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Difference between revisions of "2005 USAMO Problems/Problem 3"
(→‎Solution 2)
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''Lemma''. <math>B_1, A, C_1</math> are collinear.
 
''Lemma''. <math>B_1, A, C_1</math> are collinear.
  
−
Suppose they are not collinear. Let line <math>B_1 A</math> intersect circle <math>ABP</math> (i.e. the circumcircle of <math>ABP</math>) at <math>C_2</math> distinct from <math>C_1</math>. Because <math>\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ</math>, we have that <math>B_1 C_1 PQ</math> is cyclic. Hence <math>\angle C_1 QP = \angle C_1 B_1 P = \angle P B_1 A = \angle C</math>, so <math>C_2 Q // AC</math>. Then <math>C_1 = C_2</math>, contradiction. Hence, <math>B_1, A, C_1</math> are collinear.
+
Suppose they are not collinear. Let line <math>B_1 A</math> intersect circle <math>ABP</math> (i.e. the circumcircle of <math>ABP</math>) at <math>C_2</math> distinct from <math>C_1</math>. Because <math>\angle C_2 B_1 Q = \angle C_2 AB = \angle C_2 PB = 180^\circ - \angle C_2 PQ</math>, we have that <math>B_1 C_2 PQ</math> is cyclic. Hence <math>\angle C_2 QP = \angle C_2 B_1 P = \angle P B_1 A = \angle C</math>, so <math>C_2 Q // AC</math>. Then <math>C_2</math> must be the other intersection of the parallel to <math>AC</math> through <math>Q</math> with circle <math>ABP</math>. Then <math>C_2</math> is on segment <math>C_1 Q</math>, so <math>C_2</math> is contained in triangle <math>ABQ</math>. However, line <math>AB_1</math> intersects this triangle only at point <math>A</math> because <math>B_1</math> lies on arc <math>AC</math> not containing <math>P</math> of circle <math>APC</math>, a contradiction. Hence, <math>B_1, A, C_1</math> are collinear, as desired.
  
 
As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired.
 
As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired.

Revision as of 20:29, 19 May 2015

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 $ABP$) 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.

See also

  • <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url>
2005 USAMO (Problems • Resources)
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Problem 4
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All USAMO Problems and Solutions

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