Apian’s Problem Again
There is a fascinating observation about that problem. If you tried to solve the problem analytically, you will notice that the formula for the cosine of the angle is of fourth order. The derivative is thus of third order. Nevertheless, it has a trivial solution, namely the intersection of the line with the line of the segment. Thus it is possible to express the two other solutions (on each side of the intersection) with square roots. This confirms to the fact that the solution can be constructed by compass and ruler.
Once we know the geometric solution, we see that we have to find the intersection of a parabola and a line. Here, the parabola is the locus of all points with same distance to one endpoint of the segment and the line. Again, the analytic expression has a solution expressible in square roots.
Apian’s Problem
If I trust my reliable sources, the following problem is due to Phillip Apian: Determine the point on a line g that sees a given line segment s at the maximal angle.
The sketch vaguely indicates a geometric solution. How about an analytic solution, and how about rational geometry?
Apian posed that problem in more poetic form: How far do you have to stand away from the house to see most of your beloved on a balcony? That is a special case with s orthogonal to g.
Rational Trigonometry
I made a page about rational trigonometry, an idea proposed by N.J. Wildberger in his book about divine proportions.
Before you go to the page, try to find the height h of the following triangle, and the length of the angle bisector d, both as algebraic expressions, not approximations.
New page about installing Latex
I created a page with a description of the installation of Latex for Windows.
Math and the Press
Just recently I heard a radio interview with some German professor of math about votes and seats, and the problem to compute the distribution of the seats from the votes. The radio woman correctly identified the problem as the fractions, which occur when you multiply the percentage of votes by the total number of seats. After all, you cannot send 0.2 politician to the parliament. Funny remark, indeed.
The professor started to explain, that these fractions have to be used somehow to divide the remaining seats under the parties, threw in the previously used d’Hondt procedure, and both agreed that it would be too complicated to explain that here and now. The moderator admitted that he would not know exactly anyway.
The professor then continued that we have now a new algorithm which gives each vote the same weight, as he said, as it should be, he said. The surprising solution is to round to the next integer. That, he said, is much more fair than the d’Hondt procedure, which always rounds down. Both did not bother to confront the listener with the problem what to do if the number of seats assigned that way does not happen to be the number needed. But we learned that it is fair, and no longer hurts the small parties more than the bigger ones.
To demonstrate the problem of rounding down small versus large numbers, the professor used a striking example. If we would round down his income to the next million, it would hurt him much more than if we did the same with Bill Gate’s income. I assume the professor makes more than half a million, maybe by radio interviews. For otherwise, a better rounding can only help Bill Gates.
So, if you want to learn how useful math really is, listen to the radio!
The iterated Sine
I am a bit surprised that none of my readers like to solve problems. OK, then here is a proof. I adopt the notation 
It is a simple problem to show that 
Since all 
Now, all we need to show is for all k
But this is the same as 
The same can be done with any other monone function of the form
For the iterated function, we always get
for 
Since you are so fond of my problems, here is another one, I cannot solve yet myself. Numerical experiments show that
exists. Can you prove that? And what is that limit?
A Challenge
Find out, how
behaves. Here, I do not mean the n-th power of sine, but the sine function applied n times. Actually, I do not like to write the n-th power this way, but perfer to write the power after the bracket of the argument.
Three Cylinders, Continued
What is the volume of the intersection in the last blog posting? To compute that we decompose the intersection into 7 parts, three caps, and one cube, using PovRay again. That took me another hour.
The top cap can be integrated along its center, and the cube has side length 
This is a bit more than 4.
Three intersecting Cylinders in Povray
Can you imagine how three cylinders, each aligned along one axis, intersect? It is very difficult. So I took Povray for a the job. Here is the result.
I am absolutely no expert for Povray, more a beginner. It took me two or three hours studying the documentation and introduction pages in the net, before I got my result.
However, maybe I should use this tool more often. It might even be a good idea to add an export for Povray to the Euler Math Toolbox. If you look at my page about 3D geometry, you will find other programs to do 3D stuff. But none would produce the result as nicely as Povray.
Since you probably still cannot imagine the intersection, here is the union of the three cylinders. I hope this helps
Here is the code. It took me the most time to find out how to increase the ambient light. This is done by the assumed_gamma parameter below. I also used the 640×480 rendering with anti-aliazing. For the blog, the image was cropped and saved as PNG by Photoshop.
#include "colors.inc"
#include "stones.inc"
camera
{ location <5, 3, -3>
look_at <0, 0, 0>
angle 30 // opening angle of the lens
}
global_settings
{ ambient_light White
assumed_gamma 1.4
}
background
{ color Gray }
light_source
{ <2, 4, -3>
White
}
#declare cylobject = cylinder
{ <-2,0,0>, <2,0,0>, 1 }
intersection
{ object
{ cylobject
pigment { Red }
}
object
{ cylobject
pigment { Green }
rotate 90*y
}
object
{ cylobject
pigment { Blue }
rotate 90*z
}
}
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