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 1 (Efficient)

From a diagram, we have $\angle{RQL}=90+\angle{QCL}=90+\dfrac{C}{2}$ , and similarly $\angle{RPK}=90+\angle{PCK}=90+\dfrac{C}{2}$ . Therefore, we have $\angle{RQL}=\angle{RPK}$ . Using the area of a triangle formula $A=\dfrac{1}{2}bc\sin{\angle{A}}$ , we have $RQ \cdot QL \cdot \frac{1}{2} \cdot \sin \angle{RQL} =RP \cdot PK \cdot \frac{1}{2} \cdot \sin \angle{RPK}$ . By cancelling the sines and the constant, we now have to prove that $RQ \cdot QL=RP \cdot PK$ , or $\dfrac{PK}{QL}=\dfrac{RQ}{RP}$ . Draw line $QD$ perpendicular to BC that intersects BC at $D$ . Then $QD=QL$ because the perpendicular bisectors are congruent, (or alternatively that $\triangle QDC \cong \triangle QLC$ ) and from that we have $\dfrac{PK}{QL}=\dfrac{PK}{QD}=\dfrac{PC}{QC}$ by similar triangles. Now the problem is to prove $\dfrac{PC}{QC}=\dfrac{RQ}{RP}$ , or $RQ \cdot QC=RP \cdot PC$ .

Since $\angle{OPQ}=180-\angle{RPK}=180-\angle{RQL}=\angle{OQP}$ , we have $\triangle OPQ$ is isosceles . Draw the perpendicular from $O$ to $RC$ , intersecting at $E$ . Then $PE = QE = x$ for a real $x$ . Now, because the perpendicular from the center of a circle to a chord bisects that chord, $RE = CE$ . Let $y = RE$ . Then $RQ \cdot QC = (y+x) \cdot (y-x) = PC \cdot RP$ , proving our claim. $\boxed{QED}$

Alternative Solution (Power of a Point)

From $\angle{OPQ}=180-\angle{RPK}=180-\angle{RQL}=\angle{OQP}$ , 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 $RP*PC=RQ*QC$ . $\square$

Solution by ~KingRavi

Alternate Solution by ~mathdummy

Solution 2

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 3

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WLOG, let the diameter of $(ACBR)$ be $1.$

We see that $PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C$ and $QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C$ from right triangles $\triangle PKC$ and $\triangle QLC.$

We now look at $AR.$ By the Extended Law of Sines on $\triangle ACR,$ we get that $AR = \sin\frac{1}{2}C.$ Similarly, $BR = \sin \frac{1}{2}C.$

We now look at $CR.$ By Ptolemy's Theorem, we have $AR \cdot BC + BR \cdot AC = AB \cdot CR,$ which gives us $\sin \frac{1}{2}C (a + b) = c(CR).$ This means that $CR = \dfrac{\sin \frac{1}{2}C (a + b)}{c}.$ We now seek to relate the lengths computed with the areas.

To do this, we consider the altitude from $R$ to $PK.$ This is to find the area of $RPK.$ Finding the area of $\triangle RQL$ is similar.

We claim that $RF = \dfrac{1}{2}b.$ In order to prove this, we will prove that $\triangle RFP \cong \triangle QLC.$ In other words, we wish to prove that $PR = QC.$ This is equivalent to proving that $PC + QC = CR.$

Note that $PC = \dfrac{PK}{\sin \frac{1}{2}C}$ and $QC = \dfrac{QL}{\sin \frac{1}{2}C}.$ Therefore, we get that $PC + QC = \dfrac{PK}{\sin \frac{1}{2}C} + \dfrac{QL}{\sin\frac{1}{2}C}$ $= \dfrac{PK + QL}{\sin\frac{1}{2}C}$ $= \dfrac{PK(1 + \frac{b}{a})}{\sin\frac{1}{2}C}$ $= \dfrac{PK(\frac{a + b}{a})}{\sin\frac{1}{2}C}$ $= \dfrac{\frac{1}{2}a\tan\frac{1}{2}C \cdot (a + b)}{a\sin\frac{1}{2}C}$ $= \dfrac{\frac{1}{2}a\sin\frac{1}{2}C \cdot (a + b)}{a\sin\frac{1}{2}C\cos\frac{1}{2}C}$ $= \dfrac{\frac{1}{2}C \cdot (a + b)}{2\sin \frac{1}{2}C\cos\frac{1}{2}C}$ $= \dfrac{\frac{1}{2}C \cdot (a + b)}{\sin C}$ $= \dfrac{\frac{1}{2}C \cdot (a + b)}{c}$ $= CR.$ Thus, $RF = \dfrac{1}{2}b.$ In this way, we get that the altidude from $R$ to $QL$ has length $\dfrac{1}{2}a.$ Therefore, we see that $[RPK] = \dfrac{1}{8}ab \tan \frac{1}{2}C$ and $[RQL] = \dfrac{1}{8}ab \tan \frac{1}{2}C,$ so the two areas are equal.

Solution by Ilikeapos

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