2007 USAMO Problems/Problem 6
Problem
Let be an acute triangle with
,
, and
being its incircle, circumcircle, and circumradius, respectively. Circle
is tangent internally to
at
and tangent externally to
. Circle
is tangent internally to
at
and tangent internally to
. Let
and
denote the centers of
and
, respectively. Define points
,
,
,
analogously. Prove that

with equality if and only if triangle is equilateral.
Solution
![[asy] size(400); defaultpen(fontsize(8)); pair A=(2,8), B=(0,0), C=(13,0), I=incenter(A,B,C), O=circumcenter(A,B,C), p_a, q_a, X, Y, X1, Y1, D, E, F; real r=abs(I-foot(I,A,B)), R=abs(A-O), a=abs(B-C), b=abs(A-C), c=abs(A-B), x=(((b+c-a)/2)^2)/(r^2+4*r*R+((b+c-a)/2)^2), y=((b+c-a)/2)^2/(r^2+((b+c-a)/2)^2); p_a=x*(O-A)+A; q_a=y*(O-A)+A; X=intersectionpoint(Circle(p_a,x*abs(O-A)), A+(O-A)*0.01--O); X1=intersectionpoint(intersectionpoint(incircle(A,B,C),I--I+O-A)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+O-A)-abs(A-O)*expi(angle(A-O)-pi/2), q_a--q_a+O-A); Y=intersectionpoint(Circle(q_a,y*abs(O-A)), O--2*O-A); Y1=intersectionpoint(intersectionpoint(incircle(A,B,C),I--I+A-O)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2), O--A); draw(intersectionpoint(incircle(A,B,C),I--I+A-O)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2),blue+0.7); draw(intersectionpoint(incircle(A,B,C),I--I+O-A)+abs(A-O)*expi(angle(A-O)-pi/2)--intersectionpoint(incircle(A,B,C),I--I+O-A)-0.5*abs(A-O)*expi(angle(A-O)-pi/2),red+0.7); draw(A--B--C--A); draw(Circle(A,abs(foot(I,A,B)-A))); draw(incircle(A,B,C)); draw(I--A--O--Y); draw(Circle(p_a,x*abs(O-A)),red+0.7); draw(Circle(q_a,y*abs(O-A)),blue+0.7); label("$A$",A,(-1,1)); label("$I$",I,(-1,0));label("$O$",O,(-1,-1)); label("$P_A$",p_a,(0.5,1));label("$Q_A$",q_a,(1,0)); label("$X$",X,(1,0));label("$Y$",Y,(1,-1)); label("$X'$",X1,(1,0));label("$Y'$",Y1,(0,1)); label("$\omega$",I+r*expi(pi/12), (1,0)); label("$\omega_A$",p_a+x*abs(O-A)*expi(pi/6), (1,1)); label("$\Omega_A$",q_a+y*abs(O-A)*expi(pi/6), (1,0)); label("$\Omega_A'$",intersectionpoint(incircle(A,B,C),I--I+A-O)-abs(A-O)*expi(angle(A-O)-pi/2), (0,2)); label("$\omega_A'$",intersectionpoint(incircle(A,B,C),I--I+O-A)-0.5*abs(A-O)*expi(angle(A-O)-pi/2), (0,2)); dot(p_a^^q_a^^A^^I^^O^^X^^Y^^X1^^Y1); [/asy]](http://latex.artofproblemsolving.com/d/c/7/dc7462ed555c973a14140b29ca11f239711d0576.png)
Lemma:

Proof:
Note and
lie on
since for a pair of tangent circles, the point of tangency and the two centers are collinear.
Let touch
,
, and
at
,
, and
, respectively. Note
. Consider an inversion,
, centered at
, passing through
,
. Since
,
is orthogonal to the inversion circle, so
. Consider
. Note that
passes through
and is tangent to
, hence
is a line that is tangent to
. Furthermore,
because
is symmetric about
, so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about
, it must be perpendicular to
. Likewise,
is the other line tangent to
and perpendicular to
.
Let and
(second intersection).
Let and
(second intersection).
Evidently, and
. We want:

by inversion. Note that , and they are tangent to
, so the distance between those lines is
. Drop a perpendicular from
to
, touching at
. Then
. Then
,
=
. So

Note that . Applying the double angle formulas and
, we get


End Lemma
The problem becomes:


which is true because , equality is when the circumcenter and incenter coincide. As before,
, so, by symmetry,
. Hence the inequality is true iff
is equilateral.
Comment: It is much easier to determine by considering
. We have
,
,
, and
. However, the inversion is always nice to use. This also gives an easy construction for
because the tangency point is collinear with the intersection of
and
.
See also
2007 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Question | |
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.