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

(Solution to Problem 2 -- good ol' IM2 days)
 
m (Better placement of O)
 
Line 46: Line 46:
 
draw(P--R);
 
draw(P--R);
 
dot((27.5,32.5));
 
dot((27.5,32.5));
label("O",(27.5,32.5),E);
+
label("O",(27.5,32.5),SE);
 
</asy>
 
</asy>
  

Latest revision as of 00:27, 11 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 $SO = OQ.$

See Also

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
Invalid username
Login to AoPS