2007 Indonesia MO Problems/Problem 7

Problem

Points $A,B,C,D$ are on circle $S$ , such that $AB$ is the diameter of $S$ , but $CD$ is not the diameter. Given also that $C$ and $D$ are on different sides of $AB$ . The tangents of $S$ at $C$ and $D$ intersect at $P$ . Points $Q$ and $R$ are the intersections of line $AC$ with line $BD$ and line $AD$ with line $BC$ , respectively.

(a) Prove that $P$ , $Q$ , and $R$ are collinear.

(b) Prove that $QR$ is perpendicular to line $AB$ .

Solution

$[asy] size(225); pair a=(-50,0),b=(50,0),c=(-48,14),d=(-30,-40),o=(0,0); draw(circle(o,50)); draw(a--c--b--d--a); dot(a); label("$A$",a,NE); dot(b); label("$B$",b,E); dot(c); label("$C$",c,NE); dot(d); label("$D$",d,S); dot(o); label("$O$",o,SE); pair r=(-57.692,15.385),q=(-57.692,-53.846); draw (r--b--q); dot(r); label("$R$",r,NW); dot(q); label("$Q$",q,SW); draw(c--q); draw(r--d); draw(r--q); draw(a--b,dotted); draw(c--o--d--c,dotted); [/asy]$ Let $O$ be the center of the circle. Let $\angle CBA = b$ and $\angle DBA = a$ . To prove that $P$ is on line $QR$ , we can show that $\angle PRD = \angle QRD$ .

Because $AB$ is a diameter, we know that $\angle ACB = 90^\circ$ and $\angle ADB = 90^\circ$ . By the Vertical Angle Theorem, $\angle RAC = \angle QAD$ , so by AA Similarity, $\triangle RAC \sim \triangle QAD$ . Thus, $\frac{RA}{AC} = \frac{QA}{AD}$ . Because $\angle RAQ = \angle CAD$ , by SAS Similarity, $\triangle RAQ \sim \triangle CAD$ , so $\angle ARQ = \angle ACD$ . Since $\angle ACD$ and $\angle ABD$ are both inscribed angles with the same arc of a given circle, $\angle ABD = \angle ACD = a$ , so $\angle ARQ = a$ . $[asy] size(225); pair a=(-50,0),b=(50,0),c=(-48,14),d=(-30,-40),o=(0,0); draw(circle(o,50)); draw(a--c--b--d--a); dot(a); label("$A$",a,NE); dot(b); label("$B$",b,E); dot(c); label("$C$",c,NE); dot(d); label("$D$",d,S); dot(o); label("$O$",o,SE); pair r=(-57.692,15.385),q=(-57.692,-53.846),p=(-57.692,-19.231); dot(r); label("$R$",r,NW); dot(q); label("$Q$",q,SW); dot(p); label("$P$",p,W); draw(c--p--d); draw(r--b--q); draw(c--q); draw(r--d); draw(r--p); draw(a--b,dotted); draw(c--o--d--c,dotted); [/asy]$

Because $PC$ and $PD$ are both tangents to the circle, we must have $\angle ODP = \angle OCP$ . Therefore, $\angle RDP = \angle BDO = a$ and $\angle PCQ = \angle OCB = b$ . Additionally, from a property of inscribed angles, we must have $\angle ACD = \angle ABD = a$ and $\angle ADC = \angle ABC = b$ . Thus, since the sum of the angles in a triangle is $180^\circ$ , $\angle CPD = 180-(2a+2b)$ .

Additionally, since $\triangle RDB$ is a right triangle, we must have $\angle BRD = 90-(a+b)$ . Because $\angle CRD = \tfrac12 \cdot \angle CPD$ and $CP = PD$ , we know that $C,R,D$ are in a circle with center $P$ , so $RP = PC = PD$ . Since $RB$ is a line, $\angle RCP = 90-b$ , so by the Base Angle Theorem, $\angle CRP = \angle RCP = 90-b$ . Thus, from the Angle Addition Postulate, $\angle PRD = (90-b)-(90-a-b) = a$ .

Thus, we proved that $P$ is on line $RQ$ . Additionally, by letting $K$ be the intersection of $RP$ and $AB$ , we must have $\angle KRB = 90-b$ and $\angle RBK = b$ , so $\angle RKB = 90^\circ$ . By definition, $AB \perp RP$ , and since $P$ is on line $RQ$ , $AB \perp RQ$ .

2007 Indonesia MO (Problems)
Preceded by Problem 6	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 8
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2007 Indonesia MO Problems/Problem 7

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