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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.

Diagram

[asy] size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),R=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair K = midpoint(B--C); pair L = midpoint(A--C); pair I=incenter(A,B,C); pair O = circumcenter(A,B,C); dot(O); dot(A^^B^^C^^R^^K^^L); draw(C--R); draw(circumcircle(A,B,R)); draw(A--C--B); draw(A--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$R$",R,S); label("$K$",K,N); label("$L$",L,S); label("$O$",O,N); draw(K--O); draw(L--O); pair Q = intersectionpoint(L--O, C--R); dot(Q); label("$Q$",Q,SW); pair E = midpoint(C--R); dot(E); label("$E$",E,W); draw(O--E, dashed); pair P = intersectionpoint(O--(c/2-1.2,0), C--R); dot(P); label("$P$",P,W); draw(O--P); draw(R--L, dashed); draw(R--K, dashed); [/asy]

~KingRavi

Solution 1 (Efficient)

$\angle{RQL} = 90+\angle{QCL} = 90+\dfrac{C}{2}$, and similarly $\angle{RPK} = 90+\angle{PCK} = 90+\dfrac{C}{2}$. Therefore, $\angle{RQL} = \angle{RPK}$. Using the triangle area formula $A = \dfrac{1}{2}bc\sin{\angle{A}}$ yields $RQ \cdot QL = RP \cdot PK = \dfrac{PK}{QL} = \dfrac{RQ}{RP}$ after cancelling the sines and constant. Draw line $QD$ perpendicular to $BC$ that intersects $BC$ at $D$, then $QD=QL$ because the perpendicular bisectors are congruent, (or alternatively $\triangle QDC\cong\triangle QLC$). This presents us $\dfrac{PK}{QL}=\dfrac{PK}{QD}=\dfrac{PC}{QC}$ by similar triangles; now, we have only to prove $\dfrac{PC}{QC}=\dfrac{RQ}{RP}$, or $RQ \cdot QC=RP \cdot PC$.

Since $\angle{OPQ} =1 80-\angle{RPK} =1 80-\angle{RQL} = \angle{OQP}$, 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 $y = RE$, and then $RQ \cdot QC = (y+x) \cdot (y-x) = PC \cdot RP$, proving our claim.

Alternative END Solution (Power of a Point)

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 $RP*PC=RQ*QC$. $\square$

Solution by ~KingRavi

Alternate Solution by ~mathdummy

Edifying edits made by ~TheGrandioseGeometrician

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

(Image Link) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi85LzQ5OTJlZGNhYTQ0YjJjODcxMTBmZGNmMTdiZDdkMGRjZGUyOWQ5LnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOS0wOC0wOCBhdCAxMi4yMC4zOCBQTS5wbmc=

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 \[PC + QC = \dfrac{PK}{\sin \frac{1}{2}C} + \dfrac{QL}{\sin\frac{1}{2}C}\] \[= \dfrac{PK + QL}{\sin\frac{1}{2}C}\] \[= \dfrac{PK(1 + \frac{b}{a})}{\sin\frac{1}{2}C}\] \[= \dfrac{PK(\frac{a + b}{a})}{\sin\frac{1}{2}C}\] \[= \dfrac{\frac{1}{2}a\tan\frac{1}{2}C \cdot (a + b)}{a\sin\frac{1}{2}C}\] \[= \dfrac{\frac{1}{2}a\sin\frac{1}{2}C \cdot (a + b)}{a\sin\frac{1}{2}C\cos\frac{1}{2}C}\] \[= \dfrac{\sin\frac{1}{2}C \cdot (a + b)}{2\sin \frac{1}{2}C\cos\frac{1}{2}C}\] \[= \dfrac{\sin\frac{1}{2}C \cdot (a + b)}{\sin C}\] \[= \dfrac{\sin\frac{1}{2}C \cdot (a + b)}{c}\] \[= CR.\] 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

Solution 4

$[\triangle{RPK}]=[\triangle{RQL}], LQ*RQ*\sin\angle{LQR}=KP*PR*\sin\angle{RPK}$

Since $CR$ bisects $\angle{ACB},\angle{QCL}=\angle{PCK}$, $OL,OK$ are perpendicular to sides $AC,BC$ separately, $\angle{QLC}=\angle{PKC}=90^{\circ}, \angle{CQL}=\angle{CPK}, \angle{LQR}=\angle{RPK}$

So now, we only have to prove $RP*PK=LQ*QR$, which is $RP*(CQ+QP)*\cos\angle{CPK}=CQ*(QP+PR)*\cos\angle{CQL}$, as mentioned above, the two angles are the same, we have to prove that $RP(CQ+QP)=CQ(QP+PR)$, which is equivalent to $RP*QP=QP*CQ$, we have to prove $RP=CQ$

Now notice that $\triangle{OQP}$ is isosceles. $\angle{CQL}=\angle{OQP}=\angle{OPQ}$, construct $OJ \bot CR$, $CJ=JR,JQ=JP,CQ=PR$ as desired

~bluesoul

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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