2017 AIME II Problems/Problem 15
Problem
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
.
Official Solution (MAA)
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
.
Solution 2
Set ,
,
. Let
be the point which minimizes
.
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
.
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
.
Solution 3
Isosceles tetrahedron or Disphenoid (https://en.wikipedia.org/wiki/Disphenoid) can be inscribed in a parallelepiped
whose facial diagonals are the pares of equal edges of the tetrahedron
where
This parallelepiped is right-angled, therefore it is circumscribed and has equal diagonals. The center O of the circumscribed sphere (coincide with the centroid) has equal distance from each vertex. Tetrachedrons
and
are congruent, so point of symmetry O is point of minimum
, where
is the circumradius of parallelepiped.
Shelomovskii, vvsss, www.deoma-cmd.ru (Reconstruction)
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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