Difference between revisions of "1972 USAMO Problems/Problem 2"

Revision as of 10:44, 29 September 2011

Problem

A given tetrahedron $ABCD$ is isosceles, that is, $AB=CD, AC=BD, AD=BC$ . Show that the faces of the tetrahedron are acute-angled triangles.

Solution

Solution 1

Suppose $\triangle ABD$ is fixed. By the equality conditions, it follows that the maximal possible value of $BC$ occurs when the four vertices are coplanar, with $C$ on the opposite side of $\overline{AD}$ as $B$ . In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.

For the sake of contradiction, suppose $\angle ABD$ is non-acute. Then, $(AD)^2\geq (AB)^2+(BD)^2$ . In our optimal case noted above, $ACDB$ is a parallelogram, so $\begin{align*} 2(BD)^2 + 2(AB)^2 &= (AD)^2 + (CB)^2 \\ &= 2(AD)^2 \\ &\geq 2(BD)^2+2(AB)^2. \end{align*}$ 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 $AB\leq BC \leq CA$ . Now assume, for the sake of contradiction, that each face is non-acute; that is, right or isosceles. Consider triangles $\triangle ABC$ and $\triangle ABD$ . They share side $AB$ . Let $k$ and $l$ be the planes passing through $A$ and $B$ , respectively, that are perpendicular to side $AB$ . We have that triangles $ABC$ and $ABD$ are non-acute, so $C$ and $D$ are not strictly between planes $k$ and $l$ . Therefore the length of $CD$ is at least the distance between the planes, which is $AB$ . However, if $CD=AB$ , then the four points $A$ , $B$ , $C$ , and $D$ are coplanar, and the volume of $ABCD$ would be zero. Therefore $CD>AB$ . However, we were given that $CD=AB$ in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.

@@ Line 19: / Line 19: @@
 \end{align*}</cmath>
 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 <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==

1972 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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