A purely 3-D geometrical solution to Mathematics Magazine Problem 2065
SSolution to Mathematics Magazine Problem 2065
Yawen ZhangElizabethtown CollegeOne Alpha DriveElizabethtown, PA [email protected]
Let Q be a cube centered at the origin of R . Choose a unit vector ( a, b, c ) uniformly atrandom on the surface of the unit sphere a + b + c = 1, and let Π be the plane ax + by + cz = 0through the origin and normal to ( a, b, c ). What is the probability that the intersection of Π with Q is a hexagon?[1] Solution.
Without loss of generality, assume that Q has sides of length 2. Since Q has six faces,the intersection of Q and Π is a hexagon only if Π intersects two distinct edges on each of the sixdifferent faces of Q . There are eight symmetric octants, so we focus on the first octant and thenmultiply by 8 to account for the other octants. If a, b, c >
0, then the six edge intersections of Πand Q are ± (cid:18) b − ca , − , (cid:19) , ± (cid:18) − , a − cb , (cid:19) , and ± (cid:18) − , , a − bc (cid:19) , where a < b + c , b < a + c , c < a + b , and a + b + c = 1.The planes a = b + c , b = a + c , and c = a + b intersect a + b + c = 1 in great circles that intersectin the first octant at points A (cid:18) √ , √ , (cid:19) , B (cid:18) , √ , √ (cid:19) , and C (cid:18) √ , , √ (cid:19) , as shown in the figure. Choices of ( a, b, c ) in the first octant that lead to hexagonal intersections ofΠ and Q are in the spherical triangle ∆ ABC .Let O be the point at the origin and let α , β , and γ be the lengths of the arcs of ∆ ABC oppositethe angles at points A , B , and C , respectively. Since −→ OA · −−→ OB = −→ OA · −→ OC = −−→ OB · −→ OC = 1 / a r X i v : . [ m a t h . HO ] O c t adius of the sphere is 1, (cid:93) AOB = (cid:93) AOC = (cid:93) BOC = π/ α = β = γ = π/
3. The sphericallaw of cosines states, for example, thatcos α = cos β cos γ + sin β sin γ cos ( (cid:94) BAC ) , so cos ( (cid:94) BAC ) = cos ( (cid:94)
ABC ) = cos ( (cid:94)
ACB ) = 1 /
3. The radius of the sphere is 1, so the resultingarea of ∆
ABC is the sum of the interior angles minus π , or3 cos − (cid:18) (cid:19) − π. Since there are spherical 8 triangles like ∆
ABC and the surface area of the unit sphere is 4 π , theprobability that the intersection of Π and Q is a hexagon is84 π (cid:20) − (cid:18) (cid:19) − π (cid:21) = 6 π cos − (cid:18) (cid:19) − ≈ . . References [1] Problems and solutions.