2002 Indonesia MO Problems/Problem 7
Let be a rhombus with , and is the intersection of diagonals and . Let , , and are three points on the rhombus' perimeter. If is also a rhombus, show that exactly one of , , and is located on the vertices of rhombus .
Suppose that are all on one side of the rhombus. Then, in order for to be a parallelogram, should also be on that side. But this is not so, so this case is impossible.
Suppose that are on two sides of the rhombus; then one side is occupied by two of these points (the "majority side") and one side is occupied by only one of these points (the "minority side"). If is on the minority side, then is necessarily self-intersecting and thusly not a parallelogram. Thusly, either or is on the minority side; WLOG it is . Then is parallel to the majority side, so must also be parallel to the majority side, so that is the midpoint of the minority side. Then must be exactly half the length of the majority side.
From here, we consider cases. Based on the symmetry of , however, we only need consider two: that where the majority side is and the minority side , and that where the majority side is and the minority side . In the first case, we find that in order to satisfy and collinear, we must have or be outside of the segment , which is forbidden, so that exactly one vertex of () is also a vertex of . In the second case, we find that in order to satisfy and collinear, we must have or be the midpoint of , so that exactly one vertex of ( or , respectively) is also a vertex of .
Finally, suppose that are on three different sides. WLOG, suppose that . If one of the other vertices is on (WLOG it is ), then must be outside the parallelogram (since , where is the (signed) height of to , scaled by the height of ). This is impossible, so we know that and must not be on ; WLOG, we have . Then the midpoint of is on the line halfway between lines and . Since the midpoint of and that of are the same, is the midpoint of . Then, in order to satisfy and , we must have the midpoint of and , so that exactly one vertex of (that is, ) is also a vertex of .
All cases having been considered, we have shown that if is a rhombus, then exactly one of , , and is a vertex of , and we are done.
|2002 Indonesia MO (Problems)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8||Followed by|
|All Indonesia MO Problems and Solutions|