Difference between revisions of "1972 USAMO Problems/Problem 2"
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===Solution 2=== | ===Solution 2=== | ||
− | It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that <math>AB\leq BC \leq CA</math>. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or | + | It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that <math>AB\leq BC \leq CA</math>. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. Consider triangles <math>\triangle ABC</math> and <math>\triangle ABD</math>. They share side <math>AB</math>. Let <math>k</math> and <math>l</math> be the planes passing through <math>A</math> and <math>B</math>, respectively, that are perpendicular to side <math>AB</math>. We have that triangles <math>ABC</math> and <math>ABD</math> are non-acute, so <math>C</math> and <math>D</math> are not strictly between planes <math>k</math> and <math>l</math>. Therefore the length of <math>CD</math> is at least the distance between the planes, which is <math>AB</math>. However, if <math>CD=AB</math>, then the four points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are coplanar, and the volume of <math>ABCD</math> would be zero. Therefore <math>CD>AB</math>. However, we were given that <math>CD=AB</math> in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute. |
===Solution 3=== | ===Solution 3=== |
Revision as of 19:37, 5 July 2016
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 Congruence. Without loss of generality, assume that . Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. 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.
Solution 3
Let ,
, and
. The conditions given translate to
We wish to show that
,
, and
are all positive. WLOG,
, so it immediately follows that
and
are positive. Adding all three equations,
In addition,
Equality could only occur if
, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.
Solution 4
Suppose for the sake of contradiction that is not acute. Since all three sides of triangles
and
are congruent, those two triangles are congruent, meaning
. Construct a sphere with diameter
. Since angles
and
are both not acute,
and
both lie on or inside the sphere. We seek to make
to satisfy the conditions of the problem. This can only occur when
is a diameter of the sphere, since both points lie on or inside the sphere. However, for
to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron
degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.