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2007 IMO Problems/Problem 4

Revision as of 13:14, 24 February 2022 by Kingravi (talk | contribs) (→‎Diagram)

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] import olympiad; unitsize(150); pair A,B,C,O,I; A=origin; B=2*right; C=1.5*dir(70); O=circumcenter(A,B,C); // olympiad - circumcenter I=incenter(A,B,C); // olympiad - incenter draw(A--B--C--cycle); dot(O); dot(I); draw(circumcircle(A,B,C)); // olympiad - circumcircle draw(incircle(A,B,C)); // olympiad - incircle label("$I$",I,W); label("$O$",O,S); [/asy]

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
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