Kenmotu point

Consider a point $Q$ in the triangle $ABC$ with actual distance trilinear coordinates $(\alpha,\beta,\gamma)$ , so that the distance from $Q$ to $BC$ is $\alpha$ , the distance from $Q$ to $AC$ is $\beta$ , and the distance from $Q$ to $AB$ is $\gamma$ . Note that $a\alpha + b\beta + c\gamma \;= \; 2\Delta \hspace{1cm} (\star)$ where $\Delta$ is the area of $ABC$ . Setting up coordinate axes as shown, $Q$ has coordinates $(\xi,\gamma)$ for some $\xi$ , and it is clear that $\beta \; = \; \binom{\xi}{\gamma} \cdot \binom{\sin A}{-\cos A} \; =\; \xi \sin A - \gamma \cos A$ Thus the distance $AQ$ is such that $\begin{aligned} AQ^2\sin^2A & = \; (\xi^2 + \gamma^2)\sin^2A \; =\; (\beta + \gamma \cos A)^2 + \gamma^2\sin^2A \; = \; \beta^2 + \gamma^2 + 2\beta\gamma\cos A \\ AQ & = \; \frac{\sqrt{\beta^2 + \gamma^2 + 2\beta\gamma\cos A}}{\sin A} \end{aligned}$ The first Kenmotu point is Kimberling centre $X_{371}$ , and has triangle centre function $\alpha_{371} = \cos A + \sin A$ . This means that, using the above notation, $\alpha \; =\; \lambda(\cos A + \sin A) \hspace{1cm} \beta \; = \;\lambda(\sin B + \cos B) \hspace{1cm} \gamma \; = \; \lambda(\sin C + \cos C)$ where $\lambda$ is chosen to make $(\star)$ hold.

For this triangle, we have $a = 20$ , $b = 13$ , $c = 21$ , $\cos A = \tfrac{5}{13}$ , $\sin A = \tfrac{12}{13}$ , $\cos B = \tfrac45$ , $\sin B = \tfrac35$ , $\cos C = \tfrac{16}{65}$ , $\sin C = \tfrac{63}{65}$ and $\Delta = 126$ . These can all be determined directly using the Cosine Rule for $ABC$ , or else by spotting that $ABC$ is $(5,12,13)$ right-angled triangle and a $(12,16,20)$ right-angled triangle stuck together. Thus $\alpha \; = \; \tfrac{17}{13}\lambda \hspace{1cm} \beta \; = \; \tfrac75\lambda \hspace{1cm} \gamma \; = \; \tfrac{79}{65}\lambda$ and $(\star)$ tells us that $\lambda = \tfrac{2730}{757}$ , so that $\alpha \; = \; \tfrac{3570}{757} \hspace{1cm} \beta \; = \; \tfrac{3822}{757} \hspace{1cm} \gamma \; = \; \tfrac{3318}{757}$ Plugging these values into the original formula gives $AQ \; = \; \frac{273\sqrt{557}}{757} \; = \; \boxed{8.51127}$