2005 USAMO Problems/Problem 3

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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 $\pi$ .

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. $\blacksquare$

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: AoPS \| ML)
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4

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2005 USAMO Problems/Problem 3

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Problem

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