Nātīvitās Geoculus

Extend the sides of the bowl to make a cone, and label the diagram as follows (where the cookie is $DF$ ):

By alternate interior angles, $\alpha = \angle BFG = \angle FEG$ , and by corresponding angles, $\beta = \angle AFB = \angle FIC$ .

By properties of a conic section , the eccentricity is $e = \frac{\cos \alpha}{\cos \beta} = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{24^2}{25^2}} = \frac{7}{25}$ . Therefore, $\cos \alpha = \frac{7}{25}\cos \beta$ .

From $\triangle BFD$ , the height of the frustum is $CG = 50 \cos \alpha = 14 \cos \beta$ .

From $\triangle CDI$ , one radius of the frustum is $CD = ID \sin \beta$ and the height of the cone is $CI = ID \cos \beta$ .

Since $GB = CB - CG = ID \cos \beta - 14 \cos \beta$ , from $\triangle GHI$ , the other radius of the frustum is $GH = (ID - 14) \cos \beta$ .

By the law of sines on $\triangle FDI$ , $ID = \frac{50 \sin(180 - \alpha - \beta)}{\sin 2\beta} = \frac{25\sin \alpha}{\sin \beta} + \frac{25\cos \alpha}{\cos \beta}$ . Substituting $\frac{\cos \alpha}{\cos \beta} = \frac{7}{25}$ from above gives $ID = \frac{25\sin \alpha}{\sin \beta} + 7$ .

The volume of the frustum is $V = \frac{1}{3}\cdot \pi \cdot CG (CD^2 + GH^2 + CD \cdot GH)$ . Substituting the above values and simplifying gives $V_f = \frac{56}{3}\pi(481 \cos \beta - 49 \cos^3 \beta)$ .

Since the derivative of $481 \cos \beta - 49 \cos^3 \beta$ with respect to $\cos \beta$ is $481 - 147\cos^2 \beta$ , and $481 - 147\cos^2 \beta > 0$ for all possible values of $\cos \beta$ , the volume increases as $\cos \beta$ increases, and the maximum volume is when $\cos \beta$ is a maximum at $\cos \beta = 1$ . (In other words, when the frustum is a cylinder.)

The maximum volume is then $V_f = \frac{56}{3}\pi(481 \cdot 1 - 49 \cdot 1^3) = 8064\pi$ , so $\lfloor V \rfloor = \boxed{8064}$ .

Let $a$ denote the semi-major and $b$ the semi-minor. Consider the isosceles trapezoid, the cross-section orthogonal to the circular bases $r_{\text{top}}$ and $r_{\text{bottom}}$ , and the angle $\alpha$ formed by the diagonal and one of the circular bases. Since $b = \sqrt{r_{\text{top}}r_{\text{bottom}}}$ for the cross section to exist within the right-angle frustum (left that as the Sangaku exercise), condition $r_{\text{top}} \leq r_{\text{bottom}}$ . Then, since the minor axis of the ellipse exists as the chord of the circular cross section of the radius $\dfrac{1}{2}(r_{\text{top}} + r_{\text{bottom}})$ , by the Pythagorean Theorem ,

$\begin{aligned} s&= \sqrt{\left(\dfrac{1}{2}(r_{\text{bottom}} + r_{\text{top}})\right)^2 - \left(\sqrt{r_{\text{top}}r_{\text{bottom}}}\right)^2} \\ &= \sqrt{\dfrac{1}{4}(r_{\text{top}})^2 + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{1}{2}r_{\text{top}}r_{\text{bottom}}} \end{aligned}$

which shows that

$\begin{aligned} \text{Diameter of the bottom base} &= 2r_{\text{bottom}} &= 2a\cos\alpha + 2\sqrt{\dfrac{1}{4}(r_{\text{top}})^2 + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{1}{2}r_{\text{top}}r_{\text{bottom}}}\\ \text{Diameter of the top base} &= 2r_{\text{top}} &= 2a\cos\alpha - 2\sqrt{\dfrac{1}{4}(r_{\text{top}})^2 + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{1}{2}r_{\text{top}}r_{\text{bottom}}}\\ \end{aligned}$

Using the following volume formula of a frustum,

$\text{Volume} = \dfrac{1}{3}h \left(A_1 + A_2 + \sqrt{A_1A_2}\right)$

where

• $h = 2a\sin\alpha$ denotes the height of the trapezoid
• $A_1 = \pi (r_{\text{top}})^2$ denotes the area of the top circular base
• $A_2 = \pi (r_{\text{bottom}})^2$ denotes the area of the bottom circular base

using what we already know about the frustum and the given cross section, we obtain

$\begin{aligned} \text{Volume} &= \dfrac{2\pi}{3}a\sin\alpha \left(3a^2\cos^2\alpha + \dfrac{1}{4}(r_{\text{top}})^2 + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{1}{2}r_{\text{top}}r_{\text{bottom}}\right)\\ &= \dfrac{2\pi}{3}a\sin\alpha \left(3a^2\cos^2\alpha + \dfrac{b^4}{4(r_{\text{bottom}})^2} + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{b^2}{2}\right)\\ &= \dfrac{2\pi}{3}\sqrt{a^2 - \left(r_{\text{bottom}} - \sqrt{\dfrac{b^4}{4(r_{\text{bottom}})^2} + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{b^2}{2}}\right)^2}\left( 3\left(r_{\text{bottom}} - \sqrt{\dfrac{b^4}{4(r_{\text{bottom}})^2} + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{b^2}{2}}\right)^2 + \dfrac{b^4}{4(r_{\text{bottom}})^2} + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{b^2}{2}\right) \end{aligned}$

where $a^2 \geq \left(\dfrac{b^4}{4(r_{\text{bottom}})^2} + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{b^2}{2}\right)^2$ . But that suggests the frustum "vanishes" at the positive upper bound $r_{\text{bottom}} = a + \sqrt{a^2 - b^2}$ , following $b = \sqrt{r_{\text{bottom}}(2a - r_{\text{bottom}})}$ (illustrating the ellipse of the fixed axes overlaps both circular bases).

As $a^2 - \left(r_{\text{bottom}} - \sqrt{\dfrac{b^4}{4(r_{\text{bottom}})^2} + \dfrac{1}{4}(r_{\text{bottom}})^2 - \dfrac{b^2}{2}}\right)^2 = 0$

for $r_{\text{bottom}} = a + \sqrt{a^2 - b^2}$ , the height is decreasing and continuous at $b \leq r_{\text{bottom}} \leq a + \sqrt{a^2 - b^2}$ , involving AM-GM Inequality . With some algebraic and calculus calculation (as the derivative of $V$ is nonpositive at that interval), it achieves its maximum at $r_{\text{bottom}} = b$ , yielding a cylinder of the volume $V_{\text{max}} = 2\pi b^2 \sqrt{a^2 - b^2}$ for $a = 25$ and $b = 24$ .

In this case, $V_{\text{max}} = 8064\pi$ , where $\lfloor V \rfloor = 8064$ .

---

The graph above illustrates the idea of the volume equation in the Cartesian plane. While it has maxima at $r_{\text{bottom}} < b$ , they do not satisfy the condition $b = \sqrt{r_{\text{top}}r_{\text{bottom}}}$ needed to solve the problem.

Last, but not least, $\color{#D61F06}\text{Merry Christmas!}$