Monday, November 23, 2015
Unizor - Geometry3D - Spheres - Problems 1
Unizor - Creative Minds through Art of Mathematics - Math4Teens
Notes to a video lecture on http://www.unizor.com
Spheres - Problems 1
Explain that a sphere of radius R divides an entire three-dimensional space into two parts - points outside of a sphere that are at a distance greater than R from its center and points inside a sphere that are at a distance smaller than R from its center.
(not really a rigorous proof)
Choose any point M outside a sphere and connect it with its center O. The fact that point M is outside a sphere means that segment OM intersects a sphere at some point P lying in between points O and M.
That is, OP⊂OM.
Since the length of segment OP equals exactly to R, the length of segment OM must be greater than R.
Choose any point M inside a sphere and connect it with center O. The fact that point M is inside a sphere means that a continuation of segment OM beyond point M intersects a sphere at some point P, so point M is lying in between points O and P.
That is, OM⊂OP.
Since the length of segment OP equals exactly to R, the length of segment OM must be smaller than R.
A plane intersects a sphere, but does not pass through its center.
Prove that a base of a perpendicular from a center of this sphere to this plane is inside the sphere.
Let the base of a perpendicular from the center of our sphere O to a plane be point M.
Choose any point A on an intersection of a sphere and a plane.
Since OM is perpendicular to an entire plane, it's perpendicular to line MA on it.
Therefore, triangle ΔOMA is a right triangle with catheti OM and MA and hypotenuse OA.
Since hypotenuse is longer than cathetus, the length of OA, equaled to R, is greater than the length of OM.
According to a previous problem, point M must lie inside the sphere.
Given a sphere of radius R with center O.
Plane δ intersects this sphere.
Prove that the intersection of this plane with a sphere is a circle of a radius not greater than R and a center being a base of a perpendicular from a sphere's center O onto plane δ.
Determine the radius of this circle, if plane δ is positioned at a distance d (smaller than R) from a center of a sphere O.
Drop a perpendicular from center O onto a plane δ. Its base - point M - is inside a sphere, according to a previous problem. The length of segment OM is d.
Choose any two points A and B on intersection of our sphere with plane δ.
We know that OM⊥δ
⇒ OM⊥AM; OM⊥BM
Consider two right triangles ΔAOM and ΔBOM.
They are congruent by common cathetus OM and equal hypotenuses OA and OB, each being a radius of a sphere. Therefore, two catheti AM and BM are equal.
Since A and B are any two points on the intersection of a sphere and plane δ, all points on this intersection are equidistant from point M - a center of a circle and a base of a perpendicular from a sphere's center O onto plane δ.
The radius AM=r can be calculated by Pythagorean theorem as
r² = R² − d²
r = √R² − d²
Technically, this proof is not applicable, if the plane δ passes through a center of a sphere O because in this case we cannot form triangles used above in the proof. However, this situation does not need any special proof since what we have to prove is contained in the definitions of a sphere and a circle. All points lying on the intersection of our sphere and plane δ are lying in one plane and equidistant from one point on this plane - point O - since they belong to a sphere. Therefore, these points form a circle.
The radius of this centrally located circle is exactly R, which is greater than the radius of any other circle formed by intersection of a plane and a sphere represented by a formula above.
By the way, that formula for d=0 (which technically is not applicable) gives the radius of a circle R, which corresponds our intuition that, as a plane moves closer and closer to a center, the radius of a circle it cuts from a sphere becomes greater and greater with a limit R.
Consider a sphere of radius R and center O and a plane δ tangential to this sphere, that is the one that has only one point of intersection with this sphere - point P.
Prove that a radius OP connecting a center of a sphere O with a point of tangency of a plane with this sphere P is perpendicular to a plane δ.
All points of plane δ, except point of tangency to a sphere P, are located outside a sphere. Therefore, their distance from a center of a sphere O is greater than the radius of a sphere R (see Problem A above). Only a point of tangency P is at a distance equal to a radius of a sphere R.
We see now that segment OP is the shortest among all segments connecting center O to points on plane δ. Since we know that a perpendicular to a plane from a point is the shortest connection between them, and any other connection between center O and points on plane δ is greater than R, OP⊥δ.