Pentagon interior point

With $R$ being the circumradius of the pentagon, the 5 vertices (from A to E) are given by,

$\mathbf{p}_k = R \mathbf{u}_k \hspace{12pt} \text{ for } k = 1,...,5 \hspace{12pt} (1)$

where $\mathbf{u}_k = ( - \sin \dfrac{2\pi(k-1)}{5}, \cos \dfrac{2\pi(k-1)}{5}) \hspace{12pt} (2)$

If we now define $\mathbf{p} = (x, y)$ to be the coordinate vector of the interior point, then, for $i = 1 , 2, 3$ , we have

$(\mathbf{p} - \mathbf{p}_i)^T ( \mathbf{p} - \mathbf{p}_i) = {d_i}^2 \hspace{12pt} (3)$

where $d_1 = 16 , d_2 = 19, d_3 = 17$ .

Expanding the left hand side of equation (3), gives,

$\mathbf{p}^T \mathbf{p} - 2 \mathbf{p}^T \mathbf{p}_i + {\mathbf{p}_i}^T \mathbf{p}_i = {d_i}^2 \hspace{12pt} (4)$

Substituting equation (1) into (4), yields, for $i = 1, 2, 3$ ,

$\mathbf{p}^T \mathbf{p} - 2 R \mathbf{p}^T \mathbf{u}_i + R^2 = {d_i}^2 \hspace{12pt} (5)$

So that,

$\mathbf{p}^T \mathbf{p} - 2 R \mathbf{p}^T \mathbf{u}_1 + R^2 = {d_1}^2 \hspace{12pt} (6)$

$\mathbf{p}^T \mathbf{p} - 2 R \mathbf{p}^T \mathbf{u}_2 + R^2 = {d_2}^2 \hspace{12pt} (7)$

$\mathbf{p}^T \mathbf{p} - 2 R \mathbf{p}^T \mathbf{u}_3 + R^2 = {d_3}^2 \hspace{12pt} (8)$

subtracting (7) from (6), and (8) from (6),

$- 2 R \mathbf{p}^T (\mathbf{u}_1 - \mathbf{u}_2 ) = {d_1}^2 - {d_2}^2 \hspace{12pt} (9)$

$- 2 R \mathbf{p}^T (\mathbf{u}_1 - \mathbf{u}_3 ) = {d_1}^2 - {d_3}^2 \hspace{12pt} (10)$

Dividing (9) by (10), eliminates $R$ , and yields, after cross multiplication,

$\mathbf{p}^T \mathbf{w} = 0 \hspace{12pt} (11)$

where,

$\mathbf{w} = ({d_1}^2 - {d_3}^2 ) (\mathbf{u}_1 - \mathbf{u}_2) - ({d_1}^2 - {d_2}^2) (\mathbf{u}_1 - \mathbf{u}_3) \hspace{12pt}(12)$

Since the vector $\mathbf{w} = (w_x, w_y)$ is available, equation (11), can be solved directly, to yield,

$\mathbf{p} = t \mathbf{v} \hspace{12pt} (13)$

where $\mathbf{v} = (-w_y, w_x) \hspace{12pt} (14)$

and $t$ is a real scalar parameter to be determined. Substituting equation (13) into (9), yields,

$-2 t R \mathbf{v}^T (\mathbf{u}_1 - \mathbf{u}_2) = {d_1}^2 - {d_2}^2 \hspace{12pt} (15)$

from which,

$t R = C_0$ , where $C_0 = \dfrac{({d_1}^2 - {d_2}^2)}{(-2 \mathbf{v}^T (\mathbf{u}_1 - \mathbf{u}_2) )} \hspace{12pt} (16)$

Note that, since $R$ is positive, $t$ has the same sign as $C_0$ .

Substituting equation (13) into equation (6),

$t^2 (\mathbf{v}^T \mathbf{v}) - 2 t R \mathbf{v}^T \mathbf{u}_1 + R^2 = {d_1}^2 \hspace{12pt} (17)$

using equation (16), this becomes,

$t^2 (\mathbf{v}^T \mathbf{v}) - 2 C_0 \mathbf{v}^T \mathbf{u}_1 + \dfrac{{C_0}^2}{t^2} = {d_1}^2 \hspace{12pt} (18)$

and this can solved for $t$ quite easily (a quadratic equation in $t^2$ ), so that, from equation (13), this in turn specifies the point $\mathbf{p}$ ,

and from equation (16), this also specifies the circumradius $R$ .

The reader may note that, in general, there will be two solutions, but for the distances in this problem, the point $\mathbf{p}$ for the other solution lies outside the pentagon, and is therefore considered an extraneous solution.

Let $s$ be the side of the pentagon. Then by the law of cosines on $\triangle PBA$ , $\angle PBA = \cos^{-1} \Big( \frac{s^2 + 19^2 - 16^2}{2 \cdot 19 \cdot 16} \Big)$ , and on $\triangle PBC$ , $\angle PBC = \cos^{-1} \Big( \frac{s^2 + 19^2 - 17^2}{2 \cdot 19 \cdot 17} \Big)$ . Since the interior angle of a regular pentagon is $108°$ , $\angle PBA + \angle PBC = 108°$ . These equations solve numerically to $s \approx 5.201157564$ or $s \approx 17.17083836$ . However, if $s \approx 5.201157564$ then point $P$ is outside the pentagon, so $s \approx 17.17083836$ . The circumradius is then $R = \frac{s}{2 \sin 36°} \approx 14.60638753$ .

$A$ 's coordinates are $A(0, R)$ , and $B$ 's coordinates are $B(R \cos 162°, R \sin 162°)$ . Since $PA = 16$ , $x^2 + (y - R)^2 = 16^2$ , and since $PB = 19$ , $(x - R \cos 162°)^2 + (y - R \sin 162°)^2 = 19^2$ . When $R \approx 14.60638753$ , these equations solve to $x \approx 4.35327$ and $y \approx -0.79001$ , and $\lfloor 1000(x + y + R) \rfloor = \boxed{18169}$ .