 #1
Trysse
 38
 8
 Homework Statement:

This is not a homework assignment. It is a problem with which I came up for myself. However, I think it best belongs here.
Imagine the following figure: You have a point ##O## from which four rays point outward. The rays are arranged, so that the angle between the rays is ##arccos  \frac{1}{3}## or ##\approx 109.47°##. You get this figure for example by taking a regular tetrahedron and drawing rays from the center of the tetrahedron through the apices.
Next, imagine you define one arbitrary point on each of the rays. These four arbitrary points always describe a sphere. I can denote this sphere with the 4tuple ##(a,b,c,d)##, where ##a##, ##b##, ##c##, and ##d## are the distances of the points from the point O along each of the rays.
If I now choose ##a=b=c=d## then ##a##, ##b##, ##c##, and ##d## are the radius of the sphere. On the other side if I define only ##a>0## and ##c=b=d=0## , ##a## is the diameter ##a=d## of the sphere or twice the radius ##a=2r##.
As the surface area of a sphere is ##A=4* \pi * r^2## I wonder, if
$$A=\pi *(a^2+b^2+c^2+d^2)$$
for the above figure. What would a proof look like, that can either prove or disprove the above equation?
 Relevant Equations:
 ##A=4* \pi * r^2##
I am not very good at proofs. The only thing I have come up with is the following regularity. However, I am not sure how this can be related to the above problem.
Given a sphere ##S_a## with a center ##C## and a diameter of ##a##. I can now construct a line segment ##b## with the endpoints ##B## and ##B'## so that ##b## is tangent to the sphere and the center of ##b## touches the sphere. Next, I construct a sphere around ##C## with a radius of ##CB=CB'##. According to the theorem of Pythagoras, the surface area of the second sphere is ##\pi * (a^2+b^2)##. In the same way, I can "add" two more line segments to create a sphere where ##a^2 + b^2 + c^2 +d^2##.
Given a sphere ##S_a## with a center ##C## and a diameter of ##a##. I can now construct a line segment ##b## with the endpoints ##B## and ##B'## so that ##b## is tangent to the sphere and the center of ##b## touches the sphere. Next, I construct a sphere around ##C## with a radius of ##CB=CB'##. According to the theorem of Pythagoras, the surface area of the second sphere is ##\pi * (a^2+b^2)##. In the same way, I can "add" two more line segments to create a sphere where ##a^2 + b^2 + c^2 +d^2##.
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