Quartet of 3rds

Geometry Level 3

Given a uniform sphere partitioned by 2 parallel planes, place the following in ascending order of percentage increase in the total surface areas of the resultant solids compared to the original sphere.

A). Solids of equal volumes.

B). Solids of equal heights.

C). Solids of equal surface areas.

D). Solids of equal volume to surface area ratios.

C,A,D,B D,C,B,A A,D,B,C B,C,D,A B,D,A,C D,B,C,A A,C,B,D B,A,C,D

1 solution

K T
Jul 17, 2019

Suppose our sphere is of radius $R$ is given by $x^2+y^2+z^z=R^2$ We will intersect it by planes perpendicular to the x-axis. By symmetry, both caps must be equal in size and shape (for each of the 4 cases) so that the parallel planes are equidistant from the centre, say $x=\pm c (0 \leq c \leq R)$

The distance $r$ between the sphere surface and the $x$ -axis is given by $r(x)=\sqrt{R^2-x^2}$ , so each of the intersection areas is $π(R^2-c^2)$ . Below I list formulas for the relevant quantities as a function of R and c.

	Spherical area	planar area	total surface area	volume
Cap	$2πR(R-c)$	$π(R^2-c^2)$	$π(3R^2 -2Rc -c^2)$	$π(\frac23R^3-cR^2+\frac13c^3)$
Mid section	$4πcR$	$2π(R^2-c^2)$	$2π(2cR+R^2 - c^2)$	$2πc(R^2-\frac{1}{3}c^2)$
All three parts	$4πR^2$	$4π(R^2-c^2)$	$4π(2R^2 - c^2)$	$\frac43 πR^3$

Now I used numerical analysis to find the value for $c$ in each of the 4 scenarios. I used $R=1$ (but any fixed positive value for $R$ should do, as we are comparing areas to areas and volumes to volumes).

partition	c/R
A. equal volumes	0.22607...
B. equal heights	0.33333...
C. equal surface areas	0.17157...
D. equal volume to surface area ratios	0.25410...

Because total surface area of all three parts is a monotonously decreasing function of $c$ on the interval $0<c<R$ , we should list the scenarios in decreasing order of $c$ . This is $\boxed{ BDAC}$ .

Excellent exposition of the method you used to find the formulas to solve this problem.

W Rose - 1 year, 10 months ago

Quartet of 3rds

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