Best value: 1116.34 (two semiellipses)
This question is interesting. I would like to answer it.
The most efficient shape is the a raindrop shape type. That is: an isosceles triangle with a semicircle.
Indeed, we are looking to maximize the area in the right because of the x2 incentive
. The best way to maximize the area of a shape with a cord is a circle. We can't create a full circle (P would be outside of it otherwise), so the other side of the circle has to be transformed.
We are looking to minimize the area in the left side because of the x1 deterrent effect
. The best way to minimize an area with a cord is a segment. But the lollipop strategy shows that it reduces dramaticaly the size of a full circle in the right area. So we need 3 sides; we need a triangle. The triangle can have straight sides or outwards curved sides (see below).
Last but not least, the shape must be symmetric. If not, the area of the semicircle is grazed by that of the triangle (see this example). Of course, the symmetry is not vertical (because we have two very different shapes), but horizontal.
A triangle + a semicircle + a mandatory horizontal symmetry = this shape, as discovered by our friend.Another option: two half-ellipses side by side (see below).
Shape 1: a triangle with straight sides : 1083.63
We need to compute the best areas of the triangle and the semicircle in order to have the highest value. To do so, I would like to add some information:

- a) I add
q
at the center of the side of the triangle that is on L.
- b) I add
r
at one end of this side.
- c) Because the triangle is isosceles (horizontal symmetry),
Pq
is
the altitude of the triangle. Its lenght is 20 (Pq = 20
). Moreover, for the same reason, angle
Pqr = 90°
.
- d) I name
i
the length of the side of the triangle that is on L, at
the same time the diameter of the semicircle.
- e)
A1
is the area of the triangle; A2
the area of the semicircle.
Now, we know how to compute the area of a triangle with its altitude (20) and the base (i
), and we know how to compute the area of a semicircle with its radius (here: i/2
). If we find the optimal length i
, then we have the solution.
To find i
, I have coded a small c++ program.
#include <iostream>
#include <math.h>
#define PI 3.1415926535
int main()
{
double i, // the length we are looking for
area1, // area of the triangle
area2, // area of the circle
perimeter, // perimeter of the whole shape
value; // the value to maximize
for (i = 0; i < 200; i=i+0.001) { // 200 is an arbitrary integer way outside the maximum expected value of i
area1 = (20 * i) / 2; // classic formula of the area of a triangle
area2 = (PI*((i/2)*(i/2)))/2; // (classic formula of the area of a circle)/2
value = area1 + area2 * 2; // We compute the expected value
perimeter = (sqrt((20 * 20) + (i / 2)*(i / 2))) * 2// Perimeter = lengths of the sides minus the base (Pythagorean theorem)...
+ (i*PI) / 2; // ... plus (perimeter of a circle)/2
if (perimeter > 99.999 && perimeter < 100.001) // if the perimeter +/- = 100, then show the solution
std::cout << "i: " << i << " | p: " << perimeter << " | a1: " << area1 << " | a2 : " << area2 << " | value : "<< value << std::endl;
}
}
After having incremented i by 0.001 steps, it appears that the optimal data are (for this shape):
i: 31.32 m. | p.: 100 (+/- .001) | area 1: 313.2 | area 2: 385.215
Value: 1083.63
Shape 2: a triangle with outwards curved sides (semiellipse+semicircle) : 1113.6

We suppose that the triangle is composed of 2 quarters of ellipse. See the following image. The curved triangle is ABC
. It is composed of two quarters of an ellipse : ABO
and AOC
. Well, in fact, our triangle is more half an ellipse, but anyway. :)

I have modified my program.
First, the area of the triangle is computed with the curves of the ellipse instead of the original segments. The formula for an ellipse is pi * minor axis [i/2] * major axis [20]
.
Second, the perimeter of the triangle minus the base is computed thanks to a Ramanujan formula:

Then the new program is:
#include <iostream>
#include <math.h>
#define PI 3.1415926535
int main()
{
double i, // the length we are looking for
area1, // area of the triangle
area2, // area of the circle
perimeter, // perimeter of the whole shape
value; // the value to maximize
for (i = 0; i < 200; i=i+0.01) { // 200 is an arbitrary integer way outside the maximum expected value of i
area1 = (PI*20*(i/2))/2; // ((area of ellipse))/4)*2
area2 = (PI*((i/2)*(i/2)))/2; // (classic formula of the area of a circle)/2
value = area1 + area2 * 2; // We compute the expected value
perimeter = (PI*(3*(20 + (i / 2)) - sqrt(((3 * 20 + (i / 2))*(20 + 3 * (i / 2)))))) / 2 // Perimeter of 2*1/4 of the ellipses (Ramanujan formula)
+ (i*PI) / 2; // ... plus (perimeter of a circle)/2
if (perimeter > 99.99 && perimeter < 100.01) // if the perimeter +/- = 100, then show all the possible solutions
std::cout << "i: " << i << " | p: " << perimeter << " | a1: " << area1 << " | a2 : " << area2 << " | value : "<< value << std::endl;
}
}
The optimal data with outwards curved triangle are:
i: 28.96 m. | p.: 100 (+/- .002) | area 1: 454.903 | area 2: 329.349
Value: 1113.6
Shape 3: Two half-ellipses side by side : 1116.34
A last hypothesis. What happens if the shape on the right is, as well as that on the left, a semi-ellipse?

We have to add a new variable: the minor axis of the right ellipse. We could call this variable j
.
My new program:
#include <iostream>
#include <math.h>
#define PI 3.14159265358979323846264338327950288419716939937510
int main()
{
double i, j,// the lengths we are looking for
area1, // area of right semiellipsis
area2, // area of left semiellipsis
perimeter, // perimeter of the whole shape
value; // the value to maximize
for (i = 0; i < 200; i = i + 0.01) {
for (j = 0; j<200; j += 0.01) {// 200 is an arbitrary integer way outside the maximum expected value of i
area1 = (PI * 20* (i/2) )/ 2; // area of semiellipse
area2 = (PI * j * (i/2)) / 2; // area of right semiellipse
value = area1 + area2 * 2; // Computation of the expected value
perimeter = (PI*(3 * (20 + (i/2)) - sqrt(((3 * 20 + (i/2))*(20 + 3 * (i/2)))))) / 2 // Perimeter of left semiellipse (Ramanujan formula)
+ (PI*(3 * (j + (i/2)) - sqrt(((3 * j + (i/2))*(j + 3 * (i/2)))))) / 2; // ... plus Perimeter of right semiellipse
if (perimeter > 99.999 && perimeter < 100.001) // if the perimeter +/- = 100, then show all the possible solutions
std::cout << "i: " << i << " | j: " << j << " | p: " << perimeter << " | a1: " << area1 << " | a2 : " << area2 << " | value : " << value << std::endl;
}
}
}
And now we have the best optimal value for this problem:
i: 27.87 | j: 15.5 | p.: 100 (+/- .001) | area 1: 437.781 | area 2: 339.28
value : 1116.34
My answer is then:
