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Difference between revisions of "2007 IMO Problems/Problem 4"
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The area of <math>\triange{RQL}</math> is given by <math>\dfrac{1}{2}QL*RQ\sin{\angle{RQL}}</math> and the area of <math>\triangle{RPK}</math> is <math>\dfrac{1}{2}RP*PK\sin{\angle{RPK}}</math>. Let <math>\angle{BCA}=C</math>, <math>\angle{BAC}=A</math>, and <math>\angle{ABC}=B</math>. Now <math>\angle{KCP}=\angle{QCL}=\dfrac{C}{2}</math> and <math>\angle{PKC}=\angle{QLC}=90</math>, thus <math>\angle{RPK}=\angle{RQL}=90+\dfrac{C}{2}</math>. <math>\triangle{PKC} \sim \triangle{QLC}</math>, so <math>\dfrac{PK}{QL}=\dfrac{KC}{LC}</math>, or <math>\dfrac{PK}{QL}=\dfrac{BC}{AB}</math>. The ratio of the areas is <math>\dfrac{[RPK]}{[RQL]}=\dfrac{BC*RP}{AC*RQ}</math>. The two areas are only equal when the ratio is 1, therefore it suffices to show <math>\dfrac{RP}{RQ}=\dfrac{AC}{BC}</math>. Let <math>O</math> be the center of the circle. Then <math>\angle{ROK}=A+C</math>, and <math>\angle{ROP}=180-(A+C)=B</math>. Using law of sines on <math>\triangle{RPO}</math> we have: <math>\dfrac{RP}{\sin{B}}=\dfrac{OR}{\sin{(90+\dfrac{C}{2})}}</math> so <math>RP*\sin{(90+\dfrac{C}{2})}=OR*\sin{B}</math>. <math>OR*\sin{B}=\dfrac{1}{2}AC</math> by law of sines, and <math>\sin{(90+\dfrac{C}{2})}=\cos{\dfrac{C}{2}}</math>, thus 1) <math>2RP\cos{\dfrac{C}{2}}=AC</math>. Similarly, law of sines on <math>\triangle{ROQ}</math> results in <math>\dfrac{RQ}{\sin{(180-A)}}=\dfrac{OR}{\sin{(90-\dfrac{C}{2})}}</math> or <math>\dfrac{RQ}{\sin{A}}=\dfrac{OR}{\cos{\dfrac{C}{2}}}</math>. Cross multiplying we have <math>RQ\cos{\dfrac{C}{2}}=OR*\sin{A}</math> or 2) <math>2RQ\cos{\dfrac{C}{2}}=BC</math>. Dividing 1) by 2) we have <math>\dfrac{RP}{RQ}=\dfrac{AC}{BC}</math> <math>\square</math>  
 
The area of <math>\triange{RQL}</math> is given by <math>\dfrac{1}{2}QL*RQ\sin{\angle{RQL}}</math> and the area of <math>\triangle{RPK}</math> is <math>\dfrac{1}{2}RP*PK\sin{\angle{RPK}}</math>. Let <math>\angle{BCA}=C</math>, <math>\angle{BAC}=A</math>, and <math>\angle{ABC}=B</math>. Now <math>\angle{KCP}=\angle{QCL}=\dfrac{C}{2}</math> and <math>\angle{PKC}=\angle{QLC}=90</math>, thus <math>\angle{RPK}=\angle{RQL}=90+\dfrac{C}{2}</math>. <math>\triangle{PKC} \sim \triangle{QLC}</math>, so <math>\dfrac{PK}{QL}=\dfrac{KC}{LC}</math>, or <math>\dfrac{PK}{QL}=\dfrac{BC}{AB}</math>. The ratio of the areas is <math>\dfrac{[RPK]}{[RQL]}=\dfrac{BC*RP}{AC*RQ}</math>. The two areas are only equal when the ratio is 1, therefore it suffices to show <math>\dfrac{RP}{RQ}=\dfrac{AC}{BC}</math>. Let <math>O</math> be the center of the circle. Then <math>\angle{ROK}=A+C</math>, and <math>\angle{ROP}=180-(A+C)=B</math>. Using law of sines on <math>\triangle{RPO}</math> we have: <math>\dfrac{RP}{\sin{B}}=\dfrac{OR}{\sin{(90+\dfrac{C}{2})}}</math> so <math>RP*\sin{(90+\dfrac{C}{2})}=OR*\sin{B}</math>. <math>OR*\sin{B}=\dfrac{1}{2}AC</math> by law of sines, and <math>\sin{(90+\dfrac{C}{2})}=\cos{\dfrac{C}{2}}</math>, thus 1) <math>2RP\cos{\dfrac{C}{2}}=AC</math>. Similarly, law of sines on <math>\triangle{ROQ}</math> results in <math>\dfrac{RQ}{\sin{(180-A)}}=\dfrac{OR}{\sin{(90-\dfrac{C}{2})}}</math> or <math>\dfrac{RQ}{\sin{A}}=\dfrac{OR}{\cos{\dfrac{C}{2}}}</math>. Cross multiplying we have <math>RQ\cos{\dfrac{C}{2}}=OR*\sin{A}</math> or 2) <math>2RQ\cos{\dfrac{C}{2}}=BC</math>. Dividing 1) by 2) we have <math>\dfrac{RP}{RQ}=\dfrac{AC}{BC}</math> <math>\square</math>  
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<math>(tkhalid)</math>
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 22:45, 18 January 2015

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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