Difference between revisions of "2005 USAMO Problems/Problem 3"

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== Problem ==
 
== Problem ==
 
 
(''Zuming Feng'') Let <math>ABC</math> be an acute-angled triangle, and let <math>P</math> and <math>Q</math> be two points on side <math>BC</math>. Construct point <math>C_1 </math> in such a way that convex quadrilateral <math>APBC_1</math> is cyclic, <math>QC_1 \parallel CA</math>, and <math>C_1</math> and <math>Q</math> lie on opposite sides of line <math>AB</math>. Construct point <math>B_1</math> in such a way that convex quadrilateral <math>APCB_1</math> is cyclic, <math>QB_1 \parallel BA </math>, and <math>B_1 </math> and <math>Q </math>  lie on opposite sides of line <math>AC</math>.  Prove that points <math>B_1, C_1,P</math>, and <math>Q</math> lie on a circle.
 
(''Zuming Feng'') Let <math>ABC</math> be an acute-angled triangle, and let <math>P</math> and <math>Q</math> be two points on side <math>BC</math>. Construct point <math>C_1 </math> in such a way that convex quadrilateral <math>APBC_1</math> is cyclic, <math>QC_1 \parallel CA</math>, and <math>C_1</math> and <math>Q</math> lie on opposite sides of line <math>AB</math>. Construct point <math>B_1</math> in such a way that convex quadrilateral <math>APCB_1</math> is cyclic, <math>QB_1 \parallel BA </math>, and <math>B_1 </math> and <math>Q </math>  lie on opposite sides of line <math>AC</math>.  Prove that points <math>B_1, C_1,P</math>, and <math>Q</math> lie on a circle.
  
 
== Solution ==
 
== Solution ==
 
 
Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>.  It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>.  All our angles will be directed, and measured mod <math>\pi</math>.
 
Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>.  It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>.  All our angles will be directed, and measured mod <math>\pi</math>.
  
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It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>.  Hence <math>Q' = Q</math>.  Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>.  Hence <math>B_1' = B_1</math>, as desired.  <math>\blacksquare</math>
 
It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>.  Hence <math>Q' = Q</math>.  Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>.  Hence <math>B_1' = B_1</math>, as desired.  <math>\blacksquare</math>
 
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
== Resources ==
+
== See also ==
 +
* <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url>
  
 
{{USAMO newbox|year=2005|num-b=2|num-a=4}}
 
{{USAMO newbox|year=2005|num-b=2|num-a=4}}
 
* <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url>
 
 
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 13:06, 3 May 2008

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
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