An Old Problem

The incircle of a triangle with sides 3, 4, 5 has a radius 1.

The two circles are located as shown. The distance between their centers is $\sqrt{2^1+1^2}=\sqrt{5}\approx2.236$ .

The fact that $R=1$ can be derived from $A=R\times s$ , that is area of triangle is equal to radius of incircle times half-perimeter.

In this case $A=\frac{1}{2}\times 3 \times 4=6$ .

$s=\frac{1}{2}\times (3+4+5)=\frac{12}{2}=6$

$R=\frac{6}{6}=1$

The diagram is definitely not to scale! The circles are NOT tangent to each other.

The angle DBC is arctan(4/3) = 53.13 degrees. Let the tangent point on BC be X. Let the tangent point of that circle on BD be Y.

Since two tangents to a circle that intersect have congruent segment lengths then KXB is congruent to KYB by SSS. Therefore angle KBX

= 26.57 degrees. Let radius of circle = r.

Tan (26.57) = KC/CB = r/(3-r) = 0.5 This leads to r = 1 and diameter of each circle = 2.

Since diameters = 2 then the vertical distance from J to K is 2. The circles must overlap by 1 so that the horizontal distance from J to K is 1.

The slope distance ^2 = 2^2 + 1^2 = 5. Therefore JK = sqrt(5) = 2.236