Surface area of a sphere ##= \pi * (a^2+b^2+c^2+d^2)##

[Trysse]

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 4-tuple $(a,b,c,d)$, 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 $a^2 + b^2 + c^2 +d^2$.

 

[anuttarasammyak]

If I take it right r the radius of sphere you say is
$$r=a=b=c=d$$ so
$$4\pi r^2=\pi(a^2+b^2+c^2+d^2)$$

 

[Trysse]

View attachment 301992If I take it right r the radius of sphere you say is
$$r=a=b=c=d$$ so
$$4\pi r^2=\pi(a^2+b^2+c^2+d^2)$$

Yes, this is one of the two special cases, where the formula for area holde.

 

[pbuk]

As the surface area of a sphere is $A=4* \pi * r^2$ I wonder, if

$$A=\pi *(a^2+b^2+c^2+d^2)$$

No. As a simple way to visualize this, first set a, b, c and d equal, and note that the three points A, B and C define a unique circle: call this ABC. Now imagine decreasing d a bit: the sphere will shrink and its centre will move away from D. This keeps happening until the centre of the sphere is at the centre of the circle ABC which are now on a great circle of the sphere. Now if we decrease d any more, the sphere can't get any smaller (it can't have a radius smaller than the circle ABC) so it must get larger. If A sometimes gets larger as d gets smaller this contradicts your assertion $A=\pi *(a^2+b^2+c^2+d^2)$.

 

[pbuk]

As for the other special case you say
$$a=2r, b=0,c=0,d=0$$

No, at most one of {a, b, c, d} can be zero otherwise there are no longer 4 separate points, however it is easy to prove that in the limit as e.g. {b, c, d} all tend to zero (or alternatively a tends to infinity) the area tends to $\pi a^2$, so these are the two other special cases where the equation is true.

 

[pbuk]

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 $a^2 + b^2 + c^2 +d^2$.

You have not thought this through and there are many problems with it. I'll pick just one:

Next, I construct a sphere around $C$ with a radius of $CB=CB'$

What makes you think that this sphere will go through the point defined by $a$?

 

[PeroK]

No. As a simple way to visualize this, first set a, b, c and d equal, and note that the three points A, B and C define a unique circle: call this ABC. Now imagine decreasing d a bit: the sphere will shrink and its centre will move away from D. This keeps happening until the centre of the sphere is at the centre of the circle ABC which are now on a great circle of the sphere. Now if we decrease d any more, the sphere can't get any smaller (it can't have a radius smaller than the circle ABC) so it must get larger. If A sometimes gets larger as d gets smaller this contradicts your assertion $A=\pi *(a^2+b^2+c^2+d^2)$.

If $a = b = c$ then we can take the triangle ABC in the x-y plane, with the centroid at the origin and the apex (D) on the z-axis. Whichever point we choose on the z-axis, defines a sphere with A, B, C and D on the surface (*).

If we choose any values for $a, b, c, d$, then I can't see whether the points must lie on a sphere, or whether we have a counterexample.

(*) PS I assume that $d > 0$, which means the point must be beyond the centroid.

PPS If $d <1$, then the radius of the sphere increases as $d$ decreases. So, the OP's formula cannot hold in that case.

 

[PeroK]

Next, imagine you define one arbitrary point on each of the rays. These four arbitrary points always describe a sphere.

Can you prove that?

 

[pbuk]

If we choose any values for $a, b, c, d$, then I can't see whether the points must lie on a sphere, or whether we have a counterexample.

Provided the four points are not coplanar (ie at least one of { a, b, c, d } is non-zero) then they define a sphere. This is a well-known result.

 

[PeroK]

Provided the four points are not coplanar (ie at least one of { a, b, c, d } is non-zero) then they define a sphere. This is a well-known result.

Ah, of course, any four points lie on a sphere!

 

[pbuk]

I meant at most one is zero.

 

[pbuk]

PPS If $d <1$, then the radius of the sphere increases as $d$ decreases. So, the OP's formula cannot hold in that ccase.

No, initially the radius decreases until it reaches the critical point where ABC is a great circle as in #4.

 

[Trysse]

As a simple way to visualize this, first set a, b, c and d equal, and note that the three points A, B and C define a unique circle: call this ABC. Now imagine decreasing d a bit: the sphere will shrink and its centre will move away from D. This keeps happening until the centre of the sphere is at the centre of the circle ABC which are now on a great circle of the sphere.

I do not agree with the statement that the sphere's center will be at the center of the triangle ABC. You forgot about the point $O$ in which the four rays originate. If you decrease $d$ to $0$ I.e. you move one of the points inward as far as possible, the points $O$ and $D$ coincide. However, $O$ is not at the center of the triangle $ABC$.

I have tried to visualize this in GeoGebra. However, this is cumbersome as I cannot construct a sphere from four points in GeoGebra. As a workaround I have constructed four circles instead.
The red ray (top) has the point $A$, the orange (left front) has the point $B$, the green ray (right front) has the point $C$ and the blue ray (in the back) has the point $D$.

In the picture below I have moved the point $A$ all the way down so $AO=0$. I think you can see that the pink circle going through the points $BCD$ is not a great circle.

You can take a look at the original GeoGebra file here https://www.geogebra.org/classic/fdjwwpy7

I found, that the strange angle between the axes and the use of four axes makes this problem very counterintuitive.

I think I have to reformulate my problem...

 

[PeroK]

No, initially the radius decreases until it reaches the critical point where ABC is a great circle as in #4.

To do something useful. If we let $a = b = c = 1$, then I get $$r = \frac{d^2 + \frac 2 3 d +1}{2d + \frac 2 3} = \frac 1 2(d + \frac 1 3) + \frac 4 9(d + \frac 1 3)^{-1}$$And if we set $d = 1$ we get $r = 1$, as expected.

Setting $r'(d) = 0$ gives $(d + \frac 1 3) = \frac{\sqrt{8}}{3}$, hence:
$$r_{min} = \frac{\sqrt{8}}{3}$$And if the distance from the centre of the tetrahedron to a vertex is $1$, then that is indeed the distance from a vertex to the centroid of the associated equilateral triangle. And we have a sphere with ABC on a great circle.

 

[pbuk]

I do not agree with the statement that the sphere's center will be at the center of the triangle ABC. You forgot about the point $O$ in which the four rays originate.

No I didn't.

If you decrease $d$ to $0$ I.e. you move one of the points inward as far as possible, the points $O$ and $D$ coincide. However, $O$ is not at the center of the triangle $ABC$.

I did not suggest that you decreased $d$ to $0$, and I did not suggest that $O$ would be at the center of the triangle $ABC$: it is obvious that O, A, B and C are fixed and O is not in the same plane as triangle ABC so cannot be its centroid. The centre of the sphere is only at $O$ when ${ a, b, c, d }$ are equal.

What I said was if you decrease $d$ until A, B and C lie on a great circle then the centre of the sphere lies at the centroid of triangle ABC.

In the picture below I have moved the point $A$ all the way down so $AO=0$. I think you can see that the pink circle going through the points $BCD$ is not a great circle.

So don't move it down that far, only move it as far as you need to so it is a great circle.

I found, that the strange angle between the axes and the use of four axes makes this problem very counterintuitive.

It might help if you look at it more simply - because the disposition of the 'axes' is symmetrical there must be a position for the fourth point where the other three points (held fixed in an equilateral triangle) lie on a great circle of radius $r$ with the fourth point $r$ from the centroid and at this point the sphere can't get any smaller.

 

[PeroK]

And in general, for $a=b = c$, we have$$r = \frac 1 2(d + \frac a 3) + \frac {4a^2}{ 9}(d + \frac a 3)^{-1}$$

 

[pbuk]

hence:
$$r_{min} = \frac{\sqrt{8}}{3}$$And if the distance from the centre of the tetrahedron to a vertex is $1$, then that is indeed the distance from a vertex to the centroid of the associated equilateral triangle. And we have a sphere with ABC on a great circle.

It's nice when the numbers confirm what must be true by symmetry (or is it the other way round ).

 

[Trysse]

@PeroK, @pbuk,
Thanks a lot. Got it.

 

[WWGD]

Let's now figure out why :

SA*( $$S^2 $$) = $$ 4 \pi r^2$$= $$ (d/dr)( \frac {4}{3} \pi r^3) $$ ;).

*Surface Area ; $$S^2$$ is the 2-Sphere.