# 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

### Solution 1

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. $\blacksquare$

Motivation: One can notice that if you take such a $B_1'$ then $PQB_1'C_1$ is cyclic, and that similarly $PQB_1C_1'$ is also cyclic. One gets the intuition that only one such circle should exist where the other chord passes through A, and so sets up a ghost point for $Q$, which works. ~cocohearts

### 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.

### Solution 3

Due to the parallel lines, $m\angle C_1QB_1=m\angle A.$ Therefore, it suffices to prove that $$m\angle C_1PB_1=m\angle C_1BA+m\angle ACB_1=m\angle A.$$ Note that $m\angle BC_1A+m\angle AB_1C=180^{\circ}$ by the cyclic quadrilaterals. Now, the condition simplifies to proving $C_1,A,B_1$ collinear.

Use barycentric coordinates. Let $$A=(1,0,0), B=(0,1,0), C=(0,0,1), P=(0,p,1-p), Q=(0,q,1-q).$$ By the parallel lines, $$C_1=(1-q-z',q,z'), B_1=(q-y',y',1-q)$$ for some $y',z'.$ As $A$ is a vertex of the reference triangle, we must prove that $q(1-q)=y'z'.$

Now we find the circumcircle of $\triangle APC.$ Let its equation be $$-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+vy+wz)=0.$$ Substituting in $A,C$ gives $u=w=0.$ Substituting in $P$ yields $$-a^2p(1-p)+vp=0\implies v = a^2(1-p).$$ Substituting in $B_1$ yields $$0=-a^2y'(1-q)-b^2(q-y')(1-q)-c^2y'(q-y')+a^2(1-p)y'=0.$$ $$c^2y'^2+(a^2q-a^2p+b^2(1-q)-c^2q)y'-b^2q(1-q)=0$$

Similarly, the circumcircle of $\triangle APB$ is $$-a^2yz-b^2xz-c^2xy+(x+y+z)a^2pz.$$ Now, we substitute $C_1$ into the equation

$$0=-a^2qz'-b^2z'(1-q-z')-c^2q(1-q-z')+a^2pz'=0$$ $$b^2 z'^2+(a^2p-a^2q-b^2(1-q)+c^2q)z'-c^2q(1-q)=0$$

Let $k=a^2p-a^2q-b^2(1-q)+c^2q,$ and note that both $y',z'$ are negative. By the quadratic formula, we have $$y'=\frac{k-\sqrt{k^2+4b^2c^2q(1-q)}}{2c^2}, z'=\frac{-k-\sqrt{k^2+4b^2c^2q(1-q)}}{2b^2}$$ Multiplying these two, we have $$y'z'=\frac{4b^2c^2q(1-q)}{4b^2c^2}=q(1-q),$$ as desired. $\blacksquare$