2017 AIME II Problems/Problem 15
Tetrahedron has , , and . For any point in space, suppose . The least possible value of can be expressed as , where and are positive integers, and is not divisible by the square of any prime. Find .
Let and be midpoints of and . The given conditions imply that and , and therefore and . It follows that and both lie on the common perpendicular bisector of and , and thus line is that common perpendicular bisector. Points and are symmetric to and with respect to line . If is a point in space and is the point symmetric to with respect to line , then and , so .
Let be the intersection of and . Then , from which it follows that . It remains to minimize as moves along .
Allow to rotate about to point in the plane on the side of opposite . Because is a right angle, . It then follows that , and equality occurs when is the intersection of and . Thus . Because is the median of , the Length of Median Formula shows that and . By the Pythagorean Theorem .
Because and are right angles, It follows that . The requested sum is .
Set , , . Let be the point which minimizes .
is the gravity center .
Let and denote the midpoints of and . From and , we have , an hence is a perpendicular bisector of both segments and . Then if is any point inside tetrahedron , its orthogonal projection onto line will have smaller -value; hence we conclude that must lie on . Similarly, must lie on the line joining the midpoints of and .
The gravity center coincides with the circumcenter.
Let be the centroid of triangle ; then (by vectors). If we define , , similarly, we get and so on. But from symmetry we have , hence .
Now we use the fact that an isosceles tetrahedron has circumradius . Here so . Therefore, the answer is .
Isosceles tetrahedron is inscribed in a rectangular box, whose facial diagonals are the edges of the tetrahedron. Minimum occurs at the center of gravity, and , where is the length of the spatial diagonal of the rectangular box.
Let the three dimensions of the box be .
Add three equations, . Hence .
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