You have a closed cord of length 100 meters which touches point $P$. There is a line $L$ whose closest point is 20 meters from point $P$.
The value of a layout of cord is defined by the area enclosed on the side of the line with point $P$ plus twice the area enclosed on the opposite side.
What is the optimal shape of the cord that maximizes value?
See diagram below:
My best solution: $score = 1142.5655\ m^2$
As Jeffrey also explained, the curve is made of 3 arcs, one going from P to A, the second from A to B, the third going from B to P. The local optimality of the use of the available length implies constant curvature. It is only at boundaries, where the value of the surface changes, that the curvature can change.
The curvature is proportional to the value of the area. You can think of it as pressure. The higher the pressure, the more the lines curves out. Or to put it differently, the higher the value of the enclosed surface, the more it is worth to make an incursion in that area. That translates to higher curvature.
Anyway, the area right of the line counting twice, the curvature is double and the radius of the arc is halved. If the radius of the semi-circle on the right is $r$, then the radius of the 2 arcs left is $R = 2r$.
The problem being symmetrical, I will assume the optimal curve is also symmetrical relative to the horizontal axis. So I will only consider the upper half of the curve.
$CA$ is at an angle $\alpha$ from the horizontal.
$DP$ is at an angle $\alpha+\beta$.
From the image, we can see some relations beteen $\alpha$, $\beta$ and $r$:
The distance from $P$ to the axis is 20. It can be computed from $\alpha$, $\beta$ and $2r$:
(1) $20 = 2r\cos\alpha - 2r\cos(\alpha+\beta)$
The length of the curve is half of the available length of 100:
(2) $50 = r\alpha + 2r\beta$
The distance from A to the horizontal axis can be computed in 2 ways:
$dist = r\sin(\alpha) = 2r\sin(\alpha+\beta)$
(3) $\sin(\alpha) = 2\sin(\alpha+\beta)$
We have 3 equations, (1), (2) and (3), and 3 unknowns. We can solve it numerically:
$\alpha=1.7390005916$, $\beta=0.8871224528$, $r=14.2318548591$.
It places A at $(x_0,0)$, C at $(0,y_1)$ and D at $(2x_0,-y_1)$
where $x_0 = -r\sin(\alpha) = 2.38258653$
and $y_1 = r\cos(\alpha) = 14.03100047$.
The areas can be computed (for the whole curve):
$area\ right = \alpha r^2 + x_0 y_1 = 385.657152221015$
$area\ left = \beta (2r)^2 - (20+x_0) y_1 - x_0 y_1 = 371.25117209358604$
And the final scores evaluates to:
$area\ left + 2\ area\ right = 1142.565477$.
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).
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:
q at the center of the side of the triangle that is on L.r at one end of this side. Pq is
the altitude of the triangle. Its lenght is 20 (Pq = 20). Moreover, for the same reason, angle
Pqr = 90°.i the length of the side of the triangle that is on L, at
the same time the diameter of the semicircle.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
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
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:
For this, I did a little math and an angle at P of around 40 degrees to make this work.
The shape has to be defined by three circular arcs:
Each three having curvature away.
Ratioanle: Given any optimal shape maximizing the weighted enclosed area, one can fix the points A and B as well as the line length in each area. Then, by keeping A and B and the segment lengths, each area is maximize by using a circular arc.
So, to find the optimal shape, just do a search by fixing the three lengths (so they sum to 100 m) and picking A and B. So, finding the maximum of a well-defined function over 4 variables (the last length is a function of the other two).
Shouldn't be to hard from here. :-)
