Miscellaneous Problem Solving

$K_4= k_1+k_2+k_3 \pm 2\sqrt{(k_1k_2+k_2k_3+k_3k_1)}$ where $k$ (curvature) $= \dfrac{1}{r}$

$K_4 = \dfrac{23}{6}, r = \dfrac{6}{23} = \dfrac{p}{q}, p + q = 29$

Joining the centres of the three main triangles, we see that we have a 3,4,5 right angled triangle. Let us give it's vertices coordinates as shown. Let us also say that the small circle in the middle has radius r and its centre has coordinates (x,y) Now looking at the lengths of the three lines joining the outer vertices to the centre of the small circle, we have the equations

X^2 + y^2 = ( 1 + r)^2 (1) x^2 + (3 - y)^2 = (2+r)^2 (2) (4 - x)^2 + y^2 = (3+r) ^2 (3)

Subtracting the first one from each of the others gives

              (3-y)^2 -y^2   = (2+r)^2 - (1 +r)^2
              (4-x)^2 -x^2   = (3+ r)^2 -(1+r)^2

which simplify to 3y = 3 -r, 2x = 2-r. Substituting from these back into (1) gives (1- r/2)^2 + (1- r/3)^2 = (1+r)^2 which, upon multiplying by 36 and simplifying, gives 23r^2 + 132r - 36 = 0. Applying the usual formulae, we have

    r = -66 + √5128 ⁄23 = -66 + 72╱23 = 6╱23 = p╱q

(the other root is negative, which is impossible) so, p + q = 6 + 23 = 29