2003 Indonesia MO Problems/Problem 2

Problem

Given a quadrilateral $ABCD$ . Let $P$ , $Q$ , $R$ , and $S$ are the midpoints of $AB$ , $BC$ , $CD$ , and $DA$ , respectively. $PR$ and $QS$ intersects at $O$ . Prove that $PO = OR$ and $QO = OS$ .

Solution 1

$[asy] pair A=(0,0), B=(60,0), C=(40,80), D=(10,50), P=(30,0), Q=(50,40), R=(25,65), SX=(5,25); draw(A--B--C--D--cycle); draw(P--Q--R--SX--cycle); draw(A--C, dotted); draw(B--D, dotted); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NE); dot(D); label("D",D,NW); dot(P); label("P",P,S); dot(Q); label("Q",Q,E); dot(R); label("R",R,N); dot(SX); label("S",SX,W); [/asy]$

Draw lines $AC$ and $BD$ . By SAS Similarity, $\triangle DSR \sim \triangle DAC,$ $\triangle CRQ \sim \triangle CDB,$ $\triangle ASP \sim \triangle ADB,$ and $\triangle CRQ \sim \triangle CDB.$ That means $RQ \parallel DB \parallel SP$ and $SR \parallel AC \parallel PQ,$ making $PQRS$ a parallelogram.

$[asy] pair P=(30,0), Q=(50,40), R=(25,65), SX=(5,25); draw(P--Q--R--SX--cycle); dot(P); label("P",P,S); dot(Q); label("Q",Q,E); dot(R); label("R",R,N); dot(SX); label("S",SX,W); draw(Q--SX); draw(P--R); dot((27.5,32.5)); label("O",(27.5,32.5),SE); [/asy]$

Since $PQRS$ is a parallelogram, $RS = QP.$ In addition, by the Alternating Interior Angle Theorem, $\angle SRP = \angle RPQ$ and $\angle RSQ = \angle SQP.$ Thus, by ASA Congruency, $\triangle SRO \cong \triangle QPO.$ Finally, using CPCTC shows that $RO = OP$ and $SO = OQ.$

Solution 2

Let $A = (A_x, A_y)$ , $B = (B_x, B_y)$ , $C = (C_x, C_y)$ , and $D = (D_x, D_y)$ . Then, we have $P = (\dfrac{A_x + B_x}{2}, \dfrac{A_y + B_y}{2})$ , $Q = (\dfrac{B_x + C_x}{2}, \dfrac{B_y + C_y}{2})$ , $R = (\dfrac{C_x + D_x}{2}, \dfrac{C_y + D_y}{2})$ , and $S = (\dfrac{D_x + A_x}{2}, \dfrac{D_y + A_y}{2})$ . Note that any line that goes through two points $X$ and $Y$ also goes through their midpoint. So, the line through $P$ and $R$ also goes through $(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})$ . Similarly, any line through $Q$ and $S$ also goes through $(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})$ . That means that both $\overline{PR}$ and $\overline{QS}$ go through $(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})$ , and because two non-identical lines that intersect only intersect once, $O = (\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})$ . Since $O$ is the midpoint of both $\overline{PR}$ and $\overline{QS}$ , we have proved that $PO = OR$ and $QO = OS$ . $\blacksquare$ ~Puck_0

2003 Indonesia MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 3
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2003 Indonesia MO Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

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