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

Revision as of 17:32, 8 August 2019

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 $A=\dfrac{1}{2}bc\sin{\angle{A}}$ , 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 $RP*PC=RQ*QC$ . $\square$

$(mathdummy)$

Solution 3

[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi85LzQ5OTJlZGNhYTQ0YjJjODcxMTBmZGNmMTdiZDdkMGRjZGUyOWQ5LnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOS0wOC0wOCBhdCAxMi4yMC4zOCBQTS5wbmc=[/img]

WLOG, let the diameter of $(ACBD)$ 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 \begin{align*} &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(a + b)}{a\sin\frac{1}{2}C} \\ & = \dfrac{\frac{1}{2}a\sin\frac{1}{2}C(a + b)}{a\sin\frac{1}{2}C\cos\frac{1}{2}C} \\ & = \dfrac{\frac{1}{2}C(a + b)}{2\sin{1}{2}C\cos\frac{1}{2}C} \\ & = \dfrac{\frac{1}{2}C(a + b)}{\sin C} \\ & = \dfrac{\frac{1}{2}C(a + b)}{c} \\ &= CR. \end{align*} 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
All IMO Problems and Solutions

@@ Line 13: / Line 13: @@
 <math>(mathdummy)</math>
+==Solution 3==
+[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi85LzQ5OTJlZGNhYTQ0YjJjODcxMTBmZGNmMTdiZDdkMGRjZGUyOWQ5LnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOS0wOC0wOCBhdCAxMi4yMC4zOCBQTS5wbmc=[/img]
+WLOG, let the diameter of <math>(ACBD)</math> be <math>1.</math>
+We see that <math>PK = \dfrac{1}{2}a \tan \dfrac{1}{2}C</math> and <math>QL = \dfrac{1}{2}b \tan \dfrac{1}{2}C</math> from right triangles <math>\triangle PKC</math> and <math>\triangle QLC.</math>
+We now look at <math>AR.</math> By the Extended Law of Sines on <math>\triangle ACR,</math> we get that <math>AR = \sin\frac{1}{2}C.</math> Similarly, <math>BR = \sin \frac{1}{2}C.</math>
+We now look at <math>CR.</math> By Ptolemy's Theorem, we have <cmath>AR \cdot BC + BR \cdot AC = AB \cdot CR,</cmath> which gives us <cmath>\sin \frac{1}{2}C (a + b) = c(CR).</cmath> This means that <cmath>CR = \dfrac{\sin \frac{1}{2}C (a + b)}{c}.</cmath> We now seek to relate the lengths computed with the areas.
+To do this, we consider the altitude from <math>R</math> to <math>PK.</math> This is to find the area of <math>RPK.</math> Finding the area of <math>\triangle RQL</math> is similar.
+We claim that <math>RF = \dfrac{1}{2}b.</math> In order to prove this, we will prove that <math>\triangle RFP \cong \triangle QLC.</math> In other words, we wish to prove that <math>PR = QC.</math> This is equivalent to proving that <math>PC + QC = CR.</math>
+Note that <math>PC = \dfrac{PK}{\sin \frac{1}{2}C}</math> and <math>QC = \dfrac{QL}{\sin \frac{1}{2}C}.</math> Therefore, we get that \begin{align*}
+&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(a + b)}{a\sin\frac{1}{2}C}   \\
+& = \dfrac{\frac{1}{2}a\sin\frac{1}{2}C(a + b)}{a\sin\frac{1}{2}C\cos\frac{1}{2}C} \\
+& = \dfrac{\frac{1}{2}C(a + b)}{2\sin{1}{2}C\cos\frac{1}{2}C}          \\
+& = \dfrac{\frac{1}{2}C(a + b)}{\sin C}          \\
+& = \dfrac{\frac{1}{2}C(a + b)}{c}          \\
+&= CR.
+\end{align*} Thus, <math>RF = \dfrac{1}{2}b.</math> In this way, we get that the altidude from <math>R</math> to <math>QL</math> has length <math>\dfrac{1}{2}a.</math> Therefore, we see that <math>[RPK] = \dfrac{1}{8}ab \tan \frac{1}{2}C</math> and <math>[RQL] = \dfrac{1}{8}ab \tan \frac{1}{2}C,</math> so the two areas are equal.
+Solution by Ilikeapos
 {{alternate solutions}}

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