Difference between revisions of "1972 USAMO Problems/Problem 2"
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===Solution 2=== | ===Solution 2=== | ||
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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.