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

(Solution to Problem 2 -- good ol' IM2 days)
 
 
(One intermediate revision by one other user not shown)
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Given a quadrilateral <math>ABCD</math>.  Let <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are the midpoints of <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively.  <math>PR</math> and <math>QS</math> intersects at <math>O</math>.  Prove that <math>PO = OR</math> and <math>QO = OS</math>.
 
Given a quadrilateral <math>ABCD</math>.  Let <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are the midpoints of <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively.  <math>PR</math> and <math>QS</math> intersects at <math>O</math>.  Prove that <math>PO = OR</math> and <math>QO = OS</math>.
  
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==Solution==
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==Solution 1==
  
 
<asy>
 
<asy>
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draw(P--R);
 
draw(P--R);
 
dot((27.5,32.5));
 
dot((27.5,32.5));
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label("O",(27.5,32.5),E);
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label("O",(27.5,32.5),SE);
 
</asy>
 
</asy>
  
 
Since <math>PQRS</math> is a parallelogram, <math>RS = QP.</math>  In addition, by the Alternating Interior Angle Theorem, <math>\angle SRP = \angle RPQ</math> and <math>\angle RSQ = \angle SQP.</math>  Thus, by ASA Congruency, <math>\triangle SRO \cong \triangle QPO.</math>  Finally, using CPCTC shows that <math>RO = OP</math> and <math>SO = OQ.</math>
 
Since <math>PQRS</math> is a parallelogram, <math>RS = QP.</math>  In addition, by the Alternating Interior Angle Theorem, <math>\angle SRP = \angle RPQ</math> and <math>\angle RSQ = \angle SQP.</math>  Thus, by ASA Congruency, <math>\triangle SRO \cong \triangle QPO.</math>  Finally, using CPCTC shows that <math>RO = OP</math> and <math>SO = OQ.</math>
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==Solution 2==
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Let <math>A = (A_x, A_y)</math>, <math>B = (B_x, B_y)</math>, <math>C = (C_x, C_y)</math>, and <math>D = (D_x, D_y)</math>. Then, we have <math>P = (\dfrac{A_x + B_x}{2}, \dfrac{A_y + B_y}{2})</math>, <math>Q = (\dfrac{B_x + C_x}{2}, \dfrac{B_y + C_y}{2})</math>, <math>R = (\dfrac{C_x + D_x}{2}, \dfrac{C_y + D_y}{2})</math>, and <math>S = (\dfrac{D_x + A_x}{2}, \dfrac{D_y + A_y}{2})</math>. Note that any line that goes through two points <math>X</math> and <math>Y</math> also goes through their midpoint. So, the line through <math>P</math> and <math>R</math> also goes through <math>(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>. Similarly, any line through <math>Q</math> and <math>S</math> also goes through <math>(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>. That means that both <math>\overline{PR}</math> and <math>\overline{QS}</math> go through <math>(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>, and because two non-identical lines that intersect only intersect once, <math>O = (\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>. Since <math>O</math> is the midpoint of both <math>\overline{PR}</math> and <math>\overline{QS}</math>, we have proved that <math>PO = OR</math> and <math>QO = OS</math>. <math>\blacksquare</math> ~Puck_0
  
 
==See Also==
 
==See Also==

Latest revision as of 16:52, 22 January 2024

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

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
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