USAMO 1972 #5 August 4, 2009Posted by lumixedia in Problem-solving.
Tags: contest math, geometry, olympiad math, USAMO, USAMO 1972
USAMO 1972 #5. A given convex pentagon has the property that the area of each of the five triangles , , , , is unity. Show that every non-congruent pentagon with the above property has the same area, and that, furthermore, there are an infinite number of such non-congruent pentagons.
Solution. Let be the intersection of and . Since and have the same area and the same base, they must also have the same height, so we know that is parallel to . Similarly, is parallel to and is parallel to , so is a parallelogram and . Now let , so that . We have
so we know , or . Solving the quadratic gives
so has only one possible area, as desired.
We can create an infinite number of noncongruent pentagons with this property as follows. First we construct an arbitrary triangle with area and extend to and to so that . Then we let be the intersection of the line through parallel to and the line through parallel to .* Then we have and . To show that has the desired property, we just need to prove that . [Edit: see comments for much better ways to proceed from here.] Let be the intersection of and . We can calculate
Since and are parallel, we have . So we can write
where the last equality comes from . We conclude that
and, by symmetry, . So has the desired property.
*Note: the official solution stops here and proclaims that satisfies the desired property. This means either that the solution writers completely forgot to check that , or that there is a far more obvious reason why this is true than the proof I found. Help.