Difference between revisions of "2003 Indonesia MO Problems/Problem 2"

Revision as of 23:27, 10 August 2018

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

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

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 $SO = OQ.$

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

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Difference between revisions of "2003 Indonesia MO Problems/Problem 2"

Revision as of 23:27, 10 August 2018

Problem

Solution

See Also

@@ Line 46: / Line 46: @@
 draw(P--R);
 dot((27.5,32.5));
-label("O",(27.5,32.5),E);
+label("O",(27.5,32.5),SE);
 </asy>