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

Revision as of 01:40, 12 June 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 2

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.

@@ Line 3: / Line 3: @@
 ==Solution==
-===Solution 1===
-Proof by contradiction: Obviously, triangles ACD, DCB, and ABC are all congruent (the three sides are all equal). For the sake of clarity, triangle ABC is the "base", and D is the "pointed" vertex. Assume <DCB is obtuse (this proof will work for any other angle as well). Therefore angles <ABD and <DCA are also obtuse because of the congruence. However, this will mean that AB and AC will be extended outwards, and as these angles approach over 90 degrees, CA and AB will be parallel at 90, and then will extend outwards, making the point A non-existent, which obviously is a contradiction of the fact that ABCD is a tetrahedron.
 ===Solution 2===

1972 USAMO (Problems • Resources)
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Difference between revisions of "1972 USAMO Problems/Problem 2"

Revision as of 01:40, 12 June 2011

Contents

Problem

Solution

Solution 2

See also