Ants marching

The first thing we need to notice is that the ants cannot crawl on most of the given shapes if they're to follow the rules.

If you start walking on a straight line without turning on the surface of a sphere, the path is necessarily a diameter of said sphere (think of yourself walking in a straight line around the planet, you would end where you started). But if 2 ants were to do that, then their paths would have to cross. By the rule 2, this cannot happen and hence they aren't crawling on a sphere.

Next thing we need to notice is the length $L$ of their paths, which is the same for all the ants, and is equal to $L = \left(0.25\frac{\text{cm}}{\text{s}} \right) \cdot \left( 20\text{ s}\right) = 5 \text{ cm.}$

Consider the cylinder now. The only way we can arrange to place 100 ants crawling around such shapes the same length and trying to maximize the volume they surround, is to have them crawling in paths that are parallel to each other. But this in turns will come into contradiction with rule 3. Thing about the leftmost ant of the cylinder, when she turns to her left, she will not see another ant, ergo, the ants can't be crawling on the cylinder.

This same argument works for the rectangular prism.

The ants crawl on a torus!

If we have them crawling surrounding the tube of the torus, their paths can follow all 4 rules without problem. The separations of the ants varies, but as long as its separation at the outer side of the torus is 4 mm, we're good.

The only thing left to do is to determine the maximum volume a torus with tube of circumference equal to $L$ , and outer circumference equals to $100\times 4 \text{ mm} = 40 \text { cm}$ . The volume of a torus is given by $V = \left( \pi r^2 \right) \left( 2 \pi R \right)$ , where $R$ is the distance from the center of the tube to the center of the torus, $r$ is the radius of the tube.

It's easy to see that $L = 2 \pi r$ , therefore $r = \frac{L}{2\pi r} = \frac{5\text{cm}}{2 \pi}$ . Also, we notice that $40 \text{ cm} = 2 \pi \left(R+r \right)$ which leads to $R = \frac{35\text{cm}}{2 \pi} .$

Plugging these into the volume formula, we obtain $V= \left( \pi \left[ \frac{35\text{cm}}{2 \pi} \right]^2 \right) \left( 2 \pi \left[ \frac{35\text{cm}}{2 \pi} \right] \right),$ which simplifying it, leads to $V = \frac{5^2 \cdot 35}{4 \pi} \text{cm}^3 = 69.630 \text{ cm}^3.$

This question is rather unconventional. It consists of two parts, first, identify the type of solid the ants are crawling on, second, given the type of solid and the constraints, maximize the volume of the solid.

Property 1, in other words, means that a geodesic, the path traced out by the ant, exists on the surface. This is true for all 4 surfaces.

Property 2 and 4 implies that there must be geodesics on the surface which are "parallel". This is impossible on a sphere, as two geodesics (great circles in this case), will always cross each other.

Property 3 states that there must be at least one ant on the left side and at least one ant on the right side. At first glance, it seems that we must have an infinite row of ants marching on, say for example, the curled side of a cylinder. However, since we only have 100 ants, there must be an alternative.

If we twist the cylinder such that the two flat ends join together (to form a torus), it can satisfy the condition that the paths never cross and the condition that each ant would have one ant on the left and one on the right.

So, we have found the ants are crawling on the doughnut.

The torus is a solid of revolution of a circle. Let the major radius and the minor radius be "R and r" respectively. By the Pappus's centroid theorem, the volume of the torus is given by $V = 2 \pi^2 Rr^2$ . The length $L$ of the path is $L = 2\pi r = Tv$ , where $T$ is the period of the ant, and $v$ is the speed of the ant. Rearranging the equation yields $r = \frac{Tv}{2\pi}$ . The ants are spaced at most $4 mm$ apart. It is easy to see that they are furthest apart at the outermost circle of the torus. The circumference $C$ of this circle is $C = 2\pi(R + r)$ . Since the ants are separated by a distance $d = 4mm$ at this point in time, and there are 100 ants, $C = 2\pi(R + r) = 100d$ . Rearranging this equation gives $R = \frac{100d - 2\pi r}{2\pi}$ . Substituting R and r into $V = 2 \pi^2 Rr^2$ gives a value of $69.6 cm^3$

Any two geodesics on the sphere intersect, so we can eliminate the orange. Geodesics on the soda can can be parallel circles on the lateral surface, but the top-most ant wouldn't have neighbors on both sides, so we can eliminate the can of soda. Similarly, there can be several parallel geodesics on the sandwich, but by stepping to the left repeatedly you would reach a geodesic which intersects the previous, a contradiction. So, the ants must be crawling on a doughnut.

The largest possible volume occurs when the ants are walking abreast along circles in the plane of the axis of rotation. Then the circumference of the torus is 4mm x 100 = 40 cm, so the radius of the outer circle is 40/(2pi) cm. The 20 second condition means that the circular paths have length 5 cm, radius 5/(2pi) cm. The volume of a torus is 2 pi^2 R r^2 (http://www.mathsisfun.com/geometry/torus.html), where r is the radius of the rotated circle, and R is the distance between the axis and the center of a path. R = 40/(2pi) - 5/(2pi) = 35/(2pi). The volume is thus 875/(4pi) = 69.63 cm^3.

first speed(s=0.25 cm/s) and time period(t=20 s) of each ant is fixed, thus each ant has to travel equal distance(d=20\times 0.25=5 cm) in each cycle. this rule out the choice sphere because more than one ant cannot crawl on it satisfying the above condition.

Also,due to property 2 and 3,cylinder and prism are ruled out because - If they follow property 2,then there are always 2 ant who have neighbour at one side only. - if they follow property 3,there must be intersection of paths.

torus satisfy all the properties and each ant crawls perpendicular to its area of cross section (i.e their paths are in the direction of magnetic field if the torus is to carry a current. )

Now,let a,r and R be radius of cross-section,inner and outer radius of torus respectively.

circumference of cross section of torus=d which gives, a=d/(2 pi)=5 / (2 pi) cm.

since every ant see two other on its sides within of 0.4 cm.

outer cicumference of torus<=x \times total number of ants ,which gives 2 pi R<=x \times 100 = 40 cm R<=20 / pi cm (for volume to be maximized we take R to be its max. value )

inner radius,r=R-2 \times a= 30 / (2 pi) cm

Now, volume of torus is given by v=pi (r+ R) \times (area of cross section) v=pi (r + R) \times pi a^2

substituting the values,we get ,v=( 70 /times 5^2 )/ (8 pi )=69.6302

so required answer,v=69.6

A torus gives the best solution. For any of the "closed" (equivalent to a sphere) objects, the trajectory has to be highly concave and close to the trajectory of another ant for the maximum distance condition to be satisfied. This severely limits the maximum area and thus the maximum volume attainable.

However, the torus allows a simple way to overcome this problem with a convex trajectory. Consider a torus formed by revolving a circle with radius r about a coplanar axis A situated R away from the centre. If we divide the torus into 100 pieces with radial planes from A, each resultant cut region gives the rajectory of the ant. Since the perimeter of the cut region is $2\pi r$ and this should be equal to the ant trajectory length of 5cm, we obtain that $r=\frac{5}{2\pi}$ .

If the ants follow the trajectories identically, the maximum distance between adjacent pairs is $\frac{2\pi(R+r)}{100}$ . This should be equal to the maximum separation between ants, 4mm. Hence $R=\frac{35}{2\pi}$ .

The volume of the torus is $2\pi^2 r^2 R = \frac{875}{4\pi}$

The starting point is to recognize which piece of food the ants are on. Since the ants never turn, they are traveling on geodesics of the surface of the food. Since the paths never cross and the paths are geodesics, this immediately rules out the orange as geodesics on a sphere necessarily cross. We can cover part of the sandwich and coke can with geodesics that remain a finite distance apart and never cross, but the "ends" of each surface won't then be covered by ants and so requirement 3 will not be satisfied.

The only possibility for the food is the doughnut. For the maximum volume of the food, we want the circuits to be the shortest closed geodesics on the torus, which are the ones around the smaller circle of the doughnut (as opposed to the "bounded geodesics" on the torus, which nobody other than mathematicians probably know about). Note that we could also twist these geodesics so they progress around the big circle of the doughnut N times as they traverse the smaller circle, but this would make the doughnut smaller.

If the period is 20 seconds and the speed is $0.25~\mbox{cm/s}$ , then the circumference around the small circle of the doughnut is $5~\mbox{cm}$ . Since the ants never get more than $4~\mbox{mm}$ apart and there are $100$ of them, the circumference of the large circle of the doughnut is $40~\mbox{cm}$ . The two standard radii of the torus are therefore $r=0.796~\mbox{cm}$ , and $R=5.57~\mbox{cm}$ . The volume of the doughnut is $69.66~ \mbox{cm}^3$