The plane cuts the conic surface in an ellipse with (unknown) semimajor and semiminor axes p, q. We have to calculate the ellipse area $S = \pi p q$ and the distance d of the cone apex A from the plane. Then the volume of the cone part containing the cone apex A will be $V_A = \frac{S d}{3}$. Let O be the center of the cone circular base with radius r, h = OA the cone height, M the midpoint of OA and B the tangency point of the cutting plane with the cone circular base. Put the xyz-coordinate origin at the center O of the cone base, the positive z-axis in the direction of the ray OA and negative y-axis in the direction the ray OB. This makes the x- axis parallel and the yz-plane perpendicular to the cutting plane. The conic surface cuts the yz-plane in 2 cone elements AB, AB', with the points B, B' on the y-axis, and the cutting plane cuts the yz-plane in the line BM. Let the line BM cut the line AB' at a point C. The segment $BC = a = 2p$ is the ellipse major axis and the ellipse focal points $F_1, F_2$ are on this segment. The ellipse focal points $F_1, F_2$ are the tangency points of 2 spheres inscribed in the conic surface with the cutting plane. These 2 spheres cut the yz-plane in the incircle (I) and excircle (E) of the triangle $\triangle ABC$ touching the side BC. Let N be the midpoint of the segment BC (the ellipse center). The distances of the tangency points from the ellipse center are $ep = NF_1 = NF_2$, where e < 1 is the ellipse excentricity. The semininor axis q is then given by Pythagorean theorem for the right angle triangle $\triangle F_1NQ$, where Q, Q' are the intersections of a line in the cutting plane through the ellipse center N perpendicular to the ellipse major axis BC. That's because $QF_1 + QF_2 = 2p$ for a point on the ellipse and $QF_1 = QF_2 = p$ because of symmetry. Hence, $q = \sqrt{QF_1^2 - NF_1^2} = \sqrt{p^2 - e^2p^2}$. Working in the yz-plane, the points A, B, B', M have the yz-coordinates A = (0, h), B = (-r, 0), B' = (r, 0), M = (0, h/2). The equations of the lines BM, AB' intersecting at the point C are $BM: \ \ \ \frac{2z}{h} - \frac{y}{r} = 1$ $AB': \ \ \ \frac{z}{h} + \frac{y}{r} = 1$ Solving, we get the coordinates of the point $C = \left(\frac{r}{3}, \frac{2h}{3}\right)$ and the distance BC $BC = \sqrt{(y_C - y_B)^2 + (z_C - z_B)^2} = \sqrt{\frac{16r^2}{9} + \frac{4h^2}{9}} = \frac 2 3 \sqrt{4r^2 + h^2}$ Thus the ellipse semimajor axis is equal to $p = \frac{BC}{2} = \frac{\sqrt{4r^2 + h^2}}{3}$. Labeling the sides of the triangle $\triangle ABC$ as a = BC, b = CA, c = AB and its semiperimeter s, the distance of the tangency point $F_1$ of the incircle (I) with the side BC from the vertex C is equal to $CF_1 = s - c$ and the distance of this tangency point from the midpoint N of the side BC is equal to $NF_1 = \frac a 2 - (s - c) = \frac{c - b}{2}$. Since $z_C = \frac{2h}{3} = \frac{2z_A}{3}$ and the z-axis bisects the angle $\angle B'AB \equiv \angle CAB$, it is clear that $b = CA = \frac{AB}{3} = \frac{\sqrt{r^2 + h^2}}{3}$ and $NF_1 = \frac{c - b}{2} = \frac{\sqrt{r^2 + h^2}}{3} = b$. The ellipse semiminor axis is then equal to $q = \sqrt{QF_1^2 - NF_1^2} = \sqrt{\frac{a^2}{4} - b^2} = \sqrt{\frac{4r^2 + h^2}{9} - \frac{r^2 + h^2}{9}} = \frac{r}{\sqrt 3}$ and the ellipse area is $S = \pi pq = \frac{\pi r \sqrt{4r^2 + h^2}}{3 \sqrt 3}$. The distance d of the cone apex A from the cutting plane is equal to the A-altitude AD of the triangle $\triangle ABC$. The altitude passes through the point A = (0, h) and has the slope $m_{AD} = -\frac{1}{m_{BM}} = -\frac{2r}{h}$, being perpendicular to the line $BC \equiv BM$. Hence, the altitude equation is $AD: \ \ \ z = -\frac{2r}{h}\ y + h$ Solving the equations of the lines BM, AD, we get the coordinates of their intersection $D = \left(\frac{rh^2}{4r^2 + h^2}, \frac{2r^2h}{4r^2 + h^2} + h\right)$ and the distance d = AD $d = AD = \sqrt{(y_D - y_A)^2 + (z_D - z_A)^2} = \frac{\sqrt{r^2h^4 + 4r^4h^2}}{4r^2 + h^2} = \frac{rh}{\sqrt{4r^2 + h^2}}$ As a result, the volume $V_A$ is equal to $V_A = \frac{S d}{3} = \frac 1 3 \cdot \frac{\pi r \sqrt{4r^2 + h^2}}{3 \sqrt 3} \cdot \frac{rh}{\sqrt{4r^2 + h^2}} = \frac{\pi r^2 h}{9 \sqrt 3}$. On the other hand, the volume of the entire cone is equal to $V = \frac{\pi r^2 h}{3}$, so that $\frac{V_A}{V} = \frac{1}{3\sqrt 3}$. Clearly, $V_A = \frac{V}{3 \sqrt 3} < \frac V 2$ is the smaller volume.

Attachments: