2005 USAMO Problems/Problem 3

Revision as of 13:06, 3 May 2008 by I like pie (talk | contribs)

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. $\blacksquare$

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

See also

  • <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url>
2005 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAMO Problems and Solutions
Invalid username
Login to AoPS