##
USAMO 1972 #5 *August 4, 2009*

*Posted by lumixedia in Problem-solving.*

Tags: contest math, geometry, olympiad math, USAMO, USAMO 1972

trackback

Tags: contest math, geometry, olympiad math, USAMO, USAMO 1972

trackback

**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

but also

so we know , or . Solving the quadratic gives

and

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

and

so

from which

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.

Seemingly, it is a tradition for textbooks to leave some exercises to the reader.

From the construction, it is clear to me that and (from parallel lines). So we only need to show that which is equivalent to .

By above argument, we have

.

Thus and .

In the real contest, one should also justify that ABCDE is convex. This might be trivial but we have to make sure that the construction is valid.

Oh…I didn’t see that we can get [DEA]=[EAB]=[ABC] from the parallel lines. Yeah, that pretty much makes the stuff I added unnecessary since we have [EAB]=[EPB]=(PB/PD)[DPE]=1 by construction. Oops. Thanks for the pointer!

seems like it should be sufficient to show that ABCDE is convex, I think. Does that look right?