2002 Indonesia MO Problems/Problem 7

Let $ABCD$ be a rhombus with $\angle A = 60^\circ$ , and $P$ is the intersection of diagonals $AC$ and $BD$ . Let $Q$ , $R$ , and $S$ are three points on the rhombus' perimeter. If $PQRS$ is also a rhombus, show that exactly one of $Q$ , $R$ , and $S$ is located on the vertices of rhombus $ABCD$ .

Solution

Firstly, all rhombi are parallelograms, so that $P$ is the centroid of $ABCD$ .

Suppose that $Q, R, S$ are all on one side of the rhombus. Then, in order for $PQRS$ to be a parallelogram, $P$ should also be on that side. But this is not so, so this case is impossible.

Suppose that $Q, R, S$ 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 $R$ is on the minority side, then $PQRS$ is necessarily self-intersecting and thusly not a parallelogram. Thusly, either $Q$ or $S$ is on the minority side; WLOG it is $Q$ . Then $\vec{SR}$ is parallel to the majority side, so $\vec{PQ}$ must also be parallel to the majority side, so that $Q$ is the midpoint of the minority side. Then $\vec{PQ} = \vec{SR}$ must be exactly half the length of the majority side.

From here, we consider cases. Based on the symmetry of $ABCD$ , however, we only need consider two: that where the majority side is $AB$ and the minority side $BC$ , and that where the majority side is $AB$ and the minority side $DA$ . In the first case, we find that in order to satisfy $PQ = QR$ and $A, R, B$ collinear, we must have $R = B$ or $R$ be outside of the segment $AB$ , which is forbidden, so that exactly one vertex of $PQRS$ ( $R$ ) is also a vertex of $ABCD$ . In the second case, we find that in order to satisfy $PQ = QR$ and $A, R, B$ collinear, we must have $R = A$ or $R$ be the midpoint of $AB$ , so that exactly one vertex of $PQRS$ ( $R$ or $S$ , respectively) is also a vertex of $ABCD$ .

Finally, suppose that $Q, R, S$ are on three different sides. WLOG, suppose that $R\in AB$ . If one of the other vertices is on $CD$ (WLOG it is $Q$ ), then $S$ must be outside the parallelogram (since $h_S = h_P - h_Q + h_R = \frac{1}{2} - 1 + 0 = -\frac{1}{2}$ , where $h_X$ is the (signed) height of $X$ to $AB$ , scaled by the height of $C$ ). This is impossible, so we know that $Q$ and $S$ must not be on $CD$ ; WLOG, we have $Q\in DA, S\in BC$ . Then the midpoint of $QS$ is on the line halfway between lines $DA$ and $BC$ . Since the midpoint of $QS$ and that of $PR$ are the same, $R$ is the midpoint of $AB$ . Then, in order to satisfy $PQ = QR$ and $PS = SR$ , we must have $Q$ the midpoint of $DA$ and $S = B$ , so that exactly one vertex of $PQRS$ (that is, $S$ ) is also a vertex of $ABCD$ .

All cases having been considered, we have shown that if $PQRS$ is a rhombus, then exactly one of $Q$ , $R$ , and $S$ is a vertex of $ABCD$ , and we are done.

2002 Indonesia MO (Problems)
Preceded by Problem 6	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Last Problem
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2002 Indonesia MO Problems/Problem 7

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