Soddy Circles

If the lengths of the sides of $ABC$ are $a,b,c$ as usual, then the nonnormalised barycentric coordinates of the centres $D,H$ of the two Soddy circles are $\hat{D} = \left(a + \tfrac{\Delta}{s-a},b + \tfrac{\Delta}{s-b},c + \tfrac{\Delta}{s-c}\right) \hspace{2cm} \hat{H} = \left(a - \tfrac{\Delta}{s-a},b - \tfrac{\Delta}{s-b},c - \tfrac{\Delta}{s-c}\right)$ respectively, where $s = \tfrac12(a+b+c)$ is the semiperimeter and $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ is the area of $ABC$ .

If we normalise the barycentric coordinates of each point, so that the coefficients sum to $1$ , obtaining $P = \tfrac{1}{u+v+w}\big(u,v,w\big) \hspace{1cm} \mathrm{where} \hspace{1cm} \hat{D} = \big(u,v,w\big)$ and similarly obtaining $Q$ from $\hat{H}$ , and if we calculate $P-Q = \big(x,y,z\big)$ , then the distance $DH$ between the two Soddy circles is given by the formula $DH^2 \; = \; -a^2yz - b^2xz - c^2xy$ In this case we have $a = 6+7=13$ , $b=6+8=14$ , $c=7+8=15$ , so that $s = 21$ , $\Delta = 84$ and hence $DH \; = \; \frac{672\sqrt{37}}{1727}$ making the answer $672 + 37 + 1727 = \boxed{2436}$ .

1. Find radius for two Soddy circles - $\frac{168}{157}, \frac{168}{11}$ .

   SoddyCircles

   R m SoddyCircle

   R M SoddyCircle

2. If $A(0,0)$ and $C(13,0)$ - centers of circles. Find points H and D - SoddyCircles Centers.

   H - OuterSoddyCenter

   D - InnerSoddyCenter

3. $DH=\sqrt{(\frac{12078}{2041}-\frac{1026}{143})^2+(\frac{7896}{2041}-\frac{840}{143})^2}= \frac {672 \sqrt{37}}{1727}$

   Answer $2436$ .