In 1888, there were 13 characters: MDCCCLXXXVIII

In 2888, there will be 14 characters: MMDCCCLXXXVIII

Starting symbols are bold.

Immanuel Kant calculated the exact time of his arrival home as follows. He had wound his clock before leaving, so a glance at its face told him the amount of time that had elapsed during his absence. From this he subtracted the length of the time spent with Schmidt by checking Schmidt’s hallway clock when he arrived and again when he left. This gave him the total time spent walking. Since he returned along the same route, at the same speed, he halved the total walking time to obtain the length of time it took him to walk home. This added to the time of his departure from Schimdt’s house gave him the time of his arrival home.

Thank you to Antonio Palladino for drawing the diagram and to Gianfranco Bo for sharing it in 2005. Palladino shared a solution, which I translated and expanded upon below.

Let P be the point of intersection of the two arcs Γ1 and Γ2; let B be the center of the circle Γ3 tangent internally Γ1 and Γ2 and externally the internal square; let C be the center of the circle Γ4 externally tangent Γ1 and Γ2 and internally the unit external square. The vertical axis BPC is of symmetry throughout the figure.

Let's call the side length of the inner square x.

To find x, draw a line from the top right of the inner square, F, to the bottom left of the outer square, O. This creates right triangle OEF.

Applying the Pythagorean Theorem, we find:

OE^2 + EF^2 = OF^2

OE, thanks to vertical symmetry, is half of each the inner and outer squares' side lengths. This becomes clear if we look carefully at OH + HE.

OE = OH + HE = a/2 + x/2

EF is exactly the side length of the inner square.

EF = x

The last length, OF, is captivating. Touching the arc means it is the same as a radius of the quarter circle represented by Γ2. Because OA is also a radius, and its length is a, OF is equal to the side length of the external square.

OF = a

Plugging into the Pythagorean Theorem we get:

(a/2 + x/2)^2 + x^2 = a^2

To ease solving let's assume a = 1.

(1/2 + x/2)^2 + x^2 = 1

We obtain the quadratic equation in x.

5x^2 + 2x - 3 = 1

which, when solved, gives the side length of the internal square:

x = 3/5

We used a = 1 to ease solving. We bring it back to see the side length of the internal square in terms of a.

x = (3/5)a

Let's now find the radius b of the larger inner circle Γ3. Draw a line OG through center B.

OG = OB + BG

OG touches the arc in the same way as before, meaning it equals the external square's side length. We again assume a = 1 for ease of solving.

OG = 1

OB is the hypotenuse of the right triangle OBH, which can be obtained via the Pythagorean Theorem.

OB = sqrt(OH^2 + BH^2) = sqrt((1/2)^2 + (3/5 + b)^2)

BG is the radius we're trying to find.

BG = b

Returning to our equation for OG, we see the irrational equation:

1 = sqrt((1/2)^2 + (3/5 + b)^2) + b

Solving for b, the radius of the larger inner circle:

b = 39/320

We used a = 1 again to ease solving. Bringing it back we get the radius in terms of a:

b = (39/320)a

Finally, we want the radius c of the smaller inner circle Γ4. Draw a line from the bottom left of the external square, O, to center C.

To get the radius c, we need to find IC. This can be observed as:

IC = OC - OI

OC is the hypotenuse of right triangle OCH; it can be obtained through the Pythagorean theorem. We again assume a = 1 for ease of solving.

OC = sqrt(OH^2 + CH^2) = sqrt((1/2)^2 + (1-c)^2)

OI is touching the arc again, giving us the value of the side length of the external square.

OI = a = 1

Returning to our search for c or IC:

IC = OC - OI

c = sqrt((1/2)^2 + (1-c)^2) - 1

c = 1/16

And again bringing back to terms of the side length of the external square:

c = (1/16)a

To summarize, the side length of the internal square is (3/5)a, the radius of the larger internal circle is (39/320)a, and the radius of the smaller internal circle is (1/16)a.

The original pile had 3121 coconuts.