+180 A sphere passes through all four vertices of one face of a unit cube, and is tangent to the opposite face of the cube. Find the radius of the sphere.
Points A B and C are on a sphere whose radius is 5. If AB = BC = 6, what is the longest possible value of AC?
+1145 Problem 1
Let the sphere have radius r, and let the cube have side length 1. Let P be the center of the sphere, and let Q be the center of the square face of the cube that is opposite the face containing the vertices of the sphere.
Since P is the center of the sphere, it is equidistant to the four vertices of the face of the cube that is opposite it. Therefore, PQ=2r.
Since P is the center of the sphere, it is also tangent to the face of the cube containing the vertices of the sphere. Therefore, r=1−QP=1−2r. Solving for r, we get r= sqrt(3)/2.
+1145 Problem 2
There are two cases to consider.
(i) Triangle ABC is right-angled.
Then, by Pythagoras' theorem, AC2=62+62=72. Hence, AC=6*sqrt(2).
(ii) Triangle ABC is not right-angled.
Then, by the triangle inequality, AC
If AC<10, then dropping a perpendicular from A to BC gives a right-angled triangle with hypotenuse less than 10, and legs greater than 6. However, Pythagoras' theorem forbids this.
Therefore, the longest possible value of AC is 6*sqrt(2).
