## Greedy Clams – Solution

Here’s a solution to the Greedy Clam problem that was inspired by Luyi’s “Satiated Clam” geometry problem.

In this post, I make frequent use of the area formula $\frac{1}{2}ab \sin \theta$ for a triangle where $\theta$ is the measure of an angle sandwiched between two sides of length $a$ and $b$.

I’ll start with a solution that explains why I wanted to pose this problem in the first place. When I was thinking about Luyi’s problem, I realized that the volume of the polyhedron formed by the convex hull of the two hexagons could be seen as a nifty application of this 3D trapezoid volume formula that came about after a discussion about the area formula for trapezoids I had with some students. What an amusing coincidence!

In the figure at left, the greedy clam polyhedron (in blue) has some of its edges extended. The whole figure is an example of the generalized trapezoid described in the 3D Trapezoid blog post. The greedy clam polyhedron can be seen as this big 3D trapezoid minus two congruent 3D trapezoids (the uncolored tips on either side of the clam). Please note the “hinge,” which is the edge shared by the two hexagons. We’ll refer to this hinge after our first computation for the volume.

The larger 3D trapezoid (the entire object in the figure) has “bases” of length 1, 3, and 3. (If you don’t know what I mean by “bases,” please read 3D Trapezoids.) Its cross-sectional area is equal to $\frac{3}{2}\sin \theta$. Therefore, its volume is $\frac{7}{2}\sin \theta$.

Each of the 3D trapezoids that must be removed to leave the greedy clam has “bases” of length $\frac{h}{2}$, $h$, and $h$, where $h = 2\sqrt{3} \sin \frac{\theta}{2}$. Its cross-sectional area is equal to that of an equilateral triangle of unit side length that has been reduced in height by a factor of $\cos \frac{\theta}{2}$, and hence has area $\frac{\sqrt{3}}{4} \cos \frac{\theta}{2}$. Therefore, its volume is $\frac{5}{4} \sin \frac{\theta}{2} \cos \frac{\theta}{2} = \frac{5}{8}\sin \theta$.

So, the greedy clam has volume $\frac{7}{2} \sin \theta - 2(\frac{5}{8}\sin \theta) = \frac{9}{4} \sin \theta$. Since $\sin \theta$ achieves its maximal value of 1 at 90 degrees, we conclude that the maximum volume occurs when $\theta$ is 90 degrees and that the maximum volume is $\frac{9}{4}$ cubic units.

Even though this settles the question, let’s not stop here! The formula for the volume as a function of $\theta$ turns out to be a sine wave. Such a straightforward answer points to a more general fact which we can find if we think about the volume in another way.

When we computed the volume of the big 3D trapezoid, we had to look at its cross-section by a plane perpendicular to its “bases.” This cross-section is also a cross-section of the greedy clam, and if you slide the intersecting plane along the “bases,” you will get a variety of cross-sections of the greedy clam all of which are isosceles triangles possibly with their tips chopped off.

Let $t$ be the position of the sectioning plane along the line containing the hinge of the greedy clam. In this sectioning plane, the cross-section can be seen as an isosceles triangle (possibly with its tip removed) with an apex angle of $\theta$. Denote by $R(t)$ the length of the two equal sides of the isosceles triangle and by $L(t)$ the length of the two equal sides of the isosceles triangular tip that’s removed.  (If no tip is removed, then $L(t) = 0$.) If you graph $L(t)$ and $R(t)$, you will have drawn a picture of the bottom and top borders of a regular hexagon.

The area of the cross-section is then $\frac{1}{2}R(t)^2 \sin\theta - \frac{1}{2}L(t)^2 \sin \theta = \frac{1}{2}(R(t)^2 - L(t)^2)\sin\theta$. Since the volume of the greedy clam can be obtained by integrating this cross-sectional area along the line containing the hinge, we can see that the volume will be $\sin\theta$ times a constant that is intrinsic to the geometry of the regular hexagon.

But notice that we didn’t actually use the specific geometry of a regular hexagon to come to this last conclusion. We could have created a volume by taking any convex region in the plane, picking a line in the plane which doesn’t intersect the interior of the region as an axis of rotation, rotating the convex region about this axis by $\theta$, and forming the convex hull of the convex region and its image under the rotation. When we compute the volume of the convex hull, we’d find exactly the same form of formula:

$\frac{1}{2}\sin\theta \int (R(t)^2 - L(t)^2)dt$,

where the limits of the integral are set to just capture the desired region.

If you press further, you’ll see that this formula reduces to $CA\sin\theta$, where $C$ is the distance of the center of mass of the convex region from the axis of rotation and $A$ is its area.

In the case of the greedy clam, $A = 6(\frac{\sqrt{3}}{4})$ (6 equilateral triangles with unit side length) and $C = \frac{\sqrt{3}}{2}$ (by symmetry), and the volume does work out to be $\frac{9}{4}\sin\theta$ as we found before.

But with this new volume formula, we can quickly compute the volume of many other shapes. For example, consider a circle of radius $r$. Rotate this circle about a tangent line by 90 degrees to obtain a second circle and take the convex hull of these two circles. What is the volume of the convex hull?

From our formula, we see that it is $r(\pi r^2) = \pi r^3.$

Can you see the volume formula $CA\sin\theta$ without having to perform the integral?