Difference between revisions of "2000 USAMO Problems/Problem 5"
m (→= Solution 1) |
|||
(2 intermediate revisions by 2 users not shown) | |||
Line 3: | Line 3: | ||
== 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]]. | ||
Line 20: | Line 21: | ||
<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. | ||
+ | |||
+ | ===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 == |
Latest revision as of 09:52, 20 July 2016
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
.
![[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]](http://latex.artofproblemsolving.com/3/7/1/3717a532a6f30403d5d32173be48cce905c005af.png)
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.