Juggling Points On An Incircle

We will use barycentric coordinates.

Let $A = (1,0,0) , B=(0,1,0) , C = (0, 0, 1)$

Notice that $P$ is the Nagel Point.

It is well-known that the Nagel Point is $(s-a : s-b: s-c)$ , or when normalized, $(\frac{s-a}{s} , \frac{s-b}{s} , \frac{s-c}{s})$

We know that $D_1 = (0: s-c :s-b) = (0, \frac{s-c}{a} , \frac{s-b}{a})$ and $D_2 = (0 : s-b : s-c) = (0 , \frac{s-b}{a} , \frac{s-c}{a} )$

Let $Q_1 = (x_0 , y_0 , z_0)$ be a point on $AD_2$ such that $AQ_1 = PD_2$ .

This implies that

$x_0 + \frac{s-a}{s} = 1 \implies x_0 = \frac{a}{s}$

$y_0 + \frac{s-b}{s} = \frac{s-b}{a} \implies y_0 = \frac{(s-b)(s-a)}{sa}$

$z_0 + \frac{s-c}{s} = \frac{(s-c}{a}\implies y_0 = \frac{(s-c)(s-a)}{sa}$

$Q_1 = (\frac{a}{s} , \frac{(s-b)(s-a)}{sa} , \frac{(s-c)(s-a)}{sa} )$

Now it is well known that $Q$ is diametrically opposite to $D_1$ . (You can prove this easily using Homothety)

Let $Q = (x_1, y_1, z_1)$

It is well known that the incenter $I$ is $(a:b:c)$ or $( \frac{a}{2s} , \frac{b}{2s} , \frac{c}{2s} )$ when normalized. This implies that

$x_1 =2 \cdot \frac{a}{2s} = \frac{a}{s}$

$y_1 + \frac{s-c}{a} = 2 \cdot \frac{b}{2s} = \frac{b}{s}\implies y_1 = \frac{(s-b)(s-a)}{sa}$

$z_1 = \text{ similar computations} = \frac{(s-c)(s-a)}{sa}$

This implies that $Q$ and $Q_1$ coincide and $AQ = AQ_1 = D_2P = \boxed{5}$

Synthetic Solution: Construct the midpoints of $AC,BC$ and denote them $M,N$ . Well known parallels in the configuration gives us homothetic triangles $APB, NIM$ with ratio $2:1$ . Hence $AP=2IN=QD_2=5$