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 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. | 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 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. | ||
+ | ===Solution 3=== | ||
+ | Let <math>\vec{a} = \overrightarrow{DA}</math>, <math>\vec{b} = \overrightarrow{DB}</math>, and <math>\vec{c} = \overrightarrow{DC}</math>. The conditions given translate to | ||
+ | <cmath>\begin{align*} | ||
+ | \vec{a}\cdot\vec{a} &= \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} - 2(\vec{b}\cdot\vec{c}) \ | ||
+ | \vec{b}\cdot\vec{b} &= \vec{c}\cdot\vec{c} + \vec{a}\cdot\vec{a} - 2(\vec{c}\cdot\vec{a}) \ | ||
+ | \vec{c}\cdot\vec{c} &= \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} - 2(\vec{a}\cdot\vec{b}) | ||
+ | \end{align*}</cmath> | ||
+ | We wish to show that <math>\vec{a}\cdot\vec{b}</math>, <math>\vec{b}\cdot\vec{c}</math>, and <math>\vec{c}\cdot\vec{a}</math> are all positive. WLOG, <math>\vec{a}\cdot\vec{a}\geq \vec{b}\cdot\vec{b}, \vec{c}\cdot\vec{c}</math>, so it immediately follows that <math>\vec{a}\cdot\vec{b}</math> and <math>\vec{a}\cdot\vec{c}</math> are positive. Adding all three equations, | ||
+ | <cmath>\vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} = 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c})</cmath> | ||
+ | In addition, | ||
+ | <cmath>\begin{align*} | ||
+ | (\vec{a} - \vec{b} - \vec{c})\cdot(\vec{a} - \vec{b} - \vec{c})&\geq 0 \ | ||
+ | \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c}&\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \ | ||
+ | 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c})&\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \ | ||
+ | \vec{b}\cdot\vec{c}&\geq 0, | ||
+ | \end{align*}</cmath> | ||
+ | as desired. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 21:50, 31 March 2014
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 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.
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,
as desired.
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.