Difference between revisions of "1972 USAMO Problems/Problem 2"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
However, as stated, equality cannot be attained, so we get our desired contradiction. | However, as stated, equality cannot be attained, so we get our desired contradiction. | ||
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+ | ===Solution 2=== | ||
+ | It's not hard to see that the four faces are congruent from SSS similarity. 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 isosceles. 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. | ||
==See also== | ==See also== |
Revision as of 09:44, 29 September 2011
Contents
[hide]Problem
A given tetrahedron is isosceles, that is,
. Show that the faces of the tetrahedron are acute-angled triangles.
Solution
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.
See also
1972 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |