Difference between revisions of "2005 USAMO Problems/Problem 3"
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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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | ''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. | ||
+ | |||
+ | 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. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 20:23, 19 May 2015
Contents
[hide]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.
Solution 2
Lemma. are collinear.
Suppose they are not collinear. Let line intersect circle (i.e. the circumcircle of ) at distinct from . Because , we have that is cyclic. Hence , so . Then , contradiction. Hence, are collinear.
As a result, we have , so 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) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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