The Average Radius

The problem does not specify any particular ellipse. Hence, without loss of originality we can consider a standard ellipse with the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a=5$ and $b=4$ (As mentioned).

Since we have the lengths of the semi-major and semi-minor axes we can easily find the eccentricity of the ellipse which is given by the formula $\displaystyle e = \sqrt{1 - \frac{b^2}{a^2}}$ , implicitly assuming that $b < a$ which is anyway suggested by our assumed equation. Substituting the values of $a$ and $b$ we get, $\displaystyle e = \frac{3}{5}$

Now, having assumed the ellipse , we automatically fix the foci $S$ and $S'$ . The foci are at a distance of $ae$ from the centre of the ellipse along the major axis hence their coordinates are given by $S(ae,0)$ and $S'(-ae,0)$ . Therefore, the distance $SS' = 2ae$ .

Coming to point $P$ . SInce it is a random point on the ellipse, we can assign it some parametric coordinates. If the line segment joining the point $P$ and the origin makes an angle $\theta$ in counter clockwise sense from the positive x-axis, then coordinates of point $P$ can be represented in terms of parameter $\theta$ as $P( a\cos \theta , b\sin \theta )$ .

Now for the triangle $\Delta PSS'$ . Let us denote its area by $\Delta$ , semi-perimeter by $s$ and the radius of the incircle by $r$ .The aforementioned attributes of a triangle are related by the formula $r = \frac{\Delta}{s}$ Hence to find $r$ we will rather concentrate on finding the other two attributes.

For area, let us look back to the basics.
$\Delta = \frac{1}{2} \times base \times height$
Taking $SS'$ as the base , it can be easily observed that the height of the triangle would the perpendicular distance of the point $P$ from the x-axis, which is actually its y-coordinate ! Since we are interested only in the magnitude, we will use the absolute value of the y-coordinate. Hence,
$\Delta = \frac{1}{2} \times SS' \times |y_{P}| = \frac{1}{2} \times 2ae \times |b\sin \theta|$

$\Rightarrow \Delta = abe |\sin \theta|$

Now, by definition, $\displaystyle s = \frac{SS' + PS + PS'}{2}$ We have the value SS'. But for the distances $PS$ and $PS'$ we can apply the distance formula. Since, we are only interested in their sum it better for us to recall the definition of an ellipse. It states that an ellipse is the locus of a point whose sum of distances from two fixed points is constant and the fixed points are the foci of the ellipse and the sum of distances is equal to the length of major axis of the ellipse. Hence $PS + PS' = 2a$ .Therefore,
$s = \frac{2ae + 2a}{2} = a(1+e)$

$\displaystyle \Rightarrow r = \frac{abe |\sin \theta|}{a(1+e)}$

Therefore, $< r > = \frac{be}{1+e} \times \frac{\int\limits_0^{2\pi} |\sin \theta| d\theta}{\int\limits_0^{2\pi} d\theta}$ $\Rightarrow < r > = \frac{be}{1+e} \times \frac{4}{2\pi}$ Substitute the values of $b$ and $e$ to obtain $\boxed{< r > = \frac{3}{\pi} = 0.954929}$