Difference between revisions of "2017 AIME II Problems/Problem 15"
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Claim: <math>O</math> is the gravity center <math>\tfrac14(\vec A + \vec B + \vec C + \vec D)</math>. | Claim: <math>O</math> is the gravity center <math>\tfrac14(\vec A + \vec B + \vec C + \vec D)</math>. | ||
− | Proof | + | |
+ | Proof: Let <math>M</math> and <math>N</math> denote the midpoints of <math>AB</math> and <math>CD</math>. From <math>\triangle ABD \cong \triangle BAC</math> and <math>\triangle CDA \cong \triangle DCB</math>, we have <math>MC=MD</math>, <math>NA=NB</math> an hence <math>MN</math> is a perpendicular bisector of both segments <math>AB</math> and <math>CD</math>. Then if <math>X</math> is any point inside tetrahedron <math>ABCD</math>, its orthogonal projection onto line <math>MN</math> will have smaller <math>f</math>-value; hence we conclude that <math>O</math> must lie on <math>MN</math>. Similarly, <math>O</math> must lie on the line joining the midpoints of <math>AC</math> and <math>BD</math>. <math>\blacksquare</math> | ||
Claim: The gravity center <math>O</math> coincides with the circumcenter. | Claim: The gravity center <math>O</math> coincides with the circumcenter. |
Revision as of 01:57, 9 November 2020
Problem
Tetrahedron has
,
, and
. For any point
in space, define
. 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
.
Solution
Solution 1
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
.
Claim: is the gravity center
.
Proof: 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
.
Claim: The gravity center coincides with the circumcenter.
Proof. 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 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
.
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.