2007 IMO Problems/Problem 4

Problem

In $\triangle ABC$ the bisector of $\angle{BCA}$ intersects the circumcircle again at $R$ , the perpendicular bisector of $BC$ at $P$ , and the perpendicular bisector of $AC$ at $Q$ . The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$ . Prove that the triangles $RPK$ and $RQL$ have the same area.

Solution

The area of $\triange{RQL}$ (Error compiling LaTeX. Unknown error_msg) is given by $\dfrac{1}{2}QL*RQ\sin{\angle{RQL}}$ and the area of $\triangle{RPK}$ is $\dfrac{1}{2}RP*PK\sin{\angle{RPK}}$ . Let $\angle{BCA}=C$ , $\angle{BAC}=A$ , and $\angle{ABC}=B$ . Now $\angle{KCP}=\angle{QCL}=\dfrac{C}{2}$ and $\angle{PKC}=\angle{QLC}=90$ , thus $\angle{RPK}=\angle{RQL}=90+\dfrac{C}{2}$ . $\triangle{PKC} \sim \triangle{QLC}$ , so $\dfrac{PK}{QL}=\dfrac{KC}{LC}$ , or $\dfrac{PK}{QL}=\dfrac{BC}{AB}$ . The ratio of the areas is $\dfrac{[RPK]}{[RQL]}=\dfrac{BC*RP}{AC*RQ}$ . The two areas are only equal when the ratio is 1, therefore it suffices to show $\dfrac{RP}{RQ}=\dfrac{AC}{BC}$ . Let $O$ be the center of the circle. Then $\angle{ROK}=A+C$ , and $\angle{ROP}=180-(A+C)=B$ . Using law of sines on $\triangle{RPO}$ we have: $\dfrac{RP}{\sin{B}}=\dfrac{OR}{\sin{(90+\dfrac{C}{2})}}$ so $RP*\sin{(90+\dfrac{C}{2})}=OR*\sin{B}$ . $OR*\sin{B}=\dfrac{1}{2}AC$ by law of sines, and $\sin{(90+\dfrac{C}{2})}=\cos{\dfrac{C}{2}}$ , thus 1) $2RP\cos{\dfrac{C}{2}}=AC$ . Similarly, law of sines on $\triangle{ROQ}$ results in $\dfrac{RQ}{\sin{(180-A)}}=\dfrac{OR}{\sin{(90-\dfrac{C}{2})}}$ or $\dfrac{RQ}{\sin{A}}=\dfrac{OR}{\cos{\dfrac{C}{2}}}$ . Cross multiplying we have $RQ\cos{\dfrac{C}{2}}=OR*\sin{A}$ or 2) $2RQ\cos{\dfrac{C}{2}}=BC$ . Dividing 1) by 2) we have $\dfrac{RP}{RQ}=\dfrac{AC}{BC}$ $\square$

$(tkhalid)$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2007 IMO (Problems) • Resources
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2007 IMO Problems/Problem 4

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