KVPY 2016 SA Question 4

Geometry Level 3

Three circles with radii 1, 2 and 3 touch each other externally in a plane. Find the circumradius of the triangle formed by joining the centers of these 3 circles.


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The answer is 2.5.

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2 solutions

The triangle form by the centers will have sides (1+2,2+3,3+1)=(3,5,4).
It is well known that this is a right angled triangle with 5 as hypotenuse.
Its area =1/2 * 3 * 4=6. So R= abc/(4 * area)=60/24=2.5.

The length of the sides of this triangle formed by joining the centers of the circle,is calculated by the sum of the two radii that are tangent to each other.

In this case,the length of the sides of this triangle are: a = 1 + 2 = 3 a = 1 + 2 = 3 , b = 3 + 2 = 5 b = 3 + 2 = 5 , and c = 3 + 1 = 4 c = 3 + 1 = 4

We notice that the sides of this triangle are the sides of a right triangle, because b 2 = 25 = 16 + 9 = c 2 + a 2 b^2 = 25 = 16 + 9 = c^2 + a^2 and this implies that the circumradius R R of the triangle formed by joining the centers of the circle, is applying sine rule :

b sin 90 º = 5 = 2 R R = 2.5 \frac{b}{\sin 90º} = 5 = 2R \Rightarrow R = 2.5

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