Difference between revisions of "2000 USAMO Problems/Problem 5"
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== Solution == | == Solution == | ||
− | Let the [[circumcenter]] of <math>\triangle ABC</math> be <math>O</math>, and let the center of <math> | + | === 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]]. | ||
− | <math>O</math> is the intersection of the perpendicular bisectors of <math>\overline{A_1A_2}, \overline{A_2A_3}, \overline{A_3A_1}</math>, and each of the centers <math>O_k</math> lie on the perpendicular bisector of the side of the triangle that determines <math> | + | <math>O</math> is the intersection of the perpendicular bisectors of <math>\overline{A_1A_2}, \overline{A_2A_3}, \overline{A_3A_1}</math>, and each of the centers <math>O_k</math> lie on the perpendicular bisector of the side of the triangle that determines <math>\omega_k</math>. It follows from <math>OA_k = OA_{k+1}, O_kA_k = O_kA_{k+1}, OO_k = OO_k \Longrightarrow \triangle OA_kO_k \cong \triangle OA_{k+1}O_k</math> that <math>\angle OA_kO_k = \angle OA_{k+1}O_k</math>. |
<center><asy> | <center><asy> | ||
− | + | size(300); | |
+ | pathpen = linewidth(0.7); pen t = linetype("2 2"); | ||
+ | pair A = (0,0), B=3*expi(1), C=(3.5)*expi(0); /* arbitrary points */ | ||
+ | pair O=circumcenter(A,B,C), O1 = O + 5*( ((B+C)/2) - O ), O2 = IP(O -- O + 100*( ((A+C)/2) - O ), O1 -- O1 + 10*( C - O1 )); | ||
+ | D(MP("A_3",A,SW)--MP("A_1",B,N)--MP("A_2",C,SE)--cycle); D(MP("O",O,NW)); D(MP("O_1",O1,E)); D(MP("O_2",O2)); D(O--B--O1--C--O--A--O2--C, linetype("4 4") + linewidth(0.7)); D(O1--O--O2,linetype("4 4") + linewidth(0.6)); D(CP(O1,C),t);D(CP(O2,C),t);D(circumcircle(A,B,C),t); | ||
</asy></center> | </asy></center> | ||
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Now <math>\theta_1 = \angle OA_1O_1 = \angle OA_2O_1 = 180 - \angle OA_2O_2 = 180 - \theta_2</math> (due to collinearility). Hence, we have the recursion <math>\theta_k = 180 - \theta_{k-1}</math>, and so <math>\theta_k = \theta_{k-2}</math>. Thus, <math>\theta_{1} = \theta_{7}</math>. | Now <math>\theta_1 = \angle OA_1O_1 = \angle OA_2O_1 = 180 - \angle OA_2O_2 = 180 - \theta_2</math> (due to collinearility). Hence, we have the recursion <math>\theta_k = 180 - \theta_{k-1}</math>, and so <math>\theta_k = \theta_{k-2}</math>. Thus, <math>\theta_{1} = \theta_{7}</math>. | ||
− | <math>\theta_{1} = \theta_{7}</math> implies that <math>O_1 \equiv O_7</math>, and circles <math> | + | <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. |
+ | |||
+ | ===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 | + | == See Also == |
{{USAMO newbox|year=2000|num-b=4|num-a=6}} | {{USAMO newbox|year=2000|num-b=4|num-a=6}} | ||
[[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.