Difference between revisions of "2000 USAMO Problems/Problem 5"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
Let the [[circumcenter]] of <math>\triangle ABC</math> be <math>O</math>, and let the center of <math>\omega_k</math> be <math>O_k</math>. <math>\omega_k</math> and <math>\omega_{k-1}</math> are externally tangent at the point <math>A_k</math>, so <math>O_k, A_k, O_{k-1}</math> are [[collinear]]. | Let the [[circumcenter]] of <math>\triangle ABC</math> be <math>O</math>, and let the center of <math>\omega_k</math> be <math>O_k</math>. <math>\omega_k</math> and <math>\omega_{k-1}</math> are externally tangent at the point <math>A_k</math>, so <math>O_k, A_k, O_{k-1}</math> are [[collinear]]. | ||
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<math>\theta_{1} = \theta_{7}</math> implies that <math>O_1 \equiv O_7</math>, and circles <math>\omega_1</math> and <math>\omega_7</math> are the same circle since they have the same center and go through the same two points. | <math>\theta_{1} = \theta_{7}</math> implies that <math>O_1 \equiv O_7</math>, and circles <math>\omega_1</math> and <math>\omega_7</math> are the same circle since they have the same center and go through the same two points. | ||
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+ | ===Solution 2=== | ||
+ | Using the collinearity of certain points and the fact that <math>A_k A_{k+1} O_k</math> is isosceles, we quickly deduce that | ||
+ | <cmath>\angle{A_1 A_2 O_1} = 180^\circ - \angle{A_2} - (180^\circ - \angle{A_3}) + (180^\circ - \angle{A_1}) - (180^\circ - \angle{A_2}) + (180^\circ - \angle{A_3}) - (180^\circ - \angle{A_1}) + \angle{A_7 A_8 O_7}.</cmath> | ||
+ | From ASA Congruence we deduce that <math>A_1 A_2 O_1</math> and <math>A_7 A_8 O_7</math> are congruent triangles, and so <math>O_1 A_1 = O_7 A_7</math>, that is <math>\omega_1 = \omega_7</math>. | ||
== See Also == | == See Also == | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 08:52, 20 July 2016
Contents
[hide]Problem
Let be a triangle and let be a circle in its plane passing through and Suppose there exist circles such that for is externally tangent to and passes through and where for all . Prove that
Solution
Solution 1
Let the circumcenter of be , and let the center of be . and are externally tangent at the point , so are collinear.
is the intersection of the perpendicular bisectors of , and each of the centers lie on the perpendicular bisector of the side of the triangle that determines . It follows from that .
Since , and the perpendicular bisector of are fixed, the angle determines the position of (since lies on the perpendicular bisector). Let ; then, and together imply that .
Now (due to collinearility). Hence, we have the recursion , and so . Thus, .
implies that , and circles and are the same circle since they have the same center and go through the same two points.
Solution 2
Using the collinearity of certain points and the fact that is isosceles, we quickly deduce that From ASA Congruence we deduce that and are congruent triangles, and so , that is .
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.