $PB_{3}= 4.3,$ $PB_{4}= 5.5$ $).$

$\bullet$ Because of the points $A_{2}\equiv B_{1}C_{12}\cap B_{3}C_{23},$ $B_{2}\equiv A_{1}C_{12}\cap A_{3}C_{23}$ and $P\equiv A_{1}B_{1}\cap A_{3}B_{3}$ are collinear, by Desarques’s theorem, we conclude that the triangles $\bigtriangleup A_{1}B_{1}C_{12},$ $\bigtriangleup A_{3}B_{3}C_{23}$ are perspective and so, the segment lines $A_{1}A_{3},$ $B_{1}B_{3}$ and $C_{12}C_{23},$ are concurrent at one point, so be it $S.$

Similarly, because of the collinearity of the points $A_{4}\equiv B_{1}C_{14}\cap B_{3}C_{34},$ $B_{4}\equiv A_{1}C_{14}\cap A_{3}C_{34}$ and $P\equiv A_{1}B_{1}\cap A_{3}B_{3}$ we have that the triangles $\bigtriangleup A_{1}B_{1}C_{14},$ $\bigtriangleup A_{3}B_{3}C_{34}$ are perspective and so, the segment lines $A_{1}A_{3},$ $B_{1}B_{3}$ and $C_{14}C_{34},$ are concurrent at the point $S,$ as the intersection point of $A_{1}A_{3}$ and $B_{1}B_{3}.$

Because of now, the collinearity of the points $S\equiv B_{1}B_{3}\cap C_{34}C_{14},$ $A_{1}\equiv B_{3}C_{13}\cap B_{4}C_{14}$ and $A_{3}\equiv B_{1}C_{13}\cap B_{4}C_{34},$ we conclude that the triangles $\bigtriangleup B_{1}B_{3}C_{13},$ $\bigtriangleup C_{34}C_{14}B_{4}$ are perspective and so, the segment lines $B_{1}C_{34},$ $B_{3}C_{14},$ $C_{13}B_{4},$ are concurrent at one point, so be it $N.$

$\bullet$ We can see now, some other useful perspective triangles.

Because of the concurrency of the segment lines $C_{12}C_{23},$ $B_{1}B_{3}$ and $C_{14}C_{34},$ we conclude that the triangles $\bigtriangleup B_{1}C_{12}C_{34},$ $\bigtriangleup B_{3}C_{23}C_{14}$ are perspective and so, the points $A_{2}\equiv B_{1}C_{12}\cap B_{3}C_{23},$ $M\equiv C_{12}C_{34}\cap C_{23}C_{14}$ and $N\equiv B_{1}C_{34}\cap B_{3}C_{14},$ are collinear.

That is the point $M,$ as the intersection point of the segment lines $C_{12}C_{34},$ $C_{23}C_{14},$ lies on the segment line $A_{2}N$ where $N\equiv B_{1}C_{34}\cap B_{3}C_{14}\cap B_{4}C_{13}.$

By the same way, we can prove that the triangles $\bigtriangleup A_{4}B_{4}C_{24},$ $\bigtriangleup A_{3}B_{3}C_{23}$ are perspective $($ because of the collinearity of $A_{2},$ $B_{2},$ $P$ $)$ and so, the segment lines $A_{3}A_{4},$ $B_{3}B_{4}$ and $C_{23}C_{24},$ are concurrent at one point, so be it $T.$

Also the triangles $\bigtriangleup A_{4}B_{4}C_{14},$ $\bigtriangleup A_{3}B_{3}C_{13}$ are perspective $($ because of the collinearity of $A_{1},$ $B_{1},$ $P$ $)$ and so, the segment lines $A_{3}A_{4},$ $B_{3}B_{4}$ and $C_{13}C_{14},$ are concurrent at the point $T,$ as the intersection point of $A_{3}A_{4}$ and $B_{3}B_{4}.$

$\bullet$ Because of now, the concurrency point $T\equiv B_{3}B_{4}\cap C_{23}C_{24}\cap C_{13}C_{14},$ we conclude also that the triangles $\bigtriangleup B_{3}C_{23}C_{14},$ $B_{4}C_{24}C_{13}$ are perspective and so, the points $A_{2}\equiv B_{3}C_{23}\cap B_{4}C_{24},$ $M'\equiv C_{23}C_{14}\cap C_{24}C_{13}$ and $N\equiv B_{3}C_{14}\cap B_{4}C_{13},$ are collinear.

That is the point $M',$ as the intersection point of $C_{14}C_{23},$ $C_{13}C_{24},$ lies on the segment line $A_{2}N.$

But the common point of $A_{2}N,$ $C_{14}C_{23},$ may be only the point $M$ and so, we have that $M'\equiv M.$

Hence, we conclude that $M\equiv C_{12}C_{34}\cap C_{13}C_{24}\cap C_{14}C_{23}$ and the proof is completed.

[b][size=109]NOTE.[/size][/b] – As a remark, we can say that this problem is true, when the point $P,$ is outwardly of the given quadrilateral $A_{1}A_{2}A_{3}A_{4}$ and also when it is not in the same plane of it, as [b][size=109]''armpist''[/size][/b] mentioned. We use the same words as before, in order to prove these extended results.

Kostas Vittas.


There are 2008 red cards and 2008 white cards. 2008  players sit down in circular toward the inside of the circle in situation that 2 red cards and 2 white cards from each card are delivered to each person. Each person conducts the following procedure in one turn as follows.

$ (*)$ If you have more than one red card, then you will pass one red card to the left-neighbouring player.

If you have no red card, then you will pass one white card to the left -neighbouring player.
Find the maximum value of the number of turn required for the state such that all person will have one red card and one white card first.

Could anybody translate this problem to English? Thanks.

There are two decks of 2008 cards each. One of the decks contains only red cards, the other only white cards. 2008 players sit in a circle, facing inwards. Each player is given 2 red cards OR 2 white cards. Each 'turn', every player will do one of two things:

$ A)$ If the player has one or more red cards, he will pass one red card to his left.

$ B)$ If the player has no red cards, he will pass one (white) card to his left.

Find the maximum number of 'turns' required for each player to have one red card, and one white card.

(It could be reworded to have the same meaning as such:

Each turn, every player will pass one card to the left. If the player has a red card, it must be passed.)

The answer is: 2007.

It is obvious, that after every turn each player will have exactly two cards. We can now forget about red cards.
Now each turn every player do one of these things:
a) If the player has two white cards, he will pass one of them to his left.
b) If the player doesn't have two white cards, he will do nothing.
We have to find the maximum number of turns required for each player to have one white card.

First notice is that, if the player at some moment have one or two white cards, one of them won't move any more (if he has got two white cards, he will pass the second one). Second notice is that, unless every player have his own white card, there will be one card, that will move every turn.
From the second notice we get that the maximum number of turns is no more than 2007. There is also a situation, when we need exactly 2007 turns: first 1004 players have two white cards each. Other players get red cards.

How does the fact that unless every player has his own white card, there will be one card, that will move every turn imply there is a maximum of $2007$ turns?


Well, just apply the inequality (trivial, but hyper-useful) $9(a+b)(b+c)(c+a)$ is greater than or equal than $8(a+b+c)(ab+bc+ca)$

Are you sure , Harazi ? I can't apply your ineq to solve Kapi's problem .

Dear nttu , Do you have a solution for this problem ?
  Let $ a+b+c =1 $
 $ \to (a+b)(b+c)(c+a) = ab+ac+bc -abc $
 Applying Harazi 's ineq , we have :
 $ ab+ac+bc \geq 9abc $
 Let $ t = ab+ac+bc , u = abc $
 We need to prove :
 $ f(t)= 64t^3 - 27t^2 + 54t.u -27u^2 \geq 0 $ with $ t \geq 9u $
 Anybody helps me to finish my solution !!!

Romano, just a question: How did you manage to find such a complicated approach? I mean, with Harazi's hint, it's really easy to find the simple proof:

First note that Harazi's trivial but hyper-useful inequality,

$9\left(b+c\right)\left(c+a\right)\left(a+b\right)\geq 8\left(a+b+c\right)\left(bc+ca+ab\right)$

for nonnegative reals a, b, c, is clear, since

$9\left(b+c\right)\left(c+a\right)\left(a+b\right)-8\left(a+b+c\right)\left(bc+ca+ab\right)$
$=a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2$,

what is obviously $\geq 0$. Now, we have to prove that, for nonnegative reals a, b, c, the inequality

$\sqrt[3]{\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{8}}\geq\sqrt{\frac{bc+ca+ab}{3}}$

holds. Well, Harazi's inequality $9\left(b+c\right)\left(c+a\right)\left(a+b\right)\geq 8\left(a+b+c\right)\left(bc+ca+ab\right)$ easily transforms into

$\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{8}\geq \frac{\left(a+b+c\right)\left(bc+ca+ab\right)}{9}$.

Hence, instead of proving the required inequality

$\sqrt[3]{\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{8}}\geq\sqrt{\frac{bc+ca+ab}{3}}$,

it is enough to show the stronger inequality

$\sqrt[3]{\frac{\left(a+b+c\right)\left(bc+ca+ab\right)}{9}}\geq\sqrt{\frac{bc+ca+ab}{3}}$.

After taking this inequality into power 6, it becomes

$\left(\frac{\left(a+b+c\right)\left(bc+ca+ab\right)}{9}\right)^2\geq\left(\frac{bc+ca+ab}{3}\right)^3$.

This transforms into

$\frac{\left(a+b+c\right)^2\left(bc+ca+ab\right)^2}{81}\geq\frac{\left(bc+ca+ab\right)^3}{27}$.

This simplifies to

$\left(a+b+c\right)^2\geq 3\left(bc+ca+ab\right)$.

But this is trivial, since

$\left(a+b+c\right)^2-3\left(bc+ca+ab\right)=\frac12\cdot\left(\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\right)$,

what is trivially $\geq 0$.

  darij

Oh , I managed to prove an incorrect ineq !!! Let's look at again Kapi's problem !

Let $a+b+c=3u,$ $ab+ac+bc=3v^{2}$ and $abc=w^{3},$ where $v\geq0.$ Then
$\sqrt{\frac{ab+bc+ca}{3}}\leq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}\Leftrightarrow v\leq\sqrt[3]{\frac{9uv^{2}-w^{3}}{8}}\Leftrightarrow$
$\Leftrightarrow9uv^{2}-8v^{3}-w^{3}\geq0,$ which trivial.

Another solution is to choose $ab+bc+ca=3$, we need to prove that \[(a+b)(b+c)(c+a)\geq8.\] Using the identity \[(a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a),\] we need to prove that \[3(a+b+c)-abc\geq8.\] But this is easy.

Because of homogenity take ab+bc+ac=1. From here it s quite simple

[quote="Kapi"]Let $a$, $b$, $c$ be nonnegative reals. Prove that
$\sqrt{\frac{ab+bc+ca}{3}}\leq \sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}}$.[/quote]

This inequality is equivalent to
\[
\left(\frac{1}{8} (a+b) (a+c) (b+c)\right)^2-\left(\frac{1}{3} (a b+a c+b c)\right)^3 \geq 0
\]
and after bringing all terms to one fraction it becomes equivalent to proving that $f(a,b,c) \geq 0$ where
\[
f(a,b,c)=27 a^4 b^2+54 a^4 b c+27 a^4 c^2-10 a^3 b^3-30 a^3 b^2 c-30 a^3 b c^2-10 a^3 c^3+27 a^2 b^4-30 a^2 b^3 c-114 a^2 b^2 c^2-30 a^2 b c^3+27 a^2 c^4+54 a b^4 c-30 a b^3 c^2-30 a b^2 c^3+54 a b c^4+27 b^4 c^2-10 b^3 c^3+27 b^2 c^4
\].
Inequality is symmetric and we may assume that $a \geq b \geq c$. Then $a=c+x+y$ and $b=c+x$ for some $x,y \geq 0$ and then
\[
f(a,b,c)=f(c+x+y,c+x,c)=288 c^4 x^2+288 c^4 x y+288 c^4 y^2+800 c^3 x^3+1200 c^3 x^2 y+1104 c^3 x y^2+352 c^3 y^3+780 c^2 x^4+1560 c^2 x^3 y+1524 c^2 x^2 y^2+744 c^2 x y^3+108 c^2 y^4+312 c x^5+780 c x^4 y+840 c x^3 y^2+480 c x^2 y^3+108 c x y^4+44 x^6+132 x^5 y+159 x^4 y^2+98 x^3 y^3+27 x^2 y^4 \geq 0
\]
Done.

[quote=Kapi]Let $a$, $b$, $c$ be nonnegative reals. Prove that
$\sqrt{\frac{ab+bc+ca}{3}}\leq \sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}}$.[/quote]
[url=https://artofproblemsolving.com/community/c1068820h2091391p15109243]Austrian - Polish 1992 [/url]
[url=https://artofproblemsolving.com/community/c6h1990392p13867261]h[/url]


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

how can i resolve this triangle construction ?

 :(

[hide="Common, this is trivial."]If $ O$ is reflected in $ BC, CA, AB$ into $ O_a, O_b, O_c,$ triangles $ \triangle O_aO_bO_c \cong \triangle ABC$ are centrally congruent with similarity center at 9-point circle center $ N$ and coefficient -1. Orthocenter $ H$ of $ \triangle ABC$ is circumcenter of $ \triangle O_aO_bO_c,$ $ A$ is reflection of $ H$ in $ O_bO_c$ and $ E_a$ midpoint of $ O_bO_c.$ Quadrilateral $ AE_aM_bO_b$ with right angles at $ E_a, M_b$ is cyclic.[/hide]
$ b$ is line perpendicular to $ OM_b$ at $ M_b.$ Reflect $ O$ in $ b$ into $ O_b.$ Circle $ \odot (E_aM_bO_b)$ cuts $ b$ at $ A.$ Circumcircle $ (O)$ with center $ O$ and radius $ OA$ cuts $ b$ again at $ C.$ Perpendicular $ a$ to $ AE_a$ through $ C$ cuts $ (O)$ again at $ B.$


ratio
number theory
Diophantine equation

[quote="MCrawford"]What is the largest two-digit number which is equal to four times the sum of its digits?[/quote]

[hide="Hint"]$ 10x+y=4(x+y) $

$ 10x+y=4x+4y $

$ 6x-3y=0 $

$ 2x-y=0 $

Find the largest $ x $ and $ y $ that satisfies this Diophantine Equation such that $ x $ and $ y $ are not two digit numbers. Remember, $ x $ has a higher priority than $ y $ because it's the tens digit.[/hide]

[quote=Equinox8][hide=Solution]
Let the digits of the number have a tens digit of $L$ and a units digit of $K$. Then the number is $10L + K$, and four times the sum of the digits would be $4(L+K) = 4L + 4K$. Thus we have $10L+K = 4L+4K$ so $6L = 3K$, so $2L = K$. Because the units digit must be double the tens digit, the units digit must be even. In addition, $L < 5$ since otherwise $K \geq 10$; thus $\boxed{48}$ is our number with maximized $L$ (since tens digits take priority) and further maximized $K$.
[/hide]

Ignore the fact that this is an almost $17$ year bump o_0[/quote]

Let's also just ignore the fact that ilp has more posts than someone who existed in 2005

Oh MCrawford wrote some of the AoPS books orz

forester2015 moment


A better question is, who will come in second to Kevin Chen? But anyway here we go.

The best question is why "someone else" isn't on that list. And please don't think I'm referring to myself  :roll:. Chances are someone who wasn't on the radar makes a run for it.

 Also, in 2001, Ryan Ko won Nationals with what I believe was a 2nd place written performance. He ended up in 3rd written, 4th countdown the next year. So you NEVER KNOW. Although placing a bet, if you were forced to, on Kevin Chen would be your best best :lol:

No New Yorkers????  :(

If I was forced to take a bet, I would say

"None of the above"

[quote="frost13"]What about Bobby Shen who to the best of my knowledge outscored almost everyone on that list and bested Kevin in Countdown at States?  Could we see our first 6th grade National champ???????  Has to be an option at least :lol:[/quote]
I think he took the top ranked 7th graders from last year, then added Yuan for pwning the AIME (11), Wenyu Cao for his pwnage at practically everything, and himself.

WHOA MAN TWO OTHER PEOPLE VOTED FOR ME?
WHAT IS THIS?

Edit: OH COME ON ANOTHER ONE? HOW DO I HAVE [b]FOUR[/b] VOTES????

On your list on your blog you have you winning it :P.

This is referring to countdown, right? Or is it written? Written is much more solid and easy to predict. Countdown is... like... impossible to predict almost.

I put Kevin Chen assuming it's written. Countdown... I've never seen anyone in real live countdown action other than Kevin (I know l33t's fast on IM though I guess).

...you're lucky someone in connecticut voted twice for a redsox fan!

im not sure what the lowest written to win cd ever is...but anyone in the top 12 has a chance (ur cd skills can own your written skills)...

jorian

Kevin Chen will win

Dan Tobias (36? That barely makes nationals in Kentucky :rotfl: )

If it's countdown, I retract my vote from Kevin Chen and place it in the "this thread is absolutely useless because countdown is impossible to predict" slot :P. Given that we've only seen one of these people in countdown, it's kinda hard to measure...

[quote="Fanatic"]If it's countdown, I retract my vote from Kevin Chen and place it in the "this thread is absolutely useless because countdown is impossible to predict" slot :P. ......................[/quote]

That's why it says who do you THINK will win.

WELL DUH KEVIN CHEN!
But thats for written. If countdown... that's definitely impossible to predict.

Well yeah.
How did Allen somehow conjure up seven votes in the five hours I was gone?  :huh:

i suggest you put Richard Ni on the list, his brother won mathcounts although that doesn't mean he will.

Na, I haven't done any behind-the-scenes stuff yet.  Contrary to popular belief, mods can't change poll votes ^_^.

MysticTerminator?  How is spamming acting cool?  :lol:

Nationals is....full of surpises, so I won't pick anyone anymore.

Huh, "Peepel" owns them all with a 110 now.

Ray S in second with 105

Seth(icheated) has 100... well apparently he cheated so... lol.

whoa. who's that..? and why would they say that anyway
he knows it must be a matter of time b4 he gets caught by UBER NINJA PWNAGE SKILZ
anyhoo cheating is bad so... :yankchain: HAhAHA

FYI, Peepel Hacked. I'm 100% sure of that.

FYI, I know "peepel" in real life (no that's not his real name) and yes he cheated. He's not a kid that does math... just someone I introduced the game to.

The cheat, we think, only works on internet explorer.

(All my scores are legit, I guarantee you).

People that cheated: (icheated)seth, Peepel, nerby.

i dont think that aime scores affect mc(i got 5), but of course, they could, maybe i'm dumb.

also, who voted for me? :huh:
i doubt ill do well
countdown(cd) is not predictable e.g. i'm typically good, but i screwed up at states, luckily it dont count. kevin chen or bobby from tx might mess up(probably highly unlikely)

finally, arithmetic is not the only thing involved with cd.  you need to be able to skim the paragraph.  it only helps with those "simplify:" type questions

lol sorry, and i dont mean to be offensive here, but i think that kevin and bobby messing up is less likely than you getting a 10 total score at nationals... i was on kevin's team at st8 and chap, so i had it easy... i mean, you dont have to do anything; kevin's just so awesome like that....
lol
  :lol:
i think kevin will finally take first place this year at nats. and he deserves it, too
he's a really helpful and cool person as well as being a genius

(kevin if you read this, it's krit typing )
anyways
i think bobby is a force to be reckoned with as well

[quote="ZzZzZzZzZzZz"][quote="mchoi815"] . . . . finally, arithmetic is not the only thing involved with cd.  you need to be able to skim the paragraph.  it only helps with those "simplify:" type questions[/quote]

I agree.  Even on the simplify questions, you're asked to simplify things like 16/24, which everyone who's at nationals should know is 2/3 in a tenth of a second.

Actually, I think the most important thing for countdown is buzzer skills.  You have to know how to press the buzzer before you get the answer and use up your three seconds before you begin.  It can really make a difference.[/quote]

yeah, and if you have a long name it gets u a bit extra time  :P
like this girl at our school.. pavidhra chibadabamaram
or abhishek balakrishnan
or something long
lol u get a couple of extra seconds while
1) they figure which one you are
2) they pronounce your name
so use that time to think
^__^

[quote="ZzZzZzZzZzZz"][quote="mchoi815"] . . . . finally, arithmetic is not the only thing involved with cd.  you need to be able to skim the paragraph.  it only helps with those "simplify:" type questions[/quote]

I agree.  Even on the simplify questions, you're asked to simplify things like 16/24, which everyone who's at nationals should know is 2/3 in a tenth of a second.

Actually, I think the most important thing for countdown is buzzer skills.  You have to know how to press the buzzer before you get the answer and use up your three seconds before you begin.  It can really make a difference.[/quote]

Sometimes though it's not smart to try to do the problem in your 3 seconds. I remember at chapter in the final round Joseph buzzed in on a problem and was trying to simplify $\sqrt{40}$ in his head and his brain just...I dunno but his answer came out like : "2 root ksdjfkdlsajgsadkgf" or something...I got an extra point from it though :D . He still won, but it was tied 2-2 (first to 3) before he won...

I'm not saying you shouldn't ever buzz in and use the 3 seconds, but it might backfire on you...just a warning

Another strategy when you know you can't get the answer and have already buzzed in and there's not much time on the clock left is to say an extremely long number like 1426371079812...... and waste time.  I'd really like to see someone do that, but it might also get you disqualified.

Under pressure, it happens.

At science bowl I was simplifying like the square root of 28, I froze, and it felt like .5 seconds had passed, but somehow they all said my 3 seconds were up and I was like WHOA...

Happened to me at state against u in the first question, actually.

Then I was afraid to answer in the second one.

[quote="mathcountsfreak"]I'm not saying you shouldn't ever buzz in and use the 3 seconds, but it might backfire on you...just a warning[/quote]
actually you know, that's absolutely right.. it's a dangerous strategy, but i still think it helps. it worked for me, but i lost anyway
(my luck, i went up against bobby shen -___________- )

[quote="Fanatic"]Under pressure, it happens.

At science bowl I was simplifying like the square root of 28, I froze, and it felt like .5 seconds had passed, but somehow they all said my 3 seconds were up and I was like WHOA...

Then I was afraid to answer in the second one.[/quote]

first, i don't remember that question from the science bowl.
second, how did u guys do down there.  2nd? 1st?

btw, the answer is 2 root 7.

Honey Creek placed first in the state, and they've placed first in the nation the last two years.

The point of my post wasn't that simplifying sqrt.28 is oh so difficult, my point was that sometimes, under pressure, people freeze up on easy problems.

no, how did u guys do, i mean, with the whole "last minute thing" giong on.  i kind of thought honey creek would win.  btw, i was trying to answer your question, not show off.  i knew u were trying to make a point.


modular arithmetic
group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra theorems

I was told that the following theorem can be proven using group theory:

\[ a^{\varphi\left(n\right)}\equiv 1\pmod{n}.  \]

Is it that the collection of numbers relatively prime to $n$ form a subgroup of $\mathbb{Z}\big/n\mathbb{Z}$?  If this is the case, then by Lagrange's Theorem I could see that the size of the subgroup is $\varphi\left(n\right)$.  Since this is also the order of $a$, we have $a^{\varphi\left(n\right)}\equiv 1\pmod{n}$ again by Lagrange's Theorem.  Is this the right approach?

Your approach does work.  This is the usual other approach.  Let $I$ denote the set of invertible elements in $\mathbb{Z}/n\mathbb{Z}$.  Then if $(a,n)=1$, we have $I=aI$.  Consider the equality $\prod_{x\in I} x = \prod_{y\in I} ay$.


[quote="Allnames"]Find the greatest of value of $ k$ satisfying $ 1999^k$ is a factor of
\[ A \equal{} 1998^{{1999}^{2000}} \plus{} 2000^{{1999}^{1998}}
\]
[/quote]
I use a lemma in my proof.[b]
Lemma[/b]
If $ a$ is an even ,$ n\in Z^*$ then
\[ {(1\plus{}a)^a}^n \equal{}1\plus{}S_na^{n\plus{}1}
\]
The result is $ 1999$