Expert: Vishwas K Hajirnis Date: 11/6/2004 Subject: Circles in a triangle
Question An equalateral triangle has three circles inscribed within it. The circles have a radius of 1 and are tangent to each other and to the sides of the triangle. What is the perimeter of the triangle. How did you solve the problem? Thanks
Answer Let ABC be the given triangle.
Let D,E & F be the centers of the 3 triangles,
such that the circle with center at D touches
the sides AB & AC; the circle with center at E
touches the sides AB & BC; the circle with center at F
touches the sides BC & AC. Let us draw perpendiculars
from E & F on the side BC. Let these be EE' & FF'.
EE'=FF'=1 (=given radius).
E'F'=1+1=2 (EE'F'F is a rectangle)
Now length BC = BE'+E'F'+F'C
= BE'+2+F'C ...(I)
Triangles EBE' and FCF' are congruent
(Easy to prove:
Both triangles are right angled,
EE'=FF'
m (< EBE')= m(<FCF')
Since both EB and FC are bisectors of
<ABC and <ACB. But m(<ABC) = m(<ACB),
as they are angles of an equlateral triangle.)
So from (I), we get
length BC= 2*BE'+2 ...(II)
( '*' denotes multiplication. )
Now consider the triangle EE'B.
This is a 30-60-90 triangle.
With m(<EBE')=30
m(<BE'E)=90 and m(<BEE')=60
(There is theorem about 30-60-90 triangles that says
that the side opposite to 30 degrees is 1/2 the
hypotenuse and the side opposite to 60 degrees is
square root(3)/2 times the hypotenuse.)
From this we get the ratio of lengths of side opposite
to 30 degrees divided by side opposite 60 degrees
= 1 /(square root of 3)
So in triangle EBE'
EE' / BE' = 1 /[1/(square root of 3)]
= (square root of 3)
Therefore
BE' = EE'/ (square root of 3)
= 1 / (square root of 3) ...(III)
Substituting this value in (II),
we get
BC = 2/(square root of 3)+2 ...(IV)
So the perimeter of the triangle must be
= 3*[2/(square root of 3)+2]
= 9.464102
