Difference between revisions of "2017 AIME II Problems/Problem 15"
(→Solution 3) |
(→Solution 3) |
||
Line 30: | Line 30: | ||
==Solution 3== | ==Solution 3== | ||
[[File:2017 AIME II 15b.png|300px|right]] | [[File:2017 AIME II 15b.png|300px|right]] | ||
− | Isosceles tetrahedron <math>ABCD</math> or Disphenoid (https://en.wikipedia.org/wiki/Disphenoid) can be inscribed in a parallelepiped <math>AB'CD'C'DA'B,</math> whose facial diagonals are the pares of equal edges of the tetrahedron <math>(AC = B'D'</math> where <math>B'D' = BD).</math> 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 <math>ABCD</math> and <math>A'B'C'D'</math> are congruent, so point of symmetry O is point of minimum <math>f(X). f(O)= 4R</math>, where <math>R</math> is the circumradius of parallelepiped. | + | Isosceles tetrahedron <math>ABCD</math> or Disphenoid (https://en.wikipedia.org/wiki/Disphenoid) can be inscribed in a parallelepiped <math>AB'CD'C'DA'B,</math> whose facial diagonals are the pares of equal edges of the tetrahedron <math>(AC = B'D',</math> where <math>B'D' = BD).</math> 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 <math>ABCD</math> and <math>A'B'C'D'</math> are congruent, so point of symmetry O is point of minimum <math>f(X). f(O)= 4R</math>, where <math>R</math> is the circumradius of parallelepiped. |
− | <cmath>R = OC =\sqrt{\frac {AB^2 + AC^2 + AD^2}{8}}, | + | <cmath>8R^2 = 2 CC'^2 = 2CD'^2 + 2D'B^2 + 2BC'^2, </cmath> |
+ | <cmath>2 CC'^2 = (CD'^2 + BC'^2) + (BC'^2 + BD'^2) + (CD'^2 + BD'^2) = AC^2 + AB^2+BC^2,</cmath> | ||
+ | <cmath>R = OC =\sqrt{\frac {AB^2 + AC^2 + AD^2}{8}}, f(O)= 4R = 4\sqrt {678}.</cmath> | ||
− | + | '''Shelomovskii, vvsss, www.deoma-cmd.ru''' (Reconstruction) | |
==See Also== | ==See Also== |
Revision as of 07:23, 17 June 2022
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.