A circle is inscribed in an equilateral triangle. Smaller circles
touch it and the sides of the triangle, the process continuing
indefinitely. What is the sum of the areas of all the circles?
One side of a triangle is divided into segments of length a and b
by the inscribed circle, with radius r. Prove that the area is:
The square ABCD is split into three triangles by the lines BP and
CP. Find the radii of the three inscribed circles to these
triangles as P moves on AD.
Published December 2001,July 2001,December 2011,February 2011.
The largest circle which fits inside a triangle just touching
the three sides of the triangle is called the inscribed circle or
incircle. This article is about triangles in which the lengths of
the sides and the radii of the inscribed circles are all whole
numbers. Following on from the problem Incircles (February 2000)
about right angled triangles we now find similar results for
Before you read on, can you find the radius of the inscribed
circle of the triangle with sides of lengths 3, 4 and 5 units and
then find the radius of the inscribed circle of the triangle with
sides of lengths 5, 12 and 13 units? These numbers are Pythagorean
triples, the triangles are right angled, the inscribed circle of
the first has radius 1 unit and the second has radius 2 units. So
can we find a right angled triangle with incircle of radius 3 units
(or any other whole number) whose sides are a primitive Pythagorean
triple? You'll find the answer to this question here .
The solution to the 'Incircles' problem shows that, for any
circle whose radius is a whole number k, we are guaranteed at least
one right angled triangle containing this circle as its inscribed
circle where the lengths of the sides of the triangle are the a
primitive Pythagorean triple:
It is certainly possible to construct triangles with sides a, b
and c which give integer value to the incircle radius, but which
are not a Pythagorean triple. One such is the isosceles triangle
with sides 10, 10 and 12.
It is formed by putting two triangles back to back whose sides
are given by the Pythagorean triple 6, 8, 10. As in the solution to
the original problem, the radius r of the incircle is found by
splitting the triangle into three and finding its area
Another such triangle is the 39, 39, 30 triangle which is formed
from two 39, 15, 36 triangles (a 5, 12, 13 triangle enlarged by a
factor of 3). The inradius in this case is 10.
For a general case, take two triangles 'back to back' with sides
given by the primitive Pythagorean triple with m=x+1, n=x.
The radius of the incircle is given by:
As x, x+1 and 2x+1 are all relatively prime, if we enlarge the
triangle by a factor x+1 taking m=(x+1)(x+1) and n=x(x+1) then the
radius of the incircle of the new triangle will be the whole
As an example, for x=1, m=4, n=2, the isosceles triangle has
sides of lengths 20, 20 and 24 (made up of two right angled
triangles with sides of length 20, 16 and 12) and the inscribed
circle has radius 6 units.
For another example take x=2, m= 9, n=6, then the isosceles
triangle has sides of lengths 117, 117 and 90 (made up of two right
angled triangles with sides of length 117, 108 and 45) and the
inscribed circle has radius 30 units.
What happens for scalene triangles?