Monday, October 19, 2015
Unizor - Geometry3D - Cylinders - Not So Easy Problems
Unizor - Creative Minds through Art of Mathematics - Math4Teens
Notes to a video lecture on http://www.unizor.com
Cylinders - Not So Easy Problems
The problems presented here are steps towards the last problem and their combined solutions are supposed to constitute a complete solution to that last problem.
The Last Problem
There is a cylinder inscribed into a cube ABCDA'B'C'D' as follows.
The main axis of a cylinder, connecting centers of its circular bases, lies along a main diagonal of the cube AC'.
One base of a cylinder, a circle, is tangential to three faces of a cube, ABCD, AA'B'B and AA'D'D, that share vertex A, one end of a cube's main diagonal.
An opposite base of a cylinder, a circle, is tangential to three other faces of a cube, C'D'A'B', C'CBB' and C'CDD', that share the opposite end of the main diagonal, vertex C'.
A section of a cylinder along its main axis is a square (that is, a diameter of its base equals to its height).
Find the ratio of a volume of a cylinder to a volume of a cube.
First of all, let's think about the problem in general. There is no information about the size of a cube, which implies that the ratio of a volume of a cylinder to a volume of a cube is independent of a size. Is it true?
All the cubes are similar to each other. All the cylinders with a diameter of a base equal to a height also are similar to each other. So, it looks like changing the size of a cube might proportionally change the size of an inscribed into it cylinder, so the ratio of volumes is constant and the problem does make sense. These are logical considerations rather than proof.
The real solution would be to assign to a cube some dimensions and calculate the dimensions of an inscribed into it cylinder. Then we can calculate the ratio of the volumes and the result should be independent of the initial size of a cube.
Let's assume that the cube has all edges of a length d. Hence, its volume is d³. This is the simple part of the calculations.
With a cylinder it's not as simple. The bases of a cylinder are perpendicular to the main diagonal of a cube. From considerations of symmetry, the plane perpendicular to the main diagonal of a cube should intersect its edges on equal distance from the vertex that is the end of a diagonal. So, if the plane of the base of a cylinder that is closer to vertex A, which we will call δ, intersects edge AB at point P, edge AD at point Q and edge AA' at point R, segments AP, AQ and AR should be of the same length. To prove it is the subject of Problem 1 below.
Assume, this is proven, and continue our analysis.
Next step is to understand that a circular base tangential to three faces of a cube that share vertex A (and lying within a plane δ defined by triangle ΔPQR) is a circle inscribed into this triangle PQR. This follows from the fact that our circular base completely belongs to plane δ, so its common points with any one of three faces of a cube should belong to intersection of that face and plane δ. But this intersection is a line - a side of a triangle ΔPQR, so a circle has only one common point with a side of a triangle, that is tangential to it.
Let's assign a size x to three segments AP, AQ and AR. They completely define plane δ where a base of a cylinder lies. Using this length, we can calculate the radius of a base circle tangential to three faces that share vertex A as a radius of a circle inscribed into triangle ΔPQR - see Problem 2 below.
Another plane, the one where the other base of this cylinder (closer to vertex C') is located, intersects three other edges of a cube, C'B', C'D' and C'C at, correspondingly, points P', Q' and R'.
It should be obvious from the formula connecting the radius of an inscribed into triangle ΔPQR circle to x that C'P'=C'Q'=C'R'=x. This simple conclusion is the consequence of the fact that both circles at the bases of a cylinder are congruent.
The final component is the height of a cylinder. This might be calculated based on the length of a main diagonal of a cube and the height of a pyramid with vertex A and base triangle ΔPQR - see Problem 3 below. This height should be subtracted twice from the length of a diagonal (once for a pyramid with apex A, another - for a pyramid with apex C') to get the height of a cylinder.
Equating this height of a cylinder to a diameter of its base should give us the value of x in terms of the length of a cube's side d (see Problem 4).
With all this calculated, we can determine the volume of a cylinder in terms of d and compare it with a volume of a cube to get the ratio of a volume of a cylinder to a volume of a cube.
The variable d should be reducible from a formula.