Difference between revisions of "2005 USAMO Problems/Problem 3"
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== Problem == | == Problem == | ||
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(''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 == | ||
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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> | ||
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{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | == 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}} | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 13:06, 3 May 2008
Problem
(Zuming Feng) Let be an acute-angled triangle, and let and be two points on side . Construct point in such a way that convex quadrilateral is cyclic, , and and lie on opposite sides of line . Construct point in such a way that convex quadrilateral is cyclic, , and and lie on opposite sides of line . Prove that points , and lie on a circle.
Solution
Let be the second intersection of the line with the circumcircle of , and let be the second intersection of the circumcircle of and line . It is enough to show that and . All our angles will be directed, and measured mod .
Since points are concyclic and points are collinear, it follows that But since points are concyclic, It follows that lines and are parallel. If we exchange with and with in this argument, we see that lines and are likewise parallel.
It follows that is the intersection of and the line parallel to and passing through . Hence . Then is the second intersection of the circumcircle of and the line parallel to passing through . Hence , 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) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |