If a triangle has two particular circles as its circumcircle and incircle, there exist an infinite number of other triangles with the same circumcircle and incircle.
Find a necessary and sufficient condition for such triangles to exist.
A circle can circumscribe a triangle and a
second circle can be inscribed in the same
triangle if and only if d^2 = R^2  2Rr.
Here R is the radius of the larger circle,
r the radius of the smaller, and d is the
distance between the centers.
In the following proof we will use P(a) to
denote a circle with center P and radius a.
ONLY IF Part
Let ABC be an arbitrary triangle with O(R)
as its circumcircle, I(r) its incircle,
and d = IO. See construction below.
AFI and ECD are similar right triangles
since /FAI = /CED. Therefore,
AIDC = EDIF = (2R)(r). (1)
From the Intersecting Chords theorem,
AIDI = PIQI
= (POIO)(QO+IO)
= (R  d)(R + d)
= R^2  d^2 (2)
/DIC is an external angle of triangle ACI.
Therefore,
/DIC = /IAC + /ICA
= /DAC + /ICB
= /DAB + /BCI
= /DCB + /BCI
= /DCI
Therefore, DC = DI (3)
Equations (1)(3) imply d^2 = R^2  2Rr.
CONSTRUCTION
Let the line AI intersect O(R) again at D.
Let the line DO intersect O(R) again at E.
Let the line IO intersect O(R) at points
P and Q. (If I=O, the we have Charlie's
equilateral triangle with R = 2r).
Let F be the foot of the perpendicular
from I to side AB.
Connect point C to points D, E, and I with
line segments.
IF Part
Let O(R) and I(r) be our two circles with
d^2 = R^2  2Rr. From the equation it is
clear that I(r) lies inside O(R). Let A be
an arbitrary point on O(R) and construct
the two lines through A tangent to I(r).
These two lines intersect O(R) in points
B and C. We need to prove that BC is
tangent to I(r). We will do this by
proving that /ACI = /BCI. See
construction above.
From d^2 = R^2  2Rr and equation (2) we get
AIDI = 2Rr. (4)
Equations (1) and (4) give
DC = DI ==> /DIC = /DCI ==>
/IAC + /ICA = /DCB + /BCI ==>
/DAC + /ACI = /DAB + /BCI ==>
/ACI = /BCI.
Note that point A being arbitrary implies
that there are an infinite number of
triangles that satisfy the requirement.
QED
Edited on October 17, 2015, 7:22 pm

Posted by Bractals
on 20151017 19:19:44 