integration
logarithms
vector
geometry
geometric transformation
rotation
calculus

A motorcyclist wishes to travel in a circle of radius R on level ground. The coefficient of friction between the tires and the ground is $ \mu$. the motorcycle starts at rest. What is the minimum distance it must travel in order to achieve its maximum allowable speed, that is, the speed above which it will skid out of the circular path?

(Does the book have answers? I'm not sure at all about the below)
To find the maximum allowable speed, note that the maximum friction in the tires has to be equal to the centripetal force:

$ \mu mg \equal{} \frac {mv_{max}^2}R$
$ \mu g \equal{} \frac {v_{max}^2}R$
$ v_{max}^2 \equal{} \mu gR$

From here, I could find two possible interpretations:

1) The motorcyclist travels in a straight line until he achieves that speed.

Then we have that

$ v^2 \equal{} v_0^2 \plus{} 2as$
$ \mu gR \equal{} 2 \mu g s$
$ s \equal{} \frac {R}2$

2) The motorcyclist travels in the radius R circle until he achieves that speed.

Then we have that

$ ma_t \equal{} \mu mg \minus{} \frac {mv^2}R$
$ a_t \equal{} \mu g \minus{} \frac {v^2}R$

As

$ \Delta s \equal{} \int \frac {vdv}{a}$
$ s \equal{} \int_0^s \frac {vdv}{\mu g \minus{} \frac {v^2}R}$
$ s \equal{} \frac {R(\ln{\mu g} \minus{} \ln{(\mu g \minus{} \frac {v^2}R)})}2$

As $ v^2 \rightarrow \mu gR$, $ s \rightarrow \infty$:

$ s \equal{} \frac {R(\ln{\mu g} \minus{} \ln{(\mu g \minus{} \mu g)})}2$
$ s \equal{} \frac {R(\ln{\mu g} \minus{} \ln{0})}2$ (undefined)

I got an ugly differential equation when I did this problem, which I know is correct but I'm not sure how to solve.

you have to realize that the friction force changes direction with time; part of it is responsible for the tangential acceleration, part of it for the radial.

Oh yes, I realize something I did terribly wrong. This is (one of) the mistakes:

$ ma_t \equal{} \mu mg \minus{} \frac {mv^2}R$

I assumed the total friction acceleration was the sum of the tangential and the radial acceleration, while in fact it's a right triangle. So I think it would be correct to say that:

$ (\mu mg)^2 \equal{} (ma_t)^2 \plus{} (\frac {mv^2}R)^2$
$ \mu^2 g^2 \equal{} a_t^2 \plus{} \frac {v^4}R^2$

$ a_t^2 \equal{} \mu^2 g^2 \minus{} \frac {v^4}R^2$
$ a_t \equal{} \sqrt {\mu^2 g^2 \minus{} \frac {v^4}{R^2}}$

If we proceed in the same manner (i.e. $ s \equal{} \int \frac {v}{a}dv$), then

$ s \equal{} \int \frac {v}{\sqrt {\mu^2 g^2 \minus{} \frac {v^4}{R^2}}}dv$

This is really tough to crack. In fact, I got more than two different answers, and I don't know which one is right :) (I'll ponder about it when I have time). Also, I'd be interested in your method.

You already did most of the work, ffao, especially in noting that $ a \equal{} v\frac{dv}{dx}$.

Slightly rearranging by writing $ v_\textrm{max}^2 \equal{} R\mu g$, and then making the change of variable $ u \equal{} v^2$, gives
\[ \frac{R}{2}\int\frac{du}{\sqrt{v_\textrm{max}^2 \minus{} u^2}} \equal{} \int dx.
\]
Now this is easy to compute: $ x \equal{} \frac{R}{2}\arcsin\left(\frac{v^2}{v_\textrm{max}^2}\right)$, and taking $ v \equal{} v_\textrm{max}$, $ x \equal{} \pi R/4$.

[hide="A Solution"]$ m\frac{v_{max}^{2}}{R}$ is a radial force. It doesn't come into play here.

The force of friction ($ \mu mg$) resists motion in whatever direction a thing is moving. Furthermore, if the forward force exceeds the force of friction then the wheels will begin to slip. With that in mind, we can apply the work-energy theorem as follows:

\[ \frac{1}{2}mR^2\omega_{max}^{2} \equal{} \int_{\theta_1}^{\theta_2} \mu mg\, R\,\mathrm{d}\theta \\ \\ \Delta s \equal{} R\Delta \theta \\ \\ \therefore \Delta s \equal{} \frac{\frac{1}{2}mR^2\omega_{max}^{2}}{\mu mg} \][/hide]

[hide="Where am I wrong?"]
Since the tangential acceleration of the motor cycle cannot exceed $ \mu g$, we have $ v_{max}^2 \equal{} 2as$ where $ a \equal{} \mu g$ and $ s$ is distance travelled. Now, $ \frac {mv^2}{R} \le \mu mg$. Thus solving we get $ v_{max} \equal{} \sqrt {\mu R g}.$

Substituting, we get $ \boxed {s \equal{} \frac {R}{2}}$.
But what worries me is that this is a motorcycle and not a car, hence it can tilt/incline a lot (obviously if the rider is a skilled one!!! :winner_first:) to one side and hence the value of $ v_{max}$ is definitely greater than what I've found here. :maybe:
[/hide]

[quote="Rajiv"]Where am I wrong?
Since the tangential acceleration of the motor cycle cannot exceed $ \mu g$, we have $ v_{max}^2 \equal{} 2as$ where $ a \equal{} \mu g$ and $ s$ is distance travelled. Now, $ \frac {mv^2}{R} \le \mu mg$. Thus solving we get $ v_{max} \equal{} \sqrt {\mu R g}.$

Substituting, we get $ \boxed {s \equal{} \frac {R}{2}}$.
But what worries me is that this is a motorcycle and not a car, hence it can tilt/incline a lot (obviously if the rider is a skilled one!!! :winner_first:) to one side and hence the value of $ v_{max}$ is definitely greater than what I've found here. :maybe:
[/quote]

$ \frac{v^2}{R}$ is a radial acceleration. Balancing it with friction does nothing in this case. On an inclined track balancing friction and $ \frac{v^2}{R}$ is important as friction tends to prevent side slip.

On a level track, all $ \frac{v^2}{R}$ does is to keep the motorcycle traveling in a circle. $ \frac{v^2}{R}$ is an acceleration because the direction of the velocity vector is continually changing; however, the magnitude of the velocity vector is not. That's why it's called [i]uniform[/i] circular motion.

The thing to keep in mind is that friction resists motion, in whatever direction that motion takes place. Given that fact and that the motorcycle is moving in a circle on a level track we are not necessarily concerned with side slip. We are concerned with friction balancing the forward motion.

Let the force of friction be $ f$. Then if the torque on the motorcycle's wheels exceeds $ r\times f$, where $ r$ is the radius of the motorcycle wheel, there will be a net angular acceleration on said wheel. If this occurs then the wheels lose traction and the wheels slip.

On the other hand, if the force of friction (times $ r$) matches the torque on the motorcycle's wheels then the force of friction does work on the motorcycle as the motorcycle circles the track. Namely,

\[ \int_{v_{min}}^{v_{max}} m v\,\mathrm{d}v \equal{} \int_{\theta_{min}}^{\theta_{max}} f R\,\mathrm{d}\theta\]

\[ \Rightarrow \frac{1}{2}m(v_{max}^{2} \minus{} v_{min}^{2}) \equal{} f R (\theta_{max} \minus{}\theta_{min})\]

\[ \Delta s \equal{} R (\theta_{max} \minus{}\theta_{min})\]

\[ v_{min} \equal{} 0\]

\[ \Delta s \equal{} \frac{\frac{1}{2}m v_{max}^{2}}{f} \equal{} \frac{\frac{1}{2}m v_{max}^{2}}{\mu m g} \equal{} \frac{v_{max}^{2}}{2\mu g}\]

Friction does not balance forward motion. Friction [i]is[/i] the forward motion.

Also, the velocity vector's magnitude does change with time, otherwise the motorcycle wouldn't accelerate.

Rajiv's solution is essentially my "first interpretation" above, which I believe is not correct.

assuming nonslip conditions and the guy travels in a circle...

we all agree that the max speed $ v_{max} \equal{} \sqrt{\mu R g}$ or whatever.

since we want to minimize distance, we clearly want to maximize acceleration. the acceleration is limited only by the maximum acceleration due to force of static friction, which we can easily compute. But the thing is, a certain portion of the friction needs to be "allocated" to keeping the moto in the circle, i.e. radial acceleration. so we have \[ a_t^2\plus{}a_r^2 \equal{} constant\]
\[ a_t^2 \equal{} C \minus{} C v_t^4\]
where i am disregarding random constants cuz i am lazy
and $ v\equal{}v_t$ since the circle has const. radius
\[ a_t \equal{} \sqrt{C \minus{} C v_t^4}\]
\[ x \equal{} \int_0^{v_{max}} {\frac{v_t}{a_t} dv}\]
\[ x \equal{} \int_0^{v_{max}} {\frac{v_t}{\sqrt{C \minus{} C v_t^4}} dv}\]
which can be solved with the $ u\equal{}v^2$ and $ arcsin(u)$ trick of genericme

[quote="ffao"]Friction does not balance forward motion. Friction [i]is[/i] the forward motion.[/quote]

It's accurate enough to say that friction is "the cause of forward motion." Friction itself is the contract force between the wheels and the roadway.

That distinction notwithstanding, if friction did not balance the forward motion there would be a net torque on the wheels, the wheels would slip and the forward motion would be diminished if not curtailed altogether.

There are actually two different rotational motions in this problem. 1. There is the "orbit" (for lack of a better term) of the motorcycle about the center of the track, and 2. there is the rotation of the wheels about their axles.

It is in the second sense that friction must balance the forward motion; otherwise, the wheels will slip losing traction.

[quote="ffao"]Also, the velocity vector's magnitude does change with time, otherwise the motorcycle wouldn't accelerate.[/quote]

It is intuitively obvious that the motorcycle must accelerate to acheive its maximum speed, and thereby the magnitude of the velocity vector must change.

When I said the magnitude of the velocity vector doesn't change I was referring to why the direction of $ \frac{v^2}{R}$ is radial; it doesn't actually accelerate anything, in the sense of changing its speed.

[quote="ffao"]Rajiv's solution is essentially my "first interpretation" above, which I believe is not correct.[/quote]

Rajiv is essentially correct IMO, it just took me awhile to see it.

Considering the motion of the motorcycle around the track, we have the following Lagrangian:

\[ \mathscr{L} = \frac{1}{2} m (\dot{R}^2 + R^2\dot{\theta}^2) - mgz \]

From which we can obtain the equations of motion:

\begin{eqnarray} m(\ddot{R} -R\dot{\theta}^2) & = & F_{R} \\ mR^2\ddot{\theta} & = & \tau_{\theta} \\ mg & = & N\end{eqnarray}

The quantities on the right hand sides of (1), (2), and (3) are reaction forces.

In (1) $ F_{R} = f = \mu N$

If we now stipulate no net radial acceleration (because we don't want the motorcycle to slip) then from (1) we obtain, $ -mR\dot{\theta}^2 = f$.

We also know that $ \dot{\theta} =\frac{v}{R}.$ and therefore,

$ f = -mR\left(\frac{v}{R}\right)^2 = -m\frac{v^2}{R}$.

In my previous post I had obtained,

\[ \Delta s =\frac{\frac{1}{2}mv_{max}^{2}}{f}\]

Substituting the above expression for $ f$ into the expression for $ \Delta s$,

\[ \Delta s = \frac{\frac{1}{2}mv_{max}^{2}}{-m\frac{v^2}{R}} = -\frac{1}{2}R\left(\frac{v_{max}}{v}\right)^2\]

I believe that the original problem asked for the minimum distance that the motorcyclist can travel in order to acheive maximum speed without slipping.

It should be obvious from inspection that said minimum occurs where $ v = v_{max}$ then,

\[ \boxed{\Delta s_{min} = \frac{R}{2}}\]

[quote="Rajiv"][quote="Dr. No"]
It should be obvious from inspection that said minimum occurs where $ v \equal{} v_{max}$ then,
\[ \boxed{\Delta s_{min} \equal{} \frac {R}{2}}
\]
[/quote]
That means my answer is correct!  :wow:  :jump:  :trampoline: (although the method may be wrong!!!)[/quote]
I'm a noob at physics, but it seems the above result comes from the fact that the acceleration of the motorcycle is in the same direction as the centripetal force, which seems incorrect because if the centripetal force increases then the circular path will be drawn into a smaller radius...maybe I'm doing something wrong.

Shouldn't the motorcyclist's acceleration be perpendicular to the centripetal force, since that would create a torque and increase the tangential velocity? (As long as it doesn't slip, we can say the motorcycle stays in the circular path right? So it's kinda like tying the motorcycle down to the center)

[quote="serialk11r"][quote="Rajiv"][quote="Dr. No"]
It should be obvious from inspection that said minimum occurs where $ v \equal{} v_{max}$ then,
\[ \boxed{\Delta s_{min} \equal{} \frac {R}{2}}
\]
[/quote]
That means my answer is correct!  :wow:  :jump:  :trampoline: (although the method may be wrong!!!)[/quote]

....

Shouldn't the motorcyclist's acceleration be perpendicular to the centripetal force, since that would create a torque and increase the tangential velocity? (As long as it doesn't slip, we can say the motorcycle stays in the circular path right? So it's kinda like tying the motorcycle down to the center)[/quote]

The motorcyclist's acceleration is perpendicular to the centripetal force. The minimum he covers in accelerating to maximum speed is equal to $ \frac{R}{2}$.

The reason for this is because he doesn't slip. No slipping implies no free radial acceleration (that is his speed is not allowed to change in the direction of centripetal force). Since there is no free acceleration in the radial direction, friction is equal to the centripetal force. When that is the case, the distance traveled along the track is equal to $ \frac{R}{2}$ by consequence of the work-energy theorem.

[quote="ffao"]I'm a noob at physics too, but it wouldn't hurt to give my opinion.

If the motorcycle doesn't slip, it means that friction is balancing the backwards rotation movement of the tires, keeping the velocity of the bottommost point always equal to 0. But that's pretty much irrelevant.

About the no-slipping -> tie into the center, I don't see why would that be the case. Consider a car, for example, in a circular track. The car doesn't slip, and yet, you have to constantly turn it inwards. In fact, if the acceleration were perpendicular to the centripetal force, the tendency of the motorcycle would be to go off on a tangent.[/quote]
You turn the bike inwards to provide centripetal acceleration. If this force suddenly disappears, the bike would, as you said, move away tangentially.

I don't know what else to tell you, gents.

The equation of motion in the radial direction is:

\[ m\,a_R \minus{} mR\,\omega^2 \equal{} f\]

If there is no slipping then $ a_R \equal{} 0$. (The motorcycle is only accelerating in the direction of its velocity.)

From the above EOM, with $ a_R \equal{} 0$, $ f \equal{} \minus{}mR\,\omega^2$.

Since the velocity is $ v \equal{} R\,\omega$, (the track is a circle, and a circle has a fixed radius by definition) the kinetic energy, $ \frac{1}{2}mv^2 \equal{} \frac{1}{2}mR^2\,\omega^2$.

From the work-energy theorem: $ \Delta s \equal{} \frac{\frac{1}{2}mv^2}{f} \equal{}\frac{\frac{1}{2}mR^2\,\omega^2}{\minus{}mR\,\omega^2} \equal{} \minus{}\frac{R}{2}$.

(The minus sign comes from the direction of friction relative to the direction of motion of the motorcycle.)

I'm a physics noob too, and I have a question about that equation.

Why do we care about the EOM in the radial direction? We really want to analyze the tangential component, which is the most interesting bit.

Also, what is $ f$ in the EOM? Frictional force? But frictional force doesn't do any work in the radial direction...  :huh:

[quote="bowei"]I'm a physics noob too, and I have a question about that equation.

Why do we care about the EOM in the radial direction? We really want to analyze the tangential component, which is the most interesting bit.[/quote]

Because the EOM in the radial direction tells us that $ f \equal{} \minus{}mR\,\omega^2$, and that is a most useful thing to know.

[quote="bowei"]Also, what is $ f$ in the EOM? Frictional force? But frictional force doesn't do any work in the radial direction...  :huh:[/quote]

Yes, $ f$ in the EOM is the force of friction. I had defined it as such in a previous post. Friction does not do any work in the radial direction, but it does in the tangential direction. Indeed it is the cause of motion in the tangential direction.

The thing that most of you seem not to be seeing is that radial friction and tangential friction have the same magnitude.

\[ \left(f\right)_R \equal{} \left(f\right)_{\theta} \equal{}\mu\,N \equal{} mR\,\omega^2\]

The distance traveled, $ \Delta s$, is in the tangential ($ \theta$) direction, and its magitude is $ \frac{R}{2}$

Huh?
Okay I'm getting confused, so here's what I'm thinking:
When the object moves at the maximum velocity in the circular path, the centripetal force is at its maximum.
When an object accelerates in its circular path, the acceleration is in the direction of its initial velocity (if it had any), which is perpendicular to the centripetal force.
When the object accelerates, the centripetal force must increase as well. The only thing limiting the motorcycle's acceleration then is the limit to the centripetal force, given by the coefficient of friction.

So what is needed to be found is the effect of tangential acceleration on the centripetal acceleration, and substitute the maximum centripetal acceleration in to find the maximum tangential acceleration. (which I am not sure how to do)

It seems unlikely that that would come to the same result as simply increasing centripetal force, since the force causing tangential acceleration is in a totally different direction, and the $ f\equal{}\minus{}mR{\omega}^2$ thing you are using appears to indicate that the acceleration is in fact in the direction of the centripetal force, which I don't believe.

[quote="serialk11r"]Huh?
Okay I'm getting confused, so here's what I'm thinking:
When the object moves at the maximum velocity in the circular path, the centripetal force is at its maximum.[/quote]

Correct.

[quote="serialk11r"]When an object accelerates in its circular path, the acceleration is in the direction of its initial velocity (if it had any), which is perpendicular to the centripetal force.[/quote]

Its acceleration is in the direction of its velocity at any given instant, which is always perpendicular to the centripetal force. The acceleration is not in the direction of the object's initial velocity because the direction of the velocity vector is continuously changing, and especially in this case because the initial velocity, given in the problem statement, is zero. The motorcycle starts from rest.

[quote="serialk11r"]When the object accelerates, the centripetal force must increase as well. The only thing limiting the motorcycle's acceleration then is the limit to the centripetal force, given by the coefficient of friction.[/quote]

Well, velocity is limited by friction you got that part right. The coefficient of friction is merely a proportionality constant that relates friction to the normal force exerted by the roadway on the wheels.

[quote="serialk11r"]So what is needed to be found is the effect of tangential acceleration on the centripetal acceleration, and substitute the maximum centripetal acceleration in to find the maximum tangential acceleration. (which I am not sure how to do)[/quote]

They are related but they are also orthogonal; and as such, solving for one in terms of the other is irrelevant as far as this problem is concerned.

[quote="serialk11r"]It seems unlikely that that would come to the same result as simply increasing centripetal force, since the force causing tangential acceleration is in a totally different direction, and the $ f \equal{} \minus{} mR{\omega}^2$ thing you are using appears to indicate that the acceleration is in fact in the direction of the centripetal force, which I don't believe.[/quote]

If you don't believe me, I cannot help you. Nevertheless, it is the case that where radial acceleration is zero friction is equal to the centripetal force. Centripetal force is what keeps the bike moving in a circle -- it doesn't just magically arise because the bike [u]is[/u] moving in a circle. In this case, the centripetal force is provided by friction.

FYI, I developed the EOM using [url=http://electron6.phys.utk.edu/phys594/Tools/mechanics/summary/lagrangian/lagrangian.htm]Lagrangian Mechanics[/url], the particulars of which I prefer not to discuss with a self-professed "noob."

Notwithstanding, what the Lagrangian shows is: $ m\,a_r \minus{}m\,r\,\omega^2 \equal{} F_r$ for the radial direction. This is the EOM that I have referred to several times by now.

In this equation, $ F_r$ is any force on the body in question, not derivable from a potential energy function. $ F_r$ includes inputs from the rider to keep the bike moving in a circle, friction, and any other forces that don't arise from a change in potential energy. To keep it simple, I assumed $ F_r$ was only due to friction. When that is the case then the EOM becomes:

\[ m\,a_r \minus{}m\,r\,\omega^2 \equal{} f\]

For circular motion in general $ a_r \equal{} 0$ because a circle is defined as, "The locus of points equidistant from a given center point." With that definition in hand, the radius of any given circle is fixed and therefore the velocity and, by extension,  the acceleration in the radial direction is zero.

Our equation of motion therefore reduces to,
\[ \minus{}m\,r\,\omega^2 \equal{} f\].

The force of friction is equal to the centripetal force. QED

It is interesting to note that when there is no contraint force provided by the rider or by friction, either one ($ F_r \equal{} 0$) the EOM becomes:

\[ m\, a_r \minus{}m\,r\,\omega^2 \equal{} 0\]

indicating that $ a_r \equal{} r\,\omega^2$, which means that the motorcycle, car, or whatever is trying to maintain curvilinear motion will tend to slide to the outside of the curve it is attempting to follow.

[quote="bowei"][quote="Dr. No"]
The thing that most of you seem not to be seeing is that radial friction and tangential friction have the same magnitude.
\[ \left(f\right)_R \equal{} \left(f\right)_{\theta} \equal{} \mu\,N \equal{} mR\,\omega^2
\]
The distance traveled, $ \Delta s$, is in the tangential ($ \theta$) direction, and its magitude is $ \frac {R}{2}$[/quote]

Why does radial friction have the same magnitude as tangential friction?[/quote]

Because friction is the contact force between an object and the surface it rests upon. It arises due to a mismatch in the texture of the contact surfaces, and is created by rubbing those surfaces against one another. The coefficient of friction is a proportionality constant that relates the normal force, i.e. the reaction force with which a surface pushes on a body that rests on it, to the force of friction. That is their magnitudes are related by the coefficient of friction. The direction of friction always opposes the direction of motion or impending motion.

[quote="serialk11r"]I did not mean the initial velocity, and I do understand that the velocity is always changing.

My idea of solving this is, since centripetal acceleration depends only on tangential velocity, we can express tangential velocity as the integral of the acceleration, as a function of time, plug it into $ v^2/r$ and somehow solve for x (distance), but I can't get it to work.
Another approach I thought of was examining how the centripetal acceleration is affected by changing velocity, but I have no idea how to do that.[/quote]

Those are both good ideas, but they're also both unnecessary.

We know that friction is constant, $ |f| \equal{} \mu mg$.

By assumption, friction is the only constraint force on the RHS of the EOM:
\[ m\,a_r \minus{} m\,r\,\omega^2 \equal{} f
\]
Further if $ a_r \equal{} 0$ then $ \minus{} m\,r\,\omega^2 \equal{} f\; (\text{const.})$, it follows that $ f \equal{} \minus{} m\,r\,\omega_{max}^{2}$.

If $ \omega < \omega_{max}$ then $ F_r \equal{} F_{rider} \plus{} f$ rather than $ f$ alone; the rider must provide some input to maintain the circular path.

[quote="serialk11r"]
Another approach I thought of was examining how the centripetal acceleration is affected by changing velocity, but I have no idea how to do that.[/quote]
$ a_{centripetal} \equal{} \frac{v^2}{R}$
Differentiating with respect to time, you get:
$ \frac{da}{dt} \equal{} \frac{2v\left(a_{tangential}\right)}{R}$
However, as [b]Dr. No[/b] said, this is quite unnecessary.

$ a_r \equal{} r\omega^2$ if the motorcycle is traveling in a circle. $ a_r \ne 0$ implies the motorcycle's acceleration is [i]not[/i] in the same direction as the motorcycle's velocity.

If we discuss this in terms of classical mechanics rather than equations of motion, perhaps you'd understand. What forces are acting in the bike? In which direction?

EDIT: I'll give myself a limit. One more post and I'm out of this thread...

Yes I understand that centripetal acceleration is constant.
Here is my understanding though:
The way we get $ F \equal{} \frac {mv^2}{R}$ is by considering the amount of force pulling into the center needed to change the direction of the object such that it stays in the same circular path. I haven't ever worked through it.
I think we can all agree that to make the object move faster in its path, we need to apply some force perpendicular to the centripetal force. That means the net force is no longer perpendicular to the velocity, but it is at some angle, and one of the components is the centripetal force, which is perpendicular. Keeping the centripetal force at its maximum, it looks like acceleration changes as a function of time. What we have to find is the function of acceleration with respect to time such that the net acceleration changes the rate at which the direction changes at a constant rate with respect to position, keeping the centripetal acceleration component of the acceleration constant.

EDIT: after reading the wikipedia article on circular motion, I think I have a method, but I'll solve it during math class tomorrow or something.

[quote="ffao"]$ a_r \equal{} r\omega^2$ if the motorcycle is traveling in a circle.[/quote]

That's incorrect. (As I have been using the symbol $ a_r$, it means $ \frac {d^2 r}{dt^2}$. The radius of a circle is constant, and its derivatives are therefore all zero.) Furthermore, centripetal acceleration though pointed in the radial (or rather anti-radial) direction does not actually accelerate anything. It merely changes the direction of the velocity vector.

[quote="ffao"]If we discuss this in terms of classical mechanics rather than equations of motion, perhaps you'd understand. What forces are acting in the bike? In which direction?[/quote]

Your use of the term [i]classical mechanics[/i] is idiosyncratic. Classical mechanics refers to systems to which Newton's laws apply, as opposed to quantum mechanics that refers to systems to which they don't.

$ \vec{F} \equal{} m\vec{a}$ is an equation of motion. Lagrangian methods, I say methods instead of "mechanics" because Lagrangian methods apply to classical mechanical systems. Furthermore, they allow one to answer your question exactly as you asked it, "What forces are acting on the bike? [and] in which direction(s) [are they pointed] ?"

[quote="ffao"]I'll give myself a limit. One more post and I'm out of this thread...[/quote]
Yes, he's right. Dr. No has explained the concepts here reasonably well. I think there are other topics to argue about!! :D  :rotfl:

[quote="ffao"]Advertising topics, are you? :rotfl:

Well, as I said, nobody's changing their minds here, so I'm off to somewhere else.[/quote]
Yo! buddy, I just wanted help with that one! Nice thinking though!

[quote="Anvay"]ACTUALLY IT SHUD DEPEND UPON THE ANGLE OF INCLINATION OF THE MOTORCYCLIST W.R.T. THE GRND??[/quote]
My dear friend,
all these aspects have been discussed rather exhaustively earlier. Please read the entire thread (which in itself will be rather exhausting exercise :rotfl: )

So it appears I was indeed right ;)  (phew...)

I wasn't actually able to solve the integral, and honestly I don't get your rearranging, but that's a different matter as I never took calculus in the first place...

[quote="genericme"]http://liquids.seas.harvard.edu/oleg/competition/99/99test.pdf
http://liquids.seas.harvard.edu/oleg/competition/99/sol99.pdf for the solution.[/quote]

You should've posted that long back... :rotfl:  :P

[quote="genericme"]Dudes...

See http://liquids.seas.harvard.edu/oleg/competition/99/99test.pdf, number 3, for the problem, and

http://liquids.seas.harvard.edu/oleg/competition/99/sol99.pdf for the solution.[/quote]

genericme, since you posted the solution, or rather the link to it, I have discovered a fallacy in my own thinking about the problem, that it might be worth clearing up for everyone.

I was thinking that friction was a constant force equal to $ \mu N$. It's not. According to the Wikipedia article on [url=http://en.wikipedia.org/wiki/Friction]friction[/url], $ f\le \mu N$. That means where this problem is concerned the force of friction, $ f \equal{} \minus{}mr\omega^2$, regardless of the value of $ \omega$. (Friction is always equal to the centripetal force.) That means, my attempts to superimpose additional constraint forces other than friction were wrong headed. [i]Mea culpa, mea culpa, mea culpa.[/i] Furthermore, the tangential driving force (as we all now know) is not itself equal to the force of friction. If it were, there would exist a differential equation for $ \omega$ as a function of time that is imminently solvable but whose final result is absurd. I should have recognized this at the beginning. I apologize for not having done so.

Hmm? Friction is a force of constant magnitude though in this problem...I think everyone assumed that centripetal force was the frictional force, which was incorrect since it was the total force, which centripetal force was a component of (maybe bowei didn't, but I forgot what he wrote).

I didn't assume it, genericme didn't assume it and bowei didn't assume it. All three of us posted the same solution.


Prove that for all $ n \in \mathbb{N}$, none of the prime divisors of $ 2^{n}+1$ is of the form $ 8k+7$ with $ k \in \mathbb{N}$.

For n even 2^n+1 is a sum of 2 squares ((2^a)^2+1^2) and every prime of the form 4k+3 which divides m^2+n^2 must divide m and n (a well known result) so in our case every prime of the form 8k+7 also has form 4k+3 so it must divide 1, false. I guess this leaves the case n odd.

For the case n odd it's like this: 2^n+1=2^(2k+1)+1=2*(2^k)^2+1^2.

Now I'll prove this lemma using Legendre symbol:
Lemma 1      -2 is not a square residue (mod p) if p is a prime of the form 8k+7 (meaning that there is no number x s.t. x^2=-2 (p)).

Proof of Lemma 1 (I don't know how to prove everything I used here):
We know that the Legendre symbol of 2 with respect to p is (2/p)=(-1)^((p^2-1)/4). For -1:                            (-1/p)=(-1)^((p-1)/2). We also know that Legendre symbol is multiplicative, so (-2/p) = (2/p)*(-1/p). By replacing (2/p) and (-1/p) we can see that in the case p=8k+7 (-2/p)=-1 iff -2 is not a square residue (p).

Lemma 2 If p is a prime of form 8k+7 and p|2a^2+b^2 then p|a and p|b. Let's assume a^2=k<>0 (p). Then 2a^2=2k (p), so b^2=-2k (p). This means that both k and -2k are square residues (p). Because Legendre symbol is multiplicative it means that -2 must also be a square residue (p), contradiction from Lemma 1.

Now we apply Lemma 2 like this: if p|2*(2^k)^2+1^2 where p is a prime of the form 8k+7 then, from Lemma 2. p|2^k and p|1, which is false, so there are no primes of the form 8k+7 dividing 2^(2k+1)+1.

P.S. Legendre symbol of a with respect to the odd prime p = (a/p)= 1 if a is a square residue (p) and -1 otherwise; (p) means (mod p).


Solve in complex numbers the equation:

$ z^{n\plus{}1}\plus{}|z|^n\equal{}2$, where $ n\ge 0$ is a given integer .

[quote="Mateescu Constantin"]Solve in complex numbers the equation:

$ z^{n \plus{} 1} \plus{} |z|^n \equal{} 2$, where $ n\ge 0$ is a given integer .[/quote]

how do we tackle such questions?

any help by anyone?

[hide="Not sure"]Write $ z$ in polar form as $ r( \cos \theta \plus{} i\sin \theta)$. Then, the equation becomes

$ r^{n \plus{} 1}(\cos \theta (n \plus{} 1) \plus{} i\sin \theta(n \plus{} 1) ) \plus{} r^n \equal{} 2$

The imaginary part on the LHS must be 0, so

$ \sin \theta(n \plus{} 1) \equal{} 0$

$ \theta (n \plus{} 1) \equal{} \pi k$ for some integer $ k$

Back to our equation. We have

$ r^{n \plus{} 1} \cos \theta(n \plus{} 1) \plus{} r^n \equal{} 2$

$ r^{n \plus{} 1} \cos \pi k \plus{} r^n \equal{} 2$

$ \pm r^{n \plus{} 1} \plus{} r^n \equal{} 2$ both sides must be positive; k is even

$ r^n ( r \plus{} 1) \equal{} 2$

$ r^n (r \plus{} 1) \equal{} 2$

So doesn't this give infinately many solutions??  :huh:[/hide]

We have $ z^{n\plus{}1} \equal{} 2 \minus{} |z|^n$, Denoting $ |z| \equal{} r \geq 0$ and taking moduli on both sides we get (there was a typo in the above post)

1. $ r^{n\plus{}1} \minus{} r^n \plus{} 2 \equal{} 0$, with no non-negative real root; or
2. $ r^{n\plus{}1} \plus{} r^n \minus{} 2 \equal{} 0$, with unique non-negative root $ r\equal{}1$, when we therefore have $ z^{n\plus{}1} \equal{} 1$, so $ z$ is any root of order $ n\plus{}1$ of the unit.


quadratics
geometry
geometric transformation
reflection
algebra
polynomial
calculus

Consider the set of all quadratic tables [tex]p\times{p}[/tex] of squares [tex](p>1)[/tex], filled with natural numbers [tex]1,2,...,p^2[/tex]. Let us call such a table "right", if there are numbers [tex]1,2,...,p[/tex] in the first row (column) that are ordered in succession, in the second row (column) [tex]p+1,p+2,...,2p[/tex], also ordered in succession, and so on.

Let [tex]A[/tex] be the subset of the set of our quadratic tables, in which each table can be obtained from the "right" one by permutating columns and rows. Let [tex]B[/tex] be the subset, in which one can get a table with equal numbers by adding the number 1 to all numbers of a row or column from any table.

Prove that [tex]A=B[/tex] if and only if [tex]p[/tex] is a prime number.

First of all, it’s clear that if a table is in $B$, then all the tables obtained from it through permutations of rows/columns are also in $B$. From this and the fact that a “right” table is obviously in $B$, we conclude that $A\subseteq B,\ \forall p$. We must thus show that $B\subseteq A$ is equivalent to $p$ being a prime.

One implication is easy: assume $p$ is not a prime, and let $d|p,\ 1<d<p$ be a proper divisor of $p$. Take any right table, and divide it in $d\times d$ subsquares. Now separately transpose each one of these squares (i.e. reflect each one of them wrt its own main diagonal). The large square we get after this does not belong to $A$, but belongs to $B$ (these claims are easy to check).

For the other implication, suppose $p$ is a prime, and also assume that our table belongs to $B$. Permute the rows and columns s.t. the numbers are ordered increasingly from left to right on each row and from top to bottom on each column (it’s clear that it’s possible to do this for a table belonging to $B$). Denote the entries of this table by $a_{ij},\ i,j\in\overline{0,p^2-1}$ (we shift all the numbers one unit down, so that we’re working with the numbers from $0$ to $p^2-1$ instead of those from $1$ to $p^2$).

Put $u_i=a_{0i}-a_{00},\ \forall i\in\overline{0,p-1}$ and $v_j=a_{j0}-a_{00},\ \forall j\in\overline{0,p-1}$. It’s easy to see now that the hypothesis that our table belongs to $B$ implies that $1+x+\ldots+x^{p^2-1}=f(x)g(x)$, where $f(x)=\sum_{i=0}^{p-1}x^{u_i},\ g(x)=\sum_{j=0}^{p-1}x^{v_j}$.

If $\phi_n$ denotes the $n$’th cyclotomic polynomial, we have $f(x)g(x)=\frac{x^{p^2}-1}{x-1}=\prod_{d|p^2,\ d>1}\phi_d(x)=\phi_p(x)\cdot\phi_{p^2}(x)$. Since $f,g$ have integral coefficients and the cyclotomic polynomials also have integral coefficients and are irreducible, it means that $f,g$ are precisely the two cyclotomic polynomails $\phi_p,\phi_{p^2}$, and the conclusion follows immediately (we get the two “right” tables according to the situation: either $f=\phi_p$ and $g=\phi_{p^2}$ or vice-versa).


[quote="madness"]Exactly  :D  And even a person who goes to iit classes and doesnt listen or absents himself may not do well (Am boasting here  :P )[/quote]
Shreyas make notice of that :rotfl:  :rotfl:

Was obviously referring to him  :D  Good note  :D  And gr8 minds think alike  :P

[quote="Cyber-Shreyas"]Fools seldom differ. I actually adopt a much more logical approach to chem, just that I get suppressed by ppl like Rohit and Rakesh and I can't remember stuff. Ironical for a person who got 100 in SST. :([/quote]

Ya Ironical [b][i]NV[/i][/b] thalai like u who got 100 in SST thadavify in Chem.

[quote="Cyber-Shreyas"]I am not good at logic NV. It's either pure logic or pure NV. But then again Rohit was unable to answer a simply question like why HI bond is weaker while I was. Clearly shows this question is not present in the numerous books he reads.[/quote]
[b][size=125][color=blue]"YANAIKUM ADI SARUKUM"[/color][/size][/b]
Clearly one more example there. But he might have got that correct but think of the numerous questions he answered and you literally struggled to answer(including through pocket ones).

[quote="madness"][quote="Cyber-Shreyas"]I am not good at logic NV. It's either pure logic or pure NV. But then again Rohit was unable to answer a simply question like why HI bond is weaker while I was. Clearly shows this question is not present in the numerous books he reads.[/quote]

Yes i accept i faltered wid that ques da but not every question which ive encountered has been read by me. I agree that i am not strong in that concept (leave alone that question) while u have already nved that concept from some book  :D . And also u need to know which type of compounds show that concept which u conveniently nved da  :P . So ive got to read that concept sooner or later. And it always happen in anyone's life da bcos i can claim that i know sth which carcul sir or sundari miss or rakesh or jothishwaran or for that matter anyone of our friends may not know the answer but my cousin sister who is in 5 th std may kniw that answer so there is nothing to be proud abt mr. peacock  :P[/quote]

I NV'ed that concepta?? LOL! Good joke. As I've been telling you from time to time, I have a prob remembering stuff in chem. I have to keep on revising if I have to remember it..... And I would think it was logically obvious given H and I, the first thing would have occured to a person was the size diff (since I isn't so electronegative... ). I also like GOC and the like, cause they are very logically IMHO.

Oh! Rakesh, not for a moment did I doubt baby's prowess in chemistry, obviously he is probably going for the TC. :D Just a little kalai in chem since I am at the recieving end most of time.

Okay, lets stop this discussion and go back to the olympiad. Rohit, if you have anything to say, then post in Meta thread.

[quote="madness"]No i would like to say it here where its relevant that today u urself said that u had nved from rsm that the ortho product on nitration of benzene was 33% and ...............................................
Now what do you call that  :huh:   :D   :P[/quote]

Yes, I NV facts which are interesting. No it isnt 33% da, its 33 times. LOL! And w/o nving you can't survive in chem.

Pardesi do u also NV chemistry or is it logical like our hero shreyas :rotfl:

if i nv chemistry i also nv maths and also nv Physics....
Chemsitry is as logical as Physics or maths....but may be less interesting to me...
I luv organic and planning to take up Organic Synthesis as my summer programme(if i get thru NIUS) with HBCSE...and that says it all :)

[quote="valeriummaximum"]Thala....
I prfer organic hem to physical becuase i have dysgraphia(just joking)[/quote]

What does dysgraphia have to do with organic??

that is i don like physical because of that dysgraphia

[quote="chemrock"][quote="valeriummaximum"]Thala....
I prfer organic hem to physical becuase i have dysgraphia(just joking)[/quote]
Obviously ranga wants to show off that he is a good english NV(though I have no right to talk about it). Apparum athu 'chem' hem illa.[/quote]

Are you trying to impress some people in this forum??

OK shreyas,overa double la pesara....
w'll stop this here....
try the qualitative question given by chemrock.........

Dei straight forward sollen enna solla varenu.

[quote="chemrock"][quote="madness"]No i would like to say it here where its relevant that today u urself said that u had nved from rsm that the ortho product on nitration of benzene was 33% and ...............................................
Now what do you call that  :huh:   :D   :P[/quote]

[size=200][b][i][color=red]U HAVE NO RIGHT TO TALK ABOUT NV[/color][/i][/b][/size]
U nved the resonance energies of benzene and naphthalene which far worse than the what u mentioned above.[/quote]

Well one cant help remembering thinks after one has been told the same piece of information in class by one's professor atleast 10 times  :huh:   :D . But shreyas is fresh nv i.e. he recently read it in rsm  :P


conics
ellipse
quadratics
Vieta
symmetry
analytic geometry
graphing lines

Given an ellipse $\frac{x^2}{9}+y^2=1$. $l$ is a line meets the ellipse at two different points $M$, $N$. If $MN$ is bisected by $x=-\frac{1}{2}$, fing the inclation of $l$.

[hide]Okay, definitely not sure as to whether I got the correct answer... I'm definitely sure that there's a better way than this to do the problem  :D

Graph out the ellipse and let the equation of the line be $y = ax - \frac{1}{2}$. Plug this into the ellipse and solve for $x$ and $y$. Then solve for $a$. And now comes the horrendously evil part. SOLVING FOR $a$. I came up with three different values: $-6^{1/3}$, $-.8069$, and $3.06918$, which is technically cheating, because I used my calculator  :P[/hide]

I'd say your line equation is wrong - it should be $y=a(x+{1\over 2})$´

[hide="Solution"]Substituting $y=a(x+{1\over 2})$ into the ellipse equation, we get

$\left({1\over 9}+a^2\right)x^2+a^2x+\frac{a^2}{4}-1=0$

If the solutions of this quadratic are $x_1$ and $x_2$, we get ${{x_1+x_2}\over 2}=-{1\over 2}$ (by the formula for the midpoint of a segment). Using Vieta's formulas, we get

$-\frac{a^2}{{1\over 9}+a^2}=-1$

Since this has no solution in $a$, our line equation is wrong, and the only other possibility is the line $l: x=-{1\over 2}$, which indeed is a solution, due to the ellipse symmetry, and its inclination is $90^\circ.$[/hide]

Oh no... man I accidentally made the equation of the line $y = \frac{-1}{2}$ Lies...  :ninja: I'll work on it later...  :(

Here's my new edited solution  :)

[hide]Not sure about no solution though... it seems as if there could be one... using the CORRECT equation for the line (grr... ) :lol: and substituting into the equation $\frac{x^2}{9}+y^2=1$, we get $x^2 + a^2(x^2+x+\frac{1}{4})$. Expanding and arranging, $x^2(a^2+\frac{1}{9}) + a^2x+(\frac{1}{4}a^2-1) = 0$. Since the sum of the $x$ values must be $\frac{-1}{2}$, $\frac{-a^2}{a^2+\frac{1}{9}} = \frac{-1}{2}$ Solving, we get $a = \pm \frac{1}{3}$. Lol, it seems much easier than my [i]last [/i]answer...  :D [/hide]

[quote="Karth"]...Since the sum of the $x$ values must be $\frac{-1}{2}$...[/quote]

You're mistaken again :) [hide="The reason"]The HALF-sum of $x$ values must be $-{1\over 2}.$[/hide]

Oh yeah... my bad. Sorry about that :blush: I usually don't make such messed up mistakes lol. Guess it's pretty late. Going to bed. zzzzz :?

[hide]$M(x_M,y_M)$
$N(x_N,y_N)$

$MN: y=ax+b$

$\begin{array}{cc} x^2_M + 9y^2_M = 9\\ \\ x^2_N + 9y^2_N = 9 \end{array} \}\Rightarrow (x^2_M-x^2_N)+9(y^2_M-y^2_N)=0\Rightarrow$

$(x_M+x_N)(x_M-x_N) = -9(y_M+y_N)(y_M-y_N)$

$a=\frac{y_M-y_N}{x_M-x_N} = -\frac{1}{9} \cdot \frac{x_M+x_N}{y_M+y_N} \ \ (1)$

Let $P$ be the midpoint of $MN$. Then $P(-\frac{1}{2},m) = (\frac{x_M+x_N}{2},\frac{y_M+y_N}{2})\Rightarrow \frac{x_M+x_N}{y_M+y_N} = -\frac{1}{2m}$

$(1) \Rightarrow \boxed{a = \frac{1}{18m}}$

The point $P$ is an inner point of the ellipse, so $|m|\leq \sqrt{1-\frac{x^2}{9}}=\sqrt{1-\frac{1}{36}} = \frac{\sqrt{35}}{6} \Rightarrow$

$\frac{1}{|m|} \geq \frac{6}{\sqrt{35}} \Rightarrow$

$\frac{1}{|18m|} \geq \frac{1}{3\sqrt{35}}$

$a = \frac{1}{18m} \Rightarrow a \in (-\infty, -\frac{1}{3\sqrt{35}}) \bigcup ({\frac{1}{3\sqrt{35}},+\infty)}$

(the title of the problem asks for a [u]range[/u] of the inclination $a$)
[/hide]

I see - you took that the chord is bisected by the LINE $x=-{1\over 2}$, not the point $(-{1\over 2},0)$. Seems sensible.

[b]Theory.[/b] Given is a particular conic (from the high-school)
 $Ax^2+By^2+2Cx+2Dy+E=0$.
If $m,m'$ are the slopes of any two conjugate diameters,
then you can prove easily that  $\boxed {\ mm'=-\frac{A}{Bm}\ }\ \ (1)\ .$

[b]Remark.[/b] The geometrical locus of the midpoint for a mobile chord (in the conic) with the fixed slope $m$ is the diameter with a slope $m'$ which respects the relation $(1)$ and in this case the directions $m\ ,\ m'$ call [color=red]conjugate directions[/color] and the corresponding diameters call [color=red]conjugate diameters[/color].

I found (similarly to Pontios) that the slope $m$ of the line $MN$ is equal to
$\frac{1}{18\lambda }$, where $\left(-\frac 12\ ,\ \lambda \right)\in MN$ a.s.o.


calculus
integration
inequalities
calculus computations

Evaluate

$ \displaystyle\int^1_0 e^{x \plus{} e^{x \plus{} e^{x \plus{} e^x}}}\ dx$.

This is equivalent to evaluate the integral $ \int_{e}^{e^e} \frac {t^{t^t}}{t} \, dt$, which is obtained by the substitution $ t \equal{} e^{e^x}$. But I believe it is impossible to give a closed form of this integral.

Even its large value blocks us to attack this via numerical methods. Put $ f(t) \equal{} \frac {t^{t^t}}{t}$, then $ f$ is convex, thus Jensen's inequality shows that

$ \int_{e}^{e^e} f(t) \, dt \geq (e^e \minus{} e) f \left( \frac{\int_{e}^{e^e} t \, dt}{e^e \minus{} e} \right) \approx 3.5576 \times 10^{300638152}$


algebra
polynomial
vector
linear algebra
linear algebra unsolved

Suppose $\lambda_1,\dots,\lambda_n\in\mathbb F$ are eigenvalues of $A\in M_n(\mathbb F)$. Prove that if $p(x)\in\mathbb F[x]$ is a polynomial then $p(\lambda_1),\dots,p(\lambda_n)$ are eigenvalues of $p(A)$. (Consider multicipity of eigenvvalues)

if $v$ is an eigenvector of A with eigenvalue $c$, and $p(x) = a_0 + a_1x + \cdots + a_nx^n$, then $p(A)v = (a_0 + a_1A + \cdots + a_n A^n)v = (a_0 + a_1c + \cdots + a_n c^n)v = p(c)v$. Thus $p(c)$ is an eigenvalue of p(A). If c is of multiplicity m, we can have m different v, and all of them are eigenvectors of $p(A)$ corresponding to eigenvalue $p(c)$, thus the multiplicity remains the same.

Just make $A$ upper triangular through a similarity in the algebraic closure of $\mathbb F$.

To Soarer: You can not say that we have $m$ vectors that $Av=\lambda v$ iff $\lambda$ is a root of multicipity $n$.

[b]My Solution :[/b] WLOG, Suppose that $\mathbb F$ is an infinite algebraic closed field. Then we know that for each $t\in \mathbb F$ there exist $c,\zeta_1,\zeta_2,\dots,\zeta_n$ that $p(x)-t=c(x-\zeta_1)(x-\zeta_2)\dots(x-\zeta_k)$. You see $\det (p(A)-tI)=\det (A-\zeta_1I)\det(A-\zeta_2I)\dots\det(A-\zeta_kI)$. You see that eigenvalues of $A+sI$ are $\lambda_1+s,\dots,\lambda_n+s$. So \[ \displaystyle\det(p(A)-tI)=c^n(\prod_{i=1}^{n}(\lambda_i-\zeta_1))(\prod_{i=1}^{n}(\lambda_i-\zeta_2))\dots(\prod_{i=1}^{n}(\lambda_i-\zeta_k))\\ \displaystyle=c^n(\prod_{i=1}^{k}(\lambda_1-\zeta_i))(\prod_{i=1}^{k}(\lambda_2-\zeta_i))\dots(\prod_{i=1}^{k}(\lambda_n-\zeta_i)) \displaystyle=(p(\lambda_1)-t)(p(\lambda_2)-t)\dots(p(\lambda_n)-t) \]
So $\det (tI-p(A))=(t-p(\lambda_1))(t-p(\lambda_2))\dots(t-p(\lambda_n))$, and because this relation is true for each $t\in\mathbb F$ so $\det(xI-p(A))=\prod_{i=1}^{n}(x-p(\lambda_i))$.
QED

sorry I treated A as a diagonalizable matrix. I think grobber's way is nice.


A dietician in a hospital must prepare a meal for a patient as was prescribed by a doctor to have 800 calories, 55 g of protein, and 220 mg of vitamin C. How many servings of broccoli ($ b$), potatoes ($ p$), and steak ($ s$) are needed in order to exactly satisfy the physician's requirements? Express your answer as the ordered triple ($ b$, $ p$, $ s$).

\begin{tabular}{cccc}
 & Calories & Protein (g) & Vitamin C (mg) &
Steak & 300 & 20 & 0 &
Potato & 100 & 5 & 20 &
Broccoli & 50 & 5 & 100
\end{tabular}

From the fact that the dietician needs 220 mg of Vitamin C, we can deduce that thedietician needs one serving of potato. Then, directly from that last fact, we deduce that he needs $ 2$ servings of broccoli. Finally, from both of these facts, we can deduce that he needs $ 2$ servings of steak. Thus our ordered triple is $ \boxed{(2,1,2)}$.


geometry inradius trigonometry calculus integration

$ a$, $ b$, $ c$ are the lengths of the sides, $ r$ is the inradius of a triangle. We know that $ r\equal{}1$ and $ a,b,c\in\mathbb{Z}$. What is $ a$,  $ b$, $ c$?

[quote="mathgeniuse^ln(x)"][quote="dogcat"]$ a$, $ b$, $ c$ are the lengths of the sides, $ r$ is the inradius of a triangle. We know that $ r \equal{} 1$ and $ a,b,c\in\mathbb{Z}$. What is $ a$,  $ b$, $ c$?[/quote]

Well, if this wasn't a proof, then $ 3$, $ 4$, and $ 5$ would easily be allowed.[/quote]
Yeah, but it should be a proof. :maybe: (I've also discovered that $ 3$, $ 4$, $ 5$ is good.) Maybe we should prove that there are no other solutions. (Just guessing - maybe there are...  :huh: )

[quote="The Zuton Force"]Assume $ c > b > a$. We know that $ 2A \equal{} Pr$ and $ r \equal{} 1$.

$ ab \equal{} a \plus{} b \plus{} \sqrt {a^2 \plus{} b^2}$,   [i]by Pythagorean Theorem.[/i]
$ \implies ab < (2 \plus{} \sqrt {2})b$[i],   where a was replaced with b on the RHS.[/i]
$ \implies a < 2 \plus{} \sqrt {2} < 4$
$ \implies a \in \{ 1, 2, 3 \}$

It is clear that $ a \equal{} 1, \; 2$ yield no solutions in \mathbb{N}, hence $ a \equal{} 3$.

Using $ a \equal{} 3$, the only possible solution is $ (a,b,c) \equal{} (3,4,5)$.[/quote]

Nice, but what if the triangle is not a right triangle?

Right.

[b] [[ In progress ]] [/b]

[hide="Part 1a"]If it's a non-obtuse, then we have $ ab \sin \theta \equal{} a \plus{} b \plus{} c$, and thus, since $ \theta \geq \frac {\pi}{3}$, we have that:

$ \frac {\sqrt {3}}{2} ab < ab \sin \theta < a \plus{} b \plus{} c < (2 \plus{} \sqrt {2}) b$, since $ c \leq \sqrt {a^2 \plus{} b^2}$

Then, $ ab < \frac {2}{\sqrt {3}} (2 \plus{} \sqrt {2}) b \implies a < 4$, as before, and $ a \in \{1, 2, 3 \}$. Clearly, $ a \equal{} 3$, since $ a > D \equal{} 2r \equal{} 2$. Now proceed to part 2a.[/hide]

[hide="Part 2a"]So, now we have a (3, b, c) triangle, and $ r^2 \equal{} \frac {(a \plus{} b \minus{} 3)(a \minus{} b \plus{} 3)( \minus{} a \plus{} b \plus{} 3)}{4(a \plus{} b \plus{} 3)}$ after using heron's and squaring.

Hence, $ 4(c \plus{} b \plus{} 3) \equal{} (c \plus{} b \minus{} 3)(c \minus{} b \plus{} 3)( \minus{} c \plus{} b \plus{} 3)$. Letting $ c \plus{} b \equal{} u, \; c \minus{} b \equal{} v$, we have:

$ 4(u \plus{} 3) \equal{} (u \minus{} 3)(9 \minus{} v^2)$, and divide by $ u \minus{} 3$ to get:

$ 4 \plus{} \frac {24}{u \minus{} 3} \equal{} 9 \minus{} v^2 \implies v^2 \equal{} 5 \minus{} \frac {24}{u \minus{} 3} < 5 \implies v < \sqrt {5} < 3$, since $ u > c \equal{} 3$.
Hence, the integral solutions are $ (v,u) \equal{} (1, 9), \; (2, 27)$

Only $ (v,u) \equal{} (1,9)$ produces an integral solution in $ (b,c)$, namely $ (4,5)$.

Hence, the only non-obtuse triangle satisfying this is $ (a,b,c) \equal{} (3,4,5)$.[/hide]


[quote="me@home"][quote]2.1 (JGeneson 03') - Let $a_{0}= 1$, and $a_{n}= 2a_{n-1}\sin\frac{n\pi}{48}$ for $n \geq 1$. Compute: $\prod_{i=1}^{47}a_{i}$.[/quote]
Any nice solutions?[/quote]

there is a closed form for the sine product (this problem is $n=48$)

[hide="hint"]$e^{i}$ substitution[/hide]

I don't see how your hint helps... do you just mean complex numbers in general or representing $\sin$ as $\frac{e^{i\theta}-{e}^{-i\theta}}{2}$? :maybe:


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $A$ be a finite ring with more than 1 element and such that the only nilpotent is 0. Prove that $A$ is a direct sum of fields.

Let $M_1,M_2,\ldots,M_k$ be the maximal (bilateral) ideals of $A$, and for $x\in A$ let $\bar x_n$ be the image of $x$ through the canonical projection $A\to A/M_n$ (for $n=\overline{1,k}$).

If $r\in R=\bigcap_n M_n$, then $1-xr$ is invertible for all $x\in A$, because otherwise there would be some maximal ideal which contains $1-xr$, and if that ideal also contained $r$, then it would contain $1$ as well, giving a contradiction. Now, for every such $r\in R$, we have $r^p=r^q$ for some $q>p$, due to the finiteness of the ring. We then get $r^p(1-r^{q-p})=0$, and since $1-r^{q-p}$ is invertible, we get $r^p=0$, i.e. $r$ is nilpotent, which implies $r=0$. We have thus shown that $R=\{0\}\ (*)$.

Consider the mapping $x\mapsto (\bar x_1,\bar x_2,\ldots,\bar x_k)$. It's easy to prove now, using $(*)$, that this is an isomorphism from $A$ to $\bigoplus_n(A/M_n)$.

Thanks :).

One thing I was worried about: the ring was supposed to be unital, right?

Do :) n't worry, I forgot to mention that.

This works only for $ A$ commutative.
However assume we have $ A$ non-commutative with $ |A|$ minimal satisfying the condition.
Then the centre $ Z(A)$ is commutative, and by the above argument it must be a direct product of some fields $ F_1,\ldots,F_n$.
This gives a decomposition $ 1\equal{}e_1\plus{}\cdots\plus{}e_n$ in orthogonal central idempotents $ e_1,\ldots,e_n\in Z(A)\setminus \{0\}$.
Then each $ A_i: \equal{}Ae_i$ is a ring with neutral element $ 1_{A_i}: \equal{}e_i$ and we get
the isomorphism of rings $ A\to \times_{i\equal{}1}^nA_i, x\mapsto (xe_1,\ldots,xe_n)$ with inverse map $ (x_1,\ldots,x_n)\mapsto x_1\plus{}\cdots\plus{}x_n$.
Then by the minimal choice of $ |A|$ we must have $ n\equal{}1$ and $ Z(A)$ a field.

Assume $ x\in A\setminus Z(A)$ doesn't have a left inverse, i.e. $ Ax\subsetneq A$, and take $ |Ax|$ minimal.
If $ x^2\in Z(A)$, then $ x^2\not\equal{}0$ must have an inverse $ y$ in $ Z(A)$, which gives the left inverse $ yx$ of $ x$. Contradiction!
So $ x^2\not\in Z(A)$ and $ Ax^2\subseteq Ax$ gives $ Ax^2\equal{}Ax$ and $ x\equal{}yx^2$ for some $ y\in A$.
Then $ (1\minus{}yx)x\equal{}0$ gives $ (x(1\minus{}yx))^2\equal{}0$, and so $ x(1\minus{}yx)\equal{}0$.
So $ e: \equal{}yx$ satisfies $ (1\minus{}e)e\equal{}e(1\minus{}e)\equal{}yx(1\minus{}yx)\equal{}0$.
Then for all $ z\in A$ we get $ (ez(1\minus{}e))^2\equal{}ez(1\minus{}e)ez(1\minus{}e)\equal{}0$, so $ ez(1\minus{}e)\equal{}0\Rightarrow ez\equal{}eze$.
Analog we get $ ze\equal{}eze\equal{}ez$ for all $ z\in A$, which gives $ e\in Z(A)$.
But now $ 0\equal{}e(1\minus{}e)$ in the field $ Z(A)$ gives $ e\equal{}0$ or $ e\equal{}1$
and $ ex\equal{}yx^2\equal{}x\not\equal{}0$ gives $ 1\equal{}e\equal{}yx$ and $ y$ is left inverse of $ x$.

This contradiction shows, that every $ x\in A\setminus Z(A)$ has a left inverse, and analog also a right inverse.
But since $ Z(A)$ is a field, we get $ A$ a finite division ring, which is commutative by [url=http://en.wikipedia.org/wiki/Wedderburn_theorem]Wedderburn[/url].

you seem to show  that every finite unital ring with no nilpotent elements is commutative? but this is false for $ \mathbb{F}_2[S_3]$.

Oh, but it is true.
Take $ \sigma\equal{}(12)\in S_3$ the transposition switching $ 1$ and $ 2$.
Then $ (1\plus{}\sigma)^2\equal{}1\plus{}2\sigma\plus{}\sigma^2\equal{}1\plus{}0\plus{}1\equal{}0$ in $ \mathbb{F}_2S_3$,
so $ 0\not\equal{} 1\plus{}\sigma$ is nilpotent in $ \mathbb{F}_2S_3$.

the direct product is meant here.
@olorin: ok, thank you for the proof :-). in particular, every finite unital ring with the property $ x^n\equal{}x$ is commutative (which is a lot harder in the infinite case).

[quote="Kalle"]Is direct sum the same thing as the product, i.e. $ n$-tuples with component-wise addition and multiplication?[/quote]
If $ A_i,i\in I$ are groups, then the direct product or simply product $ \prod_{i\in I}A_i=\times_{i\in I}A_i$
is the set of all sequences $ (x_i)_{i\in I}$ with $ x_i\in A_i$, which is again a group by component-wise operation.
And the direct sum or coproduct $ \coprod_{i\in I}A_i=\sum_{i\in I}A_i=\bigoplus_{i\in I}A_i$ is the set of those sequences $ (x_i)_{i\in I}$
with only finitely many $ x_i\not= 0_{A_i}$. (the neutral element of $ A_i$)
This gives also a group by component-wise operation.
And those are simply the same for $ I$ finite.

Also if $ A_i$ are unital rings, then the direct product and sum of the additive groups $ A_i$
form again rings by component-wise operation, but only the product is unital,
as $ \coprod_{i\in I}A_i$ doesn't contain the $ 1$-element $ (1,1,\ldots)$. (unless $ A_i=0$ for almost all $ i$)

olorin, what you said about coproducts is not totally correct:
A coproduct of groups (in the category of ALL groups, not only in the category of abelian groups [i.e. $ \mathbb{Z}$-modules] ) is the [u]free product[/u] of groups.
If you work in the category of abelian groups then a coproduct of abelian groups is exactly as you said.

The coproduct of commutative rings (in the category of commutative rings) is their tensor product over $ \mathbb{Z}$.
See MacLane - "Categories for Working Mathematicians" (page 63).

In the problem, as -oo- mentioned, is meant "direct product".


What's wrong with expressing the three bases in $ \mathbb{F}^{2}_{2}$

as $ (1,1),(0,1),(1,0)$?

That's a dependent set of course. Each of these three vectors can be written as the sum of the other two.

so then for sure $ (1,0),(0,1)$ form the basis. But then what's the third one, I feel like there aren't anymore possibilities except for $ (0,0)$, but anything added with that is just itself, so I don't think that can be part of the basis.

A basis of a two-dimensional space is a two-element set. You have one basis, not two.

A basis is a maximal independent set. Any set containing the zero vector is dependent, so the zero vector cannot be a member of any basis.

$ \{(1,0),(0,1)\}$ is a basis for $ \mathbb{F}_2^2.$

$ \{(1,0),(1,1)\}$ is a basis for $ \mathbb{F}_2^2.$

$ \{(0,1),(1,1)\}$ is a basis for $ \mathbb{F}_2^2.$

You misunderstood the question. A basis isn't a vector, it's a set of vectors. You weren't asked to find three vectors, you were asked to find three sets of vectors.

Also, in your last post, you used the words "the basis." You should never use the definite article "the" in this context, since bases are extremely far from unique.

OOH. ok, thank you very much, that explains A LOT.


What does $ [\cdot]$ signify? If it's integer part, then it doesn't hold ($ n \equal{} 19$ is a counterexample), but it may be "closest integer"? Please clarify.

(I'm assuming that's a cubic root on the RHS. Is that right?)

EDIT: Even if it's "closest integer", it doesn't hold. A counterexample is $ n\equal{}3$.


calculus
integration
derivative
LaTeX
calculus computations

Hello everyone,

Let me use [ as the integral sign

if I take the [ x^2 dx it equals 3x^3, I've seen people use the dx as the same dx as when I take the derivative of 3x^3. For example

dy/dx = 3x^3
dx/dy = 1/3x^3
so dx = 1/3x^3 dy

and I've seen people use this dx as the same dx in the integral sign for [ x^2 dx, why does this work?

Thanks for help!!!

Why don't you use $\int$ as the integral sign?  (The link on the sidebar labelled "LaTeX" will teach you all about it.)

In french we say "abus de notation"
chris have a look a course about integrate differential form of degre one
see Spivak book .


For a positive integer $n$ let $S(n)$ be the sum of digits in the decimal representation of $n$.   Any positive integer obtained by removing several (at least one) digits from the right-hand end of the decimal representation of $n$ is called a [i]stump[/i] of $n$. Let $T(n)$ be the sum of all stumps of $n$. Prove that $n=S(n)+9T(n)$.

induction on the number of digits. base case is easy. assume the relation is true for numbers of $d$ digits and let $n$ be a number of $d+1$ digits, $\overline{z_d...z_2z_1z_0}$, and let $t_i$ denote the stump obtained by deleting the last $i$ digits. then by induction

$n = 10t_1+z_0 = 10(9T(t_1)+S(t_1))+z_0 = 90T(t_1)+10S(t_1)+z_0 = 90(T(n) - t_1)+9S(t_1)+S(n) = 9[10(T(n)-t_1)+S(t_1)]+S(n) = 9T(n)+S(n)$.

this last equality results from

$10(T(n)-t_1) + S(t_1) =z_d0+z_dz_{d-1}0+...+z_d...z_20+z_d+z_{d-1}+...+z_1 = z_d+z_dz_{d-1}+...+z_d...z_2z_1$

The problem is straightforward without induction.  Let $n = d_k d_{k-1} ... d_2 d_1 d_0 = \sum_{i=0}^{k} 10^i d_i$, where $d_i$ are the digits of $n$.

Now, $S(n) = \sum_{i=0}^{k} d_i$.  As to the value of $T(n)$, note that the $n-m$-digit stump - which we denote $U_m (n)$ - has the form $U_m (n) = \sum_{i=m}^{k} 10^{i-m} d_i$, and $T(n) = \sum_{i=1} U_i (n)$, so

$S(n) + 9 T(n) = \sum_{i=0}^{k} \left( 1 + 9( \sum_{j=0}^{k-1} 10^j ) \right) d_i = \sum_{i=0}^{k} 10^i d_i = n$,

as desired.

yeah, that became evident in the process of writing up the induction...but then i left the proof as it was anyway.

Nevertheless an easy problem ,just write and sum.:)

We may simply induct from $n$ to $n+1$. Let $\nu_{10} (n+1) = k$.

If $k = 0$, then $S(n+1) = S(n) + 1$ while $T(n) = T(n+1)$, so we are okay.

Otherwise, $S(n)$ decreases by $9k-1$, while $T(n)$ increases by $k$ for each stump for which the value increases by 1. Then$$S(n+1) + 9 T(n+1) = S(n) + 9T(n) -9k+1 + 9k = n+1,$$as needed.

[hide=Notation]By $m=\overline{x_1x_2\dots x_k}$, we denote the base $10$ representation of a number $m$ such that $x_1,x_2,\ldots,x_k$ are the digits of $m$.[/hide]
Let $n=\overline{a_1a_2\dots a_n}=a_1\cdot 10^{n-1}+a_2\cdot 10^{n-2}+\dots+a_n$. Then $ S(n)=a_1+a_2+\dots +a_n $.
The stumps of $n$ are $\overline{a_1a_2\dots a_{n-1}}$, $\overline{a_1a_2\dots a_{n-2}}$, ... , $\overline{a_1a_2}$, $\overline{a_1}$. So,
\[T(n)=\left(a_1\cdot 10^{n-2}+\dots+a_{n-1}\right)+\left(a_1\cdot10^{n-3}+\dots+a_{n-2}\right)+\dots+a_1\]
\[\implies T(n)=\underbrace{111\ldots 1}_{(n-1)\text{ times}}\cdot a_1+\underbrace{111\ldots 1}_{(n-2)\text{ times}}\cdot a_2+\dots+ a_{n-1}\]
\[\implies 9T(n)=\underbrace{999\ldots 9}_{(n-1)\text{ times}}\cdot a_1+\underbrace{999\ldots 9}_{(n-2)\text{ times}}\cdot a_2+\dots+ 9a_{n-1}\]
\[\implies S(n)+9T(n)=a_1\cdot 10^{n-1}+a_2\cdot 10^{n-2}+\dots+10a_{n-1}+a_n=n\]
And we are done...


I have just edited, how about this time?

Edit the second equation from the top as well, otherwise now it's just fine :)

There is a nice way to do this using linear algebra...

[quote="kunny"]I know the solution using matrix, but  I'm lazy, that solution leaves you. :lol:[/quote]

Well i am lazy too...  :D

let $<1>: a_{n+1}=2a_{n}+2b_{n}$ and $<2>: b_{n+1}=3a_{n}+b_{n}$

denote $E<1>$ as substituting $n\rightarrow n+1$,

(remember addition in this way keeps the right and left sides the way they are)

then

$\frac{1}{2}\cdot E<1>-\frac{1}{2}\cdot <1>+<2>$ eliminates $b_{n}$

at which point we use characteristic equations to solve for $b_{n}$

LaTeX note: you can use \langle and \rangle for $\langle$ and $\rangle$ (looks better than "less than" and "greater than")

Let $A=\left(\begin{array}{cc}2 & 2 \\ 3 & 1 \end{array}\right),$ we have $A\left(\begin{array}{c}2 &-3 \end{array}\right)=-\left(\begin{array}{c}2 &-3 \end{array}\right),\ A\left(\begin{array}{c}1 & 1 \end{array}\right)=4\left(\begin{array}{c}1 & 1 \end{array}\right)$

$\Longrightarrow A^{n}\left(\begin{array}{c}2 &-3 \end{array}\right)=(-1)^{n}\left(\begin{array}{c}2 &-3 \end{array}\right),\ A^{n}\left(\begin{array}{c}1 & 1 \end{array}\right)=4^{n}\left(\begin{array}{c}1 & 1 \end{array}\right).$

Rewrite the given system of recursion, $\left(\begin{array}{c}a_{n+1}& b_{n+1}\end{array}\right)=\left(\begin{array}{cc}2 & 2 \\ 3 & 1 \end{array}\right)\left(\begin{array}{c}a_{n}& b_{n}\end{array}\right)$ and  $\left(\begin{array}{c}2 & 3 \end{array}\right)=-\frac{1}{5}\left(\begin{array}{c}2 &-3 \end{array}\right)+\frac{12}{5}\left(\begin{array}{c}1 & 1 \end{array}\right),$

by the linearity of linear transformation we have

$\left(\begin{array}{c}a_{n}& b_{n}\end{array}\right)=A^{n-1}\left(\begin{array}{c}2 & 3 \end{array}\right)=-\frac{1}{5}A^{n-1}\left(\begin{array}{c}2 &-3 \end{array}\right)+\frac{12}{5}A^{n-1}\left(\begin{array}{c}1 & 1 \end{array}\right)$

$=-\frac{1}{5}(-1)^{n-1}\left(\begin{array}{c}2 &-3 \end{array}\right)+\frac{12}{5}\cdot4^{n-1}\left(\begin{array}{c}1 & 1 \end{array}\right)$

$=\frac{1}{5}\left(\begin{array}{c}3\cdot 4^{n}+2(-1)^{n}& 3\cdot 4^{n}-3(-1)^{n}\end{array}\right).$


Find the sum of the first n terms of a sequence whose nth term is
$ \frac{3}{(n)(n\plus{}1)(n\plus{}2)}$

Dah thats the point!how did it arrive to that is it by using partial fractions?

so practically everything cancels except(assuming 1 is the first term) 1/2(1/1(2)-1/(n+1)(n+2)) which is 1/4-1/2(n+1)(n+2)


If $(a,b)=(a,c)=1$ for some integers $a,b,c$ (distinct than $0$) and for $\forall m \in \mathbb{N}$ we have $a^{m}|b^{m}-c^{m}$ prove that $a=\pm 1$.

chose m=p wich divide a and you get easly the contradiction with fermat little :lol:  theorem

Simply $m=p|a$ doesn't work: $a=2$, $b=9$, $c=1$.

I think we can consider that $a$ divides $b^{m}-c^{m}$ a finite number of times, so it must either divide $b^{m}+c^{m}$, or $b^{2m}+c^{2m}$, or...
So $a|b^{k}+c^{k}$ for some $k$, but $a^{k}|b^{k}-c^{k}\implies 2c^{k}\equiv 0\mod{a}$.

Er...right, $a$ can be $2$, so we consider that case...then $b^{m}\equiv c^{m}\mod{2^{m}}$ for all $m$.  Right I don't know what to do here (or rather too lazy to finish it)

EDIT: I'm stupid.  We can in fact say $a^{i}|b^{k}+c^{k}$ for some $m>i>1$ (if we consider $k=m, 2m, 4m, ...$, then if $a$ divided each only once, we'd have a problem).  Hence we get $2c^{k}\equiv 0\mod{a^{i}}$, and there we have the contradiction.

let $a^{m}|a^{m}-b^{m}$ for$\forall m\in\mathbb{N}$ we have $(a,c)=1$ then$\exists c^{-1}$ then we have $(bc^{-1})^{m}\equiv1\mod(a^{m})$ we choose $m=p|a$ then$(bc^{-1})^{p}\equiv1\mod{p}$ wich absurd exipt $a=2$ wich we can use ordre lemma to find that $m|2^{m}$ then $a=\pm{1}$. :maybe:


Show that a square with side length 1 cannot be covered by five squares with side length less than 1/2.

Source: 2001 Iberoamerican Olympiad.

A beginning:

[hide]Consider the following 5 points: the center of the square and its four corners.  Assume that our goal can in fact be accomplished.  Note that the distance between any pair of the 5 points is greater than the largest distance between any two points in a square of side less than 1/2.  Thus, each point must be covered by a unique square.  Now, I believe the argument can follow the following form: first, note that the square at the center must cross the side of the big square, since each little square can take up at most 2s of the perimeter so together the other 4 can take up at most 8s < 4 perimeter.  When we assume such a placement for the center square, we should be able to show that there are points other places (either on the edge or near the center) that can't get covered.  Actually, I'm not so confident that this can be made rigorous, but it seems possible. [/hide]

Ravi B wrote:That's an interesting approach.  Here is my attempt at an approach:

[hide]Draw line segments to divide the square into four equal subsquares.  Together with the boundary of the original square, that gives a total perimeter of 6.  If we can show that any little square intersects these lines in length less than 1, then that would show that at least 6 (in fact at least 7) little squares would be needed to cover the entire square.  [/hide]

I'm terrible at geometry so I don't if this approach makes any sense.

Don't you mean a total perimeter of 8?

No, I meant 6, but perhaps I wasn't very clear.  I am only counting each line segment once, even if two subsquares share that line segment.

But now that I think of it, I don't this approach works.  We can put a little square at the center of the original square, rotate it by 45 degrees, and we cover something like :rt2: units of perimeter.  So...never mind.

So, I will amend my idea to just draw one of those midpoint line segments, not both.  The total perimeter is now 5.  I don't know if this amended idea will last longer than my last idea...

You can still get one rectangle to cover too much length by placing its center at the midpoint of one side, rotated 45 degrees, so that it covers 3 pieces of length [tex]\displaystyle \frac{1}{2\sqrt{2}}[/tex], which adds up to just a tad bigger than 1.  The only way to make either of our ideas work is to do case analysis: "if it does intersect x amount of edge length, then we lose.  If it doesn't intersect x amount of edge length, then we still lose."

Actually, what if we combine your idea with my idea?  Having a square covering the center might automatically limit the amount of edge that it can cover; furthermore, a square placed where I put it can't cover the corners.  I think that there's a way to put these two ideas together, but I'm not quite sure yet.

Maybe you can have the problem be equivalent to making four squares of side length < 1/2 cover all the area outside of and on the perimeter of a circle of diameter [tex]\frac{1}{\sqrt{2}}[/tex].  This should be relatively easy to prove impossible if the circle is completely within the square, but I'm not sure what to do if it's not... perhaps then one of the squares can't cover both part of the circle's perimeter and one of the square's corners.  Then 3 squares are left...

We are attempting to cover a 1x1 square with some quantity of squares of side length [tex]s=\frac{1}{2}-\epsilon[/tex], where [tex]\epsilon[/tex] is arbitrarily small.  Obviously, since the total area to be covered is 1 and the area of each square is less than 1/4, we know it will take at least 5 small squares to cover the 1x1 square.  Our question is whether or not 5 is sufficient.

My proof is subject to the proof of a (hopefully simpler) Lemma.

Figure 1 shows 12 line segments, each of length [tex]\frac{1}{2}-\epsilon[/tex].  These segments are arranged in and on the perimeter of a square of side length 1.  Each red segment touches one corner of the 1x1 square and lies on a side of the 1x1 square.  Each black segment touches the midpoint of one side of the 1x1 square and extends perpendicular to that side.  Note that each set of parallel line segments is spaced exactly [tex]\frac{1}{2}[/tex] units from the adjacent parallel line segments.

A necessary condition for covering the 1x1 square is that we must completely cover all 12 line segments.  Another necessary condition for covering the 1x1 square is that we must cover each corner of the large square.

Let's consider one corner of the larger square, along with the small square that covers this corner.  If the sides of the small square are parallel to the sides of the large square, it is evident that the small square can completely cover two of the red line segments, but none of the black segments.

Now let's consider the case where the sides of the small square are not parallel to the sides of the large square.  (See Figure 2.)

Let the point P of the small square coincide with the corner of the large square.  Clearly, no portion of the large square extends outside corner P.  Suppose we find a minimal covering of the large square, such that P does not lie on the border of the small square.  We can shift the square, parallel to one of its sides until P touches the side of the small square.  We will still have a covering of the large square.  Therefore, we can assume, without loss of generality that point P will lie on the small square.

Now, we ask our unproven Lemma... Within a square of side-length 1/2, what is the maximum total length of a 1/2 unit square grid of line segments, oriented obliquely to the sides of our square, where at least one lattice point of the oblique grid lies on a side of the 1/2 unit square.

I postulate the answer is 1, which would allow the small square to cover a maximum of 2 line segments.  [When someone proves it, we'll have the Lemma that completes this proof.  I think if we can prove that [tex]PV+YZ<\frac{1}{2}[/tex], we'll have our proof.]

Since the maximum distance across the diagonal of the small square is less than the side length of the large square, we must have a distinct small square covering each corner of the 1x1 square.

We have used 4 squares and we have covered a maximum of 8 of the line segments.

Now we are left with 4 line segments that must be covered by other small squares.  Since the remaining small squares are not constrained to cover a corner of the 1x1 square, they may cover the entire length of WY plus XZ.  (See figure 3.)  The largest possible values of WY and XZ occur when both of these line segments are a diagonal of the smaller square.  Thus the maximum segment length that can be covered is [tex]2\sqrt{2}s[/tex].

Since [tex]\displaystyle\frac{4s}{2\sqrt{2}s}\approx 1.414[/tex], we know it will take at least 2 more small squares to cover the entire large square.

Thus I conclude it takes at least 6 squares of area < 1/4 to completely cover a square of area 1.

Hmm, rcv, that's an interesting approach.  I am going to have to think about it.

Joel, thanks for pointing out a counterexample to my suggestion.  I have a second amendment to my proposal. :)  I'll go back to drawing both midpoint line segments, but this time I will give them weights of 1/2.  So, including the outer square, the total weight is 5.  If we can show that any little square has weight less than 1, then we would be done.  The only reason I like this kind of approach is that we get to reason about one little square, instead of an arrangement of 5 squares.

We are given a "large" square with grid lines painted at unit intervals.  The origin of the grid lines is at corner, C, of the square, and the grid lines are parallel to the sides of the square.

We are also given a unit square.

The question is:  What is the maximum length of grid lines that can be covered by the unit square, given the large square is oriented obliquely to the unit square, and given a corner of the large square lies on an edge of the unit square.

Given a fixed obliquity, we consider how the grid lines on the large square interact with the small square as one edge of the unit square slides along the corner, C.

Each column of Figure 1 represents a fixed obliquity, as the unit square slides across the big square.

Column 1 shows a 45-degree obliquity.  Columns 2 through 5 show 35-, 25-, 15-, and 5-degree obliquity, respectively.  Note that we only consider obliquity angles between 0 and 45.  Obliquity angles between 45 and 90 are mirror images of their complementary angle counterparts.

As the red unit square slides from right to left across the large square, the grid lines of the large square interact with the small square, as described below.

Between each pair of adjacent rows, the total covered grid-line length either increases or decreases linearly with the offset (or remains constant).

The only variable is the offset, [tex]x[/tex], from the corner.  As grid lines at the obliquity angle, [tex]\theta[/tex], enter (or leave) the coverage of the red unit square, we gain (or lose) total grid-line coverage length at a rate equal to [tex]\displaystyle\Delta l=\pm\frac{\Delta x}{\cos(\theta)}[/tex].  And as grid lines perpendicular to the obliquity angle enter (or leave) the coverage of the red unit square, we gain (or lose) total grid-line coverage length at a rate equal to [tex]\displaystyle\Delta l=\pm\frac{\Delta x}{\sin(\theta)}[/tex].  Also note that since we are only dealing with [tex]\displaystyle0 < \theta <= 45[/tex], [tex]\sin(\theta)<=\cos(\theta)[/tex], so [tex]\displaystyle\frac{1}{\cos(\theta)}<=\frac{1}{\sin(\theta)}[/tex].

The top row shows one extreme, where one corner of the small square is in contact with the corner of the large square.

As the red unit square slides to the left from row 1 to row 2, we are losing coverage of a grid line at the obliquity angle and we are gaining coverage of a grid line perpendicular to the obliquity angle.  [tex]\Delta l=(\Delta x)(-\frac{1}{\cos(\theta)}+\frac{1}{\sin(\theta)})[/tex].  The total covered grid-line length increases linearly for all positive angles less than 45 degrees, and remains constant for a 45-degree obliquity.  At the position shown on row 2, the intersection of a pair of grid lines is about to pass out the right edge of the unit square.

As the red unit square continues its slide to the left from row 2 to row 3, the total covered grid-line length decreases linearly.  We are simultaneously gaining and losing coverage of a pair of lines perpendicular to the obliquity angle, and we are losing coverage of one line at the obliquity angle.  [tex]\Delta l=(\Delta x)(-\frac{1}{\cos(\theta)})[/tex].  At the position shown on row 3, a grid line, not previously covered is about to enter the upper left corner of the red unit square.

Between rows 3 and 4, the amount of grid-line length entering coverage exactly equals the amount of grid-line length leaving coverage.  The total covered grid-line length remains constant.  At the position shown on row 4, a grid line is about to completely leave the coverage area at the upper right corner of the red unit square.

Between rows 4 and 5, the total covered grid-line length increases linearly.  We are gaining coverage of one line perpendicular to the obliquity angle, and we are gaining and losing coverage of a pair of lines at the obliquity angle.  [tex]\Delta l=(\Delta x)(+\frac{1}{\sin(\theta)})[/tex].  At the position shown on row 5, the intersection of a pair of grid lines is about to enter the left edge of the unit square.

As we continue our sojourn from row 5 to row 6, we are losing coverage of a grid line at the obliquity angle and we are gaining coverage of a grid line perpendicular to the obliquity angle.  [tex]\Delta l=(\Delta x)(-\frac{1}{\cos(\theta)}+\frac{1}{\sin(\theta)})[/tex].  The total covered grid-line length increases linearly for all positive angles less than 45 degrees, and remains constant for a 45-degree obliquity.  At the position shown on row 6, one side of the large square is located at the upper left corner of the red unit square.

Once we pass row 6, no new grid lines are entering our red square, but we are losing coverage of a grid line at the obliquity angle.  [tex]\Delta l=(\Delta x)(-\frac{1}{\cos(\theta)})[/tex].  The total covered area decreases linearly between rows 6 and 7.  At the position shown on row 7, we have completed the slide of one edge of the unit square across the corner of the large square.

The key positions shown in our diagram occur for all obliquity angles between 0 and 45 degrees.  Between successive positions the total covered grid-line length changed linearly.  Thus, if we can evaluate the covered grid-line length at every key position, for arbitrary angles from 0 to 45 degrees, we can determine, with certainty, the maximum covered grid-line length.

In fact, the positions represented at rows 2 and 6 are the only local maxima.

At row 2, we can determine a formula for the total covered grid-line length.  It is [tex]\displaystyle l=1+\frac{1-\cos{\theta}}{\sin{\theta}}+\frac{1-\sin{\theta}}{\cos{\theta}}[/tex].

The above expression has a minimum.  At [tex]\theta=45\degrees, l=2\sqrt{2}-1\approx 1.828[/tex].  And [tex]\displaystyle\lim_{\theta\to 0^+}l=2-\theta[/tex].  (Theta in radians.)

At row 6, we can also determine a formula for the total covered grid-line length.  It is a lot messier:  [tex]\displaystyle l=\frac{1}{\cos\theta}+\tan(\frac{\theta}{2})+\frac{1-\tan{\theta}}{\cos\theta}[/tex].  This is monotonically decreasing over (0, 45 degrees].  [tex]\displaystyle\lim_{\theta\to 0^+}l=2-\frac{\theta}{2}[/tex].  (Theta in radians.)

Lemma proved.  Theorem proved.  It will take at least 6 squares of area < 1/4 to cover a square of area 1.

Impressive work, rcv!  Your proof looks good to me.  I wish there were a simpler way, but I don't know of any.

Near the end, I was a little confused by your statement [tex]\lim_{\theta \rightarrow 0^+} l = 2 - \theta[/tex].  Since we are taking the limit over :theta:, there shouldn't be a :theta: on the right side.  But perhaps you mean that l is approximately 2 - :theta: as a first-order approximation.  My calculation gives 2 - :theta:/2, but it doesn't matter.

[quote="Ravi B"]Near the end, I was a little confused by your statement [tex]\lim_{\theta \rightarrow 0^+} l = 2 - \theta[/tex].  Since we are taking the limit over :theta:, there shouldn't be a :theta: on the right side.  But perhaps you mean that l is approximately 2 - :theta: as a first-order approximation.  My calculation gives 2 - :theta:/2, but it doesn't matter.[/quote]

Thanks, Ravi.  I was trying to emphasize that the length remains less than 2.  I suppose I could have stated [tex]\lim_{\theta \rightarrow 0^+} l=2^-[/tex].

And you are correct.  Both limits should have been given as 2 - :theta:/2.  My bad.

I'm satisfied with my proof of the main theorem, but I know the proof of my lemma is horrid!  [I began thinking about 4-color maps as I was working on it.]

Now that we know it's true, perhaps one of the younger generation can simplify this proof or provide a nicer, more elegant proof.  You *know* nobody who took the 2001 Iberoamerican Olympiad would have given anything like my proof!

This is a very hard problem in my opinion. :) I'll post mine soon.


hello, see here
http://www.wolframalpha.com/input/?_=1328435999548&i=sum+(1%2b(-3)%5en)%2f(4%5e(2%5e(n-1)))+for+n%3d1+to+infinity&fp=1&incTime=true
Sonnhard.


trigonometry
inequalities
limit
inequalities proposed

If $ z^n\equal{}1, n \in \mathbb{N}, n \ge 2, z \in \mathbb{C}\minus{}\{ 1 \}$ then prove that:

\[ \frac{2}{n\minus{}1}\le |z\minus{}1|\]

We can write $ z\equal{}cis(\frac{i}{n}2\pi)$ for some $ 1\leq i<n$. Then $ |z\minus{}1|^2\equal{}\left(\cos\left(\frac{2i\pi}{n})\minus{}1\right)\right)^2\plus{}\sin^2\left(\frac{2i\pi}{n}\right)\equal{}2\minus{}2\cos\left(\frac{2i\pi}{n}\right)$, so we have to prove $ \frac{4}{(n\minus{}1)^2}\leq2\minus{}2\cos\left(\frac{2i\pi}{n}\right)$. But $ \cos\left(\frac{2i\pi}{n}\right)\leq\cos\left(\frac{2\pi}{n}\right)$ with equality occuring when $ i\equal{}1$ or $ n\minus{}1$, so we prove $ \frac{2}{(n\minus{}1)^2}\leq1\minus{}\cos\left(\frac{2\pi}{n}\right)$. That is as far as I have gotten.

[quote="moldovan"]If $ z^n \equal{} 1, n \in \mathbb{N}, n \ge 2, z \in \mathbb{C} \minus{} \{ 1 \}$ then prove that:
\[ \frac {2}{n \minus{} 1}\le |z \minus{} 1|\]
[/quote]

It's been a long time I haven't seen such a beautiful inequality, especially with complex numbers. I wonder if it is yours.
Well, here is what I tried.

In the complex plane, let P be the image of 1.
To the circle with center O(0) and radius OP = 1 is inscribed a regular n-gon in which P(1) is a summit. It's known that each summit of this regular n-gon is the image of a root of the equation $ z^n \equal{} 1$.
Let M(z) be one of the two nearest summits to P(1) and notice that each side is equal to $ |z \minus{} 1|$. It's clear that it suffices to prove the inequality for M.
Consider the point P'(-1).
If n is even, then it's easy to prove that the sum of the sides belonging to one of the two semi-circles bounded by P and P' is superior to the diameter, i.e. $ \frac {n}{2}|z \minus{} 1| \geq 2$. So, since $ n \minus{} 1 \geq \frac {n}{2}$, $ (n \minus{} 1)|z \minus{} 1| \geq 2$.
If n is odd, I stand one can prove similarly that $ (n \minus{} 1)|z \minus{} 1| \geq \frac {n \plus{} 1}{2}|z \minus{} 1| \geq 2$ but I haven't proved it yet.

[quote="moldovan"]If $ z^n \equal{} 1, n \in \mathbb{N}, n \ge 2, z \in \mathbb{C} \minus{} \{ 1 \}$ then prove that:
\[ \frac {2}{n \minus{} 1}\le |z \minus{} 1|\]
[/quote]

Since $ z^{n\minus{}1}\plus{}z^{n\minus{}2}\plus{}...\plus{}z\plus{}1\equal{}0$ it follows that:

\[ (z\minus{}1)(z^{n\minus{}2}\plus{}2z^{n\minus{}3}\plus{}3z^{n\minus{}4}\plus{}...\plus{}n\minus{}1)\equal{}\minus{}n\]

So we can deduce that:

\[ n\equal{}|z\minus{}1|\cdot |z^{n\minus{}2}\plus{}2z^{n\minus{}3}\plus{}3z^{n\minus{}4}\plus{}...\plus{}n\minus{}1|\le\] \[ \le |1\minus{}z|(|z|^{n\minus{}2}\plus{}2|z|^{n\minus{}3}\plus{}...\plus{}n\minus{}1)\equal{}|1\minus{}z|\cdot \frac{n(n\minus{}1)}{2}\]

As for the source it's an old romanian problem.

Hello, I've come back with another interesting proof.

$ z^n \equal{} 1$ <==> $ z^n \equal{} e^{2i\pi}$ <==> $ z \equal{} e^{\frac {2i\pi}{n}}$.
Thus, $ |z \minus{} 1| \equal{} |e^{\frac {2i\pi}{n}} \minus{} 1|$
<==> $ |z \minus{} 1| \equal{} |e^{\frac {i\pi}{n}}(e^{\frac {i\pi}{n}} \minus{} e^{\frac { \minus{} i\pi}{n}})|$
<==> $ |z \minus{} 1| \equal{} |2e^{\frac {i\pi}{n}}sin(\frac {\pi}{n})i|$
<==> $ |z \minus{} 1| \equal{} 2sin(\frac {\pi}{n})$
<==> $ (n \minus{} 1)|z \minus{} 1| \minus{} 2 \equal{} 2((n \minus{} 1)sin(\frac {\pi}{n}) \minus{} 1)$.

Now, let $ (u_n)_{n \geq 2}$ be a sequence defined as $ u_n \equal{} (n \minus{} 1)sin(\frac {\pi}{n})$.
Then it's easy to show, thanks to the usual limit $ \lim_{x \to 0}\frac {sin(x)}{x} \equal{} 1$, that $ \lim_{n \to \plus{} \infty}\frac {sin(\frac {\pi}{n})}{sin(\frac {\pi}{n \plus{} 1})} \equal{} 1$.
But, since $ \frac {n}{n \minus{} 1} > 1$, so $ \frac {n}{n \minus{} 1} > \frac {sin(\frac {\pi}{n})}{sin(\frac {\pi}{n \plus{} 1})}$, leading to $ u_{n \plus{} 1} > u_n$.
Hence, since $ u_2 \equal{} 1$, $ (\forall n \geq 2) ; (n \minus{} 1)sin(\frac {\pi}{n}) \geq 1$.
Conclusion : $ (\forall n \geq 2) ; (n \minus{} 1)|z \minus{} 1| \minus{} 2 \geq 0$.


calculus AMC USA(J)MO USAMO Stanford college Harvard

What are my chances for getting into an ivy league in 2 years?

I am a Sophomore, and I am taking 3 AP classes:

BC Calculus (99)
AP Physics (95)
AP European History (88)

I am really studying hard for History to push it up to a 5 standard

On my PSAT I made

79 on maths
77 on reading
77 on writing

thats corresponds to about a 2330 on the SAT

Also, if I counted it correctly, my index this year for AMC/AIME will be 211.5, which should (I think) qualify me for USAMO.

I participate in several extra curricular academic activities, as well as athletic activities.

I play piano and have a Grade 8 qualification (1 level below diploma)

Next year I plan on taking 5 AP courses

Thank you!

Threads like this are better suited to [url=http://www.collegeconfidential.com/]College Confidential.[/url]  In any case, I'm sorry to say nothing stands out (with the possible exception of USAMO qualification).  Profiles like yours exist all over the place.

You would do well to learn more about two things:  1)  What selective colleges are really looking for (one word: [b]passion[/b]), and 2)  Whether going to an Ivy League college should be your only reason for living.

You are doing well -- certainly you will have a strong academic record. Probably what will hurt your application most at this point is having "Next year I plan on taking 5 AP courses" as the only statement about the future. Indeed, nothing stands out extraordinarily (although I don't know enough to judge "Grade 8 qualification" in piano... if this is of national merit, it would do it). But this is hardly a setback yet. You still have a lot of time! Do something with the next year and a half; that's plenty of time to start doing something unique or investing more into an existing hobby.

And as mentioned above, ivy league schools aren't everything -- there's also Stanford! :D On a more serious note, where you go to college can make a difference, but what you do in college can make much more of a difference. I don't want to tell you to not worry about college because then I'd be a hypocrite, but there really are more worthwhile things to do (learned from experience).

I know someone in my school whose GPA and SATs are higher than mine but got rejected from somewhere I got accepted, so GPA and SAT's aren't the whole picture. On the other hand, one of my friends does not have the ridiculous grades or SAT scores behind him but he is probably the best violist in the state (or very close to it), and he got in. You mention many extracurriculars, so make sure you show the passion for them and continue with it!
[color=#EFEFEF][size=9]an incidious badly drawn banana -- animal second same the and space the with whale killer the uses hunt puzzle evil this of page manage real the.[/color][/size]

Thanks for the advice guys!
since the word [b]passion[/b] keeps popping up I will seriously consider what I really want and pursue it to the best of my ability.

you are probably ahead of where I was in 10th grade, so you're doing fine (unless you're like, extremely bent on Harvard or Princeton or something... they weren't quite as kind to me in terms of admissions).
I'd say that taking such AP courses and doing well in them is extremely good, as is USAMO qualification. If you think about it, in your grade, there are probably less than 200 USAMO qualifiers in the whole nation. So theoretically, MIT can suck them all up and still have like.... 4/5th of a class to fill...

As I've written in other threads, make sure to do what you love, and to try to not be a stat stuffer.

It matters a lot where you come from (i.e. how many other people with a similar background to yours are applying). Hence, class rank is very important in a sense that you want to be the best at what you do in the local area where you live. Grades could be very meaningless as well as very meaningful. If there are 30 people in your school with 95 averages, that's tough. If there are 3-4, you're doing great. Main thing, do what you like to do as well as you can.

i thought i got away from college confidential.
  :maybe:
looks like you are doing really well, i wished i had waken up my sophomore year and done more things, i'm currently a junior and this is my first 3 months of competitive math- which i love- and i am just getting into doing research so i feel way behind (except maybe the research bit) but keep it up. The trite but true statement of do what you love (passion) is something you should seriously consider. Get more involved in things you enjoy, those extra curriculars aren't much, i.e volunteer, math/science organizations, debate -whatever catches your fancy.
Good luck in your endeavors and your new addiction, college admissions.

wow lol, we are like twins or something... i have like the exact same stats (except for piano and little lower SAT score)
hmm but i used to do everything for college, but now im getting away from that...i think you should do things based on what you like and youll end up doing even better than what you expected.

ugh but if you post like "what are my chances" on CC and you say your a 10th grader, people are gonig to respond with like "get a life" or something and stop worrying...


symmetry
geometry
parallelogram
calculus
trapezoid
geometry proposed

The line $ l$ parallel to the median $ AA_{1}$ of the triangle $ \triangle ABC$ intersects the sides $ BC$, $ CA$, and $ AB$ in points $ H$, $ N$, and $ D$ respectively. Show that the symmetrical points of $ H$ in respect to the midpoints of $ NC$ and $ DB$ are symmetrical of each other in respect to vertex $ A$.

P=Symmetry of H respect to AB & k=Symmetry of H respect to AC.
Y=(PH,BD),X=(KH,AC)
We Prove AC is parallel to YH.
CH/BH = BA1+HA1/BH=2HA1/BH   +1
DH is Parallel to AA1 So 2HA1/BH=2AD/BD=2AD/2BY=AD/BY
So: 1+  2HA1/BH=AD/BY+1=AD+BY/BY=AD+DY/BY=AY/BY
So: AY/BY=CH/BH SO: PH is parallel to AX
Now KX/XH=AK/AP=1 so AK=AP

E=XB∩CN     F=BA∩CY
XBHD , HCYN , EFCB are parallelograms , respectively
DH/CF=BH/BC       HC/BC=HN/BE
DH/CF+ HN/BE= DH/CF+ HC/BC=1
DH+NH=BE=CF
XB=YF   XB∥YF    =>   XBYF is a parallelogram  X,A,Y are collinear and XA=AY

Dear Mathlinkers,
I think that a synthetic solution without calculus or transformation is possible.
Any ideas?
Sincerely
Jean-Louis

Dear Mathlinkers,

1. Let B", C" the intersection of NC' and BB', of DB' and CC'.
2. By Pappus's theorem, B"C", CN and BD are concurrent at A.
3. A is the midpoint of B"C".
4. B'B"C'C" is a parallelogram
5. B', A, C' are collinear and A is the midpoint of B'C'.

Sincerely
Jean-Louis

Call P the symmetrical of H w.r.t. middle of BD, and M the symmetrical of H w.r.t. middle of CN. HDPB and HCMN are parallelograms, so PB = DH and CM = NH.
From Thales  HN/AA_1=CH/CA_1 and  HD/AA_1=BH/BA_1; by adding the 2 equalities, with
BA1 = CA1 we get HN + HD = 2 AA1, or CM + BP = 2AA1, i.e. AA1 is midline in the trapezoid MPBC.

Best regards,
sunken rock


given a coin with radius 0.5, find its thickness such that the probability of landing on one of its faces is equal to the probability of landing on the edge of the coin (ie coin standing up)

I dont think this is as simple as: surface area of edge=surface area of face
please consider this in a physics perspective

Are we taking into account the fact that when it lands on it's edge, its inertia will keep it moving?  Or should we just assume that it will stay there pefectly?

on the contest it says that you may make any assumption necessary for a real world situation