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Difference between revisions of "2007 IMO Problems/Problem 4"

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<math>(tkhalid)</math>
 
<math>(tkhalid)</math>
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==Solution 2 (Power of a point)==
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<math>\angle{RQL}=90+\angle{QCL}=90+\dfrac{C}{2}</math>, and similarly <math>\angle{RPK}=90+\angle{PCK}=90+\dfrac{C}{2}</math>, we have <math>\angle{RQL}=\angle{RPK}</math>. Using triangle area formula <math>A=bc\sin{\angle{A}}</math>, the problem is equivalent to proving <math>RQ*QL=RP*PK</math>, or <math>\dfrac{PK}{QL}=\dfrac{RQ}{RP}</math>. Draw line <math>QM</math> perpendicular to BC and intersects BC at <math>M</math>, then <math>QM=QL</math>, and <math>\dfrac{PC}{QC}=\dfrac{PK}{QM}=\dfrac{PK}{QL}</math>. Now the problem is equivalent to proving <math>\dfrac{PC}{QC}=\dfrac{RQ}{RP}</math>, or <math>RQ*QC=RP*PC</math>. Since <math>\angle{OPQ}=180-\angle{RPK}=180-\angle{RQL}=\angle{OQP}</math>, we have <math>OQ=OP=x</math>. Let the radius of the circumcircle be <math>r</math>, then the diameter through <math>P</math> is divided by point <math>P</math> into lengths of <math>r+x</math> and <math>r-x</math>. By power of point,  <math>RP*PC=(r+x)(r-x)</math>. Similarly, <math>RQ*QC=(r+x)(r-x)</math>. Therefore <math>RP*PC=RQ*QC</math>. <math>\square</math>
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<math>(mathdummy)</math>
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 02:31, 19 August 2018

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 $\triangle{RQL}$ 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 $\dfrac{RP}{RQ}=\dfrac{AC}{BC}$ $\square$

$(tkhalid)$

Solution 2 (Power of a point)

$\angle{RQL}=90+\angle{QCL}=90+\dfrac{C}{2}$, and similarly $\angle{RPK}=90+\angle{PCK}=90+\dfrac{C}{2}$, we have $\angle{RQL}=\angle{RPK}$. Using triangle area formula $A=bc\sin{\angle{A}}$, the problem is equivalent to proving $RQ*QL=RP*PK$, or $\dfrac{PK}{QL}=\dfrac{RQ}{RP}$. Draw line $QM$ perpendicular to BC and intersects BC at $M$, then $QM=QL$, and $\dfrac{PC}{QC}=\dfrac{PK}{QM}=\dfrac{PK}{QL}$. Now the problem is equivalent to proving $\dfrac{PC}{QC}=\dfrac{RQ}{RP}$, or $RQ*QC=RP*PC$. Since $\angle{OPQ}=180-\angle{RPK}=180-\angle{RQL}=\angle{OQP}$, we have $OQ=OP=x$. Let the radius of the circumcircle be $r$, then the diameter through $P$ is divided by point $P$ into lengths of $r+x$ and $r-x$. By power of point, $RP*PC=(r+x)(r-x)$. Similarly, $RQ*QC=(r+x)(r-x)$. Therefore $RP*PC=RQ*QC$. $\square$

$(mathdummy)$

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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Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
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