2005 USAMO Problems/Problem 3

Problem

(Zuming Feng) Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $BC$ . Construct point $C_1$ in such a way that convex quadrilateral $APBC_1$ is cyclic, $QC_1 \parallel CA$ , and $C_1$ and $Q$ lie on opposite sides of line $AB$ . Construct point $B_1$ in such a way that convex quadrilateral $APCB_1$ is cyclic, $QB_1 \parallel BA$ , and $B_1$ and $Q$ lie on opposite sides of line $AC$ . Prove that points $B_1, C_1,P$ , and $Q$ lie on a circle.

Solution

Let $B_1'$ be the second intersection of the line $C_1A$ with the circumcircle of $APC$ , and let $Q'$ be the second intersection of the circumcircle of $B_1' C_1P$ and line $BC$ . It is enough to show that $B_1'=B_1$ and $Q' =Q$ . All our angles will be directed, and measured mod $\pi$ .

$[asy] size(300); defaultpen(1); pair A=(2,5), B=(-1,0), C=(5,0); pair C1=(.5,5.7); path O1=circumcircle(A,B,C1); pair P=IntersectionPoint(O1,B--C,1); path O2=circumcircle(A,P,C); pair B1=IntersectionPoint(O2,C1--5A-4C1,0); path O=circumcircle(B1,C1,P); pair Q=IntersectionPoint(O,B--C,1); draw(C1--P--A--B--C--A); draw(P--B1--C1--Q--B1); draw(O1,dashed+linewidth(.7)); draw(O2,dashed+linewidth(.7)); draw(O,dotted); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,S); label("$Q'$",Q,S); label("$C_1$",C1,N); label("$B_1'$",B1,E); [/asy]$

Since points $C_1, P, Q', B_1'$ are concyclic and points $C_1, A,B_1'$ are collinear, it follows that $\angle C_1 Q' P \equiv \angle C_1 B_1' P \equiv \angle A B_1' P .$ But since points $A, B_1', P, C$ are concyclic, $\angle AB_1'P \equiv \angle ACP .$ It follows that lines $AC$ and $C_1 Q'$ are parallel. If we exchange $C$ with $B$ and $C_1$ with $B_1'$ in this argument, we see that lines $AB$ and $B_1' Q'$ are likewise parallel.

It follows that $Q'$ is the intersection of $BC$ and the line parallel to $AC$ and passing through $C_1$ . Hence $Q' = Q$ . Then $B_1'$ is the second intersection of the circumcircle of $APC$ and the line parallel to $AB$ passing through $Q$ . Hence $B_1' = B_1$ , as desired. $\blacksquare$

Solution 2

Lemma. $B_1, A, C_1$ are collinear.

Suppose they are not collinear. Let line $B_1 A$ intersect circle $ABP$ (i.e. the circumcircle of $ABP$ ) again at $C_2$ distinct from $C_1$ . Because $\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ$ , we have that $B_1 C_2 PQ$ is cyclic. Hence $\angle C_2 QP = \angle C_2 B_1 P = \angle P B_1 A = \angle C$ , so $C_2 Q // AC$ . Then $C_2$ must be the other intersection of the parallel to $AC$ through $Q$ with circle $ABP$ . Then $C_2$ is on segment $C_1 Q$ , so $C_2$ is contained in triangle $ABQ$ . However, line $AB_1$ intersects this triangle only at point $A$ because $B_1$ lies on arc $AC$ not containing $P$ of circle $APC$ , a contradiction. Hence, $B_1, A, C_1$ are collinear, as desired.

As a result, we have $\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ$ , so $B_1 C_1 PQ$ is cyclic, as desired.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2005 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
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2005 USAMO Problems/Problem 3

Contents

Problem

Solution

Solution 2

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