1972 USAMO Problems/Problem 2
Problem
A given tetrahedron is isosceles, that is,
. Show that the faces of the tetrahedron are acute-angled triangles.
Solutions
Solution 1
Suppose is fixed.
By the equality conditions, it follows that the maximal possible value of
occurs when the four vertices are coplanar, with
on the opposite side of
as
.
In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.
For the sake of contradiction, suppose is non-acute.
Then,
.
In our optimal case noted above,
is a parallelogram, so
However, as stated, equality cannot be attained, so we get our desired contradiction.
Solution 2
It's not hard to see that the four faces are congruent from SSS similarity. Without loss of generality, assume that . Now assume, for the sake of contradiction, that each face is non-acute; that is, right or isosceles. Consider triangles
and
. They share side
. Let
and
be the planes passing through
and
, respectively, that are perpendicular to side
. We have that triangles
and
are non-acute, so
and
are not strictly between planes
and
. Therefore the length of
is at least the distance between the planes, which is
. However, if
, then the four points
,
,
, and
are coplanar, and the volume of
would be zero. Therefore
. However, we were given that
in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1972 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |