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

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==Solution==
 
==Solution==
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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.
 
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.
  

Revision as of 17:20, 6 March 2010

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

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.

See also

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