The mess of common tangents to two circles: a revisit
by math_explorer, Feb 28, 2012, 7:14 AM
(USAMO 1991.5. I'm not sure how much satisfaction I'm supposed to feel now.)
Full disclosure: on first reading I thought we were drawing the common tangent to two circumcircles and was very confused.
Okay this is annoying we don't even know what arc we want to prove
is on and there isn't anything obvious. Well, we can calculate that the angle through the incenters and with vertex at 
is right, and maybe play with reflections and half-turns... no, that doesn't work
Recall the magical lemma I read about and then posted briefly/vaguely/uselessly on AoPS years ago! Wait, it's only nine months. I have a terrible sense of time. This is also why I am giving my posts nice accessible subjects now. And actually it's a close variant, not the same thing. But whatever.
Lemma. Let two nonintersecting circles
and 
be given. Let an internal common tangent touch the circles at 
and intersect the two external common tangents at 
. Then 
.
And the problem is unexpectedly trivialized!
In the diagram,
, so 
. Now, 
and 
(incircle properties), so 
, which does not depend on . Therefore lies on the arc centered at 
with radius 
inside angle 
.
As my intuition tells me and as a quick check on AoPS threads also demonstrate, solutions don't seem to have much deviation from this path.
Paranoid rigor notice: will trace out all points on that arc because will go across the whole segment and 
will cover the whole angle. Something like that.
MithsApprentice wrote:
Let 
be an arbitrary point on side 
of a given triangle 
and let 
be the interior point where 
intersects the external common tangent to the incircles of triangles 
and 
. As assumes all positions between 
and 
, prove that the point traces the arc of a circle.
be an arbitrary point on side 
of a given triangle 
and let 
be the interior point where 
intersects the external common tangent to the incircles of triangles 
and 
. As assumes all positions between 
and 
, prove that the point traces the arc of a circle.Full disclosure: on first reading I thought we were drawing the common tangent to two circumcircles and was very confused.
Okay this is annoying we don't even know what arc we want to prove
is on and there isn't anything obvious. Well, we can calculate that the angle through the incenters and with vertex at 
is right, and maybe play with reflections and half-turns... no, that doesn't workRecall the magical lemma I read about and then posted briefly/vaguely/uselessly on AoPS years ago! Wait, it's only nine months. I have a terrible sense of time. This is also why I am giving my posts nice accessible subjects now. And actually it's a close variant, not the same thing. But whatever.
Lemma. Let two nonintersecting circles
and 
be given. Let an internal common tangent touch the circles at 
and intersect the two external common tangents at 
. Then 
.
is closer to 
than 
.
.
.
,
and And the problem is unexpectedly trivialized!
In the diagram,
, so 
. Now, 
and 
(incircle properties), so 
, which does not depend on . Therefore lies on the arc centered at 
with radius 
inside angle 
.As my intuition tells me and as a quick check on AoPS threads also demonstrate, solutions don't seem to have much deviation from this path.
Paranoid rigor notice:
will trace out all points on that arc because will go across the whole segment and 
will cover the whole angle. Something like that.
May 2020
January 2020