Invisible objects resolution

We consider the following configuration;



Observe that the light from infinity strikes the first mirror at a 30 degree angle (angle of incidence)and that the angle of reflection equals the angle of incidence. This means that the ray is now at a 30 degree angle with respect to horizontal.

The rays passing between the mirrors are parallel and the distance between the rays as they hit the second mirror is the same as the distance between the incidence points when they hit the first mirror. Now assume that a ray hitting the very top of the first mirror will travel a vertical distance of sqrt(3)/4. This is one side of an right triangle with base angle 30 degrees. Since the long side of the triangle is sqrt(3)/2 the other leg of the right triangle is 3/4. This places the reflected ray at the apex of the opposite triangle.

Using this fact we see that corresponding rays also hit the opposite triangle at corresponding points. That is: if a ray hits the first triangle at a distance x from base, then the ray hits the second mirror at a point 1/4 - x from the base of the second mirror.

The reflected rays hit the third mirrored surface at precisely the same horizontal points and these points will reflect to horizontal points on the fourth mirrored surface directly below the original ray position.

Simple Railroad Car problem

Consider the simple railroad car problem

Two trains headed by engines wish to pass each other so that the trains are in the same order after reassembly. There is one siding which will hold one car at a time. The engines may disconnect and go into the siding or they may push a car into the siding but only from the right side of the diagram.

This is similar to a sliding block puzzle except that that certain blocks or cars can only move if attached to the engines.

Give a solution to the problem. Is your solution the best (shortest number of moves) possible where a move is a position where a car has been put into the siding or removed from the siding.

A partial form of a solution is given below.

Task Problem of Laraudogoitia's (See Mathematics Recreation link)

This is a class of problems that have kept philosophers busy for some time.

Paul Benacerraf is right to say that the state of the problem before t=1 (i.e. 0
One of the questions is why is there so much confusion about this. The answer is that mathematicians regularly compute the limits of sequence and series. This is the simplest form of a supertask. Indeed, Eudoxus' solution of Zeno's paradox was to show that there was a limit in the case of the arrow problem.

It is natural to accept Eudoxus' conclusion because we can see the arrow strike the target with our own eyes. On the other hand we can not see Laraudogoitia's problem played out because it deals with point masses that do not exist and can not exist.

In general the problem is that this conundrum is not well posed. The axioms that determine the outcome of the problem are not stated, but they are implied. If we take the problem to be a problem of mathematics, then the answer is that no ball reaches x=0. The real trick is that by introducing time into the problem we introduce a red herring. If we ignore time and set the problem up as a series of balls at x=1,x=2,x=3, and send a ball with speed 1 from x=0 toward the ball at x=1, we will see an infinite series of collisions none of which are outside the real line.

In the case of Laraudogoitia's conundrum, the balls are struck one after the other but the collisions never reach the end point which we could just as well have chosen to be at infinity instead of at x=0.