Difference between revisions of "1972 USAMO Problems/Problem 2"
(→Solution 1) |
(→Solution 2) |
||
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
| − | ===Solution | + | ===Solution 1=== |
Suppose <math>\triangle ABD</math> is fixed. | Suppose <math>\triangle ABD</math> is fixed. | ||
Revision as of 01:40, 12 June 2011
Contents
Problem
A given tetrahedron 
is isosceles, that is, 
. Show that the faces of the tetrahedron are acute-angled triangles.
Solution
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.
See also
| 1972 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
