Place Your Tip in the Envelope

Without loss of generality because of rotational symmetry, let $P$ be on the $x$ -axis with coordinates $(p, 0)$ .

Since $X$ is on the circle with equation $x^2 + y^2 = r^2$ , let its coordinates be $(k, \sqrt{r^2 - k^2})$ .

Then the midpoint of $XP$ is $(\frac{p + k}{2}, \frac{\sqrt{r^2-k^2}}{2})$ , the slope of $XP$ is $\frac{\sqrt{r^2-p^2}}{k - p}$ , and the perpendicular bisector of $XP$ is $y = -\frac{p - k}{\sqrt{r^2 - k^2}}(x - \frac{p + k}{2}) + \frac{\sqrt{r^2 - k^2}}{2}$ or $y = \frac{2px - 2kx - p^2 + r^2}{2\sqrt{r^2 - k^2}}$ .

The top part of the ellipse follows the minimum height of each $x$ -value over all $k$ -values, which occurs when $\frac{dy}{dk} = 0$ . Therefore, $\frac{dy}{dk} = \frac{2kpx - kp^2 + kr^2 - 2r^2x}{4(\sqrt{r^2 - k^2})^3} = 0$ , which solves to $k = \frac{2r^2x}{2px - p^2 + r^2}$ .

Substituting $k = \frac{2r^2x}{2px - p^2 + r^2}$ into $y = \frac{2px - 2kx - p^2 + r^2}{2\sqrt{r^2 - k^2}}$ and simplifying gives $y^2 = \frac{(r - 2x + p)(r + 2x - p)(r - p)(r + p)}{4r^2}$ , which can further be simplified to $\frac{(x - \frac{p}{2})^2}{(\frac{r}{2})^2} +\frac{y^2}{(\frac{r}{2})^2 - (\frac{p}{2})^2} = 1$ , an equation of an ellipse in standard form with a major axis of $a = \frac{r}{2}$ and a minor axis of $b = \sqrt{(\frac{r}{2})^2 - (\frac{p}{2})^2} = \frac{1}{2}r\sqrt{1 - (\frac{p}{r})^2}$ .

The ratio of the area of the ellipse to the area of the circle is then $\frac{\pi ab}{\pi r^2} = \frac{\pi \cdot \frac{r}{2} \cdot \frac{1}{2}r\sqrt{1 - (\frac{p}{r})^2}}{\pi r^2} = \frac{1}{4}\sqrt{1 - (\frac{p}{r})^2}$ . Setting this equal to the ratio of areas of $15\%$ gives $\frac{1}{4}\sqrt{1 - (\frac{p}{r})^2} = \frac{15}{100}$ , which solves to $\frac{p}{r} = \frac{4}{5}$ . Therefore, $m = 4$ , $n = 5$ , and $m + n = \boxed{9}$ .