1972 USAMO Problems/Problem 5
Problem
A given convex pentagon has the property that the area of each of the five triangles , , , , and is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property.
Solution
Lemma: Convex pentagon has the property that if and only if for (indices taken mod 5).
Proof: For the "only if" direction, since , and are equidistant from , and since the pentagon is convex, . The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Let be the inner pentagon, labeled so that and are opposite each other, and let be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon is similar to pentagon with and parallelogram (and cyclic) has area 2. Supposing we have Since and are parallelograms, . Triangle is similar to triangle , so , and with the requirement that , Now, we compute that , and similar computation for the other four triangles gives . From the aforementioned pentagon similarity, . Solving for , we have
To show that there are infinitely many pentagons with the given property, we start with triangle and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.
See Also
1972 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.