Difference between revisions of "1972 USAMO Problems/Problem 2"
(New page: ==Problem== A given tetrahedron <math>ABCD</math> is isosceles, that is, <math>AB=CD, AC=BD, AD=BC</math>. Show that the faces of the tetrahedron are acute-angled triangles. ==Solution== ...) |
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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. | ||
==See also== | ==See also== | ||
Revision as of 17:20, 6 March 2010
Problem
A given tetrahedron 
is isosceles, that is, 
. Show that the faces of the tetrahedron are acute-angled triangles.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it. 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 (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
