geometry
3D geometry
sphere
rotation
geometric transformation
LaTeX
analytic geometry

1. A large wedge rests on a horizontal frictionless surface, as shown. A block starts from rest and slides down the inclined surface of the wedge, which is rough. During the motion of the block, the center of mass of the block and wedge:
(a)does not move
(b)moves horizontally with constant speed
(c)moves horizontally with increasing speed
(d)moves vertically with increasing speed
(e)moves both horizontally and vertically

2. A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a fixed axis through its center. Assume the hoop is connected to the rotation axis by light spokes. With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown. Rank the objects according to their kinetic energies after a given time t, from least to greatest.

3.When the speed of a rear-drive car is increasing on a horizontal road, the direction of the frictional force on the tires is
backward on the front tires and forward on the rear tires.
forward on the front tires and backward on the rear tires.
forward on all tires.
backward on all tires.
zero.

4. A small point-like object is thrown horizontally off of a 50.0-m high building with an initial speed of 10.0 m/s. At any point along the trajectory there is an acceleration component tangential to the trajectory and an acceleration component perpendicular to the trajectory. How many seconds after the object is thrown is the tangential component of the acceleration of the object equal to twice the perpendicular component of the acceleration of the object? Ignore air resistance.
12.00 s
1.50 s
1.00 s
0.50 s
The building is not high enough for this to occur.

5. Find the period of small oscillations of a water pogo, which is a stick of mass m in the shape of a box (a rectangular parallelepiped.) The stick has a length L, a width w and a height h and is bobbing up and down in water of density ρ. Assume that the water pogo is oriented such that the length L and width w are horizontal at all times. Hint: The buoyant force on an object is given by Fbuoy = ρVg where V is the volume of the medium displaced by the object and ρ is the density of the medium. Assume that at equilibrium, the pogo is floating.

Your first two questions don't have any images.

don't know how to add images.
the first problem just has an incline with a block at the top

the second problem just has a disk, sphere, and hoop of the same size with a force being applied tangentially in the same spot on the three figures.

I'm guessing you want explanations, since there is an answer key for these problems?

1. This can be a bit hard to visualize. You have to look at the block and wedge as the system. There are only 2 unbalanced forces: gravity and the normal force, which are vertical. Since gravity acts on the center of mass, the center of mass is going to move downwards.

2. This has to do with rotation, which is taught in AP Physics C.

3. The rear tires are pushing the car foward (by pushing on the ground backwards), so frictional force has to be forwards. And the front tires are just there, so frictional force is backwards.

2. In particular, you'll want to consider moments of intertia.

Thanks to both of you for your replies.

1) So the normal force acts at the contact point between the wedge and block, not at the com so that's why the com moves downward with increasing speed? am i correct?

2)yes, i know $ I_{disk} \equal{} \frac{1}{2} MR^2$ , $ I_{hoop}\equal{} MR^2$ and $ I_{sphere} \equal{} \frac{2}{5}MR^2$ and that $ K \equal{} \frac{1}{2}I\omega^2$. But I wasn't sure how to use that.

3) i suppose i misunderstood the problem. i didn't know when you're moving backwards only your rear tires are doing any work. the front ones are just following along

[quote="mihail911"]i didn't know when you're moving backwards only your rear tires are doing any work. the front ones are just following along
[/quote]

Actually, rear-wheel drive is simply a different way of driving a car.
You're not always moving backwards...but true second part of your statement is true.

[url]http://www.rearwheeldrive.org/rwd/rwdbenefits.htm[/url]

I'm not sure if that's what you meant, but in my opinion, it looks like a misunderstanding.

1.  You can also think about it as though as the block moves, the frictionlessness of the wedge with the table will cause the wedge to move the opposite direction.  However, the normal force cancels out any downward force by the block, making the block move down while the wedge does not.

[quote="mihail911"]2)yes, i know $ I_{disk} \equal{} \frac {1}{2} MR^2$ , $ I_{hoop} \equal{} MR^2$ and $ I_{sphere} \equal{} \frac {2}{5}MR^2$ and that $ K \equal{} \frac {1}{2}I\omega^2$. But I wasn't sure how to use that.
[/quote]

So I got the problem from the exam to find the pictures haha

So same force applied at the same distance from the center of mass for the same time so $ \tau t\equal{}C$ for all three. That means that the change in angular momentum is the same. Since along the rim, $ K\equal{}\dfrac 12 I\omega^2 \equal{} \dfrac{I^2\omega^2}{2I} \equal{} \dfrac{L^2}{2I}$ because the translational component is 0, K_f is proprotional to 1/I (don't know tex for proportional to) and then it should be easy

[quote="123456789"]For the problem with the block and ramp's center of mass, I understand that the center of mass moves vertically down, but if I look at it geometrically, it doesn't seem to make sense. When the block is at the top of the ramp, the center of mass (I would think), horizontally, would be shifted towards the block. When the block is at the bottom of the ramp, I would think that the center of mass would be closer to the block at the bottom of ramp, thus a horizontal movement of the center of mass. Why is this not so?[/quote]

Based on the picture given in the test, the box moves down and to the left, while, because the horizontal surface is frictionless, the wedge moves to the right. Given this opposite movement, the center of mass should intuitively not experience horizontal motion.
Hope that makes sense

[hide="hint for #8"]
[quote="Q8"]...rider does not slide down the wall...[/quote]
It means rider is in equillibrium. Put Net vertical force on the rider zero.
What forces act in vertical direction on the rider?
[hide="solution for #8"]
$ mg = f_s$ But $ f_s \le \mu_s N = \mu_s mR \omega^2$ $ \implies \mu_s \ge \frac {g}{R \omega^2}$
[hide="calculation for #8"]
$ \omega = \frac {45(2 \pi)}{60} rads^{ - 1}$ So finally $ \mu_s \ge 0.11$
[/hide]
[/hide][/hide]

[hide=" hint for #12"]
What is the expression for Rotational Kinetic energy about a fixed axis?  (In term of MoI and angular velocity}
[hide="furthur hint and solution"]What are MoIs of a uniform disk about centre(say $ I_c$) and about end point?(say $ I_e$)
[hide="solution for #12"]
K.E = $ \frac12 I \omega^2$
$ I_c = \frac12 MR^2$ and $ I_e = I_c + MR^2$(parallel-axis theorem) = ${ \frac32 MR^2}$
So $ E_e = \frac32 E_c$
[/hide]
[/hide][/hide]

[b]Q9 latexified[/b]
9. A ball of mass $ m_1$ travels along the x-axis in the positive direction with an initial speed of $ v_0$. It collides with a ball of mass $ m_2$ that is originally at rest. After the collision, the ball of mass $ m_1$ has velocity$ v_{1_x} \hat{x} + v_{1_y} \hat{y}$ and the ball of mass $ {m_2}$ has velocity $ v_{2_x} \hat{x} + v_{2_y} \hat{y}$
Consider the following five statements:
I)   $ 0 = m_1 v_{1_x} + m_1 v_{2_x}$
II)  $ m_1 v_0 = m_1 v_{1_y} \ + m_2 v_{2_y}$
III) $ 0 = m_1 v_{1_y} + m_2 v_{2_y}$
IV) $ m_1 v_0 = m_1 v_{1_x} + m_1 v_{1_y}$
V)  $ m_1 v_0 = m_1 v_{1_x} + m_2 v_{2_x}$

[hide="hint for #9"]
Apply the principle of conservation of momentum in both $ X$-direction and $ Y$-direction.. i.e, equate initial and final momenta i.e, before and after collision
This can be done as there is no net [i]external [/i]force on the system of the two colliding bodies.
[hide="solution for #9"]
In X-direction $ m_1 v_0 = m_1 v_{1_x} + m_2 v_{2_x}$
In Y-direction $ 0 = m_1 v_{1_y} + m_2 v_{2_y}$
So $ (III)$ and $ (V)$ are correct and hence option $ B$.
[/hide][/hide]

[hide="hint for #13"]
giving certain velocity means giving certain K.E. At the maximum elongation(called amplitude) all of this K.E is converted into spring Potential Energy(U).
[hide="solution for #13"]
$ U = \frac12 kA^2$
In both cases U is same. and $ k_2 = 2k \implies A_2 = \frac {A}{\sqrt2}$
The options aren't clear. Is B this?
[/hide][/hide]

[hide="P.S."] Please learn latex here- http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About . It's easy and comfortable to use.[/hide]

14. A spaceborne energy storage device consists of two equal masses connected by a tether and rotating
about their center of mass. Additional energy is stored by reeling in the tether; no external forces
are applied. Initially the device has kinetic energy E and rotates at angular velocity !. Energy is
added until the device rotates at angular velocity 2!. What is the new kinetic energy of the device?
(a) √2E
(b) 2E
(c) 2√2E
(d) 4E
(e) 8E

An experiment consists of pulling a heavy wooden block across a level surface with a spring force
meter. The constant force for each try is recorded, as is the acceleration of the block. The data
are shown below.

10. Which is the best value for the mass of the block?
(a) 3 kg
(b) 5 kg
(c) 10 kg
(d) 20 kg
(e) 30 kg
11. Which is the best value for the coefficient of friction between the block and the surface?
(a) 0.05
(b) 0.07
(c) 0.09
(d) 0.5
(e) 0.6
[img]http://img26.imageshack.us/img26/2176/tablemw.jpg[/img]

Here're some hints:

14) $ E \equal{} 2 \times \frac{1}{2}I\omega^2$; Angular momentum $ L \equal{} mv_{1}r_{1} \equal{} mv_{2}r_{2}$ is conserved.

10) Write a net force equation. Does it look like $ y \equal{} mx \plus{}b$ or $ F \equal{} ma \plus{}$ something? What can you figure out from the table of values?

The net external force acting on the block is mg (weight) which acts in the vertical direction. Since the center of the mass's motion is impacted by the external forces acting on it, the com would move in the direction of the external force.

BTW can anyone help me with F = ma 2007, problem 26 (sled with children 25° etc)

A sled loaded with children starts from rest and slides down a snowy 25° incline traveling 85 meters in 17 seconds.  (This is like 5m/s, so i assumed that friction force cancels out with mgsintheta to result in zero acc, constant velocity)
Ignore air resistance. What is the coefficient of kinetic friction between the sled and the slope?

So eventually i end up with uk = inverse tan of 25° which is 87 wtfff.

The answer's [b].40[/b], but i can't seem to get that close using other methods, closest i've gotten is (.46)

Why would you assume that velocity is constant? Try again with non-zero acceleration.

Ahaha. Thanks man! Assuming really made a donkey out of me.

Does the energy double because the distance between the masses decreases (because the rotational inertia decreases???) ? I'm a bit confused as to what "reeling in the tether does",  I never seem to understand or notice these weird "hints" that are part of the problem.

Qualitatively, the rotational kinetic energy increases because energy was expended to "reel it in" and to increase its angular speed.

To what hints are you referring? The ones I gave aren't "part of the problem"

Haha. Not your hints, just the little statements in the problem like "Additional energy is stored by reeling in the tether; no external forces are applied."

Also, you said angular momentum was conserved. Are you treating each of the masses like particles, so you can use
$ L \equal{} mv_{1}r_{1} \equal{} mv_{2}r_{2}$ .

I tried to do, Li=Lf
and then  Iwi=Iwf, and wf=2wi
but wouldn't the rotational inertia decrease, in order to conserve angular momentum?
I think I am approaching this problem wrong; does "reeling in" the "tether refer to shortening the length of the tether between the two masses?

EDIT:
$ L \equal{} mv_{1}r_{1} \equal{} mv_{2}r_{2}$
rewriting v as wr...
$ \omega_{1} r_{1}^{2}\equal{}\omega_{2} r_{2}^{2}$
With  $ \omega_{2}\equal{}2\omega_{1}$, we get
$ r_{1}^{2}\omega_{1}\equal{}r_{2}^{2}\left(2 \omega_{1}\right)$
$ \frac{r_{1}}{\sqrt{2}}\equal{}r_{2}$
Kinetic energy...
$ K_{1}\equal{}mv^{2}\equal{}m\left(\omega r\right)^{2}\equal{}m\omega^{2}r^{2}$
So, replacing with and ,
$ K_{2}\equal{}m\left(2\omega\right)^{2}\left(\frac{r}{\sqrt{2}}\right)^{2}\equal{}m\left(4\omega^{2}\right)\left(\frac{r^{2}}{2}\right)\equal{}2m\omega^{2}r^{2}\equal{}2K_{1}$

Nvm, i guess the radius does decrease.

Yes, that's correct. For these types of problems, remember that rotational inertia can always change by a change in mass or radius or distribution to conserve angular momentum.


MATHCOUNTS ARML AMC AIME USA(J)MO USAMO HMMT

I really don't get why there is one. You get recognition sure...but no chance into another comp or something...I know I probably sound stupid and will be murdered by replies...but why is there an AMC 8?

Why is there an AMC 10? Why is there an AMC 12? Why is there Mathcounts? Why is there ARML? et cetera.

The point of math competitions is to stimulate interest in good students. The AMC 8 is just one contest for that purpose.

^The OP meant that contests such as the AMC-10/12 lead to the Aime, Usamo, MOP/IMO, while the AMC-8 does not lead to anything in particular.

I agree with worthawholebean's reasoning, math contests are intended partly for competetive purposes, but mostly to stimulate interest in mathematics.

Contests like ARML, HMMT, PUMAC etc. do not lead to anything further but are simply enjoyable experiences in themselves.

Well, for people in middle school, there's normally not very many contests to take. The Amc 8 is like the only competition (besides Mathcounts) that most schools participate in.

AMC 8, as said by many others, prepares younger problem solvers or introduces people to problem solving before they try the harder contests (although I started the AMC 8 and 10 during the same year). I think it does give people an idea of what problems will look like on the harder competitions. For my part, I used it as a benchmark to see how much I could improve, and I did improve quite a bit (increased 7 points in a year). Also, it's a place to start learning NOT TO MAKE STUPID MISTAKES! Well, we still all do...
Off topic, but I'm now a Hodge Conjecture  :) .

I think the main reason for the AMC 8 is that in middle school a lot of people are still just starting to try math contests, and so the AMC 8 is a pretty good opportunity for that -- it's not that hard, you're not really competing anything.  I think it's actually good that you're not trying to move on or anything, because it means that for the first-timers, there's nothing you can really lose.

It's just like the middle school/high school Math League contests, which is mostly unimportant in the fact it doesn't lead to anything, and the only real reward is your name/school possibly being put on a single web page once in a year.

[quote="SamE"]I got a plaque the last two years from those Math League contests! Maybe it's just my state...[/quote]My school doesn't do those and I haven't asked them to because they don't sound interesting.

Its basically so that people like me (who are new to AMC's) get a chance to start out...
except in my case i did AMC 10 too ... but it encourages people to continue with math

dont diss state mathleauge---its good for picking up easy trophies :P

Yes it is an easy way to pick up trophies. I got 24 (stupid mistake) this year and still got 1st place in my state.

I got a plaque in 6th grade. I got a 39 this year and 4th. :mad:

How do you get 39 if there are only 25 problems?

I think he's talking about the Math League thing.

Meh, I had a perfect last year in 6th grade and my new school doesn't offer.

Dude xpmath you xuck :P

My scores have been like 39, 39, 39, 39, 39... :(

I didn't hear about Math League until they were over...*sigh*

But I tried the sample test thing for my grade and it's so easy!! I made a million careless mistakes so I only got a 37 on the samples.

I never took the AMC 8, but I wish I had.  This yr was my first yr taking AMC-10, and the transition was somewhat rough, after several yrs of mathcounts.  The multiple choice format was kinda hard to get used to.  I'm sure if I had taken AMC-8, the transition would've been easier.  I still did ok (120), but I didn't feel the comfort level I felt with mathcounts, I didn't know wat kinda problems they wrote, wat traps they tried to use, etc.  My mathcounts stupid mistakes were almost 0 this yr, b/c I understood the problems so well, but that wasn't the case for AMC-10.  I think if I had taken the AMC-8 I would've done better.

yeah, i took 7th grade Math League test last year as a 6th grader, got a perfect score...


inequalities
integration
topology
advanced fields
advanced fields unsolved

Let $ \Omega$ a bounded open set with smooth border $ \Gamma: = \partial \Omega$ and let

$ V: = H^2(\Omega)\cap H_0^1(\Omega) = \{v\in H^2(\Omega): v = 0 \,\text{in}\, \Gamma \}$

Let $ v\in V$. Prove that exists $ C > 0$ such that

${ \parallel{}v\parallel{}_{H^1(\Omega)}\leq C\parallel{}\Delta v\parallel{}_{L^2(\Omega)}}$

Hint:  Let $ v\in V$ and define $ f: = - \Delta v$ and use the lax milgram lemma to the problem $ - \Delta u = f$ in $ \Omega$ and $ u = 0$ in $ \Gamma$

Dear jorgeston,

I don't know exactly what you are allowed to suppose, but I will work from the fact that the problem
\[ \minus{}\Delta u\equal{} f \quad\text{in }\Omega\]
with zero boundary conditions has a unique solution in $ H_0^1(\Omega)$ if $ f\in L^2(\Omega)$,
even if this assumption almost directly yields the result (see the book Partial Differential Equations by Evans chapter 6).
This follows immediately from the Lax-Milgram theorem and the Poincaré inequality.

To finish up, we know that the Poincaré inequality explicitely yields
\[ \left\Vert u\right\Vert_{H^1(\Omega)}^2\le C\left\Vert\nabla u\right\Vert_{L^2(\Omega)}^2 \equal{}C B[u,u]\]
where $ B$ is the bilinear form corresponding to the Laplacian. Since $ u$ is a solution in the weak sense,
\[ B[u,w]\equal{}(f,w): \equal{}\int_\Omega f w \, dx,\qquad \forall w\in H_0^1(\Omega).\]
With $ f\equal{}\minus{}\Delta v\in L^2(\Omega)$ and $ w\equal{}u$, we get
\[ B[u,u] \equal{} ( \minus{}\Delta v, u ) \equal{} \minus{} \int_\Omega \Delta v u \,dx \le \left\Vert\Delta v\right\Vert_{L^2(\Omega)} \left\Vert u\right\Vert_{L^2(\Omega)} \le \left\Vert\Delta v\right\Vert_{L^2(\Omega)} \left\Vert u\right\Vert_{H^1(\Omega)}.\]
Here, I have used the Cauchy-Schwarz inequality and the Poincaré inequality.
We use this in the first inequality:
\[ \left\Vert u\right\Vert_{H^1(\Omega)}^2\le C \left\Vert\Delta v\right\Vert_{L^2(\Omega)} \left\Vert u\right\Vert_{H^1(\Omega)}\\
\implies \left\Vert u\right\Vert_{H^1(\Omega)}\le C \left\Vert\Delta v\right\Vert_{L^2(\Omega)}.\]
Now we conclude by observing that $ u\equal{}v$ is a solution (and therefore the unique solution) to the above problem so
\[ \left\Vert v\right\Vert_{H^1(\Omega)}^2\le C \left\Vert\Delta v\right\Vert_{L^2(\Omega)}.\]

However, I don't see the reason why you cannot do the following:
\[ \left\Vert v\right\Vert_{H^1(\Omega)}^2 \le C\left\Vert \nabla v\right\Vert_{L^2(\Omega)}^2 \equal{} C\int_\Omega |\nabla v(x)|^2\,dx \equal{} \{\text{IBP}\}
\equal{}C\underbrace{\int_{\partial\Omega} v \frac{\partial v}{\partial n}\, dS}_{\equal{}0} \minus{} C\int_\Omega \Delta v v\, dx \le C  \left\Vert\Delta v\right\Vert_{L^2(\Omega)}\left\Vert v\right\Vert_{L^2(\Omega)}\le C \left\Vert\Delta v\right\Vert_{L^2(\Omega)}\left\Vert v\right\Vert_{H^1(\Omega)} \implies \left\Vert v\right\Vert_{H^1(\Omega)}^2\le C \left\Vert\Delta v\right\Vert_{L^2(\Omega)}.\]
Maybe someone can help me point out the possible error/errors in both of the solutions above. I really don't see the reason why you need a smooth boundary either. However if you do have a boundary that is at least $ C^2$, then it can be shown that the (weak) solution to the Laplace equation with homogeneous Dirichlet conditions, and $ f\in L^2(\Omega)$ not only lies in $ H_0^1(\Omega)$ but also in $ H^2(\Omega)$. In this case it seems to follow from the uniqueness of the solution and your particular $ f$.


quadratics articles parameterization algebra

1. Solve for x:

 $ \frac {1}{X \minus{} 6a}$ + $ \frac {8}{x \plus{} 3a}$+ $ \frac {3}{x \minus{} 2a}$ = $ \frac {6}{x \minus{} a}$

Thank you.

Are $ X$ and $ x$ different or the same?

If same the answers are:

Real solution:

Assuming $ a$ is a real constant the only real solution to the system is:

$ x \equal{} \left[3 \plus{} \sqrt[3]{3 \plus{} \sqrt {10}} \minus{} \frac {1}{\sqrt[3]{3 \plus{} \sqrt {10}}}\right]a\approx 4.28791 a$

Complex solutions:

Apart from the above one, there are two more:

$ x \equal{} \left[3 \plus{} \frac {1 \plus{} i \sqrt {3}}{2 \sqrt[3]{3 \plus{} \sqrt {10}}} \minus{} \frac { (1 \pm i \sqrt {3}) \sqrt[3]{3 \plus{} \sqrt {10}}}{2}\right]a\approx (2.35605\pm 2.06011 i)a$

BTW, In my opinion the problem is more likely to be :

$ \frac {1}{x \minus{} 6a} \plus{} \frac {8}{x \plus{} 3a} \equal{} \frac {3}{x \minus{} 2a} \plus{} \frac {6}{x \minus{} a}$

In which case it would be a quadratic equation actually, apart from it's looks, because $ 1 \plus{} 8 \equal{} 6 \plus{} 3$

In that case the solutions are:

$ x \equal{} \left[\frac {9\pm \sqrt {33}}{2}\right]a\approx (4.5\pm 2.87228)a$

[hide="PS"]Thanks to Mathematica 6.0  :P  :rotfl:  :rotfl: [/hide]

Ok, x and X are the same. But, this is beyond my comprehension.

And I would not be visiting the forum till at least 24th Feb due to my exams .May be I will renew the thread again after that to learn from you all.

Thank you.

Did I scare you? I'm sorry. That's what Mathematica (a software) spat out. Some cubic and quartic equations have such scary closed form solutions (check [url=http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method]here[/url] to have a look how to solve general cubics). Such things are not needed. The thing is, you copied the problem wrong. it should be
\[ \frac {1}{x \minus{} 6a} \plus{} \frac {8}{x \plus{} 3a} \equal{} \frac {3}{x \minus{} 2a} \plus{} \frac {6}{x \minus{} a}\]I believe you can solve it yourself.

Akashnil, where do you egt the sotware? it is nice but definitely scary for an 8th std guy man.

[quote="Akashnil"] Some cubic and quartic equations have such scary closed form solutions (check [url=http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method]here[/url] to have a look how to solve general cubics). [/quote]
Are non-super-geniuses supposed to understand that wiki article?  :huh:

I suppose mathematica is paid software????????  :maybe:  :maybe:

Well those of you taking Mathematica too seriously... it didn't find out all those by itself, it has the cubic formula programmed into it and it just put in the parameters and displayed the result. (you could make a C++ program to solve general cubics for fun.) Having that wiki article in front, anyone could do that by hand provided sufficient time.

And gaurav, I agree that wiki article as a whole requires quite some background (I don't have that much too)
But having background is not necessarily being genius in my opinion.
And I actually referred to the "Cardano's method" part of the article, which for some reason, doesn't seem to work.
That Cardano's method part, in my opinion isn't too advanced.

Anyway for those interested in Mathematica: here's their site: http://www.wolfram.com/

Actually there are loads of other math softwares out there (mapple or something for example). But Mathematica is quite vast, as it covers a huge range of the subject of mathematics and they say it is useful for research stuff also.

Wow! You actually bought the softwares? Isn't there a free demo lol? And thanks for all the info Akshanil.

Find the length of the side of the square....

arre.... i want the method.... how did you get it??

just use trigo

and one important thing that 3,4,5 triangle has angles 53,37,90 .........

use a bit classical and u will be done

its not that tough.........

rather its pretty easy far below the olympiad level

There are other methods, but trigo is the easiest.


Find all triplets $ (x,y,z) \in ]0,1[^3$  Such that:

$ \left(x \plus{} \frac {1}{2x} \minus{} 1\right)\left(y \plus{} \frac {1}{2y} \minus{} 1\right)\left(z \plus{} \frac {1}{2z} \minus{} 1\right) \equal{} \left(1 \minus{} \frac {xy}{z}\right)\left(1 \minus{} \frac {yz}{x}\right)\left(1 \minus{} \frac {zx}{y}\right)$

This is equivalent to
\[ (x^2\plus{}(1\minus{}x)^2)(y^2\plus{}(1\minus{}y)^2)(z^2\plus{}(1\minus{}z)^2)\equal{}(x\minus{}yz)(y\minus{}zx)(z\minus{}xy).\]The left hand side is clearly positive, so $ x\minus{}yz,y\minus{}zx,z\minus{}xy>0$ ($ x\minus{}yz,y\minus{}zx<0<z\minus{}xy$, WLOG, means that $ 0>(x\minus{}yz)\plus{}(y\minus{}zx)\equal{}(x\plus{}y)(1\minus{}z)>0$ since $ 1>z$). But
\[ (x^2\plus{}(1\minus{}x)^2)(y^2\plus{}(1\minus{}y)^2)(z^2\plus{}(1\minus{}z)^2)\ge\frac1{64}\]by QM-AM, and
\[ (x\minus{}yz)(y\minus{}zx)(z\minus{}xy)\le\frac1{64}\]by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=318626]this thread[/url].

Thus the only solution is $ (x,y,z)\equal{}(1/2,1/2,1/2)$.


calculus
integration
limit
logarithms
calculus computations

yes it's i think it's very easy with that theorem, but can you sove it without the Dominate Convergence Theorem ? :D


Prove for positive reals:

[tex]\displaystyle (a^2+2ab)^a(b^2+2bc)^b(c^2+2ca)^c \le (a^2+b^2+c^2)^{a+b+c}[/tex]

Taking logarithms and f(x) = ln(x) we want to prove:
af(a + 2ab) + bf(b + 2bc) + cf(c + 2ca)
:le: (a + b + c) f(a + b + c).

f(x) = ln(x) is convex.
Denote S = a + b + c, x = a/S, y = b/S and z = c/S.
Then x + y + z = 1.
Also let p = a + 2ab, q = b + 2bc and r = c + 2ca.
Then Jensen gives us:
xf(p) + yf(q) + zf(s) :le: f(xp + yq + zs).
This gives us:
af(a + 2ab) + bf(b + 2bc) + cf(c + 2ab)
:le: (a + b + c) f((a + b + c + 2ab + 2bc + 2ca)/(a + b + c))
So if we can prove that
(a + b + c + 2ab + 2bc + 2ca)/(a + b + c) :le: a + b + c
we are done (since f is increasing) but this is only true if
ab + bc + ca :ge: ab + bc + ca or
(a - b)(b - c)(c - a) :ge: 0.

WLOG a = max(a, b, c). Then a - b :ge: 0 and c - a :le: 0.
So if (a - b)(b - c)(c - a) :ge: 0 then b :le: c.
Thus the case b :le: c is proved.

The case where a :ge: b :ge: c remains.

...

Ok, I solved it now, it's actually easier than it looks :)

We can suppose WLOG that [tex]a + b + c = 1[/tex].
Then weighted AM-GM (used twice) gives us
[tex](a + 2b)^a(b + 2c)^b(c + 2a)^c \leq a(a + 2b) + b(b + 2c) + c(c + 2a) = 1[/tex]
and [tex]a^a b^b c^c \leq a^2 + b^2 + c^2[/tex].
Now multiply both inequalities.

Thats nice! Easier than what I did...

From weighted AM-GM,

[tex]\displaystyle \prod_{cyc}\left(\sqrt{a^2+2ab}\right)^\frac{a}{a+b+c}\le\frac{1}{a+b+c}\sum_{cyc}a\sqrt{a^2+2ab}[/tex],

and by cauchy schwarz,

[tex]\displaystyle \sum_{cyc}a\sqrt{a^2+2ab}\le\sqrt{(a^2+b^2+c^2)(a^2+2ab+b^2+2bc+c^2+2ca)}[/tex]
[tex]\displaystyle=(a+b+c)\sqrt{a^2+b^2+c^2}[/tex].

Putting those together and raising to the power of [tex]2(a+b+c)[/tex] finishes the proof.


Let P be a point inside triangle ABC such that 3<ABP = 3<ACP = <ABC + <ACB. Prove that AB/(AC + PB) = AC/(AB + PC).

Let  $CP$ cuts $AB$ at $F$ and $BP$ cuts $AC$ at $E$. Simple angle chasing gives that $\angle{FBP}=\angle{BPF}=\angle{EPC}=\angle{PCE}$ so $BF=FP, PE=EC$. After this: $AC+PB=AE+EC+PB=AE+PE+PB=AE+BE$, similarly we obtain $AB+PC=AF+FC$. So we have to prove that $\frac{AB}{AE+BE}=\frac{AC}{AF+FC}$ and this is obvious as triangles $ABE$ and $ACF$ are similar.
I hope my solution is right (I am too lazy to check it).
 :)


On a circle we take the points $A , B , D, E , C$ ( in this order)  such that $\angle ABC = \angle BCD = \angle CDE = 45$. If $M , N$ are the projections of  $B , D$  on  lines  $CA , CE$ respectively , prove that  $AM = EN$

 [u]Babis[/u]

let m<ACB=x, then AB=2Rsin(x) so AM=2Rsinx.sin(45-x) in the same way: ED=2Rsin(45-x) so EN=2Rsin(45-x)sinx. therefore AM=EN.


geometry
3D geometry
algebra
factorization
sum of cubes

Compute the value of the summation: $\sum_{i=1}^{100}\sum_{j=1}^{i}\sum_{k=1}^{j}\sum_{m=1}^{k}1$.

[quote="knexpert"][hide="Answer"]
Apply the Hockey Stick Identity four times to find that the sum is 104 C 5=91,962,520[/hide][/quote]

I got a different answer.
[hide="Method"]
I Just worked from the right to the left, using basic summation identities. The answer i got was $\frac{1}{6}\sum_{1}^{100}(i^{3})+\frac{1}{2}\sum_{1}^{100}(i^{2})+\frac{1}{3}\sum_{1}^{100}(i) = \boxed{19080291275}$ [/hide]

[quote="nutz_for2.718281828"][hide="partial solution"] $\sum_{m=1}^{k}1=k$

$\sum_{k=1}^{j}k = \frac{j(j+1)}{2}$

What we now need is

$\sum_{i=1}^{100}\sum_{j=1}^{i}\frac{j(j+1)}{2}$[/hide][/quote]

This becomes what I have in my solution above. Just split it up into seperate summations, and use the formulae $\sum_{x=1}^{n}(x^{2}) = \frac{n(n+1)(2n+1)}{6}$ and ${\sum_{x=1}^{n}(x^{3})= ( \sum_{x=1}^{n}x})^{2}$

EDIT: Kurt is right; I had my sum-of-cubes formula mixed up with another one.

[quote="t0rajir0u"][quote="knexpert"][hide="Answer"]
Apply the Hockey Stick Identity four times to find that the sum is 104 C 5=91,962,520[/hide][/quote]

While this number may not be correct, it seems to me that this problem would be tackled most easily by a nice combinatorial interpretation rather than brute-forcing out the summation identities.[/quote]

My answer isn't correct only because I used the hockey stick identity one too many times. If I had used it three times I would have gotten 103 C 4 which is the same number that Kurt got.

I agree with Kurt Gödel's answer, but nice solutions drunner2007 and knexpert.

Wait a minute, how does the hockey stick work here???


For the real numbers $ a,b$ asociate two complex numbers $ z_A\equal{}a(1\plus{}2i)$ and $ z_B\equal{}b(3\plus{}i)$ and theit geometrycal images $ A$ and $ B$.It is known that the segment $ AB$ has the color $ n$, if $ n$ is the natural number for the points $ K\in [AB]$ where $ z_k\equal{}x\plus{}yi$, with $ x,y$ integer numbers.Prove that:
[b]a)[/b] For any natural number $ n$, the segments $ AB$ with color $ n$ exist.
[b]b)[/b] if segment $ [AB]$ has the color $ n$ and $ n\geq 2$ then the numbers $ a$ and $ b$ are rationals.

[quote="balan razvan"]It is known that the segment $ AB$ has the color $ n$, if $ n$ is the natural number for the points $ K\in [AB]$ where $ z_k \equal{} x \plus{} yi$, with $ x,y$ integer numbers.[/quote]

I don't quite understand what you mean by this. What do you mean "the natural number for a point" and where does $ z_k$ come from?

$ n$ represents the number of points $ K$....and $ z_k$ represents the afix of the points $ K$.The problem seems very weird but in fact it is not so hard...I will check my english again for this problem


linear algebra
matrix
vector
combinatorics unsolved
combinatorics

[b]N contestants answered an exam with n questions. ai points are awarded for a correct answer to question i and nil for an incorrect answer. After the questions had been marked it was noticed that by a suitable choice of positive numbers ai any desired ranking of the contestants could be achieved. What is the largest possible value of N? [/b]

The largest value of N is n. Obviously we have example with n contestants when everybody has solved only one (different) problem.
It remains to prove that $N\leq n$.
Consider the matrix $A=(a_{ij})$ where $a_{ij}$ is 0 if the i-th contestant hasn't solved the j-th problem and 1 if he has. We shall prove that the row-vectors are linearly independent. Suppose they are linearly dependent. Then there exhist natural numbers $n_{1}....n_{N}$ such that:
$n_{1}v_{1}+n_{2}v_{2}+...n_{l}v_{l}=n_{l+1}v_{l+1}....n_{N}v_{N}$ where $v_{1}..v_{N}$ are the row vectors. But now if the sum of first l n-s is smaller than the sum of the rest then we cannot make the first l contestant to be the first in the ranking and if the sum of the first l of n-s is bigger we cannot make them to be the last in the ranking. In each case we cannot make any ranking we want.
So the row vectors are linearly independent but we cannot have more than n n-dimensioned linearly independent vectors so we are done.

I didn't understand the solution :( , could anyone explain the meaning of[b] "row-vectors"[/b] plz, I couldn't find in anywhere!

If you consider $(a_{ij})$matrix the row vectors are the vectors $(a_{11},a_{12}, a_{13}..a_{1n})$ and $({a_{21},a_{22}...a_{2n})}$ and so on. The numbers in i-th row form the i-th row vector.

Can any masters show me the following in this problem? Thanks :dry:

1.what is linear dependent or independent, also, why n dimension cannot have more than n linear independent?
2.what is the relationship $ n_1v_1\plus{}...\plus{}n_lv_l\equal{}n_{l\plus{}1}v_{l\plus{}1}\plus{}...\plus{}n_Nv_N$ mean? what is to multiply a row vector with a number?
3.how all these things relate to this problem.

Treat me as somebody who don't know anything about linear stuff  :oops:

[quote="greentreeroad"]Treat me as somebody who don't know anything about linear stuff  :oops:[/quote]As is the case ;)  Go read any introduction to linear algebra -- I imagine that wikipedia has some good entries on this kind of thing.  (In general, their math articles for not-too-advanced subjects (like linear algebra) are very good.)

Is there a solution without linear algebra?


prove that for all integers a,b there exists integers x,y such that
ax+by=(a,b)

Keyword: Euclidean Algorithm.
And this is not an "Open Problem" (that would mean: unsolved for all mathematicians, not just you). Please choose an appropiate subforum next time, I moved it now.


The sum of three numbers is 50. The mean of five other numbers is 14. What is the mean of all eight numbers?


[size=150]Descartes's Rule of signs Lemma 3[/size]
[hide]   Let $r_{1},r_{2},r_{3},...,r_{k}$ denote $k$ positive numbers and let $P(x)=\prod^{k}_{i=1}(x-r_{i})$.
Then the coeffiecients of $P(x)$ are all alternating in sign and this polynomial has exactly $k$ sign variation in its coefficients.[/hide]

does $r_{1},r_{2},r_{2},...,r_{k}$ has to be consecutive positive numbers or any number but positive?
example: $x^{4}-2x^{3}+x^{2}-4x+1=0$ because the variation sign is equal to 4, does it mean it's roots are four consecutive numbers?

The $r_{i}$ do not have to be consecutive, just positive.  Also, $2n$ of them can be imaginary where $n$ is a non-negative integer $\le\frac{k}{2}$.

The first two roots are not negative, but imaginary.  They just happen to have negative real portions.

The full Descartes' sign rule states that the number of positive roots that the polynomial p(x) has is equal to the number of changes of signs between the coefficients that are used (counting from the highest power of x to the lowest power of x and ignoring any powers whose coefficients are 0) or is less than this number by a multiple of two (because complex roots will come in pairs).

Similarly, the number of negative roots is the number of positive roots of p(-x) which can be obtained by reversing the signs of the coefficients of odd powers of x, while leaving the coefficients of even powers alone.

Note that Descartes' sign rule does not give any information about the signs of the real imaginary portions of complex roots.

One last piece of information that is more basic than Descartes' sign rule: if a polynomial with real coefficients has $a+bi$ as a root $(b\not=0)$, then $a-bi$ will also be a root.  This is why the positive and negative roots are less than the upper bound by a multiple of two.


abstract algebra
inequalities
group theory
combinatorics proposed
combinatorics

Is there a sequence $ a_1, a_2, a_3, \ldots$ such that each integer is contained in the sequence exactly once and each nonzero integer is contained in the sequence $ a_1 \minus{} a_2, a_2 \minus{} a_3, \ldots$ exactly once?

What if we allow our sequence to be two-ended, i.e., is there are sequence $ \ldots, a_{ \minus{} 1}, a_0, a_1, \ldots$ containing each integer exactly once such that each nonzero integer is contained in the sequence $ \ldots, a_{ \minus{} 1} \minus{} a_0, a_0 \minus{} a_1, \ldots$ exactly once?

What about if we replace $ \bf Z$ with $ \bf Q$?

(This is the natural extension of the problem raised [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199166]here[/url] to countably infinite abelian groups.)
Edit: this properly belongs to Open Questions, as far as I know -- if you care, moderators, I have no objection to its being moved (but also none to its staying put).

In reference to the first question, here's an idea for constructing such a sequence.  I henceforth denote $ b_k \equal{} a_k \minus{} a_{k \plus{} 1}$.  I also denote the set of the first $ K$ elements of the sequence by $ S_K \equal{} \{ a_k \}_{1\leq k\leq K}$ and the corresponding first $ K \minus{} 1$ differences $ D_K \equal{} \{b_k \}_{1\leq k\leq K \minus{} 1}$.  I call a finite sequence [i]good[/i] if all its elements are distinct and all the elements of the corresponding sequence of differences are distinct.

Assume that for some $ N\geq 2$ we have a good sequence $ \{a_k \}_{1\leq k\leq N}$ satisfying $ \{x\in \mathbb{Z}: n_l \leq x \leq n_r\} \subset S_N$, $ \{ x\in \mathbb{Z} \backslash \{ 0\}: m_l \leq x \leq m_r\} \subset D_N$, $ n_l,m_l < 0, n_r,m_r > 0$, $ n_l \minus{} 1,n_r \plus{} 1\notin S_N$, $ m_l \minus{} 1,m_r \plus{} 1\notin D_N$, then by adding six elements we construct a good sequence $ \{ a_k\}_{1\leq k\leq N \plus{} 8}$ such that $ \{x\in \mathbb{Z}: n_l \minus{} 1 \leq x \leq n_r \plus{} 1\} \subset S_{N \plus{} 6}$, $ \{ x\in \mathbb{Z} \backslash \{ 0\}: m_l \minus{} 1 \leq x \leq m_r \plus{} 1\} \subset D_{N \plus{} 6}$ as follows:
[hide="more notation"]
\[ a_{\min} \equal{} \min_{a\in S_N}a \\
a_{\max} \equal{} \max_{a\in S_N}a \\
b_{\min} \equal{} \min_{b\in D_N}b \\
b_{\max} \equal{} \max_{b\in D_N}b
\]
and $ c_1,c_2$ are non-negative integers to be selected (they're essentially arbitrary in some sense, clarified later).[/hide]
\[ a_{N \plus{} 1} \equal{} a_{\max} \plus{} b_{\max} \minus{} b_{\min} \plus{} c_1 > a_{\max} \\
a_{N \plus{} 2} \equal{} a_{N \plus{} 1} \minus{} (m_l \minus{} 1) > a_{\max} \\
a_{N \plus{} 3} \equal{} n_r \plus{} 1 \\
a_{N \plus{} 4} \equal{} a_{\min} \plus{} b_{\min} \minus{} b_{\max} \minus{} c_2 < a_{\min} \\
a_{N \plus{} 5} \equal{} a_{N \plus{} 4} \minus{} (m_r \plus{} 1) < a_{\min} \\
a_{N \plus{} 6} \equal{} n_l \minus{} 1
\]
the inequalities imply none of these elements are in $ S_N$, similarly
\[ b_{N} \equal{} a_n \minus{} a_{\max} \plus{} b_{\min} \minus{} b_{\max} \minus{} c_1 < b_{\min} \\
b_{N \plus{} 1} \equal{} m_l \minus{} 1 \\
b_{N \plus{} 2} \equal{} a_{\max} \plus{} b_{\max} \minus{} b_{\min} \minus{} m_l \minus{} n_r \plus{} c_1 > b_{\max} \\
b_{N \plus{} 3} \equal{} n_r \plus{} 1 \minus{} a_{\min} \minus{} b_{\min} \plus{} b_{\max} \plus{} c_2 > b_{\max} \\
b_{N \plus{} 4} \equal{} m_r \plus{} 1 \\
b_{N \plus{} 5} \equal{} a_{\min} \plus{} b_{\min} \minus{} b_{\max} \minus{} m_r \minus{} n_l \minus{} c_2 < b_{\min}
\]
so none of the differences are in $ D_N$, and therefore $ \{a_k\}_{1\leq k \leq N \plus{} 6}$ is a good sequence (it's easy to check that $ \{ a_j \}_{N \plus{} 1\leq j\leq N \plus{} 6}$ are all distinct, as are $ \{ b_j \}_{N\leq j \leq N \plus{} 5}$). [hide="remark"] If we didn't have the 'arbitrary' constants $ c_1,c_2$ it would be possible that $ b_{N} \equal{} b_{N \plus{} 5}$ or $ b_{N \plus{} 2} \equal{} b_{N \plus{} 3}$; it is to avoid this scenario that we need the degrees of freedom given by the constants.  Note that we only need one of the constants to be unequal to at most two particular values in order to guarantee that the elements are unique, so the constants are essentially arbitrary in the sense that all but a finite number of choices yield good sequences.  This suggests there is an uncountable number of sequences with the given properties. [/hide]
Furthermore, we have that $ \{x\in \mathbb{Z}: n_l \minus{} 1 \leq x \leq n_r \plus{} 1\} \subset S_{N \plus{} 6}$ and $ \{ x\in \mathbb{Z} \backslash \{ 0\}: m_l \minus{} 1 \leq x \leq m_r \plus{} 1\} \subset D_{N \plus{} 6}$ as desired.

This gives us a recurrence relation to construct the sequence we want.  Specifically, the starting values $ a_1 \equal{} 0$, $ a_2 \equal{} 1$, $ a_3 \equal{} \minus{} 1$, $ a_4 \equal{} \minus{} 2$, give $ n_l \equal{} \minus{} 2$, $ n_r \equal{} 1$, $ m_l \equal{} \minus{} 1$, $ m_r \equal{} 2$, and can be used to build a good sequence with $ 10$ elements covering at least the integers in $ [ \minus{} 3,2]$, which can then be used to build a good sequence with $ 16$ elements covering at least the integers in $ [ \minus{} 4,3]$, and so forth ad infinitum.  After $ k$ iterations, the sequence of length $ 6k \plus{} 4$ is guaranteed to contain at least the integers in $ [ \minus{} 2 \minus{} k,1 \plus{} k]$, and the sequence of differences must contain at least the nonzero integers in $ [ \minus{} 1 \minus{} k,2 \plus{} k$].

The notation isn't great and the argument isn't very polished, but hopefully the idea itself didn't get lost.  Basically we're just adding on elements to the sequence in such a way that every other element is so far out of the range covered by the previous elements that they don't interact.  It seems the argument can be used in essentially the same form for the second question, but for $ \mathbb{Q}$ it's not clear if the approach would work.

[hide="Mathematica script"]
[code]itrs = 17; a = Table[0, {i, 4 + 6 itrs}];
b = Table[0, {i, 3 + 6 itrs}];
a[[1]] = 0; a[[2]] = 1; a[[3]] = -1; a[[4]] = -2;
amax = 1; amin = -2; nl = -2; nr = 1;
b[[1]] = -1; b[[2]] = 2; b[[3]] = 1;
bmax = 2; bmin = -1; ml = -1; mr = 2;
For[k = 1, k <= itrs, k++,
  good = 0; c = -1;
  While[good == 0 ,
   c = c + 1;
   a[[6 k - 1]] = amax + bmax - bmin + c;
   a[[6 k]] = a[[6 k - 1]] - (ml - 1);
   a[[6 k + 1]] = nr + 1;
   a[[6 k + 2]] = amin + bmin - bmax;
   a[[6 k + 3]] = a[[6 k + 2]] - (mr + 1);
   a[[6 k + 4]] = nl - 1;
   For[j = 1, j <= 6, j++,
    b[[6 k + j - 3]] = a[[6 k + j - 3]] - a[[6 k + j - 2]]];
   If[b[[6 k - 2]] != b[[6 k + 3]] && b[[6 k]] != b[[6 k + 1]],
    good = 1, good = 0]];
  i = nr+1; While[Length[Intersection[a, {i}]] > 0, i = i + 1];
  nr = i - 1;
  i = mr+1; While[Length[Intersection[b, {i}]] > 0, i = i + 1];
  mr = i - 1;
  i = nl-1; While[Length[Intersection[a, {i}]] > 0, i = i - 1];
  nl = i + 1;
  i = ml-1; While[Length[Intersection[b, {i}]] > 0, i = i - 1];
  ml = i + 1;
  amax = Max[a]; amin = Min[a]; bmax = Max[b]; bmin = Min[b];];
Print[Commonest[b] == b]; Print[Commonest[a] == a];
Print[Intersection[Table[i, {i, -2 - itrs, 1 + itrs}], a] ==
   Table[i, {i, -2 - itrs, 1 + itrs}]];
Print[Intersection[Table[i, {i, -1 - itrs, 2 + itrs}],
    Union[b, {0}]] == Table[i, {i, -1 - itrs, 2 + itrs}]];[/code]
The itrs variable at the top can be changed to determine how many members of the sequence to generate (it generates the first $ 6itrs \plus{} 4$ elements).  The four print commands at the end should return True if the resulting sequence is good and covers the integers in the appropriate range.  The sequence itself is stored under list variable a, and the differences under list b.

Running it with $ itrs \equal{} 1000$ gives $ a_{6000}\approx 3.92\cdot 10^{477}$, but Mathematica still evaluates each of the four checks to True. :)

[/hide]

The fact that the group is infinite gives you a lot of freedom in the way you arrange the elements. In fact, the folowing holds:

Let $ G$ be a countable abelian group with no elements of order two. There is an enumeration $ (a_n)_{n\ge 0}$ of $ G$ such that the sequence $ (a_{n \plus{} 1} \minus{} a_n)_{n\ge 0}$ is an enumeration of $ G\setminus\{0\}$.

First, order the elements of $ G$ in an arbitrary way, so that it makes sense to speak of "smaller" and "larger" elements. I'll say that a finite sequence $ a_0,\ldots,a_n$ is "good" if its elements are different and all the differences $ a_{n \plus{} 1} \minus{} a_n$ are also different. We now place the elements of $ G$ in the sequence $ (a_n)$ recursively as follows: first put $ a_0 \equal{} 0$, and regard this as step $ 0$. In each step $ \ge 1$ we place not one element of $ G$ in the sequence, but two.

Assume we've finished taking steps $ 0,1,\ldots,2n\ (n\ge 0)$, and hence we've chosen the elements $ a_0,\ldots,a_{4n}$ such that they form a good sequence. Our goal is now to make $ a_{4n \plus{} 2}$ the smallest element not appearing among $ a_0,\ldots,a_{4n}$. Clearly, because $ G$ is infinite, it's possible to choose $ a_{4n \plus{} 1}$ in such a way that with our choice of $ a_{4n \plus{} 2}$, the new sequence $ a_0,\ldots,a_{4n \plus{} 2}$ is good. This was step $ 2n \plus{} 1$.

Now to describe the even-numbered steps. Suppose we're done with steps $ 0,1,\ldots,2n \minus{} 1\ (n\ge 1)$ so we have the good sequence $ a_0,\ldots,a_{4n \minus{} 2}$. We want to choose $ a_{4n \minus{} 1}$ and $ a_{4n}$ such that $ a_{4n} \minus{} a_{4n \minus{} 1}$ is the smallest element different from $ 1$ not appearing among the products $ a_{i \plus{} 1} \minus{} a_i,\ i \equal{} \overline{0,4n \minus{} 3}$. Again, it should be clear that we can always do this such that the resulting sequence $ a_0,\ldots,a_{4n}$ is good.

The "no elements of order two" hypothesis is needed in order to conclude that given $ a,b\in G$, there is at most one element $ x\in G$ such that $ x \minus{} a \equal{} b \minus{} x$; we need to avoid such $ x$ when we make some of the choices above. Maybe we could drop that assumption or make it weaker - I don't know.

You can even extend this statement to uncountable groups, with the right definition of "sequence". In the general case the elements should be well-ordered with the order type of the cardinal number of $ G$ (i.e. the smallest ordinal number having the cardinality of $ G$).


how do u simplify expressions like $ \sqrt{65\minus{}\sqrt{47\minus{}\sqrt{23}}}$ again? i think it was possible, with this setting equal to x or something...(haven't been working on my algebra recently, so yeah.....)  :)

[hide="hint"]square both sides and see what happens :)[/hide]

yeah, and then u move the constants over there too (and then square, move constants, etc.), and at the end, u wind up with a bunch of algebra, with like x to the power of 8 or something (i thought of that method as the "slow" method of solving this problem before i even posted this topic. thought there was some easier way). :mad: solving the equation would take a lot of time for a problem like this.

Actually, it might help to first simplify the $ \sqrt{47\minus{}\sqrt{23}}$ first. This can be done by letting $ x\equal{}a\plus{}b\sqrt{23}$, since 23 is prime.

wat. i heard of that type of solution before. then what do u do?  :)

[hide]$ \left(a \plus{} b\sqrt {23}\right)^2 \equal{} 47 \minus{} \sqrt {23}\implies47 \minus{} \sqrt {23} \equal{} a^2 \plus{} 23b^2 \plus{} 2ab\sqrt {23}$. Since $ a,b\in \mathbb{R}$, we have the two simultaneous equations
(1) $ a^2 \plus{} 23b^2 \equal{} 47$
(2) $ 2ab\sqrt {23} \equal{} \minus{} \sqrt {23}\implies{ab \equal{} \minus{} \frac12}\implies{a \equal{} \minus{} \frac1{2b}}$
Substituting (2) in (1), we have $ \frac1{4b^2} \plus{} 23b^2 \equal{} 47\implies92b^4 \minus{} 47b^2 \plus{} 1 \equal{} 0$. Using the quadratic formula, we have $ b^2 \equal{} \frac {47\pm\sqrt {1841}}{184}$. However, 1841 is not a perfect square, and $ b\in\mathbb{R}\implies{b^2\in\mathbb{R}}$, and this cannot be simplified.
However, something that works...
For example, $ (2 \plus{} \sqrt5)^2 \equal{} 9 \plus{} 4\sqrt5$. So say we have $ \sqrt {9 \plus{} 4\sqrt5}$. Again, we let $ a\sqrt {n} \plus{} b\sqrt {c} \equal{} \sqrt {9 \plus{} 4\sqrt5}$. WLOG, we let $ (n,c) \equal{} (1,5)$, since 5 is prime. Squaring both sides, we obtain $ a^2 \plus{} 5b^2 \plus{} 2ab\sqrt5 \equal{} 9 \plus{} 4\sqrt5$. Now we have two equations (as above)
(1) $ a^2 \plus{} 5b^2 \equal{} 9$
(2) $ ab \equal{} 2\implies{a \equal{} \frac2b}$
$ \frac4b^2 \plus{} 5b^2 \equal{} 9\implies5b^4 \minus{} 9b^2 \plus{} 4 \equal{} 0$. Factoring, we get $ (5b^2 \minus{} 4)(b^2 \minus{} 1) \equal{} 0\implies{b^2 \equal{} \left\{\frac45,1\right\}}$. Note that $ b\in\mathbb{R}$, though, so the only possible solution is $ b^2 \equal{} 1\implies{b \equal{} \pm1}$. Substituting in (2), we get the two solutions $ 2 \plus{} \sqrt5$ and $ \minus{} 2 \minus{} \sqrt5$. However, the square root must be positive, so our only solution is $ 2 \plus{} \sqrt5$.[/hide]

Hope this helps! :)

so basically there's some cases where it can be simplified and some that can't?

is there a trick to tell if it can be simplified?  :)

There's always a solution if the number in the radical (i.e. 1841,...) is a perfect square...
So if we have prime c, $ \sqrt{a\plus{}b\sqrt{c}}\equal{}d\plus{}e\sqrt{c}$, then
$ a\plus{}b\sqrt{c}\equal{}d^2\plus{}ce^2\plus{}2de\sqrt{c}$
If you try solving this, you'll eventually get some random stuff in the radical...that has to be a perfect square.


The altitudes $ AD, BE, CF$ of the triangle $ ABC$ meet at $ H$. Points$ P, Q, R$ are taken on the segments $ AD, BE, CF$ respectively, so that the sum of the areas of triangles $ ABR, AQC$ and $ PBC$ equals the area of $ ABC$. Show that $ P, Q, R, H$are cyclic.

[quote="Superficial"]The altitudes $ AD, BE, CF$ of the triangle $ ABC$ meet at $ H$. Points$ P, Q, R$ are taken on the segments $ AD, BE, CF$ respectively, so that the sum of the areas of triangles $ ABR, AQC and PBC$ equals the area of $ ABC$. Show that $ P, Q, R, H$are cyclic.[/quote]
[hide="Solution"]
Let the parallel through $ P$ to $ BC$ and the parallel through $ Q$ to $ AC$ meet at $ X$.  Thus,\[ [AXB] \equal{} [ABC] \minus{} [AXC] \minus{} [BXC] \equal{} [ABC] \minus{} [BPC] \minus{} [AQC] \equal{} [ABR]\]so $ XR\parallel AB$.  Now, notice that $ AB\perp CR$, so $ \angle XRH \equal{} 90$.  Similarly, $ \angle HQX \equal{} 90$, so $ \angle XQH \plus{} \angle XRH \equal{} 180$, which implies that $ HQXR$ is cyclic.  Now, we can also notice that $ \angle AFH \plus{} \angle AEH \equal{} 90 \plus{} 90 \equal{} 180$, so $ AFHE$ is cyclic.  Also, $ \angle CFB \equal{} 90 \equal{} \angle BEC$, so $ FECB$ is cyclic.  This gives us that since $ XP\parallel BC$ and $ QX\parallel AC$, we have that\[ \angle QXP \equal{} \angle ACB \equal{} \angle AFE \equal{} \angle AHE \equal{} \angle QHP\]so $ QHXP$ is cyclic.  This means that $ QHRXP$ is cyclic, so $ QHRP$ is cyclic.  [/hide]


There are two values for $a$ for which the equation $4x^{2}+ax+8x+9=0$ has only one solution for $x$. What is the sum of those values of $a$?

A. -16

B. -8

C. 0

D. 8

E. 20

[hide="solution"]For there to only be one solution, the trinomial must be a perfect square.  This can be obtained as $(2x+3)^{2}$ or $(2x-3)^{2}$. [the other two values give the same value for a] Expanding these, we get $12x$ and $-12x$ as the x terms.  so the values of $a$ are $4$ and $-20$, which have a sum of $\boxed{-16, A}$[/hide]

[hide="Hint to an alternate solution"] The discriminant is zero. [hide="So that means..."] $(a+8)^{2}-4 \cdot 4 \cdot 9 = 0$  [hide="And notice that..."] We can find the sum of the solutions to this equation by considering its linear term, which is $16a$.  The answer to this question is actually independent of the value of the constant and quadratic terms!  [/hide] [/hide] [/hide]

[hide]Discriminant $=0=a^{2}+16a+...$, from Vieta, the sum of the values of a is -16, $\boxed{A}$.[/hide]


geometry
cyclic quadrilateral
geometry unsolved

Give a triangle $ABC$ with $\angle B = 90^{\circ}$ and $D$ be a point on side $AC$. $M$ is  midpoint of side $BD$. $H$ is orthocenter of triangle $BCD$. Find the point such that triangle $ABM$ is similar to triangle $BCH$.

Let $BH$ intersect $AC$ at $P$, $CH$ intersect $BD$ at $Q$. Then we have $PHQD$ a cyclic quadrilateral so we have $\angle{BHC}=\angle{ADB}$. Also $\angle{HBC}=\angle{HBQ}+\angle{DBC}=90^{o}-\angle{PDA}+\angle{DBC}=90^{o}-\angle{DBC}=\angle{BAD}$
So we have triangles $BAD$ and $BHC$ similar so $\angle{ABM}=\angle{HCB}$
Because of the requirement of the problem, we have triangles $BAD$ and $ABM$ similar [b]and this is possible, Jan[/b]. So the only case when that is true is if $\angle{BAM}=\angle{ADB}$ and $\angle{AMB}=\angle{BAD}$
So:
$\frac{BM}{AB}=\frac{AB}{BD}$ so $2AB^{2}=BD^{2}$ so we must take $D$ on $AC$ such that $BD=AB\sqrt{2}$


When I log on with #H34N1 and my password, a blank page shows up. Also the username [b]anirudh[/b] still works.

[quote="D3m0n Shad0w"][quote="7h3.D3m0n.117"][quote="DPatrick"]Wiki usernames must begin with a letter.  You'll have to keep your old username for the wiki.[/quote]

really? I believe my username works?[/quote]

well #H34N1 has no letters while u have letters :D[/quote]

No DPatrick said that it had to START with a letter..

I'm guessing it's the # sign in your username that is tripping it up . . .


This happens to me too, it appears the media player if broken. I'd suggest googling for other free media players and find a quick download.

Media Player Classic :D
http://sourceforge.net/project/showfiles.php?group_id=82303&package_id=84358

I use BSPlayer.
[url]http://www.bsplayer.org/[/url] : it can be downloaded for free here.

The program is especially convenient when working with subtitles (drag and drop) and it also allows you to take nice snapshots of the movie.

I use VLC media player. [url=http://www.download.com/VLC-Media-Player/3000-2194_4-10267151.html]here[/url]

I use MPC and VLC (only the latter in Linux). Try getting K-Lite Codec Pack if a video still doesn't play after that.

[quote="ccy"][quote="Ignite168"]I recommend Winamp.

If you want me to buy you a license, just PM me.[/quote]

Ignite, that's illegal.  Don't do it.[/quote]

How so ?

Winamp is incompatible with this computer. I've already bought it, why can't I give the license to other people?

haha jathan you're a crazy liar

and fredbel, BSPlayer? rofl

[quote="Treething"]
and fredbel, BSPlayer? rofl[/quote]
Why is that funny? :maybe:

[quote="Silverfalcon"]For odd reason, I cannot seem to play MPG files (.mpg or .mpeg) in my Windows Media Player. It just shut down. I have Windows Media Player Version 10 and Windows XP so is there anything in configuration that I can do to fix this? I'm like novice when it comes to Windows Media Player.  :maybe:[/quote]

I had this kind of problem when I wanted to play an .avi on WMP. Eventually, I downloaded VLC Media Player, which enables you to play all sorts of extensions. (I needed .avi for Family Guy  :roll: )

[quote="Kurt Gödel"]
I had this kind of problem when I wanted to play an .avi on WMP. Eventually, I downloaded VLC Media Player, which enables you to play all sorts of extensions. (I needed .avi for Family Guy  :roll: )[/quote]
Uhm, two weird things about BSplayer though :

1. When I watch a video and double click on a music file to play it in BSPlayer (my default player) he starts playing it but with an error popup.  When I stop BSPlayer and start it again the problem is gone.  He doesn't do this when switching from movie to movie.  So it's like once I enter the "land of films" I have to stick around there.
2. One episode "Brian,Portret of a Dog" from Family Guy can't be played in any of my players, and BSPlayer really annoys my computer (makes all sorts of buttons black) when I try to play it and I have to restart my computer to make it right again???!

[quote="Kurt Gödel"]I have [b]VLC Media Player[/b], not [b]BSplayer[/b]!

That thing you've got with "Brian,Portret of a Dog" is really strange...download VLC  :wink: Maybe BSplayer is rather a music player than a movie player?[/quote]
I quoted you because you have Family Guy.
I will look into this "VLC".

Of course me and Kurt only own legal copies of Family Guy. :wink:

Everybody please abandon Windows Media Player and now that you're on it, abandon windows as well. Install a nice beautiful linux distribution, download mplayer if you dont have it already and there you are, the best player ever.

Daniel

PS have you noticed the funny, typical way of talking when you want to discuss the subject linux vs windows.

[quote="fredbel6"]
I will look into this "VLC".
[/quote]

VideoLAN (VLC) is much better than WMP. WMP freezes to much. And you can't play so many different file extensions as in VLC.

 :)
read [url=http://en.wikipedia.org/wiki/VLC_media_player]"the special features"[/url].  :P

[quote="deimos"]
VideoLAN (VLC) is much better than WMP. WMP freezes to much. And you can't play so many different file extensions as in VLC.

 :)
read [url=http://en.wikipedia.org/wiki/VLC_media_player]"the special features"[/url].  :P[/quote]

Looks nice, but I'm too much of a rookie to actually gain from those things anyway :blush: .

But I would like to point out that while BSPlayer couldn't play my Family Guy episode, BSPlayer is ver very handy (drag and drop) to try out different subtitles, correct their timing,.... :)

Deimos thanks for the link but you can even get the url in your browser of those paragraphs if you click on the link in the "Contents" box :
[url]http://en.wikipedia.org/wiki/VLC_media_player#Special_features[/url]
Wikipedia is our friend. :roll:

[quote="fredbel6"]
Deimos thanks for the link but you can even get the url in your browser of those paragraphs if you click on the link in the "Contents" box :
[url]http://en.wikipedia.org/wiki/VLC_media_player#Special_features[/url]
Wikipedia is our friend. :roll:
[/quote]

I made an wikipedia account three-four years ago, and still I don't know to use it.  :blush:

thanks, it's good to learn new things.  :P

VLC is really good...
musicmatch jukebox is also good


geometry
3D geometry
sphere
ratio
trigonometry
geometric series

Say a cone has an angle of $30^\circ$ at its vertex, and a sphere with radius $1$ is inscribed within it (as in it is pushed inside as far as possible.) This is the first sphere, and a total of $n$ are inscribed within the cone such that each sphere after the first is tangent to the sphere drawn before it and to the cone, [i]and[/i] each sphere is consecutively larger than the last. After $n$ spheres are drawn, what is the total internal volume between the spheres and the cone itself?

Well are the spheres going towards the vertex/getting smaller, or are they going away from the vertex/getting bigger?

[quote="computer_nuke"][quote="cincodemayo5590"]Oh, oops I forgot to specify. They are going away from vertex/getting bigger. Sorry about that.  :)[/quote]

Does the bottom of the cone continue forever? I mean when you ask "what is the total internal volume between the spheres and the cone itself" after n spheres are drawn, do you mean after the nth sphere is drawn, you close the cone and then calculate the area.[/quote]

The latter is exactly what I meant. (It's volume, though, not area.  :) )

Pretty sure I did this wrong because my solution is so ugly...
[hide]Take a look at a cutaway of the spheres inscribed in the cone. It will be circles inscribed inscribed in a triangle (see diagram). Let $r_{n}$ be the radius of the nth circle. Let $d$ be the distance of the triangle's tip. By similiar triangles you know the ratio of the hypotenuse to the radius is always equal, so:
\begin{eqnarray*}\frac{r_{n}+2(r_{1}+r_{2}+\cdots+r_{n-1})+d}{r_{n}}&=& \frac{r_{1}+d}{r_{1}}\\ r_{1}r_{n}+2r_{1}(r_{1}+r_{2}+\cdots+r_{n-1})+r_{1}d &=& (r_{1}+d)r_{n}\\ \end{eqnarray*}
From this similarity, you know this:
\begin{eqnarray*}r_{n}&=& \frac{2r_{1}(r_{1}+r_{2}+\cdots+r_{n-2}+r_{n-1})+r_{1}d}{d}\\ r_{n-1}&=& \frac{2r_{1}(r_{1}+r_{2}+\cdots+r_{n-2})+r_{1}d}{d}\end{eqnarray*}
Subtracting $r_{n-1}$ from $r_{n}$:
\begin{eqnarray*}r_{n}-r_{n-1}&=& \frac{2 r_{1}r_{n-1}}{d}\\ \implies r_{n}&=& (\frac{2 r_{1}}{d}+1) r_{n-1}\\ \implies r_{n}&=& (\frac{2 r_{1}}{d}+1)^{n-1}\end{eqnarray*}
This shows that the radii go in a geometric sequence. We can find $d$ using the sine of the angle:
\begin{eqnarray*}\frac{r_{1}+d}{r_{1}}&=& 1+d = \frac{1}{\sin{15^\circ}}= \sqrt{6}+\sqrt{2}\\ \implies d &=& \sqrt{6}+\sqrt{2}-1 \end{eqnarray*}
Let $k = (\frac{2}{\sqrt{6}+\sqrt{2}-1}+1)$, the constant term of the geometric sequence. The volume in between the spheres and the cone will be:
$\frac{1}{3}\pi r^{2}h-\frac{4}{3}\pi (r_{1}^{3}+r_{2}^{3}+\cdots+r_{3}^{3})$
Most of these things are geometric series:
$h = 2(r_{1}+r_{2}+\cdots+r_{n})+d = 2(k^{0}+k^{1}+\cdots+k^{n-1})+d = 2(\frac{1-k^{n}}{1-k})+d$
$\tan{15^\circ}= r/h \implies r^{2}= (h \tan{15^\circ})^{2}= h^{2}(2-\sqrt{3})^{2}$
$r_{1}^{3}+r_{2}^{3}+\cdots+r_{n}^{3}= k^{0}+k^{1}+\cdots+k^{n-1}= \frac{1-k^{3n}}{1-k}$
$V=\frac{(2-\sqrt{3})^{2}\pi}{3}(2(\frac{1-k^{n}}{1-k})+d)^{3}-\frac{4\pi}{3}((\frac{1-k^{3n}}{1-k})+d)$
I didn't feel like substituting and simplifying this expression so I used Mathematica to get:
$V = \frac{\pi}{3}(-2\sqrt{6}-2\sqrt{2}+2+(11\sqrt{6}+16\sqrt{3}-12\sqrt{2}-29)(1+\frac{2}{\sqrt{6}+\sqrt{2}-1})^{3n}))$[/hide]


In triangle $ ABC$ the points $ D, E$, and $ F$ trisect the sides so that $ BC\equal{}3BD$, $ CA\equal{}3CE$, and $ AB\equal{}3AF$. Similarly, the points $ G, H$, and $ I$ trisect the sides of triangle $ DEF$ such that $ EF\equal{}3EG$, $ FD\equal{}3FH$, and $ DE\equal{}3DI$. Prove that the sides of triangle $ GHI$ are parallel to the sides of triangle $ ABC$ and that each side of the smaller triangle is $ \frac{1}{3}$ as long as its parallel side in the larger triangle.

I used vectors to prove that each side of the smaller triangle is $ \frac{1}{3}$ as long as its parallel side in the larger triangle, but I can't figure out how to prove that the sides are parallel.

Is there any way to prove that the two sides are parallel using vectors?

Let $ AB \equal{} a, AC\equal{}b$, where $ AB$ and $ AC$ should have arrows on top and $ a$ and $ b$ tildes uderneath.

then $ GI \equal{} GE\plus{}EI \equal{} \frac{1}{3} FE \plus{} \frac{2}{3}ED$

$ \equal{}\frac{1}{3}(FA\plus{}AE) \plus{} \frac{2}{3} (EC\plus{}CD)$

$ \equal{}\frac{1}{3}(\frac{\minus{}1}{3}a\plus{}\frac{2}{3}b) \plus{} \frac{2}{3}(\frac{1}{3}b\plus{} \frac{2}{3}(a\minus{}b))$

$ \equal{}\frac{1}{3}a$

This shows that $ GI$ is parallel to $ AB$ and $ \frac{1}{3}$ of its length. Similarly for the other sides.

So I suspect you've shown that they're parallel.

thanks andrewtom.
but I just need help with proving the parellelism, I already got how to prove that $ GI$ was a third of $ AB$

Hi Demosthenes.

Once you know that the vector $ GI$ is one third of the vector $ AB$, this means that the length of $ GI$ is one third of the length of $ AB$ AND that $ GI$ is parallel to $ AB$ because if one vector is a multiple of another then the two vectors are parallel.

I hope this helps a bit more. If I haven't explained it very well, perhaps someone else will.


Could anybody please post a batch file here that Windows users could use with TeXnicCenter so that we don't have to manually compile three times to process a tex file with asy code in it?

I've wanted to do it on my own, but I don't really know which commands I use in MikTex, and the arguments and the like.

Just copy the commands into a new profile. So for example
C:\your path\bin\pdflatex.exe -interaction=nonstopmode "%pm"
is the code for generating a PDF
Then put the code for Asymptote you already have in the Postprocessor tab and repeat the pdflatex code below it.
In other words it's just a matter of copying and pasting into the profile.

But personally I'm not sure about this. Everyone (and I mean everyone) makes mistakes* and that's a good way to learn. The second time you run pdflatex it will overwrite the previous log file and you can't then easily spot the errors/warnings/bad boxes. I prefer to run pdflatex, check the log (which TexnicCenter makes easy), if everything is OK I then run Asymptote check for errors (I have put a batch file on the Wiki which puts errors into TexnicCenter), then, if OK, run pdflatex again and check for errors.
If you make a bad error on the first run you could end up with a terrible mess after the third run.

* I think someone once said it was impossible for a human being to write more than a dozen lines of code without making an error first time.

I went to Define Output Profiles, and I copied the LaTeX => PDF into a new profile. I went to the Postprocessor, and for Asymptote, I typed in [b]C:\Program Files\Asymptote\asy.exe[/b] for executable, [b]-batchView -tex "pdflatex" %tc[/b] for arguments, and [b]%dc[/b] for both input and output redirections.

Then I make a new thing under postprocessor for pdflatex, and I type in the code you provided me, separating the arguments from the executable.

But then when I compile, the asy command window doesn't even run. Why?
Even if I remove the [b]%dc[/b], it still doesn't work.

The asymptote file is not the same file as the tex one so you must change the extension of the file being procesed by asymptote. All you need do is replace %tc by %bm.asy

Still the window doesn't open up.

Also, what do %tc and %bm mean? Is there a glossary anywhere?

It's not supposed to - it does the processing in the background - there'll be a pause after LaTeX has done its bit and before the errors and warnings are displayed. You'll see the files have been created and you can view the output.

The TexnicCenter help file gives all the placeholders like %bm.

Never mind. I fixed it. Thanks! I removed the %dc and it worked.


Is there a program for including $\LaTeX$ in Yahoo and MSN Messenger????

I am sure there is, they might even be using it here in the AoPS classes... but I don't know much about it.

Read the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50046]"LATEX OVER AIM!!!!"[/url] in this forum.


trigonometry
inequalities
LaTeX
inequalities proposed

Let ABC be an acute triangle . Prove that :

sum(cosAcosB/cosC) + 4cosAcosBcosC >= 2

[quote="tranminhhoang"]Let ABC be an acute triangle . Prove that :

$\sum(\cos A \cos B/\cos C) + 4\cos A\cos B\cos C \geq 2$[/quote]

Making the substitution: $x= (b^2+c^2-a^2)/2$, $y= (c^2+a^2-b^2)/2$ and $z= (a^2+b^2-c^2)/2$, then $x,y,z>0$ since ABC is an acute triangle. And we have $a=\sqrt{y+z}$, $b=\sqrt{z+x}$ and $c=\sqrt{x+y}$. So the trig inequality is equivalent to:

\[
\frac{xy}{z(x+y)} + \frac{yz}{x(y+z)} + \frac{zx}{y(z+x)} + \frac{4xyz}{(x+y)(y+z)(z+x)} \geq 2.
\]

If my calculation is correct, this can be proved by Schurs on $xy,yz,zx$ after simplifying.

I rewrite your problem with LaTeX notation:

[i]Let ABC be an acute triangle. Then prove that

[tex]\frac{\cos B \cos C}{\cos A} + \frac{\cos C \cos A}{\cos B} + \frac{\cos A \cos B}{\cos C} + 4 \cos A \cos B \cos C \geq 2[/tex].[/i]

Here is my solution of this problem. In fact, since the triangle ABC is acute, its angles A, B, C are < 90°, so that the angles 90° - A, 90° - B, 90° - C are all positive. Hence, the angles 180° - 2A, 180° - 2B, 180° - 2C are also positive. Also, all of these angles are < 180°, but their sum is = 180°, since

  $\left( 180^{\circ} - 2A \right) + \left( 180^{\circ} - 2B \right) + \left( 180^{\circ} - 2C \right)$
  $= 3 \cdot 180^{\circ} - 2\left( A+B+C\right) = 3 \cdot 180^{\circ} - 2 \cdot 180^{\circ} = 180^{\circ}$.

Hence, these angles 180° - 2A, 180° - 2B, 180° - 2C are the angles of a triangle. If we denote X = 180° - 2A, Y = 180° - 2B, Z = 180° - 2C, then we have thus showed the existence of a triangle with the angles X, Y, Z. Call this triangle XYZ, and its sidelengths x = YZ, y = ZX and z = XY. Furthermore, let $\displaystyle s = \frac{x+y+z}{2}$ be the semiperimeter of the triangle XYZ.

Moreover, since $X = 180^{\circ} - 2A$, we have $\displaystyle A = 90^{\circ} - \frac{X}{2}$, so that $\displaystyle \cos A = \sin \frac{X}{2}$. Similarly $\displaystyle \cos B = \sin \frac{Y}{2}$ and $\displaystyle \cos C = \sin \frac{Z}{2}$. Thus, the inequality which must be proven takes the form

[tex]\frac{\sin \frac{Y}{2} \sin \frac{Z}{2}}{\sin \frac{X}{2}} + \frac{\sin \frac{Z}{2} \sin \frac{X}{2}}{\sin \frac{Y}{2}} + \frac{\sin \frac{X}{2} \sin \frac{Y}{2}}{\sin \frac{Z}{2}} + 4 \sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} \geq 2[/tex].

Now, by the half-angles theorem, we have

[tex]\sin \frac{X}{2} = \sqrt{\frac{\left(s-y\right) \left(s-z\right)}{yz}}[/tex];
[tex]\sin \frac{Y}{2} = \sqrt{\frac{\left(s-z\right) \left(s-x\right)}{zx}}[/tex];
[tex]\sin \frac{Z}{2} = \sqrt{\frac{\left(s-x\right) \left(s-y\right)}{xy}}[/tex].

Using these identities, the above inequality simplifies to

[tex]\frac{s-x}{x} + \frac{s-y}{y} + \frac{s-z}{z} + 4 \frac{\left(s-x\right) \left(s-y\right) \left(s-z\right) }{xyz} \geq 2[/tex].

Now, remembering that $\displaystyle s = \frac{x+y+z}{2}$ and that x, y, z are the sides of a triangle, we see that the numbers d, e, f defined by d = s - x, e = s - y and f = s - z are positive and satisfy the equations x = e + f, y = f + d and z = d + e. Hence we can rewrite our above inequality in the form

[tex]\frac{d}{e+f} + \frac{e}{f+d} + \frac{f}{d+e} + 4 \frac{de f}{\left(e+f\right) \left(f+d\right) \left(d+e\right) } \geq 2[/tex].

This can be proved in a straightforward way by multiplying with (e + f) (f + d) (d + e):

[tex]d \left(f+d\right) \left(d+e\right) + e \left(d+e\right) \left(e+f\right) + f \left(e+f\right) \left(f+d\right) + 4 de f \geq 2 \left(e+f\right) \left(f+d\right) \left(d+e\right) [/tex].

Expand everything and get

[tex]d^3 + e^3 + f^3 + e^2 f + f^2 e + f^2 d + d^2 f + d^2 e + e^2 d + 7de f[/tex]
[tex]\geq 2 \left( e^2 f + f^2 e + f^2 d + d^2 f + d^2 e + e^2 d \right) + 4de f[/tex].

This simplifies to

[tex]d^3 + e^3 + f^3 + 3de f \geq e^2 f + f^2 e + f^2 d + d^2 f + d^2 e + e^2 d[/tex],

what can be rewritten as

[tex]d \left(d-e\right) \left(d-f\right) + e \left(e-f\right) \left(e-d\right) + f \left(f-d\right) \left(f-e\right) \geq 0[/tex].

But this follows from Schur. Thus the inequality is proven.

Nice inequality, Tran Minh!

[b]PS.[/b] I just see Fuzzylogic posting a much simpler solution, using my favorite substitution, but well, after I've written up this one I won't throw it away...

  Darij

[quote="darij grinberg"]
[tex]\frac{d}{e+f} + \frac{e}{f+d} + \frac{f}{d+e} + 4 \frac{de f}{\left(e+f\right) \left(f+d\right) \left(d+e\right) } \geq 2[/tex].
[/quote]

[quote="fuzzylogic"]
\[
\frac{xy}{z(x+y)} + \frac{yz}{x(y+z)} + \frac{zx}{y(z+x)} + \frac{4xyz}{(x+y)(y+z)(z+x)} \geq 2.
\]
[/quote]

Interestingly, these two inequalities are exactly the same if we let $d=1/x, e=1/y, f=1/z$.  :)

I am wondering what a pure trig proof would look like.

All of you have very nice solutions. This inequality is equivalent to IG >= 0 ( I is intercenter and G is centroid of a triangle)

After all, you can call me Hoang.


[quote="SUPERMAN2"]The sequence $ (x_n)_{n \geq 0}$ is defined by $ x_{n} \equal{} (\frac {1 \plus{} \sqrt 5}{2})^n \plus{} (\frac {1 \minus{} \sqrt 5}{2})^n$ for $ n \equal{} 0,1,2,...$.
Prove that $ x_n$ is a perfect square if and only of $ n \equal{} 3$.[/quote]

Wrong : $ x_1\equal{}1$ is a perfect square.

isn't it the direct formula for $ L_n$,Lucas sequence ...????see [url=http://en.wikipedia.org/wiki/Lucas_number]here[/url]...
see this proof ...

[quote="shoki"] ...see this proof ...[/quote]

I did not read the proof but it can not be OK since $ x_1$ is a perfect square. So "$ x_n$ is a perfect square if [u][b]and only if [/b][/u]$ n\equal{}3$" is false.

[quote="SUPERMAN2"]The sequence $ (x_n)_{n \geq 0}$ is defined by $ x_{n} \equal{} (\frac {1 \plus{} \sqrt 5}{2})^n \plus{} (\frac {1 \minus{} \sqrt 5}{2})^n$ for $ n \equal{} 0,1,2,...$.
Prove that $ x_n$ is a perfect square if and only of $ n \equal{} 3$.[/quote]
Sorry,I missed the condition  for $ n>1$. :wink:
EDIT:The sequence $ (x_n)_{n \geq 0}$ is defined by $ x_{n} \equal{} (\frac {1 \plus{} \sqrt 5}{2})^n \plus{} (\frac {1 \minus{} \sqrt 5}{2})^n$ for $ n \equal{} 0,1,2,...$.
Prove that for $ n>1$, $ x_n$ is a perfect square if and only of $ n \equal{} 3$


integration
advanced fields
advanced fields unsolved

I need help with this:

Suppose that $f'(z) = 0$ for all $z$ in a path connected open subset $U$ of $\mathbb{C}$. Show that f is constant.

By the Chain Rule, $\frac{d}{dt}f(z+t\zeta)=f'(z+t\zeta)\zeta$. If the segment with endpoints $z$ and $z+\zeta$ lies in $U$, then $f(z+\zeta)-f(z)=\int_{0}^{1}\frac{d}{dt}f(z+t\zeta)dt=0$. It remains to observe that any two points of $U$ can be connected by a polygonal line.


cos2a+2sin(a+60)sin(a-60)    pls if u can help me do it

[hide]From Transformations of Products and Sums we have

$2\sin A\sin B=\cos{(A-B)}-\cos{(A+B)}$

In this case:

$A=a+60$
$B=a-60$

So,

$\cos{(2a)}+2\sin{(a+60)}\sin{(a-60)}$
$=\cos{(2a)}+\cos{(a+60-(a-60))}-\cos{(a+60+a-60)}$
$=\cos{(2a)}+\cos{120}-\cos{2a}$
$=\cos{120}=\boxed{-\frac{1}{2}}$[/hide]

EDIT: I forgot to hide my solution

thank you very much man the result is exactely correct,but i need more explanations abut that exercise because im still young ok for ex:from where came cosp/3


Hi all.  I am currently a college student majoring in Math at SF State.  My sister who is a senior in high school got an advertisement in the mail yesterday for the 17th Annual USAMTS with the questions for round 1 of this year on it.  Along with the 5 questions, it had a section called "Proofs Without Words," where it had 6 Mathematical Diagrams, and a list of 6 Mathematical Properties.  The goal is to match each property with the diagram that proves it.  I was able to match all 6 of them, and show how exactly 4 of the 6 diagrams could be used to prove their corresponding properties.  I am not sure if anyone else has looked at it very much, but one of the ones I couldn't get had me stumped for quite a while.  It is the one that goes with the property that the sum of the powers of 3 from 1 to n-1 is equal to ((3^n) - 1) / 2.  In sigma notation, it was the sum from k=1 to n-1 of 3^k on the left side, and the formula ((3^n) - 1) / 2 on the right.  I could see what the picture was saying just fine, but what was stumping me was that the picture seemed to be proving that they were equal if you added one to the left side (increasing the sigma side by one, or subtracting one from the equation on the right).  Eventually, I decided to try plugging in a value into the equation to see if it came out equal, and I was surprised to see that they were actually one off.  When I plugged in 4 for n, I got 39 for the left (3+9+27), and 40 for the right ((81 - 1) / 2).  I believe that the problem is that the sigma is supposed to go from k=0, not k=1, because changing it to 0 would add the term 3^0 to the left side, increasing it by one, and thus making it equal.

I guess my question is, is what I am saying correct, or is there something I am missing?  I am not sure how many people have looked at that sheet, but for those who did, did any of you find the same?  I must have spent an hour looking at that diagram wondering, "Why am I getting 1 extra dot in the sum?"    I just want to make sure that I am not going crazy :).  Thank you for any relief your help gives me :).

    -Chris

[quote="musicman3320"]It is the one that goes with the property that the sum of the powers of 3 from 1 to n-1 is equal to ((3^n) - 1) / 2.  In sigma notation, it was the sum from k=1 to n-1 of 3^k on the left side, and the formula ((3^n) - 1) / 2 on the right.  I could see what the picture was saying just fine, but what was stumping me was that the picture seemed to be proving that they were equal if you added one to the left side (increasing the sigma side by one, or subtracting one from the equation on the right).  Eventually, I decided to try plugging in a value into the equation to see if it came out equal, and I was surprised to see that they were actually one off.  When I plugged in 4 for n, I got 39 for the left (3+9+27), and 40 for the right ((81 - 1) / 2).  I believe that the problem is that the sigma is supposed to go from k=0, not k=1, because changing it to 0 would add the term 3^0 to the left side, increasing it by one, and thus making it equal.

I guess my question is, is what I am saying correct, or is there something I am missing?  I am not sure how many people have looked at that sheet, but for those who did, did any of you find the same?  I must have spent an hour looking at that diagram wondering, "Why am I getting 1 extra dot in the sum?"    I just want to make sure that I am not going crazy :).  Thank you for any relief your help gives me :).  [/quote]

I have not seen the sheet myself, but I would say that you have a valid proof that there is a typographical error in the summation notation (Starting the summation at 1 instead of 0 is a common typograhical mistake). You should be more confident in your own mathematical abilities. The correct summation formulae are
\[\sum_{i=0}^{n-1}3^i = 1+3+9+\cdots + 3^{n-1} = \frac{3^n-1}{2}\]
\[\sum_{i=1}^{n-1}3^i = 3+9+27+\cdots + 3^{n-1} = \frac{3^n-3}{2}\]

Erin Schram
USAMTS grader

Good catch - that is indeed a typo - should have been a 0 on the bottom index.


我已经做出来了！ :idea:


Let $ p>3$ be a prime. If $ p$ divides $ x$, prove that the equation $ x^2\minus{}1\equal{}y^p$ does not have positive integer solutions. :lol:  :lol:

Source: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229046]Oliforum contest, round 2[/url]

http://www.mathlinks.ro/viewtopic.php?t=228441   I sent you this link by email.

[quote="bboypa"]Let $ p > 3$ be a prime. If $ p$ divides $ x$, prove that the equation $ x^2 \minus{} 1 \equal{} y^p$ does not have positive integer solutions. :lol:  :lol:

Source: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229046]Oliforum contest, round 2[/url][/quote]
x>0 so let x=pk>3. From our equation:  y^p≡-1(mod x) so y^kp≡±1 mod(x), and (x,y)=1 because if (x,y)=a>1 a|1: so but by Ferma theorem y^kp=y^x≡y(mod x) so y≡±1 (mod x). If y=1: x^2=0 but x>3, y>0 so y≥x-1  p>3 so x^2= y^p+1>(x-1)^3+1=x(x^2-3x+3) or x^2-4x+3=(x+1)(x-3)<0 but x>3 so equation  does not have positive integer solutions.

we have :
$ (x \minus{} 1)(x \plus{} 1) \equal{} y^p$.
case 1) $ x$ is even :so we have $ (x \minus{} 1,x \plus{} 1) \equal{} 1$ so we have $ x \minus{} 1 \equal{} y^t$ and $ x \plus{} 1 \equal{} y^u$.but but distance between
$ y^t$ and $ y^u$ is bigger that 2 if $ y\neq 2$.but if $ y \equal{} 2$ we have $ t \equal{} 2$ and $ u \equal{} 1$ so $ p \equal{} 3$.
case 2)
$ x$ is odd.$ (x \minus{} 1,x \plus{} 1)$=2
now we have 2 case:
case 1) $ x \plus{} 1 \equal{} 2t^p$ and $ x \minus{} 1 \equal{} 2^{p \minus{} 1} u^p$;
we have $ 4(t^{2p} \minus{} (2u)^p) \equal{} (x \plus{} 1)^2 \minus{} 8(x \minus{} 1) \equal{} (x \minus{} 3)^2 \equiv l mod(p)$.
but $ l$ is not o $ mod(p)$.because we have $ p|x$(p is not 3).
so we have $ (t^2)^p \minus{} (2u)^p \equal{} (\frac {x \minus{} 3}{2})^2 \equal{} (t^2 \minus{} 2u) \frac {(t^2)^p \minus{} (2u)^p}{(t^2 \minus{} 2u)}$.
but we know that $ (\frac {(t^2)^p \minus{} (2u)^p}{(t^2 \minus{} 2u)},(t^2 \minus{} 2u)) \equal{} 1$.because its easy to show
this lemma:$ (a \minus{} b,\frac {a^p \minus{} b^p}{a \minus{} b}) \equal{} 1$ 0r $ p$.with condition$ (a,b) \equal{} 1$
prove:suppose $ z|a \minus{} b$ and $ z$ is prime.we have $ \frac {a^p \minus{} b^p}{a \minus{} b} \equal{} \sum a^{p \minus{} i} b^i\equiv pa^p mod(z)$.
so if $ z\neq p$ we have $ z|a$ so $ z|b$ which contradicts with $ (a,b) \equal{} 1$.
but because $ l$ is not zero so we have $ (\frac {(t^2)^p \minus{} (2u)^p}{(t^2 \minus{} 2u)},(t^2 \minus{} 2u)) \equal{} 1$
so $ t^2 \minus{} 2u \equal{} k^2(t > u)$so we have $ t^2 \minus{} 2u\leq (t \minus{} 1)^2$ means $ u > t$ which contradicts.
case 2)
$ x \minus{} 1 \equal{} 2t^p$ and $ x \plus{} 1 \equal{} 2^{p \minus{} 1} u^p$;again we have:
$ 4(t^{2p} \plus{} (2u)^p) \equal{} (x \minus{} 1)^2 \plus{} 8(x \plus{} 1) \equal{} (x \plus{} 3)^2 \equiv u mod(p)$.but $ u$ is not $ o mod(p)$.
because we have $ p|x(p \neq 3)$.
so we have $ (t^2)^p \plus{} (2u)^p \equal{} (\frac {x \plus{} 3}{2})^2 \equal{} (t^2 \plus{} 2u) \frac {(t^2)^p \plus{} (2u)^p}{(t^2 \plus{} 2u)}$.
the same as before we have
$ t^2 \plus{} 2u \equal{} j^2$
but again its easy that $ t > u \plus{} 1$ beacase $ p > 3$.so we have $ t^2 \plus{} 2u\geq (t \plus{} 1)^2$ but this means
$ u > t$

[quote="bboypa"]why is it equivalent?  :o[/quote]
The only solution to $ a^b \minus{} c^d \equal{} 1$ with $ a,b,c,d > 1$ is $ (a,b,c,d) \equal{} (3,2,2,3)$. (Catalans conjecture)

[quote="bboypa"]@dima_ukraine: you can see very easily that $ y \ge x \minus{} 1$ is false

@ sumita : my congrats.. :lol:

@mathias_dk: you cannot use catalan conjecture to solve the exercise! and by the way, you need to demostrate that x is a power of p to say that two exercises are equivalent (and i never said that x is a power of p).

@lambruscokid: exactly..[/quote]If y=kx±1 y>1 and y, x positive integers y>x-1 am I right?

[quote="dima ukraine"]..so but by Ferma theorem y^kp=y^x≡y(mod x) so y≡±1 (mod x)...[/quote]

are you sure that you can apply fermat?

Sorry bboypa i have a mistayke, mods delete this post

[quote="lambruscokid"]Yes of course that the problem can be solved be using the catalan´s conjeture.
But if you are asked to write a solution down in a contest you have to think in a more clever and simplier way to solve the problem, you can´t just write in the middle of the page CATALAN unless you provide a solution (that will be probably more difficult than solving the problem by elementary methods.)

Daniel[/quote]
In most competitions, you can use theorems like catalans conjecture.... :wink:

But I haven't read the rules for this one.

wow the beautiful and professional solutions . my solution is really simple. it's equal that the equation $ 2^n \equal{} a^p \plus{} 1$ for odd prime

$ (a \plus{} 1)(a^{p\minus{}1} \minus{} a^{p \minus{} 2} \plus{} ... \plus{} 1) \equal{} 2^n$

then $ a \plus{} 1 \mid a^{p \minus{} 1} \plus{} ... \plus{} 1$

we know $ a^{p \minus{} 1} \equiv 1(mod a \plus{} 1)$ and  $ \minus{} a^{p \minus{} 2} \equiv 1(mod a \plus{} 1)$

then $ a \plus{} 1 \mid p$ contradiction

[quote="t0rajir0u"][quote="bboypa"]by the way, you need to demostrate that x is a power of p to say that two exercises are equivalent [/quote]
Not at all; you might be thinking of Fermat's Last Theorem.  Mihailescu's Theorem (to give Catalan's conjecture the air of a true statement) is more than enough to solve this problem.[/quote]

i know that mihailescu is stronger, but this is a problem from a olimpic contest.. in a imo solution should you accept merely it?


real analysis
calculus
calculus computations

i was going through some of my previous year homework sheets and this problem came up.show that $f(x)=xsinx$ is not uniformly continuous on R. i did something, but i'm just not confident about the limits, can someone tell me whether this is right?

it is enough to provide 2 sequences $x_{n}$ and $y_{n}$ such that $lim(x_{n}-y_{n})=0$ but $f(x_{n})-f(y_{n})$ doesnt tend to 0.

take $x_{n}=2\pi{n}$ and $y_{n}=2\pi{n}+\frac{1}{n}$ , these two sequences solve the problem?

Yes, because $f(x_{n})-f(y_{n})=2\pi n\sin(1/n)\to 2\pi$ as $n\to\infty$.


function
logarithms
inequalities proposed
inequalities

the function $\ln \frac{1+a^{2}}{1+a}$ has a single inflection point so we can assume 4 of the variables equal.
now $f(a)=\left(a-1+\frac{2}{a+1}\right)^{4}\left(4-4a+\frac{1}{3-2a}\right)$ has a minimum at $a=1$ so we get the required inequality.

Can you prove without computer that $f(a)$ has a minimum at $a=1$?

Vasc do you have a simple proof ?

@ ehku : What do you mean ?

Using LCF-Theorem and Remark 8 (page 148), it suffices to show that
$f'(x) \geq f'(y)$ for $0\leq x\leq 1\leq y$ with $4x+y=5$ and ${f(u)=\log (u+1)-\log(u^{2}+1)}$.
This is very easy if we use $4(x^{2}+1)(y^{2}+1) \geq (x+1)^{2}(y+1)^{2}$ .

Ok i didn't think of that. Very nice. I believe it is generalisable to n variables. :)


abstract algebra
search
group theory
superior algebra
superior algebra unsolved

Let be G a p-group then G is cyclic group?How to demonstrate?
And if G/K is a group cyclic then G is abelian group?How to demonstrate?
Thanks
sastra

Do a search on the forum please for the second one :wink: and for the first one consider the Klein's four group...

for the first use lagrange theorem: if H is SG of G then card(h)/card(G).

I mentioned Klein's four group as a counterexample to the first question since it is not a cyclic group but of order $2^{2}$. A $p$-group need not be even abelian:eg consider $D_{4}$.

for the first is true in fact we use lagrange we find the only SG of G is {e} and G then is cyclic.


Prove that:
[tex] \frac{\sqrt{xy+4yz+4xz}}{x+y} + \frac{\sqrt{yz+4xy+4xz}}{y+z}+\frac{\sqrt{xz+4xy+4yz}}{x+z} \geq \frac{9}{2} [/tex]
where
[tex] i)  x,y,z \geq 0 [/tex] and [tex] min\lbrace{x,y,z}\rbrace \geq \frac{4}{5} max\lbrace{x,y,z}\rbrace [/tex].
[tex] ii) x,y,z \geq 0 [/tex].
The second question is an exercise from Mr.Nam Dung for me. That one was post here before but I don't know where it is.
Please accept my crazy proposal: wait for next week if you want to post a solution or give me a link for ii), I still want to have more time to do it.
Thank you very much. :)

Or you can post it in the hide tags: [hide="like this"]post solution here ;) [/hide]


function
integration
real analysis
real analysis unsolved

There are rational functions $R(x)$ such that for arbitrary function $f(x)$:
\[\int_{-\infty}^{\infty}f\left(R(x)\right)dx=\int_{-\infty}^{\infty}f(x)dx \]
Provided the integer on the right hand exists.
Prove that $R(x)$ such the above conditon if and only if $R(x)$ has the form:
\[R(x)=\epsilon(x-\alpha-\frac{p_{1}}{x-\alpha_{1}}-\frac{p_{2}}{x-\alpha_{2}}-\cdots-\frac{p_{n}}{x-\alpha_{n}}) \]
where $\epsilon=1$ or $-1$,$\alpha,\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$ are real numbers and $p_{1},p_{2},\cdots,p_{n}$ are positive numbers.

I think I have proved the sufficient part.Can somebody provide some help the necessary part,I want to take some particular function $f(x)$ to prove the result but failed.In fact the sufficient part is also a problem appear in AMM(american mathematical monthly in 2006.6-7)

Pick $a\in \mathbb R$ and let $f$ be the characteristic function of the interval $[a-\epsilon,a+\epsilon]$. Suppose that $b_{1},\dots,b_{n}$ are all roots of the equation $R(x)=a$. By our assumption the set $\{x\colon |R(x)-a|\le \epsilon\}$ has measure $2\epsilon$, which should imply $\sum_{i=1}^{n}|R'(b_{n})|^{-1}=1$ (*) in the limit $\epsilon\to 0$. This already tells you that $R'$ does not vanish. Perhaps a careful analysis of (*) together with a partial fraction decomposition of $R$ will lead to a solution.

Thanks for reply.
But I think $b_{1},b_{2},\cdots,b_{n}$ are just real roots of the equation $R(x)=a$.Can you also give some hint for the solution from (*)
Thank you.


geometry
trigonometry
analytic geometry
vector
linear algebra
matrix
area of a triangle

Prove the following statement.

[hide="statement"]
Given that the points $ A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ are vertices of triangle $ ABC$, the area of triangle $ ABC$ is $ \frac {1}{2} \times |(x_1y_2 \plus{} x_2 \plus{} y_3 \plus{} x_3y_1) \minus{} (y_1x_2 \plus{} y_2x_3 \plus{} y_3x_1)|$
[/hide]

Are there any more methods to find the area of a triangle?

[hide="methods"]

Some methods I know:
[list]Heron's formula[/list]
[list]$ 1/2 ab \sin C$[/list]
[list]Pick's Theorem[/list]
[list]Your usual $ ab/2$[/list]
[list]$ 1/2 nr^2 \sin \frac{2\pi}{n}$[/list]

hmm...
[/hide][/list][/hide]

Your formula is just the cross product.  Speaking of which, I believe there is a typo and it should read $ x_{2}y_{3}$. not $ x_{2}\plus{}y_{3}$.

To your second question, there are literally an unfathomably large number of ways to calculate a triangle's area.

[quote="ThetaPi"]Prove the following statement.

[hide="statement"]
Given that the points $ A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ are vertices of triangle $ ABC$, the area of triangle $ ABC$ is $ \frac {1}{2} \times |(x_1y_2 \plus{} x_2 \plus{} y_3 \plus{} x_3y_1) \minus{} (y_1x_2 \plus{} y_2x_3 \plus{} y_3x_1)|$
[/hide]
I think there may be a typo in that formula. I don't like the
"$ \plus{} x_2 \plus{} y_3$"   part. For one, the units in the added terms would not be the same. For another, if I scale up the triangle by doubling each coordinate, I don't get four times the original area.

Wait... isn't that the shoelace theorem or something? That requires vectors to prove, I thought.

[quote="ThetaPi"]Prove the following statement.

[hide="statement"]
Given that the points $ A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ are vertices of triangle $ ABC$, the area of triangle $ ABC$ is $ \frac {1}{2} \times |(x_1y_2 \plus{} x_2 \plus{} y_3 \plus{} x_3y_1) \minus{} (y_1x_2 \plus{} y_2x_3 \plus{} y_3x_1)|$
[/hide]

Are there any more methods to find the area of a triangle?

[quote="JRav"]More generally, it extends to the cross product in three dimensions.[/quote]

i wonder if we can generlisated for $ n$ dimensions   :roll:

The "volume" of any region in $ \mathbb{R}^{n}$ is equal to the determinant of the matrix of the spanning vectors.


function
functional equation
algebra
IMO Shortlist

Let $ \mathbb{R}$ be the set of real numbers. Does there exist a function $ f: \mathbb{R} \mapsto \mathbb{R}$ which simultaneously satisfies the following three conditions?

[b](a)[/b] There is a positive number $ M$ such that $ \forall x:$ $ \minus{} M \leq f(x) \leq M.$
[b](b)[/b] The value of $f(1)$ is $1$.
[b](c)[/b] If $ x \neq 0,$ then
\[ f \left(x \plus{} \frac {1}{x^2} \right) \equal{} f(x) \plus{} \left[ f \left(\frac {1}{x} \right) \right]^2
\]

Let M be minimal such that $ - M\lef(x)\le M$, although f(x) may never actually evaluate to M, there is such a number.

Now, $ f(2) = 2$ by plugging in x=1 so we know that $ M \ge 2$

By the way we chose M, there are 2 possibilities.

i) for every positive $ e$ we can chose a k s.t. $ f(k)\ge M - e$.

Then let $ f(k) = M - p$ with $ p \le e$

$ f(k + 1/k^2) = M - p + f(1/k)^2$

$ f(1/k)^2 \le p$ so $ |f(1/k)| \le \sqrt {p}$

$ f(k^2 + 1/k) = f(1/k) + M^2$

Now, by chosing e as small as we want, we can make p as small as we want, so we can get values of $ f(x)$ as close to $ M^2$ as we want. If M>1, $ M^2 > M$ so therefore we could find values of f(X) so that either $ f(x) > M^2 > M$ or $ M^2 > f(x) > M$ either way a contradiction (for example, chosing $ e = (M^2 + M)/3$ ).

Therefore, $ M \le 1$ but $ M\ge f(2) = 2 > 1\ ge M$ contradiction!

ii) for every positive $ e \le 1$ we can chose a z s.t. $ f(z)\le - M + e$.

Then let $ f(z) = - M + p$ with $ p \le e$

Now, we already know that $ M \ge 2$. As well, it is easy to see that $ x + 1/x^2$ is surjective everywhere except possibly x=1. since $ z \ne 1$ by our choice of e and our bounds of M, we can write $ z = k + 1/k^2$.

$ - M + p = f(k) + f(1/k)^2$ So therefore $ - M \le f(k) \le - M + p$ and $ f(1/k)^2 \le p$ so $ |f(1/k)| \le \sqrt {p}$

$ f(k^2 + 1/k) = f(1/k) + f(k)^2$ By choosing e small enough, we make the R.H.S. as close to $ M^2$ as we want.

M>1 means the same thing as before, so therefore M=1, but then $ f(2) > M$ contradiction!

Tell me if my proof is right please. Vying for IMO team next year.

[quote="orl"]Let $ \mathbb{R}$ be the set of real numbers. Does there exist a function $ f: \mathbb{R} \mapsto \mathbb{R}$ which simultaneously satisfies the following three conditions?

[b](a)[/b] There is a positive number $ M$ such that $ \forall x:$ $ \minus{} M \leq f(x) \leq M.$
[b](b)[/b] The value of f(1) is 1.
[b](c)[/b] If $ x \neq 0,$ then
\[ f \left(x \plus{} \frac {1}{x^2} \right) \equal{} f(x) \plus{} \left[ f \left(\frac {1}{x} \right) \right]^2
\]
[/quote]

It is obvious that a function which satisfies the conditions (b) and (c) but not condition (a),  is f(x)=x ...
If we prove that this f(x) is unique that satisfies (b) and(c) then we can say that there is no function that simultaneusly satisfies the 3 conditions... :maybe:

i think it´s the following:
suppose n is the least upper bound.
so
there is a d such that
f(d)=n
but f(d+1/d^2)=f(d)+f(1/d)^2
so f(1/d)=0
now let x=1/d
we have that
f(1/d)+f(d)^2=<n
but f(d)>=2, absurd. :wink:

Take care, because there could be a least upper bound $ n$ but there could be too a function such that gets each time more next to $ n$ but never gets this value (an example: $ f(n) \equal{} \dfrac{x^{n \plus{} 1} \minus{} 1}{x \minus{} 1}$, where $ 0 < x < 1$).

Here is my solution :
[hide]Denote by $M$ the least upper bound of the set $\{|f(x)| \ ; \  x\in \mathbb{R^*}\}$. (it exists since f is bounded)
Notice that : $f(\frac{1}{x})^2=f(x+\frac{1}{x^2})-f(x) \leq 2M$, which gives $|f(\frac{1}{x})| \leq \sqrt{2M}$, that is, $|f(x)| \leq \sqrt{2M}$ for all non zero real x.
As $M$ is the least upper bound, we need to have $\sqrt{2M} \geq M$, which directly gives $M \leq 2$.
But this gives a contradiction because we can show that $f(2)=2$ and then $f(\frac{1}{2})=0$ and then $f(\frac{1}{2}+4)=4$.[/hide]
Can someone confirm please ?

Two issues.

One minor. You should define $M = \sup \{ |f(x)| \ ; \ x\in \mathbb{R}^*\}$, otherwise it does not follow that $\sqrt{2M} \geq M$.
One major. How do you determine that $f(1/2)=0$? All you can (quickly) say is that (for $x = 1/2$) $f(9/2) = f(1/2) + f(2)^2 = f(1/2) + 4$, but that actually implies $f(1/2) = -2$ and $f(9/2) = 2$ (since $|f(x)| \leq 2$ for all $x\neq 0$).

However, let us now take $x=2$, so $f(9/4) = f(2) + f(1/2)^2 = 2+4 = 6$, absurd.

[quote="mavropnevma"]Two issues.

One minor. You should define $M = \sup \{ |f(x)| \ ; \ x\in \mathbb{R}^*\}$, otherwise it does not follow that $\sqrt{2M} \geq M$.[/quote]
Sure.

[quote="mavropnevma"]One major. How do you determine that $f(1/2)=0$? All you can (quickly) say is that (for $x = 1/2$) $f(9/2) = f(1/2) + f(2)^2 = f(1/2) + 4$, but that actually implies $f(1/2) = -2$ and $f(9/2) = 2$ (since $|f(x)| \leq 2$ for all $x\neq 0$).

However, let us now take $x=2$, so $f(9/4) = f(2) + f(1/2)^2 = 2+4 = 6$, absurd.[/quote]
Take x=2 first.

Indeed, taking $x = 2$ first, as you say, yields $2\geq f(9/4) = f(2) + f(1/2)^2 = 2+f(1/2)^2 \geq 2$, implying $f(1/2) = 0$, but[b] it needs to be said[/b]. You just said "... we can show that $f(2) = 2$ and then $f(1/2) = 0$ and then ...".

You're right, and thank you for checking my solution.
However, based on my solution, I think this problem is far easy for an IMO A5 Shortlist...

Plugging $x=1$ into the assertion gives that $f(2)=2$, and $x=2$ gives $f\left(\frac{9}{4}\right)=2+f\left(\frac{1}{2}\right)^2$. Denote $f\left(\frac{1}{2}\right)^2=2\epsilon$, and define a sequence $\{a_i\}$, where $a_1=\frac{9}{4}$, and $a_{i+1}=a_i^2+\frac{1}{a_i}$ if $\left|f\left(\frac{1}{a_i}\right)+f(a_i)^2\right|>(1+\epsilon)f(a_i)$ and $a_{i+1}=\frac{1}{a_i}$ otherwise. I claim that $f(a_i)^2\ge (1+\epsilon)^2f(a_{i-1})^2$, which we will prove using induction.

Note that $x=\frac{1}{a_i}$ gives $f\left(a_i^2+\frac{1}{a_i}\right)=f\left(\frac{1}{a_i}\right)+f(a_i)^2$. So, in the case where $a_{i+1}=a_i^2+\frac{1}{a_i}$, the claim is trivial, as we immediately obtain $f(a_{i+1})^2>(1+\epsilon)^2f(a_i)^2$. On the other hand, when $\left|f\left(\frac{1}{a_i}\right)+f(a_i)^2\right|\le(1+\epsilon)f(a_i)$, we must have that $\left|f\left(\frac{1}{a_i}\right)\right|\ge f(a_i)^2-|(1+\epsilon)f(a_i)|$. We wish to show $f(a_i)^2-|(1+\epsilon)f(a_i)|\ge |(1+\epsilon)f(a_i)|$. However, we know $f(a_i)^2\ge |(2+2\epsilon)f(a_i)|$, since $|f(a_i)|\ge 2+2\epsilon=f(a_1)$ by inductive hypothesis. Thus, the claim is proven.

Of course, this means that $|f(a_i)|$ is monotonically increasing, so $f$ cannot be bounded, as desired.

[b][u]Claim: There exists no such $f.$[/u][/b]
[i]Proof.[/i] Assume the contrary. Let $u$ be the upper bound of the set of values of $f.$
$f(2)=f(1+\frac{1}{12})=f(1)+f(1)^2=2 \implies u\geq 2.$
\begin{align*} &\text{WLOG, }\exists a: f(a)>u-\frac{1}{4} \\ &\implies u \geq f(a+\frac{1}{a^2})=f(a)+f(\frac{1}{a})^2> u-\frac{1}{4} +f(\frac{1}{a})^2\\ &\implies f(\frac{1}{a})^2 <\frac{1}{4} \implies f(\frac{1}{a})>-\frac{1}{2}.\end{align*}
\begin{align*} &\text{Also }u\geq f(\frac{1}{a}+a^2)=f(\frac{1}{a})+f(a)^2>\\ &-\frac{1}{2}+(u-\frac{1}{4})^2=u^2-\frac{u}{2}-\frac{7}{16} \implies (u-\frac{3}{4})^2<1,\end{align*} a contradiction with $u\geq 2.$ $\blacksquare$


So have any Georgians gotten phone calls? No one from Georgia is in the top 24 apparently.

Apparently I (surprise!) overestimated my score.  Actually, I was kind of surprised.  But I don't know anything yet, but even if I don't make MOP, I have a happy summer ahead. :)

And I (suprise!) overestimated my score. Actually, I was rather suprised. But I too have a happy summer ahead in my non-MOPping state.

And I (surprise!) also overestimated my score. Estimated a 9: 2/2/0/5/0/0 Actually got a 7: most likely 0/0/0/7/0/0 or 1/1/0/5/0/0. At least that was 5 more than I was shooting for.

i think i will break the trend, even though i did as well. blegh...estimated like a 12 or 13, got a 9......

I got an 18.  Not sure how this happened.  I've never gotten less than a 7 on something I was sure was a 7 before, or at least, not in two years.  Maybe they somehow docked me on number 5?  Oh well, I guess I'll just have to make do with RSI. :D

Um, I thought I actually got a pretty good score in my opinion.  Instead, I got a 2 (not surprise!). :clap2:

Yeah, you guys can start laughing now.  Red Mop next year will be something that I will only make in the wildest of my dreams, though I never had any hope of making it in the first place, so...yeah, say nothing.  Again, give a round of applause for my absolutely pathetic performance. :lol:

PS: The depressing part is that I actually got the final answer to #4 right but only got a maximum of 2 points.  The graders make me feel bad...

[quote="Twiz"]So 7 of the 8 Georgia people whose scores I know got multiples of 18 (Santhosh, you messed up the trend!). :wink:[/quote]

Do you mean factors?

You can add 1 to that... I got a 9

I suspect 7/1/0/1/0/0

Yes, factors, but it would be nice if I meant multiples. :)
And great job getting a 9.

Congrats to every Georgian USAMOer this year; you all beat me: I got a ONE. I'm still wondering how I managed to do that after I ensured my correctness of my #1's solution and my #4's answer. Yeah, I sucked.

BTW, Eric, you can add me, since 1 is definitely a factor of any number. :wink:

I bet Harrison breaks the trend as well.  Don't worry, Tim; I know someone else who got a 1 as well.

I bet I break the trend! I got an 8, which was about half of my (over)expected value...  :huh:

Well, at least you're not relatively prime to 18 like Santhosh. :wink:


$ f: N\rightarrow N$ $ n\geq 2$   $ f(f(n \minus{} 1))\equal{}f(n \plus{} 1)\minus{} f(n)$

[quote="cnyd"]$ f: N\rightarrow N n\geq 2$   $ f($f$ (n \minus{} 1))$=$ f(n \plus{} 1)$- $ f(n)$[/quote]

$ f(n\plus{}1)\minus{}f(n)\equal{}f(f(n\minus{}1))>0$, so $ f(n)$ is strictly increasing.

$ f(f(n\minus{}1))\equal{}f(n\plus{}1)\minus{}f(n)<f(n\plus{}1) \Rightarrow f(n\minus{}1) \le n$ for all $ n$.
$ f(1) \ge 1$ and $ f(n)$ is strictly increasing $ \Rightarrow f(n) \ge n$

So $ n\plus{}1 \ge f(n) \ge n$ for all $ n$

$ f(n\plus{}1)\minus{}f(n) \le (n\plus{}2)\minus{}n\equal{}2$, so $ f(f(n\minus{}1)) \le 2$ for all $ n$ , which is impossible as $ f(n)$ is strictly increasing.
So no such function exists.