Heronian Triangle Dissection B

Per the diagram below, let the triangle be $ABC$ , with $AB=195$ , $BC=225$ , $CA=210$ , and let the point the blue line meets $BC$ be $D$ .

Further, let the tangency point of the incircle of $\Delta ABD$ on $BD$ be $E$ and the tangency point of the incircle of $\Delta ACD$ on $CD$ be $F$ .

Now, the distances to the tangency points along $AD$ are equal to $DE$ and $DF$ (using this property) - that is, $DE-DF=64$ .

We have $DE=\frac12 (AD+BD-AB)$ and $DF=\frac12 (AD+CD-CA)$ .

Subtracting, we get $DE-DF=\frac12 (BD-CD+CA-AB)$ .

Substituting in, this is $64=\frac12 (BD-CD+210-195)$ , so that $BD-CD=113$ .

Also, $BD+CD=BC=225$ ; so $BD=169$ and $CD=56$ .

Using the cosine rule in triangles $\Delta ABC$ and $\Delta ABD$ , we find

$\cos B=\frac{AB^2+BC^2-CA^2}{2AB\cdot BC}=\frac{AB^2+BD^2-AD^2}{2AB\cdot BD}$

Solving, we find $AD=\boxed{182}$ . A nice bonus fact is that both $\Delta ABD$ and $\Delta ACD$ are also Heronian.

Let the $225$ side be split into lengths of $x$ and $225-x$ , and let the length of the blue line be $s$ . Then the distance from the top vertex to the meeting point of the blue line with the left incircle is $\tfrac12(195 + s - x)$ , while the distance from the top vertex to the meeting point of the blue line with the right incircle is $\tfrac12(210 + s - (225-x)) = \tfrac12(s + x - 15)$ , so that $64 \; = \; \tfrac12(s+x-15)-\tfrac12(195+s-x) \; = \; x - 105$ and hence $x = 169$ . The Cosine Rule gives the left-hand angle as having cosine $\tfrac{33}{65}$ , and hence $s = \boxed{182}$ .

Label the figure as shown. Let $\angle DAC = \alpha$ and $\angle BAD = \beta$ , the radius of the big and small circles be $R$ and $r$ respectively. We note that:

$\begin{cases} \dfrac r{\tan \frac \alpha 2} - \dfrac R{\tan \frac \beta 2} = 64 & & ...(1) \\ \dfrac r{\tan \frac \alpha 2} + \dfrac r{\tan \frac C 2} = 210 & \implies r = \dfrac {210 \tan \frac \alpha 2 \tan \frac C 2}{\tan \frac \alpha 2 + \tan \frac C 2} & ...(2) \\ \dfrac R{\tan \frac \beta 2} + \dfrac R{\tan \frac B 2} = 195 & \implies R = \dfrac {195 \tan \frac \beta 2 \tan \frac B 2}{\tan \frac \beta 2 + \tan \frac B 2} & ...(3) \end{cases}$

Then $(1)$ becomes $\dfrac {210 \tan \frac C 2}{\tan \frac \alpha 2 + \tan \frac C 2} - \dfrac {195 \tan \frac B 2}{\tan \frac \beta 2 + \tan \frac B 2} = 64 \quad ...(1a)$ .

Using cosine rule : $195^2 = 210^2 + 225^2 - 2\times 210 \times 225 \cos C$ , we get $\cos C = \dfrac 35$ . Using half-angle tangent substitution , $\dfrac {1-\tan^2 \frac C2} {1+\tan^2 \frac C2} = \dfrac 35 \implies \tan \dfrac C2 = \dfrac 12$ . Similarly, we get $\tan \dfrac B 2 = \dfrac 47$ and $\tan \dfrac A2 = \dfrac 23$ . Since $\dfrac \alpha 2 + \dfrac \beta 2 = \dfrac A 2$ , $\implies \tan \dfrac \beta 2 = \dfrac {\frac 23 - \tan \frac \alpha 2}{1+\frac 23 \tan \frac \alpha 2} = \dfrac {2-3\tan \frac \alpha 2}{3+2\tan \frac \alpha 2}$ . Let $t = \tan \dfrac \alpha 2$ ; then equation $(1a)$ becomes:

$\begin{aligned} \frac {210}{2t+1} - \frac {60(3+2t)}{2-t} & = 64 & \small \blue{\text{Rearrange}} \\ 56 t^2 + 441 - 56 & = 0 \\ (8t-1)(7t +56) & = 0 & \small \blue{\text{As }t = \tan \frac \alpha 2 > 0} \\ \implies \tan \frac \alpha 2 & = \frac 18 \end{aligned}$

By sine rule , the length of the blue line $AD$ is given by:

$\begin{aligned} \frac {AD}{\sin C} & = \frac {210}{\blue{\sin (180^\circ - \alpha - C)}} & \small \blue{\text{Note that }\sin (180^\circ - \theta) = \sin \theta} \\ \implies AD & = \frac {210 \sin C}{\blue{\sin (\alpha + C)}} & \small \blue{\text{and }\sin \alpha = \frac {2t}{1+t^2} = \frac {16}{65}} \\ & = \frac {210 \times \frac 45}{\frac {16}{65}\times \frac 35 + \frac {63}{65}\times \frac 45} \\ & = \boxed{182} \end{aligned}$

A beautiful solution of this problem was given by Chris Lewis . The only change is in the distance between the two tangent points. Following him, we get the final equation connecting the length of the blue line $l$ as $225l^2=225(195^2+169^2)-169(195^2+225^2-210^2)=7452900$ , or $l=\sqrt {\dfrac{7452900}{225}}=\boxed {182}$ .