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 Generalizing (Posted on 2015-10-16)
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.

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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,

|AI||DC| = |ED||IF| = (2R)(r).     (1)

From the Intersecting Chords theorem,

|AI||DI| = |PI||QI|
= (|PO|-|IO|)(|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

|AI||DI| = 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 2015-10-17 19:19:44

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